1,n^{\prime}\rrbracket$}.$

For each index $\pm\in\\{+,-\\}$, let
$\bm{\lambda}^{\pm}(s)=\big{(}\lambda_{1}^{\pm}(s),\lambda_{2}^{\pm}(s),\ldots,\lambda_{n^{\prime}}^{\pm}(s)\big{)}$
denote Dyson Brownian motion, run for time $s$, with initial data
$\bm{\lambda}^{\pm}$. Since we have restricted to the event $\mathscr{A}_{0}$,
we have by (8.12); (8.4) (with the definitions of $G^{-}$ and $G^{+}$); and
the fact from (8.13) that
$\partial_{y}\widetilde{G}^{\pm}\leq-\frac{\varepsilon}{2}$, that
$\bm{\lambda}^{-}-2(n^{\prime})^{3\omega/5-1}\leq\widetilde{\bm{w}}^{-}(0)\leq\bm{\lambda}^{-}+2(n^{\prime})^{3\omega/5-1}$
and
$\bm{\lambda}^{+}-2(n^{\prime})^{3\omega/5-1}\leq\widetilde{\bm{w}}^{+}(0)\leq\bm{\lambda}^{+}+2(n^{\prime})^{3\omega/5-1}$.
Thus, by 8.7, it is possible to couple the processes
$\big{(}\bm{\lambda}^{-}(s),\bm{\widetilde{w}}^{-}(s)\big{)}$ such that
$\bm{\lambda}^{-}(t)-2(n^{\prime})^{3\omega/5-1}\leq\widetilde{\bm{w}}^{-}(t)\leq\bm{\lambda}^{+}(t)+2(n^{\prime})^{3\omega/5-1}$
for each $t\geq 0$; similarly, it is possible to couple
$\big{(}\bm{\lambda}^{+}(s),\widetilde{\bm{w}}^{+}(s)\big{)}$ such that
$\bm{\lambda}^{+}(t)-2(n^{\prime})^{3\omega/5-1}\leq\widetilde{\bm{w}}^{+}(t)\leq\bm{\lambda}^{-}(t)+2(n^{\prime})^{3\omega/5-1}$
for each $t\geq 0$. Hence, it suffices to show

(8.19) $\displaystyle\begin{aligned}
\mathbb{P}\Bigg{[}\mathscr{A}_{0}\cap\bigcap_{\begin{subarray}{c}s\in[0,\tilde{s}_{0}+\tilde{s}_{1}]\\\
j\in\llbracket
n^{\prime}/2-2n^{\prime\prime},n^{\prime}/2+2n^{\prime\prime}\rrbracket\end{subarray}}\Big{\\{}\big{|}\lambda_{j}^{-}(s)-\lambda_{j}^{-}(0)\big{|}\leq(n^{\prime})^{\omega/2-1}\Big{\\}}\Bigg{]}\geq
1-c^{-1}e^{-c(\log n)^{2}};\\\
\mathbb{P}\Bigg{[}\mathscr{A}_{0}\cap\bigcap_{\begin{subarray}{c}s\in[0,\tilde{s}_{0}+\tilde{s}_{1}]\\\
j\in\llbracket
n^{\prime}/2-2n^{\prime\prime},n^{\prime}/2+2n^{\prime\prime}\rrbracket\end{subarray}}\Big{\\{}\big{|}\lambda_{j}^{+}(s)-\lambda_{j}^{+}(0)\big{|}\leq(n^{\prime})^{\omega/2-1}\Big{\\}}\Bigg{]}\geq
1-c^{-1}e^{-c(\log n)^{2}}.\end{aligned}$

We only show the first bound in (8.19), as the proof of the second is entirely
analogous. Under this notation, 2.17 applies with the $(n,\bm{\lambda})$ there
given by $(n^{\prime},\bm{\lambda}^{-})$ here and the $\mu$ there given by
$\widetilde{\mu}_{0}^{-}$ here. It therefore yields a constant $c_{1}>0$ such
that

$\displaystyle\mathbb{P}[\mathscr{E}_{1}^{-}]\geq 1-c_{1}^{-1}e^{-c_{1}(\log
n)^{2}},$

where we have defined the event

$\displaystyle\mathscr{E}_{1}^{-}=\bigcap_{j=1}^{n^{\prime}}\bigcap_{t\in[0,1]}\bigg{\\{}\widetilde{\gamma}_{j+\lfloor(\log
n)^{5}\rfloor}^{-}(t)-n^{-1}\leq\lambda_{j}^{-}(t)\leq\widetilde{\gamma}_{j-\lfloor(\log
n)^{5}\rfloor}^{-}(t)+n^{-1}\bigg{\\}}.$

Observe on $\mathscr{E}_{1}^{-}$ that

$\displaystyle\big{|}\lambda_{j}^{-}(s)-\lambda_{j}^{-}(0)\big{|}\leq\widetilde{\gamma}_{j-\lfloor(\log
n)^{5}\rfloor}^{-}(s)-\widetilde{\gamma}_{j+\lfloor(\log
n)^{5}\rfloor}^{-}(0)+2n^{-1}.$

This yields on $\mathscr{E}_{1}^{-}$ a constant
$C_{1}=C_{1}(\varepsilon,\delta,B_{0},m)>1$ such that, for each pair
$(j,s)\in\big{\llbracket}\frac{n^{\prime}}{2}-2n^{\prime\prime},\frac{n^{\prime}}{2}+2n^{\prime\prime}\big{\rrbracket}\times[0,n^{-\omega}]$,
we have

$\displaystyle\big{|}\lambda_{j}^{-}(s)-\lambda_{j}^{-}(0)\big{|}$
$\displaystyle\leq
C_{1}s\Big{(}\displaystyle\frac{n^{\prime}}{n}+\displaystyle\frac{n^{\prime\prime}}{n^{\prime}}+s\Big{)}+\widetilde{\gamma}_{j-\lfloor(\log
n)^{5}\rfloor}^{-}(0)-\widetilde{\gamma}_{j+\lfloor(\log
n)^{5}\rfloor}^{-}(0)+\displaystyle\frac{2}{n}$ $\displaystyle\leq
2C_{1}s\Big{(}\displaystyle\frac{1}{n^{\prime\prime}}+s\Big{)}+\displaystyle\frac{4(\log
n)^{5}}{\varepsilon n^{\prime}}+\displaystyle\frac{2}{n},$

where in the first inequality we used the second part of 8.6; in the second
inequality we used (8.1) and the fact that
$\widetilde{\varrho}_{0}^{-}=-\partial_{x}\widetilde{G}^{-}\geq\frac{\varepsilon}{2}$.
Together with the fact that
$s\big{(}\frac{1}{n^{\prime\prime}}+s\big{)}\leq\frac{1}{n^{\prime}}$ for
sufficiently large $n$ (since $n^{\prime\prime}=(n^{\prime})^{1/2}$,
$s\leq\widetilde{s}_{0}+\widetilde{s}_{1}\leq 2(n^{\prime})^{6\omega-1}$, and
$\omega\leq\frac{1}{20}$), this implies the first statement of (8.19) and thus
the lemma. ∎

###  8.4. Proofs of 8.6 and 8.7

In this section we establish first 8.6 and then 8.7.

###### Proof of 8.6.

We only establish the estimates in the lemma for $\widetilde{\varrho}_{t}$ and
$\widetilde{\gamma}_{j}(t)$, as the proofs of the analogous bounds for
$\widetilde{\varrho}_{t}^{\pm}$ and $\widetilde{\gamma}_{j}^{\pm}(t)$ are
entirely analogous. We begin with the first statement of the lemma. To that
end observe that, since
$\widetilde{G}\in\operatorname{Adm}_{\varepsilon/2}\big{(}\frac{n}{n^{\prime}}\cdot\mathfrak{R}\big{)}$
(by (8.13)), we have

(8.20)
$\displaystyle-\displaystyle\frac{2}{\varepsilon}\leq\widetilde{\varrho}_{0}(x)\leq-\displaystyle\frac{\varepsilon}{2},\qquad\text{for
each $x\in\big{[}\widetilde{G}(0,1),\widetilde{G}(0,0)\big{]}$}.$

Again by (8.13), we also have that
$[\widetilde{G}]_{k}\leq\frac{2B_{0}n^{\prime}}{n}$ for each integer
$k\in\llbracket 2,m\rrbracket$, from which it quickly follows from (8.17),
(8.18), and (8.20) that there exists a constant
$C_{1}=C_{1}(\varepsilon,\delta,B_{0},m)>1$ with

(8.21)
$\displaystyle[\widetilde{\varrho}_{0}]_{k}\leq\displaystyle\frac{C_{1}n^{\prime}}{n},\qquad\text{on
$\operatorname{supp}\widetilde{\varrho}_{0}$, for each $k\in\llbracket
1,m-1\rrbracket$}.$

As such, 2.11 applies to $\widetilde{\varrho}_{0}$ around any point in the
interior of $\operatorname{supp}\widetilde{\varrho}_{0}$ and yields (since
$\widetilde{G}\big{(}0,\frac{1}{2}\big{)}=0$ and
$\partial\widetilde{G}\leq-\frac{\varepsilon}{2}$, by (8.13)) a constant
$C_{2}=C_{2}(\varepsilon,\delta,B_{0},m)>1$ such that

(8.22)
$\displaystyle\displaystyle\sup_{t\in[0,1/C_{2}]}\displaystyle\sup_{|x|\leq
1/C_{2}}\big{|}\partial_{t}\widetilde{\varrho}_{t}(x)\big{|}\leq
C_{2};\qquad\displaystyle\sup_{t\in[0,1/C_{2}]}\displaystyle\sup_{|x|\leq
1/C_{2}}\displaystyle\max_{j\in\llbracket
1,m-3\rrbracket}\big{|}\partial_{t}\partial_{x}^{j}\widetilde{\varrho}_{0}(x)\big{|}\leq
C_{2}.$

The first bound in (8.22), with the first two statements of (8.13), yields the
first statement of the lemma for sufficiently large $n$ (so that
$4C_{2}n^{-\omega}\leq\varepsilon$).

To establish the second part of the lemma, first observe for
$i\in\big{\llbracket}-(n^{\prime})^{1-\omega},(n^{\prime})^{1-\omega}\big{\rrbracket}$
that

(8.23)
$\displaystyle\big{|}\widetilde{\gamma}_{n^{\prime}/2+i}(0)\big{|}=\bigg{|}\widetilde{G}\Big{(}0,\displaystyle\frac{1}{2}+\frac{i}{n^{\prime}}\Big{)}\bigg{|}\in\bigg{[}-\displaystyle\frac{2i}{\varepsilon
n^{\prime}},\displaystyle\frac{2i}{\varepsilon
n^{\prime}}\bigg{]}\subseteq\bigg{[}-\displaystyle\frac{2}{\varepsilon(n^{\prime})^{\omega}},\displaystyle\frac{2}{\varepsilon(n^{\prime})^{\omega}}\bigg{]},$

due to the first and third statements of (8.13). We then claim that there
exists a constant $C_{3}=C_{3}(\varepsilon,\delta,B_{0},m)>1$ such that
(recalling the Hilbert transform from (2.7))

(8.24) $\displaystyle\big{|}H\widetilde{\varrho}_{s}(y)\big{|}\leq
C_{3}\Big{(}\displaystyle\frac{n^{\prime}}{n}+|y|+s\Big{)},\qquad\text{for
each
$(s,y)\in[0,n^{-\omega}]\times\big{[}-(n^{\prime})^{-\omega/2},(n^{\prime})^{-\omega/2}\big{]}$}.$

Before verifying (8.24) let us confirm the second part of the lemma assuming
it. To that end, observe for any $j\in\llbracket 1,n^{\prime}\rrbracket$ that

$\displaystyle\displaystyle\int_{\tilde{\gamma}_{j}(0)}^{\infty}\widetilde{\varrho}_{t}(x)dx-\displaystyle\frac{2j-1}{n^{\prime}}=\displaystyle\int_{\tilde{\gamma}_{j}(0)}^{\infty}\widetilde{\varrho}_{t}(x)dx-\displaystyle\int_{\tilde{\gamma}_{j}(0)}^{\infty}\widetilde{\varrho}_{0}(x)dx$
$\displaystyle=\displaystyle\int_{\tilde{\gamma}_{j}(0)}^{\infty}\displaystyle\int_{0}^{t}\partial_{s}\widetilde{\varrho}_{s}(x)dsdx$
$\displaystyle=\pi\displaystyle\int_{0}^{t}\displaystyle\int_{\tilde{\gamma}_{j}(0)}^{\infty}\partial_{x}\big{(}\widetilde{\varrho}_{s}(x)H\widetilde{\varrho}_{s}(x)\big{)}dxds$
$\displaystyle=\pi\displaystyle\int_{0}^{t}\widetilde{\varrho}_{s}\big{(}\widetilde{\gamma}_{j}(0)\big{)}\cdot
H\widetilde{\varrho}_{s}\big{(}\widetilde{\gamma}_{j}(0)\big{)}ds,$

where the first statement follows from the definition of the classical
location $\widetilde{\gamma}_{j}(0)$; the second from performing the
integration in $s$; the third from integrating (2.12); and the fourth from
performing the integration in $x$ (using the fact that
$\widetilde{\varrho}_{s}(x)=0$ for $x$ sufficiently large). This with
$j=\frac{n^{\prime}}{2}+i$ for
$i\in\big{\llbracket}-(n^{\prime})^{1-\omega},(n^{\prime})^{1-\omega}\big{\rrbracket}$,
together with (8.24), (8.23), and the fact (from the first part of the lemma)
that $\big{|}\widetilde{\varrho}_{s}(x)\big{|}\leq 4\varepsilon^{-1}$ for each
$(s,x)\in[0,n^{-\omega}]\times[-n^{-\omega/8},n^{-\omega/8}]$, yields (since
$|\gamma_{j}|\leq\frac{3|i|}{\varepsilon n}$, by the second and third
statements in (8.13))

$\displaystyle\displaystyle\frac{1}{2}+\displaystyle\frac{i}{n^{\prime}}-\displaystyle\frac{8\pi
C_{3}t}{\varepsilon}\Big{(}\displaystyle\frac{n^{\prime}}{n}+\displaystyle\frac{3|i|}{\varepsilon
n^{\prime}}+t\Big{)}\leq\displaystyle\int_{\tilde{\gamma}_{n^{\prime}/2+i}(0)}^{\infty}\widetilde{\varrho}_{t}(x)dx\leq\displaystyle\frac{1}{2}+\displaystyle\frac{i}{n^{\prime}}+\displaystyle\frac{8\pi
C_{3}t}{\varepsilon}\Big{(}\displaystyle\frac{n^{\prime}}{n}+\displaystyle\frac{3|i|}{\varepsilon
n^{\prime}}+t\Big{)}.$

Together with the facts that
$\int_{\tilde{\gamma}_{n^{\prime}/2+i}(t)}^{\infty}\widetilde{\varrho}_{t}(x)dx=\frac{1}{2}+\frac{2i-1}{n^{\prime}}$
for each $j\in\llbracket 1,n^{\prime}\rrbracket$ and that
$\widetilde{\varrho}_{t}(x)\geq\frac{\varepsilon}{2}$ for $|x|\leq
n^{-\omega/8}$, it follows that

$\displaystyle\big{|}\widetilde{\gamma}_{n^{\prime}/2+i}(t)-\widetilde{\gamma}_{n^{\prime}/2+i}(0)\big{|}\leq\displaystyle\frac{30\pi
C_{3}t}{\varepsilon^{3}}\Big{(}\displaystyle\frac{n^{\prime}}{n}+\displaystyle\frac{|i|}{n^{\prime}}+t\Big{)},$

which verifies the second part of the lemma.

It then remains to establish (8.24), to which end we will bound
$\big{|}H\widetilde{\varrho}_{0}(y)\big{|}$ and
$\big{|}\partial_{t}H\widetilde{\varrho}_{s}(y)\big{|}$. First observe that
$\operatorname{supp}\widetilde{\mu}_{0}=\big{[}\widetilde{G}(0,1),\widetilde{G}(0,0)\big{]}$
and that

(8.25)
$\displaystyle-2\varepsilon^{-1}\leq\widetilde{G}(0,1)\leq-\displaystyle\frac{\varepsilon}{4}\leq\displaystyle\frac{\varepsilon}{4}\leq\widetilde{G}(0,0)\leq
2\varepsilon^{-1};\qquad\big{|}\widetilde{G}(0,0)+\widetilde{G}(0,1)\big{|}\leq\displaystyle\frac{4B_{0}n^{\prime}}{n},$

where the first statement follows from the first and third statements of
(8.13), and the second follows from the first and last statement of (8.13). We
next have that

(8.26)
$\displaystyle\big{|}H\widetilde{\varrho}_{0}(y)\big{|}=\pi^{-1}\Bigg{|}\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\widetilde{\varrho}_{0}(w)dw}{w-y}\Bigg{|}=\pi^{-1}\Bigg{|}\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\big{(}\widetilde{\varrho}_{0}(w)-\widetilde{\varrho}_{0}(y)\big{)}dw}{w-y}\Bigg{|}.$

Thus, for $y\in[-n^{-\omega/8},n^{-\omega/8}]$, we have

(8.27) $\displaystyle\begin{aligned}
\big{|}H\widetilde{\varrho}_{0}(y)\big{|}&\leq\pi^{-1}\Bigg{|}\displaystyle\int_{\tilde{G}(0,1)}^{\tilde{G}(0,0)}\displaystyle\frac{\big{(}\widetilde{\varrho}_{0}(w)-\widetilde{\varrho}_{0}(y)\big{)}dw}{w-y}\Bigg{|}+\pi^{-1}\widetilde{\varrho}_{0}(y)\Bigg{|}\displaystyle\int_{-\infty}^{\tilde{G}(0,1)}\displaystyle\frac{dw}{w-y}+\displaystyle\int_{\tilde{G}(0,0)}^{\infty}\displaystyle\frac{dw}{w-y}\Bigg{|}\\\
&\leq\displaystyle\frac{C_{1}n^{\prime}}{n}\big{|}\widetilde{G}(0,0)-\widetilde{G}(0,1)\big{|}+\varepsilon^{-1}\Big{(}\big{|}\widetilde{G}(0,0)+\widetilde{G}(0,1)\big{|}+2|y|\Big{)}\cdot\displaystyle\max_{w\notin[\tilde{G}(0,1),\tilde{G}(0,0)]}|w-y|^{-2}\\\
&\leq\displaystyle\frac{4C_{1}n^{\prime}}{\varepsilon
n}+\displaystyle\frac{256B_{0}}{\varepsilon^{3}}\Big{(}\displaystyle\frac{n^{\prime}}{n}+|y|\Big{)},\end{aligned}$

where the first bound follows from (8.26) and the fact that
$\operatorname{supp}\widetilde{\varrho}_{0}=\big{[}\widetilde{G}(0,1),\widetilde{G}(0,0)\big{]}$;
the second follows from (8.21) and (8.20); and the third follows from (8.25)
(using the fact that $|w-y|\geq\frac{\varepsilon}{8}$ for $|y|\leq
n^{-\omega}$ and $w\notin\big{[}\widetilde{G}(0,1),\widetilde{G}(0,0)\big{]}$,
due to the first statement of (8.25)).

Next, we claim that there exists a constant
$C_{4}=C_{4}(\varepsilon,\delta,B_{0},m)>1$ such that

(8.28)
$\displaystyle\displaystyle\sup_{s\in[0,n^{-\omega}]}\displaystyle\sup_{|y|\leq(n^{\prime})^{-\omega/2}}\big{|}\partial_{t}H\widetilde{\varrho}_{s}(y)\big{|}\leq
C_{4}.$

Together with (8.27), this would yield (8.24) and thus the lemma. To that end,
observe that

(8.29)
$\displaystyle\big{|}\partial_{t}H\widetilde{\varrho}_{s}(y)\big{|}=\pi^{-1}\Bigg{|}\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\partial_{t}\widetilde{\varrho}_{s}(w)dw}{w-y}\Bigg{|}=\Bigg{|}\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\partial_{x}\big{(}\widetilde{\varrho}_{s}(w)H\widetilde{\varrho}_{s}(w)\big{)}dw}{w-y}\Bigg{|},$

where in the second equality we used (2.12). Further observe for
$s\in[0,n^{-\omega}]$ that

(8.30)
$\displaystyle\operatorname{supp}\widetilde{\varrho}_{s}\subseteq\operatorname{supp}\widetilde{\mu}_{0}+\operatorname{supp}\widetilde{\mu}_{\operatorname{semci}}^{(s)}\subseteq\big{[}\widetilde{G}(0,1)-2t^{1/2},\widetilde{G}(0,0)+2t^{1/2}\big{]}\subseteq[-4\varepsilon^{-1},4\varepsilon^{-1}],$

where the first holds since
$\widetilde{\mu}_{s}=\widetilde{\mu}_{0}\boxplus\mu_{\operatorname{semci}}^{(s)}$,
and the second holds by (2.9), (2.8), and the fact that
$\operatorname{supp}\widetilde{\mu}_{0}=\big{[}\widetilde{G}(0,1),\widetilde{G}(0,0)\big{]}$;
and the third holds by (8.25). Thus, due to (8.29), (8.30), (8.20), and
(8.21), to verify (8.28) it suffices to show the existence of a constant
$C_{5}=C_{5}(\varepsilon,\delta,B_{0},m)>1$ such that

(8.31)
$\displaystyle\displaystyle\sup_{s\in[0,n^{-\omega}]}\displaystyle\sup_{|y|\leq
1/C_{5}}\big{|}\partial_{y}^{k}H\widetilde{\varrho}_{s}(y)\big{|}\leq
C_{5};\qquad\text{for each $k\in\\{0,1,2\\}$}.$

To confirm (8.31), fix $k\in\\{0,1,2\\}$, and observe that

$\displaystyle\big{|}\partial_{y}^{k}H\widetilde{\varrho}_{s}(y)\big{|}=\pi^{-1}\Bigg{|}\partial_{y}^{k}\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\widetilde{\varrho}_{s}(w)dw}{w-y}\Bigg{|}$
$\displaystyle\leq\Bigg{|}\displaystyle\int_{-1/C_{2}}^{1/C_{2}}\displaystyle\frac{\partial_{y}^{k}\widetilde{\varrho}_{s}(y+w)dw}{w}\Bigg{|}+\Bigg{|}\displaystyle\int_{|w-y|>1/C_{2}}\displaystyle\frac{\widetilde{\varrho}_{s}(w)dw}{(w-y)^{k+1}}\Bigg{|}$
$\displaystyle\leq\Bigg{|}\displaystyle\int_{-1/C_{2}}^{1/C_{2}}\displaystyle\frac{\partial_{y}^{k}\big{(}\widetilde{\varrho}_{s}(y+w)-\widetilde{\varrho}_{s}(y)\big{)}dw}{w}\Bigg{|}+(2C_{2})^{3}\leq
10C_{2}^{3},$

where in the first statement we used (2.7); in the second we changed variables
in the first integral by replacing $w$ with $y+w$; in the third we used the
facts that $k\leq 2$ and that $\widetilde{\varrho}_{t}$ is a probability
measure; and in the fourth we used (8.21) and the second bound in (8.22). This
verifies (8.31), which as mentioned above establishes the lemma. ∎

###### Proof of 8.7.

Let $\bm{G}(s)=\bm{G}_{M}(s)$ denote an $M\times M$ Hermitian Brownian motion
as described in Section 2.3, and for every $j\in\llbracket 1,k\rrbracket$ let
$\Lambda^{(j)}$ denote the $M\times M$ diagonal matrix whose $(i,i)$ diagonal
entry is given by $\lambda_{i}^{(j)}(0)$, for each $i\in\llbracket
1,M\rrbracket$. Since
$\bm{\lambda}^{(1)}\leq\bm{\lambda}^{(2)}\leq\cdots\leq\bm{\lambda}^{(k)}$, we
have $\Lambda^{(1)}\leq\Lambda^{(2)}\leq\cdots\leq\Lambda^{(k)}$, where we
write $\Lambda\leq\Lambda^{\prime}$ if $\Lambda^{\prime}-\Lambda$ is
nonnegative definite; in particular,
$\Lambda^{(1)}+\bm{G}(s)\leq\Lambda^{(2)}+\bm{G}(s)\leq\cdots\leq\Lambda^{(k)}+\bm{G}(s)$.
Together with the fact that (by the first part of 2.15)
$\bm{\lambda}^{(j)}(s)$ has the same law as the eigenvalues of
$\Lambda^{(j)}+\bm{G}(s)$ for each $j$, jointly in $s\geq 0$, this yields
$\bm{\lambda}^{(1)}(s)\leq\bm{\lambda}^{(2)}\leq\cdots\leq\bm{\lambda}^{(k)}(s)$
for each $s\geq 0$. ∎

##  9\. Regularity Properties of Solutions

In this section we state and establish various regularity properties of
solutions to the equation (1.2) governing the limit shape for non-intersecting
Brownian bridges. These include comparison principles (9.1 below); derivative
bounds (9.2 and 9.4 below); and exponential decay of discrepancy estimates
(9.5 below). We also provide a result that explains how perturbing regular
solutions to (1.2) preserves their regularity (9.6 below), thereby showing
that 1.3 is in a sense stable under smooth perturbations of boundary data.

