# Edge Rigidity of Dyson Brownian Motion with General Initial Data

Amol Aggarwal1,2 and  Jiaoyang Huang3 Department of Mathematics, Columbia
University  Clay Mathematics Institute  Department of Statistics and Data
Science, University of Pennsylvania

###### Abstract.

In this paper, we study the edge behavior of Dyson Brownian motion with
general $\beta$. Specifically, we consider the scenario where the averaged
initial density near the edge, on the scale $\eta_{*}$, is lower bounded by a
square root profile. Under this assumption, we establish that the fluctuations
of extreme particles are bounded by $(\log
n)^{\operatorname{\mathcal{O}}(1)}n^{-2/3}$ after time $C\sqrt{\eta_{*}}$. Our
result improves previous edge rigidity results from [24, 1] which require both
lower and upper bounds of the averaged initial density. Additionally,
combining with [24], our rigidity estimates are used to prove that the
distribution of extreme particles converges to the Tracy-Widom $\beta$
distribution in short time.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Edge Rigidity and Universality
  4. 4 Examples

##  1\. Introduction

###  1.1. Preface

In 1962 [17], Dyson interpreted the Gaussian orthogonal and unitary ensembles
as dynamical limit of the matrix valued Brownian motion, which is given by

(1.1) $\displaystyle{\rm d}H(t)={\rm d}B(t),$

where $B(t)$ is the Brownian motion on real symmetric or complex Hermitian
matrices. It turns out the eigenvalues of the above matrix valued Brownian
motion satisfy a system of stochastic differential equations. These equations
have been later generalized to stochastic differential equations, called the
$\beta$-Dyson Brownian motion,

(1.2) $\displaystyle{\rm d}\lambda_{i}(t)=\Big{(}\displaystyle\frac{2}{\beta
n}\Big{)}^{1/2}{\rm
d}B_{i}(t)+\frac{1}{n}\displaystyle\sum_{\begin{subarray}{c}1\leq j\leq n\\\
j\neq i\end{subarray}}\displaystyle\frac{{\rm
d}t}{\lambda_{i}(t)-\lambda_{j}(t)},\quad 1\leq i\leq n,$

where
${\bm{\lambda}}^{(n)}(t)={\bm{\lambda}}(t)=(\lambda^{(n)}_{1}(t),\lambda^{(n)}_{2}(t),\cdots,\lambda_{n}^{(n)}(t))=(\lambda_{1}(t),\lambda_{2}(t),\cdots,\lambda_{n}(t))\in\overline{\mathbb{W}}_{n}$
lies in the closure of the Weyl chamber
$\mathbb{W}_{n}=\\{\bm{y}\in\mathbb{R}^{n}:y_{1}>y_{2}>\cdots>y_{n}\\}$ and
$\\{B_{i}(t)\\}_{1\leq i\leq n}$ are independent Brownian motions. The real
symmetric and complex Hermitian matrix valued Brownian motion corresponds to
(1.2) with $\beta=1$ and $\beta=2$ respectively.

Dyson Brownian motion is a prototype model for a one-dimension interacting
particle system with singular interactions, which capture the statistical
properties of energy levels in complex systems. The understanding of Dyson
Brownian motion plays an important role in the proof of Wigner’s original
universality conjecture by Erdős, Schlein and Yau [18]. For a review about the
general framework regarding the proof of the universality conjecture, one can
read the book [19].

In this paper, we investigate the edge behavior of Dyson Brownian motion (DBM)
described by equation (1.2), considering various initial data. If the initial
data is a delta mass at zero, the law of Dyson Brownian motion at time one
corresponds to the Gaussian-$\beta$-ensemble distribution. Notably, the
largest particle fluctuates exhibits fluctuations on the scale of $n^{-2/3}$,
commonly referred to as optimal edge rigidity. Moreover, its fluctuations
converge to the Tracy-Widom $\beta$ distribution [27, 11, 22]. A natural
question arises: _when starting from general initial data, how long does it
take to observe optimal edge rigidity, characterized by $n^{-2/3}$
fluctuations?_

It has been proven in [24] ($\beta=1,2$) and [1] (general $\beta$), if the
initial data has a spectral edge with square root behavior down to a scale
$\eta_{*}$, then after time $C\sqrt{\eta_{*}}$, the extreme particle
concentrates on the scale $n^{-2/3+\operatorname{o}(1)}$. However, when the
initial data contains an outlier, the extreme eigenvalue initially experiences
fluctuations on the scale of $n^{-1/2}$, and after time
$\operatorname{\mathcal{O}}(1)$, it merges with the bulk of the spectral
distribution, resulting in $n^{-2/3+\operatorname{o}(1)}$ fluctuations. This
phenomenon is commonly referred to as the Baik–Ben Arous–Péché (BBP) phase
transition [5, 7, 8].

In this paper, we propose a general criterion that encompasses both
aforementioned cases and is believed to be necessary (up to constants).
Specifically, we demonstrate that if on the scale $\eta_{*}$, the averaged
density near the edge is lower bounded by a square root profile (without
imposing the analogous upper bound), then the fluctuations of extreme
particles are bounded by $(\log n)^{\operatorname{\mathcal{O}}(1)}n^{-2/3}$
after time $C\sqrt{\eta_{*}}$. Combining with [24], our optimal edge rigidity
estimates are used to give a proof of the local ergodicity of Dyson Brownian
motion for more general initial data, i.e., the distribution of extreme
particles converges to Tracy-Widom $\beta$ distribution in short time.

The limiting empirical particle density of Dyson Brownian motion can be
expressed as the free convolution of the initial density with the rescaled
semicircle distribution, which corresponds to the solution of a complex
Burgers equation [4]. To establish the edge rigidity, we employ a refined
comparison between the empirical particle density of Dyson Brownian motion and
a deterministic measure-valued process represented by the solution of the
complex Burgers equation. This solution can be obtained using the method of
characteristics. The Stieltjes transform of the empirical particle density
satisfies the same complex Burgers equation, with the addition of a stochastic
term and some small error terms. By plugging the characteristic flow of the
complex Burgers equation, we can analyze the differences between the two
Stieltjes transforms using estimates of Grönwall type. The method of
characteristics has previously been employed in different contexts related to
Dyson Brownian motion. In [20], it was utilized to investigate the bulk
behavior of Dyson Brownian motion. In [1], it was applied to study the edge
behavior of Dyson Brownian motion when starting from a profile exhibiting
square root behavior. Furthermore, the method of characteristics was utilized
in [9] to study the extreme gaps between eigenvalues of Wigner matrices, and
in [2], to analyze particle rigidity in unitary Dyson Brownian motion.

Our case poses additional challenges as we only assume a lower bound for the
averaged initial density near the edge, without assuming square root behavior.
As a crucial initial step, we establish that if the averaged initial density
near the edge satisfies this lower bound, it continues to hold within a finite
time frame. To prove this, we develop a coupling technique for Dyson Brownian
motions (see 2.10) with different initial data. This coupling ensures that if
one Dyson Brownian motion exhibits larger gaps between particles at the
beginning, this property persists for all subsequent times. Notably, this
coupling technique may have independent significance beyond our specific
study.

In previous work [1], a key insight relied on the observation that the
characteristic flows near the edge progress at a slower rate compared to the
movement of the spectral edge, which is easily obtained when the density has
square root behavior. In our case, we undertake a more meticulous analysis of
the equations governing the relative movement of our characteristics with
respect to the spectral edge. As a result, we establish two quantitative
versions (see 3.2 and 3.3) of these estimates when the averaged density near
the edge is lower bounded by a square root profile. These two estimates,
combined with a stopping time argument, enable us to prove that the Stieltjes
transform of the empirical particle density remains small outside the right
edge of the free-convolution density. Consequently, there are no outliers, and
optimal edge rigidity is guaranteed.

### Notation.

Let $\mathbb{H}=\\{z\in\mathbb{C}:\operatorname{Im}z>0\\}$ denote the upper
half-plane and $\overline{\mathbb{H}}$ denote its closure. For any real
numbers $a,b\in\mathbb{R}$ with $a<b$, we set $\llbracket
a,b\rrbracket=[a,b]\cap\mathbb{Z}$. We will write
$X=\operatorname{\mathcal{O}}(Y)$ to mean that $|X|\leq CY$ for some universal
constant $C>0$. We also write $a\wedge b=\min\\{a,b\\}$ and $a\vee
b=\max\\{a,b\\}$. Let
$\mathscr{P}_{\operatorname{fin}}=\mathscr{P}_{\operatorname{fin}}(\mathbb{R})$
denote the set of nonnegative measures $\mu$ on $\mathbb{R}$ with finite total
mass, $\mu(\mathbb{R})<\infty$. Further let
$\mathscr{P}=\mathscr{P}(\mathbb{R})\subset\mathscr{P}_{\operatorname{fin}}$
denote the set of probability measures on $\mathbb{R}$; the support of any
measure $\nu\in\mathscr{P}$ is denoted by $\operatorname{supp}\nu$. We say
that a probability measure $\mu\in\mathscr{P}$ has density $\varrho$ (with
respect to Lebesgue measure) if $\varrho:\mathbb{R}\rightarrow\mathbb{R}$ is a
measurable function satisfying $\mu(dx)=\varrho(x)dx$. For any real number
$x\in\mathbb{R}$, we let $\delta_{x}\in\mathscr{P}$ denote the delta function
at $x$. Given two measures $\mu,\nu\in\mathscr{P}_{\operatorname{fin}}$ of
finite total mass, the Lévy distance between them is

(1.3) $\displaystyle d_{{\rm
L}}(\mu,\nu)=\inf\Bigg{\\{}a>0:\displaystyle\int_{-\infty}^{y-a}\mu(dx)-a\leq\displaystyle\int_{-\infty}^{y}\nu(dx)\leq\displaystyle\int_{-\infty}^{y+a}\mu(dx)+a,\quad\text{for
all $y\in\mathbb{R}$}\Bigg{\\}}.$

For two sets $A,B\in{\mathbb{R}}$ we denote
$\operatorname{dist}(A,B)=\inf_{x\in A,y\in B}|x-y|$. When $A=\\{x\\}$ is a
single point, we simply write
$\operatorname{dist}(\\{x\\},B)=\operatorname{dist}(x,B)$. We denote
$\overline{{\mathbb{R}}}={\mathbb{R}}\cup\\{\pm\infty\\}$ and for any
$\lambda\in\mathbb{R}$ make the convention

(1.4)
$\displaystyle+\infty>\lambda>-\infty,\quad\infty-\lambda,\lambda-(-\infty)=+\infty,\quad(-\infty)-\lambda,\lambda-\infty=-\infty.$

### Organization.

We present the main results in the rest of Section 1. In Section 2, we collect
some preliminary results on free convolution with semicircle distributions,
previous results on Dyson Brownian motion and couplings of Dyson Brownian
motions with different initial data. In Section 3, we prove our main results
on the edge rigidity and universality of Dyson Brownian motion from general
initial data. Finally in Section 4, we discuss some examples and applications
of our optimal rigidity result.

###  1.2. Main Results

We fix $\beta\geq 1$ and consider Dyson Brownian motion (1.2) with general
initial data. We denote the empirical particle density of the $n$-particle
Dyson Brownian motion ${\bm{\lambda}}(t)$ by

(1.5)
$\displaystyle\mu_{t}^{(n)}=\frac{1}{n}\sum_{i=1}^{n}\delta_{\lambda_{i}(t)},\quad
t\geq 0.$

######  Remark 1.1.

In some cases (we will state when explicitly), we allow some particles to be
at $\pm\infty$. More precisely, we allow for some $1\leq\ell\leq m\leq n$ to
have $\lambda_{1}(t)=\lambda_{2}(t)=\cdots\lambda_{\ell-1}(t)=+\infty$ and
$\lambda_{m+1}(t)=\lambda_{m+2}(t)=\cdots\lambda_{n}(t)=-\infty$. In this
case, recalling our convention (1.4), (1.2) is a Dyson Brownian motion with
$m-\ell+1$ particles, indexed by $\llbracket\ell,m\rrbracket$. In this case,
the empirical density is a measure valued process, with total mass
$\mu_{t}^{(n)}({\mathbb{R}})=(m-\ell+1)/n$.

We study $n$-particle Dyson Brownian motion, with initial data

(1.6)
$\displaystyle\mu_{0}^{(n)}=\frac{1}{n}\sum_{i=1}^{n}\delta_{\lambda_{i}(0)}.$

######  Assumption 1.2.

Fix constants $\mathsf{T}\geq 100$ and $\mathsf{b}\in(0,1)$. Take a small
parameter $\eta_{*}=\eta_{*}(n)\in(0,1/4)$ (possibly depending on $n$). Assume
the initial data $\mu_{0}^{(n)}$ (as in (1.6)) is supported on $[-\infty,0]$,
satisfying $\mu_{0}^{(n)}([-x,0])\geq{\mathsf{b}}x^{3/2}$ for all
$\eta_{*}\leq x\leq{\mathsf{T}}^{2}$.

We denote by $\mu_{t}=\mu_{0}^{(n)}\boxplus\mu_{\operatorname{sc}}^{(t)}$ the
free convolution of $\mu_{0}^{(n)}$ and the rescaled semicircle distribution
$\mu_{\operatorname{sc}}^{(t)}$ (see (2.5)); see Section 2.1 for a more
detailed definition. Adopt 1.2. We will prove in 2.14, that for
$(30/{\mathsf{b}})\sqrt{\eta_{*}}\leq t\leq{\mathsf{T}}$, $\mu_{t}({\rm
d}x)=\varrho_{t}(x){\rm d}x$ has square root behavior close to the right edge
$E_{t}:=\sup(\operatorname{supp}\mu_{t})$ of the support of $\mu_{t}$, that
is,

(1.7) $\displaystyle\varrho_{t}(x){\rm
d}x=(1+\operatorname{\mathcal{O}}(|E_{t}-x|))\frac{\sqrt{{\mathcal{A}}(t)(E_{t}-x)}}{\pi},\quad
x\rightarrow E_{t},$

where ${\mathcal{A}}(t)={\mathcal{A}}^{\mu_{0}}(t)$ is a scalar depending on
$\mu_{0}$ and $t$. We refer to 2.14 for the explicit formula.

######  Theorem 1.3.

Adopt 1.2, and let ${\mathfrak{B}}=2^{60}{\mathsf{b}}^{-6}>0$. Then there
exists a large $C=C({\mathsf{b}},{\mathsf{T}})>1$ such that the following
holds with probability $1-Ce^{-(\log n)^{2}}$. For all
$t\in[{\mathfrak{B}}\sqrt{\eta_{*}},{\mathsf{T}}]$, we have

(1.8) $\displaystyle\lambda_{1}(t)\leq E_{t}+\frac{(\log n)^{15}}{n^{2/3}}.$

######  Theorem 1.4.

Adopt 1.2; fix real numbers $0<{\mathfrak{a}}<1\leq\Theta$ and a smooth,
compactly supported test function $F:\mathbb{R}\rightarrow\mathbb{R}$; and let
${\mathfrak{B}}=2^{60}{\mathsf{b}}^{-6}>0$ and
${\mathfrak{t}}=\max\\{2{\mathfrak{B}}\sqrt{\eta_{*}},n^{-1/3+{\mathfrak{a}}}\\}$.
There exists a constant
$\delta=\delta(\mathsf{b},\mathsf{T},\mathfrak{a},\Theta)>0$ such that the
following holds for any sufficiently large integer $n$. If for each
${\mathfrak{t}}/2\leq t\leq{\mathsf{T}}$ we have
${\mathcal{A}}(t)\in[\Theta^{-1},\Theta]$ (recall (1.7)) and
$\Theta^{-1}\sqrt{x}\leq\varrho_{t}(E_{t}-x)\leq\Theta\sqrt{x}$ for $0\leq
x\leq\Theta^{-1}$, then

(1.9)
$\displaystyle\begin{split}\mathbb{E}[F(n^{2/3}{\mathcal{A}}(t)^{1/3}(E_{t}-\lambda_{1}))]=\mathbb{E}_{\emph{Airy}_{\beta}}[F(\Lambda_{1})]+\operatorname{\mathcal{O}}\left(n^{-\delta}\right),\quad\text{for
any}\quad{\mathfrak{t}}\leq t\leq{\mathsf{T}}-\Theta,\end{split}$

where the second expectation is with respect to the Tracy–Widom $\beta$
distribution $TW_{\beta}$ of [27, Theorem 1.3].

The proofs of 1.3 and 1.4 will be given in Section 3.

### Acknowledgements

Amol Aggarwal was partially supported by a Packard Fellowship for Science and
Engineering, a Clay Research Fellowship, by the NSF grant DMS-1926686, and by
the IAS School of Mathematics. The research of Jiaoyang Huang is supported by
NSF grant DMS-2054835 and DMS-2331096. Amol Aggarwal and Jiaoyang Huang also
wish to acknowledge the NSF grant DMS-1928930, which supported their
participation in the Fall 2021 semester program at MSRI in Berkeley,
California titled, “Universality and Integrability in Random Matrix Theory and
Interacting Particle Systems.”

##  2\. Preliminaries

###  2.1. Free Convolution With Semicircle Distributions

In this section we recall various results concerning Stieltjes transforms and
free convolutions with the semicircle distribution. To that end, fix a finite
measure $\mu\in\mathscr{P}_{\operatorname{fin}}$. We define the _Stieltjes
transform_ of $\mu$ to be the function
$m=m^{\mu}:\mathbb{H}\rightarrow\mathbb{H}$ for any complex number
$z\in\mathbb{H}$ setting

(2.1) $\displaystyle
m(z)=\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\mu({\rm
d}x)}{x-z}.$

We have the following estimates on the Stieltjes transform and its
derivatives.

######  Lemma 2.1.

Let $m(z)=m_{\mu}(z)$ be the Stieltjes transform of a finite measure
$\mu\in\mathscr{P}_{\rm fin}$ with $A=\mu({\mathbb{R}})$. For any integer
$p\geq 1$, we denote its $p$-th derivative by $m^{(p)}(z)$. Then,

(2.2)
$\displaystyle|m(z)|\leq\frac{A}{\operatorname{dist}(z,\operatorname{supp}(\mu))},\quad\quad|m^{\prime}(z)|\leq\frac{{\rm
Im}[m(z)]}{{\rm
Im}[z]},\quad|m^{(p)}(z)|\leq\frac{p!A}{\operatorname{dist}(z,\operatorname{supp}(\mu))^{p+1}}$

###### Proof.

The Stieltjes transform and its derivatives are given by

$\displaystyle|m(z)|=\left|\int_{\mathbb{R}}\frac{{\rm
d}\mu(x)}{x-z}\right|\leq\int_{\mathbb{R}}\frac{{\rm
d}\mu(x)}{|z-x|}\leq\frac{A}{\operatorname{dist}(z,\operatorname{supp}(\mu))},$

and

$\displaystyle|m^{(p)}(z)|=\left|\int_{\mathbb{R}}\frac{p!{\rm
d}\mu(x)}{(x-z)^{p+1}}\right|\leq\int_{\mathbb{R}}\frac{p!{\rm
d}\mu(x)}{|z-x|^{p+1}}\leq\frac{p!A}{\operatorname{dist}(z,\operatorname{supp}(\mu))^{p+1}}.$

Finally for the middle statement of (2.2), we have

$\displaystyle|m^{\prime}(z)|=\left|\int_{\mathbb{R}}\frac{{\rm
d}\mu(x)}{(x-z)^{2}}\right|\leq\int_{\mathbb{R}}\frac{{\rm
d}\mu(x)}{|z-x|^{2}}=\frac{1}{{\rm Im}[z]}\int_{\mathbb{R}}\frac{{\rm
Im}[z]{\rm d}\mu(x)}{|z-x|^{2}}=\frac{{\rm Im}[m(z)]}{{\rm Im}[z]}.$

∎

If $\mu$ has a density with respect to Lebesgue measure, that is, $\mu({\rm
d}x)=\varrho(x){\rm d}x$ for some $\varrho\in L^{1}(\mathbb{R})$, then
$\varrho$ can be recovered from its Stieltjes transform by the identity [26,
Equation (8.14)],

(2.3) $\displaystyle\pi^{-1}\displaystyle\lim_{y\rightarrow
0}\operatorname{Im}m(x+\mathrm{i}y)=\varrho(x);\qquad\displaystyle\pi^{-1}\lim_{y\rightarrow
0}\operatorname{Re}m(x+\mathrm{i}y)=H\varrho(x),$

for any $x\in\mathbb{R}$. In the latter, $Hf$ denotes the Hilbert transform of
any function $f\in L^{1}(\mathbb{R})$, given by

(2.4) $\displaystyle
Hf(x)=\pi^{-1}\cdot\operatorname{PV}\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{f(w)dw}{w-x},$

where $\operatorname{PV}$ denotes the Cauchy principal value (assuming the
integral exists as a principal value).

The _semicircle distribution_ is a measure
$\mu_{\operatorname{sc}}\in\mathscr{P}(\mathbb{R})$ whose density
$\varrho_{\operatorname{sc}}:\mathbb{R}\rightarrow\mathbb{R}_{\geq 0}$ with
respect to the Lebesgue measure is given by

(2.5)
$\displaystyle\varrho_{\operatorname{sc}}(x)=\displaystyle\frac{(4-x^{2})^{1/2}}{2\pi}\cdot\textbf{1}_{x\in[-2,2]},\qquad\text{for
all $x\in\mathbb{R}$}.$

For any real number $t>0$, we denote the rescaled semicircle density
$\varrho_{\operatorname{sc}}^{(t)}$ and distribution
$\mu_{\operatorname{sc}}^{(t)}\in\mathscr{P}$ by

(2.6)
$\displaystyle\varrho_{\operatorname{sc}}^{(t)}(x)=t^{-1/2}\varrho_{\operatorname{sc}}(t^{-1/2}x);\qquad\mu_{\operatorname{sc}}^{(t)}=\varrho_{\operatorname{sc}}^{(t)}(x)dx.$

We next discuss the free convolution of a probability measure
$\mu\in\mathscr{P}$ with the (rescaled) semicircle distribution
$\mu_{\operatorname{sc}}^{(t)}$. For any $t>0$, denote the function
$M=M^{\mu}=M^{t;\mu}:\mathbb{H}\rightarrow\mathbb{C}$ and the set
$\Lambda_{t}=\Lambda_{t;\mu}\subseteq\mathbb{H}$ by (see Figure 1)

(2.7) $\displaystyle
M(z)=z-tm(z);\quad\Lambda_{t}=\Big{\\{}z\in\mathbb{H}:\operatorname{Im}\big{(}z-tm(z)\big{)}>0\Big{\\}}=\Bigg{\\{}z\in\mathbb{H}:\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\mu(dx)}{|z-x|^{2}}<\displaystyle\frac{1}{t}\Bigg{\\}}.$

Figure 1. $\Lambda_{t}$ as defined in (2.7) is an open subset of the upper
half plane $\mathbb{H}$. $M(z)$ is a holomorphic map from $\Lambda_{t}$ to
$\mathbb{H}$.

######  Lemma 2.2 ([6, Lemma 4]).

The function $M$ is a homeomorphism from $\overline{\Lambda}_{t}$ to
$\overline{\mathbb{H}}$. Moreover, it is a holomorphic map from $\Lambda_{t}$
to $\mathbb{H}$ and a bijection from $\partial\Lambda_{t}$ to $\mathbb{R}$.

