maximum is achieved by only one of the trees $T_{1},\ldots,T_{m}$}\\\
\max_{i=1,\dots,m}H(T_{i})+1,&\text{otherwise.}\end{array}\right.$

$H(T)$ can be thought of as a way of measuring the branching complexity of the
tree $T$. For instance, if $T$ is simply a path, then $H(T)=1$, no matter how
long this path is. On the other hand, if $T$ is a “perfect binary tree”, where
each interior vertex has two children and where all leaves have the same
depth, then $H(T)$ is equal to the depth of $T$ which is also equal to
$\log_{2}(\\#V(T)+1)$. In general, it follows directly from its recursive
definition that the Horton–Strahler number of a tree can be bounded by
$\log_{2}(\\#V(T)+1)$. See the introduction of [BDR21] for more background on
the Horton–Strahler number, especially in the context of
Bienaymé–Galton–Watson trees.

Let $\theta$ be a probability distribution on $\mathbb{R}^{d}$. Assume that
$\theta$ is centred with some finite exponential moment. Denote by $\Gamma$
the covariance matrix of $\theta$, which we suppose invertible.

######  Theorem 10.1.

There exist $c_{1},c_{2},c_{3}>0$ such that the following holds. Let $T$ be a
continuous tree with finitely many branching points and edges of unit length.
Let $S_{T}$ be a $T$-indexed random walk (with linear interpolation between
neighbouring branching points) with increments distributed according to
$\theta$ and let $B_{T}$ be a $T$-indexed Brownian motion. There exists a
coupling between $S_{T}$ and $B_{T}$ such that for all $x\geq 0$,

$\mathbb{P}\left(\sup_{t\in T}|\Gamma^{-1/2}S(t)-B(t)|\geq
c_{1}H(T)\log(1+d(T))+c_{1}\log\ell(T)+x\right)\leq c_{2}e^{-c_{3}x}$

where $d(T)$ denotes the depth of $T$ and $\ell(T)$ its number of leaves.

The error obtained in Theorem 10.1 above is fairly small if $H(T)$ is small
compared to the depth of the tree. This is the case for critical Galton–Watson
trees conditioned to survive a long time. See Corollary 10.2 and (10.2) below.
On the other hand, if $H(T)$ is large, it does not seem reasonable to expect a
very good coupling between a $T$-indexed random walk and a $T$-indexed
Brownian motion.

###### Proof.

Let $T$ be a finite tree. By a “path” in the tree $T$, we will mean a monotone
path $v_{1},\dots,v_{m}$ such that for all $i=2,\dots,m$, $v_{i}$ is a child
of $v_{i-1}$. We will moreover say that two paths are disjoint if they use
disjoint sets of edges.

To obtain a strong coupling between $S_{T}$ and $B_{T}$, one can proceed as
follows. We partition the tree $T$ into a set of disjoint paths. On each of
these paths, we use the standard (i.e., dyadic or KMT) strong coupling between
random walk and Brownian motion. We then “weld” these successive couplings to
get an overall coupling on the whole tree. It is however crucial to ensure
that on each path from the root to a leaf of the tree, we do not weld too many
couplings. Indeed, each such welding potentially increases the error in the
overall coupling. In other words, we need to partition $T$ using a special set
of paths that form a network of “highways”. By this, we mean that each path
from the root to one of the leaves will use only a few highways.

Finding such a set of highways depends very much on the branching complexity
of the tree. This is where the Horton–Strahler number $H(T)$ of $T$ will be
useful. Indeed, an equivalent definition of $H(T)$ goes as follows. Starting
with $T_{1}=T$, define recursively $T_{i+1}$ as the tree obtained from $T_{i}$
by erasing each leaf of $T_{i}$ and each path of nodes with a single child
leading to leaves of $T_{i}$. It can be proved by induction on $H(T)$ that the
minimal number of steps needed to erase all vertices of $T$ is equal to
$H(T)$. From this equivalent definition of $H(T)$, it is clear that we can
partition $T$ into a set of disjoint paths $\wp_{1},\dots,\wp_{m}$ in such a
way that each path from the root to one of the leaves is the concatenation of
at most $H(T)$ paths in $\\{\wp_{1},\dots,\wp_{m}\\}$.

By the multidimensional strong approximation of Einmahl [Ein89], for each path
$\wp_{l},(l=1,\ldots,m)$, we can find a probability space
$(\Omega_{l},\mathcal{F}_{l},\mathbb{P}_{l})$ and a random walk $S_{l}$ with
increment distribution $\theta$ and Brownian motion $B_{l}$ defined on
$(\Omega_{l},\mathcal{F}_{l},\mathbb{P}_{l})$, both starting at 0 and indexed
by the path $\wp_{l}\subset T$ (rather than the more conventional interval
$[0,\text{len}(\wp_{l})]$), such that for all $x>0$,

$\mathbb{P}_{l}\left(\sup_{t\in\wp_{l}}|\Gamma^{-1/2}S_{l}(t)-B_{l}(t)|\geq
c^{\prime}_{1}\log d(T)+x\right)\leq c^{\prime}_{2}e^{-c^{\prime}_{3}x}.$
(10.1)

Note that we simply bound the length of $\wp_{l}$ by the depth of $T$. This
rough bound will be enough for our purposes. We now consider the product space
$\Omega=\prod_{l=1}^{m}\Omega_{l}$,
$\mathcal{F}=\bigotimes_{l=1}^{m}\mathcal{F}_{l}$ and
$\mathbb{P}=\bigotimes_{l=1}^{m}\mathbb{P}_{l}$. Let $v\in T$ (recall that $T$
is a continuous tree). Denote by $\varnothing$ the root of this tree, and let
$[\varnothing,v]$ denote the unique geodesic path, parameterised by length,
between the root $\varnothing$ and the point $v\in T$. We decompose this
geodesic path according to the partition $\wp_{l},l=1,\ldots,m$ as follows:

$[\varnothing,v]=\cup_{l=1}^{m}(\wp_{l}\cap[\varnothing,v])=:\cup_{l=1}^{m}[t_{l},u_{l}]$

where $t_{l}=t_{l}(v),u_{l}=u_{l}(v)\in T$ and $[t_{l},u_{l}]$ corresponds to
the path $\wp_{l}\cap[\varnothing,v]$ (parameterised by length). (If $\wp_{l}$
does not intersect $[\varnothing,v]$ then by convention we take $t_{l},u_{l}$
to be the starting point of $\wp_{l}$.) With this definition we can now
specify a coupling between the $T$-indexed walk $(S_{T}(v))_{v\in T}$ and
$T$-indexed Brownian motion $(B_{T}(v))_{v\in T}$, as follows: namely, for
$v\in T$ we set

$S_{T}(v)=\sum_{l=1}^{m}S_{l}(u_{l})\quad\text{and}\quad
B_{T}(v)=\sum_{l=1}^{m}B_{l}(u_{l}).$

It is easy to check with the Markov property that this indeed defines a valid
coupling between the two processes. Moreover, by (10.1), for any given leaf
$v$, the maximal error in this coupling on the geodesic $[\varnothing,v]$ from
the root $\varnothing$ to $v$ satisfies

$\max_{u\in[\varnothing,v]}|\Gamma^{-1/2}S_{T}(u)-B_{T}(u)|\preceq
c^{\prime}_{1}H(T)\log d(T)+c^{\prime}_{4}\sum_{l=1}^{H(T)}E_{l}$

where $E_{l},l=1,\dots,H(T)$, are i.i.d. exponential random variables with
mean 1, and $\preceq$ stands for stochastic domination. By Chernoff’s
inequality,

$\mathbb{P}\left(\sum_{l=1}^{H(T)}E_{l}\geq 2H(T)+x\right)\leq
c^{\prime}_{5}e^{-c^{\prime}_{6}x},\quad x>0.$

We now need to take the maximum of this error over all leaves of the tree.
This is handled by the following fact, which follows by a simple union bound
argument: the maximum of $N$ (not necessarily independent) exponential random
variables with mean 1 is stochastically dominated by $\log N+X$ where the
tails of $X$ decay exponentially fast, uniformly in $N$ and the joint law of
the relevant exponential random variables. ∎

We now rephrase Theorem 10.1 in the context of Galton–Watson trees. Let $\xi$
be a nondegenerate critical offspring distribution: $\sum_{n\geq 1}\xi(n)n=1$
and $\xi(1)<1$. Let $\mathbf{T}_{n}$ be a (discrete) critical Galton-Watson
tree with offspring distribution $\xi$, conditioned to have size $n$.
Depending on the value of $n$, this conditioning might be degenerate. We will
say that $n$ is $\xi$-admissible if the event we are conditioning on has a
positive probability. As before, we will view $\mathbf{T}_{n}$ as a continuous
tree by assigning length 1 to each edge.

Given $\mathbf{T}_{n}$, and given a probability distribution $\theta$ on
$\mathbb{R}^{d}$ with vanishing mean, invertible covariance matrix $\Gamma$
and a finite exponential moment, let $S_{\mathbf{T}_{n}}$ be a
$\mathbf{T}_{n}$-indexed random walk with increment distribution $\theta$, and
let $B_{\mathbf{T}_{n}}$ be a $\mathbf{T}_{n}$-indexed Brownian motion. In
other words, $S_{\mathbf{T}_{n}}$ is simply a branching random walk
conditioned on the size of its total progeny being $n$. We stress however that
in the rest of the article we used a different notion of branching Brownian
motion, in which the branching times were not deterministically equal to 1,
but were instead random and exponentially distributed with mean 1. We explain
below the proof of Corollary 10.2 how we remedy this issue.

######  Corollary 10.2.

There exist constants $c_{1},c_{2},c_{3}$ such that the following holds. For
all $\xi$-admissible integer $n\geq 1$, there exists a coupling between
$S_{\mathbf{T}_{n}}$ and $B_{\mathbf{T}_{n}}$ such that

$\mathbb{P}\left(\sup_{t\in\mathbf{T}_{n}}|\Gamma^{-1/2}S_{\mathbf{T}_{n}}(t)-B_{\mathbf{T}_{n}}(t)|\geq
c_{1}\log(n)^{2}+x\right)\leq c_{2}e^{-c_{3}x}.$

###### Proof.

This follows directly from Theorem 10.1 and from the deterministic bounds
$H(\mathbf{T}_{n})\leq\log_{2}n$, $\log d(\mathbf{T}_{n})\leq\log n$ and
$\log\ell(\mathbf{T}_{n})\leq\log n$. Note that these bounds are not very
wasteful: under the extra assumption that the offspring distribution has
finite variance,

$\frac{H(\mathbf{T}_{n})}{\log_{2}n}\xrightarrow[n\to\infty]{\mathbb{P}}\frac{1}{2}\quad\text{and}\quad\frac{\log
d(\mathbf{T}_{n})}{\log n}\xrightarrow[n\to\infty]{\mathbb{P}}\frac{1}{2}.$
(10.2)

The first convergence is [BDR21, Theorem 1.1], while the second is a direct
consequence of Aldous’ convergence of the depth first walk rescaled by
$n^{1/2}$ to the Brownian excursion [Ald93]; see also [MM03, Theorem 1]. ∎

When we will want to compare a branching Brownian motion (with exponentially
distributed branching times) to a branching random walk, we will proceed as
follows. We will first restrict the branching Brownian motion to the branching
times. This leads to a branching random walk with increments distributed as
$\sqrt{E}G$ where $E$ is an exponential random variable with parameter 1 and
$G$ is an independent standard Gaussian random variable in $\mathbb{R}^{4}$.
We will then use Corollary 10.2 to compare it with branching Brownian motion
with deterministic branching times. Finally, we will use Corollary 10.2 once
more to compare the latter to our initial target branching random walk.

###  10.2 Thick points of branching random walk

We now have all the ingredients to prove Theorem 1.10.

