1
$\begingroup$

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?

$\endgroup$

1 Answer 1

4
$\begingroup$

$\newcommand\al\alpha\newcommand\EE{\mathcal E}\newcommand\ip[2]{\langle #1,#2\rangle}$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, $$P(X\in L^2)=P(A) \\ =P(A\cap B)+P(A\setminus B)=0+0=0.$$

$\endgroup$
6
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.