For any real number $t\geq 0$, define
$m_{t}=m_{t}^{\mu}:\mathbb{H}\rightarrow\mathbb{H}$ as follows. First set
$m_{0}(z)=m(z)$; for any real number $t>0$, define $m_{t}$ so that

(2.8) $\displaystyle m_{t}\big{(}z-tm_{0}(z)\big{)}=m_{0}(z),\qquad\text{for
any $z\in\Lambda_{t}$}.$

Since by 2.2 the function $M(z)=z-tm_{0}(z)$ is a bijection from $\Lambda_{t}$
to $\mathbb{H}$, (2.8) defines $m_{t}$ on $\mathbb{H}$. By [6, Proposition 2],
$m_{t}$ is the Stieltjes transform of a measure
$\mu_{t}\in\mathscr{P}(\mathbb{R})$. This measure is called the _free
convolution_ between $\mu$ and $\mu_{\operatorname{sc}}^{(t)}$, and we often
write $\mu_{t}=\mu\boxplus\mu_{\operatorname{sc}}^{(t)}$. By [6, Corollary 2],
$\mu_{t}$ has a density
$\varrho_{t}=\varrho_{t}^{\mu}:\mathbb{R}\rightarrow\mathbb{R}_{\geq 0}$ with
respect to Lebesgue measure for $t>0$.

In the following lemma, we collect some properties of free convolution with
rescaled semicircle distributions; it mainly follows from [6].

######  Lemma 2.3.

The following statements hold, for any real number $t>0$.

  1. (1)

Define the function $v_{t}:\mathbb{R}\rightarrow\mathbb{R}_{\geq 0}$ by for
each $u\in\mathbb{R}$ setting

(2.9) $\displaystyle v_{t}(u)=\displaystyle\inf\bigg{\\{}v\geq
0:\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\mu(dx)}{(u-x)^{2}+v^{2}}\leq
t^{-1}\bigg{\\}}.$

Then $v_{t}$ is continuous on $\mathbb{R}$. Moreover, the boundary of
$\Lambda_{t}$ is parameterized by
$\partial\Lambda_{t}=\\{E+\mathrm{i}v_{t}(E):E\in{\mathbb{R}}\\}$, and the set
$\big{\\{}E\in{\mathbb{R}}:v_{t}(E)>0\big{\\}}$ consists of countably many
open intervals $\bigcup_{i\geq 1}(a_{i},b_{i})$.

    1. (a)

For each $E\in\bigcup_{i\geq 1}(a_{i},b_{i})$, we have
$\int_{-\infty}^{\infty}\big{|}x-E-\mathrm{i}v_{t}(E)\big{|}^{-2}\mu(dx)=t^{-1}$.

    2. (b)

For each $E\in{\mathbb{R}}\setminus\bigcup_{i\geq 1}(a_{i},b_{i})$, we have
$\int_{-\infty}^{\infty}\big{|}x-E-\mathrm{i}v_{t}(E)\big{|}^{-2}\mu(dx)\leq
t^{-1}$.

  2. (2)

We have $\operatorname{supp}\mu\subseteq\bigcup_{i\geq 1}[a_{i},b_{i}]$, and
$\mu\big{(}\bigcup_{i\geq 1}\\{a_{i},b_{i}\\}\big{)}=0$.

  3. (3)

The function $M\big{(}E+\mathrm{i}v_{t}(E)\big{)}$ is (strictly) increasing in
$E\in{\mathbb{R}}$. Moreover, $v_{t}(E)$ and
$M\big{(}E+\mathrm{i}v_{t}(E)\big{)}$ are smooth for $E$ in the interior of
$\mathbb{R}\setminus\bigcup_{i\geq 1}\\{a_{i},b_{i}\\}$.

  4. (4)

A real number $y\in\mathbb{R}$ satisfies $\varrho_{t}(y)>0$ if and only if
$y=M(w)=w-tm_{0}(w)$ for some $w=w(y)\in\partial\Lambda_{t}\cap\mathbb{H}$.
Moreover, the function $w(y)$ is smooth in
$y\in\big{\\{}y^{\prime}\in\mathbb{R}:\varrho_{t}(y^{\prime})>0\big{\\}}$.

    1. (a)

We have $\varrho_{t}(y)=\pi^{-1}\operatorname{Im}m_{0}(w)=(\pi
t)^{-1}\operatorname{Im}w$.

    2. (b)

The Hilbert transform of $\varrho_{t}$ is given by
$H\varrho_{t}(y)=\pi^{-1}\operatorname{Re}m_{0}(w)=(\pi
t)^{-1}\operatorname{Re}(w-y)$.

  5. (5)

Denote $\mathfrak{e}_{+}=\mathfrak{e}_{+}(t)=\max(\operatorname{supp}\mu_{t})$
and $\mathfrak{e}_{-}=\mathfrak{e}_{-}(t)=\min\operatorname{supp}\mu_{t}$, and
let

$\displaystyle
w_{+}=\sup\Bigg{\\{}w\in\mathbb{R}:\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\mu(dx)}{(x-w)^{2}}>t^{-1}\Bigg{\\}};\qquad
w_{-}=\inf\Bigg{\\{}w\in\mathbb{R}:\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\mu(dx)}{(x-w)^{2}}>t^{-1}\Bigg{\\}}.$

Then $\mathfrak{e}_{+}=w_{+}-tm_{0}(w_{+})$ and
$\mathfrak{e}_{-}=w_{-}-tm_{0}(w_{-})$.

  6. (6)

For any real number $t>0$, both $\varrho_{t}(y)$ and its Hilbert transform
$H\varrho_{t}(y)$ are smooth in
$y\in\big{\\{}y^{\prime}\in\mathbb{R}:\varrho_{t}(y^{\prime})>0\big{\\}}$.
Moreover, we have

(2.10)
$\displaystyle\partial_{t}\varrho_{t}(y)=\pi\cdot\partial_{y}\big{(}\varrho_{t}(y)H\varrho_{t}(y)\big{)},\qquad\text{for
$y\in\big{\\{}y^{\prime}\in\mathbb{R}:\varrho_{t}(y^{\prime})>0\big{\\}}$}.$

###### Proof.

Item 1 of the lemma follows from [6, Lemma 2] and the content in [6] directly
following it (the fact that $\big{\\{}E\in\mathbb{R}:v_{t}(E)>0\big{\\}}$ is
open, implying that it consists of countably many open intervals, follows from
the definition (2.9) of $v_{t}$). To establish Item 2, assume to the contrary
that $\mu\big{(}\mathbb{R}\setminus\bigcup_{i\geq 1}(a_{i},b_{i})\big{)}>0$.
Then, there exists some constant $c>0$ and subset
$I\subseteq\mathbb{R}\setminus\bigcup_{i\geq 1}(a_{i},b_{i})$ such that
$\mu(I)\geq c$ and $I\subseteq[-c^{-1},c^{-1}]$. Denoting for any integers
$n\geq 1$ and $j\in\llbracket-n,n-1\rrbracket$ the subsets
$I_{j;n}=I\cap\big{[}j/cn,(j+1)/cn\big{]}$, it follows for each $n\geq 1$ that
there exists some $j_{n}\in\llbracket-n,n-1\rrbracket$ such that
$\mu(I_{j_{n};n})\geq c/2n$. Hence, for any real number $E_{n}\in
I_{j_{n};n}\subset\mathbb{R}\setminus\bigcup_{i\geq 1}(a_{i},b_{i})$, we have

$\displaystyle
t^{-1}\geq\displaystyle\int_{-\infty}^{\infty}|x-E_{n}|^{-2}\mu(dx)\geq\displaystyle\int_{I_{j_{n};n}}|x-E_{n}|^{-2}\mu(dx)\geq
c^{2}n^{2}\mu(I_{j_{n};n})\geq\displaystyle\frac{c^{3}n}{2},$

where in the first inequality we used Item 1 of the lemma; in the second we
restricted the integral to $I_{j_{n};n}$; in the third we used the bound
$|x-E|\leq(cn)^{-1}$ for each $x,E\in I_{j_{n};n}$; and in the fourth we used
the fact that $\mu(I_{j_{n};n})>c/2n$. Taking $n>2(tc^{3})^{-1}$ gives a
contradiction, implying Item 2 of the lemma.

To verify Item 3 of the lemma, recall from 2.2 that $M$ is a continuous
bijection from $\partial\Lambda_{t}$ to $\mathbb{R}$. Together with Item 1 of
the lemma, this implies that $M\big{(}E+\mathrm{i}v_{t}(E)\big{)}$ is either
(strictly) increasing or (strictly) decreasing. By [6, Lemma 5],
$M\big{(}E+\mathrm{i}v_{t}(E)\big{)}$ is (strictly) increasing on
$\bigcup_{i\geq 1}(a_{i},b_{i})$ (which is nonempty by Item 2 of the lemma),
meaning that $M\big{(}E+\mathrm{i}v_{t}(E)\big{)}$ is (strictly) increasing on
$\mathbb{R}$; this verifies the first statement in Item 3 of the lemma. The
fact that $v_{t}$ is smooth on $\bigcup_{i\geq 1}(a_{i},b_{i})$ follows from
[6, Lemma 2]; since $v_{t}=0$ in the interior of
$\mathbb{R}\setminus\bigcup_{i\geq 1}(a_{i},b_{i})$, it follows that $v_{t}$
is smooth in the interior of $\mathbb{R}\setminus\bigcup_{i\geq
1}\\{a_{i},b_{i}\\}$. This also implies that
$M\big{(}E+\mathrm{i}v_{t}(E)\big{)}$ is smooth in the interior of
$\mathbb{R}\setminus\bigcup_{i\geq 1}(a_{i},b_{i})$ (since, by (2.1),
$m\big{(}E+\mathrm{i}v_{t}(E)\big{)}$ is smooth in $v_{t}(E)$, as long as
$v_{t}(E)>0$ or $E\notin\operatorname{supp}\mu$), verifying Item 3 of the
lemma.

All statements in Item 4 of the lemma, except for the smoothness of $w$ in
$y$, follow from [6, Corollary 2, Corollary 3]. To show this smoothness, we
use Item 1 of the lemma to express $w=E+\mathrm{i}v_{t}(E)$ for some real
number $E\in\mathbb{R}$. Thus,
$y(E)=M(w)=M\big{(}E+\mathrm{i}v_{t}(E)\big{)}$, which is smooth in $E$ by
Item 3 of the lemma, which also indicates that
$\partial_{E}M\big{(}E+\mathrm{i}v_{t}(E)\big{)}>0$ on the domain where
$v_{t}(E)>0$. So, the implicit function theorem implies that
$w=E+\mathrm{i}v_{t}(E)$ is smooth in $y$ when $v_{t}(E)>0$ (that is, when
$w\in\mathbb{H}$, or equivalently when $\varrho_{t}(y)>0$). Then, Item 5 of
the lemma follows from Item 4 and the definition (2.9) of $v_{t}$.

The first part of Item 6 follows from [6, Corollary 3] (together with [6,
Lemma 2]). For the second part of Item 6, by taking the derivative on both
sides of (2.8) with respect to $t$, we get for any $z\in\Lambda_{t}$ that

$\displaystyle
0=\partial_{t}m_{0}(z)=\partial_{t}\Big{(}m_{t}\big{(}z-tm_{0}(z)\big{)}\Big{)}$
$\displaystyle=\partial_{t}m_{t}\big{(}z-tm_{0}(z)\big{)}-\partial_{z}m_{t}\big{(}z-tm_{0}(z)\big{)}\cdot
m_{0}(z)$
$\displaystyle=\partial_{t}m_{t}\big{(}z-tm_{0}(z)\big{)}-\partial_{z}m_{t}\big{(}z-tm_{0}(z)\big{)}\cdot
m_{t}\big{(}z-tm_{0}(z)\big{)}.$

Setting $w=z-tm_{0}(z)$ and applying 2.2, it follows for any
$w=y+\mathrm{i}\eta\in\mathbb{H}$ that

$\displaystyle\partial_{t}m_{t}(y+\mathrm{i}\eta)=\displaystyle\frac{1}{2}\cdot\partial_{z}\big{(}m_{t}(y+\mathrm{i}\eta)\big{)}^{2}.$

Sending $\eta$ to zero and applying (2.3) (with the first part of the lemma),
we find

(2.11)
$\displaystyle\pi\cdot\partial_{t}\big{(}H\varrho_{t}(y)+\mathrm{i}\varrho_{t}(y)\big{)}=\frac{\pi^{2}}{2}\partial_{y}\Big{(}\big{(}H\varrho_{t}(y)+\mathrm{i}\varrho_{t}(y)\big{)}^{2}\Big{)}.$

Then (2.10) follows from comparing the imaginary parts on both sides of
(2.11). ∎

######  Remark 2.4.

While free convolutions are typically defined between probability measures,
the relation (2.8) also defines the free convolution of any measure
$\mu\in\mathscr{P}_{\operatorname{fin}}$, satisfying
$A=\mu(\mathbb{R})<\infty$, with the rescaled semicircle distribution
$\mu_{\operatorname{sc}}^{(t)}$. Indeed, define the probability measure
$\widetilde{\mu}\in\mathscr{P}$ from $\mu$ by setting
$\widetilde{\mu}(I)=A^{-1}\cdot\mu(A^{1/2}I)$, for any interval
$I\subseteq\mathbb{R}$. Furthermore, for any real number $s\geq 0$, define the
probability measure
$\widetilde{\mu}_{s}=\widetilde{\mu}\boxplus\mu_{\operatorname{sc}}^{(s)}$,
and denote its Stieltjes transform by $\widetilde{m}_{s}$. Then, define the
free convolution $\mu_{t}=\mu\boxplus\mu_{\operatorname{sc}}^{(t)}$ and its
Stieltjes transform $m_{t}=m_{t;\mu}$ by setting

$\displaystyle\mu_{t}(I)=A\cdot\widetilde{\mu}_{t}(A^{-1/2}I),\qquad\text{for
any interval $I\subseteq\mathbb{R}$, so that}\qquad
m_{t}(z)=A^{1/2}\cdot\widetilde{m}_{t}(A^{-1/2}z),$

where the second equality follows from the first by (2.1). Then,

$\displaystyle
m_{t}\big{(}z-tm_{0}(z)\big{)}=A^{1/2}\cdot\widetilde{m}_{t}\Big{(}A^{-1/2}\big{(}z-tm_{0}(z)\big{)}\Big{)}$
$\displaystyle=A^{1/2}\cdot\widetilde{m}_{t}\big{(}A^{-1/2}z-t\widetilde{m}_{0}(A^{-1/2}z)\big{)}$
$\displaystyle=A^{1/2}\cdot\widetilde{m}_{0}(A^{-1/2}z)=m_{0}(z),$

so that (2.8) continues to hold for $m_{t}$. In particular, 2.2 and 2.3 also
hold for $\mu$.

######  Example 2.5.

If we start from the delta mass $\mu_{0}=A\delta_{0}$, then the free
convolution $\mu_{t}=\mu_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$ is
supported on $[-2\sqrt{At},2\sqrt{At}]$, and it is given explicitly by

(2.12) $\displaystyle\mu_{t}({\rm d}x)=\frac{\sqrt{4At-x^{2}}}{2\pi t}{\rm
d}x,\quad t\geq 0.$

###  2.2. Empirical measure of Dyson Brownian motion

It is known that for $n$-particle Dyson Brownian motion, if the initial data
(1.6) weakly converges to a probability measure $\mu$, then the empirical
particle density (1.5) (as a measure valued process) converges to the free
convolution of $\mu$ with the rescaled semicircle distribution. We recall the
precise statement from [4].

######  Proposition 2.6 ([4, Proposition 4.3.10, Corollary 4.3.11]).

Fix a compactly supported probability measure $\mu\in\mathscr{P}$ and real
number ${\mathsf{T}}>1$. Denote the free convolution of $\mu$ with the
rescaled semicircle distribution $\mu_{\operatorname{sc}}^{(t)}\in\mathscr{P}$
by $\mu_{t}$. For each integer $n\geq 1$, let
${\bm{u}}^{(n)}=(u^{(n)}_{1},u^{(n)}_{2},\cdots,u^{(n)}_{n})\in\overline{\mathbb{W}}_{n}$
denote sequences such that the empirical measures
$(1/n)\sum_{i=1}^{(n)}\delta_{u^{(n)}_{i}}$ converge weakly to $\mu$. Consider
$n$ particle Dyson Brownian motion
$\bm{\lambda}^{(n)}(t)=(\lambda_{1}^{(n)}(t),\lambda_{2}^{(n)}(t),\ldots,\lambda_{n}^{(n)}(t))$
starting from ${\bm{u}}^{(n)}$. We denote the empirical particle density as
$\mu_{t}^{n}=(1/n)\sum_{i=1}^{(n)}\delta_{\lambda_{i}^{(n)}(t)}$. Then, for
any real number $\varepsilon>0$, we have
$\lim_{n\rightarrow\infty}\mathbb{P}\big{[}\max_{0\leq
t\leq{\mathsf{T}}}d_{{\rm L}}(\mu_{t}^{(n)},\mu_{t})>\varepsilon\big{]}=0$,
where $d_{\text{L}}$ is the Lévy metric from (1.3).

######  Remark 2.7.

In 2.6, we can allow some particles to be at $\pm\infty$, as in Remark 1.1.
Assume the initial data $(1/n)\sum_{i=1}^{n}\delta_{u^{n}_{i}}$ converges
weakly to a positive measure $\mu$ with $\mu({\mathbb{R}})\leq 1$. Let
$\mu_{t}$ be the free convolution of $\mu$ with the rescaled semicircle
distribution $\mu_{\operatorname{sc}}^{(t)}\in\mathscr{P}$ as in 2.4. Then 2.6
still holds.

A stronger version of convergence has been proven in [20], which proves that
the optimal bulk rigidity holds for Dyson Brownian motion with general
potentials. To state it, we need some further notation.

######  Definition 2.8.

Let $\mu\in\mathscr{P}_{\operatorname{fin}}$ denote a finite measure with
$\mu({\mathbb{R}})=A$. We denote the associated inverted cumulative density
$\gamma:[0,A]\mapsto{\mathbb{R}}$, as

(2.13)
$\displaystyle\gamma(y)=\displaystyle\sup\Bigg{\\{}\gamma\in\mathbb{R}:\displaystyle\int_{\gamma}^{\infty}{\rm
d}\mu(x)\geq y\Bigg{\\}}.$

We also set $\gamma(y)=\infty$ if $y<0$, and $\gamma(y)=-\infty$ if $y>A$.
Then $\gamma(y)$ is non-increasing and left continuous, i.e.,
$\gamma(y^{-})=\gamma(y)$. For any integers $n\geq 1$ and $j\in\mathbb{Z}$, we
define the _classical location_ (also called _$n^{-1}$ -quantiles_) with
respect to $\mu$,

(2.14)
$\displaystyle\gamma_{j}=\gamma_{j}^{\mu}=\gamma_{j;n}=\gamma_{j;n}^{\mu}\in\mathbb{R}=\gamma((j-1/2)/n),$

for $j\in\llbracket 1,An\rrbracket$. We also set $\gamma_{j}=\infty$ if $j<1$
and $\gamma_{j}=-\infty$ if $j>An$.

######  Proposition 2.9 ([20, Corollary 3.2]).

For any real number $D>1$, there exists a constant $C=C(D)>1$ such that the
following holds. Consider $n$ particle Dyson Brownian motion
$\bm{\lambda}(t)=(\lambda_{1}(t),\lambda_{2}(t),\ldots,\lambda_{n}(t))$
starting from $\mu_{0}=\mu_{0}^{(n)}$ with
$\operatorname{supp}\mu_{0}\subset[-n^{D},n^{D}]$. We denote by
$\mu_{t}=\mu_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$ the free convolution of
$\mu_{0}$ with rescaled semicircle distribution, and its classical locations
by $\\{\gamma_{i}(t)\\}_{1\leq i\leq n}$ (as in (2.14)). Then, with
probability $1-Ce^{-(\log n)^{2}}$, we have for all $0\leq t\leq n^{D}$ and
$1\leq i\leq n$ that

(2.15) $\displaystyle\gamma_{i+\lfloor(\log
n)^{5}\rfloor}(t)-n^{-D}\leq\lambda_{i}(t)\leq\gamma_{i-\lfloor(\log
n)^{5}\rfloor}(t)+n^{-D}.$

We mention that, In [20], the probability on the right side of (2.15) was
written to be $1-Cn^{-D}$ for any $D>1$, but it can be seen from the proof
(see that of [21, Proposition 3.8], where $\delta$ there is $\frac{5}{4}$
here) that it can be taken to be $1-Ce^{-(\log n)^{2}}$ instead.

###  2.3. Dyson Brownian Motion Comparison

In this section, we allow some particles to be at $\pm\infty$ as in Remark
1.1. The following lemma gives $i$-th particle location and gap comparisons
between Dyson Brownian motions with different initial data. The proof utilizes
a coupling of Dyson Brownian motions and a maximum principle, which has been
used previously in the random matrix literature [14, 9].

Figure 2. Left Panel: if the initial data satisfies
$\lambda_{i}(0)\geq\widetilde{\lambda}_{i}(0)$, then we can couple the two
Dyson Brownian motions $\lambda_{i}(t)\geq\widetilde{\lambda}_{i}(t)$ (with
possibly some particles at $\pm\infty$). Right Panel: if the initial data
satisfies
$\lambda_{i}(0)-\lambda_{j}(0)\geq\widetilde{\lambda}_{i}(0)-\widetilde{\lambda}_{j}(0)$,
then we can couple the two Dyson Brownian motions
$\lambda_{i}(t)-\lambda_{j}(t)\geq\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{j}(t)$.

######  Lemma 2.10.

Take any integers $1\leq\ell\leq m\leq n$. Consider two coupled Dyson Brownian
motions (see Figure 2)

(2.16) $\displaystyle{\rm
d}\lambda_{i}(t)=\displaystyle\sum_{\begin{subarray}{c}1\leq j\leq n\\\ j\neq
i\end{subarray}}\displaystyle\frac{{\rm
d}t}{\lambda_{i}(t)-\lambda_{j}(t)}+\Big{(}\displaystyle\frac{2}{\beta
n}\Big{)}^{1/2}{\rm d}B_{i}(t),$ (2.17) $\displaystyle{\rm
d}\widetilde{\lambda}_{i}(t)=\displaystyle\sum_{\begin{subarray}{c}1\leq j\leq
n\\\ j\neq i\end{subarray}}\displaystyle\frac{{\rm
d}t}{\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{j}(t)}+\Big{(}\displaystyle\frac{2}{\beta
n}\Big{)}^{1/2}{\rm d}B_{i}(t),$

which share the same Brownian motions $\\{B_{i}(t)\\}_{1\leq i\leq n}$.

  1. (1)

Assume that $\lambda_{i}(t)=+\infty$ for $i\in\llbracket 1,\ell-1\rrbracket$
and $\lambda_{i}(t)\in{\mathbb{R}}$ for $i\in\llbracket\ell,n\rrbracket$, and
that $\widetilde{\lambda}_{i}(t)\in{\mathbb{R}}$ for $i\in\llbracket
1,m\rrbracket$ and $\widetilde{\lambda}_{i}(t)=-\infty$ for $i\in\llbracket
m+1,n\rrbracket$. If $\lambda_{i}(0)\geq\widetilde{\lambda}_{i}(0)$ for all
$\ell\leq i\leq m$, then $\lambda_{i}(t)\geq\widetilde{\lambda}_{i}(t)$ for
all $\ell\leq i\leq m$ and $t\geq 0$.

  2. (2)

Assume that $\lambda_{i}(t)=+\infty$ for $i\in\llbracket 1,\ell-1\rrbracket$;
$\lambda_{i}(t)=-\infty$ for $i\in\llbracket m+1,n\rrbracket$;
$\lambda_{i}(t)\in{\mathbb{R}}$ for $i\in\llbracket\ell,m\rrbracket$; and
$\widetilde{\lambda}_{i}(t)\in{\mathbb{R}}$ for $i\in\llbracket
1,n\rrbracket$. If
$\lambda_{i}(0)-\lambda_{j}(0)\geq\widetilde{\lambda}_{i}(0)-\widetilde{\lambda}_{j}(0)$
for all $\ell\leq i<j\leq m$, then
$\lambda_{i}(t)-\lambda_{j}(t)\geq\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{j}(t)$
for all $\ell\leq i<j\leq m$ and $t\geq 0$.

