# Large Deviations Asymptotics of Rectangular Spherical Integral

Alice Guionnet<EMAIL_ADDRESS>CNRS-ENS Lyon  Jiaoyang Huang
<EMAIL_ADDRESS>New York University

###### Abstract

In this article we study the Dyson Bessel process, which describes the
evolution of singular values of rectangular matrix Brownian motions, and prove
a large deviation principle for its empirical particle density. We then use it
to obtain the asymptotics of the so-called rectangular spherical integrals as
$m,n$ go to infinity while $m/n$ converges.

###### Contents

  1. 1 Introduction
  2. 2 Dyson Bessel Process
  3. 3 Large deviations for the Dyson Brownian motion
  4. 4 Large deviations for the Dyson Bessel process
  5. 5 Applications

##  1 Introduction

In this article we shall study the asymptotics of the so-called rectangular
spherical integrals, also called Berezin-Karpelevich type integrals in the
literature. This type of integrals arises when one studies rectangular
matrices and is the natural counterpart of the well known Harish-Chandra
-Itzykson-Zuber (HCIZ) integral. The interest in spherical integrals comes
from different fields. Harish-Chandra was motivated by Fourier analysis in
semi-simple Lie algebras. They appear in physics as the density in matrix
models such as the Ising model [25, 47, 16] or more generally matrix models
with an external field [11], including the famous Kontsevich matrix model
[43]. Their uses in random matrix theory appeared more recently. First it was
shown that spherical integral with a rank one external field gives
asymptotically the famous $R$-transform defined by Voiculescu in free
probability [29] as an analogue of Fourier transform. This approach was
generalized to the rectangular-free convolution by using rectangular spherical
integrals [27] or to the multiplicative free convolution and the $S$-transform
[10, 48].

Knowing the asymptotics of rank one spherical integrals allowed as well to
investigate the large deviations for the extreme eigenvalues of random
matrices. This approach was introduced in [33] where it was shown that the
probability that the largest eigenvalue of a Wigner matrix takes an unexpected
value is the same when the entries are Rademacher or Gaussian. This
universality phenomenon was shown to hold for random matrices with i.i.d.
entries whose Laplace transform is bounded by the Laplace transform of a
Gaussian variable with the same covariance. For more general sub-Gaussian
entries, a transition appears in the rate function between large deviations
towards a very large value with a heavy tail type rate function, and
deviations close to the bulk which are governed by the Gaussian rate function.
Such considerations were extended to unitary invariant ensembles [35], to the
joint distribution of the largest eigenvalue and its eigenvector [9], to sum
of matrices, to finitely many extreme eigenvalues [34]. Indeed, the
asymptotics of spherical integral could be extended to finite rank external
fields [34]. For small enough matrices, the same asymptotics were shown to
extend to the case where the rank goes to infinity more slowly than the
dimension [19] and to full rank matrices [18]. However, the limit differs when
the rank of both matrices are of the same order and the matrices do not have
small norms. Such a limit can as well be used to prove large deviation
principles for the empirical measure of the eigenvalues of random matrices [4]
and more generally study the asymptotics of matrix models with an external
field[11, 32].

The formula for the asymptotics of HCIZ integrals was foreseen by Matytsin
[45] and then proven rigorously in [30, 31, 32]. Matytsin used the description
of Spherical integrals as invariant eigenfunctions of the Laplacian. The
approach of [30] is kind of dual and based on a representation of spherical
integrals as the density of a Dyson Brownian motion conditioned at time one, a
representation which allows to use large deviations techniques and
martingales. In this paper, we follow the same route for the rectangular case
but prove a more general large deviation principle for conditioned Dyson
Brownian motions. In fact, the result in [30] relies on the matrix model, and
only concerns the case $\beta=1$ or $2$ whereas we can deal in this paper with
all cases $\beta\geqslant 1$. The extension of [30] to the rectangular case is
a natural step, which however posed significant difficulties for the proof of
the lower bound if one uses the methods of [30], due to additional singularity
of the drifts. We should also mention the heuristics proposed in this setting
in [26] following Matystin’s arguments. One key idea of this paper is to
improve the large deviations lower bound by obtaining better criteria for the
uniqueness of solutions to McKean-Vlasov equations with smooth fields inspired
from [44], rather than the weaker approach developed in [15]. Another novelty
in this paper is a quantitative estimate for the convergence to Dyson Brownian
motion with very general potential by a coupling argument, see Proposition
3.5. Under more restricted assumptions, i.e. the limiting profile has square
root behavior around the edge, such quantitative estimates for the convergence
has been obtained in [40, 1, 39] by using the characteristic method. The
quantitative estimate for the convergence allows us to efficiently control the
locations of each particles and extend our result to Dyson Bessel processes
which arises when one considers rectangular matrices and hence derive the
limits of rectangular spherical integrals. We now state more precisely our
main results.

The rectangular spherical integral is given by

$\displaystyle I_{n,m}(A_{n},B_{n})=\iint e^{\beta
n\mathrm{Re}[\operatorname{Tr}(A_{n}^{*}UB_{n}V^{*})]}{\rm d}U{\rm d}V,$ (1.1)

where if $\beta=1$, $U\in{\mathcal{O}}(n),V\in{\mathcal{O}}(m)$ follow the
Haar distribution over the orthogonal group, and
$A_{n}\in{\mathbb{R}}^{n\times m},B_{n}\in{\mathbb{R}}^{n\times m}$, whereas
for $\beta=2$, $U\in{\mathcal{U}}(n),V\in{\mathcal{U}}(m)$ follow the Haar
distribution over the unitary group, $A_{n}\in{\mathbb{C}}^{n\times
m},B_{n}\in{\mathbb{C}}^{n\times m}$ for $\beta=2$. We call such integrals
rectangular spherical integrals and shall study their asymptotic behavior when
$m$ and $n$ go to infinity so that the ratio $m/n$ converges towards some
$1+\alpha\in[1,\infty)$. This type of spherical integral arises when one
studies rectangular matrices and is the natural counterpart of the well known
Harish-Chandra -Itzykson-Zuber (HCIZ) integral defined when $\beta=2$ and for
two self-adjoint matrices $A_{n},B_{n}\in{\mathbb{C}}^{n\times n}$ by

$I_{n}(A_{n},B_{n})=\int e^{n\operatorname{Tr}(A_{n}UB_{n}U^{*})}{\rm d}U\,,$

where $U$ follows the Haar distribution over the unitary group. This integral
was shown by Harish-Chandra [38] and then Itzykson and Zuber [41] to be equal
to a determinant:

$I_{n}(A_{n},B_{n})=c_{n}\frac{\det\left[e^{na_{i}b_{j}}\right]_{1\leqslant
i,j\leqslant n}}{\Delta(\bm{a})\Delta(\bm{b})},$ (1.2)

where $\bm{a}=(a_{1},a_{2},\cdots,a_{n}),\bm{b}=(b_{1},b_{2},\cdots,b_{n})$
are eigenvalues of $A_{n}$ and $B_{n}$ respectively, and
$\Delta(\bm{a})=\prod_{i<j}(a_{i}-a_{j}),\Delta(\bm{b})=\prod_{i<j}(b_{i}-b_{j})$
are Vandermonde determinants. In 2003, Schlittgen and Wettig [49] considered a
generalization of the above rectangular spherical integral given by

$\displaystyle\iint\det[UV]^{\nu}e^{\tau\operatorname{Tr}(A_{n}^{*}UB_{n}V^{*}+VD_{n}^{*}U^{*}C_{n})/2}{\rm
d}U{\rm d}V,$ (1.3)

where $U,V\in{\mathcal{U}}(n)$ are $n\times n$ unitary matrices following Haar
distribution, $A_{n},B_{n},C_{n},D_{n}$ are deterministic $n\times n$
matrices, and $\nu$ is a non-negative integer. They showed that the
generalization of the above integral to the case of unequal dimensions of
$U,V$ leads to an integral which can be nonzero only if $\nu=0$, and predicted
the following formula: for $m\geqslant n$

$\displaystyle\iint
e^{\tau\operatorname{Tr}(A_{n}^{*}UB_{n}V^{*}+VD_{n}^{*}U^{*}C_{n})/2}{\rm
d}U{\rm
d}V=\frac{\tau^{n(m-1)}\prod_{i=1}^{n}(m-i)!(n-i)!}{\Delta({\bm{x}}^{2})\Delta({\bm{y}}^{2})\prod_{i=1}^{n}(x_{i}y_{i})^{m-n}}\det[I_{m-n}(2\tau
x_{i}y_{j})]_{1\leqslant i,j\leqslant n},$ (1.4)

where $U\in{\mathcal{U}}(n)$ is an $n\times n$ unitary matrix,
$V\in{\mathcal{U}}(m)$ is an $m\times m$ unitary matrix, both follow the Haar
distribution, $B_{n},C_{n}$ are deterministic $n\times m$ matrices, and
$A_{n},D_{n}$ are deterministic $m\times n$ rectangular matrices, $I_{m-n}(x)$
is the Bessel function

$I_{\kappa}(2y)=y^{\kappa}\sum_{k=0}^{\infty}\frac{y^{2k}}{k!(k+\kappa)!}\,,$

and ${\bm{x}}^{2}=(x_{1}^{2},x_{2}^{2},\cdots,x_{n}^{2})$,
${\bm{y}}^{2}=(y_{1}^{2},y_{2}^{2},\cdots,y_{n}^{2})$ are eigenvalues of the
matrices $A_{n}C_{n}^{*}$, $B_{n}D_{n}^{*}$. This formula was proven in [28].
We get the rectangular spherical integral (1.1) from (1.4) by taking
$A_{n}=C_{n}$ and $B_{n}=D_{n}$. Such formulas can be obtained by using the
character expansion method. Another approach is based on heat flows [46, 11].
Indeed, one can notice that Fourier functions $X\rightarrow
e^{i\operatorname{Tr}(AX)}$ are the eigenfunctions of the Laplacian for any
matrix $A$. Looking for eigenfunctions depending only on the eigenvalues of
$X$ one gets the spherical integral $I_{n}(A_{n},X_{n})$, which in turns has
to be an eigenfunction of the Laplace operator restricted to functions
invariant under conjugation, namely the Dyson Laplace operator
$L=-\Delta(X)^{-1}\sum_{i}\delta_{x_{i}}^{2}\Delta(X)$. Note however that
(1.2) and (1.4) are not useful to derive asymptotics as they are given in
terms of a signed sum of diverging terms.

For a rectangular $n\times m$ matrix $A_{n}$ , $m\geqslant n$, with non
trivial singular values $(s_{i})_{{1\leqslant i\leqslant n}}$, we denote
$\hat{\nu}_{A}^{n}$ its symmetrized empirical singular values

$\hat{\nu}^{n}_{A}=\frac{1}{2n}\sum_{i=1}^{n}\left(\delta_{s_{i}}+\delta_{-s_{i}}\right)\,.$

We denote by $\Sigma$ the non commutative entropy

$\Sigma(\nu)=\int\log|x-y|d\nu(x)d\nu(y)\,.$

Then, we prove the following asymptotics for the rectangular spherical
integrals:

######  Theorem 1.1.

Let $A_{n},B_{n}\in{\mathbb{R}}^{n\times m}$ and
$U\in{\mathcal{O}}(n),V\in{\mathcal{O}}(m)$ following Haar distribution over
orthogonal group for $\beta=1$; $A_{n},B_{n}\in{\mathbb{C}}^{n\times m}$ and
$U\in{\mathcal{U}}(n),V\in{\mathcal{U}}(m)$ following Haar distribution over
unitary group, for $\beta=2$, where $m\geqslant n$ and $m/n\rightarrow
1+\alpha,\alpha\geqslant 0$. We assume that the symmetrized empirical singular
values $\hat{\nu}_{A}^{n}$ and $\hat{\nu}_{B}^{n}$ of $A_{n}$ and $B_{n}$
converge weakly to $\hat{\nu}_{A}$ and $\hat{\nu}_{B}$ respectively. We
moreover assume that for $C=A$ or $B$, we have
$\sup_{n}\hat{\nu}_{C}^{n}(x^{2})<\infty$, $\Sigma(\hat{\nu}_{C})>-\infty$
and, if $\alpha\neq 0$, $\int\ln|x|d\hat{\nu}_{C}>-\infty$. Then, the
following limit of the rectangular spherical integral exists

$\lim_{n}\frac{1}{n^{2}}\log
I_{n,m}(A_{n},B_{n})=\frac{\beta}{2}I^{\alpha}(\hat{\nu}_{A},\hat{\mu}_{B}),\quad
I_{n,m}(A_{n},B_{n})=\int e^{\beta
n\mathrm{Re}[\operatorname{Tr}(A_{n}^{*}UB_{n}V^{*})]}{\rm d}U{\rm d}V.$

It is given explicitly by

$\displaystyle\begin{split}&I^{\alpha}(\hat{\nu}_{A},\hat{\nu}_{B})=-\inf_{\\{\hat{\rho}_{t}\\}_{0\leqslant
t\leqslant 1}}\left\\{\int_{0}^{1}\int u_{s}^{2}\hat{\rho}_{s}{\rm d}x{\rm
d}s+\frac{\pi^{2}}{3}\int_{0}^{1}\int\hat{\rho}^{3}_{s}{\rm d}x{\rm
d}s+\frac{\alpha^{2}}{4}\int\frac{\hat{\rho}_{s}(x)}{x^{2}}{\rm d}x{\rm
d}s\right\\}\\\
&+(\hat{\nu}_{A}(x^{2}-\alpha\log|x|)+\hat{\nu}_{B}(x^{2}-\alpha\log|x|))-(\Sigma(\hat{\nu}_{A})+\Sigma(\hat{\nu}_{B}))+\mbox{const},\end{split}$
(1.5)

where const is a constant depending on $\alpha$. The infimum is taken over
continuous symmetric measure valued processes $(\hat{\rho}_{t}(x){\rm
d}x)_{{0<t<1}}$ such that

$\displaystyle\lim_{t\rightarrow 0}\hat{\rho}_{t}(x){\rm
d}x=\hat{\nu}_{A},\quad\lim_{t\rightarrow 1}\hat{\rho}_{t}(x){\rm
d}x=\hat{\nu}_{B}.$ (1.6)

Moreover, $u$ is the weak solution of the following conservation of mass
equation

$\displaystyle\partial_{s}\hat{\rho}_{s}+\partial_{x}(\hat{\rho}_{s}u_{s})=0.$

This theorem will be proved in Section 5.1. We show in Proposition 5.1 that in
fact the non commutative law of $(A_{n},UB_{n}V^{*})$ converges when$(U,V)$
follows the Gibbs measure with free energy $I_{n,m}(A_{n},B_{n})$. As in [30],
the main point is to derive a large deviation principle for the associated
processes, namely Bessel Dyson processes. Indeed, let $G_{n}$ be an $n\times
m$ rectangular matrix with independent real ($\beta=1$) or complex ($\beta=2$)
Gaussian entries and set

$\displaystyle X_{n}=A_{n}+\frac{1}{\sqrt{n}}G_{n}.$

then, we claim that the large deviation principle for the symmetrized
empirical singular values of $X_{n}$ gives the asymptotics of spherical
integrals. In fact, denote the singular value decomposition of $X_{n}$ as
$X_{n}=UB_{n}V^{*}$. Then the joint law of $(B_{n},U,V)$ is given by

$\displaystyle\begin{split}\frac{1}{Z_{n,m}}\prod_{i}b_{i}^{\beta(m-n+1)-1}\prod_{i<j}|b^{2}_{i}-b^{2}_{j}|^{\beta}e^{-\frac{\beta
n}{2}(\sum_{i}b^{2}_{i}+\sum a^{2}_{i})+\beta
n\mathrm{Re}[\operatorname{Tr}(A_{n}^{*}UB_{n}V^{*})]}{\rm d}U{\rm d}V{\rm
d}B_{n}.\end{split}$ (1.7)

Assume that we have proven a large deviation principle for $\hat{\nu}^{n}_{X}$
with a good rate function $I_{\hat{\nu}_{A}}$ so that for any symmetric
probability measure $\hat{\nu}_{B}$

$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\hat{\nu}^{n}_{B}\in{\mathbb{B}}(\hat{\nu}_{B},\delta))=-I_{\hat{\nu}_{A}}(\hat{\nu}_{B})+o_{\delta(1)}$
(1.8)

where $\operatorname{o}_{\delta}(1)$ goes to zero as $\delta$ goes to zero. By
integrating (1.7) over the ball ${\mathbb{B}}(\hat{\nu}_{B},\delta)$, we have

$\displaystyle\phantom{{}={}}\int_{\hat{\nu}^{n}_{B}\in{\mathbb{B}}(\hat{\nu}_{B},\delta)}\frac{1}{Z_{n,m}}\prod_{i}b_{i}^{\beta(m-n+1)-1}\prod_{i<j}|b^{2}_{i}-b^{2}_{j}|^{\beta}e^{-\frac{\beta
n}{2}(\sum_{i}b^{2}_{i}+\sum a^{2}_{i})+\beta
n\mathrm{Re}[\operatorname{Tr}(A^{*}UBV^{*})]}{\rm d}U{\rm d}V{\rm d}B_{n}$
$\displaystyle=\frac{1}{Z_{n,m}}e^{\frac{\beta
n^{2}}{2}(2\alpha\int\log|x|{\rm
d}\hat{\nu}_{B}+2\Sigma(\hat{\nu}_{B})-(\hat{\nu}_{A}(x^{2})+\hat{\nu}_{B}(x^{2}))+\operatorname{o}_{\delta}(1))}\int_{\hat{\nu}^{n}_{B}\in{\mathbb{B}}(\hat{\nu}_{B},\delta)}\int
e^{\beta n\mathrm{Re}[\operatorname{Tr}(A^{*}UBV^{*})]}{\rm d}U{\rm d}V{\rm
d}B_{n}.$

By rearranging, we obtain the following asymptotics of the spherical integral
(following the standard arguments to prove large deviations for Beta-ensembles
[6]):

$\displaystyle\begin{split}&\lim_{n\rightarrow\infty}\frac{1}{n^{2}}\log
I_{n,m}(A_{n},B_{n})=-I_{\hat{\nu}_{A}}(\hat{\nu}_{B})\\\
&-\frac{\beta}{2}\left(2\alpha\int\log x{\rm
d}\hat{\nu}_{B}(x)+2\Sigma(\hat{\nu}_{B})-(\hat{\nu}_{A}(x^{2})+\hat{\nu}_{B}(x^{2}))\right)+\mbox{const}.\end{split}$

To prove (1.8), we see $X_{n}=H(1)$ as the matrix valued process
$H(t)=A+G_{n}(t)/\sqrt{n}$ at time one, where $G_{n}(t)$ is field with
independent Brownian motions. The singular values $s_{1}(t)\geqslant
s_{2}(t)\geqslant\cdots\geqslant s_{n-1}(t)\geqslant|s_{n}(t)|$ of $H(t)$
follow the Dyson Bessel process:

$\displaystyle\begin{split}\mathrm{d}s_{i}(t)&=\frac{\mathrm{d}W_{i}}{\sqrt{\beta
n}}+\left(\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{s_{i}(t)-s_{j}(t)}+\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{s_{i}(t)+s_{j}(t)}+\frac{\alpha_{n}}{2s_{i}(t)}\right)\mathrm{d}t,\quad
1\leqslant i\leqslant n,\end{split}$ (1.9)

where $W_{1},W_{2},\cdots,W_{n}$ are independent Brownian motions and

$\displaystyle\alpha_{n}=\frac{m-n}{n}+(1-\frac{1}{\beta})\frac{1}{n}.$

We denote the empirical particle density of (1.9) and its symmetrized version,
which is also the symmetrized empirical singular values of $H(t)$, as

$\displaystyle\nu_{t}^{n}=\frac{1}{n}\sum_{i=1}^{n}\delta_{s_{i}(t)},\quad\hat{\nu}_{t}^{n}=\frac{1}{2n}\sum_{i=1}^{n}(\delta_{s_{i}(t)}+\delta_{-s_{i}(t)}),$

We prove a large deviation principle for $\\{\hat{\nu}_{t}^{n}\\}_{0\leqslant
t\leqslant 1}$, in Section 4. The rate function is given by

$\displaystyle S^{\alpha}_{\hat{\mu}_{0}}(\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant 1})=\sup_{f\in{\cal
C}^{2,1}_{b}}S^{\alpha}(\\{\hat{\nu}_{t},f_{t}\\}_{0\leqslant t\leqslant 1}),$
(1.10)

where

$\displaystyle S^{\alpha}(\\{\hat{\nu}_{t},f_{t}\\}_{0\leqslant t\leqslant
1})$
$\displaystyle=\hat{\nu}_{1}(f_{1})-\hat{\mu}_{0}(f_{0})-\int_{0}^{1}\int\partial_{s}f_{s}(x){\rm
d}\hat{\nu}_{s}(x){\rm
d}s-\frac{1}{2}\int_{0}^{1}\int\frac{f_{s}^{\prime}(x)-f_{s}^{\prime}(y)}{x-y}{\rm
d}\hat{\nu}_{s}(x){\rm d}\hat{\nu}_{s}(y){\rm d}s$
$\displaystyle-\frac{\alpha}{2}\int_{0}^{1}\int\frac{f_{s}^{\prime}(x)}{x}{\rm
d}\hat{\nu}_{s}(x){\rm
d}s-\frac{1}{8\beta}\int^{1}_{0}\int(f_{s}^{\prime}(x)-f_{s}^{\prime}(-x))^{2}{\rm
d},\hat{\nu}_{s}(x){\rm d}s.$

If $\hat{\nu}_{0}\neq\hat{\mu}_{0}$,
$S^{\alpha}_{\hat{\mu}_{0}}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant
1})=\infty$. We then prove the following result

######  Theorem 1.2.

Fix a symmetric probability measure $\hat{\mu}_{0}$ and an initial condition
with symmetrized empirical measure $\hat{\nu}^{n}_{0}$ with uniformly bounded
second moment converging weakly to $\hat{\mu}_{0}$. Then, if $\alpha_{n}$
converges towards $\alpha\in[0,\infty)$ when $n$ goes to infinity so that
either $\alpha_{n}\geqslant 1/\beta n$ or $\alpha_{n}\equiv 0$, the
distribution of the empirical particle density
$\\{\hat{\nu}_{t}^{n}\\}_{0\leqslant t\leqslant 1}$ of the Dyson Bessel
process (4.3) satisfies a large deviations principle in the scale $n^{2}$ and
with good rate function $S_{\hat{\nu}_{0}}^{\alpha}$. In particular, for any
continuous symmetric measure-valued process $\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant 1}$, we have:

$\displaystyle\begin{split}&\phantom{{}={}}\lim_{\delta\rightarrow
0}\liminf_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\hat{\nu}^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant
1},\delta))\\\ &=\lim_{\delta\rightarrow
0}\limsup_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\hat{\nu}^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant
1},\delta))=-{{S^{\alpha}_{\hat{\mu}_{0}}(\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant 1}).}}\end{split}$ (1.11)

######  Remark 1.3.

In Theorem 1.2, we assumed that either $\alpha_{n}\geqslant 1/\beta n$ or
$\alpha_{n}\equiv 0$. This assumption is always true for $\beta=1,2$ and
$m\geqslant n$. If this condition is violated, i.e. $0<\alpha_{n}<1/\beta n$,
the particles $s_{n}(t)$ and $s_{-n}(t)$ in (1.9) may collapse at $0$. In this
case, to make sense of (1.9), we need to specify the boundary condition when
they collapse at $0$. We will not discuss these conditions in this paper.

