Now, we give an argument for the existence of a solution starting from any $\lambda_0 \in \mathbb{R}^N$.
Let us consider the fundamental case $A_0 = 0$ as an initial matrix, so with initial spectrum
$\lambda_0 = 0 \in \mathbb{R}^N$.

We consider a sequence of initial data $(\lambda_{0,n})_{n \in \mathbb{N}} \in (\mathbb{R}_>^N)^N$ that converges to $0 \in \mathbb{R}^N$ when $n$ goes to $+\infty$.

It remains to notice that for all $\lambda \in \mathbb{R}^N$:

$$
\det \left[ \left( \exp \left( -\frac{|\lambda^i - \lambda_{0,n}^j|^2}{2t} \right) \right)_{1 \le i,j \le N} \right] = \exp \left( -\frac{|\lambda|^2}{2t} \right) \exp \left( -\frac{|\lambda_{0,n}|^2}{2t} \right) \det \left[ \left( \exp \left( \frac{\lambda^i \lambda_{0,n}^j}{t} \right) \right)_{1 \le i,j \le N} \right]. \tag{6.27}
$$

Using a Taylor expansion of the exponential, we have that for $n$ that goes to $+\infty$:

$$
\begin{align*}
\det \left[ \left( \exp \left( \frac{\lambda^i \lambda_{0,n}^j}{t} \right) \right)_{1 \le i,j \le N} \right]
&= \det \left[ \left( \sum_{k=0}^{+\infty} \left( \frac{(\lambda^i)^k (\lambda_{0,n}^j)^k}{k! t^k} \right)_{1 \le i,j \le N} \right) \right] \\
&= \det \left[ \left( \sum_{k=0}^{N-1} \left( \frac{(\lambda^i)^k (\lambda_{0,n}^j)^k}{k! t^k} \right)_{1 \le i,j \le N} \right) \right] + o(\Delta_N(\lambda_{0,n})) \\
&= \det \left[ \left( \sum_{k=0}^{N-1} \left( \frac{(\lambda^i)^k}{k! t^k} (\lambda_{0,n}^j)^k \right)_{1 \le i,j \le N} \right) \right] + o(\Delta_N(\lambda_{0,n})) \\
&= \det \left[ \left( \frac{(\lambda^i)^{k-1}}{(k-1)! t^{k-1}} \right)_{1 \le i,k \le N} \right] \det \left[ \left( (\lambda_{0,n}^i)^{k-1} \right)_{1 \le k,i \le N} \right] + o(\Delta_N(\lambda_{0,n})) \\
&= \frac{1}{1! 2! \dots (N-1)!} \Delta_N \left( \frac{\lambda}{t} \right) \Delta_N (\lambda_{0,n}) + o(\Delta_N(\lambda_{0,n})). \tag{6.28}
\end{align*}
$$

Applying Theorem 6.13 with $\lambda_{0,n} \in \mathbb{R}_>^N$ as an initial data, we shall pass to the limit in
the formula (6.25) when $n$ goes to $+\infty$ and using (6.27) and (6.28) yields that the solution
of the Fokker Planck equation (6.20) with initial condition $\lambda_0 = 0 \in \mathbb{R}^N$ is given by:

$$
\rho(t, \lambda) = \frac{1}{(2\pi t)^{N/2} 1! 2! \dots N!} \exp\left(-\frac{|\lambda|^2}{2t}\right) \Delta_N(\lambda) \Delta_N\left(\frac{\lambda}{t}\right). \quad (6.29)
$$

To justify the passage to the limit, we can use the Dyson theorem 5.8 to interpret $\rho(t, \cdot)$ as the law of the eigenvalues of $D_N^2$ and using the Hoffman-Wielandt Lemma 4.3 to pass to the limit.

Taking $t=1$ in (6.29), we obtain that the density of the eigenvalues of a $N$ complex Wigner matrix is given by:

$$
\rho(1, \lambda) = \frac{1}{(2\pi)^{N/2} 1! 2! \dots N!} \exp\left(-\frac{|\lambda|^2}{2}\right) |\Delta_N(\lambda)|^2.
$$

This gives another proof of the formula obtained in Theorem 2.25 and Theorem 2.48.

6.4 Interpretation of this system of particle

We shall give a better physical interpretation of (6.16) revealing the important role of the
Vandermonde determinant. This section is quite formal.
This interpretation is linked with the notion of Brownian motions conditioned to live in a Weyl Chamber. Here the Weyl Chamber is $\mathbb{R}^N$. We shall not detail the notion of Weyl chamber in this section, focusing on the particular Dyson case. We refer to [41, 53] for the links between random matrices and Weyl Chambers and [69] for general results about diffusion processes conditioned to remain in a certain domain. This part is inspired by the blog article of Mufan Li: An Unusually Clean Proof: Dyson Brownian Motion via Conditioning on Non-intersection.

First we shall briefly recall the notion of h-transform (or Doob transform) of a Markov process. For a general introduction to h-transform, we refer to [34].

### 6.4.1 h-transform

**Definition 6.15.** Let $(X(t))_{t \ge 0}$ be a Markov process with value in $\mathbb{R}^d$. An event A is said to be invariant if for all $x \in \mathbb{R}^d$, for all $s > 0$ we have:

$$ \mathbb{P}((X(t))_{t \ge 0} \in A | X(0) = x) = \mathbb{P}(X(t+s))_{t \ge 0} \in A | X(s) = x). $$

For a such process X and event A, let us define:

$$ \begin{aligned} \mathbb{P}_x(\cdot) &= \mathbb{P}(\cdot | X(0) = x) \\ h(x) &= \mathbb{P}_x(A) \\ \mathbb{P}^t(x, dy) &= \mathbb{P}_x(X(t)) \\ \tilde{\mathbb{P}}^t(x, dy) &= \mathbb{P}_x(X(t) \in dy | A). \end{aligned} $$

**Proposition 6.16.** With the notations of the previous definition, we have that for all $x \in \mathbb{R}^d$ the measure $\tilde{\mathbb{P}}^t(x, \cdot)$ is absolutely continuous with respect to the measure $\mathbb{P}^t(x, \cdot)$ and the Radon-Nikodyn derivative is given by

$$ g(y) = h(y)/h(x). $$

*Proof.* We give the intuition of the proof. Thanks to the Bayes formula and invariance of A we have:

$$ \tilde{\mathbb{P}}^t(x, dy) = \mathbb{P}_x(X(t) \in dy | A) = \frac{\mathbb{P}_x(A | X(t) \in dy) \mathbb{P}_x(X(t) \in dy)}{\mathbb{P}_x(A)} = \frac{h(y)}{h(x)} \mathbb{P}^t(x, dy). $$

□

For a Markov process we can define the generator by

$$ L[f](x) = \lim_{t \to 0} \frac{\mathbb{E}_x(f(X(t)) - f(x))}{t}, $$

when this quantity exists.

We can express the generator $\tilde{L}$ under the probability $\tilde{P}$ of X. Indeed by Proposition 6.16,
we get:

$$
\begin{align*}
\tilde{L}[f](x) &= \lim_{t \to 0} \frac{\mathbb{E}_x(f(X(t))|A) - f(x)}{t} \\
&= \lim_{t \to 0} \frac{1}{t} \left( \int f(y) \tilde{\mathbb{P}}^t(x, dy) - f(x) \right) \\
&= \lim_{t \to 0} \frac{1}{t} \left( \int f(y) \frac{h(y)}{h(x)} \tilde{\mathbb{P}}^t(x, dy) - f(x) \right) \tag{6.30} \\
&= \frac{1}{h(x)} \lim_{t \to 0} \frac{1}{t} \left( \int f(y) h(y) \tilde{\mathbb{P}}^t(x, dy) - f(x) h(x) \right) \\
&= \frac{1}{h(x)} L[fh](x).
\end{align*}
$$

An important example is when $X(t)$ is the solution of a SDE. Indeed, consider $X$ as the solution of the SDE:

$$dX(t) = g(X(t))dt + dB(t),$$

with $g$ a function defined on an open subset $\Omega$ of $\mathbb{R}^d$ to $\mathbb{R}^d$ and smooth. By the Itô formula, the generator associated of $(X(t))_{t \ge 0}$ is

$$L[f](x) = \langle g(x), \nabla f(x) \rangle + \frac{1}{2} \Delta f(x),$$

with $\langle \cdot, \cdot \rangle$ the canonical scalar product on $\mathbb{R}^d$.

Now, assume that $h$ is harmonic on $\Omega$ which means $\Delta h(x) = 0$ on $\Omega$. Using the formula (6.30) we have:

$$\tilde{L}[f](x) = \langle g(x) + \nabla \log(h(x)), \nabla f(x) \rangle + \frac{1}{2} \Delta f(x). \quad (6.31)$$

We recognise the generator of the solution of the SDE

$$d\tilde{X}(t) = (g(\tilde{X}(t)) + \nabla \log h(\tilde{X}(t)))dt + dB(t).$$

So, the conditioning by an invariant event transform the law of a process solution of a SDE into the law of another process, solution of another SDE whose difference is the drift term.

### 6.4.2 Law of Brownian motions conditioned on the fact they do not intersect

We want to prove the following result.

**Theorem 6.17.** Let $(\lambda_t^i)_{t \ge 0, 1 \le i \le N}$ be the solution of (6.16) starting from $\lambda_0 \in \mathbb{R}_>^N$. Let $(B(t))_{t \ge 0}$ be a Brownian motion in $\mathbb{R}^N$ starting from $\lambda_0$. Then, we have the following equality in law:

$$(\lambda_t^i)_{t \ge 0, 1 \le i \le N} = (B_i(t))_{t \ge 0, 1 \le i \le N} |A,$$

$$\text{with } A = \{\omega \in \Omega | \forall t > 0, B(t)(\omega) \in D\}.$$

Let us notice that we use a slight abuse of notation in the previous theorem. Indeed the event $A$ is of 0 probability. But as for the Brownian bridge (for which we conditioned by the event $B_1 = 0$) the meaning of this conditional probability is that we shall construct a non increasing sequence of probability event $(A_c)_{c>0}$ which converges to $A$ but for all $c > 0$, the probability of $A_c$ is not 0.
By definition $B(t) \notin D$ is equivalent to $\Delta_N(B(t)) = 0$ and moreover we notice that $\Delta_N(\lambda) > 0$ for $\lambda \in \mathbb{R}_>^N$. The harmonic property of the Vandermonde determinant shall be used to compute the h-transform of $(B(t))_{t \ge 0}$.

For all $c > 0$, let

$$A_c = \{(\Delta_N(B(t)))_{t \ge 0} \text{ hits c before 0}\},$$

and note that

$$A = \lim_{c \to \infty} A_c.$$

Moreover, to compute $\mathbb{P}_{\lambda_0}(A_c)$ we shall use the same idea as for the gambler ruin problem for the discrete random walk.

**Lemma 6.18.** Let $\tau = \inf\{t > 0, \Delta_N(B(t)) \text{ hits c or 0}\}$, then $(\Delta_N(B(t)))_{t \wedge \tau}$ is a martingale.

*Proof.* This is a consequence of the Itô formula. Since $\Delta_N(\lambda) = 0$ by Lemma 6.10, in the Itô formula we only have the local martingale term which is this case a real martingale using the expression of the first derivative of the Vandermonde determinant. $\square$

The stopping time theorem yields that

$$\mathbb{E}_{\lambda}(\Delta_N(B(\tau))) = \Delta_N(\lambda).$$

By definition of $\tau$, we have:

$$\mathbb{E}_{\lambda}(\Delta_N(B(\tau))) = c \mathbb{P}_{\lambda}(A_c).$$

Hence it gives the expression of the h-transform with respect to the event $A_c$:

$$h_c(\lambda) = \mathbb{P}_{\lambda}(A_c) = \frac{\Delta_N(\lambda)}{c}.$$

Now thanks to the formula (6.31), for all $c > 0$ under the probability $\mathbb{P}$ conditioned by the event $A_c$, the law of $(B(t))_{t \ge 0}$ is the law of the solution of the SDE

$$d\lambda_t^c = \nabla \log h_c(\lambda_t^c) dt + dB(t).$$

Since $h_c(x) = \frac{\Delta_N(x)}{c}$, Lemma 6.10 yields:

$$\frac{\partial \log h_c}{\partial x_i}(x) = \sum_{j \neq i} \frac{1}{x_i - x_j}.$$

Hence the law of the process $(\lambda_t^c)_{t \ge 0}$ is the same for all $c > 0$ and so we can let $c$ goes to $+\infty$ so that the law of $(B(t))_{t \ge 0}$ conditioned by the event that $(B(t))_{t \ge 0}$ remains in $D$ is the law of the solution of (6.16).
# 7 Mean field limit of the system of particles

In this section, we shall deal with the Dyson case (5.9) and the Ornstein-Uhlenbeck case (5.22). More precisely, we shall focus the case $\beta = 1$ in the Dyson case to lighten the notations, but all the results also hold for the case $\beta = 2$. For the Dyson case we refer to the Section 4.3 of [5] and for the Ornstein-Uhlenbeck case to [29, 71, 61].

The method we will use is really standard when dealing with system of particles in mean field interaction. The general strategy is to prove a compactness in a certain space (Section 7.3) and then to prove the uniqueness of the limit point (Section 7.4.2). For people interested in this kind of questions, we recommend after the read of this section to see for instance [4] for a similar approach for the study of the eigenvalues of random matrices in a non confining potential or [79] for the Keller-Segel in chemotaxis which is a much more difficult model than the Dyson model because of the difficulty to prove the uniqueness of the limit point.

In all this section, we consider a family $(B^i)_{i \in \mathbb{N}}$ of independent Brownian motions defined on a same probability space $(\Omega, \mathcal{F}, \mathbb{P})$.

## 7.1 Empirical distribution

Given a solution $(\lambda^i)_{1 \le i \le N}$ of the SDE system (5.9), we define the empirical distribution of this system of particles as:

$$ \mu_N(t) = \frac{1}{N} \sum_{i=1}^{N} \delta_{\lambda_t^i}. $$

$\mu_N$ is a random variable and we shall specify in which space it is valued.

Fix $T > 0$ to be a fix time. For all $\omega \in \Omega$, $\mu_N(\omega)$ is a continuous function from $[0, T]$ to $\mathcal{P}(\mathbb{R})$ endowing $\mathcal{P}(\mathbb{R})$ with the natural weak topology which is metrizable by certain metrics. We shall detail more precisely in Section 7.2 the topology we will consider.

For a measurable function $f: \mathbb{X} \to \mathbb{R}$ and a measure $\mu$ on a measurable set $\mathbb{X}$ we write $\langle \mu, f \rangle := \int_{\mathbb{X}} f d\mu$.

We first give a key proposition that shall be used in all this section.

**Proposition 7.1.** Let $f \in C^2([0, T] \times \mathbb{R}, \mathbb{R})$ with all its derivatives bounded, then for all $t \in [0, T]$, we have:

$$ \begin{aligned} \langle \mu_N(t), f(t, \cdot) \rangle ={}& \langle \mu_N(0), f(0, \cdot) \rangle + \int_0^t \langle \partial_s f(s, \cdot), \mu_N(s) \rangle ds \\ & + \frac{1}{2} \int_0^t \int \int \frac{\partial_x f(s, x) - \partial_y f(s, y)}{x-y} d\mu_N(s)(x)d\mu_N(s)(y)ds \\ & + (2-1) \frac{1}{2N} \int_0^t \langle \mu_N(s), \partial_x^2 f(s, \cdot) \rangle ds + M_f^N(t) \end{aligned} \quad (7.1) $$

with $(M_f^N(t))_{0 \le t \le T}$ a martingale given by:

$$ M_f^N(t) := \frac{\sqrt{2}}{N^{3/2}} \sum_{i=1}^{N} \int_{0}^{t} \partial_x f(s, \lambda_t^i) dB_s^i. $$

*Proof.* By definition of $\mu_N(t)$ we have:

$$ \langle \mu_N(t), f(t, \cdot) \rangle = \frac{1}{N} \sum_{i=1}^{N} f(t, \lambda_t^i). $$
Using that $(\lambda^i)_{1\le i\le N}$ are solution of (5.9) with $\beta = 1$, the Itô formula yields:

$$ \begin{aligned} \langle \mu_N(t), f(t, \cdot) \rangle = & \langle \mu_N(0), f(0, \cdot) \rangle + \int_0^t \langle \partial_s f(s, \cdot), \mu_N(s) \rangle ds \\ & + \frac{1}{2} \int_0^t \int \int_{x \ne y} \frac{\partial_x f(s, x)}{x-y} d\mu_N(s)(x) d\mu_N(s)(y) ds \\ & + 2 \frac{1}{2N} \int_0^t \langle \mu_N(s), \partial_x^2 f(s, \cdot) \rangle ds + M_f^N(t). \end{aligned} \quad (7.2) $$

We remark that for every $s \in [0, T]$

$$ \int \int_{x \ne y} \frac{\partial_x f(s, x)}{x-y} d\mu_N(s)(x) d\mu_N(s)(y) = \frac{1}{2} \int \int_{x \ne y} \frac{\partial_x f(s, x) - \partial_y f(s, y)}{x-y} d\mu_N(s)(x) d\mu_N(s)(y). $$

The quantity

$$ \frac{\partial_x f(s, x) - \partial_y f(s, y)}{x - y} $$

can be extended to $\partial_x^2 f(s, x)$ if $x = y$.

Hence we can rewrite for $s \in [0, T]$:

$$ \begin{aligned} \int \int_{x \ne y} \frac{\partial_x f(s, x)}{x-y} d\mu_N(s)(x) d\mu_N(s)(y) &= \frac{1}{2} \int \int_{x \ne y} \frac{\partial_x f(s, x) - \partial_y f(s, y)}{x-y} d\mu_N(s)(x) d\mu_N(s)(y) \\ &= \frac{1}{2} \int \int \frac{\partial_x f(s, x) - \partial_y f(s, y)}{x-y} d\mu_N(s)(x) d\mu_N(s)(y) - \frac{1}{2} \int \int_{x=y} \partial_x^2 f(s, x) d\mu_N(s)(x) d\mu_N(s)(y). \end{aligned} \quad (7.3) $$

We then compute the last term:

$$ \begin{aligned} \int \int_{x=y} \partial_x^2 f(s, x) d\mu_N(s)(x) d\mu_N(s)(y) &= \frac{1}{N^2} \sum_{i=1}^N \sum_{j=1}^N \mathbb{1}_{\lambda_s^i = \lambda_s^j} \partial_x^2 f(s, \lambda_s^i) \\ &= \frac{1}{N} \langle \mu_N(s), \partial_x^2 f(s, \cdot) \rangle \end{aligned} \quad (7.4) $$

So (7.3) and (7.4) gives:

$$ \begin{aligned} \int_0^t \int \int_{x \ne y} \frac{\partial_x f(s, x)}{x-y} d\mu_N(s)(x) d\mu_N(s)(y) ds = & \\ \frac{1}{2} \int_0^t \int \int \frac{\partial_x f(s, x) - \partial_y f(s, y)}{x-y} d\mu_N(s)(x) d\mu_N(s)(y) ds - & \\ \int_0^t \frac{1}{2N} \langle \mu_N(s), \partial_x^2 f(s, \cdot) \rangle ds. & \end{aligned} \quad (7.5) $$

Using (7.5) in (7.2) gives (7.1).

Finally, we shall prove that $(M_f^N(t))_{0 \le t \le T}$ is not just a local martingale but a martingale.

Indeed we can compute its bracket:

$$ \langle M_f^N(t), M_f^N(t) \rangle = \frac{2}{N^3} \int_0^t \sum_{i=1}^N (\partial_x f(s, \lambda_s^i))^2 ds \le \frac{2t \sup_{s \in [0,T]} ||\partial_x f(s, \cdot)||_\infty^2}{N^2}. \quad (7.6) $$

This implies that $\mathbb{E}[\langle M_f^N(T), M_f^N(T) \rangle] < +\infty$. So $(M_f^N(t))_{0 \le t \le T}$ is a martingale. $\square$
## 7.2 Compact subsets of $C([0, T], \mathcal{P}(\mathbb{R}))$

This part can be skip for people non familiar with topology. The only result that shall be used after is Lemma 7.2 that gives some typical compact subsets of $C([0, T], \mathcal{P}(\mathbb{R}))$.

As mentioned before, we consider $\mathcal{P}(\mathbb{R})$ as a topological space with the weak topology. Since now we fix a family $\mathcal{F} := (f_i)_{i \ge 1}$ of $C^2(\mathbb{R}, \mathbb{R})$ functions such that for all $i \ge 1$, $||f_i||_\infty + ||f_i'||_\infty + ||f_i''||_\infty \le 1$ and $\text{span}(\mathcal{F})$ is dense in the set of continuous compactly supported functions. Then the distance on $\mathcal{P}(\mathbb{R})$ defined by:

$$d_{\mathcal{F}}(\mu, \nu) = \sum_{n \ge 0} 2^{-n} \left| \int f_{n+1} d\mu - \int f_{n+1} d\nu \right|$$

metrizes the weak topology on $\mathcal{P}(\mathbb{R})$.

Now we endow $C([0, T], \mathcal{P}(\mathbb{R}))$ with the uniform convergence: $d(f, g) = \sup_{0 \le t \le T} d_{\mathcal{F}}(f(t), g(t))$.

With this distance $C([0, T], \mathcal{P}(\mathbb{R}))$ is a Polish space.

**Lemma 7.2.** Let $K$ be a compact subset of $\mathcal{P}(\mathbb{R})$, and $(C_i)_{i \ge 1}$ be a family of compact subsets of $C([0, T], \mathbb{R})$. Then the set

$$K = C([0, T], \mathcal{P}(\mathbb{R})) \bigcap \{\forall t \in [0, T], \mu(t) \in K\} \bigcap_{i \ge 1} \{t \mapsto \langle \mu(t), f_i \rangle \in C_i\}$$

is a compact subset of $C([0, T], \mathcal{P}(\mathbb{R}))$.

*Proof.* We have to prove that $K$ is relatively compact and closed.

$K$ is closed as an intersection of closed subsets. It remains to prove that $K$ is relatively compact. Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of elements of $K$. By definition, for all $i \ge 1$ the sequence of functions $(t \to \langle \mu_n(t), f_i \rangle)_{n \in \mathbb{N}}$ is a sequence of functions in $C_i$. Hence by a diagonalization extraction argument, we can find a subsequence $\phi$ such that for all $i \ge 1$, $(t \to \langle \mu_{\phi(n)}(t), f_i \rangle)_{n \in \mathbb{N}}$ converges towards a function $g_i \in C_i$.

Now, let $(t_k)_{k \in \mathbb{N}}$ a dense countable sequence in $[0, T]$. By a same diagonalization extraction argument we can find sub sub sequence that shall be written $\phi$ such that for all $k \ge 0$ we have $\mu_{\phi(n)}(t_k)$ converges to $\mu(t_k) \in \mathcal{P}(\mathbb{R})$ and still for all $i \ge 1$, $(t \to \langle \mu_{\phi(n)}(t), f_i \rangle)_{n \in \mathbb{N}}$ converges towards $g_i \in C_i$. We necessarily have that

$$\langle \mu(t_k), f_i \rangle = g_i(t_k) \qquad (7.7)$$

for all $i \ge 1$ and $k \ge 0$. We then extend $\mu$ on $[0, T]$ by a continuity argument. Indeed, by construction, for all $k \ge 0$, $\mu(t_k) \in K$. Let $0 \le t \le T$, and consider a subsequence, still denoted $t_k$ that converges to $t$. The sequence of measures $(\mu(t_k))_{k \ge 0}$ is relatively compact since for all $k \ge 0$, $\mu(t_k) \in K$. Moreover, let $\nu \in \mathcal{P}(\mathbb{R})$ be a accumulation point, passing to the limit in (7.7) yields

$$\langle \nu, f_i \rangle = g_i(t)$$

for all $i \ge 1$ by continuity of $g$. Since the $(f_i)_{i \ge 1}$ are dense, these relations uniquely characterize $\nu$. So $(\mu(t_k))_{k \ge 0}$ converges to an element $\mu(t) \in \mathcal{P}(\mathbb{R})$ such that for all $i \ge 1$, $\langle \mu(t), f_i \rangle = g_i(t)$. This extend $\mu$ continuously as an element of $C([0, T], \mathcal{P}(\mathbb{R}))$. Since, $(f_i)_{i \ge 1}$ is a dense sequence and $(t \to \langle \mu_{\phi(n)}(t), f_i \rangle)_{n \in \mathbb{N}}$ converges towards a function $\langle \mu(\cdot), f_i \rangle \in C_i$ (uniformly on $[0, T]$) for all $i \ge 1$, $(\mu_{\phi(n)})_{n>0}$ converges to $\mu$ in $C([0, T], \mathcal{P}(\mathbb{R}))$. $\square$
## 7.3 Compactness of $(\mu_N)_{N \ge 1}$

The goal of this section is to prove that almost surely the sequence $(\mu_N)_{N > 0}$ is relatively compact in $C([0, T], \mathcal{P}(\mathbb{R}))$. Now we suppose that for all $N \ge 1$, $(\lambda_0^{i,N})_{1 \le i \le N} \in D$ is the initial data for the system of $N$ particles (5.9). We assume that:

$$C_0 := \sup_{N \ge 0} \frac{1}{N} \sum_{i=1}^{N} \log((\lambda_0^{i,N})^2 + 1) < +\infty.$$

**Proposition 7.3.** The sequence $(\mu_N)_{N \ge 1}$ is almost surely relatively compact in $C([0, T], \mathcal{P}(\mathbb{R}))$.

*Proof.* We follow the arguments of [5]. We shall apply Lemma 7.2 to construct a compact subset of $C([0, T], \mathcal{P}(\mathbb{R}))$ such that for $N$ large enough $\mu_N$ is in this compact almost surely. So, there is two parts in the proof. The first one will be to construct a compact subset of $\mathcal{P}(\mathbb{R})$ in which almost surely for all $t \le T$, $\mu_N(t)$ is and then compacts subsets of $C([0, T], \mathbb{R})$ as in Lemma 7.2.

Let us consider the function $f(x) := \log(1 + x^2)$ which is $C^2$ on $\mathbb{R}$ and $||f'||_\infty \le 1$, $||f''||_\infty \le 2$. We also have that:

$$\forall x, y \in \mathbb{R}, \left| \frac{f'(x) - f'(y)}{x - y} \right| \le ||f''||_\infty \le 2.$$

Hence using the Itô formula (7.1), we get the following bound:

$$\forall N > 0, \sup_{t \le T} |\langle \mu_N(t), f \rangle| \le |\langle \mu_N(0), f \rangle| + T \left(1 + \frac{1}{N}\right) + \sup_{t \le T} |M_f^N(t)|. \quad (7.8)$$

Moreover, by Proposition 7.1, $(M_f^N(t))_{0 \le t \le T}$ is a martingale with its bracket bounded by:

$$\sup_{0 \le t \le T} \langle M_f^N(t), M_f^N(t) \rangle \le \frac{2T}{N^2}, \quad (7.9)$$

as obtained in (7.6). Let

$$K_M := \{ \mu \in \mathcal{P}(\mathbb{R}), \int \log(1 + x^2) d\mu(x) \le M \}$$

for a certain $M > 0$ that will be specified later.

$K_M$ is a compact subset of $\mathcal{P}(\mathbb{R})$. Indeed, $K_M$ is closed by the Portmanteau theorem and is relatively compact by the Prokhorov theorem since the Markov inequality yields that for every $\mu \in K_M$, for every $K > 0$

$$\mu(\mathbb{R} - [-K, K]) \le \frac{\int \log(1 + x^2) d\mu(x)}{\log(1 + K^2)} \le \frac{M}{\log(1 + K^2)}.$$

Now we shall prove that almost surely, for $N$ large enough, for all $0 \le t \le T$, $\mu_N(t) \in K_M$ for a good $M$. We use the Borel-Cantelli Lemma.

Indeed, we first estimate the martingale term using (7.9), the Markov inequality and the Burkholder-Davis-Gundy inequality:

$$\mathbb{P}(\sup_{t \le T} |M_f^N(t)| \ge \varepsilon) \le \frac{C \mathbb{E}(\langle M_f^N(T), M_f^N(T) \rangle)}{\varepsilon^2} \le \frac{2TC}{N^2 \varepsilon^2}, \quad (7.10)$$
with $C > 0$ a constant independent of $\varepsilon$ and $N$ given by the Burkholder–Davis–Gundy inequality. Hence using (7.8), we obtain that for all $N > 0$

$$ \sup_{t \le T} |\langle \mu_N(t), f \rangle| \le |\langle \mu_N(0), f \rangle| + T \left(1 + \frac{1}{N}\right) + \sup_{t \le T} |M_f^N(t)| \le C_0 + T \left(1 + \frac{1}{N}\right) + \sup_{t \le T} |M_f^N(t)|. $$

So for $M > 2T + C_0$, we get

$$ \mathbb{P} \left( \sup_{t \le T} |\langle \mu_N(t), f \rangle| \ge M \right) \le \mathbb{P} \left( \sup_{t \le T} |M_f^N(t)| \ge M - 2T - C_0 \right) \le \frac{\tilde{C}(T)}{(M - 2T - C_0)^2 N^2}, \quad (7.11) $$

with $\tilde{C}(T)$ a constant independent of $M$ and $N$.

Hence the Borel Cantelli lemma yields:

$$ \mathbb{P} \left( \bigcup_{K} \bigcap_{N \ge K} \{ \forall t \le T, \mu_N(t) \in K_M \} \right) = 1. \quad (7.12) $$

This proves the first part.

Now we shall find good functions $(f_i)_{i \ge 0}$ and compact subsets $(C_i)_{i \ge 0}$ of $C([0, T])$ such that almost surely the sequence $(\mu_N)_{N > 0}$ is in

$$ \bigcap_{i>0} \{t \mapsto \langle \mu(t), f_i \rangle \in C_i\} $$

for $N$ large enough.

Let us recall that the sets of the form:

$$ C = \bigcap_n \{g \in C([0, T], \mathbb{R}), \sup_{|t-s| \le \eta_n} |g(t) - g(s)| \le \varepsilon_n, \sup_{t \le T} |g(t)| \le M\} \quad (7.13) $$

with $M > 0$ a constant, $(\varepsilon_n)_{n \ge 0}$ and $(\eta_n)_{n \ge 0}$ two sequence that converge to 0 are compact subsets of $C([0, T], \mathbb{R})$ by the Arzela-Ascoli theorem. So, we shall look at the oscillation of $t \to \langle \mu_N(t), g \rangle$ with $g \in C^2(\mathbb{R})$ and its derivatives bounded by 1.

Let $M > 2$ and $\delta \in (0, 1)$, thanks to (7.1) we have that for all $0 \le s, t \le T$,

$$ |\langle \mu_N(t), g \rangle - \langle \mu_N(s), g \rangle| \le ||g''||_\infty |s-t| + |M_g^N(t) - M_g^N(s)|. $$

Let $|s-t| \le \delta$ then we have:

$$ |\langle \mu_N(t), g \rangle - \langle \mu_N(s), g \rangle| \le \delta + |M_g^N(t) - M_g^N(s)|. \quad (7.14) $$

Now, shall prove the following estimate:

$$ \mathbb{P} \left( \sup_{|t-s| \le \delta} |\langle \mu_N(t), g \rangle - \langle \mu_N(s), g \rangle| \ge M \delta^{1/8} \right) \le \frac{a \delta^{1/2}}{N^4 M^4}, \quad (7.15) $$

with $a > 0$ a constant independent of $\delta$, $M$, $N$ and $g$.

