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Jan 26, 2023 at 20:42 comment added Drew Brady I am not sure, I fully follow what you have in mind. When/if you have a moment, if you can sketch the details, I would be greatly appreciative.
Jan 26, 2023 at 19:38 comment added Martin Hairer @DrewBrady Just stick a suitable weight on $\ell^2$ and you get a perfectly nice separable Hilbert space which is a measurable subset of full measure of $\mathbb{R}^{\mathbb{N}}$...
Jan 26, 2023 at 16:29 comment added Drew Brady @MartinHairer, I see that Prop 3.46 is related, but it is stated for a Gaussian measure on a separable Banach space. Are you saying the same argument applies if we replace Banach by Fréchet as in the case of $\mathbb{R}^\mathbb{N}$?
Jan 26, 2023 at 15:53 comment added Christophe Leuridan Since the vector $x$ is assumed to be Gaussian and centered, the components $x_i$ must be orthonormal therefore i.i.d. if you want the map $f \mapsto \sum_i f_ix_i$ from $\ell^2$ to $L^2(P)$ to be an isometry.
Jan 26, 2023 at 7:32 comment added Martin Hairer A precise statement is that one can find a Hilbert space $H$ containing $\ell^2$ such that $T$ can be extended (uniquely up to measure zero modifications) to a linear map from a measurable linear subspace of $\mathbb{R}^{\mathbb{N}}$ to $H$. In particular, the law of $Z$ is a Gaussian measure on $H$. See for example Prop. 3.46 and Thm 3.47 in my lecture notes on SPDEs at hairer.org/notes/SPDEs.pdf.
Jan 26, 2023 at 3:25 comment added Drew Brady @MattF., I tried to find a proof of this claim, and I couldn't locate it. Do you know of one?
Jan 26, 2023 at 2:39 comment added user44143 I wouldn’t expect any additional assumptions to be needed
Jan 26, 2023 at 0:10 history edited Drew Brady CC BY-SA 4.0
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Jan 26, 2023 at 0:08 comment added Drew Brady Reference: Bogachev, Vladimir I. Gaussian measures. Mathematical Surveys and Monographs, 62. American Mathematical Society, Providence, RI, 1998. xii+433 pp. ISBN: 0-8218-1054-5
Jan 26, 2023 at 0:04 history asked Drew Brady CC BY-SA 4.0