As a consequence, we deduce from the contraction principle [20] that (1.12)
holds and more precisely

######  Corollary 1.4.

For any symmetric probability measures $\hat{\nu}^{n}_{A},\hat{\nu}^{n}_{B}$
with uniformly bounded second moment converging weakly towards
$\hat{\nu}_{A},\hat{\nu}_{B}$, under the measure (1.7) we have

$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\hat{\nu}^{n}_{B}\in{\mathbb{B}}(\hat{\nu}_{B},\delta))=-I_{\hat{\nu}_{A}}(\hat{\nu}_{B})+o_{\delta(1)},$
(1.12)

where

$I_{\hat{\nu}_{A}}(\hat{\nu}_{B})=\inf_{\hat{\nu}_{1}=\hat{\nu}_{B}}S_{\hat{\nu}_{A}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant 1})\,.$

Theorem 1.1 is deduced from Theorem 1.2 in section 5.1. The main difficulty to
prove Theorem 1.2 lies in the singularity of the potential at the origin and
the repulsion between the particles. To prove it, we revisit in section 3 the
large deviation principle for the empirical measure of the Dyson Brownian
motion of [30] and extend it to to all values of $\beta$ greater or equal to
one.

Acknowledgements  The research of J.H. is supported by the Simons Foundation
as a Junior Fellow at the Simons Society of Fellows, and NSF grant
DMS-2054835. The work of A. Guionnet is partly supported by ERC Project LDRAM
: ERC-2019-ADG Project 884584. We thank O. Zeitouni for many inspiring
discussions about spherical integrals, including preliminary ideas about the
questions addressed in this article.

Notations ${\mathcal{O}}(n)$ denotes the orthogonal group in dimension $n$ and
${\mathcal{U}}(n)$ the unitary group in dimension $n$. We denote by
$d(\cdot,\cdot)$ the $2$-Wasserstein distance defined on the space
$\mathcal{P}_{2}(\mathbb{R})$ of probability measures with finite second
moment by

$d(\mu,\nu)=\inf\left\\{\int|x-y|^{2}d\pi(x,y)\right\\}^{1/2},$

where the infimum is taken over distribution on $\mathbb{R}^{2}$ with marginal
distribution $\mu$ and $\nu$. ${\cal C}_{b}^{2,1}(\mathbb{R}\times[0,1])$ is
the space of functions on $\mathbb{R}\times[0,1]$ with bounded first two
derivatives in $x$ and bounded derivative in $t$. ${\cal
C}([0,1],{\mathbb{M}}_{1}({\mathbb{R}}))$ is the space of continuous (with
respect to weak topology) measure valued process.

##  2 Dyson Bessel Process

In this section we introduce Dyson Bessel process, which is the singular value
process of rectangular matrix brownian motions. Then in section 2.2, we will
write Dyson Bessel process as a change of measure from Dyson Brownian motion
using Girsanov’s theorem.

###  2.1 Decomposition

The rectangular spherical integral (1.1) is related to real ($\beta=1$) and
complex ($\beta=2$) rectangular random matrices with nonzero mean. We consider
an $n\times m$ rectangular random matrices $X_{n}$ with nonzero mean,

$\displaystyle X_{n}=A_{n}+\frac{1}{\sqrt{n}}G_{n},$ (2.1)

where $A_{n}={\mathbb{E}}[X_{n}]$ is deterministic, and $G_{n}$ is an $n\times
m$ rectangular matrix with independent real ($\beta=1$) or complex ($\beta=2$)
Gaussian entries. We denote the singular value decomposition of
$X_{n}=UB_{n}V^{*}$, with
$B_{n}=\operatorname{diag}\\{b_{1},b_{2},\cdots,b_{n}\\}$. Then we can rewrite
the law of $X_{n}$ as

$\displaystyle\begin{split}&\phantom{{}={}}\left(\sqrt{\frac{\beta
n}{2\pi}}\right)^{\beta mn}e^{-\frac{\beta
n}{2}\operatorname{Tr}((X_{n}-A_{n})(X_{n}-A_{n})^{*})}{\rm d}X_{n}\\\
&\propto\prod_{i}b_{i}^{\beta(m-n+1)-1}\prod_{i<j}|b^{2}_{i}-b^{2}_{j}|^{\beta}e^{-\frac{\beta
N}{2}(\sum_{i}b^{2}_{i}+\sum a^{2}_{i})+\beta
n\mathrm{Re}[\operatorname{Tr}(A_{n}^{*}UB_{n}V^{*})]}{\rm d}U{\rm d}V{\rm
d}B_{n}.\end{split}$ (2.2)

Therefore, conditioning on the singular values of $X_{n}$, i.e. the matrix
$B_{n}$, the joint law of singular vectors of $X_{n}$, i.e. $U,V$ is given by
the integrand of the rectangular spherical integral (1.1)

$\displaystyle\frac{e^{\beta
N\mathrm{Re}[\operatorname{Tr}(A_{n}^{*}UB_{n}V^{*})]}}{Z^{\beta}_{m,n}}{\rm
d}U{\rm d}V.$ (2.3)

We study the random matrices $X_{n}$ as in (2.1) via a dynamical approach. By
constructing a matrix valued real/complex Brownian motions starting from
$A_{n}$, its value at time $t=1$ has the same law as $X_{n}$.

######  Theorem 2.1 (Dyson Bessel Process).

Take $\beta\geqslant 1$. Fix $m\geqslant n$, and let $H(t)$ be a $n\times m$
matrix with entries given by independent real/complex Brownian motions
starting from $A_{n}$:

$\displaystyle H(t)=A_{n}+\frac{1}{\sqrt{n}}G(t),$ (2.4)

The singular values $s_{1}(t)\geqslant s_{2}(t)\geqslant\cdots\geqslant
s_{n-1}(t)\geqslant|s_{n}(t)|$ of $H(t)$ satisfies the following stochastic
differential equations

$\displaystyle\begin{split}\mathrm{d}s_{i}(t)&=\frac{\mathrm{d}W_{i}}{\sqrt{\beta
n}}+\left(\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{s_{i}(t)-s_{j}(t)}+\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{s_{i}(t)+s_{j}(t)}+\frac{\alpha_{n}}{2s_{i}(t)}\right)\mathrm{d}t,\quad
1\leqslant i\leqslant n,\end{split}$ (2.5)

where

$\displaystyle\alpha_{n}=\frac{m-n}{n}+\left(1-\frac{1}{\beta}\right)\frac{1}{n},$

and $W_{1},W_{2},\cdots,W_{n}$ are independent Brownian motions. We denote by
$\mathbb{P}$ the law of ${\bf s}(t)=(s_{1}(t),\ldots,s_{n}(t)),0\leqslant
t\leqslant 1$.

The eigenvalues process of
$\lambda_{1}(t)\geqslant\lambda_{2}(t)\geqslant\cdots\geqslant\lambda_{n}(t)$
of $H(t)H^{*}(t)$ has been intensively studied in the literature [13, 12, 21,
22, 42], called the $\beta$-Laguerre process or $\beta$-Wishart process

$\displaystyle\mathrm{d}\lambda_{i}(t)=$ $\displaystyle
2\sqrt{\lambda_{i}}\frac{\mathrm{d}W_{i}(t)}{\sqrt{\beta
n}}+\left(\frac{1}{n}\sum_{j:j\neq
i}\frac{\lambda_{i}(t)+\lambda_{j}(t)}{\lambda_{i}(t)-\lambda_{j}(t)}+\frac{m}{n}\right)\mathrm{d}t,\quad
1\leqslant i\leqslant n,$ (2.6)

where $W_{1},W_{2},\cdots,W_{n}$ are independent Brownian motions. In [42],
the case $\beta=2$ Laguerre process was shown to correspond to squared Bessel
processes conditioned never to collide in the sense of Doob. It is known that
for $\beta\geqslant 1$ and $m\geqslant n$, (2.6) has a unique strong solution
satisfying $\lambda_{1}(t)>\lambda_{2}(t)>\cdots>\lambda_{n}(t)\geqslant 0$
for $t>0$. Then a formal calculation gives that
$s_{i}(t)=\sqrt{\lambda_{i}(t)}$ satisfies (2.5). When $n=1$, $s_{1}$ is a
Bessel process. We call the process (2.5) Dyson Bessel process. The same
argument as in [2, Lemma 4.3.3], we can show that for $\beta\geqslant 1$,
$\alpha_{n}\geqslant 1/\beta n$, and any initial condition $s_{1}(0)\geqslant
s_{2}(0)\geqslant\cdots\geqslant s_{n}(0)\geqslant 0$, the unique strong
solution of (2.5) satisfy $s_{1}(t)>s_{2}(t)>\cdots s_{n}(t)>0$ for $t>0$.
Therefore, $s_{1}(t)>s_{2}(t)>\cdots>s_{n-1}(t)>s_{n}(t)>0$ has the same law
of singular values of $H(t)$. We notice that $\alpha_{n}\geqslant 1/\beta n$
is satisfied for any $m\geqslant n$ and $\beta=2$. In the special case that
$\beta=1$, $m=n$ and $\alpha_{n}=0$, as discussed in [17, Appendix 1],
$s_{n}(t)$ can be negative, and
$s_{1}(t)>s_{2}(t)>\cdots>s_{n-1}(t)>|s_{n}(t)|>0$ has the same law of
singular values of $H(t)$. For our study of Dyson Bessel process, we restrict
ourselves to these two choices of parameters

###  2.2 Change of Measure

In this section, we relate the Dyson Bessel process (2.5) with the Dyson
Brownian motion by a change of measure using Girsanov’s theorem. We recall the
Dyson Brownian motion (DBM) is given for $\beta\geqslant 1$ by

$\displaystyle\begin{split}\mathrm{d}x_{i}(t)&=\frac{\mathrm{d}W_{i}(t)}{\sqrt{\beta
n}}+\frac{1}{2n}\sum_{j:j\neq
i}\frac{\mathrm{d}t}{x_{i}(t)-x_{j}(t)}.\end{split}$ (2.7)

We denote the law of Dyson Brownian motion (2.7) as ${\mathbb{Q}}$.

The Dyson Bessel process (2.5) can be obtained from the DBM (2.7) by a change
of measure using an exponential martingale constructed from the following
function

$\displaystyle\theta(s_{1},s_{2},\cdots,s_{n})=\frac{\beta}{2}\left(\sum_{i<j}\log(s_{i}+s_{j})+\alpha_{n}n\sum\log
s_{k}\right),\quad s_{1}\geqslant s_{2}\geqslant\cdots\geqslant s_{n}.$ (2.8)

The above function $\theta$ has logarithmic singularity when $s_{n}$ is close
to $0$. Fix a small parameter ${\mathfrak{a}}>0$, we define the stopping time
$\tau_{\mathfrak{a}}$, the first time that $s_{n}(t)$ gets too close to $0$,

$\displaystyle\tau_{{\mathfrak{a}}}=\inf\\{t\geqslant
0:s_{n}\leqslant{\mathfrak{a}}\\}.$ (2.9)

Then for $t\leqslant\tau_{\mathfrak{a}}$, we have
$s_{n}(t)\geqslant{\mathfrak{a}}$, and
$\theta(s_{1}(t),s_{2}(t),\cdots,s_{n}(t))$ is bounded below uniformly.

######  Proposition 2.2.

Let ${\cal F}_{t}$ be the $\sigma$ algebra generated by the Brownian motions
$\\{W_{i}(t)\\}$. We take ${\mathbb{Q}}$ the law of DBM

$\displaystyle{\rm d}x_{i}(t)=\frac{{\rm d}W_{i}(t)}{\sqrt{\beta
n}}+\frac{1}{2n}\sum_{j:j\neq i}\frac{{\rm d}t}{x_{i}(t)-x_{j}(t)},\quad
1\leqslant i\leqslant n,$ (2.10)

and ${\mathbb{P}}^{\mathfrak{a}}$ the law of the following modified Dyson
Bessel process

$\displaystyle\mathrm{d}s_{i}(t)=\frac{\mathrm{d}W_{i}(t)}{\sqrt{\beta
n}}+\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{s_{i}(t)-s_{j}(t)}+\bm{1}(t\leqslant\tau_{\mathfrak{a}})\left(\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{s_{i}(t)+s_{j}(t)}+\frac{\alpha_{n}}{2s_{i}(t)}\right)\mathrm{d}t,$

for $1\leqslant i\leqslant n$. Then the two laws ${\mathbb{P}}^{\mathfrak{a}}$
and ${\mathbb{Q}}$ are related by a change of measure

$\displaystyle{\mathbb{P}}^{\mathfrak{a}}=e^{{L_{1\wedge\tau_{\mathfrak{a}}}-\frac{1}{2}\langle
L,L\rangle_{1\wedge\tau_{\mathfrak{a}}}}}{\mathbb{Q}},$

where the exponent is given by

$\displaystyle\begin{split}L_{t\wedge\tau_{\mathfrak{a}}}-\frac{1}{2}\langle
L,L\rangle_{t\wedge\tau_{\mathfrak{a}}}&=\left.\theta(x_{1}({u}),\cdots,x_{n}({u}))\right|_{0}^{t\wedge\tau_{\mathfrak{a}}}-\frac{\beta
n}{2}\int_{0}^{t\wedge\tau_{\mathfrak{a}}}\sum_{i}\frac{\alpha_{n}^{2}}{4x_{i}^{2}(u)}{\rm
d}u\\\
&-\left(\frac{\beta}{2}-1\right)\int_{0}^{t\wedge\tau_{\mathfrak{a}}}\frac{1}{4n}\sum_{k\neq\ell}\frac{{\rm
d}u}{(x_{k}(u)+x_{\ell}(u))^{2}}+\int_{0}^{t\wedge\tau_{\mathfrak{a}}}\frac{\alpha_{n}}{4}\sum_{k}\frac{{\rm
d}u}{x_{k}^{2}(u)}.\end{split}$

######  Remark 2.3.

We remark that ${\mathbb{P}}^{\mathfrak{a}}$ depends on ${\mathfrak{a}}>0$.
For any event $\Omega$ of singular value DBM, we can can lower bound its
probability in the following way

$\displaystyle{\mathbb{P}}(\Omega)\geqslant{\mathbb{P}}(\Omega\cap\\{s_{n}\geqslant{\mathfrak{a}}\\})={\mathbb{P}}^{\mathfrak{a}}(\Omega\cap\\{s_{n}\geqslant{\mathfrak{a}}\\}).$

###### Proof of Proposition 2.2.

The first and second derivatives of $\theta$ are given by

$\displaystyle\begin{split}\partial_{s_{i}}\theta(s_{1},s_{2},\cdots,s_{n})&=\frac{\beta}{2}\left(\sum_{j:j\neq
i}\frac{1}{s_{i}+s_{j}}+\frac{\alpha_{n}n}{s_{i}}\right),\\\
\partial^{2}_{s_{i}}\theta(s_{1},s_{2},\cdots,s_{n})&=-\frac{\beta}{2}\left(\sum_{j:j\neq
i}\frac{1}{(s_{i}+s_{j})^{2}}+\frac{\alpha_{n}n}{s_{i}^{2}}\right),\end{split}$
(2.11)

for $1\leqslant i\leqslant n$. Since $\theta$ is ${\cal C}^{\infty}$ on sets
where it is bounded below, Itô’s lemma gives that if ${\bf
x}(t)=(x_{1}(t),\ldots,x_{n}(t))$,

$\displaystyle{\rm d}\theta({\bf x(t)})$ $\displaystyle={\rm
d}L_{t}+\frac{1}{4n}\sum_{i\neq j}\frac{\partial_{x_{i}}\theta({\bf
x}(t))-\partial_{x_{j}}\theta({\bf
x}(t))}{x_{i}(t)-x_{j}(t)}\mathrm{d}t+\sum_{i}\frac{\partial^{2}_{x_{i}}\theta({\bf
x}(t))}{2\beta n}\mathrm{d}t,$ (2.12)

where the martingale term $L_{t}$ is

$\displaystyle{\rm d}L_{t}=\sum_{i}\partial_{x_{i}}\theta({\bf
x}(t))\frac{{\rm d}W_{i}(t)}{\sqrt{\beta
n}}=\sum_{i}\left(\frac{\sqrt{\beta}}{2\sqrt{n}}\sum_{j:j\neq
i}\frac{1}{x_{i}(t)+x_{j}(t)}+\sqrt{\beta
n}\frac{\alpha_{n}}{2x_{i}(t)}\right){\rm d}W_{i}(t),$

Its quadratic variance is given by

$\displaystyle\langle
L,L\rangle_{t}=\int^{t}_{0}\sum_{i}\left(\frac{\sqrt{\beta}}{2\sqrt{n}}\sum_{j:j\neq
i}\frac{1}{x_{i}(u)+x_{j}(u)}+\sqrt{\beta
n}\frac{\alpha_{n}}{2x_{i}(u)}\right)^{2}{\rm d}u.$

For the second term on the righthand side of (2.12), using (2.11) we have

$\displaystyle\begin{split}&\phantom{{}={}}\frac{1}{4n}\sum_{i\neq
j}\frac{\partial_{x_{i}}\theta({\bf x}(t))-\partial_{x_{j}}\theta({\bf
x}(t))}{x_{i}-x_{j}}=-\frac{\beta}{8n}\sum_{i\neq j\neq
k}\frac{1}{(x_{i}+x_{k})(x_{j}+x_{k})}-\frac{\beta\alpha_{n}}{8}\sum_{i\neq
j}\frac{1}{x_{i}x_{j}}\\\ &=-\frac{\beta}{8n}\sum_{i}\left(\sum_{j:j\neq
i}\frac{1}{(x_{i}+x_{j})}\right)^{2}+\frac{\beta}{8n}\sum_{i\neq
j}\frac{1}{(x_{i}+x_{j})^{2}}-\frac{\beta\alpha_{n}}{8}\left(\sum_{i}\frac{1}{x_{i}}\right)^{2}+\frac{\beta\alpha_{n}}{8}\sum_{i}\frac{1}{x_{i}^{2}}.\end{split}$
(2.13)

For the last term on the righthand side of (2.12), using (2.11)we have

$\displaystyle\sum_{i}\frac{\partial^{2}_{x_{i}}\theta({\bf x}(t))}{2\beta
n}=-\frac{1}{4n}\sum_{i\neq
j}\frac{1}{(x_{i}+x_{j})^{2}}-\frac{\alpha_{n}}{4}\sum_{i}\frac{1}{x_{i}^{2}}.$
(2.14)

By plugging (LABEL:e:term1) and (2.14) back into (2.12), we get

$\displaystyle\begin{split}{\rm d}\theta({\bf x}(t))&={\rm
d}L_{t}-\sum_{i}\frac{\beta}{8n}\left(\sum_{j:j\neq
i}\frac{1}{(x_{i}(t)+x_{j}(t))}\right)^{2}-\frac{\beta\alpha_{n}}{8}\left(\sum_{i}\frac{1}{x_{i}(t)}\right)^{2}\\\
&+\left(\frac{\beta}{2}-1\right)\left(\frac{1}{4n}\sum_{i\neq
j}\frac{1}{(x_{i}(t)+x_{j}(t))^{2}}+\frac{\alpha}{4}\sum_{i}\frac{1}{x_{i}^{2}(t)}\right).\end{split}$
(2.15)

We recall the stopping time $\tau_{\mathfrak{a}}$ from (2.9), then

$\displaystyle\langle L,L\rangle_{t\wedge\tau_{\mathfrak{a}}}$
$\displaystyle=\int^{t\wedge\tau_{\mathfrak{a}}}_{0}\sum_{i}\left(\frac{\sqrt{\beta}}{2\sqrt{n}}\sum_{j:j\neq
i}\frac{1}{x_{i}(u)+x_{j}(u)}+\sqrt{\beta
n}\frac{\alpha_{n}}{2x_{i}(u)}\right)^{2}{\rm d}u$
$\displaystyle\leqslant\int^{t\wedge\tau_{\mathfrak{a}}}_{0}\sum_{i}\left(\frac{\sqrt{\beta}}{2\sqrt{n}}\sum_{j:j\neq
i}\frac{1}{2{\mathfrak{a}}}+\sqrt{\beta
n}\frac{\alpha_{n}}{2{\mathfrak{a}}}\right)^{2}{\rm
d}u\leqslant\left(\frac{\alpha_{n}}{2}+\frac{1}{4}\right)^{2}\frac{\beta
n^{2}{(t\wedge\tau_{\mathfrak{a}})}}{{\mathfrak{a}}^{2}},$

which is uniformly bounded. Therefore, Novikov’s theorem [2, H.10] implies the
following is an exponential martingale

$\displaystyle e^{{L_{t\wedge\tau_{\mathfrak{a}}}-\frac{1}{2}\langle
L,L\rangle_{t\wedge\tau_{\mathfrak{a}}}}}.$

Using (2.15), more explicitly, we can rewrite

$\displaystyle L_{t\wedge\tau_{\mathfrak{a}}}-\frac{1}{2}\langle
L,L\rangle_{t\wedge\tau_{\mathfrak{a}}}$
$\displaystyle=\left.\theta(x_{1}({u}),\cdots,x_{n}({u})\right|_{0}^{t\wedge\tau_{\mathfrak{a}}}-\frac{\beta
n}{2}\int_{0}^{t\wedge\tau_{\mathfrak{a}}}\sum_{i}\frac{\alpha_{n}^{2}}{4x_{i}^{2}(u)}{\rm
d}u$
$\displaystyle-\left(\frac{\beta}{2}-1\right)\int^{t\wedge\tau_{\mathfrak{a}}}\frac{1}{4n}\sum_{k\neq\ell}\frac{{\rm
d}u}{(x_{k}(u)+x_{\ell}(u))^{2}}+\int^{t\wedge\tau_{\mathfrak{a}}}\frac{\alpha_{n}}{4}\sum_{k}\frac{{\rm
d}u}{x_{k}^{2}(u)}.$

We recall that ${\mathbb{Q}}$ is the law of DBM (2.10), and denote the
rescaled Brownian motions $M$,

$\displaystyle
M_{i}(t)=x_{i}(t)-x_{i}(0)-\int_{0}^{t}\frac{1}{2n}\sum_{j:j\neq i}\frac{{\rm
d}u}{x_{i}(u)-x_{j}(u)}=\int_{0}^{t}\frac{{\rm d}W_{i}(u)}{\sqrt{\beta
n}}=\frac{W_{i}(t)}{\sqrt{\beta n}},$

then Girsanov’s theorem [2, Theorem H.11] implies that

$\displaystyle\begin{split}M_{i}(t)-\langle
M_{i},L\rangle_{t\wedge\tau_{\mathfrak{a}}}&=x_{i}(t)-x_{i}(0)-\int_{0}^{t}\frac{1}{2n}\sum_{j:j\neq
i}\frac{{\rm d}u}{x_{i}(u)-x_{j}(u)}\\\
&-\int_{0}^{t\wedge\tau_{\mathfrak{a}}}\left(\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{x_{i}(u)+x_{j}(u)}+\frac{\alpha_{n}}{2x_{i}(u)}\right){\rm
d}u,\end{split}$ (2.16)

are independent Brownian motions under the measure
${\mathbb{P}}^{{\mathfrak{a}}}$:

$\displaystyle{\mathbb{P}}^{\mathfrak{a}}=e^{{L_{1\wedge\tau_{\mathfrak{a}}}-\frac{1}{2}\langle
L,L\rangle_{1\wedge\tau_{\mathfrak{a}}}}}{\mathbb{Q}},$

Therefore, ${\mathbb{P}}^{\mathfrak{a}}$ is the unique solution of the
stochastic differential system

$\displaystyle\mathrm{d}s_{i}(t)=\frac{\mathrm{d}W_{i}(t)}{\sqrt{\beta
n}}+\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{s_{i}(t)-s_{j}(t)}+\bm{1}(t\leqslant\tau_{\mathfrak{a}})\left(\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{s_{i}(t)+s_{j}(t)}+\frac{\alpha_{n}}{2s_{i}(t)}\right)\mathrm{d}t.$

where $W_{1},W_{2},\cdots,W_{n}$ are independent Brownian motions. ∎

##  3 Large deviations for the Dyson Brownian motion

Thanks to Proposition 2.2, the law of singular value Dyson Brownian motion can
be rewritten as a change of measure from the Dyson Brownian motion. The large
deviations principle for Dyson Brownian motion has been proven in [30, 37]
when $\beta=1$ or $2$ and the initial condition has finite $5+\varepsilon$
moment for some $\varepsilon>0$. In this section we give a shorter proof for
the large deviations principle valid for any $\beta\geqslant 1$ and under the
assumption that the initial condition has finite second moment only. The main
technical improvement comes from Propositions 3.3 and 3.4 which allow to prove
the lower bound in greater generality, thanks to better approximation of our
processes by processes with smooth drifts

We denote the empirical particle density of the Dyson Brownian motion (2.10)
as

$\displaystyle\nu^{n}_{t}=\frac{1}{n}\sum_{i=1}^{n}\delta_{x_{i}(t)}.$ (3.1)

######  Assumption 1.

We assume the probability density $\mu_{0}$ has bounded second moment.
Moreover, as $n$ goes to infinite, $\nu_{0}^{n}$ converges to $\mu_{0}$ in
$2$-Wasserstein distance, i.e.
$d(\mu_{0},\nu_{0}^{n})=\operatorname{o}_{n}(1)$.

