# Tightness for Thick Points in two dimensions

Jay Rosen

(Date: March 27, 2022)

###### Abstract.

Let $W_{t}$ be Brownian motion in the plane started at the origin and let
$\theta$ be the first exit time of the unit disk $D_{1}$. Let

$\mu_{\theta}(x,\epsilon)=\frac{1}{\pi\epsilon^{2}}\int_{0}^{\theta}1_{\\{B(x,\epsilon)\\}}(W_{t})\,dt,$

and set $\mu^{\ast}_{\theta}(\epsilon)=\sup_{x\in
D_{1}}\mu_{\theta}(x,\epsilon)$. We show that

$\sqrt{\mu^{\ast}_{\theta}(\epsilon)}-\sqrt{2/\pi}\left(\log\epsilon^{-1}-\frac{1}{2}\log\log\epsilon^{-1}\right)$

is tight.

###### Key words and phrases:

Thick points. Two dimensional sphere. Barrier estimates

###### 2010 Mathematics Subject Classification:

60J65

Jay Rosen was partially supported by the Simons Foundation

##  1\. Introduction

Let $W_{t}$ be Brownian motion in the plane started at the origin and let
$\theta$ be the first exit time of the unit disk $D_{1}$. In [6] we showed
that

(1.1) $\lim_{\epsilon\to 0}\sup_{x\in
D_{1}}\frac{1}{\epsilon^{2}\log^{2}(\epsilon)}\int_{0}^{\theta}1_{\\{B(x,\epsilon)\\}}(W_{t})\,dt=2,\quad
a.s.,$

where $B(x,\epsilon)$ is the ball of radius $\epsilon$ centered at $x$. The
integral above is the occupation measure of $B(x,\epsilon)$, and points $x$
with large occupation measure are referred to as thick points. Taking square
roots we can write this as

(1.2) $\lim_{\epsilon\to 0}\frac{1}{\log(\epsilon^{-1})}\sqrt{\sup_{x\in
D_{1}}\frac{1}{\pi\epsilon^{2}}\int_{0}^{\theta}1_{\\{B(x,\epsilon)\\}}(W_{t})\,dt}=\sqrt{2/\pi},\quad
a.s.$

Let

(1.3)
$\mu_{\theta}(x,\epsilon)=\frac{1}{\pi\epsilon^{2}}\int_{0}^{\theta}1_{\\{B(x,\epsilon)\\}}(W_{t})\,dt,$

and set $\mu^{\ast}_{\theta}(\epsilon)=\sup_{x\in
D_{1}}\mu_{\theta}(x,\epsilon)$. Then (1.2) says that
$\sqrt{\mu^{\ast}_{\theta}(\epsilon)}\sim\sqrt{2/\pi}\log\epsilon^{-1}$, as
$\epsilon\to 0$. In this paper we obtain more detailed asymptotics. Let

(1.4)
$m_{\epsilon}=\sqrt{2/\pi}\left(\log\epsilon^{-1}-\frac{1}{2}\log\log\epsilon^{-1}\right).$

We will say that the thick points in $D_{1}$ are tight if
$\sqrt{\mu^{\ast}_{\theta}(\epsilon)}-m_{\epsilon}$ is a tight family of
random variables. That is,

(1.5) $\lim_{K\to\infty}\varlimsup_{\epsilon\to
0}\mathbb{P}\left(\Big{|}\sqrt{\mu^{\ast}_{\theta}(\epsilon)}-m_{\epsilon}\Big{|}>K\right)=0.$

######  Theorem 1.1.

The thick points in $D_{1}$ are tight.

In fact we obtain the following improvement on the right tail of (1.5), where
$c^{\ast}:=2\sqrt{2}$.

######  Theorem 1.2.

On $D_{1}$, for some $0<C,C^{\prime},z_{0}<\infty$ and all $z\geq z_{0}$,

(1.6) $\varlimsup_{\epsilon\to
0}\mathbb{P}\left(\sqrt{\mu^{\ast}_{\theta}(\epsilon)}-m_{\epsilon}\geq
z\right)\leq Cze^{-c^{\ast}\sqrt{\pi}\,z},$

and

(1.7) $\varliminf_{\epsilon\to
0}\mathbb{P}\left(\sqrt{\mu^{\ast}_{\theta}(\epsilon)}-m_{\epsilon}\geq
z\right)\geq C^{\prime}ze^{-c^{\ast}\sqrt{\pi}\,z}.$

It follows from Brownian scaling that Theorems 1.1 and 1.2 hold if $D_{1}$ is
replaced by any disc centered at the origin.

For reasons of symmetry it is easier to work on the sphere ${\bf S}^{2}$, and
derive our results for thick points in $D_{1}$ from results for thick points
on ${\bf S}^{2}$. We use $B_{d}(x,r)$ for the ball centered at $x$ of radius
$r$, in the spherical metric $d$.

Let $X_{t}$ be Brownian motion on ${\bf S}^{2}$ started at some point $v$ (the
‘South Pole’). For some (small) $r^{\ast}$ let $\tau$ be the first hitting
time of $\partial B_{d}(v,r^{\ast})$ (the ‘Antarctic Circle’). Let
$\omega_{\epsilon}=2\pi(1-\cos\epsilon)$, the area of $B_{d}(x,\epsilon)$, and
set

(1.8)
$\bar{\mu}_{\tau}(x,\epsilon)=\frac{1}{\omega_{\epsilon}}\int_{0}^{\tau}1_{\\{B_{d}(x,\epsilon)\\}}(X_{t})\,dt,$

With $\bar{\mu}_{\tau,\epsilon}^{\ast}=\sup_{x\in{\bf
S}^{2}}\bar{\mu}_{\tau}(x,\epsilon)$ we will say that the thick points on
${\bf S}^{2}$ are tight if
$\sqrt{\bar{\mu}_{\tau,\epsilon}^{\ast}}-m_{\epsilon}$ is a tight family of
random variables.

######  Theorem 1.3.

The thick points on ${\bf S}^{2}$ are tight.

As in Theorem 1.2 we obtain the following improvement for the right tail.

######  Theorem 1.4.

On ${\bf S}^{2}$, for some $0<C,C^{\prime},z_{0}<\infty$ and all $z\geq
z_{0}$,

(1.9) $\varlimsup_{\epsilon\to
0}\mathbb{P}\left(\sqrt{\bar{\mu}_{\tau,\epsilon}^{\ast}}-m_{\epsilon}\geq
z\right)\leq Cze^{-c^{\ast}\sqrt{\pi}\,z},$

and

(1.10) $\varliminf_{\epsilon\to
0}\mathbb{P}\left(\sqrt{\bar{\mu}_{\tau,\epsilon}^{\ast}}-m_{\epsilon}\geq
z\right)\geq C^{\prime}ze^{-c^{\ast}\sqrt{\pi}\,z}.$

Theorems 1.3 and 1.4 are stated and first proven for $r^{\ast}$ sufficiently
small. In Section 8 we show that they hold for any $0<r^{\ast}<\pi$.

In analogy with [6], rather than work directly with occupation measures, we
work with excursion counts. To define this let $h_{l}=2\arctan(r_{0}e^{-l}/2)$
with $r_{0}$ small. For some $d_{0}\leq 1/1000$ let $F_{l}$ be the centers of
a $d_{0}h_{l}$ covering of ${\bf S}^{2}$.

Let $\mathcal{T}_{x,l}^{\tau}$ be the number of excursions from $\partial
B_{d}\left(x,h_{l-1}\right)$ to $\partial B_{d}\left(x,h_{l}\right)$ prior to
$\tau$. We will obtain the following result for $\sup_{x\in
F_{L}}\mathcal{T}_{x,L}^{\tau}$.

######  Theorem 1.5.

On ${\bf S}^{2}$, for some $0<z_{0},C,C^{\prime}<\infty$, all $L$ large and
all $z_{0}\leq z\leq\log L$,

(1.11) $C^{\prime}ze^{-2z}\leq\mathbb{P}\left(\sup_{x\in
F_{L}}\sqrt{2\mathcal{T}_{x,L}^{\tau}}-(2L-\log L)\geq z\right)\leq
Cze^{-2z}.$

Equivalently

(1.12) $C^{\prime}ze^{-2z}\leq\mathbb{P}\left(\sup_{x\in
F_{L}}\mathcal{T}_{x,L}^{\tau}\geq 2L\left(L-\log L+z\right)\right)\leq
Cze^{-2z}.$

Since $L\sim\log h_{L}^{-1}$, Theorem 1.5 is then suggestive of Theorem 1.4 if
we knew that on average the occupation measure of $B_{d}\left(x,h_{L}\right)$
during an excursion from $\partial B_{d}\left(x,h_{L}\right)$ to $\partial
B_{d}\left(x,h_{L-1}\right)$ was ‘about’ $h^{2}_{L}$. While this is basically
known for our choices of $h_{L},h_{L-1}$, see [6, Lemma 6.2], it is more
delicate to get the precision necessary to show the equivalence of Theorem 1.5
with (1.9).

We now write (1.11) in a more convenient form. Set

(1.13) $\rho_{L}=2-\frac{\log L}{L}.$

We will prove the following version of Theorem 1.5.

######  Theorem 1.6.

On ${\bf S}^{2}$, for some $0<z_{0},C,C^{\prime}<\infty$, all $L$ large and
all $z_{0}\leq z\leq\log L$,

(1.14) $C^{\prime}ze^{-2z}\leq\mathbb{P}\left[\sup_{x\in
F_{L}}\sqrt{2\mathcal{T}_{x,L}^{\tau}}\,\geq\rho_{L}L+z\right]\leq Cze^{-2z}.$

###  1.1. Background

This paper is based in many ways on my work [4] with Belius and Zeituni on
tightness for the cover time of ${\bf S}^{2}$. The general approach is
similar, and whenever results of that paper could be used directly I did so.
However, the mathematics often necessitated different arguments.

The family

$\\{\bar{\mu}_{\tau}(x,\epsilon);x\in B_{d}(v,r^{\ast}),\epsilon>0\\}$

is associated with a second order Gaussian chaos $H(x,\epsilon)$, $x\in
B_{d}(v,r^{\ast}),\epsilon>0$ by an isomorphism theorem of Dynkin [12].
Intuitively,

(1.15)
$H(x,\epsilon)=\int_{B_{d}(x,\epsilon)}G^{2}_{y}\,dm(y)-E\left(\int_{B_{d}(x,\epsilon)}G^{2}_{y}\,dm(y)\right)$

where $G_{x}$ is the mean zero Gaussian process with covariance $u(x,y)$, the
Green’s function for $B_{d}(v,r^{\ast})$ and $m$ denotes the standard surface
measure on $S^{2}$. Since $u(x,x)=\infty$ for all $x$, (1.15) is not a priori
well defined. Nevertheless, this would suggest that there is a close
relationship between $\bar{\mu}_{\tau,\epsilon}^{\ast}=\sup_{x\in
R^{2}}\bar{\mu}_{\tau}(x,\epsilon)$ and the supremum of Gaussian fields. For
details on $H$ and the isomorphism theorem see [19, Section 2].

For related work see [1, 15, 16, 17, 18].

###  1.2. Open problems

1\. Based on the analogy with the extrema of Branching random walks and log-
correlated Gaussian fields, one expects that Theorem 1.1 should be replaced by
the statement

The sequence of random variables
$\sqrt{\mu^{\ast}_{\theta}(\varepsilon)}-m_{\varepsilon}$ converges in
distribution to a randomly shifted Gumbel random variable.

A key step in proving such convergence would be the improvement of the tail
estimates in Theorems 1.2 and 1.4 for $z$ large, which in turn would require a
corresponding improvement of Theorem 1.6.

2\. In [6] we also proved a conjecture of Erdős and Taylor concerning the
number $L^{\ast}_{n}$ of visits to the most visited site for simple random
walk in ${\bf Z}^{2}$ up to step $n$. It was shown there that

(1.16) $\lim_{n\to\infty}\frac{L^{\ast}_{n}}{(\log n)^{2}}=1/\pi\hskip
7.22743pt\mbox{a.s.}$

The approach in that paper was to first prove (1.1) for planar Brownian motion
and then to use strong approximation. Subsequently, in [22], we presented a
purely random walk method to prove (1.16) for simple random walk. A natural
problem is to prove tightness for $\sqrt{L^{\ast}_{n}}$.

3\. Following [6] we analyzed thick points for several other process. See [7],
[8] and [9]. One can ask about tightness or some analog for these processes.

###  1.3. Structure of the paper

In Section 2 we obtain the upper bounds for excursion counts in Theorem 1.6,
and in Section 3 we derive the lower bounds. These sections employ many of the
tools developed in [4]. In Section 4 we show how to go from results on
excursion counts to Theorems 1.3 and 1.4 which involve
$\bar{\mu}_{\tau}(x,\epsilon)$ in ${\bf S}^{2}$. Here we have to deal with a
new problem for the upper bounds: $\bar{\mu}_{\tau}(x,\epsilon)$ in ${\bf
S}^{2}$ is not in general monotone in $\epsilon$. This requires interpolation
and a continuity estimate which are developed in Sections 5 and 7. In the
short Section 8 we derive our results on thick points for the unit disc in the
plane from our results on thick points for ${\bf S}^{2}$, and use this to show
that Theorems 1.3 and 1.4 hold for any $0<r^{\ast}<\pi$. The last section is
an Appendix containing the barrier estimates we need for Sections 2 and 3.

###  1.4. Index of Notation

The following are frequently used notation, and a pointer to the location
where the definition appears.

$\begin{array}[]{ll}\mu_{\theta}(x,\epsilon)&\mbox{\eqref{plane.1}}\\\
m_{\epsilon}&\mbox{\eqref{121}}\\\ c^{\ast}&\mbox{\eqref{goal.1plane}}\\\
\bar{\mu}_{\tau}(x,\epsilon),B_{d}(x,r)&\mbox{\eqref{tp.1}}\\\
\rho_{L}&\mbox{(\ref{eq:defofts})}\\\
\mathcal{T}_{x,l}^{\tau}&\mbox{(\ref{goal.1astos})}\\\
r_{l},h_{l}&\mbox{\eqref{dr.1}}\\\ F_{l}&\mbox{\eqref{dr.2}}\\\
\mathcal{T}_{y,l}^{k\to 0}&\mbox{\rm\eqref{eq:NotHitByrL}}\\\
l_{L}&\mbox{(\ref{h1}) }\\\ \alpha_{z,+}(l)&\mbox{
(\ref{eq:AlphaBarrierDef})}\\\ k_{y}&\mbox{(\ref{label.1})}\\\
F^{\ast}_{L}&\mbox{(\ref{label.2})}\\\
F^{m}_{L},\mathcal{H}_{m,l}&\mbox{\rm\eqref{123H4}}\\\
\mathcal{B}_{m,l}&\mbox{\rm\eqref{123}}\\\
\mathcal{C}_{m,l}&\mbox{\rm\eqref{ind.34}}\\\
\mathcal{D}_{m,l}(j)&\mbox{\rm\eqref{ind.35}}\\\
\mathcal{B}_{m,l}^{\gamma,k}&\mbox{\rm\eqref{zdek}}\\\
\mathcal{T}_{y,\widetilde{r}_{l}}^{u_{m},r_{l-2},n}&\mbox{\rm\eqref{eq:to
showcent}}\\\ \beta_{z}\left(l\right)&\mbox{\rm\eqref{eq:BetaDefn2}}\\\
\alpha_{z,-}\left(l\right)&\mbox{\rm\eqref{14.1}}\\\
\mathcal{T}_{y,l}^{1},\mathcal{T}_{y,l}^{1,x^{2}}&\mbox{\rm\eqref{14.1}}\\\
F^{0}_{L}&\mbox{\rm\eqref{F0.def}}\\\
{\mathcal{W}}_{y,k}(n)&\mbox{(\ref{eq:LBTruncatedSumz})}\\\
N_{k,a}&\mbox{\rm\eqref{nkdef.2}}\\\ N_{k},I_{u}&\mbox{\rm\eqref{nkdef.2j}}\\\
{\mathcal{H}}_{k,a}&\mbox{\rm\eqref{hka.1}}\\\
k^{+},k^{++}&\mbox{\rm\eqref{eq:DefOfKPlus}}\\\
\widehat{\mathcal{I}}_{y,z}&\mbox{\rm\eqref{nkdef.2}}\\\
N_{k,a}&\mbox{\rm\eqref{nkdef.2}}\\\ N_{k}&\mbox{\rm\eqref{nkdef.2j}}\\\
{\mathcal{I}}_{y,z}&\mbox{\rm\eqref{eq:TruncatedSummandLBzz}}\\\
{\mathcal{H}}_{k,a}&\mbox{\rm\eqref{hka.1}}\\\
J_{y,k}^{\uparrow}&\mbox{\rm\eqref{eq:JDownz3}}\\\
{\mathcal{B}}_{y,k,a}&\mbox{\rm\eqref{eq:KUpz3}}\\\
\overline{\mathcal{M}}_{x,\epsilon,a,b}(n)&\mbox{\rm\eqref{occm.n}}\\\
t_{L}\left(z\right)&\mbox{\rm\eqref{clt.2}}\\\
\overline{\mathcal{M}}_{y,\bar{\epsilon}_{y},y_{0},a,b}(n)&\mbox{\rm\eqref{unif.1a}}\\\
D_{\ast}&\mbox{\rm\eqref{dast}}\\\ \end{array}$

##  2\. Upper bounds for excursions

Let

(2.1) $h(r)=2\arctan(r/2)$

and let

(2.2) $r_{l}=r_{0}e^{-l},l=0,1,\ldots,\hskip 14.45377pt\mbox{and }\hskip
14.45377pth_{l}=h(r_{l})$

for some $r_{0}<1$. We can take $r_{0}<1$ sufficiently small that for all
$0\leq x\leq r_{0}$

(2.3) $x-x^{3}\leq h(x)\leq x\hskip 14.45377pt\mbox{and}\hskip
14.45377pt|h^{\prime}(x)-1|\leq x^{2}.$

For some $d_{0}\leq 1/1000$ let $F_{l}$ be the centers of an $d_{0}h_{l}$
covering of $S^{2}$. We record for future use that

(2.4) $\left|F_{l}\right|\asymp cr_{l}^{-2}=cr^{-2}_{0}e^{2l},l\geq 0.$

Recall that $\mathcal{T}_{x,l}^{\tau}$ is the number of excursions from
$\partial B_{d}\left(x,h_{l-1}\right)$ to $\partial B_{d}\left(x,h_{l}\right)$
prior to $\tau$. In this section we will assume that $2r^{\ast}\leq h_{0}$ so
that for all $y\in B_{d}(v,r^{\ast})$ we have $B_{d}(v,r^{\ast})\subseteq
B_{d}(y,h_{0})$.

The reason for using $h(r)$ is due to the following result for ${\bf S}^{2}$,
see [4, (2.6)]. For any $u_{1}<u_{2}<u_{3}$

(2.5) $\mathbb{P}^{x\in\partial B_{d}\left(0,h(u_{2})\right)}\left[H_{\partial
B_{d}\left(0,h(u_{1})\right)}<H_{\partial
B_{d}\left(0,h(u_{3})\right)}\right]=\frac{\log\left(\frac{u_{2}}{u_{3}}\right)}{\log\left(\frac{u_{1}}{u_{3}}\right)}.$

The next Lemma provides simple bounds which will be adequate to handle points
which are close to the ‘South Pole‘ $v$.

######  Lemma 2.1.

For $L$ large, any $y\in B_{d}^{c}\left(v,h_{k}\right)$ and all $|z|\leq\log
L$,

(2.6)
$\mathbb{P}\left[\sqrt{2\mathcal{T}_{y,L}^{\tau}}\,\geq\rho_{L}L+z\right]\leq
cke^{-2L}Le^{-2z},$

for some $c<\infty$ independent of $1\leq k\leq L-1$.

If $y\in B_{d}\left(v,h_{L-1}\right)$

(2.7)
$\mathbb{P}\left[\sqrt{2\mathcal{T}_{y,L}^{\tau}}\,\geq\rho_{L}L+z\right]\leq
ce^{-2L}L^{2}e^{-2z}.$

Proof:  For $k\leq l-1$, let $\mathcal{T}_{y,l}^{k\to 0}$ be the number of
excursions from $\partial B_{d}\left(y,h_{l-1}\right)$ to $\partial
B_{d}\left(y,h_{l}\right)$ between $H_{\partial B_{d}\left(y,h_{k}\right)}$
and $H_{\partial B_{d}\left(y,h_{0}\right)}$. We first estimate probabilities
involving $\mathcal{T}_{y,l}^{k\to 0}$. Using (2.5), an excursion from
$\partial B_{d}\left(y,h_{k}\right)$ hits $B_{d}\left(y,h_{l-1}\right)$ before
exiting $B_{d}(y,h_{0})$ has probability $k/(l-1)$, and then the probability
to hit $\partial B_{d}\left(y,h_{l}\right)$ before exiting $B_{d}(y,h_{0})$ is
$1-\frac{1}{l}$. Thus, using the strong Markov property,

$\displaystyle\mathbb{P}\left[\mathcal{T}_{y,l}^{k\to 0}\geq n\right]$
$\displaystyle=$ $\displaystyle\frac{k}{l-1}\left(1-\frac{1}{l}\right)^{n}$
$\displaystyle\leq$ $\displaystyle\frac{k}{l-1}e^{-\frac{n}{l}},$

for $n$ large. Since (recall (1.13))

(2.9) $\displaystyle(\rho_{L}L+z)^{2}$ $\displaystyle=$
$\displaystyle(\rho_{L}L)^{2}+2z\rho_{L}L+z^{2}$ $\displaystyle=$
$\displaystyle 4L^{2}-4L\log L+4zL+z^{2}-2z\log L+\log^{2}L,$

it follows that for $L$ large

(2.10) $\mathbb{P}\left[\sqrt{2\mathcal{T}_{y,L}^{k\to
0}}\,\geq\rho_{L}L+z\right]\leq cke^{-2L}Le^{-2z}.$

(2.6) follows since for $y\in B_{d}^{c}\left(v,h_{k}\right)$ we have
$\mathcal{T}_{y,L}^{\tau}\leq\mathcal{T}_{y,L}^{k\to 0}$.

For (2.7) we note that for $y\in B_{d}\left(v,h_{L-1}\right)$ we have
$\mathcal{T}_{y,L}^{\tau}\leq\mathcal{T}_{y,L}^{L-1\to 0}$.

######  Lemma 2.2.

For $L$ large and all $0\leq z\leq\log L$,

(2.11) $\mathbb{P}\left[\sup_{y\in F_{L}\cap B_{d}\left(v,h_{\log
L}\right)}\sqrt{2\mathcal{T}_{y,L}^{\tau}}\,\geq\rho_{L}L+z\right]\leq
ce^{-2z},$

for some $c<\infty$.

Proof: By Lemma 2.1 the probability in (2.11) is bounded by

(2.12) $\displaystyle\sum_{k=\log L}^{\infty}\mathbb{P}\left[\sup_{y\in
F_{L}\cap B_{d}\left(v,h_{k}\right)\cap
B^{c}_{d}\left(v,h_{k+1}\right)}\sqrt{2\mathcal{T}_{y,L}^{\tau}}\,\geq\rho_{L}L+z\right]$
$\displaystyle\leq\sum_{k=\log L}^{\infty}|F_{L}\cap
B_{d}\left(v,h_{k}\right)\cap
B^{c}_{d}\left(v,h_{k+1}\right)|cke^{-2L}Le^{-2z}$ $\displaystyle\leq
cLe^{-2z}\sum_{k=\log L}^{\infty}ke^{-2k}\leq ce^{-2z}.$

Thus we only need deal with $y\in B_{d}^{c}\left(v,h_{\log L}\right)$.
However, Lemma 2.1 would give, for example, that

(2.13) $\mathbb{P}\left[\sup_{y\in F_{L}\cap
B_{d}^{c}\left(v,h_{1}\right)}\sqrt{2\mathcal{T}_{y,L}^{\tau}}\,\geq\rho_{L}L+z\right]\leq
CLe^{-2z},$

which would be disastrous if we let $L\to\infty$. To deal with this we
introduce a barrier.

