Moreover, the asymptotic estimate

$g_{\lambda}(x)\sim-\frac{2}{x}\qquad\text{as $x\to\infty$}$ (4.18)

holds for each fixed $\lambda\in(0,1)$.

A consequence of this lemma is that $g_{\lambda}$ remains roughly constant
from $x_{1}=O(\log 1/\lambda)$ until the much later time when $x$ is of order
$1/\lambda$.

###### Proof of Lemma 4.5.

We continue to write $g=g_{\lambda}$ and define
$\delta:=2g^{\prime}-g^{2}=-g^{\prime\prime}$. We first prove that

$\sup_{x\geq x_{1}}|g^{\prime\prime}(x)|\leq O(\lambda^{3})\qquad\text{as
$\lambda\downarrow 0$.}$ (4.19)

The lower bound $g^{\prime\prime}(x)\geq 0$ for all $x\geq x_{1}$ follows from
Lemma 4.2. To prove the other bound, first observe that $\delta$ is a
nonnegative function on $[x_{1},\infty)$ vanishing at $x_{1}$ and converging
to 0 at infinity (see Lemma 3.12). Therefore, it reaches its maximum at some
point $x_{2}>x_{1}$ where $\delta^{\prime}=-2\delta-2gg^{\prime}$ vanishes.
This shows that $\max_{x\geq x_{1}}\delta(x)\leq\max_{x\geq
x_{1}}|gg^{\prime}(x)|$. Since $|g|$ and $g^{\prime}$ are both non increasing
on $[x_{1},\infty)$ we have that $\max_{x\geq
x_{1}}|gg^{\prime}(x)|=|gg^{\prime}(x_{1})|$, and the claim follows from Lemma
4.4.

We now use the estimate $\sup_{x\geq
x_{1}}|g^{\prime\prime}_{\lambda}(x)|=O(\lambda^{3})$ to prove (4.17). We
begin by proving the claimed estimate under the additional assumption that
$\limsup x_{\lambda}^{2}\lambda^{3}=0$; large values of $x_{\lambda}$
satisfying $x_{\lambda}=\Omega(\lambda^{-3/2})$ will be handled afterwards,
simultaneously to the proof of (4.18). Since $g_{\lambda}$ is increasing on
$[x_{0},\infty)$ and $x_{0}\sim\frac{1}{2}\log(1/\lambda)$ as
$\lambda\downarrow 0$, it suffices to prove the claim under the stronger
assumption that $\liminf\lambda x_{\lambda}>0$. Fix $\nu_{*}<1/2$; all
implicit constants in the remainder of the proof may depend on $\nu_{*}$. Let
$0<\nu<\nu_{*}$ be a parameter and let
$x(\nu):=\inf\\{x>x_{1}:g(x)\geq-\nu\lambda\\}$. On $[x_{1},x(\nu)]$, the
absolute value $|g|$ is larger than $\nu\lambda$, so (4.19) implies that

$\frac{1}{2}g^{2}\leq
g^{\prime}\leq\frac{1}{2}g^{2}+O(\lambda^{3})=(1+O(\nu^{-2}\lambda))\frac{1}{2}g^{2}\quad\text{on~{}}[x_{1},x(\nu)].$

Integrating this relation between $x_{1}$ and $x(\nu)$ by considering the
implied inequalities on the derivative of $1/g$ yields

$\frac{1}{2}(x(\nu)-x_{1})\leq\frac{1}{g(x_{1})}-\frac{1}{g(x_{2})}\leq(1+O(\nu^{-2}\lambda))\frac{1}{2}(x(\nu)-x_{1}).$
(4.20)

Together with (4.20), this implies that

$x(\nu)-x_{1}=(1+O((\nu^{-2}\lambda)\left[\frac{2}{g(x_{1})}-\frac{2}{g(x_{2})}\right]=(1+O(\nu^{-2}\lambda))\left(\frac{2}{\nu}-4\right)\frac{1}{\lambda}$

where we safely absorbed the $1+O(\lambda)$ prefactor on $g(x_{1})$ into the
$(1+O((\nu^{-2}\lambda))$ prefactor in the second expression since $2/\nu\geq
2/\nu_{*}$ is bounded away from $4$. On the other hand, by Lemmas 4.3 and 4.4,
$x_{1}=o(1/\lambda)$, so that we can write simply

$x(\nu)=(1+O(\nu^{-2}\lambda))\left(\frac{2}{\nu}-4\right)\frac{1}{\lambda}.$

Inverting this expression yields the claim under the assumption that $\limsup
x_{\lambda}^{2}\lambda^{3}=0$. (The assumption that $\liminf\lambda
x_{\lambda}>0$, which we made at the beginning of this paragraph, lets us pick
$\nu_{*}<1/2$ and work only with $\nu<\nu_{*}$. The assumption that $\limsup
x^{2}_{\lambda}\lambda^{3}=0$ ensures that the
$O(\nu^{-2}\lambda)=O(x^{2}(\nu)\lambda^{3})$ term is negligible.)

We now prove (4.18) along with the case of (4.17) in which $\liminf
x_{\lambda}\lambda=\infty$; together with our previous analysis of the case
$\limsup x_{\lambda}^{2}\lambda^{3}$ this establishes (4.17) in all relevant
asymptotic regimes (indeed, there is a significant overlap in the regimes
treated by the two arguments). Since $g^{\prime\prime}=g^{2}-2g$ is negative
on $[x_{1},\infty)$, we can integrate the relation
$(-1/g)^{\prime}=g^{\prime}/g^{2}\geq 1/2$ to obtain that

$-\frac{1}{g(x)}+\frac{1}{g(x_{1})}\geq\frac{x-x_{1}}{2}$

for every $x\geq x_{1}$, and hence that

$g(x)\geq-\frac{2}{x-x_{1}-2/g(x_{1})}=-\frac{2}{x-O(\lambda^{-1})}$

for every $x\geq x_{1}$, which yields the desired asymptotic lower bound when
$\liminf x_{\lambda}\lambda=\infty$. We now prove the matching upper bound.
Fix $0<M<\infty$ and let $x>x_{1}\vee\lambda^{-1}M$. Recall that
$g=g_{\lambda}$ is the unique solution to $g^{\prime\prime}+2g^{\prime}=g^{2}$
with $g(0)=0$ and initial slope $g^{\prime}(0)=-\lambda$. Since $\lambda$ is
larger than $M/x$ and smaller than $1$, Lemma 3.9 implies that $g$ is bounded
above by the solution with initial slope $-M/x$, and it follows from the
$\limsup x_{\lambda}^{2}\lambda^{3}=0$ case of the lemma treated above that

$g(x)\leq g_{M/x}(x)\sim-\frac{2M}{(M+4)x}$

as $x\to\infty$. The lower bound of the claimed estimate follows since $M$ was
arbitrary. ∎

###  4.3 Upper bounds on pioneers

In this section, we combine Theorem 3.1 and the results of Section 4.2 to
establish a sharp upper bound on the tail of the number of pioneers. We then
use this bound on the tail to bound from above the number of thick points,
establishing the upper bound of Theorem 1.5. We begin by proving the following
proposition, which implies the upper bound of Theorem 1.3.

######  Proposition 4.6.

Let $x_{0}>0$. For each $\varepsilon>0$, there exist constants
$C=C(\varepsilon)>0$ and $x_{*}=x_{*}(x_{0},\varepsilon)>x_{0}$ such that

$\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{r}\geq\frac{r^{2}\log(R/r)}{2}n\right)\leq\frac{C}{(e^{-x_{0}}R)^{2}\log(R/r)}e^{-(1-\varepsilon)n}$
(4.21)

for every $1\leq r\leq e^{-x_{*}}R$ and $n\geq 1$.

To see that this implies the upper bound of Theorem 1.3, simply take
$n=\lfloor a\log(R/r)\rfloor$ and apply the hitting probability estimate (4.2)

$\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{r}>0\right)\geq\frac{c(x_{0})}{(e^{-x_{0}}R)^{2}\log(R/r)}.$
(4.22)

###### Proof of Proposition 4.6.

Let $x=\log(R/r)$ so that $r=e^{-x}R$. Let $x_{0}>0$, let
$\mu=\mu(\varepsilon)>0$ be large enough to ensure that $t:=\mu/(4+\mu)$ is at
least $1-\delta$, and let $x_{1}=x_{1}(\varepsilon,x_{0})>x_{0}$ be such that
if $x\geq x_{1}$ then $\mu/x\leq 1/2$. (The constant $1/2$ could be replaced
by anything positive and strictly less than $1$.) Applying Theorem 3.1 with
$\lambda=\mu/x$, we obtain that

$\frac{1}{(e^{-x_{0}}R)^{2}}g_{\mu/x}(x_{0})=1-\mathbb{E}_{e^{-x_{0}}R,R}\left[\left(1-\frac{g_{\mu/x}(x)}{(e^{-x}R)^{2}}\right)^{\\!N_{r}}\,\right].$

Using the asymptotics for $g_{\mu/x}(x)$ proven in Lemma 4.5 and the bound
$g_{\mu/x}(x_{0})\geq-\sup_{[0,\infty)}|g_{\mu/x}|\geq-\mu/(2x)$ from Lemma
3.12, we deduce that there exists a quantity $\delta(x,\mu)$ satisfying
$\lim_{x\to\infty}\delta(x,\mu)=0$ such that

$1-\mathbb{E}_{e^{-x_{0}}R,R}\left[\left(1+\frac{(1+\delta(x,\mu))2\mu}{(\mu+4)x(e^{-x}R)^{2}}\right)^{\\!N_{r}}\,\right]\geq-\frac{\mu}{2x(e^{-x_{0}}R)^{2}}$

Recall that $t=\mu/(4+\mu)$, so that $\mu/2=2t/(1-t)$. Rewriting the above
slightly, we obtain that there exists a quantity $\delta^{\prime}(x,\mu)$
satisfying $\lim_{x\to\infty}\delta^{\prime}(x,\mu)=0$ such that

$\mathbb{E}_{e^{-x_{0}}R,R}\left[\exp\left((1+\delta^{\prime}(x,\mu))t\frac{2N_{r}}{x(e^{-x}R)^{2}}\right)\mathbf{1}_{\left\\{N_{r}>0\right\\}}\right]\leq\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{r}>0\right)+\frac{t}{1-t}\frac{2}{x(e^{-x_{0}}R)^{2}}.$

By Lemma 4.1, the probability that $N_{r}>0$ is not larger than
$(1+o(1))2/(x(e^{-x_{0}}R)^{2})$ as $x\to\infty$. Using this and Markov’s
inequality, we can now bound the left hand side of (4.21) by

$\displaystyle
e^{-(1+o(1))tn}\mathbb{E}_{e^{-x_{0}}R,R}\left[\exp\Big{(}(1+\delta^{\prime}(x,\mu))t\frac{2N_{r}}{x(e^{-x}R)^{2}}\Big{)}\mathbf{1}_{\left\\{N_{r}>0\right\\}}\right]\leq\frac{1+o(1)}{1-t}\frac{2}{x(e^{-x_{0}}R)^{2}}e^{-t(1+o(1))n}.$

Since we took $\mu$ large enough to ensure that $t\geq 1-\delta$, this proves
the proposition. ∎

We now explain how the upper bound of Theorem 1.5 follows from Proposition
4.6.

###### Proof of Theorem 1.5 – Upper bound.

We start by arguing that it is enough to show the upper bound with the killing
on the outer sphere $\partial B(R)$ (or more precisely, on $\partial B(AR)$
for $A>0$ large). Indeed, conditionally on surviving $R^{2}$ generations, the
probability that a particle reaches $\partial B(AR)$ goes to zero as
$R\to\infty$ and $A\to\infty$. So we can a priori restrict ourselves to this
event and add a killing on $\partial B(AR)$.

We now prove the upper bound under
$\mathbb{P}_{z_{0},R}(\,\cdot\,|\zeta>R^{2})$, where we recall that $\zeta$
denotes the extinction time of the branching process. Let $a>0$. Recall the
definition (1.9) of the set $\mathcal{T}_{R}(a)$ of $a$-thick points. We
compute

$\displaystyle\mathbb{E}_{z_{0},R}\left[\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\mathcal{T}_{R}(a)\right\\}}\right]\leq\sum_{z\in\mathbb{Z}^{4}\cap
B(0,R)}\mathbb{P}_{z_{0},R}\left(N_{z,1}\geq\frac{a}{2}(\log R)^{2}\right).$

Let $z\in\mathbb{Z}^{4}\cap B(0,R)$. By killing the particles on the sphere
$\partial B(z,2R)$, which is strictly larger than the sphere $\partial B(R)$,
we reduce the problem to a spherically symmetric situation once more.
Proposition 4.6 then gives that

$\mathbb{P}_{z_{0},R}\left(N_{z,1}\geq\frac{a}{2}(\log R)^{2}\right)\leq
R^{-2-a+o(1)}.$

Because $\mathbb{P}_{z_{0},R}(\zeta>R^{2})=R^{-2+o(1)}$ (by Kolmogorov’s
theorem about the extinction probability of a critical Bienaymé Galton–Watson
tree; see e.g. [KNS66, Corollary 1]), the expectation of
$\\#\mathcal{T}_{R}(a)$ conditional on $\\{\zeta>R^{2}\\}$ is bounded by
$R^{4-a+o(1)}$. By Markov’s inequality, we deduce that for any
$\varepsilon>0$,

$\lim_{R\to\infty}\mathbb{P}_{z_{0},R}\left(\\#\mathcal{T}_{R}(a)\geq
R^{4-a+\varepsilon}\,|\,\zeta>R^{2}\right)=0.$

This gives the desired upper bound on the number of thick points. It also
gives the upper bound on $\sup_{z\in\mathbb{Z}^{4}}N_{z,1}$ since
$\mathcal{T}_{R}(a)$ is empty with high
$\mathbb{P}_{z_{0},R}(\cdot|\zeta>R^{2})$-probability as $R\to\infty$ for each
$a>4$. ∎

##  5 Asymptotic expansion of solutions (proof of Theorem 1.7)

The goal of this section is to prove the following theorem, which establishes
infinite-order asymptotic expansions for solutions to the ODEs satisfied by
the functions $\mathsf{g}_{s}$ and $\mathsf{h}_{s}$. The following theorem
immediately implies 1.7 since

$h(x)=e^{2x}\mathbb{P}_{e^{x},\infty}(N_{1}>0)$

is a non-zero solution to the ODE $h^{\prime\prime}-2h^{\prime}=h^{2}$ defined
on $[0,\infty)$.

######  Theorem 5.1.

Let $h$ be any solution to $h^{\prime\prime}-2h^{\prime}=h^{2}$ defined on
$[0,\infty)$ that is not identically zero, and let $g$ be any maximal solution
to the initial value problem

$\left\\{\begin{array}[]{l}g^{\prime\prime}+2g^{\prime}=g^{2},\\\
g(0)=0,~{}g^{\prime}(0)=-\lambda\end{array}\right.$

with $\lambda\in(-1,0)$. Then $h(x)$ is positive for all $x\geq 0$, $g(x)$ is
defined on $[0,\infty)$ and negative for all $x\geq 0$, and there exist two
sequences $(Q_{n})_{n\geq 1}$ and $(P_{n})_{n\geq 1}$ of polynomials such that
for any $N\geq 1$,

$g(x)=\sum_{n=1}^{N}\frac{Q_{n}(\log x)}{x^{n}}+o(x^{-N})\quad\text{and}\quad
h(x)=\sum_{n=1}^{N}\frac{P_{n}(\log
x)}{x^{n}}+o(x^{-N}),\quad\quad\text{as~{}}x\to\infty.$

Moreover, there exist constants $C_{\lambda},C_{*}\in\mathbb{R}$, with $C_{*}$
depending on the choice of $h$, such that $Q_{1}=-2$,
$Q_{2}(X)=2X+C_{\lambda}$, $P_{1}=2$, $P_{2}(X)=2X+C_{*}$ and for all $n\geq
3$, $Q_{n}$ and $P_{n}$ are the unique polynomials satisfying the recurrence
relations

$(2n-4)Q_{n}-2Q_{n}^{\prime}=Q_{n-1}^{\prime\prime}-(2n-1)Q_{n-1}^{\prime}+n(n-1)Q_{n-1}-\sum_{k=2}^{n-1}Q_{k}Q_{n+1-k}$
(5.1)

and

$(2n-4)P_{n}-2P_{n}^{\prime}=-P_{n-1}^{\prime\prime}+(2n-1)P_{n-1}^{\prime}-(n-1)nP_{n-1}+\sum_{k=2}^{n-1}P_{k}P_{n+1-k}.$
(5.2)

######  Remark 5.2.

As already mentioned, the main difficulty in proving this theorem comes from
the fact that the expansion $\sum_{n=1}^{\infty}\frac{P_{n}(\log x)}{x^{n}}$
diverges, i.e., is not summable for any finite $x$. Although we do not _prove_
that this series diverges, we note that this feature is already present for
the solutions of the linearised equation

$h_{\text{lin}}^{\prime\prime}-2h_{\text{lin}}^{\prime}=\frac{2}{x}h_{\text{lin}},\quad
h_{\text{lin}}(\infty)=0.$

Indeed such solutions can be explicitly written in terms of the exponential
integral $\text{Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}{t}\mathrm{d}t$:

$h_{\text{lin}}(x)=c\left(1+2xe^{2x}\text{Ei}(-2x)\right),\quad
c\in\mathbb{R},$

from which we obtain that

$h_{\text{lin}}(x)=c\sum_{n=1}^{N-1}(-1)^{n+1}\frac{n!}{(2x)^{n}}+O\left(N!(2x)^{-N}\right).$

Although the series on the right hand side is also not a summable series, it
is however a classical example of a Borel summable series. (Since we do not
need this here, we do not explain the details of this notion. See [Har00] for
background on this topic.) This suggests that the series in Theorem 5.1 may
also be summable in the Borel sense. We do not investigate this in this paper.

In particular, from Theorem 5.1, we see that there is a constant $C_{\lambda}$
such that

$g_{\lambda}(x)=-\frac{2}{x}+\frac{2\log
x+C_{\lambda}}{x^{2}}+o(x^{-2})\quad\quad\text{as}~{}x\to\infty.$ (5.3)

The second goal of this section will be to study the constant $C_{\lambda}$
appearing in this asymptotic expansion.

The following lemma gives an alternative expression for $C_{\lambda}$ and
establishes asymptotic estimates on $C_{\lambda}$ as $\lambda\downarrow 0$.

######  Lemma 5.3.

For all $\lambda\in(0,1)$ and $x_{0}>0$,

$C_{\lambda}=-\int_{x_{0}}^{\infty}\left(\frac{g_{\lambda}^{\prime\prime}(x)}{g_{\lambda}(x)^{2}}+\frac{1}{x}\right)\mathrm{d}x-\frac{2}{g_{\lambda}(x_{0})}-x_{0}-\log
x_{0}.$ (5.4)

Moreover, the function $\lambda\in(0,1)\mapsto C_{\lambda}$ is continuous,
decreasing, and satisfies $C_{\lambda}\sim 4/\lambda$ as $\lambda\downarrow
0$.

Figure 5.1: Left: Numerical plot of
$x^{2}[g_{\lambda}(x)-(-\frac{2}{x}+\frac{2\log x}{x^{2}})]$ with $x=1000$,
which is a numerical approximation to $C_{\lambda}$, as a function of
$\lambda$. Right: Numerical approximations to the difference of $g_{0.5}$ and
the three approximations $-2/x$ (blue), $-2/x+2\log x/x^{2}$ (red), and
$-2/x+(2\log x+9.209)/x^{2}$ (green), where the numerical value
$C_{0.5}\approx 9.209$ was extracted from the plot on the left.

The fact that the right hand side of (5.4) is independent of $x_{0}$ follows
from the fact that $g_{\lambda}$ satisfies the ODE
$g_{\lambda}^{\prime\prime}+2g_{\lambda}^{\prime}=g_{\lambda}^{2}$, so that
$g_{\lambda}^{\prime\prime}/g_{\lambda}^{2}=1-2g^{\prime}_{\lambda}/g_{\lambda}^{2}=(x+2/g_{\lambda})^{\prime}$.

We prove Theorem 5.1 in Sections 5.1 and 5.2 and Lemma 5.3 in Section 5.3.

###  5.1 Series expansion of $g$

This section is dedicated to the proof of the part of Theorem 5.1 concerning
the solutions of $g^{\prime\prime}+2g^{\prime}=g^{2}$. The derivation of the
asymptotic expansion of solutions to $h^{\prime\prime}-2h^{\prime}=h^{2}$ is
closely analogous except that a different argument is required for the first-
order estimate $h\sim 2/x$, and an appropriate replacement is needed for the
unimodality properties proven for $g$ in Lemma 4.2; this is discussed in more
detail in the next subsection.

We now outline the main steps of the proof. The first-order estimate
$g_{\lambda}(x)\sim-2/x$ was already proven in Lemma 4.5. In Lemma 5.4, we
warm up to the full asymptotic expansion by obtaining its second term, i.e.,
compute the asymptotic behaviour of $g$ up to an error of order $o(x^{-2})$.
We will then show in Lemma 5.5 that the proof of this preliminary estimate can
be generalised to obtain the existence of such an expansion for both
$g_{\lambda}$ and $g_{\lambda}^{\prime}$. The combinatorics of this step will
be fairly involved and the induction relation (5.1) will be difficult to
extract directly from the proof. Instead, we will show separately that if such
an expansion holds for both $g_{\lambda}$ and $g_{\lambda}^{\prime}$, then the
polynomials $Q_{n}$ have to satisfy (5.1). This final step will be
straightforward since we will have already laid the groundwork to be able to
differentiate term by term.

We now begin to carry out this plan. We begin by estimating $g_{\lambda}$ and
$g^{\prime}_{\lambda}$ up to errors of order $o(x^{-2})$ and $o(x^{-3})$
respectively.

######  Lemma 5.4.

Let $\lambda\in(0,1)$. There exists a constant $C_{\lambda}\in\mathbb{R}$ such
that

$\displaystyle g_{\lambda}(x)$ $\displaystyle=\frac{-2}{x}+\frac{2\log
x+C_{\lambda}}{x^{2}}+o\left(\frac{1}{x^{2}}\right)\qquad\text{and}$ (5.5)
$\displaystyle g_{\lambda}^{\prime}(x)$
$\displaystyle=\frac{2}{x^{2}}-\frac{4\log
x+2C_{\lambda}-2}{x^{3}}+o\left(\frac{1}{x^{3}}\right)$ (5.6)

as $x\to\infty$.

