Let $\mathcal S '(\mathbb R^d)$ be the space of Schwartz tempered distributions equipped with the weak-* topology. I need to know if this space is second countable, i.e. if this topology has a countable basis. To put this in context, I need this because I am considering random variables with values in $\mathcal S'$ and I need to know if the Borel sigma field is the same as the cylinder sigma field in this particular case and second countability would be a sufficient condition. I think the answer is yes, but I cannot find a reference.
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4 Answers
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Since $\mathscr{S}'$ is the dual of an infinite-dimensional Fréchet space, the weak-* topology in $\mathscr{S}'$ is not even first countable. What people usually do is to define probability measures in $\mathscr{S}'$ which are supported in a subspace $\mathscr{W}$ which is endowed with a stronger topology which makes $\mathscr{W}$ a Polish space, that is, a separable complete metric space (in particular, second countable). This is the natural context for cylinder set measures, as discussed in Bourbaki's "Eléments de Mathématiques, Intégration" (Hermann, 1969), Chapter 9 (there, such measures are called "promeasures", for they can be understood as "projective systems of probability measures"). It is important to notice that cylinder set measures are also supported in an extremely "small" subset $\mathscr{V}\subset\mathscr{W}$, in the sense that $\mathscr{V}$ is locally compact (therefore, essentially "finite-dimensional") with respect to the induced topology from $\mathscr{W}$.
An important special case is when $\mathscr{W}$ is the dual of a nuclear space, which is discussed in Gel'fand-Vilenkin's "Generalized Functions", Volume 4. You can also find material on this in the book of J. Glimm and A. Jaffe, "Quantum Physics - A Functional Integral Point of View" (2nd. ed., Springer-Verlag, 1988).
answered Feb 10, 2014 at 15:11
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$\begingroup$ The interesting (and mysterious?) thing is that, even though it is not first-countable, continuity for linear functionals can be checked using convergent sequences only. $\endgroup$ Feb 10, 2014 at 15:19
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$\begingroup$ Thanks to the "magic" of linearity (and, of course, the fact that the codomain of functionals is a finite-dimensional locally convex vector space). $\endgroup$ Feb 10, 2014 at 15:26
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$\begingroup$ I think that it is a bit more subtle than that. It is because the space is a countable inductive of Banach spaces with compact intertwining mappings (in this case, they are even nuclear). The portuguese mathematician Sebastião e Silva proved this fact and many others about such spaces in the 50's---hence the name "Silva space". $\endgroup$– alphaFeb 10, 2014 at 15:37
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1$\begingroup$ @Thomas: no, it's not quite the same thing. It simply means that a straightforward argument to get equality of both $\sigma$-algebras using second countability of the whole space is untenable (see, however, Jochen Wengenroth's answer for a more careful approach). The fact that cylinder set measures usually have very small supports has to do rather with the particular way they are constructed. $\endgroup$ Feb 10, 2014 at 17:28
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1$\begingroup$ @Thomas: you can find it, for instance, in Gel'fand-Vilenkin's book, "Generalized Functions - Volume IV", Chapter IV, pp. 303ff. Once again, the precise relationship is connected to Jochen Wengenroth's answer: since $\mathscr{S}$ is separable, the $\sigma$-algebra generated by the Borel cylinder sets of $\mathscr{S}'$ contains the polar $A^\circ\subset\mathscr{S}'$ of every subset $A\subset\mathscr{S}$. This implies, together with the first countability of $\mathscr{S}$, that this $\sigma$-algebra includes all weak-* Borel sets - therefore, both $\sigma$-algebras coincide. $\endgroup$ Feb 10, 2014 at 20:21
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Pedro and Alpha are right, of course. But perhaps the following remark can be helpful:
Being the dual of a Frechet space, $\mathscr S'$ (endowed with the weak* topology) is sigma-compact: if $U_n$ form a base of $0$-neighborhoods in $\mathscr S$ it is covered by the polars $U_n^\circ$ which are weak*-compact by Alaoglu's theorem. Using the separability of $\mathscr S$ (and the fact that a compact space does not have strictly coarser topologies) you get that $U_n^\circ$ are metrizable in the weak*-topology (and thus second countable).
Moreover, the Borel $\sigma$-algebra of $\mathscr S'$ is the supremum of the Borel $\sigma$-algebras of all $U_n^\circ$.
answered Feb 10, 2014 at 16:29
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A locally convex with a countable basis is metrisable, which your space is not. However it is, with its strong topology as the dual of the rapidly decreasing smooth function,s a nuclear Silva spaces and such spaces have very strong properties, in particular with respect to measurability. I would try looking up Minlos' theorem, e.g., in the book "Radon measures on topological spaces and cylindrical measures" by Laurent Schwartz.
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$\begingroup$ The above answer of Pedro was posted while I was typing---hence the duplication. $\endgroup$– alphaFeb 10, 2014 at 15:19
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Just a small bibliographical complement to the very nice answers already given. The identity of the two $\sigma$-algebras is proven in Proposition 2.1 of the article An investigation of the properties of generalized Gaussian random fields by Dobrushin and Minlos.
answered Apr 5, 2015 at 21:24