Let $H$ be a Gaussian Hilbert space defined on a probability space $(\Omega,\mathcal{F},P)$. Set $$P_n(H)=\{p(\xi_1,\ldots,\xi_m): p\;\text{polynomial of degree at most }\; n\text{and}\;\xi_1,\ldots,\xi_m\in H\}$$ and $\overline{P_n}(H)$ the closure of $P_n(H)$ in $L^2(\Omega)$.
if $H$ is finite dimensional then $P_n(H)$ is closed but if $H$ is not finite dimensional it is not.
I would like to prove that, for exemple, $$\sum_{i=1}^\infty\frac{\xi²_i}{2^i}$$ is in $\overline{P_2}(H)$ where $(\xi_i)$ is orthonormal sequence.
It clear that $\sum_{i=1}^N\frac{\xi²_i}{2^i}\in P_2(H)$ and I can compute $$E\left(\sum_{i=1}^N\frac{\xi²_i}{2^i}\right)^2=\left(\sum_{i=1}^N\frac{E(\xi²_i)}{2^i}\right)^2$$ where I used independance.
But I don't see why it converge.
- I didn't use yet that $\xi_i$ are gaussian neither the hypothesis that the sequence is a basis. So it must play a role here.
First observe that $\sum_{i=1}^{+\infty}2^{-i}\xi_i^2$ is defined as the limit of the Cauchy sequence $\left(X_n\right)_{n\geqslant 1}$ defined by $X_n=\sum_{i=1}^{n}2^{-i}\xi_i^2$. This sequence is Cauchy because for $n\geqslant m$, $$ 0\leqslant \mathbb E\left[X_n-X_m\right]=\sum_{i=m+1}^n 2^{-i} \mathbb E\left[\xi_i^2\right]= \sum_{i=m+1}^n 2^{-i}\leqslant 2^{-m}, $$ where we used in the last equality the definition of a Gaussian Hilbert space $\mathbb E\left[\xi_i\xi_j\right]=\langle e_i,e_j\rangle$ with $i=j$, where $\left(e_i\right)_{i\geqslant 1}$ is a Hilbert basis of $H$.