# Sausage Volume of the Random String and Survival in a medium of Poisson
Traps

Siva Athreya and Mathew Joseph and Carl Mueller  Siva Athreya
International centre for theoretical Sciences
Survey No. 151, Shivakote,
Hesaraghatta Hobli,
Bengaluru - 560 089
and
Statmath Unit
Indian Statistical Institute
8th Mile Mysore Road
Bangalore 560059<EMAIL_ADDRESS>or<EMAIL_ADDRESS>Mathew Joseph
Statmath Unit
Indian Statistical Institute
8th Mile Mysore Road
Bangalore 560059<EMAIL_ADDRESS>Carl Mueller
Department of Mathematics, University of Rochester Rochester, NY 14627
<EMAIL_ADDRESS>
###### Abstract.

We provide asymptotic bounds on the survival probability of a moving polymer
in an environment of Poisson traps. Our model for the polymer is the vector-
valued solution of a stochastic heat equation driven by additive spacetime
white noise; solutions take values in $\mathbb{R}^{d},d\geq 1$. We give upper
and lower bounds for the survival probability in the cases of hard and soft
obstacles. Our bounds decay exponentially with rate proportional to
$T^{d/(d+2)}$, the same exponent that occurs in the case of Brownian motion.
The exponents also depend on the length $J$ of the polymer, but here our upper
and lower bounds involve different powers of $J$.

Secondly, our main theorems imply upper and lower bounds for the growth of the
Wiener sausage around our string. The Wiener sausage is the union of balls of
a given radius centered at points of our random string, with time less than or
equal to a given value.

###### Key words and phrases:

heat equation, white noise, stochastic partial differential equations,
Poisson, hard obstacles, survival probability.

###### 2010 Mathematics Subject Classification:

Primary, 60H15; Secondary, 60G17, 60G60.

##  1\. Introduction

The model of particles performing random diffusive motion in a region
containing randomly located traps is known as the trapping problem (see
[dHW94] for review). Particle motion is typically Brownian motion in
$\mathbb{R}^{d}$ or a random walk in $\mathbb{Z}^{d}$. The traps are placed in
a Poissonian manner and the particle gets annihilated on encountering a trap.
The main question of interest in such models is the “Survival Probability” of
the particle. We refer the reader to [Szn98] and references there in for a
review of the problem of Brownian motion among Poissonian obstacles, to
[Kön16] and references there in for a review of the problem of a random walk
in a random potential and to [ADS19] for a review of Random walks among mobile
and immobile traps.

There is an extensive literature about such trapping problems, see the
references in the preceding paragraph. These results often depend on refined
estimates for the eigenvalues of the Laplacian or potential theory. On the
other hand, the process we consider takes values in function space, and we
found it impossible to analyze the situation using existing techniques.
Indeed, carrying over finite-dimensional potential theoretic arguments to the
infinite dimensional case is often difficult or impossible. We will consider a
Gaussian process, but we did not find any Gaussian tools which were relevant
to trapping problems.

In this article we will study the annealed survival probability of a random
string in a Poissonian trap environment. Let
$(\Omega,{\mathcal{F}},{\mathcal{F}}_{t},\mathbb{P}_{0})$ be a filtered
probability space on which $\dot{\mathbf{W}}=\dot{\mathbf{W}}(t,x)$ is a
$d$-dimensional random vector whose components are i.i.d. two-parameter white
noises adapted to ${\mathcal{F}}_{t}$. We consider a random string
$\mathbf{u}(t,x)\in\mathbb{R}^{d}$, which is the solution to the following
stochastic heat equation (SHE)

(1.1)
$\begin{split}\partial_{t}{\mathbf{u}}(t,x)&=\frac{1}{2}\,\partial_{x}^{2}{\mathbf{u}}(t,x)+\,\dot{\mathbf{W}}(t,x)\\\
\mathbf{u}(0,x)&=\mathbf{u}_{0}(x)\end{split}$

on the circle $x\in[0,J]$, having endpoints identified, and $t\in[0,T].$ The
initial profile $\mathbf{u}_{0}$ is assumed to be continuous. Note that we
will use boldface letters to denote vector-valued quantities.

We will be interested in the evolution of the random string in a field of
obstacles centered at points coming from an independent Poisson point process.
More precisely, let $(\Omega_{1},\mathcal{G},\mathbb{P}_{1})$ be a second
probability space on which is defined a Poisson point process $\bm{\eta}$ with
intensity $\nu$ given by

$\bm{\eta}(\omega_{1})=\sum_{i\geq
1}\delta_{\bm{\xi}_{i}(\omega_{1})},\quad\omega_{1}\in\Omega_{1},$

with points $\\{\bm{\xi}_{i}(\omega_{1})\\}_{i\geq 1}\subset\mathbb{R}^{d}$.

The obstacles will be formed via a potential
$\text{V}:\mathbb{R}^{d}\times\Omega_{1}\rightarrow[0,\infty]$

$\text{V}(\mathbf{z},\bm{\eta})=\sum_{i\geq
1}\text{H}(\mathbf{z}-\bm{\xi_{i}}),$

where $\text{H}:\mathbb{R}^{d}\rightarrow[0,\infty]$ is a non-negative,
measurable function whose support of H is contained in the closed ball
$B(\mathbf{0},a)$ of radius $0<a\leq 1$ centered at $\mathbf{0}$.

We will work in the product space
$(\Omega\times\Omega_{1},\mathcal{F}\times\mathcal{G},\mathbb{P}_{0}\times\mathbb{P}_{1})$
along with the filtration $(\mathcal{F}_{t}\times\mathcal{G})_{t\geq 0}$. We
will write $\mathbb{E}$ for the expectation with respect to
$\mathbb{P}:=\mathbb{P}_{0}\times\mathbb{P}_{1}$, and $\mathbb{E}_{i}$ for the
expectation with respect to $\mathbb{P}_{i}$ for $i=0,1$. Our main quantity of
interest is the quenched and the annealed survival probabilities given by

$\begin{split}S_{T,\eta}(\omega_{1})&=\mathbb{E}_{0}\left[\exp\left(-\int_{0}^{T}\int_{0}^{J}\text{V}\Big{(}\mathbf{u}(s,x),\bm{\eta}(\omega_{1})\Big{)}dxds\right)\right],\text{
and }\\\
S_{T}&=\mathbb{E}\left[\exp\left(-\int_{0}^{T}\int_{0}^{J}\text{V}\left(\mathbf{u}(s,x),\bm{\eta}\right)dxds\right)\right]\end{split}$

respectively. Sometimes we will write $S_{T}^{\text{H},J,\nu}$ and
$S_{T,\bm{\eta}}^{\text{H},J,\nu}$ to emphasize the dependence on
$\text{H},J,\nu$.

###  1.1. Main Result

Our first result on hard obstacles, i.e. the string is killed immediately on
contact and the only way it can survive is to avoid them.

######  Theorem 1.1 (Hard obstacles).

Consider the solution to (1.1) with $d\geq 2$ and $J\geq 1$, and let $\nu$ and
$a$ be as above. Then the following hold in the case
$\text{H}(\cdot)\equiv\infty\cdot\mathbf{1}_{B(\mathbf{0},a)}$

  1. (1)

(Lower bound) There exist positive constants $C_{0},C_{1},C_{2}$ independent
of $T,J$ such that for $T\geq C_{0}J^{2+\frac{d}{2}}$

(1.2) $S^{\textnormal{H},J,\nu}_{T}\geq
C_{1}\exp\left(-C_{2}\left(\frac{T}{J}\right)^{\frac{d}{d+2}}\right).$

  2. (2)

(Upper bound) There exist positive constants $C_{3},C_{4}$ independent of
$T,J$ such that for all $T>0,J\geq 1$

(1.3) $S^{\textnormal{H},J,\nu}_{T}\leq C_{3}\exp\left(-\frac{C_{4}}{1+|\log
J|}\left(\frac{T}{J^{2}}\right)^{\frac{d}{d+2}}\right).$

In the case of hard obstacles we immediately see that the survival of the
string is only possible if the string avoids the traps. Thus the“sausage of
radius $a$ around string up to time $T$” should be devoid of traps. Indeed it
is easy to check using standard properties of the Poisson random variable that

(1.4)
$S_{T}^{\text{H},J,\nu}=\mathbb{E}\exp\left(-\nu\left|\mathscr{S}^{J}_{T}(a)\right|\right),$

where

(1.5) $\mathscr{S}^{J}_{T}(a)=\mathop{\bigcup}_{\begin{subarray}{c}0\leq s\leq
T,\\\ 0\leq y\leq
J\end{subarray}}\left\\{\mathbf{u}(s,y)+B(\mathbf{0},a)\right\\},$

is the sausage of radius $a$ around $\mathbf{u}$. Thus Theorem 1.1 also
provides bounds on the exponential moments of the volume of the sausage of
radius $a$ around the string up to time $T$.

We next turn our attention to the case of soft obstacles, i.e. H does not take
the value $\infty$. We make the following specific assumptions on H.

######  Assumption 1.1.

There is a $\mathscr{C}>0$ such that
$\text{H}(\mathbf{x})\geq\mathscr{C}\cdot\mathbf{1}_{B(\mathbf{0},\frac{a}{2})}(\mathbf{x})$.

We note that under this assumption there is a positive probability of survival
even if the string interacts with the obstacle environment. We are now ready
to state our result in this setting.

######  Theorem 1.2 (Soft obstacles).

Consider the solution to (1.1) with $d\geq 2$ and $J\geq 1$, let $\nu>0$ be as
above and let H be a soft obstacle satisfying Assumption 1.1. Then

  1. (1)

(Lower bound) There exist positive constants $C_{0},C_{1},C_{2}$ independent
of $T,J$ such that for $T\geq C_{0}J^{2+\frac{d}{2}}$

(1.6) $S^{\textnormal{H},J,\nu}_{T}\geq
C_{1}\exp\left(-C_{2}\left(\frac{T}{J}\right)^{\frac{d}{d+2}}\right).$

  2. (2)

(Upper bound) Fix $\beta>0$. There exist positive constants $C_{3},C_{4}$
independent of $T,J$ such that for all $T>0,J\geq 1$

(1.7) $S^{\textnormal{H},J,\nu}_{T}\leq
C_{4}\exp\left(-\frac{C_{5}}{J^{3+\beta}(1+|\log
J|)}\left(\frac{T}{J^{2}}\right)^{\frac{d}{d+2}}\right).$

We conclude this sub-section with a few remarks.

######  Remark 1.1.

Though the bounds do not match with regard to the exponents of $J$, they do
match in that of $T$.

  1. (i)

If we set $J=1$ in (1.2)and (1.3) or in (1.6) and (1.7) then for large enough
$T>0$, in both the hard and soft obstacle cases we have

$C_{1}\exp\left(-C_{2}T^{\frac{d}{d+2}}\right)\leq
S^{\textnormal{H},1,\nu}_{T}\leq
C_{3}\exp\left(-C_{4}T^{\frac{d}{d+2}}\right),$

for some constants $C_{1},C_{2},C_{3},C_{4}>0.$

  2. (ii)

Due to (1.4), Theorem 1.1 immediately gives us bounds on
$\mathbb{E}\exp\left(-\nu\left|\mathscr{S}^{J}_{T}(a)\right|\right)$. Further,
as observe above at $J=1$, from Theorem 1.1 that we have for sufficiently
large $T>0$

$C_{1}\exp\left(-C_{2}T^{\frac{d}{d+2}}\right)\leq\mathbb{E}\exp\left(-\nu\left|\mathscr{S}^{1}_{T}(a)\right|\right)\leq
C_{3}\exp\left(-C_{4}T^{\frac{d}{d+2}}\ \right),$

for some constants $C_{1},C_{2},C_{3},C_{4}>0.$ The exponent of $T$ matches
that of the asymptotics of the volume of the Brownian Sausage. Since in our
case ${\mathbf{u}}$ is the solution of a stochastic PDE, this seems to be a
new result of independent interest.

  3. (iii)

The constants mentioned in Theorem 1.1 and Theorem 1.2 are all independent of
$T,J$ but do depend on $\nu,a,$ and $\mathscr{C}$. The bound $0\leq a\leq 1$
is used for technical convenience.

###  1.2. Overview of Proof

We will say that ${\mathbf{u}}$ is a solution to (1.1) if it satisfies,

(1.8)
$\mathbf{u}(t,x)=\int_{0}^{J}G^{(J)}(t,x-y)\mathbf{u}_{0}(y)dy+\int_{[0,t]\times[0,J]}G^{(J)}(t-s,x-y)\mathbf{W}(dsdy),$

where $[0,J]$ is the circle with endpoints identified and
$G^{(J)}:\mathbb{R}_{+}\times[0,J]\rightarrow\mathbb{R}$ is the fundamental
solution of the heat equation

$\begin{split}\partial_{t}G^{(J)}(t,x)&=\frac{1}{2}\,\partial_{x}^{2}G^{(J)}(t,x),\\\
G^{(J)}(0,x)&=\delta(x).\end{split}$

Furthermore, the final integral in (1.8) can be regarded as either a Wiener
integral or a white noise integral in the sense of Walsh [Wal86].

