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Let $\mathcal{E}(S^1)$ be the space of smooth functions on the circle $S^1$ and denote its dual as $\mathcal{E}'(S^1)$.

Then, by the Minlos theorem, there exists a unique probability measure $\mu$ on $\mathcal{E}'(S^1)$ such that \begin{equation} e^{-\frac{1}{2}\lVert f \rVert^2_{L^2}}=\int_{\mathcal{E}'(S^1)} e^{iT(f)}d\mu(T) \end{equation} for all $f \in \mathcal{E}(S^1)$.

Also, we know that $\mathcal{E}(S^1) \subset L^2(S^1) \subset \mathcal{E}'(S^1)$.

My question: is it true that $\mu\bigl(L^2(S^1) \bigr) \neq 0$? If so, how do I prove this?

Could anyone please help me?

Keith
  • 8,359
  • I don't know the answer. Did you try to use $\mathcal{E}'(S^1) = \cup_{n\in\mathbb{N}} \mathcal{H}^{-n}(S^1)?$ Each $\mathcal{H}^{-n}(S^1)$ is isometric to $L^2(S^1).$ – dsh Nov 02 '23 at 00:11

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