By (7.14) we have:

$$ \mathbb{P} \left( \sup_{|t-s| \le \delta} |\langle \mu_N(t), g \rangle - \langle \mu_N(s), g \rangle| \ge M \delta^{1/8} \right) \le \mathbb{P} \left( \sup_{|t-s| \le \delta} |M_g^N(t) - M_g^N(s)| \ge (M-1) \delta^{1/8} \right). \quad (7.16) $$
It remains to estimate right hand term of (7.16). By the same computation as for (7.6)
we have that for all $t \ge k\delta$,

$$
|\langle M_g^N(t) - M_g^N(k\delta), M_g^N(t) - M_g^N(k\delta) \rangle|^2 \le 4 \frac{(t - k\delta)^2 ||g'||_\infty^2}{N^4}. \quad (7.17)
$$

The idea is to truncate $[0, T]$ in interval of length $\delta$. Let $J = \lfloor T/\delta \rfloor$ and now we bound the
right hand term using as before the Markov inequality and the Burkholder–Davis–Gundy
inequality:

$$
\begin{align*}
& \mathbb{P} \left( \sup_{|t-s| \le \delta} |M_g^N(t) - M_g^N(s)| \ge (M-1)\delta^{1/8} \right) \\
& \le \sum_{k=1}^{J+1} \mathbb{P} \left( \sup_{k\delta \le t \le (k+1)\delta} |M_g^N(t) - M_g^N(k\delta)| \ge \frac{(M-1)\delta^{1/8}}{3} \right) \\
& \le \sum_{k=1}^{J+1} \frac{3^4}{\delta^{1/2}(M-1)^4} \mathbb{E} \left( \sup_{k\delta \le t \le (k+1)\delta} |M_g^N(t) - M_g^N(k\delta)|^4 \right) \tag{7.18} \\
& \le \sum_{k=1}^{J+1} \frac{3^4}{\delta^{1/2}(M-1)^4} \mathbb{E} \left( \left| \langle M_g^N((k+1)\delta) - M_g^N(k\delta) \rangle \right|^2 \right) \\
& \le \frac{C\delta^2}{(7.17) N^4 \delta^{1/2} (M-1)^4} (J+1) ||g'||_\infty^2 \le \frac{a\delta^{1/2}}{N^4 (M-1)^4} ||g'||_\infty^2,
\end{align*}
$$

with $a$ a constant independent of $N$, $M$, $\delta$ and $g$. This gives (7.15)

For $g$ a $C^2$ function with derivatives bounded by 1 and $\varepsilon > 0$ we define the subset of $C([0, T], \mathcal{P}(\mathbb{R}))$

$$
C(g, \varepsilon) := \bigcap_{n \ge 1} \{ \mu \in C([0, T], \mathcal{P}(\mathbb{R})), \sup_{|t-s| \le n^{-4}} |\langle \mu(t), g \rangle - \langle \mu(s), g \rangle| \le \frac{1}{\varepsilon\sqrt{n}} \}.
$$

As a consequence of (7.15) we obtain

$$
\begin{align}
\mathbb{P}(\mu_N \notin C(g, \varepsilon)) &\le \sum_{n \ge 1} \mathbb{P} \left( \mu_N \notin \{\mu \in C([0, T], \mathcal{P}(\mathbb{R})), \sup_{|t-s| \le n^{-4}} |\langle \mu(t), g \rangle - \langle \mu(s), g \rangle| \frac{1}{\varepsilon\sqrt{n}}\} \right) \nonumber \\
&\le \sum_{(7.15) \atop n \ge 1} \frac{a\varepsilon^4}{N^4 n^2} := \frac{\tilde{a}\varepsilon^4}{N^4}, \tag{7.19}
\end{align}
$$

with $\tilde{a}$ a constant independent of $g$, $\varepsilon$ and $N$. We recall that the family of functions
$\mathfrak{F} = (f_i)_{i \ge 1}$ is the family introduced at the beginning of Section 7.2 to define a metric on
$\mathcal{P}(\mathbb{R})$. Set $\varepsilon_k := 1/k$. Using (7.19) we get:

$$
\mathbb{P} \left( \mu_N \notin \bigcap_k C(f_k, \varepsilon_k) \right) \le \frac{b}{N^4}, \quad (7.20)
$$

with $b > 0$ a constant independent on $N$. This quantity is summable and so the Borel
Cantelli lemma yields:

$$
\mathbb{P} \left( \bigcup_{K \ge 1} \bigcap_{N \ge K} \{ \mu_N \in \bigcap_k C(f_k, \varepsilon_k) \} \right) = 1. \qquad (7.21)
$$
Set

$$ \mathcal{K} = K_M \bigcap_k C(f_k, \varepsilon_k) \subset C([0, T], \mathcal{P}(\mathbb{R})), $$

which is a compact subset of $C([0, T], \mathcal{P}(\mathbb{R}))$ by Lemma 7.2. By (7.14) and (7.21) we get that

$$ \mathbb{P}\left(\bigcup_{K \ge 1} \bigcap_{N \ge K} \{\mu_N \in \mathcal{K}\}\right) = 1. $$

Hence, almost surely the sequence $(\mu_N)_N$ is relatively compact in $C([0, T], \mathcal{P}(\mathbb{R}))$. $\square$

<figure><img src="image_12.png" /></figure>

## 7.4 Convergence of $\mu_N$

In this section we shall prove that almost surely the sequence $(\mu_N)_{N \ge 1}$ converges to an element of $C([0, T], \mathcal{P}(\mathbb{R}))$. We assume that $\mu_N(0) \in \mathcal{P}(\mathbb{R})$ converges weakly to a measure $\mu_0 \in \mathcal{P}(\mathbb{R})$.

Since the sequence is relatively compact we shall just prove that there exists a unique limit for this sequence. Fix $\omega \in \Omega$.

### 7.4.1 Characterization of the limit

First, we can give a description of a limit point of $(\mu_N(\omega))_{N \ge 1}$.

**Proposition 7.4.** Let $\mu \in C([0, T], \mathcal{P}(\mathbb{R}))$ be a limit point of the sequence $(\mu_N(\omega))_{N \ge 1}$ in the space $C([0, T], \mathcal{P}(\mathbb{R}))$, then $\mu(0) = \mu_0$ and for every $f \in C^2([0, T] \times \mathbb{R}, \mathbb{R})$ with its derivatives bounded, for every $t \ge 0$,

$$ \begin{aligned} \int f(t, x)d\mu(t)(x) = \int f(0, x)d\mu_0(x) + \int_0^t \int \partial_s f(s, x)d\mu(s)(x)ds \\ + \frac{1}{2}\int_0^t \int \int \frac{\partial_x f(s, x) - \partial_y f(s, y)}{x-y}d\mu(s)(x)d\mu(s)(y)ds. \end{aligned} \quad (7.22) $$

*Proof.* We pass to the limit in the Itô formula (7.1). The martingale term $\left( \sup_{0 \le t \le T} |M_f^N(t)| \right)_{N \ge 1}$ goes to 0 almost surely using the Markov inequality and the Burkholder-Davis-Gundy inequality as we already did for the compactness of the sequence in Proposition 7.3. $\square$

To identify the limit point we shall use the Stieltjes transform. We recall that we introduced this transformation and some standard properties in Appendix A.2.

**Definition 7.5.** Let $\mu$ be a probability measure on $\mathbb{R}$. We define its Stieltjes transform as:

$$ S_{\mu}(z) = \int_{\mathbb{R}} \frac{1}{z - x} d\mu(x), \quad z \in \mathbb{C} - \mathbb{R}. $$

**Example 7.6.** We computed in Proposition A.20 the Stieltjes transform the semicircle distribution defined by:

$$ \sigma(dx) := \frac{1}{2\pi} \sqrt{(4 - x^2)}_+ dx. $$

It is equal to:

$$ S_{\sigma}(z) = \frac{z - \sqrt{z^2 - 4}}{2}, $$

with $\sqrt{z}$ a suitable determination of the square root.
**Proposition 7.7.** Let $\mu$ be a limit point of $(\mu_N(\omega))_{N \ge 1}$ as in Proposition 7.4, then $\mu(0) = \mu_0$. Set $S_t := S_{\mu(t)}$ for the Stieltjes transform of the measure $\mu(t)$ for every $t \ge 0$ then we have

$$S_t(z) = S_0(z) - \int_0^t S_s(z) \partial_z S_s(z) ds, \forall z \in \mathbb{C} - \mathbb{R} \quad (7.23)$$

*Proof.* This is an immediate consequence of Proposition 7.4. Evaluating (7.22) with the test functions $f_z(t, x) = 1/(z-x)$ for $z \in \mathbb{C} - \mathbb{R}$, we obtain:

$$
\begin{align*}
S_t(z) &= S_0(z) + \frac{1}{2} \int_0^t \int \int \frac{(z-x)^2 - (z-y)^2}{x-y} d\mu(s)(x)d\mu(s)(y)ds \\
&= S_0(z) + \frac{1}{2} \int_0^t \int \int \frac{(z-y)^2 - (z-x)^2}{(z-x)^2(z-y)^2(x-y)} d\mu(s)(x)d\mu(s)(y)ds \\
&= S_0(z) + \frac{1}{2} \int_0^t \int \int \frac{2z-x-y}{(z-x)^2(z-y)^2} d\mu(s)(x)d\mu(s)(y)ds \\
&= S_0(z) + \frac{1}{2} \int_0^t \int \int \frac{z-x+(z-y)}{(z-x)^2(z-y)^2} d\mu(s)(x)d\mu(s)(y)ds \\
&= S_0(z) + \int_0^t \int \int \frac{z-x}{(z-x)^2(z-y)^2} d\mu(s)(x)d\mu(s)(y)ds \\
&= S_0(z) + \int_0^t \int \int \frac{1}{(z-x)(z-y)^2} d\mu(s)(x)d\mu(s)(y)ds \\
&= S_0(z) - \int_0^t S_s(z) \partial_z S_s(z) ds
\end{align*}
$$

□

### 7.4.2 Uniqueness of the limit and convergence

We fix $\mu \in C([0, T], \mathcal{P}(\mathbb{R}))$ a limit point of $(\mu_N(\omega))_{N \ge 1}$ and we recall that we defined $S_t := S_{\mu(t)}$ as the Stieljes transform of the measure $\mu(t)$ for every $t \ge 0$ which is solution of (7.23) and satisfies $\mu(0) = \mu_0$.

Equation (7.23) can be written

$$\partial_t S(t, z) + S(t, z) \partial_z S(t, z) = 0.$$

This is a Burger's equation on the complex plane. Uniqueness of a solution of a Burger equation is not immediate and sometimes it is not the case. In this case we should prove the uniqueness of a solution of this Burger equation using that we look for solutions that are the Stieljes transform of a certain measure at any time. We shall use the method of characteristic. Set $\mathbb{H} = \{z \in \mathbb{C}, \mathfrak{Im}(z) > 0\}$ and recall by Remark (A.18) that the Stieljes transform on $\mathbb{H}$ uniquely determined a real probability measure (see Proposition (A.19)). More exactly we shall show that there exists a unique solution of the system:

$$
\left\{
\begin{array}{l}
S(0, z) = S_{\mu_0}(z), \forall z \in \mathbb{H} \\
\partial_t S(t, z) + S(t, z) \partial_z S(t, z) = 0, \forall t > 0, \forall z \in \mathbb{H} \\
\text{for all } t \ge 0, S(t, \cdot) \text{ is the Stieljes transform of a real probability measure}
\end{array}
\right.
\qquad (7.24)
$$

Assume that the uniqueness of solution of (7.24) has been proved. Since the Stieljes transform on $\mathbb{H}$ of a real probability measure uniquely characterizes this measure, we
deduce from Proposition (7.7) $(\mu_N(\omega))_{N \ge 1}$ has a unique limit point.

We shall prove some general lemmas in order to apply the characteristic method as the existence of the characteristics as solutions of a ordinary differential equations and secondly the fact that we can go back up the characteristic that have been built.

**Lemma 7.8.** For every $z \in \mathbb{H}$, for every $t > 0$ there exists a unique $r \in \mathbb{H}$ (which depends on $z$ and $t$) such that:

$$z = r + tS_{\mu_0}(r).$$

*Proof.* We follow the arguments of [71]. First of all, let us notice a general property of the Stieljes transform that shall be used throughout the proof:

$$\mathfrak{I}\mathfrak{m}(S_{\nu}(z)) = -\mathfrak{I}\mathfrak{m}(z) \int_{\mathbb{R}} \frac{1}{|z-x|^2} d\nu(x) \quad \text{for } \nu \in \mathcal{P}(\mathbb{R}). \qquad (7.25)$$

Let us prove the uniqueness. Assume that we have $r_1$ and $r_2$ both in $\mathbb{H}$ that are different and such that $z = r_1 + tS_{\mu_0}(r_1)$ and $z = r_2 + tS_{\mu_0}(r_2)$. If we subtract the two identities we have:

$$0 = r_1 - r_2 + tS_{\mu_0}(r_1) - tS_{\mu_0}(r_2).$$

So it gives

$$r_1 - r_2 = t \int_{\mathbb{R}} \left( \frac{1}{r_2 - x} - \frac{1}{r_1 - x} \right) d\mu_0(x) = t \int_{\mathbb{R}} \frac{r_1 - r_2}{(r_2 - x)(r_1 - x)} d\mu_0(x).$$

Hence, we have the following relation:

$$\frac{1}{t} = \int_{\mathbb{R}} \frac{1}{(r_2 - x)(r_1 - x)} d\mu_0(x). \qquad (7.26)$$

Now we write $r_j = a_j + ib_j$ for $j = 1, 2$, $a_j \in \mathbb{R}$ and $b_j \in \mathbb{R}^*$. Taking the imaginary and real part of (7.26) yields:

$$\begin{aligned} \int_{\mathbb{R}} \frac{(x - a_1)(x - a_2) - b_1 b_2}{|x - r_1|^2 |x - r_2|^2} d\mu_0(x) &= \frac{1}{t} \\ \int_{\mathbb{R}} \frac{b_1(x - a_2) + b_2(x - a_1)}{|x - r_1|^2 |x - r_2|^2} d\mu_0(x) &= 0. \end{aligned} \qquad (7.27)$$

Thanks to the second equation of (7.27) we have:

$$\int_{\mathbb{R}} \frac{x}{|x - r_1|^2 |x - r_2|^2} d\mu_0(x) = \frac{a_2 b_1 + a_1 b_2}{b_1 + b_2} \int_{\mathbb{R}} \frac{1}{|x - r_1|^2 |x - r_2|^2} d\mu_0(x). \qquad (7.28)$$

Then using that

$$(x - a_1)(x - a_2) - b_1 b_2 = |x - r_2|^2 + (a_2 - a_1)(x - a_2) - b_2^2 - b_1 b_2,$$

the first equation of (7.27) gives that:

$$\begin{aligned} \frac{1}{t} &= \int_{\mathbb{R}} \frac{d\mu_0(x)}{|x - r_1|^2} + (a_2 - a_1) \int_{\mathbb{R}} \frac{xd\mu_0(x)}{|x - r_1|^2 |x - r_2|^2} \\ &\quad - (a_2(a_2 - a_1) + b_2^2 + b_1 b_2) \int_{\mathbb{R}} \frac{d\mu_0(x)}{|x - r_1|^2 |x - r_2|^2} \\ &= \frac{a_2(a_2 - a_1)(a_2 b_1 + a_1 b_2)}{b_1 + b_2} \int_{\mathbb{R}} \frac{d\mu_0(x)}{|x - r_1|^2 |x - r_2|^2} \\ &\quad - (a_2(a_2 - a_1) + b_2^2 + b_1 b_2) \int_{\mathbb{R}} \frac{d\mu_0(x)}{|x - r_1|^2 |x - r_2|^2}. \end{aligned} \qquad (7.29)$$
We can rewrite the last term of (7.29) by taking imaginary part of the relation $z = r_1 + tS_{\mu_0}(r_1)$:

$$ \mathfrak{Im}(z) = b_1 - tb_1 \int_{\mathbb{R}} \frac{1}{|r_1 - x|^2} d\mu_0(x). $$

Using this identity in (7.29) gives:

$$ \frac{1}{t} = \frac{1}{t} - \frac{\mathfrak{Im}(z)}{tb_1} - \frac{b_2}{b_1 + b_2}((a_2 - a_1)^2 + (b_1 + b_2)^2) \int_{\mathbb{R}} \frac{d\mu_0(x)}{|x - r_1|^2|x - r_2|^2}. $$

However the second term is clearly strictly smaller than $1/t$ since $z \in \mathbb{H}$ and $b_1$ and $b_2$ are positive which gives a contradiction.

Now let us prove the existence. The proof uses some similar ideas with the proof of the D'Alembert-Gauss theorem and the Liouville theorem for holomorphic and bounded functions on $\mathbb{C}$.

Fix $z \in \mathbb{H}$. We remark that the function defined on $\mathbb{H}$ by

$$ \phi(r) := z - tS_{\mu_0}(r) $$

is holomorphic on $\mathbb{H}$ and takes value in $\mathbb{H}$ (since $\mathfrak{Im}(\phi(r)) = \mathfrak{Im}(z) + \mathfrak{Im}(r) \int_{\mathbb{R}} \frac{1}{|r-x|^2} d\mu_0(x)$ by (7.25)).

To prove the existence in Lemma 7.8, we have to prove $r \mapsto r - \phi(r)$ has a zero on $\mathbb{H}$. Assume that is not the case. Then, the function $h$ defined on $\mathbb{H}$ by

$$ h(r) := \frac{1}{r - \phi(r)} $$

is holomorphic on $\mathbb{H}$. Let us prove that $h$ is bounded on $\mathbb{H}$.

Indeed, first we have that if $r \in \mathbb{H}$ and $0 < \mathfrak{Im}(r) < \mathfrak{Im}(z)/2$, then $\mathfrak{Im}(r - \phi(r)) \le -\mathfrak{Im}(z)/2 < 0$ which gives that for $r \in \mathbb{H}$ with $0 < \mathfrak{Im}(r) < \mathfrak{Im}(z)/2$,

$$ |h(r)| \le \frac{1}{|\mathfrak{Im}(r - \phi(r))|} \le \frac{2}{\mathfrak{Im}(z)} =: C_1. $$

Secondly, we see that for all $r \in \mathbb{H}$,

$$ |\phi(r)| \le |z| + t \frac{1}{\mathfrak{Im}(r)} $$

and so in particular if $\mathfrak{Im}(r) \ge \mathfrak{Im}(z)/2$, we get:

$$ |\phi(r)| \le |z| + t \frac{2}{\mathfrak{Im}(z)}. $$

Hence we can find constants $K, C_2$ such that for all $r \in \mathbb{H}$ such that $|r| \ge K$ and $\mathfrak{Im}(r) \ge \mathfrak{Im}(z)/2$ we have $|h(r)| \le C_2$. By continuity of $h$ on $\mathbb{H}$ we can find a constant $C_3$ such that for all $r \in \mathbb{H}$ such that $\mathfrak{Im}(r) \ge \mathfrak{Im}(z)/2$ and $|r| \le K$ we get $|h(r)| \le C_3$.

So for all $r \in \mathbb{H}$, $|h(r)| \le \max(C_1, C_2, C_3)$. It proves that $h$ is bounded on $\mathbb{H}$.

For all $\varepsilon > 0$, we define $h_{\varepsilon}(r) := h(r + i\varepsilon)$ for $r \in \mathbb{C}$ such that $\mathfrak{Im}(r) > -\varepsilon$.

From now on, fix $r \in \mathbb{H}$. The goal shall be to express the value of $h_{\varepsilon}(r)$ thanks to the value of $h_{\varepsilon}(x)$ for $x \in \mathbb{R}$. To do so, we use the residue theorem to the function

$$ g_{\varepsilon,r}(z') := \frac{h_{\varepsilon}(z')}{(z' - \mathfrak{Re}(r))^2 + \mathfrak{Im}(r)^2}, $$
with the closed curved $\gamma_R = \gamma_{1,R} \lor \gamma_{2,R}$ such that $\gamma_{1,R} = [-R, R]$ and $\gamma_{2,R}$ is the semi circle centred in 0 which joins $R$ and $-R$ in the upper half plane for $R > 0$. For $R$ large enough, there is one residue inside $\gamma_R$ at $z' = r$ with value $\frac{h_\epsilon(r)}{2i\mathfrak{Im}(r)}$. The residue theorem yields that $R > 0$ large enough:

$$ \frac{\pi h_{\epsilon}(r)}{\mathfrak{Im}(r)} = \int_{\gamma_{1,R}} g_{\epsilon,r}(z') dz' + \int_{\gamma_{2,R}} g_{\epsilon,r}(z') dz'. \qquad (7.30) $$

This $h$ is bounded on $\mathbb{H}$, thanks to the dominated convergence theorem we obtain

$$ \int_{\gamma_{2,R}} g_{\epsilon,r}(z') dz' \underset{R \to +\infty}{\longrightarrow} 0. $$

Passing to the limit in (7.30) yields

$$ \pi h_{\epsilon}(r) = \int_{\mathbb{R}} h_{\epsilon}(x) \frac{\mathfrak{Im}(r)}{(x - \mathfrak{Re}(r))^2 + \mathfrak{Im}(r)^2} dx. \qquad (7.31) $$

Now we notice that for $x \in \mathbb{R}$, $\mathfrak{Im}(x + i\epsilon - \phi(x + i\epsilon)) \le \epsilon - \mathfrak{Im}(z)$. So if we choose $\epsilon < \mathfrak{Im}(z)/2$, then $\mathfrak{Im}(x + i\epsilon - \phi(x + i\epsilon)) < 0$ for all $x \in \mathbb{R}$.

It gives that $\mathfrak{Im}(h_\epsilon(x)) > 0$ for all $x \in \mathbb{R}$. We deduce from formula (7.31) that $\mathfrak{Im}(h_\epsilon(r)) > 0$ for all $r \in \mathbb{H}$ if $\epsilon$ is small enough. Let $\epsilon$ goes to 0 to obtain $\mathfrak{Im}(h(r)) \ge 0$ for all $r \in \mathbb{H}$.

Hence, it gives that for all $r \in \mathbb{H}$, $\mathfrak{Im}(\phi(r) - r) \ge 0$.

$$ 0 \le \mathfrak{Im}(\phi(r) - r) = \mathfrak{Im}(z) + t \mathfrak{Im}(r) \int_{\mathbb{R}} \frac{1}{|r - x|^2} d\mu_0(x) - \mathfrak{Im}(r), $$

and thanks to:

$$ \int_{\mathbb{R}} \frac{1}{|r - x|^2} d\mu_0(x) \le \frac{1}{|\mathfrak{Im}(r)|^2}, $$

we obtain for $r \in \mathbb{H}$ that

$$ 0 \le \mathfrak{Im}(\phi(r) - r) \le \mathfrak{Im}(z) + t \frac{1}{\mathfrak{Im}(r)} - \mathfrak{Im}(r) $$

Let $\mathfrak{Im}(r)$ goes to $+\infty$ to have a contradiction. $\square$

A second important lemma will be the study of an ordinary differential equation to construct the characteristic.

**Lemma 7.9.** Let $z \in \mathbb{H}$ and $S(\cdot, \cdot)$ be a solution of (7.24). We can construct a solution of the ordinary differential equation in $\mathbb{C}$:

$$ \begin{cases} z(0) = z \\ z'(t) = S((t, z(t))), \end{cases} \qquad (7.32) $$

which will exist up to the time $z(t)$ hits the real axis.

*Proof.* For $a > 0$, let $I_a := \{z \in \mathbb{C}, \mathfrak{Im}(z) > a\}$. We fix $a$ such that $0 < a < \mathfrak{Im}(z)$. We notice that $(t, z) \to S(t, z)$ is uniformly in time Lipschitz for the second variable on
$\mathbb{R}^+ \times I_a$. Indeed, for all $t \ge 0$, we consider $\gamma(t) \in \mathcal{P}(\mathbb{R})$ such that $S(t, \cdot)$ is the Stieljes transform of $\mu(t)$ (by definition of a solution of (7.24)). For $t \ge 0$ and $z, z'$ in $I_a$ we have:

$$S_t(z) - S_t(z') = \int_{\mathbb{R}} \frac{1}{z-x} - \frac{1}{z'-x} d\gamma(t)(x) = \int_{\mathbb{R}} \frac{z'-z}{(z-x)(z'-x)} d\gamma(t)(x),$$

and for $x \in \mathbb{R}$,

$$a^2 \le |\mathfrak{Im}(z)||\mathfrak{Im}(z')| \le |z-x||z'-x|$$

which gives that

$$|S_t(z) - S_t(z')| \le \frac{1}{a^2}|z - z'|.$$

Hence, by the Cauchy-Lipschitz theorem, for all $a > 0$ we have a solution of the ordinary differential equation (7.32), $z_a$ which exists until the time $\tau_a$ which is the moment when $z_a$ hits $\{z \in \mathbb{C}, \mathfrak{Im}(z) = a\}$. Thanks to the uniqueness part of the Cauchy-Lipschitz theorem if $a < b$, $z_a$ and $z_b$ coincide until $\tau_b$. This allow us to define a solution of (7.32); $z_0$ until the moment when this solution hits $\{z \in \mathbb{C}, \mathfrak{Im}(z) = 0\}$ (which is possibly infinite). $\square$

Now let us prove the existence and uniqueness of a solution of (7.24).

**Proposition 7.10.** *There exists a unique solution S of (7.24).*

*Proof.* The existence of a solution of this system is obtained by Proposition 7.7 taking any almost sure limit point of $(\mu_N)_{N \ge 1}$ which is almost surely relatively compact by Proposition 7.3.

As said before we shall use the method of characteristic to prove the uniqueness. Indeed, by Lemma 7.9 let $z_r(t)$ be the solution of (7.32) starting from $r \in \mathbb{H}$ and defined until a time $\tau(r) > 0$ (since $r \in \mathbb{H}$) which is the moment when $z_r(t)$ its the real axis. By definition of $z_r(t)$ the derivative of $t \to S(t, z_r(t))$ is 0 since S is solution of (7.24). So, for all $r \in \mathbb{H}$, for all $t < \tau(r)$,

$$\left\{ \begin{array}{l} S(t, z_r(t)) = S_{\mu_0}(r) \\ z_r'(t) = S(t, z_r(t)) = S_{\mu_0}(r) \end{array} \right. \qquad (7.33)$$

Hence we deduce that for all $z \in \mathbb{H}$, for all $t < \tau(z)$,

$$\left\{ \begin{array}{l} z_r(t) = r + tS_{\mu_0}(r), \\ S(t, z_r(t)) = S(t, r + tS_{\mu_0}(r)) = S_{\mu_0}(r). \end{array} \right. \qquad (7.34)$$

Now we have an explicit expression of $z_r(t)$ and so we can have an explicit expression of $\tau(r)$ which is solution of:

$$\mathfrak{Im}(z_r(t)) = \mathfrak{Im}(r) - t\mathfrak{Im}(r) \int_{\mathbb{R}} \frac{1}{|r-x|^2} d\mu_0(x) = 0.$$

Hence, we find

$$\tau(r) = \frac{1}{\int_{\mathbb{R}} \frac{1}{|r-x|^2} d\mu_0(x)}.$$

Now fix $t > 0$ and $z \in \mathbb{H}$. By Lemma 7.8, we can find a unique $r \in \mathbb{H}$ such that $z = r + tS_{\mu_0}(r)$. So, we can write $S(t, z) = S(t, r + tS_{\mu_0}(r))$. Since $z \in \mathbb{H}$ we necessarily have that $t < \tau(r)$ because if not we would have that:

$$\mathfrak{Im}(z) = \mathfrak{Im}(r + tS_{\mu_0}(r)) = \mathfrak{Im}(r) - t\mathfrak{Im}(r) \int_{\mathbb{R}} \frac{1}{|r-x|^2} d\mu_0(x) = \mathfrak{Im}(r) \left(1 - \frac{t}{\tau(r)}\right) \le 0.$$
So using (7.34) we get:

$$S(t, z) = S(t, r + tS_{\mu_0}(r)) = S(t, z_r(t)) = S_{\mu_0}(r).$$

This proved that the value of $S(t, z)$ for every $t \ge 0$ and $z \in \mathbb{H}$ is uniquely determined by $S_{\mu_0}$ proving the uniqueness of a solution of (7.24). $\square$

Finally, we conclude this section by stating a theorem for the convergence of the system of particles which summarizes the results obtained in this section.

**Theorem 7.11.** For all $N \ge 1$, let $(\lambda_t^{i,N})_{1 \le i \le N}$ be the solution of (5.9) starting from $\lambda_0^N \in \mathbb{R}^N$. Let $(\mu_N)_{N \ge 1} \in (C([0, T], \mathcal{P}(\mathbb{R})))^{N \ge 1}$ be the sequence of empirical distributions associated to the $(\lambda_t^{i,N})_{1 \le i \le N}$. Assume that $(\mu_N(0))_{N \ge 1} \in (\mathcal{P}(\mathbb{R}))^N$ converges weakly to a measure $\mu_0 \in \mathcal{P}(\mathbb{R})$ and that

$$C_0 := \sup_{N \ge 0} \frac{1}{N} \sum_{i=1}^{N} \log((\lambda_0^{i,N})^2 + 1) < +\infty.$$

Then almost surely $(\mu_N)_{N \ge 1} \in (C([0, T], \mathcal{P}(\mathbb{R})))^N$ converges to the deterministic measure valued process $\mu \in C([0, T], \mathcal{P}(\mathbb{R}))$ whose Stieljes transform $S_t(z) = S_{\mu(t)}(z)$ is the unique solution of

$$S_t(z) = S_{\mu_0}(z) - \int_0^t S_s(z) \partial_z S_s(z) ds, \forall t \ge 0, \forall z \in \mathbb{C} - \mathbb{R} \quad (7.35)$$

*Proof.* By Proposition 7.3

$$(\mu_N)_{N \ge 1} \in (C([0, T], \mathcal{P}(\mathbb{R})))^N$$

is almost surely relatively compact in $(C([0, T], \mathcal{P}(\mathbb{R})))$.

Let $\omega \in \Omega$ such that $(\mu_N(\omega))_{N \ge 1}$ is relatively compact in $(C([0, T], \mathcal{P}(\mathbb{R})))$. Consider $\nu(\omega) \in C([0, T], \mathcal{P}(\mathbb{R}))$ a limit point of $(\mu_N(\omega))_{N \ge 1}$. By Proposition 7.7 and Proposition 7.10 $\nu(\omega)$ is uniquely determined as the unique measure valued process $\nu \in C([0, T], \mathcal{P}(\mathbb{R}))$ whose Stieljes transform $S_t(z) := S_{\nu(t)}(z)$ is the unique solution of (7.35).

