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Is the space of tempered distributions 1st countable ?

  1. Feb 15, 2014 #1

    dextercioby

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    Hi everyone,

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

    I have no idea, honestly. (The question has occured 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: [itex] \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) [/itex])
     
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  3. Feb 15, 2014 #2

    jgens

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    Assuming the weak*-topology my recollection is that this space is not first countable.
     
  4. Feb 15, 2014 #3

    dextercioby

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    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 ?
     
  5. Feb 15, 2014 #4

    jgens

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    Last edited: Feb 15, 2014
  6. Feb 15, 2014 #5

    dextercioby

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    Alright, thank you, got that. So can you show it's not first countable?
     
  7. Feb 15, 2014 #6

    jgens

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    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.
     
  8. Feb 16, 2014 #7

    dextercioby

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    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.
     
    Last edited: Feb 16, 2014
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