###### Proof of Theorem 1.10.

We start by proving the lower bound. Let $\theta$ be the distribution on
$\mathbb{Z}^{d}$ such that for any neighbour $x$ of the origin,
$\theta(\\{x\\})=1/(2d)$. Let $\theta_{1}$ and $\theta_{2}$ be the law
respectively of $W_{E}$ and $W_{1}$ where $W$ is a 4d Brownian motion and $E$
is an independent exponential random variable with mean 1. Let $\mathbf{T}$ be
a Galton–Watson tree with offspring distribution
$\frac{1}{2}\delta_{0}+\frac{1}{2}\delta_{2}$ conditioned to survive at least
$R^{2}$ generations. We will consider three $\mathbf{T}$-indexed random walks
$S_{\mathbf{T}}^{1}$, $S_{\mathbf{T}}^{2}$ and $S_{\mathbf{T}}$ with
respective increment distributions $\theta_{1}$, $\theta_{2}$ and $\theta$. To
give us some room, we will kill the particles of $S_{\mathbf{T}}^{1}$,
$S_{\mathbf{T}}^{2}$, and $S_{\mathbf{T}}$ when they reach $B(R/4)^{c}$,
$B(R/2)^{c}$, and $B(R)^{c}$ respectively. We will denote their respective
discrete local times on $B(x,r)$ by $L^{1}_{x,r}$, $L_{x,r}^{2}$, and
$L_{x,r}$. Without loss of generality, assume that each branching random walk
starts at the origin.

The strategy of the proof is as follows. We will use Theorem 1.5 and a
standard concentration result to show a thick point result for
$S_{\mathbf{T}}^{1}$ in terms of local times of _mesoscopic_ balls. These
local times of mesoscopic balls have the advantage of being very stable: if
one perturbs slightly the spatial motions of a branching process, then one
only needs to increase slightly the radii to ensure that the local time of
balls have not decreased. Note that such a property might not hold in general
for the number of pioneers. With the help of the strong coupling previously
established and this stability of local time of balls, we will then
successively transfer this thick point result to $S_{\mathbf{T}}^{2}$ and then
to $S_{\mathbf{T}}$.

Let $a>0$ be a thickness level. Let $\varepsilon>0$ be small and $\eta>0$ be
much smaller than $\varepsilon$. Let $r=R^{\eta}$ be a small mesoscopic scale.
Let us start by considering a critical Branching Brownian motion
$B_{\mathbf{T}}$ with exponentially distributed gestation time. Let us define
the discrete local time of the ball $B(x,r)$ as the sum over each branching
point $v\in\mathbf{T}$ (including the leaves) of the indicator function that
just before the branching at $v$ the particle belongs to $B(x,r)$. This
process of discrete local times has the same law as
$(L_{x,r}^{1},x\in\mathbb{R}^{d})$. We first claim that

$\mathbb{P}_{z_{0},R/4}^{\operatorname{BRW},\theta_{1}}\Bigl{(}\\#\bigl{\\{}x\in
10r\mathbb{Z}^{4}:L^{1}_{x,r}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\bigr{\\}}\geq(R/r)^{4-a-2\varepsilon}\,\Big{|}\,\zeta\geq
R^{2}\Bigr{)}\to 1$ (10.3)

as $R\to\infty$. Indeed, we can write for all $x\in\mathbb{Z}^{4}$ and $r>0$,

$L^{1}_{x,r}=\sum_{i=1}^{N_{x,r}}L_{i}$

where $N_{x,r}$ is as before the number of pioneers on $\partial B(x,r)$ and
$L_{i}$ denotes the local time of $B(x,r)$ induced by each pioneer. By a
slight variant of Theorem 1.5 (with spheres of radius $r$ instead of $1$),

$\mathbb{P}_{z_{0},R/4}\Bigl{(}\\#\bigl{\\{}x\in
10r\mathbb{Z}^{4}:N_{x,r}\geq\frac{a+\varepsilon}{2}r^{2}(\log
R/r)^{2}\bigr{\\}}\geq(R/r)^{4-a-2\varepsilon}\,\Big{|}\,\zeta\geq
R^{2}\Bigr{)}\to 1,$

as $R\to\infty$. Furthermore, and recalling the definition (1.6) of $m_{1}$,
for a fixed centre $x$, the local times $L_{i}$, $i=1,\dots,N_{x,r}$, are
i.i.d. with average $(1+o(1))m_{1}r^{2}$. Because the $L_{i}$’s are
nonnegative, it is very costly for the sum to be much smaller than its
average. Specifically, by standard large deviations for nonnegative random
variables (sometimes called Benett’s inequality, see, e.g., [BLM13, Theorem
2.9]), for any fixed $x$, the probability

$\displaystyle\mathbb{P}_{z_{0},R/4}\Bigl{(}N_{x,r}\geq\frac{a+\varepsilon}{2}r^{2}(\log
R/r)^{2}\text{ and }L^{1}_{x,r}\leq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\,\Big{|}\,\zeta\geq R^{2}\Bigr{)}$

decays faster than any polynomial in $R$. This proves (10.3).

We will now use Corollary 10.2 to transfer this result to
$S_{\mathbf{T}}^{2}$. Conditionally on surviving at least $R^{2}$ generations,
$\mathbf{T}$ (as a discrete tree) has size $O(R^{4})$, or equivalently
contains $O(R^{4})$ branch points including leaves. We therefore define
$E_{\mathbf{T}}$ to be the event that $\mathbf{T}$ contains at most $R^{5}$
such branch points. By Corollary 10.2, on this event, we can couple
$S_{\mathbf{T}}^{1}$ and $S_{\mathbf{T}}^{2}$ in such a way that
$\max_{v\in\mathbf{T}}|S_{\mathbf{T}}^{1}(v)-S_{\mathbf{T}}^{2}(v)|\leq C(\log
R)^{2}$ with large probability. Recall that particles of $S_{\mathbf{T}}^{1}$
are killed on $\partial B(R/4)$ and those of $S_{\mathbf{T}}^{2}$ are killed
on $\partial B(R/2)$. By slightly enlarging the balls $B(x,r)$ into balls
$B(x,r+C(\log R)^{2})$, we deduce that

$\mathbb{P}_{z_{0},R/2}^{\operatorname{BRW},\theta_{2}}\Bigl{(}\\#\bigl{\\{}x\in
10r\mathbb{Z}^{4}:L^{2}_{x,r+C(\log R)^{2}}\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\bigr{\\}}\geq(R/r)^{4-a-2\varepsilon}\,\Big{|}\,\zeta\geq
R^{2}\Bigr{)}\to 1.$