Let

(2.14) $l_{L}=l\wedge(L-l).$

Fix $z$ and set

(2.15)
$\alpha_{z,+}\left(l\right)=\alpha\left(l,L,z\right)=\rho_{L}l+z+l_{L}^{1/4}.$

Let

(2.16) $k_{y}=\inf\\{k\,|\,y\in B_{d}^{c}\left(v,h_{k}\right)\\},$

and

(2.17) $F^{\ast}_{L}=F_{L}\cap B_{d}^{c}\left(v,h_{\log L}\right).$

Since $\alpha_{z,+}\left(L\right)=\rho_{L}L+z$ and $k_{y}\leq\log L$ for $y\in
F^{\ast}_{L}$, our desired upper bound will follow from the next Lemma.

######  Lemma 2.3.

There exists $z_{0}>0$ such that for all $z_{0}\leq z\leq\log L$ and all $L$
large

$\displaystyle\mathbb{P}\left[\exists y\in
F^{\ast}_{L},l\in\left\\{k_{y}+1,\ldots,L\right\\}\mbox{ s.t.
}\mathcal{T}_{y,l}^{\tau}\geq\alpha_{z,+}^{2}\left(l\right)/2\right]$ (2.18)
$\displaystyle\hskip 216.81pt\leq cze^{-2z}.$

Although this formulation looks more complicated and demanding than our
desired upper bound, it will allow us to proceed level by level and to
eventually use a barrier estimate. The next Lemma will be used in our proof.

######  Lemma 2.4.

For $L$ large, any $y\in B_{d}^{c}\left(v,h_{k}\right)$ and all $0\leq
z\leq\log L$,

(2.19)
$\mathbb{P}\left[\mathcal{T}_{y,l}^{\tau}\geq\alpha_{z,+}^{2}\left(l\right)/2\right]\leq
ckle^{-2l}e^{-2(z+l_{L}^{1/4})-(z+l_{L}^{1/4})^{2}/2l},$

for some $c<\infty$ independent of $k\geq 1$ and
$l\in\left\\{k+1,\ldots,L\right\\}$.

Proof: As in (2)

(2.20)
$\mathbb{P}\left[\mathcal{T}_{y,l}^{\tau}\geq\alpha_{z,+}^{2}\left(l\right)/2\right]\leq
c\frac{k}{l-1}e^{-\frac{\alpha_{z,+}^{2}\left(l\right)}{2l}}$

and

$\displaystyle\alpha_{z,+}^{2}\left(l\right)=l^{2}\rho^{2}_{L}+2(z+l_{L}^{1/4})l\rho_{L}+(z+l_{L}^{1/4})^{2}$
$\displaystyle=l^{2}\left(4-4{\log L\over L}+{\log^{2}L\over
L^{2}}\right)+2(z+l_{L}^{1/4})l\left(2-{\log L\over L}\right)$ (2.21)
$\displaystyle\hskip 216.81pt+(z+l_{L}^{1/4})^{2}$

Hence

(2.22) $\displaystyle\alpha_{z,+}^{2}\left(l\right)/2l$ $\displaystyle=$
$\displaystyle\left(2l-2\frac{l}{L}\log
L\right)+2(z+l_{L}^{1/4})+(z+l_{L}^{1/4})^{2}/2l+o_{L}(1)$
$\displaystyle\geq\left(2l-2\log
l\right)+2(z+l_{L}^{1/4})+(z+l_{L}^{1/4})^{2}/2l+o_{L}(1),$

using the concavity of the logarithm. Our result follows.

###  2.1. Proof of (2.18) for $l$ not too large

######  Proposition 2.5.

There exists $z_{0}>0$ such that for all $z_{0}\leq z\leq\log L$ and all $L$
large

$\displaystyle\mathbb{P}\left[\exists y\in
F^{\ast}_{L},l\in\left\\{k_{y}+1,\ldots,L-\left(4\log
L\right)^{4}\right\\}\right.$ (2.23) $\displaystyle\hskip
180.67499pt\left.\mbox{ s.t.
}\mathcal{T}_{y,l}^{\tau}\geq\alpha_{z,+}^{2}\left(l\right)/2\right]\leq
ce^{-2z}.$

Proof:  We use a packing argument. Let $\phi(l)=e^{.25\,l_{L}^{1/4}}$.
Considering separately the case of $l\leq L/2$ and $L/2<l\leq L-\left(4\log
L\right)^{4}$, we see that for some $m_{0}$

(2.24) $l_{L}^{1/4}\geq 4\log l,\hskip 14.45377ptm_{0}\leq l\leq L-\left(4\log
L\right)^{4},$

so that

(2.25) ${l\over\phi(l)}={l\over e^{.25\,l_{L}^{1/4}}}\leq 1,\hskip
14.45377ptm_{0}\leq l\leq L-\left(4\log L\right)^{4}.$

We define modified radii by

(2.26) $r_{l-1}^{-}=\left(1-\frac{1}{\phi(l-1)}\right)r_{l-1}\mbox{ and
}r_{l}^{+}=\left(1+\frac{1}{\phi(l)}\right)r_{l}\mbox{ for }l\geq 1.$

Note that

(2.27)
$h(r_{l+\log\phi(l)})\overset{\eqref{dr.1c}}{\leq}r_{l+\log\phi(l)}\overset{\eqref{dr.1}}{=}\frac{r_{l}}{\phi(l)}=\frac{r_{l-1}}{e\,\phi(l)}\leq\frac{r_{l-1}}{\phi(l-1)}.$

Using this and (2.3) we have for $\phi(l)$ large enough

(2.28) $h(r_{l-1}^{-})+{1\over 10^{3}}h(r_{l+\log\phi(l)})\leq
h(r_{l-1})\mbox{ and }h(r_{l})+{1\over 10^{3}}h(r_{l+\log\phi(l)})\leq
h(r_{l}^{+}).$

For each $y\in{\bf S}^{2}$ let $y_{l}$ denote the point in $F_{l}$ closest to
$y$ (breaking ties in some arbitrary way). By the definition of
$F_{l+\log\phi(l)}$ we have

(2.29) $\hskip 14.45377ptd\left(y,y_{l+\log\phi(l)}\right)\leq{1\over
10^{3}}h(r_{l+\log\phi(l)}),$

so that using (2.28) we see that for all $y\in{\bf S}^{2}$

(2.30) $B_{d}\left(y,h_{l}\right)\subset
B_{d}\left(y_{l+\log\phi(l)},h(r_{l}^{+})\right)\subset
B_{d}\left(y_{l+\log\phi(l)},h(r_{l-1}^{-})\right)\subset
B_{d}\left(y,h_{l-1}\right).$

Now for $k\leq l-1$ set

(2.31) $r_{k,l}^{-}=\left(1-\frac{1}{\phi(l)}\right)r_{k}\mbox{ and
}r_{0,l}^{+}=\left(1+\frac{1}{\phi(l)}\right)r_{0}\mbox{ for }l\geq 0.$

As in the proof of (2.28) we have

$h(r_{k,l}^{-})+{1\over 10^{3}}h(r_{l+\log\phi(l)})\leq h(r_{k})\mbox{ and
}h(r_{0})+{1\over 10^{3}}h(r_{l+\log\phi(l)})\leq h(r_{0,l}^{+}),$

so that (2.29) also implies that

(2.32) $B_{d}\left(y_{l+\log\phi(l)},h(r_{k,l}^{-})\right)\subset
B_{d}\left(y,h(r_{k})\right)\subset B_{d}\left(y,h(r_{0})\right)\subset
B_{d}\left(y_{l+\log\phi(l)},h(r_{0,l}^{+})\right).$

For each $y\in F_{l+\log\phi(l)}$ let $\widetilde{\mathcal{T}}_{y,l}^{k\to 0}$
be the number of excursions from $\partial B_{d}\left(y,h(r_{l-1}^{-})\right)$
to $\partial B_{d}\left(y,h(r_{l}^{+})\right)$ prior to the first excursion
from $\partial B_{d}\left(y,h(r_{k,l}^{-})\right)$ to $\partial
B_{d}\left(y,h(r_{0,l}^{+})\right)$. Then define

$\widetilde{\mathcal{T}}_{y,l}^{k\to
0}=\widetilde{\mathcal{T}}_{y_{l+\log\phi(l)},l}^{k\to 0},\mbox{\,\ for
}y\in{\bf S}^{2}\backslash F_{l+\log\phi(l)}\mbox{ for all }l\geq k+1.$

It follows from (2.30) and (2.32) that
$\widetilde{\mathcal{T}}_{y,l}^{k_{y}\to 0}\geq\mathcal{T}_{y,l}^{k_{y}\to
0}\geq\mathcal{T}_{y,l}^{\tau}$ for all $l\geq k_{y}+1$. Thus

######  Lemma 2.6.

For all $y\in{\bf S}^{2},l\geq k_{y}+1$ we have that

(2.33) $\widetilde{\mathcal{T}}_{y,l}^{k_{y}\to
0}\geq\mathcal{T}_{y,l}^{\tau}.$

Because of this the probability in (2.23) is bounded above by

$\displaystyle{\displaystyle\sum_{k=1}^{\log
L}}\,\,\,{\displaystyle\sum_{l=k+1}^{L-\left(4\log
L\right)^{4}}}{\displaystyle\sum_{y\in B_{d}\left(v,h(r_{k-1})\right)\cap
F_{l+\log\phi(l)}}}\mathbb{P}\left[\widetilde{\mathcal{T}}_{y,l}^{k\to
0}\geq\alpha_{z,+}^{2}\left(l\right)/2\right]$ $\displaystyle=$
$\displaystyle{\displaystyle\sum_{k=1}^{\log
L}}\,\,\,{\displaystyle\sum_{l=k+1}^{L-\left(4\log
L\right)^{4}}}{\displaystyle\left|B_{d}\left(v,h(r_{k-1})\right)\cap
F_{l+\log\phi(l)}\right|}\mathbb{P}\left[\widetilde{\mathcal{T}}_{y,l}^{k\to
0}\geq\alpha_{z,+}^{2}\left(l\right)/2\right]$ $\displaystyle\leq$
$\displaystyle c{\displaystyle\sum_{k=1}^{\log
L}}\,\,\,{\displaystyle\sum_{l=k+1}^{L-\left(4\log
L\right)^{4}}}e^{.5\,l_{L}^{1/4}}e^{2(l-k)}\mathbb{P}\left[\widetilde{\mathcal{T}}_{y,l}^{k\to
0}\geq\alpha_{z,+}^{2}\left(l\right)/2\right],$

for some arbitrary $y\in F_{l+\log\phi(l)}$. We show below that for all $k\leq
l$

(2.35) $\mathbb{P}\left[\widetilde{\mathcal{T}}_{y,l}^{k\to
0}\geq\alpha_{z,+}^{2}(l)/2\right]\leq ce^{-2l-l_{L}^{1/4}}e^{-2z},$

and since

${\displaystyle\sum_{k=1}^{\infty}}\,\,\,{\displaystyle\sum_{l=1}^{L}}e^{.5\,l_{L}^{1/4}}e^{2(l-k)}e^{-2l-l_{L}^{1/4}}e^{-2z}\leq
ce^{-2z},$

this will complete the proof of (2.23).

We now turn to the proof of (2.35). Let

(2.36) $\displaystyle p_{l}$ $\displaystyle=$
$\displaystyle\frac{\log\left(r_{l-1}^{-}/r_{0,l}^{+}\right)}{\log\left(r_{l}^{+}/r_{0,l}^{+}\right)}=\frac{\log\left(\left(1-\frac{1}{\phi(l-1)}\right)\left(1+\frac{1}{\phi(l)}\right)^{-1}e^{-(l-1)}\right)}{\log\left(e^{-l}\right)}$
$\displaystyle=$
$\displaystyle\frac{l-1+2/\phi(l)+O(\phi(l)^{-2})}{l}=1-\frac{1-2/\phi(l)+O(\phi(l)^{-2})}{l},$

and

$\displaystyle q_{l}$ $\displaystyle=$
$\displaystyle\frac{\log\left(r_{k,l}^{-}/r_{0,l}^{+}\right)}{\log\left(r_{l-1}^{-}/r_{0,l}^{+}\right)}=\frac{\log\left(\left(1-\frac{1}{\phi(l)}\right)\left(1+\frac{1}{\phi(l)}\right)^{-1}e^{-k}\right)}{\log\left(\left(1-\frac{1}{\phi(l-1)}\right)\left(1+\frac{1}{\phi(l)}\right)^{-1}e^{-(l-1)}\right)}$
$\displaystyle=$
$\displaystyle\frac{k+2/\phi(l)+O(\phi(l)^{-2})}{l-1+2/\phi(l)+O(\phi(l)^{-2})}=\frac{k+O(\phi(l)^{-1})}{l-1}.$

Using the fact that $p_{l}<1$ together with (2.36) we can write

(2.37) $p_{l}=1-\frac{1-b_{l}/\phi(l)}{l}$

with $1-b_{l}/\phi(l)>0$. In addition, using (2.25) and possibly increasing
$m_{0}$,

(2.38) $\frac{lb_{l}}{\phi(l)}\leq 3,\qquad l\geq m_{0}.$

Since $q_{l}$ is the probability for an excursion from $\partial
B_{d}\left(y,h(r_{k,l}^{-})\right)$ to hit
$B_{d}\left(y,h(r_{l-1}^{-})\right)$ before $\partial
B_{d}\left(y,h(r_{0,l}^{+})\right)$, and $p_{l}$ is the probability for an
excursion from $B_{d}\left(y,h(r_{l-1}^{-})\right)$ to hit $\partial
B_{d}\left(y,h(r_{l}^{+})\right)$ before $\partial
B_{d}\left(y,h(r_{0,l}^{+})\right)$, we see that as in (2)

(2.39) $\mathbb{P}\left[\widetilde{\mathcal{T}}_{y,l}^{k\to
0}\geq\alpha_{z,+}^{2}(l)/2\right]\leq\frac{ck}{l}e^{-\frac{\alpha_{z,+}^{2}(l)}{2l}\left(1-\frac{b_{l}}{\phi(l)}\right)}.$

By (2.22) we have that

(2.40) $\displaystyle{\alpha_{z,+}^{2}(l)\over 2l}$ $\displaystyle\geq$
$\displaystyle\left(2l-2\log l\right)+2(z+l_{L}^{1/4})+z^{2}/2l+o_{L}(1),$

so that, for $k\leq l$

(2.41) $\mathbb{P}\left[\widetilde{\mathcal{T}}_{y,l}^{k\to
0}\geq\alpha_{z,+}^{2}(l)/2\right]\leq
cl^{2}e^{-\left(2l+2(z+l_{L}^{1/4})+z^{2}/2l\right)\left(1-\frac{b_{l}}{\phi(l)}\right)}.$

We claim that

${z^{2}\over 2l}\left(1-\frac{b_{l}}{\phi(l)}\right)-2zb_{l}/\phi(l)>0,$

that is

$z\left(1-\frac{b_{l}}{\phi(l)}\right)>4lb_{l}/\phi(l)$

for $z\geq z_{0}$ sufficiently large. For $l>m_{0}$, this follows from (2.38),
and for $l\leq m_{0}$ we can just increase $z$ further. Thus for such $z$

(2.42) $\mathbb{P}\left[\widetilde{\mathcal{T}}_{y,l}^{k\to
0}\geq\alpha_{z,+}^{2}(l)/2\right]\leq
cl^{2}e^{-\left(2l+2l_{L}^{1/4}\right)\left(1-\frac{b_{l}}{\phi(l)}\right)}e^{-2z}.$

For $k\leq l\leq m_{0}$ this already proves (2.35) with $c$ sufficiently
large. For $l>m_{0}$, using (2.38) again we now have

(2.43) $\mathbb{P}\left[\widetilde{\mathcal{T}}_{y,l}^{k\to
0}\geq\alpha_{z,+}^{2}(l)/2\right]\leq
cl^{2}e^{-\left(2l+2l_{L}^{1/4}\right)}e^{-2z}.$

and (2.24) completes the proof of (2.35).

###  2.2. Proof of (2.18) for $l$ very large

It suffices to show that for some small but fixed constant $\tilde{c}$ to be
chosen later we have that for all $L$ sufficiently large and all $z_{0}\leq
z\leq\log L$

(2.44) $\mathbb{P}\left[\begin{array}[]{c}\exists y\in F^{\ast}_{L}\cap
B_{d}\left(0,\tilde{c}h_{0}\right)\mbox{ and }k_{y}+1\leq l\leq L\\\ \mbox{
such that
}\sqrt{2\mathcal{T}_{y,l}^{\tau}}\geq\alpha_{z,+}(l)\end{array}\right]\leq
cze^{-2z}.$

Here $0$, the center of $B_{d}\left(0,\tilde{c}h_{0}\right)$, is used to
denote an arbitrary point in ${\bf S}^{2}$.

Now consider

$\mathcal{G}_{l}=\left\\{\sqrt{2\mathcal{T}_{y,l^{\prime}}^{\tau}}\leq\alpha_{z,+}(l^{\prime})\mbox{
for all }l^{\prime}=k_{y}+1,\ldots,l\mbox{ and }\forall y\in F^{\ast}_{L}\cap
B_{d}\left(0,\tilde{c}h_{0}\right)\right\\}.$

Let $L^{\prime}=L-\left(4\log L\right)^{4}$. With

(2.45) $\mathcal{H}_{l}=\left\\{\exists y\in F^{\ast}_{L}\cap
B_{d}\left(0,\tilde{c}h_{0}\right)\mbox{ s.t.
}\sqrt{2\mathcal{T}_{y,l}^{\tau}}\geq\alpha_{z,+}(l),\,k_{y}<l\right\\},$

we will prove that for all $l>L^{\prime}$

(2.46) $\mathbb{P}\left[\mathcal{H}_{l}\cap\mathcal{G}_{l-2}\right]\leq
cze^{-l_{L}^{1/4}-2z},$

so that we have

$\displaystyle\mathbb{P}\left[\mathcal{G}_{L}^{c}\right]$
$\displaystyle\leq\sum_{l=L^{\prime}+1}^{L}\mathbb{P}\left[\mathcal{G}_{l}^{c}\cap\mathcal{G}_{l-1}\right]+\mathbb{P}\left[\mathcal{G}^{c}_{L^{\prime}}\right]$
$\displaystyle\leq\sum_{l=L^{\prime}+1}^{L}\mathbb{P}\left[\mathcal{H}_{l}\cap\mathcal{G}_{l-2}\right]+\mathbb{P}\left[\mathcal{G}^{c}_{L^{\prime}}\right]$
$\displaystyle\leq\sum_{l=L^{\prime}+1}^{L}cze^{-l_{L}^{1/4}-2z}+\mathbb{P}\left[\mathcal{G}^{c}_{L^{\prime}}\right]\leq
cze^{-2z},$

by Proposition 2.5, which will prove (2.44).

Setting $F^{m}_{L}=F_{L}\cap B_{d}^{c}\left(v,h_{m}\right)\cap
B_{d}\left(v,h_{m-1}\right)$, so that $k_{y}=m$ for $y\in F^{m}_{L}$, and for
any $l>m$

(2.47) $\mathcal{H}_{m,l}=\left\\{\exists y\in F^{m}_{L}\cap
B_{d}\left(0,\tilde{c}h_{0}\right)\mbox{ s.t.
}\sqrt{2\mathcal{T}_{y,l}^{\tau}}\geq\alpha_{z,+}(l)\right\\},$

we will prove that for all $l>L^{\prime}$

(2.48) $\sum_{m=1}^{\log
L}\mathbb{P}\left[\mathcal{H}_{m,l}\cap\mathcal{G}_{l-2}\right]\leq
cze^{-l_{L}^{1/4}-2z},$

which gives (2.46).

For any $l>m$ let

(2.49) $\mathcal{B}_{m,l}=\left\\{\exists x\in F^{m}_{L}\cap
B_{d}\left(u_{m},\tilde{c}h_{l}\right)\mbox{ s.t.
}\sqrt{2\mathcal{T}_{x,l}^{\tau}}\geq\alpha_{z,+}(l)\right\\},$

where we can assume that $u_{m}\in F^{m}_{L}$. By a union bound,
$\mathbb{P}\left[\mathcal{H}_{m,l}\cap\mathcal{G}_{l-2}\right]$ is bounded
above by

(2.50)
$ce^{2(l-m)}\times\mathbb{P}\left[\mathcal{B}_{m,l}\cap\mathcal{G}_{l-2}\right].$

Hence it suffices to show that

(2.51) $\sum_{m=1}^{\log
L}e^{-2m}\mathbb{P}\left[\mathcal{B}_{m,l}\cap\mathcal{G}_{l-2}\right]\leq
cze^{-2l-l_{L}^{1/4}-2z}.$

Since $\mathcal{G}_{l-2}\subset\mathcal{C}_{m,l}$, where

(2.52)
$\mathcal{C}_{m,l}=\left\\{\sqrt{2\mathcal{T}_{u_{m},l^{\prime}}^{\tau}}\leq\alpha_{z,+}(l^{\prime})\mbox{
for all }l^{\prime}=m+1,\ldots,l-2\right\\},$

it suffices to show that

(2.53) $\sum_{m=1}^{\log
L}e^{-2m}\mathbb{P}\left[\mathcal{B}_{m,l}\cap\mathcal{C}_{m,l}\right]\leq
cze^{-2l-l_{L}^{1/4}-2z}.$

We show in the next Section that for all $l\geq L-\left(4\log L\right)^{4}$

(2.54)
$\mathbb{P}\left[\mathcal{B}_{m,l}\cap\left\\{\sqrt{2\mathcal{T}_{u_{m},l-2}^{\tau}}\leq{1\over
2}\alpha_{z,+}(l-2)\right\\}\right]\leq ce^{-c^{\prime}L^{2}}.$

It follows from this that with

(2.55) $\mathcal{D}_{m,l}(j)=\left\\{\sqrt{2\mathcal{T}_{u_{m},l-2}^{\tau}}\in
I_{\alpha_{z,+}(l-2)+j}\right\\},$

it suffices to show that

(2.56) $\displaystyle\sum_{m=1}^{\log L}e^{-2m}\sum_{j=0}^{{1\over
2}\alpha_{z,+}(l-2)}\mathbb{P}\left[\mathcal{B}_{m,l}\cap\mathcal{C}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]\leq
cze^{-2l-l^{1/4}_{L}-2z}.$

We also show in the next Section that we can find a fixed $j_{0}$ such that
for all $j_{0}\leq j\leq{1\over 2}\alpha_{+,z}(l-2)$, uniformly in $1\leq
m\leq\log L$ and $z_{0}\leq z\leq\log L$, for any
$\widetilde{\mathcal{C}}_{m,l}\in{\mathcal{F}}\left(\mathcal{T}_{u_{m},k}^{\tau},k=1,\ldots,l-2\right)$

(2.57)
$\mathbb{P}\left[\mathcal{B}_{m,l}\,\big{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]\leq
Ce^{-4j},$

by taking $\tilde{c}>0$ sufficiently small.

