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:


1. is the space $C\_K$ equipped with a norm, which extends the norm on the Cameron-Martin subspace $A\_K$?
2. if so, is this norm complete? (i.e., is $C\_K$ naturally a Banach space?)
3. 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.



Agreement of two topologies on a linear space
What do you think about this rather trivial example: Let $X=c\_0$ be the Banach space of
(real) null sequences, $X^\*=\ell\_1$ its dual (where $y=(y\_n)\_{n} \in \ell\_1$ is considered as the functional $(x\_n)\_n \mapsto \sum\_n x\_n y\_n$) and $K:X^\* \to X$
 the inclusion. Then $A\_K=\ell\_2$ is dense in $c\_0$ so that $C\_K=X$.
In this situation 3. is certainly false. 


In general, if $(X,\mathscr T)$ is a Frechet space (by the way, don't you need some completeness of $X$ in order to have $i: H\_K \to X$ well defined?), the closed graph theorem is an essential obstacle against your wishes: The Banach topology on $C\_K$ is either equal to the Frechet space topology (so that $A\_K$ would be closed in $X$ which hardly happens) or it must be so strange that you can not apply the closed graph theorem (in particular, there is no Hausdorff topology on $X$ which is coarser than the Frechet topology of $X$ and the Banach topology of $C\_K$). I doubt that such a strange topology on $C\_K$ would be of any use.


