11
$\begingroup$

Let $E=C[0,1]$, space of all real-valued continuous functions on $[0,1]$, $\mathcal{E}$ be its Borel $\sigma$-algebra and $\mu$ a Gaussian measure on $E$. Let $E^*$ be a space of all continuous linear functions on $E$. Define map $R$ on $E^*$ by $$x^* \mapsto R(x^*)=\int_E\langle x^*,x\rangle x\;\mu(dx)=\int_E x^*(x)\; x\;\mu(dx)$$ And let $H_\mu$ be the completion of $R(E^*)$ with respect a norm induced by an inner product defined as $\langle Rx^*,Ry^* \rangle=\int_Ex^*(x)y^*(y)\;\mu(dx)$.

$H_\mu$ stands for Reproducing Kernel Hilbert Space and it is dense$^1$ in $E$ if topological support$^2$ of $\mu$ is the whole space $E$. Why?

I think I understand the construction well enough, but the statement is somewhat unexpected.

$^1$ $i(H_\mu)$ to be precise, $i$ for inclusion from $H_\mu$ to $E$.

$^2$ topological support is the smallest closed set $F$ such that $\mu(F) = 1$.

Edit This is page 84 of Interest Rate Models: an Infinite Dimensional Stochastic Analysis Perspective. Read online on Springer: Link (the statement is on page 88)

$\endgroup$
4
  • $\begingroup$ What is the topology on $C[0,1]$ (sup norm I'm guessing)? Is there a particular reference this is from? $\endgroup$ Commented Jan 25, 2012 at 23:28
  • $\begingroup$ From a supremum norm. I will edit for the source. $\endgroup$ Commented Jan 25, 2012 at 23:31
  • $\begingroup$ I obviously need to show that $\mu(\overline{H}_\mu)=1$. $\endgroup$ Commented Jan 25, 2012 at 23:51
  • $\begingroup$ I am suspecting Theorem 3.3 (page 92) is the answer, just need to check the logic is not circular. $\endgroup$ Commented Jan 26, 2012 at 0:06

1 Answer 1

6
$\begingroup$

Let's prove the contrapositive.

First, you should check that $R : E^* \to H_\mu$ is the adjoint of $i : H_\mu \to E$. That is, for $h \in H_\mu$ and $x^* \in E^*$, $\langle R x^*, h \rangle = x^*(i(h))$.

Now suppose $i(H_\mu)$ is not dense in $E$. Then by the Hahn-Banach theorem there exists a nonzero $x^* \in E^*$ with $x^*(i(h)) = 0$ for all $h \in H_\mu$. Taking $h = R x^*$, we have that $0 = x^*(i(Rx^*)) = \langle R x^*, R x^* \rangle$. That is, $\int_E |x^*(x)|^2 \mu(dx) = 0$, so as a function on $E$, $x^*$ vanishes $\mu$-a.e. Hence the kernel of $x^*$ is a proper closed subset of $E$ with measure 1, so $\mu$ does not have full support.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .