Is the space of tempered distributions 1st countable ?

[dextercioby]

Hi everyone,

the question is simple: is \mathcal{S}'\left(\mathbb{R}^3\right) a first countable topological space ?

I have no idea, honestly. (The question has occurred to me from a statement of Rafael de la Madrid in his PhD thesis when discussing the general rigged Hilbert space formalism. He says that even though the space of wavefunctions is assumed 1st countable, his antidual generally isn't. So I took the simplest case of a rigged Hilbert space: \mathcal{S}\left(\mathbb{R}^3\right)\subset \mathcal{L}^2 \left(\mathbb{R}^3\right)\subset \mathcal{S}^{\times}\left(\mathbb{R}^3\right))

 

[jgens]

the question is simple: is \mathcal{S}'\left(\mathbb{R}^3\right) a first countable topological space ?

Assuming the weak*-topology my recollection is that this space is not first countable.

 

[dextercioby]

The only handy reference I have on weak*-topology is this: http://en.wikipedia.org/wiki/Weak-star_operator_topology and has to do with operator spaces and particularly the space of the trace-class operators... How's that related to the (anti)dual of the Schwartz space ?

 

[jgens]

Since the space of tempered distributions lives inside a continuous dual space the weak-* topology is often the most natural choice. Basically any decent book on functional analysis will go into detail on this but for a quick reference consult: http://en.wikipedia.org/wiki/Weak_topology#The_weak-.2A_topology

 

[dextercioby]

Alright, thank you, got that. So can you show it's not first countable?

 

[jgens]

If memory serves the argument is essentially that the space of tempered distributions with the weak-* topology is not metrisable (since for topological vector spaces the two are equivalent). I forget how exactly this is shown, but if you perform an internet search I am sure this argument will turn up somewhere.

 

[dextercioby]

It doesn't show as the proof, only as the result, thing which is quite frustrating.

I found an answer on the competitor's website, for those of you interested in the same question.
Thanks jgens for the interest shown.