1.5 Wiener’s Construction

Theorem 1.5.1 ([Theorem 8.3.1, Str24]).label Let $H$ be an infinite-dimensional, separable Hilbert space over $\real$, $E$ be a Banach space with $H$ continuously embedded as a dense subspace.

Let $\seq{h_n}\subset H$ be an orthonormal basis and $\seq{X_n}$ be mutually independent $\gamma_{0, 1}$-distributed random variables on a probability space $(\Omega, \cf, \bp)$. If

  1. (1)

    $\sum_{n = 1}^{\infty} X_{n} h_{n}$ converges in $E$ almost surely.

  2. (2)

    $S: \Omega \to E$ is given by

    \[S(\omega) = \begin{cases}\sum_{n = 1}^{\infty} X_{n}(\omega) h_{n}&\text{The series converges in }$E$ \\ 0&\text{Otherwise}\end{cases}\]

  3. (3)

    $\wien = S_{*} \bp$.

then $(H, E, \wien)$ is an abstract Wiener space.

Proof. Let $S_{n} = \sum_{k = 1}^{n} X_{k}h_{k}$, then

\[\wh \wien(x^{*}) = \limv{n}\ev^{\bp}\braks{e^{\dpb{S_n, x^*}{E}}}= \limv_{n}\prod_{k = 1}^{n} e^{-\dpb{\iota^*x^*, h_k}{H}^2/2}= e^{-\norm{\iota^*x^*}_H^2/2}\]

$\square$

Theorem 1.5.2.label Let $(H, E, \wien)$ be an abstract Wiener space and $\seq{h_n}$ be an orthonormal sequence, then

  1. (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. (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. (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 $S$.$\square$