Given a continuous measure process $\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}$
with $\nu_{0}$ satisfying Assumption 1, we define the following dynamical
entropy:

$\displaystyle\begin{split}S(\\{\nu_{t},f_{t}\\}_{0\leqslant t\leqslant
1})&=\left\\{\nu_{1}(f_{1})-\nu_{0}(f_{0})-\int_{0}^{1}\int\partial_{t}f_{t}(x){\rm
d}\nu_{t}(x){\rm d}t\right.\\\
&\left.-\frac{1}{2}\int_{0}^{1}\int\frac{f_{t}^{\prime}(x)-f_{t}^{\prime}(y)}{x-y}{\rm
d}\nu_{t}(x){\rm d}\nu_{t}(y){\rm
d}t-\frac{1}{2\beta}\int^{1}_{0}\int(f_{t}^{\prime}(x))^{2}{\rm d}\nu_{t}{\rm
d}t\right\\},\end{split}$ (3.2)

where $f_{t}(x)\in{\cal C}_{b}^{2,1}$ has bounded twice derivative in $x$ and
bounded derivative in $t$. For any measure $\mu_{0}$, if $\nu_{0}=\mu_{0}$, we
set

$\displaystyle S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1})=\sup_{f\in{\cal C}^{2,1}}S(\\{\nu_{t},f_{t}\\}_{0\leqslant t\leqslant
1}).$ (3.3)

If $\nu_{0}\neq\mu_{0}$, we set $S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant
t\leqslant 1})=\infty$. In this section we give a new proof of the following
large deviations principle for the empirical particle density of the Dyson
Brownian motion (3.1)

######  Theorem 3.1.

Fix a probability density $\mu_{0}$ and an initial condition with empirical
distribution $\nu_{0}^{n}$ satisfying Assumption 1. Then, the empirical
particle density $\\{\nu^{n}_{t}\\}_{0\leqslant t\leqslant 1}$ of the Dyson
Brownian motion (3.1) satisfies a large deviations principle in the scale
$n^{2}$ and with good rate function $S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant
t\leqslant 1})$. In particular for any continuous measure process
$\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}$, it holds

$\displaystyle\lim_{\delta\rightarrow
0}\limsup_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\nu^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1},\delta))$
$\displaystyle=\lim_{\delta\rightarrow
0}\liminf_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\nu^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1},\delta))=-S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1})\,.$

For any measure valued proces $\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}$ such
that $S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1})<\infty$, by Riesz
representation theorem, there exists a measurable function
$\partial_{x}k_{t}\in L^{2}(\mathrm{d}\nu_{t}(x)\mathrm{d}t)$, such that for
any $f\in{\cal C}_{b}^{2,1}$

$\displaystyle\nu_{1}(f_{1})-\nu_{0}(f_{0})-\int\partial_{t}f_{t}(x){\rm
d}\nu_{t}(x){\rm d}t-\frac{1}{2}\int_{0}^{1}\int
f_{t}^{\prime}(x)H(\nu_{t}){\rm d}\nu_{t}(x){\rm d}t=\int_{0}^{1}\int
f^{\prime}_{t}(x)\partial_{x}k_{t}(x){\rm d}\nu_{t}{\rm d}t.$ (3.4)

Here $H(\nu)$ denotes the Hilbert transform of $\nu$. Then we can rewrite the
rate function $S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1})$ in (3.2)
as

$\displaystyle S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1})=\sup_{f\in{\cal
C}_{b}^{2,1}}\int_{0}^{1}f^{\prime}_{t}(x)\partial_{x}k_{t}(x){\rm
d}\nu_{t}{\rm d}t-\frac{1}{2\beta}\int^{1}_{0}\int(f_{t}^{\prime}(x))^{2}{\rm
d}\nu_{t}{\rm
d}t=\frac{\beta}{2}\int_{0}^{1}\int(\partial_{x}k_{t}(x))^{2}{\rm
d}\nu_{t}{\rm d}t,$ (3.5)

where the equality is achieved when
$f^{\prime}_{t}(x)=\beta\partial_{x}k_{t}(x)$.

We collect some properties of the rate function (3.2), which were essentially
proven in [30, 32, 37].

######  Proposition 3.2.

Fix a probability measure $\mu_{0}$ with finite second moment and bounded free
entropy, i.e. $\Sigma(\mu_{0})>-\infty$. Then, $S_{\mu_{0}}$ is a good rate
function on ${\cal C}([0,1],{\mathbb{M}}_{1}({\mathbb{R}}))$. If
$S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1})<0$, then we have

  1. (i)

There exists a constant ${\mathfrak{C}}$ depending only on $\mu_{0}$ and
$S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1})$, such that the $L_{2}$
norms of $\nu_{t}$ are uniformly bounded,

$\displaystyle\int x^{2}{\rm d}\nu_{t}(x)\leqslant{\mathfrak{C}}.$ (3.6)

  2. (ii)

$\nu_{t}$ has a density for almost surely all $0\leqslant t\leqslant 1$, i.e.

$\displaystyle\frac{{\rm d}\nu_{t}(x)}{{\rm d}x}=\rho_{t}(x).$

  3. (iii)

We denote the velocity field $u_{t}(x)=H(\nu_{t})(x)/2+\partial_{x}k_{t}(x)$,
then it satisfies the conservation of mass equation

$\displaystyle\partial_{t}\rho_{t}+\partial_{x}(\rho_{t}u_{t})=0,\quad
0\leqslant t\leqslant 1,$ (3.7)

in the sense of distribution. We can rewrite the dynamical entropy (3.2) as

$\displaystyle\begin{split}S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1})&=\frac{\beta}{2}\left(\int_{0}^{1}\int(u_{t}^{2}+H(\nu_{t})^{2}/4)\rho_{t}(x){\rm
d}x{\rm d}t-\frac{1}{2}(\Sigma(\nu_{1})-\Sigma(\nu_{0}))\right)\\\
&=\frac{\beta}{2}\left(\int_{0}^{1}\int(u_{t}^{2}+\frac{\pi^{2}}{12}\rho_{t}(x)^{2})\rho_{t}(x){\rm
d}x{\rm d}t-\frac{1}{2}(\Sigma(\nu_{1})-\Sigma(\nu_{0}))\right).\end{split}$
(3.8)

###### Proof.

It is proven in [30, Theorem 1.4] that $S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant
t\leqslant 1})$ is a good rate function. If
$S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1})<\infty$, by definition
we have $\mu_{0}=\nu_{0}$. For Item (i), we take a test function
$f_{\varepsilon}(x)=x^{2}/(1+\varepsilon x^{2})$ with small $\varepsilon>0$.
Then it is easy to see that $f_{\varepsilon}^{\prime}(x)=2x/(1+\varepsilon
x^{2})^{2}$ and $|f_{\varepsilon}^{\prime\prime}(x)|\leqslant 10$. By the
definition of the dynamical free entropy (3.2), for any $0<t\leqslant 1$, we
have

$\displaystyle\begin{split}&\phantom{{}={}}\nu_{t}(f_{\varepsilon})-\nu_{0}(f_{\varepsilon})-\frac{1}{4}\int_{0}^{t}\int\frac{f_{\varepsilon}^{\prime}(x)-f_{\varepsilon}^{\prime}(y)}{y-x}{\rm
d}\nu_{s}(y){\rm d}\nu_{s}(x){\rm
d}s-\frac{1}{2\beta}\int_{0}^{t}\int(f_{\varepsilon}^{\prime}(x))^{2}{\rm
d}\nu_{s}{\rm d}s\\\ &\leqslant S_{\mu_{0}}(\\{\nu_{s}\\}_{0\leqslant
s\leqslant 1})<0.\end{split}$ (3.9)

By our assumption that $\nu_{0}=\mu_{0}$ has finite second moment, it holds
that $\sup_{\varepsilon}\nu_{0}(f_{\varepsilon})<\infty$. Using
$|f_{\varepsilon}^{\prime\prime}(x)|\leqslant 10$, we find for $t\leqslant 1$,

$\displaystyle\left|\frac{1}{4}\int_{0}^{t}\int\frac{f_{\varepsilon}^{\prime}(x)-f_{\varepsilon}^{\prime}(y)}{y-x}{\rm
d}\nu_{s}(y){\rm d}\nu_{s}(x){\rm d}s\right|\leqslant 5/2.$

Therefore, there exists a constant ${\mathfrak{C}}$ depending only on
$\mu_{0}$ and $S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1})$, such
that

$\displaystyle\nu_{t}(f_{\varepsilon})$
$\displaystyle=\int\frac{x^{2}}{1+\varepsilon x^{2}}{\rm
d}\nu_{t}\leqslant{\mathfrak{C}}+\frac{1}{2\beta}\int_{0}^{t}\int(f_{\varepsilon}^{\prime}(x))^{2}{\rm
d}\nu_{s}{\rm d}s$
$\displaystyle\leqslant{\mathfrak{C}}+\frac{2}{\beta}\int_{0}^{t}\int\frac{x^{2}}{1+\varepsilon
x^{2}}{\rm d}\nu_{s}{\rm
d}s={\mathfrak{C}}+\frac{2}{\beta}\int_{0}^{t}\nu_{s}(f_{\varepsilon}){\rm
d}s.$

Grönwall’s inequality then implies that for all $t\leqslant 1$

$\displaystyle\nu_{t}(f_{\varepsilon})\leqslant e^{2/\beta}{\mathfrak{C}}.$

The claim 3.6 follows by sending $\varepsilon$ to $0$ and monotone convergence
theorem.

It was proven in [32, Theorem 2.1] and [37, Theorem 3.3] that if
$\mu_{0}=\nu_{0}$ has bounded $5+\varepsilon$ moments, i.e.

$\displaystyle\int|x|^{5+\varepsilon}{\rm d}\nu_{0}<\infty,$ (3.10)

and $\Sigma(\mu_{0}),\Sigma(\mu_{1})$ are finite, then Item (ii) and (iii)
hold. This can be extended to the case where $\mu_{0}$ has only a finite
second moment following the arguments of the proof of [15, Lemma 5.9]. We
briefly recall the main steps of the proof. First recall that free convolution
reduces the dynamical entropy (see [15]) so that if $\sigma_{\varepsilon}$
denotes the semi-circle law with covariance $\varepsilon$

$\displaystyle
S_{\mu_{0}\boxplus\sigma_{\varepsilon}}(\\{\nu_{t}\boxplus\sigma_{\varepsilon}\\}_{0\leqslant
t\leqslant 1})\leqslant S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}).$

But on the other hand, $H(\nu_{t}\boxplus\sigma_{\varepsilon})$ is uniformly
bounded by $1/\sqrt{\varepsilon}$. Therefore if we denote by $u^{\varepsilon}$
the velocity field of
$\nu_{t}^{\varepsilon}=\nu_{t}\boxplus\sigma_{\varepsilon}$,

$\int_{0}^{1}\int(u^{\varepsilon}_{t})^{2}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t\leqslant
2\int_{0}^{1}\int(u^{\varepsilon}_{t}-H\nu_{t}^{\varepsilon})^{2}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t+2\int_{0}^{1}\int(H\nu_{t}^{\varepsilon})^{2}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t\leqslant\frac{4}{\beta}S_{\mu_{0}\boxplus\sigma_{\varepsilon}}(\\{\nu_{t}^{\varepsilon}\\}_{0\leqslant
t\leqslant 1})+\frac{2}{\varepsilon}<\infty\,.$

Hence, we can write

$S_{\mu_{0}\boxplus\sigma_{\varepsilon}}(\\{\nu_{t}^{\varepsilon}\\}_{0\leqslant
t\leqslant
1})=\frac{\beta}{2}\int_{0}^{1}\int(u^{\varepsilon}_{t})^{2}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t+\frac{\beta}{2}\int_{0}^{1}\int(H\nu^{\varepsilon}_{t})^{2}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t-\beta\int_{0}^{1}\int
H\nu^{\varepsilon}_{t}u^{\varepsilon}_{t}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t\,.$
(3.11)

For the second term we used the well known formula (recall that
$d\nu^{\varepsilon}_{t}\ll\mathrm{d}x$)

$\int_{0}^{1}\int(H\nu^{\varepsilon}_{t})^{2}\mathrm{d}\nu_{t}^{\varepsilon}=\frac{\pi^{2}}{3}\int_{0}^{1}\int(\frac{d\nu^{\varepsilon}_{t}}{\mathrm{d}x})^{3}\mathrm{d}x\,.$

Finally for the last term of (3.11), we observe following [15, Lemma 5.9] that
the continuity of $t\mapsto\nu_{t}$ implies that $t\mapsto
H\nu^{\varepsilon}_{t}$ is continuous (thanks to the explicit formulas for the
Hilbert transform of measures freely convoluted with the semi-circle laws
given by Biane [8]). Since it is bounded and $u^{\varepsilon}$ is in $L^{2}$,
we see that we can approximate the last term by Riemann sum. Then, recall that
by definition we have

$\int^{t}u^{\varepsilon}_{s}\frac{\mathrm{d}\nu_{s}^{\varepsilon}}{\mathrm{d}x}\mathrm{d}s=-\int^{x}\mathrm{d}\nu^{\varepsilon}_{t},$

to conclude that

$\int_{0}^{1}\int
H\nu^{\varepsilon}_{t}u^{\varepsilon}_{t}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t=\frac{1}{2}\int_{0}^{1}\partial_{t}\Sigma(\nu^{\varepsilon}_{t})\mathrm{d}t=\frac{1}{2}\left(\Sigma(\nu_{1}\boxplus\sigma_{\varepsilon})-\Sigma(\mu_{0}\boxplus\sigma_{\varepsilon})\right)\,.$

Hence, (3.8) holds for $\\{\nu^{\varepsilon}_{t}\\}_{0\leqslant t\leqslant
1}$. This implies that $\Sigma(\nu_{1}\boxplus\sigma_{\varepsilon})$ is
bounded since it is bounded from above as
$\nu_{1}\boxplus\sigma_{\varepsilon}$ has bounded second moment and also from
below since

$\frac{\beta}{2}\int_{0}^{1}\int(u^{\varepsilon}_{t})^{2}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t+\frac{\beta}{2}\int_{0}^{1}\int(H\nu^{\varepsilon}_{t})^{2}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t-\frac{\beta}{2}(\Sigma(\nu_{1}\boxplus\sigma_{\varepsilon})-\Sigma(\nu_{0}))\leqslant
S_{\mu_{0}\boxplus\sigma_{\varepsilon}}(\\{\nu_{t}^{\varepsilon}\\}_{0\leqslant
t\leqslant 1})\leqslant S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1})\,.$

In fact, because we could have done the same reasoning on the time interval
$[0,t]$, we also see that for all $s\leqslant 1$

$S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1})\geqslant\frac{\beta}{2}\int_{0}^{s}\int(u^{\varepsilon}_{t})^{2}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t+\frac{\beta}{2}\int_{0}^{s}\int(H\nu^{\varepsilon}_{t})^{2}\mathrm{d}\nu_{t}^{\varepsilon}\mathrm{d}t-\frac{\beta}{2}(\Sigma(\nu_{s}\boxplus\sigma_{\varepsilon})-\Sigma(\nu_{0})),$

which implies that $\Sigma(\nu_{s}\boxplus\sigma_{\varepsilon})$ is uniformly
bounded. We can finally let $\varepsilon$ going to zero to conclude. As a
consequence of (3.6), we deduce that $\nu_{t}$ has finite free entropy, i.e.
$\Sigma(\nu_{t})<+\infty$. We refer the reader to [15] for details.

∎

###  3.1 Large deviations upper bound

In this section, we prove the large deviations upper bound. We recall that the
exponential tightness was already proven in this setting in the proof of [30,
Theorem 2.4]: for the sake of completeness we will recall this proof but in
the new setting of the Bessel Dyson processes, see section 4. We next prove
the large deviations upper bound of Theorem 3.1

$\displaystyle\limsup_{\delta\rightarrow
0}\limsup_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\nu^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1},\delta))\leqslant-S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}).$
(3.12)

Take any test function $f_{t}(x)\in{\cal
C}_{b}^{2,1}([0,1]\times{\mathbb{R}})$, and use Itô’s lemma to find that

$\displaystyle\begin{split}&{\rm
d}\sum_{i}f_{t}(x_{i}(t))=\sum_{i}f^{\prime}_{t}(x_{i}(t)){\rm
d}x_{i}(t)+\sum_{i}\left(\frac{f^{\prime\prime}_{t}(x_{i}(t))}{2\beta
n}+\partial_{t}f_{t}(x_{i}(t))\right){\rm d}t\\\
&=\sum_{i}\left(\frac{f^{\prime\prime}_{t}(x_{i}(t))}{2\beta
n}+\partial_{t}f_{t}(x_{i}(t))+\frac{f^{\prime}_{t}(x_{i}(t))}{2n}\sum_{j:j\neq
i}\frac{1}{x_{i}(t)-x_{j}(t)}\right)\mathrm{d}t+\sum_{i}f^{\prime}_{t}(x_{i}(t))\frac{{\rm
d}W_{i}(t)}{\sqrt{\beta n}}\\\ &={\rm d}L^{f}_{t}+\frac{1}{4n}\sum_{i\neq
j}\frac{f^{\prime}_{t}(x_{i}(t))-f^{\prime}_{t}(x_{j}(t))}{x_{i}(t)-x_{j}(t)}\mathrm{d}t+\sum_{i}\frac{f^{\prime\prime}_{t}(x_{i}(t))}{2\beta
n}{\rm d}t+\sum_{i}\partial_{t}f_{t}(x_{i}(t)){\rm d}t,\end{split}$ (3.13)

where the martingale term is given by

$\displaystyle{\rm d}L^{f}_{t}=\sum_{i}f^{\prime}_{t}(x_{i}(t))\frac{{\rm
d}W_{i}(t)}{\sqrt{\beta n}},\quad\langle L^{f},L^{f}\rangle_{t}=\frac{1}{\beta
n}\int^{t}_{0}\sum_{i}(f_{t}^{\prime}(x_{i}(t)))^{2}{\rm d}t.$ (3.14)

We recall the empirical particle density $\\{\nu^{n}_{t}\\}_{0\leqslant
t\leqslant 1}$ from (3.1). With it, we can rewrite (LABEL:e:lin0) as

$\displaystyle\begin{split}&\phantom{{}={}}\int f_{t}(x){\rm
d}\nu^{n}_{t}-\int f_{t}(x){\rm d}\nu^{n}_{0}\\\
&=\frac{L^{f}_{t}}{n}+\frac{1}{4}\int_{0}^{t}\int_{x\neq
y}\frac{f_{s}^{\prime}(x)-f_{s}^{\prime}(y)}{x-y}{\rm d}\nu^{n}_{s}(x){\rm
d}\nu^{n}_{s}(y){\rm d}s+\frac{1}{2\beta n}\int_{0}^{t}\int
f^{\prime\prime}_{s}(x){\rm d}\nu^{n}_{s}(x){\rm
d}s+\int_{0}^{t}\int\partial_{s}f_{s}(x){\rm d}\nu^{n}_{s}(x){\rm d}s\\\
&=\frac{L^{f}_{t}}{n}+\int_{0}^{t}\left(\frac{1}{4}\int\frac{f_{s}^{\prime}(x)-f_{s}^{\prime}(y)}{x-y}{\rm
d}\nu^{n}_{s}(x){\rm d}\nu^{n}_{s}(y)+\int\partial_{s}f_{s}(x){\rm
d}\nu^{n}_{s}(x)+\frac{1}{n}\left(\frac{1}{2\beta}-\frac{1}{4}\right)\int
f^{\prime\prime}_{t}(x){\rm d}\nu^{n}_{s}(x)\right){\rm d}s.\end{split}$
(3.15)

As $L^{f}$ is bounded uniformly for $f\in C^{2,1}_{b}$, we can construct an
exponential martingale using the martingale ${\rm d}L^{f}_{t}$ from (3.14)

$\displaystyle D_{t}=e^{nL^{f}_{t}-\frac{n^{2}}{2}\langle
L^{f},L^{f}\rangle_{t}},\quad{\mathbb{E}}[D_{t}]={\mathbb{E}}[D_{0}]=1.$
(3.16)

Using (LABEL:e:df220) we can rewrite

$nL^{f}_{t}-\frac{n^{2}}{2}\langle
L^{f},L^{f}\rangle_{t}=n^{2}S^{n}_{t}(\\{\nu_{t}^{n},f_{t}\\}_{0\leqslant
t\leqslant 1}),$

where

$\displaystyle S^{n}_{t}(\\{\nu^{n}_{s},f_{s}\\}_{0\leqslant s\leqslant t})$
$\displaystyle=\int f_{t}{\rm d}\nu^{n}_{t}-\int f_{0}{\rm
d}\nu^{n}_{0}-\frac{1}{4}\int_{0}^{t}\int\frac{f_{s}^{\prime}(x)-f_{s}^{\prime}(y)}{x-y}{\rm
d}\nu^{n}_{s}(x){\rm d}\nu^{n}_{s}(y){\rm
d}s-\int_{0}^{t}\int\partial_{s}f_{s}(x){\rm d}\nu^{n}_{s}(x){\rm d}s$
$\displaystyle-\int_{0}^{t}\int\frac{1}{n}\left(\frac{1}{2\beta}+\frac{1}{4}\right)f^{\prime\prime}_{t}(x){\rm
d}\nu^{n}_{s}{\rm
d}s-\frac{1}{2\beta}\int^{t}_{0}\int(f_{t}^{\prime}(x))^{2}{\rm
d}\nu^{n}_{s}{\rm d}s.$

We also define

$\displaystyle S^{n}(\\{\nu_{t}^{n},f_{t}\\}_{0\leqslant t\leqslant 1})$
$\displaystyle=\frac{1}{n^{2}}\left(\frac{nL^{f}_{1}}{2}-\frac{n^{2}}{2}\langle
L^{f},L^{f}\rangle_{1}\right)=S^{n}_{1}(\\{\nu_{t}^{n},f_{t}\\}_{0\leqslant
t\leqslant 1}).$ (3.17)

Then for $\\{\nu^{n}_{t}\\}_{0\leqslant t\leqslant
1}\in{\mathbb{B}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1},\delta)$, we have by
uniform (in $n\geqslant 1$) continuity of $\nu\mapsto
S^{n}(\\{\nu_{t},f_{t}\\}_{0\leqslant t\leqslant 1})$ for any $f\in{\cal
C}_{b}^{2,1}([0,1]\times{\mathbb{R}})$,

$\displaystyle S^{n}(\\{\nu_{t}^{n},f_{t}\\}_{0\leqslant t\leqslant
1})=S^{n}(\\{\nu_{t},f_{t}\\}_{0\leqslant t\leqslant
1})+\operatorname{o}_{n}(1).$

We can use the exponential martingale (3.16) to obtain the large deviations
upper bound as follows.