It follows from the barrier estimate (9.5) that for $0\leq j\leq{1\over
2}\alpha_{z,+}(l-2)$,

(2.58) $\mathbb{P}\left[\mathcal{C}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]\leq
ce^{-2l-2z-2l^{1/4}_{L}+2j}\times
m^{2}\left(1+z+m+l^{1/4}_{L}\right)\left(1+j\right).$

Combining the last 2 displays we can bound the left hand side of (2.56) by

(2.59) $\displaystyle\hskip
21.68121ptCze^{-2l-2z-l^{1/4}_{L}}\sum_{j=0}^{\infty}e^{-4j1_{\\{j\geq
j_{0}\\}}+2j}(1+j),$

which proves (2.56).

###  2.3. Proof of the conditional barrier estimate (2.57) and (2.54)

We first prove that for some $j_{0}$ fixed and all $j_{0}\leq j\leq{1\over
2}\alpha_{+,z}(l)$,

(2.60)
$\mathbb{P}\left[\mathcal{B}_{m,l}\,\big{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]\leq
Ce^{-4j}.$

Proof:  For each $\gamma\in(0,1]$ and $y$, let
$\mathcal{T}_{y,\widetilde{r}_{l}}^{\tau}$ be the number of excursions from
$\partial B\left(y,h(\widetilde{r}_{l-1})\right)$ to $\partial
B\left(y,h(\widetilde{r}_{l})\right)$ prior to $\tau$, where

(2.61) $\widetilde{r}_{l-1}=r_{l-1}\left(1-\gamma\right),\hskip
14.45377pt\widetilde{r}_{l}=r_{l}\left(1+\gamma\right).$

Note that

(2.62)
$\mathcal{T}_{y^{\prime},\widetilde{r}_{l}}^{\tau}\geq\mathcal{T}_{y,l}^{\tau}\mbox{
for all }y^{\prime}\mbox{ such that }d\left(y,y^{\prime}\right)\leq{\gamma
r_{l}\over 2},$

since then

$B_{d}\left(y,h_{l-1}\right)\supset
B_{d}\left(y^{\prime},h(r_{l-1}\left(1-\gamma\right))\right)\supset
B_{d}\left(y^{\prime},h(r_{l}\left(1+\gamma\right))\right)\supset
B_{d}\left(y,h_{l}\right).$

Let

(2.63) $\mathcal{B}_{m,l}^{\gamma,k}=\left\\{\exists y\in F^{m}_{k}\cap
B_{d}\left(u_{m},\tilde{c}\,h_{l}\right)\mbox{ such that
}\sqrt{2\mathcal{T}_{y,\widetilde{r}_{l}}^{\tau}}\geq\alpha_{+,z}(l)\right\\}.$

Note $F^{m}_{k}$ not $F^{m}_{L}$. From now on we fix

(2.64) $\gamma=\frac{1}{\alpha_{+,z}(l)-j},\hskip 14.45377pt\mbox{and }\hskip
14.45377ptk=\log\left(2(\alpha_{+,z}(l)-j)\right)+l.$

We will show that with these values

(2.65)
$\mathbb{P}\left[\mathcal{B}_{m,l}^{\gamma,k}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]\leq
Ce^{-4j}.$

Using (2.62) this will imply (2.60), since for each $y\in F^{m}_{L}\cap
B_{d}\left(u_{m},\tilde{c}h_{l}\right)$ there exists a representative
$y^{\prime}\in F^{m}_{k}\cap B_{d}\left(u_{m},\tilde{c}h_{l}\right)$ such that

$d\left(y,y^{\prime}\right)\leq
r_{k}=\frac{1}{2(\alpha_{+,z}(l)-j)}\,\,r_{l}={\gamma r_{l}\over 2}.$

To show (2.65), we first show that for some $c_{3}>0$

(2.66)
$\mathbb{P}\left[\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}\geq\alpha_{+,z}(l)-\frac{j}{2}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]\leq
c^{\prime}e^{-c_{3}j^{2}}.$

Let $\mathcal{T}_{y,\widetilde{r}_{l}}^{u_{m},r_{l-2},n}$ be the number of
excursions from $\partial B\left(y,h(\widetilde{r}_{l-1})\right)$ to $\partial
B\left(y,h(\widetilde{r}_{l})\right)$ during the first $n$ excursions from
$\partial B\left(u_{m},h_{l-2}\right)$ to $\partial
B\left(u_{m},h_{l-3}\right)$. Using the Markov property we have that

$\displaystyle\mathbb{P}\left[\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}\geq\alpha_{+,z}(l)-\frac{j}{2}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]$
$\displaystyle=$
$\displaystyle\mathbb{P}\left[\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}\geq\alpha_{+,z}(l)-\frac{j}{2}\,\bigg{|}\,\mathcal{D}_{m,l}(-j)\right]$
$\displaystyle=$
$\displaystyle\mathbb{P}\left[\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}\geq\alpha_{+,z}(l)-\frac{j}{2}\,\bigg{|}\,\sqrt{2\mathcal{T}_{u_{m},l-2}^{\tau}}\in
I_{\alpha_{z,+}(l)-j}\right]$ $\displaystyle=$
$\displaystyle\mathbb{P}\left[\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{u_{m},r_{l-2},\mathcal{T}_{u_{m},l-2}^{\tau}}}\geq\alpha_{+,z}(l)-\frac{j}{2}\,\bigg{|}\,\sqrt{2\mathcal{T}_{u_{m},l-2}^{\tau}}\in
I_{\alpha_{z,+}(l)-j}\right].$

To prove (2.66) it suffices to show that show that uniformly for $s\in
I_{\alpha_{z,+}(l)-j}$

(2.68)
$\mathbb{P}\left[\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{u_{m},r_{l-2},s^{2}/2}}\geq\alpha_{+,z}(l)-\frac{j}{2}\right]\leq
c^{\prime}e^{-c_{3}j^{2}}.$

To see this, let $s=\alpha_{z,+}(l)-j+\zeta$, where $0\leq\zeta\leq 1$. It
follows from [3, Lemma 4.6], that

(2.69)
$\mathbb{P}\left[\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{u_{m},r_{l-2},s^{2}/2}}\geq\alpha_{+,z}(l)-\frac{j}{2}\right]\leq
e^{-\left(\sqrt{q}(\alpha_{+,z}(l)-j+\zeta)-\sqrt{p}(\alpha_{+,z}(l)-\frac{j}{2})\right)^{2}/2}$

where

(2.70) $\displaystyle q$ $\displaystyle=$
$\displaystyle\mathbb{P}^{u\in\partial
B_{d}\left(u_{m},h_{l-2}\right)}\left[H_{\partial
B_{d}\left(u_{m},h(r_{l}\left(1+\gamma\right))\right)}<H_{\partial
B_{d}\left(u_{m},h_{l-3}\right)}\right]$ $\displaystyle=$
$\displaystyle\frac{\log r_{l-3}-\log r_{l-2}}{\log
r_{l-3}-\log(r_{l}\left(1+\gamma\right))}=\frac{1}{3+O\left(\gamma\right)},$

and

(2.71) $\displaystyle p$ $\displaystyle=$
$\displaystyle\mathbb{P}^{u\in\partial
B_{d}\left(u_{m},h(r_{l-1}\left(1-\gamma\right))\right)}\left[H_{\partial
B_{d}\left(u_{m},h_{l-3}\right)}<H_{\partial
B_{d}\left(u_{m},h(r_{l}\left(1+\gamma\right))\right)}\right]$
$\displaystyle=$
$\displaystyle\frac{\log(r_{l-1}\left(1-\gamma\right))-\log(r_{l}\left(1+\gamma\right))}{\log
r_{l-3}-\log(r_{l}\left(1+\gamma\right))}=\frac{1+O\left(\gamma\right)}{3+O\left(\gamma\right)}.$

To apply [3, Lemma 4.6] it suffices to show that

$\alpha_{z,+}(l)-\frac{j}{2}\geq(\alpha_{+,z}(l)-j+1)\sqrt{q/p},$

and the right hand side

$=(\alpha_{+,z}(l)-j+1)(1+O\left(\gamma\right))=(\alpha_{+,z}(l)-j)+O\left(1\right),$

since $\gamma\left(\alpha_{+,z}(l)-j\right)=1$. Thus we can use (2.69) for all
$j\geq c_{3}$ for some $c_{3}<\infty$. For such $j$ we therefore have

$\mathbb{P}\left[\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{u_{m},r_{l-2},s^{2}/2}}\geq\alpha_{+,z}(l)-\frac{j}{2}\right]\leq
ce^{-\frac{1}{6}\left(-\frac{j}{2}+\zeta+O\left(\gamma\left(\alpha_{+,z}(l)-\frac{j}{2}\right)\right)\right)^{2}},$

and since $\gamma\left(\alpha_{+,z}(l)-j\right)=1$ and $j\leq{1\over
2}\alpha_{+,z}(l)$ so that $j\leq\left(\alpha_{+,z}(l)-j\right)$, it follows
that

$\gamma\left(\alpha_{+,z}(l)-\frac{j}{2}\right)=\gamma\left(\alpha_{+,z}(l)-j\right)+\gamma\frac{j}{2}\leq
2,$

so that we obtain (2.68) for all $j\geq c_{3}$. By enlarging $c^{\prime}$ we
then have (2.68) for all $j$.

We now bound

$\displaystyle\mathbb{P}\left[\mathcal{B}_{m,l}^{\gamma,k}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]\leq\mathbb{P}\left[\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}\geq\alpha_{+,z}(l)-\frac{j}{2}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]$
(2.72) $\displaystyle\hskip
72.26999pt+\mathbb{P}\left[\mathcal{B}_{m,l}^{\gamma,k}\cap\Big{\\{}\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}<\alpha_{+,z}(l)-\frac{j}{2}\Big{\\}}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right].$

Because of the bound (2.66), to prove (2.65) it suffices to show that

(2.73)
$\mathbb{P}\left[\mathcal{B}_{m,l}^{\gamma,k}\cap\Big{\\{}\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}<\alpha_{+,z}(l)-\frac{j}{2}\Big{\\}}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]\leq
Ce^{-4j}.$

Assign to each $y\in F^{m}_{l+i}\cap B_{d}\left(u_{m},\tilde{c}h_{l}\right)$ a
unique “parent” $\tilde{y}\in F^{m}_{l+i-1}\cap
B_{d}\left(u_{m},\tilde{c}h_{l}\right)$ such that
$d\left(\tilde{y},y\right)\leq r_{l+i}$. In particular, for $i=1$ we set
$\tilde{y}=u_{m}$. Let $q=q(\tilde{y},y)=d\left(\tilde{y},y\right)/r_{l}$ and
set

(2.74) $A_{i}=\Big{\\{}\underset{y\in F^{m}_{l+i}\cap
B_{d}\left(u_{m},\tilde{c}h_{l}\right)}{\sup}\left|\mathcal{T}_{y,\widetilde{r}_{l}}^{\tau}-\mathcal{T}_{\tilde{y},\widetilde{r}_{l}}^{\tau}\right|\leq
d_{0}ji\left(\alpha_{+,z}(l)-j\right)\sqrt{q}\Big{\\}},$

where $d_{0}$ will be chosen later, but small enough that $d_{0}\sum_{i\geq
1}ie^{-i/2}\leq\frac{1}{8}$. We now show that

(2.75)
$\displaystyle\bigcap_{i=1}^{k-l}A_{i}\cap\Big{\\{}\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}<\alpha_{+,z}(l)-\frac{j}{2}\Big{\\}}$
$\displaystyle\subseteq\Big{\\{}\sqrt{2\mathcal{T}_{y,\widetilde{r}_{l}}^{\tau}}<\alpha_{+,z}(l),\,\forall
y\in F^{m}_{k}\cap B_{d}\left(u_{m},\tilde{c}h_{l}\right)\Big{\\}}.$

For this, we use $q=d\left(\tilde{y},y\right)/r_{l}\leq r_{l+i}/r_{l}=e^{-i}$
for $y\in F^{m}_{l+i}$ to see that for any trajectory in the left hand side of
(2.75) and all $y\in F^{m}_{k}\cap B_{d}\left(u_{m},\tilde{c}h(r_{l})\right)$

$\mathcal{T}_{y,\widetilde{r}_{l}}^{\tau}\leq\left(\alpha_{+,z}(l)-\frac{j}{2}\right)^{2}/2+j\left(\alpha_{+,z}(l)-j\right)d_{0}\sum_{i\geq
1}ie^{-i/2},$

which, since $d_{0}\sum_{i\geq 1}ie^{-i/2}\leq\frac{1}{8}$, implies that

$\begin{array}[]{ccl}\mathcal{T}_{y,\widetilde{r}_{l}}^{\tau}&\leq&\left(\alpha_{+,z}(l)-\frac{j}{2}\right)^{2}/2+\frac{1}{8}j\left(\alpha_{+,z}(l)-j\right)\\\
&=&\alpha^{2}_{+,z}(l)/2-\alpha_{+,z}(l)j/2+(\frac{j}{2})^{2}/2+\frac{1}{8}\alpha_{+,z}(l)j-\frac{j^{2}}{8}\\\
&<&\alpha^{2}_{+,z}(l)/2.\end{array}$

This establishes (2.75) and taking complements we see that

(2.76) $\displaystyle\mathcal{B}_{m,l}^{\gamma,k}=\Big{\\{}\exists y\in
F^{m}_{k}\cap B_{d}\left(u_{m},\tilde{c}h_{l}\right)\mbox{ such that
}\sqrt{2\mathcal{T}_{y,\widetilde{r}_{l}}^{\tau}}\geq\alpha_{+,z}(l)\Big{\\}}$
$\displaystyle\hskip
108.405pt\subseteq\bigcup_{i=1}^{k-l}A^{c}_{i}\cup\Big{\\{}\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}\geq\alpha_{+,z}(l)-\frac{j}{2}\Big{\\}}.$

It follows that

(2.77)
$\displaystyle\mathcal{B}_{m,l}^{\gamma,k}\cap\Big{\\{}\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}<\alpha_{+,z}(l)-\frac{j}{2}\Big{\\}}\subseteq\bigcup_{i=1}^{k-l}A^{c}_{i}.$

We can thus bound
$\mathbb{P}\left[\mathcal{B}_{m,l}^{\gamma,k}\cap\Big{\\{}\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}<\alpha_{+,z}(l)-\frac{j}{2}\Big{\\}}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right]$
by

(2.78) $\sum_{i=1}^{k-l}\mathbb{P}\left[\underset{y\in F^{m}_{l+i}\cap
B_{d}\left(u_{m},\tilde{c}h_{l}\right)}{\sup}\left|\mathcal{T}_{y,\widetilde{r}_{l}}^{\tau}-\mathcal{T}_{\tilde{y},\widetilde{r}_{l}}^{\tau}\right|\geq
d_{0}ji\left(\alpha_{+,z}(l)-j\right)\sqrt{q}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right].$

Since $\left|F^{m}_{l+i}\cap B_{d}\left(u_{m},\tilde{c}h_{l}\right)\right|\leq
ce^{2i}$, a union bound gives that (2.78) is at most

$\displaystyle\hskip 14.45377ptc\sum_{i=1}^{k-l}e^{2i}\underset{y\in
F^{m}_{l+i}\cap B_{d}\left(u_{m},\tilde{c}h_{l}\right)}{\sup}$ (2.79)
$\displaystyle\hskip
72.26999pt\mathbb{P}\left[\left|\mathcal{T}_{y,\widetilde{r}_{l}}^{\tau}-\mathcal{T}_{\tilde{y},\widetilde{r}_{l}}^{\tau}\right|\geq
d_{0}ji\left(\alpha_{+,z}(l)-j\right)\sqrt{q}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right].$

We can write the last probability as

(2.80)
$\mathbb{P}\left[\left|\mathcal{T}_{y,\widetilde{r}_{l}}^{u_{m},r_{l-2},\mathcal{T}_{u_{m},l-2}^{\tau}}-\mathcal{T}_{\tilde{y},\widetilde{r}_{l}}^{u_{m},r_{l-2},\mathcal{T}_{0,l-2}^{\tau}}\right|\geq
d_{0}ji\left(\alpha_{+,z}(l)-j\right)\sqrt{q}\,\bigg{|}\,\widetilde{\mathcal{C}}_{m,l}\cap\mathcal{D}_{m,l}(-j)\right].$

Using [4, Lemma 5.6] with $\theta=d_{0}ji$ and $n=(\alpha_{+,z}(l)-j)^{2}/2$,
we find that for an appropriate choice of $d_{0},\tilde{c}$, the last
probability is bounded by $ce^{-4(j+i)}$. To apply [4, Lemma 5.6] we must
verify several points.

First, we need to verify that for some small $\bar{c}_{0}$ we have
$\theta\leq\bar{c}_{0}(n-1)$, that is
$d_{0}ji\leq\bar{c}^{\prime}_{0}(\alpha_{+,z}(l)-j)^{2}$. For this it suffices
to note that for $j,l$ in our range
$i/(\alpha_{+,z}(l)-j)\leq(k-l)/(\alpha_{+,z}(l)-j)=(\log
2(\alpha_{+,z}(l)-j))/(\alpha_{+,z}(l)-j)$ goes to $0$ as $L\to\infty$.

Secondly, we need to show that $\theta\leq((n-1)q)^{2}$. Since we have already
seen that $\theta\leq\bar{c}_{0}(n-1)$, it suffices to show that
$(n-1)q^{2}\geq c^{2}_{2}$ for some $c_{2}>0$, or equivalently that
$\sqrt{2n}\,\,q\geq c^{\prime}_{2}>0$. That is,
$(\alpha_{+,z}(l)-j)d\left(\tilde{y},y\right)/r_{l}\geq c^{\prime}_{2}$.
Assume that $d\left(\tilde{y},y\right)\geq c_{3}r_{k}$ for a small $c_{3}>0$,
so that, see (2.64),

$(\alpha_{+,z}(l)-j)d\left(\tilde{y},y\right)/r_{l}\geq
c_{3}(\alpha_{+,z}(l)-j)e^{-\left(k-l\right)}=c_{3}/2.$

With the $F_{l}$ constructed appropriately we can indeed assume that either
$d\left(\tilde{y},y\right)\geq c_{3}r_{k}$ for a small $c_{3}>0$, or that
$y=\tilde{y}$, in which case the corresponding term in the sum in (2.79) is
zero. Also, by taking $\tilde{c}=q_{0}/2$ we will have
$d\left(\tilde{y},y\right)/r_{l}\leq q_{0}$.

Thus we see that (2.79) is at most

(2.81) $c\sum_{i=1}^{k-l}e^{2i}e^{-4(j+i)}\leq Ce^{-4j}.$

This completes the proof of (2.73).

Proof of (2.54):  As in (2.62)

(2.82)
$\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}\geq\mathcal{T}_{y,l}^{\tau}\mbox{
for all }y\mbox{ such that }d\left(y,u_{m}\right)\leq{\gamma r_{l}\over 2},$

where we take $\gamma$ to be some fixed small number. Hence under
$\mathcal{B}_{m,l}$ we have
$\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}\geq\alpha_{z,+}(l)$. The
fact that for all $l\geq L-\left(4\log L\right)^{4}$

(2.83) $\mathbb{P}\left[\sqrt{2\mathcal{T}_{u_{m},l-2}^{\tau}}\leq{1\over
2}\alpha_{z,+}(l-2),\,\,\sqrt{2\mathcal{T}_{u_{m},\widetilde{r}_{l}}^{\tau}}\geq\alpha_{z,+}(l)\right]\leq
ce^{-c^{\prime}L^{2}}$

then follows easily as in the proof of (2.68). (In fact, the proof uses the
same ideas but is much easier).

##  3\. Lower bounds for excursions

In this section we will prove the following.

######  Lemma 3.1.

There exist $0<c_{1},c_{2}<\infty$ such that for all $L$ large and all $0\leq
z\leq\log L$,

(3.1) $\mathbb{P}\left[\sup_{y\in
F_{L}}\sqrt{2\mathcal{T}_{y,L}^{\tau}}\,\geq\rho_{L}L+z\right]\geq
c_{1}{(1+z)e^{-2z}\over(1+z)e^{-2z}+c_{2}}.$

This will immediately give the lower bounds in Theorems 1.6 and 1.5 and hence
complete the proofs of those Theorems.

Note that for any $z_{0}$ it suffices to show that (3.1) holds for all
$z_{0}\leq z\leq\log L$, since by adjusting $c_{1}$ we then get (3.1) for all
$0\leq z\leq\log L$.

Let

(3.2) $\beta_{z}\left(l\right)=\rho_{L}l+z,$

and

(3.3)
$\alpha_{z,-}\left(l\right)=\alpha_{z,-}\left(l,L,z\right)=\rho_{L}l+z-l_{L}^{1/4}.$

For each $k\geq 1$ we define $\mathcal{T}_{y,l}^{k,m}$ be the number of
excursions from $\partial B_{d}\left(y,h_{l-1}\right)$ to $\partial
B_{d}\left(y,h_{l}\right)$ during the first $m$ excursions from $\partial
B_{d}\left(y,h_{k}\right)$ to $\partial B_{d}\left(y,h_{k-1}\right)$. We
abbreviate $\mathcal{T}_{y,l}^{1}=\mathcal{T}_{y,l}^{1,x^{2}}$ with $x$ fixed.

Choose $r_{0}$ in (2.2) sufficiently small that $4h(r_{-1})\leq r^{\ast}$. Let
$\widehat{r}=h_{1}/20$, and with $F^{0}:=B_{d}(v,\widehat{r})$ we set

(3.4) $F^{0}_{L}=F^{0}\cap F_{L},\hskip 14.45377pt\mbox{so that }\hskip
14.45377pt|F^{0}_{L}|\asymp e^{2L}.$

That is, $|F^{0}_{L}|$ does not depend on $r_{0}$.

In this section we show that

######  Lemma 3.2.

There exists a $0<c<\infty$ such that for all $0<r_{0}$ sufficiently small,
$L$ large and all $0\leq z\leq\log L$,

(3.5) $\mathbb{P}\left[\sup_{y\in
F^{0}_{L}}\sqrt{2\mathcal{T}_{y,L}^{1}}\,\geq\rho_{L}L+z\right]\geq{(1+z)e^{-2z}\over(1+z)e^{-2z}+c}.$

Since the probability of $x^{2}$ excursions from $\partial
B_{d}(v,h_{1}-\widehat{r})$ to $\partial B_{d}(v,h_{0}+\widehat{r})$ before
$\tau$ is greater than $0$ and does not depend on $L$, (3.5) will imply Lemma
3.1. We note that the $r_{0}$ used in this Lemma, and hence all $h_{l}$, are
smaller than the corresponding quantities used until now. This is for
notational convenience and, as can easily be seen, does not affect Lemma 3.1
which concerns large $L$. We could have kept the original $r_{0}$ and in place
of $h_{l}$ used $h_{l+k}$ for some fixed $k$, but this would have made the
notation cumbersome.

The proof of Proposition 3.2 uses a modified second moment method and occupies
the rest of this section.

