Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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: (the statement is on page 88)

share|cite|improve this question
What is the topology on $C[0,1]$ (sup norm I'm guessing)? Is there a particular reference this is from? – Jonas Meyer Jan 25 '12 at 23:28
From a supremum norm. I will edit for the source. – Tom Artiom Fiodorov Jan 25 '12 at 23:31
I obviously need to show that $\mu(\overline{H}_\mu)=1$. – Tom Artiom Fiodorov Jan 25 '12 at 23:51
I am suspecting Theorem 3.3 (page 92) is the answer, just need to check the logic is not circular. – Tom Artiom Fiodorov Jan 26 '12 at 0:06
up vote 4 down vote accepted

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.

share|cite|improve this answer

Your Answer


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.