###### Proof of Lemma 5.4.

To lighten notation we write $g$ instead of $g_{\lambda}$. We also define
$\delta:=2g^{\prime}-g^{2}$. Although $\delta$ is equal to
$-g^{\prime\prime}$, we prefer to view it as the ‘error term’ describing the
extend to which $g$ fails to satisfy the separable ODE $2g^{\prime}=g^{2}$.
The function $\delta$ satisfies

$\delta^{\prime}=-2\delta-2gg^{\prime}=-(2-|g|)\delta+|g|^{3}.$ (5.7)

Define $x_{1}:=\inf\\{x>0:\delta(x)=0\\}$ as in Section 4.2. We showed in
Lemma 4.2 that $x_{1}<\infty$ and that $\delta$ is nonnegative on
$[x_{1},\infty)$. We will iteratively prove a sequence of estimates on $g$ and
$\delta$ of increasing strength.

Upper bound on $\delta$: Using the estimate $g(x)\sim-2/x$ in the differential
equation (5.7), we obtain that

$\delta^{\prime}(x)=-\left(2-\frac{2+o(1)}{x}\right)\delta(x)+\frac{8+o(1)}{x^{3}}\qquad\text{
as $x\to\infty$}.$ (5.8)

and hence that if $x$ is sufficiently large (so that the $(2+o(1))/x$ term has
absolute value at most $1$ and the $(8+o(1))/x^{3}$ term is at most
$16/x^{3}$) then

$\delta^{\prime}(x)+\delta(x)\leq\frac{16}{x^{3}}.$

Let $x_{2}\geq x_{1}$ be such that this estimate holds for all $x\geq x_{2}$.
After integration (Lemma 3.10) this gives for all $x>x_{2}$,

$\delta(x)\leq
e^{-x}\int_{x_{2}}^{x}e^{t}\frac{16}{t^{3}}dt+e^{-x+x_{2}}\delta(x_{2})=O(x^{-3}).$

We now plug this bound back into (5.8) to obtain that

$\delta^{\prime}(x)+2\delta(x)\leq\frac{8+o(1)}{x^{3}}+O(x^{-4})=\frac{8+o(1)}{x^{3}}.$

We integrate once more (using Lemma 3.10) to obtain that

$\delta(x)\leq
e^{-2x}\int_{x_{2}}^{x}e^{2t}\left[\frac{8+o(1)}{x^{3}}\right]dt+e^{-2x+x_{2}}\delta(x_{2})=\frac{4+o(1)}{x^{3}},$
(5.9)

where the final estimate follows by integration by parts.

Lower bound on $\delta$: We now use the estimate $g(x)\sim-2/x$ to prove a
lower bound on $\delta$. Plugging the bound (4.18) into (5.7) and using that
$|g|\delta\geq 0$ yields that

$\delta^{\prime}(x)+2\delta(x)\geq|g(x)|^{3}\geq\frac{8+o(1)}{x^{3}}$

as $x\to\infty$, and integrating this bound with Lemma 3.10 implies that

$\delta(x)\geq
e^{-2x}\int_{x_{2}}^{x}e^{2t}\left[\frac{4+o(1)}{x^{3}}\right]dt+e^{-2x+x_{2}}\delta(x_{2})=\frac{4+o(1)}{x^{3}}\qquad\text{as
$x\to\infty$},$ (5.10)

where again we use integration by parts to bound the relevant integrals.
Together with (5.9) this yields that $\delta(x)\sim 4/x^{3}$ as $x\to\infty$.

Improved lower bound on $g$: We now use the estimate $\delta(x)\sim 4/x^{3}$
to improve our lower bound on $g$. We have by (5.9) and (4.18) that

$2g^{\prime}(x)-g(x)^{2}=\delta(x)\sim\frac{4+o(1)}{x^{3}}\sim\frac{1}{x}g(x)^{2}\qquad\text{as
$x\to\infty$}$

and hence that

$-\left(\frac{1}{g(x)}+\frac{x}{2}\right)^{\prime}=\frac{g^{\prime}(x)}{g(x)^{2}}-\frac{1}{2}=\frac{1+o(1)}{2x}\qquad\text{as
$x\to\infty$.}$

It follows in particular that if $x$ is sufficiently large, then the left hand
side term is at most $1/x$ and integrating this bound yields that

$|g(x)|\geq\frac{2}{x+2\log(x)+O(1)}\qquad\text{ as $x\to\infty$.}$ (5.11)

Note that this is not yet a sharp estimate since the coefficient of $\log x$
is $2$ rather than $1$.

Completing the proof: We are now ready to conclude the proof. Consider the
telescoping sum

$\displaystyle\int_{1}^{x}\frac{\delta(t)}{g(t)^{2}}\mathrm{d}t=\log
x+\int_{1}^{x}\left(\delta(t)-\frac{4}{t^{3}}\right)\frac{1}{g(t)^{2}}\mathrm{d}t+\int_{1}^{x}\frac{4}{t^{3}}\left(\frac{1}{g(t)^{2}}-\frac{t^{2}}{4}\right)\mathrm{d}t.$

The estimates (4.18), (5.9), (5.11) and (5.10) on $|g|$ and $\delta$ show that
the two integrals on the right hand side converge absolutely as $x\to\infty$.
Therefore, there exists some $C_{0}\in\mathbb{R}$ such that

$\int_{1}^{x}\frac{\delta(t)}{g(t)^{2}}\mathrm{d}t=\log
x+C_{0}+o(1)\quad\quad\text{as~{}}x\to\infty.$ (5.12)

But by definition of $\delta$, the left hand side can also be expressed as

$\int_{1}^{x}\frac{\delta(t)}{g(t)^{2}}\mathrm{d}t=\int_{1}^{x}\left[\frac{2g^{\prime}}{g(t)^{2}}-1\right]\mathrm{d}t=-\frac{2}{g(x)}+\frac{2}{g(1)}-(x-1).$

Rearranging, we obtain that

$g(x)=\frac{-2}{x+\log x+C_{0}-2/g(1)-1+o(1)}\qquad\text{ as $x\to\infty$,}$
(5.13)

concluding the proof of (5.5). The estimate (5.6) follows and the relation
$2g^{\prime}=g^{2}+\delta$ since $\delta\sim 4/x^{3}$. ∎

We next extend this analysis to arbitrary order, without yet determining the
relevant polynomials.

######  Lemma 5.5.

Let $\lambda\in(0,1)$. There exist two sequences of polynomials
$(Q_{n})_{n\geq 1}$ and $(R_{n})_{n\geq 2}$ such that for any $N\geq 1$,

$g_{\lambda}(x)=\sum_{n=1}^{N}\frac{Q_{n}(\log x)}{x^{n}}+o(x^{-N}),\quad
g_{\lambda}^{\prime}(x)=\sum_{n=2}^{N+1}\frac{R_{n}(\log
x)}{x^{n}}+o(x^{-N-1}),\quad\quad\text{as~{}}x\to\infty.$ (5.14)

###### Proof of Lemma 5.5.

We will prove the statement by induction on $N$, the case $N=2$ having already
been treated in Lemma 5.4. Let $N\geq 2$ and suppose that there exist
polynomials $(Q_{n})_{n=1,\ldots,N}$ and $(R_{n})_{n=2,\ldots,N+1}$ such that
(5.14) holds. As in the proof of Lemma 5.4, we let
$\delta:=2g^{\prime}-g^{2}$, which satisfies

$\displaystyle\delta^{\prime}+2\delta$
$\displaystyle=-2gg^{\prime}=-2\left(\sum_{n=1}^{N}\frac{Q_{n}(\log
x)}{x^{n}}\right)\left(\sum_{n=2}^{N+1}\frac{R_{n}(\log
x)}{x^{n}}\right)+o(x^{-N-2})=\sum_{n=3}^{N+2}\frac{Q_{n}^{(1)}(\log
x)}{x^{n}}+o(x^{-N-2}).$

Here and in the rest of the proof, we write $(Q_{n}^{(1)})_{n\geq 1}$,
$(Q_{n}^{(2)})_{n\geq 1}$, and so for certain polynomials arising in our
computations. Although these polynomials can be written explicitly in terms of
$(Q_{n})_{n=1\dots N}$ and $(R_{n})_{n=2\dots N+1}$, we keep the dependence
inexplicit for now and return to their precise computation later once we know
we can differentiate term-by-term. Using that
$(e^{2x}\delta)^{\prime}=e^{2x}(\delta^{\prime}+2\delta)$, we can integrate
this equality to obtain that

$\displaystyle\delta(x)$
$\displaystyle=e^{-2x}\delta(1)+e^{-2x}\sum_{n=3}^{N+2}\int_{1}^{x}e^{2t}\frac{Q_{n}^{(1)}(\log
t)}{t^{n}}\mathrm{d}t+e^{-2x}\int_{1}^{x}e^{2t}o(t^{-N-2})\mathrm{d}t$
$\displaystyle=\sum_{n=3}^{N+2}\frac{Q_{n}^{(2)}(\log x)}{x^{n}}+o(x^{-N-2})$
(5.15)

as $x\to\infty$. It follows that

$\displaystyle\frac{\delta(x)}{g(x)^{2}}=(1+o(x^{-N+1}))\left(\sum_{n=3}^{N+2}\frac{Q_{n}^{(2)}(\log
x)}{x^{n}}\right)\left(\sum_{n=1}^{N}\frac{Q_{n}(\log x)}{x^{n}}\right)^{-2}.$

Since $Q_{1}=-2$ is a constant function, we can expand the negative binomial
to rewrite this as

$\frac{\delta(x)}{g(x)^{2}}=\sum_{n=1}^{N}\frac{Q_{n}^{(3)}(\log
x)}{x^{n}}+o(x^{-N})=\frac{1}{x}+\sum_{n=2}^{N}\frac{Q_{n}^{(3)}(\log
x)}{x^{n}}+o(x^{-N}),$ (5.16)

where we used the first-order asymptotics for $\delta$ and $g$ computed in
Lemma 5.4 to compute the first term of the asymptotic series. Note that here
it is crucial that the polynomial $Q_{1}$ is constant (equal to $-2$),
otherwise the above negative-binomial expansion would involve rational
functions of $\log x$ instead of polynomials.

The estimate (5.16) implies that

$\int_{x}^{\infty}\left(\frac{\delta(t)}{g(t)^{2}}-\frac{1}{t}\right)\mathrm{d}t=\sum_{n=2}^{N}\int_{x}^{\infty}\frac{Q_{n}^{(3)}(\log
t)}{t^{n}}\mathrm{d}t+o(x^{-N+1})=\sum_{n=1}^{N-1}\frac{Q_{n}^{(4)}(\log
x)}{x^{n}}+o(x^{-N+1})$

as $x\to\infty$, where the expansion of the integrals appearing here can be
established using iterated integration by parts. Meanwhile, the definition of
$\delta$ and the preliminary estimate we obtained in Lemma 5.4 yield that

$\lim_{y\to\infty}\int_{x}^{y}\left(\frac{\delta(t)}{g(t)^{2}}-\frac{1}{t}\right)\mathrm{d}t=\lim_{y\to\infty}\left(\frac{2}{g(x)}-\frac{2}{g(y)}-(y-x)-\log\frac{y}{x}\right)=\frac{C_{\lambda}}{2}+\frac{2}{g(x)}+x+\log
x,$

so that

$-\frac{2}{g(x)}=x+\log
x+\frac{C_{\lambda}}{2}-\sum_{n=1}^{N-1}\frac{Q_{n}^{(4)}(\log
x)}{x^{n}}+o(x^{-N+1})$

as $x\to\infty$. By using once more the power series of $(1+u)^{-1}$, this
gives exactly the estimate we wanted for $g$:

$g(x)=\sum_{n=1}^{N+1}\frac{Q_{n}^{(5)}(\log x)}{x^{n}}+o(x^{-N-1}).$

The estimate for $g^{\prime}$ is then a consequence of this estimate together
with (5.15):

$2g^{\prime}(x)=\delta(x)+g(x)^{2}=\sum_{n=1}^{N+2}\frac{Q_{n}^{(6)}(\log
x)}{x^{n}}+o(x^{-N-2}).$

This concludes our inductive proof. ∎

We conclude this section by proving the part of Theorem 5.1 concerning $g$:

###### Proof of Theorem 5.1 for the function $g$.

By using the asymptotic expansions of $g$ and $g^{\prime}$ from Lemma 5.5 and
by using the fact that $g^{\prime\prime}=g^{2}-2g^{\prime}$, we obtain for
free an asymptotic expansion for $g^{\prime\prime}$ of a similar form. Since
$g$ and $g^{\prime}$ both converge to zero as $x\to\infty$, we can write
$g^{\prime}(x)=-\int_{x}^{\infty}g^{\prime\prime}(t)\mathrm{d}t$ and
$g(x)=-\int_{x}^{\infty}g^{\prime}(t)\mathrm{d}t$, which allows us to reobtain
the asymptotic expansions of $g$ and $g^{\prime}$ by integrating the
asymptotic expansion of $g^{\prime\prime}$ term by term. Wrapping up, we have
shown that there exists a sequence of polynomials $(Q_{n})_{n\geq 1}$ such
that for any $N\geq 1$,

$\displaystyle g(x)$ $\displaystyle=\sum_{n=1}^{N}\frac{Q_{n}(\log
x)}{x^{n}}+o(x^{-N})$ $\displaystyle g^{\prime}(x)$
$\displaystyle=\sum_{n=2}^{N+1}\frac{Q^{\prime}_{n-1}(\log
x)-(n-1)Q_{n-1}(\log x)}{x^{n}}+o(x^{-N-1})$ $\displaystyle
g^{\prime\prime}(x)$
$\displaystyle=\sum_{n=3}^{N+2}\frac{Q_{n-2}^{\prime\prime}(\log
x)-(2n-3)Q_{n-2}^{\prime}(\log x)+(n-1)(n-2)Q_{n-2}(\log
x)}{x^{n}}+o(x^{-N-2}).$

Because $g$ satisfies $g^{\prime\prime}+2g^{\prime}=g^{2}$, we must have
$2Q_{1}^{\prime}-2Q_{1}=Q_{1}^{2}$ and that

$\displaystyle
Q_{n-2}^{\prime\prime}-(2n-3)Q_{n-2}^{\prime}+(n-1)(n-2)Q_{n-2}+2Q_{n-1}^{\prime}-2(n-1)Q_{n-1}=\sum_{k=1}^{n-1}Q_{k}Q_{n-k}$

for every $n\geq 3$. It follows from Lemma 5.4 that $Q_{1}=-2$ and
$Q_{2}(X)=2X+C_{\lambda}$, and rearranging the above display leads to the
recurrence (5.1). One can also easily check that this recurrence relation
uniquely determines $Q_{n}$ in terms of $Q_{1},\ldots,Q_{n-1}$ for every
$n\geq 2$, since the matrix that must be inverted to solve for the
coefficients of $Q_{n}$ is upper-triangular with diagonal terms all equal to
$2n-4$. (For $n=2$ the term $(2n-4)Q_{n}$ vanishes and the recurrence relation
does not put any constraints on the constant term of $Q_{2}$.) This concludes
the proof. ∎

###  5.2 Series expansion of $h$

In this section we explain how to modify the analysis of $g$ carried out in
the previous section to derive the series expansion for $h$. We will focus on
the first-order asymptotics $h(x)\sim 2/x$ together with the appropriate
replacement for the unimodality properties of $g$ derived in Lemma 4.2
(namely, the fact that $h$ is totally monotone), with the rest of the proof
being very similar.

We begin with the following proposition, which establishes non-negativity of
all solutions to $h^{\prime\prime}-2h^{\prime}=h^{2}$ defined on all of
$[0,\infty)$ together with the fact that these solutions are always totally
monotone. The latter property replaces the unimodality properties of $g$
established in Lemma 4.2. The fact that these properties hold for _all_
solutions defined on $[0,\infty)$ indicates an important difference between
the two ODEs (or equivalently between the forward and backward evolution of
either ODE).

######  Proposition 5.6.

Let $h$ be any solution to $h^{\prime\prime}-2h^{\prime}=h^{2}$ defined on
$[0,\infty)$. If $h$ is not identically zero then the following hold:

  1. 1.

$h(x)>0$ for all $x\geq 0$.

  2. 2.

$h(x)\to 0$ as $x\to\infty$.

  3. 3.

$h$ is totally monotone in the sense that $(-1)^{n}h^{(n)}(x)>0$ for every
$n\geq 1$.

  4. 4.

$h^{(n)}(x)\to 0$ as $x\to\infty$ for every $n\geq 1$.

We note that this proposition together with the uniqueness of non-negative
solutions to boundary value problems for the ODE
$h^{\prime\prime}-2h^{\prime}=h^{2}$ (which can be proven as in Lemma 2.6) has
the following simple corollary.

######  Corollary 5.7.

There exists a totally monotone function $h_{\infty}:(0,\infty)\to(0,\infty)$
with $h_{\infty}(x)\to\infty$ as $x\to 0$ and $h_{\infty}(x)\to 0$ as
$x\to\infty$ such that if $h$ is any solution to
$h^{\prime\prime}-2h^{\prime}=h^{2}$ defined on $[0,\infty)$ then there exists
$c>0$ such that $h(x)=h_{\infty}(x+c)$ for every $x\geq 0$.

######  Remark 5.8.

The quantity $h_{\infty}(x)$ can be written in terms of the measure given to
the set $\\{$the unit ball is hit$\\}$ by the four-dimensional superBrownian
excursion measure started at $e^{x}$.

We break the proof of Proposition 5.6 into a few steps, beginning with the
following lemma.

######  Lemma 5.9.

Let $h$ be any solution to $h^{\prime\prime}-2h^{\prime}=h^{2}$ defined on
$[0,\infty)$. If $h$ is not identically zero then $h^{\prime}(x)<0$ for all
$x\geq 0$.

###### Proof of Lemma 5.9.

Suppose not, so that there exists $x_{0}\geq 0$ with $h^{\prime}(x_{0})\geq
0$. Since $h^{\prime\prime}(x_{0})=2h^{\prime}(x_{0})+h(x_{0})^{2}\geq 0$ and
$h^{\prime}$ cannot vanish on any non-trivial interval (by uniqueness of
forward and backward solutions to the IVP associated to the ODE), there must
exist $x_{1}>x_{0}$ with $h^{\prime}(x_{1})>0$. Using that
$h^{\prime\prime}-2h^{\prime}\geq 0$, it follows by Lemma 3.10 that
$h^{\prime}(x)\geq e^{2(x-x_{1})}$ for every $x\geq x_{1}$. Integrating this
inequality, it follows that there exists $x_{2}\geq x_{1}$ such that
$h^{\prime}(x)$ and $h(x)$ are both bounded below by $1$ for all $x\geq
x_{2}$, and hence that $h^{\prime\prime}(x)\geq 0$ for all $x\geq x_{2}$. Now,
let $y_{0}=x_{2}$ and for each $n\geq 1$ let $y_{n}\geq y_{n-1}$ be minimal
such that $h^{\prime}(y_{n})=2^{n}$. For each $x\geq y_{n}$ we have that

$h(x)=h(y_{n})+\int_{y_{n}}^{x}h^{\prime}(t)\mathrm{d}t\geq 2^{n}(x-y_{n})$

and hence by Lemma 3.10 that

$h^{\prime}(x)\geq
e^{2(x-y_{n})}h(y_{n})+2^{2n}\int_{y_{n}}^{x}(t-y_{n})^{2}\mathrm{d}t\geq\frac{1}{3}2^{2n}(x-y_{n})^{3}$

for every $x\geq y_{n}$. This implies that $y_{n+1}-y_{n}\leq(3\cdot
2^{-n})^{1/3}$ for every $n\geq 0$, and since $\sum(3\cdot
2^{-n})^{1/3}<\infty$ this shows that $h^{\prime}$ blows up in finite time and
contradicts the assumption that $h$ is defined on all of $[0,\infty)$. ∎

######  Lemma 5.10.

Let $h$ be any solution to $h^{\prime\prime}-2h^{\prime}=h^{2}$ defined on
$[0,\infty)$. If $h$ is not identically zero then $h(x)>0$ for all $x\geq 0$
and $h(x)\to 0$ as $x\to\infty$.

###### Proof of Lemma 5.10.

Since $h^{\prime}(x)<0$ for all $x\geq 0$, $h(x)$ converges to some
$c\in[-\infty,\infty)$ as $x\to\infty$, and it suffices to prove that $c=0$.
We will rule out the cases $c>0$ and $c<0$ separately.

First suppose for contradiction that $c$ is positive, in which case $h(x)$ is
positive for all $x\geq 0$. We first claim that $h^{\prime\prime}(x)\geq 0$
for every $x\geq 0$. Suppose that this is not the case, so that there exists
$x_{0}\geq 0$ with $h^{\prime\prime}(x)<0$. The second derivative
$h^{\prime\prime}$ satisfies the differential inequality
$h^{\prime\prime\prime}-2h^{\prime\prime}=2hh^{\prime}\leq 0$, so that (by
Lemma 3.10) $h^{\prime\prime}(x)\leq e^{2(x-x_{0})}h^{\prime\prime}(x_{0})$
for every $x\geq x_{0}$. Integrating this inequality twice implies that $h(x)$
is eventually negative, contradicting the assumption that $c\geq 0$. This
completes the proof that the second derivative is non-negative. Now, since
$h(x)\to c$ and $h^{\prime}(x)$ is monotone, we must have that
$h^{\prime}(x)\to 0$ as $x\to\infty$. But the ODE then implies that
$h^{\prime\prime}(x)\to c^{2}>0$ as $x\to\infty$, a contradiction. This rules
out the case that $c$ is positive.

Now suppose that $c$ is negative, and let $x_{0}\geq 0$ be such that $h(x)\leq
0$ for all $x\geq x_{0}$. In this case, the second derivative satisfies the
differential inequality
$h^{\prime\prime\prime}-2h^{\prime\prime}=2hh^{\prime}\geq 0$ for every $x\geq
x_{0}$, and it follows by a similar argument to above that
$h^{\prime\prime}(x)<0$ for all $x>x_{0}$. As such, $h^{\prime}(x)\leq
h^{\prime}(x_{0})<0$ for all $x\geq x_{0}$, so that $c$ cannot be finite and
negative.

