# Tracy-Widom Law: Vague convergence and Ledoux Bound

This is a continuation of the previous post available here. In the previous post, we develop the ingredients required for the vague convergence proof.  Let us now return to the random matrix scenario.

Proof of Theorem 4: $X_n$ be a sequence of $n\times n$ GUE matrices. Let $\lambda_1,\ldots, \lambda_n$ be the eigenvalues of $X_n$. Fix $-\infty < t< t' < \infty$. Let us quickly evaluate the limit

$\displaystyle \lim_{n\to \infty} P\left[n^{2/3}\left(\frac{\lambda_i}{\sqrt{n}}-2\right) \not\in (t,t'), \ \ i=1,2, \ldots, n\right]$

Observe that by using Theorem 3, we have

\displaystyle \begin{aligned} & P\left[n^{2/3}\left(\frac{\lambda_i}{\sqrt{n}}-2\right) \not\in (t,t'), \ \ i=1,2, \ldots, n\right] \\ & = P\left[\lambda_i \not\in \left(2\sqrt{n}+\frac{t}{n^{1/6}},2\sqrt{n}+\frac{t'}{n^{1/6}}\right), \ \ i=1,2, \ldots, n \right] \\ & = 1+\sum_{k=1}^{\infty} \frac{(-1)^k}{k!}\int\limits_{2\sqrt{n}+\frac{t}{n^{1/6}}}^{2\sqrt{n}+\frac{t'}{n^{1/6}}}\cdots\int\limits_{2\sqrt{n}+\frac{t}{n^{1/6}}}^{2\sqrt{n}+\frac{t'}{n^{1/6}}} \det_{i,j=1}^k K^{(n)}(x_i,x_j)\prod_{i=1}^{k}dx_i \\ & = 1+\sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \int_{t}^{t'}\cdots\int_{t}^{t'} \det_{i,j=1}^k \left[\frac1{n^{1/6}}K^{(n)}\left(2\sqrt{n}+\frac{x_i}{n^{1/6}},2\sqrt{n}+\frac{x_j}{n^{1/6}}\right)\right]\prod_{i=1}^k dx_i \end{aligned}

where in the last line we use change of variable formula. Let us define

$\displaystyle A^{(n)}(x,y):=\frac1{n^{1/6}}K^{(n)}\left(2\sqrt{n}+\frac{x}{n^{1/6}},2\sqrt{n}+\frac{y}{n^{1/6}}\right)$

Note that $A^{(n)}$ are kernels since $K^{(n)}$ is a kernel. Let us also define $\displaystyle \phi_n(x) = n^{1/12}\psi_n\left(2\sqrt{n}+\frac{x}{n^{1/6}}\right)$. Then using proposition 1 (c) we have

$\displaystyle A^{(n)}(x,y)=\frac{\phi_n(x)\phi_n'(y)-\phi_n'(x)\phi_n'(y)}{x-y}-\frac1{2n^{1/3}}\phi_n(x)\phi_n(y)$

Note that Proposition 4 implies that  for every $C>1$, $\phi_n \to Ai$ uniformly over a ball of radius C in the complex plane. $\phi_n$ are entire functions. Hence $\phi_n' \to Ai'$ and $\phi_n'' \to Ai''$. Hence for $t\le x,y \le t'$ we have

$\displaystyle ||A^{(n)}(x,y)-A(x,y)|| \to 0$

Hence by continuity lemma for fredholm determinants we have

\displaystyle \begin{aligned} & \lim_{n\to \infty} P\left[n^{2/3}\left(\frac{\lambda_i}{\sqrt{n}}-2\right) \not\in (t,t'), \ \ i=1,2, \ldots, n\right] \\ & = \lim_{n\to\infty} \Delta(A^{(n)}) \\ & = \lim_{n\to\infty} \Delta(A) \\ & = 1+\sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \int_{t}^{t'}\cdots\int_{t}^{t'} \det_{i,j=1}^k A(x_i,x_j) \prod_{i=1}^k dx_i \end{aligned}

where the measure is the lebesgue measure on the bounded interval $\displaystyle (t,t')$. This completes the proof of Theorem 4

$\square$