###### Proof of 2.10.

By taking difference of (2.16) and (2.17), we have

(2.18) $\displaystyle{\rm
d}(\lambda_{i}(t)-\widetilde{\lambda}_{i}(t))=-\displaystyle\sum_{\begin{subarray}{c}1\leq
j\leq n\\\ j\neq
i\end{subarray}}\displaystyle\frac{(\lambda_{i}(t)-\widetilde{\lambda}_{i}(t))-(\lambda_{j}(t)-\widetilde{\lambda}_{j}(t)){\rm
d}t}{(\lambda_{i}(t)-\lambda_{j}(t))(\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{j}(t))},\quad\ell\leq
i\leq m.$

Let $i_{*}(t):=\operatorname{argmin}_{\ell\leq i\leq
m}\\{\lambda_{i}(t)-\widetilde{\lambda}_{i}(t)\\}$ (if $i_{*}(t)$ is not
unique, we can simply take the smallest one). By our assumption for
$i\leq\ell-1$ or $i\geq m+1$,
$\lambda_{i}(t)-\widetilde{\lambda}_{i}(t)=\infty$. Thus we also have
$i_{*}(t)=\operatorname{argmin}_{1\leq i\leq
n}\\{\lambda_{i}(t)-\widetilde{\lambda}_{i}(t)\\}.$ Since
$\lambda_{i}(t)-\widetilde{\lambda}_{i}(t)$ are continuous, $i_{*}(t)$ is
piecewise constant. Either $i_{*}(t)$ is continuous at $t$ or $i_{*}(t)$ has a
jump at $t$. If $i_{*}(t)$ is continuous at $t$, then ${\rm d}i_{*}(t)=0$ and

(2.19) $\displaystyle\frac{{\rm
d}(\lambda_{i_{*}(t)}(t)-\widetilde{\lambda}_{i_{*}(t)}(t))}{{\rm
d}t}=-\displaystyle\sum_{\begin{subarray}{c}1\leq j\leq n\\\
j\neq{i_{*}(t)}\end{subarray}}\displaystyle\frac{(\lambda_{i_{*}(t)}(t)-\widetilde{\lambda}_{i_{*}(t)}(t))-(\lambda_{j}(t)-\widetilde{\lambda}_{j}(t))}{(\lambda_{i_{*}(t)}(t)-\lambda_{j}(t))(\widetilde{\lambda}_{i_{*}(t)}(t)-\widetilde{\lambda}_{j}(t))}\geq
0,$

where in the last inequality we used the definition of $i_{*}(t)$ that
$(\lambda_{i_{*}(t)}(t)-\widetilde{\lambda}_{i_{*}(t)}(t))-(\lambda_{j}(t)-\widetilde{\lambda}_{j}(t))\leq
0$, and
$(\lambda_{i_{*}(t)}(t)-\lambda_{j}(t))(\widetilde{\lambda}_{i_{*}(t)}(t)-\widetilde{\lambda}_{j}(t))>0$.
If $i_{*}(t)$ has a jump at $t$, then we have
$\lambda_{i_{*}(t^{-})}(t)-\widetilde{\lambda}_{i_{*}(t^{-})}(t)=\lambda_{i_{*}(t^{+})}(t)-\widetilde{\lambda}_{i_{*}(t^{+})}(t)$,
and as in (2.19) we have

(2.20) $\displaystyle\frac{{\rm
d}(\lambda_{i_{*}(t^{-})}(t^{-})-\widetilde{\lambda}_{i_{*}(t^{-})}(t^{-}))}{{\rm
d}t},\frac{{\rm
d}(\lambda_{i_{*}(t^{+})}(t^{+})-\widetilde{\lambda}_{i_{*}(t^{+})}(t^{+}))}{{\rm
d}t}\geq 0.$

We conclude from (2.19) and (2.20) that
$\lambda_{i_{*}(t)}(t)-\widetilde{\lambda}_{i_{*}(t)}(t)$ is non-decreasing in
$t$. Moreover by our assumption
$\lambda_{i_{*}(0)}(0)-\widetilde{\lambda}_{i_{*}(0)}(0)\geq 0$. We conclude
that for all $\ell\leq i\leq m$,
$\lambda_{i}(t)-\widetilde{\lambda}_{i}(t)\geq\lambda_{i_{*}(t)}(t)-\widetilde{\lambda}_{i_{*}(t)}(t)\geq
0$ for any $t\geq 0$. This finishes the first claim.

To prove the second claim, we further take the difference of (2.18) with $i$
replaced by $i+1$,

(2.21) $\displaystyle\begin{split}&\frac{{\rm
d}((\lambda_{i}(t)-\lambda_{i+1}(t))-(\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{i+1}(t)))}{{\rm
d}t}=-2\displaystyle\frac{(\lambda_{i}(t)-\lambda_{i+1}(t))-(\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{i+1}(t))}{(\lambda_{i}(t)-\lambda_{i+1}(t))(\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{i+1}(t))}\\\
&-\displaystyle\sum_{\begin{subarray}{c}1\leq j\leq n\\\ j\neq
i,i+1\end{subarray}}\frac{\lambda_{i}(t)-\lambda_{i+1}(t)}{(\lambda_{i}(t)-\lambda_{j}(t))(\lambda_{i+1}(t)-\lambda_{j}(t))}-\frac{\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{i+1}(t)}{(\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{j}(t))(\widetilde{\lambda}_{i+1}(t)-\widetilde{\lambda}_{j}(t))}.\end{split}$

Let $i_{*}(t)=\operatorname{argmin}_{\ell\leq i\leq
m-1}\\{(\lambda_{i}(t)-\lambda_{i+1}(t))-(\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{i+1}(t))\\}$
(if $i_{*}(t)$ is not unique, we can simply take the smallest one). By our
assumption for $i\leq\ell-1$ or $i\geq m$,
$(\lambda_{i}(t)-\lambda_{i+1}(t))-(\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{i+1}(t))=\infty$.
Thus we also have $i_{*}(t)=\operatorname{argmin}_{1\leq i\leq
n}\\{(\lambda_{i}(t)-\lambda_{i+1}(t))-(\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{i+1}(t))\\}.$
Since
$(\lambda_{i}(t)-\lambda_{i+1}(t))-(\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{i+1}(t))$
are continuous, $i_{*}(t)$ is piecewise constant. Either $i_{*}(t)$ is
continuous at $t$ or $i_{*}(t)$ has a jump at $t$. If $i_{*}(t)$ has a jump at
$t$, then we have
$(\lambda_{i_{*}(t^{-})}(t)-\lambda_{{i_{*}(t^{-})}+1}(t))-(\widetilde{\lambda}_{i_{*}(t^{-})}(t)-\widetilde{\lambda}_{{i_{*}(t^{-})}+1}(t))=(\lambda_{i_{*}(t^{+})}(t)-\lambda_{{i_{*}(t^{+})}+1}(t))-(\widetilde{\lambda}_{i_{*}(t^{+})}(t)-\widetilde{\lambda}_{{i_{*}(t^{+})}+1}(t))$.
In both cases
$(\lambda_{i_{*}(t)}(t)-\lambda_{{i_{*}(t)}+1}(t))-(\widetilde{\lambda}_{i_{*}(t)}(t)-\widetilde{\lambda}_{{i_{*}(t)}+1}(t))$
is continuous in $t$.

By our assumption
$(\lambda_{i_{*}(0)}(0)-\lambda_{i_{*}(0)+1}(0))-(\widetilde{\lambda}_{i_{*}(0)}(0)-\widetilde{\lambda}_{i_{*}(0)+1}(0))\geq
0$. We denote the stopping time

(2.22) $\displaystyle\tau=\inf_{t\geq
0}\\{t:(\lambda_{i_{*}(t)}(t)-\lambda_{i_{*}(t)+1}(t))-(\widetilde{\lambda}_{i_{*}(t)}(t)-\widetilde{\lambda}_{i_{*}(t)+1}(t))<0\\}.$

Then either $\tau=\infty$, and our claim follows or $\tau<\infty$. Next we
prove by contradiction that $\tau<\infty$ is impossible. Otherwise

(2.23)
$\displaystyle\begin{split}&(\lambda_{i_{*}(\tau)}(\tau)-\lambda_{i_{*}(\tau)+1}(\tau))-(\widetilde{\lambda}_{i_{*}(\tau)}(\tau)-\widetilde{\lambda}_{i_{*}(\tau)+1}(\tau))=0,\\\
&\lambda_{i}(\tau)-\lambda_{j}(\tau)\geq\widetilde{\lambda}_{i}(\tau)-\widetilde{\lambda}_{j}(\tau),\quad\ell\leq
i\leq j\leq m,\end{split}$

and there exists arbitrarily small $\varepsilon>0$ such that

(2.24)
$\displaystyle(\lambda_{i_{*}(\tau+\varepsilon)}(\tau+\varepsilon)-\lambda_{i_{*}(\tau+\varepsilon)+1}(\tau+\varepsilon))-(\widetilde{\lambda}_{i_{*}(\tau+\varepsilon)}(\tau+\varepsilon)-\widetilde{\lambda}_{i_{*}(\tau+\varepsilon)+1}(\tau+\varepsilon))<0.$

By plugging $i=i_{*}(\tau)$ (if $i_{*}(t)$ has a jump at $\tau$, we interpret
this as $i=i_{*}(\tau^{+})$) and $t=\tau$ in (LABEL:e:gapdiff), and using the
first statement in (LABEL:e:diff0), we get

(2.25) $\displaystyle\begin{split}\frac{{\rm
d}((\lambda_{i_{*}(\tau)}(\tau)-\lambda_{i_{*}(\tau)+1}(\tau))-(\widetilde{\lambda}_{i_{*}(\tau)}(\tau)-\widetilde{\lambda}_{i_{*}(\tau)+1}(\tau)))}{{\rm
d}t}=-(\lambda_{i_{*}(\tau)}(\tau)-\lambda_{{i_{*}(\tau)}+1}(\tau))\times\\\
\sum_{\begin{subarray}{c}1\leq j\leq n\\\
j\neq{i_{*}(\tau)},{i_{*}(\tau)}+1\end{subarray}}\frac{1}{(\lambda_{i_{*}(\tau)}(\tau)-\lambda_{j}(\tau))(\lambda_{{i_{*}(\tau)}+1}(\tau)-\lambda_{j}(\tau))}-\frac{1}{(\widetilde{\lambda}_{i_{*}(\tau)}(\tau)-\widetilde{\lambda}_{j}(\tau))(\widetilde{\lambda}_{{i_{*}(\tau)}+1}(\tau)-\widetilde{\lambda}_{j}(\tau))}\geq
0,\end{split}$

where we also used that
$(\lambda_{i_{*}(\tau)}(\tau)-\lambda_{{i_{*}(\tau)}+1}(\tau))>0$ and
$0<(\widetilde{\lambda}_{i_{*}(\tau)}(\tau)-\widetilde{\lambda}_{j}(\tau))(\widetilde{\lambda}_{{i_{*}(\tau)}+1}(\tau)-\widetilde{\lambda}_{j}(\tau))\leq(\lambda_{i_{*}(\tau)}(\tau)-\lambda_{j}(\tau))(\lambda_{{i_{*}(\tau)}+1}(\tau)-\lambda_{j}(\tau))$
from (LABEL:e:diff0). If (2.25) is postive, then this contradicts (2.24).

Otherwise (2.25) equals zero, which is only possible if
$(\widetilde{\lambda}_{i_{*}(\tau)}(\tau)-\widetilde{\lambda}_{j}(\tau))(\widetilde{\lambda}_{{i_{*}(\tau)}+1}(\tau)-\widetilde{\lambda}_{j}(\tau))-(\lambda_{i_{*}(\tau)}(\tau)-\lambda_{j}(\tau))(\lambda_{{i_{*}(\tau)}+1}(\tau)-\lambda_{j}(\tau))=0$
for all $j\neq i_{*}(\tau),i_{*}(\tau)+1$. Together with (LABEL:e:diff0), this
implies that
$\lambda_{i_{*}(\tau)}(\tau)-\lambda_{j}(\tau)=\widetilde{\lambda}_{i_{*}(\tau)}(\tau)-\widetilde{\lambda}_{j}(\tau)$
for all $1\leq j\leq n$. In particular,
$\\{\widetilde{\lambda}_{j}(\tau)\\}_{1\leq j\leq
n}=\\{\lambda_{j}(\tau)+c\\}_{1\leq j\leq n}$ for
$c:=\widetilde{\lambda}_{i_{*}(\tau)}(\tau)-\lambda_{i_{*}(\tau)}(\tau)\in{\mathbb{R}}$.
Then the coupling (2.16) (notice that DBM only depends on the gaps
$\lambda_{i}(t)-\lambda_{j}(t)$) implies that for any $t\geq\tau$, we have
$\\{\widetilde{\lambda}_{j}(t)\\}_{1\leq j\leq
n}=\\{\lambda_{j}(t)+c\\}_{1\leq j\leq n}$, which again contradicts the
definition of $\tau$ in (2.22). ∎

2.6 and 2.10 together give a continuous version of comparison for the free
convolution with a rescaled semicircle distribution.

######  Lemma 2.11.

Fix a real number $A\in(0,1]$, and finite measures
$\mu,\widetilde{\mu}\in\mathscr{P}_{\rm fin}$ such that $\mu(\mathbb{R})=A$
and $\widetilde{\mu}(\mathbb{R})=1$. Let
$\mu_{t}=\mu\boxplus\mu_{\operatorname{sc}}^{(t)}$ and
$\widetilde{\mu}_{t}=\mu\boxplus\mu_{\operatorname{sc}}^{(t)}$ denote the free
convolutions of $\mu$ and $\widetilde{\mu}$ with the rescaled semicircle
distribution respectively. Also denote the associated inverted cumulative
density functions (recall (2.13)) by $\gamma_{t}:[0,A]\rightarrow\mathbb{R}$
and $\widetilde{\gamma}_{t}:[0,1]\rightarrow\mathbb{R}$ respectively.

  1. (1)

Assume for each $y\in[0,A]$ that $\gamma_{0}(y)\leq\widetilde{\gamma}_{0}(y)$.
Then, $\gamma_{t}(y)\leq\widetilde{\gamma}_{t}(y)$ holds for each
$(t,y)\in[0,\infty]\times[0,A]$.

  2. (2)

Assume for each $y,y^{\prime}\in[0,A]$ that
$|\gamma_{0}(y)-\gamma_{0}(y^{\prime})|\geq|\widetilde{\gamma}_{0}(y)-\widetilde{\gamma}_{0}(y^{\prime})|$.
Then,
$|\gamma_{t}(y)-\gamma_{t}(y^{\prime})|\geq|\widetilde{\gamma}_{t}(y)-\widetilde{\gamma}_{t}(y^{\prime})|$
holds for each $t\in[0,\infty]$ and $y,y^{\prime}\in[0,A]$.

###### Proof of 2.11.

Using 2.6 and 2.10 as input, the proofs of the two statements are similar. So
we will only prove the second statement. We take two sets of initial data,
given by $\\{\lambda_{i}(0)=\gamma_{0}((i-1/2)/n)\\}_{1\leq i\leq\lfloor
An\rfloor}$, $\\{\lambda_{i}(0)=-\infty\\}_{\lfloor An\rfloor<i\leq n}$ and
$\\{\widetilde{\lambda}_{i}(0)=\widetilde{\gamma}_{0}((i-1/2)/n)\\}_{1\leq
i\leq n}$. Then for any index $1\leq i\leq n$, by our assumption for $1\leq
i\leq\lfloor An\rfloor$,

(2.26)
$\displaystyle\begin{split}\lambda_{i}(0)-\lambda_{i+1}(0)&=\gamma_{0}((i-1/2)/n)-\gamma_{0}((i+1/2)/n)\\\
&\geq\widetilde{\gamma}_{0}((i-1/2)/n)-\widetilde{\gamma}_{0}((i+1/2)/n)=\widetilde{\lambda}_{i}(0)-\widetilde{\lambda}_{i+1}(0).\end{split}$

Then (2.26) verifies the assumptions in the second statement of 2.10, so we
can couple the two associated Dyson Brownion motions such that

(2.27)
$\displaystyle\lambda_{i}(t)-\lambda_{i+1}(t)\geq\widetilde{\lambda}_{i}(t)-\widetilde{\lambda}_{i+1}(t),\quad
1\leq i\leq\lfloor An\rfloor.$

By 2.6 (recalling the definitions of Lévy distance from (1.3) and the inverted
cumulative density from (2.13)), we have for any $\varepsilon>0$, $0\leq a\leq
A$, and sufficiently large $n\geq 1$ that

(2.28) $\displaystyle\gamma_{t}(a+\varepsilon)-\varepsilon\leq\lambda_{\lfloor
an\rfloor}(t)\leq\gamma_{t}(a-\varepsilon)+\varepsilon,\quad\widetilde{\gamma}_{t}(a+\varepsilon)-\varepsilon\leq\widetilde{\lambda}_{\lfloor
an\rfloor}(t)\leq\widetilde{\gamma}_{t}(a-\varepsilon)+\varepsilon,$

In particular if $\gamma_{t}$ and $\widetilde{\gamma}_{t}$ are both continuous
at $a$, then we have

(2.29) $\displaystyle\gamma_{t}(a)=\lim_{n\rightarrow\infty}\lambda_{\lfloor
an\rfloor}(t),\quad\widetilde{\gamma}_{t}(a)=\lim_{n\rightarrow\infty}\widetilde{\lambda}_{\lfloor
an\rfloor}(t),$

and for any $0\leq a<b\leq A$ such that $\gamma_{t}$ and
$\widetilde{\gamma}_{t}$ are both continuous at $a,b$ we have

(2.30)
$\displaystyle\gamma_{t}(a)-\gamma_{t}(b)=\lim_{n\rightarrow\infty}\lambda_{\lfloor
an\rfloor}(t)-\lambda_{\lfloor
bn\rfloor}(t)\geq\lim_{n\rightarrow\infty}\widetilde{\lambda}_{\lfloor
an\rfloor}(t)-\widetilde{\lambda}_{\lfloor
bn\rfloor}(t)=\widetilde{\gamma}_{t}(a)-\widetilde{\gamma}_{t}(b).$

Recall $\gamma_{t}$ and $\widetilde{\gamma}_{t}$ are non-increasing and left
continuous, by from the discussion after (2.13). They are thus continuous at
all but countably many points. So for any $0<a<b\leq A$, we can take sequences
$\\{a_{k}\\}_{k\geq 1},\\{b_{k}\\}_{k\geq 1}$ which converge to $a,b$
respectively from below, such that $\gamma_{t}$ and $\widetilde{\gamma}_{t}$
are both continuous at each $a_{k},b_{k}$. Then it follows from (2.30) and the
fact the $\gamma_{t}$ and $\widetilde{\gamma}_{t}$ are left continuous

(2.31)
$\displaystyle\gamma_{t}(a)-\gamma_{t}(b)=\lim_{k\rightarrow\infty}\gamma_{t}(a_{k})-\gamma_{t}(b_{k})\geq\lim_{k\rightarrow\infty}\widetilde{\gamma}_{t}(a_{k})-\widetilde{\gamma}_{t}(b_{k})=\widetilde{\gamma}_{t}(a)-\widetilde{\gamma}_{t}(b).$

If $a=0$, then $\gamma_{t}(a)=\widetilde{\gamma}_{t}(a)=\infty$ and we have
the trivial statement
$\gamma_{t}(a)-\gamma_{t}(b)=\infty=\widetilde{\gamma}_{t}(a)-\widetilde{\gamma}_{t}(b)$.
This together with (2.31) finishes the proof of the second statement in
(2.11). ∎

###  2.4. Comparisons with Free Convolutions

In this section we discuss comparisons between Dyson Brownian motions with the
classical locations of the free convolution, which will be used in Section
4.2.

######  Proposition 2.12.

Fix ${\mathsf{T}}>0$. Given a probability measure $\mu_{0}\in\mathscr{P}$, we
denote by $\mu_{t}=\mu_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$ the free
convolution of $\mu$ with the rescaled semicircle distribution
$\mu_{\operatorname{sc}}^{(t)}$. We denote by
$\gamma_{i}(t)=\gamma_{i;n}^{\mu_{t}}(t)$ the classical locations of
$\mu_{t}({\rm d}x)=\varrho_{t}(x){\rm d}x$ (recall from (2.14)). We assume
there exists a small constant $c>0$ such that the following two statements
hold for any $0\leq t\leq{\mathsf{T}}$.

  1. (1)

$\mu_{t}=\varrho_{t}(x){\rm d}x$ has a bounded density:
$\sup_{x\in\mathbb{R}}\varrho_{t}(x)\leq(2c)^{-1}$.

  2. (2)

The derivative of $\varrho_{t}(x)$ is bounded in the following way: when
$\operatorname{dist}(x,\\{\gamma_{1}(t),\gamma_{2}(t),\cdots,\gamma_{n}(t)\\})\leq
c/2n$, we have $|\varrho_{t}^{\prime}(x)|\leq n/c$.

Then there exists $C=C(c,{\mathsf{T}})>1$ such that the following holds for
any $0\leq t\leq{\mathsf{T}}$

(2.32) $\displaystyle\mathbb{P}\left(\sup_{0\leq s\leq t,\atop 1\leq i\leq
n}|\lambda_{i}(s)-\gamma_{i}(s)|\leq\max_{1\leq j\leq
n}|\lambda_{j}(0)-\gamma_{j}(0)|+C\left(t+\frac{(\log
n)\sqrt{t}}{\sqrt{n}}\right)\right)\geq 1-Ce^{-(\log n)^{2}}.$

###### Proof of 2.12.

We denote ${\mathfrak{e}}_{+}(t)=\max(\operatorname{supp}\mu_{t})$ and
${\mathfrak{e}}_{-}(t)=\min\operatorname{supp}\mu_{t}$, as in Item 5 of 2.3.
By our assumption $\mu_{t}=\varrho_{t}(x){\rm d}x$ has a bounded density, so
the inverted cumulative density (recall from (2.13)) satisfies

(2.33) $\displaystyle y=\int_{\gamma_{t}(y)}^{\infty}\varrho_{t}(x){\rm d}x.$

By taking the time derivative of (2.33), and using Item 6 in 2.3, we get

(2.34)
$\displaystyle\operatorname{\partial}_{t}\gamma_{t}(y)\varrho_{t}(\gamma_{t}(y))=\int_{\gamma_{t}(y)}^{\infty}\operatorname{\partial}_{t}\varrho_{t}(x){\rm
d}x=\int_{\gamma_{t}(y)}^{\infty}\pi\operatorname{\partial}_{x}(\varrho_{t}(x)H\varrho_{t}(x)){\rm
d}x=-\pi\varrho_{t}(\gamma_{t}(y))H\varrho_{t}(\gamma_{t}(y)).$

It follows that $\operatorname{\partial}_{t}\gamma_{t}(y)=-\pi
H\varrho_{t}(\gamma_{t}(y))$. In particular, the classical locations of
$\varrho_{t}$ (recall from (2.14)) satisfy
$\operatorname{\partial}_{t}\gamma_{i}(t)=-\pi H\varrho_{t}(\gamma_{i}(t))$.
Next we show that the Hilbert transform on the right side of this equation can
be approximated by

(2.35) $\displaystyle\operatorname{\partial}_{t}\gamma_{i}(t)=-\pi
H\varrho_{t}(\gamma_{i}(t))=\frac{1}{n}\sum_{j:j\neq
i}\frac{1}{\gamma_{i}(t)-\gamma_{j}(t)}+\operatorname{\mathcal{O}}\left(1\right),\quad
1\leq i\leq n.$

where $\operatorname{\mathcal{O}}(1)$ depends on $c$.

