Convergence of Expectation of norm of sub-gaussian random vector 
*

*We know that if $X=(X_1,...,X_n)$ be a random vector with independent sub-gaussian coordinates $X_i$ that satisfy $EX_i^2=1$, then


$$\left\lVert \left\lVert X\right\rVert_2-\sqrt{n}\right\rVert_{\psi_2}\leq CK^2$$
where $K=\max||X_i||_{\psi_2}$. Additionally, $-CK^2\leq \mathbb{E}||X||_2-\sqrt{n}\leq CK^2$.
Can we conclude that $\mathbb{E}||X||_2-\sqrt{n}$ converges to $0$ as $n \rightarrow \infty$?
I know that $\frac{||X||_2}{\sqrt{n}}-1$ converges to $0$ in probability. But I don't know if it converges in expectation or not.


*I want to show that $\operatorname{Var}\left(||X||_2\right)\leq CK^4$.


It suffices to show that $(\mathbb{E}||X||_2)^2\geq n-CK^4$. I want to exploit above inequalities to drive the desired bound.
 A: The first question is addressed here, by a positive answer.
For the second question, we can follow the following steps:

*

*$\mathbb E\left[\left(\lVert X\rVert_2^2-n\right)^2\right]
=\mathbb E\left[\left(\sum_{i=1}^n\left(X_i^2-\mathbb E\left[X_i^2\right]\right)\right)^2\right]\leqslant K^4n.$


*Then using
$$ \left(\lVert X\rVert_2^2-n\right)^2=\left(\lVert X\rVert_2-\sqrt n\right)^2 \left(\lVert X\rVert_2+\sqrt n\right)^2\geqslant\left(\lVert X\rVert_2-\sqrt n\right)^2 n,  $$
we get $$\tag{*}\mathbb E\left[\left(\lVert X\rVert_2-\sqrt n\right)^2\right]\leqslant K^4.$$


*Using $(a+b)^2\leqslant 2a^2+2b^2$, we derive by (*),
$$
\operatorname{Var}\left(\lVert X\rVert_2\right)= \mathbb E\left[\left(\lVert X\rVert_2-\sqrt{n}-\mathbb E\left[\lVert X\rVert_2\right]+\sqrt n\right)^2\right]
\leqslant 2K^4+2  \left(\mathbb E\left[\lVert X\rVert_2\right]-\sqrt n\right)^2.$$
Moreover, $(*)$ gives
$$K^4\geqslant \left\lVert  \lVert X\rVert_2-\sqrt n \right\rVert_{L^2}^2\geqslant \left\lVert  \lVert X\rVert_2-\sqrt n \right\rVert_{L^1}^2$$
hence
$$
\operatorname{Var}\left(\lVert X\rVert_2\right)\leqslant 3K^4.
$$
