# Reproducing Kernel Hilbert Space is dense?

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)

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

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.