Lemma 1.1.4 ([Lemma 8.1.1, Str24]).label Let $(\Omega, \cf, \bp)$ be a probability space and $(B_{t}, \bracs{\cf_t: t \ge 0})$ be a standard Brownian motion, then
- (1)
Almost every sample path of $B$ is in $\sps$.
- (2)
$\ev[\norm{B}_{\sps}^{2}] \le 32d$.
Proof. $(1)$: By the Brownian Strong Law of Large Numbers, $B(\omega) \in \sps$ for almost every $\omega$.
$(2)$: By Doob’s maximal inequality, for any $r \ge 0$,
\begin{align*}\ev\braks{\norm{B}_{u, [0, 2]}^2}^{1/2}&\le 2\ev\braks{\norm{B_r}_{\real^d}^2}^{1/2}\\ \ev\braks{\norm{B}_{u, [0, 2]}^2}&\le 4\ev\braks{\norm{B_r}_{\real^d}^2}\end{align*}
and for any $\theta \in \sps$,
\begin{align*}\norm{\theta}_{\sps}^{2}&\le \sum_{n = 0}^{\infty} \sup_{t \in [0, 2^n]}\frac{\abs{\theta(t)}^{2}}{(1 + t)^{2}}\le \sum_{n = 0}^{\infty} \frac{1}{2^{2(n-1)}}\norm{\theta}_{u, [0, 2^n]}^{2}\\&\le \sum_{n = 0}^{\infty} 4^{1 - n}\norm{\theta}_{u, [0, 2^n]}^{2}\end{align*}
Thus
\begin{align*}\ev[\norm{B}_{\sps}^{2}]&\le \sum_{n = 0}^{\infty} 4^{1 - n}\ev\braks{\norm{\theta}_{u, [0, 2^n]}^2}\le \sum_{n = 0}^{\infty} 4^{1 - n}\ev\braks{\norm{B_{2^n}}_{\real^d}^2}\le 32d\end{align*}
$\square$