We now argue as in the proof of Lemma 5.9 that if $c=-\infty$ then in fact
$h^{\prime}$ blows up to negative infinity in finite time. Let $x_{1}\geq
x_{0}$ be such that $h(x)\leq-1$ for all $x\geq x_{1}$ and, for each $n\geq 0$
let $y_{n}\geq x_{0}$ be minimal such that $h^{\prime}(y_{n})\leq-2^{n}$.
Integrating the differential inequality $h^{\prime\prime}-2h^{\prime}\geq 1$
on $[x_{1},\infty)$ yields that $y_{n}<\infty$ for every $n\geq 0$. Since
$h^{\prime}$ is decreasing on $[x_{1},\infty)$, it follws as in the proof of
Lemma 5.9 that

$h(x)\leq-2^{n}(x-y_{n})\text{ and that
}h^{\prime}(x)\leq-2^{2n}\int_{y_{n}}^{x}(t-y_{n})^{3}$

for every $n\geq 0$ and $x\geq y_{n}$, so that $y_{n+1}\leq(3\cdot
2^{-n})^{1/3}$ for every $n\geq 0$. It follows as before that $h^{\prime}$
blows up to negative infinity in finite time, contradicting the assumption
that $h$ was defined on $[0,\infty)$. ∎

###### Proof of Proposition 5.6.

We have already shown that $h(x)>0$ for all $x\geq 0$, $h^{\prime}(x)<0$ for
all $x\geq 0$, and that $h(x)\to 0$ as $x\to\infty$. We now prove by induction
on $n$ that $(-1)^{n}h^{(n)}(x)>0$ for all $x\geq 0$ and that $h^{(n)}(x)\to
0$. Suppose that $n\geq 2$ and that $(-1)^{n}h^{(k)}(x)>0$ and $h^{(k)}(x)\to
0$ for all $0\leq k<n$ and $x\geq 0$. The $n$th derivative $h^{(n)}$ satisfies

$h^{(n+1)}-2h^{(n)}=(h^{2})^{(n-1)}=\sum_{k=0}^{n-1}\binom{n-1}{k}h^{(k)}h^{(n-1-k)}.$
(5.17)

By the induction hypothesis, the right hand side is positive if $n$ is odd and
negative if $n$ is even. Thus, it follows by Lemma 3.10 that if
$(-1)^{n}h^{(n)}(x_{0})\leq 0$ for some $x_{0}\geq 0$ then
$(-1)^{n}h^{(n)}(x)\leq 0$ for all $x\geq x_{0}$. This is inconsistent with
the induction hypothesis that $(-1)^{n-1}h^{(n-1)}(x)$ converges to $0$ from
above as $x\to\infty$, so that we must have $(-1)^{n}h^{(n)}(x)>0$ for all
$x\geq 0$ as claimed.

It remains to prove that $h^{(n)}(x)\to 0$ as $x\to\infty$. Since
$h^{(n-1)}(x)\to 0$ as $x\to\infty$, we must have that
$\liminf_{n\to\infty}|h^{(n)}(x)|=0$. By (5.17) and the induction hypothesis,
there exists $x_{0}<\infty$ such that

$|h^{(n+1)}(x)-2h^{(n)}(x)|\leq\varepsilon$

for all $x\geq x_{0}$. Applying Lemma 3.10, we deduce that

$|h^{(n)}(x)-e^{2(x-x_{1})}h^{(n)}(x_{1})|\leq\varepsilon\int_{x_{1}}^{x}e^{2x-2t}\mathrm{d}t\leq\frac{\varepsilon}{2}e^{2(x-x_{1})}$

for every $x\geq x_{1}\geq x_{0}$. Thus, if $x_{1}\geq x_{0}$ is such that
$|h^{(n)}(x_{1})|\geq\varepsilon$ then
$|h^{(n)}(x)|\geq\frac{\varepsilon}{2}e^{2(x-x_{1})}$ for every $x\geq x_{1}$,
contradicting the fact that $\liminf_{n\to\infty}|h^{(n)}(x)|=0$. This
completes the proof of the induction step and hence of the proposition. ∎

We now compute the first-order asymptotics of $h$.

######  Lemma 5.11.

Let $h$ be any solution to $h^{\prime\prime}-2h^{\prime}=h^{2}$ defined on
$[0,\infty)$. If $h$ is not identically zero then $h(x)\sim 2/x$ as
$x\to\infty$.

###### Proof.

Define $\delta=h^{\prime\prime}=2h^{\prime}+h^{2}$. Since $h$ is totally
monotone and $\delta=h^{\prime\prime}$, we have that $\delta(x)\geq 0$ for all
$x\geq 0$ and hence that $h$ satisfies the differential inequality
$2h^{\prime}\geq-h^{2}$. Integrating the resulting inequality on the
derivative of $1/h$ yields that

$\frac{1}{h(x)}-\frac{1}{h(0)}=\int_{0}^{x}-\frac{h^{\prime}(t)}{h^{2}(t)}\mathrm{d}t\leq\frac{x}{2}$

and hence that $h(x)\geq\frac{2+o(1)}{x}$ as $x\to\infty$.

We now prove the matching upper bound. We first claim that
$\delta(x)=|h^{\prime\prime}(x)|=o(|h^{\prime}(x)|)$ as $x\to\infty$. Since
$h$ is totally monotone, $|h^{\prime\prime}(x)/h^{\prime}(x)|$ is decreasing
and hence converges from above to some limit $c\geq 0$. If $c>0$ then we have
that $h^{\prime\prime}(x)\geq-ch^{\prime}(x)$ for all $x\geq 0$ and hence by
Grönwall (Lemma 3.10) that $|h^{\prime}(x)|\leq e^{-cx}|h^{\prime}(0)|$ for
every $x\geq 0$. Since $h(x)\to 0$ as $x\to\infty$, this inequality is
inconsistent with the sub-exponential lower bound $h(x)\geq(2+o(1))/x$, so
that we must have $c=0$ as claimed. Using that
$h^{\prime\prime}(x)=o(|h^{\prime}(x)|)$ as $x\to\infty$, we can approximately
simplify the ODE $h^{\prime\prime}+2h^{\prime}=h^{2}$ by writing

$h^{\prime}(x)\sim-\frac{1}{2}h^{2}.$

The claim follows by integrating the resulting asymptotic estimate
$(1/h)^{\prime}\sim 1/2$. ∎

######  Remark 5.12.

It should be possible to adapt this proof to give an alternative derivation of
the first-order asymptotics of $g_{\lambda}$ also.

Given Propositon 5.6 and Lemma 5.11, the rest of the proof of the part of
Theorem 5.1 concerning $h$ is very similar to the part concerning $g$, and we
omit the details.

###  5.3 Proof of Lemma 5.3

In this section we prove Lemma 5.3, providing an integral expression for the
constant $C_{\lambda}$ which allows us to deduce continuity of $\lambda$ and
the asymptotic expression $C_{\lambda}\sim 4/\lambda$ as $\lambda\downarrow
0$.

###### Proof of Lemma 5.3.

As usual, we will denote
$\delta_{\lambda}=2g_{\lambda}^{\prime}-g_{\lambda}^{2}=-g_{\lambda}^{\prime\prime}$.
By (5.12) and (5.13), $C_{\lambda}$ can be written as

$C_{\lambda}=\int_{1}^{\infty}\left(\frac{\delta_{\lambda}(x)}{g_{\lambda}(x)^{2}}-\frac{1}{x}\right)\mathrm{d}x-\frac{2}{g_{\lambda}(1)}-1.$
(5.18)

The first integration bound was chosen in an arbitrary way, meaning that for
any $x_{0}>0$, $C_{\lambda}$ can be written as

$C_{\lambda}=\int_{x_{0}}^{\infty}\left(\frac{\delta_{\lambda}(x)}{g_{\lambda}(x)^{2}}-\frac{1}{x}\right)\mathrm{d}x-\frac{2}{g_{\lambda}(x_{0})}-x_{0}-\log
x_{0}.$ (5.19)

This corresponds to the formula stated in (5.4). We emphasise that the term
$\log x_{0}$ was not apparent when $x_{0}$ was chosen to be equal to 1 since
$\log 1=0$. (Note that the equation
$g_{\lambda}^{\prime\prime}+2g^{\prime}_{\lambda}=g_{\lambda}^{2}$ easily
implies that the expression on the right hand side of the above equation does
not depend on $x_{0}$.)

We now prove continuity of $C_{\lambda}$. We use (5.18). The maps
$\lambda\mapsto g_{\lambda}(1)$ and
$\lambda\mapsto\frac{\delta_{\lambda}(x)}{g_{\lambda}(x)^{2}}-\frac{1}{x}$,
$x>1$, are continuous since solutions to initial value problems depend
continuously on their initial conditions. By (5.4) and the dominated
convergence theorem, to conclude that $C_{\lambda}$ is continuous it is enough
to check that for each $\lambda\in(0,1)$ there is a neighbourhood of $\lambda$
on which $\frac{\delta_{\lambda}(x)}{g_{\lambda}(x)^{2}}-\frac{1}{x}$ is
dominated by some integrable function. Let $0<\lambda_{1}<\lambda_{2}<1$. By
Theorem 3.1, $g_{\lambda}(x)$ is monotone in $\lambda$, so for all
$\lambda\in[\lambda_{1},\lambda_{2}]$,
$|g_{\lambda_{1}}(x)|\leq|g_{\lambda}(x)|\leq|g_{\lambda_{2}}(x)|,$ meaning
that we have uniform control on $g_{\lambda}(x)$ in the sense that

$\sup_{\lambda\in[\lambda_{1},\lambda_{2}]}|g_{\lambda}(x)|=|g_{\lambda_{2}}(x)|=\frac{2}{x}+O\left(\frac{\log
x}{x^{2}}\right)$

and

$\inf_{\lambda\in[\lambda_{1},\lambda_{2}]}|g_{\lambda}(x)|=|g_{\lambda_{1}}(x)|=\frac{2}{x}+O\left(\frac{\log
x}{x^{2}}\right).$

Here and below we use the notation that $f(x)=O(g(x))$ to mean that there
exists there exists a constant $C>0$ such that $|f(x)|\leq C|g(x)|$. In
particular the reader should keep in mind that the left hand side might well
be negative.

Since we obtained the bounds (5.9) and (5.10) on $\delta_{\lambda}$ using only
first-order estimates on $|g_{\lambda}|$, we deduce that the same bounds on
$\delta$ hold uniformly in $\lambda\in[\lambda_{1},\lambda_{2}]$ in the sense
that

$\inf_{\lambda\in[\lambda_{1},\lambda_{2}]}\delta_{\lambda}(x)=\frac{4}{x^{3}}+O\left(\frac{\log
x}{x^{4}}\right)\qquad\text{ and
}\qquad\sup_{\lambda\in[\lambda_{1},\lambda_{2}]}\delta_{\lambda}(x)=\frac{4}{x^{3}}+O\left(\frac{\log
x}{x^{4}}\right)$

as $x\to\infty$. It follows that

$\left|\frac{\delta_{\lambda}(x)}{g_{\lambda}(x)^{2}}-\frac{1}{x}\right|=O\left(\frac{\log
x}{x^{2}}\right)$

uniformly in $\lambda\in[\lambda_{1},\lambda_{2}]$. This provides the required
domination by an integrable function and concludes the proof of continuity of
$C_{\lambda}$. The fact that $C_{\lambda}$ is decreasing on $(0,1)$ follows
immediately from the fact that $g_{\lambda}(x)$ is a decreasing function of
$\lambda\in(0,1)$ as proven in Lemma 3.9.

We now show that $C_{\lambda}\sim 4/\lambda$ as $\lambda\downarrow 0$. Roughly
speaking, we will check that the asymptotic expansion (5.3) is a good
approximation as soon as $x\geq 1/\lambda$ and conclude using Lemma 4.5. The
following line of argument is close to the proof of Lemma 5.4. Let
$\lambda\in(0,1/2]$. All the big-O written in the rest of the proof are
uniform with respect to $\lambda$. Let $x_{\lambda}=1/\lambda$ and $x\geq
x_{\lambda}$. As noticed above, the upper bounds on $|g_{\lambda}|$ and
$|\delta_{\lambda}|$ are uniform in $\lambda(0,1/2]$:

$|g_{\lambda}(x)|\leq|g_{1/2}(x)|\leq\frac{2}{x}+O(1)\frac{\log x}{x}$ (5.20)

and

$\delta(x)\leq\frac{4}{x^{3}}+O(1)\frac{\log x}{x^{4}}.$ (5.21)

Obtaining uniform lower bounds require more care. (Indeed, we recall from
Lemma 3.12 that $g_{\lambda}\to 0$ pointwise as $\lambda\to 0$.) First, notice
that since $1/x\leq\lambda$,

$|g_{\lambda}(x)|\geq|g_{\lambda=1/x}(x)|\sim\frac{2}{5x}\quad\quad\text{as}~{}x\to\infty$

by Lemma 4.5. Hence, there exists $\lambda_{0}\in(0,1/2]$ small enough, such
that for all $\lambda\in(0,\lambda_{0}]$ and $x\geq x_{\lambda}$,
$|g_{\lambda}(x)|\geq 1/(5x)$. Injecting this estimate in (5.21) shows that

$-\left(\frac{1}{g_{\lambda}(x)}+\frac{x}{2}\right)^{\prime}=\frac{g_{\lambda}^{\prime}(x)}{g(x)^{2}}-\frac{1}{2}=\frac{\delta_{\lambda}(x)}{2g_{\lambda}(x)^{2}}\leq\frac{O(1)}{x},\quad\quad
x\geq x_{\lambda},$

where we stress again that all big-$O$ estimates are uniform in
$\lambda\in(0,1/2]$ and $x\geq x_{\lambda}$. Integrating between $x_{\lambda}$
and some $x\geq x_{\lambda}$ shows that

$|g_{\lambda}(x)|\geq\frac{2}{x+O(1)\log(x/x_{\lambda})+2/|g_{\lambda}(x_{\lambda})|-x_{\lambda}}.$

By Lemma 4.5, the difference $2/|g_{\lambda}(x_{\lambda})|-x_{\lambda}$ is
asymptotic to $4/\lambda$ as $\lambda\to 0$. Using in particular that it is
bounded by $O(1/\lambda)$, we can inject this improved lower bound on
$|g_{\lambda}|$ back in (5.21) to obtain that

$-\left(\frac{1}{g_{\lambda}(x)}+\frac{x}{2}\right)^{\prime}=\frac{\delta_{\lambda}(x)}{2g_{\lambda}(x)^{2}}\leq\frac{1}{2x}+\frac{O(1)}{\lambda
x^{2}}+O(1)\frac{\log x}{x^{2}},\quad\quad x\geq x_{\lambda}.$

Integrating this inequality finally shows that

$|g_{\lambda}(x)|\geq\frac{2}{x+\log(x/x_{\lambda})+2/|g_{\lambda}(x_{\lambda})|-x_{\lambda}+O(1)/(\lambda
x)+O(1)\log x/x}.$

In particular,

$C_{\lambda}=\lim_{x\to\infty}\frac{2}{|g_{\lambda}(x)|}-x-\log
x\leq\frac{2}{|g_{\lambda}(x_{\lambda})|}-x_{\lambda}-\log x_{\lambda}.$

We can now use this lower bound on $|g_{\lambda}|$ to get a lower bound on
$\delta_{\lambda}$ using the differential inequality
$\delta_{\lambda}^{\prime}+2\delta_{\lambda}=|g_{\lambda}|\delta+|g_{\lambda}|^{3}\geq|g_{\lambda}|^{3}$,
which follows from (5.7). We get that

$\delta_{\lambda}(x)\geq\frac{4}{x^{3}}+O(1)\frac{1}{\lambda
x^{4}}+O(1)\frac{\log x}{x^{4}},\quad\quad x\geq 2x_{\lambda}.$

See the paragraph around (5.10) for details. This lower bound on
$\delta_{\lambda}$ together with the upper bound (5.20) on $|g_{\lambda}|$
shows that

$\int_{2x_{\lambda}}^{\infty}\left(\frac{\delta_{\lambda}(x)}{g_{\lambda}(x)^{2}}-\frac{1}{x}\right)\mathrm{d}x\geq
O(1)\int_{2x_{\lambda}}^{\infty}\left(\frac{1}{\lambda x^{2}}+\frac{\log
x}{x^{2}}\right)\mathrm{d}x\geq O(1)\frac{1}{\lambda
x_{\lambda}}+O(1)\frac{\log x_{\lambda}}{x_{\lambda}}\geq O(1).$

Applying the formula (5.4) for $C_{\lambda}$ to $x_{0}=2x_{\lambda}$, we get
that

$\displaystyle C_{\lambda}$
$\displaystyle=\frac{2}{|g_{\lambda}(2x_{\lambda})|}-2x_{\lambda}-\log
2x_{\lambda}+\int_{2x_{\lambda}}^{\infty}\left(\frac{\delta_{\lambda}(x)}{g_{\lambda}(x)^{2}}-\frac{1}{x}\right)\mathrm{d}x\geq\frac{2}{|g_{\lambda}(2x_{\lambda})|}-2x_{\lambda}-\log
2x_{\lambda}+O(1).$

Wrapping up, we have proved that

$\frac{2}{|g_{\lambda}(2x_{\lambda})|}-2x_{\lambda}-\log 2x_{\lambda}+O(1)\leq
C_{\lambda}\leq\frac{2}{|g_{\lambda}(x_{\lambda})|}-x_{\lambda}-\log
x_{\lambda}.$

By Lemma 4.5, both sides of the above inequalities are asymptotically
equivalent to $4/\lambda$ as $\lambda\to 0$. This concludes the proof. ∎

##  6 Tail estimates with killing II: Lower bounds (proof of Theorem 1.3)

The aim of this section is to prove the lower bound of Theorems 1.3,
completing the proof of that theorem. The proof relies on the following
precise asymptotic of the Laplace transform of the number of pioneers, which
we prove in this section using Theorem 1.7 as an input.

######  Theorem 6.1.

Let $x_{0}>0$. For each $\delta>0$, there exist constants $T_{0},R_{0}>0$ and
$x_{*}>x_{0}$ depending only on $x_{0}$ and $\delta$ such that

$e^{-\delta}\frac{2(e^{2x_{0}}-1)}{R^{2}x}\frac{x}{T}\leq\mathbb{E}_{e^{-x_{0}}R,R}\left[\exp\left\\{\left(1-\frac{T}{x}\right)\frac{2N_{e^{-x}R}}{(x+\log
x)(e^{-x}R)^{2}}\right\\}-1\right]\leq
e^{\delta}\frac{2(e^{2x_{0}}-1)}{R^{2}x}\frac{x}{T}$ (6.1)

for all $T\in[T_{0},2T_{0}]$, $R\geq R_{0}$, and $x\in[x_{*},\log R]$.

######  Remark 6.2.

As we will see in Lemma 4.1 below, the term $\frac{2(e^{2x_{0}}-1)}{R^{2}x}$
corresponds to the asymptotic behaviour of the hitting probability of
$\partial B(0,e^{-x}R)$ starting from a single particle on $\partial
B(0,e^{-x_{0}}R)$. Therefore, Theorem 6.1 says in a very strong and precise
way that the law of the number of pioneers $N_{e^{-x}R}$ conditioned to be
positive is close to that of an exponential random variable with mean
$\frac{1}{2}(x+\log x)(e^{-x}R)^{2}$.

The logarithmic shift in the mean (which corresponds to shifting by $\log\log
R$ in the original scale) is important in the sense that the theorem would not
be true without it. This logarithmic shift will eventually come from the
second order term in the expansion in Theorem 5.1. To see why it matters, one
simply needs to expand

$\left(1-\frac{T}{x}\right)\frac{1}{x+\log x}=\frac{1}{x}-\frac{\log
x}{x^{2}}-\frac{T}{x^{2}}+\cdots.$

The dependence in $T$, which is the key feature of this theorem, appears only
from the third term onward; if we replaced $\log x$ by any other quantity
differing by a divergent term, this effect would overwhelm that of changing
$T$.

Before proving Theorem 6.1, we will first explain how we will use it to deduce
Theorem 1.3. The idea is of course to convert information on the Laplace
transform into information on the tail. This is usually achieved by the means
of the classical Tauberian theorem of Hardy–Littlewood. However, the
assumptions of that theorem are too restrictive to apply given only the
information we obtain in Theorem 6.1. Roughly speaking, in order to use such a
theorem we would require asymptotics of the Laplace transform for values of
$1-\tfrac{T}{x}$ arbitrarily close to $1$. In our case, even if we take $x$ as
large as possible (namely, take $x=\log R$ so that we consider the pioneers on
the ball of radius $1$) our estimates on the Laplace transform hold only on an
interval of the type $[1-t_{1}/\log R,1-t_{2}/\log R]$. The Hardy–Littlewood
theorem also requires an assumption of regular or slow variation, something
which we do not have at our disposal. To proceed, we therefore require a
finitary improvement of the classical Tauberian theorem. Fortunately, we
manage to prove such a finitary improvement using fairly soft arguments in
Section 6.2; we anticipate that this will be useful in other problems too.
This is an improvement over the classical theorem in the sense that the
assumptions are less restrictive, but of course this is reflected in the
conclusion too, which is not as strong.

We prove Theorem 6.1 in Section 6.1, formulate and prove our finitary
Tauberian theorem in Section 6.2, and then deduce Theorem 1.2 in Section 6.3.

###  6.1 Estimates on the Laplace transform

In this section we prove Theorem 6.1. The proof will rely on the results of
Sections 4 and 5.

###### Proof of Theorem 6.1.

To ease notations, we will prove the statement only for $x=L$ so that
$e^{-x}R=1$. The same line of argument proves the result in general: Simply
replace the number of pioneers $N_{1}$ on the unit sphere by the renormalised
number of pioneers $N_{e^{-x}R}/(e^{-x}R)^{2}$ on the sphere $\partial
B(e^{-x}R)$ and each occurrence of $L$ by $x$. Throughout the proof, we will
write $e^{\pm\delta}$ for a quantity that is bounded between $e^{-\delta}$ and
$e^{\delta}$, the precise value of which may vary from line to line.