###  9.1. Regularity Estimates

In this section we state estimates for solutions to (1.2). In what follows, we
recall the sets of functions $\operatorname{Adm}(\mathfrak{R})$ and
$\operatorname{Adm}_{\varepsilon}(\mathfrak{R})$ from Section 1.3. For any
real number $r>0$ and point $z\in\mathbb{R}^{2}$, we recall that
$\mathcal{B}_{r}(z)=\big{\\{}z^{\prime}\in\mathbb{R}:|z^{\prime}-z|<r\big{\\}}$
denotes the open disk of radius $r$ centered at $z$; if $z=(0,0)$, we
abbreviate $\mathcal{B}_{r}(z)=\mathcal{B}_{r}$. We begin with the below
maximum and comparison principles for solutions of (1.2), which are quick
consequences of the analogous principles for solutions to nonlinear elliptic
partial differential equations; it is established in Section E.1 below.

######  Lemma 9.1.

Fix an open set $\mathfrak{R}\subset\mathbb{R}$, and let
$F_{1},F_{2},F\in\operatorname{Adm}(\mathfrak{R})\cap\mathcal{C}^{2}(\mathfrak{R})$
be solutions to (1.2) on $\mathfrak{R}$.

  1. (1)

If $F_{1}(z)\leq F_{2}(z)$ for each $z\in\partial\mathfrak{R}$, then
$F_{1}(z)\leq F_{2}(z)$ for each $z\in\mathfrak{R}$.

  2. (2)

We have
$\sup_{z\in\mathfrak{R}}\big{|}F_{1}(z)-F_{2}(z)\big{|}\leq\sup_{z\in\partial\mathfrak{R}}\big{|}F_{1}(z)-F_{2}(z)\big{|}$.
In particular,
$\sup_{z\in\mathfrak{R}}\big{|}F(z)\big{|}=\sup_{z\in\partial\mathfrak{R}}\big{|}F(z)\big{|}$.

We next have the following lemma indicating boundedness of the derivatives for
a solution to (1.2); it is established in Section E.1 below, again as a
consequence of its analog in the context of elliptic partial differential
equations. In what follows, $\mathfrak{R}^{\prime}\subseteq\mathfrak{R}$ is a
subset that accounts for boundary regularity, in that $F$ is regular in the
interior of $\mathfrak{R}$ but up to the boundary of $\mathfrak{R}^{\prime}$;
in particular, if $\mathfrak{R}^{\prime}$ is empty, the below lemma is an
interior regularity estimate and, if $\mathfrak{R}^{\prime}=\mathfrak{R}$,
then it is a global one. Here, for any sets $\mathcal{S}$ and
$\mathcal{S}^{\prime}$, we let
$\mathcal{S}\setminus\mathcal{S}^{\prime}=\mathcal{S}\setminus(\mathcal{S}\cap\mathcal{S}^{\prime})$.
See the left side of Figure 9.

######  Lemma 9.2.

For any integer $m\geq 1$; real numbers $r>0$, $\varepsilon\in(0,1)$, and
$B>1$; and bounded open subsets
$\mathfrak{R}^{\prime}\subseteq\mathfrak{R}\subset\mathbb{R}^{2}$ such that
$\mathfrak{R}^{\prime}$ has a smooth boundary, there exists a constant
$C=C(\varepsilon,r,B,m,\mathfrak{R}^{\prime})>1$ such that the following
holds. Let $f\in\mathcal{C}(\partial\mathfrak{R})$ be a function satisfying
$\|f\|_{0}\leq B$, and let
$F\in\operatorname{Adm}_{\varepsilon}(\mathfrak{R})\cap\mathcal{C}^{2}(\mathfrak{R})$
be a solution to (1.2) on $\mathfrak{R}$ with boundary data
$F|_{\partial\mathfrak{R}}=f$. Assume that there exists a function
$\varphi\in\mathcal{C}^{m+1}(\overline{\mathfrak{R}}^{\prime})$ with
$\|\varphi\|_{\mathcal{C}^{m+1}(\mathfrak{R}^{\prime})}\leq B$, such that
$f(z)=\varphi(z)$ for each
$z\in\partial\mathfrak{R}^{\prime}\cap\partial\mathfrak{R}$. Letting

(9.1)
$\displaystyle\mathfrak{D}_{r}=\big{\\{}z\in\mathfrak{R}:\operatorname{dist}(z,\partial\mathfrak{R}\setminus\partial\mathfrak{R}^{\prime})>r\big{\\}},$

we have $\|F\|_{\mathcal{C}^{m}(\mathfrak{D}_{r})}\leq C$.

The following lemma states that the solutions of (1.2) are real analytic (that
is, they have an absolutely convergent power series expansion at each point in
the interior of their domains); it is shown in Section E.1 below, as a quick
consequence of classical results stating real analyticity for solutions of
elliptic partial differential equations.

######  Lemma 9.3.

Fix a real number $\varepsilon\in(0,1)$ and some open set
$\mathfrak{R}\subset\mathbb{R}^{2}$; let
$F\in\operatorname{Adm}_{\varepsilon}(\mathfrak{R})\cap\mathcal{C}^{2}(\mathfrak{R})$
denote a solution to (1.2) on $\mathfrak{R}$. Then, $F$ is real analytic
$\mathfrak{R}$.

Figure 9. Shown on the left is depiction for 9.2. Shown in the middle is a
depiction for 9.5 stating, if $F_{1}(z)=F_{2}(z)$ for each $z$ on the blue
sides of $\mathfrak{R}$, then $F_{1}-F_{2}$ is exponentially small in the
shaded region. Shown on the right is a depiction for 9.6.

We next have the following lemma bounding the effect of a $\mathcal{C}^{0}$
boundary perturbation on the derivatives of a solution to (1.2); its proof is
similar to [2, Proposition 2.13] and is given in Section E.2 below.

######  Lemma 9.4.

For any integer $m\geq 1$; real numbers $r>0$, $\varepsilon\in(0,1)$, and
$B>1$; and bounded open subsets
$\mathfrak{R}^{\prime}\subseteq\mathfrak{R}\subset\mathbb{R}^{2}$ such that
$\mathfrak{R}^{\prime}$ has a smooth boundary, there exists a constant
$C=C(\varepsilon,r,B,m,\mathfrak{R}^{\prime})>1$ such that the following
holds. For each index $i\in\\{1,2\\}$, let
$F_{i}\in\operatorname{Adm}_{\varepsilon}(\mathfrak{R})\cap\mathcal{C}^{2}(\mathfrak{R})$
denote a solution to (1.2) such that
$\sup_{z\in\partial\mathfrak{R}}\big{|}F_{i}(z)\big{|}\leq B$. Assume that
there exist functions
$\varphi_{i}\in\mathcal{C}^{m+5}(\overline{\mathfrak{R}})$ such that
$\|\varphi_{i}\|_{\mathcal{C}^{m+5}(\mathfrak{R})}<B$ and
$F_{i}(z)=\varphi_{i}(z)$ for each
$z\in\partial\mathfrak{R}^{\prime}\cap\partial\mathfrak{R}$. Then, letting
$\mathfrak{D}_{r}\subseteq\mathfrak{R}$ be as in (9.1), we have

$\displaystyle\|F_{1}-F_{2}\|_{\mathcal{C}^{m}(\mathfrak{D}_{r})}\leq
C\big{(}\|F_{1}-F_{2}\|_{\mathcal{C}^{0}(\mathfrak{R})}+\|\varphi_{1}-\varphi_{2}\|_{\mathcal{C}^{m+2}(\mathfrak{R})}\big{)}.$

The following result is established in Section 9.2 below. It states that,
given two solutions $F_{1},F_{2}$ to (1.2) on a tall rectangle of height $1$
and width $L^{-1}$ (with $L$ bounded below), whose boundary data match on its
west and east boundaries, $|F_{1}-F_{2}|$ decays exponentially in $L$ in the
middle of the rectangle. As such, it can be viewed as indicating that the
“discrepancy” between $F_{1}$ and $F_{2}$ (originating from their different
boundary data along the north and south sides of the rectangle) decay
exponentially around the middle of the rectangle. See the middle of Figure 9.

######  Proposition 9.5.

For any real numbers $\varepsilon,r\in\big{(}0,\frac{1}{4}\big{)}$ and $B>1$,
there exists a constant $c=c(\varepsilon,r,B)>0$ such that the following
holds. Let $L>0$ be a real number, and define the open rectangle
$\mathfrak{R}=(0,L^{-1})\times(-1,1)\subset\mathbb{R}^{2}$. Let
$F_{1},F_{2}\in\operatorname{Adm}_{\varepsilon}(\mathfrak{R})\cap\mathcal{C}^{5}(\overline{\mathfrak{R}})$
be two solutions to (1.2) such that
$\|F_{i}\|_{\mathcal{C}^{5}(\mathfrak{R})}\leq B$ for each $i\in\\{1,2\\}$. If
$F_{1}(t,x)=F_{2}(t,x)$ for any $(t,x)\in\partial\mathfrak{R}$ with
$t\in\\{0,L^{-1}\\}$, then

$\displaystyle\big{|}F_{1}(t,x)-F_{2}(t,x)\big{|}\leq
c^{-1}e^{-cL^{1/8}},\qquad\text{for any $(t,x)\in[0,L^{-1}]\times[r-1,1-r]$}.$

We conclude this section with the following lemma, which is a consequence of
the above bounds and is established in Section E.3 below; it essentially
states the following. Fix a solution $F$ to (1.2), bounded in
$\mathcal{C}^{m}$ for some integer $m$ on a rectangle $\mathfrak{R}$, as well
some boundary data $g_{0}$ and $g_{1}$ on the two vertical sides on the
rectangle; assume that $g_{0}$ and $g_{1}$ are small perturbations of $F$
(restricted to those boundaries). Then, it is possible to find a solution $G$
to (1.2) on a slightly shorter rectangle $\mathfrak{S}$, whose boundary data
on the vertical sides of the rectangle are given by $g_{0}$ and $g_{1}$ (the
first condition in the below lemma), and that is close to $F$ (quantified
through the second and third conditions of the below lemma). The second part
of the lemma states that $F$ and $G$ are close in any $\mathcal{C}^{k}$ norm
in the interior of $\mathfrak{S}$, and the third part states that $F$ and $G$
are close in $\mathcal{C}^{m-5}$ (that is, fewer derivatives than the original
assumed bound on $F$) up to the boundary of $\mathfrak{S}$. See the right side
of Figure 9.

######  Lemma 9.6.

For any integers $m,k\geq 7$, and real numbers $\varepsilon>0$;
$r\in\big{(}0,\frac{1}{4}\big{)}$; and $B>1$, there exist constants
$\delta=\delta(\varepsilon,B)>0$, $C_{1}=C_{1}(\varepsilon,r,B,k)>1$, and
$C_{2}=C_{2}(\varepsilon,B,m)>1$ such that the following holds. Fix a real
number $L>2$, and define the open rectangles

$\displaystyle\mathfrak{R}=\Big{(}0,\displaystyle\frac{1}{L}\Big{)}\times(-1,1);\qquad\mathfrak{S}_{r}=\Big{(}\displaystyle\frac{r}{L},\displaystyle\frac{1-r}{L}\Big{)}\times\Big{(}\displaystyle\frac{r+1}{L}-1,1-\displaystyle\frac{r+1}{L}\Big{)};\qquad\mathfrak{S}=\mathfrak{S}_{0}.$

Let
$F\in\operatorname{Adm}_{\varepsilon}(\mathfrak{R})\cap\mathcal{C}^{m}(\overline{\mathfrak{R}})$
denote a solution to (1.2) such that
$\|F\|_{\mathcal{C}^{m}(\mathfrak{R})}\leq B$, and define the functions
$f_{0},f_{1}:[-1,1]\rightarrow\mathbb{R}$ by setting $f_{i}(x)=F(iL^{-1},x)$
for each $(i,x)\in\\{0,1\\}\times[-1,1]$. Further let
$g_{0},g_{1}:[-1,1]\rightarrow\mathbb{R}$ denote two functions such that
$\|g_{i}\|_{\mathcal{C}^{m}(-1,1)}\leq B$ and
$\big{|}g_{i}(x)-f_{i}(x)\big{|}\leq\delta L^{-1}$ for each
$(i,x)\in\\{0,1\\}\times[-1,1]$. Then, there exists a solution
$G\in\operatorname{Adm}_{\varepsilon/2}(\mathfrak{S})\cap\mathcal{C}^{m-5}(\overline{\mathfrak{S}})$
to (1.2) satisfying the following three properties.

  1. (1)

For each $i\in\\{0,1\\}$ and $x\in\big{[}L^{-1}-1,1-L^{-1}\big{]}$, we have
$G(iL^{-1},x)=g_{i}(x)$.

  2. (2)

We have $\|F-G\|_{\mathcal{C}^{k}(\mathfrak{S}_{r})}\leq
C_{1}L^{k}\cdot\big{(}\|f_{0}-g_{0}\|_{\mathcal{C}^{0}}+\|f_{1}-g_{1}\|_{\mathcal{C}^{0}}\big{)}$.

  3. (3)

We have $\|F-G\|_{\mathcal{C}^{m-5}(\mathfrak{S})}\leq
C_{2}L^{m-5}\cdot\big{(}\|f_{0}-g_{0}\|_{\mathcal{C}^{m-3}}+\|f_{1}-g_{1}\|_{\mathcal{C}^{m-3}}\big{)}$
and $\|F-G\|_{\mathcal{C}^{m-5}(\mathfrak{S})}\leq
C_{2}L^{m-5}\cdot\big{(}\|f_{0}-g_{0}\|_{\mathcal{C}^{0}}^{3/m}+\|f_{1}-g_{1}\|_{\mathcal{C}^{0}}^{3/m}\big{)}$.

###  9.2. Proof of 9.5

In this section we establish 9.5. In what follows, we fix real numbers
$\varepsilon>0$, $L>1$, and $L_{0}\in\big{[}\frac{L}{2B},2BL\big{]}$. We also
denote the rectangular open set $\mathfrak{R}_{0}=(0,1)\times(-L_{0},L_{0})$
and let
$F_{1},F_{2}\in\operatorname{Adm}_{\varepsilon}(\overline{\mathfrak{R}_{0}})\cap\mathcal{C}^{2}(\overline{\mathfrak{R}_{0}})$
denote two solutions to the equation (1.2).

The next lemma bounds $\big{|}F_{1}(z)-F_{2}(z)\big{|}$, for a point
$z\in\mathfrak{R}_{0}$ such that $F_{1}=F_{2}$ on the part of
$\partial\mathfrak{R}_{0}$ near $z$; we establish it in Section 9.3 below.

######  Lemma 9.7.

For any real number $B>1$, there exist constants $c=c(\varepsilon,B)>0$ and
$C=C(\varepsilon,B)>1$ such that the following holds. Fix a point
$z=(t_{0},x_{0})\in\mathfrak{R}_{0}$, such that
$[0,1]\times[x_{0}-2L^{1/2},x_{0}+2L^{1/2}]\subseteq\mathfrak{R}_{0}$. Let
$A\leq B$ be a real number, and assume the below three statements hold.

  1. (1)

For any indices $i\in\\{1,2\\}$ and $j,k\in\\{t,x\\}$, and point
$w\in\mathfrak{R}_{0}$, we have $\big{|}\nabla F_{i}(w)\big{|}\leq B$ and
$\big{|}\partial_{j}\partial_{k}F_{i}(w)\big{|}\leq BL^{-1}$; we also have
$\|F_{i}\|_{\mathcal{C}^{5}(\mathfrak{R}_{0})}-\|F_{i}\|_{\mathcal{C}^{0}(\mathfrak{R}_{0})}\leq
B$.

  2. (2)

For any point $w\in\partial\mathfrak{R}_{0}$ with $|w-z|\leq 2L^{1/2}$, we
have $F_{1}(w)=F_{2}(w)$.

  3. (3)

For any point $w\in\partial\mathfrak{R}_{0}$, we have
$\big{|}F_{1}(w)-F_{2}(w)\big{|}\leq A$.

Then, $\big{|}F_{1}(z)-F_{2}(z)\big{|}\leq C(L^{-1/8}A+e^{-cL^{1/8}})$.

We can quickly prove 9.5 by repeated application of 9.7.

###### Proof of 9.5.

We may assume throughout this proof that $L$ is sufficiently large (in a way
only dependent on $\varepsilon$ and $B$), for otherwise the lemma holds by
altering the values of $c$ and $C$ (using the fact that
$\big{|}F_{1}(t,x)-F_{2}(t,x)\big{|}\leq 2B$). Define the rescaled domain
$\widehat{\mathfrak{R}}=L\cdot\mathfrak{R}=(0,1)\times(-L,L)$ and, for each
index $i\in\\{1,2\\}$, set
$\widehat{F}_{i}\in\mathcal{C}^{2}(\widehat{\mathfrak{R}})$ by for each
$w\in[0,1]\times[-L,L]$ letting

$\displaystyle\widehat{F}_{i}(w)=L\cdot
F_{i}(L^{-1}w),\quad\text{so}\quad\nabla\widehat{F}_{i}(w)=\nabla\widehat{F}_{i}(L^{-1}w),\quad\text{and}\quad\partial_{j}\partial_{k}\widehat{F}_{i}(w)=L^{-1}\cdot\partial_{i}\partial_{j}F_{i}(L^{-1}w),$

for any indices $j,k\in\\{t,x\\}$. In particular, the $\widehat{F}_{i}$
continue to solve (1.2).

It therefore suffices to show that
$\big{|}\widehat{F}_{i}(t,x)-\widehat{F}_{2}(t,x)\big{|}\leq Ce^{-cL^{1/8}}$
for each $(t,x)\in[0,1]\times\big{[}(r-1)L,(1-r)L\big{]}$. Denoting the open
rectangle $\widehat{\mathfrak{R}}_{m}=(0,1)\times(mL^{2/3}-L,L-mL^{2/3})$ for
any integer $m\geq 1$, we will more generaly show by induction on
$m\in\llbracket 0,L^{1/8}\rrbracket$ that there exist constants
$c=c(\varepsilon,B)>0$ and $C=C(\varepsilon,B)>1$ such that

(9.2)
$\displaystyle\big{|}\widehat{F}_{1}(t,x)-\widehat{F}_{2}(t,x)\big{|}\leq
2BL^{-m/9}+Cme^{-cL^{1/8}},\quad\text{for any
$(t,x)\in\widehat{\mathfrak{R}}_{m}$},$

from which the lemma would follow upon taking $m=\lfloor L^{1/8}\rfloor$; see
the left side of Figure 10.

Let us first show that the $\widehat{F}_{i}$ satisfy the first and third
assumptions of 9.7 (with the $A$ there equal to $2B$ here). Since
$\|F_{i}\|_{\mathcal{C}^{5}(\overline{\mathfrak{R}})}\leq B$, we have that
$\big{|}\nabla\widehat{F}_{i}(w)\big{|}\leq B$ and
$\big{|}\partial_{j}\partial_{k}\widehat{F}_{i}(w)\big{|}\leq BL^{-1}$ for
each $w\in\widehat{\mathfrak{R}}$ and that
$\|\widehat{F}_{i}\|_{\mathcal{C}^{5}(\hat{\mathfrak{R}}_{m})}-\|\widehat{F}_{i}\|_{\mathcal{C}^{0}(\hat{\mathfrak{R}}_{m})}\leq
B$; this verifies the first assumption. Moreover, since
$F_{1}(t,x)=F_{2}(t,x)$ for $t\in\\{0,L^{-1}\\}$, since $\big{|}\nabla
F_{i}(w)\big{|}\leq B$, and since $\mathfrak{R}=(0,L^{-1})\times(-1,1)$ we
have $\big{|}F_{1}(w)-F_{2}(w)\big{|}\leq 2BL^{-1}$ for each
$w\in\overline{\mathfrak{R}}$. Hence,
$\big{|}\widehat{F}_{1}(w)-\widehat{F}_{2}(w)\big{|}\leq 2B$ for each
$w\in\widehat{\mathfrak{R}}$, verifying the third assumption of 9.7. This in
particular confirms (9.2) at $m=0$.

We therefore assume in what follows that (9.2) holds at some value of
$m\in\llbracket 0,L^{1/8}-1\rrbracket$ and verify that it also holds with $m$
replaced by $m+1$. For any integer $k\in\llbracket 0,L^{1/8}\rrbracket$, set

$\displaystyle\varsigma_{k}=\sup_{w\in\mathfrak{R}_{k}}\big{|}\widehat{F}_{1}(w)-\widehat{F}_{2}(w)\big{|}.$

Then the second assumption of 9.7 applies, with the
$(\mathfrak{R}_{0},F_{1},F_{2},z)$ there given by the
$(\widehat{\mathfrak{R}}_{m},\widehat{F}_{1},\widehat{F}_{2},z)$ with any
$z\in\widehat{\mathfrak{R}}_{m+1}$. Since
$L-(m+1)L^{2/3}\geq\frac{L}{2B}+2L^{1/2}$ for $m\in\llbracket
0,L^{1/8}\rrbracket$, 9.7 therefore yields constants
$c_{1}=c_{1}(\varepsilon,B)>0$ and $C_{1}=C_{1}(\varepsilon,B)>1$ such that

$\displaystyle\varsigma_{m+1}=\displaystyle\sup_{w\in\widehat{\mathfrak{R}}_{m+1}}\big{|}\widehat{F}_{1}(w)-\widehat{F}_{2}(w)\big{|}$
$\displaystyle\leq
C_{1}L^{-1/8}\cdot\displaystyle\sup_{w\in\widehat{\mathfrak{R}}_{m}}\big{|}\widehat{F}_{1}(w)-\widehat{F}_{2}(w)\big{|}+C_{1}e^{-c_{1}L^{1/8}}$
$\displaystyle=C_{1}L^{-1/8}\varsigma_{m}+C_{1}e^{-c_{1}L^{1/8}}\leq
L^{-1/9}\varsigma_{m}+C_{1}e^{-c_{1}L^{1/8}},$

where in the last inequality we used the fact that $L$ is sufficiently large.
Applying this with (9.2), with the $(c,C)$ there equal to $(c_{1},C_{1})$ here
yields

$\displaystyle\varsigma_{m+1}\leq
2BL^{-(m+1)/9}+C_{1}mL^{-1/9}e^{-c_{1}L^{1/8}}+C_{1}e^{-c_{1}L^{1/8}}\leq
2BL^{-(m+1)/9}+C_{1}(m+1)e^{-c_{1}L^{1/8}}.$

This verifies (9.2) (with the $m$ there replaced by $m+1$ here), establishing
the proposition. ∎

To establish 9.7, we use a probabilistic interpretation of the $F_{i}$. To
that end, for any real numbers $u\in\mathbb{R}$ and
$v\in[-\varepsilon^{-1},\varepsilon]$, we recall from 3.19 the $2\times 2$
matrix $\bm{D}=\bm{D}(u,v)$. Let $W:\mathbb{R}_{\geq
0}\rightarrow\mathbb{R}^{2}$ denote a standard two-dimensional Brownian
motion. For any point $z\in\mathfrak{R}_{0}$ and index $i\in\\{1,2\\}$, define
the process $Y_{i}=Y_{i}^{z}:[0,\tau_{i}]\rightarrow\mathbb{R}^{2}$ through
the stochastic differential equation

(9.3) $\displaystyle dY_{i}(s)=\bm{D}\Big{(}\nabla
F_{i}\big{(}Y_{i}(s)\big{)}\Big{)}^{1/2}\cdot dW(s),\qquad\text{with initial
data $Y_{i}(0)=z$},$

so that $Y_{1}$ and $Y_{2}$ are coupled through the same Brownian motion
$W(s)$; see the right side of Figure 10 for a depiction. Here, the hitting
time $\tau_{i}=\tau_{i;z}$ is given by

$\displaystyle\tau_{i}=\inf\big{\\{}t\geq
0:Y_{i}(s)\in\partial\mathfrak{R}_{0}\big{\\}}.$

That the stochastic differential equation (9.3) is well-defined follows from
the fact that the coefficients $\bm{D}\big{(}\nabla F_{i}(t,x)\big{)}^{1/2}$
of the equation are uniformly Lipschitz (as
$F_{i}\in\mathcal{C}^{2}(\overline{\mathfrak{R}_{0}})$).

