Random Matrix Theory, semicircular law, and Bai-Yin Theorem I am trying to find a proof for the following lower bound Bai-Yin theorem.
Theorem: For an ensemble of Wigner matrices $(W_n)_{n\in\mathbb{Z}_+}$, we have for any $\epsilon>0$, almost surely,
\begin{align*}
||W_n||>(2-\epsilon)\sqrt{n}
\end{align*}
for large enough $n$. Here we denote the operator norm of matrix $A$ with $||A||$. In fact, we have almost surely
\begin{align*}
\lim_{n\rightarrow\infty}\inf||W_n||/\sqrt{2}\geq 2
\end{align*}
My question: Only thing I found on the internet was the following proof fragment, which looks kind of strange for me. For instance, the used function $g$ seems to be not defined for some arguments ($x=2-\epsilon$) and overall I am not able to follow the argumentation.
Proof:
By the semicircular law, we know that the empirical spectral distribution of $(1/\sqrt{n})W_n$ converges to a semicircular distribution almost surely. Fix $\epsilon>0$. Let $g:\mathbb{R\rightarrow}[0,1]$ be a continuous function with $g(x)=1$ if $|x|<2-\epsilon$ and $g(x)=0$ if $|x|>2-\epsilon/2$.
If $||W_n||<(2-\epsilon)\sqrt{2}$, then
\begin{align*}
\int_{\mathbb{R}}g(x)d\mu_{\frac{1}{\sqrt{n}}W_n}(x)=1.
\end{align*}
But we also see that
\begin{align*}
\int_{\mathbb{R}}g(x)d\mu_{sc}(x)<1.
\end{align*}
This means that we can almost surely find $N$ such that
\begin{align*}
||W_n||>(2-\epsilon)\sqrt{n}
\end{align*}
for any $n>N$.
 A: Fix $\varepsilon>0$. First consider the (real-valued) random variable
$$X_n=\int_\mathbb{R} h \,d\mu_{W_n/\sqrt{n}}.$$
where $h(x)=1$ if $|x|\le 2-\varepsilon$ and $h(x)=0$ if $|x|> 2-\varepsilon$. $X_n$ is counting the fraction of eigenvalues of $W_n$ inside the interval $[(-2+\varepsilon)\sqrt{n},(2-\varepsilon)\sqrt{n}]$.
Now the following events all happen with probability 1. From the semicircular law, $X_n$ converges to the random variable equal to the mass of the semicircular distribution on the interval $[-2+\varepsilon,2-\varepsilon]$. Since it is a (deterministic) number strictly less than 1, $X_n<1$ for all but finitely many $n$. Recalling the interpretation of $X_n$, this means that for all but finitely many $n$, $W_n$ has an eigenvalue of magnitude larger than $(2-\varepsilon)\sqrt{n}$, i.e. $\lVert W_n\rVert>(2-\varepsilon)\sqrt{n}$.

There is one flaw in our proof: We need to take a continuous test function $h$ to apply the convergence results of the semicircular law. However, the same idea works by picking some continuous approximation $g$ of $h$ that satisfies $g(x)=1$ if $|x|\le 2-\varepsilon$, $g(x)=0$ if $|x|>2-\varepsilon/2$ and $g(x)\in[0,1]$ inbetween. Now $X_n$ only provides an upper bound on the fraction of eigenvalues inside $[(-2+\varepsilon)\sqrt{n},(2-\varepsilon)\sqrt{n}]$. However, you can check that $X_n$ still converges a.s. to a number strictly less than 1, so this still means that a.s. $\lVert W_n\rVert>(2-\varepsilon)\sqrt{n}$ for all but finitely many $n$.

Note that if you do not have the semicircular law, there are also more direct proofs under additional assumptions on the boundedness of the entries (see e.g. Proposition 32 of Tao's set of notes).
