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. Show that $$\left|\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) - \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\widehat{\Lambda}_{i,n}\right)\right)\right| \leqslant \frac{K}{n} \mathbb{E}\left(\left\|M_{n} - \widehat{M}_{n}(C)\right\|_{F}\right).$$
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.
Show that
$$\left|\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\Lambda_{i,n}\right)\right) - \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^{n} f\left(\widehat{\Lambda}_{i,n}\right)\right)\right| \leqslant \frac{K}{n} \mathbb{E}\left(\left\|M_{n} - \widehat{M}_{n}(C)\right\|_{F}\right).$$