We first introduce some parameters. Let $x_{0}>0$ be fixed, and let $\delta>0$
be small. Let us first pick $\mu_{0}=\mu_{0}(\delta)$ small enough so that,
defining $A:=1/\mu_{0}$,

$\frac{1}{\delta}e^{-\delta A/2}\leq\delta\quad\text{and}\quad A\geq
8e^{10\delta}.$ (6.2)

By Lemmas 4.5 and 5.3, for all $\mu>0$, we have that

$\lim_{\xi\to\infty}\xi
g_{\mu/\xi}(\xi)=\frac{-2}{1+4/\mu}\quad\text{and}\quad\lim_{\xi\to\infty}\frac{1}{\xi}C_{\mu/\xi}=\frac{4}{\mu}$

uniformly with respect to $\mu$ on compact subsets of $(0,\infty)$. Thus, by
choosing $\mu_{0}$ smaller if necessary, we can find
$\xi=\xi(\delta,\mu_{0})\geq\max\\{x_{0},3\mu_{0}\\}$ large such that if
$\mu\in[\mu_{0},3\mu_{0}]$ then

$\frac{e^{-\delta}}{C_{\lambda=\mu/\xi}}\leq-\frac{1}{2}g_{\lambda=\mu/\xi}(\xi)\leq\frac{e^{\delta}}{C_{\lambda=\mu/\xi}}\quad\text{and}\quad
4e^{-\delta}\frac{\xi}{\mu}\leq C_{\lambda=\mu/\xi}\leq
4e^{\delta}\frac{\xi}{\mu}.$ (6.3)

Moreover, by Proposition 4.6 applied with $\varepsilon=1/2$, we may assume
that $\xi$ is large enough (depending only on $x_{0}$ and $\delta$) so that
for all $n\geq 1$,

$\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{e^{-\xi}R}\geq\frac{\xi(e^{-\xi}R)^{2}}{2}n\right)\leq\frac{C(x_{0})}{\xi(e^{-x_{0}}R)^{2}}e^{-n/2};\quad\text{
and }\quad e^{-2\xi}\leq 1-e^{-\delta}$ (6.4)

for some constant $C(x_{0})$ depending only $x_{0}$.

Now that our parameters are defined, we can start more concretely the proof.
Since $\xi\geq 3\mu_{0}$, it follows from Theorem 3.1 that

$1-\mathbb{E}_{e^{-\xi}R,R}\left[\left(1-g_{\lambda}(L)\right)^{N}\right]=\frac{1}{(e^{-\xi}R)^{2}}g_{\lambda}(\xi)$

for all $\lambda\in[\mu_{0}/\xi,3\mu_{0}/\xi]$. Applying Theorem 5.1 (more
precisely (5.3)), it follows that

$g_{\lambda}(L)=\frac{-2}{L+\log L+C_{\lambda}+o(1)}$

as $L\to\infty$, where, here and in what follows, all uses of asymptotic
notation are taken to be uniform in $\lambda\in[\mu_{0}/\xi,3\mu_{0}/\xi]$
(the fact that the estimate is uniform in this range follows by continuity of
$C_{\lambda}$). Writing

$N\log\left(1+\frac{2}{L+\log
L+C_{\lambda}+o(1)}\right)=\left(1-\frac{C_{\lambda}+o(1)}{L}\right)\frac{2N}{L+\log
L}$

and using (6.3), we deduce that there exists a quantity $\eta(\lambda,L)$
which is jointly continuous in $\lambda$ and $L$ with $\eta(\lambda,L)\to 0$
as $L\to\infty$ uniformly over $\lambda\in[\mu_{0}/\xi,3\mu_{0}/\xi]$, such
that

$1-\mathbb{E}_{e^{-\xi}R,R}\left[\exp\left\\{\left(1-\frac{C_{\lambda}+\eta(\lambda,L)}{L}\right)\frac{2N}{L+\log
L}\right\\}\right]=-\frac{2e^{\pm\delta}}{(e^{-\xi}R)^{2}C_{\lambda}}$

for every $\lambda\in[\mu_{0}/\xi,3\mu_{0}/\xi]$. Starting now from a point on
the sphere $\partial B(e^{-x_{0}}R)$, we have

$\displaystyle\mathbb{E}_{e^{-x_{0}}R,R}\left[1-\exp\left\\{\left(1-\frac{C_{\lambda}+\eta(\lambda,L)}{L}\right)\frac{2N}{L+\log
L}\right\\}\right]$ $\displaystyle\hskip
56.9055pt=\sum_{n=1}^{\infty}\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{e^{-\xi}R}=n\right)\left(1-\mathbb{E}_{e^{-\xi}R,R}\left[\exp\left\\{\left(1-\frac{C_{\lambda}+\eta(\lambda,L)}{L}\right)\frac{2N}{L+\log
L}\right\\}\right]^{n}\right)$ $\displaystyle\hskip
56.9055pt=\sum_{n=1}^{\infty}\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{e^{-\xi}R}=n\right)\left(1-\left(1+\frac{2e^{\pm\delta}}{(e^{-\xi}R)^{2}C_{\lambda}}\right)^{n}\right),$

where $e^{\pm\delta}$ may represent a different quantity bounded between
$e^{-\delta}$ and $e^{\delta}$ in each term of the sum. The linearisation
$(1+u)^{n}\approx 1+nu$ is a good approximation as long as $nu$ is small.
Using the estimate (6.3) on $C_{\lambda}$ and recalling that
$A=1/\mu_{0},\lambda=\mu/\xi$, we deduce after some elementary computations
that for $n\leq n_{1}:=\delta(e^{-\xi}R)^{2}A\xi$,

$1-\left(1+\frac{2e^{\pm\delta}}{(e^{-\xi}R)^{2}C_{\lambda}}\right)^{n}=-\frac{2e^{\pm
10\delta}}{(e^{-\xi}R)^{2}C_{\lambda}}n.$

This implies that

$\displaystyle\mathbb{E}_{e^{-x_{0}}R,R}\left[1-\exp\left\\{\left(1-\frac{C_{\lambda}+\eta(\lambda,L)}{L}\right)\frac{2N}{L+\log
L}\right\\}\right]$ (6.5) $\displaystyle\hskip 85.35826pt=-e^{\pm
10\delta}\mathbb{E}_{e^{-x_{0}}R,R}\left[N_{e^{-\xi}R}\mathbf{1}_{\left\\{N_{e^{-\xi}R}\leq
n_{1}\right\\}}\right]\frac{2}{(e^{-\xi}R)^{2}C_{\lambda}}$
$\displaystyle\hskip
170.71652pt+\sum_{n=n_{1}+1}^{\infty}\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{e^{-\xi}R}=n\right)\left(1-\left(1+\frac{2e^{\pm\delta}}{(e^{-\xi}R)^{2}A\xi}\right)^{n}\right).$

It remains to argue that the contribution of the sum above is small and that
the restriction to the event $\\{N_{e^{-\xi}R}\leq n_{1}\\}$ leaves the
expectation of $N_{e^{-\xi}R}$ almost unchanged. This follows from (6.4): for
all $k\geq 1$,

$\displaystyle\Big{|}\sum_{n=kn_{1}+1}^{(k+1)n_{1}}\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{e^{-\xi}R}=n\right)\left(1-\left(1+\frac{2e^{\pm\delta}}{(e^{-\xi}R)^{2}A\xi}\right)^{n}\right)\Big{|}$
$\displaystyle\hskip
85.35826pt\leq\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{e^{-\xi}R}\geq
kn_{1}+1\right)\left(1+\frac{2e^{\pm\delta}}{(e^{-\xi}R)^{2}A\xi}\right)^{(k+1)n_{1}}$
$\displaystyle\hskip
85.35826pt\leq\frac{C(x_{0})}{\xi(e^{-x_{0}}R)^{2}}\exp\left(-k\delta
A+2(k+1)e^{\pm
10\delta}\delta\right)\leq\frac{C(x_{0})}{\xi(e^{-x_{0}}R)^{2}}\exp\left(-k\delta
A/2\right)$

since $A\geq 8e^{10\delta}$. This proves that the sum on the right hand side
of (6.5) is at most

$\frac{C^{\prime}(x_{0})}{\xi(e^{-x_{0}}R)^{2}}\frac{1}{\delta
A}\exp\left(-\delta A/2\right)\leq\frac{\delta
C^{\prime}(x_{0})}{(e^{-x_{0}}R)^{2}C_{\lambda}},$

where we used (6.2) and the fact that $C_{\lambda}$ is of order $A\xi$ (see
(6.3)) in the last inequality. Similarly, one can show that the expectation on
the right hand side of (6.5) equals $e^{\pm
10\delta}\mathbb{E}_{e^{-x_{0}}R,R}\left[N_{e^{-\xi}R}\right]$. The
expectation $\mathbb{E}_{e^{-x_{0}}R,R}\left[N_{e^{-\xi}R}\right]$ is equal to
the probability that a Brownian trajectory starting on the sphere $\partial
B(e^{-x_{0}}R)$ hits $\partial B(e^{-\xi}R)$ before $\partial B(R)$ which is
equal to

$\frac{e^{2x_{0}}-1}{e^{2\xi}-1}=e^{\pm\delta}e^{-2\xi}(e^{2x_{0}}-1),$

by (6.4). Wrapping things up, we have obtained that

$\mathbb{E}_{e^{-x_{0}}R,R}\left[1-\exp\left\\{\left(1-\frac{C_{\lambda}+\eta(\lambda,L)}{L}\right)\frac{2N}{L+\log
L}\right\\}\right]=-(1\pm
C^{\prime\prime}(x_{0})\delta)\frac{2(e^{2x_{0}}-1)}{R^{2}C_{\lambda}}.$

By continuity of $\lambda\mapsto C_{\lambda}+\eta(\lambda,L)$ for each fixed
$L$, the intermediate value theorem and (6.3),
$\\{C_{\lambda}+\eta(\lambda,L),\lambda\in[A/\xi,3A/\xi]\\}$ contains the
interval $[T_{0},2T_{0}]$ with $T_{0}=C_{2.99A/\xi}$ if $L$ is large enough.
The above estimate thus implies the claim. ∎

###  6.2 A finitary Tauberian theorem

In this section we prove the following finitary Tauberian theorem, which will
be applied when we deduce Theorem 1.3 from Theorem 6.1.

######  Theorem 6.3.

For each $0<a<1$ and $0<c<C<\infty$ there exists $\delta>0$ such that if $T>1$
and $X$ is a non-negative random variable satisfying

$\frac{e^{-\delta}}{1-\lambda}\leq\mathbb{E}\left[e^{\lambda
X}\right]\leq\frac{e^{\delta}}{1-\lambda}$

for every $1-C/T\leq\lambda\leq 1-c/T$ then

$\mathbb{P}\Bigl{(}(1-a)T\leq X\leq(1+a)T\Bigr{)}\geq\delta
aT\exp\left[-(1+a)T\right].$

######  Remark 6.4.

The proof of this theorem also yields an upper bound of the same form. We
focus on the lower bound since upper bounds on the tail are easy to obtain
from generating function estimates via Markov’s inequality.

We begin the proof of this theorem with the following lemma.

######  Lemma 6.5.

For each $0<\varepsilon<a<1$, $0<c_{1}<c_{2}<1$, and $A<\infty$ there exists
$\delta>0$ such that if $\mu$ is a probability measure on $[0,\infty)$
satisfying

$\frac{e^{-\delta}}{1-\lambda}\leq\int_{0}^{\infty}e^{\lambda
x}\mathrm{d}\mu(x)\leq\frac{e^{\delta}}{1-\lambda}$

for every $c_{1}\leq\lambda\leq c_{2}$ then

$\mu([y-a,y+a])\geq(1-\varepsilon)(1-e^{-2a})e^{-y+a}$

for every $0\leq y\leq A$.

Note that $e^{-y+a}(1-e^{-2a})$ is equal to the probability that an
Exponential$(1)$ random variable belongs to the interval $[y-a,y+a]$.

###### Proof of Lemma 6.5.

Suppose not. Then there exists $0<\varepsilon<a<1$, $0<c_{1},c_{2}<1$,
$A<\infty$, a sequence of measures $(\mu_{n})_{n\geq 1}$ and a sequence of
numbers $(y_{n})_{n\geq 0}$ in $[0,A]$ such that

$\frac{e^{-1/n}}{1-\lambda}\leq\int_{0}^{\infty}e^{\lambda
x}\mathrm{d}\mu_{n}(x)\leq\frac{e^{1/n}}{1-\lambda}$

for every $n\geq 1$ and $c_{1}\leq\lambda\leq c_{2}$ but for which
$\mu_{n}([y_{n}-a,y_{n}+a])\leq(1-\varepsilon)(1-e^{-2a})e^{-y_{n}+a}$ for
each $n\geq 1$. Taking a subsequence if necessary, we may take $y_{n}$ to
converge to a limit $y\in[0,A]$. Letting $0<\eta<1$ be such that
$(1-e^{-2(a-\eta)})e^{-y+a-\eta}>(1-\varepsilon)(1-e^{-2a})e^{-y+a-\varepsilon}$,
it follows that
$\mu_{n}([y-a+\eta,y+a-\eta])\leq\mu_{n}([y_{n}-a,y_{n}+a])\leq(1-\varepsilon)(1-e^{-2a})e^{-y_{n}+a}$
for all sufficiently large $n$. If $\mu$ is the law of an Exp(1) random
variable then
$\mu([y-a+\eta,y+a-\eta])=(1-e^{-2(a-\eta)})e^{-y+a-\eta}>(1-\varepsilon)(1-e^{-2a})e^{-y+a-\varepsilon}$,
so to reach a contradiction it suffices to prove that $\mu_{n}$ converges
weakly to $\mu$ as $n\to\infty$; this is clear since the moment generating
function of any subsequential limit of the $\mu_{n}$’s agrees with the moment
generating function of $\mu$ on the interval $[c_{1},c_{2}]$. ∎

###### Proof of Theorem 6.3.

We will prove the theorem by applying Lemma 6.5 to a certain biased and
rescaled version of $X$. Fix $T>1$ and suppose that $X$ satisfies

$\frac{e^{-\delta}}{1-\lambda}\leq\mathbb{E}\left[e^{\lambda
X}\right]\leq\frac{e^{\delta}}{1-\lambda}$

for every $1-C/T\leq\lambda\leq 1-c/T$. Let $s=1-(c+C)/2T$ and let $Z_{s}$ be
a random variable whose law is obtained by biasing the law of $X$ by $e^{sX}$
and multiplying the resulting random variable by $(1-s)$. The moment
generating function of $Z_{s}$ satisfies

$\frac{e^{-2\delta}}{1-\lambda}\leq\mathbb{E}\left[e^{\lambda
Z_{s}}\right]=\frac{\mathbb{E}\left[e^{sX+(1-s)\lambda
X}\right]}{\mathbb{E}\left[e^{sX}\right]}\leq\frac{e^{2\delta}}{1-\lambda}$

for every $\lambda$ satisfying $1-C/T\leq s+(1-s)\lambda\leq 1-c/T$. This
condition holds if and only if $1-2C/(c+C)\leq\lambda\leq 1-2c/(c+C)$. Thus,
it follows from Lemma 6.5 (applied with “$y$” equal to $T$, “$a$” equal to
$(1-s)Ta/2=(c+C)a/4$, and “$\varepsilon$” equal to $1/2$) that if $\delta>0$
is sufficiently small as a function of $a$, $c$, and $C$ then

$\mathbb{P}\left(1-a\leq\frac{Z_{s}}{(1-s)T}\leq
1+a\right)\geq(1-\varepsilon)\left(1-\exp\left(-\frac{(1-s)Ta}{2}\right)\right)e^{-1+a}$

and hence that

$\displaystyle\mathbb{P}\left((1-a)T\leq X\leq(1+a)T\right)$
$\displaystyle=\mathbb{E}[e^{sX}]\mathbb{E}\left[e^{-\frac{sZ_{s}}{1-s}}\mathbf{1}_{\left\\{1-a\leq\frac{Z_{s}}{(1-s)T}\leq
1+a\right\\}}\right]$
$\displaystyle\geq\frac{1}{2}\frac{e^{-\delta}}{1-s}\left(1-\exp\left(-\frac{(1-s)Ta}{2}\right)\right)\exp\left[-s(1+a)T-1+a\right].$

Since $s\leq 1$, the claim follows by using the approximation
$1-\exp[-\frac{1}{2}(1-s)Ta]\geq c^{\prime}(1-s)Ta$ (where $c^{\prime}>0$
depends only on $c$ and $C$) and noting that the two factors of $1-s$ cancel.
∎

###  6.3 Proof of Theorem 1.3

This section combines the estimates on the Laplace transform (Theorem 6.1)
together with the Tauberian theorem from Section 6.2 to prove Theorem 1.3.

###### Proof of Theorem 1.3.

Let $\varepsilon>0$. We apply Theorem 6.3 with ${a=\varepsilon}$, $c=1$, and
$C=2$, letting $\delta>0$ be small enough that the statement made in Theorem
6.3 holds. Now, by Theorem 6.1, there exists $u_{0}>0$ (that we referred to as
$T_{0}$ in Theorem 6.1), $R_{0}>0$ and $x_{*}>x_{0}$ large enough such that
for all $u_{0}<u<2u_{0}$, $R>R_{0}$, $x>x_{*}$,

$\mathbb{E}_{e^{-x_{0}}R,R}\left[1-\exp\left\\{\left(1-\frac{u}{x}\right)\frac{2N_{e^{-x}R}}{(x+\log
x)(e^{-x}R)^{2}}\right\\}\right]=-e^{\pm\delta/3}\frac{2(e^{2x_{0}}-1)}{R^{2}x}\frac{x}{u},$

where, as before, we write $e^{\pm\delta/3}$ for a quantity bounded between
$e^{-\delta/3}$ and $e^{\delta/3}$ whose precise value may vary from line to
line. Also, by Lemma 4.1, if $x$ and $R$ are large enough,

$\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{e^{-x}R}>0\right)=e^{\pm\delta/3}\frac{2(e^{2x_{0}}-1)}{R^{2}x}.$
(6.6)

These two estimates together imply that

$\mathbb{E}_{e^{-x_{0}}R,R}\left[\left.\exp\left\\{\left(1-\frac{u}{x}\right)\frac{2N_{e^{-x}R}}{(x+\log
x)(e^{-x}R)^{2}}\right\\}\right|N_{e^{-x}R}>0\right]=1+e^{\pm
2\delta/3}\frac{x}{u}.$

Fix $u\in[u_{0},2u_{0}]$. If we take $x$ to be large enough that the first
term on the right hand side is much smaller than the second one, we obtain
that

$\mathbb{E}_{e^{-x_{0}}R,R}\left[\left.\exp\left\\{\left(1-\frac{u}{x}\right)\frac{2N_{e^{-x}R}}{(x+\log
x)(e^{-x}R)^{2}}\right\\}\right|N_{e^{-x}R}>0\right]=e^{\pm\delta}\frac{x}{u}$

whenever $x$ and $R$ are sufficiently large. Applying Theorem 6.3 to
$T=(1-\varepsilon)^{-1}x/u_{0}$ and the random variable
$X=\frac{2N_{e^{-x}R}}{(x+\log x)(e^{-x}R)^{2}}$ conditioned to be positive,
we obtain that

$\mathbb{P}_{e^{-x_{0}}R,R}\left(\left.N_{e^{-x}R}\geq\frac{1}{2}\frac{x}{u_{0}}(x+\log
x)(e^{-x}R)^{2}\right|N_{e^{-x}R}>0\right)\geq\frac{\delta\varepsilon
x}{(1-\varepsilon)u_{0}}\exp\left[-\frac{1+\varepsilon}{1-\varepsilon}\cdot\frac{x}{u_{0}}\right]$

Using (6.6) to rewrite this estimate slightly, we have shown for each
$\varepsilon>0$ that there exists $\delta^{\prime}>0$ such that

$\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{e^{-x}R}\geq\frac{x(e^{-x}R)^{2}}{2}\frac{x}{u_{0}}\right)\geq\frac{\delta^{\prime}}{R^{2}x}\exp\left[-\frac{1+\varepsilon}{1-\varepsilon}\cdot\frac{x}{u_{0}}\right]$
(6.7)

for all $u_{0}>0$ large enough (depending on $\varepsilon$) and for all
$x>x_{0}$ large enough (depending on $\varepsilon$ and $u_{0}$). This shows
that the lower bound of the desired form holds _for all sufficiently small
values of $a$._

We now bootstrap this result to obtain any thickness level $a>0$ instead of
small levels $1/u_{0}$ as follows. Let $x_{0}^{\prime}=x_{0}+1$ be an
intermediate level. Starting from a single particle on the sphere of radius
$e^{-x_{0}}R$, we can first condition on the event that the branching process
emanating from this particle contains at least $\left\lceil
au_{0}\right\rceil(e^{-x_{0}^{\prime}}R)^{2}$ pioneers on the sphere of radius
$e^{-x_{0}^{\prime}}R$. This occurs with some probability larger than
$c(a,u_{0},x_{0},x_{0}^{\prime})R^{-2}$. Conditionally on this event, to
create at least $ax^{2}(e^{-x}R)^{2}/2$ pioneers on the sphere $\partial
B(e^{-x}R)$, we can ask each of the $\left\lceil au_{0}\right\rceil$ sets of
$(e^{-x_{0}^{\prime}}R)^{2}$ pioneers on $\partial B(e^{-x_{0}^{\prime}}R)$ to
create at least $\frac{1}{u_{0}}x^{2}(e^{-x}R)^{2}/2$ pioneers. By (6.7)
applied to $x_{0}^{\prime}$ instead of $x_{0}$, this probability is at least

$\left(\frac{C}{x}e^{-(1+\varepsilon)x/u_{0}}\right)^{\left\lceil
au_{0}\right\rceil}\geq e^{-(1+2\varepsilon)\frac{\left\lceil
au_{0}\right\rceil}{u_{0}}x}.$

Because $\varepsilon$ and $u_{0}$ can be chosen arbitrarily small and large
respectively, this concludes the proof. ∎

##  7 Tail estimates without killing (proof of Theorem 1.2)

In this we complete the proof of Theorem 1.2. Since we have already proved
above Theorem 1.3, which implies directly the part of Theorem 1.2 with
killing, it suffices to prove (1.3), which is the part of that concerns the
process in infinite volume, with no killing. In fact we will prove the
following more general theorem which allows for a killing boundary at an
arbitrary distance.

