I posted this question on ME as "A Gaussian measure on $\mathcal{E}'(S^1)$ by Minlos Theorem and its value for $L^2(S^1)$", but it seems much more nontrivial than I expected... so, I post an extended version here.

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 $\mathcal{E}(S^1) \subset H^{\alpha}(S^1) \subset \mathcal{E}'(S^1)$ for all $\alpha \in [0,\infty)$.

My question: is it true that $\mu\bigl(H^{\alpha}(S^1) \bigr) \neq 0$ for any $\alpha \in [0,\infty)$? Also, can we get an exact value of $\mu\bigl(\mathcal{E}(S^1) \bigr)$?

Could anyone please help me?


1 Answer 1


$\newcommand\al\alpha\newcommand\EE{\mathcal E}\newcommand\ip[2]{\langle #1,#2\rangle}$The answer is $$\mu(\EE(S^1))=\mu(H^\al(S^1))=0$$ for all real $\al\ge0$.

Indeed, since $\EE(S^1))\subseteq H^\al(S^1)\subseteq H^0(S^1)=L^2(S^1)=:L^2$ for all real $\al\ge0$, it is enough to show that $\mu(L^2)=0$.

Let $X$ be a random vector in $\EE'(S^1)$ with distribution $\mu$. Let $(e_1,e_2,\dots)$ be an orthonormal basis of $L^2$, with $e_n\in\EE(S^1)$ for each $n$. Then $X_1:=X(e_1),X_2:=X(e_2),\dots$ are independent standard normal random variables and hence on the event $X\in L^2$ we have $\|X\|_{L^2}^2=\sum_{n=1}^\infty X_n^2=\infty$ almost surely (a.s.) -- say, by the strong law of large numbers. So, $P(X\in L^2)=0$; that is, $\mu(L^2)=0$. $\quad\Box$

Details on the latter two sentences: By the strong law of large numbers, $$\sum_{n=1}^\infty X_n^2 =\lim_{N\to\infty}N\frac1N\,\sum_{n=1}^N X_n^2 \\ =\lim_{N\to\infty}N \; \lim_{N\to\infty}\frac1N\,\sum_{n=1}^N X_n^2 =\lim_{N\to\infty}N \times 1=\infty\text{ a.s.}$$

Introducing now the events $A:=\{X\in L^2\}$ and $B:=\{\sum_{n=1}^\infty X_n^2=\infty\}$, we see that $A\cap B=\emptyset$ and $P(B)=1$. So, $$P(X\in L^2)=P(A) \\ =P(A\cap B)+P(A\setminus B)=0+0=0.$$

  • $\begingroup$ I do not understand. Why is the norm of $X$ necessarily infinite a.s.? If $X \in L^2$, then the norm must be finite. $\endgroup$
    – Isaac
    Commented Nov 1, 2023 at 14:51
  • $\begingroup$ @Isaac : As I said, by the strong law of large numbers: $\sum_{n=1}^\infty X_n^2=\lim_{N\to\infty}N\frac1N\,\sum_{n=1}^N X_n^2=\lim_{N\to\infty}N \lim_{N\to\infty}\frac1N\,\sum_{n=1}^N X_n^2=\lim_{N\to\infty}N \times 1=\infty$ almost surely. $\endgroup$ Commented Nov 1, 2023 at 14:58
  • $\begingroup$ But, how is the mean equal to $1$ for $L^2$? I think the mean is equal to $1$ only for $E'(S^1)$.. $\endgroup$
    – Isaac
    Commented Nov 1, 2023 at 15:02
  • 2
    $\begingroup$ @Isaac : Consider the events $A:=\{X\in L^2\}$ and $B:=\{\sum_{n=1}^\infty X_n^2=\infty\}$. Then $A\cap B=\emptyset$ and $P(B)=1$. So, $P(X\in L^2)=P(A)=P(A\cap B)+P(A\setminus B)=0+0=0$. $\endgroup$ Commented Nov 1, 2023 at 15:10
  • $\begingroup$ I have came across the following question: mathoverflow.net/questions/457725/… $\endgroup$
    – Isaac
    Commented Nov 4, 2023 at 12:05

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