We introduce the events ${\mathcal{I}}_{y,z}$, beginning with a barrier event.
Let

(3.6)
$\widehat{\mathcal{I}}_{y,z}=\left\\{\sqrt{2\mathcal{T}_{y,l}^{1}}\leq\alpha_{z,-}\left(l\right)\mbox{
for }l=1,\ldots,L-1\mbox{ and
}\sqrt{2\mathcal{T}_{y,L}^{1}}\geq\rho_{L}+z\right\\},$

for $y\in F_{L}$. As discussed in [4], we need to augment
$\widehat{\mathcal{I}}_{y,z}$ by information on the angular increments of the
excursions. Instead of keeping track of individual excursions, we track the
empirical measure of the increments, by comparing it in Wasserstein distance
to a reference measure. Recall that the Wasserstein $L^{1}$-distance between
probability measures on ${\bf R}$ is given by

(3.7) $d^{1}_{\mbox{\rm
Wa}}(\mu,\nu)=\inf_{\xi\in{\mathcal{P}}^{2}(\mu,\nu)}\Big{\\{}\int|x-y|\,d\xi(x,y)\Big{\\}},$

where ${\mathcal{P}}^{2}(\mu,\nu)$ denotes the set of probability measures on
${\bf R}\times{\bf R}$ with marginals $\mu,\nu$. If $\mu$ is a probability
measure on ${\bf R}$ with finite support and if $\theta_{i},\,1\leq i\leq n$
denotes a sequence of i.i.d $\mu$-distributed random variables then it follows
from [13, Theorem 2] that for some $c_{0}=c_{0}(\mu)$

(3.8) $\mbox{\rm Prob}\left\\{d^{1}_{\mbox{\rm
Wa}}\left(\frac{1}{n}\sum_{i=1}^{n}\delta_{\theta_{i}},\mu\right)>\frac{c_{0}x}{\sqrt{n}}\right\\}\leq
2e^{-x^{2}}.$

Let $W_{t}$ be Brownian motion in the plane. For each $k$ let $\nu_{k}$ be the
probability measure on $[0,2\pi]$ defined by

(3.9) $\nu_{k}(dx)=P^{(r_{k},0)}\left(\mbox{arg }W_{H_{\partial
B(0,r_{k-1})}}\in\,dx\right),$

where $\mbox{arg }x$ for $x\in{\bf R}^{2}$ is the argument of $x$ measured
from the positive $x$-axis and $P^{w}$ is the law of $W_{\cdot}$ started from
$w$.

Returning to $X_{t}$, our Brownian motion on the sphere, and using isothermal
coordinates, see [4, Section 2], let $0\leq\theta_{k,i}\leq 2\pi$,
$i=1,2,\ldots$ be the angular increments centered at $y$, mod $2\pi$, from
$X_{H^{i}_{\partial B(y,h_{k})}}$ to $X_{H^{i}_{\partial B(y,h_{k-1})}}$, the
endpoints of the $i$’th excursion between $\partial B(y,h_{k})$ and $\partial
B(y,h_{k-1})$. By the Markov property the $\theta_{k,i}$, $i=1,2,\ldots$ are
independent, and using [4, Section 2] we see that each $\theta_{k,i}$ has
distribution $\nu_{k}$. We set, for $n$ a positive integer,

(3.10) ${\mathcal{W}}_{y,k}(n)=\left\\{d^{1}_{\mbox{\rm
Wa}}\left(\frac{1}{n}\sum_{i=1}^{n}\delta_{\theta_{k,i}},\nu_{k}\right)\leq\frac{c_{0}\log(L-k)}{2\sqrt{n}}\right\\}.$

We are ready to define the good events ${\mathcal{I}}_{y,z}$. For $a\in{\bf
Z}_{+}$ let

(3.11) $N_{k,a}=[(\rho_{L}k+z-a+1)^{2}/2].$

We set

(3.12) $N_{k}=N_{k,a}\hskip 7.22743pt\mbox{ if }\hskip
7.22743pt\sqrt{2\mathcal{T}_{y,k}^{1}}\in I_{\rho_{L}k+z-a},$

where $I_{s}=[s,s+1]$. With $L_{+}=L-(500\,logL)^{4}$ and $d^{*}$ a constant
to be determined below, let

(3.13)
${\mathcal{I}}_{y,z}=\widehat{\mathcal{I}}_{y,z}\cap_{k=L_{+}}^{L-d^{\ast}}{\mathcal{W}}_{y,k}\left(N_{k}\right),$

and define the count

(3.14) $J_{z}=\sum_{y\in F^{0}_{L}}{\bf 1}_{{\mathcal{I}}_{y,z}}.$

To obtain (3.5), we need a control on the first and second moments of $J_{z}$,
which is provided by the next two lemmas. In fact, (3.5) will follow directly
from these two Lemmas as in the proof of [4, Proposition 4.2], taking into
account that $|F^{0}_{L}|$ does not depend on $r_{0}$. Most of this section is
devoted to their proof. We emphasize that in the statements of the lemma, the
implied constants are uniform in $r_{0}$ smaller than a fixed small threshold.

######  Lemma 3.3 (First moment estimate).

There is a large enough $d^{\ast}$, such that for all $L$ sufficiently large,
all $0\leq z\leq\log L$, and all $y\in F^{0}_{L}$,

(3.15) $\mathbb{P}\left({\mathcal{I}}_{y,z}\right)\asymp(1+z)e^{-2L}e^{-2z}.$

Let

$\displaystyle G_{0}$ $\displaystyle=$
$\displaystyle\left\\{\left(y,y^{\prime}\right):y,y^{\prime}\in F_{L}\mbox{
s.t. }d\left(y,y^{\prime}\right)>2h_{0}\right\\},$ $\displaystyle G_{k}$
$\displaystyle=$
$\displaystyle\left\\{\left(y,y^{\prime}\right):y,y^{\prime}\in F_{L}\mbox{
s.t. }2h_{k}<d\left(y,y^{\prime}\right)\leq 2h_{k-1}\right\\}\mbox{ for }1\leq
k<L,$ $\displaystyle G_{L}$ $\displaystyle=$
$\displaystyle\left\\{\left(y,y^{\prime}\right):y,y^{\prime}\in F_{L}\mbox{
s.t. }0<d\left(y,y^{\prime}\right)\leq 2h_{L-1}\right\\}.$

######  Lemma 3.4 (Second moment estimate).

There are large enough $d^{*},c^{\prime}$, such that for all $L$ sufficiently
large, all $0\leq z\leq\log L$ and all $(y,y^{\prime})\in G_{k}$, $1\leq k\leq
L$,

(3.17)
$\mathbb{P}\left({\mathcal{I}}_{y,z}\cap{\mathcal{I}}_{y^{\prime},z}\right)\leq
c^{\prime}(1+z)e^{-4L+2k}e^{-2z}e^{-ck_{L}^{1/4}}.$

Before turning to the proofs, we introduce some notation and record some
simple estimates that will be useful in calculations. Recall (3.2),
(3.11)-(3.12) and for $a\in{\bf Z}_{+}$ let

(3.18) ${\mathcal{H}}_{k,a}=\left\\{\sqrt{2\mathcal{T}_{y,k}^{1}}\in
I_{\rho_{L}k+z-a}\right\\}=\left\\{\sqrt{2\mathcal{T}_{y,k}^{1}}\in
I_{\beta_{z}(k)-a}\right\\}.$

Note that on ${\mathcal{H}}_{k,a}$ we have $N_{k}=N_{k,a}$.

Before proceeding we need to state a deviation inequality of Gaussian type for
the Galton-Watson process $T_{l},l\geq 0$ under $P^{\mbox{\rm GW}}_{n}$, the
law of a critical Galton-Watson process with geometric offspring distribution
with initial offspring $n$. The proof is very similar to [3, Lemma 4.6], and
is therefore omitted.

######  Lemma 3.5.

For all $n=1,2,3,\ldots$,

(3.19) $P^{\mbox{\rm
GW}}_{n}\left(\left|\sqrt{2T_{l}}-\sqrt{2T_{0}}\right|\geq\theta\right)\leq
ce^{-\frac{\theta^{2}}{2l}},\quad\theta\geq 0.$

In the proof of our moment estimates we will need the following.

######  Lemma 3.6.

For any $a,b\leq L/\log L$

(3.20)
$\displaystyle\mathbb{P}\left[\sqrt{2\mathcal{T}_{y,L}^{k,(\beta_{z}(k)-a)^{2}/2}}\geq\rho_{L}L+z-b\right]$
$\displaystyle\hskip 72.26999pt\leq ce^{-2(L-k)-2(a-b)-{(a-b)^{2}\over
2(L-k)}}L^{2{(L-k)\over L}}.$

Proof:  By (3.19) we have that for all $\theta>n\geq 1$

(3.21)
$\mathbb{P}\left[\mathcal{T}_{y,L}^{k,n^{2}/2}\geq\theta^{2}/2\right]\leq
c\exp\left(-{\left(\theta-n\right)^{2}\over 2(L-k)}\right).$

We apply this with $\theta=\rho_{L}L+z-b$ and

$n=\beta_{z}(k)-a=\rho_{L}k+z-a$

so that

$\theta-n=\rho_{L}\left(L-k\right)+a-b,$

and hence

${\left(\theta-n\right)^{2}\over 2(L-k)}\geq 2\left(L-k\right)-2{(L-k)\over
L}(\log L)+2(a-b)+{(a-b)^{2}\over 2(L-k)}+O_{L}(1).$

This gives (3.20).

###  3.1. First moment estimate

In this subsection we prove Lemma 3.3.

For the lower bound we have that

(3.22) $\displaystyle\mathbb{P}\left[{\mathcal{I}}_{y,z}\right]\geq
P\left[\widehat{I}_{y,z}\right]-\sum_{k=L_{+}}^{L-d^{\ast}}\mathbb{P}\left[\widehat{\mathcal{I}}_{y,z}\bigcap\,W^{c}_{y,k}\left(N_{k}\right)\right]$
$\displaystyle\geq
c(1+z)e^{-2L}e^{-2z}-\sum_{k=L_{+}}^{L-d^{\ast}}\mathbb{P}\left[\widehat{\mathcal{I}}_{y,z}\bigcap\,W^{c}_{y,k}\left(N_{k}\right)\right],$

where for $P\left[\widehat{I}_{y,z}\right]$ we have used the barrier estimate
(9.12) of Appendix I. We note that

(3.23)
$\mathbb{P}\left[\widehat{\mathcal{I}}_{y,z}\bigcap\,W^{c}_{y,k}\left(N_{k}\right)\right]\leq\sum_{a\geq
k_{L}^{1/4}}\mathbb{P}\left(\widehat{I}_{y,z}^{k,a}\right),$

where

(3.24)
$\widehat{I}_{y,z}^{k,a}=\widehat{\mathcal{I}}_{y,z}\bigcap{\mathcal{H}}_{k,a}\,\bigcap\,W^{c}_{y,k}\left(N_{k,a}\right).$

We show below that for all $L_{+}\leq k\leq L-d^{\ast}$ and $0\leq z\leq\log
L$,

(3.25) $\sum_{a\geq
k_{L}^{1/4}}\mathbb{P}\left(\widehat{I}_{y,z}^{k,a}\right)\leq
c^{\prime}(1+z)\left(e^{-2L}e^{-2z}\right)e^{-c\log^{2}(L-k)},$

which will finish the proof of the lower bound for (3.15) for $d^{\ast}$
sufficiently large.

Furthermore, it is easily seen using (3.20) and the fact that $L-k\leq(500\log
L)^{4}$ that the sum in (3.25) over $a\geq L^{3/4}$ is much smaller than the
right hand side of (3.25), hence it suffices to show that

(3.26) $\sum_{a\geq
k_{L}^{1/4}}^{L^{3/4}}\mathbb{P}\left(\widehat{I}_{y,z}^{k,a}\right)\leq
c^{\prime}(1+z)\left(e^{-2L}e^{-2z}\right)e^{-c\log^{2}(L-k)},$

We now turn to the proof of (3.26). Let

(3.27)
$J_{y,k}^{\uparrow}=\left\\{\sqrt{2\mathcal{T}_{y,l}^{1}}\leq\rho_{L}l+z\mbox{
for }l=1,\ldots,k\right\\},$

and

(3.28)
${\mathcal{B}}_{y,k,a}=\left\\{\sqrt{2\mathcal{T}_{y,L}^{k,(\beta_{z}(k)-a)^{2}/2}}\geq\rho_{L}L+z\right\\}.$

Then with

$\mathcal{K}_{k,p,a}=J_{y,k-3}^{\uparrow}\bigcap{\mathcal{H}}_{k-3,p}\bigcap{\mathcal{H}}_{k,a}\,\bigcap\,W^{c}_{y,k}\left(N_{k,a}\right)\bigcap{\mathcal{B}}_{y,k,a}$

we have

(3.29)
$\mathbb{P}\left(\widehat{I}_{y,z}^{k,a}\right)\leq\sum_{p\geq(k-3)_{L}^{1/4}}^{L^{3/4}}\mathbb{P}\left(\mathcal{K}_{k,p,a}\right),$

plus a term which is much smaller than the right hand side of (3.25).

Let

(3.30) ${\mathcal{W}}^{\in x}_{y,k}(n)=\left\\{d^{1}_{\mbox{\rm
Wa}}\left(\frac{1}{n}\sum_{i=1}^{n}\delta_{\theta_{k,i}},\nu_{k}\right)\in\frac{c_{0}}{\sqrt{n}}I_{x}\right\\},$

so that

(3.31)
${\mathcal{W}}^{c}_{y,k}(N_{k,a})\subseteq\cup_{m=\log(L-k)}^{\infty}{\mathcal{W}}^{\in
m}_{y,k}(N_{k,a}),$

and consequently, setting

(3.32)
${\mathcal{L}}_{k,m,p,a}={\mathcal{K}}_{k,p,a}\cap{\mathcal{W}}_{y,k}^{\in
m}(N_{k,a}),$

we have

(3.33)
$\mathbb{P}\left({\mathcal{K}}_{k,p,a}\right)\leq\sum_{m=\log(L-k)}^{\infty}\mathbb{P}\left({\mathcal{L}}_{k,m,p,a}\right).$

Let

${\mathcal{L}}^{\prime}_{k,m,p,a}=:J_{y,k-3}^{\uparrow}\bigcap{\mathcal{H}}_{k-3,p}\bigcap{\mathcal{H}}_{k,a}\,\bigcap\,W^{\in
m}_{y,k}\left(N_{k,a}\right).$

To prove (3.25) it suffices to prove that for all $m\geq\log(L-k)$,

(3.34) $\sum_{a\geq
k_{L}^{1/4}}^{L^{3/4}}\sum_{p\geq(k-3)_{L}^{1/4}}^{L^{3/4}}\mathbb{P}\left({\mathcal{B}}_{y,k,a}\cap{\mathcal{L}}^{\prime}_{k,m,p,a}\right)\leq
c^{\prime}(1+z)\left(e^{-2L}e^{-2z}\right)e^{-cm^{2}}.$

######  Lemma 3.7.

(3.35)
$\displaystyle\mathbb{P}\left({\mathcal{L}}^{\prime}_{k,m,p,a}\right)=\mathbb{P}\left(J_{y,k-3}^{\uparrow}\bigcap{\mathcal{H}}_{k-3,p}\bigcap{\mathcal{H}}_{k,a}\,\bigcap\,W^{\in
m}_{y,k}\left(N_{k,a}\right)\right)$ $\displaystyle\leq
C(1+z)\left(1+p\right)e^{-2k-2(z-p)}e^{-c(p-a)^{2}}e^{-m^{2}}.$

Proof:  By (3.8)

(3.36) $\mathbb{P}\left(W^{\in
m}_{y,k}\left(N_{k,a}\right)\,|\,{\mathcal{H}}_{k,a}\right)\leq e^{-m^{2}}.$

By (3.20)

(3.37)
$\mathbb{P}\left({\mathcal{H}}_{k,a}\,|\,{\mathcal{H}}_{k-3,p}\right)\leq
ce^{-c(p-a)^{2}},$

and by (9.14) we see that

(3.38)
$\mathbb{P}\left(J_{y,k-3}^{\uparrow}\bigcap{\mathcal{H}}_{k-3,p}\right)\leq
C(1+z)\left(1+p\right)e^{-2k-2(z-p)}.$

It follows as in the proof of [4, Lemma 4.7] that for some $M_{0}<\infty$

(3.39)
$\displaystyle\mathbb{P}\left({\mathcal{B}}_{y,k,a}\cap{\mathcal{L}}^{\prime}_{k,m,p,a}\right)\leq\mathbb{P}\left\\{\sqrt{2\mathcal{T}_{y,L}^{k,(\beta_{z}(k)-a)^{2}/2}}\geq\rho_{L}L+z-M_{0}m\right\\}$
$\displaystyle\hskip
144.54pt\times\mathbb{P}\left({\mathcal{L}}^{\prime}_{k,m,p,a}\right)+e^{-4L}.$

We note that by (3.20)

(3.40)
$\mathbb{P}\left\\{\sqrt{2\mathcal{T}_{y,L}^{k,(\beta_{z}(k)-a)^{2}/2}}\geq\rho_{L}L+z-M_{0}m\right\\}\leq
ce^{-2(L-k)-2(a-M_{0}m)-{(a-M_{0}m)^{2}\over 2(L-k)}},$

Putting this all together with (3.35), and using $|a-p|\leq 1+(p-a)^{2}$ we
find that

(3.41)
$\mathbb{P}\left({\mathcal{B}}_{y,k,a}\cap{\mathcal{L}}^{\prime}_{k,m,p,a}\right)\leq
C(1+z)e^{-2L-2z}e^{-m^{2}/2}\left(1+p\right)e^{-c(p-a)^{2}}e^{-{a^{2}\over
2(L-k)}}.$

Summing first over $p$ and then over $a$ it is easy to see, using a fraction
of the exponent $m^{2}/2$, that (3.34) holds for all $m\geq\log(L-k).$ This
completes the proof of the lower bound in (3.15) .

Since ${\mathcal{I}}_{y,z}\subseteq\widehat{\mathcal{I}}_{y,z}$ the upper
bound in (3.15) follows from the barrier estimate (9.11) of Appendix I.

###  3.2. Second moment estimate: branching in the bulk

We prove the second moment estimate for $y,y^{\prime}\in F^{0}_{L}$ with

$2h_{k-1}<d(y,y^{\prime})\leq 2h_{k-2}.$

In this subsection we prove Lemma 3.4 for

(3.42) $(500\log L)^{4}<k\leq L-(500\log L)^{4}.$

We need to “give ourselves a bit of space” , and we therefore define

(3.43) $k^{+}=k+\lceil 100\log L\rceil.$

Let

$\widehat{\mathcal{I}}_{y,z;k\pm
3}=\left\\{\sqrt{2\mathcal{T}_{y,l}^{1}}\leq\rho_{L}l+z;\,l=1,\ldots,k-4,k+4,\ldots,L-1\right\\}$
(3.44) $\hskip
216.81pt\cap\left\\{\sqrt{\mathcal{T}_{y,L}^{1}}\geq\rho_{L}L+z\right\\},$

where we have skipped the barrier condition for $k-3,\ldots,k+3$. To obtain
the two point bound for the range (3.42) we will bound the probability of

(3.45) $\widehat{\mathcal{I}}_{y,z;k\pm
3}\cap\left\\{\sqrt{2\mathcal{T}_{y^{\prime},L}^{k^{+},\alpha_{z,-}^{2}(k^{+})/2}}\geq\rho_{L}L+z\right\\},$

which contains the event
${\mathcal{I}}_{y,z}\cap{\mathcal{I}}_{y^{\prime},z}$.

Let $\mathcal{G}^{y^{\prime}}$ denote the $\sigma$-algebra generated by the
excursions from $\partial B_{d}(y^{\prime},h_{k-1})$ to $\partial
B_{d}(y^{\prime},h(r_{k^{+}}))$. Note that $\widehat{\mathcal{I}}_{y,z;k\pm
3}\in\mathcal{G}^{y^{\prime}}$. Since

$\left\\{\sqrt{2\mathcal{T}_{y^{\prime},L}^{k^{+},\alpha_{z,-}^{2}(k^{+})/2}}\geq\rho_{L}L+z\right\\}$

is measurable with respect to the first $\alpha_{z,-}^{2}(k^{+})$ excursions
from $\partial B_{d}(y^{\prime},h(r_{k^{+}}))$ to $\partial
B_{d}(y^{\prime},h(r_{k^{+}-1}))$, it follows from the basic ideas in [3, sub-
section 6.2] that

(3.46) $\displaystyle\mathbb{P}\left[\widehat{\mathcal{I}}_{y,z;k\pm
3},\,\sqrt{2\mathcal{T}_{y^{\prime},L}^{k^{+},\alpha_{z,-}^{2}(k^{+})/2}}\geq\rho_{L}L+z\right]$
$\displaystyle\leq c\mathbb{P}\left[\widehat{\mathcal{I}}_{y,z;k\pm
3}\right]P\left[\sqrt{2\mathcal{T}_{y^{\prime},L}^{k^{+},\alpha_{z,-}^{2}(k^{+})/2}}\geq\rho_{L}L+z\right].$

By Lemma 9.3

(3.47) $\mathbb{P}\left[\widehat{\mathcal{I}}_{y,z;k\pm 3}\right]\leq
c(1+z)e^{-2L}e^{-2z}.$

Using (3.20) for the last term in (3.46) together with the fact that in the
range (3.42) we have $(k^{+})_{L}^{1/4}\geq 500\log L$, we find that (3.46) is
bounded by

(3.48) $\displaystyle
c(1+z)e^{-2L}e^{-2z}e^{-2(L-k^{+})-2(k^{+})_{L}^{1/4}}L^{2},$
$\displaystyle\leq c(1+z)e^{-2L}L^{202}e^{-2(L-k)-2(k^{+})_{L}^{1/4}}e^{-2z}.$
$\displaystyle\leq c(1+z)e^{-2L}e^{-2(L-k)-k_{L}^{1/4}}e^{-2z}.$

###  3.3. Second moment estimate: early branching

In this subsection we prove Lemma 3.4 for

$1\leq k<(500\log L)^{4}.$

Since we no longer have $k_{L}^{1/4}\geq\log L$ we will have to use barrier
estimates to control the factors of $L$ such as arise in the first line of
(3.48). On the other hand, since the number of excursions at lower levels is
not so great we don’t need such a large separation. Let

(3.49) $\widetilde{k}=k+\lceil 100\log k\rceil,\hskip 14.45377ptk_{z}=k+\lceil
100\log z\rceil.$

For $v\in\left\\{y,y^{\prime}\right\\}$

$\displaystyle J_{v,s,\widetilde{k}}^{\downarrow}$ $\displaystyle=$
$\displaystyle\left\\{\sqrt{2\mathcal{T}_{v,l}^{\widetilde{k},s^{2}/2}}\leq\rho_{L}l+z\mbox{
for }l=\widetilde{k}+1,\ldots,L-1;\right.$ $\displaystyle\left.\hskip
144.54pt\,\,\sqrt{2\mathcal{T}_{v,L}^{\widetilde{k},s^{2}/2}}\geq\rho_{L}L+z\right\\},$

with the barrier condition applied only for $l\geq\widetilde{k}$.