So, the sequence $(\mu_N(\omega))_{N \ge 1}$ has a unique limit point and so converges towards it. $\square$

### 7.4.3 Application: the Wigner theorem

In this section, for all $N \ge 0$ we consider the unique strong solution of (5.9), $(\lambda_t^{i,1})_{t \in \mathbb{R}^+, 1 \le i \le N}$ starting from $(\lambda_0^{i,N})_{1 \le i \le N} = 0 \in \mathbb{R}^N$ as it was explained in Remark 6.6. We shall use the notations of Theorem 7.11 as for instance that $S_t$ is the Stieljes transform of $\mu(t)$ the almost sure limit point of $(\mu_N)_{N \ge 1}$.

We recall that by the Dyson Theorem for all $N \ge 1$, if we denote $(\lambda_t^{i,1,Dyson})_{t \in \mathbb{R}^+, 1 \le i \le N}$ the spectrum of $(D_N^1(t))_{t \in \mathbb{R}^+}$, then $(\lambda_t^{i,1})_{t \in \mathbb{R}^+, 1 \le i \le N}$ and $(\lambda_t^{i,1,Dyson})_{t \in \mathbb{R}^+, 1 \le i \le N}$ have the same law.

**Lemma 7.12.** The Stieljes transform of $\mu$ satisfies the following identity: for all $t \ge 0$, for all $z \in \mathbb{C} - \mathbb{R}$:

$$S_t(z) = t^{-1/2} S_1(zt^{-1/2}).$$

*Proof.* Let $t > 0$. We denote $S_t^N(z)$ the Stieljes transform of $\mu_N(t)$ (let us mention that this Stieljes transform is a random variable since $\mu_N$ is). Since $\mu_N(t)$ converges almost
surely towards $\mu(t)$ in $\mathcal{P}(\mathbb{R})$ (which is deterministic), for all $z \in \mathbb{C} - \mathbb{R}$ the dominated convergence theorem yields

$$ \mathbb{E}(S_t^N(z)) \underset{N \to +\infty}{\longrightarrow} \mathbb{E}(S_t(z)) = S_t(z). \qquad (7.36) $$

We denote $\mu_N^{\text{Dyson}}(t)$ the empirical distribution of the Dyson particles $(\lambda_t^{i,1,\text{Dyson}})_{t \in \mathbb{R}^+, 1 \le i \le N}$ and $S_t^{N,\text{Dyson}}$ the Stieljes transform of $\mu_N^{\text{Dyson}}(t)$. By the Dyson theorem, for all $N \ge 1$, for all $z \in \mathbb{C} - \mathbb{R}$,

$$ \mathbb{E}(S_t^N(z)) = \mathbb{E}(S_t^{N,\text{Dyson}}(z)) \qquad (7.37) $$

By the scaling property of the Brownian motion: $B(t) \stackrel{\text{Law}}{=} \sqrt{t}B(1)$, $D_N(t)$ and $\sqrt{t}D_N(1)$ have the same law and so $(\lambda_t^{i,1,\text{Dyson}})_{1 \le i \le N}$ and $(\sqrt{t}\lambda_1^{i,1,\text{Dyson}})_{1 \le i \le N}$ also. This gives that for all $z \in \mathbb{C} - \mathbb{R}$:

$$ \begin{aligned} \mathbb{E}(S_t^{N,\text{Dyson}}(z)) &= \mathbb{E}\left[\frac{1}{N}\sum_{i=1}^N \frac{1}{z - \lambda_t^{i,1,\text{Dyson}}}\right] = \mathbb{E}\left[\frac{1}{N}\sum_{i=1}^N \frac{1}{z - \sqrt{t}\lambda_1^{i,1,\text{Dyson}}}\right] \quad (7.38) \\ &= t^{-1/2}\mathbb{E}(S_1^{N,\text{Dyson}}(zt^{-1/2})) \end{aligned} $$

Using (7.37) we obtain:

$$ \mathbb{E}(S_t^N(z)) = t^{-1/2}\mathbb{E}(S_1^N(zt^{-1/2})) \qquad (7.39) $$

Passing to the limit in (7.39) and using (7.36) we get that for all $z \in \mathbb{C} - \mathbb{R}$

$$ S_t(z) = t^{-1/2}S_1(zt^{-1/2}). \qquad \square $$

Thanks to this lemma, to compute the Stieljes transform at any time of $\mu$ it suffices to compute it at the time $t = 1$.

**Lemma 7.13.** The Stieljes transform of $\mu$ at time $t = 1$ is given by $S_1 = S_\sigma$ where $\sigma$ is the semicircle distribution introduced in Example 7.6. As a consequence $\mu(1) = \sigma$.

*Proof.* Using the identity obtained in Lemma 7.12 and Proposition 7.7, for all $z \in \mathbb{C} - \mathbb{R}$, for all $t > 0$, we get

$$ S_1(z) = \sqrt{t}S_t(\sqrt{t}z) = \frac{1}{z} - \sqrt{t} \int_0^t s^{-1/2}S_1(zt^{1/2}s^{-1/2})s^{-1}S_1'(zt^{1/2}s^{-1/2})ds. \quad (7.40) $$

Doing the change of variable $t^{1/2}s^{-1/2} = u$ in the integral yields

$$ \begin{aligned} \sqrt{t} \int_0^t s^{-1/2} S_1(zt^{1/2}s^{-1/2}) s^{-1} S_1'(zt^{1/2}s^{-1/2}) ds &= \int_1^{+\infty} S_1(zu) S_1'(zu) u^3 \frac{2}{u^3} du \quad (7.41) \\ &= 2 \int_1^{+\infty} S_1(zu) S_1'(zu) du. \end{aligned} $$

However we have

$$ \int_1^{+\infty} S_1(zu)S_1'(zu)du = \frac{1}{2z}\int_1^{+\infty} \partial_u((S_1(zu))^2)du = -\frac{1}{2z}S_1(z)^2. \quad (7.42) $$

So (7.40) and (7.42) give

$$ S_1(z) = \frac{1}{z} + \frac{1}{z}S_1(z)^2. $$

We recognize the functional equation satisfied by $S_\sigma$. By Proposition A.20 we deduce that $S_1 = S_\sigma$.

<figure><img src="image_13.png" /></figure>
Let state the Wigner theorem in the real and complex case.

**Theorem 7.14** (Wigner). For all $N \ge 1$, let $X_N$ be a real Wigner matrix of size $N$ and $\mu_N^{\text{Wigner}}$ be the spectral measure of $X_N/\sqrt{N}$:

$$ \mu_N^{\text{Wigner}} = \frac{1}{N} \sum_{i=1}^{N} \delta_{\frac{\lambda_i^N}{\sqrt{N}}}, $$

with $(\lambda_i^N)_{1 \le i \le N}$ the eigenvalues of $X_N$. Then for all $f: \mathbb{R} \to \mathbb{R}$ continuous and bounded:

$$ \mathbb{E} \left[ \int f d\mu_N^{\text{Wigner}} \right] \underset{N \to +\infty}{\longrightarrow} \int f d\sigma_1, $$

where $\sigma_1(dx)$ is given by

$$ \sigma_1(dx) := \frac{1}{\pi} \sqrt{(2-x^2)_+} dx. $$

*Proof.* We use the notations of Lemma 7.12.

We recall that for all $N \ge 1$, $X_N/\sqrt{N}$ and $D_N^1(1/2)$ have the same law. Moreover, by Lemma 7.12 and Lemma 7.13, we have $\mu(1/2) = \sigma_1$. Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous bounded function. Then for all $N \ge 1$

$$ \mathbb{E} \left[ \int f d\mu_N^{\text{Wigner}} \right] = \mathbb{E} \left[ \int f d\mu_N^{\text{Dyson}}(1/2) \right]. \qquad (7.43) $$

By the Dyson theorem for every $N \ge 1$ we also have:

$$ \mathbb{E} \left[ \int f d\mu_N^{\text{Dyson}}(1/2) \right] = \mathbb{E} \left[ \int f d\mu_N(1/2) \right]. \qquad (7.44) $$

Moreover, by Theorem 7.11, the dominated convergence theorem yields:

$$ \mathbb{E} \left[ \int f d\mu_N(1/2) \right] \underset{N \to +\infty}{\longrightarrow} \mathbb{E} \left[ \int f d\mu(1/2) \right] = \int f d\mu(1/2) = \int f d\sigma_1. \qquad (7.45) $$

(7.43), (7.44) and (7.45) gives the result. $\square$

**Theorem 7.15** (Wigner). For all $N \ge 1$, let $X_N$ be a complex Wigner matrix of size $N$ and $\mu_N^{\text{Wigner}}$ be the spectral measure of $X_N/\sqrt{N}$:

$$ \mu_N^{\text{Wigner}} = \frac{1}{N} \sum_{i=1}^{N} \delta_{\frac{\lambda_i^N}{\sqrt{N}}}, $$

with $(\lambda_i^N)_{1 \le i \le N}$ the eigenvalues of $X_N$. Then for all $f: \mathbb{R} \to \mathbb{R}$ continuous and bounded:

$$ \mathbb{E} \left[ \int f d\mu_N^{\text{Wigner}} \right] \underset{N \to +\infty}{\longrightarrow} \int f d\sigma, $$

where $\sigma(dx)$ is the semicircle law.

*Proof.* We sketch the proof because it is similar to the real case. In this section we detailed the case $\beta = 1$ to lighten the notations but all the computations are also valid for $\beta = 2$. We notice that when we pass to the limit in Proposition 7.4, the coefficient in front the Brownian motions in the system of particles (5.9) does not play any role. Indeed this coefficient appears in the martingale term and in the term with second derivatives in the
Itô formula in Proposition 7.1 which disappear when we pass to the limit when $N$ goes to $+\infty$. Hence in the real case, the exact same Theorem 7.11 holds with the same equation at the limit for the measure valued process $\mu \in C([0, T], \mathcal{P}(\mathbb{R}))$. Since in the complex case for all $N \ge 1$, $X_N/\sqrt{N}$ and $D_N^2(1)$ have the same law, the exact same proof yields that for all $f: \mathbb{R} \to \mathbb{R}$ continuous and bounded:

$$ \mathbb{E} \left[ \int fd\mu_N^{\text{Wigner}} \right] \underset{N \to +\infty}{\longrightarrow} \int fd\mu(1) = \int fd\sigma. \quad \square $$

**Remark 7.16.** An interpretation of the equation (7.22) comes from the theory of free probability. The free probability were introduced by Voiculescu [83] and appears when we deal with non commutative random variables as random matrices. We choose in these notes to not use the language of free probability but the free probability represents an important topic in the study of random matrices as it gives amazing results and interpretations for some results that can be obtained in random matrix theory. We refer to [75, 5, 45] for general introductions on this topic. The Dyson Brownian motion can be considered as the free Brownian motion [13, 15] and (7.22) can be in this sense considered as a free version of the heat equation. As we saw in this section, the semicircle distribution can be viewed as the fundamental solution of this partial differential equation which means the solution with initial condition $\delta_0(dx)$. As for the heat equation, thanks to the free probability interpretation we can give an "explicit" formula of the solution starting from $\mu_0 \in \mathcal{P}(\mathbb{R})$. The solution is given by what is called the free convolution of $\mu_0$ and the semicircle distribution. Thanks to this representation of the solutions one can study some analytic properties of solutions of (7.22) as the regularity. We refer to [14] for such results.

## 7.5 Case of Ornstein-Uhlenbeck process

In this section, we consider the particles which are solutions of the Ornstein Uhlenbeck equation (5.22). First we explain the convergence of the system of particles to obtain the counterpart of Theorem 7.11 for the ornstein-Uhlenbeck case. We shall not give many proof but just explain how to adapt the proof of the Dyson case (that corresponds to the case $\theta = 0$). Then, we study the long time behaviour of solutions of the Ornstein-Uhlenbeck case. We shall follow the arguments of [71, 29] for this case.

### 7.5.1 Convergence of the system of particles

For all $N \ge 1$, we consider the family of $N$ particles solution of the Ornstein-Uhlenbeck system of stochastic differential equations.

$$ \left\{ \begin{array}{l} d\lambda_i(t) = \frac{1}{N} \sum_{j \neq i} \frac{1}{\lambda_i(t) - \lambda_j(t)} dt + \sqrt{\frac{2}{N}} dB_t^i - \theta \lambda_i(t) dt \quad \forall 1 \le i \le N \\ \\ \lambda_i(0) = \lambda_0^i \quad \forall 1 \le i \le N \end{array} \right. \qquad (7.46) $$

with $\lambda_0 \in D$ an initial data. One can prove a counterpart of Theorem (6.5) that is the existence of a strong solution for this system of SDE and such that the particles never collide for $t > 0$ almost surely (see Lemma 1 of [71]). As for the Dyson case, for all $N \ge 1$, we denote $\mu_N \in C([0, T], \mathcal{P}(\mathbb{R}))$ the empirical distribution of a solution $(\lambda_i)_{1 \le i \le N}$ of (7.46) which means that for all $N$, for all $t \ge 0$ we have

$$ \mu_N(t) = \frac{1}{N} \sum_{i=1}^{N} \delta_{\lambda_i(t)}. $$
We can prove that almost surely the sequence $(\mu_N)_{N \ge 1}$ is relatively compact (see Section 3 of [29, 71]). We can state the counterpart of Proposition (7.4).

**Proposition 7.17.** Fix $\omega \in \Omega$. Suppose that $(\mu_N(0))_{N \ge 1}$ converges in $\mathcal{P}(\mathbb{R})$ towards $\mu_0 \in \mathcal{P}(\mathbb{R})$. Let $\mu \in C([0, T], \mathcal{P}(\mathbb{R}))$ be a limit point of the sequence $(\mu_N(\omega))_{N \ge 1}$ in the space $C([0, T], \mathcal{P}(\mathbb{R}))$, then $\mu(0) = \mu_0$ and for every $f \in C^2(\mathbb{R}, \mathbb{R})$ with its derivatives bounded, for every $t \ge 0$,

$$ \int_{\mathbb{R}} f(x)d\mu(t)(x) = \int_{\mathbb{R}} f(x)d\mu_0(x) - \theta \int_0^t \int_{\mathbb{R}} xf'(x)d\mu(s)(x)ds \\ + \frac{1}{2}\int_0^t \int \int \frac{f'(x) - f'(y)}{x-y}d\mu(s)(x)d\mu(s)(y)ds \quad (7.47) $$

We introduce the Stieljes transform $S_t(z) = S_{\mu(t)}(z)$ of $\mu(t)$ where $\mu$ is a limit point as in Proposition 7.17. The counterpart of Proposition 7.7 is the following result.

**Proposition 7.18.** Let $\mu$ a limit point as in Proposition 7.17, then $\mu(0) = \mu_0$ and $(S_t)_{t \ge 0}$ satisfies:

$$ S_t(z) = S_0(z) - \int_0^t S_s(z)\partial_z S_s(z)ds + \int_0^t [\theta z \partial_z S_s(z) + \theta S_s(z)]ds, \forall z \in \mathbb{C} - \mathbb{R}. \quad (7.48) $$

The counterpart of (7.24) is

$$ \left\{ \begin{array}{l} S(0, z) = S_{\mu_0}(z), \forall z \in \mathbb{H} \\ \partial_t S(t, z) + (S(t, z) - \theta z) \partial_z S(t, z) = \theta S(t, z), \forall t > 0, \forall z \in \mathbb{H} \\ \text{for all } t \ge 0, S(t, \cdot) \text{ is the Stieljes transform of a real probability measure} \end{array} \right. \quad (7.49) $$

To prove that there is a unique solution for this system we again use the characteristic method. We follow the arguments of the Dyson case. We consider the solution of the ordinary differential equation

$$ \left\{ \begin{array}{l} \partial_t z_r(t) = S(t, z_r(t)) - \theta z_r(t) \\ z(0) = r \in \mathbb{H} \end{array} \right. \quad (7.50) $$

Using that $S(\cdot, \cdot)$ is a solution of (7.49), we obtain:

$$ \partial_t^2 z_r(t) = \theta^2 z_r(t). $$

Solving this equation yields:

$$ z_r(t) = r \exp(-\theta t) + \frac{1}{\theta} S_0(r) \sinh(\theta t) \quad (7.51) $$

(if $\theta$ converges to 0 it gives that $z(t) = r + S_0(r)t$ which is the result obtained in Dyson case). In particular for every $t \ge 0$,

$$ S(t, z_r(t)) = z'_r(t) + \theta z_r(t) = \exp(\theta t)S_0(r). \quad (7.52) $$

Hence to prove the uniqueness of the solution of (7.48) we prove that for all $t > 0$, for all $z \in \mathbb{H}$ there exists a unique $z_0(t, z) \in \mathbb{H}$ such that:

$$ z_0(t, z) \exp(-\theta t) + \frac{1}{\theta} S_0(z_0(t, z)) \sinh(\theta t) = z. \quad (7.53) $$
This result is a consequence of Lemma 7.8.

Thanks to (7.51) and (7.53) we can follow the exact same proof as for Proposition 7.10 to obtain that for all $t \ge 0$ and $z \in \mathbb{H}$, $S(t, z)$ is uniquely determined by $S_{\mu_0}$ giving the uniqueness of a solution of (7.49). We refer to [71] for a detailed proof.

We now state a theorem for the convergence of the particles associated to an Ornstein-Uhlenbeck process.

**Theorem 7.19.** For all $N \ge 1$, let $(\lambda_t^{i,N})_{1\le i\le N}$ be the solution of (5.9) starting from $\lambda_0^N \in \mathbb{R}^N$. Let $(\mu_N)_{N \ge 1} \in (C([0,T], \mathcal{P}(\mathbb{R}))^{N \ge 1})$ be the sequence of empirical distributions associated to the $(\lambda_t^{i,N})_{1\le i\le N}$. Assume that $(\mu_N(0))_{N \ge 1} \in (\mathcal{P}(\mathbb{R}))^N$ converges weakly to a measure $\mu_0 \in \mathcal{P}(\mathbb{R})$ and that

$$C_0 := \sup_{N \ge 0} \frac{1}{N} \sum_{i=1}^{N} \log((\lambda_0^{i,N})^2 + 1) < +\infty.$$

Then almost surely $(\mu_N)_{N \ge 1} \in (C([0,T], \mathcal{P}(\mathbb{R}))^N)$ converges to the element $\mu^\theta \in C([0,T], \mathcal{P}(\mathbb{R}))$ whose Stieljes transform $S_t(z) := S_{\mu^\theta(t)}(z)$ is the unique solution of

$$S_t(z) = S_{\mu_0}(z) - \int_0^t S_s(z) \partial_z S_s(z) ds + \int_0^t \theta z \partial_z S_s(z) + \theta S_s(z) ds, \forall z \in \mathbb{C} - \mathbb{R}.$$

### 7.5.2 Long time behaviour and stationary solutions

In this section we consider $\theta > 0$ and let $\mu^\theta \in C([0,T], \mathcal{P}(\mathbb{R}))$ be the almost sure limit of the system of particles for the Ornsetin Uhlenbeck process obtained in Theorem 7.19. We want to look at the convergence of the process of measures $(\mu^\theta(t))_{t \ge 0}$ when $t$ goes to $\infty$. Actually we shall prove that it convergences to a semi-circle distribution.

**Theorem 7.20.** The measure valued process $(\mu_\theta(t))_{t \ge 0}$ converges weakly towards the measure $\sigma_\theta \in \mathcal{P}(\mathbb{R})$ with density:

$$\sigma_\theta(x) = \frac{\theta}{\pi} \sqrt{\frac{2}{\theta} - x^2} \mathbf{1}_{|x| \le \sqrt{2/\theta}}.$$

We shall prove that the Stieljes transform of $\mu^\theta(t)$ converges to the Stieljes transform of $\sigma_\theta$ when $t$ goes to $\infty$. This implies the result by Proposition A.19.

*Proof.* We first recall the expression of $S_t(z)$, the Stieljes transform of $\mu_\theta(t)$ for $z \in \mathbb{H}$ obtained thanks to the characteristic method. Using the notations of (7.51), (7.53) and (7.52) we have:

$$ \begin{aligned} S(t, z) &= S \left( t, z_0(t, z) \exp(-\theta t) + \frac{1}{\theta} S_0(z_0(t, z)) \sinh(\theta t) \right) \\ &= S(t, z_{z_0(t,z)}(t)) \\ &= \exp(\theta t) S_0(z_0(t, z)). \end{aligned} \quad (7.54) $$

So for all $z \in \mathbb{H}$, $\forall t \ge 0$, we have:

$$S_t(z) = \exp(\theta t) S_0(z_0(t, z)), \quad (7.55)$$

with $z_0(t, z)$ the unique element of $\mathbb{H}$ such that:

$$z_0(t, z) \exp(-\theta t) + \frac{1}{\theta} S_0(z_0(t, z)) \sinh(\theta t) = z. \quad (7.56)$$
To look at the convergence of $S(t, z)$ when $t$ goes to $\infty$, we shall look at the behaviour of $z_0(t, z)$.

We fix $z \in \mathbb{H}$ and write $\exp(-\theta t)z_0(t, z) := a(t) + ib(t)$. Now let us study $b$ and $a$.

By taking the imaginary part of (7.56), we have:

$$b(t) - \frac{\sinh(\theta t)b(t)\exp(\theta t)}{\theta} \int_{\mathbb{R}} \frac{1}{|z_0(t, z) - x|^2} d\mu_0(x) = \mathfrak{Im}(z). \quad (7.57)$$

From (7.57), we deduce that $b(t) \ge \mathfrak{Im}(z)$ and also using that

$$\int_{\mathbb{R}} \frac{1}{|z_0(t, z) - x|^2} d\mu_0(x) \le \frac{1}{|\mathfrak{Im}(z_0(t, z))|^2} = \frac{1}{\exp(2\theta t)b(t)^2},$$

we get

$$b(t) - \frac{1}{b(t)\theta} \le \mathfrak{Im}(z).$$

Solving this quadratic inequality we obtain that for all $t \ge 0$,

$$\mathfrak{Im}(z) \le b(t) \le \frac{\mathfrak{Im}(z) + \sqrt{\mathfrak{Im}(z)^2 + \frac{4}{\theta}}}{2}. \quad (7.58)$$

We shall now prove that $\sup_{t>0} |a(t)| < +\infty$. Indeed if it is not the case we can find a sequence $t_n$ that goes to $+\infty$ such that $|a(t_n)|$ goes to $+\infty$ when $n$ goes to $+\infty$.

Now we take the real part of (7.56) to have that:

$$\mathfrak{Re}(z) = a(t_n) + \frac{1}{\theta} \mathfrak{Re} \left( \sinh(\theta t_n) \int_{\mathbb{R}} \frac{1}{z_0(t_n, z) - x} d\mu_0(x) \right). \quad (7.59)$$

We remark that:

$$\sinh(\theta t_n) \frac{1}{z_0(t_n, z) - x} = \frac{1}{\frac{z_0(t_n, z)}{\sinh(\theta t_n)} - \frac{x}{\sinh(\theta t_n)}}, \quad (7.60)$$

and

$$\mathfrak{Re} \left( \frac{z_0(t_n, z)}{\sinh(\theta t_n)} \right) = a(t_n) \frac{\exp(\theta t_n)}{\sinh(\theta t_n)}, \quad (7.61)$$

which converges in absolute value to $+\infty$ by hypothesis on $a(t_n)$. As a consequence of (7.60) and (7.61) we obtain:

$$\begin{aligned} \left| \sinh(\theta t_n) \frac{1}{z_0(t_n, z) - x} \right| &= \frac{1}{\left| \frac{z_0(t_n, z)}{\sinh(\theta t_n)} - \frac{x}{\sinh(\theta t_n)} \right|} \\ &\le \frac{1}{\left| \mathfrak{Re} \left( \frac{z_0(t_n, z)}{\sinh(\theta t_n)} - \frac{x}{\sinh(\theta t_n)} \right) \right|} \quad (7.62) \\ &= \frac{1}{\left| \mathfrak{Re} \left( \frac{z_0(t_n, z)}{\sinh(\theta t_n)} \right) - \frac{x}{\sinh(\theta t_n)} \right|} \xrightarrow{n \to +\infty} 0 \end{aligned}$$
Moreover we also have:

$$ \left| \sinh(\theta t_n) \frac{1}{z_0(t_n, z) - x} \right| \leq \left| \sinh(\theta t_n) \right| \frac{1}{\mathfrak{Im}(z_0(t_n, z))} = \frac{|\sinh(\theta t_n)|}{\exp(\theta t_n) b(t_n)} \leq \frac{1}{b(t_n)} \leq \frac{1}{\mathfrak{Im}(z)}, $$

by bound (7.58) on $b(t)$ for all $t \geq 0$. The dominated convergence theorem yields:

$$ \sinh(\theta t_n) \int_{\mathbb{R}} \frac{1}{z_0(t_n, z) - x} d\mu_0(x) \underset{n \to \infty}{\longrightarrow} 0, \qquad (7.63) $$

which is in contradiction with the equality (7.59).

At this point we obtained $t \geq 0$,

$$ \mathfrak{Im}(z) \leq b(t) \leq \frac{\mathfrak{Im}(z) + \sqrt{\mathfrak{Im}(z)^2 + 4/\theta}}{2}, \quad \sup_{t \geq 0} |a(t)| < +\infty. $$

We now want to prove that $(a(t))_{t \geq 0}$, $(b(t))_{t \geq 0}$ converges when $t$ goes to $\infty$. Since they are bounded we only have to prove that there is only one limit point for $((a(t), b(t))_{t \geq 0})$. So let $(a, b) \in \mathbb{R} \times \mathbb{R}^{+*}$ be a limit point of $((a(t), b(t))_{t \geq 0})$ and a sequence $t_n$ that goes to $+\infty$ such that $\exp(-\theta t_n)z_0(t_n, z) = a(t_n) + ib(t_n)$ converges to $w := a + ib \in \mathbb{H}$. Using a similar dominated convergence theorem as for (7.63), passing to the limit in

$$ \begin{aligned} z &= z_0(t_n, z) \exp(-\theta t_n) + \frac{1}{\theta} S_0(z_0(t_n, z)) \sinh(\theta t_n) \\ &= z_0(t_n, z) \exp(-\theta t_n) + \frac{\sinh(\theta t_n)}{\theta} \int_{\mathbb{R}} \frac{1}{z_0(t_n, z) - x} d\mu_0(x) \end{aligned} \qquad (7.64) $$

we obtain

$$ z = w + \frac{1}{2\theta w}. $$

The only solution in $\mathbb{H}$ is:

$$ w := \frac{z + \sqrt{z^2 - \frac{2}{\theta}}}{2} = \frac{1}{\theta} \frac{1}{z - \sqrt{z^2 - \frac{2}{\theta}}}. $$

Since $(a(t), b(t))_{t \geq 0}$ has only one limit point, this proves that $\exp(-\theta t)z_0(t, z)$ converges to $w$ when $t$ goes to $+\infty$.

Using (7.55) and again dominated convergence theorem we get

$$ S_t(z) = \exp(\theta t) \int_{\mathbb{R}} \frac{1}{z_0(t, z) - x} d\mu_0(x) \underset{t \to +\infty}{\longrightarrow} \frac{1}{w} = \theta \left( z - \sqrt{z^2 - \frac{2}{\theta}} \right) = \int_{\mathbb{R}} \frac{1}{z - x} d\sigma_{\theta}(x). $$

Hence, for all $z \in \mathbb{C} - \mathbb{R}$, $S_{\mu^{\theta}(t)}(z) \underset{t \to \infty}{\longrightarrow} S_{\sigma_{\theta}}(z)$. It proves that $(\mu^{\theta}(t))_{t \geq 0}$ converges weakly to the measure $\sigma_{\theta} \in \mathcal{P}(\mathbb{R})$. $\square$

We shall now link the long time behaviour and the notion of stationary solutions.

**Definition 7.21.** $\nu \in \mathcal{P}(\mathbb{R})$ is said to be a stationary solution of (7.47) if and only if for every $f \in C^2(\mathbb{R}, \mathbb{R})$ with its derivatives bounded,

$$ 0 = -\theta \int_{\mathbb{R}} x f'(x) d\nu(x) + \frac{1}{2} \int_{\mathbb{R}} \int_{\mathbb{R}} \frac{f'(x) - f'(y)}{x - y} d\nu(x) d\nu(y) \qquad (7.65) $$
**Proposition 7.22.** $\nu \in \mathcal{P}(\mathbb{R})$ is a stationary solution if and only if for all $x \in \text{supp}(\nu)$,

$$ \theta x = \int_{\mathbb{R}} \frac{d\nu(y)}{x-y}. $$

*Proof.* By definition of a stationary measure we have that $\mu \in \mathcal{P}(\mathbb{R})$ is a stationary measure if and only if for every $f \in C^2(\mathbb{R}, \mathbb{R})$ with its derivatives bounded,

$$ 0 = \int_{\mathbb{R}} \left( -\theta x + \int_{\mathbb{R}} \frac{d\nu(y)}{x-y} \right) f'(x) d\nu(x). \qquad (7.66) $$

Since this identity should hold for all $f \in C^2(\mathbb{R}, \mathbb{R})$ with its derivatives bounded it gives the result. $\square$

For $\nu \in \mathcal{P}(\mathbb{R})$, let us specify the definition of the term

$$ \int_{\mathbb{R}} \frac{d\nu(y)}{x-y} $$

for $x \in \text{supp}(\nu)$ which is a priori not defined. This term must be viewed as a distribution. We recall that the principal value of $1/x$: $P.V.(1/x)$ is a distribution defined by:

$$ P.V.(1/x)(\phi) = \lim_{\epsilon \to 0} \int_{|y| \ge \epsilon} \frac{\phi(y)}{y} dy, $$

for $\phi$ a smooth function with compact support.

**Definition 7.23.** Let $\nu \in \mathcal{P}(\mathbb{R})$, we define the Hilbert transform of $\nu$ as the distribution

$$ \mathcal{H}(\mu) = P.V.(1/x) * \mu. $$

We can express the Hilbert transform of $\mu$ as

$$ \mathcal{H}(\nu)(x) = \lim_{\epsilon \to 0} \int_{|x-y| \ge \epsilon} \frac{\nu(dy)}{x-y} $$

if this quantity is well defined. This quantity is linked with the Stieljes transform of $\nu$ by the following formula:

$$ \mathcal{H}(\nu)(x) = \lim_{\epsilon \to 0} \frac{S_{\nu}(x + i\epsilon) + S_{\nu}(x - i\epsilon)}{2} = \lim_{\epsilon \to 0} \mathfrak{Re}(S_{\nu}(x + i\epsilon)). $$

With this definition we can reformulate the equation for a stationary solution $\nu \in \mathcal{P}(\mathbb{R})$ writing that $\nu \in \mathcal{P}(\mathbb{R})$ is a stationary solution if and only if for all $x \in \text{supp}(\nu)$:

$$ \mathcal{H}(\nu)(x) = \theta x. \qquad (7.67) $$

**Example 7.24.** We can compute the Hilbert transform of the Wigner semi circle law using its Stieljes transform. For all $x \in [-\sqrt{2}/\theta, \sqrt{2}/\theta]$:

$$ \mathcal{H}(\sigma_{\theta})(x) = \theta x, $$

which proves that $\sigma_{\theta}$ is a stationary solution. This is coherent with Theorem 7.20. In other words, a solution of (7.47) converges in long time to a stationary solution of the equation.
To conclude this section let us mention an heuristic to solve (7.67). We focus on the case $\theta = 1/2$ to lighten the notations.