The same procedure can now be done between $S_{\mathbf{T}}$ and
$S_{\mathbf{T}}^{2}$. The only difference comes from the fact that the
covariance matrix $\Gamma$ of $\theta$ is equal to $\frac{1}{4}I$ instead of
$I$, so that the balls $B(x,r+C(\log R)^{2})$ should be replaced by

$C_{x,r}:=\\{z\in\mathbb{Z}^{4}:|\Gamma^{-1/2}z-x|<r+C(\log R)^{2}\\}.$

Denoting $L(C_{x,R})$ the local time accumulated by $S_{\mathbf{T}}$ in
$C_{x,r}$, we obtain that

$\mathbb{P}_{z_{0},R}^{\operatorname{BRW},\theta}\Bigl{(}\\#\bigr{\\{}x\in
10r\mathbb{Z}^{4}:L(C_{x,r})\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\bigr{\\}}\geq(R/r)^{4-a-\varepsilon}\,\Big{|}\,\zeta\geq
R^{2}\Bigr{)}\to 1.$ (10.4)

Since the number of points of $C_{x,r}$ is at most
$(1-\varepsilon)^{-1}\frac{\pi^{2}}{2}\sqrt{\det\Gamma}r^{4}$ (where
$\pi^{2}/2$ is the volume of the unit ball in $\mathbb{R}^{4}$), if
$L(C_{x,r})\geq\frac{am_{1}}{2}r^{4}(\log R/r)^{2}$ then there must be a point
$z\in C_{x,r}$ such that
$\ell_{z}\geq(1-\varepsilon)\frac{am_{1}}{\pi^{2}\sqrt{\det\Gamma}}(\log
R/r)^{2}$. In other words, we have shown that

$\mathbb{P}_{z_{0},R}^{\operatorname{BRW},\theta}\Bigl{(}\\#\bigl{\\{}z\in\mathbb{Z}^{4}:\ell_{z}\geq(1-\varepsilon)\frac{am_{1}}{\pi^{2}\sqrt{\det\Gamma}}(\log
R/r)^{2}\bigr{\\}}\geq(R/r)^{4-a-\varepsilon}\,\Big{|}\,\zeta\geq
R^{2}\Bigr{)}\to 1.$

Recall that $\sqrt{\det(\Gamma)}=1/16$. Since $r=R^{\eta}$ and $\eta$ and
$\varepsilon$ can be arbitrary small, this concludes the proof.

We now move to the proof of the upper bound. Let $a\in(0,a_{0})$, where
$a_{0}>0$ is from Theorem 1.4. As before, let $\varepsilon>0$, $\eta>0$ be
much smaller than $\varepsilon$ and $r=R^{\eta}$. Let

$\widetilde{C}_{x,r}:=\\{z\in\mathbb{Z}^{4}:|\Gamma^{-1/2}z-x|<r\\}=\\{z\in\mathbb{Z}^{4}:|2z-x|<r\\}.$

Using the exact same strategy as above, we can show that

$\mathbb{P}_{z_{0},R}^{\operatorname{BRW},\theta}\Bigl{(}\\#\bigl{\\{}x\in\mathbb{Z}^{4}:L(\widetilde{C}_{x,r})\geq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\bigr{\\}}\leq(R/r)^{4-a+\varepsilon}\,\Big{|}\,\zeta\geq
R^{2}\Bigr{)}\to 1,$ (10.5)

where $L(\widetilde{C}_{x,R})$ stands for the local time accumulated by
$S_{\mathbf{T}}$ in $\widetilde{C}_{x,r}$. In the proof of the lower bound, we
then transferred the analogous estimate (10.4) to an estimate about local time
of points simply by using the fact that for all $x$, there must exist a point
$y\in C_{x,r}$ such that $\ell_{y}\geq L(C_{x,r})/\\#C_{x,r}$. This argument
does not work directly for the upper bound. We now explain how to circumvent
this issue. As in the branching Brownian motion case, let us denote by
$N_{x,r}$ the number of pioneers of $\widetilde{C}_{x,r}$ of our branching
random walk. Because of the lack of rotational invariance, we will need to
consider the sigma algebra $\mathcal{F}_{x,r}$ generated by the number of
pioneers together with their locations. From (10.5), one can deduce an upper
bound on $N_{x,r}$ of the type

$\mathbb{P}_{z_{0},R}^{\operatorname{BRW},\theta}\Bigl{(}\\#\bigl{\\{}x\in\mathbb{Z}^{4}:N_{x,r}\geq\frac{a+\varepsilon}{2}r^{2}(\log
R/r)^{2}\bigr{\\}}\leq(R/r)^{4-a+\varepsilon}\,\Big{|}\,\zeta\geq
R^{2}\Bigr{)}\to 1.$ (10.6)

Indeed, for each $x\in\mathbb{Z}^{4}$, we can decompose

$L(\widetilde{C}_{x,r})=\sum_{i=1}^{N_{x,r}}L_{i},$

where $L_{i}$ is the local time of $\widetilde{C}_{x,r}$ accumulated by the
$i$-th pioneer on $\widetilde{C}_{x,r}$. Conditionally on $\mathcal{F}_{x,r}$,
the $L_{i}$’s are independent, nonnegative with mean equal to
$(1+o(1))m_{1}r^{2}$. By Benett’s inequality, we deduce that for all
$x\in\mathbb{Z}^{4}$,

$\mathbb{P}_{z_{0},R}^{\operatorname{BRW},\theta}\Bigl{(}L(\widetilde{C}_{x,r})\leq\frac{am_{1}}{2}r^{4}(\log
R/r)^{2}\,\Big{|}\,N_{x,r}\geq\frac{a+\varepsilon}{2}r^{2}(\log R/r)^{2}\text{
and }\zeta\geq R^{2}\Bigr{)}$

decays faster than any polynomial in $R$. Together with (10.5), this leads to
(10.6). As already alluded to, transferring directly an upper bound on the
number of pioneers from the continuous to the discrete without relying on
local time of balls might not be an easy task. This is why we took this
convoluted approach to get (10.6).