$\displaystyle\begin{split}&\phantom{{}={}}{\mathbb{P}}(\\{\nu^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1},\delta))={\mathbb{E}}\left[\bm{1}(\\{\nu^{n}_{t}\\}_{0\leqslant t\leqslant
1}\in{\mathbb{B}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1},\delta))\frac{e^{n^{2}S^{n}(\\{\nu_{t}^{n},f_{t}\\}_{0\leqslant t\leqslant
1})}}{e^{n^{2}S^{n}(\\{\nu_{t}^{n},f_{t}\\}_{0\leqslant t\leqslant
1})}}\right]\\\ &={\mathbb{E}}\left[\bm{1}(\\{\nu^{n}_{s}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\nu_{s}\\}_{0\leqslant t\leqslant
1},\delta))e^{n^{2}S^{n}(\\{\nu_{t}^{n},f_{t}\\}_{0\leqslant t\leqslant
1})}\right]\frac{e^{\operatorname{o}(n^{2})}}{e^{n^{2}S^{n}(\\{\nu_{t},f_{t}\\}_{0\leqslant
t\leqslant 1})}}\\\
&\leqslant{\mathbb{E}}\left[e^{n^{2}S^{n}(\\{\nu_{t}^{n},f_{t}\\}_{0\leqslant
t\leqslant
1})}\right]\frac{e^{\operatorname{o}(n^{2})}}{e^{n^{2}S^{n}(\\{\nu_{t},f_{t}\\}_{0\leqslant
t\leqslant 1})}}=e^{-n^{2}S^{n}(\\{\nu_{t},f_{t}\\}_{0\leqslant t\leqslant
1})+\operatorname{o}_{n}(1))}.\end{split}$ (3.18)

The large deviations upper bound (3.12) follows from rearranging
(LABEL:e:upp), and taking the infimum over $f\in{\cal C}^{2,1}_{b}$.

###  3.2 Large deviations Lower Bound

In the rest of this section, we prove the large deviations lower bound of
Theorem 3.1, namely we show that for any continuous measure-valued process
$\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}$, we have

$\displaystyle\liminf_{\delta\rightarrow
0}\liminf_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\nu^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1},\delta))\geqslant-S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}).$
(3.19)

The proof itself will be used to derive the large deviations for Dyson Bessel
processes as it allows to control the positions of the extreme particules, see
Proposition 3.5, key to control the singularity at the origin of the Dyson
Bessel process.

The proof consists of two steps, in the first step we approximate
$\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}$ by a sequence of measure-valued
process with benign properties.

######  Proposition 3.3.

Fix a probability measure $\mu_{0}$ with finite second moment. Then, any
measure-valued process $\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}$ with
$S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1})<\infty$ can be
approximated by a sequence of measure-valued processes
$\\{\nu^{\varepsilon}_{t}\\}_{0\leqslant t\leqslant 1}$ satisfying

  * •

$\nu_{t}^{\varepsilon}$ has uniformly bounded density
$\rho_{t}^{\varepsilon}$, $\operatorname{supp}(\nu_{t}^{\varepsilon})$ is a
single interval for all times $t\in[0,1]$, and

$\displaystyle\lim_{\varepsilon\rightarrow 0}\sup_{0\leqslant t\leqslant
1}d(\nu_{t},\nu^{\varepsilon}_{t})=0.$

  * •

The dynamical entropy satisfies

$\displaystyle\lim_{\varepsilon\rightarrow
0}S_{\nu_{0}^{\varepsilon}}(\\{\nu_{t}^{\varepsilon}\\}_{0\leqslant t\leqslant
1})=S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}).$ (3.20)

  * •

The density $\\{\rho^{\varepsilon}_{t}(x)\\}_{0\leqslant t\leqslant 1}$ of the
measure-valued process $\\{\nu^{\varepsilon}_{t}\\}_{0\leqslant t\leqslant 1}$
is smooth in both $x,t$, and the corresponding drift
$\partial_{x}k^{\varepsilon}_{t}(x)$ as defined by

$\displaystyle\partial_{t}\rho^{\varepsilon}_{t}+\partial_{x}(\rho^{\varepsilon}_{t}u^{\varepsilon}_{t})=0,\quad
u_{t}^{\varepsilon}(x)=\frac{1}{2}H(\nu_{t}^{\varepsilon})(x)+\partial_{x}k_{t}^{\varepsilon}(x),$

is also smooth in both $x,t$.

Above, smooth means differentiable and with continuous derivative (we shall
not need more the proof yields eventually the existence of more derivatives).

######  Proposition 3.4.

Fix a probability measure $\mu_{0}$ satisfying Assumption 1 and $\delta>0$.
Let $\tilde{\mu}_{0}$ be a compactly supported probability measure such that
$d(\mu_{0},\tilde{\mu}_{0})\leqslant\delta/3$. Let
$\\{\tilde{\nu}_{t}(x)\\}_{0\leqslant t\leqslant 1}$ be a compactly supported
measure-valued process with a smooth density $\tilde{\rho}$ in both $x,t$ such
that $\tilde{\nu}_{0}=\tilde{\mu}_{0}$. Assume that the corresponding drift
$\partial_{x}\tilde{k}_{t}$ defined by

$\displaystyle\partial_{t}\tilde{\rho}_{t}(x)+\partial_{x}(\tilde{\rho}_{t}\tilde{u}_{t})=0,\quad\tilde{u}_{t}(x)=\frac{1}{2}H(\tilde{\nu}_{t})(x)+\partial_{x}\tilde{k}_{t}(x),$
(3.21)

is also smooth in both $x,t$. Then, the following large deviations lower bound
holds

$\displaystyle\begin{split}&\liminf_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\nu^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\tilde{\nu}_{t}\\}_{0\leqslant t\leqslant
1},\delta))\geqslant-S_{\tilde{\mu}_{0}}(\\{\tilde{\nu}_{t}\\}_{0\leqslant
t\leqslant 1})+\operatorname{o}_{\delta}(1).\end{split}$

###### Proof of Large deviations lower bound (3.19).

We approximate $\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}$ by
$\\{\nu^{\varepsilon}_{t}\\}_{0\leqslant t\leqslant 1}$ as in Proposition 3.3
and take $\\{\tilde{\nu}_{t}(x)\\}_{0\leqslant t\leqslant 1}$ equal
$\\{\nu^{\varepsilon}_{t}\\}_{0\leqslant t\leqslant 1}$ in Proposition 3.4
with sufficiently small $\varepsilon$. Then it follows

$\displaystyle\begin{split}&\phantom{{}={}}\lim_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\nu^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1},\delta))\geqslant\lim_{\varepsilon\rightarrow
0}\lim_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\nu^{n}_{t}\\}\in{\mathbb{B}}(\\{\nu^{\varepsilon}_{t}\\}_{0\leqslant
t\leqslant 1},\delta/2))\\\ &\geqslant\lim_{\varepsilon\rightarrow
0}-S_{\nu_{0}^{\varepsilon}}(\\{\nu_{t}^{\varepsilon}\\}_{0\leqslant
t\leqslant 1}))+\operatorname{o}_{\delta}(1)\geqslant-
S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1}))+\operatorname{o}_{\delta}(1),\end{split}$

where in the last inequality we used (3.20). The large deviations lower bound
follows by taking $\delta\rightarrow 0$. ∎

###### Proof of Proposition 3.3.

We fix three parameters
$\varepsilon_{3}\ll\varepsilon_{2}\ll\varepsilon_{1}\ll 1$. The construction
of $\nu_{t}^{\varepsilon}$ consists of the following three steps. Note that
$S$ is lower semi-continuous hence we only need to show that

$\limsup_{\varepsilon\rightarrow
0}S_{\nu^{\varepsilon}_{0}}(\\{\nu_{t}^{\varepsilon}\\}_{0\leqslant t\leqslant
1})\leqslant S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}).$

_Step 1 (Free Convolution)._ We replace $\nu_{t}$ by
$\nu_{t}^{(1)}=\nu_{t}\boxplus\sigma_{\varepsilon_{1}}$, its free convolution
with a small semi-circle distribution of size $\varepsilon_{1}$. Then we have
$d(\nu_{t},\nu_{t}^{(1)})=\operatorname{o}_{\varepsilon_{1}}(1)$. More
importantly, $\nu_{t}^{(1)}(dx)=\rho_{t}^{(1)}(x)\mathrm{d}x$ has density
bounded by $\operatorname{O}(1/\sqrt{{\varepsilon_{1}}})$, and it is proven in
[15] that

$\displaystyle S_{\nu_{0}^{(1)}}(\\{\nu_{t}^{(1)}\\}_{0\leqslant t\leqslant
1})\leqslant S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}).$ (3.22)

By our assumption $S_{\mu_{0}}(\\{\nu_{t}\\}_{0\leqslant t\leqslant
1})<\infty$, Proposition 3.2 implies that $\nu_{1}$ has bounded second moment
and finite free entropy $-\infty<\Sigma(\nu_{1})<\infty$. The same bound holds
for its free convolution with semi-circle distribution, i.e.
$-\infty<\Sigma(\nu_{1}^{(1)})<\infty$. Moreover, the second moments of
$\rho^{(1)}$ and $u^{(1)}$ under $\rho^{(1)}_{t}(x)\mathrm{d}x\mathrm{d}t$ are
bounded independently of $\varepsilon_{1}$ by Proposition 3.2 and (3.22).

_Step 2 (Truncation)._ If $\\{\nu^{(1)}_{t}\\}_{0\leqslant t\leqslant 1}$ is
not compactly supported, in this step we truncate it to have compact support.
Let ${\mathfrak{a}}(t),{\mathfrak{b}}(t)$ be such that

$\displaystyle\int_{\infty}^{{\mathfrak{a}}(t)}\rho_{t}^{(1)}(x){\rm
d}x={\varepsilon_{1}}/2,\quad\int^{\infty}_{{\mathfrak{b}}(t)}\rho_{t}^{(1)}(x){\rm
d}x={\varepsilon_{1}}/2\,.$

Observe that because $t\rightarrow\nu^{(1)}_{t}$ is weakly continuous and with
bounded density, $t\rightarrow{\mathfrak{a}}(t)$ and
$t\rightarrow{\mathfrak{b}}(t)$ are continuous. Moreover, because the second
moments of $\nu^{(1)}_{t}$ are uniformly bounded we see that
${\mathfrak{a}}(t)$ and ${\mathfrak{b}}(t)$ are at most of order
$1/\sqrt{\varepsilon_{1}}$. Then we restrict $\rho^{(1)}_{t}$ on
$[{\mathfrak{a}}(t),{\mathfrak{b}}(t)]$ by setting

$\displaystyle\rho^{(2)}_{t}=\frac{\bm{1}([{\mathfrak{a}}(t),{\mathfrak{b}}(t)])}{1-{\varepsilon_{1}}}\rho^{(1)}_{t}(x).$

and let $\nu^{(2)}_{t}(\mathrm{d}x)=\rho^{(2)}_{t}(x)\mathrm{d}x$. From the
construction, we have
$d(\rho^{(1)}_{t},\rho_{t}^{(2)})=\operatorname{o}_{\varepsilon_{1}}(1)$, and
$\rho_{t}^{(2)}$ has bounded $L_{2}$ norm. The corresponding $u^{(2)}_{t}(x)$
is the restriction of $u^{(1)}_{t}(x)$ to
$[{\mathfrak{a}}(t),{\mathfrak{b}}(t)]$. Hence, we get

$\displaystyle S_{\nu_{0}^{(2)}}(\\{\nu_{t}^{(2)}\\}_{0\leqslant t\leqslant
1})=\frac{\beta}{2}\left(\int_{0}^{1}\int((u_{t}^{(2)})^{2}+\frac{\pi^{2}}{12}(\rho^{(2)}_{t}(x))^{2})\rho^{(2)}_{t}(x){\rm
d}x{\rm d}t-\frac{1}{2}(\Sigma(\nu^{(2)}_{1})-\Sigma(\nu^{(2)}_{0}))\right)$
$\displaystyle=\frac{\beta}{2}\left(\int_{0}^{1}\frac{1}{1-\varepsilon_{1}}\int_{{\mathfrak{a}}(t)}^{{\mathfrak{b}}(t)}((u_{t}^{(1)})^{2}+\frac{\pi^{2}}{12}(\frac{\rho^{(1)}_{t}(x)}{1-\varepsilon_{1}})^{2})\rho^{(1)}_{t}(x){\rm
d}x{\rm d}t-\frac{1}{2}(\Sigma(\nu^{(2)}_{1})-\Sigma(\nu^{(2)}_{0}))\right)$
(3.23)
$\displaystyle\rightarrow\frac{\beta}{2}\left(\int_{0}^{1}\int((u_{t}^{(1)})^{2}+\frac{\pi^{2}}{12}(\rho^{(1)}_{t}(x))^{2})\rho^{(1)}_{t}(x){\rm
d}x{\rm
d}t-\frac{1}{2}(\Sigma(\nu^{(1)}_{1})-\Sigma(\nu^{(1)}_{0}))\right)=S_{\nu_{0}^{(1)}}(\\{\nu_{t}^{(1)}\\}_{0\leqslant
t\leqslant 1}),$

as ${\varepsilon_{1}}\rightarrow 0$ by monotone convergence theorem.

_Step 3. (Smoothing)_ We first extend $\rho_{t}^{(2)}$ by $\rho_{0}^{(2)}$ for
$t\leqslant 0$ and $\rho^{(2)}_{1}$ for $t\geqslant 1$. Then we replace it by
its convolution with a bump function $\varphi_{\varepsilon_{3}}$ on the scale
$\varepsilon_{3}$, i.e a smooth function with $L^{1}$ norm equal to one,
supported on $[-\varepsilon_{3},\varepsilon_{3}]^{2}$, with
$\varepsilon_{3}\ll\varepsilon_{1}$:

$\displaystyle\rho_{t}^{(3)}(x)=\int\rho^{(2)}_{t-s}(x-y)\varphi_{\varepsilon_{3}}(y,s){\rm
d}y{\rm d}s.$

Then $\nu^{(3)}(\mathrm{d}x)=\rho_{t}^{(3)}(x)\mathrm{d}x$ is supported on
$[{\mathfrak{a}}(t)-\varepsilon_{3},{\mathfrak{b}}(t)+\varepsilon_{3}]$. Next
we replace this smoothed density by its average with a smooth characteristic
function $\chi_{t}(x)$ constructed in the following way. Take a smoothed step
function $\phi(x)$ such that $\phi(x)=0$ for $x\leqslant 0$, $\phi(x)=1$ for
$x\geqslant{\varepsilon_{1}}$. Let
$\tilde{\mathfrak{a}}(t),\tilde{\mathfrak{b}}(t)$ be two smooth functions such
that for all times
$[{\mathfrak{a}}(t)-\varepsilon_{3},{\mathfrak{b}}(t)+\varepsilon_{3}]\subset[\tilde{\mathfrak{a}}(t),\tilde{\mathfrak{b}}(t)]$.
Then we let

$\displaystyle\chi_{t}(x)=\left\\{\begin{array}[]{cc}\phi(x-\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}}),&x\in[\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}},\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}],\\\
\Omega(1),&x\in[\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}},\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}],\\\
\phi(\tilde{\mathfrak{b}}(t)+2{\varepsilon_{1}}-x),&x\in[\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}},\tilde{\mathfrak{b}}(t)+2{\varepsilon_{1}}].\end{array}\right.$

Moreover, we construct $\chi_{t}(x)$ such that $\int\chi_{t}(x){\rm d}x=1$ for
all $0\leqslant t\leqslant 1$. Because ${\mathfrak{a}}(t),{\mathfrak{b}}(t)$
are at most of order $1/\sqrt{\varepsilon_{1}}$, we can choose
$\tilde{\mathfrak{a}}(t)$ and $\tilde{\mathfrak{b}}(t)$ such that $\chi_{t}$
is lower bounded by $\operatorname{O}(\sqrt{\varepsilon_{1}})$ on
$[\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}},\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}]$.
We can also make sure that it is upper bounded by one, and such that
$\int^{x}\partial_{t}\chi_{t}$ is uniformly bounded. We replace
$\rho_{t}^{(2)}$ by

$\displaystyle\rho_{t}^{\varepsilon}(x)={\varepsilon_{2}}\chi_{t}(x)+(1-{\varepsilon_{2}})\rho_{t}^{(3)}(x).$

Then $\rho_{t}^{\varepsilon}(x)$ is smooth, and for
$x\in[\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}},\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}]$,
$\rho_{t}^{\varepsilon}$ is uniformly lower bounded by
$\Omega({\varepsilon_{2}}\sqrt{\varepsilon_{1}})$. For
$x\in[\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}},\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}]$,
$\rho_{t}^{\varepsilon}(x)={\varepsilon_{2}}\phi(x-\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}})$
and
$x\in[\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}},\tilde{\mathfrak{b}}(t)+2{\varepsilon_{1}}]$,
$\rho_{t}^{\varepsilon}(x)={\varepsilon_{2}}\phi(\tilde{\mathfrak{b}}(t)+2{\varepsilon_{1}}-x)$.
From the construction, we have that
$d(\rho^{\varepsilon}_{t},\rho_{t}^{(2)})=\operatorname{o}_{\varepsilon_{1}}(1)$.

For the second term in the dynamical entropy (3.8), by Young’s convolution
inequality,

$\displaystyle\begin{split}\int(\rho_{t}^{\varepsilon}(x))^{3}{\rm d}x{\rm
d}t&=\int\left({\varepsilon_{2}}\chi_{t}(x)+(1-{\varepsilon_{2}})\rho_{t}^{(3)}(x)\right)^{3}{\rm
d}x{\rm d}t\\\
&=(1+\operatorname{O}({\varepsilon_{2}}))\int\left(\int\rho^{(2)}_{t-s}(x-y)\varphi_{\varepsilon_{3}}(y,s){\rm
d}y{\rm d}s\right)^{3}{\rm d}x{\rm d}t+\operatorname{O}({\varepsilon_{2}})\\\
&\leqslant(1+\operatorname{O}({\varepsilon_{2}}))\int(\rho_{t}^{(2)}(x))^{3}{\rm
d}x{\rm d}t+\operatorname{O}({\varepsilon_{2}}),\end{split}$ (3.24)

where we used that $\chi_{t}$ was uniformly bounded above and with bounded
expectation. In the following we study the first term in (3.8),

$\displaystyle\int_{0}^{1}\int\frac{\left(\partial_{t}\int^{x}\rho^{\varepsilon}_{t}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm
d}t=\int_{0}^{1}\int\frac{\left(\partial_{t}\int^{x}({\varepsilon_{2}}\chi_{t}(y)+(1-{\varepsilon_{2}})\rho_{t}^{(3)}(y)){\rm
d}y\right)^{2}}{{\varepsilon_{2}}\chi_{t}(x)+(1-{\varepsilon_{2}})\rho_{t}^{(3)}(x)}{\rm
d}x{\rm d}t.$

For
$x\in[\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}},\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}]$,
$\rho_{t}^{\varepsilon}(x)={\varepsilon_{2}}\phi(x-\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}})$,
the integrand simplifies and is of order
${\varepsilon_{2}}(\tilde{\mathfrak{a}}^{\prime}(t))^{2}\phi(x-\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}})$,
the total contribution is $\operatorname{O}(\varepsilon_{1}\varepsilon_{2})$.
Similarly for
$x\in[\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}},\tilde{\mathfrak{b}}(t)+2{\varepsilon_{1}}]$,
$\rho_{t}^{\varepsilon}(x)={\varepsilon_{2}}\phi(\tilde{\mathfrak{b}}(t)+2{\varepsilon_{1}}-x)$,
the total contribution is $\operatorname{O}(\varepsilon_{1}\varepsilon_{2})$.
We get

$\displaystyle\int_{0}^{1}\int\frac{\left(\partial_{t}\int^{x}\rho^{\varepsilon}_{t}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm
d}t=\int_{0}^{1}\int_{{\mathfrak{a}}(t)-{\varepsilon_{1}}}^{{\mathfrak{b}}(t)+{\varepsilon_{1}}}\frac{\left(\partial_{t}\int^{x}\rho^{\varepsilon}_{t}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm
d}t+\operatorname{O}({\varepsilon_{1}}\varepsilon_{2}).$ (3.25)

For
$x\in[\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}},\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}]$,
we have $\rho_{t}^{\varepsilon}(x)$ is lower bounded by
$\Omega(\varepsilon_{2}({\varepsilon_{1}})^{1/2})$. Moreover, from Step 1, we
know that $\rho^{(1)}_{t}(x)$ is at most of order
$\operatorname{O}(1/{\sqrt{{\varepsilon_{1}}}})$, and so are the densities
$\rho_{t}^{(2)}(x)$ and $\rho^{\varepsilon}_{t}(x)$. The boundedness of
$S(\\{\nu_{t}^{(2)}\\}_{0\leqslant t\leqslant 1})$ implies that
$\partial_{t}\int^{x}\rho_{t}^{(2)}(y){\rm d}y$ is in $L^{2}$:

$\int_{0}^{1}\int\left(\partial_{t}\int^{x}\rho_{t}^{(2)}(y){\rm
d}y\right)^{2}{\rm d}x{\rm
d}t\lesssim\frac{1}{\sqrt{\varepsilon_{1}}}\int_{0}^{1}\int(u_{t}^{(2)}(y))^{2}\rho_{t}^{(2)}(y){\rm
d}y{\rm
d}t\leqslant\frac{2}{\beta\sqrt{\varepsilon_{1}}}S_{\nu_{0}^{(2)}}(\\{\nu_{t}^{(2)}\\}_{0\leqslant
t\leqslant 1})\,.$

Therefore the convolution density $\partial_{t}\int^{x}\rho_{t}^{(3)}(y){\rm
d}y=\int^{x}\partial_{t}(\rho_{\cdot}^{(2)}*\varphi_{\varepsilon_{3}})(y,t){\rm
d}y$ converges to $\int^{x}\partial_{t}\rho_{t}^{(2)}(y){\rm d}y$ in $L^{2}$
norm as $\varepsilon_{3}\rightarrow 0$ faster than $\varepsilon_{1}$ .