Let $\nu \in \mathcal{P}(\mathbb{R})$ with a density with respect to Lebesgue measure such that for all $x \in \text{supp}(\mu)$

$$x = 2\mathcal{H}(\nu)(x).$$

We can rewrite this equality as saying that for all $x \in \mathbb{R}$ we have

$$(x - 2\mathcal{H}(\nu)(x))\nu(x) = 0 \qquad (7.68)$$

Using the properties of the Stieljes transform (see Proposition A.19, for all $x \in \mathbb{R}$, we have

$$\lim_{\epsilon \to 0} \mathfrak{Im}(S_{\nu}(x + i\epsilon)) = -\pi\nu(x), \lim_{\epsilon \to 0} \mathfrak{Re}(S_{\nu}(x + i\epsilon)) = \mathcal{H}(\nu)(x).$$

We can rewrite (7.68) as

$$\lim_{\epsilon \to 0} [(x - 2\mathfrak{Re}(S_{\nu}(x + i\epsilon)))\mathfrak{Im}(S_{\nu}(x + i\epsilon))] = 0 \qquad (7.69)$$

for all $x \in \mathbb{R}$. This gives that for all $x \in \mathbb{R}$:

$$\lim_{\epsilon \to 0} \mathfrak{Im} \left[ (x + i\epsilon)S_{\nu}(x + i\epsilon) - S_{\nu}(x + i\epsilon)^2 \right] = 0. \qquad (7.70)$$

Let $f(z) := zS_{\mu}(z) - S_{\mu}(z)^2$. By the properties of the Stieljes transform, $f$ is holomorphic on $\mathbb{C} - \mathbb{R}$, satisfies $\bar{f}(z) = f(\bar{z})$ for all $z \in \mathbb{C}$ and for all $x \in \mathbb{R}$, $\mathfrak{Im}[\lim_{\epsilon \to 0} f(x + i\epsilon)] = 0$. This allows us to extend $f$ as an analytic function on $\mathbb{C}$. Moreover, since for all $z \in \mathbb{C} - \mathbb{R}$, we have

$$S_{\mu}(z) = \int_{\mathbb{R}} \frac{1}{z-t} d\nu(t) = \frac{1}{z} \int_{\mathbb{R}} \frac{1}{1-\frac{t}{z}} d\nu(t) \underset{z \to +\infty}{\sim} \frac{1}{z},$$

we obtain $f(z) \underset{z \to +\infty}{\longrightarrow} 1$. The Liouville theorem gives that $f$ is constant equal to 1. So for all $z \in \mathbb{C}$,

$$zS_{\mu}(z) - S_{\mu}(z)^2 = 1.$$

We recover the functional equation that characterizes the Stieljes transform of the semi-circle distribution obtained in Proposition A.20.
# Part III

## Large deviations

The goal of large deviations is to understand how small is the probability that rare events append. The most usual example is the deviations of the empirical mean of iid random variables from its mean. By the law of large numbers we know that if $(X_i)_{i \in \mathbb{N}_{\ge 1}}$ are iid real random variables in $L^1$ then $(X_1 + \dots + X_n)/n$ converges almost surely towards $\mathbb{E}(X_1)$. The large deviations theorem associated to the law of large numbers is the Cramer theorem.

**Theorem 7.25 (Cramer [33]).** Let $(X_i)_{i \in \mathbb{N}_{\ge 1}}$ be a family of real random variables and $\Lambda(t) = \log(\mathbb{E}[\exp(tX_1)])$ be the logarithmic moment generating function of $X_1$. Assume that for all $t \in \mathbb{R}$, $\Lambda(t) < +\infty$. Let $\Lambda^*(x) = \sup_{t \in \mathbb{R}}(tx - \Lambda(t))$ be the Legendre transform of $\Lambda$. Then for all $x > \mathbb{E}(X_1)$ we have:

$$ \lim_{n \to \infty} \frac{1}{n} \log \left[ \mathbb{P} \left( \sum_{i=1}^{n} X_i \ge nx \right) \right] = -\Lambda^*(x). $$

In the language of large deviations (the complete definitions of large deviations shall be given in the Section 8.3), we can reformulate the Cramer theorem saying that the family $(\mathbb{P}_n)_{n \in \mathbb{N}_{\ge 1}}$ where for all $n \ge 1$, $\mathbb{P}_n$ is the law of $\sum_{i=1}^n X_i/n$, satisfies a large deviations principle at speed $n$ and with good rate function $\Lambda^*$.

Another important theorem in large deviations theory is the Sanov theorem [30, 33]. Again by the law of large numbers if $(X_i)_{i \in \mathbb{N}_{\ge 1}}$ are iid real random variables with law $\mathcal{L}$ in $L^1$ then $(\delta_{X_1} + \dots + \delta_{X_n})/n$ converges almost surely in $\mathcal{P}(\mathbb{R})$ endowed with the weak convergence towards $\mathcal{L}$. The Sanov theorem states that for $(X_i)_{i \in \mathbb{N}_{\ge 1}}$ a family of independent random variables of law $\mathcal{L} \in \mathcal{P}(\mathbb{R})$, defining $(\mathbb{P}_n)_{n \in \mathbb{N}_{\ge 1}}$ the family of law of $(\delta_{X_1} + \dots + \delta_{X_n})/n$ (viewed as random variables in $\mathcal{P}(\mathbb{R})$), then this sequence of probabilities satisfies a large deviations principle at speed $n$ with good rate function $D(P||\mathcal{L}) := \int \log(dP/d\mathcal{L})dP$ if $P \ll \mathcal{L}$ and $+\infty$ otherwise. The function $D$ is called the Kullback-Leibler divergence and appears in statistics and also in information theory where it is called relative entropy. The goal of this section is to present the theory of large deviations through the large deviations of the empirical mean of eigenvalues of random matrices. The theorem is similar with the Sanov theorem but here the eigenvalues are not at all independent and identically distributed. The good rate function shall be linked with the so-called free entropy of Voiculescu [83] instead of the entropy. In 1997, G.Ben Arous and A.Guionnet proved this large deviations principle for the eigenvalues of so-called $\beta$-ensembles as the $GUE_N$ and $GOE_N$ [6]. Then it was extended to other models of random matrices as Ginibre ensemble [7] and the circular ensemble [51]. The method they used is actually quite general and has been generalised to Coulomb and Riesz gases in any dimensions [26, 74, 73, 27].

# 8 Notations and motivations

## 8.1 Recalls and notations

First we recall some general notations and classical results about random matrices that have been seen in the two first parts.
**Definition 8.1.** For a finite family of real numbers $(\lambda_i)_{1\le i\le N}$ we write $\overline{\mu_N}$ its empirical distribution, which the is a probability measure on $\mathbb{R}$ defined by

$$ \overline{\mu_N} = \frac{1}{N} \sum_{i=1}^{N} \delta_{\lambda_i}. $$

Furthermore if a matrix $M \in M_N(\mathbb{C})$ with real eigenvalues $(\lambda_i)_{1\le i\le N}$, we define its spectral measure as the empirical distribution of its eigenvalues.

**Definition 8.2.** For $\beta > 0$, let $\beta$ be the semicircular distribution defined as the probability measure on $\mathbb{R}$ with density

$$ \sigma_{\beta}(x) := \frac{1}{\beta\pi} \mathbf{1}_{[-\sqrt{2\beta}, \sqrt{2\beta}]} \sqrt{2\beta - x^2}. $$

**Definition 8.3.** The space of probability measure on $\mathbb{R}$ will be denoted $\mathcal{P}(\mathbb{R})$. We endow it with its usual weak topology which is metrizable by the Kantorovich-Rubinstein distance.

We do not explicit the Kantorovich-Rubinstein distance for now. The important point is to take a distance that metrizes the weak topology of $\mathcal{P}(\mathbb{R})$. Let us recall that we defined the Gaussian Unitary Ensemble and the Gaussian Orthogonal Ensemble in Definition 2.1. The Wigner theorem which can be viewed as the law of large numbers for the eigenvalues of random matrices (see Theorem 4.9).

**Theorem 8.4 (Wigner).** If $X_N$ is a $N$ complex Wigner matrix (resp. $N$ real Wigner matrix) then the spectral measure of $X_N/\sqrt{N}$ almost surely converges to $\sigma_2$ (resp. $\sigma_1$).

In Part I, we computed the distribution of the eigenvalues of a $N$ complex Wigner matrix (see the Theorem 2.25). A similar result holds for a matrix which is a $N$ real Wigner matrix.

**Proposition 8.5.** Let $X_N$ be a $N$ real Wigner matrix then the probability distribution of the eigenvalues of $X_N/\sqrt{N}$ is

$$ Q_1^N(d\lambda_1, \dots, d\lambda_N) := \frac{1}{Z_1^N} \prod_{1\le i<j\le N} |\lambda_i - \lambda_j| \exp \left( -\frac{1}{2} N \sum_{i=1}^N \lambda_i^2 \right) \prod_{i=1}^N d\lambda_i $$

with $Z_1^N$ a normalizing constant.

**Proposition 8.6.** Let $X_N$ be a $N$ complex Wigner matrix then the probability distribution of the eigenvalues of $X_N/\sqrt{N}$ is

$$ Q_2^N(d\lambda_1, \dots, d\lambda_N) := \frac{1}{Z_2^N} \prod_{1\le i<j\le N} |\lambda_i - \lambda_j|^2 \exp \left( -\frac{1}{2} N \sum_{i=1}^N \lambda_i^2 \right) \prod_{i=1}^N d\lambda_i $$

with $Z_2^N$ a normalizing constant.

These two distributions can be viewed as particular cases of so-called $\beta$ Hermite ensembles.

**Definition 8.7.** Let $\beta > 0$, we define

$$ Q_{\beta}^{N}(d\lambda_{1}, \dots, d\lambda_{N}) := \frac{1}{Z_{\beta}^{N}} \prod_{1 \le i < j \le N} |\lambda_{i} - \lambda_{j}|^{\beta} \exp \left( -\frac{1}{2} N \sum_{i=1}^{N} \lambda_{i}^{2} \right) \prod_{i=1}^{N} d\lambda_{i} $$
with $Z_{\beta}^N$ a normalizing constant to obtain a probability measure on $\mathbb{R}^N$. The measure $Q_{\beta}^N$ defined the $\beta$ Hermite ensemble.

We also define

$$ \overline{Q}_{\beta}^{N}(d\lambda_{1}, \dots, d\lambda_{N}) := \prod_{1 \le i < j \le N} |\lambda_{i} - \lambda_{j}|^{\beta} \exp \left( -\frac{1}{2} N \sum_{i=1}^{N} \lambda_{i}^{2} \right) \prod_{i=1}^{N} d\lambda_{i}, $$

the measure without normalization.

**Remark 8.8.** It is quite natural to ask ourselves if for all $\beta > 0$ we can find a law on random matrices such that if a random matrix is distributed with this law, then its eigenvalues are distributed like the $\beta$ Hermite ensemble. This is the case with $\beta = 1$ and $\beta = 2$ by the Proposition 8.5 and the Proposition 8.6. The answer is true for all $\beta > 0$ with random tridiagonal hermitian matrices [36].

Now we extend $Q_{\beta}^N$ on $\mathcal{P}(\mathbb{R})$ to view $Q_{\beta}^N$ as a probability measure on $\mathcal{P}(\mathbb{R})$ and not only $\mathbb{R}^N$ in view of proving a large deviations principle in $\mathcal{P}(\mathbb{R})$ as for the Sanov theorem.

**Definition 8.9.** For all $N \ge 1$, let $i_N : \mathbb{R}^N \to \mathcal{P}(\mathbb{R})$ be the map defined by $i_N(\lambda_1, \dots, \lambda_n) := \overline{\mu_N}$. For all $N \ge 1$, for all $\beta > 0$, we extend $Q_{\beta}^N$ as a probability measure on $\mathcal{P}(\mathbb{R})$ by considering the push forward measure $i_N \sharp Q_{\beta}^N$ defined by

$$ (i_N \sharp Q_{\beta}^N)(A) = Q_{\beta}^N(i_N^{-1}(A)), $$

for A a measurable subset of $\mathcal{P}(\mathbb{R})$. By abuse of notation we still call $Q_{\beta}^N$ these measures on $\mathcal{P}(\mathbb{R})$.

To avoid confusions this means in practice that for a measurable function $F : \mathcal{P}(\mathbb{R}) \to \mathbb{R}^+$ we have

$$ \int_{\mathcal{P}(\mathbb{R})} F(\mu) Q_{\beta}^{N}(d\mu) = \int_{\mathbb{R}^{N}} F\left(\frac{1}{N}\sum_{i=1}^{N} \delta_{\lambda_{i}}\right) Q_{\beta}^{N}(d\lambda_{1}, \dots, d\lambda_{N}) = \int_{\mathbb{R}^{N}} F(\overline{\mu_{N}}) Q_{\beta}^{N}(d\lambda_{1}, \dots, d\lambda_{N}), $$

where the notation $\overline{\mu_N}$ is introduced in Definition 8.1. In particular if $F$ is just the indicator of a ball $B(\mu, \varepsilon)$ we get

$$ Q_{\beta}^{N}(B(\mu, \varepsilon)) = \int_{\mathbb{R}^{N}} \mathbb{1}_{\overline{\mu_{N}} \in B(\mu, \varepsilon)}(\lambda_{1}, \dots, \lambda_{N}) Q_{\beta}^{N}(d\lambda_{1}, \dots, d\lambda_{N}) = Q_{\beta}^{N}(\overline{\mu_{N}} \in B(\mu, \varepsilon)), $$

where the $Q_{\beta}^N$ on the left refers to the measure on $\mathcal{P}(\mathbb{R})$ and the $Q_{\beta}^N$ on the right refers to the measure on $\mathbb{R}^N$.

## 8.2 Motivation

The Wigner theorem could be viewed as a law of large numbers. We want to obtain as the Cramer or the Sanov theorem an estimation of the probability that the empirical mean of the eigenvalues of random matrices is different from its almost sure limit.

As explained at the beginning of this section the goal of large deviations is to find a "good" function $I$ on $\mathcal{P}(\mathbb{R})$ such that

$$ \forall \mu \in \mathcal{P}(\mathbb{R}), \mathbb{P}(\overline{\mu_N} \cong \mu) \underset{N \to +\infty}{\sim} \exp(-N^2 I(\mu)), $$

where $\overline{\mu_N}$ is the spectral measure of a random matrix in the $GUE_N$ for instance. We can naturally expect a term in $N^2$ because a random matrix has $N^2$ entries and we want that $I(\mu) > 0$ for every $\mu \in \mathcal{P}(\mathbb{R})$ different from $\sigma_2$ and $I(\sigma_2) = 0$ in the case of the $GUE_N$. Hence it shall imply an exponential decrease of the probability for the spectral measure to be far from $\sigma_2$ in this case.
## 8.3 Few definitions of large deviations

We give some standard definitions of the theory of large deviations, for a more general approach of this topic see the textbook [33] or for instance [80].

**Definition 8.10.** Let $X$ be a metric space. A function $I: X \to [-\infty, \infty]$ is a lower semi-continuous (lsc) function if for all $M \in \mathbb{R}$, $\{x, I(x) \le M\}$ is a closed subset of $X$.

We have a sequential characterization for the lsc functions.

**Proposition 8.11.** Let $X$ be a metric space. A function $I: X \to [-\infty, \infty]$ is a lsc function if and only if for all sequences $(x_n)_{n \in \mathbb{N}} \in X^{\mathbb{N}}$ that converge to $x \in X$ we have $\liminf_n I(x_n) \ge I(x)$.

*Proof.* For the direct sense, we suppose that $I$ is lsc and we consider a sequence $(x_n)_{n \in \mathbb{N}}$ that converges to an $x \in X$. Assume by contradiction that $\liminf_n I(x_n) < I(x)$. We can find $\varepsilon > 0$ and a subsequence of $(x_n)_{n \in \mathbb{N}}$ that we call $(x_{\phi(n)})_{n \in \mathbb{N}}$ such that for $n$ large enough $I(x_{\phi(n)}) \le I(x) - \varepsilon$. Since $\{y, I(y) \le I(x) - \varepsilon\}$ is closed and $(x_n)_{n \in \mathbb{N}}$ converges to $x$, we deduce that $x \in \{y, I(y) \le I(x) - \varepsilon\}$. So $I(x) \le I(x) - \varepsilon$ which is a contradiction. Conversely, let $M \in \mathbb{R}$ and $(x_n)_{n \in \mathbb{N}}$ a sequence that converges to $x \in X$ such that for all $n \in \mathbb{N}$, $x_n \in \{y, I(y) \le M\}$. It yields $M \ge \liminf_n I(x_n) \ge I(x)$. So $x \in \{y, I(y) \le M\}$. This proves the result. □

**Lemma 8.12.** Let $X$ be a compact metric space and $I: X \to [-\infty, \infty]$ a lsc function. Then $I$ reaches its infimum.

*Proof.* Let $m = \inf_X I(x)$ the infimum of $I$. By definition of the infimum, we can find a sequence $(x_n) \in X^{\mathbb{N}}$ such that $(I(x_n))_{n \in \mathbb{N}}$ converges to $m$. Since $X$ is compact, we can find a subsequence $(x_{\phi(n)})_{n \in \mathbb{N}}$ that converges to a $x \in X$. By Proposition 8.11, $m \ge I(x)$. Hence by definition of the infimum, it gives that $m = I(x)$ and so $I$ reaches his infimum in $x$. □

**Definition 8.13.** Let $X$ be a metric space. A rate function is a lsc function $I: X \to [-\infty, +\infty]$. A rate function $I$ is called a good rate function if its sub-level sets $\{x, I(x) \le M\}$ are compact.

**Lemma 8.14.** Let $X$ be a metric space and $I: X \to [0, \infty]$ a good rate function that has a unique minimizer $x^*$ such that $I(x^*) = 0$. Let $F$ be a closed subset of $X$ that does not contain $x^*$. Then $\inf_F I > 0$.

*Proof.* Assume by contradiction that $\inf_F I = 0$. We can find a sequence $x_n \in F$ such that $I(x_n)$ converges to 0. Since for all $M \in \mathbb{R}$, $\{I \le M\}$ is a compact subset of $X$, we can find a sub sequence $x_{\phi(n)}$ that converges to an element $x_0 \in F$ since $F$ is closed. Since $I$ is lower semi continuous, Proposition 8.11 yields

$$0 = \liminf_n I(x_{\phi(n)}) \ge I(x_0).$$

By uniqueness of the minimizer of $I$, we get that $x_0 = x^*$ which contradicts that $x^* \notin F$. □

We can now give a definition for a large deviations principle.
**Definition 8.15.** Let $(P_N)_{N \in \mathbb{N}}$ be a sequence of Borel probability measures on a metric space $X$ and $(a_N)_{N \in \mathbb{N}}$ be a sequence of positive real numbers diverging to $+\infty$. Let also $I$ be a good rate function on $X$. The sequence $(P_N)_{N \in \mathbb{N}}$ is said to satisfy a large deviations principle (LDP) at speed $a_N$ with good rate function $I$ if for every Borel set $E \subset X$ the following inequalities hold:

$$ - \inf_{x \in \mathring{E}} I(x) \le \lim_{N \to +\infty} \frac{1}{a_N} \log(P_N(\mathring{E})) \le \overline{\lim_{N \to +\infty}} \frac{1}{a_N} \log(P_N(\overline{E})) \le - \inf_{x \in \overline{E}} I(x). $$

**Proposition 8.16.** Let $(P_N)_{N \in \mathbb{N}}$ be a sequence of Borel probability measures on a metric space $X$ that satisfies a large deviations principle with good rate function $I$, then $\inf_X I = 0$.

*Proof.* Taking $E$ equal to $X$ in Definition 8.15 gives:

$$ - \inf_{X} I \le 0 \le 0 \le - \inf_{X} I, $$

since $P_N(X) = 1$ for all $N \in \mathbb{N}$. This gives the result. $\square$

We finally justify how we usually go from a LDP to an almost sure convergence using the Borel-Cantelli Lemma.

**Proposition 8.17 (From large deviations to almost sure convergence).** Let $(X_N)_{N \in \mathbb{N}_{\ge 1}}$ be a family of random variables defined on a same probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with value in a metric space $X$. For all $N \ge 1$, let $P_N$ be the law of $X_N$. Assume that $(P_N)_{N \in \mathbb{N}_{\ge 1}}$ satisfies a LDP with good rate function $I$ at speed $N^s$ with $s > 0$. First, $I$ reaches its infimum. Assume that this infimum is reached in a unique $x^* \in X$. Then, we have the following convergence

$$ X_N \xrightarrow[N \to +\infty]{\text{Almost surely}} x^*. $$

*Proof.* By Proposition 8.16, $\inf_X I = 0$. Let $K = \{x \in X, I(x) \le 1\}$. Since $\inf_X I = 0$ and $I$ is a good rate function, $K$ is a non empty compact subset of $X$. Moreover as $\inf_X I = \inf_K I$, Lemma 8.12 yields that $I$ reaches its infimum. We now assume that this infimum is uniquely reached in $x^* \in X$. The LDP gives that for all $\varepsilon > 0$

$$ \overline{\lim}_{N \to +\infty} \frac{1}{N^s} \log(P_N(X \setminus B(x^*, \varepsilon))) \le - \inf_{x \in X \setminus B(x^*, \varepsilon)} I(x), \quad (8.1) $$

with $B(x^*, \varepsilon)$ the open ball of center $x^*$ and of radius $\varepsilon > 0$ and $X \setminus B(x^*, \varepsilon)$ its complement in $X$.

By Lemma 8.14, for all $\varepsilon > 0$ we have $0 < \inf_{x \in X \setminus B(x^*, \varepsilon)} I(x) =: 2c_\varepsilon$. We deduce from (8.1) that for all $\varepsilon > 0$ there exists $N(\varepsilon)$ such that for all $N \ge N(\varepsilon)$

$$ \mathbb{P}(X_N \notin B(x^*, \varepsilon)) \le \exp(-N^s c_\varepsilon). $$

The Borel-Cantelli lemma yields that for all $\varepsilon > 0$, for almost all $\omega \in \Omega$ there exists $N^*(\omega, \varepsilon)$ such that for all $N \ge N^*(\omega, \varepsilon)$ $X_N(\omega) \in B(x^*, \varepsilon)$. Using a countable dense sequence of $\varepsilon$, we deduce that for almost all $\omega \in \Omega$, for all $\varepsilon > 0$, there exists $N^*(\omega, \varepsilon)$ such that for all $N \ge N^*(\omega, \varepsilon)$ $X_N(\omega) \in B(x^*, \varepsilon)$.

This proves that $X_N$ converges almost surely towards $x^*$. $\square$
# 9 Free entropy and rate function

In this section we define the function that shall be the good rate function for the large deviations principle. In this section $\beta$ is a fixed positive real number.

## 9.1 General study of the rate function

**Definition 9.1.** Let $H_\beta$ the function defined on $\mathcal{P}(\mathbb{R})$ by:

$$H_{\beta}(\mu) := \frac{1}{2} \int_{\mathbb{R}^2} \left[ \frac{x^2}{2} + \frac{y^2}{2} - \beta \log(|x - y|) \right] \mu(dx)\mu(dy).$$

We write $f_\beta(x, y) := \frac{1}{2}(x^2 + y^2) - \beta \log(|x - y|)$ if $x \neq y$ and $f_\beta(x, x) = +\infty$ such that $H_\beta(\mu) = \frac{1}{2} \int_{\mathbb{R}^2} f_\beta(x, y)\mu(dx)\mu(dy)$.

Let us explain formally why this function is a priori the good function to look at as a good rate function for the large deviations principle of the sequence of probability $(Q_\beta^N)_N$. Indeed we can write

$$Q_{\beta}^{N}(\lambda_{1}, \dots, \lambda_{N}) = \exp(-N^{2}H_{\beta}^{N}(\lambda_{1}, \dots, \lambda_{N})),$$

where

$$H_{\beta}^{N}(\lambda_{1}, \dots, \lambda_{N}) = -\frac{\beta}{2N^{2}} \sum_{1 \le i \ne j \le N} \log(|\lambda_{i} - \lambda_{j}|) + \frac{1}{2N} \sum_{i=1}^{N} \lambda_{i}^{2} + \frac{\log(Z_{\beta}^{N})}{N^{2}}. \quad (9.1)$$

Formally, let us suppose that the empirical mean $\sum_{i=1}^N \delta_{\lambda_i}/N$ converges to $\mu$ and that $\frac{\log(Z_\beta^N)}{N^2}$ converges to a finite quantity $Z$, then $H_\beta^N(\lambda_1, \dots, \lambda_N)$ converges to $H_\beta(\mu) + Z$. So formally, we get that

$$Q_{\beta}^{N}(\lambda_{1}, \dots, \lambda_{N}) \approx \exp(-N^{2}H_{\beta}(\mu)).$$

This makes a link with Coulomb and Riesz gases [74] as the formula (9.1) corresponds to particles in interaction through a logarithmic potential that corresponds in dimension 1 to a Riesz gas and an external potential $V(x) = x^2/2$. We shall give some definitions of Riesz and Coulomb gases in Section 11.2.

We first prove that $H_\beta(\mu) > -\infty$ for all $\mu \in \mathcal{P}(\mathbb{R})$.

**Lemma 9.2.** $H_\beta$ is defined on $\mathcal{P}(\mathbb{R})$ and takes values in $[\frac{\beta}{4}(1 - \log(2\beta)), +\infty]$.

*Proof.* We shall first justify that

$$-\beta \log(|x - y|) + \frac{x^2}{2} + \frac{y^2}{2} \ge \frac{\beta}{2} (1 - \log(2\beta)). \quad (9.2)$$

We write

$$-\beta \log(|x - y|) + \frac{x^2}{2} + \frac{y^2}{2} = \frac{1}{2}((x^2 + y^2) - \beta \log(x^2 + y^2)) + \beta \log\left(\frac{\sqrt{x^2 + y^2}}{|y - x|}\right).$$

Since we have that for all $t \in \mathbb{R}^+$, $t - \beta \log(t) \ge \beta(1 - \log(\beta))$ by an easy study of function and for all $(x, y) \in \mathbb{R}^2$, $\frac{\sqrt{x^2 + y^2}}{|y - x|} \ge \frac{1}{\sqrt{2}}$.

Integrating (9.2) against $\mu(dx)\mu(dy)$ gives the result. $\square$
**Remark 9.3.** Let us also points out that since $f_β$ is bounded from below by (9.2) we have that for $\mu \in \mathcal{P}(\mathbb{R})$, $H_β(\mu)$ is finite if and only if $f_β \in L^1(\mu \otimes \mu)$.

When $H_β$ is finite we can rewrite it in another form.

**Lemma 9.4.** For all $\mu \in \mathcal{P}(\mathbb{R})$ such that $H_β(\mu) < +\infty$ we have $\int_{\mathbb{R}} x^2\mu(dx) < +\infty$ and $\int_{\mathbb{R}^2} |\log|x-y||\mu(dx)\mu(dy) < +\infty$. So if $H_β(\mu) < +\infty$ we can rewrite $H_β(\mu)$ as

$$H_β(\mu) = -\frac{\beta}{2} \int_{\mathbb{R}^2} \log|x-y|\mu(dx)\mu(dy) + \int_{\mathbb{R}} \frac{x^2}{2}\mu(dx) =: -\frac{\beta}{2} \Sigma(\mu) + \int_{\mathbb{R}} \frac{x^2}{2}\mu(dx).$$

*Proof.* Since the quadratic term dominates the logarithmic term we shall justify that we can find a constant $C_β$ such that for all $x, y \in \mathbb{R}^2$

$$-\beta \log(|x-y|) + \frac{x^2}{4} + \frac{y^2}{4} \geq C_\beta. \qquad (9.3)$$

Doing the same computations as in Lemma 9.2 yields

$$-\beta \log(|x-y|) + \frac{x^2}{4} + \frac{y^2}{4} = \frac{1}{4}((x^2 + y^2) - 2\beta \log(x^2 + y^2)) + \beta \log\left(\frac{\sqrt{x^2 + y^2}}{|y - x|}\right).$$

Since we have that for all $t \in \mathbb{R}^+$, $t - 2\beta \log(t) \geq 2\beta(1 - \log(2\beta))$ by an easy study of function and for all $(x, y) \in \mathbb{R}^2$, $\frac{\sqrt{x^2 + y^2}}{|y - x|} \geq \frac{1}{\sqrt{2}}$ we obtain (9.3).

Let $\mu \in \mathcal{P}(\mathbb{R})$ such that $H_β(\mu) < +\infty$. From (9.3) we have

$$+\infty > H_β(\mu) \geq \frac{1}{2} \int_{\mathbb{R}^2} \left[ C_β + \frac{x^2}{4} + \frac{y^2}{4} \right] \mu(dx)\mu(dy) = \frac{C_β}{2} + \int_{\mathbb{R}} \frac{x^2}{4}\mu(dx).$$

So we get that $\int_{\mathbb{R}} x^2 \mu(dx) < +\infty$. Moreover, again by (9.3) for all $(x, y) \in \mathbb{R}^2$ we have

$$C_β - \frac{x^2}{4} - \frac{y^2}{4} \leq -\beta \log|x-y| \leq f_β(x, y).$$

Since the two extremal terms of the inequalities are in $L^1(\mu \otimes \mu)$, this implies that $\int_{\mathbb{R}^2} |\log|x-y||\mu(dx)\mu(dy) < +\infty$.

For $\mu$ is a probability measure such that $H_β(\mu) < +\infty$ we get

$$H_β(\mu) = \frac{1}{2} \int_{\mathbb{R}^2} f_β(x, y)\mu(dx)\mu(dy) = \frac{1}{2} \left( \int_{\mathbb{R}} \frac{x^2}{2}\mu(dx) + \int_{\mathbb{R}} \frac{y^2}{2}\mu(dy) \right) - \frac{\beta}{2}\Sigma(\mu).$$

□

**Remark 9.5.** For $\mu \in \mathcal{P}(\mathbb{R})$, the free entropy of $\mu$ is defined as

$$\Sigma(\mu) = \int_{\mathbb{R}^2} \log|x-y|\mu(dx)\mu(dy).$$

This notion of free entropy was introduced by Voiculescu in his work on free probability [83, 82].

Let us state some properties on the function $H_β$.

**Proposition 9.6.**
1. $H_β$ is defined on $\mathcal{P}(\mathbb{R})$ and takes values in $[\frac{\beta}{4}(1 - \log(2\beta)), +\infty]$.

2. $H_β(\mu)$ is infinite if one of the conditions is satisfied:

(a) $\int_{\mathbb{R}} x^2 \mu(dx) = +\infty$.

(b) There exists a set $A$ with positive mass for $\mu$ but null logarithmic capacity, which means

$$ \mu(A) > 0, \quad \mathrm{cap}(A) := \exp \left( - \inf_{\nu \in \mathcal{P}(A)} \int \log(1/|x-y|) \nu(dx) \nu(dy) \right) = 0. $$

3. If $\mu$ has an atom, then $H_β(\mu) = +\infty$. If $\mu$ has no atom,

$$ H_β(\mu) = \int_{x<y} f_β(x, y)\mu(dx)\mu(dy). $$

4. For all $M \in \mathbb{R}$, $\{H_β \le M\}$ is a compact subset of $\mathcal{P}(\mathbb{R})$.

5. $H_β$ is convex on $\mathcal{P}(\mathbb{R})$.