Figure 10. Shown to the left is the rectangle $\widehat{\mathfrak{R}}_{m}$
used in the proof of 9.5. Shown to the right are the coupled stochastic
processes $Y_{1}(s)$ and $Y_{2}(s)$ as in (9.3).

######  Lemma 9.8.

For each index $i\in\\{1,2\\}$, the process $F_{i}\big{(}Y_{i}(s)\big{)}$ is a
martingale. In particular,

(9.4) $\displaystyle dF_{i}\big{(}Y_{i}(s)\big{)}=\nabla
F_{i}\big{(}Y_{i}(s)\big{)}\cdot\bm{D}\Big{(}\nabla
F_{i}\big{(}Y_{i}(s)\big{)}\Big{)}^{1/2}\cdot dW(s).$

###### Proof.

Since $\tau_{i}=0$ if $z\in\partial\mathfrak{R}_{0}$, we may assume that
$z\in\mathfrak{R}_{0}$. Then, denoting the Hessian of $F_{i}$ by

$\displaystyle\bm{H}(z)=\left[\begin{array}[]{cc}\partial_{tt}F_{i}(z)&\partial_{tx}F_{i}(z)\\\
\partial_{xt}F_{i}(z)&\partial_{xx}F_{i}(z)\end{array}\right],$

we have by Itô’s lemma that

$\displaystyle dF_{i}\big{(}Y_{i}(s)\big{)}$ $\displaystyle=\nabla
F_{i}\big{(}Y_{i}(s)\big{)}\cdot\bm{D}\Big{(}\nabla
F_{i}\big{(}Y_{i}(s)\big{)}\Big{)}^{1/2}\cdot
dW(s)+\frac{1}{2}\operatorname{Tr}\bigg{(}\bm{D}\Big{(}\nabla
F_{i}\big{(}Y_{i}(s)\big{)}\Big{)}\bm{H}\Big{(}F_{i}\big{(}Y_{i}(s)\big{)}\Big{)}\bigg{)}$
$\displaystyle=\nabla F_{i}\big{(}Y_{i}(s)\big{)}\cdot\bm{D}\Big{(}\nabla
F_{i}\big{(}Y_{i}(s)\big{)}\Big{)}^{1/2}\cdot dW(s),$

where in the last equality we used the fact that $F_{i}$ satisfies (1.2) (with
the $(t,y)$ there equal to $Y_{i}(s)$ here). This yields (9.4) and thus that
$F_{i}\big{(}Y_{i}(s)\big{)}$ is a martingale. ∎

Hence, we must analyze properties of the process $Y_{i}$, to which end we have
the following lemma.

######  Lemma 9.9.

There exists a constant $c=c(\varepsilon)>0$ such that, for any real number
$A>1$, index $i\in\\{1,2\\}$, and point $z=(t_{0},x_{0})\in\mathfrak{R}_{0}$,
we have

(9.5) $\displaystyle\begin{aligned} &\mathbb{P}[\tau_{i;z}\geq A]\leq
c^{-1}e^{-cA};\qquad\mathbb{P}\bigg{[}\displaystyle\sup_{s\in[0,\tau_{i;z}]}\big{|}Y_{i}(s)-z\big{|}\geq
A\Big{]}\leq c^{-1}e^{-cA}.\end{aligned}$

###### Proof.

Denote $W(s)=\big{(}W_{1}(s),W_{2}(s)\big{)}\in\mathbb{R}$, so that
$W_{1},W_{2}:\mathbb{R}_{\geq 0}\rightarrow\mathbb{R}$ are two independent
standard Brownian motions; also let
$Y_{i}(s)=\big{(}Y_{i;1}(s),Y_{i;2}(s)\big{)}\in\mathbb{R}^{2}$. Then, by
(9.3) and the fact that $\bm{D}(u,v)=\operatorname{diag}(1,\pi^{2}v^{-4})$, we
have $Y_{i;1}(s)-t_{0}=W_{1}(s)$, for each $s\in[0,\tau_{i;z}]$. In particular
we have for any $s\leq\min\\{\tau_{1;z},\tau_{2;z}\\}$ that

(9.6) $\displaystyle Y_{1;1}(s)=Y_{2;1}(s)=W_{1}(s)+t_{0}.$

Since $\mathfrak{R}_{0}\subset[0,1]\times\mathbb{R}$, it follows that
$\tau_{i;z}\geq A$ can only hold if $W_{i}(s)=Y_{i;1}(s)\in[-1,1]$, for each
$s\in[0,A]$. Thus,

(9.7) $\displaystyle\begin{aligned} \mathbb{P}[\tau_{i;z}\geq
A]&\leq\mathbb{P}\Bigg{[}\bigcap_{s\in[0,A]}\big{\\{}W_{1}(s)\in[-1,1]\big{\\}}\Bigg{]}=\mathbb{P}\Bigg{[}\bigcap_{k=1}^{\lfloor
A\rfloor}\Big{\\{}\big{|}W_{1}(k)-W_{1}(k-1)\big{|}\leq
2\Big{\\}}\Bigg{]}.\end{aligned}$

Now observe that there exists a constant $c_{1}>1$ such that
$\mathbb{P}\big{[}W_{1}(k)-W_{1}(k-1)\leq 2\big{]}<1-c_{1}$, for any $k\geq
1$. Hence,

$\displaystyle\mathbb{P}\Bigg{[}\bigcap_{k=1}^{\lfloor A\rfloor}$
$\displaystyle\Big{\\{}\big{|}W_{1}(k)-W_{1}(k-1)\big{|}\leq
2\Big{\\}}\Bigg{]}=\displaystyle\prod_{k=1}^{\lfloor
A\rfloor}\mathbb{P}\Big{[}\big{|}W_{1}(k)-W_{1}(k-1)\big{|}\leq
2\Big{]}\leq(1-c_{1})^{\lfloor A\rfloor},$

which together with (9.7) yields the first bound in (9.5).

Next, since $F_{i}\in\operatorname{Adm}_{\varepsilon}(\mathfrak{R}_{0})$, we
have $\partial_{x}F_{i}(t,x)\in[-\varepsilon^{-1},-\varepsilon]$ for each
$(t,x)\in\mathfrak{R}_{0}$; it follows that $\big{|}\bm{D}\big{(}\nabla
F_{i}(t,x)\big{)}^{1/2}\big{|}\leq 5\varepsilon^{-2}$ for each
$(t,x)\in\mathfrak{R}_{0}$. By the definition (9.3) of $Y_{i}$ (and the fact
that $Y_{i}$ is a time-changed Brownian motion by [48, Theorem 3.4.6]), this
yields

$\displaystyle\mathbb{P}\bigg{[}\displaystyle\sup_{s\in[0,\tau_{i}]}\big{|}Y_{i}(s)-z\big{|}\geq
A\bigg{]}$
$\displaystyle\leq\mathbb{P}\bigg{[}\displaystyle\sup_{s\in[0,A]}\big{|}W(s)\big{|}\geq
10\varepsilon^{-2}A\bigg{]}+\mathbb{P}[\tau_{i}\geq A]$
$\displaystyle\leq\mathbb{P}\bigg{[}\displaystyle\sup_{s\in[0,1]}\big{|}W(s)\big{|}\geq
10\varepsilon^{-2}A^{1/2}\bigg{]}+c^{-1}e^{-cA},$

where in the last inequality we used the first bound in (9.5). This, together
with the fact (see [48, Chapter 4, Equation (3.40)]) that there is a constant
$c_{2}=c_{2}(\varepsilon)>0$ such that $\sup_{s\in[0,1]}\big{|}W(s)\big{|}\leq
10\varepsilon^{-2}A^{1/2}$ holds with probability at least
$1-c_{2}^{-1}e^{-c_{2}A}$, implies the second bound in (9.5). ∎

###  9.3. Proof 9.7

In this section we establish 9.7.

###### Proof of 9.7.

We may assume throughout this proof that $L>10$ (for otherwise taking the
constant $C=2$ in the lemma gives
$\big{|}F_{1}(z)-F_{2}(z)\big{|}\leq\sup_{z^{\prime}\in\partial\mathfrak{R}_{0}}\big{|}F_{1}(z^{\prime})-F_{2}(z^{\prime})\big{|}\leq
A\leq CL^{-1/8}A$, by 9.1). Fix $M=L^{1/4}$; abbreviate $\tau_{i}=\tau_{i;z}$
for each $i\in\\{1,2\\}$; set $\tau=\min\\{\tau_{1},\tau_{2}\\}$; and define
the event

(9.8) $\displaystyle\mathscr{A}=\\{\tau\leq
M\\}\cap\bigg{\\{}\displaystyle\max_{i\in\\{1,2\\}}\displaystyle\sup_{s\in[0,\tau_{i}]}\big{|}Y_{i}(s)-z\big{|}\leq
M\bigg{\\}},\qquad\text{so that
$\mathbb{P}\big{[}\mathscr{A}^{\complement}\big{]}\leq
c_{1}^{-1}e^{-c_{1}M}$},$

for some constant $c_{1}=c_{1}(\varepsilon)>0$, by 9.9. Denote
$W(s)=\big{(}W_{1}(s),W_{2}(s)\big{)}$ and
$Y_{i}(s)=\big{(}Y_{i;1}(s),Y_{i;2}(s)\big{)}$, for each index $i\in\\{1,2\\}$
and real number $s\geq 0$. Recall from (9.6) that we have
$Y_{1;1}(s)=W_{1}(s)+t_{0}=Y_{2;1}(s)$ for each $s\in[0,\tau]$. Since on the
event $\mathscr{A}$ we have $Y_{i_{0}}(\tau)\in\partial\mathfrak{R}_{0}$ for
some $i_{0}\in\\{1,2\\}$ while $\big{|}Y_{i_{0};2}(\tau)-x_{0}\big{|}\leq
L^{1/4}\leq L^{1/2}$, and since
$(0,1)\times[x_{0}-2L^{1/2},x_{0}+2L^{1/2}]\subseteq\mathfrak{R}_{0}$, it
follows that $Y_{i_{0};1}(\tau)\in\\{0,1\\}$ (that is, the $t$-coordinate of
$Y_{i_{0}}(\tau)$ causes it to be on $\partial\mathfrak{R}_{0}$). Hence,

(9.9) $\displaystyle Y_{1,1}(\tau)=Y_{2,1}(\tau)\in\\{0,1\\},\quad\text{and
thus}\quad\tau_{1}=\tau_{2},\quad\text{on the event $\mathscr{A}$}.$

By (9.3), we have

(9.10) $\displaystyle d\big{(}Y_{1}(t)-Y_{2}(t)\big{)}$
$\displaystyle=\bigg{(}\bm{D}\Big{(}\nabla
F_{1}\big{(}Y_{1}(t)\big{)}\Big{)}^{1/2}-\bm{D}\Big{(}\nabla
F_{2}\big{(}Y_{2}(t)\big{)}\Big{)}^{1/2}\bigg{)}\cdot dW(t).$

Thus, for any real numbers $t>0$ and $p>1$, the Burkholder–Davis–Gundy
inequality [13, Theorem 3.2] (see also [14, Theorem 11.2.1]) yields

(9.11)
$\displaystyle\mathbb{E}\bigg{[}\displaystyle\sup_{s\in[0,t]}\big{|}Y_{1}(s)-Y_{2}(s)\big{|}^{2p}\bigg{]}\leq(72p)^{2p}\cdot\mathbb{E}\Bigg{[}\bigg{(}\displaystyle\int_{0}^{t}\Big{\|}\bm{D}\big{(}\nabla
F_{1}(Y_{1}(s)\big{)}^{1/2}-\bm{D}\big{(}\nabla
F_{2}(Y_{2}(s))\big{)}^{1/2}\Big{\|}ds\bigg{)}^{2p}\Bigg{]}.$

To bound the integral on the right side of (9.11), first observe that since
$\partial_{x}F_{1}(w),\partial_{x}F_{2}(w)\in[-\varepsilon^{-1},-\varepsilon]$
that the $\bm{D}(\nabla F_{i})$ are uniformly smooth (in a way dependent on
$\varepsilon$) in $\nabla F_{i}$. Thus, there exists a constant
$C_{1}=C_{1}(\varepsilon)>0$ such that

(9.12) $\displaystyle\begin{aligned} \bigg{\|}\bm{D}\Big{(}\nabla
F_{1}\big{(}&Y_{1}(t)\big{)}\Big{)}^{1/2}-\bm{D}\Big{(}\nabla
F_{2}\big{(}Y_{2}(t)\big{)}\Big{)}^{1/2}\bigg{\|}\\\ &\leq C_{1}\Big{|}\nabla
F_{1}\big{(}Y_{1}(t)\big{)}-\nabla F_{2}\big{(}Y_{2}(t)\big{)}\Big{|}\\\ &\leq
C_{1}\bigg{(}\Big{|}\nabla F_{1}\big{(}Y_{1}(t)\big{)}-\nabla
F_{1}\big{(}Y_{2}(t)\big{)}\Big{|}+\Big{|}\nabla
F_{1}\big{(}Y_{2}(t)\big{)}-\nabla
F_{2}\big{(}Y_{2}(t)\big{)}\Big{|}\bigg{)}\\\ &\leq
C_{1}BL^{-1}\big{|}Y_{1}(t)-Y_{2}(t)\big{|}+C_{1}\Big{|}\nabla
F_{1}\big{(}Y_{2}(t)\big{)}-\nabla
F_{2}\big{(}Y_{2}(t)\big{)}\Big{|},\end{aligned}$

where in the last bound we used the first assumption of the lemma
$\big{|}\partial_{j}\partial_{k}F_{i}(w)\big{|}\leq BL^{-1}$. Since we
deterministically have $Y_{1}(t)-Y_{2}(t)\leq 4L+1\leq 5L$ (as
$\mathfrak{R}_{0}\subseteq[0,1]\times[-2L,2L]$) and $\nabla
F_{i}(w)\in[-\varepsilon^{-1},-\varepsilon]$, we therefore have that

(9.13) $\displaystyle\begin{aligned}
\mathbb{E}\Bigg{[}\textbf{1}_{\mathscr{A}^{\complement}}\cdot\bigg{(}\displaystyle\int_{0}^{t}\Big{\|}\bm{D}\big{(}\nabla
F_{1}(Y_{1}(s))\big{)}-\bm{D}\big{(}\nabla
F_{2}(Y_{2}(s))\big{)}\Big{\|}ds\bigg{)}^{2p}\Bigg{]}&\leq\mathbb{P}\big{[}\mathscr{A}^{\complement}\big{]}\cdot(C_{1}t)^{2p}(5B+\varepsilon^{-1})^{2p}\\\
&\leq(C_{2}t)^{2p}e^{-c_{1}M},\end{aligned}$

for some constant $C_{2}=C_{2}(\varepsilon,B)>1$, where in the last inequality
we applied (9.8).

To bound the second term on the right side of (9.12), fix some neighborhood
$\mathfrak{R}_{0}^{\prime}\subseteq\mathfrak{R}_{0}$ of $Y_{2}(t)$ with smooth
boundary such that
$[0,1]\times\big{[}Y_{2}(t)-1,Y_{2}(t)+1\big{]}\subseteq\mathfrak{R}_{0}^{\prime}\subseteq[0,1]\times\big{[}Y_{2}(t)-2,Y_{2}(t)+2\big{]}$.
On the event $\mathscr{A}$, we have (since $M=L^{1/2}$) that
$Y_{2}(t)\in[0,1]\times[x_{0}-2L^{1/2},x_{0}+2L^{1/2}]\subseteq\mathfrak{R}_{0}$,
and so we may take $\mathfrak{R}_{0}^{\prime}$ to be of some fixed shape (for
example, a rectangle of uniformly bounded height and width, with smoothened
corners) up to translation, independently of $\varepsilon$, $A$, $B$, and $L$.
Then,

(9.14) $\displaystyle\begin{aligned}
\textbf{1}_{\mathscr{A}}\cdot\Big{|}\nabla F_{1}\big{(}Y_{2}(t)\big{)}-\nabla
F_{2}\big{(}Y_{2}(t)\big{)}\Big{|}&\leq\displaystyle\sup_{w\in\mathfrak{R}_{0}^{\prime}}\big{|}\nabla
F_{1}(w)-\nabla F_{2}(w)\big{|}\\\ &\leq
C_{3}\displaystyle\sup_{w\in\partial\mathfrak{R}_{0}^{\prime}}\big{|}F_{1}(w)-F_{2}(w)\big{|}\leq
C_{3}\displaystyle\sup_{w\in\partial\mathfrak{R}_{0}}\big{|}F_{1}(w)-F_{2}(w)\big{|}\leq
C_{3}A,\end{aligned}$

for some some constant $C_{3}=C_{3}(\varepsilon,B)>1$, where in the second
inequality we applied (the $m=1$ case of) 9.4 (applied with both
$\varphi_{i}=F_{1}|_{\mathfrak{R}_{0}^{\prime}}$, which satisfy
$\varphi_{1}-\varphi_{2}=0$ and
$\|\varphi_{i}\|_{\mathcal{C}^{5}(\mathfrak{R}_{0}^{\prime})}-\|\varphi_{i}\|_{\mathcal{C}^{0}(\mathfrak{R}_{0}^{\prime})}\leq\|F_{1}\|_{\mathcal{C}^{5}(\mathfrak{R}_{0})}-\|F_{1}\|_{\mathcal{C}^{0}(\mathfrak{R}_{0})}\leq
B$), in the third we applied 9.1, and in the fourth we applied the third
assumption of the lemma. Inserting this, (9.12), and (9.13) into (9.11) gives

$\displaystyle\mathbb{E}\bigg{[}\displaystyle\sup_{s\in[0,t]}\big{|}$
$\displaystyle Y_{1}(s)-Y_{2}(s)\big{|}^{2p}\bigg{]}$
$\displaystyle\leq(72p)^{2p}\Bigg{(}(C_{2}t)^{2p}e^{-c_{1}M}+(2C_{1}C_{3}At)^{2p}+(2C_{1}BL^{-1})^{2p}\bigg{(}\displaystyle\int_{0}^{t}\big{|}Y_{1}(s)-Y_{2}(s)\big{|}ds\bigg{)}^{2p}\Bigg{)}.$

By Markov’s inequality, we deduce the existence of a constant
$C_{4}=C_{4}(\varepsilon,B)>1$ such that

$\displaystyle\mathbb{P}\Bigg{[}\displaystyle\sup_{s\in[0,t]}\big{|}Y_{1}(s)-Y_{2}(s)\big{|}\geq
C_{4}pt(A+e^{-c_{1}M/2p})+C_{4}pL^{-1}\displaystyle\int_{0}^{t}\big{|}Y_{1}(s)-Y_{2}(s)\big{|}ds\Bigg{]}\leq
4^{-p}.$

Taking $p=L^{1/8}$ gives (after replacing $c_{1}$ by $\frac{c_{1}}{2}$) with
probability at least $1-4^{-L^{1/8}}$ that

(9.15)
$\displaystyle\displaystyle\sup_{s\in[0,t]}\big{|}Y_{1}(s)-Y_{2}(s)\big{|}\leq
C_{4}L^{3/8}(A+e^{-c_{1}L^{1/8}})+C_{4}L^{-7/8}\displaystyle\int_{0}^{t}\big{|}Y_{1}(s)-Y_{2}(s)\big{|}ds,$

for any $t\in[0,L^{1/4}]$. Together with Gronwall’s inequality,101010Here, we
implicitly take a union bound so that (9.15) holds for all $t$ in a
$2^{-L^{1/8}}$-mesh of $[0,L^{1/8}]$, and apply the discrete Gronwall
inequality [39, Equation (8)]. this yields a constant
$C_{5}=C_{5}(\varepsilon,B)>1$ such that

(9.16)
$\displaystyle\mathbb{P}\bigg{[}\displaystyle\sup_{s\in[0,L^{1/2}]}\big{|}Y_{1}(s)-Y_{2}(s)\big{|}\geq
C_{5}L^{3/8}(A+e^{-c_{1}L^{1/8}})\bigg{]}\leq 2^{-L^{1/8}}.$

Since by 9.8 the process $F_{i}\big{(}Y_{i}(s)\big{)}$ is a martingale, we
have

(9.17) $\displaystyle
F_{1}(z)-F_{2}(z)=\mathbb{E}\Big{[}F_{1}\big{(}Y_{1}(\tau)\big{)}-F_{2}\big{(}Y_{2}(\tau)\big{)}\Big{]}.$

Setting $z_{0}=z-(t_{0},0)=(0,x_{0})\in\partial\mathfrak{R}_{0}$, Taylor
expanding $F_{1}$ and $F_{2}$, and using the fact that
$\big{|}\partial_{j}\partial_{k}F_{i}(w)\big{|}\leq BL^{-1}$, we find that

$\displaystyle\bigg{|}\mathbb{E}\Big{[}F_{1}\big{(}Y_{1}($
$\displaystyle\tau)\big{)}-F_{2}\big{(}Y_{2}(\tau)\big{)}\Big{]}-\mathbb{E}\Big{[}F_{1}\big{(}Y_{1}(\tau)\big{)}-F_{2}\big{(}Y_{1}(\tau)\big{)}\Big{]}-\mathbb{E}\Big{[}\nabla
F_{2}(z_{0})\cdot\big{(}Y_{1}(\tau)-Y_{2}(\tau)\big{)}\Big{]}$
$\displaystyle\quad-\mathbb{E}\bigg{[}\Big{(}\nabla
F_{2}\big{(}Y_{2}(\tau)\big{)}-\nabla
F_{2}(z_{0})\Big{)}\cdot\big{(}Y_{1}(\tau)-Y_{2}(\tau)\big{)}\bigg{]}\bigg{|}\leq\displaystyle\frac{B}{2L}\cdot\mathbb{E}\Big{[}\big{|}Y_{1}(\tau)-Y_{2}(\tau)\big{|}^{2}\Big{]}.$