The first reduction in the proof is to reduce to the case $J=1.$ We will do
this by deriving a scaling relation for $S_{T}^{\text{H},J,\nu}$. Consider

$\mathbf{v}(t,x):=J^{-\frac{1}{2}}\mathbf{u}(J^{2}t,Jx),$

defined for $x\in[0,1]$ with endpoints identified, and $t\in[0,TJ^{-2}]$. The
initial profile is
$\mathbf{v}(0,x)=\mathbf{v}_{0}(x)=J^{-1/2}\mathbf{u}_{0}(Jx)$. It was proved
in Lemma 2.2 of [AJM21] that $\mathbf{v}$ satisfies

$\begin{split}\partial_{t}\mathbf{v}&=\frac{1}{2}\partial_{x}^{2}\mathbf{v}+\dot{\widetilde{\mathbf{W}}},\quad
t\in[0,TJ^{-2}]\\\ \mathbf{v}(0,x)&=\mathbf{v}_{0}(x),\quad
x\in[0,1]\end{split}$

for some other white noise $\dot{\widetilde{\mathbf{W}}}$. Now it is easily
checked

$\displaystyle S_{T}^{\text{H},J,\nu}$
$\displaystyle=\mathbb{E}\left[\exp\left(-\int_{0}^{T}\int_{0}^{J}\sum_{i\geq
1}\text{H}\left(\mathbf{u}(s,x)-\bm{\xi}_{i}\right)dxds\right)\right]$
$\displaystyle=\mathbb{E}\left[\exp\left(-\int_{0}^{\frac{T}{J^{2}}}\int_{0}^{1}\sum_{i\geq
1}J^{3}\text{H}\left(J^{\frac{1}{2}}\left(\mathbf{v}(\tilde{s},\tilde{x})-\frac{\bm{\xi}_{i}}{J^{\frac{1}{2}}}\right)\right)d\tilde{x}d\tilde{s}\right)\right]$

Define

(1.9)
$\widetilde{\text{H}}(\cdot):=J^{3}\text{H}(J^{\frac{1}{2}}\cdot),\qquad\tilde{\bm{\xi}_{i}}:=\frac{\bm{\xi}_{i}}{J^{\frac{1}{2}}},\qquad\mbox{
and }\qquad\widetilde{\nu}:=\nu J^{\frac{d}{2}}.$

It is easily seen that $\widetilde{\text{H}}$ is supported in the ball
$B(\mathbf{0},aJ^{-\frac{1}{2}})$. The points $\tilde{\bm{\xi}}$ form a
Poisson point process of intensity $\widetilde{\nu}$. Setting
$\widetilde{T}=TJ^{-2}$ we obtain

(1.10)
$S_{T}^{\text{H},J,\nu}=S_{\widetilde{T}}^{\widetilde{\text{H}},1,\widetilde{\nu}}.$

######  Remark 1.2 (Important).

For the rest of article will focus on $J=1$, so (1.8) becomes

$\mathbf{u}(t,x)=\int_{0}^{1}G(t,x-y)\mathbf{u}_{0}(y)\,dy+\mathbf{N}(t,x),$

where

$\mathbf{N}(t,x)=\int_{[0,t]\times[0,1]}G(t-s,x-y)\mathbf{W}(dsdy)$

is the noise term. For simplicity of notation, we have also removed the
superscipt in $G^{(1)}$. We will work with $S_{T}^{\textnormal{H},1,\nu}$ and
finally use the scaling relation (1.10) to obtain the bounds for
$S_{T}^{\textnormal{H},J,\nu}$

The strategy for proving the lower bound for survival probability in Theorem
1.1 Theorem 1.2 is the same as that of Random walks or Brownian motions moving
in a field of Poisson Traps. This is obtained by identifying an optimal
configuration for the traps ${\mathbf{\xi}}$. The configuration being one
which has an area free of traps in a ball of radius $\alpha_{T}$ around the
origin and the string under this potential is forced to stay inside this ball
till time $T$. The probability of first event is of order
$\exp(-C_{1}(\alpha_{T}+a)^{d})$. To calculate the second probability we
decompose the string in to two components namely center of mass and radius of
$\mathbf{u}$ respectively. More precisely

(1.11) $\begin{split}\mathbf{X}_{t}&=\int_{0}^{1}\mathbf{u}(t,x)\,dx,\mbox{
(Center of Mass) }\\\
\mathbf{R}_{t}&=\sup_{x\in[0,1]}\left|\mathbf{u}(t,x)-\mathbf{X}_{t}\right|,\mbox{
(Radius). }\end{split}$

We show that $\mathbf{X}_{t}$ and $\mathbf{R}_{t}$ are independent. Then
separately we compute the probabilities of
$|\mathbf{X}_{t}|\leq\frac{\alpha_{T}}{2},\;t\leq T$ and
$\mathbf{R}_{t}\leq\frac{\alpha_{T}}{2},\;t\leq T$, and show that these are
bounded by $\exp(-C_{2}(\frac{T}{\alpha_{T}^{2}})).$ Optimising over choice of
$\alpha_{T}$ and the scaling relations discussed above yield the results. We
present the details in Section 2.

Unlike the lower bound, the proof of upper bound differs from the classical
setting of random walks or that of Brownian motion. Following Remark 1.2 we
first obtain an upper bound for $S_{T}^{\text{H},1,\nu}$. Recall

(1.12)
$S_{T}^{\text{H},1,\nu}=\mathbb{E}\exp\left(-\nu\left|\mathscr{S}^{1}_{T}(a)\right|\right),$

where $\mathscr{S}^{1}_{T}(a)$ is the sausage of radius $a$ around
$\mathbf{u}$, that is

$\mathscr{S}^{1}_{T}(a)=\mathop{\bigcup}_{\begin{subarray}{c}0\leq s\leq T,\\\
0\leq y\leq 1\end{subarray}}\left\\{\mathbf{u}(s,y)+B(\mathbf{0},a)\right\\}.$

We will explain the strategy for the proof in the case of hard obstacles, the
argument for the soft obstacles not being very different. Due to (1.12), an
upper bound on the partition function $S_{T}^{\text{H},1,\nu}$ essentially
boils down to obtaining a lower bound on the volume of the sausage
$\mathscr{S}_{T}^{1}(a)$ around $\mathbf{u}$. For this, we consider the
sausage around $\mathbf{u}(t)=\mathbf{u}(t,\cdot)$, that is

(1.13) $\mathscr{S}^{1}(a;t)=\mathop{\bigcup}_{0\leq y\leq
1}\left\\{\mathbf{u}(t,y)+B(\mathbf{0},a)\right\\},$

so that

$\mathscr{S}^{1}_{T}(a)=\bigcup_{0\leq t\leq T}\mathscr{S}^{1}(a;t).$

We will identify times at which $\mathscr{S}^{1}(a;t)$ do not intersect, so
the sum of the volumes of these fixed time sausages will provide the desired
lower bound.

We will consider a set of stopping times $\tau_{i}$ (see (3.3)) such that the
center of mass at these time points, ${\mathbf{X}}_{\tau_{i}}$ are separated
by at least $4\Lambda$ from each other (where $\Lambda$ is suitably chosen see
Lemma 3.10). Using known results on the volume of the Wiener sausage, we show
that the number of $\tau_{i}$ before time $T$ should be of order
$T^{\frac{d}{d+2}}$ (see Lemma 3.2). Now, let

$\mathbf{N}(s,t;x):=\int_{[s,t]\times[0,1]}G(t-r,x-y)\mathbf{W}(drdy),$

which represents the noise term from time $s$ to $t$. Then $\mathbf{N}(s,t)$
will represent the function from $x\in[0,1]$ to $\mathbb{R}^{d}$. For $s<t$,
we use the Markov property for $\mathbf{u}$ to write

$\mathbf{u}(t)=G_{t-s}*\mathbf{u}(s)+\mathbf{N}(s,t).$

If $s\ll t$ then the first term is almost a constant because of the
smoothening effect of the Laplacian. The volume of the sausage around
$G_{t-s}*\mathbf{u}(s)$ will then be approximately $a^{d}$. We show in Lemma
3.11, using the independence of $\mathbf{X}_{t}$ and $\mathbf{R}_{t}$, that
with probability $\geq\frac{1}{2}$ the range (see (3.8) for precise
definition) of $\mathbf{N}(s,t)$ is at most $\Lambda$ and the volume of the
sausage of radius $a$ around $\mathbf{N}(s,t)$ is of order at least
$a^{d-2+\epsilon}$. Consequently there is a subset $\\{T_{i}\\}$ of
$\\{\tau_{i}\\}$ of positive density, such that the sausages around
$\mathbf{u}(T_{i})$ are disjoint and have volume $a^{d-2+\epsilon}$ each. This
gives a lower bound of $a^{d-2+\epsilon}T^{\frac{d}{d+2}}$ on the volume
$\mathscr{S}_{T}^{1}(a)$.

As mentioned earlier, our strategy for the lower bound of the annealed
survival probability is the same as for a random walk or Brownian motion
surviving in a field of random traps. The upper bound is different, since the
usual strategies depend on potential theory and eigenvalues of the Laplacian,
and both of these are much harder to study for infinite dimensional processes
such as the random string. We are forced to go back to first principles, which
perhaps explains the fact that our upper and lower bounds do not completely
match. It is also important to note that while the upper and lower bounds
match for the case $J=1$, the scaling relations in (1.9) imply that they don’t
carry over to the general case via space-time scaling.

The quenched survival probability is of keen interest and will be the focus of
future work. Here the geometry of the string and its topology will come into
play. We did not explore large deviations for the volume of the sausage as
there was no immediate ergodicity to establish a limiting value of a Lyapunov
exponent.

Convention We will use $C$ to denote constants whose value might change from
line to line. Sometimes we will indicate dependence of constants on parameters
by putting the parameters in parentheses, for example $C(d),C(\nu,d)$ etc. The
notation $C_{1},C_{2},\cdots$ will be used to denote constants whose value
remain fixed throughout a lemma, proposition, theorem etc. Such constants
might be used later in which case it will be clear from the context.

Acknowledgement: S.A. research was partially supported by the CPDA grant from
the Indian Statistical Institute and the Knowledge Exchange grant from the
International Centre for Theoretical Sciences, C.M. research was partially
supported by Simons Collaboration Grant 513424, and M.J. research was
partially supported by Serb Matrics grant MTR/2020/000453 and a CPDA grant
from Indian Statistical Institute.

##  2\. Proof of the lower bound in Theorems 1.1 and 1.2

As indicated above we will use the same strategy for the lower bound for the
survival probability for both the hard and soft obstacle case. We need a
couple of technical results before we begin the proof.

The following lemma is crucial.

######  Lemma 2.1.

With $\mathbf{X}_{t}$ and $\mathbf{R}_{t}$ as in (1.11), we have

  * (a)

$\mathbf{X}_{t}$ is a standard Brownian motion starting at
$\int_{0}^{1}\mathbf{u}_{0}(x)dx$.

  * (b)

$\mathbf{X}_{t}$ and $\mathbf{R}_{t}$ are independent.

###### Proof.

It is easily checked that

$\mathbf{X}_{t}=\int_{0}^{1}\mathbf{u}_{0}(x)dx+\int_{0}^{t}\mathbf{W}(dyds)$

is a standard Brownian motion, and

$\mathbf{u}(t,x)-\mathbf{X}_{t}=\int_{0}^{1}\left[G(t,x-y)-1\right]\mathbf{u}_{0}(y)dy+\int_{[0,t]\times[0,1]}\left[G(t-s,x-y)-1\right]\mathbf{W}(dsdy),$

where $G=G^{(1)}$ is the heat kernel on the unit circle. Both these processes
are Gaussian. The components of $\mathbf{X}_{t}$ and
$\mathbf{u}(t,x)-\mathbf{X}_{t}$ are uncorrelated since

$\int_{0}^{t}\int_{0}^{1}\left[G(t-s,x-y)-1\right]\,dyds=0.$

The second part of the lemma immediately follows. ∎

We will also need

######  Proposition 2.1.

Assume $\sup_{x}|\mathbf{u}_{0}(x)|\leq\frac{\alpha}{2}$. Then there are
constants $0<C_{0}<1$ and $K_{0}>0$ such that for all $\alpha\geq K_{0}$

(2.1) $\mathbb{P}_{0}\left(\sup_{\stackrel{{\scriptstyle
s\leq\alpha^{2}}}{{x\in[0,1]}}}\left|\mathbf{u}(s,x)\right|\leq\alpha,\;\sup_{x\in[0,1]}\left|\mathbf{u}(\alpha,x)\right|\leq\frac{\alpha}{2}\right)\geq
C_{0}.$

###### Proof.

Let us consider first the case $\mathbf{u}_{0}\equiv{\mathbf{0}}$. From the
previous lemma

$\mathbb{P}_{0}\left(\sup_{\stackrel{{\scriptstyle
s\leq\alpha^{2}}}{{x\in[0,1]}}}\left|\mathbf{u}(s,x)\right|\leq\frac{\alpha}{2}\right)\geq\mathbb{P}_{0}\left(\sup_{s\leq\alpha^{2}}|\mathbf{R}_{s}|\leq\frac{\alpha}{4}\right)\mathbb{P}_{0}\left(\sup_{s\leq\alpha^{2}}|\mathbf{X}_{s}|\leq\frac{\alpha}{4}\right).$

Since the last term is a positive constant independent of $T$, it is enough to
show that there is a $K_{0}>0$ such that

$\sup_{\alpha\geq
K_{0}}\;\mathbb{P}_{0}\left(\sup_{s\leq\alpha^{2}}|\mathbf{R}_{s}|>\frac{\alpha}{4}\right)<1.$

Now

(2.2)
$\displaystyle\mathbb{P}_{0}\left(\sup_{s\leq\alpha^{2}}|\mathbf{R}_{s}|>\frac{\alpha}{4}\right)=\mathbb{P}_{0}\left(\sup_{\stackrel{{\scriptstyle
s\leq\alpha^{2}}}{{x\in[0,1]}}}|\mathbf{u}(s,x)-\mathbf{X}_{s}|>\frac{\alpha}{4}\right).$

By splitting the time interval into subintervals of length $1$ we have the
bound

(2.3) $\begin{split}&\mathbb{P}_{0}\left(\sup_{\stackrel{{\scriptstyle
s\leq\alpha^{2}}}{{x\in[0,1]}}}|\mathbf{u}(s,x)-\mathbf{X}_{s}|>\frac{\alpha}{4}\right)\\\
&\leq\sum_{k=0}^{[\alpha^{2}]+1}\mathbb{P}_{0}\left(\sup_{\stackrel{{\scriptstyle
s\in[k,k+1]}}{{x\in[0,1]}}}|\mathbf{u}(s,x)-\mathbf{X}_{s}|>\frac{\alpha}{4}\right)\\\
&\leq\sum_{k=0}^{[\alpha^{2}]+1}\mathbb{P}_{0}\left(|\mathbf{u}(k,0)-\mathbf{X}_{k}|>\frac{\alpha}{8}\right)\\\
&\qquad\qquad+\sum_{k=0}^{[\alpha^{2}]+1}\mathbb{P}_{0}\left(\sup_{\stackrel{{\scriptstyle
s\in[i,i+1]}}{{x\in[0,1]}}}|\left[\mathbf{u}(k,0)-\mathbf{X}_{k}\right]-\left[\mathbf{u}(s,x)-\mathbf{X}_{s}\right]|>\frac{\alpha}{8}\right)\end{split}$

Using the standard Fourier decomposition of $G(t,x)$ (see Section 3 of
[AJM21]) one obtains that each coordinate of $\mathbf{u}(i,0)-\mathbf{X}_{i}$
has variance

$\displaystyle\int_{0}^{k}\int_{0}^{1}\left[G(s,y)-1\right]^{2}\,dyds$
$\displaystyle=\int_{0}^{k}\int_{0}^{1}G^{2}(s,y)\,dyds-k$
$\displaystyle=\int_{0}^{k}\sum_{l\geq 1}\exp\left(-(2\pi l)^{2}s\right)\,ds$
$\displaystyle\leq C.$

See Lemma 3.1 of [AJM21] for details. Therefore for large $\alpha$

$\sum_{k=0}^{[\alpha^{2}]+1}\mathbb{P}_{0}\left(|\mathbf{u}(k,0)-\mathbf{X}_{k}|>\frac{\alpha}{8}\right)\leq\exp\left(-C\alpha^{2}\right).$

Now we turn to the last term in (2.3). Consider the process

$\mathbf{M}(s,x)=\left[\mathbf{u}(k,0)-\mathbf{X}_{k}\right]-\left[\mathbf{u}(s,x)-\mathbf{X}_{s}\right],\quad
s\in[k,k+1],\;x\in[0,1].$

Note that $\mathbf{M}(k,0)=0$. A quick calculation gives

$\displaystyle\mathbf{M}(s,x)-\mathbf{M}(s,\tilde{x})$
$\displaystyle=\int_{[0,s]\times[0,1]}\left[G(s-r,\tilde{x}-y)-G(s-r,x-y)\right]\mathbf{W}(drdy)$

whose components have variance less than $C|x-y|$ (see Lemma 3.1 of [AJM21]
for details). Similarly for $k\leq s<\tilde{s}\leq k+1$ we obtain

$\displaystyle\mathbf{M}(s,x)-\mathbf{M}(\tilde{s},x)$
$\displaystyle=\int_{[0,s]\times[0,1]}\left[G(\tilde{s}-r,x-y)-G(s-r,x-y)\right]\mathbf{W}(drdy)$
$\displaystyle\quad+\int_{[s,\tilde{s}]\times[0,1]}G(\tilde{s}-r,x-y)\mathbf{W}(drdy)+\left[\mathbf{X}_{s}-\mathbf{X}_{\tilde{s}}\right].$

Following Lemma 3.1 of [AJM21] we obtain that the components have variance
less than $C\sqrt{\tilde{s}-s}$. Note that $\tilde{s}\leq s+1$ so that the
variance of the components of $\mathbf{X}_{\tilde{s}}-\mathbf{X}_{s}$ are also
bounded by $C\sqrt{\tilde{s}-s}$.

