# A Gaussian measure $\mu$ on $\mathcal{E}'(S^1)$ by Minlos theorem and its value for Sobolev spaces $H^{\alpha}(S^1)$

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 $$$$e^{-\frac{1}{2}\lVert f \rVert^2_{L^2}}=\int_{\mathcal{E}'(S^1)} e^{iT(f)}d\mu(T)$$$$ for all $$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 $$\mu\bigl(\mathcal{E}(S^1) \bigr)$$?

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

• 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. Commented Nov 1, 2023 at 14:51
• @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. Commented Nov 1, 2023 at 14:58
• But, how is the mean equal to $1$ for $L^2$? I think the mean is equal to $1$ only for $E'(S^1)$.. Commented Nov 1, 2023 at 15:02
• @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, $P(X\in L^2)=P(A)=P(A\cap B)+P(A\setminus B)=0+0=0$. Commented Nov 1, 2023 at 15:10
• I have came across the following question: mathoverflow.net/questions/457725/… Commented Nov 4, 2023 at 12:05