Together with (9.17), the fact that
$\mathbb{E}\big{[}Y_{1}(\tau)\big{]}=z=\mathbb{E}\big{[}Y_{2}(\tau)\big{]}$
(as $Y_{1}$ and $Y_{2}$ are martingales by (9.3)), and again the bound
$\big{|}\partial_{j}\partial_{k}F_{i}(w)\big{|}\leq BL^{-1}$, this yields

$\displaystyle\big{|}F_{1}(z)-F_{2}(z)\big{|}$
$\displaystyle=\bigg{|}\mathbb{E}\Big{[}F_{1}\big{(}Y_{1}(\tau)\big{)}-F_{2}\big{(}Y_{2}(\tau)\big{)}\Big{]}\bigg{|}$
$\displaystyle\leq\bigg{|}\mathbb{E}\Big{[}F_{1}\big{(}Y_{1}(\tau)\big{)}-F_{2}\big{(}Y_{1}(\tau)\big{)}\Big{]}\bigg{|}+\displaystyle\frac{B}{L}\cdot\mathbb{E}\Big{[}\big{|}Y_{2}(\tau)-z_{0}\big{|}\cdot\big{|}Y_{2}(\tau)-Y_{1}(\tau)\big{|}\Big{]}$
$\displaystyle\qquad+\displaystyle\frac{B}{2L}\cdot\mathbb{E}\Big{[}\big{|}Y_{1}(\tau)-Y_{2}(\tau)\big{|}^{2}\Big{]},$

and hence

(9.18) $\displaystyle\begin{aligned}
\big{|}F_{1}(z)-F_{2}(z)\big{|}&\leq\bigg{|}\mathbb{E}\Big{[}F_{1}\big{(}Y_{1}(\tau)\big{)}-F_{2}\big{(}Y_{1}(\tau)\big{)}\Big{]}\bigg{|}\\\
&\qquad+\displaystyle\frac{2B}{L}\cdot\mathbb{E}\bigg{[}\Big{(}\big{|}z_{0}-Y_{1}(\tau)\big{|}+\big{|}Y_{2}(\tau)-z_{0}\big{|}\Big{)}\cdot\big{|}Y_{1}(\tau)-Y_{2}(\tau)\big{|}\bigg{]}.\end{aligned}$

Since $F_{1}\big{(}Y_{1}(\tau)\big{)}=F_{2}\big{(}Y_{1}(\tau)\big{)}$ on the
event $\mathscr{A}$ (by (9.9)), and since
$\big{|}F_{1}\big{(}Y_{1}(\tau)\big{)}-F_{2}\big{(}Y_{2}(\tau)\big{)}\big{|}\leq
10BL$ (as $\big{|}\nabla F_{1}(w)\big{|}\leq B$, $\big{|}\nabla
F_{2}(w)\big{|}\leq B$, and the diameter of $\mathfrak{R}_{0}$ is at most
$5L$), it follows from (9.8) that

(9.19)
$\displaystyle\bigg{|}\mathbb{E}\Big{[}F_{1}\big{(}Y_{1}(\tau)\big{)}-F_{2}\big{(}Y_{1}(\tau)\big{)}\Big{]}\bigg{|}=\mathbb{E}\bigg{[}\textbf{1}_{\mathscr{A}^{\complement}}\cdot\Big{|}F_{1}\big{(}Y_{1}(\tau)\big{)}-F_{2}\big{(}Y_{1}(\tau)\big{)}\Big{|}\bigg{]}\leq
10c_{1}^{-1}BLe^{-c_{1}L^{1/8}}$

Moreover, from (9.16), (9.8), and the fact that
$\big{|}Y_{i}(\tau)-z_{0}\big{|}\leq M+1\leq L^{1/2}$ for each $i\in\\{1,2\\}$
on $\mathscr{A}$ (and again the fact that the diameter of $\mathfrak{R}_{0}$
is at most $5L$), we deduce the existence of constants
$c_{2}=c_{2}(\varepsilon,B)>0$ and $C_{6}=C_{6}(\varepsilon,B)>1$ such that

(9.20) $\displaystyle\begin{aligned}
\mathbb{E}&\bigg{[}\Big{(}\big{|}Y_{1}(\tau)-z_{0}\big{|}+\big{|}Y_{2}(\tau)-z_{0}\big{|}\Big{)}\cdot\big{|}Y_{1}(\tau)-Y_{2}(\tau)\big{|}\bigg{]}\\\
&\leq 2L^{1/2}\cdot
C_{5}L^{3/8}(A+e^{-c_{1}L^{1/8}})+\big{(}c_{1}^{-1}e^{-c_{1}L^{1/8}}+2^{-L^{1/8}})\cdot
2(5L)^{2}\leq C_{6}L^{7/8}A+C_{6}e^{-c_{2}L^{1/8}},\end{aligned}$

where we have used the fact that, on $\mathscr{A}$, we have
$|Y_{1}(\tau)-z_{0}\big{|}+\big{|}Y_{2}(\tau)-z_{0}\big{|}\leq 2M=2L^{1/2}$.
Inserting (9.19) and (9.20) into (9.18) yields

$\displaystyle\big{|}F_{1}(z)-F_{2}(z)\big{|}\leq
2C_{6}BL^{-1/8}A+2C_{6}BL^{-1}e^{-c_{2}L^{1/8}}+10c_{1}^{-1}BLe^{-c_{1}L^{1/8}},$

which implies the lemma. ∎

##  Appendix A Concentration Bounds for the Height Function

In this section we establish a concentration bound for non-intersecting
Brownian bridges; it (and its proof) is analogous to the one that appears in
the context of random tilings (see [16, Theorem 21], for example). For any
family of continuous curves $\bm{x}=(x_{1},x_{2},\ldots,x_{n})\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[a,b]\big{)}$, we recall from (3.6) that
the associated height function
$\mathsf{H}=\mathsf{H}^{\bm{x}}:[a,b]\times\mathbb{R}\rightarrow\mathbb{R}$ is
defined by $\mathsf{H}(t,w)=\\#\big{\\{}j\in\llbracket
1,n\rrbracket:x_{j}(t)>w\big{\\}}$. In what follows, we also recall the
measure $\mathfrak{Q}_{f;g}^{\bm{u};\bm{v}}$ from Definition 1.1.

######  Lemma A.1.

Let $n\geq 1$ be an integer; $T>0$ and $r,B\geq 1$ be real numbers;
$\bm{u},\bm{v}\in\overline{\mathbb{W}}_{n}$ be $n$-tuples; and
$f,g:[0,T]\rightarrow\overline{\mathbb{R}}$ be measurable functions with
$f\leq g$. Sample non-intersecting Brownian bridges $\bm{x}\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[0,T]\big{)}$ from the measure
$\mathfrak{Q}_{f;g}^{\bm{u};\bm{v}}$. Fix real numbers $t\in[0,T]$ and
$w\in\big{[}f(t),g(t)\big{]}$. Denoting the event
$\mathscr{E}=\big{\\{}\mathsf{H}(t,w)\leq B\big{\\}}$, there exists a
deterministic number $\mathfrak{Y}=\mathfrak{Y}(\bm{u};\bm{v};f;g;T;t;w;B)\geq
0$ such that

(A.1)
$\displaystyle\mathbb{P}\Big{[}\big{|}\mathsf{H}(t,w)-\mathfrak{Y}\big{|}\geq
rB^{1/2}\Big{]}\leq
2e^{-r^{2}/4}+2\cdot\mathbb{P}\big{[}\mathscr{E}^{\complement}\big{]}.$

In particular, setting $B=n$, we have
$\mathbb{P}\big{[}\big{|}\mathsf{H}(t,w)-\mathfrak{Y}\big{|}\geq
rn^{1/2}\big{]}\leq 2e^{-r^{2}/4}$.

To establish A.1, we proceed through a discretization. For any integer
$\mathsf{S}\geq 0$, a _discrete path_ is a function $\mathsf{p}:\llbracket
0,\mathsf{S}\rrbracket\rightarrow\mathbb{Z}$ such that
$\mathsf{p}(t)-\mathsf{p}(t-1)\in\\{0,1\\}$, for each $t\in\llbracket
1,\mathsf{S}\rrbracket$. A _discrete path ensemble_ on $\llbracket
0,\mathsf{S}\rrbracket$ is an ordered sequence of discrete paths
$\bm{\mathsf{p}}=(\mathsf{p}_{1},\mathsf{p}_{2},\ldots,\mathsf{p}_{k})$, with
$\mathsf{p}_{j}:\llbracket 0,\mathsf{S}\rrbracket\rightarrow\mathbb{Z}$ for
each $j\in\llbracket 1,k\rrbracket$. Given integer sequences
$\bm{u}=(u_{1},u_{2},\ldots,u_{k})\in\mathbb{Z}^{k}$ and
$\bm{v}=(v_{1},v_{2},\ldots,v_{k})\in\mathbb{Z}^{k}$, we say that
$\bm{\mathsf{p}}$ _starts_ at $\bm{u}$ and _ends_ at $\bm{v}$ if
$\mathsf{p}_{j}(0)=u_{j}$ and $\mathsf{p}_{j}(\mathsf{S})=v_{j}$ for each
$j\in\llbracket 1,k\rrbracket$, respectively; observe that this is only
possible if $u_{j}\leq v_{j}\leq u_{j}+\mathsf{S}$ for each $j$. Given two
functions $f,g:\llbracket 0,\mathsf{S}\rrbracket\rightarrow\mathbb{Z}$, we
further say that the ensemble $\bm{\mathsf{p}}$ _remains above $f$_ and
_remains below $g$_ if $f(s)<\mathsf{p}_{j}(s)$ and $\mathsf{p}_{j}(s)<g(s)$
for each $(j,s)\in\llbracket 1,n\rrbracket\times\llbracket
0,\mathsf{S}\rrbracket$, respectively. See the left side of Figure 11 for a
depiction.

Figure 11. Shown to the left is a discrete path ensemble
${\bm{\mathsf{p}}}=({\mathsf{p}}_{1},{\mathsf{p}}_{2},{\mathsf{p}}_{3})$.
Shown to the right is a level set decomposition
$\mathscr{D}=\mathcal{S}_{0}\cup\mathcal{S}_{1}\cup\cdots\cup\mathcal{S}_{3}$;
there, $(3,2,0)$ is an exit sequence of $\mathcal{S}_{3}$.

Denoting
$\bm{\mathsf{p}}(t)=\big{(}\mathsf{p}_{1}(t),\mathsf{p}_{2}(t),\ldots,\mathsf{p}_{k}(t)\big{)}$
for any $t$, we additionally say that the ensemble $\bm{\mathsf{p}}$ is _non-
intersecting_ if $\bm{\mathsf{p}}(t)\in\mathbb{W}_{k}$, for each
$t\in\llbracket 0,\mathsf{S}\rrbracket$. Given a non-intersecting path
ensemble
$\bm{\mathsf{p}}=(\mathsf{p}_{1},\mathsf{p}_{2},\ldots,\mathsf{p}_{k})$ on
$\llbracket 0,\mathsf{S}\rrbracket$, we define its height function
$\mathsf{H}^{\bm{\mathsf{p}}}:\llbracket
0,\mathsf{S}\rrbracket\times\mathbb{Z}\rightarrow\mathbb{Z}_{\geq 0}$ by, for
any $(t,w)\in\llbracket 0,\mathsf{S}\rrbracket\times\mathbb{Z}$, setting

$\displaystyle\mathsf{H}^{\bm{\mathsf{p}}}(t,w)=\\#\big{\\{}j\in\llbracket
1,k\rrbracket:\mathsf{p}_{j}(t)>w\big{\\}}.$

Given the above notation, we can state the discrete variant of A.1 that
quickly implies it. Observe that, in the below concentration bound, the error
estimates and probabilities do not depend on the size $\mathsf{S}$ of the
domain; this is in contrast to analogous statements in the context of tilings
(for example, in [16, Theorem 21], the error in the height function grows at
least as $\mathsf{S}^{1/2}$), though the proof is similar.

######  Lemma A.2.

Let $n,\mathsf{S}\geq 1$ be integers; $r,B\geq 1$ be real numbers;
$\bm{u},\bm{v}\in\mathbb{W}_{n}\cap\mathbb{Z}^{n}$ be two $n$-tuples of
integers; and $f,g:\llbracket 0,\mathsf{S}\rrbracket\rightarrow\mathbb{Z}$ be
functions with $f\leq g$. Let
$\bm{\mathsf{p}}=(\mathsf{p}_{1},\mathsf{p}_{2},\ldots,\mathsf{p}_{n})$ denote
a uniformly random discrete path ensemble conditioned to start at $\bm{u}$;
end at $\bm{v}$; remain above $f$ and below $g$; and to be non-intersecting
(assuming at least one such ensemble exists). Fixing $(t,w)\in\llbracket
0,\mathsf{S}\rrbracket\times\mathbb{Z}$, and denoting the event
$\mathscr{E}(\bm{\mathsf{p}})=\big{\\{}\mathsf{H}^{\bm{\mathsf{p}}}(t,w)\leq
B\big{\\}}$, there exists a deterministic real number
$\mathfrak{Y}=\mathfrak{Y}(\bm{u};\bm{v};f;g;t;w;B)\in\llbracket
0,n\rrbracket$ such that

$\displaystyle\mathbb{P}\Big{[}\big{|}\mathsf{H}^{\bm{\mathsf{p}}}(t,w)-\mathfrak{Y}\big{|}\geq
rB^{1/2}\Big{]}\leq
2e^{-r^{2}/4}+2\cdot\mathbb{P}\big{[}\mathscr{E}(\bm{\mathsf{p}})^{\complement}\big{]}.$

###### Proof.

This proof follows the discussion in [3, Section 3.2]. Let
$\bm{\mathsf{p}}^{\prime}=(\mathsf{p}_{1}^{\prime},\mathsf{p}_{2}^{\prime},\ldots,\mathsf{p}_{n}^{\prime})$
denote a non-intersecting discrete path ensemble with the same law as, but
independent from, $\bm{\mathsf{p}}$. Denote
$\mathscr{D}=\mathscr{D}(\mathsf{S})=\llbracket
0,\mathsf{S}\rrbracket\times\mathbb{Z}\subset\mathbb{Z}^{2}$; for any subset
$\mathcal{S}\subseteq\mathscr{D}$, the boundary $\partial\mathcal{S}$ of
$\mathcal{S}$ denotes the set of vertices $(s,y)\in\mathcal{S}$ that are
adjacent111111Here, two vertices $(s,y)$ and $(s^{\prime},y^{\prime})$ are
adjacent if
$(s-s^{\prime},y-y^{\prime})\in\big{\\{}(0,1),(1,0),(1,1),(0,-1),(-1,0),(-1,-1)\big{\\}}$
(in this way, we view vertices as on a triangular lattice, instead of on a
square one). to a vertex in $\mathbb{Z}^{2}\setminus\mathcal{S}$. Define the
function $F:\mathscr{D}\rightarrow\mathbb{Z}$ by setting
$F(s,y)=\mathsf{H}^{\bm{\mathsf{p}}}(s,y)-\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(s,y)$,
for each $(s,y)\in\mathscr{D}$.

A family
$\bm{\mathcal{S}}=(\mathcal{S}_{0},\mathcal{S}_{2},\ldots,\mathcal{S}_{k})$ of
mutually disjoint subsets of $\mathscr{D}$ satisfying
$\bigcup_{j=0}^{k}\mathcal{S}_{j}=\mathscr{D}$ is a _level set decomposition_
of $\mathscr{D}$ with respect to $F$ if the following two properties hold.
First, for any two vertices $(s,y),(s^{\prime},y^{\prime})\in\mathcal{S}_{j}$
of the same subset, $F(s,y)=F(s^{\prime},y^{\prime})$. Second, for any two
adjacent vertices $(s,y)\in\mathcal{S}_{j}$ and
$(s^{\prime},y^{\prime})\in\mathcal{S}_{j^{\prime}}$ with $j\neq j^{\prime}$
(namely, in different subsets), $F(s,y)\neq F(s^{\prime},y^{\prime})$; observe
in this case that $\big{|}F(s,y)-F(s^{\prime},y^{\prime})\big{|}=1$, as we
either have that both
$\mathsf{H}^{\bm{\mathsf{p}}}(s,y)-\mathsf{H}^{\bm{\mathsf{p}}}(s^{\prime},y^{\prime})\in\\{0,1\\}$
and
$\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(s,y)-\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(s^{\prime},y^{\prime})\in\\{0,1\\}$
hold or that both
$\mathsf{H}^{\bm{\mathsf{p}}}(s,y)-\mathsf{H}^{\bm{\mathsf{p}}}(s^{\prime},y^{\prime})\in\\{0,-1\\}$
and
$\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(s,y)-\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(s^{\prime},y^{\prime})\in\\{0,-1\\}$
hold. We set $F(\mathcal{S}_{j})=F(s,y)$, for any $(s,y)\in\mathcal{S}_{j}$.
In the below, we condition on the level set decomposition $\bm{\mathcal{S}}$
of $\mathscr{D}$ with respect to $F$.

Further observe that $F(s,y)=0$ for any $(s,y)\in\llbracket
0,\mathsf{S}\rrbracket\times\mathbb{Z}$ that either satisfies
$(s,y)\in\partial\mathscr{D}$ (since $\bm{\mathsf{p}}$ and
$\bm{\mathsf{p}}^{\prime}$ both start and end at the same $n$-tuples $\bm{u}$
and $\bm{v}$, respectively), or that satisfies $y<f(s)$ (as then
$\mathsf{H}^{\bm{\mathsf{p}}}(s,y)=n=\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(s,y)$)
or $y\geq g(s)$ (as then
$\mathsf{H}^{\bm{\mathsf{p}}}(s,y)=0=\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(s,y)$).
Since this family of vertices is connected, there is one infinite set in
$\bm{\mathcal{S}}$, which we let be $\mathcal{S}_{0}$. It has the property
that $\mathscr{D}\setminus\mathcal{S}_{0}$ is finite, so the number of
elements $k+1$ in $\bm{\mathcal{S}}$ is finite.

Next, given $i,j\in\llbracket 1,n\rrbracket$ with $i\neq j$, we say that
$\mathcal{S}_{i}$ is _adjacent_ to $\mathcal{S}_{j}$ if there exists a vertex
in $\mathcal{S}_{i}$ that is adjacent to a vertex in $\mathcal{S}_{j}$. In
this case, we say that $\mathcal{S}_{i}$ is _exterior adjacent_ to
$\mathcal{S}_{j}$ if $\mathcal{S}_{i}$ is contained in the infinite connected
component of $\mathscr{D}\setminus\mathcal{S}_{j}$; we otherwise call
$\mathcal{S}_{i}$ _interior adjacent_ to $\mathcal{S}_{j}$. In particular,
$\mathcal{S}_{0}$ is exterior adjacent to any $\mathcal{S}_{j}$ adjacent to
it. It is quickly verified (see [3, Lemma 12]) that, for every $j\in\llbracket
1,k\rrbracket$, there is at most one element of $\bm{\mathcal{S}}$ that is
exterior adjacent to $\mathcal{S}_{j}$. See the right side of Figure 11 for a
depiction.

An _exit sequence_ of any $\mathcal{S}_{j}\in\bm{\mathcal{S}}$ is an ordered
sequence $(i_{0},i_{1},\ldots,i_{m})$ of integers in $\llbracket
0,k\rrbracket$ so that $i_{0}=j$, $i_{m}=0$, and $\mathcal{S}_{i_{a}}$ is
exterior adjacent to $\mathcal{S}_{i_{a-1}}$, for each $a\in\llbracket
1,m\rrbracket$; here, $m=m(j)=m(j;\bm{\mathcal{S}})$ depends on $j$ and
$\bm{\mathcal{S}}$. Then, it is quickly verified (see the proof of [3, Lemma
14]) that $F(\mathcal{S}_{j})$ has the same law as $\sum_{a=1}^{m}\xi_{a}$,
where the $\xi_{a}$ are mutually independent $\\{-1,1\\}$-Bernoulli random
variables with $\mathbb{P}[\xi_{a}=1]=\frac{1}{2}=\mathbb{P}[\xi_{a}=-1]$.
Hence, by Chernoff’s inequality,

(A.2)
$\displaystyle\mathbb{P}\bigg{[}\Big{|}F(\mathcal{S}_{j})-\mathbb{E}\big{[}F(\mathcal{S}_{j})\big{]}\Big{|}\geq
rm^{1/2}\bigg{|}\bm{\mathcal{S}}\bigg{]}\leq 2e^{-r^{2}/2}.$

Now fix $j\in\llbracket 0,m\rrbracket$ to be such that
$(t,w)\in\mathcal{S}_{j}$. Observe on the event
$\mathscr{E}=\mathscr{E}(\bm{\mathsf{p}})\cap\mathscr{E}(\bm{\mathsf{p}}^{\prime})=\big{\\{}\mathsf{H}^{\bm{\mathsf{p}}}(t,w)\leq
B\big{\\}}\cap\big{\\{}\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(t,w)\leq
B\big{\\}}$ that $m(j;\bm{\mathcal{S}})\leq 2B$. Indeed, define a sequence
$(w_{0},w_{1},\ldots,w_{m})$ by first setting $w_{0}=w$ and then defining
$w_{i+1}$ for $i\in\llbracket 0,m-1\rrbracket$ inductively as follows. Let
$j_{i}\in\llbracket 0,k\rrbracket$ be such that
$(t,w_{i})\in\mathcal{S}_{j_{i}}$ (so $j_{0}=j$), and let $w_{i+1}>w_{i}$ be
the maximum integer such that $(t,w_{i+1})\notin\mathcal{S}_{j_{i}}$ but
$(t,w_{i+1}-1)\in\mathcal{S}_{j_{i}}$ (assuming $j_{i}\neq 0$; if instead
$j_{i}=0$ then the sequence stops). Then, $(j_{0},j_{1},\ldots,j_{m})$ is an
exit sequence for $\mathcal{S}_{j}$. Moreover, for each $i\in\llbracket
0,m-1\rrbracket$, we either have
$\mathsf{H}^{\bm{\mathsf{p}}}(t,w_{i+1})=\mathsf{H}^{\bm{\mathsf{p}}}(t,w_{i})-1$
or
$\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(t,w_{i+1})=\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(t,w_{i})-1$,
since $F(t,w_{i-1})\neq F(t,w_{i})$ and since
$\mathsf{H}^{\bm{\mathsf{p}}}(t,y)$ and
$\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}(t,y)$ are nonincreasing in $y$. On the
event $\mathscr{E}$, at most $B$ such decrements can occur for
$\mathsf{H}^{\bm{\mathsf{p}}}$ and for
$\mathsf{H}^{\bm{\mathsf{p}}^{\prime}}$, meaning that $m\leq 2B$.

Combining this estimate $\textbf{1}_{\mathscr{E}}\cdot
m(j;\bm{\mathcal{S}})\leq 2B$ with (A.2) and a union bound, we obtain

$\displaystyle\mathbb{P}\bigg{[}\Big{|}F(\mathcal{S}_{j})-\mathbb{E}\big{[}F(\mathcal{S}_{j})\big{]}\Big{|}\geq
r(2B)^{1/2}\bigg{]}$
$\displaystyle\leq\mathbb{P}\bigg{[}\textbf{1}_{\mathscr{E}}\cdot\Big{|}F(\mathcal{S}_{j})-\mathbb{E}\big{[}F(\mathcal{S}_{j})\big{]}\Big{|}\geq
r(2B)^{1/2}\bigg{|}\bm{\mathcal{S}}\bigg{]}+\mathbb{P}\big{[}\mathscr{E}^{\complement}\big{]}$
$\displaystyle\leq
2e^{-r^{2}/2}+2\cdot\mathbb{P}\big{[}\mathscr{E}(\bm{\mathsf{p}})^{\complement}\big{]},$

which gives the lemma. ∎

###### Proof of A.1.

The second statement follows from the first, since $\mathsf{H}(t,w)\leq n$
holds deterministically. The first statement follows from A.2 and the fact
that non-intersecting random discrete paths converge to non-intersecting
Brownian motions (namely, the
$\big{(}T^{1/2}\mathsf{S}^{-1/2}\cdot\mathsf{p}_{j}(\lfloor
sT^{-1}\mathsf{S}\rfloor)\big{)}$ from A.2 over $(j,s)\in\llbracket
1,n\rrbracket\times[0,T]$ converges in law to the $\big{(}x_{j}(s)\big{)}$
over $(j,s)\in\llbracket 1,n\rrbracket\times[0,T]$ from A.1). ∎

##  Appendix B Proofs of Results From Section 2

###  B.1. Proof of 2.7

In this section we establish 2.7. We begin with the following lemma that
bounds non-intersecting Brownian bridges constrained to lie below a linear
lower boundary; see the left side of Figure 12 for a depiction. In the below,
we let $\bm{0}_{n}=(0,0,\ldots,0)\in\overline{\mathbb{W}}_{n}$ (where $0$
appears with multiplicity $n$).