Since the density $\varrho_{t}$ is bounded by $1/(2c)$, we have
$|\gamma_{i}(t)-\gamma_{i\pm 1}(t)|\geq c/n$, and
$|\gamma_{i}(t)-{\mathfrak{e}}_{\pm}(t)|\geq c/n$ (which follows from (2.14),
as it implies that
$\int_{{\mathfrak{e}}_{-}(t)}^{\gamma_{i}(t)}\varrho_{t}(x){\rm
d}x,\int_{\gamma_{i}(t)}^{{\mathfrak{e}}_{+}(t)}\varrho_{t}(x){\rm d}x\geq
1/(2n)$). We consider the integral

(2.36)
$\displaystyle\begin{split}\operatorname{PV}&\int_{\gamma_{i-1}(t)}^{{\gamma_{i+1}}(t)}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx\\\
&=\operatorname{PV}\int_{\gamma_{i}(t)-c/2n}^{{\gamma_{i}(t)}+c/2n}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx+\int_{\gamma_{i}(t)+c/2n}^{{\gamma_{i-1}}(t)}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx+\int_{\gamma_{i+1}(t)}^{{\gamma_{i}}(t)-c/2n}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx\\\
&=\operatorname{PV}\int_{\gamma_{i}(t)-c/2n}^{{\gamma_{i}(t)}+c/2n}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx+\operatorname{\mathcal{O}}\left(1\right)\\\
&=\operatorname{PV}\int_{\gamma_{i}(t)-c/2n}^{{\gamma_{i}(t)}+c/2n}\frac{\varrho_{t}(x)-\varrho_{t}(\gamma_{i}(t))}{\gamma_{i}(t)-x}dx+\operatorname{\mathcal{O}}\left(1\right)=\operatorname{\mathcal{O}}\left(\frac{\max_{x:|x-\gamma_{i}(t)|\leq
c/2n}|\varrho_{t}^{\prime}(x)|}{n}+1\right)=\operatorname{\mathcal{O}}\left(1\right),\end{split}$

where the implicit error $\operatorname{\mathcal{O}}(1)$ depends only on $c$.
We used in the third line that on $[\gamma_{i}(t)+c/2n,\gamma_{i-1}(t)]$ and
$[\gamma_{i+1}(t),\gamma_{i}(t)-c/2n]$ we have $|1/(\gamma_{i}(t)-x)|\leq
1/(c/2n))=2n/c$, so

$\displaystyle\phantom{{}={}}\left|\int_{\gamma_{i}(t)+c/2n}^{{\gamma_{i-1}}(t)}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx\right|+\left|\int_{\gamma_{i+1}(t)}^{{\gamma_{i}}(t)-c/2n}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx\right|$
$\displaystyle\leq\frac{2n}{c}\left|\int_{\gamma_{i}(t)+c/2n}^{{\gamma_{i-1}}(t)}\varrho_{t}(x)dx\right|+\frac{2n}{c}\left|\int_{\gamma_{i+1}(t)}^{{\gamma_{i}}(t)-c/2n}\varrho_{t}(x)dx\right|\leq\frac{2n}{c}\cdot\frac{1}{n}+\frac{2n}{c}\cdot\frac{1}{n}=\frac{4}{c}.$

where we also used the definition of classical locations (recall from (2.14));
in the fourth line we used that
$\operatorname{PV}\int_{\gamma_{i}(t)-c/2n}^{{\gamma_{i}(t)}+c/2n}(\gamma_{i}(t)-x)^{-1}dx=0$;
in the last line, we used that for any
$\operatorname{dist}(x,\\{\gamma_{1}(t),\gamma_{2}(t),\cdots,\gamma_{n}(t)\\})\leq
c/2n$, it holds $|\varrho_{t}^{\prime}(x)|\leq n/c$ from our assumption. For
the integral outside the interval $[\gamma_{i-1}(t),\gamma_{i+1}(t)]$, we have
the trivial bounds

$\displaystyle\int_{-\infty}^{\gamma_{i+1}(t)}\frac{\varrho_{t}(x)dx}{\gamma_{i}(t)-x}\leq\sum_{j=i+1}^{n}\int_{\gamma_{j+1}(t)}^{\gamma_{j}(t)}\frac{\varrho_{t}(x)dx}{\gamma_{i}(t)-\gamma_{j}(t)}\leq\sum_{j=i+1}^{n}\frac{1}{n(\gamma_{i}(t)-\gamma_{j}(t))},$

and

$\displaystyle\int_{-\infty}^{\gamma_{i+1}(t)}\frac{\varrho_{t}(x)dx}{\gamma_{i}(t)-x}\geq\sum_{j=i+1}^{n}\int_{\gamma_{j+1}(t)}^{\gamma_{j}(t)}\frac{\varrho_{t}(x)dx}{\gamma_{i}(t)-\gamma_{j+1}(t)}\geq\sum_{j=i+2}^{n}\frac{1}{n(\gamma_{i}(t)-\gamma_{j}(t))}.$

Thus we conclude that

(2.37)
$\displaystyle\left|\int_{-\infty}^{\gamma_{i+1}(t)}\frac{\varrho_{t}(x)dx}{\gamma_{i}(t)-x}-\sum_{j=i+1}^{n}\frac{1}{n(\gamma_{i}(t)-\gamma_{j}(t))}\right|\leq\frac{1}{n(\gamma_{i}(t)-\gamma_{i+1}(t))}\leq\frac{1}{c}.$

We have the same estimate for the integral from $\gamma_{i-1}(t)$ to $\infty$.
The claim (2.35) follows from combining (2.36) and (2.37).

Next we show (2.32) using (2.35) as input. By taking the difference of (1.2)
and (2.35), we get

(2.38) $\displaystyle\begin{split}{\rm
d}(\lambda_{i}(t)-\gamma_{i}(t))&=\Big{(}\displaystyle\frac{2}{\beta
n}\Big{)}^{1/2}{\rm d}B_{i}(t)-\frac{1}{n}\sum_{j:j\neq
i}\frac{(\lambda_{i}(t)-\gamma_{i}(t))-(\lambda_{j}(t)-\gamma_{j}(t))}{(\lambda_{i}(t)-\lambda_{j}(t))(\gamma_{i}(t)-\gamma_{j}(t))}{\rm
d}t+\operatorname{\mathcal{O}}\left(1\right),\quad 1\leq i\leq n.\end{split}$

We take $i_{*}(t)=\arg\max_{1\leq i\leq n}(\lambda_{i}(t)-\gamma_{i}(t))$ (if
$i_{*}(t)$ is not unique, we can simply take the smallest one). Since the
$\lambda_{i}(t)-\gamma_{i}(t)$ are continuous, $i_{*}(t)$ is piecewise
constant. By plugging $i=i_{*}(t)$ (if $i_{*}$ has a jump at $t$, we interpret
this as either $i_{*}(t^{-})$ or $i_{*}(t^{+})$) in (2.38), noticing that the
second term on the righthand side of (2.38) is nonpositive, we deduce (through
a very similar argument to the one used to obtain (2.19) and (2.20)) that

(2.39) $\displaystyle{\rm
d}(\lambda_{i_{*}(t)}(t)-\gamma_{i_{*}(t)}(t))\leq\Big{(}\displaystyle\frac{2}{\beta
n}\Big{)}^{1/2}{\rm
d}B_{i_{*}(t)}(t)+\operatorname{\mathcal{O}}\left(1\right){\rm d}t.$

The martingale term ${\rm d}B_{i_{*}(t)}(t)$ has the same law as a standard
Brownian motion. Thus by the reflection principle we have

(2.40) $\displaystyle\mathbb{P}\left(\sup_{0\leq s\leq t}\int_{0}^{s}{\rm
d}B_{i_{*}(\tau)}(\tau)\geq a\sqrt{t}\right)=\mathbb{P}\left(\int_{0}^{t}{\rm
d}B_{i_{*}(\tau)}(\tau)\geq a\sqrt{t}\right)\leq 2e^{-a^{2}/2}.$

By integrating both sides of (2.39) (and taking $a=\log n$ in the above tail
estimate), we deduce that there exists a large constant
$C=C(c,{\mathsf{T}})>0$ such that, the following holds with probability
$1-Ce^{-(\log n)^{2}}$ for any $0\leq s\leq t$,

$\displaystyle\lambda_{i_{*}(s)}(s)-\gamma_{i_{*}(s)}(s)$
$\displaystyle\leq\lambda_{i_{*}(0)}(0)-\gamma_{i_{*}(0)}(0)+\Big{(}\displaystyle\frac{2}{\beta
n}\Big{)}^{1/2}\int_{0}^{s}{\rm
d}B_{i_{*}(\tau)}(\tau)+\operatorname{\mathcal{O}}(s)$
$\displaystyle\leq\max_{i}|\lambda_{i}(0)-\gamma_{i}(0)|+C\left(t+\frac{(\log
n)\sqrt{t}}{\sqrt{n}}\right),$

It follows that for any $0\leq t\leq{\mathsf{T}}$ and $n$ large enough,

(2.41) $\displaystyle\mathbb{P}\left(\sup_{0\leq s\leq t,\atop 1\leq i\leq
n}\lambda_{i}(s)-\gamma_{i}(s)\leq\max_{i}|\lambda_{i}(0)-\gamma_{i}(0)|+C\left(t+\frac{(\log
n)\sqrt{t}}{\sqrt{n}}\right)\right)\geq 1-Ce^{-(\log n)^{2}}.$

By the same argument, we have a similar lower bound by considering
$i_{*}(t)=\arg\min_{1\leq i\leq n}(\lambda_{i}(t)-\gamma_{i}(t))$, that is, we
have for any $0\leq t\leq{\mathsf{T}}$ and $n$ large enough,

(2.42) $\displaystyle\mathbb{P}\left(\sup_{0\leq s\leq t,\atop 1\leq i\leq
n}\gamma_{i}(s)-\lambda_{i}(s)\leq\max_{i}|\lambda_{i}(0)-\gamma_{i}(0)|+C\left(t+\frac{(\log
n)\sqrt{t}}{\sqrt{n}}\right)\right)\geq 1-Ce^{-(\log n)^{2}},$

2.12 follows from combining (2.41) and (2.42). ∎

We also need the following special case of 2.12, which compares Dyson Brownian
motion starting from a delta mass with the quantiles of semicircle
distribution.

######  Proposition 2.13.

Fix a real number ${\mathsf{T}}>0$ and integers $1\leq m\leq n$; let
$A=m/n\in(0,1]$. Define the measure $\mu_{0}=A\delta_{0}\in\mathscr{P}$, and
denote by $\mu_{t}=\mu_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$ the free
convolution of $\mu$ with the rescaled semicircle distribution
$\mu_{\operatorname{sc}}^{(t)}$. We denote by
$\gamma_{i}(t)=\gamma_{i;n}^{\mu_{t}}(t)$ the classical locations (recall from
(2.14)) of $\mu_{t}({\rm d}x)=\varrho_{t}(x){\rm d}x$ for $1\leq i\leq m$. We
consider the $m$-particle Dyson Brownian motion (1.2) starting from
$\lambda_{1}(0)=\lambda_{2}(0)=\cdots=\lambda_{m}(0)=0$ (and
$\lambda_{m+1}(0)=\cdots=\lambda_{n}(0)=-\infty$). Then there exists a
constant $C=C({\mathsf{T}})>0$ such that the following holds for any $0\leq
t\leq{\mathsf{T}}$

(2.43) $\displaystyle\begin{split}&\mathbb{P}\left(\sup_{0\leq s\leq t,\atop
1\leq i\leq n}|\lambda_{i}(s)-\gamma_{i}(s)|\leq C\left(\sqrt{At}+\frac{(\log
n)\sqrt{t}}{\sqrt{n}}\right)\right)\geq 1-Ce^{-(\log n)^{2}},\\\
&\mathbb{P}\left(\sup_{0\leq s\leq t,\atop 1\leq i\leq n}|\lambda_{i}(s)|\leq
C\left(\sqrt{At}+\frac{(\log n)\sqrt{t}}{\sqrt{n}}\right)\right)\geq
1-Ce^{-(\log n)^{2}}.\end{split}$

###### Proof of 2.13.

The proof follows essentially the same argument as in 2.12, so we will only
sketch the proof. We recall from Example 2.5 that the free convolution
$\mu_{t}=\mu_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$ is supported on
$[-2\sqrt{At},2\sqrt{At}]$, and given explicitly by the rescaled semicircle
distribution

(2.44) $\displaystyle\mu_{t}({\rm d}x)=\varrho_{t}(x){\rm
d}x=\frac{\sqrt{4At-x^{2}}}{2\pi t}{\rm d}x,\quad t\geq 0.$

One can directly check that $\varrho_{t}(x)$ satisfies for $0\leq x\leq
2\sqrt{At}$ the bounds

(2.45)
$\displaystyle\varrho_{t}(x)\leq\varrho_{t}(0)=\frac{\sqrt{A/t}}{\pi}\leq\frac{\sqrt{A/t}}{2},\quad|\varrho_{t}^{\prime}(x)|\leq(2\sqrt{At}-x)^{-1/2}\frac{A^{1/4}}{t^{3/4}}.$

Using the above estimates as input, similarly to (2.35), next we show

(2.46) $\displaystyle\operatorname{\partial}_{t}\gamma_{i}(t)=-\pi
H(\varrho_{t})(\gamma_{i}(t))=\frac{1}{n}\sum_{1\leq j\leq m\atop j\neq
i}\frac{1}{\gamma_{i}(t)-\gamma_{j}(t)}+\operatorname{\mathcal{O}}\left(\sqrt{\frac{A}{t}}\right),\quad
1\leq i\leq m.$

where the implicit constant in $\operatorname{\mathcal{O}}(\cdot)$ depends
only on ${\mathsf{T}}$. Since the density $\varrho_{t}$ is bounded by
$\sqrt{A/t}/2$ (from (2.45), we have $|\gamma_{i}(t)-\gamma_{i\pm
1}(t)|\geq\sqrt{t/A}/n$, and $|\gamma_{i}(t)-(\pm
2\sqrt{At})|\geq\sqrt{t/A}/n$ (which follows from (2.14) that
$\int_{-2\sqrt{At}}^{\gamma_{i}(t)}\varrho_{t}(x){\rm
d}x,\int_{\gamma_{i}(t)}^{2\sqrt{At}}\varrho_{t}(x){\rm d}x\geq 1/(2n)$). We
consider the integral

(2.47)
$\displaystyle\begin{split}&\phantom{{=}}\operatorname{PV}\int_{\gamma_{i-1}(t)}^{{\gamma_{i+1}}(t)}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx\\\
&=\operatorname{PV}\int_{\gamma_{i}(t)-\sqrt{t/A}/(2n)}^{{\gamma_{i}(t)}+\sqrt{t/A}/(2n)}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx+\int_{\gamma_{i}(t)+\sqrt{t/A}/(2n)}^{{\gamma_{i-1}}(t)}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx+\int_{\gamma_{i+1}(t)}^{{\gamma_{i}}(t)-\sqrt{t/A}/(2n)}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx\\\
&=\operatorname{PV}\int_{\gamma_{i}(t)-\sqrt{t/A}/(2n)}^{{\gamma_{i}(t)}+\sqrt{t/A}/(2n)}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx+\operatorname{\mathcal{O}}\left(\sqrt{\frac{A}{t}}\right)\\\
&=\operatorname{PV}\int_{\gamma_{i}(t)-\sqrt{t/A}/(2n)}^{{\gamma_{i}(t)}+\sqrt{t/A}/(2n)}\frac{\varrho_{t}(x)-\varrho_{t}(\gamma_{i}(t))}{\gamma_{i}(t)-x}dx+\operatorname{\mathcal{O}}\left(\sqrt{\frac{A}{t}}\right)\\\
&\leq\frac{\sqrt{t/A}}{n}\max_{x:|x-\gamma_{i}(t)|\leq\sqrt{t/A}/(2n)}|\varrho_{t}^{\prime}(x)|+\operatorname{\mathcal{O}}\left(\sqrt{\frac{A}{t}}\right)\leq\frac{1}{\sqrt{tn}}+\operatorname{\mathcal{O}}\left(\sqrt{\frac{A}{t}}\right)=\operatorname{\mathcal{O}}\left(\sqrt{\frac{A}{t}}\right).\end{split}$

We used in the third line that on
$[\gamma_{i}(t)+\sqrt{t/A}/(2n),\gamma_{i+1}(t)]$ and
$[\gamma_{i-1}(t),\gamma_{i}(t)-\sqrt{t/A}/(2n)]$ we have
$|1/(\gamma_{i}(t)-x)|\leq 1/(\sqrt{t/A}/(2n)))=2n\sqrt{A/t}$. So

$\displaystyle\phantom{{}={}}\left|\int_{\gamma_{i}(t)+\sqrt{t/A}/(2n)}^{{\gamma_{i-1}}(t)}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx\right|+\left|\int_{\gamma_{i+1}(t)}^{{\gamma_{i}}(t)-\sqrt{t/A}/(2n)}\frac{\varrho_{t}(x)}{\gamma_{i}(t)-x}dx\right|$
$\displaystyle\leq
2n\sqrt{\frac{A}{t}}\int_{\gamma_{i}(t)+\sqrt{t/A}/(2n)}^{{\gamma_{i-1}}(t)}\varrho_{t}(x)dx+2n\sqrt{\frac{A}{t}}\int_{\gamma_{i+1}(t)}^{{\gamma_{i}}(t)-\sqrt{t/A}/(2n)}\varrho_{t}(x)dx\leq
4\sqrt{\frac{A}{t}},$

where we also used the definition of classical locations from (2.14); in the
fourth line we used that
$\operatorname{PV}\int_{\gamma_{i}(t)-\sqrt{t/A}/(2n)}^{{\gamma_{i}(t)}+\sqrt{t/A}/(2n)}(\gamma_{i}(t)-x)^{-1}dx=0$;
in the last line, we used that for
$x\in[\gamma_{m}(t)-\sqrt{t/A}/(2n),\gamma_{1}(t)+\sqrt{t/A}/(2n)]\subset[-2\sqrt{At}+\sqrt{t/A}/(2n),2\sqrt{At}-\sqrt{t/A}/(2n)]$,
it holds that $|\varrho_{t}^{\prime}(x)|\leq\sqrt{An}/t$ from (2.45); in the
last equality we used that $A=m/n\geq 1/n$. For the integral outside the
interval $[\gamma_{i-1}(t),\gamma_{i+1}(t)]$, we have the trivial bounds

$\displaystyle\int_{-\infty}^{\gamma_{i+1}(t)}\frac{\varrho_{t}(x)dx}{\gamma_{i}(t)-x}\leq\sum_{j=i+1}^{n}\int_{\gamma_{j+1}(t)}^{\gamma_{j}(t)}\frac{\varrho_{t}(x)dx}{\gamma_{i}(t)-\gamma_{j}(t)}\leq\sum_{j=i+1}^{m}\frac{1}{n(\gamma_{i}(t)-\gamma_{j}(t))},$

and

$\displaystyle\int_{-\infty}^{\gamma_{i+1}(t)}\frac{\varrho_{t}(x)dx}{\gamma_{i}(t)-x}\geq\sum_{j=i+1}^{n}\int_{\gamma_{j+1}(t)}^{\gamma_{j}(t)}\frac{\varrho_{t}(x)dx}{\gamma_{i}(t)-\gamma_{j+1}(t)}\geq\sum_{j=i+2}^{m}\frac{1}{n(\gamma_{i}(t)-\gamma_{j}(t))}.$

Together with the fact that $\gamma_{i}(t)-\gamma_{i+1}(t)\geq\sqrt{t/A}/n$,
this gives

(2.48)
$\displaystyle\left|\int_{-\infty}^{\gamma_{i+1}(t)}\frac{\varrho_{t}(x)dx}{\gamma_{i}(t)-x}-\sum_{j=i+1}^{n}\frac{1}{n(\gamma_{i}(t)-\gamma_{j}(t))}\right|\leq\frac{1}{n(\gamma_{i}(t)-\gamma_{i+1}(t))}\leq\sqrt{\frac{A}{t}}.$

We have the same estimate for the integral from $\gamma_{i-1}(t)$ to $\infty$.
The claim (2.46) follows from combining (LABEL:e:diffA) and (2.48).

Using (2.46) as input, following the derivations of (2.41) and (2.42), we
conclude there exists a large constant $C>0$ such that for any $0\leq
t\leq{\mathsf{T}}$ and $n$ large enough, the following holds with probability
$1-Ce^{-(\log n)^{2}}$,

(2.49) $\displaystyle\sup_{0\leq s\leq t,\atop 1\leq i\leq
n}|\lambda_{i}(s)-\gamma_{i}(s)|\leq C\left(\int_{0}^{t}\frac{A{\rm
d}\tau}{\sqrt{\tau}}+\frac{(\log
n)\sqrt{t}}{\sqrt{n}}\right)=C\left(2\sqrt{tA}+\frac{(\log
n)\sqrt{t}}{\sqrt{n}}\right).$

This finishes the proof of the first statement in (2.43). The second statement
follows from the first one and the bound $|\gamma_{i}(s)|\leq 2\sqrt{As}\leq
2\sqrt{At}$. ∎

###  2.5. Free Convolution Density Estimates

Adopt 1.2. Denote the free convolution of $\mu_{0}:=\mu_{0}^{(n)}$ with the
rescaled semicircle distribution $\mu_{\operatorname{sc}}^{(t)}$ by
$\mu_{t}=\mu_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$. Denote the Stieltjes
transform of $\mu_{t}$ by $m_{t}(z)$ as in (2.1), and its $p$-th derivative by
$m_{t}^{(p)}(z)$. The free convolution $\mu_{t}=\varrho_{t}(x){\rm d}x$ has a
density $\varrho_{t}(x)$. Thanks to Item 5, the right edge of $\mu_{t}$,
denoted by $E_{t}=\max(\operatorname{supp}\mu_{t})$, satisfies
$E_{t}=\xi(t)-tm_{0}(\xi(t))$, where $\xi(t)$ is given by

(2.50)
$\displaystyle\xi(t)=\sup\Bigg{\\{}w\in\mathbb{R}:\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{\mu_{0}({\rm
d}x)}{(x-w)^{2}}=\displaystyle\frac{1}{t}\Bigg{\\}}.$

The following two lemmas collects some properties of $\mu_{t}$, i.e., square
root edge behavior and averaged density estimates. They will be used in
Section 3 to prove our main result 1.3.

######  Lemma 2.14.

Adopt 1.2. Denote the free convolution of $\mu_{0}:=\mu_{0}^{(n)}$ with the
rescaled semicircle distribution $\mu_{\operatorname{sc}}^{(t)}$ by
$\mu_{t}=\mu_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$, and its right edge by
$E_{t}=\max(\operatorname{supp}\mu_{t})$. We denote the Stieltjes transform of
$\mu_{t}$ as $m_{t}(z)$, recall $\xi(t)$ from (2.50), and introduce the
following two parameters:

(2.51)
$\displaystyle{\mathcal{A}}(t)=\frac{2}{-t^{3}m^{(2)}_{0}(\xi(t))},\quad{\mathfrak{c}}(t)=-2^{-9}tm^{(2)}_{0}(\xi(t))\xi(t)^{2}.$

Then for any $(30/{\mathsf{b}})\sqrt{\eta}_{*}\leq t\leq{\mathsf{T}}$ we have

(2.52)
$\displaystyle\sqrt{t}\geq\xi(t)\geq\left(\frac{{\mathsf{b}}t}{20}\right)^{2},\quad{\mathcal{A}}(t)\geq
2^{-10}{\mathsf{b}}^{2},\quad{\mathfrak{c}}(t)\geq
2^{-20}({\mathsf{b}}t)^{2},$

and for all $x\in[E_{t}-{\mathfrak{c}}(t),E_{t}]$ we have

(2.53)
$\displaystyle\varrho_{t}(x)=(1+{\mathcal{E}}(E_{t}-x))\frac{\sqrt{{\mathcal{A}}(t)(E_{t}-x)}}{\pi},\qquad\text{where
$\mathcal{E}$ satisfies
$|{\mathcal{E}}(x)|\leq\frac{|x|}{{\mathfrak{c}}(t)}$}.$

######  Lemma 2.15.

Adopt 1.2. Denote the free convolution of $\mu_{0}:=\mu_{0}^{(n)}$ with the
rescaled semicircle distribution $\mu_{\operatorname{sc}}^{(t)}$ by
$\mu_{t}=\mu_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$, and its right edge by
$E_{t}=\max(\operatorname{supp}\mu_{t})$. We denote the Stieltjes transform of
$\mu_{t}$ by $m_{t}$. Then there exists constants
${\mathfrak{C}}=2^{21}/{\mathsf{b}}^{2}>0$ and
${\mathfrak{c}}={\mathsf{b}}{\mathfrak{C}}^{-3/2}>0$ such that the following
hold.

