1.3 Abstract Wiener Spaces

Proof. $(1)$: Since $\iota: H \to E$ is continuous, $\iota^{*}: E^{*} \to H$ is continuous with respect to the strong and weak* topologies. As $H \subset E$ is dense, $\iota^{*}: E^{*} \to H^{*}$ is injective.

Let $h \in H$ such that $h \perp \iota^{*}(L)$, then by the weak*-density of $L \subset E^{*}$, $\angles{h, x^*}_{E} = 0$ for all $x^{*} \in E$. Therefore $h = 0$, and $\iota^{*}(L)$ is dense in $H$.

$(2)$: By $(1)$, the image $\iota^{*}(E^{*}) \subset H$ is dense. Hence if $x \in H$, then

On the other hand, if $\sup_{x^* \in E^*, \norm{\iota^*x^*}_H = 1}\angles{x, x^*}_{E} < \infty$ then the map $\iota^{*}(E^{*}) \to \real$ with $\iota^{*}x^{*} \mapsto \angles{x, x^*}_{E}$ is well-defined as $\iota^{*}$ is injective, and continuous by assumption. By density of $\iota^{*}(E^{*}) \subset H$, there exists $h \in H$ such that $\anglesn{h, \iota^*x^*}_{H} = \angles{h, x^*}_{E}= \angles{x, x^*}_{E}$ for all $x \in E^{*}$. In which case, $x = h \in H$.

$(3)$: Since $E^{*}$ is separable with respect to the weak* topology, there exists a linearly independent set $\seq{x_n^*}\subset L$ whose span is dense in $E$. By $(1)$, $\seq{\iota^*x_n^*}\subset H$ has dense span in $H$. Using the Gram-Schmidt process, there exists an orthonormal basis $\seq{h_n}\subset \text{span}\bracs{\iota^*x_n^*: n \in \nat}$ for $H$. Since $\iota^{*}$ is injective and $L$ is a subspace, $\bracsn{(\iota^*)^{-1}(h_n)}_{1}^{\infty} \subset L$ is the desired sequence.$\square$

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

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

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 $x^{*} \in E^{*}$,

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

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

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.$\square$