######  Theorem 7.1.

Let $x_{0}>0$ and consider a thickness level $a>0$. For each $s\in[1,\infty]$,
we have that

$\displaystyle\mathbb{P}_{e^{-x_{0}}R,R^{s}}\left(\left.N_{1}\geq\frac{a}{2}(\log
R)^{2}\right|N_{1}>0\right)$ $\displaystyle=R^{-\psi_{s}(a)+o(1)}$ (7.1)

as $R\to\infty$, where we set $\psi_{s}(a):=\inf_{1\leq t\leq
s}(2(t-1)+at^{-1})$ (note that $s=\infty$ is allowed, in which case
$\psi_{s}(a)$ coincides with the function $\psi$ in Theorem 1.2).

To this end, we first state and prove the following intermediate result:

######  Lemma 7.2.

For each fixed $t>1$ and $a>0$, the estimate

$\mathbb{P}_{R,R^{t}}\left(\mathrm{hit}~{}\partial B(R^{t})\text{ and
}N_{1}\geq\frac{a}{2}(\log R)^{2}\right)=R^{-2t-a/t+o(1)}$ (7.2)

holds as $R\to\infty$.

###### Proof of Lemma 7.2.

The lower bound follows quickly from the estimate (1.4) of Theorem 1.2,
concerning the process with killing on scale $R$. Indeed, starting with one
initial particle on $\partial B(R)$, the probability that it reaches $\partial
B(R^{t}/2)$ with $2R^{2t}$ pioneers is equal to $R^{-2t+o(1)}$. We then divide
these pioneers into two conditionally independent groups of $R^{2t}$
particles. The probability that the first group reaches $\partial B(R^{t})$ is
of constant order. Meanwhile, by (1.4), the probability that the second group
generates $\frac{a}{2}(\log R)^{2}=\frac{a}{2t^{2}}(\log R^{t})^{2}$ pioneers
on $\partial B(1)$ before reaching $\partial B(R^{t})$ is equal to
$(R^{t})^{-a/t^{2}+o(1)}=R^{-a/t+o(1)}$. Altogether, this shows that the left
hand side of (7.2) is at least $R^{-2t-a/t+o(1)}$.

The upper bound requires more work. For this, we will need to rely again on
the PDE satisfied by the Laplace transform. Fix $u\in(0,1/t)$ and let
$s=-2u/\log R$. By Theorem 3.1 and Lemma 4.5, there exists a positive constant
$C(u)$ such that

$1-\mathbb{E}_{R,R^{t}}\left[\left(1-s\right)^{N_{1}}\right]\geq-\frac{C(u)}{R^{2}\log
R}.$ (7.3)

Let us define for $x\in[0,t\log R]$,

$\mathsf{g}(x):=(R^{t}e^{-x})^{2}\mathbb{E}_{R^{t}e^{-x},R^{t}}\left[1-(1-s)^{N_{1}}\right],\quad\tilde{\mathsf{g}}(x):=(R^{t}e^{-x})^{2}\mathbb{E}_{R^{t}e^{-x},R^{t}}\left[1-(1-s)^{N_{1}}\mathbf{1}_{\left\\{N_{R^{t}}=0\right\\}}\right].$

We are interested in the function
$\mathsf{f}:=\tilde{\mathsf{g}}-\mathsf{g}\geq 0$, which is given by

$\mathsf{f}(x)=(R^{t}e^{-x})^{2}\mathbb{E}_{R^{t}e^{-x},R^{t}}\left[(1-s)^{N_{1}}\mathbf{1}_{\left\\{N_{R^{t}}>0\right\\}}\right].$

The two functions $\mathsf{g}$ and $\tilde{\mathsf{g}}$ satisfy

$\left\\{\begin{array}[]{ll}\mathsf{g}^{\prime\prime}+2\mathsf{g}^{\prime}=\mathsf{g}^{2},\\\
\mathsf{g}(0)=0,~{}\mathsf{g}(t\log
R)=s,\end{array}\right.\quad\text{and}\quad\left\\{\begin{array}[]{ll}\tilde{\mathsf{g}}^{\prime\prime}+2\tilde{\mathsf{g}}^{\prime}=\tilde{\mathsf{g}}^{2},\\\
\tilde{\mathsf{g}}(0)=1,~{}\tilde{\mathsf{g}}(t\log R)=s,\end{array}\right.$

while $\mathsf{f}$ satisfies the ODE
$\mathsf{f}^{\prime\prime}+2\mathsf{f}^{\prime}=(\mathsf{g}+\tilde{\mathsf{g}})\mathsf{f}$.

Let $\eta>0$. Since $\sup_{[0,\infty)}|g_{\lambda}|\leq\lambda/2$ (Lemma 3.12)
and the slope at the origin of $\mathsf{g}$ goes to zero as $R\to\infty$
(Lemma 4.5), we have that $\mathsf{g}\geq-\eta/2$ for all sufficiently large
$R$. We now fix such a value of $R$ and claim that

$\mathsf{f}(x)\leq e^{-(2-\varepsilon)x},\quad\quad x\in[0,t\log
R],\quad\quad\text{where}\quad\varepsilon=1-\sqrt{1-\eta}.$ (7.4)

Once the claim (7.4) is proven, we may apply this estimate with $x=(t-1)\log
R$ to deduce that

$\mathbb{E}_{R,R^{t}}\left[(1+2u/\log
R)^{N_{1}}\mathbf{1}_{\left\\{N_{R^{t}}>0\right\\}}\right]\leq R^{-2t+o(1)}$

as $R\to\infty$ (recalling that $u\in(0,1/t)$ was fixed), from which the
claimed upper bound of (7.2) may be deduced via Markov’s inequality and
optimising over $u\in(0,1/t)$ by taking $u$ as close to $1/t$ as desired.

We now prove the claimed inequality (7.4). Since
$\tilde{\mathsf{g}}\geq\mathsf{g}$, we have
$\mathsf{g}+\tilde{\mathsf{g}}\geq-\eta$ and
$\mathsf{f}^{\prime\prime}+2\mathsf{f}^{\prime}\geq-\eta\mathsf{f}.$ As in the
proof of Lemma 3.11, if we let $c_{\pm}=1\pm\sqrt{1-\eta}$ we have that
$c_{+}+c_{-}=2$ and $c_{+}c_{-}=\eta$ and hence that

$(\mathsf{f}^{\prime}+c_{+}\mathsf{f})^{\prime}+c_{-}(\mathsf{f}^{\prime}+c_{+}\mathsf{f})=\mathsf{f}^{\prime\prime}+2\mathsf{f}^{\prime}+\eta\mathsf{f}\geq
0.$

Integrating between $x$ and $t\log R$ for some $x\in[0,t\log R]$ (using Lemma
3.10) leads to

$(\mathsf{f}^{\prime}(x)+c_{+}\mathsf{f}(x))e^{c_{-}x}\leq(\mathsf{f}^{\prime}(t\log
R)+c_{+}\mathsf{f}(t\log R))e^{c_{-}t\log R}.$

Because $\mathsf{f}(t\log R)=0$ and $\mathsf{f}^{\prime}(t\log R)\leq 0$, this
implies that $\mathsf{f}^{\prime}+c_{+}\mathsf{f}\leq 0$ in $[0,t\log R]$.
Integrating further between $0$ and some $x\in[0,\log R]$ (using Lemma 3.10 a
second time) leads to (7.4), which concludes the proof. ∎

We can now prove Theorem 7.1.

###### Proof of Theorem 7.1.

Fix $a>0$ and $s\in[1,\infty]$. The lower bound is a direct consequence of
(1.4) when $a\leq 2$. If $a>2$, letting $t_{*}=\min\\{s,\sqrt{a/2}\\}$, we
have by Lemma 7.2 that

$\mathbb{P}_{R,R^{s}}\left(N_{1}\geq\frac{a}{2}(\log
R)^{2}\right)\geq\mathbb{P}_{R,R^{t_{*}}}\left(N_{1}\geq\frac{a}{2}(\log
R)^{2},\mathrm{hit~{}}\partial B(R^{t_{*}})\right)=R^{-2t_{*}-a/t_{*}+o(1)}.$

The exponent $t_{*}$ was chosen to minimise the above exponent, which is equal
to $\psi_{s}(a)$. This concludes the proof of the lower bound.

For the upper bound, let $\varepsilon>0$ be the inverse of a large integer,
let $t_{k}=1+\varepsilon k$ for each $k\geq 0$, and let $k_{\max}$ be the
largest integer such that $t_{k}\leq t_{*}=\min\\{s,\sqrt{a/2}\\}$. Let
$K:=\max\\{k\geq 0:N_{R^{t_{k}}}>0\\}$, where $N_{R^{t_{k}}}$ is the number of
pioneers on the large sphere $\partial B(R^{t_{k}})$ when particles are frozen
both on this sphere and the unit sphere. We have by a union bound that

$\displaystyle\mathbb{P}_{R,R^{s}}\left(N_{1}\geq\frac{a}{2}(\log
R)^{2}\right)\leq\sum_{k=0}^{k_{\max}}\mathbb{P}_{R,R^{s}}\left(N_{1}\geq\frac{a}{2}(\log
R)^{2},K=k\right)+\mathbb{P}_{R,R^{s}}\left(K\geq k_{\max}+1\right).$

If $t_{*}=s$ then the last probability in the above display is equal to zero.
Otherwise, this probability is

$\mathbb{P}_{R,R^{s}}\left(N_{R^{t_{k_{\max}+1}}}>0\right)=R^{-2t_{k_{\max}+1}+o(1)}\leq
R^{-2\sqrt{a/2}+o(1)}=R^{-2-\psi_{s}(a)+o(1)}$

by definition of $k_{\max}$. In either case, we have that

$\mathbb{P}_{R,R^{s}}\left(N_{R^{t_{k_{\max}+1}}}>0\right)\leq
R^{-2-\psi_{s}(a)+o(1)}.$ (7.5)

We now take $k\in\\{0,\dots,k_{\max}\\}$ and work on the event $\\{K=k\\}$. We
can decompose $N_{1}$ as $N_{1}=N_{1}^{(1)}+N_{1}^{(2)}$ where $N_{1}^{(1)}$
is the number of pioneers on $\partial B(1)$ that _do not_ have an ancestor in
$\partial B(R^{t_{k}})$, and $N_{1}^{(2)}=N_{1}-N_{1}^{(1)}$ is the number of
pioneers on $\partial B(1)$ that _do_ have an ancestor in $\partial
B(R^{t_{k}})$. We have by a union bound that

$\displaystyle\mathbb{P}_{R,\infty}\left(N_{1}\geq\frac{a}{2}(\log
R)^{2},K=k\right)\leq\mathbb{P}_{R,\infty}\left(N_{R^{t_{k}}}\geq
R^{2t_{k}}(\log R)^{3/2}\right)$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+\mathbb{P}_{R,R^{t_{k+1}}}\left(N_{1}^{(1)}+N_{1}^{(2)}\geq\frac{a}{2}(\log
R)^{2},0<N_{R^{t_{k}}}<R^{2t_{k}}(\log R)^{3/2}\right).$ (7.6)

By Lemma 2.7, the first probability on the right hand side decays faster than
any polynomial in $R$. Meanwhile, the second probability is at most

$\displaystyle\mathbb{P}_{R,R^{t_{k+1}}}\left(N_{1}^{(1)}\geq\frac{a}{2}(\log
R)^{2},0<N_{R^{t_{k}}}\right)+\sum_{n=1}^{1/\varepsilon}\mathbb{P}_{R,R^{t_{k+1}}}\Big{(}N_{1}^{(1)}\in[1-n\varepsilon,1-(n-1)\varepsilon]\frac{a}{2}(\log
R)^{2},$ $\displaystyle N_{1}^{(2)}\geq\frac{(n-1)\varepsilon a}{2}(\log
R)^{2},0<N_{R^{t_{k}}}<R^{2t_{k}}(\log R)^{3/2}\Big{)}.$

Conditionally on $N_{R^{t_{k}}}$, the random variables $N_{1}^{(1)}$ and
$N_{1}^{(2)}$ are independent. Moreover, by (7.3) and the conditional
independence of particles after hitting $\partial B(R^{t_{k}})$, we have for
all $v\in(0,1)$,

$\mathbb{E}_{R,R^{t_{k+1}}}\left[\left.\left(1+\frac{2v}{\log
R^{t_{k+1}}}\right)^{N_{1}^{(2)}}\right|N_{R^{t_{k}}}<R^{2t_{k}}(\log
R)^{3/2}\right]\leq\exp(C(v)(\log R)^{1/2})=R^{o(1)}.$

By Markov’s inequality, this implies that, conditionally on
$N_{R^{t_{k}}}<R^{2t_{k}}(\log R)^{3/2}$, the probability (for the branching
Brownian motion with killing at $R^{t_{k+1}}$) that $N_{1}^{(2)}$ is at least
$\frac{(n-1)\varepsilon a}{2}(\log R)^{2}$ is at most $R^{-(n-1)\varepsilon
a/t_{k+1}+o(1)}$. Moreover, by Lemma 7.2,

$\mathbb{P}_{R,R^{t_{k+1}}}\Big{(}N_{1}^{(1)}\geq(1-n\varepsilon)\frac{a}{2}(\log
R)^{2},0<N_{R^{t_{k}}}\Big{)}\leq R^{-2t_{k}-(1-n\varepsilon)a/t_{k}+o(1)}.$

This shows that the second probability in the right hand side of (7.6) is at
most

$\displaystyle
R^{-2t_{k}-a/t_{k}+o(1)}+\sum_{n=1}^{1/\varepsilon}R^{-2t_{k}-(1-n\varepsilon)a/t_{k}-(n-1)\varepsilon
a/t_{k+1}+o(1)}\leq R^{-2t_{k}-a/t_{k}+2a\varepsilon+o(1)},$

where in the last inequality we made elementary simplifications and, in
particular, used the bound

$\frac{1}{t_{k+1}}=\frac{1}{t_{k}+\varepsilon}\geq\frac{1}{t_{k}}\Big{(}1-\frac{\varepsilon}{t_{k}}\Big{)}\geq\frac{1}{t_{k}}-\frac{\varepsilon}{t_{k}^{2}}\geq\frac{1}{t_{k}}-\varepsilon$

(as $t_{k}\geq 1$). Wrapping up, we have shown that

$\mathbb{P}_{R,\infty}\left(N_{1}\geq\frac{a}{2}(\log R)^{2},K=k\right)\leq
R^{-2t_{k}-a/t_{k}+2a\varepsilon+o(1)}$

for each $0\leq k\leq k_{\mathrm{max}}$ and hence that

$\mathbb{P}_{R,\infty}\left(N_{1}\geq\frac{a}{2}(\log R)^{2},K\leq
k_{\mathrm{max}}\right)\leq k_{\mathrm{max}}\cdot R^{-\min_{0\leq k\leq
k_{\mathrm{max}}}(2t_{k}+at_{k}^{-1})+2a\varepsilon+o(1)}\leq
R^{-\psi_{s}(a)+2a\varepsilon+o(1)}.$ (7.7)

The claimed upper bound follows from (7.5) and (7.7) since $\varepsilon>0$ was
arbitrary. ∎

##  8 Thick points (proof of Theorem 1.5)

In this section we apply the main results of Sections 4 and 6 to prove Theorem
1.5. The upper bound in this theorem has already been proved at the end of
Section 4.3, so that it remains only to prove the (more difficult) lower
bound.

###  8.1 Strategy of the proof

We now explain the overall strategy behind the proof of the lower bound of
(1.11) in Theorem 1.5. In particular, we will state three key propositions
(Propositions 8.1, 8.2 and 8.3), and explain how these propositions imply
Theorem 1.5. The rest of Section 8 is then dedicated to the proofs of these
three key propositions.

As is common in the study of analogous questions for other models, we use a
truncated second moment approach. The truncation we use will be described in
terms of _non-backtracking trajectories_ , a notion we now introduce. Let
$\eta_{1}>0$ be the inverse of a large integer, let $z\in\mathbb{Z}^{4}$, and
define $r_{k}=R^{1-k\eta_{1}}$ for each $k\in\\{1,\dots,1/\eta_{1}\\}$. We say
that a pioneer in $\partial B(z,r)$ is non-backtracking if, for each $1\leq
k\leq 1/\eta_{1}$, the trajectory taken by the ancestors of the pioneer does
not revisit the sphere $\partial B(z,r_{k}\log R)$ after first visiting the
sphere $\partial B(z,r_{k})$. (Note that this makes sense even if $r$ is
larger than some of the $r_{k}$.) We denote the number of non-backtracking
pioneers on the sphere $\partial B(z,r)$ by $\hat{N}_{z,r}$, noting that
$\hat{N}_{z,r}$ depends implicitly on the choice of $\eta_{1}$ and $R$.

In the first key proposition, we show that restricting the branching process
to non-backtracking trajectories does not reduce the exponent for a single
point to be thick (as can be seen by comparing this result with Theorem 1.3).
The constants $1/2$ and $1/10$ appearing in this proposition are somewhat
arbitrary.

######  Proposition 8.1.

Let $z,z_{0}\in B(0,R/2)$ with $|z-z_{0}|\geq R/10$. We have

$\mathbb{P}_{z_{0},R}\left(\hat{N}_{z,1}\geq\frac{a}{2}(\log R)^{2}\right)\geq
R^{-2-a+o(1)}\quad\quad\text{as}\quad R\to\infty.$

The next key proposition will state that the exponent for a point to be thick
remains the same if, in addition to restricting to non-backtracking
trajectories, we also impose a “good event” stating that the number of non-
backtracking pioneers on each mesoscopic scale is of order consistent with the
desired level of thickness at the unit scale. Let $b>a$ be a parameter close
to $a$ and define the good event $G_{R}(z)$ by

$G_{R}(z):=\left\\{\forall
k=1,\dots,1/\eta_{1},\hat{N}_{z,r}\leq\frac{b}{2}r_{k}^{2}\left(\log\frac{R}{r_{k}}\right)^{2}\right\\}.$
(8.1)

The definition of this good event is inspired by the Gaussian and Brownian
multiplicative chaos theories; see e.g. [Ber17, Jeg20a]. In the next result,
we show that adding this second layer of good event (the first layer being the
restriction to non-backtracking particles) keeps the first moment almost
unchanged (rather than merely keeping the exponent the same).

######  Proposition 8.2.

Let $z_{0},z\in B(0,R)$ with $|z-z_{0}|\geq R/10$. There exists
$\varepsilon_{1}=\varepsilon_{1}(a,b,\eta_{1})>0$ such that for all $R$ large
enough,

$\mathbb{P}_{z_{0},R}\left(\hat{N}_{z,1}\geq\frac{a}{2}(\log
R)^{2},G_{R}(z)^{c}\right)\leq R^{-2-a-\varepsilon_{1}}.$ (8.2)

Finally, we bound the relevant truncated two-point function, where the
truncation both restricts to non-backtracking trajectories and imposes that
the relevant good events hold.

######  Proposition 8.3.

Let $\eta_{3}>0$ be a small parameter, $z_{0}\in B(R/2)\setminus B(R/4)$ and
$z,w\in B(0,R^{1-\eta_{3}})$. There exists
$\varepsilon_{2}=\varepsilon_{2}(a,b,\eta_{1},\eta_{3})>0$ such that

$\mathbb{P}_{z_{0},R}\left(\hat{N}_{z,1},\hat{N}_{w,1}\geq\frac{a}{2}(\log
R)^{2},G_{R}(z),G_{R}(w)\right)\leq
R^{-2-2a+o(1)}\left(\frac{R}{|z-w|}\right)^{a+\varepsilon_{2}}.$ (8.3)

Moreover, for $a$ and $\eta_{3}$ fixed,
$\varepsilon_{2}(a,b,\eta_{1},\eta_{3})\to 0$ as $b\to a^{+}$ and $\eta_{1}\to
0$.

Assuming the three propositions above, we are now ready to prove the lower
bound in Theorem 1.5. In a nutshell, an application of Paley–Zygmund
inequality will show that the probability of having a large number of thick
points decays slower than any power of $R$. We will then take advantage of the
numerous independent branching Brownian motions present in a branching
Brownian motion conditioned to hit $\partial B(R/2)$.

