Convergence of edge eigenvalues for Gaussian matrices

Question

I am reading this lecture note. I have a difficulty in understanding the third section in chapter 6. Particularly, in Theorem 4.1, they claimed that

Let $X$ be a Gaussian Wigner matrix satisfying assumptions (A). We have a.s. $$ \lim _{n \rightarrow \infty} \lambda_1\left(\frac{X}{\sqrt{n}}\right)=-\lim _{n \rightarrow \infty} \lambda_n\left(\frac{X}{\sqrt{n}}\right)=2 . $$ Where $\lambda_1$ is the largest eigenvalue of $\frac{X}{\sqrt{n}}$.

But in the proof, I don't understand why they said that

Note also that Wigner semicircle law implies that a.s. $$ \liminf _{n \rightarrow \infty} \lambda_1 \geq 2 \text {. } $$

Could you please help me by explaining that line? Thank you for your help.

Answer 1

Let $\mu_{\mathsf{sc}}$ be the probability measuere of the semicircle law. For any $0<\varepsilon <2$, by Wigner's semicircle law, we have $$ \#\left\{\lambda_i \left(\frac{X}{\sqrt{n}}\right)\in[2-\varepsilon,2]\right\} \to n\cdot \mu_{\mathsf{sc}}\left( [2-\varepsilon,2] \right) . $$ That is, there are nearly $n\cdot \mu_{\mathsf{sc}}\left( [2-\varepsilon,2] \right)$ eigenvalues in $[2-\varepsilon,2]$. This implies that $$ \liminf_{n\to\infty} \; \lambda_1\left(\frac{X}{\sqrt{n}}\right) \geqslant 2 - \varepsilon. $$ Taking $\varepsilon \to 0$ on both sides yields that $$ \liminf_{n\to\infty} \; \lambda_1\left(\frac{X}{\sqrt{n}}\right) \geqslant 2. $$