$\displaystyle\begin{split}&\phantom{{}={}}\left|\int_{0}^{1}\int_{\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}}^{\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}}\frac{\left(\partial_{t}\int^{x}\rho^{(3)}_{t}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm
d}t-\int_{0}^{1}\int\frac{\left(\partial_{t}\int^{x}\rho^{(2)}_{t}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm d}t\right|\\\
&\lesssim\int_{0}^{1}\int_{\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}}^{\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}}\frac{1}{\varepsilon_{2}(\varepsilon_{1})^{\frac{1}{2}}}\left|\left(\int^{x}\partial_{t}\rho^{(3)}(y){\rm
d}y\right)^{2}-\left(\int^{x}\partial_{t}\rho^{(2)}_{t}(x){\rm
d}y\right)^{2}\right|{\rm d}x{\rm d}t\rightarrow 0,\end{split}$ (3.26)

provided we choose $\varepsilon_{3}$ going to zero fast enough with respect to
$\varepsilon_{1}$ and $\varepsilon_{2}$. Now we can estimate the right hand
side of (3.25) as

$\displaystyle\begin{split}&\int_{0}^{1}\int_{\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}}^{\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}}\frac{\left(\partial_{t}\int^{x}\rho^{\varepsilon}_{t}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm
d}t=\int_{0}^{1}\int_{\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}}^{\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}}\frac{\left({\varepsilon_{2}}\int^{x}\partial_{t}\chi_{t}(y)+(1-{\varepsilon_{2}})\int^{x}\partial_{t}\rho^{(3)}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm d}t\\\
&=\operatorname{O}(\frac{{\varepsilon_{2}}}{\varepsilon_{1}^{1/2}})+\operatorname{O}(\varepsilon_{2})\int_{0}^{1}\int_{\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}}^{\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}}\frac{|\int^{x}\partial_{t}\rho^{(3)}(y){\rm
d}y|}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm
d}t+(1-{\varepsilon_{2}})^{2}\int_{0}^{1}\int_{\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}}^{\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}}\frac{\left(\int^{x}\partial_{t}\rho^{(3)}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm d}t\\\
&\leqslant\operatorname{O}(\frac{{\varepsilon_{2}}}{\varepsilon_{1}^{1/2}})+\operatorname{O}(\varepsilon_{2})\int_{0}^{1}\int_{\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}}^{\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}}\frac{({\varepsilon_{2}})^{1/2}+(1/{\varepsilon_{2}})^{1/2}\left(\int^{x}\partial_{t}\rho^{(3)}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm d}t\\\
&+(1-{\varepsilon_{2}})^{2}\int_{0}^{1}\int_{\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}}^{\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}}\frac{\left(\int^{x}\partial_{t}\rho^{(3)}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm d}t\\\
&=\operatorname{O}(\frac{{\varepsilon_{2}}}{\varepsilon_{1}^{1/2}})+\operatorname{O}(\frac{{\varepsilon_{2}^{1/2}}}{\varepsilon_{1}^{1/2}})+(1+\operatorname{O}(({\varepsilon_{2}})^{1/2}))\int_{0}^{1}\int_{\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}}^{\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}}\frac{\left(\int^{x}\partial_{t}\rho^{(3)}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm d}t,\end{split}$ (3.27)

where we used that
$\rho_{t}^{\varepsilon}(x)\gtrsim{\varepsilon_{2}}\sqrt{\varepsilon_{1}}$ on
$x\in[\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}},\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}]$.
Moreover, for the second term in (LABEL:e:r3) we have

$\displaystyle\begin{split}&\int_{0}^{1}\int\frac{\left(\partial_{t}\int^{x}\rho^{(3)}_{t}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm
d}t=\int_{0}^{1}\int\frac{\left(\partial_{t}\int^{x}\rho^{(3)}_{t}(y){\rm
d}y\right)^{2}}{{\varepsilon_{2}}\chi_{t}(x)+(1-{\varepsilon_{2}})\rho_{t}^{(3)}(x)}{\rm
d}x{\rm d}t\\\
&\rightarrow\int_{0}^{1}\int\frac{\left(\partial_{t}\int^{x}\rho^{(2)}_{t}(y){\rm
d}y\right)^{2}}{{\varepsilon_{2}}\chi_{t}(x)+(1-{\varepsilon_{2}})\rho_{t}^{(2)}(x)}{\rm
d}x{\rm
d}t\leqslant\frac{1}{1-{\varepsilon_{2}}}\int_{0}^{1}\int\frac{\left(\partial_{t}\int^{x}\rho^{(2)}_{t}(y){\rm
d}y\right)^{2}}{\rho_{t}^{(2)}(x)}{\rm d}x{\rm d}t.\end{split}$ (3.28)

when $\varepsilon_{3}$ goes to zero and we used (LABEL:e:r3r2). Combining the
estimates (3.25), (LABEL:e:r3r2), (LABEL:e:r3) and (LABEL:e:r2) all together,
we get

$\displaystyle\int_{0}^{1}\int\frac{\left(\partial_{t}\int^{x}\rho^{\varepsilon}_{t}(y){\rm
d}y\right)^{2}}{\rho^{\varepsilon}_{t}(x)}{\rm d}x{\rm
d}t=(1+\operatorname{o}_{\varepsilon_{3},\varepsilon_{2},{\varepsilon_{1}}}(1))\int_{0}^{1}\int\frac{\left(\partial_{t}\int^{x}\rho^{(2)}_{t}(y){\rm
d}y\right)^{2}}{\rho_{t}^{(2)}(x)}{\rm d}x{\rm
d}t+\operatorname{o}_{\varepsilon_{3},\varepsilon_{2},{\varepsilon_{1}}}(1).$
(3.29)

where $\operatorname{o}_{\varepsilon_{3},\varepsilon_{2},{\varepsilon_{1}}}$
is small when $\varepsilon_{3}\ll\varepsilon_{2}\ll\varepsilon_{1}$. Moreover,
since the densities $\rho_{0}^{(2)}(x),\rho_{1}^{(2)}(x)$ are bounded by
$\operatorname{O}(1/\sqrt{{\varepsilon_{1}}})$, and are compactly supported,
it is easy to see that

$\displaystyle\frac{1}{2}\left(\Sigma(\nu_{1}^{(3)})-\Sigma(\nu_{0}^{(3)})\right)\rightarrow\frac{1}{2}\left(\Sigma(\nu_{1}^{(2)})-\Sigma(\nu_{0}^{(2)})\right),$

as $\varepsilon_{2},\varepsilon_{3}$ go to zero. We conclude from combining
(3.24) and (3.29) that

$\displaystyle\limsup_{\varepsilon\downarrow
0}S(\\{\nu_{t}^{\varepsilon}\\}_{0\leqslant t\leqslant 1})\leqslant
S(\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}).$

From the construction, $\rho_{t}^{\varepsilon}$ has uniformly bounded density,
i.e. $\rho_{t}^{\varepsilon}=\operatorname{O}(1/\sqrt{\varepsilon_{1}})$,
$\operatorname{supp}(\nu_{t}^{\varepsilon})$ is a single interval, and

$\displaystyle\lim_{\varepsilon\rightarrow 0}\sup_{0\leqslant t\leqslant
1}d(\nu_{t},\nu^{\varepsilon}_{t})=0.$

For the density $\rho^{\varepsilon}_{s}(x)$, it satisfies

$\displaystyle\partial_{t}\rho^{\varepsilon}_{t}+\partial_{x}(\rho^{\varepsilon}_{t}u^{\varepsilon}_{t})=0.$

The drift is given by

$\displaystyle\partial_{x}k^{\varepsilon}_{t}(x)=-\frac{\int^{x}\partial_{t}\rho^{\varepsilon}_{t}(y)\mathrm{d}y}{\rho_{t}^{\varepsilon}(x)}-\frac{1}{2}H(\nu^{\varepsilon}_{t})(x).$

Since $\rho^{\varepsilon}_{t}(x)$ is smooth, i.e. in ${\cal C}^{\infty}$, then
$H(\nu^{\varepsilon}_{t})$ is also smooth (see Remark 3.6). For the regularity
of the drift term $\partial_{x}k^{\varepsilon}_{t}(x)$, we need to understand
the regularity of
$(\int^{x}\partial_{t}\rho^{\varepsilon}_{t})/\rho^{\varepsilon}_{t}$. By our
construction, $\rho_{t}^{\varepsilon}$ is supported on
$[\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}},\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}}]$,
and has positive smooth density. Thus
$(\int^{x}\partial_{t}\rho^{\varepsilon}_{t})/{\rho^{\varepsilon}_{t}}$ is
smooth inside the support of $\rho_{t}^{\varepsilon}$. Close to the boundary
of the support, on
$[\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}},\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}]$,
$\rho_{t}^{\varepsilon}=\phi(x-\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}})$,
where $\phi(x)$ is smooth, and $\phi(x)=0$ for $x\leqslant 0$. In this way

$\displaystyle\partial_{x}k^{\varepsilon}_{t}(x)=-\frac{\int^{x}\partial_{t}\rho^{\varepsilon}_{t}(y){\rm
d}y}{\rho^{\varepsilon}_{t}(x)}-\frac{1}{2}H(\nu^{\varepsilon}_{t})(x)=\frac{\partial_{t}\int^{x}\phi(y-\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}}){\rm
d}y}{\phi(x-\tilde{\mathfrak{a}}(t))}-\frac{1}{2}H(\nu^{\varepsilon}_{t})(x)=\tilde{\mathfrak{a}}^{\prime}(t)-\frac{1}{2}H(\nu^{\varepsilon}_{t})(x),$
(3.30)

which is smooth in a neighborhood of
$[\tilde{\mathfrak{a}}(t)-2{\varepsilon_{1}},\tilde{\mathfrak{a}}(t)-{\varepsilon_{1}}]$.
The same argument holds in a neighborhood of the right edge
$[\tilde{\mathfrak{b}}(t)+{\varepsilon_{1}},\tilde{\mathfrak{b}}(t)+2{\varepsilon_{1}}]$.
So $\partial_{x}k^{\varepsilon}_{t}(x)$ is a smooth drift.

∎

In the following proposition, we show that for the Dyson Brownian motion with
smooth drift, the locations of its particles are close to the quantiles of the
limiting profile. Proposition 3.4 will then be an easy consequence.

######  Proposition 3.5.

Let $\\{\nu_{t}\\}_{0\leqslant t\leqslant 1}$ be a measure valued process with
bounded support and with smooth density $\rho_{t}(x)$ in both $x,t$, and such
that the drift $\partial_{x}k_{t}(x)$ such that

$\displaystyle\partial_{t}\rho_{t}(x)+\partial_{x}(\rho_{t}(\frac{1}{2}H(\nu_{t})(x)+\partial_{x}k_{t}(x)))=0,$
(3.31)

is uniformly Lipschitz : $|\partial_{x}k_{t}(x)-\partial_{x}k_{t}(y)|\leqslant
K|x-y|$ for all real numbers $x,y$. We denote by $\gamma_{i}(t)$ the $ith$
$(1/n)$-quantiles of $\rho_{t}$ given by

$\displaystyle\frac{i-1/2}{n}=\int_{\infty}^{\gamma_{i}(t)}\rho_{t}(x){\rm
d}x,\quad 1\leqslant i\leqslant n.$ (3.32)

For $\beta\geqslant 1$, we consider the Dyson Brownian motion with drift
$\partial_{x}k_{t}$ which is the unique strong solution of

$\displaystyle{\rm d}x_{i}(t)=\frac{{\rm d}W_{i}(t)}{\sqrt{\beta
n}}+\frac{1}{2n}\left(\sum_{j:j\neq
i}\frac{1}{x_{i}(t)-x_{j}(t)}\right)\mathrm{d}t+\partial_{x}k_{t}(x_{i}(t))\mathrm{d}t.$
(3.33)

starting from $(x_{j}(0))_{1\leqslant i\leqslant n}$. Let $M$ be a positive
real number. Then it holds

$\displaystyle\sup_{t\in[0,1]}\max_{1\leqslant i\leqslant
n}|x_{i}(t)-\gamma_{i}(t)|\leqslant e^{Kt}\left(\max_{1\leqslant j\leqslant
n}(|x_{j}(0)-\gamma_{j}(0)|)+\frac{M}{\sqrt{n}}\right),$

with probability going to one as $M$ goes to infinity.

######  Remark 3.6.

Note that if $\nu$ is a probability measure with density $\rho$ which is
${\cal C}^{k}_{b}$ and with compact support, then $H\nu$ is ${\cal
C}^{k-1}_{b}$. Indeed, for $\rho$ with support in $[-A,A]$ and $|x|\leqslant
M$

$H\nu(x)=P.V.\int\frac{\rho(y)}{x-y}\mathrm{d}y=P.V.\int_{|x-y|\leqslant
A+M}\frac{\rho(y)}{x-y}\mathrm{d}y=\int_{|x-y|\leqslant
A+M}\frac{\rho(y)-\rho(x)}{x-y}\mathrm{d}y,$

where we noticed that $P.V.\int_{|x-y|\leqslant
A+M}\frac{1}{x-y}\mathrm{d}y=0$.

###### Proof of Proposition 3.5.

We first show that the $(1/n)$-quantiles of $\rho_{t}$ approximately satisfy
the equations of Dyson Brownian motion:

$\displaystyle\partial_{t}\gamma_{i}(t)=\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{\gamma_{i}(t)-\gamma_{j}(t)}+\partial_{x}k_{t}(\gamma_{i}(t))+\operatorname{O}\left(\frac{1}{\sqrt{n}}\right),\quad
1\leqslant i\leqslant n.$ (3.34)

In the following, we denote in short $H(\rho)$ for $H(\rho(x)\mathrm{d}x)$. By
taking derivative with respect to $t$ on both sides of (3.32), we have

$\displaystyle\begin{split}0&=\partial_{t}\gamma_{i}(t)\rho_{t}(\gamma_{i}(t))+\int^{\gamma_{i}(t)}\partial_{t}\rho_{t}(x){\rm
d}x\\\
&=\partial_{t}\gamma_{i}(t)\rho_{t}(\gamma_{i}(t))-\int^{\gamma_{i}(t)}\partial_{x}(\rho_{t}(H(\rho_{t})/2+\partial_{x}k_{t}(x))){\rm
d}x\\\
&=\partial_{t}\gamma_{i}(t)\rho_{t}(\gamma_{i}(t))-\rho_{t}(\gamma_{i}(t))(H(\rho_{t}(\gamma_{i}(t)))/2+\partial_{x}k_{t}(\gamma_{i}(t))).\end{split}$

By rearranging the above equality, we obtain the following differential
equation for the quantiles of $\rho_{t}(x)$

$\displaystyle\partial_{t}\gamma_{i}(t)=\frac{1}{2}H(\rho_{t})(\gamma_{i}(t))+\partial_{x}k_{t}(\gamma_{i}(t)).$
(3.35)

In the following, we prove that we can approximate the Hilbert transform by a
discrete sum

$\displaystyle H(\rho_{t})(\gamma_{i}(t))=\frac{1}{n}\sum_{j:j\neq
i}\frac{1}{\gamma_{i}(t)-\gamma_{j}(t)}+\operatorname{O}\left(\frac{1}{\sqrt{n}}\right).$
(3.36)

Then the proof of (3.34) follows from combining (3.35) and (3.36). We fix a
large constant ${\mathfrak{C}}>0$ (which will be chosen later), and divide
(3.36) into two cases: either
$\rho_{t}(\gamma_{i}(t))\leqslant{\mathfrak{C}}/\sqrt{n}$ or
$\rho_{t}(\gamma_{i}(t))\geqslant{\mathfrak{C}}/\sqrt{n}$.

In the following we first discuss the case that
$\rho_{t}(\gamma_{i}(t))\leqslant{\mathfrak{C}}/\sqrt{n}$. Since our density
$\rho_{t}(x)$ is smooth, in particular its first derivative is uniformly
bounded $\|\rho_{t}^{\prime}\|_{\infty}\lesssim 1$. For
${\mathfrak{c}}\leqslant\|\rho_{t}^{\prime}\|_{\infty}^{-1}{\mathfrak{C}}$,
$\rho_{t}(x)\leqslant 2{\mathfrak{C}}/\sqrt{n}$ on
$[\gamma_{i}(t)-{\mathfrak{c}}/\sqrt{n},\gamma_{i}(t)+{\mathfrak{c}}/\sqrt{n}]$.
If we take also ${\mathfrak{c}}\leqslant 1/4{\mathfrak{C}}$, we deduce that
$\gamma_{i-1}(t)\leqslant\gamma_{i}(t)-{\mathfrak{c}}/\sqrt{n}$ and
$\gamma_{i+1}(t)\geqslant\gamma_{i}(t)+{\mathfrak{c}}/\sqrt{n}$ (we made the
convention $\gamma_{0}=-\infty$ and $\gamma_{n+1}=+\infty$). Then

$\displaystyle\begin{split}P.V.\int_{\gamma_{i-1}(t)}^{{\gamma_{i+1}}(t)}\frac{\rho_{t}(x)}{\gamma_{i}(t)-x}{\rm
d}x&=P.V.\int_{\gamma_{i}(t)-{\mathfrak{c}}/\sqrt{n}}^{{\gamma_{i}(t)}+{\mathfrak{c}}/\sqrt{n}}\frac{\rho_{t}(x)}{\gamma_{i}(t)-x}{\rm
d}x+\operatorname{O}\left(\frac{1}{{\mathfrak{c}}\sqrt{n}}\right)\\\
&=P.V.\int_{\gamma_{i}(t)-{\mathfrak{c}}/\sqrt{n}}^{{\gamma_{i}(t)}+{\mathfrak{c}}/\sqrt{n}}\frac{\rho_{t}(x)-\rho_{t}(\gamma_{i}(t))}{\gamma_{i}(t)-x}{\rm
d}x+\operatorname{O}\left(\frac{1}{{\mathfrak{c}}\sqrt{n}}\right)\\\
&=\operatorname{O}\left(\frac{{\mathfrak{c}}\|\rho_{t}^{\prime}\|_{\infty}}{\sqrt{n}}+\frac{1}{{\mathfrak{c}}\sqrt{n}}\right)=\operatorname{O}\left(\frac{1}{\sqrt{n}}\right),\end{split}$
(3.37)

where we used in the second line that
$P.V.\int_{\gamma_{i}(t)-{\mathfrak{c}}/\sqrt{n}}^{{\gamma_{i}(t)}+{\mathfrak{c}}/\sqrt{n}}(\gamma_{i}(t)-x)^{-1}{\rm
d}x=0$. For the integral outside the interval
$[\gamma_{i-1}(t),\gamma_{i+1}(t)]$, we have the trivial bounds

$\displaystyle\int_{-\infty}^{\gamma_{i-1}(t)}\frac{\rho_{t}(x){\rm
d}x}{\gamma_{i}(t)-x}\leqslant\sum_{j=1}^{i-1}\int_{\gamma_{j-1}(t)}^{\gamma_{j}(t)}\frac{\rho_{t}(x){\rm
d}x}{\gamma_{i}(t)-\gamma_{j}(t)}\leqslant\sum_{j=1}^{i-1}\frac{1}{n(\gamma_{i}(t)-\gamma_{j}(t))},$

and the lower bound

$\displaystyle\int_{-\infty}^{\gamma_{i-1}(t)}\frac{\rho_{t}(x){\rm
d}x}{\gamma_{i}(t)-x}\geqslant\sum_{j=1}^{i-1}\int_{\gamma_{j-1}(t)}^{\gamma_{j}(t)}\frac{\rho_{t}(x){\rm
d}x}{\gamma_{i}(t)-\gamma_{j-1}(t)}\geqslant\sum_{j=1}^{i-2}\frac{1}{n(\gamma_{i}(t)-\gamma_{j}(t))}.$

Thus we conclude that

$\displaystyle\left|\int_{-\infty}^{\gamma_{i-1}(t)}\frac{\rho_{t}(x){\rm
d}x}{\gamma_{i}(t)-x}-\sum_{j=1}^{i-1}\frac{1}{n(\gamma_{i}(t)-\gamma_{j}(t))}\right|\leqslant\frac{1}{n(\gamma_{i}(t)-\gamma_{i-1}(t))}\leqslant\frac{1}{{\mathfrak{c}}\sqrt{n}}.$
(3.38)

We have the same estimate for the integral from $\gamma_{i+1}(t)$ to $\infty$.
The claim (3.36) follows from combining (3.37) and (3.38).