Therefore the conditions of Lemma 3.4 of [AJM21] are satisfied, and we obtain
for large $\alpha$

$\sum_{k=0}^{[\alpha^{2}]+1}\mathbb{P}_{0}\left(\sup_{\stackrel{{\scriptstyle
s\in[k,k+1]}}{{x\in[0,1]}}}\left|\left[\mathbf{u}(k,0)-\mathbf{X}_{k}\right]-\left[\mathbf{u}(s,x)-\mathbf{X}_{s}\right]\right|>\frac{\alpha}{8}\right)\leq\exp\left(-C\alpha^{2}\right).$

Returning to (2.2) we obtain

$\mathbb{P}_{0}\left(\sup_{s\leq\alpha^{2}}|\mathbf{R}_{s}|>\frac{\alpha}{2}\right)\leq\exp\left(-C\alpha^{2}\right)<1,$

uniformly in $\alpha\geq K_{0}$ for some $K_{0}>0$. This completes the proof
of the proposition in the case that $\mathbf{u}_{0}\equiv{\mathbf{0}}$.

In the general case $\sup_{x}|\mathbf{u}_{0}(x)|\leq\frac{\alpha}{2}$, we
apply a Girsanov change of measure argument. Consider the measure
$\mathbb{Q}_{0}$ given by

$\frac{d\mathbb{Q}_{0}}{d\mathbb{P}_{0}}=\exp\left(-\int_{[0,\alpha^{2}]\times[0,1]}\frac{(G_{s}*\mathbf{u}_{0})(y)\cdot\mathbf{W}(dsdy)}{\alpha^{2}}-\frac{1}{2}\int_{0}^{\alpha^{2}}\int_{0}^{1}\frac{\left|G_{s}*\mathbf{u}_{0})(y)\right|^{2}}{\alpha^{4}}\right),$

where $G_{s}*\mathbf{u}_{0}$ is the convolution of $G(s,\cdot)$ with
$\mathbf{u}_{0}$. Under the measure $\mathbb{Q}_{0}$,

$\widetilde{\mathbf{W}}(dsdy)=\mathbf{W}(dsdy)+\frac{(G_{s}*\mathbf{u}_{0})(y)}{\alpha^{2}}dsdy$

is a white noise (see [All98]). Moreover

(2.4)
$\mathbf{u}(t,x)=\left(1-\frac{t}{\alpha^{2}}\right)(G_{t}*\mathbf{u}_{0})(x)+\int_{[0,t]\times[0,1]}G(t-s,x-y)\widetilde{\mathbf{W}}(dsdy),\quad
0\leq t\leq\alpha^{2},$

and the first term is $0$ at time $t=\alpha^{2}$. The case of
$\mathbf{u}_{0}\equiv{\mathbf{0}}$ shows

$\mathbb{Q}_{0}\left(\sup_{\stackrel{{\scriptstyle
t\leq\alpha^{2}}}{{x\in[0,1]}}}\left|\int_{[0,t]\times[0,1]}G(t-s,x-y)\widetilde{\mathbf{W}}(dsdy)\right|\leq\frac{\alpha}{2}\right)\geq\tilde{C}_{0},$

for some $\tilde{C}_{0}>0$. An application of the Cauchy-Schwarz inequality
gives

$\displaystyle\mathbb{P}_{0}\left(\sup_{\stackrel{{\scriptstyle
t\leq\alpha^{2}}}{{x\in[0,1]}}}\left|\int_{[0,t]\times[0,1]}G(t-s,x-y)\widetilde{\mathbf{W}}(dsdy)\right|\leq\frac{\alpha}{2}\right)$
$\displaystyle\geq\mathbb{Q}_{0}\left(\sup_{\stackrel{{\scriptstyle
t\leq\alpha^{2}}}{{x\in[0,1]}}}\left|\int_{[0,t]\times[0,1]}G(t-s,x-y)\widetilde{\mathbf{W}}(dsdy)\right|\leq\frac{\alpha}{2}\right)^{1/2}\cdot\mathbb{E}_{0}\left[\left(\frac{d\mathbb{Q}_{0}}{d\mathbb{P}_{0}}\right)^{2}\right]^{-1/2}$
$\displaystyle\geq\tilde{C}_{0}^{1/2}\exp\left(-\frac{1}{4}\int_{0}^{\alpha^{2}}\int_{0}^{1}\frac{\alpha^{2}}{4\alpha^{4}}dyds\right)$
$\displaystyle=\tilde{C}_{0}^{1/2}\exp\left(-\frac{1}{16}\right).$

The first term in (2.4) is at most $\frac{\alpha}{2}$, and so the above lower
bound is also a lower bound for the probability in (2.1). This completes the
proof of the proposition. ∎

###### Proof of lower bound in Theorems 1.1 and 1.2:.

Following Remark 1.2 we find the lower bound for $S_{T}^{\text{H},1,\nu}$. As
indicated earlier for the string to survive in a hard obstacle environment, it
must avoid the obstacles. We will use the same strategy of survival for the
soft obstacle as well. Let

$\mathcal{O}=\bigcup_{i\geq 1}B(\bm{\xi}_{i},a)$

be the obstacle set. For $T>0$, let $\alpha\equiv\alpha_{T}>0$ be a parameter
which will be chosen to devise the optimal strategy. Due to the support of H
being in a ball of radius $a$ one obtains

(2.5)
$\begin{split}S_{T}^{\text{H},1,\nu}&\geq\mathbb{P}(\mathcal{B}_{T}\cap\mathcal{C}_{T})\\\
&=\mathbb{P}_{0}(\mathcal{B}_{T})\mathbb{P}_{1}(\mathcal{C}_{T}),\end{split}$

where

$\displaystyle\mathcal{B}_{T}$
$\displaystyle=\left\\{\sup_{\stackrel{{\scriptstyle
s\in[0,T]}}{{x\in[0,1]}}}\left|\mathbf{u}(s,x)\right|\leq\alpha\right\\},$
$\displaystyle\mathcal{C}_{T}$ $\displaystyle=\Big{\\{}\text{there are no
}\bm{\xi}_{i}\text{ in the ball }B\left(\mathbf{0},\alpha+a\right)\Big{\\}}.$

It is important to observe here that the above argument does not depend on
whether the obstacles are hard or soft. Clearly

(2.6) $\mathbb{P}_{1}(\mathcal{C}_{T})=\exp\left(-\nu
c_{d}(\alpha+a)^{d}\right)$

for some dimension dependent constant $c_{d}$.

We will next estimate $\mathbb{P}_{0}(\mathcal{B}_{T})$ by using Proposition
2.1. Indeed, an application of the Markov property and (2.1) yields

(2.7)
$\mathbb{P}_{0}(\mathcal{B}_{T})\geq\mathbb{P}_{0}\left(\sup_{\stackrel{{\scriptstyle
s\leq\alpha^{2}}}{{x\in[0,1]}}}\left|\mathbf{u}(s,x)\right|\leq\alpha,\;\sup_{x\in[0,1]}\left|\mathbf{u}(\alpha,x)\right|\leq\frac{\alpha}{2}\right)^{\frac{T}{\alpha^{2}}}\leq\exp\left(\frac{T}{\alpha^{2}}\log
C_{0}\right),$

for $\alpha^{2}\ll T.$

Using (2.6) and (2.7) in (2.5) we have for $\alpha\geq K_{0}$

$\begin{split}S_{T}^{\text{H},1,\nu}&\geq\exp\left(-\nu
c_{d}(\alpha+a)^{d}\right)\exp\left(\frac{T}{\alpha^{2}}\log C_{0}\right)\\\
&\geq\exp\left(-\nu c_{d}2^{d}a^{d}\right)\exp\left(-\nu
c_{d}2^{d}\alpha^{d}+\frac{T}{\alpha^{2}}\log C_{0}\right).\end{split}$

For general $J\geq 1$ we use (1.10), as well as the fact that
$\widetilde{a}=aJ^{-\frac{1}{2}},\,\widetilde{\nu}=\nu J^{\frac{d}{2}}$ to
obtain

$\begin{split}S_{T}^{\text{H},J,\nu}&\geq\exp\left(-\nu
c_{d}2^{d}a^{d}\right)\exp\left(-\nu
J^{\frac{d}{2}}c_{d}2^{d}\alpha^{d}+\frac{T}{J^{2}\alpha^{2}}\log
C_{0}\right).\end{split}$

A simple calculus computation shows that the maximum of the exponent in the
second term is attained at
$\alpha=C(d,\nu)\left(\frac{T}{J^{2+\frac{d}{2}}}\right)^{\frac{1}{d+2}}$ so
that

(2.8) $S_{T}^{\text{H},J,\nu}\geq\exp\left(-\nu
c_{d}2^{d}a^{d}\right)\exp\left(-C_{1}(d,\nu)\left(\frac{T}{J}\right)^{\frac{d}{d+2}}\right),$

for a constant $C_{1}(d,\nu)$ independent of $J,\,T$, as long as $\alpha\geq
K_{0}$ or equivalently $T\geq C_{2}(d,\nu)J^{2+\frac{d}{2}}$. ∎

##  3\. Preliminaries for the upper bounds in Theorems 1.1 and 1.2

In this section we prove several preliminary lemmas required for the proof. We
begin with Section 3.1, where we define the stopping times $\\{\tau_{i}\\}$
precisely and show that there are order of $T^{\frac{d}{d+2}}$ such times in
$[0,T]$ with very high probability. In Section 3.2, we define the crucial
stopping times $\\{T_{i}\\}$ at which we will consider the volume of the
sausage around $\mathbf{u}(T_{i},\cdot)$. We will choose the $\\{T_{i}\\}$
from the $\\{\tau_{i}\\}$ so that

  1. (1)

$\mathbf{u}(T_{i},\cdot)$ has a larger volume than the sausage of radius $a/2$
around $\mathbf{N}\left(T_{i-1},T_{i}\right)$,

  2. (2)

the volume of the sausage around $\mathbf{N}(T_{i-1},T_{i})$ of radius
$\frac{a}{2}$ is at least ${\mathcal{C}}_{\gamma}a^{d-2+\gamma}$, and

  3. (3)

the range of $\mathbf{N}(T_{i-1},T_{i})$ has volume less than or equal to
$\Lambda$.

Then in Section 3.4 we show is that there are sufficiently many times
$\\{T_{i}\\}$, and they are far enough apart to ensure that the sausages at
these times do not intersect and the gaps between these times have finite
mean. Finally we conclude with Section 3.5 where we prove some estimates
needed for the soft obstacle case.

###  3.1. Using Estimates of the Wiener Sausage

For $\Lambda>1$ (to be chosen specifically in Lemma 3.10), it is useful to
consider the sausage of radius $4\Lambda$ around the center of mass
$\mathbf{X}_{T}$:

(3.1) $\mathcal{X}_{T}(4\Lambda):=\bigcup_{0\leq t\leq
T}\left\\{\mathbf{X}_{s}+B(\mathbf{0},4\Lambda)\right\\}.$

This is the well studied Wiener sausage.

######  Lemma 3.1 ([vdBBdH01], [Bol90], [DV75], [Szn90]).

Let $d\geq 2$ and $\Lambda>1$. There exists $C(d,\Lambda)>0$ such that for
$T>0$ we have

(3.2) $\mathbb{P}_{0}\left(\left|\mathcal{X}_{T}(4\Lambda)\right|\leq
T^{\frac{d}{d+2}}\right)\leq\exp\left(-C(d,\Lambda)T^{\frac{d}{d+2}}\right).$

Let $\tau_{0}=0$ and consider consecutive stopping times $\tau_{i}$ defined as

(3.3)
$\tau_{i+1}=\inf\left\\{t>\tau_{i}:\text{dist}\left(\mathbf{X}_{t},\bigcup_{k=0}^{i}\mathbf{X}_{\tau_{k}}\right)\geq
4\Lambda\right\\}.$

Note that $\mathbf{X}_{\tau_{i}}$ is on the boundary of the region
$\bigcup_{k=0}^{i-1}B(\mathbf{X}_{\tau_{k}},4\Lambda)$. Let

(3.4) $\\#(T):=\left|\left\\{i\geq 1:\tau_{i}\leq T\right\\}\right|$

be the number of $\tau_{i}$’s with $i\geq 1$ before time $T$.

######  Lemma 3.2.

Let $d\geq 2$ and $\Lambda>1$. There exists $C_{d}>0$ such that for $T>0$ we
have

(3.5)
$\mathbb{P}_{0}\left(\\#(T)\leq\frac{C_{d}T^{\frac{d}{d+2}}}{\Lambda^{d}}\right)\leq\exp\left(-C(d,\Lambda)T^{\frac{d}{d+2}}\right),$

where $C(d,\Lambda)$ is the same as in Lemma 3.1.

###### Proof.

We first claim that

(3.6)
$\bigcup_{i=0}^{\\#(T)+1}B\left(\mathbf{X}_{\tau_{i}},8\Lambda\right)\supset\mathcal{X}_{T}(4\Lambda).$

Clearly we need to just consider the behavior of the sausage for the time
points strictly between $\tau_{i}$ and $\tau_{i+1}$ for any fixed $i$. By the
definition of $\tau_{i+1}$ the path $\mathbf{X}_{t},\,\tau_{i}\leq
t\leq\tau_{i+1}$ is inside
$\bigcup_{k=0}^{i}B(\mathbf{X}_{\tau_{k}},4\Lambda)$. Therefore any such
$\mathbf{X}_{t}$ is within $4\Lambda$ distance of some
$\mathbf{X}_{\tau_{k}},k\leq i$. It then follows that
$B(\mathbf{X}_{t},4\Lambda)\subset B(\mathbf{X}_{\tau_{k}},8\Lambda)$. The
claim (3.6) then follows immediately.