We first consider the case where $z\leq 100k$. Then

(3.50) $\mathbb{P}\left(I_{y,z}\cap
I_{y^{\prime},z}\right)\leq\sum_{n=1}^{\alpha_{z,-}\left(\widetilde{k}\right)}\mathbb{P}\left(J_{y,n,\widetilde{k}}^{\downarrow}\cap\widehat{\mathcal{I}}_{y^{\prime},z;k\pm
3}\right).$

Let $\mathcal{G}^{y}$ denote the $\sigma$-algebra generated by the excursions
from $\partial B_{d}(y,h_{k-1})$ to $\partial B_{d}(y,h(r_{\widetilde{k}}))$.
Note that $\widehat{\mathcal{I}}_{y^{\prime},z;k\pm 3}\in\mathcal{G}^{y}$.
Since, under our assumption that $z\leq 100k$, the number of excursions from
$\partial B_{d}(y,h_{k-1})$ to $\partial B_{d}(y,h(r_{\widetilde{k}}))$ is
dominated by $n=O(k^{2})$, it follows as before that

(3.51)
$\displaystyle\mathbb{P}\left(J_{y,n,\widetilde{k}}^{\downarrow}\cap\widehat{\mathcal{I}}_{y^{\prime},z;k\pm
3}\right)\leq
cP\left(J_{y,n,\widetilde{k}}^{\downarrow}\right)\mathbb{P}\left(\widehat{\mathcal{I}}_{y^{\prime},z;k\pm
3}\right).$

By the barrier estimate (9.16) of Appendix I, with
$n=\beta_{z}(\widetilde{k})-t$

(3.52) $\mathbb{P}\left(J_{y,n,\widetilde{k}}^{\downarrow}\right)\leq
ct\,n^{1/2}e^{-2(L-\widetilde{k})-2t}\leq ck^{202}e^{-2(L-k)-2k_{L}^{1/4}}.$

where the last step followed from the fact that
$k_{L}^{1/4}\leq\widetilde{k}_{L}^{1/4}\leq t\leq\beta_{z}(\widetilde{k})\leq
ck$. Since, under our assumption that $z\leq 100k$, the number of terms in
(3.50) is $\leq ck^{2}$, and using (3.47), we find that (3.50) is bounded by

$\displaystyle ck^{204}e^{-2(L-k)-2k_{L}^{1/4}}(1+z)e^{-2L}e^{-2z}$
$\displaystyle\leq$ $\displaystyle c(1+z)e^{-4L+2k-k_{L}^{1/4}}e^{-2z}.$

Thus we can assume that

(3.54) $z\geq 100k.$

We have

(3.55) $\displaystyle\mathbb{P}\left(I_{y,z}\cap I_{y^{\prime},z}\right)$
$\displaystyle=\sum_{n=1}^{\alpha_{z,-}\left(k_{z}\right)}1_{\\{n=\beta_{z}(k_{z})-t;\,t\geq
z/2\\}}\mathbb{P}\left(\Big{\\{}\sqrt{2\mathcal{T}_{y,k_{z}}^{1}}=n\Big{\\}}\cap
I_{y,z}\cap I_{y^{\prime},z}\right)$
$\displaystyle+\sum_{n=1}^{\alpha_{z,-}\left(k_{z}\right)}1_{\\{n=\beta_{z}(k_{z})-t;\,t<z/2\\}}\mathbb{P}\left(\Big{\\{}\sqrt{2\mathcal{T}_{y,k_{z}}^{1}}=n\Big{\\}}\cap
I_{y,z}\cap I_{y^{\prime},z}\right)$

Since in the above sums $n\leq cz$ in view of (3.54), we can bound the first
sum in (3.55) by

(3.56)
$\displaystyle\sum_{n=1}^{\alpha_{z,-}\left(k_{z}\right)}1_{\\{n=\beta_{z}(k_{z})-t;\,t\geq
z/2\\}}\mathbb{P}\left(J_{y,n,k_{z}}^{\downarrow}\cap\widehat{\mathcal{I}}_{y^{\prime},z;k\pm
3}\right)$ $\displaystyle\leq
c\sum_{n=1}^{\alpha_{z,-}\left(k_{z}\right)}1_{\\{n=\beta_{z}(k_{z})-t;\,t\geq
z/2\\}}\mathbb{P}\left(J_{y,n,k_{z}}^{\downarrow}\right)\mathbb{P}\left(\widehat{\mathcal{I}}_{y^{\prime},z;k\pm
3}\right)$ $\displaystyle\leq
c\sum_{n=1}^{\alpha_{z,-}\left(k_{z}\right)}1_{\\{n=\beta_{z}(k_{z})-t;\,t\geq
z/2\\}}\mathbb{P}\left(J_{y,n,k_{z}}^{\downarrow}\right)(1+z)e^{-2L}e^{-2z},$

as before. Instead of (3.52) we now have

(3.57) $\mathbb{P}\left(J_{y,n,k_{z}}^{\downarrow}\right)\leq
ct\,z^{1/2}e^{-2(L-k_{z})-2t}\leq cz^{202}e^{-2(L-k)-z/2},$

where the last inequality used $t\geq z/2$. In view of (3.54) and the fact
that the number of terms in the sum is $\leq cz$, this gives the desired bound
for the first sum in (3.55).

Note next that if $t<z/2$ then we must have $n=\beta_{z}(k_{z})-t\geq z/2$,
(but we still have $n\leq cz$ by (3.54)). Thus we can bound the second sum in
(3.55) by

(3.58)
$\displaystyle\sum_{n,n^{\prime}=1}^{\alpha_{z,-}\left(k_{z}\right)}1_{\\{n\geq
z/2\\}}\mathbb{P}\left(\Big{\\{}\sqrt{2\mathcal{T}_{y,k_{z}}^{1}}=n\Big{\\}}\cap
J_{y,n,k_{z}}^{\downarrow}\cap
J_{y^{\prime},n^{\prime},k_{z}}^{\downarrow}\right)$ $\displaystyle\leq
c\sum_{n,n^{\prime}=1}^{\alpha_{z,-}\left(k_{z}\right)}1_{\\{n\geq
z/2\\}}\mathbb{P}\left(\Big{\\{}\sqrt{2\mathcal{T}_{y,k_{z}}^{1}}=n\Big{\\}}\cap
J_{y,n,k_{z}}^{\downarrow}\right)\mathbb{P}\left(J_{y^{\prime},n^{\prime},k_{z}}^{\downarrow}\right).$

as before. Then by the Markov property, this is bounded by

(3.59) $c\sum_{n,n^{\prime}=1}^{\alpha_{z,-}\left(k_{z}\right)}1_{\\{n\geq
z/2\\}}\mathbb{P}\left(\Big{\\{}\sqrt{2\mathcal{T}_{y,k_{z}}^{1}}=n\Big{\\}}\right)\mathbb{P}\left(J_{y,n,k_{z}}^{\downarrow}\right)\mathbb{P}\left(J_{y^{\prime},n^{\prime},k_{z}}^{\downarrow}\right).$

By (3.19) with $n\geq z/2$ and then (3.54)

(3.60)
$\mathbb{P}\left(\Big{\\{}\sqrt{2\mathcal{T}_{y,k_{z}}^{1}}=n\Big{\\}}\right)\leq
e^{-z^{2}/4k_{z}}\leq e^{-10z},$

while now, instead of (3.57), we use

(3.61) $\mathbb{P}\left(J_{y,n,k_{z}}^{\downarrow}\right)\leq
ct\,z^{1/2}e^{-2(L-k_{z})-2t}\leq cz^{202}e^{-2(L-k)},$

and a similar bound for
$\mathbb{P}\left(J_{y^{\prime},n^{\prime},k_{z}}^{\downarrow}\right).$ Thus
(3.59) is bounded by

$c\sum_{n,n^{\prime}=1}^{\alpha_{z,-}\left(k_{z}\right)}e^{-10z}z^{404}e^{-4(L-k)}\leq
ce^{-10z}z^{408}e^{2k}e^{-4L+2k}.$

In view of (3.54), this gives the desired bound for the second sum in (3.55).

###  3.4. Second moment estimate: late branching

In this subsection we prove Lemma 3.4 for $L-(500\log L)^{4}\leq k<L-1$.

Consider first the case

$L-(500\log L)^{4}\leq k<L-d^{\ast}.$

We will bound the probability of

(3.62)
$\mathcal{A}=\left\\{\sqrt{2\mathcal{T}_{y,L}^{k,\alpha_{z,-}^{2}(k)/2}}\geq\rho_{L}L+z\right\\}\cap{\mathcal{W}}_{y,k}\left(N_{k}\right)\cap\widehat{\mathcal{I}}_{y^{\prime},z;k\pm
3},$

(which contains the event
${\mathcal{I}}_{y,z}\cap{\mathcal{I}}_{y^{\prime},z}$).

It follows as in the proof of [4, Lemma 4.7] that for some $M_{0}<\infty$

(3.63)
$\displaystyle\mathbb{P}\left(\mathcal{A}\right)\leq\mathbb{P}\left\\{\sqrt{2\mathcal{T}_{y,L}^{k,\alpha_{z,-}^{2}(k)/2}}\geq\rho_{L}L+z-M_{0}\log(L-k)\right\\}$
$\displaystyle\hskip
144.54pt\times\mathbb{P}\left(\widehat{\mathcal{I}}_{y^{\prime},z;k\pm
3}\right)+e^{-4L}.$

Using (3.20) and (3.47) this shows that

(3.64) $\mathbb{P}\left(\mathcal{A}\right)\leq
ce^{-2(L-k)+2M_{0}\log(L-k)-2k_{L}^{1/4}}(1+z)e^{-2L}e^{-2z}+e^{-4L}$

By taking $d^{\ast}$ sufficiently large we will have $M_{0}\log(L-k)\leq
k_{L}^{1/4}/2$, which then gives (3.15).

For $L-d^{\ast}\leq k<L-1$ we simply bound the term
$\mathbb{P}\left({\mathcal{I}}_{y,z}\cap{\mathcal{I}}_{y^{\prime},z}\right)$
by $\mathbb{P}\left({\mathcal{I}}_{y,z}\right)$ and obtain from (3.15) the
following upper bound

(3.65)
$\mathbb{P}\left({\mathcal{I}}_{y,z}\cap{\mathcal{I}}_{y^{\prime},z}\right)\leq
c(1+z)e^{-2L}e^{-2z}\leq c_{d^{\ast}}(1+z)e^{-(4L-2k)-ck^{1/4}_{L}}e^{-2z}.$

##  4\. Excursion counts and occupation measure on ${\bf S}^{2}$.

In this section we prove Theorems 1.3 and 1.4.

For $0<\epsilon<a<b<\pi$, let ${\mathcal{M}}_{x,\epsilon,a,b}(n)$ be the total
occupation measure of $B_{d}(x,\epsilon)$ until the end of the first $n$
excursions from $\partial B_{d}\left(x,a\right)$ to $\partial
B_{d}\left(x,b\right)$. With $\omega_{\epsilon}=2\pi(1-\cos(\epsilon))$, the
area of $B_{d}\left(x,\epsilon\right)$, let

(4.1)
$\overline{\mathcal{M}}_{x,\epsilon,a,b}(n)={1\over\omega_{\epsilon}}{\mathcal{M}}_{x,\epsilon,a,b}(n).$

In particular, when starting from $\partial B_{d}(x,a)$,

(4.2)
$\overline{\mathcal{M}}_{x,\epsilon,a,b}(1)={1\over\omega_{\epsilon}}\int_{0}^{H_{\partial
B_{d}(x,b)}}1_{\\{B_{d}(x,\epsilon)\\}}(X_{t})\,dt.$

The following Lemma is proven in Section 6.

######  Lemma 4.1.

For some $c>0$, uniformly in $x\in S^{2}$, and $h_{k}/100\leq\epsilon\leq
h_{k}$,

(4.3)
$\mathbb{P}\left(\overline{\mathcal{M}}_{x,\epsilon,h_{k},h_{k-1}}(n)\leq{1\over\pi}(1-\delta)\,\,n\right)\leq
e^{-c\delta^{2}n}$

and

(4.4)
$\mathbb{P}\left(\overline{\mathcal{M}}_{x,\epsilon,h_{k},h_{k-1}}(n)\geq{1\over\pi}(1+\delta)\,\,n\right)\leq
e^{-c\delta^{2}n}$

Recall $\bar{\mu}_{\tau}\left(y,\epsilon\right)$ from (1.8) and set

(4.5) $t_{L}\left(z\right)=2L\left(L-\log L+z\right).$

######  Lemma 4.2.

We can find $0<c,c^{\prime},z_{0}<\infty$ such that for $L$ large, all
$z_{0}\leq z\leq\log L$, and all $h_{L}/100\leq\epsilon_{y}\leq h_{L}$

(4.6) $c^{\prime}ze^{-2z}\leq\mathbb{P}\left(\exists y\in F_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)\leq
cze^{-2z}.$

For the sphere, it suffices to take $\epsilon_{y}=\epsilon$ independent of
$y$. The present formulation is needed for the plane, as we will see in
Section 8. To clarify the connection with (1.9) we note that for some
$0<c_{\ast}=c_{\ast}(r_{0})<\infty$,

(4.7)
$\left(m_{h_{L}}+z\right)^{2}={1\over\pi}t_{L}\left(\sqrt{2\pi}z+c_{\ast}+o_{L}(1)\right).$

###  4.1. The upper bound

We first show that, with $F^{+}_{L}=F_{L}\cap B_{d}\left(v,h_{\log L}\right)$,

(4.8) ${\mathcal{P}}_{1}=:\mathbb{P}\left(\exists y\in F^{+}_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)\leq
cze^{-2z}.$

If

$\widehat{\mathcal{A}}_{L,z}=\Bigg{\\{}\sup_{y\in
F^{+}_{L}}\sqrt{2\mathcal{T}_{y,L}^{\tau}}\,\geq\rho_{L}L+z\Bigg{\\}},$

then by (2.11)

$\displaystyle{\mathcal{P}}_{1}$ $\displaystyle\leq$
$\displaystyle\mathbb{P}\left(\widehat{\mathcal{A}}_{L,z}\right)+\mathbb{P}\left(\widehat{\mathcal{A}}^{c}_{L,z},\exists
y\in F^{+}_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq$ $\displaystyle
ce^{-2z}+\mathbb{P}\left(\widehat{\mathcal{A}}^{c}_{L,z},\exists y\in
F^{+}_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right).$

Recalling the notation $F^{m}_{L}=F_{L}\cap B_{d}^{c}\left(v,h_{m}\right)\cap
B_{d}\left(v,h_{m-1}\right)$, we then bound

(4.9) $\displaystyle\mathbb{P}\left(\widehat{\mathcal{A}}^{c}_{L,z},\exists
y\in F^{+}_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq\sum_{m=\log L}^{L}ce^{2(L-m)}$ $\displaystyle\hskip
36.135pt\sup_{y\in
F^{m}_{L}}\mathbb{P}\left(\sqrt{2\mathcal{T}_{y,L}^{\tau}}\,\leq\rho_{L}L+z,\,\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right).$

We can write

(4.10)
$\displaystyle\mathbb{P}\left(\sqrt{2\mathcal{T}_{y,L}^{\tau}}\,\leq\rho_{L}L+z,\,\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle=\sum_{j=1}^{z+ML^{1/2}}P\left(\,\sqrt{2\mathcal{T}_{y,L}^{\tau}}\in
I_{\rho_{L}L+z-j}\mbox{ and
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle+\mathbb{P}\left(\,\sqrt{2\mathcal{T}_{y,L}^{\tau}}\leq\rho_{L}L-ML^{1/2}\mbox{
and
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right).$

######  Lemma 4.3.

For all $y\in F^{m}_{L}$, $\log L\leq m\leq L$ and $j\leq z+ML^{1/2}$

(4.11) $\displaystyle P\left(\,\sqrt{2\mathcal{T}_{y,L}^{\tau}}\in
I_{\rho_{L}L+z-j}\mbox{ and
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq cme^{-2L}Le^{-2(z-j)}e^{-c^{\prime}j^{2}},$

and

(4.12)
$\displaystyle\mathbb{P}\left(\,\sqrt{2\mathcal{T}_{y,L}^{\tau}}\leq\rho_{L}L-ML^{1/2}\mbox{
and
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq cme^{-2L}Le^{-2(z-j)}e^{-4L}.$

Proof of Lemma 4.3: By (2.9)

(4.13) $\left(\rho_{L}L+z-j\right)^{2}/2\leq t_{L}\left(z-j+2M^{2}\right)$

for all $j\leq z+ML^{1/2}$. Hence for such $j$

$\displaystyle\mathbb{P}\left(\sqrt{2\mathcal{T}_{y,L}^{\tau}}\in
I_{\rho_{L}L+z-j}\mbox{ and
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq\mathbb{P}\left(\sqrt{2\mathcal{T}_{y,L}^{\tau}}\in
I_{\rho_{L}L+z-j},\,\,\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right).$

Using the Markov property and then (2.6), we have for $y\in F^{m}_{L}$ this is

$\displaystyle=\mathbb{P}\left(\sqrt{2\mathcal{T}_{y,L}^{\tau}}\in
I_{\rho_{L}L+z-j}\right)\mathbb{P}\left(\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq
cme^{-2L}Le^{-2(z-j)}\mathbb{P}\left(\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right).$

Consider first the case of $4M^{2}\leq j$. We now apply (4.4) with

$n=t_{L}\left(z-j+2M^{2}\right)=t_{L}\left(z\right)-2(j-2M^{2})L\sim L^{2}$

and

$\delta=2(j-2M^{2})L/t_{L}\left(z-j+2M^{2}\right)<<1$

for $4M^{2}\leq j\leq z+ML^{1/2}$ to see that

(4.14) $\displaystyle\hskip
72.26999pt\mathbb{P}\left(\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle=\mathbb{P}\left(\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\right.$
$\displaystyle\left.\hskip
108.405pt\geq{1\over\pi}\left(t_{L}\left(z-j+2M^{2}\right)+2(j-2M^{2})L\right)\right)$
$\displaystyle=\mathbb{P}\left(\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\right.$
$\displaystyle\left.\hskip 72.26999pt\geq{1\over\pi}\left(1+{2(j-2M^{2})L\over
t_{L}\left(z-j+2M^{2}\right)}\right)t_{L}\left(z-j+2M^{2}\right)\right)$
$\displaystyle\leq e^{-c{\left((j-2M^{2})L\right)^{2}\over
t_{L}\left(z-j+2M^{2}\right)}}\leq e^{-c^{\prime}j^{2}}.$

Similarly, for (4.12) we use

(4.15)
$\displaystyle\mathbb{P}\left(\sqrt{2\mathcal{T}_{y,L}^{\tau}}\leq\rho_{L}L-ML^{1/2}\mbox{
and
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq\mathbb{P}\left(\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(-ML^{1/2}+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right)\leq
e^{-4L}$

by (4.14) for $M$ sufficiently large.

For $j<4M^{2}$ we simply bound the probability in (4.14) by $1$ which we can
bound by $Ce^{-c^{\prime}j^{2}}$ for $C$ sufficiently large.

Then using (4.9) and Lemma 4.3 we see that

(4.16) $\displaystyle\hskip
14.45377pt\mathbb{P}\left(\widehat{\mathcal{A}}^{c}_{L,z},\exists y\in
F^{+}_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq C\sum_{m=\log
L}^{L}cmLe^{2(L-m)}\sum_{j=1}^{z+ML^{1/2}}e^{-2L}e^{-2(z-j)}e^{-c^{\prime}j^{2}}$
$\displaystyle+\sum_{m=\log L}^{L}cmLe^{2(L-m)}e^{-4L}.$

This is easily seen to be bounded by the right hand side of (4.8).

Recalling the notation $F^{\ast}_{L}=F_{L}\cap B_{d}^{c}\left(v,h_{\log
L}\right)$ from (2.17), to complete the proof of the upper bound it remains to
show that

(4.17) ${\mathcal{P}}_{2}=:\mathbb{P}\left(\exists y\in F^{\ast}_{L}\mbox{
s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)\leq
cze^{-2z}.$

Note that with $k_{y}$ as in (2.16), if

(4.18) ${\mathcal{A}}_{L,z}=\Big{\\{}\exists y\in
F^{\ast}_{L},l\in\left\\{k_{y}+1,\ldots,L\right\\}\mbox{ s.t.
}\mathcal{T}_{y,l}^{\tau}\geq\alpha_{z,+}^{2}\left(l\right)/2\Big{\\}},$

then

$\displaystyle{\mathcal{P}}_{2}$ $\displaystyle\leq$
$\displaystyle\mathbb{P}\left({\mathcal{A}}_{L,z}\right)+\mathbb{P}\left({\mathcal{A}}^{c}_{L,z},\exists
y\in F^{\ast}_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq$ $\displaystyle
cze^{-c^{\ast}z}+\mathbb{P}\left({\mathcal{A}}^{c}_{L,z},\exists y\in
F^{\ast}_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right),$

by (2.18). Recalling again the notation $F^{m}_{L}=F_{L}\cap
B_{d}^{c}\left(v,h_{m}\right)\cap B_{d}\left(v,h_{m-1}\right)$, we have that

(4.19) $\displaystyle\mathbb{P}\left({\mathcal{A}}^{c}_{L,z},\exists y\in
F^{\ast}_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle=\sum_{m=1}^{\log
L}\mathbb{P}\left({\mathcal{A}}^{c}_{L,z},\exists y\in F^{m}_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$

Since

(4.20) ${\mathcal{A}}^{c}_{L,z}=\Big{\\{}\sup_{y\in
F^{\ast}_{L}}\mathcal{T}_{y,l}^{\tau}\leq\alpha_{z,+}^{2}\left(l\right)/2,k_{y}+1\leq
l\leq L\Big{\\}}$

and $k_{y}=m$ for $y\in F^{m}_{L}$, we see that

(4.21) $\displaystyle\hskip
14.45377pt\mathbb{P}\left({\mathcal{A}}^{c}_{L,z},\exists y\in F^{m}_{L}\mbox{
s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq ce^{2(L-m)}$ $\displaystyle\sup_{y\in
F^{m}_{L}}\mathbb{P}\left(\mathcal{T}_{y,l}^{\tau}\leq\alpha_{z,+}^{2}\left(l\right)/2,m+1\leq
l\leq L\mbox{ and
}\bar{\mu}_{\tau}\left(y,h_{L}\right)\geq{1\over\pi}t_{L}\left(z\right)\right).$

With

(4.22)
${\mathcal{B}}^{y}_{L,m,z}=\Big{\\{}\mathcal{T}_{y,l}^{\tau}\leq\alpha^{2}_{z,+}\left(l\right)/2,m+1\leq
l\leq L-1\Big{\\}}$

we have for $y\in F^{m}_{L}$,

(4.23) $\displaystyle\hskip
21.68121pt\mathbb{P}\left(\mathcal{T}_{y,l}^{\tau}\leq\alpha_{z,+}^{2}\left(l\right)/2,m+1\leq
l\leq L\mbox{ and
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle=\sum_{j=1}^{z+ML^{1/2}}P\left({\mathcal{B}}^{y}_{L,m,z},\,\sqrt{2\mathcal{T}_{y,L}^{\tau}}\in
I_{\alpha_{z,+}\left(L\right)-j}\mbox{ and
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle+\mathbb{P}\left({\mathcal{B}}^{y}_{L,m,z},\,\sqrt{2\mathcal{T}_{y,L}^{\tau}}\leq\alpha_{z,+}\left(L\right)-z-ML^{1/2}\mbox{
and
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right).$

Here, $M\geq 1$ is a fixed constant to be chosen shortly.

Recalling, see (2.15), that $\alpha_{z,+}\left(L\right)=\rho_{L}L+z$, and
using (4.13) we see that

$\displaystyle\left(\alpha_{z,+}\left(L\right)-j\right)^{2}/2\leq
t_{L}\left(z-j+2M^{2}\right)$

for all $j\leq z+ML^{1/2}$. It follows that for such $j$

$\displaystyle
P\left({\mathcal{B}}^{y}_{L,m,z},\,\sqrt{2\mathcal{T}_{y,L}^{\tau}}\in
I_{\alpha_{z,+}\left(L\right)-j}\mbox{ and
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\leq\mathbb{P}\left({\mathcal{B}}^{y}_{L,m,z},\,\sqrt{2\mathcal{T}_{y,L}^{\tau}}\in
I_{\alpha_{z,+}\left(L\right)-j},\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle=\mathbb{P}\left(\sqrt{2\mathcal{T}_{y,l}^{\tau}}\leq\alpha_{z,+}\left(l\right),m+1\leq
l\leq L-1,\,\sqrt{2\mathcal{T}_{y,L}^{\tau}}\in
I_{\alpha_{z,+}\left(L\right)-j}\right)$ $\displaystyle\hskip
130.08621pt\times\mathbb{P}\left(\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right),$

by the Markov property. Using the barrier estimate (9.5) of Appendix I, and
recalling that $m=k_{y}<\log L$, this is bounded by

(4.24) $ce^{-2L}e^{-2(z-j)}\times
m^{2}j\left(z+m\right)\mathbb{P}\left(\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right).$

The rest of the proof of (4.17) follows as in the proof of (4.8). This
completes the proof of the upper bound in Lemma 4.2.

