For every integer $n$ such that $n \geqslant 1$, we denote by $\widehat{M}_{n}(C) = \left(\widehat{X}_{ij}(C)\right)_{1 \leqslant i,j \leqslant n}$. For every $\omega \in \Omega$, we denote by $\widehat{\Lambda}_{1,n}(\omega) \geqslant \ldots \geqslant \widehat{\Lambda}_{n,n}(\omega)$ the eigenvalues of $\frac{1}{\sqrt{n}} \widehat{M}_{n}(C)(\omega)$ arranged in decreasing order. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a $K$-Lipschitz function.
Assume furthermore that $f$ is bounded. Show $$\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) \xrightarrow{n \rightarrow +\infty} \frac{1}{2\pi} \int_{-2}^{2} f(x) \sqrt{4 - x^{2}} \, \mathrm{d}x.$$

Show the semicircle law in the general case:
For every function $f : \mathbb{R} \rightarrow \mathbb{R}$, continuous and bounded, $$\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) \xrightarrow{n \rightarrow +\infty} \frac{1}{2\pi} \int_{-2}^{2} f(x) \sqrt{4 - x^{2}} \, \mathrm{d}x$$

For every integer $n$ such that $n \geqslant 1$, we denote by $\widehat{M}_{n}(C) = (\widehat{X}_{ij}(C))_{1 \leqslant i,j \leqslant n}$. For all $\omega \in \Omega$, we denote by $\widehat{\Lambda}_{1,n}(\omega) \geqslant \ldots \geqslant \widehat{\Lambda}_{n,n}(\omega)$ the eigenvalues of $\frac{1}{\sqrt{n}} \widehat{M}_{n}(C)(\omega)$ arranged in decreasing order. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a $K$-Lipschitz function.
Assume further that $f$ is bounded. Show $$\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) \xrightarrow{n \rightarrow +\infty} \frac{1}{2\pi} \int_{-2}^{2} f(x) \sqrt{4 - x^{2}} \, \mathrm{d}x.$$

Show the semicircle law in the general case:
For every function $f : \mathbb{R} \rightarrow \mathbb{R}$, continuous and bounded, $$\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) \xrightarrow{n \rightarrow +\infty} \frac{1}{2\pi} \int_{-2}^{2} f(x) \sqrt{4 - x^{2}} \, \mathrm{d}x$$