Figure 12. The left panel depicts B.1. The right panel depicts its proof by
coupling $\bm{x}$ with the non-intersecting Brownian bridges $\bm{z}$.

######  Lemma B.1.

There exist constants $c>0$ and $C>1$ such that the following holds. Fix an
integer $n\geq 1$; real numbers $A\geq 0$, $T>0$, and $v\geq AT$; and define
$g:[0,T]\rightarrow\mathbb{R}$ by setting $g(x)=-Ax$, for each $x\in[0,T]$.
Let $\bm{v}=(-v,-v,\ldots,-v)\in\overline{\mathbb{W}}_{n}$ (where $-v$ appears
with multiplicity $n$), and sample non-intersecting Brownian bridges
$\bm{x}=(x_{1},x_{2},\ldots,x_{n})\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[0,T]\big{)}$ under the measure
$\mathfrak{Q}_{-\infty;g}^{\bm{0}_{n};\bm{v}}$. Then, for any real numbers
$t\in[0,T]$ and $B\geq 1$, we have

$\displaystyle\mathbb{P}\Big{[}x_{n}(t)\leq-
vtT^{-1}-Bt^{1/2}\log|9t^{-1}T|\Big{]}\leq Ce^{Cn-cB^{2}n}.$

###### Proof.

In view of the invariance of non-intersecting Brownian bridges under affine
transformations and diffusive scaling (recall 2.1 and 2.2), we may assume that
$T=1$. Then, set $\theta=\frac{B}{8}\cdot t^{1/2}\log|9t^{-1}|$ and
$v^{\prime}=(t^{-1}+1)\theta+v$; denote
$\bm{u}=(-\theta,-\theta,\ldots,-\theta)\in\overline{\mathbb{W}}_{n}$ and
$\bm{v}^{\prime}=(-v^{\prime},-v^{\prime},\ldots,-v^{\prime})\in\overline{\mathbb{W}}_{n}$.
Sample non-intersecting Brownian bridges
$\bm{y}=(y_{1},y_{2},\ldots,y_{n})\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[0,T]\big{)}$ and
$\bm{z}=(z_{1},z_{2},\ldots,z_{n})\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[0,T]\big{)}$ from the measures
$\mathfrak{Q}_{-\infty;g}^{\bm{u};\bm{v}^{\prime}}$ and
$\mathfrak{Q}^{\bm{u};\bm{v}^{\prime}}$, respectively; see the right side of
Figure 12. By 2.4, we may couple $\bm{x}$ and $\bm{y}$ so that

(B.1)
$\displaystyle\big{|}y_{j}(s)-x_{j}(s)\big{|}\leq(t^{-1}s+1)\theta,\qquad\text{for
each $(j,s)\in\llbracket 1,n\rrbracket\times[0,1]$}.$

Now, let us compare $\bm{y}$ and $\bm{z}$. To that end, first observe from 2.6
that there exist constants $c\in(0,1)$ and $C_{1}>1$ such that

(B.2) $\displaystyle\begin{aligned}
&\mathbb{P}\Bigg{[}\bigcap_{j=1}^{n}\bigcap_{s\in[0,1]}\Big{\\{}z_{j}(s)\geq-\displaystyle\frac{B}{8}\cdot
s^{1/2}\log|9s^{-1}|-(t^{-1}s+1)\theta-vs\Big{\\}}\Bigg{]}\geq
1-C_{1}e^{C_{1}n-cB^{2}n};\\\
&\mathbb{P}\Bigg{[}\bigcap_{j=1}^{n}\bigcap_{s\in[0,1]}\Big{\\{}z_{j}(s)\leq\displaystyle\frac{B}{8}\cdot
s^{1/2}\log|9s^{-1}|-(t^{-1}s+1)\theta-vs\Big{\\}}\Bigg{]}\geq
1-C_{1}e^{C_{1}n-cB^{2}n}.\end{aligned}$

Observe that $\frac{B}{8}\cdot s^{1/2}\log|9s^{-1}|-(t^{-1}s+1)\theta-vs\leq-
As$, for each $s\in[0,1]$. Indeed, if $s\leq t$, this follows from the facts
that $v\geq A$ and $t^{1/2}\log|9t^{-1}|\geq s^{1/2}\log|9s^{-1}|$ when
$0<s\leq t\leq 1$; for $s>t$, this follows from the facts that $v\geq A$ and
$t^{-1/2}\log|9t^{-1}|\geq s^{-1/2}\log|9s^{-1}|$ when $0<t\leq s\leq 1$.
Hence, the second statement of (B.2) indicates that $z_{j}(s)$ remains below
$g$ with probability at least $1-C_{1}e^{C_{1}n-cB^{2}}$. It follows that we
may couple $\bm{z}$ and $\bm{y}$ to coincide on an event of this probability
$1-C_{1}e^{C_{1}n-cB^{2}}$. Together with (B.1), the first part of (B.2), and
a union bound, this gives

$\displaystyle\mathbb{P}\Big{[}x_{n}(t)\geq-4\theta-
vt-\displaystyle\frac{B}{8}\cdot t^{1/2}\log|9t^{-1}|\Big{]}\geq
1-2C_{1}e^{C_{1}n-cB^{2}n},$

which yields the lemma, since $\theta=\frac{B}{8}\cdot t^{1/2}\log|9t^{-1}|$.
∎

We then deduce the following variant of 2.7 that proves high probability
Hölder bounds for a single choice of $(j,s,t)$, which quickly implies 2.7. We
detail the proof of the first part and outline that of the second, which is
similar.

######  Lemma B.2.

Adopt the notation of 2.7.

  1. (1)

Under the assumption of part 1 of 2.7, we have

$\displaystyle\mathbb{P}\Big{[}\big{|}x_{j}(t+s)-x_{j}(t)-sT^{-1}(v_{j}-u_{j})\big{|}\geq
s^{1/2}\big{(}B\log|2s^{-1}T|+\kappa^{-1}(A+B)\big{)}\Big{]}\leq
Ce^{Cn^{2}-cB^{2}n}.$

  2. (2)

Under the assumption of part 2 of 2.7, we have

$\displaystyle\mathbb{P}\Big{[}\big{|}x_{j}(t+s)-x_{j}(t)-sT^{-1}(v_{j}-u_{j})\big{|}\geq
s^{1/2}\big{(}B\log|9s^{-1}T|+2\big{(}A+B)\big{)}\Big{]}\leq
Ce^{Cn^{2}-cB^{2}n}.$

###### Proof of B.2(1).

By 2.2 and 2.1, we may assume that $(a,b)=(0,1)$ and that
$(u_{j},v_{j})=(0,0)$. Since $(sT^{-1})^{1/2}\leq 1$ (as $s\leq T$), it
suffices to show that

(B.3) $\displaystyle\begin{aligned} &\mathbb{P}\big{[}x_{j}(t+s)-x_{j}(t)\geq
Bs^{1/2}\log|2s^{-1}|+\kappa^{-1}s(A+B)\big{]}\leq Ce^{Cn^{2}-cB^{2}n};\\\
&\mathbb{P}\big{[}x_{j}(t+s)-x_{j}(t)\leq-
Bs^{1/2}\log|2s^{-1}|-\kappa^{-1}s(A+B)\big{]}\leq
Ce^{Cn^{2}-cB^{2}n}.\end{aligned}$

We only establish the first bound in (B.3), as then that of the second would
follow by symmetry (namely, by reflecting the bridges into the line
$\big{\\{}t=\frac{1}{2}\big{\\}}$). To that end, we first claim that there
exist constants $c_{1}\in(0,1)$ and $C_{1}>1$ such that

(B.4)
$\displaystyle\mathbb{P}\Bigg{[}\displaystyle\sup_{r\in[0,1]}x_{j}(r)\leq
A+B\Bigg{]}\geq 1-C_{1}e^{C_{1}n-c_{1}B^{2}n}.$

To see this, define
$\bm{u}^{\prime},\bm{v}^{\prime}\in\overline{\mathbb{W}}_{n}$ by setting

$\displaystyle u_{k}^{\prime}=u_{k}+\displaystyle\frac{B}{2}+A;\qquad
v_{k}^{\prime}=v_{k}+\displaystyle\frac{B}{2}+A,$

for each $k\in\llbracket 1,n\rrbracket$. Then, sample non-intersecting
Brownian bridges $\bm{y}=(y_{1},y_{2},\ldots,y_{n})\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[0,1]\big{)}$ and
$\bm{y}^{\prime}=(y_{1}^{\prime},y_{2}^{\prime},\ldots,y_{n}^{\prime})\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[0,1]\big{)}$ according to the measures
$\mathfrak{Q}_{f}^{\bm{u}^{\prime};\bm{v}^{\prime}}$ and
$\mathfrak{Q}^{\bm{u}^{\prime};\bm{v}^{\prime}}$, respectively. Then, by 2.3,
we may couple $\bm{x}$ and $\bm{y}$ so that

(B.5) $\displaystyle x_{k}(r)\leq y_{k}(r),\qquad\text{for each
$(k,r)\in\llbracket 1,n\rrbracket\times[0,1]$}.$

Moreover, by 2.6, there exist constants $c_{1}\in(0,1)$ and $C_{2}>1$ such
that

(B.6) $\displaystyle\begin{aligned}
&\mathbb{P}\Bigg{[}\displaystyle\inf_{r\in[0,1]}y_{n}^{\prime}(r)\geq\displaystyle\sup_{r\in[0,1]}f(r)+\displaystyle\frac{B}{4}\Bigg{]}\geq
1-C_{2}e^{C_{2}n-c_{1}B^{2}n};\\\
&\mathbb{P}\Bigg{[}\displaystyle\sup_{r\in[0,1]}y_{j}^{\prime}(r)\leq
A+B\Bigg{]}\geq 1-C_{2}e^{C_{2}n-c_{1}B^{2}n},\end{aligned}$

where in the first inequality we used that
$\min\\{u_{n}^{\prime},v_{n}^{\prime}\\}=\min\\{u_{n},v_{n}\\}+A+\frac{B}{2}\geq\sup_{r\in[0,1]}f(r)+\frac{B}{2}$,
and in the second inequality we used the fact that
$u_{j}^{\prime}=A+\frac{B}{2}=v_{j}^{\prime}$ (recalling that
$u_{j}=v_{j}=0$).

The first bound in (B.6) indicates that $\bm{y}^{\prime}$ lies above $f$ with
probability at least $1-C_{2}e^{C_{2}n-c_{1}B^{2}n}$. Hence, we may couple
$\bm{y}$ and $\bm{y}^{\prime}$ so that they coincide with probability at least
$1-C_{2}e^{C_{2}n-c_{1}B^{2}n}$. This, together with the second bound in
(B.6), (B.5), and a union bound then yields (B.4).

Figure 13. The left panel depicts the proof of B.2(1). The right panel depicts
that of B.2(2).

Now, to show the first bound in (B.3), we condition on $x_{k}(r)$ for
$(k,r)\in\llbracket j+1,n\rrbracket\times[0,1]$ and for $(k,r)\in\llbracket
1,n\rrbracket\times[t+s,1]$, and set
$\bm{u}^{\prime\prime},\bm{v}^{\prime\prime}\in\overline{\mathbb{W}}_{n-j+1}$
by letting $u_{k}^{\prime\prime}=x_{j}(0)=0$ and
$v_{k}^{\prime\prime}=x_{j}(t+s)$, for each $k\in\llbracket 1,j\rrbracket$.
Sample non-intersecting Brownian bridges
$\bm{z}=(z_{1},z_{2},\ldots,z_{j})\in\llbracket
1,j\rrbracket\times\mathcal{C}\big{(}[0,t+s]\big{)}$ from the measure
$\mathfrak{Q}^{\bm{u}^{\prime\prime};\bm{v}^{\prime\prime}}(n^{-1})$. Then,
2.3 yields a coupling between $\bm{x}$ and $\bm{z}$ such that $x_{k}(r)\geq
z_{k}(r)$, for each $(k,r)\in\llbracket 1,j\rrbracket\times[0,1]$; see the
left side of Figure 13. Thus,

$\displaystyle\mathbb{P}\bigg{[}x_{j}(t)\leq x_{j}($ $\displaystyle
t+s)-Bs^{1/2}\log|2s^{-1}|-\displaystyle\frac{s(A+B)}{\kappa}\bigg{]}$
$\displaystyle\leq\mathbb{P}\bigg{[}z_{j}(t)\leq
z_{j}(t+s)-Bs^{1/2}\log|2s^{-1}|-\displaystyle\frac{s(A+B)}{\kappa}\bigg{]}\leq
2C_{1}e^{C_{1}n-c_{1}B^{2}n},$

where in the last equality we used 2.6 and the facts that $s+t\geq
t\geq\kappa$ and that

$\displaystyle\displaystyle\frac{v_{j}^{\prime\prime}-u_{j}^{\prime\prime}}{s+t}=(s+t)^{-1}\cdot
z_{j}(t+s)\leq(s+t)^{-1}\cdot\max_{r\in[0,1]}x_{j}(r)\leq\displaystyle\frac{A+B}{\kappa},$

on the event $\big{\\{}\max_{r\in[0,1]}x_{j}(r)\leq A+B\big{\\}}$. This, with
(B.4), establishes the first bound in (B.3) and thus the first statement of
the lemma. ∎

###### Proof of B.2(2).

Since the proof of the second part of the lemma is similar to that of the
first part, we only outline it. We may again assume that $(a,b)=(0,1)$ and
$(u_{j},v_{j})=(0,0)$. We may further assume by symmetry that
$t\leq\frac{1}{2}$. It suffices (again since $(sT^{-1})^{1/2}\leq 1$) to show
that

(B.7) $\displaystyle\begin{aligned} &\mathbb{P}\big{[}x_{j}(t+s)-x_{j}(t)\geq
Bs^{1/2}\log|9s^{-1}|+2s(A+B)\big{]}\leq Ce^{Cn^{2}-cB^{2}n};\\\
&\mathbb{P}\big{[}x_{j}(t+s)-x_{j}(t)\leq-
Bs^{1/2}\log|9s^{-1}|-2s(A+B)\big{]}\leq Ce^{Cn^{2}-cB^{2}n},\end{aligned}$

and by symmetry it suffices to show the latter. Following the derivation of
(B.4), we again find that there exist constants $c_{1}\in(0,1)$ and $C_{1}>1$
such that

(B.8)
$\displaystyle\mathbb{P}\Bigg{[}\displaystyle\sup_{r\in[0,1]}x_{j}(r)\leq
A+B\Bigg{]}\geq 1-C_{1}e^{C_{1}n-c_{1}B^{2}n}.$

Now condition on $x_{k}(r)$ for $(k,r)\in\llbracket
j+1,n\rrbracket\times[0,1]$ and for $(k,r)\in\llbracket
1,n\rrbracket\times[0,t]$; set $\widetilde{g}:[t,1]\rightarrow\mathbb{R}$ by
letting $\widetilde{g}(r)=x_{j}(t)+(t-r)A$, for each $r\in[t,1]$; and set
$\bm{u}^{\prime},\bm{v}^{\prime}\in\overline{\mathbb{W}}_{j}$ by letting
$u_{k}^{\prime}=x_{j}(t)$ and
$v_{k}^{\prime}=\min\\{x_{j}(1),\widetilde{g}(1)\\}=\min\\{0,\widetilde{g}(1)\\}$
(where we used that $x_{j}(1)=v_{j}=0$) for each $k\in\llbracket
1,j\rrbracket$. Then sample non-intersecting Brownian bridges
$\bm{y}=(y_{1},y_{2},\ldots,y_{j})\in\llbracket
1,j\rrbracket\times\mathcal{C}\big{(}[t,1]\big{)}$ from the measure
$\mathfrak{Q}_{-\infty;\tilde{g}}^{\bm{u}^{\prime};\bm{v}^{\prime}}(n^{-1})$;
see the right side of Figure 13. Since $\big{|}\partial_{t}g(r)\big{|}\leq A$
for each $r\in[t,1]$, and since $g(t)\geq x_{j}(t)=u_{j}^{\prime}$, we have
$g\geq\widetilde{g}$. Thus, 2.3 yields a coupling between $\bm{x}$ and
$\bm{y}$ such that

(B.9) $\displaystyle x_{k}(r)\geq y_{k}(r),\qquad\text{for each
$(k,r)\in\llbracket 1,j\rrbracket\times[t,1]$}.$

Applying B.1 to $\bm{y}$ yields constants $c_{1}\in(0,1)$ and $C_{2}>1$ such
that

(B.10) $\displaystyle\mathbb{P}\bigg{[}y_{j}(t+s)\geq
y_{j}(t)-\displaystyle\frac{(u_{j}^{\prime}-v_{j}^{\prime})s}{1-t}-Bs^{1/2}\log|9s^{-1}|\bigg{]}\geq
1-C_{2}e^{C_{2}n-c_{1}B^{2}n}.$

If $v_{j}^{\prime}=\widetilde{g}(1)=u_{j}^{\prime}-A(1-t)$, then
$u_{j}^{\prime}-v_{j}^{\prime}=A(1-t)\leq A$. If instead $v_{j}^{\prime}=0$,
and (B.8) implies $\mathbb{P}[u_{j}^{\prime}\leq A+B]\geq
1-C_{1}e^{C_{1}n-c_{1}B^{2}n}$; since $t\leq\frac{1}{2}$, we have in either
case that

$\displaystyle\mathbb{P}\bigg{[}\displaystyle\frac{u_{j}^{\prime}-v_{j}^{\prime}}{1-t}\leq
2(A+B)\bigg{]}\geq 1-C_{1}e^{C_{1}n-cB^{2}n}.$

Together with (B.10), (B.9), and a union bound, this gives the second bound in
(B.7), which implies the second statement of the lemma. ∎

###### Proof of 2.7 (Outline).

The proof follows quickly from B.2 and (the proof of) [19, Lemma 3.3], very
similarly to the proof of [19, Proposition 3.5] given [19, Lemmas 3.2 and
3.3], so further details are omitted. ∎

###  B.2. Proofs of 2.10, and 2.11

In this section we establish first 2.10, and next 2.11.

###### Proof of 2.10.

By [8, Lemma 3], we have $\big{|}m_{t}^{\mu}(z)\big{|}\leq t^{-1/2}$ for any
$z\in\overline{\Lambda}_{t;\mu}$. By (2.6), this yields
$\varrho_{t}(x)\leq\pi^{-1}t^{-1/2}$, for any $x\in\mathbb{R}$, confirming the
first statement of the lemma.

To verify the latter, fix $x\in\mathbb{R}$ and set $z\in\partial\Lambda_{t}$
such that $M(z)=z-tm_{0}(z)=x\in\mathbb{R}$ (where such a $z$ exists by 2.8).
Then, (2.6) and (2.11) give

(B.11)
$\displaystyle\varrho_{t}(x)=\pi^{-1}\operatorname{Im}m_{t}\big{(}z-tm_{0}(z)\big{)}=\pi^{-1}\operatorname{Im}m_{0}(z).$

Let $z_{0}\in\partial\Lambda_{t}$ be such that $M(z_{0})=x_{0}$. Repeatedly
applying the relation
$\partial_{x}=\big{(}\frac{\partial_{z}}{\partial_{x}}\big{)}^{-1}\partial_{z}=(1-tm_{0}^{\prime})^{-1}\partial_{z}$,
it follows that there exists a constant $C=C(k)>1$ such that

(B.12) $\displaystyle\begin{aligned}
\big{|}\partial_{x}^{k}\varrho_{t}(x_{0})\big{|}&\leq
C\cdot\displaystyle\max_{k^{\prime}\in\llbracket
1,k\rrbracket}\big{|}\partial_{z}^{k^{\prime}}m_{0}(z_{0})\big{|}^{2k/k^{\prime}}\cdot\displaystyle\max_{k^{\prime\prime}\in\llbracket
0,2k\rrbracket}\big{|}1-tm_{0}^{\prime}(z_{0})\big{|}^{-k^{\prime\prime}}.\end{aligned}$

To bound the first term on the right side, observe since
$|z_{0}-y|\geq\operatorname{Im}z_{0}$ for any $y\in\mathbb{R}$ that

(B.13) $\displaystyle\big{|}\partial_{z}^{k}m_{0}(z_{0})\big{|}\leq
k!\displaystyle\int_{-\infty}^{\infty}\frac{\varrho_{0}(y)dy}{|z_{0}-y|^{k+1}}\leq\frac{k!}{(\operatorname{Im}z_{0})^{k-1}}\displaystyle\int_{-\infty}^{\infty}\frac{\varrho_{0}(y)dy}{|z_{0}-y|^{2}}=k!\cdot\frac{\operatorname{Im}m_{0}(z_{0})}{(\operatorname{Im}z_{0})^{k}}\leq\frac{k!}{t(t\delta\pi)^{k-1}},$

where in the last two statements we used the facts that

(B.14) $\displaystyle\begin{aligned}
&\qquad\operatorname{Im}m_{0}(z_{0})=\pi\varrho_{t}(x_{0})\geq\pi\delta;\qquad\operatorname{Im}z_{0}=t\operatorname{Im}m_{0}(z_{0});\\\
&\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\varrho_{0}(y)dy}{|z_{0}-y|^{2}}=\displaystyle\frac{1}{\operatorname{Im}z_{0}}\cdot\operatorname{Im}\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\varrho_{0}(y)dy}{z_{0}-y}=\displaystyle\frac{\operatorname{Im}m_{0}(z_{0})}{\operatorname{Im}z_{0}}=t^{-1},\end{aligned}$

Here, the first statement follows from (B.11) and the assumption
$\varrho_{t}(x_{0})\geq\delta$; and the second follows from the fact that
$M(z_{0})=z_{0}-tm_{0}(z_{0})=x_{0}\in\mathbb{R}$; and the third follows from
the definition of $m_{0}$.

