We had the following definition of a Gaussian measure on a Banach space: A Gaussian probabilty measure $\mu$ on a Banach space $B$ is a Borel measure such that $\ell^*\mu$ is a real Gaussian probability measure on $\mathbb R$ for every linear functional $\ell: B\rightarrow \mathbb R$. Where we used the push-forward of a measure $(f^*\mu)(A)=\mu(f^{-1}(A))$.
Now, let $\{\xi_n\}$ be a sequence of i.i.d. $N(0,1)$ random variables and let $\{a_n\}$ be a sequence of real numbers. Show that the law of $(a_0\xi_0,a_1\xi_1,...)$ determines a Gaussian measure on $\ell^2$ if and only if $\sum_{n\geq0}a_n^2<\infty$.
I guess, here $\ell^2$ is the space of square summable sequences.
I don't really know how to prove this. The law of $(a_0\xi_0,a_1\xi_1,...)$ would be something like $\mathbb P(\{(c_n)_n:(a_0\xi_0(c_0),a_1\xi_1(c_1),....)\in A\})$ right? As a linear functional that appears in the definition one might take $\ell((c_n)_n)=\sum c_n^2$. However, it must be for every linear functional. I also don't see the other direction. Can somebody help me with this?
This is exercise 3.5 from Hairer's notes https://www.hairer.org/notes/SPDEs.pdf.
I've wondered about this problem too. Well, this might help. My approach is like this: if $(a_0\xi_0, a_1\xi_1, ...)$ is the (probability) law on $\ell^2$, then it should be the case that the series $$\sum_{n=1}^\infty a_n^2\xi_n^2$$ converge almost surely.
First, intuitively, $\lim \sup |a_n| = 0$. To see this, assume otherwise, and we might obtain $$\sum_{n=1}^\infty a_n^2\xi_n^2 \geq \displaystyle C\sum_{k=1}^\infty \xi_{n_k}^2$$ for some constant $C>0$ and subindex $(n_k)$. By Kolmogorov's three series law using the property of convergent series of means (with some technicalities because the means are truncated), the right-hand side cannot converge a.s., making it impossible for the left-hand side to converge a.s.
Then, we use this $\lim \sup$ of $|a_n|$ and property of convergent series of means again to obtain the convergence of $\displaystyle \sum_{n=1}^\infty a_n^2$.