Asymptotic order of the $L_\infty$ norm of asymptotically normally distributed random variables Let $\mathbf{X}_n \in R^p$ be a random variable and $\mathbf{s} \in \mathcal{S} = \{x \in R^p \ s.t. \ ||x||_2 = 1\}$. Then, suppose that $p$ and $n$ are allowed to diverge and that we have
$$\sqrt{n} \mathbf{s}^T \mathbf{X}_n \to \mathcal{N}(0,1),\;\; \text{(in distribution as } n \to \infty \text{)},$$
for all $\mathbf{s} \in \mathcal{S}$. Is there anything that can be concluded about the order of $||\mathbf{X}_n||_{\infty}$?
My guess is that we should have $||\mathbf{X}_n||_{\infty} = \mathcal{O}_p(\log(p)/\sqrt{n})$ but I can't show this formally.
 A: I don't think it's possible to prove any statement of the form $||\mathbf{X}_{p,n}||_{\infty} = O_p(f(p, n))$. I haven't seen $O_p$ notation with more than one variable (here $n$ and $p$) before, so I'm going to assume the definition is:

For any $\epsilon > 0$, there exist $M, N, P > 0$ such that for all $n > N, p> P$,
$$
P(||\mathbf{X}_{p,n}||_{\infty} > M |f(p, n)|) < \epsilon
$$

But the only constraint given in the problem is that for each $p$, for each $\mathbf s$,
$$
\sqrt{n} \mathbf{s}^T \mathbf{X}_{p,n} \to \mathcal{N}(0,1),\;\; \text{(in distribution)}
$$
The problem is that these convergences in distribution may not happen "in sync" for different $p$. For example, I don't see anything in the problem statement that rules out the behaviour
$$\mathbf{X}_{p,n}^{(1)} \sim \mathcal N(n^{p!},p^{n!}) \text{ whenever p > n}$$
where $\mathbf{X}_{p,n}^{(i)}$ is the $i$-th component of $\mathbf{X}_{p,n}$. This behaviour would rule out any reasonable $O_p(f(n,p))$ convergence result, because it would mean that no matter how big you choose your $N, P$, it will still be the case that $\mathbf{X}_{P+N+2,P+N+1}$ behaves ridiculously.
(Note that I am not assuming that $\mathbf{X}_{p,n}^{(i)} = \mathbf{X}_{p+1,n}^{(i)}$.)