For the case that $\rho_{t}(\gamma_{i}(t))\geqslant{\mathfrak{C}}/\sqrt{n}$,
we have for any integer $k$,

$\displaystyle\begin{split}\frac{k}{n}=\int_{\gamma_{i}(t)}^{\gamma_{i+k}(t)}\rho_{t}(x){\rm
d}x&=\int_{\gamma_{i}(t)}^{\gamma_{i+k}(t)}(\rho_{t}(\gamma_{i}(t))+\operatorname{O}(\|\rho^{\prime}_{t}\|_{\infty}|x-\gamma_{i}(t)|)){\rm
d}x\\\
&=(\gamma_{i+k}(t)-\gamma_{i}(t))\rho_{t}(\gamma_{i}(t))+\operatorname{O}(\|\rho^{\prime}_{t}\|_{\infty}|(\gamma_{i+k}(t)-\gamma_{i}(t))|^{2}).\end{split}$

By rearranging, we get

$\displaystyle\begin{split}\gamma_{i+k}(t)-\gamma_{i}(t)&=\frac{k}{n\rho_{t}(\gamma_{i}(t))}\frac{1}{1+\operatorname{O}(\|\rho^{\prime}_{t}\|_{\infty}|(\gamma_{i+k}(t)-\gamma_{i}(t))|/\rho_{t}(\gamma_{i}(t))}\\\
&=\frac{k}{n\rho_{t}(\gamma_{i}(t))}\left(1+\operatorname{O}\left(\frac{k\|\rho^{\prime}_{t}\|_{\infty}}{n\rho_{t}(\gamma_{i}(t))^{2}}\right)\right),\end{split}$
(3.39)

provided that $k\|\rho^{\prime}_{t}\|_{\infty}\leqslant
n\rho_{t}(\gamma_{i}(t))^{2}/2$. We have exactly the same estimates for
$\gamma_{i}(t)-\gamma_{i-k}(t)$, and

$\displaystyle\frac{1}{n}\left|\frac{1}{\gamma_{i}(t)-\gamma_{i-k}(t)}+\frac{1}{\gamma_{i}(t)-\gamma_{i+k}(t)}\right|\lesssim\frac{1}{n}\frac{\operatorname{O}\left(\frac{k\|\rho^{\prime}_{t}\|_{\infty}}{n\rho_{t}(\gamma_{i}(t))^{2}}\right)}{\frac{k}{n\rho_{t}(\gamma_{i}(t))}}\lesssim\frac{\|\rho^{\prime}_{t}\|_{\infty}}{n\rho_{t}(\gamma_{i}(t))}.$
(3.40)

We take

$\displaystyle
d=\left\lfloor\frac{\sqrt{n}\rho_{t}(\gamma_{i}(t))}{{\mathfrak{C}}}\right\rfloor+1,$
(3.41)

then $d\geqslant 1$ and

$\displaystyle
d\|\rho^{\prime}_{t}\|_{\infty}\leqslant\|\rho^{\prime}_{t}\|_{\infty}+\frac{\|\rho^{\prime}_{t}\|_{\infty}\sqrt{n}\rho_{t}(\gamma_{i}(t))}{{\mathfrak{C}}}\leqslant
n\rho_{t}(\gamma_{i}(t))^{2}/2,$

provided we take ${\mathfrak{C}}$ large enough. By summing over (3.40) from
$k=1$ to $k=d-1$, we have

$\displaystyle\sum_{k=1}^{d-1}\frac{1}{n}\left(\frac{1}{\gamma_{i}(t)-\gamma_{i-k}(t)}+\frac{1}{\gamma_{i}(t)-\gamma_{i+k}(t)}\right)\lesssim\frac{d\|\rho^{\prime}_{t}\|_{\infty}}{n\rho_{t}(\gamma_{i}(t))}\lesssim\frac{1}{\sqrt{n}}.$
(3.42)

Moreover, by taking $k=d$ in (3.39), we have
$|\gamma_{i+d}(t)-\gamma_{i}(t)|,|\gamma_{i}(t)-\gamma_{i-d}(t)|\asymp
1/\sqrt{n}.$ The same argument as for (3.37) gives

$\displaystyle\begin{split}P.V.\int_{\gamma_{i-d}(t)}^{{\gamma_{i+d}}(t)}\frac{\rho_{t}(x)}{\gamma_{i}(t)-x}{\rm
d}x=\operatorname{O}\left(\frac{1}{\sqrt{n}}\right).\end{split}$ (3.43)

Similarly, following the proof of (3.38), we get

$\displaystyle\left|\int_{-\infty}^{\gamma_{i-d}(t)}\frac{\rho_{t}(x){\rm
d}x}{\gamma_{i}(t)-x}-\sum_{j=1}^{i-d}\frac{1}{n(\gamma_{i}(t)-\gamma_{j}(t)}\right|\leqslant\frac{1}{n(\gamma_{i}(t)-\gamma_{i-d}(t))}\lesssim\frac{1}{\sqrt{n}}.$
(3.44)

The claim (3.36) follows from combining (3.42), (3.43) and (3.44).

By taking the difference of (3.33) and (3.34), we get

$\displaystyle\begin{split}{\rm d}(x_{i}(t)-\gamma_{i}(t))&=\frac{{\rm
d}W_{i}(t)}{\sqrt{\beta n}}-\frac{1}{2n}\sum_{j:j\neq
i}\frac{(x_{i}(t)-\gamma_{i}(t))-(x_{j}(t)-\gamma_{j}(t))}{(x_{i}(t)-x_{j}(t))(\gamma_{i}(t)-\gamma_{j}(t))}\mathrm{d}t\\\
&+(\partial_{x}k_{t}(x_{i}(t))-\partial_{x}k_{t}(\gamma_{i}(t)))\mathrm{d}t,\quad
1\leqslant i\leqslant n.\end{split}$ (3.45)

We take $i_{*}(t)=\arg\max_{i\in[\\![{1,n}]\\!]}(x_{i}(t)-\gamma_{i}(t))$.
Since $i_{*}(t)$ is piecewise constant, ${\rm d}i_{*}(t)$ is almost surely
zero. By plugging $i=i_{*}(t)$ in (3.45), and noticing that the second term on
the righthand side is nonpositive, we get

$\displaystyle{\rm d}(x_{i_{*}(t)}(t)-\gamma_{i_{*}(t)}(t))\leqslant\frac{{\rm
d}W_{i_{*}(t)}(t)}{\sqrt{\beta
n}}+K(x_{i_{*}(t)}(t)-\gamma_{i_{*}(t)}(t))\mathrm{d}t.$

The martingale term ${\rm d}W_{i_{*}(t)}(t)$ has the same law as a standard
Brownian motion. By Gronwall’s inequality, we deduce that

$\displaystyle x_{i_{*}(t)}(t)-\gamma_{i_{*}(t)}(t)$ $\displaystyle\leqslant
e^{Kt}\left(x_{i_{*}(0)}(0)-\gamma_{i_{*}(0)}(0)+\int_{0}^{t}{{e^{-Ks}}}\frac{{\rm
d}W_{i_{*}(s)}(s)}{\sqrt{\beta n}}\right)$ $\displaystyle\leqslant
e^{Kt}\left(\max_{i}|x_{i}(0)-\gamma_{i}(0)|+\frac{M}{\sqrt{n}}\right),$

where $M$ is a stochastically bounded random variable, uniformly in time (by
Doob’s martingale inequality). It follows that uniformly for any
$i\in[\\![{1,n}]\\!]$,

$\displaystyle x_{i}(t)-\gamma_{i}(t)\lesssim
e^{Kt}\left(\max_{i}|x_{i}(0)-\gamma_{i}(0)|+\frac{M}{\sqrt{n}}\right).$

By the same argument, we have a similar lower bound by considering
$i_{*}(t)=\arg\min_{i\in[\\![{1,n}]\\!]}(x_{i}(t)-\gamma_{i}(t))$. The
following holds

$\displaystyle\gamma_{i}(t)-x_{i}(t)\lesssim
e^{Kt}\left(\max_{i}|x_{i}(0)-\gamma_{i}(0)|+\frac{M}{\sqrt{n}}\right),$

where $M$ is stochastically bounded. This finishes the proof of Proposition
3.5. ∎

###### Proof of Proposition 3.4.

We first show that changing slightly the initial condition of the Dyson
Brownian motion will not change much the large deviations lower bound. This
will enable us to consider an initial measure $\tilde{\mu}_{0}$ with compact
support and finite free entropy. Let $\mu_{0}$ be a probability measure and
$\tilde{\mu}_{0}$ a compactly supported approximation so that
$d(\mu_{0},\tilde{\mu}_{0})\leqslant\delta/3$. Denote
$\tilde{F}_{0}(x)=\tilde{\mu}_{0}((-\infty,x])$. We construct a new family of
initial data

$\displaystyle\tilde{\nu}_{0}^{n}=\frac{1}{n}\sum_{i=1}^{n}\delta_{\tilde{x}_{i}(0)},\quad\tilde{x}_{i}(0)=\tilde{F}_{0}^{-1}((i-1/2)/n).$

In this way $\tilde{x}_{i}(0)$ is the $(i-1/2)/n$ quantile of
$\tilde{\rho}_{0}$. Thanks to Assumption 1, we have that

$\displaystyle
d(\tilde{\nu}_{0}^{n},\nu_{0}^{n})=\sqrt{\frac{1}{n}\sum_{1\leqslant
i\leqslant
n}|\tilde{x}_{i}(0)-x_{i}(0)|^{2}}\leqslant\frac{\delta}{3}+\operatorname{o}_{n}(1).$

We consider the Dyson Brownian motion starting from $\tilde{\nu}_{0}^{n}$,

$\displaystyle{\rm d}\tilde{x}_{i}(t)=\frac{{\rm d}W_{i}(t)}{\sqrt{\beta
n}}+\frac{1}{2n}\left(\sum_{j:j\neq
i}\frac{1}{\tilde{x}_{i}(t)-\tilde{x}_{j}(t)}\right)\mathrm{d}t,$ (3.46)

which shares the same Brownian motions $W_{i}$ as (2.7). By taking the
difference between (2.10) and (3.46), we get

$\displaystyle\partial_{t}(\tilde{x}_{i}(t)-x_{i}(t))^{2}=\frac{(\tilde{x}_{i}(t)-x_{i}(t))}{n}\sum_{j:j\neq
i}\frac{-(\tilde{x}_{i}(t)-x_{i}(t))+(\tilde{x}_{j}(t)-x_{j}(t))}{(\tilde{x}_{i}(t)-\tilde{x}_{j}(t))(x_{i}(t)-x_{j}(t))}.$
(3.47)

Averaging over all the indices $i\in[\\![{1,n}]\\!]$, we find

$\displaystyle\frac{1}{n}\partial_{t}\sum_{i}(\tilde{x}_{i}(t)-x_{i}(t))^{2}=-\frac{1}{n^{2}}\sum_{i<j}\frac{((\tilde{x}_{i}(t)-x_{i}(t))-(\tilde{x}_{j}(t)-x_{j}(t)))^{2}}{(\tilde{x}_{i}(t)-\tilde{x}_{j}(t))(x_{i}(t)-x_{j}(t))}\leqslant
0.$

It follows that if we let
$\tilde{\nu}_{t}^{n}=\frac{1}{n}\sum_{i=1}^{n}\delta_{\tilde{x}_{i}(t)}$, we
have

$\displaystyle{d(\tilde{\nu}_{t}^{n},\nu_{t}^{n})}=\sqrt{\frac{1}{n}\sum_{i}(\tilde{x}_{i}(t)-x_{i}(t))^{2}}\leqslant\sqrt{\frac{1}{n}\sum_{i}(\tilde{x}_{i}(0)-x_{i}(0))^{2}}\leqslant\delta/2,$

provided $n$ is large enough. As a consequence, we deduce that for any
compactly supported measure-valued process
$\\{\tilde{\nu}_{t}(x)\\}_{0\leqslant t\leqslant 1}$ with a smooth density
such that $\tilde{\nu}_{0}=\tilde{\mu}_{0}$ as in Proposition 3.4

$\displaystyle\begin{split}&\phantom{{}={}}\lim_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\nu^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\tilde{\nu}_{t}\\}_{0\leqslant t\leqslant
1},\delta))\geqslant\lim_{n\rightarrow\infty}\frac{1}{n^{2}}\log{\mathbb{P}}(\\{\tilde{\nu}^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\tilde{\nu}_{t}\\}_{0\leqslant t\leqslant
1},\delta/2)).\end{split}$ (3.48)

Let $\partial_{x}\tilde{k}_{t}$ denote the drift (3.21) and
${\mathbb{Q}}^{\beta\tilde{k}}$ be the distribution

$\displaystyle{\mathbb{Q}}^{\beta\tilde{k}}=e^{n^{2}S^{n}(\\{\tilde{\nu}^{n}_{t},\beta\tilde{k}_{t})\\}_{0\leqslant
t\leqslant 1}}{\mathbb{Q}},$

where $S^{n}(\\{\tilde{\nu}^{n}_{t},\beta\tilde{k}_{t})\\}_{0\leqslant
t\leqslant 1})$

$\displaystyle S^{n}(\\{\tilde{\nu}^{n}_{t},\beta\tilde{k}_{t})\\}_{0\leqslant
t\leqslant 1})$
$\displaystyle=\frac{1}{n^{2}}\left(\frac{nL_{1}^{\beta\tilde{k}}}{2}-\frac{n^{2}}{2}\langle
L^{\beta\tilde{k}},L^{\beta\tilde{k}}\rangle_{1}\right),\quad{\rm
d}L^{\beta\tilde{k}}_{t}=\sum_{i}\beta\partial_{x}\tilde{k}_{t}(x_{i}(t))\frac{{\rm
d}W_{i}(t)}{\sqrt{\beta n}},$ (3.49)

is defined in (3.14) and (3.17). Because $\partial_{x}\tilde{k}_{t}(x)$ is
continuously differentiable and bounded, this is a well defined change of
measure. By Girsanov’s formula, under ${\mathbb{Q}}^{\tilde{k}}$, the measure
valued process $\\{\tilde{\nu}^{n}_{t}\\}_{0\leqslant t\leqslant 1}$ has the
same law as the empirical distribution of

$\displaystyle\begin{split}{\rm d}\tilde{x}_{i}(t)&=\frac{{\rm
d}W_{i}(t)}{\sqrt{\beta n}}+\frac{1}{2n}\left(\sum_{j:j\neq
i}\frac{1}{\tilde{x}_{i}(t)-\tilde{x}_{j}(t)}\right)\mathrm{d}t+\langle{\rm
d}L^{\beta\tilde{k}},{\rm d}\tilde{x}_{i}\rangle_{t}\\\ &=\frac{{\rm
d}W_{i}(t)}{\sqrt{\beta n}}+\frac{1}{2n}\left(\sum_{j:j\neq
i}\frac{1}{\tilde{x}_{i}(t)-\tilde{x}_{j}(t)}\right)\mathrm{d}t+\partial_{x}\tilde{k}_{t}(\tilde{x}_{i})\mathrm{d}t.\end{split}$
(3.50)

Thanks to Proposition 3.5, it holds that

$\displaystyle\sup_{t\in[0,1]}\max_{1\leqslant k\leqslant
n}|\tilde{x}_{k}(t)-\tilde{\gamma}_{k}(t)|\lesssim
e^{Kt}\left(\max_{1\leqslant k\leqslant
n}(|\tilde{x}_{k}(0)-\tilde{\gamma}_{k}(0)|)+\frac{M}{\sqrt{n}}\right),$
(3.51)

where the constant $K$ depends on the Lipschitz constant of
$\tilde{\partial}_{x}k_{t}$, and $M$ is stochastically bounded. Especially,
(3.51) implies that
${\mathbb{Q}}^{\tilde{k}}(\\{\tilde{\nu}^{n}_{t}\\}_{0\leqslant t\leqslant
1}\in{\mathbb{B}}(\\{\tilde{\nu}_{t}\\}_{0\leqslant t\leqslant 1},\delta/2))$
with probability $1-\operatorname{o}(1)$. We conclude that

$\displaystyle\begin{split}{\mathbb{Q}}(\\{\tilde{\nu}^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\tilde{\nu}_{t}\\}_{0\leqslant t\leqslant
1},\delta/2))&={\mathbb{Q}}^{\beta\tilde{k}}(e^{-n^{2}S^{n}(\\{\tilde{\nu}^{n}_{t},\beta\tilde{k}_{t}\\}_{0\leqslant
t\leqslant 1})}\bm{1}(\\{\tilde{\nu}^{n}_{t}\\}_{0\leqslant t\leqslant
1}\in{\mathbb{B}}(\\{\tilde{\nu}_{t}\\}_{0\leqslant t\leqslant
1},\delta/2)))\\\
&=\exp\\{-n^{2}(S_{\tilde{\mu}_{0}}(\tilde{\nu}_{t})+\operatorname{o}_{\delta}(1))\\}{\mathbb{Q}}^{\beta\tilde{k}}(\\{\tilde{\nu}^{n}_{t}\\}_{0\leqslant
t\leqslant 1}\in{\mathbb{B}}(\\{\tilde{\nu}_{t}\\}_{0\leqslant t\leqslant
1},\delta/2)\\\
&=\exp\\{-n^{2}(S_{\tilde{\mu}_{0}}(\tilde{\nu}_{t})+\operatorname{o}_{\delta}(1))\\}(1-\operatorname{o}(1)),\end{split}$
(3.52)

where in the second line, we used that
$S^{n}(\\{\tilde{\nu}^{n}_{t},\beta\tilde{k}_{t}\\}_{0\leqslant t\leqslant
1})=S_{\tilde{\mu}_{0}}(\tilde{\nu}_{t})+\operatorname{o}_{\delta}(1)$ for
$\\{\tilde{\nu}^{n}_{t}\\}_{0\leqslant t\leqslant 1}$ in
${\mathbb{B}}(\\{\tilde{\nu}_{t}\\}_{0\leqslant t\leqslant 1},\delta/2)$ by
continuity of $\nu\rightarrow
S^{n}(\\{\nu_{t},\beta\tilde{k}_{t}\\}_{0\leqslant t\leqslant 1})$ and (3.5).
This and (LABEL:e:replace1) together finish the proof of (3.19). ∎

##  4 Large deviations for the Dyson Bessel process

In this section, we prove the large deviation principle, Theorem 1.2, for the
symmetrized empirical particle density of the Dyson Bessel process

$\displaystyle\begin{split}\mathrm{d}s_{i}(t)&=\frac{\mathrm{d}W_{i}}{\sqrt{\beta
n}}+\left(\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{s_{i}(t)-s_{j}(t)}+\frac{1}{2n}\sum_{j:j\neq
i}\frac{1}{s_{i}(t)+s_{j}(t)}+\frac{\alpha_{n}}{2s_{i}(t)}\right)\mathrm{d}t,\quad
1\leqslant i\leqslant n,\end{split}$ (4.1)

We can symmetrize the Dyson Bessel process (4.1), by setting
$s_{-i}(t)=-s_{i}(t),W_{-i}(t)=-W_{i}(t)$ for $1\leqslant i\leqslant n$, then
for $i\in[\\![{-n,n}]\\!]\setminus\\{0\\}$, we have

$\displaystyle\begin{split}\mathrm{d}s_{i}(t)&=\frac{\mathrm{d}W_{i}(t)}{\sqrt{\beta
n}}+\left(\frac{1}{2n}\sum_{j:j\neq\pm
i}\frac{1}{s_{i}(t)-s_{j}(t)}+\frac{\alpha_{n}}{2s_{i}(t)}\right)\mathrm{d}t,\\\
&=\frac{\mathrm{d}W_{i}(t)}{\sqrt{\beta n}}+\frac{1}{2n}\sum_{j:j\neq
i}\frac{{\rm
d}t}{s_{i}(t)-s_{j}(t)}+\frac{\alpha_{n}-1/(2n)}{2s_{i}(t)}\mathrm{d}t\end{split}$
(4.2)

where in the last line we added a term $(1/2n){\rm d}t/(s_{i}(t)-s_{j}(t))$
with $j=i$, and replaced $\alpha_{n}$ by $\alpha_{n}-(1/2n)$.

We denote the law of the Dyson Bessel process by ${\mathbb{P}}$, and the
empirical particle density and its symmetrized version as

$\displaystyle\nu^{n}_{t}=\frac{1}{n}\sum_{i=1}^{n}\delta_{s_{i}(t)},\quad\hat{\nu}^{n}_{t}=\frac{1}{2n}\sum_{i=1}^{n}\left(\delta_{s_{i}(t)}+\delta_{-s_{i}(t)}\right),\quad
0\leqslant t\leqslant 1.$ (4.3)

More generally, for a probability measure $\nu$ on the real line we set
$\hat{\nu}$ to be its symmetrized version
$\hat{\nu}(f)=\int(f(x)+f(-x))/2\mathrm{d}\nu$. Reciprocally, if $\nu$ is a
probability on $(0,\infty)$, we can retrieve $\nu$ from $\hat{\nu}$ by setting
$\nu=2\hat{\nu}|_{(0,+\infty)}$. If $\alpha_{n}\geqslant 1/n\beta$, the
solution of (4.1) for $t>0$ is non negative almost surelyand thus
$\nu_{t}^{n}=2\hat{\nu}^{n}_{t}|_{(0,+\infty)}$. We denote by
$\mathbb{M}_{1}^{s}(\mathbb{R})$ the set of symmetric probability measures on
the real line and observe that it is a closed subset of
$\mathbb{M}_{1}(\mathbb{R})$.

We recall from (1.10) that given a symmetric measure $\hat{\mu}_{0}$ and a
continuous symmetric measure-valued process $\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant 1}$ with $\hat{\nu}_{0}=\hat{\mu}_{0}$, we define the following
dynamical free entropy:

$S_{\hat{\mu}_{0}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant
1})=\sup_{f\in{\cal
C}^{2,1}_{b}}S^{\alpha}(\\{\hat{\nu}_{t},f_{t}\\}_{0\leqslant t\leqslant 1}),$
(4.4)

where

$\displaystyle S^{\alpha}(\\{\hat{\nu}_{t},f_{t}\\}_{0\leqslant t\leqslant
1})$ $\displaystyle=L^{\alpha}_{1}(\\{\hat{\nu}_{t},f_{t}\\}_{0\leqslant
t\leqslant
1})-\frac{1}{8\beta}\int^{1}_{0}\int(f_{s}^{\prime}(x)-f_{s}^{\prime}(-x))^{2}{\rm
d}\hat{\nu}_{s}(x){\rm d}s$ $\displaystyle
L^{\alpha}_{u}(\\{\hat{\nu}_{t},f_{t}\\}_{0\leqslant t\leqslant 1})$
$\displaystyle=\hat{\nu}_{u}(f_{u})-\hat{\nu}_{0}(f_{0})-\int_{0}^{u}\int\partial_{s}f_{s}(x){\rm
d}\hat{\nu}_{s}(x){\rm
d}s-\frac{1}{2}\int_{0}^{u}\int\frac{f_{s}^{\prime}(x)-f_{s}^{\prime}(y)}{x-y}{\rm
d}\hat{\nu}_{s}(x){\rm d}\hat{\nu}_{s}(y){\rm d}s$
$\displaystyle-\frac{\alpha}{2}\int_{0}^{u}\int\frac{f_{s}^{\prime}(x)}{x}{\rm
d}\hat{\nu}_{s}(x){\rm d}s.$

The supremum is taken over $f_{t}(x)\in{\cal
C}_{b}^{2,1}(\mathbb{R}\times[0,1])$ which has bounded twice derivative in $x$
and bounded derivative in $t$. We notice that since the measure process
$\hat{\nu}_{t}$ is symmetric, for any $f_{t}(x)\in{\cal
C}_{b}^{2,1}(\mathbb{R}\times[0,1])$, it holds

$\displaystyle
S^{\alpha}(\\{\hat{\nu}_{t},(f_{t}(x)+f_{t}(-x))/2\\}_{0\leqslant t\leqslant
1})=S^{\alpha}(\\{\hat{\nu}_{t},f_{t}(x)\\}_{0\leqslant t\leqslant 1}).$

Therefore, the optimization problem (4.4) can be restricted to the set of even
functions, i.e. $f_{t}(x)=f_{t}(-x)$. If $\hat{\nu}_{t}$ is not symmetric, or
$\hat{\nu}_{0}\neq\hat{\mu}_{0}$, we simply set
$S_{\hat{\mu}_{0}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant
1})=\infty$.

