I'm dealing with the formalism of an abstract Wiener space, and I'm not sure if two relevant topologies coincide.

Let $X$ be a topological vector space, and let $X^*$ be its dual space of continuous linear functionals. (if $X$ is not locally convex then $X^*$ is trivial, but that's fine for these purposes). Let $K : X^* → X$ be a covariance operator, that is, a non-negative-definite and symmetric operator.

It is not hard to see that $K$ generates an inner product on $X^*$, defined by $\langle \psi | \varphi \rangle_K := \psi[K\varphi]$. Let $H_K$ denote the Hilbert-space completion of $X^*$ with respect to this inner product. (that is, $H_K \cong \overline{X^*/\ker K}$. It is not hard to see that there exist maps $i^* : X^* \to H_K$ and $i : H_K \to X$ so that $K = i \circ i^*$. That is, the following diagram commutes: $$\begin{array}[ccccc] ~X^* && \longrightarrow && X \\ & \searrow && \nearrow & \\ && H_K && \end{array}$$

The image $A_K := iH_K$ is called the Cameron-Martin space corresponding to the covariance operator $K$. I refer to the space $C_K := \overline{KX^*} = \overline{A_K}$. as the "outer core" for the operator. This closed linear subspace is of great interest in probability theory. If $\mathbb P$ is any Radon probability measure on $X$ with covariance operator $K$, then $\operatorname{supp}(\mathbb P) \subseteq C_K$; if $\mathbb P$ is a Gaussian measure, then $\operatorname{supp}(\mathbb P) = C_K$.

Here are my questions:

is the space $C_K$ equipped with a norm, which extends the norm on the Cameron-Martin subspace $A_K$?

if so, is this norm complete? (i.e., is $C_K$ naturally a Banach space?)

does this Banach topology agree with the subspace topology on $C_K \subseteq X$?

Let me know if this doesn't make sense, and I'll be happy to expand on any point.

Gaussian Measuresrelevant? $\endgroup$ – Nate Eldredge Jun 26 '13 at 3:35Theorem 3.6.5.(p. 121) Let $\mu$ be a Radon measure on a Fréchet space $X$. There exists a linear subspace $E \subseteq X$ of full $\mu$-measure which is a reflexive, separable Banach space, and its unit ball is compact in the topology of $X$. $\endgroup$ – Tom LaGatta Jul 2 '13 at 6:28