###### Proof of Theorem 1.5 – Lower bound.

Without loss of generality, we change the conditioning: rather than
conditioning on surviving at least $R^{2}$ generations, we will condition on
the event that at least one particle reaches the outer sphere $\partial B(R)$.
We will denote by $\mathbb{P}^{*}_{z_{0},R}$ this conditional law. (Any
conditioning that guarantees that the BGW tree survives at least $R^{2+o(1)}$
generations with high probability would do.) Note that it suffices to prove
(1.11), since this trivially implies the lower bound on the maximal thickness.
Let $\eta_{3}>0$ be as in Proposition 8.3 (small, but much larger than
$\eta_{1}$) and define the following subset of $\mathcal{T}_{R}(a)$:

$\widehat{\mathcal{T}}_{R}(a):=\left\\{z\in
B(0,R^{1-\eta_{3}})\cap\mathbb{Z}^{4}:\hat{N}_{z,1}\geq\frac{a}{2}(\log
R)^{2}\right\\}.$

Let $\delta>0$. By Propositions 8.1 and 8.2,

$\displaystyle\mathbb{E}_{z_{0},R}\left[\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\widehat{\mathcal{T}}_{R}(a)\right\\}}\mathbf{1}_{G_{R}(z)}\right]\geq
R^{-2+4(1-\eta_{3})-a+o(1)}.$ (8.4)

Under $\mathbb{P}^{*}_{z_{0},R}$ there is at least one particle hitting
$\partial B(R)$. Pick such a particle uniformly at random, and let $E$ be the
event that the ancestors of this particle gave birth to at least $R^{2}$
particles, each starting in $B(R/2)\setminus B(R/4)$. It is easy to see that
the conditional probability of $E$ is uniformly bounded away from 0:
$\inf_{R}\mathbb{P}_{z_{0},R}^{*}(E)>0$, and that conditionally on $E$ under
$\mathbb{P}^{*}_{z_{0},R}$, and conditionally on the location of these
branching events, the descendants form $R^{2}$ independent branching Brownian
motions starting from the given locations in $B(R/2)\setminus B(R/4)$. We
deduce from (8.4) that

$\displaystyle\mathbb{E}_{z_{0},R}^{*}\left[\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\widehat{\mathcal{T}}_{R}(a)\right\\}}\mathbf{1}_{G_{R}(z)}\right]$
$\displaystyle\geq\mathbb{P}_{z_{0},R}^{*}(E)R^{2}\inf_{z_{0}^{\prime}\in
B(R/2)\setminus
B(R/4)}\mathbb{E}_{z_{0}^{\prime},R}\left[\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\widehat{\mathcal{T}}_{R}(a)\right\\}}\mathbf{1}_{G_{R}(z)}\right]$
$\displaystyle\geq R^{4(1-\eta_{3})-a+o(1)}.$

Meanwhile, by Proposition 8.3,

$\displaystyle\mathbb{E}_{z_{0},R}\left[\left(\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\widehat{\mathcal{T}}_{R}(a)\right\\}}\mathbf{1}_{G_{R}(z)}\right)^{2}\right]\leq
R^{-2-2a+o(1)}\sum_{z,w\in\mathbb{Z}^{4}\cap
B(0,R^{1-\eta_{3}})}\left(\frac{R}{|z-w|}\right)^{a+\varepsilon_{2}}.$

If $b$ is close enough to $a$ and $\eta_{1}$ is sufficiently small, then
$a+\varepsilon_{2}<4$, so that the above sum can be bounded by
$O(R^{8(1-\eta_{3})+\eta_{3}(a+\varepsilon_{2})})$ and the above second moment
is bounded above by
$R^{-2+8(1-\eta_{3})-2a+\eta_{3}(a+\varepsilon_{2})+o(1)}.$ To deduce a
similar bound regarding the second moment under the conditioned probability
$\mathbb{P}_{z_{0},R}^{*}$, we simply use the fact that the probability of
hitting $\partial B(R)$ is of order $R^{2}$ so that

$\displaystyle\mathbb{E}_{z_{0},R}^{*}\Bigg{[}\Big{(}\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\widehat{\mathcal{T}}_{R}(a)\right\\}}\mathbf{1}_{G_{R}(z)}\Big{)}^{2}\Bigg{]}$
$\displaystyle\leq\frac{1}{\mathbb{P}_{z_{0},R}\left(\text{hit~{}}\partial
B(R)\right)}\mathbb{E}_{z_{0},R}\Bigg{[}\Big{(}\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\widehat{\mathcal{T}}_{R}(a)\right\\}}\mathbf{1}_{G_{R}(z)}\Big{)}^{2}\Bigg{]}$
$\displaystyle\leq R^{8(1-\eta_{3})-2a+\eta_{3}(a+\varepsilon_{2})+o(1)}.$

Using the Paley–Zygmund inequality, we deduce that

$\displaystyle\mathbb{P}^{*}_{z_{0},R}\left(\\#\mathcal{T}_{R}(a)\geq
R^{4(1-\eta_{3})-a-\delta}\right)$ $\displaystyle\hskip
56.9055pt\geq\mathbb{P}^{*}_{z_{0},R}\left(\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\widehat{\mathcal{T}}_{R}(a)\right\\}}\mathbf{1}_{G_{R}(z)}\geq\frac{1}{2}\mathbb{E}^{*}_{z_{0},R}\left[\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\widehat{\mathcal{T}}_{R}(a)\right\\}}\mathbf{1}_{G_{R}(z)}\right]\right)$
$\displaystyle\hskip
56.9055pt\geq\frac{1}{4}\mathbb{E}^{*}_{z_{0},R}\left[\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\widehat{\mathcal{T}}_{R}(a)\right\\}}\mathbf{1}_{G_{R}(z)}\right]^{2}\left/\mathbb{E}^{*}_{z_{0},R}\left[\left(\sum_{z\in\mathbb{Z}^{4}}\mathbf{1}_{\left\\{z\in\widehat{\mathcal{T}}_{R}(a)\right\\}}\mathbf{1}_{G_{R}(z)}\right)^{2}\right]\right.\geq
R^{-\eta_{3}(a+\varepsilon_{2})-o(1)}.$

Rephrasing slightly, we have proved that for any $\delta>0$,

$\mathbb{P}^{*}_{z_{0},R}\left(\\#\mathcal{T}_{R}(a)\geq
R^{4-a-\delta}\right)\geq R^{-o(1)}.$ (8.5)

Using again the fact that $\mathbb{P}_{z_{0},R}\left(\text{hit~{}}\partial
B(R)\right)=R^{-2+o(1)}$, this implies the following estimate for the
unconditioned probability: for any $\delta>0$,

$\mathbb{P}_{z_{0},R}\left(\\#\mathcal{T}_{R}(a)\geq R^{4-a-\delta}\right)\geq
R^{-2+o(1)}.$ (8.6)

To conclude the proof, it remains to replace the $R^{o(1)}$ on the right hand
side of (8.5) by some quantity that goes to 1 as $R\to\infty$. Let $\delta>0$
be small and let $\varepsilon>0$ be small enough so that
$(4-a-\delta)/(1-\varepsilon)<4-a$. Let $E$ be the event that for all
$k=\left\lfloor R^{\varepsilon}/4\right\rfloor,\left\lfloor
R^{\varepsilon}/4\right\rfloor+1,\dots,\left\lfloor
R^{\varepsilon}/2\right\rfloor-1$, the number of pioneers on the sphere
$\partial B(R/2+kR^{1-\varepsilon})$ is at least $R^{2-\varepsilon/2}$. For
each $k$ and each such pioneer, consider the number of thick points generated
by that pioneer where particles are killed on the next sphere $\partial
B(R/2+(k+1)R^{1-\varepsilon})$. These numbers of thick points stochastically
dominate $\\#\mathcal{T}_{2R^{1-\varepsilon}}(a)$ under
$\mathbb{P}_{w_{0},2R^{1-\varepsilon}}$, where $w_{0}\in\partial
B(R^{1-\varepsilon})$. If $\\#\mathcal{T}_{R}(a)\leq R^{4-a-\delta}$, then all
these subsets of $\mathcal{T}_{R}(a)$ contain at most
$R^{4-a-\delta}=(R^{1-\varepsilon})^{(4-a-\delta)/(1-\varepsilon)}$ elements
(recall that $(4-a-\delta)/(1-\varepsilon)<4-a$). On the event $E$, there are
at least $cR^{\varepsilon}R^{2-\varepsilon/2}=cR^{2+\varepsilon/2}$ of these
subsets. By a repeated application of the Markov property and by (8.6) applied
to $2R^{1-\varepsilon}$ and to a starting point on $\partial
B(R^{1-\varepsilon})$ instead of $R$ and $z_{0}$, we deduce that

$\displaystyle\mathbb{P}^{*}_{z_{0},R}\left(E,\\#\mathcal{T}_{R}(a)\leq
R^{4-a-\delta}\right)$ $\displaystyle\leq
R^{2+o(1)}\mathbb{P}_{z_{0},R}\left(E,\\#\mathcal{T}_{R}(a)\leq
R^{4-a-\delta}\right)$ $\displaystyle\leq
R^{2+o(1)}(1-R^{-2+o(1)})^{cR^{2+\varepsilon/2}}=R^{2+o(1)}e^{-R^{\varepsilon/2+o(1)}}\to
0$

as $R\to\infty$. Together with the fact that $\mathbb{P}^{*}_{z_{0},R}(E)\to
1$ as $R\to\infty$, this concludes the proof of the fact that
$\mathbb{P}^{*}_{z_{0},R}(\\#\mathcal{T}_{R}(a)\leq R^{4-a-\delta})\to 0$ as
$R\to\infty$. ∎

###  8.2 One-point estimate in a non-backtracking scenario

The goal of this section is to prove Proposition 8.1. We start with a
preliminary lemma, which is essentially just a restatement of some of our
earlier results. Recall that $\eta_{1}$ stands for the inverse of a large
integer and that the decreasing sequence of radii $(r_{k})_{k}$ is defined by
$r_{k}=R^{1-k\eta_{1}}$, for $k=1,\dots,1/\eta_{1}$.

######  Lemma 8.4.

Let $\eta_{1}>0$ be the inverse of an integer. Let
$[\nu_{-},\nu_{+}]\subset(0,1)$ be a compact interval of $(0,1)$. For all
$\nu\in[\nu_{-},\nu_{+}]$ and $k\in\\{1,\dots,1/\eta_{1}\\}$,

$\mathbb{E}_{r_{k},r_{k}\log
R}\left[\exp\left(\frac{2\nu}{r_{k+1}^{2}\log\frac{r_{k}}{r_{k+1}}}N_{r_{k+1}}\right)\right]=1+(1+o(1))\frac{2}{r_{k}^{2}\log(r_{k}/r_{k+1})}\frac{\nu}{1-\nu}$
(8.7)

where $o(1)\to 0$ as $R\to\infty$, uniformly in $\nu\in[\nu_{-},\nu_{+}]$ and
$k\in\\{1,\dots,1/\eta_{1}\\}$.

###### Proof.

We will use the function $g_{\lambda}$ as defined in (3.1) and studied
extensively in Sections 3, 4, and 5. Let $\mu=4\nu/(1-\nu)$. Because $\nu$
belongs to a compact subspace of $(0,1)$, $\mu$ belongs to a compact subspace
of $(0,\infty)$. Letting $x_{0}=\log\log R$ and $x=\eta_{1}\log R+\log\log R$,
we have $r_{k}=(r_{k}\log R)e^{-x_{0}}$ and $r_{k+1}=(r_{k}\log R)e^{-x}$ and,
by Theorem 3.1, that

$1-\mathbb{E}_{r_{k},r_{k}\log
R}\left[\left(1-\frac{1}{r_{k+1}^{2}}g_{\lambda}(x)\right)^{N_{r_{k+1}}}\right]=\frac{1}{r_{k}^{2}}g_{\lambda}(x_{0})$

for all $\lambda\in(0,1)$. Taking $\lambda=\mu/x$, we have by Lemma 4.5 that

$g_{\lambda}(x)=-(1+o(1))\frac{2\mu}{\mu+4}\frac{1}{x}=-(1+o(1))2\nu\frac{1}{\eta_{1}\log
R}.$

Also, by (4.9) and because $1\ll x_{0}\ll\frac{1}{\lambda}$, we have

$g_{\lambda}(x_{0})=-(1+o(1))\frac{\lambda}{2}=(1+o(1))\frac{2\nu}{1-\nu}\frac{1}{\eta_{1}\log
R}.$

Putting things together, we obtain the desired claim. ∎

Let $\eta_{2}>0$ be much smaller than $\eta_{1}$ and define for all $k=1\dots
1/\eta_{1}$,
$I_{k}=\frac{a}{2}(1-\eta_{2})^{k}r_{k}^{2}\log(\frac{R}{r_{k}})^{2}$. The
main intermediate step in the proof of Proposition 8.1 is summarised by the
following lemma.

######  Lemma 8.5.

Let $k\in\\{1,\dots,1/\eta_{1}-1\\}$. For each $i=1,\ldots,I_{k}$, let
$N^{(i)}$ be i.i.d. random variables distributed according to the number of
pioneers on $\partial B(r_{k+1})$, starting from a single particle on
$\partial B(r_{k})$ and where particles are killed on $\partial B(r_{k}\log
R)$. We have

$\mathbb{P}\left(\sum_{i=1}^{I_{k}}N^{(i)}\geq I_{k+1}\right)\geq
R^{-a\eta_{1}(1+O(\eta_{2}/\eta_{1}))}$

where the implicit constants in $O(\eta_{2}/\eta_{1})$ are uniform provided
that $R$ is sufficiently large.

###### Proof.

The idea of the proof can be easily explained with the following toy model.
Let $n$ be a large integer and $E_{1},\dots,E_{n}$ be i.i.d. exponential
variables with mean 1. We are interested in the probability that
$E_{1}+\dots+E_{n}$ exceeds $tn$, where $t$ is some fixed real number larger
than $1$. To achieve this atypical behaviour, an optimal strategy consists in
shifting the mean of each individual exponential variable to $t$, i.e. to
shift the probability measure by $t^{-n}e^{(1-1/t)(E_{1}+\dots+E_{n})}.$ After
this shift, the sum $E_{1}+\dots+E_{n}$ exceeds $tn$ with a fairly high
probability (asymptotic to 1/2 by the central limit theorem, but more
importantly larger than any exponential in $n$). Since the numbers of pioneers
behave in many ways like exponential random variables, we will be able to use
such a strategy as well. Moreover, since we have a good control on the Laplace
transform of the number of pioneers, we will be able to control the cost of
such a shift.

We now define the change of measure that we will use in the proof. Let
$k\in\\{1,\dots,1/\eta_{1}\\}$, let $\nu=1/(1+k)$, let

$t=\frac{2\nu}{r_{k+1}^{2}\log\frac{r_{k}}{r_{k+1}}}$

and let $\mathbb{Q}$ be the probability law

$\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}=\left.\exp\left(t\sum_{i=1}^{I_{k}}N^{(i)}\right)\right/\mathbb{E}\left[\exp\left(t\sum_{i=1}^{I_{k}}N^{(i)}\right)\right].$

By (8.7) and by the special choice of $\nu$, we have

$\displaystyle\mathbb{E}\left[\exp\left(t\sum_{i=1}^{I_{k}}N^{(i)}\right)\right]=\left(1+\frac{2(1+o(1))}{r_{k}^{2}\log(r_{k}/r_{k+1})}\frac{\nu}{1-\nu}\right)^{I_{k}}$
(8.8) $\displaystyle\hskip
56.9055pt=\exp\left((1+o(1))(1-\eta_{2})^{k}a\frac{\nu}{1-\nu}\log^{2}\left(\frac{R}{r_{k}}\right)\Big{/}\log\left(\frac{r_{k}}{r_{k-1}}\right)\right)=R^{(1+o(1))(1-\eta_{2})^{k}a\eta_{1}k}.$

Define $I_{k+1}^{+}=\frac{1+\eta_{2}}{1-\eta_{2}}I_{k+1}$. We have

$\displaystyle\mathbb{P}\left(\sum_{i=1}^{I_{k}}N^{(i)}\geq I_{k+1}\right)$
$\displaystyle\geq\mathbb{P}\left(I_{k+1}\leq\sum_{i=1}^{I_{k}}N^{(i)}\leq
I_{k+1}^{+}\right)$
$\displaystyle=R^{(1+o(1))(1-\eta_{2})^{k}a\eta_{1}k}\mathbb{E}_{\mathbb{Q}}\left[\mathbf{1}_{\left\\{I_{k+1}\leq\sum_{i=1}^{I_{k}}N^{(i)}\leq
I_{k+1}^{+}\right\\}}\exp\left(-t\sum_{i=1}^{I_{k}}N^{(i)}\right)\right]$

When the sum of the $N^{(i)}$ is smaller than $I_{k+1}^{+}$, we can bound from
below the exponential in the expectation above by
$\exp\left(-tI_{k+1}^{+}\right)$. This leads to

$\displaystyle\mathbb{P}\left(\sum_{i=1}^{I_{k}}N^{(i)}\geq I_{k+1}\right)\geq
R^{((1+o(1))(1-\eta_{2})^{k}k-(1-\eta_{2})^{k-1}(1+\eta_{2})(k+1))\eta_{1}a}\mathbb{Q}\left(I_{k+1}\leq\sum_{i=1}^{I_{k}}N^{(i)}\leq
I_{k+1}^{+}\right)$ $\displaystyle\hskip
199.16928pt=R^{-(1+O(\eta_{2}/\eta_{1}))(1+o(1))\eta_{1}a}\mathbb{Q}\left(I_{k+1}\leq\sum_{i=1}^{I_{k}}N^{(i)}\leq
I_{k+1}^{+}\right).$

To conclude the proof it is enough to show that the above probability
appearing on the second line is equal to $(1+o(1))$. To do this, we will show
separately that $\mathbb{Q}\left(\sum_{i=1}^{I_{k}}N^{(i)}\leq I_{k+1}\right)$
and $\mathbb{Q}\left(\sum_{i=1}^{I_{k}}N^{(i)}\geq I_{k+1}^{+}\right)$ vanish
as $R\to\infty$. We start by bounding the first probability and we let
$\mu=\nu+\frac{k}{k+1}\frac{1}{\sqrt{1-\eta_{2}}}-1$ (which is positive and
smaller than $\nu$). By Markov’s inequality, we have

$\displaystyle\mathbb{Q}\left(\sum_{i=1}^{I_{k}}N^{(i)}\leq
I_{k+1}\right)\leq\exp\left(\frac{\mu}{\nu}tI_{k+1}\right)\mathbb{E}_{\mathbb{Q}}\left[\exp\left(-\frac{\mu}{\nu}t\sum_{i=1}^{I_{k}}N^{(i)}\right)\right]$
$\displaystyle\hskip
113.81102pt=R^{-(1+o(1))(1-\eta_{2})^{k}a\eta_{1}k}R^{\mu(1-\eta_{2})^{k+1}(k+1)^{2}a\eta_{1}}\mathbb{E}\left[\exp\left(\frac{\nu-\mu}{\nu}t\sum_{i=1}^{I_{k}}N^{(i)}\right)\right]$

where the first power of $R$ on the right hand side comes from the
normalisation of the Radon–Nikodym derivative
$\mathrm{d}\mathbb{Q}/\mathrm{d}\mathbb{P}$; see (8.8). By (8.7), this is
further equal to

$\displaystyle\exp\left[(1+o(1))\left(-(1-\eta_{2})^{k}a\eta_{1}k+\mu(1-\eta_{2})^{k+1}(k+1)^{2}a\eta_{1}+(1-\eta_{2})^{k}a\eta_{1}\frac{\nu-\mu}{1-(\nu-\mu)}k^{2}\right)\log
R\right]$
$\displaystyle=\exp\left[(1+o(1))a\eta_{1}(1-\eta_{2})^{k}\left(-k+\mu(1-\eta_{2})(k+1)^{2}+\frac{\nu-\mu}{1-(\nu-\mu)}k^{2}\right)\log
R\right].$

Our specific choice for $\mu$ leads to

$\displaystyle\mathbb{Q}\left(\sum_{i=1}^{I_{k}}N^{(i)}\leq I_{k+1}\right)$
$\displaystyle\leq\exp\left[-(1+o(1))a\eta_{1}(1-\eta_{2})^{k}(2\sqrt{1-\eta_{2}}-2+\eta_{2})k(k+1)\log
R\right]$
$\displaystyle\leq\exp\left[-\frac{1}{5}a\eta_{1}\eta_{2}^{2}k(k+1)\log
R\right]\to 0$

as $R\to\infty$. To bound the other probability, we proceed similarly. We
define this time $\mu=\left(1-\frac{1}{\sqrt{1+\eta_{2}}}\right)\frac{k}{k+1}$
(which is positive and smaller than $1-\nu$). By Markov’s inequality and
(8.7),

$\displaystyle\mathbb{Q}\left(\sum_{i=1}^{I_{k}}N^{(i)}\geq
I_{k+1}^{+}\right)\leq\exp\left(-\frac{\mu}{\nu}tI_{k+1}^{+}\right)\mathbb{E}_{\mathbb{Q}}\left[\exp\left(\frac{\mu}{\nu}t\sum_{i=1}^{I_{k}}N^{(i)}\right)\right]$
$\displaystyle=R^{-(1-\eta_{2})^{k}a\eta_{1}k}R^{-\mu(1-\eta_{2})^{k}(1+\eta_{2})(k+1)^{2}a\eta_{1}}\mathbb{E}\left[\exp\left(\frac{\nu+\mu}{\nu}t\sum_{i=1}^{I_{k}}N^{(i)}\right)\right]$
$\displaystyle=\exp\left[\left(-(1-\eta_{2})^{k}a\eta_{1}k-\mu(1-\eta_{2})^{k}(1+\eta_{2})(k+1)^{2}a\eta_{1}+(1-\eta_{2})^{k}a\eta_{1}\frac{\nu+\mu}{1-(\nu+\mu)}k^{2}\right)\log
R\right]$
$\displaystyle=\exp\left[{\left(a\eta_{1}(1-\eta_{2})^{k}\left(-k-\mu(1+\eta_{2})(k+1)^{2}+\frac{\nu+\mu}{1-(\nu+\mu)}k^{2}\right)\right)}\log
R\right].$

Our choice for $\mu$ leads to

$\displaystyle\mathbb{Q}\left(\sum_{i=1}^{I_{k}}N^{(i)}\geq
I_{k+1}^{+}\right)$
$\displaystyle\leq\exp\left[-a\eta_{1}(1-\eta_{2})^{k}\left(2+\eta_{2}-2\sqrt{1+\eta_{2}}\right)k(k+1)\log
R\right]$
$\displaystyle\leq\exp\left[-\frac{1}{5}a\eta_{1}\eta_{2}^{2}k(k+1)\log
R\right]\to 0$

as $R\to\infty$. This concludes the proof of Lemma 8.5. ∎

We can now prove Proposition 8.1.

###### Proof of Proposition 8.1.

With the same notations as in Lemma 8.5, we have

$\displaystyle\mathbb{P}_{e^{-x_{0}}R,R}\left(\hat{N}_{1}\geq
I_{1/\eta_{1}}\right)\geq\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{r_{1}}\geq
I_{1}\right)\prod_{k=1}^{1/\eta_{1}-1}\mathbb{P}_{r_{k},r_{k}\log
R}\left(\sum_{i=1}^{I_{k}}N_{r_{k+1}}^{(i)}\geq I_{k+1}\right).$

By Theorem 1.3, the first probability on the right hand side is at least
$R^{-2-a\eta_{1}(1-\eta_{2})+o(1)}$. By Lemma 8.5, each probability appearing
in the product is at least $R^{-a\eta_{1}(1+O(\eta_{2}/\eta_{1}))}$. Using the
definition of $I_{1/\eta_{1}}$, this gives

$\mathbb{P}_{e^{-x_{0}}R,R}\left(\hat{N}_{1}\geq\frac{a}{2}(1-\eta_{2})^{1/\eta_{1}}(\log
R)^{2}\right)\geq R^{-2-a(1+O(\eta_{2}/\eta_{1}))}.$

Because $\eta_{2}$ can be as small as desired, this concludes the proof. ∎

###  8.3 Truncated first moment

We now prove Proposition 8.2, which states that we can add the good event
$G_{R}(z)$ to our non-backtracking thick point event without significantly
changing its probability.