We can now transfer this upper bound on the pioneers to an upper bound on
local times at points. Indeed, for all $x\in\mathbb{Z}^{4}$, we can write

$\ell_{x}=\sum_{i=1}^{N_{x,r}}\ell_{i}$

where $\ell_{i}$ is the local time at $x$ generated by the $i$-th pioneer on
$B(x,r)$. In the proof of Theorem 1.4, we transferred the estimate we had on
the number of pioneers to estimates about local time. We can proceed in the
exact same way here. Note in particular that the discrete analogue of (9.1) is
known [AHJ21, AS22]. ∎

##  11 Hitting probabilities in dimension $d\neq 4$ (proof of Theorem 1.9)

In this section, we assume that $d\neq 4$ and we will prove the expansion of
the hitting probability stated in Theorem 1.9. We recall the definitions of
$\beta$ and $(\alpha_{\ell})_{\ell\geq 0}$ from (1.13) and (1.14):

$\beta=\beta(d)=\begin{cases}\displaystyle\frac{d-6+\sqrt{d^{2}-20d+68}}{2}&\text{
if }d\leq 3,\\\ d-4&\text{ if }d\geq 5,\end{cases}$ (11.1)

and

$\alpha_{0}=(8-2d)_{+},\quad\alpha_{1}=1\quad\text{and}\quad\alpha_{\ell}=\frac{1}{(\beta^{2}\ell+|2d-8|)(\ell-1)}\sum_{k=1}^{\ell-1}\alpha_{k}\alpha_{\ell-k},\quad\ell\geq
2$ (11.2)

where for $x\in\mathbb{R}$, $x_{+}=\max(x,0)$. We want to prove that (when
$d\neq 4$) there exists a constant $\mu_{1}=\mu_{1}(d)\in\mathbb{R}$ such that

$\mathbb{P}_{r,\infty}(N_{1}>0)=\sum_{\ell=0}^{\infty}\alpha_{\ell}\mu_{1}^{\ell}r^{-2-\beta\ell}$

for every $r$ such that this sequence converges absolutely. We also want to
prove that the constant $\mu_{1}$ is negative when $d\in\\{1,2,3\\}$, is
strictly between $0$ and $1$ when $d\geq 5$, and is the unique solution to
$\sum_{\ell=1}^{\infty}\alpha_{\ell}\mu^{\ell}=1$ when $d\geq 5$.

###  11.1 High dimensions

We start with the easier case in which $d\geq 5$. For each $\ell\geq 1$,
$\alpha_{\ell}\in(0,1]$, so the function $\mu\in[0,1]\mapsto\sum_{\ell\geq
1}\alpha_{\ell}\mu^{\ell}$ is continuous, increasing, vanishes at 0 and is
larger than $\alpha_{1}=1$ at $\mu=1$. Therefore, for all $s\in[0,1]$, there
exists a unique $\mu_{s}\in[0,1)$ such that

$\sum_{\ell\geq 1}\alpha_{\ell}\mu_{s}^{\ell}=s.$ (11.3)

The high-dimensional case of Theorem 1.9 follows by taking $s=1$ in the
following lemma:

######  Lemma 11.1.

Let $d\geq 5$. For each $s\in[0,1]$ and $r\geq 1$,

$1-\mathbb{E}_{r,\infty}\left[(1-s)^{N_{1}}\right]=\sum_{\ell\geq
1}\alpha_{\ell}\mu_{s}^{\ell}r^{-(d-4)\ell-2}.$ (11.4)

###### Proof.

Let $w(r)$ be the right hand side of (11.4). Using the recurrence equation
(11.2), one can easily verify that
$w^{\prime\prime}+\frac{d-1}{r}w^{\prime}=w^{2}$ for every $r\geq 1$.
Moreover, we also have that $w(r)\to 0$ as $r\to\infty$ and that $w(1)=s$ by
definition of $\mu_{s}$. By (2.3), the left hand side of (11.4) also satisfies
this boundary value problem, and we conclude by uniqueness of nonnegative
solutions to this problem as established in Lemma 2.6. ∎

######  Remark 11.2.

Since both sides of (11.4) are analytic in $s$, the identity (11.4) extends to
at least some negative values of $s$. This could prove useful in the study of
large deviations of ${N_{1}}$ when $d\geq 5$; see the proof of Lemma 2.7 for
arguments in this direction.

Lemma 11.1 has the following immediate corollary, which is related to the
existence of an “excursion from infinity” (or “branching” interlacements
[Szn10]) measure for high-dimensional BBM.

######  Corollary 11.3.

The law of ${N_{1}}$ under $\mathbb{P}_{r}(\cdot|{N_{1}}>0)$ converges weakly
as $r\to\infty$ to some law $\mathbb{P}_{\infty}$ satisfying

$\mathbb{E}_{\infty}\left[(1-s)^{N_{1}}\right]=1-\frac{\mu_{s}}{\mu_{1}}$
(11.5)

for all $s\in(0,1)$.

###  11.2 Low dimensions

Let us now consider the low-dimensional case $d\in\\{1,2,3\\}$. The main
result of this section is the following proposition, the $s=1$ case of which
implies Theorem 1.9 by Lemma 2.1.

######  Proposition 11.4.

Let $d\in\\{1,2,3\\}$. There exists a continuous increasing function
$s\in(0,8-2d]\mapsto\mu_{s}\in(-\infty,0]$ such that the following holds. For
each $s\in(0,8-2d]$, there exists a unique nonnegative solution $v_{s}$ to the
boundary value problem

$\left\\{\begin{array}[]{l}\Delta
v_{s}=v_{s}^{2}\quad\text{in}\quad\mathbb{R}^{d}\setminus\overline{B(1)},\\\
v_{s}=s\quad\text{on}\quad\partial B(1),\\\ v_{s}(y)\to 0\quad\text{as}\quad
y\to\infty.\end{array}\right.$ (11.6)

Moreover, the function $v_{s}$ is equal to

$v_{s}(y)=\sum_{\ell=0}^{\infty}\alpha_{\ell}\mu_{s}^{\ell}\left\|y\right\|^{-2-\beta\ell}$

for each $y\in\mathbb{R}^{d}$ with
$\left\|y\right\|>(|\mu_{s}|/R_{\alpha}^{1/\beta})$, where $R_{\alpha}>0$ is
the radius of convergence of
$x\mapsto\sum_{\ell=0}^{\infty}\alpha_{\ell}x^{\ell}$.