Using (3.6) and the formula for the volume of a $d$ sphere of radius
$8\Lambda$

(3.7)
$(\\#(T)+2)\frac{\pi^{\frac{d}{2}}(8\Lambda)^{d}}{\Gamma(\frac{d}{2}+1)}\geq\left|\mathcal{X}_{T}(4\Lambda)\right|.$

Therefore

$\displaystyle\mathbb{P}_{0}\left(\\#(T)\leq\frac{C_{d}T^{\frac{d}{d+2}}}{\Lambda^{d}}\right)$
$\displaystyle\leq\mathbb{P}_{0}\left((\\#(T)+2)\frac{\pi^{\frac{d}{2}}(8\Lambda)^{d}}{\Gamma(\frac{d}{2}+1)}\leq\frac{C_{d}16^{d}\pi^{\frac{d}{2}}T^{\frac{d}{d+2}}}{\Gamma(\frac{d}{2}+1)}\right)$
$\displaystyle\leq\mathbb{P}_{0}\left(\left|\mathcal{X}_{T}(4\Lambda)\right|\leq\frac{C_{d}16^{d}\pi^{\frac{d}{2}}T^{\frac{d}{d+2}}}{\Gamma(\frac{d}{2}+1)}\right)$
$\displaystyle\leq\exp\left(-C(d,\Lambda)T^{\frac{d}{d+2}}\right),$

if we choose

$C_{d}^{-1}=\frac{16^{d}\pi^{\frac{d}{2}}}{\Gamma\left(\frac{d}{2}+1\right)}.$

This completes the proof of the lemma. ∎

###  3.2. Stopping times for string

For a $\mathbb{R}^{d}$ valued function $\mathbf{f}$ defined on $[0,1]$, we
denote the range of $\mathbf{f}$ by

(3.8)
$\mathcal{R}(\mathbf{f}):=\sup_{x,y\in[0,1]}|\mathbf{f}(x)-\mathbf{f}(y)|$

We will need the following lemma

######  Lemma 3.3.

Let $\mathbf{f}:[0,1]\to\mathbb{R}^{d}$. We have for $t\geq 1$

$\displaystyle\mathcal{R}(G_{t}*\mathbf{f})\leq
4de^{-4\pi^{2}t}\left\|\mathbf{f}-\int_{0}^{1}\mathbf{f}(x)dx\right\|_{2}\leq
4de^{-4\pi^{2}t}\mathcal{R}(\mathbf{f})$

###### Proof.

We expand each component $\mathbf{f}$ in the Fourier basis

$f_{j}(x)=\sum_{k}a^{(j)}_{k}e^{i2\pi kx}.$

Then

$\displaystyle\sup_{x,y}\Big{|}(G_{t}*f_{j})(x)-(G_{t}*f_{j})(y)\Big{|}$
$\displaystyle=\sup_{x,y}\left|\sum_{k\neq
0}e^{-4\pi^{2}k^{2}t}a_{k}^{(j)}\left[e^{i2\pi kx}-e^{i2\pi ky}\right]\right|$
$\displaystyle\leq 4e^{-4\pi^{2}t}\left(\sum_{k\neq
0}\left[a_{k}^{(j)}\right]^{2}\right)^{1/2}$

Now observe that $\int_{0}^{1}f_{j}(x)dx$ is the zeroth Fourier coefficient of
$f_{j}$ so that

$\left(\sum_{k\neq
0}\left[a_{k}^{(j)}\right]^{2}\right)^{1/2}=\left\|f_{j}-\int_{0}^{1}f_{j}(x)dx\right\|_{2}\leq\left\|\mathbf{f}-\int_{0}^{1}\mathbf{f}(x)dx\right\|_{2}.$

Clearly $\mathcal{R}(\mathbf{f})$ is an upper bound for the right hand side.
Finally note that

$\mathcal{R}(G_{t}*\mathbf{f})\leq\sum_{i=1}^{d}\mathcal{R}(G_{t}*f_{j})$

which gives the factor $d$ in the bound. ∎

For the rest of this article, we let
(3.9) $\begin{split}\delta&=\frac{a}{100},\\\ L&=E+3|\log a|,\end{split}$
where $E$ is a large enough constant.

######  Remark 3.1.

The constant $E$ chosen above is independent of any of the parameters in the
model, e.g. $T,J,a,\nu$. It is chosen so that $L$ satisfies the following:

  * •

$4de^{-4\pi^{2}L}\leq\delta$ (see Lemma 3.3).

  * •

$L\geq\textnormal{D}+2|\log a|$, where D is the constant appearing in Lemma
3.9.

  * •

$e^{4\pi^{2}L}\geq\frac{8C_{0}d^{\frac{3}{2}}}{\delta}$, where $C_{0}$ is the
constant appearing in (3.21).

  * •

$L$ is large enough so that the last inequality in (3.24) holds.

Recall the stopping times $\tau_{i}$ defined in (3.3). Let $T_{0}=0$ and
define the sequence of stopping times

(3.10) $\begin{split}S_{1}&=\left\\{t\geq
T_{0}+L:\;\mathcal{R}\left(G_{t-T_{0}}*\mathbf{u}_{0}\right)\leq\delta\right\\}\\\
T_{1}&=\min\left\\{\tau_{j}:\;\tau_{j}\geq S_{1}\right\\},\\\
S_{2}&=\inf\left\\{t\geq
T_{1}+L:\;\mathcal{R}\Big{(}G_{t-T_{1}}*\mathbf{N}(T_{0},T_{1})\Big{)}\leq\delta\right\\},\\\
T_{2}&=\min\left\\{\tau_{j}:\;\tau_{j}\geq S_{2}\right\\},\\\
S_{3}&=\inf\left\\{t\geq
T_{2}+L:\;\mathcal{R}\Big{(}G_{t-T_{2}}*\mathbf{N}(T_{1},T_{2})\Big{)}\leq\delta\right\\},\\\
T_{3}&=\min\left\\{\tau_{j}:\;\tau_{j}\geq S_{3}\right\\},\\\
\vdots&\qquad\vdots\qquad\vdots\end{split}$

The reason for introducing the stopping times $T_{i}$ will be clear below.
Inductively

$\displaystyle\mathbf{u}(T_{i})$
$\displaystyle=G_{T_{i}-T_{i-1}}*\mathbf{u}\big{(}T_{i-1}\big{)}+\mathbf{N}(T_{i-1},T_{i})$
$\displaystyle=G_{T_{i}-T_{i-1}}*\left[G_{T_{i-1}-T_{i-2}}*\mathbf{u}\big{(}T_{i-2}\big{)}+\mathbf{N}(T_{i-2},T_{i-1})\right]+\mathbf{N}(T_{i-1},T_{i})$
$\displaystyle=G_{T_{i}-T_{i-2}}*\mathbf{u}\big{(}T_{i-2}\big{)}+G_{T_{i}-T_{i-1}}*\mathbf{N}(T_{i-2},T_{i-1})+\mathbf{N}(T_{i-1},T_{i})$
$\displaystyle\vdots\qquad\vdots\qquad\vdots$
$\displaystyle=G_{T_{i}-T_{0}}*\mathbf{u}(T_{0})+G_{T_{i}-T_{1}}*\mathbf{N}(T_{0},T_{1})+G_{T_{i}-T_{2}}*\mathbf{N}(T_{1},T_{2})+\cdots+\mathbf{N}(T_{i-1},T_{i})$

######  Definition 3.1.

For a $\mathbb{R}^{d}$ valued function $\mathbf{f}$ on $[0,1]$ we denote

$\mathscr{S}(a;\mathbf{f}):=\bigcup_{0\leq y\leq
1}\left\\{\mathbf{f}(y)+B(\mathbf{0},a)\right\\}$

to be the sausage of radius $a$ around $\mathbf{f}$.

The following lemma is crucial for the upper bound.

######  Lemma 3.4.

We have

(3.11)
$\left|\mathcal{R}\big{(}\mathbf{u}(T_{i})\big{)}-\mathcal{R}\big{(}\mathbf{N}(T_{i-1},T_{i})\big{)}\right|\leq
2\delta.$

and

(3.12)
$\left|\mathscr{S}^{1}\left(a;T_{i}\right)\right|\geq\left|\mathscr{S}\Big{(}a/2;\;\mathbf{N}\left(T_{i-1},T_{i}\right)\Big{)}\right|.$

###### Proof.

Denote by

$\bm{\mathscr{G}}:=G_{T_{i}-T_{0}}*\mathbf{u}(T_{0})+G_{T_{i}-T_{1}}*\mathbf{N}(T_{0},T_{1})+G_{T_{i}-T_{2}}*\mathbf{N}(T_{1},T_{2})+\cdots+G_{T_{i}-T_{i-1}}*\mathbf{N}(T_{i-2},T_{i-1}).$

It is easily checked that when $\mathbf{f}=\mathbf{g}+\mathbf{h}$ then
$|\mathcal{R}(\mathbf{f})-\mathcal{R}(\mathbf{g})|\leq\mathcal{R}(\mathbf{h})$.
Therefore, with our choice of $\delta$ and $L$, and by using Lemma 3.3, we
obtain

$\left|\mathcal{R}\big{(}\mathbf{u}(T_{i})\big{)}-\mathcal{R}\big{(}\mathbf{N}(T_{i-1},T_{i})\big{)}\right|\leq\mathcal{R}(\bm{\mathscr{G}})\leq\sum_{i\geq
1}\delta^{i}\leq 2\delta.$

To prove (3.12) , note that

$\mathbf{u}(T_{i},x)=[\bm{\mathscr{G}}(0)+\mathbf{N}\left(T_{i-1},T_{i};x\right)]+[\bm{\mathscr{G}}(x)-\bm{\mathscr{G}}(0)],$

and so the ball of radius $a/2$ around
$[\bm{\mathscr{G}}(0)+\mathbf{N}\left(T_{i-1},T_{i};x\right)]$ is contained in
the ball of radius $a$ around $\mathbf{u}(T_{i},x)$ (note
$\mathcal{R}(\bm{\mathscr{G}})\leq a/2$ by our choice of $\delta$).
Consequently the sausage of radius $a/2$ around
$\mathbf{N}\left(T_{i-1},T_{i}\right)$ has a smaller volume than the sausage
of radius $a$ around $\mathbf{u}(T_{i})$. ∎

######  Remark 3.2.

Equations (3.11) and (3.12) show that the range of $\mathbf{u}(T_{i})$ is
close to that of $\mathbf{N}(T_{i-1},T_{i})$, and a lower bound on the volume
of the sausage around $\mathbf{u}(T_{i})$ is given by the volume of a smaller
sausage around $\mathbf{N}(T_{i-1},T_{i})$ . As we will see later
$\mathbf{N}(T_{i-1},T_{i})$ form a weakly dependent sequence. In particular,
using a weak form of law of large numbers, we will see that there is a subset
of $O(T^{\frac{d}{d+2}})$ many $T_{i}$’s where the volume of the sausage
around $\mathbf{N}(T_{i-1},T_{i})$ is large and where the range of
$\mathbf{N}(T_{i-1},T_{i})$ is small. Because of (3.11) and (3.12) the same
holds for $\mathbf{u}(T_{i})$ on this subset. The small range guarantees that
the sausages at these $T_{i}$’s are disjoint, and thus a lower bound for
$\mathscr{S}_{T}^{1}(a)$ is obtained by adding the volumes of the sausages at
the $T_{i}$’s on this subset.

###  3.3. Volume of the sausage around $\mathbf{N}(T_{i-1},T_{i})$ and Range
of $\mathbf{N}(T_{i-1},T_{i})$

Our first objective will be to give a lower bound on the probability that the
volume of the sausage of radius $a/2$ around $\mathbf{N}(0,t)$ is at least
$Ca^{d-2+\epsilon}$ (see Lemma 3.9). As indicated in Remark 3.2, we will show
later that there are sufficiently many $i$ such that the sausages around
$\mathbf{N}(T_{i-1},T_{i})$ have at least this volume. Let

(3.13) $\mathbf{N}(t;x,y)=\mathbf{N}(t,x)-\mathbf{N}(t,y),$

and write

$\mathbf{N}(t;x,y)=\mathbf{N}^{(1)}(t;x,y)-\mathbf{N}^{(2)}(t;x,y),$

where

$\displaystyle\mathbf{N}^{(1)}(t;x,y)$
$\displaystyle=\int_{(-\infty,t]\times[0,1]}\left[G(t-s,x,z)-G(t-s,y,z)\right]\mathbf{W}(dsdz),\text{
and }$ $\displaystyle\mathbf{N}^{(2)}(t,x,y)$
$\displaystyle=\int_{(-\infty,0]\times[0,1]}\left[G(t-s,x,z)-G(t-s,y,z)\right]\mathbf{W}(dsdz)$

It is easy to see that $\mathbf{N}^{(1)}(t;\cdot,\cdot)$ is a stationary
process in $t$. The reason for considering the differences
$\mathbf{N}(t,x)-\mathbf{N}(t,y)$ instead of $\mathbf{N}(t,x)$ is that the
term

$\int_{(-\infty,t]\times[0,1]}G(t-s,x,z)\mathbf{W}(dzds)$

would not be convergent. Note also that the volume of a sausage around
$\mathbf{N}(0,t)$ is the same as the volume of the sausage around
$\mathbf{N}(t;\cdot,0)$. The lemma below shows that $\mathbf{N}^{(2)}$ becomes
smaller with increasing $t$, and hence the main contribution of
$\mathbf{N}(t,x,y)$ comes from $\mathbf{N}^{(1)}(t;x,y)$ for large $t$. The
process $\mathbf{N}^{(1)}(t;\cdot,0)$ behaves locally like Brownian motion
(see Lemma 3.6, and so we can use some techniques (see for example Lemma 3.7)
which work for Brownian motion, to obtain a lower bound on the sausage around
$\mathbf{N}^{(1)}(t;\cdot,0)$.

######  Lemma 3.5.

There exists a constant $C_{1}>0$ such that for any $\lambda>0$

(3.14)
$\mathbb{P}_{0}\left(\sup_{x,y\in[0,1]}\left|\mathbf{N}^{(2)}(t;x,y)\right|>\lambda\right)\leq
2\exp\left(-\frac{e^{t}\lambda^{2}}{C_{1}}\right).$

###### Proof.

We first observe

$\displaystyle\mathbb{E}_{0}\left[\left|\mathbf{N}^{(2)}(t;x,y)\right|^{2}\right]$
$\displaystyle=\int_{-\infty}^{0}\int_{0}^{1}\left[G(t-s,x,z)-G(t-s,y,z)\right]^{2}dzds$
$\displaystyle=\int_{-\infty}^{0}ds\sum_{k\geq
1}e^{-k^{2}(t-s)}\left|1-e^{i\cdot 2\pi k(x-y)}\right|^{2}$
$\displaystyle\leq\sum_{k\geq
1}\frac{e^{-k^{2}t}}{k^{2}}\left|1\wedge(k|x-y|)\right|^{2}$
$\displaystyle\leq Ce^{-t}|x-y|.$

An application of the Burkholder-Davis-Gundy inequality then gives us

$\mathbb{E}_{0}\left[\left|\mathbf{N}^{(2)}(t;x,y)\right|^{2k}\right]\leq
C^{k}(2\sqrt{2})^{2k}k^{k}e^{-kt}|x-y|^{k}$

for all positive integers $k\geq 1$. From the argument leading up to
inequality (6.9) in [CJK13] one then obtains

$\mathbb{E}_{0}\left[\sup_{x,y\in[0,1]}\left|\mathbf{N}^{(2)}(t;x,y)\right|^{2k}\right]\leq
C^{k}e^{-kt}k^{k},\quad k\geq 2.$

Therefore there exists a constant $C_{1}>0$ such that

$\mathbb{E}_{0}\left[\sup_{x,y\in[0,1]}\exp\left(\frac{\exp(t)\cdot|\mathbf{N}^{(2)}(t;x,y)|^{2}}{C_{1}}\right)\right]\leq
2.$

The inequality (3.14) follows immediately from this. ∎

The following lemma indicates that the process $\mathbf{N}^{(1)}(t;\cdot,0)$
behaves locally like Brownian motion. It will be used in Lemma 3.7 below.