We now remove the restriction that $y\in F_{L}$ in the upper bound, subject to
a continuity restriction on $\epsilon_{y}$.

######  Lemma 4.4.

We can find $0<c,C,z_{0}<\infty$ such that for $L$ large, all $z_{0}\leq
z\leq\log L$, and all $h_{L}/20\leq\epsilon_{y}\leq h_{L+1}$ such that
$|\epsilon_{y}-\epsilon_{y^{\prime}}|\leq C\,d(y,y^{\prime})/L$ for all
$y,y^{\prime}\in{\bf S}^{2}$,

(4.25) $\mathbb{P}\left(\exists y\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)\leq
cze^{-2z}.$

Proof of Lemma 4.4:  Let $F^{\prime}_{L}$ be the centers of a
$\frac{d_{0}}{L}h_{L}$ covering of ${\bf S}^{2}$ which contains $F_{L}$. For
any $y\in{\bf S}^{2}$ we can find $y^{\prime}\in F^{\prime}_{L}$ such that
$d(y,y^{\prime})\leq\frac{d_{0}}{L}h_{L}$, so that by our assumptions
$|\epsilon_{y}-\epsilon_{y^{\prime}}|\leq C\,\frac{d_{0}}{L^{2}}h_{L}$. If we
set $\bar{\epsilon}_{y}=\left(1+\frac{1}{L}\right)\epsilon_{y}$ for all
$y\in{\bf S}^{2}$ we see that for $L$ large
$h_{L}/30\leq\bar{\epsilon}_{y}\leq 2h_{L+1}$ and
$|\bar{\epsilon}_{y}-\bar{\epsilon}_{y^{\prime}}|\leq\,\frac{d_{0}}{L}h_{L}$.
It follows from Lemma 5.1 below that it suffices to prove that

(4.26) $\mathbb{P}\left(\exists y\in F^{\prime}_{L}\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\bar{\epsilon}_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)\leq
cze^{-2z}.$

For $0<\epsilon<a<b<\pi$, let
${\mathcal{M}}_{y,\bar{\epsilon}_{y},y_{0},a,b}(n)$ be the total occupation
measure of $B_{d}(y,\bar{\epsilon}_{y})$ during the first $n$ excursions from
$\partial B_{d}\left(y_{0},a\right)$ to $\partial B_{d}\left(y_{0},b\right)$.
With $\omega_{\epsilon}=2\pi(1-\cos(\epsilon))$, the area of
$B_{d}\left(y,\epsilon\right)$, let

(4.27)
$\overline{\mathcal{M}}_{y,\bar{\epsilon}_{y},y_{0},a,b}(n)={1\over\omega_{\bar{\epsilon}_{y}}}{\mathcal{M}}_{y,\bar{\epsilon}_{y},y_{0},a,b}(n).$

For $y_{0}\in F_{L}$ let

(4.28) $D_{y_{0}}=\\{y\in F^{\prime}_{L}\,|\,d\left(y,y_{0}\right)\leq
d_{0}h_{L}/2\\}.$

Following the proof of the upper bound for Lemma 4.2, to prove (4.26) it
suffices to show that

(4.29) $\mathbb{P}\left(\sup_{y\in
D_{y_{0}}}\overline{\mathcal{M}}_{y,\bar{\epsilon}_{y},y_{0},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right)\leq
ce^{-c^{\prime}j^{2}}$

for $j\leq z+ML^{1/2}$ sufficiently large. Setting $\epsilon=\sup_{y\in
D_{y_{0}}}\bar{\epsilon}_{y}$ and using our condition on
$|\bar{\epsilon}_{y}-\bar{\epsilon}_{y^{\prime}}|$ to control the denominator
in (4.27), we see that it suffices to show that

(4.30) $\mathbb{P}\left(\sup_{y\in
D_{y_{0}}}\overline{\mathcal{M}}_{y,\epsilon,y_{0},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z-M^{2}/2\right)\right)\leq
ce^{-c^{\prime}j^{2}}.$

Abbreviating
$Y^{(n)}_{y}=\overline{\mathcal{M}}_{y,\epsilon,y_{0},h_{L},h_{L-1}}\left(n\right)$
where $n=t_{L}\left(z-j+2M^{2}\right)$ we have that

(4.31) $\displaystyle\mathbb{P}\left(\sup_{y\in
D_{y_{0}}}\overline{\mathcal{M}}_{y,\epsilon,y_{0},h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z-M^{2}/2\right)\right)$
$\displaystyle\leq\mathbb{P}\left(\overline{\mathcal{M}}_{y_{0},\epsilon,h_{L},h_{L-1}}\left(t_{L}\left(z-j+2M^{2}\right)\right)\geq{1\over\pi}t_{L}\left(z-j/2-M^{2}/2\right)\right)$
$\displaystyle+\mathbb{P}\left(\sup_{y\in
D_{y_{0}}}|Y^{(n)}_{y}-Y^{(n)}_{y_{0}}|\geq jL/2\right).$

As in the proof of Lemma 4.3, the first term on the right hand side is bounded
by $ce^{-c^{\prime}j^{2}}$ for $j\leq z+ML^{1/2}$ sufficiently large. We then
bound

(4.32) $\displaystyle\mathbb{P}\left(\sup_{y\in
D_{y_{0}}}|Y^{(n)}_{y}-Y^{(n)}_{y_{0}}|\geq jL/2\right)$
$\displaystyle\leq\sum_{l=1}^{\log_{2}L}\mathbb{P}\left(\sup_{y,y^{\prime}\in
D_{y_{0}},\,d(y,y^{\prime})\approx
2^{-l}d_{0}h_{L}}|Y^{(n)}_{y}-Y^{(n)}_{y^{\prime}}|\geq jL/2l^{2}\right)$
$\displaystyle\leq\sum_{l=1}^{\log_{2}L}2^{2l}\sup_{y,y^{\prime}\in
D_{y_{0}},\,d(y,y^{\prime})\approx
2^{-l}d_{0}h_{L}}\mathbb{P}\left(|Y^{(n)}_{y}-Y^{(n)}_{y^{\prime}}|\geq
jL/2l^{2}\right).$

It follows from Lemma 7.2 with $n=t_{L}\left(z-j+2M^{2}\right)\sim 2L^{2}$ as
above and $\theta=j/2^{3/2}l^{2},\bar{d}(y,y^{\prime})=2^{-l}d_{0}$ that for
some $C_{0}>0$

(4.33) $\displaystyle 2^{2l}\sup_{y,y^{\prime}\in
D_{y_{0}},\,d(y,y^{\prime})\approx
2^{-l}d_{0}h_{L}}\mathbb{P}\left(|Y^{(n)}_{y}-Y^{(n)}_{y^{\prime}}|\geq
jL/2l^{2}\right)$ $\displaystyle\leq
2^{2l}\exp\left(-C_{0}j^{2}2^{l/2}/8d^{1/2}_{0}l^{4}\right)$

whose sum over $l$ is bounded by $ce^{-c^{\prime}j^{2}}$. In order to apply
Lemma 7.2 we have to verify that $\theta\leq\sqrt{\bar{d}(y,y^{\prime})n}/2$.
In our situation this means that $j/2^{3/2}l^{2}\leq 2^{-l/2}d^{1/2}_{0}L/2$,
for all $j\leq 2ML^{1/2}$. Thus we have to verify that
$2^{1/2}M2^{l/2}/l^{2}\leq d^{1/2}_{0}L^{1/2}$, which follows from the fact
that $l\leq\log_{2}L$, $L$ is large and $d_{0},M$ are fixed.

###  4.2. The lower bound

Recall the notation $\mathcal{T}_{y,l}^{1}=\mathcal{T}_{y,l}^{x^{2},1}$ from
the beginning of Section 3. Let $\tau_{y}$ be the time needed to complete
$x^{2}$ excursions from $\partial B_{d}\left(y,h_{1}\right)$ to $\partial
B_{d}\left(y,h_{0}\right)$, and set

(4.34)
$\bar{\mu}_{\tau_{y}}(y,\epsilon)=\frac{1}{\omega_{\epsilon}}\int_{0}^{\tau_{y}}1_{\\{B_{d}(y,\epsilon)\\}}(X_{t})\,dt.$

Recall $F^{0}_{L}$ from (3.4). We will prove the following analogue of Lemma
3.2.

######  Lemma 4.5.

There exists a $0<c<\infty$ such that for all $0<r_{0}$ sufficiently small,
$L$ large, all $0\leq z\leq\log L$, and all $h_{L}/100\leq\epsilon_{y}\leq
h_{L}$

(4.35) $\mathbb{P}\left[\sup_{y\in
F^{0}_{L}}\bar{\mu}_{\tau_{y}}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right]\geq{(1+z)e^{-2z}\over(1+z)e^{-2z}+c}.$

As before, the lower bound in (4.6) will follow from this, and hence combined
with (4.25) we see that for some $0<z_{0}$, and all $z_{0}\leq z\leq\log L$

(4.36) $c^{\prime}ze^{-2z}\leq\mathbb{P}\left(\exists y\mbox{ s.t.
}\bar{\mu}_{\tau}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)\leq
cze^{-2z}.$

Combined with (4.7) it is easy to check that this implies Theorem 1.4.

To prove (4.35) set

(4.37)
$\widetilde{\mathcal{I}}_{y,z+d}={\mathcal{I}}_{y,z+d}\,\cap\\{\bar{\mu}_{\tau_{y}}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\\}$

for some $d<\infty$ to be chosen shortly. We use the second moment method used
in the proof of Lemma 3.2. Indeed, since
$\widetilde{\mathcal{I}}_{y,z+d}\subseteq{\mathcal{I}}_{y,z+d}$ all upper
bounds needed follow from those used in the proof of Lemma 3.2, and it only
remains to prove the appropriate lower bound for
$\widetilde{\mathcal{I}}_{y,z+d}$.

As in (3.22)-(3.23) we have

(4.38)
$\displaystyle\mathbb{P}\left(\widetilde{\mathcal{I}}_{y,z+d}\right)\geq\mathbb{P}\left(\widehat{\mathcal{I}}_{y,z+d},\,\bar{\mu}_{\tau_{y}}\left(y,h_{L}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\hskip 72.26999pt-\sum_{k=L_{+}}^{L-d^{\ast}}\sum_{a\geq
k_{L}^{1/4}}\mathbb{P}\left[\widehat{\mathcal{I}}_{y,z+d}\bigcap{\mathcal{H}}_{k,a}\,\bigcap\,W^{c}_{y,k}\left(N_{k,a}\right)\right].$

Using the Markov property and then the barrier estimate (9.12) of Appendix I,

(4.39) $\displaystyle\hskip
21.68121pt\mathbb{P}\left(\widehat{\mathcal{I}}_{y,z+d},\,\bar{\mu}_{\tau_{y}}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\geq\mathbb{P}\left(\widehat{\mathcal{I}}_{y,z+d}\mbox{ and
}\,\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z+d\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle=\mathbb{P}\left(\widehat{\mathcal{I}}_{y,z+d}\right)\mathbb{P}\left(\,\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z+d\right)\right)\geq{1\over\pi}t_{L}\left(z\right)\right)$
$\displaystyle\geq\bar{c}(1+z)e^{-2L}e^{-2(z+d)}$ $\displaystyle\hskip
36.135pt\mathbb{P}\left(\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z+d\right)\right)\geq{1\over\pi}\left(t_{L}\left(z+d\right)-dL\right)\right)$
$\displaystyle=\bar{c}(1+z)e^{-2L}e^{-2(z+d)}$ $\displaystyle\hskip
36.135pt\mathbb{P}\left(\overline{\mathcal{M}}_{y,\epsilon_{y},h_{L},h_{L-1}}\left(t_{L}\left(z+d\right)\right)\geq{1\over\pi}\left(1-{dL\over
t_{L}\left(z+d\right)}\right)t_{L}\left(z+d\right)\right)$
$\displaystyle\geq\bar{c}(1+z)e^{-2L}e^{-2(z+d)}(1-e^{-c^{\prime\prime}{\left(dL\right)^{2}\over
t_{L}\left(z+d\right)}}),$

where the last line used (4.3). It should be clear from the structure of
$t_{L}\left(z+d\right)$ that we can choose some $d<\infty$ so that
$e^{-c^{\prime\prime}{\left(dL\right)^{2}\over t_{L}\left(z+d\right)}}\leq
1/2$ uniformly in $0\leq z\leq\log L$. Finally, after fixing such a $d$, we
can show as in the proof of the first moment estimate in Section 3.1, that for
$d^{\ast}$ large enough, the last line in (4.38) is much smaller than the last
line of (4.39).

###  4.3. The left tail

######  Lemma 4.6.

There exists a $0<c<\infty$ such that for all $0<r_{0}$ sufficiently small,
$L$ large, all $0\leq z\leq\log L$, and all $h_{L}/100\leq\epsilon_{y}\leq
h_{L}$

(4.40) $\mathbb{P}\left[\sup_{y\in
F^{0}_{L}}\bar{\mu}_{\tau_{y}}\left(y,\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(-z\right)\right]\geq{e^{2z}\over
e^{2z}+c}.$

This will complete the proof of Theorem 1.3 since, as discussed right after
the statement of Lemma 3.2, the probability of completing $x^{2}$ excursions
from $\partial B_{d}\left(y,h_{1}\right)$ to $\partial
B_{d}\left(y,h_{0}\right)$ before time $\tau$ for all $y\in F^{0}_{L}$ is a
strictly positive function of $r_{0}$ which goes to $1$ as $r_{0}\to 0$.

The proof of Lemma 4.6 is very similar to our proof of the lower bound on the
right tail, except we now have to change the upper barrier to allow for
negative $z$. Fix $|z|\leq\log L$. We fix $\widehat{x}>0$ once and for all. We
abbreviate,

(4.41)
$\widehat{\beta}_{z}\left(l\right)=f_{\widehat{x},\rho_{L}L+z}\left(l;L\right)=\widehat{x}\left(1-\frac{l}{L}\right)+\left(\rho_{L}l+z\frac{l}{L}\right),$

and

(4.42)
$\widehat{\gamma}_{z,-}\left(l\right)=\widehat{\gamma}_{z,-}\left(l,L,z\right)=\widehat{\beta}_{z}\left(l\right)-l_{L}^{1/4}.$

The barrier estimates needed are given in Lemma 9.6. We point out that the
factors $(1+z)$ which appear on the right hand side of (4.35) but not (4.40)
come from the difference in the initial points of the barriers.

##  5\.  Interpolation

Recall, (1.8), that

(5.1)
$\bar{\mu}_{\tau}(y,\epsilon_{y})=\frac{1}{\omega_{\epsilon_{y}}}\int_{0}^{\tau}1_{\\{B_{d}(y,\epsilon_{y})\\}}(X_{t})\,dt,$

where $\omega_{\epsilon_{y}}=2\pi(1-\cos\epsilon_{y})$, the area of
$B_{d}(y,\epsilon_{y})$, and, (4.5),

(5.2) $t_{L}\left(z\right)=2L\left(L-\log L+z\right).$

######  Lemma 5.1.

Assume that $d(y,y^{\prime})\leq a{h_{L}\over L}$,
$|\epsilon_{y}-\epsilon_{y^{\prime}}|\leq b\,{h_{L}\over L}$, and
$h_{L}/30\leq\epsilon_{y},\epsilon_{y^{\prime}}\leq 2h_{L+1}$. We can find a
$d_{1}<\infty$ such that for all $L$ large and $z\leq\log L$, if

(5.3)
$\bar{\mu}_{\tau}\left(y^{\prime},\epsilon_{y^{\prime}}\right)\geq{1\over\pi}t_{L}\left(z\right),$

then for any $c_{1}\geq 30(a+b)$,

(5.4)
$\bar{\mu}_{\tau}\left(y,\left(1+c_{1}/L\right)\epsilon_{y}\right)\geq{1\over\pi}t_{L}\left(z-d_{1}\right).$

Proof:  Under our assumptions, for any $z\in
B_{d}(y^{\prime},\epsilon_{y^{\prime}})$ we have $d(z,y)\leq
d(z,y^{\prime})+d(y,y^{\prime})\leq\epsilon_{y^{\prime}}+a{h_{L}\over
L}\leq\left(1+{c_{1}\over L}\right)\epsilon_{y}$ so that

(5.5) $B_{d}(y^{\prime},\epsilon_{y^{\prime}})\subseteq
B_{d}\left(y,\left(1+c_{1}/L\right)\epsilon_{y}\right).$

It follows that

$\displaystyle\bar{\mu}_{\tau}(y^{\prime},\epsilon_{y^{\prime}})$
$\displaystyle=$
$\displaystyle\frac{1}{\omega_{\epsilon_{y^{\prime}}}}\int_{0}^{\tau}1_{\\{B_{d}(y^{\prime},\epsilon_{y^{\prime}})\\}}(X_{t})\,dt$
$\displaystyle\leq$
$\displaystyle\frac{1}{\omega_{\epsilon_{y^{\prime}}}}\int_{0}^{\tau}1_{\\{B_{d}(y,\left(1+c_{1}/L\right)\epsilon_{y})\\}}(X_{t})\,dt$
$\displaystyle=$
$\displaystyle\frac{\omega_{\left(1+c_{1}/L\right)\epsilon_{y}}}{\omega_{\epsilon_{y^{\prime}}}}\bar{\mu}_{\tau}(y,\left(1+c_{1}/L\right)\epsilon_{y}).$

Hence

(5.7)
$\bar{\mu}_{\tau}\left(y^{\prime},\epsilon_{y^{\prime}}\right)\geq{1\over\pi}t_{L}\left(z\right)$

implies that

(5.8)
$\bar{\mu}_{\tau}(y,\left(1+c_{1}/L\right)\epsilon_{y})\geq\frac{\omega_{\epsilon_{y^{\prime}}}}{\omega_{\left(1+c_{1}/L\right)\epsilon_{y}}}{1\over\pi}t_{L}\left(z\right).$

But under our assumptions

(5.9)
$\frac{\omega_{\epsilon_{y^{\prime}}}}{\omega_{\left(1+c_{1}/L\right)\epsilon_{y}}}=1+O\left(1/L\right).$

This gives (5.4).

##  6\. Green’s functions and proof of Lemma 4.1

Let $G_{a}(x,y)$ denote the potential density for Brownian motion killed the
first time it leaves $B_{e}(0,a)$, that is, the Green’s function for
$B_{e}(0,a)$. We have, see [9, Section 2] or [11, Chapter 2, (1.1)],

(6.1)
$G_{a}(x,y)=-\frac{1}{\pi}\log|x-y|+\frac{1}{\pi}\log\left(\frac{|y|}{a}|x-y_{a}^{\ast}|\right),\hskip
14.45377pty\neq 0,$

where

(6.2) $y_{a}^{\ast}=\frac{a^{2}y}{|y|^{2}},$

and

(6.3) $G_{a}(x,0)=-\frac{1}{\pi}\log|x|+\frac{1}{\pi}\log a.$

Let $v$ denote the south pole of ${\bf S}^{2}$. We claim that in the
isothermal coordinates induced by stereographic projection $\sigma$, the
Green’s function for $\sigma\left(B_{d}(v,h(a))\right)=B_{e}(0,a)$ is just
$G_{a}(x,y)$. To see this we must show that if $\Delta_{{\bf S}^{2}}$ is the
Laplacian for ${\bf S}^{2}$ in isothermal coordinates and $dV(y)$ is the
volume measure, then

(6.4) $\frac{1}{2}\Delta_{{\bf S}^{2}}\int G_{a}(x,y)f(y)\,dV(y)=-f(x)$

for all continuous $f$ compactly supported in $B_{e}(0,a)$.

For $x=\left(x_{1},x_{2}\right)$, let

(6.5) $g(x)=\frac{1}{(1+{\frac{1}{4}}(x_{1}^{2}+x_{2}^{2}))^{2}}.$

As shown in [23, Chapter 7, p. 6-9], the stereographic projection $\sigma$ is
an isometry if we give $R^{2}$ the metric

(6.6) $g(x)\left(dx_{1}\otimes dx_{1}+dx_{2}\otimes dx_{2}\right).$

Because of (6.6) the Laplace-Beltrami operator takes the form

(6.7) ${1\over g(x)}\left(\partial^{2}_{x_{1}}+\partial^{2}_{x_{2}}\right).$

Thus, $\Delta_{{\bf S}^{2}}=\frac{1}{g}\Delta$ and $dV(y)=g(y)\,dy$, so that
(6.4) holds.

Proof of Lemma 4.1:  Let $\epsilon=h(\alpha)$ so that $h(\alpha)\leq h_{k}$.
lf $\tau_{h_{k-1}}$ is the first exit time of $B_{d}(v,h_{k-1})$ and
$\rho_{h_{k}}$ is uniform measure on $\partial B_{d}(v,h_{k})$, then by
symmetry, for any $z\in\partial B_{d}(v,h_{k})$

$\displaystyle J_{1}$ $\displaystyle=:$
$\displaystyle\mathbb{E}^{z}\left(\int_{0}^{\tau_{h_{k-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t})\,dt\right)$
$\displaystyle=$
$\displaystyle\mathbb{E}^{\rho_{h_{k}}}\left(\int_{0}^{\tau_{h_{k-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t})\,dt\right).$

Since uniform measure $\rho_{h_{k}}$ on $\partial B_{d}(v,h_{k})$ goes over to
uniform measure $\gamma_{r_{k}}$ on $\partial B_{e}(0,r_{k})$, using the
discussion at the beginning of this section we have

(6.9) $J_{1}=\int_{B_{e}(0,\alpha)}\int_{\partial
B_{e}(0,r_{k})}G_{r_{k-1}}(x,y)\,d\gamma_{r_{k}}(x)g(y)\,dy.$

We recall, [20, Chapter 2, Prop. 4.9] or [11, Chapter 1, (5.4), (5.5)], that

(6.10) $\int_{\partial
B_{e}(0,b)}\log\left(|x-y|\right)\,d\gamma_{b}(x)=\log\left(b\vee|y|\right).$

This shows that for $y\in B_{e}(0,r_{k})$

(6.11) $\displaystyle\int_{\partial
B_{e}(0,r_{k})}G_{r_{k-1}}(x,y)\,d\gamma_{r_{k}}(x)$
$\displaystyle=\frac{1}{\pi}\int_{\partial
B_{e}(0,r_{k})}\left(-\log|x-y|+\log\left(\frac{|y|}{r_{k-1}}|x-y_{r_{k-1}}^{\ast}|\right)\right)\,d\gamma_{r_{k}}(x)$
$\displaystyle=\frac{1}{\pi}\left(-\log
r_{k}+\log\left(\frac{|y|}{r_{k-1}}|y_{r_{k-1}}^{\ast}|\right)\right)$
$\displaystyle=\frac{1}{\pi}\left(-\log r_{k}+\log
r_{k-1}\right)=\frac{1}{\pi}\log(r_{k-1}/r_{k})=\frac{1}{\pi}.$

Thus

(6.12)
$J_{1}=\frac{1}{\pi}\int_{B_{e}(0,\alpha)}g(y)\,dy=\frac{1}{\pi}\mbox{Area
}(B_{d}(v,h(\alpha)))=\frac{1}{\pi}\omega_{h(\alpha)}=\frac{1}{\pi}\omega_{\epsilon}.$

It follows that for any $z\in\partial B_{d}(v,h_{k})$

(6.13)
$\mathbb{E}^{z}\left(\overline{\mathcal{M}}_{v,\epsilon,h_{k},h_{k-1}}(1)\right)=\frac{1}{\pi}.$

By the Kac moment formula, for any $z\in\partial B_{d}(v,h_{k})$, with
$x=\sigma(z)$

(6.14)
$\displaystyle\mathbb{E}^{z}\left(\left(\int_{0}^{\tau_{h_{k-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t})\,dt\right)^{n}\right)$
$\displaystyle=n!\int_{B^{n}_{e}(0,\alpha)}G_{r_{k-1}}(x,y_{1})\prod_{j=2}^{n}G_{r_{k-1}}(y_{j-1},y_{j})\prod_{i=1}^{n}g(y_{i})\,dy_{i}$
$\displaystyle\leq
c^{n}n!\int_{B^{n}_{e}(0,\alpha)}G_{r_{k-1}}(x,y_{1})\prod_{j=2}^{n}G_{r_{k-1}}(y_{j-1},y_{j})\prod_{i=1}^{n}\,dy_{i}$
$\displaystyle\leq
c^{n}n!\alpha^{2n}\left(\log\left(r_{k-1}/\alpha\right)+c_{0}\right)^{n}$

where the last inequality follows as in the proof of [9, Lemma 2.1]. It
follows that for any $z\in\partial B_{d}(v,h_{k})$

(6.15)
$\mathbb{E}^{z}\left(\left(\overline{\mathcal{M}}_{v,\epsilon,h_{k},h_{k-1}}(1)\right)^{n}\right)\leq
c^{n}n!\left(\log\left(r_{k-1}/\alpha\right)+c_{0}\right)^{n}.$

By (2.3), our assumption that $h_{k}/100\leq\epsilon\leq h_{k}$ implies that
$e\leq r_{k-1}/\alpha\leq 200e$. Using (6.13) and (6.15), our Lemma then
follows as in the proof of [10, Lemma 2.2]

##  7\. Continuity Estimates

For fixed $u\in{\bf S}^{2}$, let $\tau_{a}$ be the first exit time of
$B_{d}(u,a)$ and let $\rho_{m}$ be uniform measure on $\partial B_{d}(u,m)$.