Note that this expansion holds for large $\left\|y\right\|$, in contrast to
the high-dimensional case. More importantly, the existence of the function
$s\in(0,8-2d]\mapsto\mu_{s}\in(-\infty,0]$ is nontrivial. Heuristically
(letting aside the possible divergence issues), one needs to know the possible
values that

$y\mapsto\sum_{\ell\geq
0}\alpha_{\ell}\mu^{\ell}\left\|y\right\|^{-2-\beta\ell}$

can take when $y\in\partial B(1)$, by letting $\mu$ vary. If we are targeting
smaller values than $\alpha_{0}=8-2d$, we need to take $\mu<0$ (possibly very
negative). This has a different flavour than the high-dimensional case and
requires extra work that we now explain.

We are concerned with rotationally invariant solutions to $\Delta v=v^{2}$. As
before, expressing the Laplacian in spherical coordinates leads to the ODE
$v^{\prime\prime}(r)+(d-1)v^{\prime}(r)/r=v(r)^{2}$. If one makes the change
of variable $v(r)=r^{-2}f(r^{-\beta})$, one obtains the following equation for
$f$:

$\beta^{2}x^{2}f^{\prime\prime}(x)+\beta(\beta+6-d)xf^{\prime}(x)+(8-2d)f(x)=f(x)^{2}.$
(11.7)

The exponent $\beta$ was chosen exactly so that the multiplicative constants
in front of $xf^{\prime}(x)$ and $f(x)$ agree. To see why this is natural, we
can first neglect $x^{2}f^{\prime\prime}(x)$ and look at the solutions to
$c_{1}xf^{\prime}(x)+c_{2}f(x)=f(x)^{2}$, which are of the form $x\mapsto
Cc_{2}/(C+x^{c_{2}/c_{1}})$ for some constant $C$. These solutions are
therefore analytic at $x=0$ only when $c_{2}/c_{1}$ is an integer. We chose to
fix $c_{2}/c_{1}=1$.

######  Remark 11.5.

This qualitative behaviour is quite robust: if the branching mechanism were
different but still critical, this would change $f(x)^{2}$ into a more general
function of $f(x)$, but would not change the local behaviour of solutions to
(11.7) near $x=0$. This means that for other critical branching mechanism, we
would still want $\beta$ to satisfy $\beta(\beta+6-d)=8-2d$, indicating that
$\beta$ is indeed a universal exponent.

To proceed, we will fix two positive real numbers $p$ and $q$ and study the
solutions to $px^{2}f^{\prime\prime}(x)+qxf^{\prime}(x)+qf(x)=f(x)^{2}$ on
$[0,\infty)$. We will later specify our result to $p=\beta^{2}$ and $q=8-2d$.
See Figure 11.1 for numerical approximations of solutions to this equation.
Let $\alpha_{0}(p,q)=q$, $\alpha_{1}(p,q)=1$ and for all $\ell\geq 2$,

$\alpha_{\ell}(p,q)=\frac{1}{(p\ell+q)(\ell-1)}\sum_{k=1}^{\ell-1}\alpha_{k}(p,q)\alpha_{\ell-k}(p,q).$

Let $R(p,q)$ be the radius of convergence of the power series
$x\mapsto\sum_{\ell=0}^{\infty}\alpha_{\ell}(p,q)x^{\ell}$. It can be shown by
induction that for all $\ell\geq 1$, $\alpha_{\ell}(p,q)\leq(2p+q)^{1-\ell}$,
so $R(p,q)\geq(2p+q)$.

We now state and prove a lemma concerning solutions to
$px^{2}f^{\prime\prime}(x)+qxf^{\prime}(x)+qf(x)=f(x)^{2}$, which we will use
to prove Proposition 11.4.

######  Lemma 11.6.

Let $p,q>0$ be such that $q/p\geq 3+2\sqrt{2}$. For each $\mu\leq 0$, there
exists a unique nonnegative function $f_{\mu}$ in $C^{\infty}([0,\infty))$
solving the initial value problem

$\left\\{\begin{array}[]{l}px^{2}f^{\prime\prime}(x)+qxf^{\prime}(x)+qf(x)=f(x)^{2},\quad
x\geq 0,\\\ f(0)=q,\quad f^{\prime}(0)=\mu.\end{array}\right.$

The function $f_{0}$ is the constant function $x\mapsto q$, while for each
$\mu<0$, the function $f_{\mu}$ is positive, decreasing, convex, and equal to
$x\mapsto\sum_{\ell=0}^{\infty}\alpha_{\ell}(p,q)\mu^{\ell}x^{\ell}$ on
$[0,R_{\alpha}(p,q)/|\mu|)$. This family of functions also has the following
properties:

  1. 1.

$(x,\mu)\in[0,\infty)\times(-\infty,0]\mapsto f_{\mu}(x)$ is analytic;

  2. 2.

For all $\mu_{1}<\mu_{2}\leq 0$, $f_{\mu_{1}}<f_{\mu_{2}}$ on $(0,\infty)$;

  3. 3.

For all $x>0$, $f_{\mu}(x)\to 0$ as $\mu\to-\infty$.

$f_{\mu}(x)$
---
$x$
---
$8-2d$
---
Figure 11.1: Plot of numerical approximations of $f_{\mu}(x),x\in[0,15]$, for
different values of $\mu$, when $p=\beta^{2}$, $q=8-2d$ and $d=2$. From top to
bottom, $\mu=-1,-2,-3$ and $-4$.

Before we start the proof of this lemma we make a few comments. As explained
above, our analysis is predicated on the fact that the term containing
$x^{2}f^{\prime\prime}(x)$ is negligible compared to the other terms in
(11.7). It is thus reasonable to make an assumption that $p/q$ is sufficiently
small; the threshold $q/p=3+2\sqrt{2}$ is sufficient for the argument in the
proof below. Moreover, numerical simulations suggest that if the initial slope
$\mu$ is a sufficiently large negative number and $q/p<3+2\sqrt{2}$, then the
associated solution does not stay nonnegative. As such, the lemma is likely to
be false when this condition is violated.

