I'm trying to prove the following:
Let X be a fixed matrix of dimension $d*n$ and let $g$ be a standard gaussian vector in $\mathbb{R}^n$.
Then: $\mathbb{E} \||Xg\|| \leq \operatorname{Tr}\left(\left[X X^T\right]^{1/2}\right)$.
This is taken from https://arxiv.org/pdf/1610.05200.pdf (proof of Lemma 5.3).
By taking the power and the square root of the norm inside of the expected value, writing the squared norm in terms of trace, and using Jensen's inequality with the square root I managed to prove that $\mathbb{E} \||Xg\|| \leq \sqrt{\operatorname{Tr}\left(X X^T\right)}$.
How can I prove, instead the result with the square root inside of the argument of the trace? I'm having some troubles.
Thanks in advance
Best Answer
I don't see where the author uses that fact in the proof. In his final lines he says $$ \begin{aligned} \mathbf{E}_{\tilde{X}}\left[\sup _{v, w \in B} Z_{v, w}\right] \leq \mathbf{E}_{\tilde{X}}\left[\sup _{v, w \in B} Y_{v, w}\right] & \leq\|\tilde{X}\| \mathbf{E}\left\|\Sigma^{1 / 2} g\right\|+\|\Sigma\|^{1 / 2} \mathbf{E}_{\tilde{X}}\|\tilde{X} g\| \\ & \leq\|\tilde{X}\| \sqrt{\operatorname{Tr}[\Sigma]}+\|\Sigma\|^{1 / 2} \operatorname{Tr}\left[\tilde{X} \tilde{X}^*\right]^{1 / 2} \end{aligned} $$ by the Slepian-Fernique inequality. Taking the expectation with respect to $\tilde{X}$ and using that $\mathbf{E}\|\tilde{X}\|=\mathbf{E}\|X\|$ and $\mathbf{E}\left[\operatorname{Tr}\left[\tilde{X} \tilde{X}^*\right]^{1 / 2}\right] \leq \sqrt{n \operatorname{Tr}[\Sigma]}$ yields the conclusion.
And the conclusion he wishes to establish is with $r(\Sigma):=\operatorname{Tr}[\Sigma] /\|\Sigma\|$ $$ \mathbf{E}\|Z-\Sigma\| \lesssim \mathbf{E}\|X\| \frac{\sqrt{\operatorname{Tr}[\Sigma]}}{n}+\|\Sigma\| \sqrt{\frac{r(\Sigma)}{n}} $$ where $$ \mathbf{E}\|Z-\Sigma\| \leq \frac{2}{n} \mathbf{E}\left[\sup _{v, w \in B} Z_{v, w}\right]. $$ So all that needs to happen is that he takes the expectation of the list line of the first expression, and $\mathbf{E}\left[\operatorname{Tr}\left[\tilde{X} \tilde{X}^*\right]^{1 / 2}\right] \leq \sqrt{n \operatorname{Tr}[\Sigma]}$ indeed follows from Jensen's.