  1. (1)

The right edge satisfies $E_{t}\leq 2\sqrt{t}$.

  2. (2)

For any $t\in[0,{\mathsf{T}}]$ and $x\in[0,{\mathsf{T}}^{2}]$ satisfying
either $t\geq{\mathfrak{C}}\sqrt{\eta}_{*}$ or $x\geq{\mathfrak{C}}\eta_{*}$,
we have $\mu_{t}([E_{t}-x,E_{t}])\geq{\mathfrak{c}}x^{3/2}$.

  3. (3)

For any $t\in[0,{\mathsf{T}}]$ and $\kappa,\eta\geq 0$ with
$\kappa+\eta\leq{\mathsf{T}}^{2}$ satisfying either
$t\geq{\mathfrak{C}}\sqrt{\eta}_{*}$ or $\kappa\geq{\mathfrak{C}}\eta_{*}$, we
have

(2.54) $\displaystyle{\rm
Im}[m_{t}(E_{t}+\kappa+\mathrm{i}\eta)]\geq\frac{{\mathfrak{c}}\eta}{8\sqrt{\kappa+\eta}}.$

###### Proof of 2.14.

For simplicity of notation, in the rest of the proof, we will write
$\kappa=\xi(t)$ as defined in (2.50). Let $F(x)=\mu_{0}([-x,0])$, which
satisfies $F(x)\geq{\mathsf{b}}x^{3/2}$ for $x\in[\eta_{*},\mathsf{T}^{2}]$.
Then we have

(2.55) $\displaystyle\frac{1}{t}=\int_{-\infty}^{0}\frac{\mu_{0}({\rm
d}x)}{(\kappa-x)^{2}}=-\int_{0}^{\infty}\frac{{\rm
d}F(x)}{(\kappa+x)^{2}}\geq\int_{0}^{\infty}\frac{2F(x){\rm
d}x}{(\kappa+x)^{3}}\geq\int_{\eta_{*}}^{1}\frac{2{\mathsf{b}}x^{3/2}{\rm
d}x}{(\kappa+x)^{3}},$

where the first statement is from (2.50); the second statement is from the
definition $F(x)=\mu_{0}([-x,0])$; the third statement is by integration by
parts; and the last statement is by restricting the integration domain to
$[\eta_{*},1]$, and using that $F(x)\geq{\mathsf{b}}x^{3/2}$ on
$[\eta_{*},{\mathsf{T}}]$ and ${\mathsf{T}}\geq 1$ from 1.2.

If $\kappa\leq 2\eta_{*}$, then (2.55) gives

(2.56)
$\displaystyle\frac{1}{t}\geq\int_{\eta_{*}}^{1}\frac{2{\mathsf{b}}x^{3/2}{\rm
d}x}{27x^{3}}=\frac{4{\mathsf{b}}}{27}\left(\frac{1}{\sqrt{\eta}_{*}}-1\right)\geq\frac{{\mathsf{b}}}{27\sqrt{\eta}_{*}},$

where the second statement is from performing the integral, and the last
statement uses $\eta_{*}\leq 1/4$ from 1.2. This contradicts our assumption
$t\geq(30/{\mathsf{b}})\sqrt{\eta_{*}}>27\sqrt{\eta}_{*}/{\mathsf{b}}$, so we
have $\kappa\geq 2\eta_{*}$, and (2.55) gives

(2.57)
$\displaystyle\frac{1}{t}\geq\int_{\eta_{*}}^{\kappa}\frac{2{\mathsf{b}}x^{3/2}{\rm
d}x}{8\kappa^{3}}=\frac{{\mathsf{b}}(\kappa^{5/2}-\eta_{*}^{5/2})}{10\kappa^{3}}\geq\frac{{\mathsf{b}}}{20\kappa^{1/2}},\qquad\text{so}\qquad\kappa\geq\left(\frac{{\mathsf{b}}t}{20}\right)^{2},$

where the first inequality follows from restricting the integral (2.55) to
$[\eta_{*},\kappa]$ and the fact that $\kappa+x\leq 2\kappa$ on this region;
the second statement follows from performing the integral; and the last
inequality follows from the bound $\kappa\geq 2\eta_{*}$. This gives the lower
bound in the first statement in (2.52). The upper bound follows from the
following estimate

$\displaystyle\frac{1}{t}=\int_{-\infty}^{0}\frac{\mu_{0}({\rm
d}x)}{(\kappa-x)^{2}}\leq\int_{-\infty}^{0}\frac{\mu_{0}({\rm
d}x)}{\kappa^{2}}=\frac{1}{\kappa^{2}}$

where the first statement is from (2.50); the second statement uses
$\kappa-x\geq\kappa$; and the last statement uses $\mu_{0}({\mathbb{R}})=1$.

We next bound the derivatives of $m_{0}$, in order to Taylor expand it. In the
following, we recall that the $p$-th derivative of $m_{0}$ is denoted by
$m_{0}^{(p)}$. Since $m_{0}(w)$ is analytic outside
$\operatorname{supp}\mu_{0}\subset(-\infty,0]$, it is in particular analytic
on $\\{w\in{\mathbb{C}}:|w-\kappa|<\kappa/2\\}$. To bound
$m_{0}^{(2)}(\kappa)$ observe that, similarly to (2.55) and (2.57), we have

(2.58)
$\displaystyle\begin{split}-m_{0}^{(2)}(\kappa)&=\displaystyle\int_{0}^{\infty}\displaystyle\frac{6F(x)\mathrm{d}x}{(\kappa+x)^{4}}\geq\int_{\eta_{*}}^{\kappa}\frac{6{\mathsf{b}}x^{3/2}{\rm
d}x}{(\kappa+x)^{4}}\geq\int^{\kappa}_{\eta_{*}}\frac{4{\mathsf{b}}x^{3/2}{\rm
d}x}{(2\kappa)^{4}}=\frac{{\mathsf{b}}(\kappa^{5/2}-\eta_{*}^{5/2})}{10\kappa^{4}}\geq\frac{{\mathsf{b}}}{20\kappa^{3/2}},\end{split}$

where in the second statement we restricted the integral to
$[\eta_{*},\kappa]$, and used
$\kappa\leq\sqrt{t}\leq\sqrt{\mathsf{T}}\leq{\mathsf{T}}^{2}$ from the first
statement in (2.52), and $F(x)=\mu_{0}([-x,0])\geq{\mathsf{b}}x^{3/2}$ from
1.2; in the third statement we used $\kappa+x\leq 2\kappa$; and in the last
two statements used that $\kappa\geq 2\eta_{*}$. We also have the upper bounds
of $-m_{0}^{(2)}$

(2.59) $\displaystyle-m_{0}^{(2)}(\kappa)=\int_{\mathbb{R}}2!\frac{{\rm
d}\mu_{0}(x)}{(\kappa-x)^{3}}\leq\int_{\mathbb{R}}2!\frac{{\rm
d}\mu_{0}(x)}{(\kappa-x)^{2}\kappa}=\frac{2}{t\kappa},$

and

(2.60) $\displaystyle m_{0}^{(3)}(\kappa)=\int_{\mathbb{R}}3!\frac{{\rm
d}\mu_{0}(x)}{(\kappa-x)^{4}}\leq 6\int_{\mathbb{R}}\frac{{\rm
d}\mu_{0}(x)}{\kappa(\kappa-x)^{3}}=\frac{-3m_{0}^{(2)}(\kappa)}{\kappa}.$

where we bounded $|\kappa-x|\geq\kappa$ for
$x\in\operatorname{supp}\mu_{0}\subset(-\infty,0]$ (to obtain both estimates)
and used (2.50) (in the last statement of (2.59)). For any $w\in{\mathbb{C}}$
such that $|w-\kappa|<\kappa/2$, we have

(2.61) $\displaystyle|m_{0}^{(4)}(w)|\leq\int_{\mathbb{R}}4!\frac{{\rm
d}\mu_{0}(x)}{|x-w|^{5}}\leq 4!\int_{\mathbb{R}}\frac{2^{3}{\rm
d}\mu_{0}(x)}{(\kappa/2)^{2}(x-\kappa)^{3}}=\frac{-3\times
2^{7}m_{0}^{(2)}(\kappa)}{\kappa^{2}},$

where we used that $|x-w|\geq\kappa/2$ and $2|x-w|\geq|x-\kappa|$ for
$x\in\operatorname{supp}\mu_{0}\subseteq(-\infty,0]$ for the second
inequality.

We can thus Taylor expand the Stieltjes transform $m_{0}(w)$ of $\mu_{0}$
around $\kappa$ when $|w-\kappa|<\kappa/2$:

(2.62)
$\displaystyle\begin{split}m_{0}(w)&=m_{0}(\kappa)+m_{0}^{\prime}(\kappa)(w-\kappa)+m_{0}^{(2)}(\kappa)\frac{(w-\kappa)^{2}}{2}+m_{0}^{(3)}(\kappa)\frac{(w-\kappa)^{3}}{3!}+\widetilde{\mathcal{E}}(w)\frac{(w-\kappa)^{4}}{4!}\\\
&=m_{0}(\kappa)+\frac{(w-\kappa)}{t}+m_{0}^{(2)}(\kappa)\frac{(w-\kappa)^{2}}{2}+m_{0}^{(3)}(\kappa)\frac{(w-\kappa)^{3}}{3!}+\widetilde{\mathcal{E}}(w)\frac{(w-\kappa)^{4}}{4!},\end{split}$

where in the second line we used that $\kappa=\xi(t)$ and
$m_{0}^{\prime}(\xi(t))=1/t$ from the definition of $\xi(t)$ (2.50). Moreover,
thanks to (2.61), the Taylor remainder is bounded by

(2.63)
$\displaystyle|\widetilde{\mathcal{E}}(z)|\leq\max_{w\in{\mathbb{C}}:|w-\kappa|\leq\kappa/2}|m_{0}^{(4)}(w)|\leq\frac{-3\times
2^{7}m_{0}^{(2)}(\kappa)}{\kappa^{2}}.$

By Item 4 in 2.3, any $x\in\mathbb{R}$ satisfies $\varrho_{t}(E_{t}-x)>0$ if
and only if there exists a
$w\in\operatorname{\partial}\Lambda_{t}\cap\mathbb{H}$ such that
$E_{t}-x=w-tm_{0}(w)$. Recalling that $E_{t}=\kappa-tm_{0}(\kappa)$ (from
$\kappa=\xi(t)$ and Item 5 in 2.3), rearranging (2.62) gives

(2.64) $\displaystyle\begin{split}{\mathbb{R}}\ni x&=(\kappa-
tm_{0}(\kappa))-(w-tm_{0}(w))\\\
&=tm_{0}^{(2)}(\kappa)\frac{(w-\kappa)^{2}}{2}+tm_{0}^{(3)}(\kappa)\frac{(w-\kappa)^{3}}{3!}+t\widetilde{\mathcal{E}}(w)\frac{(w-\kappa)^{4}}{4!}\\\
&=\left(1+{\mathcal{E}}(w)\right)\frac{tm_{0}^{(2)}(\kappa)(w-\kappa)^{2}}{2},\end{split}$

where

(2.65)
$\displaystyle{\mathcal{E}}(w):=\frac{m_{0}^{(3)}(\kappa)}{m_{0}^{(2)}(\kappa)}\frac{(w-\kappa)}{3}+\frac{\widetilde{\mathcal{E}}(w)}{m_{0}^{(2)}(\kappa)}\frac{(w-\kappa)^{2}}{12}.$

When $|w-\kappa|\leq 2^{-3}\kappa$, from (2.65) we have

(2.66)
$\displaystyle|{\mathcal{E}}(w)|\leq\frac{|w-\kappa|}{\kappa}+\frac{2^{5}|w-\kappa|^{2}}{\kappa^{2}}\leq\frac{5|w-\kappa|}{\kappa}\leq\frac{5}{8}<1,$

where we used (2.60) and (2.63) for the first inequality, and $|w-\kappa|\leq
2^{-3}\kappa$ for the second. Next we show that the map
$w\mapsto\left(1+{\mathcal{E}}(w)\right)tm_{0}^{(2)}(\kappa)(w-\kappa)^{2}/2$
is a surjection from $|w-\kappa|\leq 2^{-3}\kappa$ onto
$\\{u\in{\mathbb{C}}:|u|\leq-2^{-9}tm_{0}^{(2)}(\kappa)\kappa^{2}\\}$.

Fix any $|u|\leq-2^{-9}tm_{0}^{(2)}(\kappa)\kappa^{2}$; we will use Rouché’s
theorem to compare the two holomorphic functions
$tm_{0}^{(2)}(\kappa)(w-\kappa)^{2}/2-u$ and
$\left(1+{\mathcal{E}}(w)\right)tm_{0}^{(2)}(\kappa)(w-\kappa)^{2}/2-u$ on the
domain $\\{w\in{\mathbb{C}}:|w-\kappa|\leq 2^{-3}\kappa\\}$. First we notice
that $tm_{0}^{(2)}(\kappa)(w-\kappa)^{2}/2-u$ has a root with $|w-\kappa|\leq
2^{-3}\kappa$. Next, for any $w$ such that $|w-\kappa|=2^{-3}\kappa$, we can
upper bound the difference

(2.67)
$\displaystyle\begin{split}&\phantom{{}={}}\left|\left(\left(1+{\mathcal{E}}(w)\right)tm_{0}^{(2)}(\kappa)(w-\kappa)^{2}/2-u\right)-\left(tm_{0}^{(2)}(\kappa)(w-\kappa)^{2}/2-u\right)\right|\\\
&=-tm_{0}^{(2)}(\kappa)|w-\kappa|^{2}|{\mathcal{E}}(w)|/2\leq-(5/8)\cdot
tm_{0}^{(2)}(\kappa)|w-\kappa|^{2}/2\leq\left|tm_{0}^{(2)}(\kappa)(w-\kappa)^{2}/2-u\right|,\end{split}$

where in the second inequality we used (2.66) indicating that
$|{\mathcal{E}}(w)|\leq 5/8$, and in the last inequality we used that, when
$|w-\kappa|=2^{-3}\kappa$,

(2.68)
$\displaystyle-\frac{3}{8}\frac{tm_{0}^{(2)}(\kappa)|w-\kappa|^{2}}{2}=-\frac{3}{8}\frac{tm_{0}^{(2)}(\kappa)2^{-6}\kappa^{2}}{2}>-2^{-9}tm_{0}^{(2)}(\kappa)\kappa^{2}\geq|u|.$

The relation (2.67) verifies the conditions of Rouché’s theorem, and hence
$\left(1+{\mathcal{E}}(w)\right)tm_{0}^{(2)}(\kappa)(w-\kappa)^{2}/2-u$ also
has a root with $|w-\kappa|\leq 2^{-3}\kappa$.

In particular for any $0\leq x\leq-2^{-9}tm_{0}^{(2)}(\kappa)\kappa^{2}$,
(2.64) has a solution $w=w(x)$ with $|w-\kappa|\leq 2^{-3}\kappa$:

(2.69) $\displaystyle
w-\kappa=\mathrm{i}\sqrt{\frac{2x}{-tm_{0}^{(2)}(\kappa)}}\left(1+{\mathcal{E}}(w)\right)^{-1/2},\quad|{\mathcal{E}}(w)|\leq
5/8.$

To further simplify (2.69), we will use the following elementary inequalities:
for any $|\varepsilon|\leq 5/8$.

(2.70) $\displaystyle\left|\frac{1}{\sqrt{1+\varepsilon}}-1\right|\leq
2\varepsilon,\quad\left|\frac{1}{\sqrt{1+\varepsilon}}-1+\frac{\varepsilon}{2}\right|\leq\frac{8\varepsilon^{2}}{3},$

Using them we can rewrite ${\mathcal{E}}(w)$ from (2.65) as

(2.71)
$\displaystyle\begin{split}&\phantom{{}={}}\left|\left(1+{\mathcal{E}}(w)\right)^{-1/2}-1+\frac{m_{0}^{(3)}(\kappa)}{6m_{0}^{(2)}(\kappa)}\mathrm{i}\sqrt{\frac{2x}{-tm_{0}^{(2)}(\kappa)}}\right|\\\
&\leq\left|\left(1+{\mathcal{E}}(w)\right)^{-1/2}-1+\frac{{\mathcal{E}}(w)}{2}\right|+\left|\frac{{\mathcal{E}}(w)}{2}-\frac{m_{0}^{(3)}(\kappa)}{m_{0}^{(2)}(\kappa)}\frac{(w-\kappa)}{6}\right|+\frac{m_{0}^{(3)}(\kappa)}{6|m_{0}^{(2)}(\kappa)|}\left|w-\kappa-\mathrm{i}\sqrt{\frac{2x}{-tm_{0}^{(2)}(\kappa)}}\right|\\\
&\leq\frac{8|{\mathcal{E}}(w)|^{2}}{3}+\frac{|\widetilde{\mathcal{E}}(w)|}{m_{0}^{(2)}(\kappa)}\frac{|w-\kappa|^{2}}{24}+\frac{3}{\kappa}\sqrt{\frac{2x}{-tm_{0}^{(2)}(\kappa)}}\frac{|{\mathcal{E}}(w)|}{3}\\\
&\leq\frac{|w-\kappa|^{2}}{\kappa^{2}}\frac{200}{3}+\frac{16|w-\kappa|^{2}}{\kappa^{2}}+\frac{5|w-\kappa|}{\kappa^{2}}\sqrt{\frac{2x}{-tm_{0}^{(2)}(\kappa)}}\\\
&\leq\frac{2x}{-t\kappa^{2}m_{0}^{(2)}(\kappa)}\left(\frac{248}{3|1+{\mathcal{E}}(w)|}+\frac{5}{|1+{\mathcal{E}}(w)|^{1/2}}\right)\leq\frac{2^{9}x}{-t\kappa^{2}m_{0}^{(2)}(\kappa)},\end{split}$

where in the second line we used the triangle inquality; in the third line we
used (2.60), the definition of ${\mathcal{E}}(w)$ from (2.65), (2.69) and
(2.70); in the fourth line we used (2.66) for the first and third terms and
(2.63) for the second term; in the last line we used (2.69) and
$1/|1+{\mathcal{E}}(w)|\leq 8/3$ from (2.66).

By plugging (LABEL:e:error) into (2.69) and using (4a) in 2.3, we conclude
that for any $0\leq x\leq-2^{-9}tm_{0}^{(2)}(\kappa)\kappa^{2}$

$\displaystyle\phantom{{}={}}\varrho_{t}(E_{t}-x)=\frac{{\rm
Im}[w]}{t\pi}=\frac{1}{\pi}\sqrt{\frac{2x}{-t^{3}m_{0}^{(2)}(\kappa)}}{\rm
Re}\left[\left(1+{\mathcal{E}}(w)\right)^{-1/2}\right]$
$\displaystyle=\frac{1}{\pi}\sqrt{\frac{2x}{-t^{3}m_{0}^{(2)}(\kappa)}}{\rm
Re}\left[1-\frac{m_{0}^{(3)}(\kappa)}{6m_{0}^{(2)}(\kappa)}\mathrm{i}\sqrt{\frac{2x}{-tm_{0}^{(2)}(\kappa)}}+\widehat{\mathcal{E}}(x)\right]=\frac{1}{\pi}\sqrt{\frac{2x}{-t^{3}m_{0}^{(2)}(\kappa)}}(1+{\rm
Re}[\widehat{\mathcal{E}}(x)])$

where for the last equality, we used that the middle term inside ${\rm
Re}[\cdot]$ is purely imaginary, and introduced the notation

(2.72)
$\displaystyle\widehat{\mathcal{E}}(x)=\left(1+{\mathcal{E}}(w(x))\right)^{-1/2}-1+\frac{m_{0}^{(3)}(\kappa)}{6m_{0}^{(2)}(\kappa)}\mathrm{i}\sqrt{\frac{2x}{-tm_{0}^{(2)}(\kappa)}},\quad|\operatorname{Re}\widehat{\mathcal{E}}(x)|\leq|\widehat{\mathcal{E}}(x)|\leq\frac{2^{9}|x|}{-t\kappa^{2}m_{0}^{(2)}(\kappa)},$

where the bound of $\widehat{\mathcal{E}}(x)$ is from (LABEL:e:error). This
gives the statement (2.53). Using (2.59) and the lower bound
$\kappa\geq({\mathsf{b}}t/20)^{2}$ from (2.57), get
${\mathcal{A}}(t)=-2/(t^{3}m_{0}^{(2)}(\kappa))\geq\kappa/t^{2}\geq{\mathsf{b}}^{2}/400\geq
2^{-10}{\mathsf{b}}^{2}$. Using the lower bound (2.58) and (2.57), we have
${\mathfrak{c}}(t)=-2^{-9}tm_{0}^{(2)}(\kappa)\kappa^{2}\geq
2^{-9}t{\mathsf{b}}\kappa^{1/2}/20\geq 2^{-9}t^{2}{\mathsf{b}}^{2}/400\geq
2^{-20}({\mathsf{b}}t)^{2}$. These two estimates give the last two statements
in (2.52). ∎

###### Proof of 2.15.

We denote the inverted cumulative density (recall from (2.13)) associated with
$\\{\mu_{t}\\}_{0\leq t\leq{\mathsf{T}}}$, by
$\gamma_{t}:[0,1]\mapsto{\mathbb{R}}$ for $0\leq t\leq{\mathsf{T}}$. Let
$F(x)=\mu_{0}([-x,0])\geq{\mathsf{b}}x^{3/2}$ for $\eta_{*}\leq
x\leq{\mathsf{T}}^{2}$.

The first statement follows from a comparison argument using 2.11. Denote
$\breve{\mu}_{0}:=\delta_{0}\in\mathscr{P}$, and let
$\breve{\mu}_{t}=\breve{\mu}_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$ be the
free convolution of $\breve{\mu}_{0}$ with the rescaled semicircle
distribution $\mu_{\operatorname{sc}}^{(t)}$. Then recall from Example 2.5
(with $A=1$) that $\breve{\mu}_{t}$ is given explicitly by

$\displaystyle{\rm d}\breve{\mu}_{t}=\frac{\sqrt{4t-x^{2}}}{2\pi t}{\rm d}x.$

For the measure valued process $\\{\breve{\mu}_{t}\\}_{0\leq
t\leq{\mathsf{T}}}$, we denote the associated inverted cumulative density
function by $\breve{\gamma}_{t}:[0,1]\mapsto{\mathbb{R}}$. From construction,
for $y\in[0,1]$ we have $\breve{\gamma}_{0}(y)=0\geq\gamma_{0}(y)$. Thus Item
1 in 2.11 implies that for any $t\geq 0$,

$\displaystyle E_{t}=\gamma_{t}(0)\leq\breve{\gamma}_{t}(0)=2\sqrt{t}.$

This gives the first statement in 2.15.

For the second claim, we analyze the two scenarios; the first is if
$t\leq\sqrt{{\mathfrak{C}}x}$, and the second is if
$t\geq\sqrt{{\mathfrak{C}}x}$. If $t\leq\sqrt{{\mathfrak{C}}x}$, then we
either have ${\mathfrak{C}}\sqrt{\eta}_{*}\leq t\leq\sqrt{{\mathfrak{C}}x}$ or
$x\geq{\mathfrak{C}}\eta_{*}$. In both cases $x\geq{\mathfrak{C}}\eta_{*}$. We
will prove the claim $\mu_{t}(E_{t}-x,E_{t})\geq{\mathfrak{c}}x^{3/2}$ by
using the comparison results from 2.11.