In the concrete cases where we will apply this lemma, we will have
$p=\beta^{2}$ and $q=8-2d$, where $\beta$ is as in (11.1). Interestingly, as
$d$ varies in the interval $d\in[0,4)$, the minimum value of $q/p$ is attained
at $d=2$ and is precisely equal to $3+2\sqrt{2}$. As such, the condition
$q/p\geq 3+2\sqrt{2}$ holds for all $d=1,2,3$, but $d=2$ seems to be a
borderline case for uniqueness. We do not know the reason why $d=2$ seems to
be critical from that point of view, or what interpretations this should have
for the behaviour of $2d$ BBM.

###### Proof of Lemma 11.6..

Let $\mu<0$ and let

$\tilde{f}:x\in[0,R_{\alpha}(p,q)/|\mu|)\mapsto\sum_{\ell=0}^{\infty}\alpha_{\ell}(p,q)\mu^{\ell}x^{\ell}.$

Using the recursive definition of $(\alpha_{n}(p,q))_{n\geq 0}$, one can check
that $\tilde{f}$ satisfies

$\tilde{f}(0)=q,\quad\tilde{f}^{\prime}(0)=\mu\qquad\text{and}\qquad
px^{2}\tilde{f}^{\prime\prime}(x)+qx\tilde{f}^{\prime}(x)+q\tilde{f}(x)=\tilde{f}(x)^{2}\quad\text{
for all $x\in[0,R_{\alpha}(p,q)/|\mu|)$}.$

Let $x_{0}$ be any positive real number in $(0,R_{\alpha}(p,q)/|\mu|)$. Since
we are away from the singularity at $x=0$, we can safely consider the unique
maximal solution $\hat{f}$ to
$px^{2}\hat{f}^{\prime\prime}(x)+qx\hat{f}^{\prime}(x)+q\hat{f}(x)=\hat{f}(x)^{2}$
in $[x_{0},x_{*})$ with $\hat{f}(x_{0})=\tilde{f}(x_{0})$ and
$\hat{f}^{\prime}(x_{0})=\tilde{f}^{\prime}(x_{0})$. Finally, let
$f_{\mu}:[0,x_{*})\to\mathbb{R}$ be the function defined by:
$f_{\mu}(x)=\tilde{f}(x)$ if $x\in[0,x_{0}]$ and $f_{\mu}(x)=\hat{f}(x)$ if
$x\in[x_{0},x_{*})$. Notice that this definition does not depend on the choice
of $x_{0}$ by the uniqueness of forward solutions when $x_{0}>0$.

We are going to show that $f_{\mu}$ is defined on $[0,\infty)$ (i.e. that
$x_{*}=+\infty$) and that it satisfies all the desired properties. In the rest
of the proof we will denote by $\gamma=q/p$ and we will simply write $f$
instead of $f_{\mu}$ when there is no ambiguity.

$f$ is positive: Let $b_{1}$ and $b_{2}$ be the roots of the polynomial
$\gamma+(1-\gamma)b+b^{2}$:

$b_{1}=\frac{\gamma-1-\sqrt{1-6\gamma+\gamma^{2}}}{2}\quad\text{and}\quad
b_{2}=\frac{\gamma-1+\sqrt{1-6\gamma+\gamma^{2}}}{2}.$ (11.8)

The assumption that the ratio $\gamma=q/p$ is at least $3+2\sqrt{2}$ is used
to ensure that $b_{1}$ and $b_{2}$ are real and positive. Since $b_{1}$ and
$b_{2}$ satisfy $b_{1}+b_{2}+1=\gamma$ and $b_{1}b_{2}=\gamma$, we get that

$\frac{\mathrm{d}(x^{b_{1}+1}f^{\prime}(x))}{\mathrm{d}x}+b_{2}\frac{\mathrm{d}(x^{b_{1}}f(x))}{\mathrm{d}x}=x^{b_{1}+1}f^{\prime\prime}(x)+(b_{1}+b_{2}+1)x^{b_{1}}f^{\prime}(x)+b_{1}b_{2}x^{b_{1}-1}f(x)\\\
=\frac{1}{p}x^{b_{1}-1}(px^{2}f^{\prime\prime}(x)+qxf^{\prime}(x)+qf(x))=\frac{1}{p}x^{b_{1}-1}f(x)^{2}\geq
0.$

Integrating this relation between $0$ and $x$ leads to
$x^{b_{1}+1}f^{\prime}(x)+b_{2}x^{b_{1}}f(x)\geq 0.$ Since $x^{b_{2}}f(x)$ has
zero derivative at $x=0$, this implies by Lemma 3.10 that the derivative of
$x\mapsto x^{b_{2}}f(x)$ is nonnegative. Since $f>0$ in a neighbourhood of the
origin, this shows that $f$ stays positive as desired.

$f$ is strictly decreasing: Since $f^{\prime}(0)<0$ there exists $x_{1}>0$
such that $f^{\prime}(x)<0$ for all $x\in[0,x_{1}]$. Let $x_{2}:=\inf\\{x\geq
x_{1}:f(x)\geq q\\}$. Using the nonnegativity of $f$, we see that on
$[x_{1},x_{2})$, $f^{2}-qf\leq 0$. This implies that
$px^{2}f^{\prime\prime}(x)+qxf^{\prime}(x)\leq 0$ on $[x_{1},x_{2})$.
Integrating this relation (Lemma 3.10) shows that for all $x\in[x_{1},x_{2})$,
$x^{1+\gamma}f^{\prime}(x)\leq x_{1}^{1+\gamma}f^{\prime}(x_{1})<0$. The
derivative $f^{\prime}$ is therefore negative on $[x_{1},x_{2})$ and $x_{2}$
must be equal to $x_{*}$ (where we recall that, by definition, $f$ is defined
on $[0,x_{*})$). This shows that $f$ is strictly decreasing on its entire
domain and also shows that $f<q$ on its entire domain.

$f$ is convex: Using the ODE satisfied by $f$ and then the fact, established
in the previous paragraph, that $f^{\prime}(f-q)\geq 0$, we have

$p\frac{\mathrm{d}(x^{2}f^{\prime\prime}(x))}{\mathrm{d}x}=2f^{\prime}(x)(f(x)-q)-qxf^{\prime\prime}(x)\geq-
qxf^{\prime\prime}(x).$

This implies that the derivative of $x^{\gamma+2}f^{\prime\prime}(x)$ is
nonnegative and hence that $f^{\prime\prime}(x)\geq 0$.

