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Let $W$ be a separable infinite-dimensional Banach space and let $W^*$ denote its continuous dual. Suppose that $\mu$ is a Borel measure on $W$ such that $W^*\subseteq L^2(W,\mu)$. Let $K$ denote the closure of $W^*$ in $L^2(W,\mu)$. What is an example of an element of $K$ that is not in $W^*$? I am especially interested in the special case $W=C([0,1])$ when $\mu$ is Wiener measure.


Some context: Elements of $L^2(W,\mu)$ are square-integrable random variables on $W$, and elements of $W^*$ are random variables represented by bounded linear functionals. In the special case above, elements of $W^*$ are centered Gaussian random variables. $K$ is the dual of the Cameron-Martin space. I believe it consists of all centered Gaussian random variables in $L^2(W,\mu)$, but I am not sure how to construct a Gaussian random variable on $W$ that is not given by a bounded linear functional.

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    $\begingroup$ Sorry, there seems to be a more basic misunderstanding here. Any element of $K$ will be representable by some (a.s. defined) linear functional, just not a bounded one. The point is that any $L^2$ limit of bounded linear functionals is itself an (a.s. defined) linear functional, just not a necessarily bounded one. This is because convergence in $L^2$ implies convergence a.s. along a subsequence. $\endgroup$ – Shalop Mar 19 '17 at 3:14
  • $\begingroup$ Thanks for pointing this out, this answers my question. To prevent further confusion, I also added the worded "bounded" in the final sentence of the "some context" paragraph in my question above. $\endgroup$ – pre-kidney Mar 19 '17 at 3:21
  • $\begingroup$ Okay, but there is an explicit answer to your question. Elaborating on my first (deleted) comment, the functionals in $K$ which are not in $W^*$ are precisely those of the form $\phi(g) = \int_0^1 g(x)f''(dx)$ where $f \in H_0^1[0,1]$ and $f''$ exists as a distribution which is not a finite (signed) measure. $\endgroup$ – Shalop Mar 19 '17 at 4:36

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