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It is known, and not too difficult to prove that if a dual of a normed space is separable, it is separable itself. Does the same hold for Frechet spaces?

(By the dual I mean the dual endowed with the strong topology.)

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  • $\begingroup$ Did you try to transfer the usual proof in case of a normed vector space? $\endgroup$ – gerw Apr 9 '18 at 6:49
  • $\begingroup$ @gerw The proof that I know relies on the norm heavily: you should be able to choose "small" functionals and "large" vectors. If you know a transferable proof, could you please give a reference? $\endgroup$ – erz Apr 9 '18 at 7:02
  • $\begingroup$ Do you consider the weak*-topology or the strong topology on the dual space? Note that $F^*$ with the strong dual is in general not metrizable. In fact, this happens only if $F$ is already normable, i.e. a banach space. In this situation we can apply the result for banach spaces. $\endgroup$ – p4sch Apr 9 '18 at 8:39
  • $\begingroup$ @p4sch I mean the strong topology. I know that $F^*$ is not always metrizable, but in my specific case it is separable, and I want to conclude that $F$ is then separable too, without assuming that $F$ is normed. $\endgroup$ – erz Apr 9 '18 at 8:48
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I think so (no books at hand for a reference). All we need for the normed proof to go through, is some sort of Hahn-Banach theorem that gives us enough functionals to works with. And Fréchet spaces have enough functionals.

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  • $\begingroup$ we also need to choose "small" functionals and "large" vectors (at least in the proof that I know). Perhaps you could elaborate? $\endgroup$ – erz Apr 9 '18 at 7:03

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