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On Zabczyk & Da Prato book about infinite dimensional SDEs they introduce the idea of Gaussian measures in infinite dimensional Banach spaces. They do so by means of Fernique theorem. In the White Noise framework, we also deal with Gaussian measures defined on infinite dimensional spaces, in the more classical case we work with the space of tempered distributions $S'(\mathbb R)$, and they are introduced by means of the Bochner-Minlos theorem.

My question is, can we "exchange" those approaches, namely can we use Bochner-Minlos (or some modification) to define a Gaussian measure on Banach space? and is there another way to introduce Gaussian measures on white noise spaces without using B-M. ?

Any idea, suggestion or comment will be greatly appreciated.

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    $\begingroup$ What is your reason for not wanting to work with $S'(\mathbb{R}^n)$? If it is because topological vector spaces scare you look up arxiv.org/abs/1706.09326 You can construct the measure on the bigger space $S'$ and then pick your favorite function space e.g. (weighted) Sobolev or Besov and then prove the measure gives that subspace (Borel set) of $S'$ probability one (for white noise you need the exponent $s$ to be less than $-d/2$). This will give you a measure on that Banach space. See Prop. 3.4 of arxiv.org/abs/1706.09326 Also useful is... $\endgroup$
    – Abdelmalek Abdesselam
    Commented Feb 2, 2021 at 23:47
  • $\begingroup$ ...the reference arxiv.org/abs/1502.07335 Finally, if you want an introduction purely with Banach spaces see arxiv.org/abs/1607.03591 $\endgroup$
    – Abdelmalek Abdesselam
    Commented Feb 2, 2021 at 23:48
  • $\begingroup$ In my first comment I mean Prop 3.4 of arxiv.org/abs/1801.09245 instead. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Feb 2, 2021 at 23:56
  • $\begingroup$ @AbdelmalekAbdesselam Thank you so much for the references! Actually I am more used to work on $S'(\mathbb R)$, this was just a curiosity that I had while reading Da Prato's book. I wanted to know if both approaches were somehow equivalent. Thanks again! $\endgroup$
    – Chaos
    Commented Feb 3, 2021 at 9:56

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