######  Lemma 7.1.

If $d(u,v),d(u,\widetilde{v})\leq d_{0}h_{L}/2$,
$d_{0}/L\leq\bar{d}=:d(v,\widetilde{v})/h_{L}\leq d_{0}$, and
$h_{L}/20\leq\epsilon\leq h_{L+1}$, then

(7.1)
$\mathbb{E}^{\rho_{h_{L}}}\left(\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t})\,dt-\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(\widetilde{v},\epsilon)\\}}(X_{t})\,dt\right)=0,$

(7.2)
$\mathbb{E}^{\rho_{h_{L}}}\left(\left(\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t})\,dt-\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(\widetilde{v},\epsilon)\\}}(X_{t})\,dt\right)^{2}\right)\leq
c\epsilon^{4}\bar{d}^{2},$

and

(7.3) $\sup_{x\in\partial
B_{d}(u,h_{L})}\mathbb{E}^{x}\left(\left(\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t})\,dt-\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(\widetilde{v},\epsilon)\\}}(X_{t})\,dt\right)^{2}\right)\leq
c\epsilon^{4}\bar{d}^{2}.$

Proof of Lemma 7.1:  As in (6)-(6.9) we have

(7.4) $\displaystyle J_{2}$ $\displaystyle=$
$\displaystyle\mathbb{E}^{\rho_{h_{L}}}\left(\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t})\,dt-\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(\widetilde{v},\epsilon)\\}}(X_{t})\,dt\right)$
$\displaystyle=$ $\displaystyle\int\int
G_{r_{L-1}}(x,y)\,d\gamma_{r_{L}}(x)\,d\mu_{v,\widetilde{v}}(y),$

where

(7.5)
$d\mu_{v,\widetilde{v}}(y)=\left(1_{\\{\sigma(B_{d}(v,\epsilon))\\}}-1_{\\{\sigma(B_{d}(\widetilde{v},\epsilon))\\}}\right)(y)g(y)\,dy.$

Then by (6.11)-(6.12) we have that

$\displaystyle J_{2}$ $\displaystyle=$ $\displaystyle\frac{1}{\pi}\int
d\mu_{v,\widetilde{v}}(y)$ $\displaystyle=$
$\displaystyle\frac{1}{\pi}\left(\mbox{Area }(B_{d}(v,\epsilon))-\mbox{Area
}(B_{d}(\widetilde{v},\epsilon))\right)=0,$

since all balls of radius $\epsilon$ on the sphere have area
$\omega_{\epsilon}=2\pi(1-\cos\epsilon)$. This completes the proof of (7.1).

We next observe that

(7.7)
$\displaystyle\mathbb{E}^{\rho_{h_{L}}}\left(\left(\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t})\,dt-\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(\widetilde{v},\epsilon)\\}}(X_{t})\,dt\right)^{2}\right)$
$\displaystyle=2\int\int\int
G_{r_{L-1}}(x,y)G_{r_{L-1}}(y,z)\,d\gamma_{r_{L}}(x)\,d\mu_{v,\widetilde{v}}(y)\,d\mu_{v,\widetilde{v}}(z)$
$\displaystyle=\frac{2}{\pi}\int\int
G_{r_{L-1}}(y,z)\,d\mu_{v,\widetilde{v}}(y)\,d\mu_{v,\widetilde{v}}(z)$

as above.

We note that for $b<a$

(7.8)
$G_{a}(bx,by)=-\frac{1}{\pi}\log\left(b\,|x-y|\right)+\frac{1}{\pi}\log\left(b\,\frac{|y|}{a/b}|x-y_{a/b}^{\ast}|\right)=G_{a/b}(x,y),$

since

(7.9) $(by)_{a}^{\ast}=\frac{a^{2}by}{b^{2}|y|^{2}}=by_{a/b}^{\ast}.$

Using this to scale by $r_{L}$ we see that

(7.10) $\int\int
G_{r_{L-1}}(y,z)\,d\mu_{v,\widetilde{v}}(y)\,d\mu_{v,\widetilde{v}}(z)=r_{L}^{4}\int\int
G_{e}(y,z)\,d\mu_{L,v,\widetilde{v}}(y)\,d\mu_{L,v,\widetilde{v}}(z),$

where

(7.11) $\displaystyle d\mu_{L,v,\widetilde{v}}(y)$ $\displaystyle=$
$\displaystyle\left(1_{\\{\sigma(B_{d}(v,\epsilon))\\}}-1_{\\{\sigma(B_{d}(\widetilde{v},\epsilon))\\}}\right)(r_{L}y)g(r_{L}y)\,dy$
$\displaystyle=$
$\displaystyle\left(1_{\\{\frac{1}{r_{L}}\sigma(B_{d}(v,\epsilon))\\}}-1_{\\{\frac{1}{r_{L}}\sigma(B_{d}(\widetilde{v},\epsilon))\\}}\right)(y)g(r_{L}y)\,dy.$

For $y$ in our range we have $g(r_{L}y)=1+O(\epsilon)$, and it is easy to
check that up to errors of order $\epsilon$,
$\frac{1}{r_{L}}\sigma(B_{d}(v,\epsilon))$ and
$\frac{1}{r_{L}}\sigma(B_{d}(\widetilde{v},\epsilon))$ can be replaced by
$B_{e}(v^{\prime},\epsilon/r_{L})$ and
$B_{e}(v^{\prime}-(0,\delta\epsilon/r_{L}),\epsilon/r_{L})$ for some
$v^{\prime}$ with $|v^{\prime}|\leq d_{0}$ and $0<\delta\leq c_{2}\bar{d}$.

Hence with

(7.12)
$d\nu_{v^{\prime}}(y)=\left(1_{\\{B_{e}(v^{\prime},\epsilon/r_{L})\\}}-1_{\\{B_{e}(v^{\prime}-(0,\delta\epsilon/r_{L}),\epsilon/r_{L})\\}}\right)(y)\,dy,$

it remains to show that

(7.13) $\int\int G_{e}(y,z)\,d\mu_{v^{\prime}}(y)\,d\mu_{v^{\prime}}(z)\leq
C\delta^{2}.$

The symmetric difference of $B_{e}(v^{\prime},s)$ and
$B_{e}(v^{\prime}-(0,\delta s),s)$ consist of two disjoint pieces we denote by
$A,B$. They have the same area

(7.14) $\mbox{Area}\left(A\right)=2s^{2}\arcsin\left({\delta\over
2}\right)+{\delta s\over 2}\sqrt{(4-\delta^{2})s^{2}}\asymp\delta s^{2}.$

We observe that

(7.15) $\lim_{y\to 0}\frac{|y|}{a}|z-y_{a}^{\ast}|=\lim_{y\to
0}\frac{|y|}{a}\left(|y_{a}^{\ast}|+O(1)\right)=\lim_{y\to
0}\frac{|y|}{a}\left(\frac{a^{2}|y|}{|y|^{2}}+O(1)\right)=a.$

It follows that for $y,z$ in our range,
$\log\left(\frac{|y|}{e}|z-y_{e}^{\ast}|\right)$ is bounded, hence to prove
(7.13) it suffices to show that

(7.16)
$\int\int\big{|}\log|y-z|\big{|}\,d\mu_{v^{\prime}}(y)\,d\mu_{v^{\prime}}(z)\leq
C\delta^{2}.$

It is then easy to see that we need only show that

(7.17) $\int_{A}\int_{A}\big{|}\log|y-z|\big{|}\,dy\,dz\leq C\delta^{2}.$

It is also clear that we only need to consider $|y-z|\leq 1/2$. Writing
$y=(y_{1},y_{2}),z=(z_{1},z_{2})$ we see that

(7.18) $\displaystyle\int_{A}\int_{A}\big{|}\log|y-z|\big{|}1_{\\{|y-z|\leq
1/2\\}}\,dy\,dz$
$\displaystyle\leq\int_{[0,1]\times[0,\delta]}\int_{[0,1]\times[0,\delta]}\big{|}\log|y-z|\big{|}1_{\\{|y-z|\leq
1/2\\}}\,dy_{1}\,dy_{2}\,dz_{1}\,dz_{2}$
$\displaystyle\leq\int_{[0,1]\times[0,\delta]}\int_{[0,1]\times[0,\delta]}\big{|}\log|y_{1}-z_{1}|\big{|}\,dy_{1}\,dy_{2}\,dz_{1}\,dz_{2}\leq
C\delta^{2},$

which completes the proof of (7.17).

To obtain (7.3), arguing as before we need to show that

(7.19) $K_{1}=\sup_{x\in\partial B_{e}(0,r_{L})}\int\int
G_{r_{L-1}}(x,y)G_{r_{L-1}}(y,z)\,d\mu_{v,\widetilde{v}}(y)\,d\mu_{v,\widetilde{v}}(z)\leq
c\epsilon^{4}\delta^{2}.$

Scaling in $r_{L}$ as before shows that

(7.20) $K_{1}=r_{L}^{4}\sup_{x\in\partial B_{e}(0,1)}\int\int
G_{e}(x,y)G_{e}(y,z)\,d\mu_{L,v,\widetilde{v}}(y)\,d\mu_{L,v,\widetilde{v}}(z)$

But for $y$ in our range, $G_{e}(x,y)$ is bounded uniformly in $x\in\partial
B_{e}(0,1)$, so that (7.3) follows as before.

The same proof shows that

$\displaystyle\sup_{x\in\partial
B_{d}(u,h_{L})}\mathbb{E}^{x}\left(\left(\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t})\,dt-\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(\widetilde{v},\epsilon)\\}}(X_{t})\,dt\right)^{2n}\right)$
(7.21) $\displaystyle\hskip
252.94499pt\leq(2n)!c_{1}^{2n}\epsilon^{4n}\bar{d}^{2n}.$

and hence by the Cauchy-Schwarz inequality

$\displaystyle\sup_{x\in\partial
B_{d}(u,h_{L})}\mathbb{E}^{x}\left(\Big{|}\frac{1}{\omega_{\epsilon}}\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t})\,dt-\frac{1}{\omega_{\epsilon}}\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(\widetilde{v},\epsilon)\\}}(X_{t})\,dt\Big{|}^{n}\right)$
(7.22) $\displaystyle\hskip 252.94499pt\leq n!c_{2}^{n}\bar{d}^{n}.$

Recall (4.27) and set

$Y^{(n)}_{y}=\overline{\mathcal{M}}_{y,\epsilon,u,h_{L},h_{L-1}}\left(n\right)$

######  Lemma 7.2.

For some $d_{0}>0$ we can find $C_{0}>0$ such that, if
$d(u,v),d(u,\widetilde{v})\leq d_{0}h_{L}/2$,
$d_{0}/L\leq\bar{d}(v,\widetilde{v})=:d(v,\widetilde{v})/h_{L}\leq d_{0}$,
$h_{L}/20\leq\epsilon\leq h_{L+1}$, and
$\theta\leq\sqrt{\bar{d}(v,\widetilde{v})n}/2$, then

(7.23)
$\mathbb{P}\left(|Y^{(n)}_{v}-Y^{(n)}_{\widetilde{v}}|\geq\theta\sqrt{n}\right)\leq
e^{-C_{0}\theta^{2}/\bar{d}^{1/2}(v,\widetilde{v})}.$

Proof of Lemma 7.2: We follow the proof of [4, Lemma 5.1].

Let $T_{i}$ denote the successive excursion times $T_{\partial
B_{d}(u,h_{L})}\circ\theta_{T_{\partial B_{d}(u,h_{L-1})}}$ from $\partial
B_{d}(u,h_{L-1})$ to $\partial B_{d}(u,h_{L})$ and set

(7.24)
$Y_{v,i}=\frac{1}{\omega_{\epsilon}}\int_{0}^{\tau_{h_{L-1}}}1_{\\{B_{d}(v,\epsilon)\\}}(X_{t+T_{i}})\,dt,$

so that

(7.25) $Y^{(n)}_{v}=\sum_{i=1}^{n}Y_{v,i}.$

Let $J$ be a geometric random variable with success parameter $p_{3}>0$,
independent of $\\{Y_{v,i},Y_{\widetilde{v},i}\\}$. It follows from (7.22) and
the proof of [4, Corollary 5.3] that, abbreviating
$\bar{d}=\bar{d}(v,\widetilde{v})$, if $c_{2}\bar{d}\lambda\leq p_{3}/2$ then
for some $c_{4}$

(7.26) $\displaystyle\sup_{x\in\partial
B_{d}(u,h_{L})}\mathbb{E}^{x}\left(\exp\left(\lambda\sum_{i=1}^{J-1}\left|\left(Y_{v,i}-Y_{\widetilde{v},i}\right)\right|\right)\right)\leq
e^{c_{4}\bar{d}\lambda/p_{3}},$

and from (7.22) together with (7.1) and the proof and notation of [4, Lemma
5.5] it follows that, after perhaps enlarging $c_{4}$

(7.27)
$\displaystyle\mathbb{E}\left(\exp\left(\lambda\sum_{i=J_{1}}^{J_{2}-1}\left(Y_{v,i}-Y_{\widetilde{v},i}\right)\left(X_{\cdot}^{i}\right)\right)\right)\leq
e^{c_{4}\left(\bar{d}\lambda/p_{3}\right)^{2}}.$

Instead of [4, (5.33)] we set

$\delta=\theta/\sqrt{\bar{d}n}\leq 1/2.$

It then follows from the proof of [4, Lemma 5.1] that for
$c_{2}\bar{d}\lambda\leq p_{3}/2$

(7.28)
$\displaystyle\mathbb{P}\left(|Y^{(n)}_{v}-Y^{(n)}_{\widetilde{v}}|\geq\theta\sqrt{n}\right)$
$\displaystyle\leq
e^{-\bar{c}\theta^{2}/\bar{d}}+\exp\left(c_{4}\lambda^{2}\bar{d}^{2}n/p_{3}+2c_{4}\lambda\theta\sqrt{\bar{d}n}-\lambda\theta\sqrt{n}\right)$

By taking $d_{0}$ sufficiently small we can be sure that $c_{2}\bar{d}\leq 1$,
so the above holds for any $\lambda\leq p_{3}/2$. If we set

$\lambda=p_{3}\theta/\sqrt{\bar{d}n}\leq p_{3}/2$

we see that

$\displaystyle\exp\left(c_{4}\lambda^{2}\bar{d}^{2}n/p_{3}+2c_{4}\lambda\theta\sqrt{\bar{d}n}-\lambda\theta\sqrt{n}\right)$
$\displaystyle=\exp\left(c_{4}p_{3}\theta^{2}\bar{d}+2c_{4}p_{3}\theta^{2}-p_{3}\theta^{2}/\sqrt{\bar{d}}\right),$

which completes the proof of (7.23) for $d_{0}$ sufficiently small.

##  8\.  From the sphere to the plane, and back

Using (6.7) it follows from [21, Chapter 5, Theorem 1.9], that we can find a
planar Brownian motion $W_{t}$ such that in the isothermal coordinates induced
by stereographic projection,

(8.1) $X_{t}=W_{U_{t}},\hskip 7.22743pt\mbox{ where }\hskip
7.22743ptU_{t}=\int_{0}^{t}{1\over g(X_{s})}\,ds.$

We take the $v$ of this paper to be $v=(0,0,0)$. Let

(8.2)
$D_{\ast}=\sigma\left(B_{d}\left(v,r^{\ast}\right)\right)=B_{e}\left((0,0),2\tan(r^{\ast}/2)\right).$

For the last equality see [4, (2.4)]. If $\theta$ is the first hitting time of
$\partial D_{\ast}$ by $W_{t}$, then under the coupling (8.1) we see that
$\theta=U_{\tau}$. Set

(8.3)
$\mu_{\theta}(x,\epsilon)=\frac{1}{\pi\epsilon^{2}}\int_{0}^{\theta}1_{\\{B_{e}(x,\epsilon)\\}}(W_{t})\,dt.$

######  Lemma 8.1.

For some $-\infty<d_{1},d_{2},d_{3},d_{4}<\infty$, all $x\in D_{\ast}$ and all
$\epsilon$ sufficiently small

(8.4)
$\mu_{\theta}(x,\epsilon)\leq(1+d_{1}\,\,\epsilon)\,\,\bar{\mu}_{\tau}(x,g^{1/2}(x)\epsilon(1+d_{2}\epsilon)),$

and

(8.5)
$\mu_{\theta}(x,\epsilon)\geq(1+d_{3}\,\,\epsilon)\,\,\bar{\mu}_{\tau}(x,g^{1/2}(x)\epsilon(1+d_{4}\epsilon)).$

Proof of Lemma 8.1:  We first note that for $\epsilon$ sufficiently small, we
can find $c_{1}<c_{2}$ such that uniformly in $x^{\prime}\in
B_{e}(x,2\epsilon)$ and $x\in D_{\ast}$

(8.6) $g(x)(1+c_{1}\epsilon)\leq g(x^{\prime})\leq g(x)(1+c_{2}\epsilon).$

For $x^{\prime}\in B_{e}(x,\epsilon)$, with $x_{t}=x+t(x^{\prime}-x)$

(8.7) $d(x,x^{\prime})\leq\int_{0}^{1}g^{1/2}(x_{t})|x^{\prime}-x|\,dt\leq
g^{1/2}(x)|x^{\prime}-x|(1+c_{3}\epsilon).$

Hence

(8.8) $B_{e}(x,\epsilon)\subseteq
B_{d}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon)).$

Similarly, for some $c_{4}<c_{3}$

(8.9) $B_{e}(x,\epsilon)\supseteq
B_{d}(x,g^{1/2}(x)\epsilon(1+c_{4}\epsilon)).$

Consider

(8.10)
$\int_{0}^{\tau}1_{\\{B_{d}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon))\\}}(W_{U_{t}})\,dt.$

By the nature of $U_{t}$ in (8.1) it follows that whenever the path
$W_{U_{t}}$ enters $B_{d}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon))$ it is slowed
by a variable factor between ${1\over g(x)(1+c_{5}\epsilon)}$ and ${1\over
g(x)(1+c_{6}\epsilon)}$. Hence the amount of time spent in
$B_{d}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon))$ during each incursion is
multiplied by a variable factor between $g(x)(1+c_{7}\epsilon)$ and
$g(x)(1+c_{8}\epsilon)$. Thus

(8.11)
$\int_{0}^{\tau}1_{\\{B_{d}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon))\\}}(W_{U_{t}})\,dt\geq
g(x)(1+c_{7}\epsilon)\int_{0}^{\theta}1_{\\{B_{d}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon))\\}}(W_{t})\,dt,$

and

(8.12)
$\int_{0}^{\tau}1_{\\{B_{d}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon))\\}}(W_{U_{t}})\,dt\leq
g(x)(1+c_{8}\epsilon)\int_{0}^{\theta}1_{\\{B_{d}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon))\\}}(W_{t})\,dt,$

It follows from (8.8) and (8.11) that

(8.13) $\displaystyle\int_{0}^{\theta}1_{\\{B_{e}(x,\epsilon)\\}}(W_{t})\,dt$
$\displaystyle\leq\int_{0}^{\theta}1_{\\{B_{d}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon))\\}}(W_{t})\,dt$
$\displaystyle\leq{1\over
g(x)(1+c_{7}\epsilon)}\int_{0}^{\tau}1_{\\{B_{d}(x,g^{1/2}(z)\epsilon(1+c_{3}\epsilon))\\}}(W_{U_{t}})\,dt.$

Since $\omega_{\delta}=2\pi\left(1-\cos\left(\delta\right)\right)$, we see
that if we set $\delta_{x}=g^{1/2}(x)\epsilon(1+c_{3}\epsilon)$, then
uniformly in $x\in D_{\ast}^{2}$ and sufficiently small $\epsilon$

$(1+f^{\prime}_{0}\epsilon)\leq\frac{\omega_{\delta_{x}}}{\pi
g\left(x\right)\epsilon^{2}}\leq(1+f_{0}\epsilon),$

so that by (8.8) and (8.13)

(8.14) $\displaystyle\mu_{\theta}(x,\epsilon)$ $\displaystyle=$
$\displaystyle\frac{1}{\pi\epsilon^{2}}\int_{0}^{\theta}1_{\\{B_{e}(x,\epsilon)\\}}(W_{t})\,dt$
$\displaystyle\leq$
$\displaystyle\frac{1}{\pi\epsilon^{2}g(x)(1+c_{7}\epsilon)}\int_{0}^{\tau}1_{\\{B_{d}(x,g^{1/2}(z)\epsilon(1+c_{3}\epsilon))\\}}(W_{U_{t}})\,dt$
$\displaystyle=$
$\displaystyle\frac{1}{(1+c_{7}\epsilon)}\,\,\frac{\omega_{\delta_{x}}}{\pi
g\left(x\right)\epsilon^{2}}\,\,\bar{\mu}_{\tau}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon))$
$\displaystyle\leq$
$\displaystyle(1+\widehat{d}\,\,\epsilon)\,\,\bar{\mu}_{\tau}(x,g^{1/2}(x)\epsilon(1+c_{3}\epsilon)),$

where

$(1+\widehat{d}\,\,\epsilon)={1+f_{0}\epsilon\over 1+c_{7}\epsilon}.$

This completes the proof of (8.4).

The lower bound (8.5) is proven similarly using (8.9) and (8.12).