To estimate the last term on the right side of (B.12), observe that

(B.15) $\displaystyle\big{|}1-tm^{\prime}_{0}(z_{0})\big{|}$
$\displaystyle\geq 1-t\operatorname{Re}m^{\prime}_{0}(z_{0}),$

and that

$\displaystyle\operatorname{Re}m^{\prime}_{0}(z_{0})=\operatorname{Re}\displaystyle\int_{-\infty}^{\infty}\frac{\varrho_{0}(y)dy}{(y-z_{0})^{2}}$
$\displaystyle=\displaystyle\int_{-\infty}^{\infty}\big{(}|y-z_{0}|^{2}-2(\operatorname{Im}z_{0})^{2}\big{)}\cdot\displaystyle\frac{\varrho_{0}(y)dy}{|y-z_{0}|^{4}}$
$\displaystyle=\frac{1}{t}-2(\operatorname{Im}z_{0})^{2}\displaystyle\int_{-\infty}^{\infty}\frac{\varrho_{0}(y)dy}{|y-z_{0}|^{4}},$

where in the last equality we used the third statement of (B.14). Inserting
this into (B.15), we find

(B.16) $\displaystyle\begin{split}\big{|}1-tm^{\prime}_{0}(z_{0})\big{|}&\geq
2t(\operatorname{Im}z_{0})^{2}\displaystyle\int_{-\infty}^{\infty}\frac{\varrho_{0}(y)dy}{|y-z_{0}|^{4}}\geq
2t(\operatorname{Im}z_{0})^{2}\left(\displaystyle\int_{-\infty}^{\infty}\frac{\varrho_{0}(y)dy}{|y-z_{0}|^{2}}\right)^{2}\geq\frac{2(\operatorname{Im}z_{0})^{2}}{t}\geq
2t(\pi\delta)^{2},\end{split}$

where in the second statement we used the fact that $\varrho_{0}$ is a
probability density; in the third we used the third statement of (B.14); and
in the fourth we used the first two statements of (B.14). The lemma then
follows from inserting (B.13) and (B.16) into (B.12). ∎

###### Proof of 2.11.

Denote the Stieltjes transform of $\mu_{0}$ by $m_{0}=m_{\mu_{0}}$. We will
use (B.12) to estimate the $x$-derivatives of $\varrho_{t}$, to which end we
must upper bound derivatives of $m_{0}$ and lower bound $|1-tm_{0}^{\prime}|$.
To implement this, we claim that there exists a constant $C_{1}=C_{1}(k)>1$
such that, for any integer $j\in\llbracket 0,k-1\rrbracket$,

(B.17) $\displaystyle\begin{aligned}
&\big{|}\partial_{z}^{j}m_{0}(z)\big{|}\leq
C_{1}B^{j+2}(\delta\mathfrak{c})^{-j-1},\qquad\text{for any
$z\in\overline{\mathbb{H}}$ such that
$|\operatorname{Re}z|\leq\displaystyle\frac{\mathfrak{c}}{2}$};\\\
&\big{|}1-tm_{0}^{\prime}(z)\big{|}\geq\displaystyle\frac{t\delta\mathfrak{c}}{2\operatorname{Im}z},\qquad\qquad\quad\text{for
any $z\in\overline{\mathbb{H}}$ such that
$|\operatorname{Re}z|\leq\displaystyle\frac{\mathfrak{c}}{2}$ and
$\operatorname{Im}z\in[0,\mathfrak{c}]$}.\end{aligned}$

To see this, fix $j\in\llbracket 0,k-1\rrbracket$ and observe that (2.5)
implies for $|\operatorname{Re}z|<\frac{\mathfrak{c}}{2}$ that

(B.18) $\displaystyle\begin{aligned}
\big{|}\partial_{z}^{j}m_{0}(z)\big{|}=j!\Bigg{|}\int_{-\infty}^{\infty}\frac{\varrho_{0}(x)dx}{(x-z)^{j+1}}\Bigg{|}&\leq
j!\Bigg{|}\int_{-\mathfrak{c}}^{\mathfrak{c}}\frac{\varrho_{0}(x)dx}{(x-z)^{j+1}}\Bigg{|}+2^{j+1}j!\mathfrak{c}^{-j-1}\\\
&\leq\bigg{|}\int_{-\mathfrak{c}}^{\mathfrak{c}}\frac{\partial_{x}^{j}\varrho_{0}(x)dx}{x-z}\bigg{|}+2^{j+1}j!\mathfrak{c}^{-j-1}+2^{j+1}(j+1)!B^{j+1}(\delta\mathfrak{c})^{-j}.\end{aligned}$

Here, to obtain the second statement we used the facts that
$|\operatorname{Re}z|\leq\frac{\mathfrak{c}}{2}$ and that $\varrho_{0}$ is the
density of a probability measure. To obtain the third, we integrated by parts
$j$ times, using the estimate on the boundary terms,

$\displaystyle\displaystyle\sum_{i=0}^{j}i!\bigg{|}\frac{\partial_{x}^{j-i}\varrho_{0}(\mathfrak{c})}{(z-\mathfrak{c})^{i}}-\frac{\partial_{x}^{j-i}\varrho_{0}(-\mathfrak{c})}{(z+\mathfrak{c})^{i}}\bigg{|}\leq\displaystyle\frac{2^{j}}{\mathfrak{c}^{j}}\displaystyle\sum_{i=0}^{j}i!\Big{(}\big{|}\partial_{x}^{j-i}\varrho_{0}(\mathfrak{c})\big{|}+\big{|}\partial_{x}^{j-i}\varrho_{0}(-\mathfrak{c})\big{|}\Big{)}\leq\displaystyle\frac{2^{j+1}(j+1)!B^{j+1}}{(\delta\mathfrak{c})^{j}},$

where the first bound holds again since
$|\operatorname{Re}z|\leq\frac{\mathfrak{c}}{2}$ and the second holds since
$\big{|}\partial_{x}^{i}\varrho_{0}(x)\big{|}\leq B^{i+1}\delta^{-i}$ for any
$i\in\llbracket 0,k\rrbracket$. Then,

$\displaystyle\big{|}\partial_{z}^{j}m_{0}(z)\big{|}$
$\displaystyle\leq\int_{-\mathfrak{c}}^{\mathfrak{c}}\bigg{|}\frac{\partial_{x}^{j}\varrho_{0}(x)-\partial_{x}^{j}\varrho_{0}(\operatorname{Re}z)}{z-x}\bigg{|}dx+\big{|}\partial_{x}^{j}\varrho_{0}(\operatorname{Re}z)\big{|}\cdot\Bigg{|}\displaystyle\int_{-\mathfrak{c}}^{\mathfrak{c}}\displaystyle\frac{dx}{x-z}\Bigg{|}+\mathcal{O}\big{(}B^{j+1}(\delta\mathfrak{c})^{-j}\big{)}$
$\displaystyle\leq\displaystyle\int_{-\mathfrak{c}}^{\mathfrak{c}}\bigg{|}\displaystyle\frac{\partial_{x}^{j}\varrho_{0}(x)-\partial_{x}^{j}\varrho_{0}(\operatorname{Re}z)}{x-z}\bigg{|}dx+\mathcal{O}\big{(}B^{j+1}(\delta\mathfrak{c})^{-j}\big{)}$
$\displaystyle\leq
2\mathfrak{c}\cdot\displaystyle\sup_{|x|\leq\mathfrak{c}}\big{|}\partial_{x}^{j+1}\varrho_{0}(x)\big{|}+\mathcal{O}\big{(}B^{j+1}\big{(}\mathfrak{c}\delta)^{-j-1}\big{)}\leq\mathcal{O}\big{(}B^{j+2}(\mathfrak{c}\delta)^{-j-1}\big{)},$

where in the first and second estimates we used (B.18), the bound
$\big{|}\partial_{x}^{j}\varrho_{0}(x)\big{|}\leq B^{j}\delta^{-j}$ for
$|x|\leq\mathfrak{c}$ (as $|\operatorname{Re}z|\leq\mathfrak{c}$), and the
fact that
$\big{|}\operatorname{PV}\int_{-\mathfrak{c}}^{\mathfrak{c}}(x-z)^{-1}dx\big{|}\leq
2$ (as $|\operatorname{Re}z|\leq\frac{\mathfrak{c}}{2}$); in the fourth, we
used the same bound on derivatives of $\varrho_{0}$, but with $j$ replaced by
$j+1$. Here, the implicit constant only depends on $j$, verifying the first
statement of (B.17).

To verify the second, observe for
$|\operatorname{Re}z|\leq\frac{\mathfrak{c}}{2}$ and
$\operatorname{Im}z\in[0,\mathfrak{c}]$ that

$\displaystyle\big{|}1-tm^{\prime}_{0}(z)\big{|}\geq
2t(\operatorname{Im}z)^{2}\int_{-\infty}^{\infty}\frac{\varrho_{0}(x)dx}{|x-z|^{4}}$
$\displaystyle\geq
2t\delta(\operatorname{Im}z)^{2}\int_{-\mathfrak{c}}^{\mathfrak{c}}\frac{dx}{|x-z|^{4}}$
$\displaystyle\geq
2t\delta(\operatorname{Im}z)^{2}\displaystyle\int_{-\mathfrak{c}/2}^{\mathfrak{c}/2}\displaystyle\frac{dx}{\big{(}|x|+\operatorname{Im}z\big{)}^{4}}$
$\displaystyle=\displaystyle\frac{4}{3}\cdot
t\delta(\operatorname{Im}z)^{2}\bigg{(}(\operatorname{Im}z)^{-3}-\Big{(}\displaystyle\frac{\mathfrak{c}}{2}+\operatorname{Im}z\Big{)}^{-3}\bigg{)}\geq\frac{t\delta\mathfrak{c}}{2\operatorname{Im}z},$

where in the first statement we used the first bound of in (B.16); in the
second we used the fact that $\varrho_{0}(x)\geq\delta$ for
$|x|\leq\mathfrak{c}$; in the third we changed variables from $x$ to
$x-\operatorname{Re}z$ and used the bound
$|\operatorname{Re}z|\leq\frac{\mathfrak{c}}{2}$; in the fourth we performed
the integration; and in the fifth we used the fact that
$\operatorname{Im}z\leq\mathfrak{c}$. This verifies the second statement of
(B.17).

To apply (B.17), let $C_{2}=C_{2}(k)>10$ denote a sufficiently large constant
(to be determined later) and fix real numbers

(B.19) $\displaystyle x_{0},t\in\mathbb{R},\qquad\text{such
that}\quad|x_{0}|\leq\frac{\mathfrak{c}}{4},\quad\text{and}\quad
t\in\Big{[}0,\frac{(\delta\mathfrak{c})^{20}}{C_{2}B^{20}}\Big{]}.$

By 2.8, there exists a complex number
$z_{0}=z_{0}(x_{0},t)\in\partial\Lambda_{t}$ such that
$z_{0}-tm_{0}(z_{0})=x_{0}$. We claim that
$|\operatorname{Re}z_{0}|\leq\frac{\mathfrak{c}}{2}$. To see this, first
observe that $z_{0}=z_{0}(x_{0},t)$ is continuous in $t$. Indeed, since the
map $M(z)=z-tm_{0}(z)$ is a homeomorphism from $\overline{\Lambda}_{t}$
(recall (2.10)) to $\overline{\mathbb{H}}$ by 2.8 that is uniformly continuous
in $t$, its inverse $M^{-1}$ is also continuous in $t$. Thus,
$z_{0}(x_{0},t)=M^{-1}(x_{0})$ is continuous in $t$. Now, assume to the
contrary that
$\big{|}\operatorname{Re}z_{0}(x,t)\big{|}>\frac{\mathfrak{c}}{2}$. Then since
$z_{0}(x_{0},0)=x_{0}\in\big{[}-\frac{\mathfrak{c}}{4},\frac{\mathfrak{c}}{4}\big{]}$,
there exists some $s\in[0,t]$ such that
$\big{|}\operatorname{Re}z_{0}(x_{0},s)\big{|}=\frac{\mathfrak{c}}{2}$. Since
$|x_{0}|\leq\frac{\mathfrak{c}}{4}$, it follows that
$\big{|}tm_{0}(z_{0})\big{|}=|z_{0}-x_{0}|\geq\frac{\mathfrak{c}}{4}$ at
$z_{0}=z_{0}(x_{0},s)$. This contradicts the fact (from the $j=0$ case of
(B.17) and (B.19)) that

$\displaystyle\big{|}tm_{0}(z_{0})\big{|}\leq\frac{(\delta\mathfrak{c})^{20}}{C_{2}B^{20}}\cdot
C_{1}B^{2}(\delta\mathfrak{c})^{-1}\leq\frac{\mathfrak{c}}{8},$

for $C_{2}>C_{1}$, and so
$|\operatorname{Re}z_{0}|\leq\frac{\mathfrak{c}}{2}$. Further observe that
$\operatorname{Im}z_{0}\leq\frac{\mathfrak{c}}{2}$, since otherwise the third
statement of (B.14) would yield (as $C_{2}>4$)

$\displaystyle\frac{1}{t}=\int_{-\infty}^{\infty}\frac{\varrho_{0}(x)dx}{|x-z_{0}|^{2}}<4\mathfrak{c}^{-2}\int_{-\infty}^{\infty}\varrho_{0}(x)dx=4\mathfrak{c}^{-2}<\displaystyle\frac{C_{2}B^{20}}{(\delta\mathfrak{c})^{20}},$

which contradicts (B.19).

Therefore,

$\displaystyle\frac{1}{t}=\int_{-\infty}^{\infty}\frac{\varrho_{0}(x)dx}{|x-z_{0}|^{2}}\leq
B\int_{-\mathfrak{c}}^{\mathfrak{c}}\frac{dx}{|x-z_{0}|^{2}}+4\mathfrak{c}^{-2}\leq\frac{B\pi}{\operatorname{Im}z}+4\mathfrak{c}^{-2},$

where in the first statement we used the third part of (B.14); in the second
we used the facts that $\sup_{|x|\leq\mathfrak{c}}\varrho(x)\leq B$, that
$\varrho$ is a probability density, and that
$|\operatorname{Re}z_{0}|\leq\frac{\mathfrak{c}}{2}$; and in the third we used
the equality
$\int_{-\infty}^{\infty}|x-z|^{-2}dx=\pi(\operatorname{Im}z)^{-1}$. Since
$t\leq\frac{\mathfrak{c}^{2}}{8}$, it follows that
$t(\operatorname{Im}z)^{-1}\geq(2B\pi)^{-1}$. Inserting this into the second
bound of (B.17) (which applies as
$|\operatorname{Re}z_{0}|\leq\frac{\mathfrak{c}}{2}$ and
$\operatorname{Im}z_{0}\leq\frac{\mathfrak{c}}{2}$) gives

(B.20)
$\displaystyle\big{|}1-tm_{0}^{\prime}(z_{0})\big{|}\geq\displaystyle\frac{\delta\mathfrak{c}}{4\pi
B}.$

Inserting the first bound of (B.17) (at $z=z_{0}$, which applies since
$|\operatorname{Re}z_{0}|\leq\frac{\mathfrak{c}}{2}$) and (B.20) into (B.12),
it follows that there exists a constant $C_{3}=C_{3}(k)>1$ such that

(B.21) $\displaystyle\big{|}\partial_{x}^{j}\varrho_{t}(x_{0})\big{|}\leq
C_{3}(B^{8}\delta^{-6}\mathfrak{c}^{-6})^{j+1},\qquad\text{for any
$j\in\llbracket 0,k-1\rrbracket$}.$

Before proceeding to bound the $t$-derivative of $\varrho_{t}$, let us quickly
establish the first statement of (2.14). To that end, observe that

$\displaystyle\big{|}\varrho_{t}(x_{0})-\varrho_{0}(x_{0})\big{|}$
$\displaystyle=\pi^{-1}\big{|}\operatorname{Im}m_{t}(x_{0})-\operatorname{Im}m_{0}(x_{0})\big{|}$
$\displaystyle=\pi^{-1}\Big{|}\operatorname{Im}m_{t}\big{(}z_{0}-tm_{0}(z_{0})\big{)}-\operatorname{Im}m_{0}(x_{0})\Big{|}$
$\displaystyle=\pi^{-1}\big{|}\operatorname{Im}m_{0}(z_{0})-\operatorname{Im}m_{0}(x_{0})\big{|}$
$\displaystyle\leq
C_{3}B^{16}\delta^{-12}\mathfrak{c}^{-12}\cdot|z_{0}-x_{0}|\leq
C_{3}B^{16}\delta^{-12}\mathfrak{c}^{-12}\cdot
t\big{|}m_{0}(z_{0})\big{|}\leq\displaystyle\frac{\delta}{4},$

where in the first statement, we applied (2.6); in the second, the definition
of $z_{0}$; in the third statement, (2.11); in the fourth, the first bound in
(B.17) at $j=1$; in the fifth, the fact that $z_{0}-x_{0}=tm_{0}(z_{0})$; and
in the sixth, the bound on $t$ from (B.19) (and making $C_{2}>4C_{3}$
sufficiently large) with first bound in (B.17) at $j=0$. Since
$\varrho_{0}(x_{0})\geq\delta$ (as
$|x|\leq\frac{\mathfrak{c}}{4}\leq\mathfrak{c}$), this implies the first
statement of (2.14).

To prove the second statement of (2.14), observe by (B.21) that it suffices to
bound the $t$-derivatives of $\varrho_{t}$. Recall from (2.12) that
$\partial_{t}\varrho_{t}(x)=\pi\partial_{x}\big{(}\varrho_{t}(x)H\varrho_{t}(x)\big{)}$,
where $H\varrho_{t}(x)$ is the Hilbert transform of $\varrho_{t}$ (recall
(2.7)). For any real number
$x\in\big{[}-\frac{\mathfrak{c}}{4},\frac{\mathfrak{c}}{4}\big{]}$ and integer
$j\in\llbracket 0,k-2\rrbracket$, we can then bound the derivatives of Hilbert
transform by

(B.22)
$\displaystyle\begin{split}\big{|}\partial_{x}^{j}H\varrho_{t}(x)\big{|}&=\pi^{-1}\Bigg{|}\partial_{x}^{j}\bigg{(}\operatorname{PV}\displaystyle\int_{-\infty}^{\infty}\frac{\varrho_{t}(x-y)dy}{y}\bigg{)}\Bigg{|}\\\
&=\pi^{-1}\Bigg{|}\partial_{x}^{j}\bigg{(}\operatorname{PV}\displaystyle\int_{-\mathfrak{c}/2}^{\mathfrak{c}/2}\displaystyle\frac{\varrho_{t}(x-y)dy}{y}\bigg{)}+\partial_{x}^{j}\displaystyle\int_{|y|>\mathfrak{c}/2}\displaystyle\frac{\varrho_{t}(x-y)dy}{y}\Bigg{|}\\\
&=\pi^{-1}\Bigg{|}\partial_{x}^{j}\bigg{(}\operatorname{PV}\int_{-\mathfrak{c}/2}^{\mathfrak{c}/2}\frac{\big{(}\varrho_{t}(x-y)-\varrho_{t}(x)\big{)}dy}{y}\bigg{)}+\partial_{x}^{j}\displaystyle\int_{|y-x|>\mathfrak{c}/2}\displaystyle\frac{\varrho_{t}(y)dy}{x-y}\Bigg{|}\\\
&\leq\mathfrak{c}\cdot\displaystyle\sup_{|y|\leq\mathfrak{c}}\big{|}\partial_{x}^{j+1}\varrho_{t}(y)\big{|}+2^{j}j!\mathfrak{c}^{-j}=\mathcal{O}\big{(}(B^{8}\delta^{-6}\mathfrak{c}^{-6})^{j+2}\big{)},\end{split}$

where in the first and second statements we used (2.7); in the third we used
the fact that $\operatorname{PV}(a)=0$ for any constant $a$ in the first
integral, and we changed variables from $y$ to $x-y$ in the second; in the
fourth we used the fact that $\big{|}\partial_{x}^{j}(x-y)^{-1}\big{|}\leq
j!2^{j+1}\mathfrak{c}^{-j-1}$ for $|x-y|\geq\frac{\mathfrak{c}}{2}$ and that
$\varrho_{t}$ is a probability measure; and in the fifth we applied (B.21).
Combining this with (B.21) and (2.12) then yields the second bound in (2.14).
∎

###  B.3. Proof of 2.4

In this section we establish 2.4, to which end we discretize in time (leaving
space continuous). For any integer $T\geq 1$, a _( $T$-step) Gaussian walk_
starting at $u\in\mathbb{R}$ is a probability measure on $(T+1)$-tuples
$\big{(}x(0),x(1),\ldots,x(T)\big{)}\in\mathbb{R}^{T}$ with $x(0)=u$ such
that, for each $j\in\llbracket 1,T\rrbracket$, the jump $x(j)-x(j-1)$ is a
centered Gaussian random variable of variance $1$. A _( $T$-step) Gaussian
bridge_ from $u$ to $v$ is a Gaussian walk starting at $u$, conditioned to end
at $v$ (so $x(T)=v$). The following definition, similar to Definition 1.1,
provides notation for non-intersecting Gaussian bridges.

######  Definition B.3.

Fix integers $T,n\geq 1$; two $n$-tuples
$\bm{u},\bm{v}\in\overline{\mathbb{W}}_{n}$; and two functions $f,g:\llbracket
0,T\rrbracket\rightarrow\overline{\mathbb{R}}$. Let
$\mathfrak{G}_{f;g}^{\bm{u};\bm{v}}$ denote the law on sequences
$\bm{x}(t)=\big{(}x_{1}(t),x_{2}(t),\ldots,x_{n}(t)\big{)}$, with
$t\in\llbracket 0,T\rrbracket$, given by $n$ independent $T$-step Gaussian
walks, conditioned on satisfying the following three properties.

  1. (1)

The $x_{j}$ do not intersect, that is, $\bm{x}(t)\in\mathbb{W}_{n}$ for each
$t\in\llbracket 1,T-1\rrbracket$.

  2. (2)

The $x_{j}$ start at $u_{j}$ and end at $v_{j}$, that is, $x_{j}(0)=u_{j}$ and
$x_{j}(T)=v_{j}$ for each $j\in\llbracket 1,n\rrbracket$.

  3. (3)

The $x_{j}$ are bounded below by $f$ and above by $g$, that is, $f<x_{j}<g$
for each $j\in\llbracket 1,n\rrbracket$.

We assume $f(0)\leq u_{n}\leq u_{1}\leq g(0)$ and $f(T)\leq v_{n}\leq
v_{1}\leq g(T)$, even when not stated explicitly.

Analogous to 2.1, the following states that non-intersecting Gaussian walks
are invariant under affine transformations.

######  Remark B.4.

Nonintersecting Gaussian walks satisfy the following invariance property under
affine transformations. Adopt the notation of Definition B.3, and fix real
numbers $\alpha,\beta\in\mathbb{R}$. Define the $n$-tuples
$\widetilde{\bm{u}},\widetilde{\bm{v}}\in\mathbb{W}_{n}$ and functions
$\widetilde{f},\widetilde{g}:\llbracket
0,n\rrbracket\rightarrow\overline{\mathbb{R}}$ by setting

$\displaystyle\widetilde{u}_{j}=u_{j}+\alpha,\quad\text{and}\quad\widetilde{v}_{j}=v_{j}+T\beta+\alpha,\qquad\qquad\qquad\quad\text{for
each $j\in\llbracket 0,n\rrbracket$};$
$\displaystyle\widetilde{g}(t)=f(t)+t\beta+\alpha,\quad\text{and}\quad\widetilde{g}(t)=g(t)+t\beta+\alpha,\qquad\text{for
each $t\in\llbracket 0,T\rrbracket$}.$

Sampling
$\widetilde{\bm{x}}=\big{(}\widetilde{x}_{1}(t),\widetilde{x}_{2}(t),\ldots,\widetilde{x}_{n}(t)\big{)}$
under $\mathfrak{G}_{\tilde{f},\tilde{g}}^{\tilde{\bm{u}};\tilde{\bm{v}}}$,
there is a coupling between $\widetilde{\bm{x}}$ and $\bm{x}$ such that
$\widetilde{x}_{j}(t)=x_{j}(t)+\beta t+\alpha$ for each $t\in\llbracket
0,T\rrbracket$ and $j\in\llbracket 1,n\rrbracket$.

As in 2.1, this follows from the analogous affine invariance of a single
Gaussian random bridge, together with the fact that affine transformations do
not affect the non-intersecting property.

The following lemma states height monotonicity (the analog of 2.3) for non-
intersecting Gaussian bridges. Its proof is very similar to that of 2.3 (from
[18]) and is thus omitted.