6. $H_β$ is actually strict convex on $\mathcal{P}(\mathbb{R})$.

*Proof.* In this proof let $\Delta = \{(x, x), x \in \mathbb{R}\}$ be the diagonal of $\mathbb{R}^2$.

1. This point corresponds to Lemma 9.2. To fix the notations for this proof, (9.2) yields

$$ \forall (x, y) \in \mathbb{R}^2, f_β(x, y) \ge \frac{\beta}{2}(1 - \log(2\beta)) =: m_β. \quad (9.4) $$

Integrating against $\mu(dx)\mu(dy)$ gives the result.

2. If the first condition holds, $H_β(\mu) = +\infty$ by Lemma 9.4.

Then, we can notice using the inequality (9.4) that for all Borelian subsets $B$ of $\mathbb{R}^2$

$$ H_β(\mu) \ge \int_B f_β(x, y)\mu(dx)\mu(dy) + (1 - \mu \otimes \mu(B))m_β. $$

But if $B = A \times A$, we also have

$$ \int_B f_β(x, y)\mu(dx)\mu(dy) \ge -\beta \int_{A \times A} \log(|x - y|)\mu(dx)\mu(dy) \ge -\beta\mu(A)^2 \log(\mathrm{cap}(A)). $$

So under the second condition we also have $H_β(\mu) = +\infty$.

3. If $\mu$ has an atom $a$, then $\mu(\{a\}) > 0$ and $\mathrm{cap}(\{a\}) = 0$, then $H_β(\mu) = +\infty$. Assume that $\mu$ has no atom, then by Fubini-Tonelli

$$ \mu \times \mu(\Delta) = \int_{\mathbb{R}} \int_{\mathbb{R}} \mathbb{1}_{x=y} \mu(dx) \mu(dy) = \int_{\mathbb{R}} \mu(\{y\}) \mu(dy) = 0. $$

Since $f_β$ is symmetric we have the result.
4. Firstly, let us define for $M > 0$, $f_{\beta}^{M}(x, y) = \min(f_{\beta}(x, y), M)$ in order to truncate $f_{\beta}$ and to have a bounded function (since $f_{\beta}$ is bounded from below by $m_{\beta}$). We naturally associate to $f_{\beta}^{M}$ the quantity:

$$H_{\beta}^{M}(\mu) := \frac{1}{2} \int_{\mathbb{R}^2} f_{\beta}^{M}(x, y) \mu(dx) \mu(dy). \qquad (9.5)$$

Since $f_{\beta}^{M}$ is bounded, $\mu \mapsto H_{\beta}^{M}(\mu)$ is continuous on $\mathcal{P}(\mathbb{R})$.

Moreover, as $(f_{\beta}^{M})_{M>0}$ increases, converges pointwise to $f_{\beta}$ and for all $M > 0$ $f_{\beta}^{M}$ is bounded from below by $m_{\beta}$ we can apply monotone convergence theorem to have that for every $\mu \in \mathcal{P}(\mathbb{R})$, $H_{\beta}^{M}(\mu)$ converges to $H_{\beta}(\mu)$.

Of course it does not imply that $H_{\beta}$ is continuous but at least $H_{\beta}$ is lower semi continuous. Indeed, since for all $\mu \in \mathcal{P}(\mathbb{R})$ $H_{\beta}^{M}(\mu)$ increases to $H_{\beta}(\mu)$, it gives that for all $L \in \mathbb{R}$ $\{H_{\beta} \le L\} = \bigcap_{M>0} \{H_{\beta}^{M} \le L\}$. Since for all $M > 0$ $H_{\beta}^{M}$ is continuous on $\mathcal{P}(\mathbb{R})$, for all $L \in \mathbb{R}$, $\{H_{\beta} \le L\}$ is an intersection of closed subsets of $\mathcal{P}(\mathbb{R})$.

It remains to prove that for all $L \in \mathbb{R}$, $\{H_{\beta} \le L\}$ is a relatively compact subset of $\mathcal{P}(\mathbb{R})$. Fix $L \in \mathbb{R}$. Let $(\mu_n)_{n \in \mathbb{N}} \in \mathcal{P}(\mathbb{R})^{\mathbb{N}}$ be a family of probability measures such that for all $n \in \mathbb{N}$, $H_{\beta}(\mu_n) \le L$. For a constant $C > 0$ fixed we can find a compact subset $K \subset \mathbb{R}$ (which depend on $C$) such that

$$\min_{(K \times K)^c} \left( \frac{x^2}{2} + \frac{y^2}{2} - \beta \log |x - y| \right) \ge C.$$

Using Lemma 9.4 yields

$$L \ge H_{\beta}(\mu_n) \ge \frac{m_{\beta}}{2} \mu_n(K)^2 + \frac{C}{2} (\mu_n \otimes \mu_n)((K \times K)^c) \\ \ge -\frac{|m_{\beta}|}{2} + \frac{C}{2} \mu_n(K^c)$$

Hence for all $\varepsilon > 0$, we can find $K$ a compact subset of $\mathbb{R}$ such that for all $n \in \mathbb{N}$, $\mu_n(K^c) \le \varepsilon$ by choosing $C$ large enough. The Prokohorov theorem implies that for all $L \in \mathbb{R}$, $\{H_{\beta} \le L\}$ is a relatively compact subset of $\mathcal{P}(\mathbb{R})$.

5. In order to prove that $H_{\beta}$ is convex we only have to prove that $\Sigma$ is concave. Let state a lemma to have another expression of $\Sigma$.

**Lemma 9.7.** Let $a, b > 0$ then

$$\int_{\mathbb{R}^+} \frac{\exp(-at) - \exp(-bt)}{t} dt = \log(b/a).$$

*Proof.* Since the problem of this integral is essentially at 0 let us focus on this part.

$$\int_{\mathbb{R}^+} \frac{\exp(-at) - \exp(-bt)}{t} dt = \lim_{\varepsilon \to 0} \int_{\varepsilon}^{+\infty} \frac{\exp(-at) - \exp(-bt)}{t} dt.$$

As the two parts of the integral are converging we can cut this integral in two parts, we have:

$$\int_{\varepsilon}^{+\infty} \frac{\exp(-at) - \exp(-bt)}{t} dt = \int_{\varepsilon}^{+\infty} \frac{\exp(-at)}{t} dt - \int_{\varepsilon}^{+\infty} \frac{\exp(-bt)}{t} dt.$$
Now, on both integral we do the change of variables $t' = at$ and $t' = bt$. Hence

$$ \int_{\epsilon}^{+\infty} \frac{\exp(-at) - \exp(-bt)}{t} dt = \int_{a\epsilon}^{+\infty} \frac{\exp(-t)}{t} dt - \int_{b\epsilon}^{+\infty} \frac{\exp(-t)}{t} dt = \int_{a\epsilon}^{b\epsilon} \frac{\exp(-t)}{t} dt $$

Now by continuity of exp at 0 we have:

$$ \lim_{\epsilon \to 0} \int_{a\epsilon}^{b\epsilon} \frac{\exp(-t)}{t} dt = \lim_{\epsilon \to 0} \int_{a\epsilon}^{b\epsilon} \frac{1}{t} dt = \log(b/a). $$

It gives the result.

<figure><img src="image_14.png" /></figure>

Thanks to this formula we can write:

$$ \forall (x, y) \in \mathbb{R}^2, \log|x-y|^2 = \int_{\mathbb{R}^+} \frac{\exp\left(-\frac{t}{2}\right) - \exp\left(-\frac{|x-y|^2}{2}\right)}{t} dt $$

Then the change of variables $t' = 1/t$, yields

$$ \forall (x, y) \in \mathbb{R}^2, \log|x-y| = \int_{\mathbb{R}^+} \frac{\exp\left(-\frac{1}{2t}\right) - \exp\left(-\frac{|x-y|^2}{2t}\right)}{2t} dt \quad (9.6) $$

We shall use (9.6) to have another expression of $\Sigma$ and view it as a limit of concave functions.

Indeed for all $T > 1$, $\mu \in \mathcal{P}(\mathbb{R})$, we define

$$ \Sigma_T(\mu) = \int_{\mathbb{R}^2} \int_{1/T}^T \frac{\exp\left(-\frac{1}{2t}\right) - \exp\left(-\frac{|x-y|^2}{2t}\right)}{2t} dt \mu(dx) \mu(dy). $$

We can cut this integral in two parts: one part with the $(x, y)$ such that $|x - y| \le 1$ and the complement of this part.

On these parts the sign of the function we integrate is constant and so the Fubini-Tonelli theorem implies that for all $\mu \in \mathcal{P}(\mathbb{R})$

$$ \lim_{T \to +\infty} \Sigma_T(\mu) = \Sigma(\mu). $$

Hence, we only have to prove that for all $T > 1$, $\Sigma_T$ is concave. Fix $T > 1$.

We can first isolate the dependence in $\mu$ of $\Sigma_T$:

$$ \Sigma_T(\mu) = \int_{1/T}^{T} \frac{\exp\left(-\frac{1}{2t}\right)}{2t} dt - \int_{1/T}^{T} \frac{1}{2t} \int_{\mathbb{R}^2} \exp\left(-\frac{|x-y|^2}{2t}\right) \mu(dx) \mu(dy) dt. $$

So to prove that $\Sigma_T$ is concave we only have to prove that

$$ \mu \mapsto \int_{1/T}^{T} \frac{1}{2t} \int_{\mathbb{R}^2} \exp\left(-\frac{|x-y|^2}{2t}\right) \mu(dx) \mu(dy) dt $$

is convex on $\mathcal{P}(\mathbb{R})$. The key argument is to express $\Sigma_T$ using the Fourier transform of $\mu$.
**Lemma 9.8.** Let $\nu$ be a finite signed measure. Then for all $t > 0$,

$$ 
\begin{aligned}
& \int_{\mathbb{R}^2} \exp\left(-\frac{(x-y)^2}{2t}\right) d\nu(x)d\nu(y) \\
& = \sqrt{\frac{t}{2\pi}} \int_{\mathbb{R}} \left|\int_{\mathbb{R}} \exp(i\lambda x)d\nu(x)\right|^2 \exp(-t\lambda^2/2)d\lambda.
\end{aligned}
\qquad (9.7) $$

*Proof.* Let us recall the Fourier transform of the Gaussian

$$ \forall z \in \mathbb{C}, \quad \int_{\mathbb{R}} \exp(z\lambda) \exp(-\lambda^2)d\lambda = \sqrt{\pi} \exp(z^2/4). $$

Using this formula with $z = \sqrt{2i(x-y)/\sqrt{t}}$, applying the Fubini theorem and doing an affine change of variables gives the result. $\square$

With this lemma the convexity of $\Sigma_T$ easily follows since

$$ \mu \mapsto \left|\int_{\mathbb{R}} \exp(i\lambda x) d\mu(x)\right|^2 $$

is convex on $\mathcal{P}(\mathbb{R})$.

Indeed if $\mu \in \mathcal{P}(\mathbb{R})$, we have:

$$ \left|\int_{\mathbb{R}} \exp(i\lambda x) d\mu(x)\right|^2 = \left(\int_{\mathbb{R}} \cos(\lambda x) d\mu(x)\right)^2 + \left(\int_{\mathbb{R}} \sin(\lambda x) d\mu(x)\right)^2 $$

and the two terms are convex using the convexity of $t \mapsto t^2$.

It concludes about the concavity of $\Sigma$ and so the convexity of $H_\beta$.

6. About the strict convexity we shall use that $\Sigma$ is a quadratic form on $\mathcal{P}(\mathbb{R})$. Let state a general lemma.

**Lemma 9.9.** Let $E$ be a vector space, $C$ be a convex subset of $E$ and $q$ a quadratic form defined on $E$ with polar form $\phi$. Assume that for all $(x, y) \in C$, such that $x \neq y$ we have $q(x - y) < 0$.

Then $q$ is strictly concave on $C$.

*Proof.* Let $x \neq y$ both in $C$ and $t \in (0, 1)$, then we have

$$ q((1-t)x + ty) = (1-t)^2q(x) + t^2q(y) + 2t(1-t)\phi(x,y). $$

Moreover, since $q(x - y) < 0$ and $q(x - y) = q(x) + q(y) - 2\phi(x, y) < 0$ it gives

$$ q(x) + q(y) < 2\phi(x, y). $$

Hence as $t \in (0, 1)$, with the previous statements we have

$$ 
\begin{aligned}
q((1-t)x + ty) &> (1-t)^2q(x) + t^2q(y) + t(1-t)(q(x) + q(y)) \\
&= (1-t)q(x) + tq(y).
\end{aligned}
$$

$\square$
We now apply this lemma with $E = \{\mu \text{ finite signed measure such that: } \Sigma(\mu) > -\infty\}$, $C = \mathcal{P}(\mathbb{R}) \cap E$ and $q(\mu) = \Sigma(\mu)$.

By definition, $\Sigma$ is a quadratic form on $E$ with $\phi(\mu, \nu) = \int_{\mathbb{R}^2} \log|x-y|d\mu(x)d\nu(y)$ its polar form.

Moreover if $(\mu, \nu) \in \mathcal{C}^2$ we can rewrite $\Sigma(\mu, \nu)$ using (9.6):

$$ \Sigma(\mu - \nu) = - \int_{\mathbb{R}^+} \frac{1}{2t} \int_{\mathbb{R}^2} \exp\left(-\frac{|x-y|^2}{2t}\right) d(\mu - \nu)(x)d(\mu - \nu)(y)dt. $$

By (9.7), for $(\mu, \nu) \in \mathcal{C}^2$ we have

$$ \Sigma(\mu - \nu) = \int_{\mathbb{R}^+} \frac{1}{2\sqrt{2\pi t}} \int_{\mathbb{R}} \left| \int_{\mathbb{R}} \exp(i\lambda x) d(\mu - \nu)(x) \right|^2 \exp(-t\lambda^2/2) d\lambda dt. $$

This quantity is non positive and is equal to 0 if and only if for every $\lambda \in \mathbb{R}$ we have $\int_{\mathbb{R}} \exp(i\lambda x)d(\mu - \nu)(x) = 0$ (by continuity in $\lambda$) if and only if for every $\lambda \in \mathbb{R}$, $\int_{\mathbb{R}} \exp(i\lambda x)d\mu(x) = \int_{\mathbb{R}} \exp(i\lambda x)d\nu(x)$.

Since the characteristic function characterizes a probability measure, $\Sigma(\mu - \nu) \ge 0$ with equality if and only if $\mu = \nu$.

Hence, the assumptions of Lemma 9.9 are checked. $\Sigma$ is strictly concave of $\mathcal{C}$. So $H_\beta$ is strictly convex on $\mathcal{C}$.

□

**Remark 9.10.** Let us explain the deep link between the formula of Lemma 9.8 and Fourier representation. This link can be used for more general Coulomb and Riesz gases as we will explain in the Section 11.2 [74]. Since the Fourier transform of $x \to -\log|x|$ is $\xi \to 1/|\xi|$, if $\mu = \mu_1 - \mu_2$ where $\mu_1$ and $\mu_2$ are probability measures on $\mathbb{R}$ we get

$$ \int_{\mathbb{R}^2} -\log|x-y|\mu(dx)\mu(dy) = \int_{\mathbb{R}} \frac{|\mathcal{F}(\mu)(\xi)|^2}{|\xi|} d\xi, \qquad (9.8) $$

where $\mathcal{F}(\mu)$ is the Fourier transform of $\mu$ (this formula can be obtained by first assuming that $\mu$ has a density in the Schwartz space and then using a density argument). Thanks to the Fourier representation (9.8), if $\mu$ is the difference of two probabilities measures

$$ -\Sigma(\mu) = ||\mu||_{\dot{H}^{-1/2}}, $$

where $||\cdot||_{\dot{H}^{-1/2}}$ is the usual Sobolev semi norm on $H^{-1/2}$. This can be directly used to prove Proposition 9.6 thanks to Lemma 9.9.

The notion of logarithmic capacity is linked with the notion of capacity in electrostatic and play an important role to have a condition that ensures $\inf_{\mu \in \mathcal{P}(\mathbb{R})} H_\beta(\mu) < +\infty$ (see for instance [59, 72] for a more detailed introduction to the notion of capacity). The only basic properties about the logarithmic capacity that shall be used is that a set of zero logarithmic capacity has also zero Lebesgue measure and for a borelian $E$, $\mathrm{cap}(E) = \sup_{K \subset E, K \text{ is compact}} \mathrm{cap}(K)$.

Let state a lemma to prove that $\inf_{\mu \in \mathcal{P}(\mathbb{R})} H_\beta(\mu)$ is finite.

**Lemma 9.11.** The quantity $\inf_{\mu \in \mathcal{P}(\mathbb{R})} H_\beta(\mu)$ is finite.
Proof. We follow the arguments of Lemma 2.4 of [74]. Let $\epsilon > 0$ and let us define $\mathcal{V}_\epsilon := \{x | V(x) \le \epsilon^{-1}\}$ where $V(x) = x^2/2$ is the external potential that appears in the potential $H_\beta$. $\mathcal{V}_\epsilon$ is a compact subset of $\mathbb{R}$ with positive Lebesgue measure and so has a positive logarithmic capacity. By definition of the logarithmic capacity, we can find a measure $\nu_\epsilon$ with a support included in $\mathcal{V}_\epsilon$ such that

$$- \int \log(|x - y|) \nu_{\epsilon}(dx) \nu_{\epsilon}(dy) < +\infty.$$

By definition of $\mathcal{V}_\epsilon$, $\int V d\nu_\epsilon < +\infty$. So $H_\beta(\nu_\epsilon) < +\infty$. $\square$

**Remark 9.12.** One can also directly prove by computations that $H_\beta(\mu)$ is finite if $\mu$ is compactly supported with a bounded density with respect to the Lebesgue measure or for instance if $\mu(dx) = \exp(-x^2/2)/\sqrt{2\pi} dx$.

Thanks to this lemma we can now define a good rate function.

**Definition 9.13.** Let $I_\beta : \mathcal{P}(\mathbb{R}) \to [0, +\infty]$ the function defined by:

$$I_{\beta}(\mu) = H_{\beta}(\mu) - \inf_{\mu \in \mathcal{P}(\mathbb{R})} H_{\beta}(\mu).$$

As a consequence of Proposition 9.6 we get the following result.

**Theorem 9.14.** $I_\beta$ is a good rate function.

## 9.2 Equilibrium measure.

We first introduce the notion of equilibrium measure.

**Proposition 9.15.** $I_\beta$ admits a minimum on $\mathcal{P}(\mathbb{R})$ and this minimum is reached in a unique element $\mu_\beta \in \mathcal{P}(\mathbb{R})$ which is called the equilibrium measure of $I_\beta$.

Proof. By Lemma 9.11, we can consider $\nu \in \mathcal{P}(\mathbb{R})$ such that $I_\beta(\nu) < +\infty$. Since $\{\mu \in \mathcal{P}(\mathbb{R}) | I_\beta(\mu) \le I_\beta(\nu)\}$ is compact and $I_\beta$ is lsc by Proposition 9.6, $I_\beta$ reaches its infimum by Lemma 8.12. Moreover, by strict convexity of $I_\beta$, this minimum is reached in a unique element. $\square$

We shall use a general result of Frostman that characterizes the unique minimizer of $I_\beta$ (called the Frostman equilibrium measure) as solution of an Euler-Lagrange equation. The following theorem is general as we will mention in Section 11.2 but we only give a proof for $I_\beta$ (but the proof would be similar for Coulomb and Riesz gases, see for instance [74]). Intuitively the equilibrium measure corresponds to the distribution of the particles that minimizes the energy of the system. Here $V(x) = x^2/2$ is the external potential and satisfies $\lim_{|x| \to +\infty} (V(x) - \log|x|) = +\infty$. This formally forces at equilibrium the particles to be not to far from the origin because if not the energy of the system would be to high. This explains why the support of the equilibrium measure shall be compact as we will see.

**Theorem 9.16 (Frostman).**

1. There exists a unique minimizer of $I_\beta$: $\mu_\beta \in \mathcal{P}(\mathbb{R})$ which satisfies

$$I_{\beta}(\mu_{\beta}) = 0.$$

2. The support $S_\beta$ of $\mu_\beta$ is compact and with positive logarithmic capacity.
3. We can find A with null logarithmic capacity such that for all $x \notin A$, we have:

$$ \beta \int_{\mathbb{R}} \log |x - y| d\mu_{\beta}(y) \leq \frac{1}{2} x^2 - 2 \inf H_{\beta} + \frac{1}{2} \int_{\mathbb{R}} y^2 d\mu_{\beta}(y). \quad (9.9) $$

4. For all $x \in S_{\beta}$ we have:

$$ \beta \int_{\mathbb{R}} \log |x - y| d\mu_{\beta}(y) = \frac{1}{2} x^2 - 2 \inf H_{\beta} + \frac{1}{2} \int_{\mathbb{R}} y^2 d\mu_{\beta}(y). \quad (9.10) $$

Proof.

1. This is Proposition 9.15.

2. Let $K$ be a compact such that $f_{\beta}(x, y) \geq 2(H_{\beta}(\mu_{\beta}) + 1)$ for $(x, y) \notin K \times K$ and $\mu_{\beta}(K) > 0$.

Assume by contradiction that $\mu_{\beta}(K) < 1$ and let us define the conditional probability

$$ \mu_{\beta,K} := \frac{\mu_{\beta}|_{K}}{\mu_{\beta}(K)}. $$

The goal is to show that $H_{\beta}(\mu_{\beta,K}) < H_{\beta}(\mu_{\beta})$ to have a contradiction. We compute $H_{\beta}(\mu_{\beta})$ by distinguishing $(x, y)$ that are in $K \times K$ and not in:

$$ \begin{aligned} H_{\beta}(\mu_{\beta}) &= \frac{1}{2} \int_{\mathbb{R} \times \mathbb{R}} f_{\beta}(x, y) \mu_{\beta}(dx) \mu_{\beta}(dy) \\ &= \frac{1}{2} \int_{K \times K} f_{\beta}(x, y) \mu_{\beta}(dx) \mu_{\beta}(dy) + \frac{1}{2} \int_{(K \times K)^c} f_{\beta}(x, y) \mu_{\beta}(dx) \mu_{\beta}(dy) \quad (9.11) \\ &= \mu_{\beta}(K)^2 H_{\beta}(\mu_{\beta,K}) + (H_{\beta}(\mu_{\beta}) + 1) \mu_{\beta} \otimes \mu_{\beta}((K \times K)^c) \\ &\geq \mu_{\beta}(K)^2 H_{\beta}(\mu_{\beta,K}) + (H_{\beta}(\mu_{\beta}) + 1)(1 - \mu_{\beta}(K)^2) \end{aligned} $$

It yields that

$$ H_{\beta}(\mu_{\beta,K}) \leq \frac{H_{\beta}(\mu_{\beta})}{\mu_{\beta}(K)^2} + \frac{\mu_{\beta}(K)^2 - 1}{\mu_{\beta}(K)^2} (H_{\beta}(\mu_{\beta}) + 1) < H_{\beta}(\mu_{\beta}), $$

since $\mu_{\beta}(K) < 1$. This is a contradiction with the minimality of $\mu_{\beta}$. Hence, the support of $\mu_{\beta}$ is compact. Moreover by Proposition 9.6, since $H_{\beta}(\mu_{\beta})$ is finite by Lemma 9.11, the logarithmic capacity of the support of $\mu$ is positive.

3. Now we shall prove the two last points. We use the standard method of variations to obtain the Euler-Lagrange equation satisfies by the minimizer of $H_{\beta}$.

Let $\mu \in \mathcal{P}(\mathbb{R})$ such that $H_{\beta}(\mu) < +\infty$. Since $\mu_{\beta}$ minimizes $H_{\beta}$, for $t \in [0, 1]$, we have:

$$ H_{\beta}((1-t)\mu_{\beta} + t\mu) \geq H_{\beta}(\mu_{\beta}). $$

Using the expression of $H_{\beta}$, we get

$$ H_{\beta}(\mu_{\beta}) + t \left[ \int f_{\beta}(x, y) \mu_{\beta}(dx)(\mu - \mu_{\beta})(dy) \right] + O(t^2) \geq H_{\beta}(\mu_{\beta}). $$
Simplifying the previous inequality, dividing by $t > 0$ and letting $t \to 0$ gives:

$$ 
\begin{gathered}
\int f_{\beta}(x, y)\mu_{\beta}(dx)(\mu - \mu_{\beta})(dy) \geq 0, \\
\int f_{\beta}(x, y)\mu_{\beta}(dx)\mu(dy) \geq \int f_{\beta}(x, y)\mu_{\beta}(dx)\mu(dy) = 2H_{\beta}(\mu_{\beta}) = 2\inf H_{\beta} \qquad (9.12)
\end{gathered}
$$

Set

$$ h_{\beta}(x) := -\beta \int \log(|x - y|)\mu_{\beta}(dy) + \frac{x^2}{2} $$

and use the expression of $f_{\beta}$ to obtain:

$$ \int h_{\beta}(x)\mu(dx) \geq 2\inf H_{\beta} - \frac{1}{2}\int x^2\mu_{\beta}(dx) = 2H_{\beta}(\mu_{\beta}) - \frac{1}{2}\int x^2\mu_{\beta}(dx) =: c_{\beta} \quad (9.13) $$

Hence for all $\mu$ such that $H_{\beta}(\mu) < +\infty$

$$ \int h_{\beta}(x)\mu(dx) \geq c_{\beta}. $$

Let us notice that (9.9) is equivalent to $h_{\beta}(x) \geq c_{\beta}$.

Formally if we could chose dirac measures $\mu(dx) = \delta_y(dx)$ we would have (9.9) for every $y \in \mathbb{R}$. However we cannot use such measures since $H_{\beta}(\delta_y(dx)) = +\infty$. We prove (9.9) by contradiction. Assume that there exists a set $E$ of positive capacity such that (9.9) is false. By properties of the capacity of a set, we can suppose that $E$ is in fact compact. We can find a probability measure $\nu$ supported in $E$ such that

$$ -\int \log(|x - y|)\nu(dx)\nu(dy) < +\infty. $$

Since $\nu$ is supported on a compact, it implies that $H_{\beta}(\nu) < +\infty$ and (9.13) gives $\int h_{\beta}(x)\nu(dx) \geq c_{\beta}$ which is in contradiction with the facts that (9.9) does not hold on $E$. This proves the third point of Theorem 9.16.

Now we prove (9.10). Since $H_{\beta}(\mu_{\beta}) < +\infty$, $\mu_{\beta}$ does not charge sets of null logarithmic capacity. Hence, we deduce that on $S_{\beta}$ the inequality (9.9) is true:

$$ h_{\beta}(x) \geq c_{\beta}, \mu_{\beta}\text{-a.e} \qquad (9.14) $$

Let us notice that by definition of $h_{\beta}$ and $c_{\beta}$, $\int h_{\beta}\mu_{\beta}(dx) = c_{\beta}$. The inequality in (9.14) is actually an equality on $S_{\beta}$.

□

This theorem gives us the following important characterization of the minimizer of $I_{\beta}$.

**Proposition 9.17.** Let $\mu \in \mathcal{P}(\mathbb{R})$. Then $\mu$ minimizes $I_{\beta}$ if and only if for $\mu$-almost all $x$ we have:

$$ \beta \int_{\mathbb{R}} \log |x - y| d\mu(y) = \frac{1}{2} x^2 - 2 \inf H_{\beta} + \frac{1}{2} \int_{\mathbb{R}} y^2 d\mu(y). \qquad (9.15) $$

*Proof.* The direct sense is a consequence of Theorem 9.16.

For the reverse, integrating (9.15) with respect to $\mu$ yields:

$$ \beta\Sigma(\mu) = \int_{\mathbb{R}} x^2 d\mu(x) - 2\inf H_{\beta}. $$

Hence by definition of $H_{\beta}$, we have: $H_{\beta}(\mu) = \inf H_{\beta}$.

□
We will use this equation to compute explicitly the equilibrium measure of $I_β$ in Section 10.4. Formally, if we differentiate (9.10) on $S_β$, we obtain that for all $x \in S_β$

$$ \beta \int_{\mathbb{R}} \frac{1}{x-y} d\mu_{\beta}(y) = x. $$

This is exactly the equation of a stationary measure obtained in (7.67) when studying the partial differential equation (7.47) of the Ornstein-Uhlenbeck process in Section 7.5. We justified why the stationary measure of such process should be the semicircle distribution. In Section 10.4 we shall give another proof using the large deviations principle.

# 10 Large deviations

In this section we present the large deviations of the family of measures $(Q_β^N)_{N \in \mathbb{N}}$ for all $β > 0$. We fix $β > 0$.

## 10.1 Large deviations principle of the normalizing factor

In Statistical mechanics, the study of the normalizing constant of the system is a first important step as it contains many information of the system and its called the partition function of the system.

We first study the behaviour when $N$ goes large of the normalizing factor.

**Proposition 10.1.** We have:

$$ \lim_{N \to \infty} \frac{1}{N^2} \log Z_{\beta}^N = \frac{\beta}{4} \log \left( \frac{\beta}{2} \right) - \frac{3}{8} =: F(\beta). $$

*Proof.* To do this, let us state an important result of Selberg [39] about the value of $Z_β^N$. This result is very important to compute explicitly the normalizing constant of the density of eigenvalues of random matrices. A proof can be found in [5] in Section 2.5.3. Let us mention that we computed $Z_β^N$ in the case $β = 2$ which corresponds to the complex Wigner case in Theorem 2.48.

**Lemma 10.2** (Admitted). For all $β > 0$, and $N > 0$, we have the following equality:

$$ Z_{\beta}^{N} = (2\pi)^{N/2} N^{-N/2 - \beta N(N-1)/4} \left[ \Gamma\left(1 + \beta/2\right) \right]^{-N} \prod_{j=1}^{N} \Gamma\left(1 + \beta j/2\right), $$

with $\Gamma$ the Euler function.

Using Lemma 10.2 we have:

$$ \frac{1}{N^2} \log(Z_{\beta}^N) = \lim_{N \to \infty} \left( o(1) - \frac{\beta}{4} \log(N) + o(1) + \frac{1}{N^2} \sum_{j=1}^{N} \log \left( \Gamma \left( 1 + \beta j/2 \right) \right) \right). $$

Moreover, Stirling's formula yields:

$$ \log(\Gamma(x)) \underset{x \to \infty}{=} x \log(x/e) + o(x). $$
Hence, we obtain: $\sum_{j=1}^{N} \log(\Gamma(1 + \beta j/2)) = \sum_{N \to \infty} \frac{\beta j}{2} \log \frac{\beta j}{2e} + o(N^2)$.