From the discussion above (and our assumption in 2.15), in the case that
$t\leq\sqrt{{\mathfrak{C}}x}$, we have ${\mathsf{T}}^{2}\geq
x\geq{\mathfrak{C}}\eta_{*}$. By 1.2 we have
$1\geq\mu_{0}([-x/{\mathfrak{C}},0])\geq{\mathsf{b}}(x/{\mathfrak{C}})^{3/2}$
for any $x/{\mathfrak{C}}\leq{\mathsf{T}}^{2}$. We denote
$A:={\mathsf{b}}(x/{\mathfrak{C}})^{3/2}\leq 1$. Then
$\gamma_{0}(A)\geq-x/{\mathfrak{C}}$. Let $\widetilde{\gamma}_{0}$ be the
restriction of $\gamma_{0}$ to $[0,A]$; then $\widetilde{\gamma}_{0}$ is the
inverted height function of a finite measure
$\widetilde{\mu}_{0}\in\mathscr{P}_{\rm fin}$ (by the definition (2.13),
$\widetilde{\mu}_{0}$ is the pushforward of the Lebesgue measure on $[0,A]$ by
$\widetilde{\gamma}_{0}$), and $\widetilde{\mu}_{0}({\mathbb{R}})=A$. We also
denote by
$\widetilde{\mu}_{t}=\widetilde{\mu}_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$
the free convolution of $\widetilde{\mu}_{0}$ with the rescaled semicircle
distribution $\mu_{\operatorname{sc}}^{(t)}$. Then
$\\{\widetilde{\mu}_{t}\\}_{t\geq 0}$ is a measure-valued process with
constant total mass $A$. We denote the associated inverted cumulative density
function (recall from (2.13)) of $\\{\widetilde{\mu}_{t}\\}_{0\leq
t\leq{\mathsf{T}}}$ by $\widetilde{\gamma}_{t}:[0,A]\mapsto{\mathbb{R}}$. Then
from construction we have $\widetilde{\gamma}_{0}(y)=\gamma_{0}(y)$ for
$y\in[0,A]$. Thus for $y,y^{\prime}\in[0,A]$ we have
$|\gamma_{0}(y)-\gamma_{0}(y^{\prime})|\leq|\widetilde{\gamma}_{0}(y)-\widetilde{\gamma}_{0}(y^{\prime})|$,
and Item 2 in 2.11 implies that, for any $t\geq 0$,

(2.73)
$\displaystyle\gamma_{t}(0)-\gamma_{t}(A)\leq\widetilde{\gamma}_{t}(0)-\widetilde{\gamma}_{t}(A).$

Next we use Item 1 in 2.11, by comparing to delta masses at $0$ and
$-x/{\mathfrak{C}}$, respectively, to show that $\widetilde{\gamma}_{t}(0)\leq
2\sqrt{tA}$ and $\widetilde{\gamma}_{t}(A)\geq-x/{\mathfrak{C}}-2\sqrt{tA}$.
The proofs of the two statements are very similar, so we will only prove the
first one. Denote $\widehat{\mu}_{0}:=A\delta_{0}$, and let
$\widehat{\mu}_{t}=\widehat{\mu}_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$ be
the free convolution of $\widehat{\mu}_{0}$ with the rescaled semicircle
distribution $\mu_{\operatorname{sc}}^{(t)}$. Then recall from Example 2.5
that $\widehat{\mu}_{t}$ is given explicitly by

$\displaystyle{\rm d}\widehat{\mu}_{t}=\frac{\sqrt{4At-x^{2}}}{2\pi t}{\rm
d}x.$

For the measure valued process $\\{\widehat{\mu}_{t}\\}_{0\leq
t\leq{\mathsf{T}}}$, we denote the associated inverted cumulative density
function by $\widehat{\gamma}_{t}:[0,A]\mapsto{\mathbb{R}}$. From
construction, for $y\in[0,A]$ we have $\widetilde{\gamma}_{0}(y)\leq
0=\widehat{\gamma}_{0}(y)$. Thus Item 1 in 2.11 implies that for any $t\geq
0$,

$\displaystyle\widetilde{\gamma}_{t}(0)\leq\widehat{\gamma}_{t}(0)=2\sqrt{At}.$

The same argument also gives that
$\widetilde{\gamma}_{t}(A)\geq-x/{\mathfrak{C}}-2\sqrt{At}$. Together with
(2.73), we conclude that for $t\leq\sqrt{\mathfrak{C}x}$ we have

$\displaystyle\gamma_{t}(0)-\gamma_{t}(A)\leq\widetilde{\gamma}_{t}(0)-\widetilde{\gamma}_{t}(A)\leq
x/{\mathfrak{C}}+4\sqrt{At}\leq
x\left(\frac{1}{{\mathfrak{C}}}+4\sqrt{\frac{{\mathsf{b}}}{{\mathfrak{C}}}}\right)\leq
x,$

where in the third inequality we used $A={\mathsf{b}}(x/{\mathfrak{C}})^{3/2}$
and $t\leq\sqrt{{\mathfrak{C}}x}$; in the last inequality we used
${\mathfrak{C}}=2^{21}/{\mathsf{b}}^{2}\geq\max\\{2,2^{6}{\mathsf{b}}\\}$.
Hence, for $t\leq\sqrt{{\mathfrak{C}}x}$, we have

(2.74) $\displaystyle\mu_{t}([E_{t}-x,E_{t}])\geq
A={\mathsf{b}}(x/{\mathfrak{C}})^{3/2}.$

For $t\geq\sqrt{{\mathfrak{C}}x}$, either $t\geq{\mathfrak{C}}\sqrt{\eta_{*}}$
or $x\geq{\mathfrak{C}}\eta_{*}$. In both cases we have
$t\geq{\mathfrak{C}}\sqrt{\eta_{*}}\geq(30/{\mathsf{b}})\sqrt{\eta_{*}}$.
Recall $\xi(t)$, ${\mathcal{A}}(t)$, $\mathcal{E}(t)$, and ${\mathfrak{c}}(t)$
from 2.14; for $(30/{\mathsf{b}})\sqrt{\eta_{*}}\leq t\leq{\mathsf{T}}$, we
have that

(2.75)
$\displaystyle\varrho_{t}(y)=(1+{\mathcal{E}}(E_{t}-y))\frac{\sqrt{{\mathcal{A}}(t)(E_{t}-y)}}{\pi}\geq\frac{1}{2}\cdot\frac{{\mathsf{b}}}{2^{5}}\frac{\sqrt{E_{t}-y}}{\pi},$

provided that $|y|\leq 2^{-21}{\mathsf{b}}^{2}t^{2}\leq{\mathfrak{c}}(t)/2$.
Moreover, for $t\geq\sqrt{{\mathfrak{C}}x}$, we have $x\leq
t^{2}/{\mathfrak{C}}\leq 2^{-21}{\mathsf{b}}^{2}t^{2}\leq{\mathfrak{c}}(t)/2$,
where we used that ${\mathfrak{C}}\geq 2^{21}/{\mathsf{b}}^{2}$. Then
integrating (2.75) gives

(2.76) $\displaystyle\mu_{t}([E_{t}-x,E_{t}])$
$\displaystyle=\int_{0}^{x}\varrho_{t}(E_{t}-y){\rm
d}y\geq\int_{0}^{x}\frac{{\mathsf{b}}}{2^{6}\pi}\sqrt{y}{\rm
d}y=\frac{{\mathsf{b}}}{96\pi}x^{3/2}.$

The estimates (2.74) and (2.76) together give the first statement in 2.15.

Next we prove (2.54). Recall our assumption that we have either
$t\geq{\mathfrak{C}}\sqrt{\eta}_{*}$, or $\kappa\geq{\mathfrak{C}}\eta_{*}$,
implying that

$\displaystyle{\rm Im}[m_{t}(E_{t}+\kappa+\mathrm{i}\eta)]$
$\displaystyle=\int_{{\mathbb{R}}}\frac{\eta\varrho_{t}(x){\rm
d}x}{(E_{t}+\kappa-x)^{2}+\eta^{2}}\geq\int_{E_{t}-(\kappa+\eta)}^{E_{t}}\frac{\eta\varrho_{t}(x){\rm
d}x}{(E_{t}+\kappa-x)^{2}+\eta^{2}}$
$\displaystyle\geq\int_{E_{t}-(\kappa+\eta)}^{E_{t}}\frac{\eta\varrho_{t}(x){\rm
d}x}{(2\kappa+\eta)^{2}+\eta^{2}}\geq\frac{{\mathfrak{c}}\eta(\kappa+\eta)^{3/2}}{2(2\kappa+2\eta)^{2}}=\frac{{\mathfrak{c}}\eta}{8\sqrt{\kappa+\eta}},$

where in the second statement we restricted the integration on
$[E_{t}-\kappa-\eta,E_{t}]$; in the third statement we used that for
$x\in[E_{t}-\kappa-\eta,E_{t}]$, we have $E_{t}+\kappa-x\leq 2\kappa+\eta$; in
the fourth statement inequality we used $\kappa+\eta\leq{\mathsf{T}}^{2}$ from
our assumption, and
$\mu_{t}([E_{t}-(\kappa+\eta),E_{t}])\geq{\mathfrak{c}}(\kappa+\eta)^{3/2}$
from the second statement in 2.15. ∎

##  3\. Edge Rigidity and Universality

In this section we study Dyson Brownian motion, with initial data
$\mu_{0}=\mu_{0}^{(n)}=(1/n)\sum_{i=1}^{n}\delta_{\lambda_{i}(0)}$ satisfying
1.2, and prove 1.3 and 1.4. Recall $\mu_{t}=\mu_{0}\boxplus\mu_{\rm
sc}^{(t)}$, $E_{t}=\max(\operatorname{supp}\mu_{t})$, and the constants
${\mathfrak{c}}$ and ${\mathfrak{C}}$ from 2.15. We will first study the case
that

(3.1) $\displaystyle{\mathfrak{C}}^{2}\eta_{*}\geq(\log n)^{15}/n^{2/3}.$

Under the assumption (3.1), define the function

(3.2) $\displaystyle
f(t)=\max\\{{\mathfrak{C}}\sqrt{\eta_{*}}-{\mathfrak{c}}\max\\{0,t-{\mathfrak{C}}\sqrt{\eta_{*}}\\}/8,(\log
n)^{15/2}n^{-1/3}\\}^{2}.$

Then for $t\leq{\mathfrak{C}}\sqrt{\eta_{*}}$, we have by (3.1) that
$f(t)={\mathfrak{C}}^{2}\eta_{*}$; for $t\geq{\mathfrak{C}}\sqrt{\eta}_{*}$,
$f(t)$ decreases to $(\log n)^{15}/n^{2/3}$, and

(3.3)
$\displaystyle(f(s))^{1/2}-(f(t))^{1/2}\leq\frac{{\mathfrak{c}}(t-s)}{8},\quad{\mathfrak{C}}\sqrt{\eta_{*}}\leq
s\leq t.$

The following proposition states that $\lambda_{1}(t)\leq E_{t}+f(t)$ for
$0\leq t\leq{\mathsf{T}}$ with overwhelming probability. 1.3 is an easy
consequence of it. The proof of 3.1 will be given in Section 3.2.

######  Proposition 3.1.

Consider Dyson Brownian motion $\bm{\lambda}(t)$ as in (1.2). We assume its
initial data
$\mu_{0}=\mu_{0}^{(n)}=(1/n)\sum_{i=1}^{n}\delta_{\lambda_{i}(0)}$ satisfies
1.2. We further assume that (3.1) holds and
${\mathsf{T}}\geq{\mathfrak{B}}\sqrt{\eta_{*}}=2^{60}{\mathsf{b}}^{-6}\sqrt{\eta_{*}}$.
Recall $\mu_{t}=\mu_{0}\boxplus\mu_{\rm sc}^{(t)}$ and
$E_{t}=\max(\operatorname{supp}\mu_{t})$ from 2.15. There exists a large
constant $C=C({\mathsf{b}},{\mathsf{T}})>1$, such that, with probability
$1-Ce^{-(\log n)^{2}}$, we have for any $0\leq t\leq{\mathsf{T}}$ that

(3.4) $\displaystyle\lambda_{1}(t)\leq E_{t}+f(t),$

where $f(t)$ is as defined in (3.2).

###### Proof of 1.3.

Recall ${\mathfrak{c}}$ and ${\mathfrak{C}}$ from 2.15. We first discuss the
case when (3.1) holds. From the construction of $f(t)$ in (3.2) and discussion
after it, for $t\geq
2^{60}{\mathsf{b}}^{-6}\sqrt{\eta_{*}}\geq(1+8/{\mathfrak{c}}){\mathfrak{C}}\sqrt{\eta_{*}}$,
we have $f(t)=\max\\{0,(\log n)^{15/2}n^{-1/3}\\}^{2}=(\log n)^{15}n^{-2/3}$.
The claim (1.8) thus follows from (3.4). When (3.1) does not hold, then 1.2
still holds by replacing $\eta_{*}$ with $(\log
n)^{15}/({\mathfrak{C}}^{2}n^{2/3})$. If $\eta_{*}=(\log
n)^{15}/({\mathfrak{C}}^{2}n^{2/3})$, we can construct $f(t)$ as in (3.2),
then $f(t)=(\log n)^{15}n^{-2/3}$ for any $t\geq 0$. Thus 3.1 gives that with
probability $1-Ce^{-(\log n)^{2}}$, $\lambda_{1}(t)\leq E_{t}+(\log
n)^{15}/n^{2/3}$ for any $0\leq t\leq{\mathsf{T}}$, and the claim (1.8)
follows. ∎

###  3.1. Characteristic flow

In the remainder of Section 3, we will use the notations
${\mathsf{T}},{\mathsf{b}},\eta_{*}$ and the measure $\mu_{0}=\mu_{0}^{(n)}$
from 1.2. We recall the functions ${\mathcal{A}}(t),{\mathfrak{c}}(t)$, and
constants ${\mathfrak{c}}$ and ${\mathfrak{C}}$ from 2.15. We also recall the
functions $f(t)$ from (3.2). We denote the free convolution of $\mu_{0}$ with
rescaled semicircle distribution by as $\mu_{t}$ (recall from Section 2.1). We
let $E_{t}=\max(\operatorname{supp}\mu_{t})$ and denote the Stieltjes
transform of $\mu_{t}$ by $m_{t}(z)$. We recall the set $\Lambda_{t}$ from
(2.7).

We can reformulate the defining relation (2.8) of the free convolution by
introducing the characteristic flow:

(3.5) $\displaystyle m_{t}(z_{t}(u))=m_{0}(u),\quad
z_{t}(u)=u-tm_{0}(u)=:E_{t}+\kappa_{t}(u)+\mathrm{i}\eta_{t}(u),\quad
u\in\Lambda_{t}.$

When the context is clear, we will simply write
$z_{t}(u),\kappa_{t}(u),\eta_{t}(u)$ as $z_{t},\kappa_{t},\eta_{t}$. It
follows from (3.5) that $z_{t}=u-tm_{t}(z_{t})\in\mathbb{H}$, since
$z_{t}:\Lambda_{t}\rightarrow\mathbb{H}$ is a bijection, by 2.2. The
characteristic flow satisfies the fundamental property that

(3.6) $\displaystyle m_{t}(z_{t})\qquad\text{is constant in $t\geq 0$},$

which we will often use without comment.

In the following we collect some basic facts about the Stieltjes transform and
the characteristic flows, which will be used repeatedly in the rest of this
section. Take any point $u\in\mathbb{H}$, and time $t\geq 0$, such that at
time $t$ we have
$z_{t}=z_{t}(u)=E_{t}+\kappa_{t}+\mathrm{i}\eta_{t}\in\Lambda_{t}$. Then for
any time $0\leq s\leq t$, using $z_{s}=u-sm_{0}(u)=u-sm_{t}(z_{t})$ (from
(3.5)) we have

(3.7) $\displaystyle
z_{s}=z_{t}+(t-s)m_{t}(z_{t}),\quad\eta_{s}=\eta_{t}+(t-s){\rm
Im}[m_{t}(z_{t})]\geq\eta_{t},$

where the last inequality follows from ${\rm Im}[m_{t}(z_{t})]={\rm
Im}[m_{0}(u)]\geq 0$.

We collect some further properties of the characteristic flow in the following
two lemmas. 3.2 states that if $\kappa_{t}\geq 0$, then $\kappa_{s}\geq 0$ for
all $0\leq s\leq t$; it also gives lower bounds on $\kappa_{s}-\kappa_{t}$.
3.3 states that if $\kappa_{t}\geq f(t)$ (recall from (3.2)), then
$\kappa_{s}\geq f(s)$ for all $0\leq s\leq t$; it also gives lower bounds on
$\kappa_{s}-f(s)$.

######  Lemma 3.2.

Adopt 1.2, and recall the constants ${\mathfrak{c}}$ and ${\mathfrak{C}}$ from
2.15. Take $u\in\Lambda_{t}$, and fix real numbers $t>0$ and $0\leq s\leq t$.
Denote $z_{s}=z_{s}(u)=E_{s}+\kappa_{s}+\mathrm{i}\eta_{s}$, and assume that
$\kappa_{s}\leq{\mathsf{T}}^{2}$ (recall ${\mathsf{T}}$ from 1.2). If
$\kappa_{t}\geq 0$ then, for $0\leq s\leq t$, we have

(3.8)
$\displaystyle\kappa_{s}-\kappa_{t}\geq\left(\frac{\eta_{s}}{\eta_{t}}-1\right)\kappa_{t}=(t-s){\rm
Im}[m_{t}(z_{t})]\frac{\kappa_{t}}{\eta_{t}}\geq 0.$

Moreover, if $\kappa_{t}\geq 0$ then, for ${\mathfrak{C}}\sqrt{\eta_{*}}\leq
s\leq t$, we have

(3.9)
$\displaystyle\kappa_{s}^{1/2}-\kappa_{t}^{1/2}\geq\frac{{\mathfrak{c}}(t-s)}{4}.$

######  Lemma 3.3.

Adopt 1.2, and recall the constants ${\mathfrak{c}}$ and ${\mathfrak{C}}$ from
2.15. Take $u\in\Lambda_{t}$, and fix real numbers $t>0$ and $0\leq s\leq t$.
Denote $z_{s}=z_{s}(u)=E_{s}+\kappa_{s}+\mathrm{i}\eta_{s}$, and assume that
$\kappa_{s}\leq{\mathsf{T}}^{2}$ (recall ${\mathsf{T}}$ from 1.2). If
$\kappa_{t}\geq f(t)$ (recall from (3.2)) then, for
${\mathfrak{C}}\sqrt{\eta_{*}}\leq s\leq t$, we have

(3.10) $\displaystyle\kappa_{s}-f(s)\geq\frac{(t-s)}{2}{\rm
Im}[m_{t}(z_{t})]\frac{\kappa_{t}}{\eta_{t}};$

Moreover, if $\kappa_{t}\geq f(t)$ then, for $0\leq
s\leq\widetilde{t}:=\min\\{t,{\mathfrak{C}}\sqrt{\eta_{*}}\\}$, we have

(3.11) $\displaystyle\kappa_{s}-f(s)\geq(\widetilde{t}-s){\rm
Im}[m_{t}(z_{t})]\frac{\kappa_{\widetilde{t}}}{\eta_{\widetilde{t}}}$

###### Proof of 3.2.

We will first show that, if $\kappa_{t}\geq 0$, then $\kappa_{s}\geq 0$ for
$0\leq s\leq t$, and we will also estimate its derivative. Notice

(3.12) $\displaystyle\operatorname{\partial}_{s}{\rm
Re}[z_{s}]=\operatorname{\partial}_{s}{\rm Re}[u-sm_{0}(u)]=-{\rm
Re}[m_{0}(u)]=-{\rm Re}[m_{s}(z_{s})],$

where the first statement is from the definition (3.5) that
$z_{s}=u-sm_{0}(u)$; the second statement is from taking derivative with
respect to $s$; the last statement follows from (3.6). Further recall the
construction from Item 5 in 2.3 that there exists $w_{s}\in{\mathbb{R}}$ such
that $m_{0}^{\prime}(w_{s})=1/s$ and $E_{s}=w_{s}-sm_{0}(w_{s})=z_{s}(w_{s})$.
By taking the time derivative on both sides, we get a similar equation as
(3.12), namely,

(3.13)
$\displaystyle\begin{split}\operatorname{\partial}_{s}E_{s}&=\operatorname{\partial}_{s}w_{s}-sm_{0}^{\prime}(w_{s})\operatorname{\partial}_{s}w_{s}-m_{0}(w_{s})\\\
&=\operatorname{\partial}_{s}w_{s}(1-sm_{0}^{\prime}(w_{s}))-m_{0}(w_{s})=-m_{0}(w_{s})=-m_{s}(E_{s}),\end{split}$

where the first statement follows from taking the derivative with respect to
$s$ on both sides of $E_{s}=w_{s}-sm_{0}(w_{s})$; the second and third
statements use $m_{0}^{\prime}(w_{s})=1/s$; and the last statement uses (3.6)
and $E_{s}=z_{s}(w_{s})$. Then by taking difference between (3.12) and (3.13),
we get

(3.14)
$\displaystyle\begin{split}\operatorname{\partial}_{s}\kappa_{s}&=-({\rm
Re}[m_{s}(z_{s})]-m_{s}(E_{s}))=-(({\rm Re}[m_{s}(z_{s})]-m_{s}({\rm
Re}[z_{s}])+(m_{s}({\rm Re}[z_{s}])-m_{s}(E_{s})))\\\
&=-\eta_{s}^{2}\int_{\mathbb{R}}\frac{\varrho_{s}(x){\rm
d}x}{((E_{s}+\kappa_{s}-x)^{2}+\eta_{s}^{2})(E_{s}+\kappa_{s}-x)}-\kappa_{s}\int_{\mathbb{R}}\frac{\varrho_{s}(x){\rm
d}x}{(E_{s}-x)(E_{s}+\kappa_{s}-x)},\end{split}$

where the first statement follows from $\kappa_{s}={\rm Re}[z_{s}]-E_{s}$; the
second statement is from separating the difference into two; in the third
statement we used ${\rm Re}[z_{s}]=E_{s}+\kappa_{s}$ and ${\rm
Re}[m_{s}(z_{s})]=\int_{\mathbb{R}}(x-E_{s}-\kappa_{s})\varrho_{s}(x){\rm
d}x/((E_{s}+\kappa_{s}-x)^{2}+\eta_{s}^{2})$. By our assumption, we have
$\kappa_{t}\geq 0$. Next we will show that $\kappa_{s}\geq 0$ for all $0\leq
s\leq t$. Otherwise, there exists $\tau=\sup\\{0\leq s\leq t:\kappa_{s}<0\\}$.
Since $\kappa_{s}$ is continuous, we have that $\tau\in[0,t)$, $\kappa_{s}\geq
0$ for $s\in[\tau,t]$ and $\kappa_{\tau}=0$. Thus for $s\in[\tau,t]$, (3.14)
gives that

(3.15) $\displaystyle\operatorname{\partial}_{s}\kappa_{s}\leq 0,$

since $E_{s}+\kappa_{s}\geq E_{s}=\max\operatorname{supp}\varrho_{s}$, and
$\kappa_{s}\geq 0$. By integrating (3.15) from $\tau$ to $t$, we conclude that
$\kappa_{\tau}\geq\kappa_{t}\geq 0$, which contradicts with our choice of
$\tau$. Thus we conclude

(3.16) $\displaystyle\kappa_{s}\geq 0\quad\text{and $\kappa_{s}$ is decreasing
for}\quad 0\leq s\leq t,$

where the second statement follows from
$\operatorname{\partial}_{s}\kappa_{s}<0$ (from (3.14)).