######  Lemma B.5.

Fix integers $T,n\geq 1$; four $n$-tuples
$\bm{u},\widetilde{\bm{u}},\bm{v},\widetilde{\bm{v}}\in\mathbb{W}_{n}$; and
functions $f,\widetilde{f},g,\widetilde{g}:\llbracket
0,T\rrbracket\rightarrow\overline{\mathbb{R}}$ with $f\leq g$ and
$\widetilde{f}\leq\widetilde{g}$. Sample non-intersecting Gaussian bridges
$\bm{x}(t)$ and $\widetilde{\bm{x}}(t)$ from the measures
$\mathfrak{G}_{f;g}^{\bm{u};\bm{v}}$ and
$\mathfrak{G}_{\tilde{f};\tilde{g}}^{\tilde{\bm{u}};\tilde{\bm{v}}}$,
respectively. If

$\displaystyle f\leq\widetilde{f};\quad
g\leq\widetilde{g};\quad\bm{u}\leq\widetilde{\bm{u}};\quad\bm{v}\leq\widetilde{\bm{v}},$

then there exists a coupling between $\bm{x}$ and $\widetilde{\bm{x}}$ such
that $\bm{x}(t)\leq\widetilde{\bm{x}}(t)$, for each $t\in\llbracket
0,T\rrbracket$.

The following lemma is a discrete (in time) variant of 2.4 and quickly implies
it.

######  Lemma B.6.

Adopt the notation and assumptions of B.5, and fix some real number $B\geq 0$.
If $u_{j}\leq\widetilde{u}_{j}\leq u_{j}+B$ and
$v_{j}\leq\widetilde{v}_{j}\leq v_{j}+B$ for all $j\in\llbracket
1,n\rrbracket$, then the following two statements hold.

  1. (1)

If $f(t)\leq\widetilde{f}(t)\leq f(t)+B$ and $g(t)\leq\widetilde{g}(t)\leq
g(t)+B$ for each $t\in\llbracket 0,T\rrbracket$, then there is a coupling
between $\bm{x}$ and $\widetilde{\bm{x}}$ so that
$x_{j}(t)\leq\widetilde{x}_{j}(t)\leq x_{j}(t)+B$ for each $(t,j)\in\llbracket
0,T\rrbracket\times\llbracket 1,n\rrbracket$.

  2. (2)

If $\bm{u}=\widetilde{\bm{u}}$, and $f(t)\leq\widetilde{f}(t)\leq
f(t)+\frac{tB}{T}$ and $g(t)\leq\widetilde{g}(t)\leq g(t)+\frac{tB}{T}$ for
each $t\in\llbracket 0,T\rrbracket$, then there is a coupling between $\bm{x}$
and $\widetilde{\bm{x}}$ so that $x_{j}(t)\leq\widetilde{x}_{j}(t)\leq
x_{j}(t)+\frac{tB}{T}$ for each $(t,j)\in\llbracket
0,T\rrbracket\times\llbracket 1,n\rrbracket$.

###### Proof of 2.4.

This follows from B.6 with the convergence of Gaussian walks to Brownian
bridges; we omit further details. ∎

Similarly to the proof [18, Lemma 2.6 and Lemma 2.7], that of B.6 makes use of
a local Glauber dynamic, which in our setting is defined as follows.

######  Definition B.7.

Fix integers $T,n\geq 1$ and two functions $f,g:\llbracket
0,T\rrbracket\rightarrow\mathbb{R}$. For $t\in\llbracket 0,T\rrbracket$, let
$\bm{y}(t)=\big{(}y_{1}(t),y_{2}(t),\ldots,y_{n}(t)\big{)}\in\mathbb{W}_{n}$
be a family of $n$ non-intersecting $T$-step walks. The _Glauber dynamics_ is
the discrete-time Markov chain whose state
$\mathcal{P}^{k}\bm{y}(t)=\big{(}\mathcal{P}^{k}y_{1}(t),\mathcal{P}^{k}y_{2}(t),\ldots,\mathcal{P}^{k}y_{n}(t)\big{)}$
at time $k\geq 0$ is determined as follows. Below, for each $t\in\llbracket
0,T\rrbracket$, let $y_{0}(t)=g(t)$ and $y_{n+1}(t)=f(t)$.

Set $\mathcal{P}^{0}\bm{y}=\bm{y}$. For $k\geq 1$, let $(a,b,d)\in\llbracket
0,T-2\rrbracket\times\llbracket 1,n\rrbracket\times\mathbb{Z}_{\geq 0}$ be
such that $k=dn(T-1)+an+b$. For $(t,j)\neq(a+1,b)$, set
$\mathcal{P}^{k}y_{j}(t)=\mathcal{P}^{k-1}y_{j}(t)$. For $(t,j)=(a+1,b)$, set
$\mathcal{P}^{k}y_{b}(a+1)$ to be the middle ($t=1$) value of a $2$-step
Gaussian random bridge, conditioned to start at $\mathcal{P}^{k-1}y_{b}(a)$;
end at $\mathcal{P}^{k-1}y_{b}(a+2)$; be above
$\mathcal{P}^{k-1}y_{b+1}(a+1)$; and be below $\mathcal{P}^{k-1}y_{b-1}(a+1)$.

######  Remark B.8.

It follows from the Gibbs property (for non-intersecting Gaussian bridges)
that $\mathfrak{G}_{f;g}^{\bm{y}(0);\bm{y}(T)}$ is a stationary measure for
the Glauber dynamics $\mathcal{P}$.

The following lemma indicates that the Glauber dynamics converge to the
stationary measure $\mathfrak{G}_{f;g}^{\bm{y}(0);\bm{y}(T)}$ in the long-time
limit; its proof is given in Section B.4 below.

######  Lemma B.9.

Adopt the notation of Definition B.7, and assume that $\max_{t\in\llbracket
0,T\rrbracket}\big{(}|f(t)|+|g(t)|\big{)}<\infty$. Then, the law of
$\mathcal{P}^{2n(T-1)k}\bm{y}$ converges as $k$ tends to $\infty$ to
$\mathfrak{G}_{f;g}^{\bm{y}(0);\bm{y}(T)}$.

Now let us briefly outline the proof of B.6.

###### Proof of B.6 (Outline).

We only discuss the second statement of the lemma, as the proof of the first
is entirely analogous. We will assume throughout that
$\big{|}f(t)\big{|}+\big{|}g(t)\big{|}<\infty$ for each $t\in\llbracket
1,T-1\rrbracket$, as the cases when $f(s)=-\infty$ or $g(s)=\infty$ for some
$s\in\llbracket 1,T-1\rrbracket$ can then be obtained from a limiting
procedure (in particular, one can consider a sequence $f_{k}$ of lower
boundaries such that $f_{k}(s)$ always remains finite but tends to $-\infty$
as $k$ tends to $\infty$, and similarly for $g$).

Let us first confirm the lemma when $(T,n)=(2,1)$, in which case we will
verify the following more general result. Fix a real number $A\geq 0$ and
assume that

$\displaystyle x_{1}(0)\leq\widetilde{x}_{1}(0)\leq x_{1}(0)+A;\qquad
x_{1}(0)\leq\widetilde{x}_{1}(2)\leq x_{1}(2)+A+B;$ $\displaystyle
f(1)\leq\widetilde{f}(1)\leq f(1)+A+\displaystyle\frac{B}{2};\quad
g(1)\leq\widetilde{g}(1)\leq g(1)+A+\displaystyle\frac{B}{2}.$

Then, there exists a coupling between $x_{1}$ and $\widetilde{x}_{1}$ such
that

(B.23) $\displaystyle x_{1}(1)\leq\widetilde{x}_{1}(1)\leq
x_{1}(1)+A+\displaystyle\frac{B}{2}.$

Indeed, this is a consequence of the fact (which follows quickly from B.5 and
B.4, or can also be verified directly) that $\mathbb{P}\big{[}x_{1}(1)\geq
r\big{]}\leq\mathbb{P}\big{[}\widetilde{x}_{1}(1)\geq
r\big{]}\leq\mathbb{P}\big{[}x_{1}(1)+A+\frac{B}{2}\geq r\big{]}$, for any
real number $r\in\mathbb{R}$. This shows (B.23), whose $A=0$ special case
yields the second part of B.6 at $(T,n)=(2,1)$.

Thus, we assume in what follows that either $T\geq 3$ or $n\geq 2$. Fix two
ensembles
$\bm{y}(t)=\big{(}y_{1}(t),y_{2}(t),\ldots,y_{n}(t)\big{)}\in\mathbb{W}_{n}$
and
$\widetilde{\bm{y}}(t)=\big{(}\widetilde{y}_{1}(t),\widetilde{y}_{2}(t),\ldots,\widetilde{y}_{n}(t)\big{)}\in\mathbb{W}_{n}$
of $n$ non-intersecting $T$-step walks, with starting points
$\bm{y}(0)=\bm{u}=\widetilde{\bm{y}}(0)$, and ending points $\bm{y}(T)=\bm{v}$
and $\widetilde{\bm{y}}(T)=\widetilde{\bm{v}}$, such that, for each
$t\in\llbracket 0,T\rrbracket$,

$\displaystyle\bm{y}(t)\leq\widetilde{\bm{y}}(t)\leq\bm{y}(t)+\displaystyle\frac{tB}{T};\qquad
f(t)\leq y_{n}(t)\leq y_{1}(t)\leq
g(t);\qquad\widetilde{f}(t)\leq\widetilde{y}_{1}(t)\leq\widetilde{y}_{n}(t)\leq\widetilde{g}(t).$

Such initial data exist due to the assumptions of the second part of the
lemma.

We next run the Glauber dynamics on $\bm{y}$ (with boundaries $(f,g)$) and
$\widetilde{\bm{y}}$ (with boundaries $(\widetilde{f},\widetilde{g})$),
coupled in the following way. If we update the pair
$\big{(}\mathcal{P}^{k-1}y_{j}(t),\mathcal{P}^{k-1}\widetilde{y}_{j}(t)\big{)}$,
then using (B.23) (with the $(A;B)$ there given by
$\big{(}\frac{(t-1)B}{T};\frac{2B}{T}\big{)}$ here), we couple
$\mathcal{P}^{k}y_{j}(t)\leq\mathcal{P}^{k}\widetilde{y}_{j}(t)\leq\mathcal{P}^{k}y_{j}(t)+\frac{tB}{T}$.
By induction on $k$, this ensures that
$\mathcal{P}^{k}y_{j}(t)\leq\mathcal{P}^{k}\widetilde{y}_{j}(t)\leq\mathcal{P}^{k}y_{j}(t)+\frac{tB}{T}$
holds for each $(t,j)\in\llbracket 0,T\rrbracket\times\llbracket
1,n\rrbracket$ and $k\geq 0$. Letting $k$ tend to $\infty$ and taking any
limit point of this coupling, the lemma then follows from B.9. ∎

###  B.4. Proof of B.9

In this section we establish B.9. We begin by recalling a general statement on
convergence to stationarity of infinite-dimensional Markov chains.

Let $\Omega$ be a measurable space with $\sigma$-algebra $\mathcal{F}$; let
$\mathscr{P}(\Omega)$ denote the space of probability measures on
$(\Omega,\mathcal{F})$. Given any two probability measures
$\nu_{1},\nu_{2}\in\mathscr{P}(\Omega)$, recall that the total variation
distance between them is defined by

(B.24) $\displaystyle
d_{\operatorname{TV}}(\nu_{1},\nu_{2})=\displaystyle\sup_{A\in\mathcal{F}}\big{|}\nu_{1}(A)-\nu_{2}(A)\big{|}.$

######  Assumption B.10.

Adopting the above notation, let
$\mathsf{K}:\Omega\times\mathcal{F}\rightarrow\mathbb{R}_{\geq 0}$ be a Markov
transition kernel. For any function $\varphi:\Omega\rightarrow\mathbb{R}_{\geq
0}$ and measure $\mu$ on $\Omega$, define the function
$\mathfrak{\mathsf{K}}\varphi:\Omega\rightarrow\mathbb{R}_{\geq 0}$ and
measure $\mathsf{K}\mu$ on $\Omega$ by setting

$\displaystyle\mathsf{K}\varphi(x)=\displaystyle\int_{\Omega}\varphi(y)\mathsf{K}(x,dy);\qquad\mathsf{K}\mu(A)=\displaystyle\int_{\Omega}\mathsf{K}(x,A)\mu(dx),$

for any $x\in\Omega$ and measurable set $A\in\mathcal{F}$. Assume that there
exist constants $\alpha\in(0,1)$, $\gamma\in(0,1)$, $B\geq 0$, and
$R>\frac{2B}{1-\gamma}$; a potential function
$V:\Omega\rightarrow\mathbb{R}_{\geq 0}$; and a probability measure $\nu$ on
$\Omega$, such that the following two conditions hold.

  1. (1)

For each $x\in\Omega$, we have $\mathsf{K}V(x)\leq\gamma V(x)+B$, for each
$x\in\Omega$.

  2. (2)

For each $x\in\Omega$ with $V(x)\leq R$, and any measurable set
$A\in\mathcal{F}$, we have $\mathsf{K}(x,A)\geq\alpha\nu(A)$.

The following result provides a convergence theorem for such Harris chains
[38]. It appears in [57], though it is stated as written below in [37].

######  Lemma B.11 ([37, Theorem 1.2]).

Adopt B.10, and fix a measure $\mu$ on $\Omega$. Then, the Markov process
defined by $\mathsf{K}$ has a unique stationary measure $\mu_{0}$, and
$\lim_{m\rightarrow\infty}\|\mathsf{K}^{m}\mu-\mu_{0}\|_{\operatorname{TV}}=0$.

We next apply B.11 to the Glauber dynamics $\mathcal{P}$ of Definition B.7.
Throughout the remainder of this section, we adopt the notation of that
definition. These include the functions $f,g:\llbracket
0,T\rrbracket\rightarrow\mathbb{R}$; the family $\bm{y}$ of $n$ non-
intersecting $T$-step walks
$\bm{y}(t)=\big{(}y_{1}(t),y_{2}(t),\ldots,y_{n}(t)\big{)}\in\mathbb{W}_{n}$
(over $t\in\llbracket 0,T\rrbracket$), and the associated Markov operator
$\mathcal{P}$ (which we also interpret as a kernel) for the Glauber dynamics.
We also set $\bm{u}=\bm{y}(0)$ and $\bm{v}=\bm{y}(T)$, which are fixed
throughout the dynamics. Then the state space for the Glauber dynamics
$\mathcal{P}$ can be viewed as

$\displaystyle\Omega_{0}=\Big{\\{}\big{(}\bm{y}(t)\big{)}_{t\in\llbracket
1,T-1\rrbracket}\in(\mathbb{W}_{n})^{T-1}:\displaystyle\min_{t\in\llbracket
1,T-1\rrbracket}\big{(}y_{n}(t)-f(t)\big{)}>0,\displaystyle\min_{t\in\llbracket
1,T-1\rrbracket}\big{(}g(t)-y_{1}(t)\big{)}>0\Big{\\}}.$

The associated potential function $V_{0}$ is defined by

$\displaystyle V_{0}(\bm{y})=\displaystyle\max_{j\in\llbracket
1,n\rrbracket}\displaystyle\max_{t\in\llbracket
1,T-1\rrbracket}\Big{(}\big{|}y_{j}(t)\big{|}+1\Big{)}.$

We then have the following two lemmas verifying B.10 for the Glauber dynamics.

######  Lemma B.12.

There exist constants $\gamma\in(0,1)$ and $B=B(\bm{u},\bm{v},f,g)\geq 0$ such
that, for any family $\bm{y}$ of $n$ non-intersecting $T$-step walks with
$\bm{y}(0)=\bm{u}$ and $\bm{y}(T)=\bm{v}$, we have
$\mathcal{P}V_{0}(\bm{y})\leq\gamma V_{0}(\bm{y})+B$.

###### Proof.

This follows from taking $\gamma=\frac{1}{2}$ and $B=\max_{t\in\llbracket
0,T\rrbracket}\big{(}|f(t)|+|g(t)|\big{)}+1$, which applies since $f(t)\leq
y_{j}(t)\leq g(t)$ holds for any $(t,j)\in\llbracket
0,T\rrbracket\times\llbracket 1,n\rrbracket$. In particular,
$\mathcal{P}V_{0}(\bm{y})\leq B\leq\gamma V_{0}(\bm{y})+B$ holds
deterministically. ∎

######  Lemma B.13.

There exists a constant $\alpha=\alpha(\bm{u},\bm{v},f,g)>0$ such that the
following holds. Letting $\nu_{0}$ denote the Lebesgue measure on the set
$\Omega_{0}$, we have $\mathcal{P}^{2n(T-1)}(\bm{y},A)\geq\alpha\nu_{0}(A)$,
for each $\bm{y}\in\Omega_{0}$ and any measurable subset
$A\subseteq(\mathbb{W}_{n})^{T-1}$.

###### Proof (Outline).

For any integer $T^{\prime}\geq 2$; two $n$-tuples
$\bm{u}^{\prime},\bm{v}^{\prime}\in\mathbb{W}_{n}$; and two functions
$f^{\prime},g^{\prime}:\llbracket
0,T^{\prime}\rrbracket\rightarrow\overline{\mathbb{R}}$, the density of the
measure
$\mathfrak{G}_{f^{\prime};g^{\prime}}^{\bm{u}^{\prime};\bm{v}^{\prime}}$ on
sequences
$\bm{\mathsf{x}}(t)=\big{(}\mathsf{x}_{1}(t),\mathsf{x}_{2}(t),\ldots,\mathsf{x}_{n}(t)\big{)}$
is

(B.25) $\displaystyle
C\cdot\textbf{1}_{\bm{\mathsf{x}}\in\Omega_{0}}\cdot\displaystyle\prod_{j=1}^{n}\Bigg{(}\textbf{1}_{\mathsf{x}_{j}(0)=u_{j}^{\prime}}\textbf{1}_{\mathsf{x}_{j}(T^{\prime})=v_{j}^{\prime}}\displaystyle\prod_{t=1}^{T^{\prime}}\exp\bigg{(}-\displaystyle\frac{1}{2}\big{(}\mathsf{x}_{j}(t)-\mathsf{x}_{j}(t-1)\big{)}^{2}\bigg{)}\displaystyle\prod_{t=1}^{T^{\prime}-1}d\mathsf{x}_{j}(t)\Bigg{)},$

for some normalization constant
$C=C(f^{\prime},\bm{u}^{\prime},\bm{v}^{\prime})>0$. Observe that there exist
some constant
$c_{1}=c_{1}(f^{\prime},g^{\prime},\bm{u}^{\prime},\bm{v}^{\prime})>0$ such
that $C>c_{1}$ (that is, the normalization constant is bounded below), since
the property $\bm{\mathsf{x}}\in\Omega_{0}$ is an open condition. Further
observe that the density (B.25) is uniformly bounded (since
$\big{|}f(t)\big{|}+\big{|}g(t)\big{|}$ is). Thus, there exists a constant
$c_{1}=c_{1}(f^{\prime};\bm{u}^{\prime};\bm{v}^{\prime})>0$ such that the
measure $\mathfrak{G}_{f^{\prime}}^{\bm{u}^{\prime};\bm{v}^{\prime}}$ is
absolutely continuous with respect to the Lebesgue measure on this set, and
its Radon–Nikodym derivative is bounded above by $c_{1}^{-1}$ and below by
$c_{1}$.

Adopting the notation of Definition B.7, we fix an integer $k\geq 1$ and apply
this in the case $(n^{\prime},T^{\prime})=(1,2)$;
$(\bm{u}^{\prime},\bm{v}^{\prime})=(u_{1}^{\prime},v_{1}^{\prime})=\big{(}\mathcal{P}^{k-1}y_{b}(a),\mathcal{P}^{k-1}y_{b}(a+2)\big{)}$;
and
$(f^{\prime},g^{\prime})=\big{(}f^{\prime}(1),g^{\prime}(1)\big{)}=\big{(}\mathcal{P}^{k-1}y_{b+1}(a+1),\mathcal{P}^{k-1}y_{b-1}(a+1)\big{)}$.
We then deduce (using the fact that the density (B.25) is bounded below by a
constant multiple of the Lebesgue measure on $\Omega_{0}$) that there is a
constant $c_{2}=c_{2}(\bm{u},\bm{v},f,g,R)>0$ such that

(B.26) $\displaystyle\mathbb{P}\big{[}\mathcal{P}^{k}y_{b}(a+1)\in
S_{k}\big{]}\geq c_{2}\displaystyle\int_{S_{k}}dy,$

for any measurable set
$S_{k}\subseteq\big{\\{}x\in\mathbb{R}:\mathcal{P}^{k-1}y_{b+1}(a+1)\leq
x\leq\mathcal{P}^{k-1}y_{b-1}(a+1)\big{\\}}$. Applying this $n(T-1)$ times
yields a constant $c_{3}=c_{3}(\bm{u},\bm{v},f,g,R)>0$ such that

(B.27) $\displaystyle\mathbb{P}\big{[}\mathcal{P}^{k+n(T-1)}\bm{y}\in
S\big{]}\geq c_{3}\cdot\nu_{0}(S),$

for any measurable set

$\displaystyle
S\subseteq\Big{\\{}\bm{x}=\big{(}\bm{x}(t)\big{)}_{t\in\llbracket
1,T-1\rrbracket}\in(\mathbb{W}_{n})^{T-1}:\mathcal{P}^{k}y_{j+1}(t)\leq
x_{j}(t)\leq g(t),\quad\text{for all $(t,j)$}\Big{\\}}.$

Applying (B.27) twice, first at $k=0$ and then at $k=n(T-1)$, yields the
lemma. ∎

Given these lemmas, we can quicky establish B.9.

###### Proof of B.9.

That $\mathfrak{G}_{f}^{\bm{u};\bm{v}}$ is stationary for $\mathcal{P}$
follows from B.8. Thus, the lemma follows from B.11, using B.12 and B.13 to
verify B.10. ∎

###  B.5. Proof of 2.18

In this section we establish 2.18. We begin with the following concentration
bound, which follows quickly from combining results of [61, 10], for the
eigenvalues of a GUE random matrix.

######  Lemma B.14 ([61, Corollary 5],[10, Corollary 1.6]).

There exists a constant $C>1$ such that the following holds. Letting $n\geq 1$
be an integer and
$(\lambda_{1},\lambda_{2},\ldots,\lambda_{n})\in\overline{\mathbb{W}}_{n}$
denote the eigenvalues of an $n\times n$ GUE matrix $\bm{G}=\bm{G}_{n}$, we
have

$\displaystyle\mathbb{P}\Bigg{[}\bigcup_{j=1}^{n}\Big{\\{}\big{|}\lambda_{j}-\gamma_{\operatorname{semci};n}(j)\big{|}\geq(\log
n)^{8}\cdot n^{-2/3}\cdot\min\\{$ $\displaystyle
j,n-j+1\\}^{-1/3}\Big{\\}}\Bigg{]}\leq Ce^{-(\log n)^{5}}.$

Now we can establish 2.18.

###### Proof of 2.18.

By 2.1, we may assume throughout that $u=v=0$. We may also assume by shifting
that $a=-b=T$, so that $(a,b)=(-T,T)$ for some $T\leq n^{D}$.

We next equate the law of $\bm{x}(t)$ with the eigenvalues of a suitably
scaled GUE matrix. To that end, let
$\bm{\lambda}(t)=\big{(}\lambda_{1}(t),\lambda_{2}(t),\ldots,\lambda_{n}(t)\big{)}\in\mathbb{W}_{n}$
denote Dyson Brownian motion with starting data $\bm{0}_{n}$. From the second
part of 2.15, $\bm{\lambda}(t)$ prescribes the law of $n$ non-intersecting
Brownian motions (of variance $n^{-1}$) run for time $t$, conditoned to never
intersect; from the first part of 2.15, it also prescribes the eigenvalues of
an $n\times n$ GUE matrix, scaled by $t^{1/2}$.