Now let us notice that:

$$ \frac{1}{N^2} \sum_{j=1}^{N} \frac{\beta j}{2} \log \frac{\beta j}{2e} - \frac{\beta}{4} \log N = \frac{\beta}{2N} \sum_{j=1}^{N} \frac{j}{N} \log \frac{\beta j}{2eN} + o(1). $$

But the last sum is a Riemann sum. It gives:

$$ \frac{\beta}{2N} \sum_{j=1}^{N} \frac{j}{N} \log \frac{\beta j}{2eN} = \frac{\beta}{2} \int_{0}^{1} x \log \frac{\beta x}{2e} + o(1) = F(\beta) + o(1). $$

Hence we have:

$$ \frac{1}{N^2} \sum_{j=1}^{N} \frac{\beta j}{2} \log \frac{\beta j}{2e} = \frac{\beta}{4} \log N + F(\beta) + o(1). $$

This gives:

$$ \frac{1}{N^2} \log(Z_{\beta}^N) = F(\beta) + o(1). $$

□

**Remark 10.3.** Let us mention that in the general case of Coulomb or Riesz gases if for instance we consider another external potential than $V(x) = x^2/2$ we cannot compute the exact value of the normalizing factor. However, we are still able to prove large deviations as we shall see in next sections and the large deviations of the normalizing factor. Indeed, similar computations as we shall do to prove the upper bound (10.8) and the lower bound (10.7) can be done to show that the normalizing factor satisfies a large deviation principles. We refer to Theorem 3.3 of [74] or Theorem 2.6.1 [5]. We choose to not present this approach seems the computations should have been very similar with what we shall do and because in the particular case of $\beta$ ensembles of this section we think it is worth mentioning the Selberg integral formula (which was computed in previous sections for the case $\beta = 2$). We shall mention in Remark 10.8 and Remark 10.12 how we can obtain a such large deviation principles for the normalizing factor without computing the exact value of $Z_{\beta}^N$.

## 10.2 Weak large deviations principle

The main goal of this section is to prove a so-called weak large deviations principle, that means a deviations principle for compact and open subsets of $\mathcal{P}(\mathbb{R})$. Let $\widehat{I}_{\beta} = H_{\beta} + F(\beta)$.

**Theorem 10.4.** $(Q_{\beta}^N)_{N \in \mathbb{N}}$ satisfies a weak deviations principle with good rate function $\widehat{I}_{\beta}$. More exactly we have:

1. For any open subsets $O$ of $\mathcal{P}(\mathbb{R})$,

$$ \lim_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^N (\overline{\mu_N} \in O) \geq - \inf_{O} \widehat{I}_{\beta}. \qquad (10.1) $$
2. For any compact subsets K of P($\mathbb{R}$),

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^{N}(\overline{\mu_N} \in K) \leq - \inf_{K} \widehat{I}_{\beta}. \quad (10.2) $$

We first state an elementary lemma that plays a key role in large deviations theory as we will see.

**Lemma 10.5.** Let $(u_{j,N})_{j,N \in \mathbb{N}^2}$ be a family of non negative real numbers and $k \in \mathbb{N}_{\ge 1}$, then

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \left( \sum_{j=1}^{k} u_{j,N} \right) \leq \max_{1 \leq j \leq k} \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log(u_{j,N}). $$

*Proof.* The proof is just a use of the definition of the superior limit. Fix $\varepsilon > 0$. By definition of the superior limit, for all $j$, we can find $N(j) \in \mathbb{N}$ such that for all $N \ge N(j)$,

$$ \frac{1}{N^2} \log(u_{j,N}) \leq \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log(u_{j,N}) + \varepsilon. $$

Let $\tilde{N} = \max_{1 \le j \le k} N(j)$. For all $1 \le j \le k$, for all $N \ge \tilde{N}$, we have

$$ \frac{1}{N^2} \log(u_{j,N}) \leq \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log(u_{j,N}) + \varepsilon, $$

which implies that for all $1 \le j \le k$, for all $N \ge \tilde{N}$,

$$ u_{j,N} \le \exp \left( N^2 \left( \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log(u_{j,N}) + \varepsilon \right) \right) \le \exp \left( N^2 \left( \max_{1 \le j \le k} \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log(u_{j,N}) + \varepsilon \right) \right). $$

Summing these inequalities and taking the logarithm yields

$$ \frac{1}{N^2} \log \left( \sum_{j=1}^{k} u_{j,N} \right) \leq \frac{\log(k)}{N^2} + \max_{1 \leq j \leq k} \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log(u_{j,N}) + \varepsilon, $$

for all $N \ge \tilde{N}$. Taking the superior limit and letting $\varepsilon$ goes to 0 gives the result. $\square$

Thanks to this lemma we can reduce the weak deviations principle to a "local" large deviations principle.

**Lemma 10.6.** If for all $\mu \in P(\mathbb{R})$ we have:

$$ -\widehat{I}_{\beta}(\mu) = \lim_{\delta \to 0} \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^{N}(\overline{\mu_N} \in B(\mu, \delta)) \quad (10.3) $$

and

$$ -\widehat{I}_{\beta}(\mu) = \lim_{\delta \to 0} \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^{N}(\overline{\mu_N} \in B(\mu, \delta)), \quad (10.4) $$

then $(Q_{\beta}^{N})_{N \in \mathbb{N}}$ satisfies a weak deviations principle with rate function $\widehat{I}_{\beta}$.
Proof. Let us prove that the hypothesis (10.4) implies the weak deviations principle for compact subsets.

Let $K$ be a compact of $\mathcal{P}(\mathbb{R})$. Fix $\eta > 0$. By hypothesis (10.4), for all $\mu \in K$ we can find $\delta_\mu > 0$ such that: $\overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \sum_{i=1}^{l_\delta} Q_\beta^N(\overline{\mu_N} \in B(\mu, \delta_\mu)) \le -I_\beta(\mu) + \eta \le -\inf_K I_\beta + \eta$. Since $K$ is compact, we can find a finite number $l_\eta$ of $\mu \in K$ such that $K \subset \bigcup_{i=1}^{l_\eta} B(\mu_i, \delta_{\mu_i})$. By Lemma 10.5 we have:

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_\beta^N(\overline{\mu_N} \in K) \le \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \sum_{i=1}^{l_\eta} Q_\beta^N(\overline{\mu_N} \in B(\mu_i, \delta_{\mu_i})) \quad (10.5) $$

$$ \le \max_{i \le l_\eta} \left( \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_\beta^N(\overline{\mu_N} \in B(\mu_i, \delta_{\mu_i})) \right) \quad (10.6) $$

Hence, thanks to the choice $\delta_\mu$ we have:

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_\beta^N(\overline{\mu_N} \in K) \le -\inf_K \widehat{I_\beta} + \eta. $$

Let $\eta$ converges to 0 to conclude.

Let us prove that the hypothesis (10.3) implies the weak deviations principle for open subsets of $\mathcal{P}(\mathbb{R})$.

Let $O$ be an open subset of $\mathcal{P}(\mathbb{R})$. Fix $\varepsilon > 0$, we take $\mu \in O$ such that $\widehat{I_\beta}(\mu) \le \inf_O \widehat{I_\beta} + \varepsilon$. Hence for all $\delta > 0$ small enough we have $B(\mu, \delta) \subset O$ and so

$$ \lim_{N \to \infty} \frac{1}{N^2} \log Q_\beta^N(\overline{\mu_N} \in O) \ge \lim_{N \to \infty} \frac{1}{N^2} \log Q_\beta^N(\overline{\mu_N} \in B(\mu, \delta)). $$

Letting $\delta$ converges to 0 and using the hypothesis (10.4) yields:

$$ \lim_{N \to \infty} \frac{1}{N^2} \log Q_\beta^N(\overline{\mu_N} \in O) \ge -\widehat{I_\beta}(\mu) \ge -\inf_O \widehat{I_\beta} - \varepsilon. $$

Let $\varepsilon$ converges to 0 to conclude. $\square$

**Remark 10.7.** As we shall explain after in the proof of the lower bound, we only have to prove the lower bound (10.7) for "regular" measures. See for instance Corollary 2.7 and Theorem 1.1 of [26].

Thanks to the large deviations principle for the normalizing constant stated in Proposition 10.1, to prove Theorem 10.4 it remains to show that for every $\mu \in \mathcal{P}(\mathbb{R})$:

$$ -H_\beta(\mu) \le \lim_{\delta \to 0} \lim_{N \to \infty} \frac{1}{N^2} \log \overline{Q_\beta^N}(\overline{\mu_N} \in B(\mu, \delta)) \quad (10.7) $$

$$ \lim_{\delta \to 0} \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \overline{Q_\beta^N}(\overline{\mu_N} \in B(\mu, \delta)) \le -H_\beta(\mu), \quad (10.8) $$

where we recall that $\overline{Q_\beta^N}$ was defined in the Definition 8.7 as the non normalize version of $Q_\beta^N$.
### 10.2.1 Upper bound (10.8)

By definition we have:

$$ \overline{Q}_{\beta}^{N}(\overline{\mu}_{N} \in B(\mu, \delta)) = \int_{\overline{\mu}_{N} \in B(\mu, \delta)} \prod_{1 \le i < j \le N} |\lambda_{i} - \lambda_{j}|^{\beta} \exp \left( -\frac{1}{2} N \sum_{i=1}^{N} \lambda_{i}^{2} \right) \prod_{i=1}^{N} d\lambda_{i}. $$

We want to make appear $H_{\beta}(\mu)$ in the previous expression. We recall that

$$ H_{\beta}(\mu) = \frac{1}{2} \int_{\mathbb{R}^2} f_{\beta}(x, y) \mu(dx) \mu(dy). $$

Hence, we write

$$ \begin{aligned} \overline{Q}_{\beta}^{N}(\overline{\mu}_{N} \in B(\mu, \delta)) &= \int_{\overline{\mu}_{N} \in B(\mu, \delta)} \prod_{\lambda_{i} < \lambda_{j}} |\lambda_{i} - \lambda_{j}|^{\beta} \exp \left( -\frac{1}{2} \sum_{\lambda_{i} < \lambda_{j}} (\lambda_{i}^{2} + \lambda_{j}^{2}) - \frac{1}{2} \sum_{i=1}^{N} \lambda_{i}^{2} \right) \prod_{i=1}^{N} d\lambda_{i} \\ &= \int_{\overline{\mu}_{N} \in B(\mu, \delta)} \exp \left( -N^{2} \int_{x<y} f_{\beta}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) \right) \exp \left( -\frac{1}{2} \sum_{i=1}^{N} \lambda_{i}^{2} \right) \prod_{i=1}^{N} d\lambda_{i}. \end{aligned} \quad (10.9) $$

Let $(\lambda_1^N, \dots, \lambda_N^N)$ be different real numbers then:

$$ \int_{\Delta} \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) = \int_{\mathbb{R}} \int_{\mathbb{R}} \mathbb{1}_{x=y} \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) = \frac{1}{N} \int_{\mathbb{R}} |\{i; \lambda_{i}^{N} = y\}| \overline{\mu}_{N}(dy). $$

Since the $(\lambda_1^N, \dots, \lambda_N^N)$ are different real numbers, for all $y$ the cardinal is either 1 if $y$ is one of these numbers or 0 otherwise. Hence:

$$ \overline{\mu}_{N} \otimes \overline{\mu}_{N}(\Delta) = \int_{\Delta} \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) = \frac{1}{N}. $$

The technical point is now that since $H_{\beta}$ is not continuous we must pay attention to the passage to the limit. So the idea is to look at $H_{\beta}^{M}$ for a certain $M$ which is continuous and which converges pointwise to $H_{\beta}$ which was defined during the proof of Proposition 9.6 by the formula (9.5). Let us notice that:

$$ H_{\beta}^{M}(\overline{\mu}_{N}) = \frac{1}{2} \int_{\mathbb{R}^2} f_{\beta}^{M}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) = \int_{x<y} f_{\beta}^{M}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) + \frac{1}{2} \int_{\Delta} f_{\beta}^{M}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy). $$

So if $\overline{\mu}_{N} \in B(\mu, \delta)$ then:

$$ \inf_{\nu \in B(\mu, \delta)} H_{\beta}^{M}(\nu) \le \int_{x<y} f_{\beta}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) + \frac{M}{2N}. \quad (10.10) $$

We can now bound (10.9) using (10.10). It gives:

$$ \overline{Q}_{\beta}^{N}(\overline{\mu}_{N} \in B(\mu, \delta)) \le \exp \left( -N^2 \inf_{\nu \in B(\mu, \delta)} H_{\beta}^{M}(\nu) + \frac{MN}{2} \right) \int_{\mathbb{R}^N} \exp \left( -\frac{1}{2} \sum_{i=1}^{N} \lambda_i^2 \right) \prod_{i=1}^{N} d\lambda_i. $$

Hence we have the following upper bound:

$$ \overline{Q}_{\beta}^{N}(\overline{\mu}_{N} \in B(\mu, \delta)) \le \exp \left( -N^2 \inf_{\nu \in B(\mu, \delta)} H_{\beta}^{M}(\nu) + \frac{MN}{2} \right) (2\pi)^{N/2}. $$
So, we get that for any $M > 0$,

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \overline{Q}_{\beta}^{N}(\overline{\mu}_{N} \in B(\mu, \delta)) \leq - \inf_{\nu \in B(\mu, \delta)} H_{\beta}^{M}(\nu). $$

Since, for any $M > 0$, $H_{\beta}^{M}$ is continuous:

$$ \lim_{\delta \to 0} \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \overline{Q}_{\beta}^{N}(\overline{\mu}_{N} \in B(\mu, \delta)) \leq -H_{\beta}^{M}(\mu). $$

Let $M$ goes to $+\infty$:

$$ \lim_{\delta \to 0} \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \overline{Q}_{\beta}^{N}(\overline{\mu}_{N} \in B(\mu, \delta)) \leq -H_{\beta}(\mu). $$

### 10.2.2 Lower bound (10.7)

We will only prove the result for a family of "regular" measures as mentioned in Remark 10.7. We consider measures $\mu \in \mathcal{P}(\mathbb{R})$ such that $\mu$ is absolutely continuous with respect to Lebesgue measure (we still write $\mu$ its density), such that $\int_{\mathbb{R}} x^2 d\mu$, $\int_{\mathbb{R}} \log(\mu) d\mu$ are integrable and for which there exists $C \in \mathbb{R}$ such that for all $x \in \mathbb{R}$

$$ \frac{x^2}{2} + \log(\mu(x)) \geq C. \qquad (10.11) $$

In this section we fix a such measure $\mu \in \mathcal{P}(\mathbb{R})$ and a constant $C$. We start as for the proof of the upper bound:

$$ \begin{aligned} \overline{Q}_{\beta}^{N}(\overline{\mu}_{N} \in B(\mu, \delta)) &= \int_{\{\overline{\mu}_{N} \in B(\mu, \delta)\}} \prod_{1 \leq i < j \leq N} |\lambda_i - \lambda_j|^{\beta} \exp \left( -\frac{1}{2} N \sum_{i=1}^{N} \lambda_i^2 \right) \prod_{i=1}^{N} d\lambda_i \\ &= \int_{\{\overline{\mu}_{N} \in B(\mu, \delta)\}} \prod_{\lambda_i < \lambda_j} |\lambda_i - \lambda_j|^{\beta} \exp \left( -\frac{1}{2} \sum_{\lambda_i < \lambda_j} (\lambda_i^2 + \lambda_j^2) - \frac{1}{2} \sum_{i=1}^{N} \lambda_i^2 \right) \prod_{i=1}^{N} d\lambda_i \\ &= \int_{\{\overline{\mu}_{N} \in B(\mu, \delta)\}} \exp \left( -N^2 \int_{x<y} f_{\beta}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) \right) \exp \left( -\frac{1}{2} \sum_{i=1}^{N} \lambda_i^2 \right) \prod_{i=1}^{N} d\lambda_i \\ &= \int_{\{\overline{\mu}_{N} \in B(\mu, \delta)\}} \exp \left( -N^2 \int_{x<y} f_{\beta}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) \right) \exp \left( -\frac{1}{2} \sum_{i=1}^{N} \lambda_i^2 - \sum_{i=1}^{N} \log(\mu(\lambda_i)) \right) \prod_{i=1}^{N} \mu(d\lambda_i). \end{aligned} \qquad (10.12) $$

Taking the logarithm and using the Jensen inequality yields:

$$ \begin{aligned} \log \left( \overline{Q}_{\beta}^{N} \left( \overline{\mu}_{N} \in B(\mu, \delta) \right) \right) &\geq \\ & \int_{\{\overline{\mu}_{N} \in B(\mu, \delta)\}} \log \left[ \exp \left( -N^2 \int_{x<y} f_{\beta}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) \right) \exp \left( -\frac{1}{2} \sum_{i=1}^{N} \lambda_i^2 - \sum_{i=1}^{N} \log(\mu(\lambda_i)) \right) \right] \prod_{i=1}^{N} \mu(d\lambda_i) \\ &= \int_{\{\overline{\mu}_{N} \in B(\mu, \delta)\}} -N^2 \int_{x<y} f_{\beta}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) - \frac{1}{2} \sum_{i=1}^{N} \lambda_i^2 - \sum_{i=1}^{N} \log(\mu(\lambda_i)) \prod_{i=1}^{N} \mu(d\lambda_i) \\ &=: I_{1,N} + I_{2,N}, \end{aligned} \qquad (10.13) $$
where $I_{1,N}$ and $I_{2,N}$ are defined by:

$$ egin{cases} I_{1,N} = \int_{\mathbb{R}^N} -N^2 \int_{x<y} f_{\beta}(x,y) \overline{\mu_N}(dx) \overline{\mu_N}(dy) - \frac{1}{2} \sum_{i=1}^N \lambda_i^2 - \sum_{i=1}^N \log(\mu(\lambda_i)) \prod_{i=1}^N \mu(d\lambda_i) \\ I_{2,N} = \int_{\{\overline{\mu_N} \notin B(\mu, \delta)\}} N^2 \int_{x<y} f_{\beta}(x,y) \overline{\mu_N}(dx) \overline{\mu_N}(dy) + \frac{1}{2} \sum_{i=1}^N \lambda_i^2 + \sum_{i=1}^N \log(\mu(\lambda_i)) \prod_{i=1}^N \mu(d\lambda_i) \end{cases} \quad (10.14) $$

We first look at $I_{2,N}$. Using (9.4) and (10.11) we get:

$$ I_{2,N} \geq \left( N^2 m_{\beta} \frac{N(N-1)}{2N^2} + CN \right) \int_{\{\overline{\mu_N} \notin B(\mu, \delta)\}} \prod_{i=1}^{N} \mu(d\lambda_i). \quad (10.15) $$

By the law of large numbers, we have

$$ \lim_{N \to \infty} \int_{\{\overline{\mu_N} \notin B(\mu, \delta)\}} \prod_{i=1}^{N} \mu(d\lambda_i) = 0. $$

So we obtain that:

$$ \frac{I_{2,N}}{N^2} \geq o_N(1). \quad (10.16) $$

We now focus on $I_{1,N}$. Using that $\mu \in \mathcal{P}(\mathbb{R})$ and the symmetry of $f_{\beta}$ yields

$$ egin{align} I_{1,N} &= \int_{\mathbb{R}^N} -N^2 \int_{x<y} f_{\beta}(x,y) \overline{\mu_N}(dx) \overline{\mu_N}(dy) - \frac{1}{2} \sum_{i=1}^N \lambda_i^2 - \sum_{i=1}^N \log(\mu(\lambda_i)) \prod_{i=1}^N \mu(d\lambda_i) \nonumber \\ &= -\frac{1}{2} \int_{\mathbb{R}^N} \sum_{i \neq j} f_{\beta}(\lambda_i, \lambda_j) \prod_{i=1}^N \mu(d\lambda_i) - \frac{N}{2} \int_{\mathbb{R}} \lambda^2 \mu(d\lambda) - N \int_{\mathbb{R}} \log(\mu(\lambda)) \mu(d\lambda) \nonumber \\ &= -\frac{1}{2} \sum_{i \neq j} \int_{\mathbb{R}^2} f_{\beta}(x,y) \mu(dx) \mu(dy) - \frac{N}{2} \int_{\mathbb{R}} \lambda^2 \mu(d\lambda) - N \int_{\mathbb{R}} \log(\mu(\lambda)) \mu(d\lambda) \nonumber \\ &= -\frac{N(N-1)}{2} \int_{\mathbb{R}^2} f_{\beta}(x,y) \mu(dx) \mu(dy) - \frac{N}{2} \int_{\mathbb{R}} \lambda^2 \mu(d\lambda) - N \int_{\mathbb{R}} \log(\mu(\lambda)) \mu(d\lambda) \nonumber \\ &= -N(N-1)H_{\beta}(\mu) - \frac{N}{2} \int_{\mathbb{R}} \lambda^2 \mu(d\lambda) - N \int_{\mathbb{R}} \log(\mu(\lambda)) \mu(d\lambda) \tag{10.17} \end{align} $$

Hence we get

$$ \lim_{N \to \infty} \frac{I_{1,N}}{N^2} = -H_{\beta}(\mu). \quad (10.18) $$

Using (10.13), (10.16) and (10.18), for all $\delta > 0$,

$$ -H_{\beta}(\mu) \leq \lim_{N \to \infty} \frac{1}{N^2} \log \overline{Q_{\beta}^{N}}(\overline{\mu_N} \in B(\mu, \delta)). $$

Let $\delta$ goes to 0 to have the lower bound:

$$ -H_{\beta}(\mu) \leq \lim_{\delta \to 0} \lim_{N \to \infty} \frac{1}{N^2} \log \overline{Q_{\beta}^{N}}(\overline{\mu_N} \in B(\mu, \delta)). $$
Approximating the measures with a density with respect to the Lebesgue measure by "regular" measures defined in this part, one can prove that for all $O$ open subset of $\mathcal{P}(\mathbb{R})$, we have the lower bound

$$ \lim_{N \to \infty} \frac{1}{N^2} \log(\overline{Q_{\beta}^{N}}(\mu_{N} \in O)) \geq -\inf\{H_{\beta}(\mu), \mu \in O \text{ is absolutely continuous}\}. $$

This result is similar with the one obtain in Lemma 10.6 and can be found in [26] (see Corollary 2.7). Then, thanks to the properties of $H_{\beta}$, we can extend the previous result for open subsets to obtain the large deviations principle for open subsets (10.1). See the hypothesis (H4) and the end of the proof of Theorem 1 of [26] to have a complete proof of this point.

**Remark 10.8.** In view of Remark 10.3, let us mention how this computation can be used to obtain a part of the large deviations principle of the normalizing constant factor without knowing its exact value.

Since $Z_{\beta}^{N}$ is the normalizing factor, we have that for all $\mu \in \mathcal{P}(\mathbb{R})$

$$ \lim_{N \to \infty} \frac{1}{N^2} \log Z_{\beta}^{N} \geq \lim_{\delta \to 0} \lim_{N \to \infty} \frac{1}{N^2} \log \overline{Q_{\beta}^{N}}(\mu_{N} \in B(\mu, \delta)) \geq -H_{\beta}(\mu). $$

This gives that

$$ \lim_{N \to \infty} \frac{1}{N^2} \log Z_{\beta}^{N} \geq - \inf_{\mathcal{P}(\mathbb{R})} H_{\beta}(\mu). $$

## 10.3 Exponential tightness and complete large deviations

First, we shall see why the exponential tightness of a family of measures is important to go from large deviations on compact sets to large deviations on closed sets.

**Definition 10.9.** The family $(Q_{\beta}^{N})_{N \in \mathbb{N}}$ is said to be exponentially tight if for every $L$ there exists a compact subset $K_{L}$ of $\mathcal{P}(\mathbb{R})$ such that $\overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^{N}(\mu_{N} \notin K_{L}) \leq -L$.

**Proposition 10.10.** Assume $(Q_{\beta}^{N})_{N \in \mathbb{N}}$ is exponentially tight, then $(Q_{\beta}^{N})_{N \in \mathbb{N}}$ satisfies a large deviations principle for closed sets, that means: for any closed sets $F$ of $\mathcal{P}(\mathbb{R})$ we have:

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^{N}(\mu_{N} \in F) \leq - \inf_{F} I_{\beta}. $$

*Proof.* Let $F$ be a closed subset of $\mathcal{P}(\mathbb{R})$ and fix $L > 0$ and $K_{L}$ such that:

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^{N}(\mu_{N} \notin K_{L}) \leq -L. $$

Then we have that:

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^{N}(\mu_{N} \in F) \leq \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log(Q_{\beta}^{N}(\mu_{N} \in F \cap K_{L}) + Q_{\beta}^{N}(\mu_{N} \notin K_{L})). $$

Using Lemma 10.5 and the compactness of $F \cap K_{L}$, the large deviations principle for compact subsets gives:

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^{N}(\mu_{N} \in F) \leq \max\{-\inf_{F \cap K_{L}} I_{\beta}; -L\} \leq \max\{-\inf_{F} I_{\beta}; -L\}. $$
Now let $L$ tends to $+\infty$ to obtain that:

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^{N}(\overline{\mu}_{N} \in F) \leq - \inf_{F} I_{\beta}. \quad \square $$

Hence to prove the complete large deviations principle it remains to prove the exponential tightness property.

**Proposition 10.11.** The sequence $(Q_{\beta}^N)_{N \in \mathbb{N}}$ is exponentially tight.

*Proof.* Let $E \subset \mathcal{P}(\mathbb{R})$ be a measurable set, doing the same computation as in the upper bound proof to obtain 10.9 we have:

$$ \begin{aligned} Q_{\beta}^{N}(\overline{\mu}_{N} \in E) &= \frac{1}{Z_{\beta}^{N}} \int_{\overline{\mu}_{N} \in E} \exp \left( -N^2 \int_{x<y} f_{\beta}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) \right) \exp \left( -\frac{1}{2} \sum_{i=1}^{N} \lambda_{i}^{2} \right) \prod_{i=1}^{N} d\lambda_{i} \\ &= \frac{1}{Z_{\beta}^{N}} \int_{\overline{\mu}_{N} \in E} \exp \left( -\frac{N^2}{2} \int_{x \neq y} f_{\beta}(x, y) \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) \right) \exp \left( -\frac{1}{2} \sum_{i=1}^{N} \lambda_{i}^{2} \right) \prod_{i=1}^{N} d\lambda_{i}. \end{aligned} \quad (10.19) $$

As a consequence of the large deviations principle for the partition function obtained in Proposition 10.1, we can find a constant $C \in \mathbb{R}$ such that for all $N \geq 1$,

$$ \log(Z_{\beta}^{N}) \geq -CN^{2}. \quad (10.20) $$

Moreover, as for the inequality 9.4, we can find a constant $\tilde{m}_{\beta} > -\infty$ such that for all $(x, y) \in \mathbb{R}^2$,

$$ f_{\beta}(x, y) \geq \frac{x^2}{4} + \frac{y^2}{4} + \tilde{m}_{\beta}. \quad (10.21) $$

For all $L > 0$, let $K_L := \{\mu \in \mathcal{P}(\mathbb{R}) | \int_{\mathbb{R}} x^2 \mu(dx) \leq L\}$. For all $L > 0$, $K_M$ is closed as a consequence of the Portmanteau theorem and is relatively compact. Indeed, the Markov inequality implies that for all $\mu \in K_L$, for all $\varepsilon > 0$,

$$ \mu(\{|x| \geq \varepsilon\}) \leq \frac{\int_{\mathbb{R}} x^2 \mu(dx)}{\varepsilon^2} \leq \frac{L}{\varepsilon^2}, $$

giving the relatively compactness of $K_L$ by the Prokhorov theorem. Using (10.21) and (10.20) in (10.19) yields

$$ \begin{aligned} Q_{\beta}^{N}(\overline{\mu}_{N} \notin K_{L}) &\leq e^{CN^{2} - \frac{\tilde{m}_{\beta}N(N-1)}{2}} \int_{\overline{\mu}_{N} \notin K_{L}} e^{-\frac{N^{2}}{4}} \int_{x \neq y} x^{2} \overline{\mu}_{N}(dx) \overline{\mu}_{N}(dy) e^{-\frac{1}{2} \sum_{i=1}^{N} \lambda_{i}^{2}} \prod_{i=1}^{N} d\lambda_{i} \\ &= e^{CN^{2} - \frac{\tilde{m}_{\beta}N(N-1)}{2}} \int_{\overline{\mu}_{N} \notin K_{L}} e^{-\frac{N(N-1)}{4}} \int_{\mathbb{R}} x^{2} \overline{\mu}_{N}(dx) e^{-\frac{1}{2} \sum_{i=1}^{N} \lambda_{i}^{2}} \prod_{i=1}^{N} d\lambda_{i} \\ &\leq e^{CN^{2} - \frac{\tilde{m}_{\beta}N(N-1)}{2} - \frac{N(N-1)L}{4}} \left( \int_{\mathbb{R}} e^{-\frac{x^{2}}{2}} dx \right)^{N} \end{aligned} \quad (10.22) $$

We deduce that for all $L \in \mathbb{R}$,

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_{\beta}^{N}(\overline{\mu}_{N} \notin K_{L}) \leq C - \frac{\tilde{m}_{\beta}}{2} - \frac{L}{4}. \quad (10.23) $$

Since $L$ is arbitrary, $(Q_{\beta}^N)_{N \in \mathbb{N}}$ is exponentially tight. $\square$
**Remark 10.12.** If we look at the previous computations, we proved in particular an exponential tightness for the family $Q_{\beta}^N$ which means that for all $L > 0$ we can find a compact $K_L$ of $\mathcal{P}(\mathbb{R})$ such that

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \overline{Q}_{\beta}^{N}(\overline{\mu_N} \notin K_L) \leq -L. \qquad (10.24) $$

Combining this property with the weak large deviations principle on compact subsets obtained as a consequence of (10.8), we have:

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Z_{\beta}^{N} = \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \overline{Q}_{\beta}^{N}(\mathcal{P}(\mathbb{R})) \leq - \inf_{\mu \in \mathcal{P}(\mathbb{R})} H_{\beta}(\mu). \qquad (10.25) $$

Indeed, for $L > 0$ fix a compact $K_L$ as in (10.24). Then using Lemma 10.5 we get:

$$ \begin{aligned} \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Z_{\beta}^N &= \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \overline{Q}_{\beta}^N(\mathcal{P}(\mathbb{R})) \\ &= \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log [\overline{Q}_{\beta}^N(K_L) + \overline{Q}_{\beta}^N((K_L)^c)] \\ &\leq \max \left( \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \overline{Q}_{\beta}^N(K_L), \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \overline{Q}_{\beta}^N((K_L)^c) \right) \\ &\leq \max(-\inf_{K_L} H_{\beta}, -L) \\ &\leq \max(-\inf_{\mathcal{P}(\mathbb{R})} H_{\beta}, -L). \end{aligned} $$

Let $L$ goes to $+\infty$ to get (10.25).

Using this bound with Remark 10.8 this gives the large deviations principle of the normalizing factor:

$$ \lim_{N \to \infty} \frac{1}{N^2} \log Z_{\beta}^N = - \inf_{\mathcal{P}(\mathbb{R})} H_{\beta}. $$

This gives a more robust approach for the large deviations principle of Section 10.1 which was mentioned in Remark 10.3.

## 10.4 Consequences of the large deviations principle

**Proposition 10.13.** For all $\beta > 0$,

$$ \inf_{\mu \in \mathcal{P}(\mathbb{R})} H_{\beta}(\mu) = -F(\beta), \quad I_{\beta} = \widehat{I}_{\beta}. $$

*Proof.* This is an immediate consequence of Proposition 8.16. Indeed, it gives that $\inf_{\mathcal{P}(\mathbb{R})} \widehat{I}_{\beta} = 0$ which implies the result. $\square$

As a consequence of Section 10.2 and Section 10.3 we can state the following result.

