The completion of $R(E^*)$ Let $E$ be a real separable Banach space and $\mu$ be a Gaussian measure and define the operator $R$ on $E^*$ by $$R(x^*)=\int_E x^*(x)x\mu (dx)$$
$R:E^*\to E$ is clearly a bounded linear operator and let $H$ be the completion of $R(E^*)$ for the inner product $$<Rx^*,Ry^*>=\int_E x^*(x)y^*(y)\mu (dx)$$
The goal is to show that $H$ is a subset of $E$
It can be easily shown that if $\{x_n^*\}$ is a Cauchy sequence in $E^*$ then $\{Rx_n^*\}$ is a Cauchy sequence for the norm given by the above inner product. I have few questions I need answers for:
Is the inclusion $H\hookrightarrow E$ continuous? If so, why is this important to show that $\{Rx_n^*\}$ is a Cauchy sequence for the norm in $E$?
Any help is appreciated!
 A: For $x = Rx^*, y = Ry^* \in R(E^*)$, I will write $\langle x,y \rangle_\mu = \int_{B} x^*(z) y^*(z) \mu(dz)$ for the inner product on $R(E^*)$ and $\|\cdot\|_\mu$ for its associated norm. 
Unless I miss something, noting that Cauchy sequences in $E^*$ get mapped to Cauchy sequences in $R(E^*)$ by $R$ is the wrong thing to check to see that we can realise $H$ as a subspace of $E$ such that the natural embedding $H \hookrightarrow E$ is continuous.
Let me give an explicit construction of $H$ as a subset of $E$. 
First notice that by Fernique's theorem, $E^* \hookrightarrow L^2(E,\mu)$ and that $\|\cdot\|_\mu$ is nothing but the unique norm such that $R:(E^*, \|\cdot\|_{L^2(E,\mu)}) \to R(E^*)$ is an isometry. This means that the reason that $R(E^*)$ isn't complete is that $E^*$ isn't closed in $L^2(E,\mu)$, so let's just fix that.
Let $E_\mu^*$ be the closure of $E^*$ in $L^2(E,\mu)$. We can extend $R$ to a map $R':E_\mu^* \to E$ by the same formula (the integral for $x^* \in L^2(E,\mu)$ is absolutely convergent by Cauchy-Schwarz and Fernique's theorem). Let $H = \operatorname{Im}R'$ and equip $H$ with the unique norm such that $R'$ is an isometry. It is immediate from the construction that $H$ is a Hilbert space containing $R(E^*)$ as a dense subset and hence is the completion. 
Finally, to see continuity of the embedding $H \hookrightarrow E$, note that for $h = R'h^* \in H$,
\begin{align}
\|h\|_E^2 = \sup_{x^* \in E^* \setminus \{0\}} \frac{x^*(h)^2}{\|x^*\|_{E^*}^2} &= \sup_{x^* \in E^* \setminus \{0\}} \frac{1}{\|x^*\|_{E^*}^2} \bigg( \int_E x^*(z) h^*(z) \mu(dz) \bigg)^2
\\& \leq \sup_{x^* \in E^* \setminus \{0\}} \frac{1}{\|x^*\|_{E^*}^2} \bigg( \int_E x^*(z)^2 \mu(dz) \bigg)\bigg( \int_E h^*(z)^2 \mu(dz) \bigg)
\\& \leq C \|h^*\|_{L^2(E,\mu)}^2 = C \|h\|_\mu^2
\end{align}
where $C$ is the second moment of $\mu$ (which is finite by Fernique's theorem since $\mu$ is Gaussian).