######  Lemma 3.6.

There are constants $C_{1},C_{2}>0$ such that for all $x,y\in[0,1]$ and all
$t\geq 1$ one has

$C_{1}d(x,y)\leq\int_{0}^{t}\int_{0}^{1}\left[G(s,x,z)-G(s,y,z)\right]^{2}dzds\leq
C_{2}d(x,y),$

where $d(x,y)$ is the distance between $x$ and $y$ on the torus
$\mathbb{T}=[0,1)$.

###### Proof.

Using the Fourier decomposition of $G(s,x)$ we obtain

(3.15)
$\begin{split}&\int_{0}^{t}ds\int_{0}^{1}dz\left[G(s,x,z)-G(s,y,z)\right]^{2}\\\
&=C\int_{0}^{t}ds\sum_{k\geq 1}\exp\left(-(2\pi
k)^{2}s\right)\big{|}1-\exp\left[i(2\pi k)d(x,y)\right]\big{|}^{2}\\\
&=C\sum_{k\geq 1}\frac{\left[1-\exp(-(2\pi
k)^{2}t)\right]}{k^{2}}\big{|}1-\exp\left[i(2\pi
k)d(x,y)\right]\big{|}^{2}.\end{split}$

Now use $|1-e^{iz}|\leq 1\wedge|z|$ and $|1-e^{-4\pi^{2}k^{2}t}|\leq 1$ to
obtain that the above is less than

$C\sum_{k\geq 1}\frac{1}{k^{2}}\big{|}1\wedge[kd(x,y)]\big{|}\leq Cd(x,y).$

The final inequality is obtained by splitting the sum according to whether
$k\leq d(x,y)^{-1}$ or $k>d(x,y)^{-1}$.

Next we turn to the lower bound. For this we only consider the sum in
(LABEL:eq:parseval) for $1\leq k\leq\frac{1}{2d(x,y)}$. Now $|1-e^{iz}|\geq
C|z|$ for all $z\in[0,\pi]$ and some $C>0$. Therefore since $t\geq 1$

$\int_{0}^{t}ds\int_{0}^{1}dz\left[G(s,x,z)-G(s,y,z)\right]^{2}\geq
C\sum_{k=1}^{[2d(x,y)]^{-1}}\frac{k^{2}d(x,y)^{2}}{k^{2}}\geq Cd(x,y).$

This completes the proof of the lemma. ∎

Before proceeding we recall

######  Definition 3.2.

The lower Minkowski dimension of a set $A$ is

$\underline{\textnormal{dim}}_{M}(A):=\liminf_{\epsilon\to 0}\frac{\log
N_{\epsilon}(A)}{\log(\epsilon^{-1})},$

where $N_{\epsilon}(A)$ is the minimum number of balls of radius $\epsilon$
needed to cover $A$.

We will need

######  Lemma 3.7.

For $t\geq 1$

$\underline{\textnormal{dim}}_{M}\left[\textnormal{Range}\big{(}\mathbf{N}^{(1)}(t;\cdot,0)\big{)}\right]\geq
2\;\;\text{a.s.}$

###### Proof.

We first recall that for any set $A$ we have

$\underline{\textnormal{dim}}_{M}(A)\geq{\textnormal{dim}}_{H}(A),$

where ${\textnormal{dim}}_{H}$ is the Hausdorff dimension (see page 115 of
[MP10]). We use the energy method (Theorem 4.27 in [MP10]) to get a lower
bound on the Hausdorff dimension of the range. Let $\mu_{t}$ be the occupation
measure of $\mathbf{N}^{(1)}(t;\cdot,0)$:

$\int_{\mathbb{R}^{d}}f(\mathbf{x})d\mu_{t}(\mathbf{x})=\int_{0}^{1}f\left(\mathbf{N}^{(1)}(t;x,0)\right)dx.$

We just need to show that for any $0<\alpha<2$

$\mathbb{E}_{0}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\frac{d\mu_{t}(\mathbf{x})d\mu_{t}(\mathbf{y})}{|\mathbf{x}-\mathbf{y}|^{\alpha}}=\mathbb{E}_{0}\int_{0}^{1}\int_{0}^{1}\frac{dxdy}{|\mathbf{N}^{(1)}(t;x,0)-\mathbf{N}^{(1)}(t;y,0)|^{\alpha}}<\infty.$

Now $\mathbf{N}^{(1)}(t,x,0)-\mathbf{N}^{(1)}(t,y,0)$ is a Gaussian random
variable with mean $0$ and variance

$\int_{0}^{\infty}\int_{0}^{1}\left[G(s,x,z)-G(s,y,z)\right]^{2}dzds.$

From Lemma 3.6 this expression is bounded above and below by a constant
multiple of $d(x,y)$. Therefore it follows that

$\mathbb{E}_{0}\int_{0}^{1}\int_{0}^{1}\frac{dxdy}{|\mathbf{N}^{(1)}(t;x,0)-\mathbf{N}^{(1)}(t;x,0)|^{\alpha}}\leq
C\int_{0}^{1}\int_{0}^{1}\frac{dxdy}{d(x,y)^{\alpha/2}}<\infty,$

as required. Consequently

$\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\frac{d\mu_{t}(\mathbf{x})d\mu_{t}(\mathbf{y})}{|\mathbf{x}-\mathbf{y}|^{\alpha}}<\infty\quad\text{a.s.}$

It thus follows from Theorem 4.27 in [MP10] that
$\underline{\textnormal{dim}}_{M}\left[\textnormal{Range}\big{(}\mathbf{N}^{(1)}(t;\cdot,0)\big{)}\right]\geq
2\;\text{a.s.}$ as required. ∎

The lower bound in the lower Minkowski dimension gives a lower bound on the
number of balls of radius $a$ which intersect the range of
$\mathbf{N}^{(1)}(t;\cdot,0)$, and consequently a lower bound on the volume of
the sausage of radius $a$ around $\mathbf{N}^{(1)}(t;\cdot,0)$.

######  Lemma 3.8.

Fix an arbitrary $0<\gamma<1$. There exists a $\tilde{\mathcal{C}}_{\gamma}>0$
such that for any $t\in\mathbb{R}$ and any $0<a\leq 1$

$\mathbb{P}_{0}\left(|\mathscr{S}\left(a;\;\mathbf{N}^{(1)}(t;\cdot,0)\right)|\geq\tilde{\mathcal{C}}_{\gamma}a^{d-2+\gamma}\right)\geq\frac{4}{5}.$

###### Proof.

First note that the process $\mathbf{N}^{(1)}(t;\cdot,0)$ is stationary in
$t$. Partition $\mathbb{R}^{d}$ into cubes of side length $a$. Let
$\tilde{N}_{a}(\mathbf{N}^{(1)},t)$ be the number of cubes through which
$\mathbf{N}^{(1)}(t;\cdot,0)$ passes. By Lemma 3.7 we obtain almost surely

$\tilde{N}_{a}(\mathbf{N}^{(1)},t)\geq\left(\frac{1}{a}\right)^{2-\gamma}\text{
for all $a$ small enough}.$

Therefore there exists a positive random variable $A(\omega)$ which is finite
almost surely such that

$\tilde{N}_{a}(\mathbf{N}^{(1)},t)\geq
A(\omega)\left(\frac{1}{a}\right)^{2-\gamma}\text{ for all }0<a\leq 1.$

We now choose the largest possible subcollection of these
$\tilde{N}_{a}(\mathbf{N}^{(1)},t)$ cubes such that no two cubes are adjacent
(that is, share a common edge). Let $N_{a}^{*}(\mathbf{N}^{(1)},t)$ be the
number of cubes in this subcollection. Clearly there exists a constant $C_{d}$
such that almost surely

$N_{a}^{*}(\mathbf{N}^{(1)},t)\geq
A(\omega)C_{d}\left(\frac{1}{a}\right)^{2-\gamma}\text{ for all }0<a\leq 1.$

From each of these $N_{a}^{*}(\mathbf{N}^{(1)},t)$ cubes choose any point in
the range of $\mathbf{N}^{(1)}(t,\cdot,0)$. The union of the balls of radius
$a$ around these points is contained in the sausage of radius $a$ around
$\mathbf{N}^{(1)}(t,\cdot,0)$, so that

$|\mathscr{S}\left(a;\;\mathbf{N}^{(1)}(t,\cdot,0)\right)|\geq
A(\omega)\tilde{C}_{d}a^{d-2+\gamma}\quad\text{ for all }0<a\leq 1,$

for some other constant $\tilde{C}_{d}>0$. The constant
$\mathcal{C}_{\gamma}>0$ is chosen so that

$\mathbb{P}_{0}\left(A(\omega)\tilde{C}_{d}\geq\tilde{\mathcal{C}}_{\gamma}\right)\geq\frac{4}{5}.$

This completes the proof of the lemma. ∎

We now use Lemma 3.5 on the smallness of $\mathbf{N}^{(2)}(t;\cdot,0)$ to
control the volume of the sausage around $\mathbf{N}(0,t)$.

######  Lemma 3.9.

Fix an arbitrary $0<\gamma<1$. There are constants ${\mathcal{C}}_{\gamma}>0$
and $\textnormal{D}>0$ such that for any $0<a\leq 1$ and
$t\geq\textnormal{D}+2|\log a|$

$\mathbb{P}_{0}\left(|\mathscr{S}\left(\frac{a}{2};\;\mathbf{N}(0,t)\right)|\geq{\mathcal{C}}_{\gamma}a^{d-2+\gamma}\right)\geq\frac{3}{4}.$

###### Proof.

We shall use (3.14).

$\mathbb{P}_{0}\left(\sup_{x,y\in[0,1]}\left|\mathbf{N}^{(2)}(t;x,y)\right|>\frac{a}{4}\right)\leq
2\exp\left(-\frac{e^{t}a^{2}}{16C_{1}}\right).$

If we choose D large enough the right hand side above is at most
$\frac{1}{20}$. Therefore,

$\mathbb{P}_{0}\left(\mathcal{R}\left(\mathbf{N}^{(2)}(t;\cdot,0)\right)\leq\frac{a}{4}\right)\geq\frac{19}{20}.$

As a consequence of this we obtain by a similar argument as in Lemma 3.4 that

$\big{|}\mathscr{S}\left(\frac{a}{2};\;\mathbf{N}(0,t)\right)\big{|}\geq\big{|}\mathscr{S}\left(\frac{a}{4};\;\mathbf{N}^{(1)}(t;\cdot,0)\right)\big{|}.$

We now use Lemma 3.8 to complete the proof. ∎

Our second objective in this subsection is (following Remark 3.2) analyze the
probability that the range of $\mathbf{N}(0,t)$ is small (see Lemma 3.10).

######  Lemma 3.10.

There exists $\Lambda>1$ such that for all $t\geq L$

(3.16)
$\mathbb{P}_{0}\left(\mathcal{R}(\mathbf{N}(0,t))\leq\Lambda\right)\geq\frac{3}{4}.$

###### Proof.

Since $\mathscr{\mathbf{N}}^{(1)}(t;\cdot,0)$ is stationary in $t$, we can
choose $\Lambda>1$ such that

$\mathbb{P}_{0}\left(\mathcal{R}\left(\mathbf{N}^{(1)}(t;\cdot,0)\right)\leq\frac{\Lambda}{2}\right)\geq\frac{4}{5}.$

In the proof of Lemma 3.9 we have seen that with probability at least
$\frac{19}{20}$ one has

$\mathcal{R}\left(\mathbf{N}^{(2)}(t;\cdot,0)\right)\leq\frac{a}{4}.$

Therefore with probability at least $\frac{3}{4}$ we have

$\mathcal{R}\Big{(}\mathbf{N}(0,t)\Big{)}\leq\mathcal{R}\Big{(}\mathbf{N}^{(1)}(t;\cdot,0)\Big{)}+\mathcal{R}\Big{(}\mathbf{N}^{(2)}(t;\cdot,0)\Big{)}\leq\Lambda.$

This completes the proof. ∎

######  Remark 3.3.

We fix and use a $\Lambda$ as in Lemma 3.10 for the rest of the article. In
particular the $\tau_{i}$’s defined in (3.3) are defined in terms of this
particular choice of $\Lambda$.

Our final objective is to show that there are sufficiently many $T_{i}$’s such
that $\mathcal{R}\left(\mathbf{N}(T_{i-1},T_{i})\right)\leq\Lambda$ and
$\mathscr{S}\Big{(}\frac{a}{2};\mathbf{N}\left(T_{i-1},T_{i}\right)\Big{)}\geq{\mathcal{C}}_{\gamma}a^{d-2+\gamma}$
(See Lemma 3.11). Before we proceed, we will need

######  Definition 3.3.

Let $\tilde{\mathcal{F}}_{t}$ be the filtration generated by white noise

$\tilde{\mathcal{F}}_{t}=\sigma\left\\{\mathbf{W}(A\times[r,s]);\;A\subset[0,1],\,0\leq
r,s\leq t\right\\}.$

Let $\mathcal{G}_{i}$ denote the $\sigma$-algebra generated by the white noise
up to time $T_{i}$:

$\mathcal{G}_{i}=\left\\{\textnormal{A}\in\mathcal{F}:\textnormal{A}\cap\left\\{T_{i}\leq
t\right\\}\in\tilde{\mathcal{F}}_{t}\right\\}$

Let

$\mathcal{H}_{i}=\mathcal{G}_{i}\vee\sigma\left\\{\mathbf{X}_{t};\,t\geq
0\right\\}$

the $\sigma$-algebra generated by the white noise up to time $T_{i}$ and the
center of mass process.