We can also lower bound the second integral in (3.14) by

(3.17) $\displaystyle\int_{\mathbb{R}}\frac{\varrho_{s}(x){\rm
d}x}{(E_{s}-x)(E_{s}+\kappa_{s}-x)}\geq\int_{\mathbb{R}}\frac{\varrho_{s}(x){\rm
d}x}{|z_{s}-x|^{2}}=\frac{1}{\eta_{s}}{\rm
Im}\int_{\mathbb{R}}\frac{\varrho_{s}(x){\rm d}x}{x-z_{s}}=\frac{{\rm
Im}[m_{s}(z_{s})]}{\eta_{s}},$

where in the first statement we used
$z_{s}=E_{s}+\kappa_{s}+\mathrm{i}\eta_{s}$ with $\kappa_{s}\geq 0$ from
(3.16) and $\max\operatorname{supp}\varrho_{s}=E_{s}$, thus $|z_{s}-x|\geq
E_{s}+\kappa_{s}-x\geq E_{s}-x$; the last two statements are from the
definition of Stieltjes transform (2.1). Combining (3.14) and (3.17), we get

(3.18)
$\displaystyle\operatorname{\partial}_{s}\kappa_{s}\leq-\frac{\kappa_{s}}{\eta_{s}}{\rm
Im}[m_{s}(z_{s})].$

Recall from the definition of the characteristic flow (3.5) and (3.7) that the
derivative for the imaginary part $\operatorname{\partial}_{s}\eta_{s}$
satisfies $\operatorname{\partial}_{s}\eta_{s}=-{\rm Im}[m_{s}(z_{s})].$ By
plugging this relation to (3.18), we conclude

$\displaystyle\operatorname{\partial}_{s}\kappa_{s}\leq\frac{\kappa_{s}}{\eta_{s}}\operatorname{\partial}_{s}\eta_{s},\qquad\text{so
that}\qquad\frac{\operatorname{\partial}_{s}\eta_{s}}{\eta_{s}}\geq\frac{\operatorname{\partial}_{s}\kappa_{s}}{\kappa_{s}},$

and hence for $0\leq s\leq t$ we obtain

(3.19) $\displaystyle\kappa_{s}\geq\frac{\eta_{s}}{\eta_{t}}\kappa_{t}.$

Recall from (3.7) that $(\eta_{s}-\eta_{t})/\eta_{t}=(t-s){\rm
Im}[m_{t}(z_{t})]/\eta_{t}$. Hence, the claim (3.8) follows from (3.19):

(3.20)
$\displaystyle\kappa_{s}-\kappa_{t}\geq\left(\frac{\eta_{s}}{\eta_{t}}-1\right)\kappa_{t}=(t-s){\rm
Im}[m_{t}(z_{t})]\frac{\kappa_{t}}{\eta_{t}}.$

If we moreover have that $s\geq{\mathfrak{C}}\sqrt{\eta_{*}}$, then we can
also bound the integral in (3.14) by

(3.21)
$\displaystyle\begin{split}-\kappa_{s}\int_{\mathbb{R}}\frac{\varrho_{s}(x){\rm
d}x}{(E_{s}-x)(E_{s}+\kappa_{s}-x)}\leq-\kappa_{s}\int_{E_{s}-\kappa_{s}}^{E_{s}}\frac{\varrho_{s}(x){\rm
d}x}{2\kappa_{s}^{2}}=-\frac{\mu([E_{s}-\kappa_{s},E_{s}])}{2\kappa_{s}}\leq-\frac{1}{2}{\mathfrak{c}}\kappa_{s}^{1/2},\end{split}$

where we in the first inequality we restricted the integral on
$[E_{s}-\kappa_{s},E_{s}]$ and used $\kappa_{s}\geq E_{s}-x$ and
$2\kappa_{s}\geq E_{s}+\kappa_{s}-x$ on this interval; in the last two
inequalities, we used $\kappa_{s}\leq{\mathsf{T}}^{2}$ and
$\mu_{s}([E_{s}-\kappa_{s},E_{s}])\geq{\mathfrak{c}}\kappa_{s}^{3/2}$ from
2.15. Plugging (3.21) into (3.14) we get

(3.22)
$\displaystyle\operatorname{\partial}_{s}\kappa_{s}\leq-\frac{{\mathfrak{c}}}{2}\kappa_{s}^{1/2},\quad{\mathfrak{C}}\sqrt{\eta_{*}}\leq
s\leq t.$

Solving it we obtain

(3.23)
$\displaystyle\kappa_{s}^{1/2}-\kappa_{t}^{1/2}\geq\frac{{\mathfrak{c}}(t-s)}{4},\quad{\mathfrak{C}}\sqrt{\eta_{*}}\leq
s\leq t.$

This finishes the proof of (3.9) ∎

###### Proof of 3.3.

For $t\geq s\geq{\mathfrak{C}}\sqrt{\eta_{*}}$, the estimates (3.8) and (3.9)
imply

(3.24) $\displaystyle\kappa_{s}-\kappa_{t}\geq(t-s){\rm
Im}[m_{t}(z_{t})]\frac{\kappa_{t}}{\eta_{t}},\quad\kappa_{s}-\kappa_{t}\geq\frac{{\mathfrak{c}}(t-s)}{4}(\sqrt{\kappa_{s}}+\sqrt{\kappa_{t}}),$

From our construction of $f(t)$, using (3.3) and (3.9), and recalling that
$\kappa_{t}\geq f(t)$, we have

(3.25)
$\displaystyle\sqrt{f(s)}\leq\frac{{\mathfrak{c}}(t-s)}{8}+\sqrt{f(t)}\leq\frac{{\mathfrak{c}}(t-s)}{8}+\sqrt{\kappa_{t}}\leq\sqrt{\kappa_{s}},\quad{\mathfrak{C}}\sqrt{\eta_{*}}\leq
s\leq t,$

Rearranging (3.3), together with (3.25), gives $\kappa_{s}\geq f(s)\geq f(t)$.
Thus,

(3.26) $\displaystyle
f(s)-f(t)\leq\frac{{\mathfrak{c}}}{8}(t-s)(\sqrt{f(s)}+\sqrt{f(t)})\leq\frac{{\mathfrak{c}}(t-s)}{8}(\sqrt{\kappa_{s}}+\sqrt{\kappa_{t}}),\quad{\mathfrak{C}}\sqrt{\eta_{*}}\leq
s\leq t,$

Put everything together, we get

$\displaystyle\phantom{{}={}}\kappa_{s}-f(s)$
$\displaystyle\geq\frac{1}{2}(\kappa_{s}-\kappa_{t})+\frac{1}{2}(\kappa_{s}-\kappa_{t})-(f(s)-f(t))$
$\displaystyle\geq\frac{(t-s)}{2}{\rm
Im}[m_{t}(z_{t})]\frac{\kappa_{t}}{\eta_{t}}+\frac{{\mathfrak{c}}(t-s)}{8}(\sqrt{\kappa_{s}}+\sqrt{\kappa_{t}})-\frac{{\mathfrak{c}}(t-s)}{8}(\sqrt{\kappa_{s}}+\sqrt{\kappa_{t}})$
$\displaystyle=\frac{(t-s)}{2}{\rm
Im}[m_{t}(z_{t})]\frac{\kappa_{t}}{\eta_{t}},$

where in the first inequality we used our assumption $\kappa_{t}\geq f(t)$ in
3.3; in the second inequality we used (3.24) to bound the first two terms and
(3.26) to bound the last term. This gives (3.10).

Next we prove (3.11), by considering two cases. If $t=\widetilde{t}$, then
$f(\widetilde{t})=f(t)\leq\kappa_{t}=\kappa_{\widetilde{t}}$. If
$t>\widetilde{t}$, then $\widetilde{t}={\mathfrak{C}}\sqrt{\eta_{*}}$ and
(3.10) (applied with the $s$ there equal to $\widetilde{t}$ here) implies
$f(\widetilde{t})\leq\kappa_{\widetilde{t}}$. In both cases we have
$f(\widetilde{t})\leq\kappa_{\widetilde{t}}$. For $s\leq\widetilde{t}$ we have
$f(s)=f(\widetilde{t})\leq\kappa_{\widetilde{t}}$, and (3.8) (applied with the
$t$ there equal to $\widetilde{t}$ here, observing that
$\kappa_{\tilde{t}}\geq\kappa_{t}\geq f(t)\geq 0$ holds) implies

(3.27)
$\displaystyle\kappa_{s}-f(s)\geq\kappa_{s}-\kappa_{\widetilde{t}}\geq({\widetilde{t}}-s){\rm
Im}[m_{s}(z_{s})]\frac{\kappa_{\widetilde{t}}}{\eta_{\widetilde{s}}}=({\widetilde{t}}-s){\rm
Im}[m_{t}(z_{t})]\frac{\kappa_{\widetilde{t}}}{\eta_{\widetilde{t}}}.$

This finishes the proof of (3.11). ∎

Figure 3. The spectral domain ${\mathcal{D}}_{t}$ constructed in (3.28).

Fix the real number
${\mathfrak{M}}=6({\mathsf{T}}+{\mathfrak{C}}^{2}\eta_{*})$, and recall $f(t)$
from (3.2). For any real number $t\in[0,\mathsf{T}]$, we define the spectral
domain ${\mathcal{D}}_{t}={\mathcal{D}}_{t}(\mu_{t})$ (see Figure 3) by

(3.28)
$\displaystyle{\mathcal{D}}_{t}\operatorname{:=}\left\\{z\in\mathbb{H}:E_{t}+f(t)\leq{\rm
Re}[z]\leq{\mathfrak{M}}-2t,\frac{1}{(\log n)n{\rm Im}[m_{t}(z)]}\leq{\rm
Im}[z]\leq{\mathfrak{M}}-2t\right\\}$

Roughly speaking, the information of the Stieltjes transform $m_{t}(z)$ on
${\mathcal{D}}_{t}$ can be used to detect particles of distance $f(t)$ from
the spectral edge $E_{t}$. The following lemma collects some properties of the
domain ${\mathcal{D}}_{t}$.

######  Lemma 3.4.

Adopt 1.2. Recall the domain ${\mathcal{D}}_{t}$ from (3.28). For $n$ large
enough and any time $0\leq t\leq{\mathsf{T}}$, the following hold.

  1. (1)

${\mathcal{D}}_{t}$ is non-empty. For any
$z=E_{t}+\kappa+\mathrm{i}\eta\in{\mathcal{D}}_{t}$, we have

(3.29) $\displaystyle{\mathfrak{M}}\geq\kappa\geq
f(t)\geq\frac{1}{n^{2/3}},\quad{\mathfrak{M}}\geq\eta\geq\frac{1}{n^{2}},\quad\kappa+\eta\leq{\mathsf{T}}^{2},\quad{\rm
Im}[m_{t}(z)]\geq\frac{{\mathfrak{c}}\eta}{8\sqrt{\kappa+\eta}},$

where ${\mathfrak{c}}$ is from 2.14, and

(3.30) $\displaystyle\frac{(\log
n)^{7/2}}{n\sqrt{(\kappa+\eta)\eta}}\leq\frac{{\rm Im}[m_{t}(z)]}{\log n}.$

  2. (2)

If $z_{t}(u)=E_{t}+\kappa_{t}+\mathrm{i}\eta_{t}\in{\mathcal{D}}_{t}$, then
$z_{s}(u)\in{\mathcal{D}}_{s}$ for any $0\leq s\leq t$.

By the second part of 3.4, if for any $u=z_{0}(u)\in{\mathcal{D}}_{0}$ we
define

(3.31) $\displaystyle{\mathfrak{t}}(u):={\mathsf{T}}\wedge\sup\\{t\geq
0:z_{t}(u)\in{\mathcal{D}}_{t}\\},$

then $z_{s}(u)\in{\mathcal{D}}_{s}$ for any $0\leq s\leq{\mathfrak{t}}(u)$.

###### Proof of 3.4.

For Item 1, ${\mathfrak{M}}=6({\mathsf{T}}+{\mathfrak{C}}^{2}\eta_{*})\geq
2{\mathsf{T}}+2\sqrt{{\mathsf{T}}}+{\mathfrak{C}}^{2}\eta_{*}\geq
2t+E_{t}+f(t)$, where we used that $t\leq{\mathsf{T}}$, $E_{t}\leq
2\sqrt{t}\leq 2\sqrt{{\mathsf{T}}}$ from the first statement in 2.15, and
$f(t)\leq f(0)={\mathfrak{C}}^{2}\eta_{*}$ from (3.2). Moreover for any
$z\in\mathbb{H}$ with $E_{t}+f(t)\leq{\rm Re}[z]\leq{\mathfrak{M}}-2t$ and
${\rm Im}[z]\leq{\mathfrak{M}}-2t$ we have

(3.32) $\displaystyle{\rm Im}[m_{t}(z)]=\int_{\mathbb{R}}\frac{{\rm
d}\mu_{t}(x)}{|x-z|^{2}}\geq\int^{E_{t}}_{E_{t}-1}\frac{{\rm
d}\mu_{t}(x)}{(2{\mathfrak{M}})^{2}}\geq\frac{{\mathfrak{c}}}{4{\mathfrak{M}}^{2}},$

where the first inequality is from $|z-x|\leq 2{\mathfrak{M}}$ for
$x\in[E_{t},E_{t}-1]$; and the last inequality follows from the second
statement of 2.15 (with $x=1$). Thus for $n$ large enough,
${\mathfrak{M}}-2t>1>1/(n{\rm Im}[m_{t}(z)]\log n)$, and ${\mathcal{D}}_{t}$
is not empty.

If $E_{t}+\kappa+\mathrm{i}\eta\in{\mathcal{D}}_{t}$, then we have
$\eta,\kappa\leq{\mathfrak{M}}$ from the construction of ${\mathcal{D}}_{t}$
in (3.28); $\kappa\geq f(t)\geq 1/n^{2/3}$ from the definition (3.2); and
$\eta\geq 1/((\log n)n{\rm Im}[m_{t}(z)])\geq 1/((\log n)n(1/\kappa))\geq
1/n^{2}$ (where the first statement is from the definition of
${\mathcal{D}}_{t}$ (3.28); the second statement is from
$\max(\operatorname{supp}\mu_{t})=E_{t}$ and 2.1 that $|m_{t}(z)|\leq
1/\operatorname{dist}(z,\operatorname{supp}\mu_{t})\leq 1/\kappa$; the last
statement follows from $\kappa\geq n^{-2/3}$) provided $n$ is large enough.
This gives the first two statement in (3.29). For the third statement, we
recall from our assumption from (3.1) that
${\mathsf{T}}\geq(2^{60}/{\mathsf{b}}^{6})\sqrt{\eta}_{*}\geq
2^{10}{\mathfrak{C}}\sqrt{\eta_{*}}$ (we recall that
${\mathfrak{C}}=2^{21}/{\mathsf{b}}^{2}$ from (2.15) and
${\mathsf{b}},\eta_{*}<1$ from 1.2). Then $\kappa+\eta\leq
2{\mathfrak{M}}=12({\mathsf{T}}+{\mathfrak{C}}^{2}\eta_{*})\leq{\mathsf{T}}^{2}$
(where we also used ${\mathsf{T}}\geq 100$ from 1.2.) The last relation in
(3.29) follows from (2.54) and $\kappa+\eta\leq{\mathsf{T}}^{2}$.

For (3.30), by the definition of ${\mathcal{D}}_{t}$, we have

(3.33) $\displaystyle\eta{\rm
Im}[m_{t}(E_{t}+\kappa+\mathrm{i}\eta)]\geq\frac{1}{(\log n)n}.$

There are two cases; we either have $\eta\geq(\log n)^{4}n^{-2/3}$ or
$\eta\leq(\log n)^{4}n^{-2/3}$. If $\eta\geq(\log n)^{4}n^{-2/3}$, then

(3.34) $\displaystyle\sqrt{(\kappa+\eta)\eta}{\rm
Im}[m_{t}(E_{t}+\kappa+\mathrm{i}\eta)]\geq\sqrt{(\kappa+\eta)\eta}\frac{{\mathfrak{c}}\eta}{8\sqrt{\kappa+\eta}}=\frac{{\mathfrak{c}}}{8}\eta^{3/2}\geq\frac{(\log
n)^{9/2}}{n},$

provided $n$ is large enough. If $\eta\leq(\log n)^{4}n^{-2/3}$, then
$\kappa\geq f(t)\geq(\log n)^{15}n^{-2/3}\geq(\log n)^{11}\eta$, and

(3.35) $\displaystyle\sqrt{(\kappa+\eta)\eta}{\rm
Im}[m_{t}(E_{t}+\kappa+\mathrm{i}\eta)]\geq\sqrt{\frac{\kappa}{\eta}}\cdot\eta{\rm
Im}[m_{t}(E_{t}+\kappa+\mathrm{i}\eta)]\geq\sqrt{\frac{\kappa}{\eta}}\cdot\frac{1}{n\log
n}\geq\frac{(\log n)^{9/2}}{n}.$

This finishes the proof of (3.30).

For Item 2, for observe that 3.3 gives if $\kappa_{t}\geq f(t)$ then
$\kappa_{s}\geq f(s)$ for $s\leq t$. Since $\eta_{s}\geq\eta_{t}$ from (3.7),
if $\eta_{t}\geq 1/((\log n)n{\rm Im}[m_{t}(z_{t})])$, then
$\eta_{s}\geq\eta_{t}\geq 1/((\log n)n{\rm Im}[m_{t}(z_{t})])=1/((\log n)n{\rm
Im}[m_{s}(z_{s})])$. Thus, $\operatorname{Re}z_{s}(u)$ and
$\operatorname{Im}z_{s}(u)$ satisfy the lower bounds required for $z_{s}(u)$
to be in $\mathcal{D}_{s}$.

Next we prove by contradiction that ${\rm Re}[z_{s}(u)]\leq{\mathfrak{M}}-2s$
and ${\rm Im}[z_{s}(u)]\leq{\mathfrak{M}}-2s$. Otherwise, assume $\max\\{{\rm
Re}[z_{s}(u)],{\rm Im}[z_{s}(u)]\\}>{\mathfrak{M}}-2s$. Recall our choice
${\mathfrak{M}}=6({\mathsf{T}}+{\mathfrak{C}}^{2}\eta_{*})\geq 6{\mathsf{T}}$
and $E_{s}\leq 2\sqrt{s}\leq 2\sqrt{\mathsf{T}}$ from the first statement of
2.15. Then
${\mathfrak{M}}-2{\mathsf{T}}-E_{s}\geq{\mathfrak{M}}-3{\mathsf{T}}\geq{\mathfrak{M}}/2$
(we used that ${\mathsf{T}}\geq 100$ from 1.2 and
${\mathfrak{M}}-2{\mathsf{T}}\geq{\mathfrak{M}}/2$). Thus
$\operatorname{dist}(z_{s}(u),(-\infty,E_{s}])\geq{\mathfrak{M}}/2$; so,
$|m_{s}(z_{s})|\leq 1/(\operatorname{dist}(z_{s}(u),(-\infty,E_{t}])\leq
2/{\mathfrak{M}}$ (from 2.1), and $|z_{s}-z_{t}|=|t-s||m_{s}(z_{s})|\leq
2(t-s)/{\mathfrak{M}}$ (here we used the definition of characteristic flow
(3.7) and (3.6)). Then ${\rm Re}[z_{t}(u)]\geq{\rm
Re}[z_{s}(u)]-2(t-s)/{\mathfrak{M}}$ and ${\rm Im}[z_{t}(u)]\geq{\rm
Im}[z_{s}(u)]-2(t-s)/{\mathfrak{M}}$. Therefore, $\max\\{{\rm
Re}[z_{t}(u)],{\rm Im}[z_{t}(u)]\\}\geq\max\\{{\rm Re}[z_{s}(u)],{\rm
Im}[z_{s}(u)]\\}-2(t-s)/{\mathfrak{M}}>{\mathfrak{M}}-2s-2(t-s)/{\mathfrak{M}}>{\mathfrak{M}}-2t$
(we used our hypothesis $\max\\{{\rm Re}[z_{s}(u)],{\rm
Im}[z_{s}(u)]\\}>{\mathfrak{M}}-2s$ and ${\mathfrak{M}}\geq 1$), which
contradicts the fact that $z_{t}(u)\in{\mathcal{D}}_{t}$. It follows that
$\operatorname{Re}z_{s}(u)$ and $\operatorname{Im}z_{s}(u)$ satisfy the upper
bounds required for $z_{s}(u)$ to be in $\mathcal{D}_{s}$, and
$z_{s}(u)\in{\mathcal{D}}_{s}$. ∎

We next define the lattice on the upper half-plane $\mathbb{H}$ given by

(3.36)
$\displaystyle\mathcal{L}=\left\\{E+\mathrm{i}\eta\in\operatorname{\mathcal{D}}_{0}:E\in\mathbb{Z}/n^{8},\eta\in\mathbb{Z}/n^{8}\right\\}.$

######  Lemma 3.5.

Adopt 1.2. For any $t\in[0,{\mathsf{T}}]$ and
$w\in\operatorname{\mathcal{D}}_{t}$, there exists some lattice point
$u\in\mathcal{L}\cap z_{t}^{-1}(\operatorname{\mathcal{D}}_{t})$, such that
$|z_{t}(u)-w|\leq n^{-5},$ provided $n$ is large enough.

