I've come across the following statement (unfortunately w/o proof, or any details). The statement is as follows:
Lemma. Let $X$ be a random variable, and recall that the common definition of a subgaussian random variable. That is, if there exists a finite $c$ such that the moment generating function $\mathbb{E}\left[\exp\left(c^{-1}X^2\right)\right] < \infty$, then we say $X$ is subgaussian. Equivalently, $X$ is subgaussian when $\sup_{1 \leq p < \infty}p^{-1/2}||X||_p < \infty.$
So rewriting, I would like to prove that:
$$\exists c < \infty~\text{such that}~\mathbb{E}\left[\exp\left(c^{-1}X^2\right)\right] < \infty \quad\Leftrightarrow\quad \sup_{1 \leq p < \infty}p^{-1/2}||X||_p < \infty.$$
So far.
$\Leftarrow$: Since $X$ is a random variable we have that
$$\sup_{1 \leq p < \infty}p^{-1/2}||X||_p = \sup_{1 \leq p < \infty}p^{-1/2}\left(\mathbb{E}|X|^p\right)^{1/p} < \infty.$$
Not exactly sure how to proceed from here, but the idea I believe is to get the following
$$\sup_{1 \leq p < \infty} p^{-1/2}\left(\mathbb{E}|X|^p\right)^{1/p} = \sup_{1 \leq p < \infty} \int |x|^pd\mu = \dots = \int_0^{\infty} P\left[\exp\left(c^{-1}X^2 \geq x \right)\right]dx = \mathbb{E}\left[\exp\left(c^{-1}X^2\right)\right].$$
$\Rightarrow:$ Not quite sure, but I think we could invoke something like Markov's inequality to start. But I'm not really seeing it.