In this section we prove Theorem 1.2. We study the properties of the rate
function (4.4) and its relation to rectangular free convolution in Sections
4.1 and 4.2. In Section 4.3, we derive dynamical equations of linear
statistics of Dyson Bessel process for general test function using Itô’s
formula. We prove the large deviations upper bound in Section 4.4, and the
large deviations lower bound in Section 4.5. Finally we give the proof of
Theorem 1.2 in Section 4.6.

###  4.1 Study of the Rate Function

In this section, we study the rate function
$S_{\hat{\mu}_{0}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant 1})$ as
defined in (4.4). If
$S^{\alpha}_{\hat{\mu}_{0}}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant
1})<\infty$, then $\hat{\nu}_{0}=\hat{\mu}_{0}$, and by Riesz representation
theorem, for any measure process $\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant
1}$, there exists a measurable function $\partial_{x}k_{t}(x)$, such that for
any $f\in{\cal C}_{b}^{2,1}([0,1]\times{\mathbb{R}})$,

$\displaystyle\begin{split}\int_{0}^{1}\int
f^{\prime}_{s}(x)\partial_{x}k_{s}(x){\rm d}\hat{\nu}_{s}{\rm
d}s=L_{1}^{\alpha}(\\{\hat{\nu}_{t},f_{t}\\}_{0\leqslant t\leqslant
1})&=\hat{\nu}_{1}(f_{1})-\hat{\nu}_{0}(f_{0})-\int\partial_{s}f_{s}(x){\rm
d}\hat{\nu}_{s}(x){\rm d}s\\\ &-\int_{0}^{1}\int
f_{s}^{\prime}(x)(H(\hat{\nu}_{s})+(\alpha/2)H(\delta_{0})){\rm
d}\hat{\nu}_{s}(x){\rm d}s.\end{split}$ (4.5)

Since $\hat{\nu}_{s}$ is symmetric, it is necessary that
$\partial_{x}k_{s}(x)$ is an odd function. With this notation, we can rewrite
the rate function as

$\displaystyle\begin{split}S_{\hat{\mu}_{0}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant 1})&=\sup_{f\in{\cal
C}^{2,1}}\left\\{\int_{0}^{1}f^{\prime}_{s}(x)\partial_{x}k_{s}(x){\rm
d}\hat{\nu}_{s}(x){\rm
d}s-\frac{1}{8\beta}\int^{1}_{0}\int(f_{s}^{\prime}(x)-f_{s}^{\prime}(-x))^{2}{\rm
d}\hat{\nu}_{s}(x){\rm d}s\right\\}\\\
&=\frac{\beta}{2}\int_{0}^{1}\int\partial_{x}k_{s}(x)^{2}{\rm
d}\hat{\nu}_{s}(x){\rm d}s,\end{split}$ (4.6)

where the equality is achieved when
$f^{\prime}_{t}(x)-f^{\prime}_{t}(-x)=2\beta\partial_{x}k_{t}(x)$.

We next see as in Proposition 3.2 that when the rate function
$S_{\hat{\mu}_{0}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant 1})$ is
finite, the measure valued process $\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant
1}$ satisfies nice properties.

######  Proposition 4.1.

Fix a symmetric probability density $\hat{\mu}_{0}$ with bounded second
moment. $S^{\alpha}_{\hat{\mu}_{0}}$ is a good rate function on ${\cal
C}([0,1],{\mathbb{M}}^{s}_{1}({\mathbb{R}}))$. If
$S^{\alpha}_{\hat{\mu}_{0}}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant 1})$ is
finite, and $\int\log|x|{\rm d}\hat{\nu}_{0}$, and the free entropy
$\Sigma(\hat{\nu}_{0})$ are finite, then we have

  1. (i)

There exists universal constant ${\mathfrak{C}}>0$ depends only on
$\hat{\mu}_{0}$ and $S_{\hat{\mu}_{0}}(\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant 1})$ such that the $L_{2}$ norms of $\hat{\nu}_{t}$ are uniformly
bounded,

$\displaystyle\int x^{2}{\rm d}\hat{\nu}_{t}(x)\leqslant{\mathfrak{C}},$ (4.7)

and if $\alpha>0$

$\displaystyle\int_{0}^{1}\int\frac{{\rm d}\hat{\nu}_{t}(x)}{x^{2}}{\rm
d}t\leqslant{\mathfrak{C}},\quad\quad\int-\log|x|{\rm
d}\hat{\nu}_{t}\leqslant{\mathfrak{C}}.$ (4.8)

  2. (ii)

$\hat{\nu}_{t}$ has a density for almost all $0\leqslant t\leqslant 1$, i.e.

$\displaystyle\frac{{\rm d}\hat{\nu}_{t}(x)}{{\rm d}x}=\hat{\rho}_{t}(x).$

  3. (iii)

We denote the velocity field
$u_{t}=(H(\hat{\nu}_{t})+(\alpha/2)H(\delta_{0}))+\partial_{x}k_{t}$, then it
satisfies the conservation of mass equation

$\displaystyle\partial_{t}\hat{\rho}_{t}+\partial_{x}(\hat{\rho}_{t}u_{t})=0,\quad
0\leqslant t\leqslant 1,$ (4.9)

in the sense of distribution. We can rewrite the dynamical entropy (4.4) as

$\displaystyle\begin{split}S_{\hat{\mu}_{0}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant 1})&=\frac{\beta}{2}\left(\int_{0}^{1}\int
u_{s}^{2}\hat{\rho}_{s}(x){\rm d}x{\rm
d}s+\frac{\pi^{2}}{3}\int_{0}^{1}\int\hat{\rho}^{3}_{s}{\rm
d}s+\frac{\alpha^{2}}{4}\int\frac{\hat{\rho}_{s}(x)}{x^{2}}{\rm d}x{\rm
d}s\right.\\\ &-\left.\left.\left(\Sigma(\hat{\nu}_{t})+\alpha\int\log|x|{\rm
d}\hat{\nu}_{t}(x)\right)\right|_{t=0}^{1}\right).\end{split}$ (4.10)

######  Remark 4.2.

Under assumptions of Proposition 4.1, if $\hat{\mu}_{0}$ and $\hat{\nu}_{t}$
are symmetrization of measures $\mu_{0}$ and $\nu_{t}$ supported on
$[0,\infty)$ respectively,

$\displaystyle\hat{\mu}_{0}(x)=(\mu_{0}(x)+\mu_{0}(-x))/2,\quad\hat{\nu}_{t}(x)=(\nu_{t}(x)+\nu_{t}(-x))/2,\quad
0\leqslant t\leqslant 1,$ (4.11)

then for almost all $0\leqslant t\leqslant 1$, we have
$\nu_{t}(x)=\rho_{t}(x){\rm d}x$, and it satisfies
$\hat{\rho}_{t}(x)=(\rho_{t}(x)+\rho_{t}(-x))/2$ and

$\displaystyle\partial_{t}\rho_{t}(x)+\partial_{x}(u_{t}(x)\rho_{t}(x))=0.$

In particular we have for almost all $0\leqslant t\leqslant 1$,
$\hat{\rho}_{t}(x)=(\rho_{t}(x)+\rho_{t}(-x))/2$. With $\rho_{t}$, we can
rewrite (3.8) as

$\displaystyle S_{\hat{\mu}_{0}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant 1})$ $\displaystyle=\frac{\beta}{2}\left(\int_{0}^{1}\int
u_{s}^{2}{\rm d}\rho_{s}{\rm
d}s+\frac{\pi^{2}}{12}\int_{0}^{1}\int\rho^{3}_{s}{\rm
d}s+\frac{\alpha^{2}}{4}\int\frac{\rho_{s}(x)}{x^{2}}{\rm d}x{\rm d}s\right.$
$\displaystyle-\left.\left.\left(\Sigma((\nu_{t}(x)+\nu_{t}(-x))/2)+\alpha\int\log|x|{\rm
d}\nu_{t}(x)\right)\right|_{t=0}^{1}\right)=:S_{\mu_{0}}^{\alpha}(\\{\nu_{t}\\}_{0\leqslant
t\leqslant 1}).$

###### Proof.

The claim that $S^{\alpha}_{\hat{\mu}_{0}}$ is a good rate function follows
from essentially the same arguments as in [30, Theorem 1.4]. The fact that
$S^{\alpha}_{\hat{\mu}_{0}}$ is lower semi-continuous comes from the
continuity of $\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant 1}\rightarrow
S^{\alpha}(\\{\hat{\nu}_{t},f_{t}\\}_{0\leqslant t\leqslant 1})$ for any
${\cal C}^{2,1}_{b}(\mathbb{R}\times[0,1])$ function. In fact, the only
difference lies in the new term

$-\frac{\alpha}{2}\int_{0}^{1}\int\frac{f_{s}^{\prime}(x)}{x}{\rm
d}\hat{\nu}_{s}(x){\rm d}s,$

which are continuous since $\hat{\nu}_{s}$ is even we can rewrite it as

$\displaystyle-\frac{\alpha}{4}\int_{0}^{1}\int\frac{f_{s}^{\prime}(x)-f_{s}^{\prime}(-x)}{x}{\rm
d}\hat{\nu}_{s}(x){\rm d}s,$

and it no long has a singularity at $x=0$. If
$S^{\alpha}_{\hat{\mu}_{0}}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant 1})$ is
finite, by definition we have $\hat{\nu}_{0}=\hat{\mu}_{0}$. To check that the
level set $\\{S^{\alpha}_{\hat{\mu}_{0}}\leqslant M\\}$ is included in a
compact set, one first checks that $\int x^{2}{\rm d}\hat{\nu}_{t}(x)$ is
uniformly bounded as in Proposition 3.2. Moreover if $f$ is in ${\cal
C}_{b}^{2,1}([0,1],{\mathbb{R}})$, (4.5) implies that

$\displaystyle|\hat{\nu}_{t}(f)-\hat{\nu}_{s}(f)|$ $\displaystyle\leqslant$
$\displaystyle
C\|f^{\prime\prime}\|_{\infty}|t-s|+\int_{s}^{t}\int|\partial_{x}k_{s}||f^{\prime}(x)|{\rm
d}\hat{\nu}_{s}(x)\mathrm{d}s$ $\displaystyle\leqslant$ $\displaystyle
C\|f^{\prime\prime}\|_{\infty}|t-s|+\|f^{\prime}\|_{\infty}\left(\int_{0}^{1}\int{\rm
d}\hat{\nu}_{s}(x)\int|\partial_{x}k_{s}|^{2}{\rm d}s\right)^{1/2}\sqrt{t-s}$
$\displaystyle\leqslant$ $\displaystyle
C\|f^{\prime\prime}\|_{\infty}|t-s|+\|f^{\prime}\|_{\infty}\left(2M\right)^{1/2}\sqrt{t-s}\,.$

This implies that $t\rightarrow\hat{\nu}_{t}(f)$ is tight by Arzela-Ascoli
theorem. The conclusion follows.

The estimate (4.7) can be proven in the same way as (3.6). In the following we
prove (4.8). We take a test function
$f(x)=-(\alpha\beta/4)\ln(\varepsilon+x^{2})$ in (4.4),

$\displaystyle\begin{split}&+\infty>S_{\hat{\mu}_{0}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant
1})=\hat{\nu}_{1}(f)-\hat{\nu}_{0}(f)+\frac{\alpha\beta}{4}\int_{0}^{1}\int\frac{x}{\varepsilon+x^{2}}H(\hat{\nu}_{s}){\rm
d}\hat{\nu}_{s}(x){\rm d}s\\\
&+\frac{\alpha^{2}\beta}{4}\int_{0}^{1}\int\frac{1}{\varepsilon+x^{2}}{\rm
d}\hat{\nu}_{s}(x){\rm
d}s-\frac{\alpha^{2}\beta}{8}\int^{1}_{0}\int\left(\frac{x}{\varepsilon+x^{2}}\right)^{2}{\rm
d}\hat{\nu}_{s}{\rm d}s\\\
&\geqslant\hat{\nu}_{1}(f)-\hat{\nu}_{0}(f)+\frac{\alpha\beta}{4}\int_{0}^{1}\int\frac{x}{\varepsilon+x^{2}}H(\hat{\nu}_{s}){\rm
d}\hat{\nu}_{s}(x){\rm
d}s+\frac{\alpha^{2}\beta}{8}\int_{0}^{1}\int\frac{1}{\varepsilon+x^{2}}{\rm
d}\hat{\nu}_{s}(x){\rm d}s.\end{split}$ (4.12)

By our assumption on $\hat{\mu}_{0}$, it holds that
$\hat{\mu}_{0}(\log|x|)>-\infty$. Using the fact that $\hat{\nu}_{s}$ is
symmetric, i.e. $\hat{\nu}_{s}(y)=\hat{\nu}_{s}(-y)$, we can rewrite the first
integral on the righthand side of (LABEL:e:tets) as

$\displaystyle\begin{split}&\phantom{{}={}}\int_{0}^{1}\int\frac{x}{\varepsilon+x^{2}}H(\hat{\nu}_{s}){\rm
d}\hat{\nu}_{s}(x){\rm
d}s=\int_{0}^{1}\int\frac{x}{\varepsilon+x^{2}}\int\frac{{\rm
d}\hat{\nu}_{s}(y)}{x-y}{\rm d}\hat{\nu}_{s}(x){\rm d}s\\\
&=\frac{1}{2}\int_{0}^{1}\int\left(\frac{x}{\varepsilon+x^{2}}-\frac{y}{\varepsilon+y^{2}}\right)\frac{{\rm
d}\hat{\nu}_{s}(y)}{x-y}{\rm d}\hat{\nu}_{s}(x){\rm
d}s=\frac{1}{2}\int_{0}^{1}\int\frac{\varepsilon-
xy}{(\varepsilon+x^{2})(\varepsilon+y^{2})}{\rm d}\hat{\nu}_{s}(y){\rm
d}\hat{\nu}_{s}(x){\rm d}s\\\
&=\frac{\varepsilon}{2}\int_{0}^{1}\left(\int\frac{1}{\varepsilon+x^{2}}{\rm
d}\hat{\nu}_{s}(x)\right)^{2}{\rm d}s\geqslant 0.\end{split}$ (4.13)

By plugging (LABEL:e:2t) into (LABEL:e:tets), and rearranging, we conclude
that there exists a constant ${\mathfrak{C}}$ depending only on $\rho_{0}$ and
$S_{\hat{\mu}_{0}}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant 1})$, such that

$\displaystyle\frac{\alpha^{2}\beta}{8}\int_{0}^{1}\int\frac{1}{\varepsilon+x^{2}}{\rm
d}\hat{\nu}_{s}(x){\rm
d}s\leqslant{\mathfrak{C}}+\int\ln(\varepsilon+x^{2}){\rm
d}\hat{\nu}_{1}(x)\lesssim{\mathfrak{C}},$

where we used (4.7) for the last inequality. Moreover, we also have that
$\int\ln(\varepsilon+x^{2}){\rm d}\hat{\nu}_{t}(x)\gtrsim-{\mathfrak{C}}$. The
claim (4.8) follows by sending $\varepsilon$ to $0$.

For Item (ii), let
$u_{t}=(H(\hat{\nu}_{t})+(\alpha/2)H(\delta_{0}))+\partial_{x}k_{t}(x)$, then
(4.6) implies that

$\displaystyle S^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant t\leqslant
1})=\frac{\beta}{2}\int_{0}^{1}\partial_{x}k_{s}(x)^{2}{\rm
d}\hat{\nu}_{s}{\rm
d}s=\frac{\beta}{2}\int_{0}^{1}(u_{s}-H(\hat{\nu}_{s})+(\alpha/2)H(\delta_{0}))^{2}{\rm
d}\hat{\nu}_{s}{\rm d}s<+\infty.$ (4.14)

In the following we show that

$\displaystyle\int_{0}^{1}\int(u_{s}-H(\hat{\nu}_{s}))^{2}{\rm
d}\hat{\nu}_{s}{\rm d}s<+\infty.$ (4.15)

If $\alpha=0$, then (4.15) is the same as (4.14). We assume $\alpha>0$, then
we can write (4.14) as

$\displaystyle\begin{split}+\infty&>\int_{0}^{1}\int\left((u_{s}-H(\hat{\nu}_{s}))^{2}-(u_{s}-H(\hat{\nu}_{s}))\frac{\alpha}{x}+\frac{\alpha^{2}}{4x^{2}}\right){\rm
d}\hat{\nu}_{s}{\rm d}s\\\
&\geqslant\int_{0}^{1}\left(\frac{1}{2}(u_{s}-H(\hat{\nu}_{s}))^{2}-\frac{\alpha^{2}}{4x^{2}}\right){\rm
d}\hat{\nu}_{s}{\rm d}s\\\
&\geqslant\int_{0}^{1}\frac{1}{2}(u_{s}-H(\hat{\nu}_{s}))^{2}{\rm
d}\hat{\nu}_{s}{\rm d}s-{\mathfrak{C}},\end{split}$ (4.16)

where we used (4.8) for the last inequality. The claim (4.15) follows by
rearranging (4.16).

The measure process $\hat{\nu}_{t}$ satisfies (4.15), which verifies the
assumption in Proposition 3.2. Item (ii) in Proposition 3.2 implies that
$\hat{\nu}_{t}$ has a density for almost surely all $0\leqslant t\leqslant 1$,
$\hat{\nu}_{t}=\hat{\rho}_{t}(x){\rm d}x$, and (4.9) holds. Moreover, we have

$\displaystyle\begin{split}&\phantom{{}={}}\int_{0}^{1}(u_{s}-H(\hat{\nu}_{s}))^{2}{\rm
d}\hat{\nu}_{s}{\rm d}s\\\ &=\int_{0}^{1}u_{s}^{2}\hat{\rho}_{s}{\rm d}x{\rm
d}s+\int_{0}^{1}\int H(\hat{\rho}_{s})^{2}\hat{\rho}_{s}{\rm d}x{\rm
d}s-\left(\Sigma(\hat{\nu}_{1})-\Sigma(\hat{\nu}_{0})\right)\\\
&=\int_{0}^{1}u_{s}^{2}\hat{\rho}_{s}{\rm d}x{\rm
d}s+\frac{\pi^{2}}{3}\int_{0}^{1}\int\hat{\rho}_{s}^{3}{\rm d}x{\rm
d}s-\left(\Sigma(\hat{\nu}_{1})-\Sigma(\hat{\nu}_{0})\right).\end{split}$
(4.17)

In the following we prove (4.10). The same as in (4.16), we have

$\displaystyle\begin{split}\int_{0}^{1}\int\partial_{x}k_{s}(x)^{2}\hat{\rho}_{s}{\rm
d}x{\rm
d}s=\int_{0}^{1}\left((u_{s}-H(\hat{\rho}_{s}))^{2}-(u_{s}-H(\hat{\rho}_{s}))\frac{\alpha}{x}+\frac{\alpha^{2}}{4x^{2}}\right)\hat{\rho}_{s}(x){\rm
d}x{\rm d}s.\end{split}$ (4.18)

For the second term on the righthand side of (4.18), we have

$\displaystyle\int_{0}^{1}(u_{s}-H(\hat{\rho}_{s}))\frac{\alpha}{x}\hat{\rho}_{s}{\rm
d}x{\rm d}s=\int_{0}^{1}\frac{\alpha}{x}u_{s}\hat{\rho}_{s}{\rm d}x{\rm
d}s-\int_{0}^{1}\frac{\alpha}{x}H(\hat{\rho}_{s})\hat{\rho}_{s}{\rm d}x{\rm
d}s.$ (4.19)

For the first term on the righthand side of (4.19), we have

$\displaystyle\begin{split}\partial_{s}\int\log x\hat{\rho}_{s}{\rm
d}x&=\int\log|x|\partial_{s}\hat{\rho}_{s}{\rm
d}x=-\int\log|x|\partial_{x}(u_{s}\hat{\rho}_{s}){\rm
d}x=\int\frac{u_{s}}{x}\hat{\rho}_{s}{\rm d}x.\end{split}$ (4.20)

For the second term on the righthand side of (4.19), we notice the following
equality

$\displaystyle\begin{split}\int
H(\hat{\rho}_{s})\frac{1}{x}\hat{\rho}_{s}(x){\rm
d}x&=\iint\frac{\hat{\rho}_{s}(y)}{x-y}\frac{\hat{\rho}_{s}(x)}{x}{\rm d}x{\rm
d}y=\frac{1}{2}\iint\left(\frac{1}{x(x-y)}-\frac{1}{y(x-y)}\right)\hat{\rho}_{s}(x)\hat{\rho}_{s}(y){\rm
d}x{\rm d}y\\\
&=-\frac{1}{2}\iint\frac{1}{xy}\hat{\rho}_{s}(x)\hat{\rho}_{s}(y){\rm d}x{\rm
d}y=0,\end{split}$

where we used that $\hat{\rho}_{s}$ is symmetric, i.e.
$\hat{\rho}_{s}(x)=\hat{\rho}_{s}(-x)$.

