For what kind of random vectors do we have $\sup_{p \ge 1}\|X\|_p < \infty$? Let $X$ be a random vector on $\mathbb R^m$ (assumed to have zero mean, for simplicity). For $p \in [1,\infty)$, define $e_p(X):=\mathbb E\sum_{j=1}^m|X_j|^p \in [0,\infty]$. Finally, define $\|X\|_p \in [0,\infty]$ by
$$
\|X\|_p = \begin{cases}e_p(X)^{1/p},&\mbox{ if }e_p(X) < \infty,\\\infty,&\mbox{else.}\end{cases}
$$
Note that $\|X\|_p$ is a increasing function of $p$ and define $\|X\|_\infty := \lim_{p \to \infty}\|X\|_p \in [0,\infty]$.

Question. For what kind of random vectors do we have $\|X\|_\infty < \infty$ ?

 A: Sorry for answering my own question. The solution presented below is inspired by a remark of user Nate Eldredge.

Define $\vert\vert\vert X\vert\vert\vert_\infty:=\max_{j=1}^k\|X_j\|_\infty \in [0, \infty]$, with $\|X_j\|_\infty := \lim_{p \to \infty}\|X_j\|_p$ and $\|X_j\|_p := (\mathbb E|X_j|^p)^{1/p}$ for all $j$. A well-known alternative representation of $\|X_j\|_\infty$ is $\|X_j\|_\infty = \inf\{B \ge 0\mid \mathbb P(|X_j| \le B) = 1\}$. Thus, $\|X_j\|_\infty < \infty$ is equivalent to saying that $X_j$ is essentially bounded. I will prove the following:

Claim. $\|X\|_\infty = \vert\vert\vert X\vert\vert\vert_\infty$.

Proof. For every $p \in [1,\infty)$, one computes
$$
e_p(X) := \mathbb E \sum_j|X_j|^p = \sum_j\mathbb E|X_j|^p = \sum_j((\mathbb E|X_j|^p)^{1/p})^p=:\sum_j\|X_j\|_p^p.
$$
Therefore in the limit $p \to \infty$, we have $e_p(X)^{1/p} = (\sum_j\|X_j\|_p^p)^{1/p} \overset{(*)}{\longrightarrow} (\sum_j\|X_j\|_\infty^p)^{1/p} \longrightarrow \vert\vert\vert X\vert\vert\vert_\infty$, where (*) follows from this post https://math.stackexchange.com/a/242792/168758. $\quad\quad\Box$
Now, it is clear that
$\|X\|_\infty < \infty \iff \vert\vert\vert X\vert\vert\vert_\infty \iff \|X_j\| < \infty\;\forall j$. Thus $\|X\|_\infty < \infty$ is equivalent to saying each coordinate of $X_j$ (a random variable!) of $X$ is essentially bounded.