######  Lemma 3.11.

Let $\Lambda$ be as in Lemma 3.10. We have

(3.17)
$\mathbb{P}_{0}\left(\mathcal{R}\left(\mathbf{N}(T_{i-1},T_{i})\right)\leq\Lambda,\;\left|\mathscr{S}\Big{(}\frac{a}{2};\,\mathbf{N}(T_{i-1},T_{i}))\Big{)}\right|\geq{\mathcal{C}}_{\gamma}a^{d-2+\gamma}\,\bigg{|}\,\mathcal{H}_{i-1}\right)\geq\frac{1}{2}.$

###### Proof.

It is enough to show

(3.18)
$\displaystyle\mathbb{P}_{0}\left(\mathcal{R}\left(\mathbf{N}\left(T_{i-1},T_{i}\right)\right)\leq\Lambda\,\bigg{|}\,\mathcal{H}_{i-1}\right)$
$\displaystyle\geq\frac{3}{4},$ (3.19)
$\displaystyle\mathbb{P}_{0}\left(\left|\mathscr{S}\Big{(}\frac{a}{2};\,\mathbf{N}\left(T_{i-1},T_{i}\right)\Big{)}\right|\geq{\mathcal{C}}_{\gamma}a^{d-2+\gamma}\,\bigg{|}\,\mathcal{H}_{i-1}\right)$
$\displaystyle\geq\frac{3}{4}.$

We have from Lemma 3.10

$\mathbb{P}_{0}\Big{(}\mathcal{R}\left(\mathbf{N}\left(0,t\right)\right)\leq\Lambda\Big{)}\geq\frac{3}{4},$

uniformly in $t\geq L$. Then observe

$\begin{split}&\mathbb{P}_{0}\Big{(}\mathcal{R}\left(\mathbf{N}\left(T_{i-1},T_{i}\right)\right)\leq\Lambda\,\bigg{|}\,\mathcal{H}_{i-1}\Big{)}\\\
&=\int_{0}^{\infty}\int_{s+L}^{\infty}\mathbb{P}_{0}\left(T_{i-1}\in
ds,\;T_{i}\in
dt,\;\mathcal{R}\Big{(}\mathbf{N}(s,t)\Big{)}\leq\Lambda\,\bigg{|}\,\mathcal{H}_{i-1}\right)\\\
&=\int_{0}^{\infty}\int_{s+L}^{\infty}\mathbf{1}\left\\{T_{i-1}\in
ds,\;T_{i}\in
dt\right\\}\cdot\mathbb{P}_{0}\left(\mathcal{R}\Big{(}\mathbf{N}(s,t)\Big{)}\leq\Lambda\right).\end{split}$

The second equality follows from an argument similar to Lemma 2.1, In fact the
event $\left\\{T_{i-1}\in ds,\;T_{i}\in dt\right\\}$ is measurable with
respect to the sigma field $\mathcal{H}_{i-1}$, while
$\mathcal{R}(\mathbf{N}(s,t))$ depends on

$\sigma\left(\int_{s}^{\tilde{t}}\int_{0}^{1}\left[G_{t-r}(x,z)-G_{t-r}(0,z)\right]\mathbf{W}(dzdr),\;x\in[0,1],\;\tilde{t}\geq
s\right),$

which is independent of $\mathcal{H}_{i-1}$. From this we obtain (3.18).
Similarly, to show (3.19) we use Lemma 3.9, and integrate over the
realizations of $T_{i-1}$ and $T_{i}$. This completes the proof of the lemma.
∎

Consequently, due to (3.11) and (3.12) and using Lemma 3.11, there are
sufficiently many $T_{i}$ such that
$\mathcal{R}\left(\mathbf{u}(T_{i})\right)\leq\Lambda+2\delta$ and
$\left|\mathscr{S}^{1}\left(a;T_{i}\right)\right|\geq{\mathcal{C}}_{\gamma}a^{d-2+\gamma}.$

###  3.4. Sufficiently many $T_{i}$ far apart

######  Lemma 3.12.

There exists $C_{2}>0$ such that for all $t\geq L$

$\mathbb{P}_{0}\left[S_{i+1}-T_{i}>t\;\Big{|}\mathcal{H}_{i-1}\right]\leq\frac{C_{2}e^{-8\pi^{2}t}}{\delta^{2}}.$

In particular for any $\eta<8\pi^{2}$ there exists $C_{3}(\eta)>0$ such that

(3.20)
$\mathbb{E}_{0}\left[\exp\left(\eta(S_{i+1}-T_{i})\right)\;\bigg{|}\mathcal{H}_{i-1}\right]\leq
e^{\eta L}+\frac{C_{3}(\eta)}{\delta^{2}}$

###### Proof.

Recall from (3.10)

$S_{i+1}=\inf\left\\{t\geq
T_{i}+L:\mathcal{R}\Big{(}G_{t-T_{i}}*\mathbf{N}(T_{i-1},T_{i})\Big{)}\leq\delta\right\\}.$

The event $\\{S_{i+1}-T_{i}>t\\}$ implies that
$\mathcal{R}\Big{(}G_{t}*\mathbf{N}(T_{i-1},T_{i})\Big{)}>\delta$, and in
light of Lemma 3.3, it further implies

$\left\|\mathbf{N}(T_{i-1},T_{i})-\int_{0}^{1}\mathbf{N}(T_{i-1},T_{i};\,x)dx\right\|_{2}>\frac{\delta}{4d}e^{4\pi^{2}t}.$

Using the subscript $j$ to denote the components of
$\mathbf{N}(T_{i-1},T_{i})$ it follows that there exists a $1\leq j\leq d$
such that

$\left\|\mathbf{N}_{j}(T_{i-1},T_{i})-\int_{0}^{1}\mathbf{N}_{j}(T_{i-1},T_{i};\,x)dx\right\|_{2}>\frac{\delta}{4d^{\frac{3}{2}}}e^{4\pi^{2}t}.$

We can formally write each component $\mathbf{W}_{j}$ of the white noise as
$\mathbf{W}_{j}(dydr)=\sum_{k\in\mathbb{Z}}e^{\imath(2\pi ky)}dB_{k}(r)dy$,
where the $B_{k}$’s are independent standard complex Brownian motions (that is
$B_{k}=\frac{R_{k}}{\sqrt{2}}+\imath\frac{C_{k}}{\sqrt{2}}$ where
$R_{k},C_{k}$ are standard real Brownian motions) with $\bar{B_{k}}=B_{-k}$.
This can be seen by integrating both sides with test functions and computing
the second moments. Since

$G_{t}(x,y)=\sum_{l\in\mathbb{Z}}e^{-2\pi^{2}l^{2}t}e^{\imath 2\pi l(x-y)},$

the $k$th Fourier coefficient of $\mathbf{N}_{j}(s,\tilde{s}),\;k\neq 0$ is

$a_{k}=\int_{s}^{\tilde{s}}e^{-2\pi^{2}k^{2}(\tilde{s}-r)}dB_{k}(r)$

Furthermore we have $\bar{a_{k}}=a_{-k}$ and $a_{k}$ is independent of
$a_{\tilde{k}}$ if $\tilde{k}\neq k,-k$. Now

$\text{E}\left[\sum_{k\neq 0}|a_{k}|^{2}\right]=\sum_{k\neq
0}\frac{1}{2\pi^{2}k^{2}}\left(1-e^{-2\pi^{2}k^{2}(\tilde{s}-s)}\right),$

and so there exist positive constants $C_{0},\,C_{1}$ such that

$\text{E}\left[\left(\sum_{k\neq 0}|a_{k}|^{2}\right)^{\frac{1}{2}}\right]\leq
C_{0}\quad\text{and}\quad\text{Var}\left[\left(\sum_{k\neq
0}|a_{k}|^{2}\right)^{\frac{1}{2}}\right]\leq C_{1}$

uniformly in $s$ and $\tilde{s}$. Therefore

(3.21)
$\begin{split}&\mathbb{P}_{0}\left[S_{i+1}-T_{i}>t\;\Big{|}\;\mathcal{H}_{i-1}\right]\\\
&\leq
d\cdot\mathbb{P}_{0}\left[\left\|\mathbf{N}_{j}(T_{i-1},T_{i})-\int_{0}^{1}\mathbf{N}_{j}(T_{i-1},T_{i};x)dx\right\|_{2}>\frac{\delta}{4d^{\frac{3}{2}}}e^{4\pi^{2}t}\bigg{|}\;\mathcal{H}_{i-1}\right]\\\
&\leq\frac{d\cdot
C_{1}}{\left(\frac{\delta}{4d^{\frac{3}{2}}}e^{4\pi^{2}t}-C_{0}\right)^{2}}.\end{split}$

The second part of the lemma follows from

$\displaystyle\mathbb{E}_{0}\left[\exp\left(\eta(S_{i+1}-T_{i})\right)\;\bigg{|}\;\mathcal{H}_{i-1}\right]$
$\displaystyle\leq e^{\eta L}-\int_{L}^{\infty}e^{\eta
t}\mathbb{P}_{0}\left[S_{i+1}-T_{i}>t\;\Big{|}\;\mathcal{H}_{i-1}\right]\,dt,$

and the above tail bound. ∎

We conclude this section regarding the spacings of $S_{i}$ and $T_{i}$. This
will be crucially used on a specific subset of $T_{i}$’s to show the upper
bound.

######  Lemma 3.13.

There is a constant $\widetilde{C}>0$ such that for any $A_{4}>0$ and $L$ as
in (3.9) we have

$\mathbb{P}_{0}\left(\sum_{i=1}^{A_{4}\frac{T^{\frac{d}{d+2}}}{L}}(S_{i}-T_{i-1})>\widetilde{C}A_{4}T^{\frac{d}{d+2}}\right)\leq\exp\left(-\frac{\widetilde{C}A_{4}}{2}T^{\frac{d}{d+2}}\right).$

###### Proof.

With the choice of $\eta=1$ in (3.20) we obtain

$\displaystyle\mathbb{P}_{0}\left(\sum_{i=1}^{A_{4}\frac{T^{\frac{d}{d+2}}}{L}}(S_{i}-T_{i-1})>\widetilde{C}A_{4}T^{\frac{d}{d+2}}\right)$
$\displaystyle\leq\mathbb{E}_{0}\exp\left(\sum_{i=1}^{A_{4}\frac{T^{\frac{d}{d+2}}}{L}}(S_{i}-T_{i-1})-\widetilde{C}A_{4}T^{\frac{d}{d+2}}\right)$
$\displaystyle\leq\left(e^{L}+\frac{C_{3}}{\delta^{2}}\right)^{A_{4}\frac{T^{\frac{d}{d+2}}}{L}}\exp\left(-\widetilde{C}A_{4}T^{\frac{d}{d+2}}\right).$

The lemma follows by a large choice of the constant $\widetilde{C}$ above. ∎

###  3.5. Estimates for Soft obstacles

We will need a few lemmas which lead up Proposition 3.2. This is a key
proposition that will be used in the proof of the upper bound in Theorem 1.2.

######  Proposition 3.1.

There is a $C_{3}>0$ such that for all $s_{0}\leq 1$

$\mathbb{P}_{0}\left(\sup_{s\leq
s_{0}}\sup_{x\in[0,1]}\left|\mathbf{N}(0,t+s;x)-\mathbf{N}(0,t;x)\right|>\lambda\right)\leq\exp\left(-\frac{C_{3}^{2}\lambda^{2}}{\sqrt{s_{0}}}\right)$

uniformly in $t$.

The proof of the above proposition follows from a sequence of lemmas. Define
for $s,\,t\geq 0$ and $x,y\in[0,1]$

$\mathbf{Z}(t,s;x,y)=\Big{\\{}\mathbf{N}(0,t+s;x)-\mathbf{N}(0,t;x)\Big{\\}}-\Big{\\{}\mathbf{N}(0,t+s;y)-\mathbf{N}(0,t;y)\Big{\\}}$

######  Lemma 3.14.

There is a constant $C_{1}>0$ such that

$\mathbb{P}_{0}\left(\left|\mathbf{Z}(t,s;x,y)\right|>\lambda\right)\leq\exp\left(-\frac{C_{1}^{2}\lambda^{2}}{\sqrt{s}\wedge|x-y|}\right)$

uniformly in $t$.

###### Proof.

We first give an upper bound on
$\mathbb{E}_{0}\left[\mathbf{Z}_{i}^{2}(t,s;x,y)\right]$, for any fixed
coordinate $\mathbf{Z}_{i}$ of $\mathbf{Z}$. This is easily seen to be equal
to

(3.22)
$\displaystyle\int_{0}^{t}\int_{0}^{1}\left[G(t+s-r,x,z)-G(t+s-r,y,z)-G(t-r,x,z)+G(t-r,y,z)\right]^{2}dzdr$
(3.23)
$\displaystyle\qquad+\int_{t}^{t+s}\int_{0}^{1}\left[G(t+s-r,x,z)-G(t+s-r,y,z)\right]^{2}dzdr$

Let us first look at (3.23). This is bounded by

$\displaystyle\int_{0}^{s}dr\sum_{k\geq 1}e^{-(2\pi
k)^{2}r}\Big{|}1-\exp\left(i(2\pi k)(x-y)\right)\Big{|}^{2}$
$\displaystyle\leq C\sum_{k\geq 1}\frac{1-e^{-(2\pi
k)^{2}s}}{k^{2}}\Big{[}1\wedge|2\pi k(x-y)|\Big{]}^{2}$ $\displaystyle\leq
C\sum_{k\geq 1}\frac{1\wedge(2\pi k)^{2}s}{k^{2}}\Big{[}1\wedge|2\pi
k(x-y)|\Big{]}^{2}$

In the case that $\sqrt{s}\leq|x-y|$, the above is bounded by

$C\sum_{k=1}^{\frac{1}{2\pi|x-y|}}\frac{k^{2}s}{k^{2}}k^{2}|x-y|^{2}+C\sum_{k=\frac{1}{2\pi|x-y|}+1}^{\frac{1}{2\pi\sqrt{s}}}s+C\sum_{k=\frac{1}{2\pi\sqrt{s}}+1}^{\infty}\frac{1}{k^{2}}\leq
C\sqrt{s}.$

In the case that $|x-y|\leq\sqrt{s}$ we obtain a bound

$C\sum_{k=1}^{\frac{1}{2\pi\sqrt{s}}}\frac{k^{2}s}{k^{2}}k^{2}|x-y|^{2}+C\sum_{k=\frac{1}{2\pi\sqrt{s}}+1}^{\frac{1}{2\pi|x-y|}}\frac{k^{2}|x-y|^{2}}{k^{2}}+C\sum_{k=\frac{1}{2\pi|x-y|}+1}^{\infty}\frac{1}{k^{2}}\leq
C|x-y|.$

Let us next consider the term (3.22). This is bounded by

$\displaystyle\int_{0}^{t}dr\sum_{k\geq 1}\left[\exp\left(-\frac{(2\pi
k)^{2}(t+s-r)}{2}\right)-\exp\left(-\frac{(2\pi
k)^{2}(t-r)}{2}\right)\right]^{2}\left[1\wedge\left|2\pi
k(x-y)\right|\right]^{2}$ $\displaystyle\leq\int_{0}^{t}dr\sum_{k\geq
1}\exp\left(-\pi^{2}k^{2}r\right)\left|1-\exp\left(-2\pi^{2}k^{2}s\right)\right|^{2}\left[1\wedge\left|2\pi
k(x-y)\right|\right]^{2}$ $\displaystyle\leq C\int_{0}^{t}dr\sum_{k\geq
1}\frac{\left[1\wedge 2\pi^{2}k^{2}s\right]^{2}}{k^{2}}\left[1\wedge\left|2\pi
k(x-y)\right|\right]^{2}$

Therefore a similar bound as that for (3.23) holds for (3.22). The conclusion
of our arguments is that

$\mathbb{E}_{0}\left[\mathbf{Z}_{i}^{2}(t,s;x,y)\right]\leq
C\left[\sqrt{s}\wedge|x-y|\right]$

Since $\mathbf{Z}(t,s;x,y)$ is Gaussian we obtain the lemma by standard
arguments. ∎

By similar arguments (see Lemma 3.3 in [AJM21]) one has

######  Lemma 3.15.

There is a constant $\tilde{C}_{1}>0$ such that for all $s\leq 1$

$\mathbb{P}_{0}\left(\left|\mathbf{N}(0,t+s;0)-\mathbf{N}(0,t;0)\right|>\lambda\right)\leq\exp\left(-\frac{\tilde{C}_{1}^{2}\lambda^{2}}{\sqrt{s}}\right)$

uniformly in $t$.

