Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
You should be able to do some abstract nonsense construction with the axiom of choice to achieve 1 and probably 2. But I doubt this is what you want, so maybe you want to add some other conditions. By any chance is the discussion of "linear supports" in section 3.6 of Bogachev's Gaussian Measures relevant? –  Nate Eldredge Jun 26 '13 at 3:35
    
@NateEldredge, actually the abstract nonsense construction is what I was looking for, except that Jochen's counterexample below means that it won't agree with the subspace topology, so it doesn't really matter. –  Tom LaGatta Jul 2 '13 at 6:22
    
@NateEldredge, thanks for the tip of Bogachev's book, it's very relevant. In particular, here is a nice theorem which I had been looking for. Theorem 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 $X$. –  Tom LaGatta Jul 2 '13 at 6:28
add comment

1 Answer

up vote 4 down vote accepted

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.

share|improve this answer
    
thanks for the nice counterexample. You're probably right about the completeness comment. Thanks for spotting it. –  Tom LaGatta Jul 2 '13 at 6:19
    
do you think that being a Cauchy space would suffice for well-definedness of the inclusion map? nlab.mathforge.org/nlab/show/Cauchy+space –  Tom LaGatta Jul 2 '13 at 6:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.