$f$ is defined and analytic on $[0,\infty)$: $f$ is defined on $[0,\infty)$
since $0\leq f\leq q$ and $f^{\prime}(0)\leq f^{\prime}\leq 0$ (which are
consequences of the previous properties); here we are using that maximal
solutions to locally Lipschitz second order ODEs must either have
$(x,f,f^{\prime})$ unbounded or accumulating to a point where the ODE is not
defined as $x$ approaches an endpoint of the domain. The fact that
$(x,\mu)\mapsto f_{\mu}(x)$ is analytic follows quickly from the definition of
$f_{\mu}$ and from Cauchy–Kovalevskaya theorem.

$f_{\mu}$ is a strictly increasing function of $\mu\in(-\infty,0]$: Let
$g=f_{\mu_{2}}-f_{\mu_{1}}$. It follows immediately from the series expansion
of $f_{\mu_{1}}$ and $f_{\mu_{2}}$ around $0$ that there exists $x_{1}>0$ such
that for all $x\in(0,x_{1}]$, $g(x)>0$. Let $x_{2}:=\inf\\{x\geq
x_{1}:g(x)=0\\}$. On $[x_{1},x_{2})$, we have that
$f_{\mu_{2}}^{2}-f_{\mu_{1}}^{2}=g(f_{\mu_{2}}+f_{\mu_{1}})\geq 0$ and, using
the equation satisfied by $f_{\mu_{1}}$ and $f_{\mu_{2}}$, we deduce that
$px^{2}g^{\prime\prime}(x)+qxg^{\prime}(x)+qg(x)\geq 0$ on $[x_{1},x_{2})$.
With the same line of argument as in the proof of the fact that solutions stay
positive, this implies that $x^{b_{2}}g(x)\geq x_{1}^{b_{2}}g(x_{1})>0$ for
all $x\in[x_{1},x_{2})$, where $b_{2}$ is defined in (11.8). This shows that
$x_{2}=+\infty$, establishing the desired monotonicity property.

$f_{\mu}$ converges to zero pointwise as $\mu\to-\infty$: We fix a small
positive real number $\varepsilon$ and let $\mu<0$ be very negative. Since
$f_{\mu}$ is decreasing and has initial value $q$, the function
$x\mapsto(q-f_{\mu}(x))f_{\mu}(x)$ increases until $f_{\mu}$ reaches $q/2$ and
then decreases. Let $x_{1}=\inf\\{x\geq
0:(q-f_{\mu}(x))f(x)\geq\varepsilon\\}$ and
$x_{2}=\inf\\{x>x_{1}:(q-f_{\mu}(x))f(x)\leq\varepsilon\\}$. On
$[x_{2},\infty)$, $f_{\mu}$ is at most

$\frac{q}{2}\left(1-\sqrt{1-4\varepsilon/q^{2}}\right)=\frac{\varepsilon}{q}+O(\varepsilon^{2}).$

It is therefore enough to show that for $\varepsilon>0$ fixed, $x_{2}\to 0$ as
$\mu\to-\infty$. Using that $f^{\prime\prime}\geq 0$, we have on
$[x_{1},x_{2}]$,

$-f^{\prime}(x)=\frac{p}{q}f^{\prime\prime}(x)+\frac{(q-f(x))f(x)}{x}\geq\frac{\varepsilon}{x}.$

Integrating between $x_{1}$ and $x_{2}$ leads to $q\geq
f(x_{1})-f(x_{2})\geq\varepsilon\log(x_{2}/x_{1})$, i.e. $x_{2}\leq
e^{q/\varepsilon}x_{1}$. Finally, using the explicit power series for
$x<R_{\alpha}(p,q)/|\mu|$, one can check that $x_{1}$ is of order $1/|\mu|$.
This concludes the proof that $x_{2}\to 0$ as $\mu\to-\infty$. ∎

We are now ready to prove Proposition 11.4.

###### Proof of Proposition 11.4.

Let $d\in\\{1,2,3\\}$ and recall the definition (11.1) of $\beta$. We apply
Lemma 11.6 with $p=\beta^{2}$ and $q=8-2d$. As already mentioned, the
assumption that $q/p$ is at least $3+2\sqrt{2}$ is satisfied for
$d\in\\{1,2,3\\}$ with equality when $d=2$. By Lemma 11.6, the function
$\mu\in(-\infty,0]\mapsto f_{\mu}(1)\in(0,8-2d]$ is a continuous decreasing
bijection. Let us denote by $s\in(0,8-2d]\mapsto\mu_{s}\in(-\infty,0]$ its
inverse. Let $s\in(0,8-2d]$ and define for all $r>0$,
$w_{s}(r)=r^{-2}f_{\mu_{s}}(r^{-\beta})$. The fact that $f_{\mu_{s}}$
satisfies $\beta^{2}x^{2}f^{\prime\prime}+(8-2d)(xf^{\prime}+f)=f^{2}$ implies
that

$w_{s}^{\prime\prime}(r)+\left(5+\beta-(8-2d)/\beta\right)\frac{w_{s}^{\prime}(r)}{r}+(4+2\beta+(8-2d)(1-2/\beta))\frac{w_{s}(r)}{r^{2}}=w_{s}(r)^{2},\quad\quad
r>0.$

The quadratic equation satisfied by $\beta$ implies that the multiplicative
constant in front of $w_{s}^{\prime}$ equals $d-1$ and the one in front of
$w_{s}$ vanishes, i.e.
$w_{s}^{\prime\prime}(r)+(d-1)w_{s}^{\prime}(r)/r=w_{s}(r)^{2}$ for all $r>0$.
By definition of $\mu_{s}$, $w_{s}(1)=s$. Wrapping up, the function
$y\in\mathbb{R}^{d}\setminus
B(1)\mapsto\left\|y\right\|^{-2}f_{\mu_{s}}(\left\|y\right\|^{-\beta})$ is a
nonnegative solution to the boundary value problem (11.6). Uniqueness of
nonnegative solutions (see Lemma 2.6) together with Lemma 11.6 concludes the
proof. ∎

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