Let $\mathbb{D}_{n}$ denote the collection of dyadic points of the form
$\frac{k}{2^{n}}$ in $[0,1]$. For any dyadic point $x\in\mathbb{D}_{n}$, we
can find a sequence $0=p_{0},p_{1},\cdots,p_{m}=x$ of points such that
$p_{i},p_{i+1}$ are nearest neighbors in some $\mathbb{D}_{k},\,k\leq n$, and
there are at most $2$ points in any $\mathbb{D}_{k}$. Now

$\displaystyle\mathbf{N}(0,t+s;x)-\mathbf{N}(0,t;x)$
$\displaystyle=\Big{[}\mathbf{N}(0,t+s;0)-\mathbf{N}(0,t;0)\Big{]}+\sum_{i=1}^{m}\mathbf{Z}(t,s;p_{i+1},p_{i})$

From this and a chaining argument, similar to that of Lemma 3.4 in [AJM21] we
obtain

######  Lemma 3.16.

There is a $C_{2}>0$ such that for all $s\leq 1$

$\mathbb{P}_{0}\left(\sup_{x\in[0,1]}\left|\mathbf{N}(0,t+s;x)-\mathbf{N}(0,t;x)\right|>\lambda\right)\leq\exp\left(-\frac{C_{2}^{2}\lambda^{2}}{\sqrt{s}}\right).$

uniformly in $t$.

We clearly have for $s,\,\tilde{s}\leq 1$

$\displaystyle\Big{|}\sup_{x\in[0,1]}\left|\mathbf{N}(0,t+s;x)-\mathbf{N}(0,t;x)\right|-\sup_{x\in[0,1]}\left|\mathbf{N}(0,t+\tilde{s};x)-\mathbf{N}(0,t;x)\right|\Big{|}$
$\displaystyle\leq\sup_{x\in[0,1]}\left|\mathbf{N}(0,t+s;x)-\mathbf{N}(0,t+\tilde{s};x)\right|,$

and just as in the above lemma we have for $\tilde{s}\leq s\leq 1$

$\mathbb{P}_{0}\left(\sup_{x\in[0,1]}\left|\mathbf{N}(0,t+s,x)-\mathbf{N}(0,t+\tilde{s},x)\right|>\lambda\right)\leq\exp\left(-\frac{C_{2}^{2}\lambda^{2}}{\sqrt{s-\tilde{s}}}\right).$

Therefore a chaining argument gives us Proposition 3.1. ∎

Now we return to the case of soft obstacles. We will need the following

######  Proposition 3.2.

Fix any $\eta>0$. There are constant $0<C_{6}(\eta)<1$ and $C_{7}(\eta)>0$
such that for $t\geq L$ we have

$\displaystyle\mathbb{P}_{0}\bigg{(}\mathcal{R}\left(\mathbf{N}(0,t)\right)\leq\frac{a}{8},\;\sup_{s\leq
C_{6}a^{4+\eta}}\sup_{x\in[0,1]}\left|\mathbf{N}(0,t+s,x)-\mathbf{N}(0,t,x)\right|\leq\frac{a}{16},$
$\displaystyle\hskip 227.62204pt\text{and }\sup_{s\leq
C_{6}a^{4+\eta}}|\mathbf{X}_{t+s}-\mathbf{X}_{t}|\leq\frac{a}{16}\bigg{)}\geq\frac{a^{2}}{C_{7}}.$

###### Proof.

We first give a lower bound on
$\mathbb{P}_{0}\left(\mathcal{R}(\mathbf{N}(0,t))\leq\frac{a}{2}\right)$.
Clearly $\mathcal{R}(\mathbf{N}(0,t))=\sup_{x,y}|\mathbf{N}(t;x,y)|$, where
$\mathbf{N}(t;x,y)$ is defined in (3.13). We have

$\mathbf{N}(t;x,y)=\mathbf{N}^{(1)}(t;x,y)-\mathbf{N}^{(2)}(t;x,y).$

Lemma 3.5 gives

$\mathbb{P}_{0}\left(\sup_{x,y\in[0,1]}|\mathbf{N}^{(2)}(t;x,y)|\geq\frac{a}{16}\right)\leq
2\exp\left(-\frac{e^{t}a^{2}}{256C_{1}}\right).$

We next obtain an upper bound on the tail probabilities of
$\sup_{x,y\in[0,1]}|\mathbf{N}^{(1)}(t;x,y)|$. Using ideas analogous to
Proposition 3.1, but now we use Lemma 3.6 instead of the bounds on (3.22) and
(3.23), we obtain

$\mathbb{P}_{0}\left(\sup_{x,y\in[0,1]}|\mathbf{N}^{(1)}(t;x,y)|\geq\frac{a}{16}\right)\leq\exp\left(-\frac{a^{2}}{256C_{4}}\right)$

for some $C_{4}>0$. Therefore

(3.24)
$\begin{split}\mathbb{P}_{0}\left(\mathcal{R}\left(\mathbf{N}(0,t)\right)\leq\frac{a}{2}\right)&=1-\mathbb{P}_{0}\left(\mathcal{R}\left(\mathbf{N}(0,t)\right)>\frac{a}{4}\right)\\\
&\geq
1-\mathbb{P}_{0}\left(\sup_{x,y\in[0,1]}|\mathbf{N}^{(1)}(t;x,y)|\geq\frac{a}{16}\right)\\\
&\qquad-\mathbb{P}_{0}\left(\sup_{x,y\in[0,1]}|\mathbf{N}^{(2)}(t;x,y)|\geq\frac{a}{16}\right)\\\
&\geq\frac{a^{2}}{C_{5}}.\end{split}$

for some $C_{5}>0$. By Proposition 3.1 and standard results on Brownian motion
the quantities

$\mathbb{P}_{0}\left(\sup_{s\leq
C_{6}a^{4+\eta}}\sup_{x\in[0,1]}\left|\mathbf{N}(0,t+s;x)-\mathbf{N}(0,t;x)\right|\leq\frac{a}{16}\right)$

and

$\mathbb{P}_{0}\left(\sup_{s\leq
C_{6}a^{4+\eta}}|\mathbf{X}_{t+s}-\mathbf{X}_{t}|\leq\frac{a}{16}\right)$

are both at least
$1-\exp\left(-\frac{C}{\sqrt{C_{6}}a^{\frac{\eta}{2}}}\right)$, uniformly in
$t$. The proposition is proved by combining the above with (3.24). ∎

##  4\. The proof of upper bound in Theorems 1.1 and 1.2

As explained earlier (in Remark 1.2) due to the scaling relations, we will
first obtain an upper bound for $S_{T}^{H,1,\nu}$. We will consider the hard
obstacle case first and then modify its proof suitably to handle the soft
obstacle case.

###### Proof of Upper bound in Theorem 1.1.

Recall from (1.4) that

$S_{T}^{\text{H},1,\nu}=\mathbb{E}_{0}\exp\left(-\nu\left|\mathscr{S}^{1}_{T}(a)\right|\right)$

where $\mathscr{S}^{1}_{T}(a)$ is the sausage of radius $a$ around
$\mathbf{u}$, that is

$\mathscr{S}^{1}_{T}(a)=\mathop{\bigcup}_{\begin{subarray}{c}0\leq s\leq T,\\\
0\leq y\leq 1\end{subarray}}\left\\{\mathbf{u}(s,y)+B(\mathbf{0},a)\right\\}.$

The upper bound on $S_{T}^{H,1,\nu}$ essentially involves finding a lower
bound on the volume of the sausage.

Recall from (3.4) that $\\#(T)=\left|\left\\{i\geq 1:\tau_{i}\leq
T\right\\}\right|,$ counts the number of $\tau_{i}$’s before time $T$, and for
$\Lambda>1$ from Lemma 3.2 that there are positive constants
$A_{1}(d,\Lambda),\,B_{1}(d,\Lambda)$ such for all $T>0$

(4.1) $\mathbb{P}_{0}\left(\\#(T)\leq
A_{1}T^{\frac{d}{d+2}}\right)\leq\exp\left(-B_{1}T^{\frac{d}{d+2}}\right).$

Therefore

$\displaystyle\mathbb{E}_{0}\exp\left(-\nu\left|\mathscr{S}^{1}_{T}(a)\right|\right)$
$\displaystyle\leq\exp\left(-B_{1}T^{\frac{d}{d+2}}\right)+\mathbb{E}_{0}\left[\exp\left(-\nu\left|\mathscr{S}^{1}_{T}(a)\right|\right)\cdot\mathbf{1}\left\\{\\#(T)>A_{1}T^{\frac{d}{d+2}}\right\\}\right].$

Now let

$\\#_{1}(T):=\left|\left\\{i\leq
A_{1}T^{\frac{d}{d+2}}:\,\tau_{i+1}-\tau_{i}\geq\Lambda^{2}\right\\}\right|.$

Clearly $\tau_{i+1}-\tau_{i}$ is more than the time it takes for the Brownian
motion $\mathbf{X}_{t}$ to leave a ball of radius $4\Lambda$ centered at
$\mathbf{X}_{\tau_{i}}$. Therefore the sequence $\tau_{i+1}-\tau_{i}$
stochastically dominates an i.i.d. sequence $\mathcal{T}_{i}$, where
$\mathcal{T}_{i}$ is distributed as the time it takes for a Brownian motion
starting at $\mathbf{0}$ to leave a ball of radius $4\Lambda$. Moreover

$\mathbb{P}\left(\mathcal{T}_{i}\geq\Lambda^{2}\right)=p>0,$

where $p$ is independent of any of the parameters. Therefore by standard large
deviation theory, there are positive constants
$A_{2}(p,d,\Lambda),B_{2}(p,d,\Lambda)$ such that

(4.2)
$\mathbb{P}_{0}\left(\\#_{1}(T)<A_{2}T^{\frac{d}{d+2}}\right)\leq\exp\left(-B_{2}T^{\frac{d}{d+2}}\right).$

Consider the event

$\mathscr{A}_{1}:=\left\\{\\#(T)>A_{1}T^{\frac{d}{d+2}},\quad\\#_{1}(T)>A_{2}T^{\frac{d}{d+2}}\right\\}.$

Equations (4.1) and (4.2) imply that there is a positive $B_{3}(d,\Lambda,p)$
such that

(4.3)
$\mathbb{P}_{0}\left(\mathscr{A}_{1}^{c}\right)\leq\exp\left(-B_{3}T^{\frac{d}{d+2}}\right).$

To get a good lower bound on the volume of the sausage, we need a good control
on the number of $T_{i}$’s up to time $T$. So let

$\\#_{2}(T):=\left|\left\\{T_{i}:T_{i}\leq T\right\\}\right|.$

Our next task is to show that on the event $\mathscr{A}_{1}$ we must have that
$\\#_{2}(T)$ is sufficiently large.

On the event

$\mathscr{A}_{2}=\left\\{\sum_{i=1}^{A_{4}\frac{T^{\frac{d}{d+2}}}{L}}(S_{i}-T_{i-1})\leq\widetilde{C}A_{4}T^{\frac{d}{d+2}}\right\\},$

the number of $\tau_{i}$ with $\tau_{i+1}-\tau_{i}\geq\Lambda^{2}$ in the
union of all the intervals
$[T_{i-1},S_{i}],\,i=1,2\cdots,A_{4}\frac{T^{\frac{d}{d+2}}}{L}$ is less than
$\frac{\widetilde{C}A_{4}T^{\frac{d}{d+2}}}{\Lambda^{2}}$. On the event
$\mathscr{A}_{1}$ we have $\\#_{1}(T)>A_{2}T^{\frac{d}{d+2}}$. Therefore with
the choice of $A_{4}$ such that

(4.4) $\frac{\widetilde{C}A_{4}}{\Lambda^{2}}=\frac{A_{2}}{4},$

the intersection of $\mathscr{A}_{1}$ and $\mathscr{A}_{2}$ contains at least
$\frac{A_{2}}{2}T^{\frac{d}{d+2}}$ many $\tau_{i}$’s before time $T$ outside
of the union of intervals
$[T_{i-1},S_{i}],\,i=1,2\cdots,A_{4}\frac{T^{\frac{d}{d+2}}}{L}$. Note
carefully that any $T_{j}$ is the smallest $\tau_{i}$ immediately following
$S_{j}$, and therefore this guarantees that there are at least
$A_{4}\frac{T^{\frac{d}{d+2}}}{L}$ many $T_{j}$’s up to time $T$. Then using
Lemma 3.13 and (4.3) we have that

(4.5)
$\mathbb{P}_{0}\left(\\#_{2}(T)<A_{4}\frac{T^{\frac{d}{d+2}}}{L}\right)\leq\mathbb{P}_{0}(\mathscr{A}_{1}^{c})+\mathbb{P}_{0}(\mathscr{A}_{2}^{c})\leq\exp\left(-B_{4}T^{\frac{d}{d+2}}\right),$

for some $B_{4}(d,\Lambda,p)>0$.

Therefore on the event $\mathscr{A}_{3}:=\mathscr{A}_{1}\cap\mathscr{A}_{2}$
we have that $\\#_{2}(T)\geq A_{4}\frac{T^{\frac{d}{d+2}}}{L}.$ We now count
the $T_{i}$’s before time $T$ with large sausage volumes at $T_{i}^{\prime}s$,
as in Lemma 3.11. Namely,

$\\#_{3}(T):=\left|\left\\{i\leq
A_{4}\frac{T^{\frac{d}{d+2}}}{L}:\,\mathcal{R}\left(\mathbf{N}(T_{i-1},T_{i})\right)\leq\Lambda,\;\left|\mathscr{S}\Big{(}\frac{a}{2};\,\mathbf{N}(T_{i-1},T_{i}))\Big{)}\right|\geq{\mathcal{C}}_{\gamma}a^{d-2+\gamma}\right\\}\right|.$

It now follows from (3.17), for $\Lambda>1$ as in Lemma 3.10, that for any
$A_{5}>0$

$\displaystyle\mathbb{P}_{0}\left(\\#_{3}(T)\leq
A_{5}\frac{T^{\frac{d}{d+2}}}{L}\right)$
$\displaystyle=\mathbb{P}_{0}\left(\exp\left(-\\#_{3}(T)\right)\geq\exp\left(-A_{5}\frac{T^{\frac{d}{d+2}}}{L}\right)\right)$
$\displaystyle\leq\left(\frac{1}{2}+e^{-1}\right)^{A_{4}\frac{T^{\frac{d}{d+2}}}{L}}\exp\left(A_{5}\frac{T^{\frac{d}{d+2}}}{L}\right)$
$\displaystyle\leq\exp\left(A_{4}\frac{T^{\frac{d}{d+2}}}{L}\log\left(\frac{1}{2}+e^{-1}\right)+A_{5}\frac{T^{\frac{d}{d+2}}}{L}\right).$

We now choose

$A_{5}=-\frac{A_{4}}{2}\log\left(\frac{1}{2}+e^{-1}\right),$

to obtain

(4.6) $\mathbb{P}_{0}\left(\\#_{3}(T)\leq
A_{5}\frac{T^{\frac{d}{d+2}}}{L}\right)\leq\exp\left(-B_{5}\frac{T^{\frac{d}{d+2}}}{L}\right),$

for a positive constant $B_{5}(d,\Lambda,p)$.