**Theorem 10.14.** For all $\beta > 0$, $(Q_{\beta}^N)_{N \in \mathbb{N}}$ satisfies a LDP at speed $N^2$ with good rate function $I_{\beta}$.
## 10.5 Computation of the equilibrium measure

We proved that for all $I_β$ reaches its infimum in a unique $μ_β ∈ P(\mathbb{R})$. The goal of this section is to compute it and accordingly with Wigner theorem, to prove that this measure is the semicircle distribution. First let state a lemma to justify that we can reduce to the case $β = 1$.

**Lemma 10.15.** $μ_β$ minimizes $I_β$ if and only if the measure $\tilde{μ}_β(A) := μ_β(\sqrt{β}A)$ minimizes $I_1$.

*Proof.* Thanks to the Proposition 9.17 we have that $μ_β$ minimizes $I_β$ if and only if for $\tilde{μ}_β$ almost all $x$:

$$ \beta \int_{\mathbb{R}} \log |\sqrt{\beta}x - \sqrt{\beta}y| d\mu_{\beta}(y) = \frac{\beta}{2} x^2 - 2 \inf H_{\beta} + \frac{\beta}{2} \int_{\mathbb{R}} y^2 d\mu_{\beta}(y). $$

Since $\inf H_β = -F(β)$ (by Proposition 10.13), and $-2F(β) + \log \sqrt{β} = -2βF(1)$, we have $\frac{2}{β}\inf H_β + \log \sqrt{β} = 2\inf H_1$.

Hence $μ_β$ minimizes $I_β$ if and only if the measure $\tilde{μ}_β$ minimizes $I_1$.

□

Thanks to Proposition 9.17 it remains to prove that for any $x \in [-\sqrt{2}, +\sqrt{2}]$

$$ \int_{\mathbb{R}} \log |x - y| d\sigma_1(y) = \frac{1}{2} x^2 - 2 \inf H_1 + \frac{1}{2} \int_{\mathbb{R}} y^2 d\sigma_1(y). $$

It is the object of the next proposition.

**Proposition 10.16.** For all $x \in [-\sqrt{2}, +\sqrt{2}]$,

$$ \int_{\mathbb{R}} \log |x - y| d\sigma_1(y) = \frac{x^2 - \log 2 - 1}{2}. $$

*Proof.* Notice that if we view $x \mapsto \int_{\mathbb{R}} \log |x - y| d\sigma_1(y)$ as a distribution, we can write $\int_{\mathbb{R}} \log |x - y| d\sigma_1(y) = \log *\sigma_1(x)$ with log the distribution associated to the function $x \mapsto \log |x|$ and $\sigma_1$ the distribution $f \mapsto \int_{\mathbb{R}} f(x) d\sigma_1(x)$.

It is well known that the distribution $P.V.(1/x)(\phi) = \lim_{\epsilon \to 0} \int_{|x|>\epsilon} \frac{\phi(x)}{x} dx$ satisfies $P.V.(1/x) = \log'$ in the sense of distributions.

By the properties of the convolution of distribution we also have that $P.V.(1/x) * \sigma_1 = (\log * \sigma_1)'$.

We first compute $P.V.(1/x) * \sigma_1(x) = \lim_{\epsilon \to 0} \int_{|x-y|>\epsilon} \frac{1}{x-y} \sigma_1(dx)$ for $x \in [-\sqrt{2}, \sqrt{2}]$. This quantity is called the Hilbert transform of the measure $\sigma_1$ and was introduced in Definition 7.23.

We can compute the Hilbert transform of $\sigma_1$ thanks to the Stieltjes transform of $\sigma_1$ which is well known. We recall that for all $z \in \mathbb{H}$, we have

$$ S_{\sigma_1}(z) = \int_{\mathbb{R}} \frac{1}{z-y} \sigma_1(dy) = z - \sqrt{z^2 - 2}, $$
with $\sqrt{z'}$ the determination of the square root on $\mathbb{C} - \mathbb{R}^+$ such that $\sqrt{-1} = i$ (see Proposition A.20).

Indeed we have:

$$ \lim_{\epsilon \to 0} \int_{|x-y| > \epsilon} \frac{1}{x-y} \sigma_1(dx) = \lim_{\epsilon \to 0} \frac{S_{\sigma_1}(x+i\epsilon) + S_{\sigma_1}(x-i\epsilon)}{2}. $$

$$ \text{Hence, for all } x \in [-\sqrt{2}, \sqrt{2}], \text{ P.V.}(1/x) * \sigma_1(x) = x. $$

So, we can find a constant $C \in \mathbb{R}$ such that for all $x \in [-\sqrt{2}, \sqrt{2}]$:

$$ \int_{\mathbb{R}} \log |x - y| d\sigma_1(y) = \frac{x^2}{2} + C. \qquad (10.26) $$

It remains to compute the constant $C$.

We write the equality (10.26) at $x = 0$, $x = \sqrt{2}$ and $x = -\sqrt{2}$. We have the three following equalities:

$$ \left\{ \begin{aligned} C &= \int_{\mathbb{R}} \log |y| d\sigma_1(y) \\ C &= -1 + \int_{\mathbb{R}} \log |\sqrt{2} - y| d\sigma_1(y) \\ C &= -1 + \int_{\mathbb{R}} \log | -\sqrt{2} - y| d\sigma_1(y) \end{aligned} \right. \qquad (10.27) $$

We make a polar change of variable $y = \sqrt{2} \sin(\theta)$. It gives:

$$ \left\{ \begin{aligned} C &= \log \sqrt{2} + \frac{4}{\pi} \int_{0}^{\pi/2} \log(\sin(\theta)) \cos^2(\theta) d\theta \\ C &= \log \sqrt{2} - 1 + \frac{4}{\pi} \int_{0}^{\pi/2} \log(1 - \sin(\theta)) \cos^2(\theta) d\theta \\ C &= \log \sqrt{2} - 1 + \frac{4}{\pi} \int_{0}^{\pi/2} \log(1 + \sin(\theta)) \cos^2(\theta) d\theta \end{aligned} \right. \qquad (10.28) $$

Now if we sum the two last equations of (10.28), we get:

$$ C = \log \sqrt{2} - 1 + \frac{2}{\pi} \int_{0}^{\pi/2} \log(1 - \sin^2(\theta)) \cos^2(\theta) d\theta = \log \sqrt{2} - 1 + \frac{4}{\pi} \int_{0}^{\pi/2} \log(\cos(\theta)) \cos^2(\theta) d\theta. $$

Doing the change of variable $\theta' = \pi/2 - \theta$ and summing this equality with the first one of (10.28) yields:

$$ C = \frac{\log(2) - 1}{2} + \frac{2}{\pi} \int_{0}^{\pi/2} \log(\sin(\theta)) d\theta. $$

It remains to compute $\int_{0}^{\pi/2} \log(\sin(\theta)) d\theta$.

**Lemma 10.17.**

$$ \int_{0}^{\pi/2} \log(\sin(\theta)) d\theta = -\pi \frac{\log(2)}{2}. $$

*Proof.* We write $I = \int_{0}^{\pi/2} \log(\sin(\theta)) d\theta$. Doing the change of variable $\theta' = \pi/2 - \theta$ gives $I = \int_{0}^{\pi/2} \log(\cos(\theta)) d\theta$. Summing these two equalities yields

$$ 2I = \int_{0}^{\pi/2} \log(\sin(\theta) \cos(\theta)) d\theta. $$
Since

$$ \int_{0}^{\pi/2} \log(\sin(\theta) \cos(\theta)) d\theta = \int_{0}^{\pi/2} \log(\sin(2\theta)/2) d\theta = \int_{0}^{\pi/2} \log(\sin(2\theta)) d\theta - \log(2) \frac{\pi}{2} $$

and

$$ \int_{0}^{\pi/2} \log(\sin(2\theta)) d\theta = \int_{0}^{\pi} \log(\sin(\theta)) \frac{d\theta}{2} = \int_{0}^{\pi/2} \log(\sin(\theta)) d\theta = I, $$

we have the following relation:

$$ 2I = I - \pi \frac{\log(2)}{2}. $$

It yields $I = -\pi \frac{\log(2)}{2}$. $\square$

As a consequence of (10.26) and Lemma 10.17, for all $x \in [-\sqrt{2}, +\sqrt{2}]$,

$$ \int_{\mathbb{R}} \log |x - y| d\sigma_1(y) = \frac{x^2 - \log 2 - 1}{2}. $$

Thanks to Lemma 10.15 and 10.16 obtain the main theorem of this section.

**Theorem 10.18.** The unique minimizer of $I_β$ is $\sigma_β$.

As a consequence of this theorem we get another proof of the Wigner theorem.

**Corollary 10.19 (Wigner's theorem).** For all $N \ge 1$, let $X_N$ be a $N$ complex (resp. real) Wigner matrix of size $N$. Let $(\lambda_i^N)_{1 \le i \le N}$ be the $N$ real eigenvalues of $X_N$ and $\mu_N$ be the spectral measure of $X_N/\sqrt{N}$:

$$ \mu_N = \frac{1}{N} \sum_{i=1}^{N} \delta_{\frac{\lambda_i^N}{\sqrt{N}}}. $$

Then, almost surely

$$ \mu_N \xrightarrow[N \to +\infty]{} \sigma_2 \text{ (resp. } \sigma_1). $$

*Proof.* This is an immediate consequence of the LDP of $(Q_β^N)_{N \ge 1}$ for $\beta = 1$ or $\beta = 2$ and Proposition 8.17. $\square$

# 11 Applications and further results

## 11.1 Convergence of the smallest and the largest eigenvalue

In this section we focus on the smallest and the largest eigenvalue of real Wigner matrices but we could have stated all the results of this part for complex Wigner matrices.

**Definition 11.1.** Let $M \in M_n(\mathbb{R})$ be a symmetric matrix. We denote $\lambda_1^n(M) \ge \dots \ge \lambda_n^n(M)$ its eigenvalues.
Let us explicit the Kantorovich-Rubinstein distance which metrizes the weak topology of $\mathcal{P}(\mathbb{R})$. We will denote $d_{KR}$ this distance between two elements of $\mathcal{P}(\mathbb{R})$. We recall that for all $\mu, \nu$ in $\mathcal{P}(\mathbb{R})$,

$$d_{KR}(\mu, \nu) = \sup_{f \in BL(\mathbb{R}); ||f||_{BL} \le 1} \left\{ \left| \int_{\mathbb{R}} f d\mu - \int_{\mathbb{R}} f d\nu \right| \right\},$$

with $BL(\mathbb{R})$ the space of bounded and Lipschitz functions endows with the natural norm: $||f||_{BL} = ||f||_{\infty} + ||f||_{lip}$.

**Proposition 11.2.** Let $\lambda_1^n \ge \dots \ge \lambda_n^n$ be a family of $n$ real numbers. Assume that there exists $\varepsilon > 0$ such that $\sqrt{2} - \varepsilon > \lambda_1^n$. Then we can find $c_\varepsilon > 0$ such that:

$$d_{KR}(\bar{\mu}_n, \sigma_1) \ge c_\varepsilon.$$

*Proof.* Let $f_\varepsilon$ the continuous function which is null on $] -\infty, \sqrt{2} - \varepsilon]$, constant to $1/2$ on $[\sqrt{2} - \varepsilon + 1, +\infty[$ and linear on $[\sqrt{2} - \varepsilon, \sqrt{2} - \varepsilon + 1]$. We clearly have that $f_\varepsilon$ is in $BL(\mathbb{R})$ and that $||f||_{BL} \le 1$.

Hence, by definition of the Kantorovich-Rubinstein distance, we have:

$$d_{KR}(\bar{\mu}_n, \sigma_1) \ge \left| \int_{\mathbb{R}} f_\varepsilon d\bar{\mu}_n - \int_{\mathbb{R}} f_\varepsilon d\sigma_1(y) \right|.$$

Since $f_\varepsilon$ is supported on $[\sqrt{2} - \varepsilon, +\infty[$, by hypothesis on the $\lambda_i^n$ we have that $\int_{\mathbb{R}} f_\varepsilon d\bar{\mu}_n = 0$.

Since $f_\varepsilon$ is not null on $[-\sqrt{2}, +\sqrt{2}]$ we have: $\int_{\mathbb{R}} f_\varepsilon d\sigma_1(y) > 0$. Let $c_\varepsilon := \int_{\mathbb{R}} f_\varepsilon d\sigma_1(y) > 0$, we get:

$$d_{KR}(\bar{\mu}_n, \sigma_1) \ge c_\varepsilon.$$

□

**Proposition 11.3.** Let $\varepsilon > 0$ and $X_N$ be a $N$ real Wigner matrix for all $N \ge 1$. Then, we can find $C_\varepsilon > 0$ such that

$$\overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \mathbb{P} \left( \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \varepsilon \right) \le -C_\varepsilon.$$

*Proof.* Using Proposition 11.2 and the fact that the law of the eigenvalues of $X_N/\sqrt{N}$ is $Q_1^N$ we have:

$$\mathbb{P} \left( \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \varepsilon \right) \le \mathbb{P} \left( d_{KR} \left( \frac{1}{N} \sum_{i=1}^N \delta_{\lambda_i^N \left( \frac{X_N}{\sqrt{N}} \right)}, \sigma_1 \right) \ge c_\varepsilon \right) \quad (11.1) \\ = Q_1^N (\{ \mu \in \mathcal{P}(\mathbb{R}) \mid d_{KR}(\mu, \sigma_1) \ge c_\varepsilon \}).$$

By the large deviations theorem obtained in Section 10, we get:

$$\overline{\lim}_{N \to \infty} \frac{1}{N^2} \log Q_1^N (\{ \mu \in \mathcal{P}(\mathbb{R}) \mid d_{KR}(\mu, \sigma_1) \ge c_\varepsilon \}) \le -\inf_{F_\varepsilon} I_1,$$

with $F_\varepsilon = \{ \mu \in \mathcal{P}(\mathbb{R}) : d_{KR}(\mu, \sigma_1) \ge c_\varepsilon \}$.

Since $I_1$ is lsc and $c_\varepsilon > 0$, $C_\varepsilon := \inf_{F_\varepsilon} I_1 > 0$ by Lemma 8.14. Hence, we have

$$\overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \mathbb{P} \left( \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \varepsilon \right) \le -C_\varepsilon.$$

□
**Remark 11.4.** We clearly also have the same result with $\lambda_N^N$ and $-\sqrt{2}$ instead of $\lambda_1^N$ and $\sqrt{2}$.

Let us mention two applications. The first one can also be viewed as a consequence of the Wigner theorem. The second one can be viewed as a first step in order to obtain a large deviations principle for the largest eigenvalue of a $N$ real Wigner matrix.

**Corollary 11.5.** For all $N \ge 1$ let $X_N$ be a $N$ real Wigner matrix, we have $\mathbb{P}$ almost surely

$$ \liminf_{N \to \infty} \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) \ge \sqrt{2}. $$

*Proof.* We compute $\mathbb{P} \left( \liminf_{N \to \infty} \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} \right)$. By definition of the $\liminf$ we have:

$$ \mathbb{P} \left( \liminf_{N \to \infty} \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} \right) = \lim_{\epsilon \to 0} \mathbb{P} \left( \text{there exists an infinite number of } N \text{ such that: } \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \epsilon \right). \quad (11.2) $$

Let $\epsilon > 0$, by Proposition 11.3, there exists $N(\epsilon) \in \mathbb{N}$ and $C_\epsilon > 0$ such that

$$ \forall N \ge N(\epsilon), \mathbb{P} \left( \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \epsilon \right) \le \exp \left( -N^2 \frac{C_\epsilon}{2} \right), $$

which gives that $\sum_N \mathbb{P} \left( \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \epsilon \right) < +\infty$.

Hence the Borel-Cantelli lemma yields that for all $\epsilon > 0$:

$$ \mathbb{P} \left( \text{there exists an infinite number of } N \text{ such that: } \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \epsilon \right) = 0. $$

It gives:

$$ \mathbb{P} \left( \liminf_{N \to \infty} \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} \right) = 0. $$

□

**Corollary 11.6.** For all $N \ge 1$, let $X_N$ be a $N$ real Wigner matrix, then for all $\epsilon > 0$ we have

$$ \overline{\lim}_{N \to \infty} \frac{1}{N} \log \mathbb{P} \left( \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \epsilon \right) = -\infty. $$

*Proof.* By Proposition 11.3 we have $C_\epsilon > 0$ such that

$$ \overline{\lim}_{N \to \infty} \frac{1}{N^2} \log \mathbb{P} \left( \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \epsilon \right) \le -C_\epsilon. $$

So there exists $N(\epsilon)$ such that for all $N \ge N(\epsilon)$

$$ \frac{1}{N^2} \log \mathbb{P} \left( \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \epsilon \right) \le -\frac{C_\epsilon}{2}. $$
This gives that for all $N \ge N(\varepsilon)$

$$ \frac{1}{N} \log \mathbb{P} \left( \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) < \sqrt{2} - \varepsilon \right) \le -\frac{C_\varepsilon}{2} N. $$

Passing to the limit $N \to +\infty$ gives the result. $\square$

Let us mention the complete result for the large deviations of the largest eigenvalue of a real Wigner matrices. We refer to the section 2.6.2 of [5] for a proof of this theorem.

**Theorem 11.7.** For all $N \ge 1$ let $X_N$ be a $N$ real Wigner matrix. Then $(\lambda_1^N(\frac{X_N}{\sqrt{N}}))_{N \ge 1}$ satisfies a LDP at speed $N$ with good rate function

$$ \mathcal{I}(x) = \begin{cases} +\infty & \text{if } x < \sqrt{2} \\ \frac{x^2}{2} - \int_{\mathbb{R}} \log|x-y| d\sigma_1(dy) - \frac{\log(2)+1}{2} & \text{if } x \ge \sqrt{2}. \end{cases} $$

The function $\mathcal{I}$ is a good rate function since it is increasing on $[\sqrt{2}, +\infty)$, strictly convex on $[\sqrt{2}, +\infty)$, continuous on $[\sqrt{2}, +\infty)$ (by the dominated convergence theorem on $] \sqrt{2}, +\infty)$ and by the monotone convergence for the right continuity in $\sqrt{2}$) and goes to $+\infty$ when $x$ goes to $+\infty$.

Moreover, $I(x) \ge 0$ by the Frostman theorem 9.16 with an equality if and only if $x = \sqrt{2}$. Hence $I$ has a unique minimizer which is $\sqrt{2}$. Using Proposition 8.17 the previous theorem implies the following result.

**Theorem 11.8.** For all $N \ge 1$ let $X_N$ be a $N$ real Wigner matrix. Then

$$ \lambda_1^N \left( \frac{X_N}{\sqrt{N}} \right) \xrightarrow[N \to +\infty]{\text{Almost surely}} \sqrt{2}. $$

## 11.2 Extensions of the result

### 11.2.1 Coulomb and Riesz gases

Let $N \ge 1$ and let $X_N$ be the $N$-tuple of points $(x_1, ..., x_N) \in (\mathbb{R}^d)^N$, where $d$ is the dimension. We define the following energies $\mathcal{H}_N : (\mathbb{R}^d)^N \to (-\infty, +\infty)$ by:

$$ \mathcal{H}_N(X_N) := \frac{1}{2} \sum_{1 \le i \ne j \le N} g(x_i - x_j) + N \sum_{i=1}^N V(x_i), \quad (11.3) $$

where $g : \mathbb{R}^d \to (-\infty, +\infty]$ is called the pair interaction potential and $V : \mathbb{R}^d \to (-\infty, +\infty]$ is called the external field.

We focus on the cases where $g$ is given by

$$ g_{d,s}(x) = \begin{cases} \frac{1}{s}|x|^{-s} & s \ne 0 \\ -\log|x| & s = 0, \end{cases} \quad (11.4) $$

The first case is called the Riesz case and the second one the logarithmic case.

The case $s = 0$ is obtained as the formal limit when $s \to 0$ and so we will name it as a Riesz case also.

The particular case $s = d - 2$ corresponds to the Coulomb case. Indeed, in this case, $g$ is the Coulomb kernel (up to a multiplicative constant); the fundamental solution to The Laplace-Poisson equation

$$ -\Delta g_{d,d-2} = c_d \delta_0, \quad (11.5) $$
where $\delta_0$ is the Dirac mass at the point 0 and $c_d$ is explicitly given by

$$c_d = \begin{cases} 2\pi & d=2 \\ |\mathbb{S}^{d-1}| & d \ge 3 \end{cases} \qquad (11.6)$$

with $\mathbb{S}^{d-1}$ the unit sphere of $\mathbb{R}^d$. For instance in dimension 3, with $s = 1$ $g$ corresponds to the gravitational potential.

More generally, the cases $d-2 \le s < d$ are much more well known. Indeed for $s \ge d$, the potential $g$ is not integrable near 0 what requires a different study. Moreover, as for the case $s = d-2$, in the cases $d-2 \le s < d$, $g_{d,s}$ is the fundamental solution of an operator (up to a multiplicative constant) called the fractional Laplacian:

$$(-\Delta)^{\frac{d-s}{2}} g_{d,s} = c_{d,s} \delta_0, \qquad (11.7)$$

where $c_{d,s}$ is explicitly given by

$$c_{d,s} = \begin{cases} 2\pi & s=0, d=1, d=2 \\ |\mathbb{S}^{d-1}| & s=d-2 > 0 \\ \frac{2^{d-s}\pi^{d/2}\Gamma\left(\frac{d-s}{2}\right)}{\Gamma\left(\frac{s}{2}\right)} & s > \max(0, d-2) \end{cases} \qquad (11.8)$$

The main difference between the fractional Laplacian and the Laplacian is the fact that the fractional Laplacian is a non local operator. Many equivalent definitions exist but we can for instance define this operator using Fourier representation by

$$\mathcal{F}((-\Delta)^{\alpha} f)(\xi) = |\xi|^{2\alpha} \mathcal{F}(f), \alpha \in (0,1),$$

or by an integral representation

$$(-\Delta)^{\alpha} f(x) = C_{d,\alpha} \int_{\mathbb{R}^d} \frac{f(x) - f(y)}{|x-y|^{d+2\alpha}} dy, \alpha \in (0,1),$$

with $C_{d,\alpha} > 0$ an explicit constant. From the Fourier definition of the fractional Laplacian, we deduce that the Fourier transform of $g_{d,s}$ is proportional to $\xi \mapsto |\xi|^{s-d}$ making link with Remark 9.10.

We can associate to the energy $H_N$ the Gibbs measure which is a probability measure on $(\mathbb{R}^d)^N$ whose density with respect to the Lebesgue measure on $(\mathbb{R}^d)^N$, $dX_N := dx_1...dx_N$ is

$$d\mathbb{P}_{N,\beta}(X_N) := \frac{1}{Z_{N,\beta}} \exp(-\beta N^{-s/d} H_N(X_N)) dX_N, \qquad (11.9)$$

where $\beta$ physically corresponds to the inverse of the temperature and $Z_{N,\beta}$ is a normalizing constant.

The particular case $s = 0$ and $d = 1$ corresponds to the case of $\beta$ Hermite ensemble studied in this part. Thanks to a similar approach, one can prove a large deviations principle for the sequence of probability measures $(P_{N,\beta})_{N \in \mathbb{N}}$ extended on $\mathcal{P}(\mathbb{R})$ as in the Definition 8.9 at speed $\beta N^{2-\frac{s}{d}}$. See chapter 2 and 3 of [74] for the complete and rigorous derivation of this large deviations principle.
Let us mention the particular Coulomb case $d = 2, s = 0$ with the potential $V(z) = |z|^2/2$ with $\beta = 2$. In this case,

$$d\mathbb{P}_{N,2}(z_1, \dots, z_N) = \frac{1}{Z_{N,2}} \exp \left( 2 \sum_{i<j} \log |z_i - z_j| - N \sum_{1 \le i \le N} |z_i|^2 \right) dz_1 \dots dz_N \\ = \frac{1}{Z_{N,2}} \prod_{i<j} |z_i - z_j|^2 \exp \left( -N \sum_{1 \le i \le N} |z_i|^2 \right) dz_1 \dots dz_N.$$

We recognize the density of the eigenvalues of $M_N/\sqrt{N}$ where $M_N$ is a $N$ complex Ginibre matrix obtained in Theorem 2.31. As explained $\mathbb{P}_{N,2}$ satisfies a large deviations principle at speed $2N^2$. The counterpart of $H_\beta$ that was studied in this section is

$$H_{\mathbb{C}}(\mu) := - \int_{\mathbb{C}^2} \log |z - z'| \mu(dz) \mu(dz') + \frac{1}{2} \int_{\mathbb{C}} |z|^2 \mu(dz).$$

By similar arguments as for the $\beta$ ensembles case studied in this section, one can prove that the good rate function for the large deviations of $\mathbb{P}_{N,2}$ is $I_{\mathbb{C}} := H_{\mathbb{C}} - \inf_{\mu \in \mathcal{P}(\mathbb{C})} H_{\mathbb{C}}(\mu)$. We can show that the unique minimizer of $I_{\mathbb{C}}$ is the uniform distribution on the circle. Indeed, the exact same proof as we did for the Frostman theorem 9.16 implies that the unique minimizer of $I_{\mathbb{C}}$ satisfies the counterpart of (9.9) and (9.10). Let $\mu_{\mathbb{C}}$ be the unique minimizer of $I_{\mathbb{C}}$ and $S_{\mu_{\mathbb{C}}}$ its support. We obtained that $z \mapsto -\int_{\mathbb{C}} \log |z - z'| \mu_{\mathbb{C}}(dz') + \frac{|z|^2}{2}$ is constant on $S_{\mu_{\mathbb{C}}}$. Since $\Delta(\log(\cdot)) = 2\pi\delta_0(dz)$, computing the Laplacian yields that

$$\mu_{\mathbb{C}}(dz) = \frac{\Delta|z|^2}{4\pi} \mathbb{1}_{S_{\mu_{\mathbb{C}}}}(z)dz = \frac{1}{\pi} \mathbb{1}_{S_{\mu_{\mathbb{C}}}}(z)dz. \qquad (11.10)$$

By uniqueness of the minimizer, one can show that $\mu_{\mathbb{C}}$ is invariant by rotation, so its support is a ball of radius $r > 0$. Using (11.10), the unique possibility such that $\mu_{\mathbb{C}}$ is a probability measure is $r = 1$. So $\mu_{\mathbb{C}}$ is the uniform measure of the disk, giving another proof of the circular law proved in Section 3.

For people interested in this topic we recommend the paper of Chafaï & al. [26] for a very rigorous derivation of a large deviation principles for particles in singular repulsion and a study of the equilibrium measure for radial external potential and the review of Serfaty [74] for a more recent overview on this topic.

### 11.2.2 Extension to other random matrices models and large deviations for the largest and smallest eigenvalue

After the pionner paper of A.Guionnet and G.Ben Arous [6], the large deviations principle of the empirical mean has been studied for many other classical models of random matrices. For instance a large deviations principle has been obtained for the circular model [7] and for the unitary one [51].

More generally a goal would be to obtain a large deviations principle for random matrices even if we do not know the density of the distribution of eigenvalues of the model (or even if this density does not exist). An interesting example would be random matrices with independent and identically distributed entries that have Bernoulli or Rademacher distributions. These questions are linked with random graph theory since we can in the Bernoulli case interpret the random matrix as the adjacent matrix of an Erdős–Rényi random graph [8]. Recently many works for the Bernoulli and Rademacher cases have
been been done to study the large deviations of the largest and smallest eigenvalue of such
matrices. Indeed, as explained in Section 11.1, the behaviour of the largest eigenvalues
seems to also obey to a large deviations principle. Indeed for random matrices whose law
of eigenvalues are given by $Q_{\beta}^{N}$, the largest eigenvalue satisfies a large deviations principle
at speed $N$ with an explicit good rate function. See the section 2.6.2 of [5] for a complete
proof. For the study of Rademacher and Bernoulli cases, let us mention the recent works
of F.Augeri, A.Guionnet, J.Husson [47, 9].
# Appendix

## A The Wigner semicircle distribution

The goal of this section is to introduce some standard tools that are used in random matrix theory/probability and to present the ideas of how they are used to prove the Wigner theorem. The goal is not to make the proofs of the Wigner theorem but to present these tools for readers that are non familiar with random matrices. For more detailed introductions to random matrices we refer to [5, 75].

### A.1 Moments

**Definition A.1.** The Wigner semicircle distribution is the probability distribution on $\mathbb{R}$ with density function

$$ \sigma(x) := 1_{[-2,2]}(x) \frac{\sqrt{4-x^2}}{2\pi}. $$

We shall write $\sigma$ for the probability measure and its density. In this section, for all $k \in \mathbb{N}$ let $m_k := \int_{\mathbb{R}} x^k d\sigma(x)$ its moment of order $k$.

**Definition A.2.** We say that a probability measure $\mu$ is characterized by its moments if for all probability measures $\nu$ such that for all $k \in \mathbb{N}$, $\int_{\mathbb{R}} x^k d\mu(x) = \int_{\mathbb{R}} x^k d\nu(x)$ we have $\mu = \nu$.

We say that a real random variable $X$ is characterized by its moments if its law is a probability measure characterized by its moments.

Let us recall that compact supported measures are determined by their moments.

**Proposition A.3.** A probability measure $\mu$ on $\mathbb{R}$ with a compact support is characterized by its moments.

*Proof.* A real probability measure $\mu$ is characterized by its characteristic function

$$ \phi_{\mu}(t) = \int_{\mathbb{R}} \exp(itx) d\mu(x). $$

Fix $R > 0$ such that $\text{supp}(\mu) \subset [-R, R]$. Then its moments $\mu_k := \int_{-R}^{R} x^k d\mu(x)$ satisfy that

$$ |\mu_k| \le R^k. \qquad (A.1) $$

Hence, using the Taylor expansion of the exponential function: $\exp(itx) = \sum_{k=0}^{\infty} \frac{(itx)^k}{k!}$ and changing the integral and the sum (which is justified by the bound (A.1)), we get that

$$ \forall t \in \mathbb{R}, \phi_{\mu}(t) = \sum_{k=0}^{+\infty} \frac{\mu_k}{k!} i^k t^k. $$

So if a probability measure has the same moments as $\mu$, it necessarily has the same characteristic function and so it is equal to $\mu$. $\square$

We first recall some basic definitions about the convergence of probability measures.
**Definition A.4.** Let $\mathcal{X}$ be a separable Banach space. We endow the set of probability measures on $\mathcal{X}$; $\mathcal{P}(\mathcal{X})$; with the weak topology which means that a sequence of probability measures $(\mu_n) \in \mathcal{P}(\mathcal{X})^\mathbb{N}$ converges weakly to $\mu \in \mathcal{P}(\mathcal{X})$ if and only if for all $f: \mathcal{X} \to \mathbb{R}$ continuous and bounded, we have $\int_{\mathcal{X}} f d\mu_n \xrightarrow{n \to +\infty} \int_{\mathcal{X}} f d\mu$.