The estimates (LABEL:e:nee) and (4.18) together give us the new formula of the
dynamical entropy

$\displaystyle\begin{split}S_{\hat{\mu}_{0}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{0\leqslant
t\leqslant 1})&=\frac{\beta}{2}\left(\int_{0}^{1}\int u_{s}^{2}{\rm
d}\hat{\rho}_{s}{\rm
d}s+\frac{\pi^{2}}{3}\int_{0}^{1}\int\hat{\rho}^{3}_{s}{\rm
d}s+\frac{\alpha^{2}}{4}\int\frac{\hat{\rho}_{s}(x)}{x^{2}}{\rm d}x{\rm
d}s\right.\\\
&-\left.\left.\left(\Sigma(\hat{\nu}_{t})+\alpha\int\log|x|\hat{\nu}_{t}(x){\rm
d}x\right)\right|_{t=0}^{1}\right).\end{split}$ (4.21)

This finishes the proof of Item (iii). ∎

###  4.2 Free Rectangular Convolution

The minimizer of the dynamical entropy (4.21) is characterized by the free
rectangular convolution as introduced in [7, 5]. For any
$\lambda=1/(1+\alpha)\in[0,1]$, the rectangular free convolution denoted by
$\boxplus_{\lambda}$ can be defined in terms of the rectangular $R$-transform.
For any symmetric measure $\hat{\nu}$ on ${\mathbb{R}}$, its Stieltjes
transform is given by

$\displaystyle G_{\hat{\nu}}(z)=\int\frac{{\rm d}\hat{\nu}(x)}{z-x}.$

The rectangular $R$-transform $C_{\hat{\nu}}(w)$ with ratio $\lambda$ of
$\hat{\nu}$ is defined on a neighborhood of zero by

$\displaystyle zG_{\hat{\nu}}(z)-1=C_{\hat{\nu}}(w),\quad
w=G_{\hat{\nu}}(z)(\lambda G_{\hat{\nu}}(z)+(1-\lambda)/z).$ (4.22)

When the symmetric measure $\hat{\nu}$ is the delta mass $\delta_{0}$ at zero,
we have $G_{\delta_{0}}(z)=1/z$, and the $R$-transform $C_{\delta_{0}}(w)=0$.
Let $W_{n}$ be a sequence of $n\times m$ matrices with entries given by
independent real/complex Gaussian random variables with mean zero and variance
one, where $m\geqslant n$ and $n/m\rightarrow\lambda\in[0,1]$. Then the
empirical eigenvalues of $(\sigma W_{n}/\sqrt{n})(\sigma W_{n}/\sqrt{n})^{*}$
converges to the rescaled Marchenko-Pastur law

$\displaystyle\mu_{\sigma
W}=\frac{\sqrt{(\sigma^{2}(1/\sqrt{\lambda}+1)^{2})-x)(x-\sigma^{2}(1/\sqrt{\lambda}-1)^{2}})}{\sqrt{2\pi}\sigma
x},$

with the Stieltjes transform given by

$\displaystyle zm_{\mu_{\sigma
W}}(z)-1=\frac{\sigma^{2}}{\lambda}m_{\mu_{\sigma W}}(z)(\lambda
zm_{\mu_{\sigma W}}(z)+1-\lambda),\quad m_{\mu_{\sigma W}}=\int\frac{{\rm
d}\mu_{\sigma W}(x)}{z-x}.$ (4.23)

We denote the limiting symmetrized singular value distribution of $\sigma
W_{n}$ by $\hat{\nu}_{\sigma W}$, we call it the square root Marchenko-Pastur
law. Then

$\displaystyle G_{\hat{\nu}_{\sigma W}}(z)=\int\frac{{\rm d}\hat{\nu}_{\sigma
W}}{z-x}=\frac{1}{2}\left(\int\frac{{\rm d}\hat{\nu}_{\sigma
W}}{z-x}+\int\frac{{\rm d}\hat{\nu}_{\sigma W}}{z-x}\right)=\int\frac{z{\rm
d}\hat{\nu}_{\sigma W}}{z^{2}-x^{2}}=\int\frac{z{\rm d}\mu_{\sigma
W}}{z^{2}-x}=zm_{\mu_{\sigma W}}(z^{2}),$ (4.24)

and we can rewrite the relation (4.23) as

$\displaystyle zG_{\hat{\nu}_{\sigma
W}}(z)-1=\frac{\sigma^{2}}{\lambda}G_{\hat{\nu}_{\sigma
W}}(z)(G_{\hat{\nu}_{\sigma W}}+(1-\lambda)/z).$

Comparing with the defining relation of rectangular $R$-transform (4.22), we
conclude that the rectangular $R$-transform of the square root Marchenko-
Pastur law is given by

$\displaystyle C_{\hat{\nu}_{\sigma W}}(w)=\frac{\sigma^{2}}{\lambda}w.$
(4.25)

######  Theorem 4.3.

Let $A_{n},B_{n}\in{\mathbb{R}}^{n\times m}$ and
$U\in{\mathcal{O}}(n),V\in{\mathcal{O}}(m)$ following Haar distribution over
orthogonal group for $\beta=1$; $A_{n},B_{n}\in{\mathbb{C}}^{n\times m}$ and
$U\in{\mathcal{U}}(n),V\in{\mathcal{U}}(m)$ following Haar distribution over
unitary group, for $\beta=2$, where $m\geqslant n$ and
$n/m\rightarrow\lambda\in[0,1]$. We assume that the symmetrized empirical
singular values $\hat{\nu}_{A}^{n}$ and $\hat{\nu}_{B}^{n}$ of $A_{n}$ and
$B_{n}$ converge to $\hat{\nu}_{A}$ and $\hat{\nu}_{B}$ respectively. Then the
symmetrized empirical singular values of $A_{n}+U_{n}B_{n}V^{*}$ converges
weakly in probability to $\hat{\nu}_{A\boxplus_{\lambda}B}$, the free
rectangular convolution

$\displaystyle\hat{\nu}_{A\boxplus_{\lambda}B}=\hat{\nu}_{A}\boxplus_{\lambda}\hat{\nu}_{B},$

which is characterized by

$\displaystyle
C_{\hat{\nu}_{A\boxplus_{\lambda}B}}(w)=C_{\hat{\nu}_{A}}(w)+C_{\hat{\nu}_{B}}(w).$
(4.26)

To use (4.26) to solve for the measure $\hat{\nu}_{A\boxplus_{\lambda}B}$, we
need to solve

$\displaystyle w=G_{\hat{\nu}_{A\boxplus_{\lambda}B}}(z)(\lambda
G_{\hat{\nu}_{A\boxplus_{\lambda}B}}(z)+(1-\lambda)/z)=G_{\hat{\nu}_{A}}(z^{\prime})(\lambda
G_{\hat{\nu}_{A}}(z^{\prime})+(1-\lambda)/z^{\prime}).$ (4.27)

where $z,z^{\prime}$ belongs to some neighborhood of infinity. We formally
show below how to deduce a closed equation for
$G_{\hat{\nu}_{A\boxplus_{\lambda}B}}(z)$ from (4.27) and (4.26) and leave the
reader check that we can take $z,z^{\prime}$ in such a neighborhood. We notice
that the defining relation (4.22) gives that

$\displaystyle\frac{G_{\hat{\nu}_{A}}(z^{\prime})}{z^{\prime}}=\frac{1+C_{\hat{\nu}_{A}}(w)}{(z^{\prime})^{2}}=\frac{w}{\lambda
C_{\hat{\nu}_{A}}(w)+1},$ (4.28)

and from (4.26) we get

$\displaystyle\frac{G_{\hat{\nu}_{A\boxplus_{\lambda}B}}(z)}{z}$
$\displaystyle=\frac{1+C_{\hat{\nu}_{A\boxplus_{\lambda}B}}(w)}{z^{2}}=\frac{w}{\lambda
C_{\hat{\nu}_{A\boxplus_{\lambda}B}}(w)+1}$ (4.29)
$\displaystyle=\frac{w}{\lambda C_{\hat{\nu}_{A}}(w)+1+\lambda
C_{\hat{\nu}_{B}}(w)}=\frac{1}{\frac{z^{\prime}}{G_{\hat{\nu}_{A}}(z^{\prime})}+\lambda\frac{C_{\hat{\nu}_{B}}(w)}{w}}.$

In particular we can rearrange (4.28) and (4.29) as

$\displaystyle\frac{z^{\prime}}{G_{\hat{\nu}_{A}}(z^{\prime})}=\frac{z}{G_{\hat{\nu}_{A\boxplus_{\lambda}B}}(z)}-\frac{\lambda
C_{\hat{\nu}_{B}}(w)}{w}.$ (4.30)

Thanks to (4.27), we have the expression of $w$ in terms of
$G_{\hat{\nu}_{A\boxplus_{\lambda}B}}(z)$ and $z$. We can then solve
$z^{\prime}$ using (4.27) and (4.30) in terms of
$G_{\hat{\nu}_{A\boxplus_{\lambda}B}}(z)$ and $z$. Plugging them into (4.30),
we finally get a self-consistent equation for
$G_{\hat{\nu}_{A\boxplus_{\lambda}B}}(z)$ and $z$ which has a unique solution
in a neighborhood of infinity, which determines
$\hat{\nu}_{A\boxplus_{\lambda}B}$.

The empirical distribution of eigenvalues of large dimensional information-
plus-noise type matrices [23, 24, 3] can also be characterized by rectangular
free convolution. This model is of particular interest because of its
applications in statistics. The following Theorem is a special case of [23,
24], which deals with more general noise.

######  Theorem 4.4.

Let $A_{n}$ be an sequence of $n\times m$ matrices and $W_{n}$ be a sequence
of $n\times m$ matrices with entries given by independent real or complex
Gaussian random variables with mean zero and variance $1/n$, where $m\geqslant
n$ and $n/m\rightarrow\lambda\in[0,1]$. If the eigenvalue distributions of
$A_{n}A_{n}^{*}$ converge to $\mu_{A}$. Then the empirical eigenvalue
distributions of $(A_{n}+\sigma W_{n})(A_{n}+\sigma W_{n})^{*}$ converge to a
deterministic measure $\mu_{A\boxplus_{\lambda}\sigma W}$ with Stieltjes
transform $m(z)$ given by

$\displaystyle\int\frac{{\rm
d}\mu_{A}(x)}{(1-\sigma^{2}m(z))\left((1-\lambda\sigma^{2}m(z))z-(1-\lambda)\sigma^{2})\right)-x}=\frac{m(z)}{1-\sigma^{2}m(z)},\quad
m(z)=\int\frac{{\rm d}\mu_{A\boxplus_{\lambda}\sigma W}(x)}{z-x}.$ (4.31)

The limit $\lim_{\eta\rightarrow 0+}{\mathrm{Im}}[m(x+\eta\rm i)]$ exists, it
is analytic when it is positive and away from $0$, and

$\displaystyle|m(z)|\leqslant\left(\frac{1}{\sigma^{2}|z|}\right)^{1/2}.$
(4.32)

We can reformulate Theorem 4.4 in terms of rectangular $R$-transform. We
recall the rectangular $R$-transform of the square root Marchenko-Pastur law
from (4.25) $C_{\hat{\nu}_{\sigma W}}(w)=\sigma^{2}w/\lambda$. We denote the
limiting symmetrized empirical singular value distribution of $A_{n}$ and
$A_{n}+\sigma W_{n}$ as $\hat{\nu}_{A}$ and
$\hat{\nu}_{A\boxplus_{\lambda}\sigma W}$ respectively, then (4.31) is
equivalent to

$\displaystyle C_{\hat{\nu}_{A\boxplus_{\lambda}\sigma
W}}(w)=C_{\hat{\nu}_{A}}(w)+\sigma^{2}w/\lambda.$

The Stieltjes transform $m(z)$ can be expressed in terms of the Stieltjes
transform of $\hat{\nu}_{A\boxplus_{\lambda}\sigma W}$, and the bound (4.32)
becomes

$\displaystyle\frac{1}{z}G_{\hat{\nu}_{A\boxplus_{\lambda}\sigma
W}}(z)=m(z^{2}),\quad|G_{\hat{\nu}_{A\boxplus_{\lambda}\sigma
W}}(z)|\leqslant\frac{1}{\sigma}.$ (4.33)

By letting $z$ approach the support of $\hat{\nu}_{A\boxplus_{\lambda}\sigma
W}$ in (4.33), we conclude that $\hat{\nu}_{A\boxplus_{\lambda}\sigma W}$ has
a density bounded by $\operatorname{O}(1/\sigma)$, and it is analytic on its
support.

For later purpose, we show that free convolution reduces the dynamical
entropy.

######  Lemma 4.5.

Let $\hat{\nu}\in\mathcal{C}([0,1],\mathbb{M}_{1}^{s}(\mathbb{R}))$ and
$p\in\mathbb{M}_{1}^{s}(\mathbb{R}))$. Then for all $\lambda\geqslant 0$,

$S^{\alpha}_{\hat{\nu}_{0}\boxplus_{\lambda}p}(\\{\hat{\nu}_{t}\boxplus_{\lambda}p\\}_{t\in[0,1]})\leqslant
S^{\alpha}(\\{\hat{\nu}_{t}\\}_{t\in[0,1]})\,.$

######  Remark 4.6.

If we take $p$ to be the square root Marchenko-Pastur law on scale
$\varepsilon$, i.e. $\sigma_{\varepsilon}=\hat{\nu}_{\varepsilon W}$ as in
(4.24), then Theorem 4.4 implies
$\hat{\nu}_{t}^{\varepsilon}=\hat{\nu}_{t}\boxplus_{\lambda}\sigma_{\varepsilon}$
has an analytic density, which is bounded by
$\operatorname{O}(1/\varepsilon)$, and Lemma 4.5 implies

$\displaystyle
S^{\alpha}_{\hat{\nu}^{\varepsilon}_{0}}(\\{\hat{\nu}^{\varepsilon}_{t}\\}_{t\in[0,1]})\leqslant
S_{\hat{\mu}_{0}}^{\alpha}(\\{\hat{\nu}_{t}\\}_{t\in[0,1]})\,.$

###### Proof.

With the notation of rectangular free convolution, we can construct the
limiting object of the matrix Brownian motions $A_{n}$ and $G(t)/\sqrt{n}$
from (2.4),

$\displaystyle{\mathsf{a}}=\left(\begin{array}[]{cc}0&a\\\
a^{*}&0\end{array}\right),\quad{\mathsf{g}}(t)=\left(\begin{array}[]{cc}0&g(t)\\\
g(t)^{*}&0\end{array}\right),$

where $g(t)$ and $a$ are the limit in $*$-moments of $G(t)/\sqrt{n}$ and
$A_{n}$ respectively. Then ${\mathsf{a}}$ and ${\mathsf{g}}(t)$ are free and
${\mathsf{g}}(t)$ has independent increment in terms of rectangular free
convolution, with $\lambda=1/(1+\alpha)$.

For any measurable odd function $\\{\partial_{x}k_{t}(x)\\}_{0\leqslant
t\leqslant 1}$, we consider the following noncommutative stochastic process,

$\displaystyle{\rm d}{\mathsf{h}}(t)={\rm
d}{\mathsf{g}}(t)+\partial_{x}k_{t}({\mathsf{h}}(t)){\rm
d}t,\quad{\mathsf{h}}(0)={\mathsf{a}}.$

Then the nonzero part of the spectral measure $\hat{\rho}_{t}$ of
${\mathsf{h}}(t)$, satisfies the equation (4.9)

$\displaystyle\partial\hat{\rho}_{t}+\partial_{x}((H(\hat{\rho}_{t})+\alpha
H(\delta_{0})/2+\partial_{x}k_{t})\hat{\rho}_{t})=0.$ (4.34)

Then for any symmetric probability measure $\hat{p}$ on ${\mathbb{R}}$, the
process $\hat{\rho}^{p}_{t}=\hat{\rho}_{t}\boxplus_{\lambda}\hat{p}$ satisfies
the same differential equation but with $\partial_{x}k_{t}$ replaced by
$\partial_{x}k^{p}_{t}$ defined by, if ${\mathsf{h}}(t)$ and ${\mathsf{p}}$
are two free random variables with respect to the rectangular free convolution
in a non-commutative probability space $\tau$ with distribution
$\hat{\rho}_{t}$ and $\hat{p}_{t}$ respectively, then

$\displaystyle\partial_{x}k^{p}_{t}(x)=\tau(\partial_{x}k_{t}({\mathsf{h}}(t))|{\mathsf{h}}(t)+{\mathsf{p}}).$

As a consequence, we find

$\displaystyle
S^{\alpha}(\\{\hat{\rho}_{t}\boxplus_{\lambda}\hat{p}\\}_{0\leqslant
t\leqslant 1})$
$\displaystyle=\frac{\beta}{2}\int_{0}^{1}\hat{\rho}_{s}\boxplus_{\lambda}\hat{p}(\partial_{x}(k^{p}_{s})^{2}){\rm
d}s=\frac{\beta}{2}\int_{0}^{1}\tau(\tau(\partial_{x}k_{s}({\mathsf{h}}(t))|{\mathsf{h}}(t)+{\mathsf{p}}))^{2}){\rm
d}s$
$\displaystyle\leqslant\frac{\beta}{2}\int_{0}^{1}\tau(\partial_{x}k_{s}({\mathsf{h}}(t))^{2}){\rm
d}s=\frac{\beta}{2}\int_{0}^{1}\int\partial_{x}k_{s}(x)^{2}\hat{\rho}_{s}{\rm
d}s=S^{\alpha}(\\{\hat{\rho}_{t}\\}_{0\leqslant t\leqslant 1}).$

∎

###  4.3 Itô’s calculus for the Dyson Bessel process

In this section, we derive dynamical equations of linear statistics of Dyson
Bessel process for general test functions using Itô’s formula. Take any test
function $f_{t}(x)\in{\cal C}_{b}^{2,1}([0,1]\times{\mathbb{R}})$. Using to
Itô’s lemma, (4.2) gives

$\displaystyle\begin{split}&\phantom{{}={}}{\rm
d}\sum_{i\in[\\![{-n,n}]\\!]\setminus\\{0\\}}f_{t}(s_{i}(t))=\sum_{i\in[\\![{-n,n}]\\!]\setminus\\{0\\}}f^{\prime}_{t}(s_{i}(t)){\rm
d}s_{i}(t)+\partial_{t}f_{t}(s_{i}(t)){\rm
d}t+\frac{f^{\prime\prime}_{t}(s_{i}(t))}{2\beta n}{\rm d}t\\\ &={\rm
d}L^{f}_{t}+\frac{1}{4n}\sum_{i\neq
j\in[\\![{-n,n}]\\!]\setminus\\{0\\}}\frac{f^{\prime}_{t}(s_{i}(t))-f^{\prime}_{t}(s_{j}(t))}{s_{i}(t)-s_{j}(t)}\mathrm{d}t+\left(\alpha_{n}-\frac{1}{2n}\right)\sum_{i\in[\\![{-n,n}]\\!]\setminus\\{0\\}}\frac{f_{t}^{\prime}(s_{i}(t))}{2s_{i}(t)}{\rm
d}t\\\
&+\sum_{i\in[\\![{-n,n}]\\!]\setminus\\{0\\}}\partial_{t}f_{t}(s_{i}(t)){\rm
d}t+\sum_{i\in[\\![{-n,n}]\\!]\setminus\\{0\\}}\frac{f^{\prime\prime}_{t}(s_{i}(t))}{2\beta
n}{\rm d}t,\end{split}$ (4.35)

where the martingale term is given by

$\displaystyle{\rm
d}L^{f}_{t}=\sum_{i\in[\\![{-n,n}]\\!]\setminus\\{0\\}}f^{\prime}_{t}(s_{i}(t))\frac{{\rm
d}W_{i}(t)}{\sqrt{\beta n}},$ (4.36)

for ${i\in[\\![{-n,n}]\\!]\setminus\\{0\\}}$.

We recall the empirical particle density $\\{\hat{\nu}^{n}_{t}\\}_{0\leqslant
t\leqslant 1}$ from (4.3). With it, we can rewrite (LABEL:e:lin) as

$\displaystyle\begin{split}&\phantom{{}={}}\int f_{t}(x){\rm
d}\hat{\nu}^{n}_{t}-\int f_{t}(x){\rm d}\hat{\nu}^{n}_{0}\\\
&=\frac{L^{f}_{t}}{2n}+\frac{1}{2}\int_{0}^{t}\int_{x\neq
y}\frac{f_{s}^{\prime}(x)-f_{s}^{\prime}(y)}{x-y}{\rm
d}\hat{\nu}^{n}_{s}(x){\rm d}\hat{\nu}^{n}_{s}(y){\rm
d}s+\left(\alpha_{n}-\frac{1}{2n}\right)\int_{0}^{t}\int\frac{f_{s}^{\prime}(x)}{2x}{\rm
d}\hat{\nu}^{n}_{s}(x){\rm d}s\\\ &+\frac{1}{2\beta
n}\int_{0}^{t}\int(f^{\prime\prime}_{s}(x)){\rm d}\hat{\nu}^{n}_{s}(x){\rm
d}s+\int_{0}^{t}\int\partial_{s}f_{s}(x){\rm d}\hat{\nu}^{n}_{s}(x){\rm d}s\\\
&=\frac{L^{f}_{t}}{2n}+\frac{1}{2}\int_{0}^{t}\int\frac{f_{s}^{\prime}(x)-f_{s}^{\prime}(y)}{x-y}{\rm
d}\hat{\nu}^{n}_{s}(x){\rm d}\hat{\nu}^{n}_{s}(y){\rm
d}s+\left(\alpha_{n}-\frac{1}{2n}\right)\int_{0}^{t}\int\frac{f_{s}^{\prime}(x)}{2x}{\rm
d}\hat{\nu}^{n}_{s}(x){\rm d}s\\\ &+\left(\frac{1}{2\beta
n}-\frac{1}{4n}\right)\int_{0}^{t}\int f^{\prime\prime}_{s}(x){\rm
d}\hat{\nu}^{n}_{s}(x){\rm d}s+\int_{0}^{t}\int\partial_{s}f_{s}(x){\rm
d}\hat{\nu}^{n}_{s}(x){\rm d}s\\\
&=\frac{L^{f}_{t}}{2n}+\int_{0}^{t}\int\partial_{s}f_{s}(x){\rm
d}\hat{\nu}^{n}_{s}(x){\rm d}s+\int_{0}^{t}\int
f_{s}^{\prime}(x)H(\hat{\nu}^{n}_{s}){\rm d}\hat{\nu}^{n}_{s}(x){\rm d}s\\\
&+\left(\frac{\alpha_{n}}{2}-\frac{1}{4n}\right)\int_{0}^{t}\int
f_{s}^{\prime}(x)H(\delta_{0}){\rm d}\hat{\nu}^{n}_{s}(x){\rm
d}s+\frac{1}{n}\left(\frac{1}{2\beta}-\frac{1}{4}\right)\int_{0}^{t}\int
f^{\prime\prime}_{t}(x){\rm d}\hat{\nu}^{n}_{s}(x){\rm d}s.\\\
&=\frac{L^{f}_{t}}{2n}+\int_{0}^{t}\int\partial_{s}f_{s}(x){\rm
d}\hat{\nu}^{n}_{s}(x){\rm d}s+\int_{0}^{t}\int
f_{s}^{\prime}(x)H(\hat{\nu}^{n}_{s}){\rm d}\hat{\nu}^{n}_{s}(x){\rm d}s\\\
&+\frac{\alpha}{2}\int_{0}^{t}\int f_{s}^{\prime}(x)H(\delta_{0}){\rm
d}\hat{\nu}^{n}_{s}(x){\rm d}s+\varepsilon_{t}^{n}(\\{f_{s}\\}_{0\leqslant
s\leqslant 1}),\end{split}$ (4.37)

where

$\varepsilon^{n}_{t}(\\{f_{s}\\}_{0\leqslant s\leqslant
t})=\frac{1}{n}\left(\frac{1}{2\beta}-\frac{1}{4}\right)\int_{0}^{t}\int
f^{\prime\prime}_{t}(x){\rm d}\hat{\nu}^{n}_{s}(x){\rm
d}s+\left(\frac{\alpha_{n}-\alpha}{2}-\frac{1}{4n}\right)\int_{0}^{t}\int
f_{s}^{\prime}(x)H(\delta_{0}){\rm d}\hat{\nu}^{n}_{s}(x){\rm d}s,$

goes to zero uniformly when $\sup_{t}\|f^{\prime\prime}_{t}\|_{\infty}$ is
finite. Since $s_{-i}(t)=-s_{i}(t)$, we can rewrite the martingale
$\\{L_{t}^{f}\\}_{0\leqslant t\leqslant 1}$ from (4.37) as

$L^{f}_{t}=\sum_{i=1}^{n}\frac{1}{\sqrt{\beta
n}}\int_{0}^{t}(f^{\prime}(s_{i}(u))-f^{\prime}(-s_{i}(u)))\mathrm{d}W_{i}(u),$
(4.38)

and its quadratic variation is given by,

$\displaystyle\langle L^{f},L^{f}\rangle_{t}=\frac{1}{\beta