Let $n_{1}<n_{2}<\cdots$ be the indices $j\leq
A_{4}\frac{T^{\frac{d}{d+2}}}{L}$ such that $T_{j}\leq T$ and

(4.7)
$\,\mathcal{R}\left(\mathbf{N}(T_{i-1},T_{i})\right)\leq\Lambda,\;\left|\mathscr{S}\Big{(}\frac{a}{2};\,\mathbf{N}(T_{i-1},T_{i}))\Big{)}\right|\geq{\mathcal{C}}_{\gamma}a^{d-2+\gamma}.$

Thanks to (4.1), (4.2), (4.5) and (4.6), outside of a set of probability
$\exp\left(-B_{6}\frac{T^{\frac{d}{d+2}}}{L}\right)$, where $B_{6}$ is a
positive constant depending only on $d,\Lambda,p$, there are at least
$A_{5}\frac{T^{\frac{d}{d+2}}}{L}$ many such $n_{i}$’s. Further, from (3.11)
we have that the fixed-time sausages $\mathscr{S}^{1}(a;T_{n_{j}})$ are
disjoint. Therefore, using (3.12) we have

$\displaystyle S_{T}^{\text{H},1,\nu}$
$\displaystyle=\mathbb{E}_{0}\exp\left(-\nu\left|\mathscr{S}^{1}_{T}(a)\right|\right)$
$\displaystyle\leq\exp\left(-B_{6}\frac{T^{\frac{d}{d+2}}}{L}\right)+\mathbb{E}_{0}\left[\exp\left(-\nu\left|{\bigcup}_{j=0}^{A_{5}\frac{T^{\frac{d}{d+2}}}{L}}\mathscr{S}^{1}(a;T_{n_{j}})\right|\right)\right]$
$\displaystyle\leq\exp\left(-B_{6}\frac{T^{\frac{d}{d+2}}}{L}\right)+\mathbb{E}_{0}\left[\exp\left(-\nu\left|{\bigcup}_{j=0}^{A_{5}\frac{T^{\frac{d}{d+2}}}{L}}\mathscr{S}(\frac{a}{2};{\mathbf{N}}(T_{n_{j-1}},T_{n_{j}})\right|\right)\right]$

Now applying (4.7),

$\displaystyle S_{T}^{\text{H},1,\nu}$
$\displaystyle\leq\exp\left(-B_{6}\frac{T^{\frac{d}{d+2}}}{L}\right)+\exp\left(-\nu
A_{5}C_{\gamma}a^{d-2+\gamma}\frac{T^{\frac{d}{d+2}}}{L}\right).$

Finally we apply the scaling (1.10) to get an upper bound for
$S_{T}^{H,J,\nu}$. We thus obtain

$\displaystyle S_{T}^{\text{H},J,\nu}$
$\displaystyle\leq\exp\left(-\frac{B_{6}(T/J^{2})^{\frac{d}{d+2}}}{E+3|\log(a/J^{\frac{1}{2}})|}\right)+\exp\left(-\nu
A_{5}C_{\gamma}a^{d-2+\gamma}\frac{(T/J^{2})^{\frac{d}{d+2}}J^{1-\frac{\gamma}{2}}}{E+3|\log(a/J^{\frac{1}{2}})|}\right).$

It is clear that the first term is the leading term. ∎

We now turn to the case of soft obstacles. As before we first find an upper
bound for $S_{T}^{H,1,\nu}$. We will explain how the argument differs from the
case of hard obstacles.

###### The proof of upper bound in Theorem 1.2 .

We will now look for a subsequence $T_{n_{j}}$ of the $T_{i}$’s such that the
entire string stays in a small ball of radius $\frac{3a}{8}$ during the time
interval $[T_{n_{j}},T_{n_{j}}+a^{4+\epsilon}]$, and there exists a Poisson
point within distance $\frac{a}{8}$ of the center of mass (See Definition
4.8). Assumption 1.1 then guarantees there is a contribution of at least
$\mathscr{C}a^{4+\epsilon}$ to the integral in $S_{T}$ during this time
period.

We then follow the argument of Proof of upper bound in Theorem 1.1. The only
difference now is in the definition of $\\#_{3}(T)$, which in the present case
becomes

(4.8) $\begin{split}&\\#_{3}(T)=\bigg{|}\bigg{\\{}i\leq
A_{4}\frac{T^{\frac{d}{d+2}}}{L}:\mathcal{R}(\mathbf{N}(T_{j-1},T_{j}))\leq\frac{a}{8},\\\
&\qquad\qquad\sup_{s\leq
C_{6}a^{4+\eta}}\sup_{x\in[0,1]}\left|\mathbf{N}(T_{j-1},T_{j}+s,x)-\mathbf{N}(T_{j-1},T_{j},x)\right|\leq\frac{a}{16},\\\
&\qquad\qquad\sup_{s\leq
C_{6}a^{4+\eta}}|\mathbf{X}_{T_{j}+s}-\mathbf{X}_{T_{j}}|\leq\frac{a}{16},\;\\\
&\qquad\qquad\text{and there is a Poisson point within distance $\frac{a}{8}$
of $\mathbf{X}_{T_{j}}$ }\bigg{\\}}\bigg{|}\end{split}$

Therefore $\\#_{3}(T)$ is a sum of Bernoulli random variables with probability
of success $\tilde{p}$ satisfying

$\tilde{p}\geq\frac{a^{2}}{C_{7}}\left(1-\exp\left(-\nu\frac{a^{d}}{8^{d}}\right)\right)\geq
C_{8}\nu a^{d+2}$

for some constant $0<C_{8}<1$.

Let $Z_{1},Z_{2},\cdots$ be i.i.d. Bernoulli random variables with success
probability $p_{*}=C_{8}\nu a^{d+2}$. By standard large deviation theory

(4.9) $\begin{split}\mathbb{P}_{0}\left(\\#_{3}(T)\leq
A_{4}\frac{T^{\frac{d}{d+2}}}{L}\cdot\frac{p_{*}}{2}\right)&\leq\mathbb{P}_{0}\left(\sum_{i=1}^{\frac{A_{4}T^{\frac{d}{d+2}}}{L}}Z_{i}\leq
A_{4}\frac{T^{\frac{d}{d+2}}}{L}\cdot\frac{p_{*}}{2}\right)\\\
&\leq\exp\left(-B_{7}\frac{T^{\frac{d}{d+2}}}{L}\cdot\nu
a^{d+2}\right),\end{split}$

for some constant $B_{7}(d,\Lambda,p)>0$.

Let $n_{1}<n_{2}<\cdots$ be the indices $j$ such that the event in the right
hand side of (4.8) occurs. Note that at these times $T_{n_{k}}$ it follows
from the proof of (3.11)

$\displaystyle\sup_{s\leq
C_{6}a^{4+\eta}}\mathcal{R}(\mathbf{u}(T_{n_{k}}+s))$
$\displaystyle\leq\sup_{s\leq
C_{6}a^{4+\eta}}\mathcal{R}(\mathbf{N}(T_{n_{k}-1},T_{n_{k}}+s))+2\delta$
$\displaystyle\leq\mathcal{R}(\mathbf{N}(T_{n_{k}-1},T_{n_{k}}))+\frac{a}{8}+2\delta$
$\displaystyle\leq\frac{a}{4}+2\delta$

Moreover we have that the center of mass satisfies $\sup_{s\leq
C_{6}a^{4+\eta}}\left|\mathbf{X}_{T_{n_{k}}+s}-\mathbf{X}_{T_{n_{k}}}\right|\leq\frac{a}{16}$.
Therefore

$\displaystyle\sup_{s\leq C_{6}a^{4+\eta}}$
$\displaystyle\left|\mathbf{u}(T_{n_{k}}+s,x)-\mathbf{X}_{T_{n_{k}}}\right|$
$\displaystyle\leq\sup_{s\leq
C_{6}a^{4+\eta}}\left|\mathbf{u}(T_{n_{k}}+s,x)-\mathbf{X}_{T_{n_{k}}+s}\right|+\sup_{s\leq
C_{6}a^{4+\eta}}\left|\mathbf{X}_{T_{n_{k}}+s}-\mathbf{X}_{T_{n_{k}}}\right|$
$\displaystyle\leq\sup_{s\leq
C_{6}a^{4+\eta}}\mathcal{R}(\mathbf{u}(T_{n_{k}}+s))+\sup_{s\leq
C_{6}a^{4+\eta}}\left|\mathbf{X}_{T_{n_{k}}+s}-\mathbf{X}_{T_{n_{k}}}\right|$
$\displaystyle\leq\frac{5a}{16}+2\delta$ $\displaystyle\leq\frac{3a}{8}.$

Thus the entire string lies within a ball of radius $\frac{3a}{8}$ centered at
$\mathbf{X}_{T_{n_{k}}}$ for the duration
$[T_{n_{k}},T_{n_{k}}+C_{6}a^{4+\eta}]$. Since there is a Poisson point within
distance $\frac{a}{8}$ of $\mathbf{X}_{T_{n_{k}}}$ the string will be entirely
contained within distance $\frac{a}{2}$ of the Poisson point during the time
interval $[T_{n_{k}},T_{n_{k}}+C_{6}a^{4+\eta}]$. Therefore using the bounds
in the proof of the upper bound in Theorem 1.1 along with (4.9) we obtain for
$T>0$

$\displaystyle S_{T}^{\text{H},1,\nu}$
$\displaystyle=\mathbb{E}\left[\exp\left(-\int_{0}^{T}\int_{0}^{1}\text{V}(\mathbf{u}(s,x),\text{\Large$\eta$})dsdx\right)\right]$
$\displaystyle\leq\exp\left(-B_{7}\frac{T^{\frac{d}{d+2}}}{L}\cdot\nu
a^{d+2}\right)+\exp\left(-B_{8}\frac{T^{\frac{d}{d+2}}}{L}\right)$
$\displaystyle\quad+\mathbb{E}\left[\exp\left(-\sum_{j=1}^{\frac{A_{5}}{L}T^{\frac{d}{d+2}}\nu
a^{d+2}}\int_{T_{n_{j}}}^{T_{n_{j}}+C_{6}a^{4+\eta}}\int_{0}^{1}\text{V}(\mathbf{u}(s,x),\text{\Large$\eta$})dsdx\right)\cdot\right],$

where $A_{5}=\frac{A_{4}C_{8}}{2}$. Now we use Assumption 1.1 to obtain an
upper bound

$\displaystyle S_{T}^{\text{H},1,\nu}$
$\displaystyle\leq\exp\left(-B_{7}\frac{T^{\frac{d}{d+2}}}{L}\cdot\nu
a^{d+2}\right)+\exp\left(-B_{8}\frac{T^{\frac{d}{d+2}}}{L}\right)$
$\displaystyle\qquad+\exp\left(-\mathscr{C}A_{5}C_{6}\nu
a^{d+6+\eta}\frac{T^{\frac{d}{d+2}}}{L}\right)$

Finally we use (1.10) to get an upper bound for $S_{T}^{\text{H},J,\nu}$. We
obtain for $T>0$

$\displaystyle S_{T}^{\text{H},J,\nu}$
$\displaystyle\leq\exp\left(-\frac{B_{7}\nu
a^{d+2}(T/J^{2})^{\frac{d}{d+2}}}{J\left(E+3|\log(a/J^{\frac{1}{2}})|\right)}\right)+\exp\left(-\frac{\mathscr{C}A_{5}C_{6}\nu
a^{d+6+\eta}(T/J^{2})^{\frac{d}{d+2}}}{J^{3+\frac{\eta}{2}}\left(E+3|\log(a/J^{\frac{1}{2}})|\right)}\right).$

It is clear that the second term dominates the first term for large $J$.

∎

## References

  * [ADS19] Siva Athreya, Alexander Drewitz, and Rongfeng Sun, _Random walk among mobile/immobile traps: a short review_ , Sojourns in Probability Theory and Statistical Physics-III (2019), 1–22.
  * [AJM21] Siva Athreya, Mathew Joseph, and Carl Mueller, _Small ball probabilities and a support theorem for the stochastic heat equation_ , Ann. Probab. 49 (2021), no. 5, 2548–2572. MR 4317712
  * [All98] H. Allouba, _Different types of SPDEs in the eyes of Girsanov’s theorem_ , Stochastic Anal. Appl. 16 (1998), no. 5, 787–810. MR 1643116
  * [Bol90] E. Bolthausen, _On the volume of the Wiener sausage_ , Ann. Probab. 18 (1990), no. 4, 1576–1582. MR 1071810
  * [CJK13] Daniel Conus, Mathew Joseph, and Davar Khoshnevisan, _On the chaotic character of the stochastic heat equation, before the onset of intermittency_ , The Annals of Probability 41 (2013), no. 3B, 2225–2260.
  * [dHW94] Frank den Hollander and George H Weiss, _Aspects of trapping in transport processes_ , Contemporary problems in statistical physics, SIAM, 1994, pp. 147–203.
  * [DV75] M. D. Donsker and S. R. S. Varadhan, _Asymptotics for the Wiener sausage_ , Comm. Pure Appl. Math. 28 (1975), no. 4, 525–565. MR 397901
  * [Kön16] Wolfgang König, _The parabolic anderson model: Random walk in random potential_ , Birkhäuser, 2016.
  * [MP10] Peter Mörters and Yuval Peres, _Brownian motion_ , Cambridge Series in Statistical and Probabilistic Mathematics, vol. 30, Cambridge University Press, Cambridge, 2010, With an appendix by Oded Schramm and Wendelin Werner. MR 2604525
  * [Szn90] Alain-Sol Sznitman, _Long time asymptotics for the shrinking Wiener sausage_ , Comm. Pure Appl. Math. 43 (1990), no. 6, 809–820. MR 1059329
  * [Szn98] by same author, _Brownian motion, obstacles and random media_ , Springer Science & Business Media, 1998.
  * [vdBBdH01] M. van den Berg, E. Bolthausen, and F. den Hollander, _Moderate deviations for the volume of the Wiener sausage_ , Ann. of Math. (2) 153 (2001), no. 2, 355–406. MR 1829754
  * [Wal86] John B. Walsh, _An introduction to stochastic partial differential equations_ , École d’été de probabilités de Saint-Flour, XIV—1984, Lecture Notes in Math., vol. 1180, Springer, Berlin, 1986, pp. 265–439. MR 876085 (88a:60114)