######  Lemma 8.2.

We can find $0<c,c^{\prime},z_{0}<\infty$ such that for $L$ large and all
$\frac{1}{12}h_{L}\leq\epsilon\leq\frac{1}{3}h_{L}$ and $z_{0}\leq z\leq\log
L$,

(8.15)
$c^{\prime}ze^{-c^{\ast}\sqrt{\pi}z}\leq\mathbb{P}\left(\sqrt{\sup_{y}\mu_{\theta}(y,\epsilon)}\geq
m_{\epsilon}+z\right)\leq cze^{-c^{\ast}\sqrt{\pi}z}.$

Proof of Lemma 8.2: We consider the upper bound. By (8.4) it suffices to show
that

(8.16)
$\mathbb{P}\left(\sqrt{\sup_{y}\bar{\mu}_{\tau}(y,g^{1/2}(y)\epsilon(1+d_{2}\epsilon))}\geq
m_{\epsilon}+z\right)\leq cze^{-c^{\ast}\sqrt{\pi}z}.$

Since for $\epsilon$ in our range

$\left(m_{\epsilon}+z\right)^{2}={1\over\pi}t_{L}\left(\sqrt{2\pi}\,\,z+O(1)\right),$

(compare (4.7)), (8.16) follows from Lemma 4.4 once we verify the condition
that $|\epsilon_{y}-\epsilon_{y^{\prime}}|\leq C\,d(y,y^{\prime})/L$, where
now $\epsilon_{y}=g^{1/2}(y)\epsilon(1+d_{2}\epsilon))$. This follows easily
since $g$ is smooth and we can assume that $\frac{4}{5}\leq g^{1/2}(y)\leq 1$.
We also point out that for $\frac{1}{12}h_{L}\leq\epsilon\leq\frac{1}{3}h_{L}$
and $L$ large we have $\frac{1}{20}h_{L}\leq\epsilon_{y}\leq h_{L+1}$.

The lower bound is similar.

We note that $\frac{1}{3}h_{L+1}=\frac{1}{3e}h_{L}\geq\frac{1}{12}h_{L}$, so
all $\epsilon$ are covered by Lemma 8.2.

Lemma 8.2 is the analog of Theorem 1.2, but where now $\theta$ is the first
hitting time of $\partial D_{\ast}$, see (8.2). Theorem 1.2 then follows by
Brownian scaling. To spell this out for later use, let $\theta_{a}$ be the
first hitting time of $\partial B_{e}(0,a)$ and set

(8.17)
$\mu_{a}(x,\epsilon)=\frac{1}{\pi\epsilon^{2}}\int_{0}^{\theta_{a}}1_{\\{B(x,\epsilon)\\}}(W_{t})\,dt.$

Then it follows from Brownian scaling that for any $a,b>0$,

(8.18)
$\\{\mu_{a}(x,\epsilon_{x});\,x,\epsilon_{x}\\}\overset{\mbox{law}}{=}\\{\mu_{ba}(bx,b\epsilon_{x});\,x,\epsilon_{x}\\}.$

The left tail and then Theorem 1.1 can be proven similarly.

###  8.1. $0<r^{\ast}<\pi$

We first note the following extension of Lemma 8.2.

######  Lemma 8.3.

We can find $0<c,c^{\prime},z_{0}<\infty$ such that for $L$ large and all
$\frac{1}{12}h_{L}\leq\epsilon_{y}\leq\frac{1}{3}h_{L}$ with
$|\epsilon_{y}-\epsilon_{y^{\prime}}|\leq C|y-y^{\prime}|/L$ and $z_{0}\leq
z\leq\log L$,

(8.19)
$c^{\prime}ze^{-2z}\leq\mathbb{P}\left(\sup_{y}\mu_{\theta}(y,\epsilon_{y})\geq{1\over\pi}t_{L}\left(z\right)\right)\leq
cze^{-2z}.$

This follows as in the proof of Lemma 8.2, once we observe that in Lemma 8.1
we can allow the $\epsilon$ to depend on $x$.

It follows from (8.18) that for any fixed $a>0$, Lemma 8.3 holds with $\theta$
replaced by $\theta_{a}$.

We now show that Theorem 1.4 holds for any $0<r^{\ast}<\pi$. This is done by
using Lemma 8.1. That is, with $a=2\tan(r^{\ast}/2)$ we have that for some
$-\infty<d_{1},d_{2},d_{3},d_{4}<\infty$, all $x\in D_{a}$ and all $\epsilon$
sufficiently small

(8.20)
$\bar{\mu}_{\tau}(x,\epsilon)\leq(1+d_{1}\,\,\epsilon)\,\,\mu_{\theta_{a}}(x,g^{-1/2}(x)\epsilon(1+d_{2}\epsilon)),$

and

(8.21)
$\bar{\mu}_{\tau}(x,\epsilon)\geq(1+d_{3}\,\,\epsilon)\,\,\mu_{\theta_{a}}(x,g^{-1/2}(x)\epsilon(1+d_{4}\epsilon)),$

Theorem 1.4 then follows from Lemma 8.3 just as Lemma 8.2 followed from Lemma
4.4. Theorem 1.3 can be proven similarly.

##  9\. Appendix I: Barrier estimates

In what follows, we use the notation $H_{y,\delta}=[y,y+\delta]$ from [5]. The
following is a variant of [5, Theorem 1.1], which can be proven similarly. Re
set

(9.1) $f_{a,b}\left(l;L\right)=a+(b-a)\frac{l}{L}.$

######  Theorem 9.1.

a) For all fixed $\delta>0,C\geq 0$, $\eta>1$ and
$\varepsilon\in\left(0,\frac{1}{2}\right)$ we have, uniformly in $\sqrt{2}\leq
x,y\leq\eta L$ such that ${x^{2}}/{2}\in\mathbb{N}$, any $0\leq x\leq
a,\,\,0\leq y\leq b$, that

$\displaystyle P^{\mbox{\rm GW}}_{x^{2}/2}\left(\sqrt{2T_{l}}\leq
f_{a,b}\left(l;L\right)+Cl_{L}^{\frac{1}{2}-\varepsilon},l=1,\ldots,L-1,\sqrt{2T_{L}}\in
H_{y,\delta}\right)$ (9.2) $\displaystyle\hskip 144.54pt\;\leq
c\frac{\left(1+a-x\right)\left(1+b-y\right)}{L}\sqrt{\frac{x}{yL}}e^{-\frac{\left(x-y\right)^{2}}{2L}}.$

b) Let
$\mbox{Tube}_{C,\tilde{C}}(l;L)=[f_{x,y}\left(l;L\right)-\tilde{C}l_{L}^{\frac{1}{2}+\varepsilon},f_{a,b}\left(l;L\right)-Cl_{L}^{\frac{1}{2}-\varepsilon}]\
$. If, in addition to the conditions in part a), we also have
$\left(1+a-x\right)\left(1+b-y\right)\leq\eta L$, $\max(xy,|y-x|)\geq L/\eta$
and $\left[y,y+\delta\right]\cap\sqrt{2\mathbb{Z}}\neq\emptyset$, and
$\mbox{Tube}_{C,\tilde{C}}(l;L)\cap\sqrt{2\mathbb{N}}\neq\emptyset$ for all
l=1,…,L-1, then

$\displaystyle P^{\mbox{\rm
GW}}_{x^{2}/2}\left(\sqrt{2T_{l}}\in\mbox{Tube}_{C,\tilde{C}}(l;L),l=1,\ldots,L-1,\sqrt{2T_{L}}\in
H_{y,\delta}\right)$ (9.3) $\displaystyle\hskip 72.26999pt\;\geq
c\frac{\left(1+a-x\right)\left(1+b-y\right)}{L}\times\left(\sqrt{\frac{x}{yL}}\wedge
1\right)e^{-\frac{\left(x-y\right)^{2}}{2L}},$

and the estimate is uniform in such $x,y,a,b$ and all $L$.

Similar results hold if we delete the barrier condition on some fixed finite
interval.

For the last statement, we simply note that following the proof of [5, Lemma
2.3] we can show that the analogue of [5, Theorem 1.1] holds where we skip
some fixed finite interval.

Recall that

(9.4) $\rho_{L}=2-\frac{\log L}{L},\hskip
14.45377pt\alpha_{z,\pm}\left(l\right)=\alpha\left(l,L,z\right)=\rho_{L}l+z\pm
l_{L}^{1/4}.$

######  Lemma 9.2.

Let $m=k_{y}+1\leq\log L$. For any $k\geq L-\left(\log L\right)^{4}$, $0\leq
j\leq\alpha_{z,+}(k)/2$ and $z\leq\log L$

(9.5)
$\displaystyle\mathbb{P}\left[\sqrt{2T_{y,l}^{\tau}}\leq\alpha_{z,+}(l),\,l=m,\ldots,k-1;\,\sqrt{2T_{y,k}^{\tau}}\in
I_{\alpha_{z,+}(k)-j}\right]$ $\displaystyle\leq
ce^{-2k-2z-2k^{1/4}_{L}+2j}\times
m^{2}\left(1+z+m+k^{1/4}_{L}\right)\left(1+j\right).$

Proof of Lemma (9.2):  Using the Markov property, the probability in (9.5) is
bounded by

$\displaystyle\sum_{s=0}^{\alpha_{z,+}(m)}\mathbb{P}\left[\sqrt{2T_{y,m}^{\tau}}\in
I_{s}\right]$ $\displaystyle\hskip 72.26999pt\times\sup_{x\in
I_{s}}\mathbb{P}\left[\sqrt{2T_{y,l}^{m,x^{2}/2}}\leq\alpha_{z,+}(l),\,l=m+1,\ldots,k-1;\,\right.$
(9.6) $\displaystyle\left.\hskip 202.35622pt\sqrt{2T_{y,k}^{m,x^{2}/2}}\in
I_{\alpha_{z,+}(k)-j}\right],$

and using the fact that $T_{y,m}^{\tau}\leq T_{y,m}^{m,0}$ and (2), we see
that (9.6) is bounded by

$\displaystyle\sum_{s=0}^{\alpha_{z,+}(m)}e^{-s^{2}/2m}\sup_{x\in
I_{s}}\mathbb{P}\left[\sqrt{2T_{y,l}^{m,x^{2}/2}}\leq\alpha_{z,+}(l),\,l=m+1,\ldots,k-1;\,\right.$
(9.7) $\displaystyle\left.\hskip 202.35622pt\sqrt{2T_{y,k}^{m,x^{2}/2}}\in
I_{\alpha_{z,+}(k)-j}\right].$

Recall that $\alpha_{z,+}(l)=\rho_{L}l+z+l^{1/4}_{L}$. Using this we can write
the last probability as

$\displaystyle
K_{1,s}:=\mathbb{P}\left[\sqrt{2T_{y,l}^{m,x^{2}/2}}\leq\rho_{L}l+z+l^{1/4}_{L}\mbox{\,\
for }l=m+1,\ldots,k-1;\right.$ (9.8) $\displaystyle\left.\hskip
180.67499pt\sqrt{2T_{y,k}^{m,x^{2}/2}}\in
I_{\rho_{L}k+z+k^{1/4}_{L}-j}\right].$

Using the fact that for all $1\leq l\leq k-1$

(9.9) $l_{L}^{1/4}\leq l_{k}^{1/4}+k_{L}^{1/4},$

see [4], it follows that

$\displaystyle K_{1,s}\leq P^{\mbox{\rm
GW}}_{x^{2}/2}\left[\sqrt{2T_{l}}\leq\rho_{L}\left(m+l\right)+z+\left(m+l\right)^{1/4}_{k}+k_{L}^{1/4},\right.$
(9.10) $\displaystyle\left.\hskip 14.45377pt\mbox{\,\ for
}l=1,\ldots,k-m-1;\sqrt{2T_{k-m}}\in I_{\rho_{L}k+z+k^{1/4}_{L}-j}\right].$

Thus using (9.2), with $a=\rho_{L}m+z+(m^{1/4}_{k}+k_{L}^{1/4})$ and
$b=\rho_{L}k+z+k^{1/4}_{L}$, $y=\rho_{L}k+z+k^{1/4}_{L}-j$,

$K_{1,s}\leq
c\frac{\left(1+a-x\right)\left(1+j\right)}{k-m}\sqrt{\frac{x}{y\left(k-m\right)}}e^{-\frac{\left(\rho_{L}k+z+k^{1/4}_{L}-j-x\right)^{2}}{2\left(k-m\right)}}.$

We have

$\displaystyle
e^{-\frac{\left(\rho_{L}k+z+k^{1/4}_{L}-j-x\right)^{2}}{2\left(k-m\right)}}$
$\displaystyle\leq$ $\displaystyle
ce^{-\frac{\left(\rho_{L}k\right)^{2}}{2k(1-m/k)}}\,e^{-2\left(z+k_{L}^{1/4}-j-x\right)}$
$\displaystyle\leq$ $\displaystyle ce^{2k\frac{\log
L}{L}}e^{-2(k+m)-2z-2k_{L}^{1/4}+2j+2x},$

$\frac{1}{k-m}\sqrt{\frac{x}{y\left(k-m\right)}}\,\,e^{2k\frac{\log
L}{L}}\asymp\sqrt{x}$, and by assumption,

$a-x\leq c\left(k_{L}^{1/4}+m+z\right).$

Hence we can bound (9.7) by

$c\left(1+k_{L}^{1/4}+m+z\right)\left(1+j\right)e^{-2k-2z-2k_{L}^{1/4}+2j}\sum_{s=0}^{\alpha_{z,+}(m)}\sqrt{s}e^{-(s-2m)^{2}/2m}.$

Our Lemma follows.

######  Lemma 9.3.

For all $L$ sufficiently large, and all $0\leq z\leq\log L$,

$\displaystyle\mathbb{P}\left[\sqrt{2T_{y,l}^{1}}\leq\alpha_{z,-}\left(l\right)\mbox{\,\
for }l=1,\ldots,L-1;\,\sqrt{2T_{y,L}^{1}}\geq\rho_{L}+z\right]$
$\displaystyle\leq\mathbb{P}\left[\sqrt{2T_{y,l}^{1}}\leq\rho_{L}l+z\mbox{\,\
for }l=1,\ldots,L-1;\,\sqrt{2T_{y,L}^{1}}\geq\rho_{L}L+z\right]$ (9.11)
$\displaystyle\hskip 180.67499pt\leq c(1+z)e^{-2L-2z-z^{2}/4L}.$

and

$\displaystyle\mathbb{P}\left[\sqrt{2T_{y,l}^{1}}\leq\alpha_{z,-}\left(l\right)\mbox{\,\
for }l=1,\ldots,L-1;\,\sqrt{2T_{y,L}^{1}}\in I_{\rho_{L}L+z}\right]$ (9.12)
$\displaystyle\hskip 144.54pt\geq c(1+z)e^{-2L-2z-z^{2}/4L}.$

Similar results hold if we delete the barrier condition on some fixed finite
interval.

Proof of Lemma 9.3: The first inequality in (9.11) is obvious. Theorem 9.1
requires that $y\leq b$ which we will not have if we go all the way to $L$.
Instead, using the Markov property at $l=L-1$ and (3.19) we bound

$\displaystyle\mathbb{P}\left[\sqrt{2T_{y,l}^{1}}\leq\rho_{L}l+z\mbox{\,\ for
}l=1,\ldots,L-1;\,\sqrt{2T_{y,L}^{1}}\geq\rho_{L}+z\right]$ $\displaystyle\leq
c\sum_{j=1}^{\rho_{L}(L-1)+z}\mathbb{P}\left[\sqrt{2T_{y,l}^{1}}\leq\rho_{L}l+z\mbox{\,\
for }l=1,\ldots,L-2;\right.$ (9.13) $\displaystyle\left.\hskip
72.26999pt\sqrt{2T_{y,L-1}^{1}}\in I_{\rho_{L}(L-1)+z-j}\right]e^{-j^{2}/2}.$

If $j\geq L/2$, then $e^{-j^{2}/2}\leq e^{-L^{2}/8}$ so we get a bound much
smaller than (9.11). Thus we need only bound the sum over $1\leq j\leq L/2$.

It follows from (9.2), with $a=z$, $b=\rho_{L}(L-1)+z,\,y=\rho_{L}(L-1)+z-j$
that the last probability is bounded by

$\displaystyle
c\frac{(1+z)\left(1+j\right)}{L}\sqrt{\frac{1}{L^{2}}}e^{-\left(\rho_{L}(L-1)+z-j\right)^{2}/2(L-1)}$
$\displaystyle\hskip 72.26999pt\leq
c(1+z)\left(1+j\right)e^{-2L-2(z-j)-(z-j)^{2}/4L},$

and our upper bound follows after summing over $j$.

The lower bound follows similarly using (9.3). The last statement in our Lemma
comes from the last statement in Theorem 9.1.

######  Lemma 9.4.

If $k\geq L/2,\,\,0\leq z\leq\log L$ and $L$ is sufficiently large, then
uniformly in $0\leq p\leq k$,

(9.14)
$\displaystyle\mathbb{P}\left[\sqrt{2\mathcal{T}_{y,l}^{1}}\leq\rho_{L}l+z\mbox{\,\
for }l=1,\ldots,k-1;\,\,\sqrt{2\mathcal{T}_{y,k}^{1}}\in
I_{\rho_{L}k+z-p}\right]$ $\displaystyle\leq
C(1+z)\left(1+p\right)e^{-2k-2(z-p)-(z-p)^{2}/4k}.$

Proof of Lemma 9.4:  Using Theorem 9.1 with $a=z$,
$y=\rho_{L}k+z-p,b=\rho_{L}k+z$ this is bounded by

(9.15) $\displaystyle
c\frac{(1+z)\left(1+p\right)}{k}\sqrt{\frac{1}{k^{2}}}e^{-\left(\rho_{L}k+z-p\right)^{2}/2k}$
$\displaystyle\leq
C\frac{(1+z)\left(1+p\right)}{k^{2}}e^{2\log(L)k/L}e^{-2k-2(z-p)-(z-p)^{2}/4k}.$

(9.14) follows since by the convexity of $\log$ we have $e^{2\log(L)k/L}\leq
e^{2\log(k)}=k^{2}$.

######  Lemma 9.5.

If $k\leq\log^{5}L,\,\,0\leq z\leq\log L$ and $L$ is sufficiently large, then
uniformly in $m=\rho_{L}k+z-t$,

$\displaystyle\mathbb{P}\left[\sqrt{2\mathcal{T}_{y,l}^{k,m^{2}/2}}\leq\rho_{L}l+z\mbox{\,\
for
}l=k+1,\ldots,L-1;\,\,\sqrt{2\mathcal{T}_{y,L}^{k,m^{2}}}\geq\rho_{L}L+z\right]$
(9.16) $\displaystyle\hskip 72.26999pt\leq
c(1+t)m^{1/2}e^{-2(L-k)-2t-t^{2}/4(L-k)}.$

Proof of Lemma 9.5:  As before, we can bound the probability by

(9.17)
$\displaystyle\sum_{j=0}^{\rho_{L}(L-1)+z}\mathbb{P}\left[\sqrt{2\mathcal{T}_{y,l}^{k,m^{2}/2}}\leq\rho_{L}l+z\mbox{\,\
for }l=k+1,\ldots,L-2;\right.$ $\displaystyle\hskip
108.405pt\left.\,\,\sqrt{2T_{y,L-1}^{k,m^{2}/2}}\in
I_{\rho_{L}(L-1)+z-j}\right]e^{-j^{2}/2}.$

Also, as before, we need only bound the sum over $1\leq j\leq L/2$.

It follows from (9.2), with the $x$ of that estimate given by $m=a-t$ and
$a=\rho_{L}k+z,b=\rho_{L}(L-1)+z,y=\rho_{L}(L-1)+z-j$ that the last
probability is bounded by

(9.18)
$c\frac{\left(1+t\right)\left(1+j\right)}{L-k}\sqrt{\frac{m}{L(L-k)}}e^{-\frac{\left(\rho_{L}\left(L-k-1\right)+t-j\right)^{2}}{2\left(L-k-1\right)}},$

and

$\displaystyle
e^{-\frac{\left(\rho_{L}\left(L-k-1\right)+t-j\right)^{2}}{2\left(L-k-1\right)}}$
$\displaystyle\leq
ce^{-\frac{\left(\rho_{L}\left(L-k-1\right)\right)^{2}}{2(L-k-1)}-2(t-j)-(t-j)^{2}/2(L-k)}$
(9.19) $\displaystyle\leq
Ce^{2\log(L)(L-k)/L}e^{-2\left(L-k\right)-2(t-j)-(t-j)^{2}/2(L-k)}.$

This gives

$\displaystyle\mathbb{P}\left[\sqrt{2\mathcal{T}_{y,l}^{k,m^{2}/2}}\leq\rho_{L}l+z\mbox{\,\
for
}l=k+1,\ldots,L-1;\,\,\sqrt{2\mathcal{T}_{y,L}^{k,m^{2}/2}}\geq\rho_{L}L+z\right]$
$\displaystyle\leq
C\sum_{j=0}^{\rho_{L}(L-1)+z}\frac{\left(1+t\right)\left(1+j\right)}{L-k}\sqrt{\frac{m}{L(L-k)}}L^{\frac{2(L-k)}{L}}$
$\displaystyle\hskip
144.54pte^{-2\left(L-k\right)-2(t-j)-(t-j)^{2}/2(L-k)}e^{-j^{2}/2}.$

(9.16) follows since by the convexity of $\log$, $e^{2\log(L)(L-k)/L}\leq
e^{2\log((L-k))}=(L-k)^{2}$.

The following Lemma states the barrier estimates needed for the proof of the
left tail estimates in Theorem 1.3. For notation see Section 4.3. The proof of
this Lemma is similar to the proofs of Lemmas 9.3 -9.5.

######  Lemma 9.6.

For all $L$ sufficiently large, and all $|z|\leq\log L$,

$\displaystyle\mathbb{P}\left[\sqrt{2T_{y,l}^{1}}\leq\widehat{\gamma}_{z,-}\left(l\right)\mbox{\,\
for }l=1,\ldots,L-1;\,\sqrt{2T_{y,L}^{1}}\geq\rho_{L}L+z\right]$
$\displaystyle\leq\mathbb{P}\left[\sqrt{2T_{y,l}^{1}}\leq\widehat{\beta}_{z}(l)\mbox{\,\
for }l=1,\ldots,L-1;\,\sqrt{2T_{y,L}^{1}}\geq\rho_{L}L+z\right]$ (9.20)
$\displaystyle\hskip 216.81pt\leq ce^{-2L-2z-z^{2}/4L}.$

and

$\displaystyle\mathbb{P}\left[\sqrt{2T_{y,l}^{1}}\leq\widehat{\gamma}_{z,-}\left(l\right)\mbox{\,\
for }l=1,\ldots,L-1;\,\sqrt{2T_{y,L}^{1}}\in I_{\rho_{L}L+z}\right]$ (9.21)
$\displaystyle\hskip 216.81pt\geq ce^{-2L-2z-z^{2}/4L}.$

Similar results hold if we delete the barrier condition on some fixed finite
interval.

If $k\geq L/2,\,\,|z|\leq\log L$, and $L$ is sufficiently large, then
uniformly in $p\leq k$,

(9.22)
$\displaystyle\mathbb{P}\left[\sqrt{2\mathcal{T}_{y,l}^{1}}\leq\widehat{\beta}_{z}(l)\mbox{\,\
for }l=1,\ldots,k-1;\,\,\sqrt{2\mathcal{T}_{y,k}^{1}}\in
I_{\widehat{\beta}_{z}(k)-p}\right]$ $\displaystyle\leq
C\left(1+p\right)e^{-2k-2(z-p)-(z-p)^{2}/4k}.$

If $k\leq\log^{5}L,\,\,|z|\leq\log L$, and $L$ is sufficiently large, then
uniformly in $m=\widehat{\beta}_{z}(k)-t$,

(9.23)
$\displaystyle\mathbb{P}\left[\sqrt{2\mathcal{T}_{y,l}^{k,m^{2}}}\leq\widehat{\beta}_{z}(l)\mbox{\,\
for
}l=k+1,\ldots,L-1;\,\,\sqrt{2\mathcal{T}_{y,L}^{k,m^{2}}}\geq\rho_{L}+z\right]$
$\displaystyle\leq c(1+t)m^{1/2}e^{-2(L-k)-2t-t^{2}/4(L-k)}.$

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