###### Proof of 3.5.

It follows from 3.4 that if $z_{t}(u)\in\operatorname{\mathcal{D}}_{t}$, then
$u\in{\mathcal{D}}_{0}$, and $z_{t}(u)=u-tm_{0}(u)$ with
$u\in\operatorname{\mathcal{D}}_{0}$. In particular $|u-E_{0}|\geq
f(0)=\mathfrak{C}^{2}\eta_{*}\geq 1/n^{2/3}$ (by (3.1)), and
$\operatorname{\partial}_{u}z_{t}(u)=1-tm_{0}^{\prime}(u)\leq
1+t/|u-E_{0}|^{2}\leq 1+{\mathsf{T}}n^{4/3}$ (using 2.1). Thus $z_{t}(u)$ is
Lipschitz in $u$ with Lipschitz constant bounded by $(1+{\mathsf{T}}n^{4/3})$.
Thus for any $w\in\operatorname{\mathcal{D}}_{t}$, there exists some lattice
point $u\in\mathcal{L}\cap z_{t}^{-1}(\operatorname{\mathcal{D}}_{t})$, such
that $|z_{t}(u)-w|\leq(1+{\mathsf{T}}n^{4/3})/n^{8}\leq n^{-5},$ provided $n$
is large enough. ∎

###  3.2. Proof of 3.1

To prove 3.1, we denote the Stieltjes transform of the empirical particle
density by

(3.37)
$\displaystyle\widetilde{m}_{t}(z)=\frac{1}{n}\sum_{i=1}^{n}\frac{1}{\lambda_{i}(t)-z},\quad
z\in\mathbb{H}.$

By Itô’s formula, $\widetilde{m}_{s}(z)$ satisfies a Burgers type stochastic
differential equation,

(3.38) $\displaystyle\begin{split}{\rm
d}\widetilde{m}_{s}(z)&=-\frac{1}{n}\sum_{i=1}^{n}\frac{{\rm
d}\lambda_{i}(s)}{(\lambda_{i}(s)-z)^{2}}+\frac{1}{n}\sum_{i=1}^{n}\frac{\langle{\rm
d}\lambda_{i}(s),{\rm d}\lambda_{i}(s)\rangle}{(\lambda_{i}(s)-z)^{3}}\\\
&=-\sqrt{\frac{2}{\beta n^{3}}}\sum_{i=1}^{n}\frac{{\rm
d}B_{i}}{(\lambda_{i}(s)-z)^{2}}-\frac{1}{n^{2}}\sum_{i=1}^{n}\frac{1}{(\lambda_{i}(s)-z)^{2}}\sum_{j:j\neq
i}\frac{1}{\lambda_{i}(s)-\lambda_{j}(s)}+\frac{2}{\beta
n^{2}}\sum_{i=1}^{n}\frac{{\rm d}s}{(\lambda_{i}(s)-z)^{3}}\\\
&=-\sqrt{\frac{2}{\beta n^{3}}}\sum_{i=1}^{n}\frac{{\rm
d}B_{i}(s)}{(\lambda_{i}(s)-z)^{2}}+\widetilde{m}_{s}(z)\operatorname{\partial}_{z}\widetilde{m}_{s}(z){\rm
d}s+\frac{2-\beta}{\beta n^{2}}\sum_{i=1}^{n}\frac{{\rm
d}s}{(\lambda_{i}(s)-z)^{3}},\end{split}$

where the last line follows from the fact that

$\displaystyle\frac{1}{n^{2}}$
$\displaystyle\sum_{i=1}^{n}\frac{1}{(\lambda_{i}(s)-z)^{2}}\sum_{j:j\neq
i}\frac{1}{\lambda_{i}(s)-\lambda_{j}(s)}$
$\displaystyle=\frac{1}{2n^{2}}\sum_{i\neq
j}^{n}\bigg{(}\frac{1}{(\lambda_{i}(s)-z)^{2}}\frac{1}{\lambda_{i}(s)-\lambda_{j}(s)}+\frac{1}{(\lambda_{j}(s)-z)^{2}}\frac{1}{\lambda_{j}(s)-\lambda_{i}(s)}\bigg{)}$
$\displaystyle=-\frac{1}{2n^{2}}\sum_{i\neq
j}\frac{\lambda_{i}(s)-z+\lambda_{j}(s)-z}{(\lambda_{i}(s)-z)^{2}(\lambda_{j}(s)-z)^{2}}$
$\displaystyle=-\frac{1}{n^{2}}\sum_{i\neq
j}\frac{1}{(\lambda_{i}(s)-z)(\lambda_{j}(s)-z)^{2}}=\widetilde{m}_{s}(z)\operatorname{\partial}_{z}\widetilde{m}_{s}(z)+\frac{1}{n^{2}}\sum_{i=1}^{n}\frac{1}{(\lambda_{i}(s)-z)^{3}}.$

Plugging the characteristic flow (3.5) into (3.38), and using the chain rule,
we have

(3.39) $\displaystyle\begin{split}{\rm
d}\widetilde{m}_{s}(z_{s})=\partial_{s}\widetilde{m}_{s}(z_{s})+\partial_{z}\widetilde{m}_{s}(z_{s})\partial_{s}z_{s}=-&\sqrt{\frac{2}{\beta
n^{3}}}\sum_{i=1}^{n}\frac{{\rm d}B_{i}(s)}{(\lambda_{i}(s)-z_{s})^{2}}\\\
&+\operatorname{\partial}_{z}\widetilde{m}_{s}(z_{s})\left(\widetilde{m}_{s}(z_{s})-m_{s}(z_{s})\right){\rm
d}s+\frac{2-\beta}{\beta n^{2}}\sum_{i=1}^{n}\frac{{\rm
d}s}{(\lambda_{i}(s)-z_{s})^{3}}.\end{split}$

Let $\mu_{t}=\mu_{0}\boxplus\mu_{\operatorname{sc}}^{(t)}$ denote the free
convolution between $\mu_{0}$ with the rescaled semicircle distribution, and
let $m_{t}$ denote its Stieltjes transform. Then, we can integrate both sides
of (3.39) from $0$ to $t$ to obtain

(3.40)
$\displaystyle\widetilde{m}_{t}(z_{t})-m_{t}(z_{t})=\widetilde{m}_{t}(z_{t})-m_{0}(z_{0})=\widetilde{m}_{t}(z_{0})-\widetilde{m}_{0}(z_{0})=\displaystyle\int_{0}^{t}\mathrm{d}\widetilde{m}_{s}(z_{s})ds=\int_{0}^{t}\left({\mathcal{E}}_{1}(s){\rm
d}s+{\rm d}{\mathcal{E}}_{2}(s)\right),$

where the error terms are

(3.41) $\displaystyle{\mathcal{E}}_{1}(s)=$
$\displaystyle\left(\widetilde{m}_{s}(z_{s})-m_{s}(z_{s})\right)\operatorname{\partial}_{z}\widetilde{m}_{s}(z_{s})$
(3.42) $\displaystyle{\rm d}{\mathcal{E}}_{2}(s)=$
$\displaystyle\frac{2-\beta}{\beta n^{2}}\sum_{i=1}^{n}\frac{{\rm
d}s}{(\lambda_{i}(s)-z_{s})^{3}}-\sqrt{\frac{2}{\beta
n^{3}}}\sum_{i=1}^{n}\frac{{{\rm d}}B_{i}(s)}{(\lambda_{i}(s)-z_{s})^{2}}.$

We remark that $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$ implicitly depend on
$u$, the initial value of the flow $z_{0}=u$.

We define the stopping time

(3.43)
$\displaystyle\begin{split}\sigma\operatorname{:=}{\mathsf{T}}\wedge\inf_{s\geq
0}\left\\{\exists
w\in\operatorname{\mathcal{D}}_{s}:\left|\widetilde{m}_{s}(w)-m_{s}(w)\right|\geq\frac{(\log
n)^{7/2}}{n\sqrt{{\rm Im}[w]({\rm Re}[w-E_{s}]+{\rm
Im}[w])}}\right\\}.\end{split}$

For $u\in{\mathcal{D}}_{0}$ and $s\leq{\mathfrak{t}}(u)\wedge\sigma$, by the
definition of the stopping time (3.43), ${\rm Im}[\widetilde{m}_{s}(w)]$ and
${\rm Im}[m_{s}(w)]$ are close. A more precise estimate is given in the
following lemma.

######  Lemma 3.6.

Adopt the notation and assumptions of 3.1, and assume $n$ is large enough.
Given any $u\in{\mathcal{D}}_{0}$, recall ${\mathfrak{t}}(u)$ from (3.31). For
any $0\leq s\leq{\mathfrak{t}}(u)$, abbreviate $z_{s}=z_{s}(u)$; we have

(3.44) $\displaystyle\begin{split}&\left|{\rm
Im}[\widetilde{m}_{s\wedge\sigma}(z_{s\wedge\sigma})]-{\rm
Im}[m_{s\wedge\sigma}(z_{s\wedge\sigma})]\right|\leq\frac{{\rm
Im}[m_{s\wedge\sigma}(z_{s\wedge\sigma})]}{\log n},\\\ &{\rm
Im}[\widetilde{m}_{s\wedge\sigma}(z_{s\wedge\sigma})]\leq 2{\rm
Im}[m_{s\wedge\sigma}(z_{s\wedge\sigma})].\end{split}$

###### Proof of 3.6.

From the definition of ${\mathfrak{t}}(u)$ as in (3.31),
$z_{s}=z_{s}(u):=E_{s}+\kappa_{s}+\mathrm{i}\eta_{s}\in{\mathcal{D}}_{s}$ for
$0\leq s\leq t(u)$. From the definition (3.43) of $\sigma$, we have that

$\displaystyle|\widetilde{m}_{s\wedge\sigma}(z_{s\wedge\sigma})-{m}_{s\wedge\sigma}(z_{s\wedge\sigma})|\leq\frac{(\log
n)^{7/2}}{n\sqrt{(\kappa_{s\wedge\sigma}+\eta_{s\wedge\sigma})\eta_{s\wedge\sigma}}}\leq\frac{\text{Im}[{m}_{s\wedge\sigma}(z_{s\wedge\sigma})]}{\log
n},$

where we used (3.30) in the last inequality. The last statement in
(LABEL:e:Im-Im) also follows for $n$ large enough. ∎

The following lemma states that before the stopping time $\sigma$, the largest
particle is bounded by $E_{t}+f(t)$.

######  Lemma 3.7.

Adopt the notation and assumptions in 3.1, and assume $n$ is large enough. For
any $t\leq\sigma$ (from (3.43)), we have

(3.45) $\displaystyle\lambda_{1}(t)\leq E_{t}+f(t).$

###### Proof of 3.7.

Fix a time $t\geq 0$ and recall ${\mathcal{D}}_{t}$ from (3.28). We notice
that

(3.46) $\displaystyle\eta{\rm
Im}[m_{t}(E_{t}+f(t)+\eta)]=\int_{\mathbb{R}}\frac{\eta^{2}\varrho_{t}(x){\rm
d}x}{(x-E_{t}-f(t))^{2}+\eta^{2}},$

is increasing in $\eta\geq 0$. When $\eta\rightarrow+\infty$, (3.46) goes to
one; when $\eta\rightarrow 0$, (3.46) goes to zero. Therefore, there exists a
(unique) $\eta_{t}>0$ satisfying

(3.47) $\displaystyle\frac{1}{(\log n)n}=\eta_{t}{\rm
Im}[m_{t}(E_{t}+f(t)+\mathrm{i}\eta_{t})].$

Then let $w_{t}=E_{t}+f(t)+\mathrm{i}\eta_{t}$. Recalling the definition of
${\mathcal{D}}_{t}$ from (3.28), (3.47) and the monotonicity (3.46) imply that
$w_{t}$ is on the south-west corner of ${\mathcal{D}}_{t}$.

Next we claim that the bound

(3.48) $\displaystyle|\widetilde{m}_{t}(w_{t})-m_{t}(w_{t})|\leq\frac{(\log
n)^{7/2}}{n\sqrt{{\rm Im}[w_{t}]({\rm Re}[w_{t}-E_{t}]+{\rm Im}[w_{t}])}},$

implies that no $\lambda_{i}(t)$ exists in the interval
$[E_{t}+f(t)-\eta_{t},E_{t}+f(t)+\eta_{t}]$. Indeed, the estimate (3.48)
implies

(3.49) $\displaystyle|\widetilde{m}_{t}(w_{t})-m_{t}(w_{t})|\leq\frac{(\log
n)^{7/2}}{n\sqrt{{\rm Im}[w_{t}]({\rm Re}[w_{t}-E_{t}]+{\rm
Im}[w_{t}])}}\leq{\rm Im}[m_{t}(w_{t})]=\frac{1}{(\log n)n{\rm Im}[w_{t}]},$

where the second inequality is from $w_{t}\in{\mathcal{D}}_{t}$ and (3.30);
the last equality is from our construction (3.47). Together with our choice of
$\eta_{t}$ from (3.47), it follows that

(3.50) $\displaystyle{\rm Im}[\widetilde{m}_{t}(w_{t})]\leq{\rm
Im}[m_{t}(w_{t})]+|\widetilde{m}_{t}(w_{t})-m_{t}(w_{t})|\leq\frac{2}{(\log
n)n{\rm Im}[w_{t}]}.$

If there exists some $\lambda_{i}(t)$ such that
$\lambda_{i}(t)\in[E_{t}+f(t)-\eta_{t},E_{t}+f(t)+\eta_{t}]$, then the left-
hand side of (3.50) is at least

$\displaystyle{\rm Im}[\widetilde{m}_{t}(w_{t})]={\rm
Im}\left[\sum_{i=1}^{n}\frac{1}{\lambda_{i}(t)-w_{t}}\right]=\frac{1}{n}\sum_{i=1}^{n}\frac{{\rm
Im}[w_{t}]}{|\lambda_{i}(t)-w_{t}|^{2}}\geq\frac{{\rm
Im}[w_{t}]}{n|w_{t}-\lambda_{i}(t)|^{2}}\geq\frac{1}{2n{\rm Im}[w_{t}]},$

where in the first inequality we used that ${\rm Im}[w_{t}]=\eta_{t}>0$, so
each term in the summation is positive; in the last inequality we used that
$|w_{t}-\lambda_{i}(t)|^{2}\leq 2\eta_{t}^{2}$ since
$\lambda_{i}(t)\in[E_{t}+f(t)-\eta_{t},E_{t}+f(t)+\eta_{t}]$. This leads to a
contradiction with (3.50), provided $n$ is large enough.

For $t\leq\sigma$, the bound (3.48) holds (by the definition 3.43), and so
there is no $\lambda_{j}(t)$ in the interval
$[E_{t}+f(t)-\eta_{t},E_{t}+f(t)+\eta_{t}]$. At time $t=0$, we have
$\lambda_{1}(0)\leq 0\leq f(0)$. Since $\lambda_{1}(t)$ is continuous, and
also $f(t)$ and $E_{t}$ are continuous (the first follows from the definition
(3.2), and the second follows from the facts $E_{t}=\xi(t)-tm_{0}(\xi(t))$ and
$\xi(t)$ is continuous from (2.50)) we conclude that $\lambda_{1}(t)\leq
E_{t}+f(t)$ for all $0\leq t\leq\sigma$. ∎

3.1 follows from the following statement and 3.7. Its proof will occupy the
remainder of this section.

######  Proposition 3.8.

Adopt the notation and assumptions in 3.1. There exists a constant
$C=C({\mathsf{b}},{\mathsf{T}})>1$, with probability $1-Ce^{-(\log n)^{2}}$,
we have $\sigma={\mathsf{T}}$.

###### Proof of 3.1.

On the event $\sigma={\mathsf{T}}$, 3.7 implies that $\lambda_{1}(t)\leq
E_{t}+f(t)$ for $0\leq t\leq{\mathsf{T}}$. Thus 3.1 follows from 3.8. ∎

In the rest of this section we prove 3.8 by analyzing the stochastic
differential equation (3.40). The following proposition will be used to upper
bound the error terms ${\mathcal{E}}_{1}(s)$ and ${\mathcal{E}}_{2}(s)$.

######  Proposition 3.9.

There exists a constant $C=C({\mathsf{b}},{\mathsf{T}})>1$ and an event
$\Omega$, measurable with respect to the Brownian paths
$\\{B_{1}(s),B_{2}(s),\cdots,B_{n}(s)\\}_{0\leq s\leq{\mathsf{T}}}$, such that
$\mathbb{P}[\Omega]\geq 1-Ce^{-(\log n)^{2}}$ and the following holds. On
$\Omega$, for any $u\in{\mathcal{L}}$ (recall this lattice from (3.36)),
denote $z_{s}(u)=E_{s}(u)+\kappa_{s}(u)+\mathrm{i}\eta_{s}(u)$. Then, for any
$0\leq s\leq{\mathfrak{t}}(u)$ (recall (3.31)), we have

(3.51) $\int_{0}^{s\wedge\sigma}\sqrt{\frac{2}{\beta
n^{3}}}\sum_{i=1}^{n}\frac{{\rm
d}B_{i}(\tau)}{|z_{\tau}(u)-\lambda_{i}(\tau)|^{2}}{\rm d}\tau\leq\frac{C(\log
n)^{2}}{n\sqrt{(\kappa_{s\wedge\sigma}(u)+\eta_{s\wedge\sigma}(u))\eta_{s\wedge\sigma}(u)}}.$

and

(3.52)
$\displaystyle\frac{|2-\beta|}{\beta}\int_{0}^{s\wedge\sigma}\frac{1}{n^{2}}\sum_{i=1}^{n}\frac{1}{|\lambda_{i}(\tau)-z_{\tau}(u)|^{3}}{\rm
d}\tau\leq\frac{C\log
n}{n(\kappa_{s\wedge\sigma}(u)+\eta_{s\wedge\sigma}(u))},$

###### Proof of 3.9.

For simplicity of notation, we write
$\mathfrak{t}(u),z_{\tau}(u),\kappa_{\tau}(u),\eta_{\tau}(u)$ as
$\mathfrak{t},z_{\tau},\kappa_{\tau},\eta_{\tau}$, respectively. By the
definition of $\sigma$, for $\tau\leq\sigma$, 3.7 gives that

(3.53) $\displaystyle\lambda_{1}(\tau)\leq E_{\tau}+f(\tau).$

To prove (3.51), we notice by 3.4 that $z_{\tau}(u)\in{\mathcal{D}}_{\tau}$
for any $0\leq{\tau}\leq{\mathfrak{t}}$. We define a series of stopping times
$0=t^{(0)}<t^{(1)}<t^{(2)}<\cdots<t^{(m)}={\mathfrak{t}}$, as follows:

(3.54)
$t^{(k)}={\mathfrak{t}}\wedge\inf\\{{\tau}>t^{(k-1)}:(\kappa_{{\tau}}+\eta_{{\tau}})\eta_{{\tau}}<(\kappa_{t^{(k-1)}}+\eta_{t^{(k-1)}})\eta_{t^{(k-1)}}/2\\},\quad
k=1,2,3,\cdots,m,$

where $m=m(u)$ might depend on $u$. From (3.29), $\kappa_{t}$ and $\eta_{t}$
are finite and cannot be smaller than $1/n^{2}$; in particular,
$\eta_{t}(\kappa_{t}+\eta_{t})\geq 2n^{-4}$, and so any $u\in\mathcal{L}$ must
satisfy $m=m(u)\leq 10\log n$. We now apply the Burkholder–Davis–Gundy
inequality to our stochastic integral. This allows us to bound the supremum of
our stochastic integral over a time range by its quadratic variation with
overwhelming probability. We refer to [15, Theorem 3.2] (see also [16, Theorem
11.2.1]) for a detailed introduction to the Burkholder–Davis–Gundy inequality.
The quadratic variation of the left side of (3.51) at $s=t^{(k)}$ is given by

(3.55)
$\displaystyle\begin{split}&\phantom{{}={}}\int_{0}^{t^{(k)}\wedge\sigma}\frac{2}{\beta
n^{3}}\sum_{k=1}^{n}\frac{{\rm
d}{\tau}}{|z_{\tau}-\lambda_{k}({\tau})|^{4}}\leq\frac{\operatorname{\mathcal{O}}(1)}{n^{3}}\int_{0}^{t^{(k)}\wedge\sigma}\sum_{i=1}^{n}\frac{1}{|z_{\tau}-E_{\tau}-f(\tau)|^{2}}\frac{{\rm
d}\tau}{|z_{\tau}-\lambda_{i}(\tau)|^{2}}\\\
&=\frac{\operatorname{\mathcal{O}}(1)}{n^{2}}\int_{0}^{t^{(k)}\wedge\sigma}\frac{\text{Im}[\widetilde{m}_{\tau}(z_{\tau})]}{\eta_{\tau}((\kappa_{\tau}-f({\tau}))^{2}+(\eta_{\tau})^{2})}{\rm
d}{\tau}\leq\frac{\operatorname{\mathcal{O}}(1)}{n^{2}}\int_{0}^{t^{(k)}\wedge\sigma}\frac{\text{Im}[m_{\tau}(z_{\tau})]}{\eta_{\tau}((\kappa_{\tau}-f({\tau}))^{2}+(\eta_{\tau})^{2})}{\rm
d}{\tau},\end{split}$

where in the first inequality we used that for $\tau\leq\sigma$ we have
$\lambda_{1}(\tau)\leq E_{\tau}+f(\tau)\leq\operatorname{Re}z_{\tau}$; in the
second inequality we used the definition of $\widetilde{m}_{t}$ from (3.37);
and in the last inequality, we used (LABEL:e:Im-Im) indicating that, for
${\tau}\leq\sigma$, we have ${\rm Im}[\widetilde{m}_{\tau}(z_{\tau})]\leq
2{\rm Im}[m_{\tau}(z_{\tau})]$.

Let $s=t^{(k)}\wedge\sigma$ and
$\widetilde{s}=\min\\{s,\mathfrak{C}\sqrt{\eta_{*}}\\}$. We can rewrite the
integral in (LABEL:e:quadraticv) as

(3.56)
$\displaystyle\begin{split}\int_{0}^{s}\frac{\text{Im}[m_{\tau}(z_{\tau})]}{\eta_{\tau}((\kappa_{\tau}-f({\tau}))^{2}+(\eta_{\tau})^{2})}{\rm
d}{\tau}&\leq\int_{0}^{\widetilde{s}}\frac{\text{Im}[m_{\tau}(z_{\tau})]}{\eta_{\tau}(((\widetilde{s}-{\tau}){\rm
Im}[m_{s}(z_{s})]\kappa_{\widetilde{s}}/\eta_{\widetilde{s}})^{2}+(\eta_{\tau})^{2})}{\rm
d}{\tau}\\\
&\qquad+\int_{\widetilde{s}}^{s}\frac{\text{Im}[m_{\tau}(z_{\tau})]}{\eta_{\tau}(((s-{\tau}){\rm
Im}[m_{s}(z_{s})]\kappa_{s}/2\eta_{s})^{2}+(\eta_{\tau})^{2})}{\rm
d}{\tau},\end{split}$

where we used (3.11) to get the first term on the righthand of (3.56), and
(3.10) to get the second term. The two terms on the righthand side of (3.56)
can be bounded in the same way, so we will only study the second one. Let
$A:={\rm Im}[m_{\tau}(z_{\tau})]={\rm Im}[m_{s}(z_{s})]$; then
$\eta_{\tau}=\eta_{s}+(s-\tau)A$ (from (3.7)). Applying a change of variables
replacing $s-\tau$ by $\tau$, and then one replacing $A\tau$ by $\tau$, we
find

$\displaystyle\int_{\widetilde{s}}^{s}\frac{\text{Im}[m_{\tau}(z_{\tau})]{\rm
d}{\tau}}{\eta_{\tau}(((s-{\tau}){\rm
Im}[m_{s}(z_{s})]\kappa_{s}/2\eta_{s})^{2}+(\eta_{\tau})^{2})}{\rm d}{\tau}$
$\displaystyle=\int_{0}^{\infty}\frac{A{\rm
d}{\tau}}{(\eta_{s}+{\tau}A)(({\tau}A\kappa_{s}/2\eta_{s})^{2}+(\eta_{s}+{\tau}A)^{2})}$
$\displaystyle=\int_{0}^{\infty}\frac{{\rm
d}{\tau}}{(\eta_{s}+{\tau})(({\tau}\kappa_{s}/2\eta_{s})^{2}+(\eta_{s}+{\tau})^{2})}.$

If $\eta_{s}\geq\kappa_{s}$, then

(3.57) $\displaystyle\int_{0}^{\infty}\frac{{\rm
d}{\tau}}{(\eta_{s}+{\tau})(({\tau}\kappa_{s}/2\eta_{s})^{2}+(\eta_{s}+{\tau})^{2})}\leq\int_{0}^{\infty}\frac{{\rm
d}{\tau}}{(\eta_{s}+{\tau})^{3}}\leq\frac{1}{2\eta_{s}^{2}}\leq\frac{1}{(\kappa_{s}+\eta_{s})\eta_{s}}.$

If $\kappa_{s}\geq\eta_{s}$, we need to decompose the integral into three
parts:

(3.58) $\displaystyle\begin{split}&\phantom{{}={}}\int_{0}^{\infty}\frac{{\rm
d}{\tau}}{(\eta_{s}+{\tau})(({\tau}\kappa_{s}/2\eta_{s})^{2}+(\eta_{s}+{\tau})^{2})}=\int_{0}^{\eta^{2}_{s}/\kappa_{s}}(\cdots)+\int_{\eta^{2}_{s}/\kappa_{s}}^{\eta_{s}}(\cdots)+\int_{\eta_{s}}^{\infty}(\cdots)\\\
&\leq\int_{0}^{\eta^{2}_{s}/\kappa_{s}}\frac{{\rm
d}{\tau}}{\eta_{s}^{3}}+\int_{\eta^{2}_{s}/\kappa_{s}}^{\eta_{s}}\frac{{\rm
d}{\tau}}{\eta_{s}({\tau}\kappa_{s}/2\eta_{s})^{2}}+\int_{\eta_{s}}^{\infty}\frac{{\rm
d}{\tau}}{{\tau}({\tau}\kappa_{s}/2\eta_{s})^{2}}\leq\frac{1}{\kappa_{s}\eta_{s}}+\frac{4}{\kappa_{s}\eta_{s}}+\frac{2}{\kappa_{s}^{2}}\leq\frac{16}{(\kappa_{s}+\eta_{s})\eta_{s}}.\end{split}$

By the same argument, the first term on the righthand side of (3.56) can be
bounded by
$16/((\kappa_{\widetilde{s}}+\eta_{\widetilde{s}})\eta_{\widetilde{s}})\leq
16/((\kappa_{s}+\eta_{s})\eta_{s})$ (where the inequality is from the facts