###### Proof of Proposition 8.2.

Since the left hand side of (8.2) increases as the value of $b$ approaches
$a$, we can assume that $b$ is close enough to $a$ to ensure that
$\sqrt{b}(1-\eta_{1})<\sqrt{a}$. Let $\varepsilon>0$ be small enough so that

$\frac{a+\varepsilon}{b}\frac{a+\varepsilon}{a}<1-\varepsilon,\quad\sqrt{\frac{b}{a}}(1-\eta_{1})<1-\varepsilon\quad\text{and}\quad\eta_{1}^{2}(\sqrt{b}-\sqrt{a})^{2}>\varepsilon.$

Now let $\delta>0$ be small enough such that for all $p\geq 0$,

$\frac{1+(p+1)\delta}{(1+p\delta)^{2}}\frac{a+\varepsilon}{b}\frac{a+\varepsilon}{a}<1-\varepsilon,\quad\sqrt{\frac{b}{a}(1+\delta)}(1-\eta_{1})<1-\varepsilon\quad\text{and}\quad\eta_{1}(-\delta
b+\eta_{1}(\sqrt{b}-\sqrt{a})^{2})>\varepsilon.$ (8.9)

Let $J_{k}=\frac{b}{2}r_{k}^{2}\left(\log\frac{R}{r_{k}}\right)^{2}$. By a
union bound, the left hand side of (8.2) is at most

$\displaystyle\sum_{k=1}^{1/\eta_{1}-1}\mathbb{P}_{e^{-x_{0}}R,R}\left(\hat{N}_{1}\geq\frac{a}{2}(\log
R)^{2},\hat{N}_{r_{k}}\geq J_{k}\right).$

Fix $k\in\\{1,\dots,1/\eta_{1}-1\\}$. Let $N^{(i)}_{1},i\geq 1$, be i.i.d.
random variables having the same law as the number of pioneers on the unit
sphere with one initial particle on $\partial B(r_{k})$ and where particles
are killed on $\partial B(r_{k}\log R)$. Conditionally on $\hat{N}_{r_{k}}=n$,
the random variable $\hat{N}_{1}$ is stochastically dominated by
$\sum_{i=1}^{n}N_{1}^{(i)}$ (there are additional killings in the variable
$\hat{N}_{1}$, so that their distributions are not identical). Therefore, we
can bound

$\displaystyle\mathbb{P}_{e^{-x_{0}}R,R}\left(\hat{N}_{1}\geq\frac{a}{2}(\log
R)^{2},\hat{N}_{r_{k}}\geq J_{k}\right)$
$\displaystyle\leq\sum_{p=0}^{\infty}\mathbb{P}_{e^{-x_{0}}R,R}\left(1+p\delta\leq\frac{\hat{N}_{r_{k}}}{J_{k}}<1+(p+1)\delta\right)\mathbb{P}\Bigg{(}\sum_{i=1}^{(1+(p+1)\delta)J_{k}}N_{1}^{(i)}\geq\frac{a}{2}(\log
R)^{2}\Bigg{)}.$ (8.10)

By Proposition 4.6, the first probability on the right hand side is at most

$\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{r_{k}}\geq(1+p\delta)J_{k}\right)\leq
R^{-2}e^{-(1+o(1))(1+p\delta)b\log\frac{R}{r_{k}}}=R^{-2-(1+o(1))(1+p\delta)b\eta_{1}k}.$

When $(1+p\delta)b\eta_{1}k\geq a+\varepsilon$, we simply bound the second
probability in (8.10) by 1. The contribution of the sum over $p$ large enough
so that $(1+p\delta)b\eta_{1}k\geq a+\varepsilon$ is then at most
$R^{-2-a-\varepsilon+o(1)}$. We now control the remaining part of the sum and
take $p\geq 0$ such that $(1+p\delta)b\eta_{1}k<a+\varepsilon$. Let

$\nu=1-\sqrt{\frac{b}{a}(1+(p+1)\delta)}\eta_{1}k.$

The first two inequalities in (8.9) ensure that
$\nu\in(\varepsilon,1-\varepsilon)$. Applying Markov’s inequality and
recalling that $r_{k}=R^{1-k\eta_{1}}$, we have

$\displaystyle\mathbb{P}\Bigg{(}\sum_{i=1}^{(1+(p+1)\delta)J_{k}}N_{1}^{(i)}\geq\frac{a}{2}(\log
R)^{2}\Bigg{)}\leq R^{-\frac{a\nu}{1-k\eta_{1}}}\mathbb{E}_{r_{k},r_{k}\log
R}\left[\exp\left(\frac{2\nu N_{1}}{\log
r_{k}}\right)\right]^{(1+(p+1)\delta)J_{k}}.$

Similarly to (8.7),

$\mathbb{E}_{r_{k},r_{k}\log R}\left[\exp\left(\frac{2\nu N_{1}}{\log
r_{k}}\right)\right]=1+(1+o(1))\frac{2}{r_{k}^{2}\log
r_{k}}\frac{\nu}{1-\nu}.$

After simplifications, this leads to

$\mathbb{P}\Bigg{(}\sum_{i=1}^{(1+(p+1)\delta)J_{k}}N_{1}^{(i)}\geq\frac{a}{2}(\log
R)^{2}\Bigg{)}\leq\exp\left(-\frac{1+o(1)}{1-k\eta_{1}}\left(\sqrt{(1+(p+1)\delta)b}\eta_{1}k-\sqrt{a}\right)^{2}\log
R\right).$ (8.11)

Overall this gives that

$\displaystyle\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{r_{k}}\geq(1+p\delta)J_{k}\right)\mathbb{P}\Bigg{(}\sum_{i=1}^{(1+(p+1)\delta)J_{k}}N_{1}^{(i)}\geq\frac{a}{2}(\log
R)^{2}\Bigg{)}$ $\displaystyle\leq
R^{-2}\exp\left(-(1+o(1))\left((1+p\delta)b\eta_{1}k+\frac{1}{1-k\eta_{1}}\left(\sqrt{(1+(p+1)\delta)b}\eta_{1}k-\sqrt{a}\right)^{2}\right)\log
R\right)$ $\displaystyle=R^{-2}\exp\left(-(1+o(1))\left(a-\delta
b\eta_{1}k+\frac{\eta_{1}k}{1-\eta_{1}k}\left(\sqrt{(1+(p+1)\delta)b}-\sqrt{a}\right)^{2}\right)\log
R\right).$

By the last inequality in (8.9), we can bound from below the exponent

$-\delta
b\eta_{1}k+\frac{\eta_{1}k}{1-\eta_{1}k}\left(\sqrt{(1+(p+1)\delta)b}-\sqrt{a}\right)^{2}\geq\eta_{1}(-\delta
b+\eta_{1}(\sqrt{b}-\sqrt{a})^{2})>\varepsilon.$

We have shown that when $p$ is small enough so that
$(1+p\delta)b\eta_{1}k<a+\varepsilon$, the probability in (8.10) is at most
$R^{-2-a-\varepsilon+o(1)}$. Since we have already bounded the remaining
contribution of the sum, this concludes the proof. ∎

######  Remark 8.6.

We note that the appearance of a quadratic exponent in (8.11) after taking the
square root of the thickness parameters $a$ and $b$ is highly suggestive of an
isomorphism-type picture, similar to the isomorphism relating the occupation
of random walk trajectories to square of Gaussian free field [LJ11, MR06].

###  8.4 Truncated second moment

We now prove Proposition 8.3, which upper bounds the probability for two
points to be thick points under the nonbacktracking condition, and each
satisfying the good event.

###### Proof of Proposition 8.3.

Let $z,w\in B(0,R^{1-\eta_{3}})$, and recall that $\eta_{3}$ is fixed while
$\eta_{1}\to 0$. If $2|z-w|\leq R^{\eta_{1}}\log R$, then we can simply bound
the left hand side of (8.3) by the probability that a single point is thick
which is at most $R^{-2-a+o(1)}$ and so (after elementary algebra) this
probability is smaller than the right hand side of (8.3) with
$\varepsilon_{2}=a\eta_{1}/(1+\eta_{1})$.

Thus we suppose now that $2|z-w|\geq R^{\eta_{1}}\log R$, hence we can find
$k\in\\{k_{0},\dots,1/\eta_{1}-1\\}$ such that

$r_{k}\log R\leq 2|z-w|\leq r_{k-1}\log R.$ (8.12)

Here $k_{0}=k_{0}(\eta_{3},\eta_{1})$ can be chosen arbitrarily large if
$\eta_{1}$ is small enough for any given $\eta_{3}>0$. By the definition of
the good event, the left hand side of (8.3) is at most

$\displaystyle\mathbb{P}_{z_{0},R}\Big{(}\hat{N}_{z,1},\hat{N}_{w,1}\geq\frac{a}{2}(\log
R)^{2},\hat{N}_{z,r_{k}},\hat{N}_{w,r_{k}}\leq\frac{b}{2}r_{k}^{2}\Big{(}\log\frac{R}{r_{k}}\Big{)}^{2}\Big{)}.$

Let $\delta>0$ be the inverse of a large integer and denote by
$J_{k}=\frac{b}{2}r_{k}^{2}\left(\log\frac{R}{r_{k}}\right)^{2}$. The above
probability is equal to

$\displaystyle\sum_{0\leq
p_{1},p_{2}<1/\delta}\mathbb{P}_{z_{0},R}\Big{(}\hat{N}_{z,1},\hat{N}_{w,1}\geq\frac{a}{2}(\log
R)^{2},p_{1}\delta<\frac{\hat{N}_{z,r_{k}}}{J_{k}}\leq(p_{1}+1)\delta,p_{2}\delta<\frac{\hat{N}_{w,r_{k}}}{J_{k}}\leq(p_{2}+1)\delta\Big{)}.$

Let $p_{1},p_{2}\in\\{0,\dots,1/\delta-1\\}$. Without loss of generality, let
us assume that $p_{2}\geq p_{1}$. Using the nonbacktracking condition, the
descendants of the particles counted in $\hat{N}_{z,r_{k}}$ and
$\hat{N}_{w,r_{k}}$ respectively, and which contribute to $\hat{N}_{z,1}$ and
$\hat{N}_{w,1}$ respectively, are independent from one another. Hence we may
bound the probability appearing in the above sum by

$\displaystyle\mathbb{P}_{z_{0},R}\left(N_{w,r_{k}}>p_{2}\delta
J_{k}\right)\prod_{j=1}^{2}\mathbb{P}\Bigg{(}\sum_{i=1}^{(p_{j}+1)\delta
J_{k}}N^{(i)}_{1}\geq\frac{a}{2}(\log R)^{2}\Bigg{)}$ (8.13)

where $N^{(i)}_{1},i\geq 1$, are i.i.d. random variables distributed according
to the number of pioneers on the unit sphere starting from a single particle
on $\partial B(r_{k})$ and where particles are killed on $\partial B(r_{k}\log
R)$. By (8.11), for $j=1,2$,

$\displaystyle\mathbb{P}\Bigg{(}\sum_{i=1}^{(p_{j}+1)\delta
J_{k}}N^{(i)}_{1}\geq\frac{a}{2}(\log
R)^{2}\Bigg{)}\leq\exp\left(-\frac{1+o(1)}{1-k\eta_{1}}\left(\sqrt{(p_{j}+1)\delta
b}\eta_{1}k-\sqrt{a}\right)^{2}\log R\right)$

and by Proposition 4.6,

$\displaystyle\mathbb{P}_{z_{0},R}\left(N_{w,r_{k}}>p_{2}\delta J_{k}\right)$
$\displaystyle\leq
R^{-2+o(1)}\exp\left(-bp_{2}\delta\log\frac{R}{r_{k}}\right)$
$\displaystyle\leq R^{-2+o(1)-bp_{2}\delta\eta_{1}k}\leq
R^{-2+o(1)-\frac{1}{2}\sum_{j=1}^{2}bp_{j}\delta\eta_{1}k}.$

In the last inequality above, we used the fact that $p_{2}\geq p_{1}$. Putting
things together, and combining terms suitably, (8.13) is at most

$\displaystyle
R^{-2-2a+o(1)}\exp\Bigg{(}-\frac{k\eta_{1}}{1-k\eta_{1}}\sum_{j=1}^{2}\left(a-2\sqrt{(p_{j}+1)\delta
ab}+(p_{j}+1)\delta b\eta_{1}k+\frac{1-k\eta_{1}}{2}p_{j}\delta b\right)\log
R\Bigg{)}$
$\displaystyle=R^{-2-2a+o(1)}\exp\Bigg{(}-\frac{k\eta_{1}}{1-k\eta_{1}}\sum_{j=1}^{2}\left(\left(\sqrt{a}-\sqrt{(p_{j}+1)\delta
b}\right)^{2}-(1-k\eta_{1})\left((p_{j}+1)\delta b-\frac{1}{2}p_{j}\delta
b\right)\right)\log R\Bigg{)}$ $\displaystyle\leq
R^{-2-2a+o(1)}\exp\Bigg{(}\frac{1}{2}k\eta_{1}\sum_{j=1}^{2}(p_{j}+2)\delta
b\log R\Bigg{)}.$

By (8.12),

$\log\frac{R}{2|z-w|}\geq(k-1)\eta_{1}\log R-\log\log
R=(1+o(1))\left(1-\frac{1}{k}\right)k\eta_{1}\log R.$

So we have obtained that (8.13) is at most

$R^{-2-2a+o(1)}\left(\frac{R}{|z-w|}\right)^{\frac{1}{2}\left(1-\frac{1}{k}\right)^{-1}\sum_{j=1}^{2}(p_{j}+2)\delta
b}.$

Since $p_{1},p_{2}\leq 1/\delta-1$, the exponent
$\frac{1}{2}(1-\frac{1}{k})^{-1}\sum_{j=1}^{2}(p_{j}+2)\delta b$ is bounded by
$(1-\frac{1}{k_{0}})^{-1}(1+\delta)b$ which can be made arbitrarily close to
$(1-\frac{1}{k_{0}})^{-1}b$ by choosing $\delta$ small enough. Recall further
that $k_{0}=k_{0}(\eta_{1},\eta_{3})$ can be chosen arbitrarily large by
taking $\eta_{1}$ small enough for any given $\eta_{3}>0$. Hence the exponent
$\frac{1}{2}(1-\frac{1}{k_{0}})^{-1}\sum_{j=1}^{2}(p_{j}+2)\delta b$ can be
made as close to $a$ as desired. This concludes the proof. ∎

##  9 Occupation measure of branching Brownian motion (proof of Theorem 1.4)

In this section, we prove Theorem 1.4, which concerns the probability that a
small ball has a large occupation measure for branching Brownian motion. We
start with the lower bound, which follows quickly from Theorem 1.3:

###### Proof of Theorem 1.4 – Lower bound.

Let $x_{0}>0$, $\varepsilon>0$. By Theorem 1.3, there exists $x_{*}>0$ and
$R_{0}>0$ such that for all $R\geq R_{0}$, $r\leq e^{-x_{*}}R$, $a>0$,

$\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{r}\geq\frac{a}{2}r^{2}(\log
R/r)^{2}|N_{r}>0\right)\geq(R/r)^{-(1+\varepsilon)a}.$

The local time of $B(r)$ can be decomposed as $\sum_{i=1\dots N_{r}}L_{i}$
where $L_{i}$ is the occupation measure accumulated by the progeny of the
$i$-th pioneer of $\partial B(r)$. The $L_{i}$’s are independent of $N_{r}$
and are i.i.d. and distributed according to the occupation measure of $B(r)$
under $\mathbb{P}_{z,R}$ where $z$ is any point of $\partial B(r)$. Letting
$n=\frac{(1+\varepsilon)^{2}a}{2}r^{2}(\log R/r)^{2}$, we have

$\displaystyle\mathbb{P}_{e^{-x_{0}}R,R}\left(L(B(r))\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}|N_{r}>0\right)\geq(R/r)^{-(1+\varepsilon)^{3}a}\mathbb{P}\left(\sum_{i=1}^{n}L_{i}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\right).$

Recall that since the branching mechanism is critical,
$\mathbb{E}\left[L_{i}\right]$ is equal to the expected value of the local
time in $B(r)$ of a single Brownian motion that starts on $\partial B(r)$ and
that is killed on $\partial B(R)$. If $x_{*}$ is large enough, this
expectation is close to the one in infinite volume, more precisely it is
larger than $(1+\varepsilon)^{-1}m_{1}r^{2}$ (recall the definition (1.6) of
$m_{1}$). The probability on the right hand side of the above display is
therefore at least

$\mathbb{P}\left(\sum_{i=1}^{n}L_{i}\geq(1+\varepsilon)^{-1}\mathbb{E}\left[L_{1}\right]n\right).$

Because the $L_{i}$’s are bounded from below by 0, this probability is
extremely close to 1. This is for instance a consequence of standard large
deviations for i.i.d. nonnegative random variables, sometimes called Benett’s
inequality (see e.g. [BLM13, Theorem 2.9]) or [Dur19, Theorem (9.5)]). This
concludes the proof. ∎

We now move on to the proof of the upper bound (1.8) in Theorem 1.4. We first
explain the difficulty of its proof. Let $r\geq 1$. As in the proof of the
lower bound, we can decompose the occupation measure of the ball $B(r)$ as
$\sum_{i=1\dots N_{r}}L_{i}$ where $L_{i}$ is the occupation measure
accumulated by the progeny of the $i$-th pioneer of $\partial B(r)$. We want
to argue that, given $N_{r}$, the sum $\sum_{i=1\dots N_{r}}L_{i}$ is
concentrated around $\mathbb{E}\left[L_{1}\right]N_{r}$. In the lower bound,
this step was straightforward and we only used the nonnegativity of the
$L_{i}$’s. For the more subtle upper bound, we need to guarantee a certain
decay of their upper tail, which is the content of the following theorem.

######  Theorem 9.1.

There exists $\lambda_{0}>0$ such that for all $\lambda\in[0,\lambda_{0}]$,

$\mathbb{E}_{r,R}\left[\exp\left(\frac{\lambda}{r^{4}\log(R/r)}L(B(r))\right)\right]=1+\frac{\lambda
m_{1}+o(1)}{r^{2}\log(R/r)}$ (9.1)

where $o(1)\to 0$ as $r/R\to 0$ and $R\to\infty$.

###### Proof.

We first claim that there exists $\lambda_{0}>0$ and $C>0$ such that for all
$\lambda\in[0,2\lambda_{0}]$, $R$ large enough and $R/r$ large enough,

$\mathbb{E}_{r,R}\left[\exp\left(\frac{\lambda}{r^{4}\log(R/r)}L(B(r))\right)\right]\leq
1+\frac{C}{r^{2}\log(R/r)}.$ (9.2)

This is a reformulation of a result obtained in [AS22] in the context of
branching random walk; see in particular (9.2) therein. Their proof is based
on the following estimate which is the content of [AS22, Proposition 9.1]:
there exists $c,C>0$ such that for all $r\geq 1$ and $t>r^{4}$,

$\sup_{x\in
B(2r)}\mathbb{P}_{x,2r}\left(L(B(2r))>t\right)\leq\frac{C}{r^{2}}\exp\left(-ct/r^{4}\right).$
(9.3)

To get (9.2) from this estimate, Asselah and Schapira decompose $L(B(r))$ as a
sum over successive “waves” (or excursions) between $\partial B(r)$ and
$\partial B(2r)$ of the occupation measure of $B(r)$ accumulated during the
wave. Each such elementary occupation measure is handled thanks to (9.3)
wheareas the total number of terms involved in the sum is handled thanks to
their knowledge on the number of pioneers. See [AS22, Section 9] for more
details. The exact same approach can be done in the continuum (in fact more
precise estimates on these pioneers were already obtained here in the previous
sections). This leads to (9.2); we omit the details.

We will need in addition the following moment estimate: for all $p\geq 1$,
there exists $C_{p}>0$ such that for all $r\geq 1$,

$\mathbb{E}_{r,\infty}\left[L(B(r))^{p}\right]\leq C_{p}r^{-2+4p}.$ (9.4)

This type of statement is relatively to prove by induction on $p\geq 1$. As
already explained, by criticality of the branching mechanism, the first moment
coincides with the first moment of the local time of a single Brownian motion,
which is equal to $m_{1}r^{2}$. For the second moment, one needs to handle the
local times generated by two particles. One can decompose over the generation
of the most recent common ancestor. The remaining trajectories are independent
and the second moment factorises into the square of the first moment.
Integrating out the randomness of the most recent common ancestor, one gets
(9.4) for $p=2$. This procedure can be iterated to control moments of
arbitrary order $p\geq 1$ by considering the first branching event, leading to
a diagrammatic sum representing the possible genealogical structure of the $p$
particles. It has been carried out in great detail in the context of branching
random walk in [AHJ21, Section 3]. We emphasise that we do not require any
control on the growth of the constants $C_{p}$, which would require a careful
analysis of these diagrammatic sums as was done in [AHJ21]. Here we will
actually only use (9.4) with $p=2$ and $4$, and thus omit the details.

We now explain how we obtain (9.1). Let us denote by
$X=L(B(r))/(r^{4}\log(R/r))$. Let $\lambda\in[0,\lambda_{0}]$. Using the mean
value theorem we can bound

$\displaystyle 0\leq\mathbb{E}_{r,R}\left[e^{\lambda X}-1-\lambda X\right]$
$\displaystyle\leq\frac{\lambda_{0}^{2}}{2}\mathbb{E}_{r,R}\left[e^{\lambda_{0}X}X^{2}\right]=\frac{\lambda_{0}^{2}}{2}\mathbb{E}_{r,R}\left[X^{2}\right]+\frac{\lambda_{0}^{2}}{2}\mathbb{E}_{r,R}\left[(e^{\lambda_{0}X}-1)X^{2}\right]$
$\displaystyle\leq\frac{\lambda_{0}^{2}}{2}\mathbb{E}_{r,R}\left[X^{2}\right]+\frac{\lambda_{0}^{2}}{2}\mathbb{E}_{r,R}\left[(e^{\lambda_{0}X}-1)^{2}\right]^{1/2}\mathbb{E}_{r,R}\left[X^{4}\right]^{1/2}$

where we used Cauchy–Schwarz in the last inequality. By (9.4),

$\mathbb{E}_{r,R}\left[X^{2}\right]\leq\frac{C}{r^{2}(\log(R/r))^{2}}\quad\text{and}\quad\mathbb{E}_{r,R}\left[X^{4}\right]^{1/2}\leq\frac{C}{r(\log(R/r))^{2}}$

and by (9.2),

$\mathbb{E}_{r,R}\left[(e^{\lambda_{0}X}-1)^{2}\right]\leq\mathbb{E}_{r,R}\left[e^{2\lambda_{0}X}\right]-1\leq\frac{C\lambda_{0}^{2}}{r^{2}\log(R/r)}.$

Thus (we absorb the dependence on $\lambda_{0}$ into the constant $C$),

$0\leq\mathbb{E}_{r,R}\left[e^{\lambda X}-1-\lambda
X\right]\leq\frac{C}{r^{2}(\log(R/r))^{2}}.$

Since $\mathbb{E}_{r,R}\left[L(B(r))\right]=(1+o(1))m_{1}r^{2}$, this
concludes the proof of (9.1). ∎

Using directly the estimate (9.1) as well as Theorem 6.1, we find that for all
$\lambda<\min(\lambda_{0},2/m_{1})$,

$\displaystyle\mathbb{P}_{e^{-x_{0}}R,R}\left(L(B(r))\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\right)\leq R^{-2+o(1)}(R/r)^{-\frac{am_{1}\lambda}{2}}.$

If we were allowed to choose $\lambda$ as close to $2/m_{1}$ as possible, this
would already imply the upper bound of Theorem 1.4 (without any restriction on
$a\in(0,a_{0})$). However, we have no control over the $\lambda_{0}$ in (9.1)
at this stage, and hence we will need to proceed in a more complicated way,
using once more the notion of “non-backtracking” particles from Section 8.