Next, recall that a Brownian bridge $B(t)$ on the interval $[-T,T]$,
conditioned to start and end at $0$ (that is, $B(-T)=B(T)=0$), has the same
law as $\frac{T-t}{(2T)^{1/2}}\cdot W\big{(}\frac{T+t}{T-t}\big{)}$, where
$W:[0,\infty)\rightarrow\mathbb{R}$ is a Brownian motion. This, implies (see
also [19, Equation (2)]) that

(B.28) $\displaystyle\text{$x_{j}(t)$ has the same law as
$(2T)^{-1/2}(T^{2}-t^{2})^{1/2}\cdot\lambda_{j}$};$

Given (B.28), the proofs of both parts of the lemma are very similar, obtained
by applying B.14 on a suitable mesh of $t$; we therefore only detail the proof
of the first statement of the lemma. To that end, let $C_{0}>1$ be the
constant from B.14. For any integer $j\in\llbracket 1,n\rrbracket$ and real
number $t\in[-T,T]$, define the events

$\displaystyle\mathscr{G}(t)=\bigcup_{j=1}^{n}\bigg{\\{}\Big{|}x_{j}(t)-(2T)^{-1/2}(T^{2}-t^{2})^{1/2}\cdot\gamma(j)\Big{|}\geq(\log
n)^{8}\cdot T^{1/2}n^{-2/3}\cdot\min\\{j,n-j+1\\}^{-1/3}\bigg{\\}};$
$\displaystyle\mathscr{G}_{0}=\bigcup_{j=1}^{n}\bigcup_{a\leq t<t+s\leq
b}\Big{\\{}\big{|}x_{j}(t+s)-x_{j}(t)\big{|}\geq
s^{1/2}n^{5}\log|4s^{-1}T|\Big{\\}};\quad\mathscr{G}_{1}=\bigcup_{j=-\lfloor
Tn^{20(D+1)}\rfloor}^{\lfloor
Tn^{20(D+1)}\rfloor}\mathscr{G}\Big{(}\displaystyle\frac{j}{n^{20(D+1)}}\Big{)},$

where we have abbreviated $\gamma(j)=\gamma_{\operatorname{semci};n}(j)$. By
(B.28), B.14, (the $B=n^{5}$ case of) 2.6, a union bound, and the fact that
$T\leq n^{D}$, there exist constants $c>0$ and $c_{1}=c_{1}(D)>1$ such that

(B.29) $\displaystyle\mathbb{P}[\mathscr{G}_{0}\cap\mathscr{G}_{1}]\leq
3C_{0}n^{20(D+1)}Te^{-c_{1}(\log n)^{5}}+c^{-1}e^{n/c-cn^{10}}\leq
2c_{1}^{-1}e^{-c(\log n)^{5}},$

for sufficiently large $n$. Moreover observe, since the function
$(T-t^{2}T^{-1})^{1/2}$ is $\frac{1}{2}$-Hölder on $[-T,T]$ with constant
$2^{1/2}$, on the event
$\mathscr{G}_{0}^{\complement}\cap\mathscr{G}_{1}^{\complement}$ we have for
each $j\in\llbracket 1,n\rrbracket$ that

(B.30) $\displaystyle\begin{aligned}
&\displaystyle\sup_{t\in[a,b]}\bigg{|}x_{j}(t)-\Big{(}\displaystyle\frac{T}{2}-\frac{t^{2}}{2T}\Big{)}^{1/2}\cdot\gamma(j)\bigg{|}\\\
&\quad\leq\displaystyle\max_{n^{20(D+1)}t\in\llbracket-
Tn^{20(D+1)},Tn^{20(D+1)}\rrbracket}\bigg{|}x_{j}(t)-(2T)^{-1/2}(T^{2}-t^{2})^{1/2}\cdot\gamma(j)\bigg{|}\\\
&\qquad+\displaystyle\sup_{t\in[a,b]}\Bigg{(}\sup_{s\in[0,n^{-20(D+1)}]}\big{|}x_{j}(t+s)-x_{j}(t)\big{|}\Bigg{)}\\\
&\qquad+\displaystyle\sup_{t\in[a,b]}\Bigg{(}\displaystyle\sup_{s\in[0,n^{-20(D+1)}]}\Big{|}\big{(}T^{2}-(t+s)^{2}\big{)}^{1/2}-(T^{2}-t^{2})^{1/2}\Big{|}\Bigg{)}\\\
&\quad\leq(\log n)^{8}\cdot
T^{1/2}n^{-2/3}\cdot\min\\{j,n-j+1\\}^{-1/3}+2n^{-10D-5}\cdot\log\big{(}10(T+1)Tn^{20(D+1)}\big{)}\\\
&\quad\leq 2(\log n)^{8}\cdot
T^{1/2}n^{-2/3}\cdot\min\\{j,n-j+1\\}^{-1/3}+n^{-D},\end{aligned}$

for sufficiently large $n$. The lemma then follows from (B.29) and (B.30). ∎

##  Appendix C Proofs of Results From Section 3

###  C.1. Proofs of 3.3, 3.5, and 3.7

In this section we establish first 3.3, then 3.5, and next 3.7.

###### Proof of 3.3.

Denote the bridge-limiting measure process on $[\widetilde{a},\widetilde{b}]$
with boundary data $(\mu_{\tilde{a}},\mu_{\tilde{b}})$ by
$(\widetilde{\mu}_{t}^{\star})_{t\in[\tilde{a},\tilde{b}]}$. Recalling the
ensemble $\bm{x}^{n}\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[a,b]\big{)}$, condition on $\bm{x}(t)$
for $t\notin[\widetilde{a},\widetilde{b}]$. Then, the law of the restriction
$\bm{y}^{n}=\bm{x}^{n}|_{[\tilde{a},\tilde{b}]}\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[\widetilde{a},\widetilde{b}]\big{)}$ of
$\bm{x}$ to $[\widetilde{a},\widetilde{b}]$ is given by
$\mathfrak{Q}^{\bm{x}^{n}(\tilde{a});\bm{x}^{n}(\tilde{b})}(An^{-1})$. Then
3.1 implies that, with probability $1-o(1)$ (that is, tending to $1$ as $n$
tends to $\infty$),

(C.1) $\displaystyle\text{the measure
$A\cdot\operatorname{emp}\big{(}\bm{y}^{n}(t)\big{)}$ converges under the
L\'{e}vy metric to $\mu_{t}^{\star}$, uniformly in
$t\in[\widetilde{a},\widetilde{b}]$}.$

In particular, (C.1) holds for $t\in\\{\widetilde{a},\widetilde{b}\\}$, and so
(with probability $1-o(1)$), the ensemble $\bm{y}$ satisfies the conditions of
3.1, with $(a,b,\mu_{a},\mu_{b})$ there equal to
$(\widetilde{a},\widetilde{b},\mu_{\tilde{a}},\mu_{\tilde{b}})$ here. Hence,
with probability $1-o(1)$, the measure
$A\cdot\operatorname{emp}\big{(}\bm{y}(t)\big{)}$ converges to
$\widetilde{\mu}_{t}^{\star}$, uniformly in
$t\in[\widetilde{a},\widetilde{b}]$. Together with (C.1), this yields the
lemma. ∎

###### Proof of 3.5.

If $\bm{x}^{n}=(x_{1}^{n},x_{2}^{n},\ldots,x_{n}^{n})\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[a,b]\big{)}$ is as in 3.1, then by 2.2
the ensemble
$\widetilde{\bm{x}}^{n}=(\widetilde{x}_{1}^{n},\widetilde{x}_{2}^{n},\ldots,\widetilde{x}_{n}^{n})\in\llbracket
1,n\rrbracket\times\mathcal{C}\big{(}[0,1]\big{)}$ defined by setting

$\displaystyle\widetilde{x}_{j}(t)=A^{-1/2}(b-a)^{-1/2}\cdot
x_{j}\big{(}a+(b-a)t\big{)},\qquad\text{for each $(j,t)\in\llbracket
1,n\rrbracket\times[0,1]$},$

is as in 3.1, with the $(a,b;A;\mu_{a},\mu_{b})$ there equal to
$(0,1;1;\widetilde{\mu}_{0},\widetilde{\mu}_{1})$ here. Then, recalling that
$\nu_{t}^{n}=\operatorname{emp}\big{(}\bm{x}^{n}(t)\big{)}$ and denoting
$\widetilde{\nu}_{t}=\operatorname{emp}\big{(}\widetilde{\bm{x}}^{n}(t)\big{)}$
for each $t\in[0,1]$, we have

$\displaystyle\widetilde{\nu}_{t}^{n}(I)=\nu_{a+t(b-a)}^{n}\big{(}A^{1/2}(b-a)^{1/2}\cdot
I\big{)},\qquad\text{for any interval $I\subseteq\mathbb{R}$}.$

Letting $n$ tend to $\infty$ and using 3.1, this verifies the first statement
of (3.5). Together with (3.3) and (3.4), this also quickly implies the second
and third statements of (3.5). ∎

###### Proof of 3.7.

The first statement of the lemma is due to [35, Corollary 2.8(b)], and the
second to [35, Theorem 2.7] (see also [7, Proposition 2.6(6)]). For the third,
the estimates on $\varrho_{t}^{\star}$ and $\partial_{x}\varrho_{t}^{\star}$
follow from the second part of the lemma, together with 2.10.

For the fourth, let us first show that $\varrho_{t}(x)$ is
$\frac{1}{7}$-Hölder in $x$. In particular, we claim that there exists a
constant $C_{1}>1$ such that, for any $t\in(0,1)$ and
$x,x^{\prime}\in\mathbb{R}$ with $|x-x^{\prime}|$ sufficiently small (in a way
only dependent on $t$), we have

(C.2)
$\displaystyle\big{|}\varrho_{t}^{\star}(x^{\prime})-\varrho_{t}^{\star}(x)\big{|}\leq
C_{1}|x-x^{\prime}|^{1/7}\big{(}t(1-t)\big{)}^{-1}.$

Assume to the contrary that this is false, so there exist $t\in(0,1)$ and
$x,x^{\prime}\in\mathbb{R}$ such that (C.2) does not hold; we may suppose by
symmetry that $x^{\prime}>x$. Then, since $\varrho_{t}^{\star}(x)\geq 0$ and
$\varrho_{t}^{\star}(x^{\prime})\geq 0$, we either have
$\varrho_{t}^{\star}(x)>C_{1}(x^{\prime}-x)^{1/7}(t-t^{2})^{-1}$ or
$\varrho_{t}^{\star}(x^{\prime})\geq C_{1}(x^{\prime}-x)^{1/7}(t-t^{2})^{-1}$;
let us assume the former holds. By the third part of the lemma, we have for
some constant $C_{2}>1$ that

$\displaystyle\big{|}\partial_{x}\varrho_{t}^{\star}(y)\big{|}\leq
C_{2}(x^{\prime}-x)^{-6/7}\big{(}t(1-t)\big{)}^{2},\qquad\text{whenever
$\varrho_{t}^{\star}(y)\geq\big{(}t(1-t)\big{)}^{-1}(x^{\prime}-x)^{1/7}$}.$

Integrating this bound over a maximal subset of $[x,x^{\prime}]$ on which
$\varrho_{t}^{\star}(y)\geq(t-t^{2})^{-1}|x-x^{\prime}|^{1/7}$, we find for
any such $y_{0}$ that

(C.3)
$\displaystyle\big{|}\varrho_{t}^{\star}(y_{0})-\varrho_{t}^{\star}(x)\big{|}\leq
C_{2}|y_{0}-x|\cdot(x^{\prime}-x)^{-6/7}\big{(}t(1-t)\big{)}^{2}\leq
C_{2}(x^{\prime}-x)^{1/7}\big{(}t(1-t)\big{)}^{2}.$

This implies for $C_{1}>C_{2}+2$ that
$\varrho_{t}^{\star}(y)\geq|x-x^{\prime}|^{1/7}(t-t^{2})^{-1}$ for any
$y\in[x,x^{\prime}]$, so (C.3) applies at $y_{0}=x^{\prime}$. This yields
(C.2), which is a contradiction, so (C.2) holds for any
$(t,x)\in(0,1)\times\mathbb{R}$.

Now fix any point $(t_{0},x_{0})\in(0,1)\times\mathbb{R}$ and sequence
$(t_{1},x_{1}),(t_{2},x_{2}),\ldots\in(0,1)\times\mathbb{R}$ with
$\lim_{j\rightarrow\infty}(t_{j},x_{j})=(t_{0},x_{0})$. For any real number
$\delta>0$, we have by the continuity of the measures $\mu_{t}^{\star}$ in $t$
that

$\displaystyle\lim_{j\rightarrow\infty}\int_{x_{0}-\delta}^{x_{0}+\delta}\varrho_{t_{j}}(x)dx=\int_{x_{0}-\delta}^{x_{0}+\delta}\varrho_{t_{0}}(x)dx.$

By (C.2), this implies for sufficiently large $j$ (dependent on $\delta$) that

$\displaystyle\big{|}\varrho_{t_{0}}(x_{0})-\varrho_{t_{j}}(x_{j})\big{|}$
$\displaystyle\leq(2\delta)^{-1}\Bigg{(}\displaystyle\int_{x_{0}-\delta}^{x_{0}+\delta}\Big{(}\big{|}\varrho_{t_{0}}(x)-\varrho_{t_{0}}(x_{0})\big{|}+\big{|}\varrho_{t_{j}}(x_{j})-\varrho_{t_{j}}(x_{0})\big{|}\Big{)}dx\Bigg{)}$
$\displaystyle\qquad+(2\delta)^{-1}\Bigg{|}\displaystyle\int_{x_{0}-\delta}^{x_{0}+\delta}\varrho_{t_{0}}(x)dx-\displaystyle\int_{x_{0}-\delta}^{x_{0}+\delta}\varrho_{t_{j}}(x)dx\Bigg{|}\leq
4C_{1}\big{(}t_{0}(1-t_{0})\big{)}^{-1}\delta^{1/7}+\delta,$

which upon letting $\delta$ tend to $0$ yields the continuity of
$\varrho_{t}(x)$ in $(t,x)$. ∎

###  C.2. Proofs of 3.12, 3.13, and 3.23

In this section we establish first 3.12, then 3.13, and next 3.23.

###### Proof of 3.12.

Expanding the second and first equations in (3.17) yields

(C.4)
$\displaystyle\partial_{t}\varrho+u\partial_{x}\varrho+\varrho\partial_{x}u=0;\qquad
u\partial_{t}\varrho+\varrho\partial_{t}u+2u\varrho\partial_{x}u+u^{2}\partial_{x}\varrho-\pi^{2}\varrho^{2}\partial_{x}\varrho=0.$

The first equation of (C.4) is equivalent to the imaginary part of (3.19).
Moreover, subtracting the second equation of (C.4) from the first one
multiplied by $u$ then yields

$\displaystyle\varrho(\partial_{t}u+u\partial_{x}u-\pi^{2}\varrho\partial_{x}\varrho)=0.$

Since $\varrho_{t}(x)>0$ for $(t,x)\in\Omega$, this implies
$\partial_{t}u+u\partial_{x}u-\pi^{2}\varrho\partial_{x}\varrho=0$ on
$\Omega$, which is equivalent to the real part of (3.19); this verifies
(3.19). ∎

###### Proof of 3.13.

Let us first show that $H^{\star}$ is locally Lispchitz; that this holds in
directions parallel to the $x$-axis follows from the definition (3.3) and the
bound on $\varrho_{t}^{\star}$ from the third part of 3.7. It remains to show
that $H^{\star}$ is locally Lipschitz in directions parallel to the $t$-axis.

To that end, fix points $(t_{1},x_{0}),(t_{2},x_{0})\in(0,1)\times\mathbb{R}$
with $t_{1}<t_{2}$; by the second part of 3.7, there exists some real number
$x_{1}>x_{0}$ such that $\varrho_{t}(x)=0$ for each $x>x_{1}-1$ and
$t\in(0,1)$. For any real number $\varepsilon>0$, let
$\varphi_{\varepsilon}:(0,1)\rightarrow\mathbb{R}$ denote a smooth
approximation of the indicator function
$\textbf{1}_{t\in[t_{1},t_{2}]}\cdot\textbf{1}_{x\in[x_{0},x_{1}]}$.
Explicitly, fix a smooth function
$\psi:\mathbb{R}^{2}\rightarrow\mathbb{R}_{\geq 0}$ with
$\operatorname{supp}\psi\subseteq[-1,1]\times[-1,1]$ and with total integral
$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi(t,x)dtdx=1$, and define
$\psi_{\varepsilon}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by setting
$\psi_{\varepsilon}(t,x)=\varepsilon^{-2}\psi(\varepsilon^{-1}t,\varepsilon^{-1}x)$
for each $(t,x)\in\mathbb{R}^{2}$. Then, set

$\displaystyle\varphi_{\varepsilon}(t,x)=\displaystyle\int_{t_{1}}^{t_{2}}\displaystyle\int_{x_{0}}^{x_{1}}\psi_{\varepsilon}(t-s,x-w)dwds.$

Since $(u^{\star},\varrho^{\star})$ weakly solves the equation
$\partial_{t}\varrho^{\star}=\partial_{x}(u^{\star}\varrho^{\star})$, we have

(C.5)
$\displaystyle\displaystyle\int_{0}^{1}\displaystyle\int_{-\infty}^{\infty}\varrho_{t}^{\star}(x)\partial_{t}\varphi_{\varepsilon}(t,x)dxdt+\displaystyle\int_{0}^{1}\displaystyle\int_{-\infty}^{\infty}u_{t}^{\star}(x)\varrho_{t}^{\star}(x)\partial_{x}\varphi_{\varepsilon}(t,x)dxdt=0.$

The continuity of $\varrho$ (due to the fourth part of 3.7), and the facts
that $\psi_{\varepsilon}$ has total integral equal to $1$ and that
$\operatorname{supp}\psi_{\varepsilon}\subseteq[-\varepsilon,\varepsilon]\times[-\varepsilon,\varepsilon]$,
imply

(C.6) $\displaystyle\begin{aligned} \displaystyle\lim_{\varepsilon\rightarrow
0}&\displaystyle\int_{0}^{1}\displaystyle\int_{-\infty}^{\infty}\varrho_{t}^{\star}(x)\partial_{t}\varphi_{\varepsilon}(t,x)dxdt\\\
&=\displaystyle\lim_{\varepsilon\rightarrow
0}\displaystyle\int_{0}^{1}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{x_{0}}^{x_{1}}\varrho_{t}^{\star}(x)\big{(}\psi_{\varepsilon}(t-t_{1},x-w)-\psi_{\varepsilon}(t-t_{2},x-w)\big{)}dwdxdt\\\
&=\displaystyle\int_{x_{0}}^{x_{1}}\big{(}\varrho_{t_{1}}^{\star}(x)-\varrho_{t_{2}}^{\star}(x)\big{)}dx=H_{t_{1}}^{\star}(x_{0})-H_{t_{2}}^{\star}(x_{0}),\end{aligned}$

where we used the fact that $H_{t_{j}}^{\star}(x_{1})=0$ for each
$j\in\\{1,2\\}$, since $\varrho_{t_{j}}^{\star}(x)=0$ for $x>x_{1}$.

Next, [35, Corollary 2.8(c)] states that, for any constant
$\omega\in\big{(}0,\frac{1}{2}\big{)}$, there exists a constant
$C=C(\omega,\mu_{0},\mu_{1})>1$ satisfying

(C.7)
$\displaystyle\displaystyle\sup_{x\in\mathbb{R}}\displaystyle\sup_{t\in[\omega,1-\omega]}\big{|}u_{t}^{\star}(x)\varrho_{t}^{\star}(x)\big{|}<C.$

We further have

(C.8) $\displaystyle\begin{aligned} \displaystyle\lim_{\varepsilon\rightarrow
0}&\displaystyle\int_{0}^{1}\displaystyle\int_{-\infty}^{\infty}u_{t}^{\star}(x)\varrho_{t}^{\star}(x)\partial_{x}\varphi_{\varepsilon}(t,x)dxdt\\\
&=-\displaystyle\lim_{\varepsilon\rightarrow
0}\displaystyle\int_{0}^{1}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{t_{1}}^{t_{2}}\displaystyle\int_{x_{0}}^{x_{1}}u_{t}^{\star}(x)\varrho_{t}^{\star}(x)\partial_{w}\psi_{\varepsilon}(t-s,x-w)dwdsdxdt\\\
&=\displaystyle\lim_{\varepsilon\rightarrow
0}\displaystyle\int_{0}^{1}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{t_{1}}^{t_{2}}u_{t}^{\star}(x)\varrho_{t}^{\star}(x)\big{(}\psi_{\varepsilon}(t-s,x-x_{0})-\psi_{\varepsilon}(t-s,x-x_{1})\big{)}dsdxdt\\\
&=\displaystyle\lim_{\varepsilon\rightarrow
0}\displaystyle\int_{0}^{1}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{t_{1}}^{t_{2}}u_{t+s}^{\star}(x+x_{0})\varrho_{t+s}^{\star}(x+x_{0})\psi_{\varepsilon}(t,x)dsdxdt,\end{aligned}$

where in first equality we used the fact that
$\partial_{x}\psi_{\varepsilon}(t-s,x-w)=-\partial_{w}\psi_{\varepsilon}(t-s,x-w)$;
in the second we performed the integration; and in the third we used the fact
that $u_{t}^{\star}(x)\varrho_{t}^{\star}(x)\psi_{\varepsilon}(t-s,w-x_{1})=0$
for sufficiently small $\varepsilon$ (as $\varrho_{t}^{\star}(x)=0$ for $x\geq
x_{1}-1$ and
$\operatorname{supp}\psi_{\varepsilon}\subseteq[-\varepsilon,\varepsilon]\times[-\varepsilon,\varepsilon]$)
and changed variables from $(t,x)$ to $(t+s,x+x_{0})$. Now, since
$\psi_{\varepsilon}$ is nonnegative with total integral equal to $1$,
inserting (C.8) and (C.6) into (C.5), and using (C.7) yields
$\big{|}H_{t_{1}}^{\star}(x_{0})-H_{t_{2}}^{\star}(x_{0})\big{|}\leq C_{1}$
for some constant $C_{1}=C_{1}(t_{1},t_{2},\mu_{1},\mu_{2})>1$. In particular,
$H_{t}^{\star}$ is locally Lipschitz.

It remains to verify (3.20); its first statement follows from (3.3), so it
remains to confirm its second. To that end, the Lebesgue differentiation
theorem yields, for almost all $(s,x_{0})\in(0,1)\times\mathbb{R}$,

$\displaystyle\displaystyle\lim_{\varepsilon\rightarrow
0}\displaystyle\int_{0}^{1}\displaystyle\int_{-\infty}^{\infty}u_{t+s}^{\star}(x+x_{0})\varrho_{t+s}^{\star}(x+x_{0})\psi_{\varepsilon}(t,x)dxdt=u_{s}^{\star}(x_{0})\varrho_{s}^{\star}(x_{0}).$

With (C.7), this implies for almost every $x_{0}\in\mathbb{R}$ that

$\displaystyle\displaystyle\lim_{\varepsilon\rightarrow 0}$
$\displaystyle\displaystyle\int_{0}^{1}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{t_{1}}^{t_{2}}u_{t+s}^{\star}(x+x_{0})\varrho_{t+s}^{\star}(x+x_{0})\psi_{\varepsilon}(t,x)dsdxdt=\displaystyle\int_{t_{1}}^{t_{2}}u_{s}^{\star}(x_{0})\varrho_{s}^{\star}(x_{0})ds.$

Together this, (C.5), (C.6), and (C.8) gives for almost all
$x_{0}\in\mathbb{R}$ that
$H_{t_{2}}^{\star}(x_{0})-H_{t_{1}}^{\star}(x_{0})=\int_{t_{1}}^{t_{2}}u_{s}^{\star}(x_{0})\varrho_{s}^{\star}(x_{0})ds$,
which establishes the second part of (3.20). ∎

###### Proof of 3.23.

We only establish the first part of the lemma, as the proofs of the second and
third are entirely analogous. To do this, it suffices to show that $H^{\star}$