Theorem 1.3.3: {{\cite[Theorem 8.3.2]{stroock}}} | Brownian Motion and Stochastic Calculus

Theorem 1.3.3 ([Theorem 8.3.2, Str24]).label Let $E$ be a separable Banach space over $\real$, and $\wien: \cb(E) \to [0, 1]$ be a Borel probability measure, then the following are equivalent:

(1)
$\wien$ is a non-degenerate centred Gaussian measure.
(2)
There exists a Hilbert space $(H, \inp_{H})$ contained in $E$ such that $(H, E, \wien)$ is an abstract Wiener space.

If $(G, \inp_{G})$ is another Hilbert space in $E$ such that $(G, E, \wien)$ is an abstract Wiener space, then $G = H$ and $\inp_{G} = \inp_{H}$.

Proof of equivalence.$(1) \Rightarrow (2)$: The requirement on the characteristic function of $\wien$ defines the desired inner product for us, thus the work lies in finding an appropriate subspace and completing it as a Hilbert space. To this end, observe that since each $x^{*} \in E^{*}$ viewed as a random variable is Gaussian, the inner product can be taken in $L^{2}(\wien; \real)$, where we can take the completion, and then recover an element of $E$ by integrating against the identity.

We now begin the construction. By Fernique’s theorem, $\int_{E} \norm{x}_{E}^{2} d\wien(x) < \infty$. Thus there exists a bounded linear map $R: E^{*} \to L^{2}(\wien; \real)$ defined by identifying each $x^{*} \in E^{*}$ as a $L^{2}$-random variable on $E$. In addition, since $\text{Id}_{E}$ itself is an $E$-valued random variable, there exists a bounded linear map

\[I: L^{2}(\wien; \real) \to E \quad Y \mapsto \int_{E} Y(x)xd\wien(x)\]

then for any $x^{*} \in E^{*}$ and $Y \in L^{2}(\wien; \real)$,

\[\angles{IY, x^*}_{E} = \int_{E} Y(x)\angles{x, x^*}d\wien(x) = \angles{Y, Rx^*}_{L^2(\wien; \real)}\]

In particular, $IY = 0$ if and only if $Y \perp R(E^{*})$, and $I|_{\overline{R(E^*)}}$ is injective.

Let $H = I(\overline{R(E^*)})$ and define $\angles{x, y}_{H} = \anglesn{I^{-1}x, I^{-1}y}_{L^2(\wien; \real)}$. Since $\wien$ is non-degenerate, for any $x^{*} \in E^{*}$, $\angles{IRx^*, x^*}_{E} = \angles{Rx^*, Rx^*}_{L^2(\wien; \real)}> 0$. Thus $H$ is dense in $E$ with $IR = \iota^{*}: E^{*} \to H$ corresponding to the adjoint of the inclusion.

Lastly, since $\wien$ is a centred Gaussian measure, for any $x^{*} \in E^{*}$, $\wh \wien (x^{*}) = e^{-\norm{Rx^*}_{L^2(\wien; \real)}^2/2}= e^{-\normn{\iota^*x^*}_{H}^2/2}$.

$(2) \Rightarrow (1)$: Since $\wh \wien(x^{*}) = e^{-\normn{\iota^*x^*}_H^2/2}$ and $\iota^{*}$ is injective, $x^{*}$ is a non-degenerate centred Gaussian. Thus $\wien$ is non-degenerate.$\square$

Proof of uniqueness.Let $R: E^{*} \to L^{2}(\wien; \real)$ and $(H, \inp_{H})$ be as in the proof of $(1) \Rightarrow (2)$, and $\iota^{*}_{G}: E^{*} \to G$ be the adjoint of the inclusion $\iota_{G}: G \to E$. Since $\wh \wien(x^{*}) = e^{-\norm{\iota_G^*x^*}_G^2/2}$ for all $x^{*} \in E^{*}$,

\[\angles{\iota_G^*x^*, \iota_G^*x^*}_{G} = \angles{Rx^*, Rx^*}_{L^2(\wien; \real)}\]

and by polarisation, for any $y^{*} \in E^{*}$,

\[\angles{\iota_G^*x^*, y^*}_{E} = \angles{\iota_G^*x^*, \iota_G^*y^*}_{G} = \angles{Rx^*, Ry^*}_{L^2(\wien; \real)}= \angles{IRx^*, y^*}_{E}\]

Therefore $\iota_{G}^{*}x^{*} = IRx^{*}$ for all $x^{*} \in E^{*}$, and the following diagram commutes

\[\begin{CD}E^* @>R>> \overline{R(E^*)} \\ @A{(\iota_G^*)^{-1}}AA @VV{I}V \\ \iota_G^*(E^*) @>{\iota}>> H\end{CD}\]

By density of $\iota^{*}(E^{*})$, the composition $IR(\iota_{G}^{*})^{-1}: \iota_{G}^{*}(E^{*}) \to H$ admits a unique extension to $G$ as an isometry. Therefore $G = H$ with the same inner product.$\square$