Law of the iterated logarithm of a Brownian motion in $\mathbb{R}^q$

Question

If $W=(W^1,...,W^q)$ is a Brownian motion in $\mathbb{R}^q$ and $\|\cdot\|$ is the Euclidean norm on $\mathbb{R}^q,$ we want to prove that $$\limsup_{u \to +\infty}\frac{\|W_u\|}{\sqrt{2u\ln(\ln(u))}}=1 \text{ a.s.}$$

An answer was given here: Law of the iterated logarithm in higher dimensions, with the property that there exists a finite $I_ϵ \subset S^{q−1}$ for $ϵ>0$ such that for all $x \in \mathbb{R}^q$ and for all $u \in S^{q−1},$ there exists $y \in I_ϵ$ such that $$\langle x,u\rangle \leq (1+\epsilon)\langle x,y\rangle$$

The above property, why is it true? Also, is there a law for an arbitrary norm $\|\cdot\|$?

Answer 1

This is a nice question. Actually, using the fact that any two norms on $\mathbb R^q$ are equivalent, it is easy to argue that for any norm $||\cdot||$, $$ \limsup_{u\to\infty} \frac{||W_u||}{\sqrt{u \ln \ln u}} \in (0,\infty). $$

However, this is not enough, as the limit might be random. I will show that it is not.

Proposition For any norm $||\cdot||$ on $\mathbb R^q$, $$ \limsup_{u\to\infty} \frac{||W_u||}{\sqrt{u \ln \ln u}} = K_{||\cdot||}, $$ where $$ K_{||\cdot||} = \sqrt2 \max_{x: ||x||_2 = 1} ||x||, $$ and $||\cdot||_2$ is the Euclidean norm.

Proof. Recall that $||x|| = \sup_{||y||^* = 1} \langle x,y\rangle$, where $||\cdot||^*$ is the dual norm. Observe that $$ \sup_{||y||^* = 1} ||y||_2 = \sup_{||y||^* = 1}\sup_{||x||_2 = 1} \langle x,y\rangle = \sup_{||x||_2 = 1}\sup_{||y||^* = 1} \langle x,y\rangle = \sup_{||x||_2 = 1}||x|| = \frac{K_{||\cdot||}}{\sqrt 2}. $$ Denote $y_0 = \operatorname{argmax}_{||y||^* = 1} ||y||_2$ (it is attained thanks to compactness). Then, $$ \limsup_{u\to\infty} \frac{||W_u||}{\sqrt{u \ln \ln u}} = \limsup_{u\to\infty} \frac{\sup_{||y||^* = 1}\langle W_u, y\rangle}{\sqrt{u \ln \ln u}}\\ \ge \limsup_{u\to\infty} \frac{\langle W_u, y_0\rangle}{\sqrt{u \ln \ln u}} = \sqrt{2}||y_0||_2 = K_{||\cdot||} $$ almost surely.

To prove the opposite inequality, it is possible to use the result for Euclidean norm. I will still give the proof for completenss and since OP also asks about the displayed property (however, I won't prove the latter, but rather use a slightly weaker result which is enough).

For arbitrary $\epsilon\in (0,1)$, let $y_1,\dots, y_n$ be the $\epsilon$-net for the unit sphere in $||\cdot||^*$; in other words, for any $y\in \mathbb R^q$, there exists $i$ such that $||y-y_i||^*<\epsilon$. Then, for any $x\in \mathbb R^q$, $$ \langle x,y\rangle = \langle x , y_i\rangle + \langle x , y-y_i\rangle \le \langle x , y_i\rangle + \epsilon ||x||. $$ Taking supremum over $y$ with $||y||^* = 1$, we get $$ ||x|| \le (1-\epsilon)^{-1}\max_{i=1,\dots,n} \langle x , y_i\rangle. $$ As a result, $$ \limsup_{u\to\infty} \frac{||W_u||}{\sqrt{u \ln \ln u}} \le (1-\epsilon)^{-1} \limsup_{u\to\infty} \frac{\max_{i = 1,\dots,n}\langle W_u, y_i\rangle}{\sqrt{u \ln \ln u}}\\ = (1-\epsilon)^{-1} \max_{i = 1,\dots,n}\limsup_{u\to\infty} \frac{\langle W_u, y_i\rangle}{\sqrt{u \ln \ln u}} = (1-\epsilon)^{-1} \max_{i = 1,\dots,n}\sqrt{2}||y_i||_2 \le (1-\epsilon)^{-1} K_{||\cdot||} $$ almost surely. Letting $\epsilon\to 0$ finishes the proof.