When we deal with the law of random variables we shall identify a random variable with its law. For random variables, we said that a sequence of random variables $(X_n)_{n \in \mathbb{N}}$ with values in $\mathcal{X}$ converges in law towards $X$ if $(\mu_n)_{n \in \mathbb{N}}$ the sequence of law of $(X_n)_{n \in \mathbb{N}}$ converges weakly towards the law of $X$.

If $(\mu_n)_{n \in \mathbb{N}}$ (resp. $(X_n)_{n \in \mathbb{N}}$) converges weakly (resp. in law) to $\mu$ (resp. $X_n \xrightarrow{n \to +\infty} X$).

Now, let us recall a result about convergence in distribution of random variables that is called method of moments which is based on Prokhorov's theorem. We first recall the definition of tightness.

**Definition A.5.** Let $\mathcal{X}$ be a separable Banach space. A sequence of measure $(\mu_n)_{n \in \mathbb{N}} \in \mathcal{P}(\mathcal{X})^\mathbb{N}$ is said to be tight if for all $\varepsilon > 0$, there exists a compact $K_\varepsilon \subset \mathcal{X}$ such that for all $n \in \mathbb{N}$ $\mu_n(K_\varepsilon) > 1 - \varepsilon$.

**Example A.6.** A sequence of real random variables $(X_n)_{n \in \mathbb{N}}$ such that there exists $k \in \mathbb{N}_{\ge 1}$ and $M \ge 0$ such that for all $n \ge 0$, $\mathbb{E}(|X_n|^k) \le M$ is tight by the Markov inequality.

We can now state the main theorem about the convergence in law of random variable (equivalently we could have stated it for the weak convergence of probability measures). We refer to the Chapter 1 of [16] for a proof of this standard result.

**Theorem A.7 (Prokhorov).** Let $(X_n)_{n \in \mathbb{N}}$ and $X$ be random variables with values in separable Banach metric space.

Then $(X_n)_{n \in \mathbb{N}}$ is relatively compact for the topology of the convergence in law if and only if $(X_n)_{n \in \mathbb{N}}$ is tight.

Furthermore $X_n \xrightarrow{n \to +\infty} X$ if and only if $(X_n)_{n \in \mathbb{N}}$ is tight and for all subsequences $\phi$ such that $X_{\phi(n)} \xrightarrow{n \to +\infty} Y$ where $Y$ is a random variable then $X$ and $Y$ have the same law (this second point shall be called uniqueness of the limit).

As an application of the Prokhorov theorem, let us prove the so-called method of moments

**Proposition A.8 (Method of moments).** Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of real random variables and $X$ be a real random variable characterized by its moments. We assume that for all $k \in \mathbb{N}$ $\lim_{n \to \infty} \mathbb{E}(X_n^k) = \mathbb{E}(X^k)$.

Then $X_n \xrightarrow{n \to +\infty} X$.

First let us prove a lemma that shall give the uniqueness of the limit in order to apply the Prokhorov theorem.

**Lemma A.9.** Let $(X_n)_{n \in \mathbb{N}}$ and $X$ be real random variables such that $(X_n)_{n \in \mathbb{N}}$ is uniformly integrable and $X_n \xrightarrow{n \to +\infty} X$, then $\mathbb{E}(X_n) \xrightarrow{n \to +\infty} \mathbb{E}(X)$.

*Proof.* Since $(X_n)_{n \in \mathbb{N}}$ converges to $X$ in law, we know that by the Skorokhod representation theorem (see [16]) we can find random variables $(X'_n)_{n \in \mathbb{N}}$ and $X'$ defined on the same
probability space, such that for all $n \ge 0$, $X'_n$ and $X_n$ have the same law, $X$ and $X'$ have the same law and $X'_n$ converges almost surely to $X'$. So $(X'_n)_{n \in \mathbb{N}}$ is uniformly integrable and converges almost surely and so in probability towards $X'$. Hence $(X'_n)_{n \in \mathbb{N}}$ converges in $L^1$ to $X'$.

We get that

$$ \mathbb{E}(X_n) = \mathbb{E}(X'_n) \underset{n \to +\infty}{\longrightarrow} \mathbb{E}(X') = \mathbb{E}(X). \quad \square $$

We now prove Proposition A.8.

*Proof.* We shall use the Prokhorov theorem to show the result. Firstly, since $(\mathbb{E}(X_n^2))_{n \in \mathbb{N}}$ converges, this sequence is bounded and so $(X_n)_{n \in \mathbb{N}}$ is tight by Example A.6. We want to show the uniqueness of the limit. Let $(X_{\phi(n)})_{n \in \mathbb{N}}$ be a subsequence that converges in law to a random variable $Y$.

Fix $k \in \mathbb{N}$. $(X_{\phi(n)}^k)_{n \in \mathbb{N}}$ is uniformly integrable since bounded in $L^2$ by the hypothesis of Proposition A.8 and converges in law to $Y^k$. Hence, by Lemma A.9, we have that for all $k \in \mathbb{N}$, $\mathbb{E}(X_{\phi(n)}^k) \underset{n \to +\infty}{\longrightarrow} \mathbb{E}(Y^k)$. As for all $k \in \mathbb{N}$ $\lim_{n \to +\infty} \mathbb{E}(X_n^k) = \mathbb{E}(X^k)$, we get that for all $k \in \mathbb{N}$, $\mathbb{E}(Y^k) = \mathbb{E}(X^k)$. Since $X$ is characterized by its moments, $X$ is equal to $Y$ in law. This proves the uniqueness of the limit.

<figure><img src="image_15.png" /></figure>

In order to apply the method of moments we have to compute the moments of Wigner semicircle distribution. Before let us introduce the Catalan numbers that are linked with these moments as we shall see.

**Definition A.10.** For all $p \in \mathbb{N}$, we call $C_p$ the $p^{th}$ Catalan number which is the number of plane trees with $p$ edges.

**Proposition A.11.** We have the following properties for $(C_p)_{p \in \mathbb{N}}$:

* $(C_p)_{p \in \mathbb{N}}$ is characterized by $C_0 = 1$ and $C_{p+1} = \sum_{k=0}^{p} C_k C_{p-k}$.

* Let $C(Z) = \sum_{k=0}^{+\infty} C_p Z^p \in \mathbb{C}[[Z]]$ its generative function (where $\mathbb{C}[[Z]]$ is the set of formal series in the variable $Z$). $C(Z)$ satisfies

$$ ZC(Z^2) - C(Z) + 1 = 0, \quad C(Z) = \frac{1 - \sqrt{1 - 4Z}}{2Z}. $$

* The explicit values of the Catalan numbers are given by:

$$ \forall p \in \mathbb{N}, C_p = \frac{1}{p+1} \binom{2p}{p}. $$

*Proof.* • $C_0$ is clearly equal to 1. For the inductive formula, consider a plane tree with $p+1 \ge 1$ edges, if we look at the leftmost edge and we isolate it, we have two plane trees with respectively $k \le p$ edges and $p-k$ with $k$ the number of edges on leftmost plane tree. Hence, if we denote $\mathcal{T}_p$ the set of plane trees with $p$ edges, we have a bijection from $\mathcal{T}_{p+1}$
to $\bigcup_{k=0}^{p} \mathcal{T}_k \times \mathcal{T}_{p-k}$. Taking the cardinality gives the result.

• Doing a Cauchy product and using the inductive formula yields

$$C(Z)^2 = \sum_{k=0}^{+\infty} \sum_{p=0}^{p} C_k C_{p-k} Z^p = \sum_{k=0}^{+\infty} C_{p+1} Z^p, \quad ZC(Z)^2 - C(Z) + 1 = 0.$$

Solving this equation in $C(Z)$, we find that $C(Z) = \frac{1 - \sqrt{1 - 4Z}}{2Z}$ using the fact $C(0) = 1$ to choose the good solution.

• Denoting $c(z) := \frac{1 - \sqrt{1 - 4z}}{2z}$, we have that $c$ is holomorphic on $D(0, 1/4)$, satisfies the equation $zc(z)^2 - c(z) + 1 = 0$ for all $z \in D(0, 1/4)$. We can do its Taylor expansion at 0

$$\forall z \in D(0, 1/4), c(z) = \sum_{p=0}^{+\infty} \frac{1}{p+1} \binom{2p}{p} z^p.$$

By uniqueness of the coefficients of Taylor expansion we deduce that

$$\forall p \in \mathbb{N}, C_p = \frac{1}{p+1} \binom{2p}{p}.$$

□

**Proposition A.12.** The moments of the Wigner semicircle law are given by:

$$\forall p \in \mathbb{N}, m_{2p+1} = 0, m_{2p} = C_p$$

*Proof.* Let us notice that $\sigma$ is even, so for all $p \in \mathbb{N}$, $m_{2p+1} = 0$. Fix $p \in \mathbb{N}$, we write

$$m_{2(p+1)} = \int_{-2}^{2} x^{2(p+1)} \frac{\sqrt{4-x^2}}{2\pi} dx = \int_{2}^{2} x^{2p+1} x \frac{\sqrt{4-x^2}}{2\pi} dx.$$

Noticing that $g(x) = x\sqrt{4-x^2}$ is the derivative of $h(x) = -\frac{1}{3}(4-x^2)^{\frac{3}{2}}$, by integrating by part yields

$$m_{2p+2} = \frac{2p+1}{3} \int_{-2}^{2} x^{2(p+1)} (4-x^2) \frac{\sqrt{4-x^2}}{2\pi} dx = \frac{2p+1}{3} (4m_{2p} - m_{2p+2}).$$

Hence, we get that $\frac{m_{2p+2}}{m_{2p}} = \frac{2p+1}{2p+4} = \frac{C_{p+1}}{C_p}$.

Since $C_0 = m_0 = 1$, we get that $C_p = m_{2p}$ for all $p \in \mathbb{N}$.

□

As a consequence of the results of this part, we can state the following corollary.

**Corollary A.13.** Let $(\mu_n)_{n \in \mathbb{N}} \in \mathcal{P}(\mathbb{R})^{\mathbb{N}}$ be a sequence of probability measures such that for all $k \in \mathbb{N}$, $\left(\int_{\mathbb{R}} x^k d\mu_n(x)\right)_{n \in \mathbb{N}}$ converges to 0 if $k$ is odd and to $C_{k/2}$ if $k$ is even then $(\mu_n)_{n \in \mathbb{N}}$ converges weakly to $\sigma$.
This Corollary is used in random matrix theory to prove the so-called mean Wigner theorem. We defined a $N$ complex Wigner matrix in Definition 2.1. For a matrix $M \in \mathcal{H}_N(\mathbb{C})$ and $f: \mathbb{R} \to \mathbb{R}$, one can define $f(M)$ by diagonalizing $M = ODO^*$ with $O \in \mathcal{U}_N(\mathbb{C})$ and $D$ a real diagonal matrix and defining $f(M) = Of(D)O^*$, where $f(D)$ is the diagonal matrix on which we apply $f$ on the diagonal elements of $D$.

Let us mention that this definition of $f(M)$ does not actually depend on the choice of basis to diagonalize $M$. Indeed, if we write $M = P_1D_1P_1^{-1} = P_2D_2P_2^{-1}$ with $D_1$ and $D_2$ diagonal matrix and $(P_1, P_2) \in GL_N(\mathbb{C})$ and denote $\{\lambda_1(M), ..., \lambda_N(M)\}$ the spectrum of $M$. Then we can define by Lagrange interpolation a polynomial $P \in \mathbb{C}[X]$ such that for all $1 \le i \le N$, $P(\lambda_i(M)) = f(\lambda_i(M))$. We have that $P(M) = P_1P(D_1)P_1^{-1} = P_2P(D_2)P_2^{-1}$ (by addition and multiplication of matrices) and so $P(M) = P_1f(D_1)P_1^{-1} = P_2f(D_2)P_2^{-1}$. This proves that the definition $f(M) = P_1f(D_1)P_1^{-1} = P_2f(D_2)P_2^{-1}$ does not depend on the basis in which we diagonalize $M$.

**Definition A.14 (Mean spectral measure).** Let $X_N$ be a $N$ complex Wigner matrix, we define the mean spectral measure $\mathbb{E}(\mu_N)$ of $X_N/\sqrt{N}$ as the unique measure in $\mathcal{P}(\mathbb{R})$ such that for all $f: \mathbb{R} \to \mathbb{R}$ continuous and bounded:

$$ \mathbb{E}\left[\frac{1}{N} \operatorname{Tr} f\left(\frac{X_N}{\sqrt{N}}\right)\right] = \int_{\mathbb{R}} f d\mathbb{E}(\mu_N) $$

Using the method of moments one can prove the Wigner theorem in mean.

**Theorem A.15 (Mean Wigner theorem).** For all $N \in \mathbb{N}$, let $X_N$ be a $N$ complex Wigner matrix and $\mathbb{E}(\mu_N)$ be the mean spectral measure of $X_N/\sqrt{N}$. Then $(\mathbb{E}(\mu_N))_{N \in \mathbb{N}}$ converges weakly to $\sigma$.

The starting point is to write

$$ \int_{\mathbb{R}} x^k d\mathbb{E}(\mu_N) = \mathbb{E}\left[\frac{1}{N} \operatorname{Tr}\left(\frac{X_N}{\sqrt{N}}\right)^k\right] = \frac{1}{N^{1+k/2}} \mathbb{E}[\operatorname{Tr}(X_N)^k)]. $$

Then, we express $\operatorname{Tr}(X_N^k)$ as $\sum_{i=1}^N \sum_{1 \le i_1, ..., i_{k-1} \le N} (X_N)_{i,i_1} (X_N)_{i_1, i_2} ... (X_N)_{i_{k-1}, i}$. The next part of the proof consists in doing combinatorics to simplify the expectation of the previous sum. We refer to Theorem 2.1.1 of [6] for a complete proof. Let us mention that this proof does not use that the entries of the Wigner matrix are Gaussian it only uses hypothesis on the moments of the entries of the matrix.

Finally, let us also point that in the case of Wigner matrices with Gaussian entries the Wigner theorem can be considered as the first term of a so-called topological expansion. This is obtained thanks to the Dyson-Schwinger equations. This makes a link between the random matrix theory and the enumerative geometry and combinatorics. See for instance [46, 48].

## A.2 Stieltjes transformation

**Definition A.16.** Let $\mu$ be a real probability measure. We define the Stieltjes transform of $\mu$ by

$$ \forall z \in \mathbb{C} - \mathbb{R}, \quad S_{\mu}(z) := \int_{\mathbb{R}} \frac{1}{z-t} d\mu(t). $$

**Example A.17.** For $\mu = \frac{1}{2}(\delta_1 + \delta_{-1})$, we have $S_{\mu}(z) = \frac{z}{z^2 - 1}$.
**Remark A.18.** For $\mu \in \mathcal{P}(\mathbb{R})$, for all $z \in \mathbb{C} - \mathbb{R}$, $\overline{S_\mu(z)} = S_\mu(\bar{z})$. Hence we only have to know the Stieljes transform on $\mathbb{H} = \{z \in \mathbb{C} | \mathfrak{Im}(z) > 0\}$ to know it on $\mathbb{C} - \mathbb{R}$.

**Proposition A.19.** Let $\mu$ be a real probability measure, we have the following properties:

1. For all $z \in \mathbb{C} - \mathbb{R}$, $|S_\mu(z)| \le \frac{1}{|\mathfrak{Im}(z)|}$.

2. If the support of $\mu$ is included in $[-A, A]$, then for all $z$ such that $|z| > A$, we have

$$S_{\mu}(z) = \sum_{k=0}^{+\infty} \int_{\mathbb{R}} t^k d\mu(t) \frac{1}{z^{k+1}}.$$

3. There exists an inversion formula. Let $I = [a, b]$ with $\mu(\partial I) = 0$, then

$$\mu(I) = \lim_{\epsilon \to 0} -\frac{1}{\pi} \mathfrak{Im} \left( \int_I S_\mu(x + i\epsilon) dx \right). \qquad (A.2)$$

As a consequence, if $\nu \in \mathcal{P}(\mathbb{R})$ is such that $S_\mu = S_\nu$ on $\mathbb{C} - \mathbb{R}$ then $\nu = \mu$.

4. A sequence $(\mu_n)_{n \in \mathbb{N}}$ of real probability measures converges weakly to $\mu \in \mathcal{P}(\mathbb{R})$ if and only if for all $z \in \mathbb{C} - \mathbb{R}$, $S_{\mu_n}(z)$ converges to $S_\mu(z)$.

5. $S_\mu$ is holomorphic on $\mathbb{C} - \text{supp}(\mu)$ where $\text{supp}(\mu)$ is the support of the measure $\mu$.

*Proof.*

1. Notice that if $z \in \mathbb{C} - \mathbb{R}$ and $t \in \mathbb{R}$, $|\mathfrak{Im}(z)| = |\mathfrak{Im}(z - t)| \le |z - t|$.

2. If $|z| > A$ and $|t| \le A$, then $\frac{1}{z-t} = \frac{1}{z} \frac{1}{1-t/z} = \frac{1}{z} \sum_{k=0}^{+\infty} t^k z^{-k}$. Exchanging the integral and the sum gives the result.

3. Since the atoms of $\mu$ are at most countable the formula A.2 implies that the measure $\mu$ is characterized by its Stieljes transform.

Let $I = [a, b]$ with $\mu(\partial I) = 0$. We have that

$$\mathfrak{Im} \left( \int_I S_\mu(x + i\epsilon) dx \right) = \mathfrak{Im} \left( \int_I \int_{\mathbb{R}} \frac{1}{x + i\epsilon - t} d\mu(t) dx \right) = - \int_{\mathbb{R}^2} \mathbb{1}_{[a,b]}(x) \frac{\epsilon}{(x-t)^2 + \epsilon^2} d\mu(t) dx.$$

Changing the variable and using Fubini-Tonelli theorem, yields

$$-\frac{1}{\pi} \mathfrak{Im} \left( \int_I S_\mu(x + i\epsilon) dx \right) = \int_{\mathbb{R}^2} \mathbb{1}_{[a,b]}(\epsilon x + t) \frac{1}{\pi(x^2 + 1)} d\mu(t) dx.$$

Let $C$ be a random variable with Cauchy distribution and $X$ be random variable with law $\mu$ and independent from $C$. We can write that

$$-\frac{1}{\pi} \mathfrak{Im} \left( \int_I S_\mu(x + i\epsilon) dx \right) = \mathbb{P}(X + \epsilon C \in [a, b]). \qquad (A.3)$$

Since $\mu(\partial I) = 0$ then $\mathbb{1}_{X+\epsilon C \in [a,b]}$ converges almost surely to $\mathbb{1}_{X \in [a,b]}$ when $\epsilon$ converges to 0. By the dominated convergence theorem we get

$$-\frac{1}{\pi} \mathfrak{Im} \left( \int_I S_\mu(x + i\epsilon) dx \right) = \mathbb{P}(X + \epsilon C \in [a, b]) \xrightarrow{\epsilon \to 0} \mathbb{P}(X \in [a, b]) = \mu(I).$$
4. If $(\mu_n)_{n \in \mathbb{N}}$ converges weakly to $\mu$ then, since $t \mapsto \mathfrak{R}\mathfrak{e}(1/(z-t))$ and $t \mapsto \mathfrak{I}\mathfrak{m}(1/(z-t))$ are bounded continuous functions for all $z \in \mathbb{C} - \mathbb{R}$, we have that for all $z \in \mathbb{C} - \mathbb{R}$, $S_{\mu_n}(z)$ converges to $S_{\mu}(z)$.

Conversely, let $(X_n)_{n \in \mathbb{N}}$, $X$, $C$ be independent random variables whose laws are respectively $(\mu_n)_{n \in \mathbb{N}}$, $\mu$ and Cauchy distribution as in the previous point. Let $(a, b) \in \mathbb{R}^2$ such that $\mu(\{a, b\}) = 0$ then for all $\varepsilon > 0$, for all $n \in \mathbb{N}$,

$$ \mathbb{P}(X_n + \varepsilon C \in [a, b]) = -\frac{1}{\pi} \mathfrak{I}\mathfrak{m} \left( \int_I S_{\mu_n}(x + i\varepsilon) dx \right) $$

converges to

$$ -\frac{1}{\pi} \mathfrak{I}\mathfrak{m} \left( \int_I S_{\mu}(x + i\varepsilon) dx \right) = \mathbb{P}(X + \varepsilon C \in [a, b]). $$

By the Portemanteau theorem (see [16]) we deduce that for all $\varepsilon > 0$, $X_n + \varepsilon C \xrightarrow{n \to +\infty} X + \varepsilon C$. Taking $\varepsilon = 1$, the Lévy continuity theorem gives that for all $t \in \mathbb{R}$,

$$ \mathbb{E}(\exp(it(X_n + C))) \xrightarrow{n \to +\infty} \mathbb{E}(\exp(it(X + C))). $$

By independence and since the characteristic function of the Cauchy distribution does not vanish, we get that for all $t \in \mathbb{R}$, $\mathbb{E}(\exp(itX_n)) \xrightarrow{n \to +\infty} \mathbb{E}(\exp(itX))$. Again by the Lévy continuity theorem $X_n \xrightarrow{n \to +\infty} X$.

5. For $K$ a compact included in $\mathbb{C} - \text{supp}(\mu)$ we have that for all $z \in K$ and $t \in \text{supp}(\mu)$

$$ \left| \frac{1}{z-t} \right| \le \frac{1}{d(K, \text{supp}(\mu))} \text{ with } d(K, \text{supp}(\mu)) \text{ the distance between } K \text{ and } \text{supp}(\mu). $$
$d(K, \text{supp}(\mu))$ is positive since $K$ is compact, $\text{supp}(\mu)$ is closed and $K \subset \mathbb{C} - \text{supp}(\mu)$. Hence, by the holomorphic parameter integral theorem $S_{\mu}$ is holomorphic on $\mathbb{C} - \text{supp}(\mu)$.

□

We can explicitly compute the Stieljes transform of the semicircle distribution.

**Proposition A.20.** $S_{\sigma}$ is characterized by the fact that it is the unique holomorphic function on $\mathbb{C} - \mathbb{R}$ such that for all $z \in \mathbb{C} - \mathbb{R}$, $S(z)^2 - zS(z) + 1 = 0$ and which converges to 0 when $|\mathfrak{I}\mathfrak{m}(z)|$ converges to $+\infty$. So we get

$$ \forall z \in \mathbb{H}, S_{\sigma}(z) = \frac{z - \sqrt{z^2 - 4}}{2} $$

where $\sqrt{z'}$ is the determination of the square root on $\mathbb{C} - \mathbb{R}^+$ such that $\sqrt{-1} = i$.

*Proof.* Since $\sigma$ is supported on the compact $[-2, 2]$ and using its moments given in Proposition A.12 we have:

$$ \forall |z| > 2, S_{\sigma}(z) = \frac{1}{z} \sum_{k=0}^{\infty} \frac{m_k}{z^k} = \frac{1}{z} C \left( \frac{1}{z^2} \right) $$

where $C$ is the generative function of the Catalan numbers defined in A.11. As explained in Remark A.18 we only have to prove the result on $\mathbb{H}$ and not $\mathbb{C} - \mathbb{R}$.

Since $ZC(Z)^2 - C(Z) + 1 = 0$ we have that for all $z \in \mathbb{H}$ such that $|z| > 2$, $S_{\sigma}^2(z) - zS_{\sigma}(z) + 1 = 0$. So there is two possibilities: for all $z \in \mathbb{H}$ with $|z| > 2$, $S_{\sigma}(z) = \frac{z - \sqrt{z^2 - 4}}{2}$ or for
all $z \in \mathbb{H}, |z| > 2$, $S_{\sigma}(z) = \frac{z + \sqrt{z^2 - 4}}{2}$ where $\sqrt{z}'$ is the square root on $\mathbb{C} - \mathbb{R}^+$ such that $\sqrt{-1} = i$.

By holomorphy and since $\mathbb{H}$ is connected, we deduce that for all $z \in \mathbb{H}$, $S_{\sigma}(z) = \frac{z - \sqrt{z^2 - 4}}{2}$ or for all $z \in \mathbb{H}$, $S_{\sigma}(z) = \frac{z + \sqrt{z^2 - 4}}{2}$.

However since for all $|z| > 2$, $|S_{\sigma}(z)| \le \frac{1}{|z| - 2}$, $S_{\sigma}$ converges to 0 when $|z|$ converges to $+\infty$. So the unique possible solution is given by $S_{\sigma}(z) = \frac{z - \sqrt{z^2 - 4}}{2}$ for all $z \in \mathbb{H}$. $\square$

In the end of this section, we briefly explain how we usually use the Stieljes transform to prove the mean Wigner theorem as we did in the last section with the method of moments.

Let us recall the notion of convergence for holomorphic functions and the Montel theorem which can be viewed as an Ascoli's theorem for holomorphic function.

**Definition A.21.** Let $U$ be an open set of $\mathbb{C}$, we write $\mathcal{H}(U)$ the set of holomorphic functions on $U$ and denote $(K_p)_{p \in \mathbb{N}}$ a sequence of compacts subsets of $U$ such that for all $p$, $K_p \subset K_{p+1}^\circ$ and $\bigcup_{p \in \mathbb{N}} K_p = U$ (we shall call a such sequence of compact sets a covering of $U$).

Let $f, g \in \mathcal{H}(U)$, we introduce the distance

$$d_{\mathcal{H}(U)}(f, g) := \sum_{p=0}^{+\infty} \frac{\min(1, ||f - g||_{\infty, K_p})}{2^{p+1}},$$

where $||f||_{\infty, K_p}$ is the infinite norm of $f$ viewed as function defined on the compact set $K_p$.

The topology defined by this distance $d_{\mathcal{H}(U)}$ does not depend on the covering $(K_p)_{p \in \mathbb{N}}$. A sequence $(f_n)_{n \in \mathbb{N}}$ in $\mathcal{H}(U)$ converges to $f \in \mathcal{H}(U)$ for this distance if and only if for all compact sets $K$ included in $U$, $||f_n - f||_{\infty, K} \xrightarrow{n \to +\infty} 0$.

Moreover $(\mathcal{H}(U), d_{\mathcal{H}(U)})$ is a complete metric space.

**Remark A.22.** Given an open set $U \subset \mathbb{C}$ a covering of $U$ always exists. Indeed we can consider the sequence of compacts $K_p := \{z \in U \mid |z| \le p\} \cap \{z \in U \mid d(z, \partial U) > 1/p\}$.

We now state the main result about relative compact subsets of $\mathcal{H}(U)$ for this topology.

**Theorem A.23 (Montel).** Let $U$ be an open subset of $\mathbb{C}$, and $A \subset \mathcal{H}(U)$. A is relatively compact in $(\mathcal{H}(U), d_{\mathcal{H}(U)})$ if and only if for all $K$ compact included in $U$, $\{f|_K, f \in A\}$ is bounded for $||.||_{\infty, K}$.

This result can be surprising at first sight when we compare it to the Ascoli theorem because there is no counterpart to the hypothesis of uniform uniform continuity for elements of $A \subset \mathcal{H}(U)$. Indeed, thanks to the Cauchy formula the uniform boundedness assumption of the elements of $A \subset \mathcal{H}(U)$ in the Montel theorem implies a uniform uniform continuity for elements of $A$.

As an application of the Montel theorem we have the following result that can be used to prove the mean Wigner theorem.

**Corollary A.24.** Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of real probability measures. If for all $z \in \mathbb{C} - \mathbb{R}$, $S_{\mu_n}(z)^2 - zS_{\mu_n}(z) + 1 \xrightarrow{n \to +\infty} 0$ then $(\mu_n)_{n \in \mathbb{N}}$ converges weakly to $\sigma$.
*Proof.* Let $K$ be a compact included in $\mathbb{C}-\mathbb{R}$, since $z \mapsto \mathfrak{Im}(z)$ is continuous on $K$ it has a minimum on $K$, which will be denoted $M_K$. For all compact $K$ included in $\mathbb{C}-\mathbb{R}$, we have that for all $n$, $||S_{\mu_n}||_{\infty,K} \le 1/M_K$ by Proposition A.19. By the Montel theorem, $(S_{\mu_n})_{n \in \mathbb{N}}$ has an accumulation point for the distance $d_{\mathcal{H}(\mathbb{C}-\mathbb{R})}$.

Let $S$ be a accumulation point of $(S_{\mu_n})_{n \in \mathbb{N}}$. Taking the limit, we get that for all $z \in \mathbb{C}-\mathbb{R}$, $S(z)^2 - zS(z) + 1 = 0$. We first notice that $\bar{S}(z) = S(\bar{z})$ by Remark A.18 and passing to the limit. So we only have to identify $S$ on $\mathbb{H}$. As in Proposition A.20 we have two possibilities: either for all $z \in \mathbb{H}$, $S(z) = \frac{z - \sqrt{z^2 - 4}}{2}$ or for all $z \in \mathbb{H}$, $S(z) = \frac{z + \sqrt{z^2 - 4}}{2}$.

However, since for all $n \in \mathbb{N}$, for all $t \in \mathbb{R}^+$, $|S_{\mu_n}(it)| \le \frac{1}{t}$ by Proposition A.12, we get that $|S(it)| \le \frac{1}{t}$. So we necessarily have that $z \in \mathbb{H}$, $S(z) = \frac{z - \sqrt{z^2 - 4}}{2} = S_{\sigma}(z)$.

So $(S_{\mu_n})_{n \in \mathbb{N}}$ is a relatively compact sequence of $(\mathbb{C}-\mathbb{R}, d_{\mathcal{H}(\mathbb{C}-\mathbb{R})})$ with a unique accumulation point $S_{\sigma}$. So $(S_{\mu_n})_{n \in \mathbb{N}}$ converges to $S_{\sigma}$. $\square$

In particular to prove the Wigner theorem in mean, with the notations of Theorem A.15, we classically first prove that for all $z \in \mathbb{C}-\mathbb{R}$, $\mathbb{E}[S_{\mu_N}^2(z)] - z\mathbb{E}[S_{\mu_N}(z)] + 1 = 0$. Then using some concentration inequalities as those obtained in Section 4.2, we prove that for all $z \in \mathbb{C}-\mathbb{R}$, $\mathbb{E}[S_{\mu_N}^2(z)] - \mathbb{E}[S_{\mu_N}(z)]^2 \underset{N \to +\infty}{\longrightarrow} 0$. So we deduce that for all $z \in \mathbb{C}-\mathbb{R}$, $\mathbb{E}[S_{\mu_N}(z)]^2 - z\mathbb{E}[S_{\mu_N}(z)] + 1 \underset{N \to +\infty}{\longrightarrow} 0$. Thanks to Corollary A.24 we obtain that for all $z \in \mathbb{C}-\mathbb{R}$, $\mathbb{E}[S_{\mu_N}(z)] = S_{\mathbb{E}[\mu_N]}(z) \underset{N \to +\infty}{\longrightarrow} S_{\sigma}(z)$. By Proposition A.19, $(\mathbb{E}[\mu_N])_{N \in \mathbb{N}}$ converges weakly towards $\sigma$. See Section 2.4 of [5] for a complete proof.

## Acknowledgments

I acknowledge Charles Bertucci for its encouragement for the writing of these notes and Edouard Maurel-Segala for its course on random matrices that was a motivation for working on this topic. Finally, I also acknowledge the seminar MEGA for the hospitality and for the diversity of talks that made me discover the richness of this topic.
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