Q:

Inequality between 2p-norm and p-norm for random variables

Recently I was studying something about random matrix theory, and class of sub-guassian / sub-exponential random variables is of interest. In the literature it gave an inequality as following:
$\sup_{p\geq 1} \frac{\|X^2\|_p}{p} \leq 2\sup_{p\geq 1} (\frac{\|X\|_p}{\sqrt{p}})^2$
which gives a sufficient condition such that:
$\|X\|_{2p} \leq \sqrt{2}\|X\|_p$
The proof of this inequality is not provided in the literature. Suppose the random variable $X\in\mathbb{L}^\infty(\mathbb{R})$ therefore the previous defined sub-gaussian norm and sub-exponential norm both exist, then how do we proof the sufficient condition? 
Any comment is greatly appreciated :)  

A:

This is a direct consequence of the general fact that $\|X^2\|_p=E[|X|^{2p}]^{1/p}=\left(\|X\|_{2p}\right)^2$ for every $p\gt0$ and every  random variable $X$ in $L^{2p}$. Hence,
$$
\sup_{p\geqslant 1} \frac{\|X^2\|_p}{p}=2\,\sup_{p\geqslant 2} \left(\frac{\|X\|_p}{\sqrt{p}}\right)^2\leqslant 2\,\sup_{p\geqslant 1} \left(\frac{\|X\|_p}{\sqrt{p}}\right)^2.
$$
The rôle of the (not always true) inequality $\|X\|_{2p} \leqslant \sqrt{2}\,\|X\|_p$ is not clear.

