4
$\begingroup$

Consider the space $\mathcal D'(M)$ of distributions on a manifold $M$.

Is there a ready reference for the fact that the Borel $\sigma$-algebra (for the strong dual topology) coincides with the weak cylindrical $\sigma$-algebra (generated by the $\{f:\langle f,\phi\rangle\in U\}$ for $U\subset\mathbb R$ open and $\phi\in\mathcal D(M)$)?

My opinion is that the Borel $\sigma$-algebra coming from the strong dual topology on $\mathcal D'(M)$ is the biggest reasonable one, while the weak cylindrical one is the smallest. This would serve, at least to me, as a strong argument for the fact that there is only one reasonable $\sigma$-agebra on $\mathcal D’(M)$.

Definitions. Here $M$ is a manifold with or without boundary, compact or not. $\mathcal D(M)=\mathcal C^\infty_c(M)$ is the space of smooth compactly supported functions on $M$, with the usual topology, i.e. the finest locally convex topology that makes every inclusion $\mathcal C^\infty(K)\to\mathcal D(M)$ of the Fréchet space of smooth functions with support in a compact set $K\subset M$ continuous. Its topological dual $\mathcal D'(M)$ is endowed with the strong dual topology. I am willing to take it as folklore that those two spaces are complete, nuclear, Montel, bornological and separable.

For the purposes of this question, if we are given a subspace (not necessarily closed) $V_0\subset E'$ of the topological dual of a topological vector space $E$, then the corresponding cylindrical $\sigma$-algebra on $E$ is the one generated by $\{x:u(x)\in U\}$, where $U\subset\mathbb R$ is open and $u\in V_0$. The strong cylindrical $\sigma$-algebra is the one corresponding to all continuous forms, i.e. $V_0=E'$, while in the case where $E$ is already the topological dual of some $V$, the weak cylindrical $\sigma$-algebra corresponds to $V_0=V$. They coincide when $V$ is reflexive, which is the case for us but I am interested in adjacent questions where it is less clear. I believe those are Borel $\sigma$-algebras, generated respectively by the weak and weak* topologies $\sigma(E,E')$ and $\sigma(V',V)$, in most cases of interest; it is obvious that the Borel $\sigma$-algebras contain the cylindrical ones, but for the reverse direction I am not 100% certain the arbitrary unions of the topology are not too much for the countable unions of the $\sigma$-algebras.

Related questions. I am also interested in partial results, for instance when $M$ is an open subset of $\mathbb R^d$ or a closed half-space; I believe any proof will generalize immediately to my case. I think the fact that the Borel $\sigma$-algebra for the weak dual topology coincides with the weak cylindrical $\sigma$-algebra is Proposition 2.1 in [DM], although I cannot read Russian so I am not entirely sure. I am also interested in the case of the Sobolev spaces $H^s(M)$ when $M$ is compact (with or without boundary) and $s\in\mathbb R$, comparing the cylindrical $\sigma$-algebra with respect to the inclusion $\mathcal D(M)\to H^s(M)'$ with the Borel $\sigma$-algebra for the Hilbert structure.

I am assuming that a statement of the form “if $V_0\subset E'$ satisfy this and that property, then the cylindrical $\sigma$-algebra of $E$ with respect to $V_0$ and the Borel $\sigma$-algebra of the strong dual topology on $V'$ coincide”; I would be most interested in this result.

Proof by hand. Fixing a countable dense subset $S\subset\mathcal D(M)$, define the topology $\tau_S$ on $\mathcal D'(M)$ inherited from the identification $\mathcal D'(M)\to\mathbb R^S$ with a subspace of the Polish space $\mathbb R^S$. Clearly the Borel $\sigma$-algebra for $\tau_S$ is contained in the weak cylindrical $\sigma$-algebra, which is contained in the Borel $\sigma$-algebra for the strong dual topology. If we show that the Borel $\sigma$-algebras of $\tau_S$ and of the strong dual topology coincide, then this will conclude. In [S], we learn there that a Hausdorff topological space is Suslin if it is the continuous image of a Polish space (some variant of an analytic set, I suppose) [S Part I/Chap II/Definition 3], and that if two Suslin topologies on the same base space are comparable, then they generate the same $\sigma$-algebra [S Part I/Chap II/Corollary 2]. The strong dual topology is Suslin as shown in Example (C) page 115 of [S] (the proof given works for $M\neq\mathbb R^d$ without changing a word). For $\tau_S$, one can write the image of $\mathcal D'(M)$ in $\mathbb R^S$ as a Borel subset by hand, and Borel subsets of Suslin spaces stay Suslin [S Part I/Chap II/Theorem 3].

A similar proof works for $H^s(M)$: the Hilbert topology is Polish so the space is Suslin, and its image under the map $H^s(M)\to\mathbb R^S$ is also Borel.

[S] L. Schwartz, Radon measures on arbitrary topological spaces and cylindrical measures (1973).

[DM] R. L. Dobrushin, R. A. Minlos, Исследование свойств обобщенных гауссовских случайных полей, in Задачи механики и математической физики (1976). My translation, based on automatic translation: A study of properties of generalized Gaussian random fields, in Problems of mechanics and mathematical physics.

$\endgroup$
3
  • $\begingroup$ You may want to check out mathoverflow.net/a/157219/11211 and the discussion in the comments... TL&DR: Gel'fand-Vilenkin's book, "Generalized Functions - Volume IV", Chapter IV, pp. 303ff.. $\endgroup$ Commented Nov 29 at 19:11
  • 1
    $\begingroup$ @PedroLauridsenRibeiro Thank you for your comment! I should have linked that question, that is where I learned about those two references... I am still a bit puzzled by all this. I think the closest result I could find in your reference is the the beginning of section IV-2.4, p330, where it is left as an exercise to show that, in my terminology, the Borel $\sigma$-algebra of the weak* topology coincides with the weak cylindrical $\sigma$-algebra (to be continued). $\endgroup$
    – Pierre PC
    Commented Nov 30 at 15:07
  • 1
    $\begingroup$ @PedroLauridsenRibeiro But frankly, I cannot do the exercise myself; what is clear is that the weak* topology and the weak cylindrical $\sigma$-algebra are generated by a common basis, but this only shows the direction "cylindrical implies weak*-Borel", and for the reverse direction I cannot seem to be able to find enough countability leverage since there is no first countability. Also I don't see how the argument applies to the strong topology (the weak topology here is the same as the weak* topology). $\endgroup$
    – Pierre PC
    Commented Nov 30 at 15:19

0

You must log in to answer this question.