######  Remark 9.2.

In fact, we believe that the optimal value for $\lambda_{0}$ in (9.1) is equal
to $2/m_{1}$. However, it seems unlikely that the approach of [AS22] can
directly achieve this level of precision.

###### Proof of Theorem 1.4 – Upper bound.

In this proof, we will assume that $a<\min(2,\lambda_{0}m_{1})$ where
$\lambda_{0}>0$ is the constant appearing in (9.1). Let $x_{0}>0$,
$k_{\max}\geq 1$ be a large integer and $\eta_{1}=1/(k_{\max}+1)$. Let
$\eta_{2}>0$ (inverse of a large integer) and $\varepsilon>0$ be both much
smaller than $\eta_{1}$. Let $R_{0}>0$ and $x_{*}>0$ be large enough so that
Theorems 1.3 and 6.1 and the estimate (9.1) apply with errors at most
$\varepsilon$. Let $R>R_{0}^{1/\eta_{1}}$ and $r\in[1,e^{-x_{*}/\eta_{1}}R]$.

For all $k=0,\dots,k_{\max}+1$, let $r_{k}=r(R/r)^{k\eta_{1}}$. By definition,
$r_{0}=r$ and $r_{k_{\max}+1}=R$. We have tuned the parameters to ensure that
$r_{1}\geq R_{0}$ and $r_{k+1}/r_{k}\geq e^{x_{*}}$ for all $k$. For each
$k\in\\{1,\dots,k_{\max}\\}$, we let $\mathbf{P}_{r_{k}}$ be the set of
pioneers on $\partial B(r_{k})$ emanating from the pioneers on $\partial
B(r)$. By convention, we also set $\mathbf{P}_{r}$ to be the set of pioneers
on $\partial B(r)$. For each $k\in\\{0,\dots,k_{\max}\\}$, let
$\mathbf{L}_{k}$ be the local time in $B(r)$ induced by $\mathbf{P}_{r_{k}}$
where particles are killed on $\partial B(r_{k+1})$. In words,
$\mathbf{L}_{k}$ captures the local time in $B(r)$ of particles that
backtracked exactly to level $r_{k}$. By construction, the total local time of
$B(r)$ is equal to $\sum_{k=0}^{k_{\max}}\mathbf{L}_{k}.$ Let
$(\mathcal{F}_{k})_{k=0,\dots,k_{\max}}$ be the filtration:

$\mathcal{F}_{k}=\sigma(\\#\mathbf{P}_{r_{\ell}},\ell=0,\dots,k),\quad
k=0,\dots,k_{\max}.$ (9.5)

Let $k\in\\{1,\dots,k_{\max}\\}$. Note that conditionally on
$\mathcal{F}_{k}$,
$(\mathbf{L}_{k},\mathbf{L}_{k+1},\dots,\mathbf{L}_{k_{\max}})$ is independent
from $(\mathbf{L}_{0},\mathbf{L}_{1},\dots,\mathbf{L}_{k-1})$ and is
distributed as follows. Consider a branching Brownian motion with
$\\#\mathbf{P}_{r_{k}}$ initial particles located on $\partial B(r_{k})$. Let
$\ell_{k}$ be the total local time of $B(r)$ where particles are killed on
$\partial B(r_{k+1})$. For $i=1,\dots,k_{\max}-k$, let $\ell_{k+i}$ be the
local time of $B(r)$ generated by the pioneers of $\partial B(r_{k+i})$ where
particles are killed on $\partial B(r_{k+i+1})$. Then the law of
$(\mathbf{L}_{k},\mathbf{L}_{k+1},\dots,\mathbf{L}_{k_{\max}})$ conditionally
given $\mathcal{F}_{k}$ is the same as the law of
$(\ell_{k},\dots,\ell_{k_{\max}})$.

For each $\mathbf{k}\in\\{0,\dots,k_{\max}\\}$, we define the events

$E_{\mathbf{k}}:=\bigl{\\{}\max\\{k=0,\dots,k_{\max}:\\#\mathbf{P}_{r_{k}}\geq
1\\}=\mathbf{k}\bigr{\\}},\quad F_{\mathbf{k}}:=\biggl{\\{}\forall
k=0,\dots,\mathbf{k},\\#\mathbf{P}_{r_{k}}\leq r_{k}^{2}\frac{(\log
R/r)^{2}}{\log\log R/r}\biggr{\\}}.$

Note that $F_{\mathbf{k}}$ is measurable with respect to
$\mathcal{F}_{\mathbf{k}}$. By Lemma 2.7, the probability
$\mathbb{P}\left(F_{k_{\max}}^{c}|N_{r}>0\right)$ decays faster than any
polynomial in $R/r$ (here we use that $\sup_{R}s_{*}(R)<0$). In the rest of
the proof, we will work on the event $F_{k_{\max}}\cap\\{N_{r}>0\\}$. We first
keep track of the contribution of $\mathbf{L}_{0}$:

$\displaystyle\mathbb{P}\Big{(}\sum_{k=0}^{k_{\max}}\mathbf{L}_{k}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2},F_{k_{\max}}\Big{)}\leq\mathbb{P}\left(\mathbf{L}_{0}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\right)$ (9.6) $\displaystyle\hskip
28.45274pt+\sum_{p=0}^{1/\eta_{2}-1}\mathbb{P}\Big{(}\mathbf{L}_{0}\in[p,p+1)\frac{am_{1}\eta_{2}}{2}r^{4}(\log
R/r)^{2},\sum_{k=1}^{k_{\max}}\mathbf{L}_{k}\geq(1-(p+1)\eta_{2})\frac{am_{1}}{2}r^{4}(\log
R/r)^{2},F_{k_{\max}}\Big{)}.$

We first claim that the killing on $\partial B(r_{1})$ and the estimate (9.1)
are enough to show that (recall that $\varepsilon$ and $\eta_{2}$ are much
smaller than $\eta_{1}$ so that $\varepsilon^{\prime}$ defined below is small)

$\mathbb{P}\left(\mathbf{L}_{0}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\,\Big{|}\,N_{r}>0\right)\leq(R/r)^{-a+\varepsilon^{\prime}}\quad\text{where}\quad\varepsilon^{\prime}=m_{1}\lambda_{0}\frac{\varepsilon+\eta_{2}}{2\eta_{1}}+\varepsilon.$
(9.7)

Indeed, we decompose $\mathbf{L}_{0}=\sum_{i=1}^{N_{r}}L_{i}$ where $L_{i}$,
$i\geq 1$, are i.i.d. and distributed as the local time of $B(r)$ starting
with one particle on $\partial B(r)$ and where particles are killed on
$\partial B(r_{1})$. We have

$\displaystyle\mathbb{P}\left(\mathbf{L}_{0}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\,\Big{|}\,N_{r}>0\right)\leq\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{r}\geq\frac{a}{2}r^{2}(\log
R/r)^{2}\,\Big{|}\,N_{r}>0\right)$ $\displaystyle\hskip
49.79231pt+\sum_{p=0}^{1/\eta_{2}-1}\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{r}\in[n_{p},n_{p+1})\,\Big{|}\,N_{r}>0\right)\mathbb{P}\left(\sum_{i=1}^{n_{p+1}}L_{i}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\right)$ (9.8)

where $n_{p}=p\eta_{2}\frac{a}{2}r^{2}(\log R/r)^{2}$, $p\geq 0$. By Markov’s
inequality and (9.1), the last probability in the above display is at most

$\displaystyle(R/r)^{-\frac{am_{1}\lambda_{0}}{2\eta_{1}}}\mathbb{E}_{r,r_{1}}\left[\exp\left(\frac{\lambda_{0}}{r^{4}(\log
r_{1}/r)}L(B(r))\right)\right]^{n_{p+1}}$
$\displaystyle=(R/r)^{-\frac{am_{1}\lambda_{0}}{2\eta_{1}}}\left(1+\frac{m_{1}\lambda_{0}}{r^{2}\eta_{1}(\log
R/r)}(1+\varepsilon)\right)^{n_{p+1}}$
$\displaystyle=\exp\left(-\frac{am_{1}\lambda_{0}}{2\eta_{1}}(1-(p+1)\eta_{2}(1+\varepsilon))\log(R/r)\right).$

Moreover, by Theorem 1.3, $\mathbb{P}_{e^{-x_{0}}R,R}\left(N_{r}\geq
n_{p}|N_{r}>0\right)\leq(R/r)^{-ap\eta_{2}+\varepsilon}$. This shows that the
sum in (9.8) is at most

$\displaystyle(R/r)^{-a+o(1)}\sum_{p=0}^{1/\eta_{2}-1}\exp\left(\left((1-p\eta_{2})a(1-\frac{m_{1}\lambda_{0}}{2\eta_{1}})+m_{1}\lambda_{0}\frac{\varepsilon(p+1)\eta_{2}+\eta_{2}}{2\eta_{1}}+\varepsilon\right)\log\frac{R}{r}\right).$

We simply bound $1-\frac{m_{1}\lambda_{0}}{2\eta_{1}}\leq 0$ and
$(p+1)\eta_{2}\leq 1$ to get that the sum in (9.8) is at most
$(R/r)^{-a+\varepsilon^{\prime}+o(1)}$ where $\varepsilon^{\prime}$ is defined
in (9.7). Together with Theorem 1.3 this establishes the claim (9.7), which
handles the first term on the right hand side of (9.6).

It remains to bound the sum over $p$ appearing on the right hand side of
(9.6). By using that $\mathbf{L}_{0}$ and
$(\mathbf{L}_{1},\dots,\mathbf{L}_{k_{\max}})$ are independent conditionally
on $\mathcal{F}_{1}$, we can further bound the left hand side of (9.6) by

$\displaystyle\mathbb{P}\left(\mathbf{L}_{0}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\right)$
$\displaystyle+\sum_{p=0}^{1/\eta_{2}-1}\mathbb{E}\Bigg{[}\mathbb{P}\left(\mathbf{L}_{0}\in[p,p+1)\frac{am_{1}\eta_{2}}{2}r^{4}(\log
R/r)^{2}\,\Big{|}\,\mathcal{F}_{1}\right)\times$ (9.9)
$\displaystyle\times\mathbb{P}\left(\sum_{k=1}^{k_{\max}}\mathbf{L}_{k}\geq(1-(p+1)\eta_{2})\frac{am_{1}}{2}r^{4}(\log
R/r)^{2},F_{k_{\max}}\,\Bigg{|}\,\mathcal{F}_{1}\right)\Bigg{]}.$

Let $p\in\\{0,\dots,1/\eta_{2}-1\\}$. We now focus on bounding the last
probability above and denote by $\alpha=(1-(p+1)\eta_{2})a$. The event inside
the probability can occur only on $\bigcup_{1}^{k_{\max}}E_{\mathbf{k}}\cap
F_{\mathbf{k}}$. Let $\mathbf{k}\in\\{1,\dots,k_{\max}\\}$. By definition of
$E_{\mathbf{k}}$ and then by Cauchy–Schwarz,

$\displaystyle\mathbb{P}\left(\sum_{k=1}^{k_{\max}}\mathbf{L}_{k}\geq\frac{\alpha
m_{1}}{2}r^{4}(\log R/r)^{2},E_{\mathbf{k}}\cap
F_{\mathbf{k}}\,\Bigg{|}\,\mathcal{F}_{1}\right)\leq\mathbb{P}\left(\sum_{k=1}^{\mathbf{k}}\mathbf{L}_{k}\geq\frac{\alpha
m_{1}}{2}r^{4}(\log
R/r)^{2},\\#\mathbf{P}_{r_{\mathbf{k}}}>0,F_{\mathbf{k}}\,\Bigg{|}\,\mathcal{F}_{1}\right)$
$\displaystyle\leq\mathbb{P}\left(\sum_{k=1}^{\mathbf{k}}\mathbf{L}_{k}\geq\frac{\alpha
m_{1}}{2}r^{4}(\log
R/r)^{2},F_{\mathbf{k}}\,\Bigg{|}\,\mathcal{F}_{1}\right)^{1/2}\mathbb{P}\left(\\#\mathbf{P}_{r_{\mathbf{k}}}>0\,\Bigg{|}\,\mathcal{F}_{1}\right)^{1/2}.$
(9.10)

On the event that $\\#\mathbf{P}_{1}\leq r_{1}^{2}(\log R/r)^{2}/\log\log R/r$
(which is measurable w.r.t. $\mathcal{F}_{1}$), we bound crudely,

$\mathbb{P}\left(\\#\mathbf{P}_{r_{\mathbf{k}}}>0\,\big{|}\,\mathcal{F}_{1}\right)^{1/2}\mathbf{1}_{F_{1}}\leq(r_{1}/r_{\mathbf{k}})(R/r)^{\varepsilon}=(R/r)^{-({\mathbf{k}}-1)\eta_{1}+\varepsilon}=(R/r)^{-\frac{\mathbf{k}-1}{k_{\max}+1}+\varepsilon}.$
(9.11)

We now bound the first probability in (9.10). For this purpose, let

$\mu_{0}=\min(\lambda_{0},\frac{2}{(1+\varepsilon)m_{1}})\quad\text{and}\quad\lambda=\mu_{0}\frac{k_{\max}+1}{\mathbf{k}+1}.$
(9.12)

By Markov’s inequality and by independence of $\mathbf{L}_{\mathbf{k}}$ and
$(\mathbf{L}_{0},\dots,\mathbf{L}_{\mathbf{k}-1})$ conditionally on
$\mathcal{F}_{\mathbf{k}}$,

$\displaystyle\mathbb{P}\left(\sum_{k=1}^{\mathbf{k}}\mathbf{L}_{k}\geq\frac{\alpha
m_{1}}{2}r^{4}(\log
R/r)^{2},F_{\mathbf{k}}\big{|}\mathcal{F}_{1}\right)\leq(R/r)^{-\frac{\lambda\alpha
m_{1}}{2}}\mathbb{E}\left[\mathbf{1}_{F_{\mathbf{k}}}\exp\left(\frac{\lambda}{r^{4}(\log
R/r)}\sum_{k=1}^{\mathbf{k}}\mathbf{L}_{k}\right)\big{|}\mathcal{F}_{1}\right]$
$\displaystyle=(R/r)^{-\frac{\lambda\alpha
m_{1}}{2}}\mathbb{E}\left[\mathbf{1}_{F_{\mathbf{k}}}\mathbb{E}\left[\exp\left(\frac{\lambda}{r^{4}(\log
R/r)}\sum_{k=1}^{\mathbf{k}-1}\mathbf{L}_{k}\right)\big{|}\mathcal{F}_{\mathbf{k}}\right]\mathbb{E}\left[\exp\left(\frac{\lambda}{r^{4}(\log
R/r)}\mathbf{L}_{\mathbf{k}}\right)\big{|}\mathcal{F}_{\mathbf{k}}\right]\big{|}\mathcal{F}_{1}\right].$
(9.13)

Moreover, by the explicit description of the conditional law of
$\mathbf{L}_{k}$ given $\mathcal{F}_{k}$ (see below (9.5)) and by (9.1) and
Proposition 4.6,

$\displaystyle\mathbb{E}\left[\exp\left(\frac{\lambda}{r^{4}(\log
R/r)}\mathbf{L}_{\mathbf{k}}\right)\big{|}\mathcal{F}_{\mathbf{k}}\right]=\mathbb{E}_{r_{\mathbf{k}},r_{{\mathbf{k}}+1}}\left[\mathbb{E}_{r,r_{{\mathbf{k}}+1}}\left[\exp\left(\frac{\lambda}{r^{4}(\log
R/r)}L(B(r))\right)\right]^{N_{r}}\right]^{\\#\mathbf{P}_{r_{\mathbf{k}}}}$
$\displaystyle=\mathbb{E}_{r_{\mathbf{k}},r_{{\mathbf{k}}+1}}\left[\left(1+\frac{(1+\varepsilon)m_{1}\lambda}{r^{2}(\log
R/r)}\right)^{N_{r}}\right]^{\\#\mathbf{P}_{r_{\mathbf{k}}}}=\left(1+\frac{2}{r_{\mathbf{k}}^{2}\log(r_{\mathbf{k}}/r)}\frac{\frac{(1+\varepsilon)m_{1}\lambda}{2}\frac{{\mathbf{k}}+1}{k_{\max}+1}}{1-\frac{(1+\varepsilon)m_{1}\lambda}{2}\frac{{\mathbf{k}}+1}{k_{\max}+1}}\right)^{\\#\mathbf{P}_{r_{\mathbf{k}}}}.$

In the second equality we used that
$\lambda/\log(R/r)\leq\lambda_{0}/\log(r_{{\mathbf{k}}+1}/r)$ and in the third
equality we used that
$(1+\varepsilon)m_{1}\lambda/\log(R/r)<2/\log(r_{{\mathbf{k}}+1}/r)$. On the
event $F_{\mathbf{k}}$, the right hand side of the above display is bounded by
$(R/r)^{2\varepsilon}$. We have obtained that

$\displaystyle\mathbb{E}\left[\mathbf{1}_{F_{\mathbf{k}}}\exp\left(\frac{\lambda}{r^{4}(\log
R/r)}\sum_{k=1}^{\mathbf{k}}\mathbf{L}_{k}\right)\big{|}\mathcal{F}_{1}\right]\leq(R/r)^{2\varepsilon}\mathbb{E}\left[\mathbf{1}_{F_{\mathbf{k}-1}}\exp\left(\frac{\lambda}{r^{4}(\log
R/r)}\sum_{k=1}^{\mathbf{k}-1}\mathbf{L}_{k}\right)\big{|}\mathcal{F}_{1}\right].$
(9.14)

By iterating, we find that the left hand side of the above display is at most
$(R/r)^{2\varepsilon k_{\max}}$. Putting together (9.11), (9.12), (9.13),
(9.14), we obtain that (9.10) is at most

$\exp\left(-\left(\frac{\mu_{0}\alpha
m_{1}}{4}\frac{k_{\max}+1}{{\mathbf{k}}+1}+\frac{{\mathbf{k}}-1}{k_{\max}+1}-(k_{\max}+1)\varepsilon\right)\log\frac{R}{r}\right)\leq(R/r)^{2\eta_{1}+\varepsilon/\eta_{1}}(R/r)^{-\sqrt{\mu_{0}m_{1}\alpha}}$

using that for all $u\in(0,1)$, $c_{1}u+c_{2}/u\geq 2\sqrt{c_{1}c_{2}}$
(applied to $u=\frac{{\mathbf{k}}+1}{k_{\max}+1}$). Since we assumed at the
very beginning of the proof that $a<\min(2,\lambda_{0}m_{1})=\mu_{0}m_{1}$, we
have $\alpha\leq\mu_{0}m_{1}$ and we can bound
$\sqrt{\mu_{0}m_{1}\alpha}\geq\alpha$. Injecting this estimate back in (9.9)
and recalling that we denoted $\alpha=(1-(p+1)\eta_{2})a$, we obtain that

$\displaystyle\mathbb{P}\left(\sum_{k=0}^{k_{\max}}\mathbf{L}_{k}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2},F_{k_{\max}}\right)\leq\mathbb{P}\left(\mathbf{L}_{0}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\right)$
$\displaystyle+(R/r)^{2\eta_{1}+\varepsilon/\eta_{1}}\sum_{p=0}^{1/\eta_{2}-1}(R/r)^{-(1-(p+1)\eta_{2})a}\mathbb{P}\left(\mathbf{L}_{0}\in[p,p+1)\frac{am_{1}\eta_{2}}{2}r^{4}(\log
R/r)^{2}\right).$

By (9.7) (where $\varepsilon^{\prime}$ is also defined), this leads to

$\mathbb{P}\left(\sum_{k=0}^{k_{\max}}\mathbf{L}_{k}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2},F_{k_{\max}}|N_{r}>0\right)\leq(R/r)^{2\eta_{1}+\eta_{2}/\eta_{1}a+\varepsilon/\eta_{1}+\varepsilon^{\prime}}(R/r)^{-a}.$

Because
${2\eta_{1}+\eta_{2}/\eta_{1}a+\varepsilon/\eta_{1}+\varepsilon^{\prime}}$ can
be arbitrary small, this concludes the proof. ∎

##  10 Strong coupling and local time of branching random walk (proof of
Theorem 1.10)

###  10.1 Strong coupling of tree-indexed random walk and Brownian motion

In this section, we state and prove our strong coupling between tree-indexed
random walk and tree-indexed Brownian motion. This is the analogue for tree-
indexed processes of the celebrated dyadic or KMT coupling giving a strong
approximation of Brownian motion by random walk. This subject is in itself one
with a long and distinguished history; we refer for instance to [LL10, Chapter
7] for a summary of some results on this topic.

All the trees considered in section will be continuous trees with finitely
many branching points, i.e. each edge will carry a length. Tree-indexed random
walks will then also be defined on the edges by linear interpolation. Our
result will follow from a novel connection between this problem and the so-
called Horton–Strahler number of a tree, which we now define:

#### Horton–Strahler number

The Horton–Strahler number $H(T)$ of a finite tree $T$ is defined recursively
as follows. If the tree $T$ consists of a single vertex (the root), we set
$H(T)=1$. Otherwise, let $T_{1},\dots,T_{m}$ be the subtrees hanging off the
children of the root. We set

$H(T)=\left\\{\begin{array}[]{ll}\max_{i=1,\dots,m}H(T_{i}),&\text{if this