Upper bound of sub-gaussian norm of bounded random variable? I am reading the High-Dimensional Probability by Dr.Roman Vershynin , where I stuck on some statement at page 28. where state as below:
Any bounded random variable $X$ is sub-gaussian with:  $$\newcommand\norm[1]{\left\lVert#1\right\rVert}
\norm{X}_{\psi_2}\leq \frac{\norm{X}_{\infty}}{\sqrt{\log2}}
$$
where $\newcommand\norm[1]{\left\lVert#1\right\rVert}
\norm{X}_{\psi_2}$ is the sub-gaussian norm define as:
$$\newcommand\norm[1]{\left\lVert#1\right\rVert}
\norm{X}_{\psi_2} =\inf \left\{ t>0 : \mathbb{E} \left[\exp{\left(\frac{X^2}{t^2}\right)} \right]   \leq 2 \right\}
$$
where $\newcommand\norm[1]{\left\lVert#1\right\rVert}
\norm{X}_{\infty} :=( \mathbb{E} |X|^p)^{1/p}$ as $p \to \infty$
I can see how why the  bounded random variable is sub-gaussian (hoeffing lemma ),but How could I see this upper bound of sub-gaussian  norm?

If $t=(\sqrt {\log 2})^{-1} \|X\|_{\infty}$ then $Ee^{\frac {X^{2}} {t^{2}}} \leq e^{\log 2}=  2$ and hence $\|X\|_{\psi_2} \leq t$. I have used the fact that $\|X\|_{\infty}$ is nothing but the essential supremum of $X$.
