Theorem 1.5.2.label Let $(H, E, \wien)$ be an abstract Wiener space and $\seq{h_n}$ be an orthonormal sequence, then
- (1)
For any $p \in [1, \infty)$,
\[\ev^{\wien}\braks{\sup_{n \in \nat}\norm{\sum_{k = 1}^n \ci(h_k)h_k}_E^p}< \infty\] - (2)
For $\wien$-almost every $x \in E$,
\[\sum_{n = 1}^{\infty} \ci(h_{k})(x)h_{k} = \ev^{\wien}\braks{x|\sigma(\bracs{\ci(h_n): n \in \nat})}\] - (3)
$\sum_{n = 1}^{\infty} \ci(h_{n})h_{n}$ is $\wien$-independent of $x - \sum_{n = 1}^{\infty} \ci(h_{n})h_{m}$.
Proof. Let $n \in \nat$, $\cf_{n} = \sigma(\bracs{\ci(h_k): 1 \le k \le n})$, and $S_{n} = \sum_{k = 1}^{n} \ci(h_{k})h_{k}$, then for any $x^{*} \in E^{*}$, $\dpb{x - S_n(x), x^*}{E}\perp h_{k}$ in $L^{2}(\wien; E)$ for all $k \le n$. Since $\bracs{\ci(h_n): n \in \nat}$ is a Gaussian family, $x - S_{n}(x)$ is independent of $\seqf{h_k}$ and $\cf_{n}$.
(1): By Theorem 1.2.3, $x \in L^{p}(\wien; E)$ for all $p \in [1, \infty)$, so (1) holds.
(2): $S_{n} = \ev^{\wien}[x|\cf_{n}]$, and $S_{n} \to \ev^{\wien}[x|\sigma(\bracs{\ci(h_n): n \in \nat})]$ $\wien$-almost surely.
(3): Since $x - S_{n}(x)$ is independent of $\seq{h_k}$ for each $n \in \nat$, $x - S$ is independent of $S$.$\square$