Is the weak$^*$ topology on space of all finite Borel measures defined on a compact set metrizable?

Is this true for any bounded domain instead of compact set? I know a result from Kesavan that

A Banach space $X$ is separable if and only if the weak$^*$ topology on the closed unit ball in $X^*$ is metrizable

In order to use this result I need $\textbf{Space of all finite Borel measures defined on a compact set}$ is separable. Is this true? $\textbf{OR}$ Is there any other way to prove this? Any references is greatly appreciated. Thanks!!

  • $\begingroup$ @KyleGannon But I know that Space of all bounded variations is not separable en.wikipedia.org/wiki/Separable_space . can you give me some references $\endgroup$ – Arun Aug 21 '17 at 6:47
  • $\begingroup$ If $\Omega$ has uncountable subset $S$, then you can consider Dirac delta measure $\{\delta_s: s\in S\}\subset M(\Omega)$. Now note that $\Vert\delta_s-\delta_t\Vert=2$ for $s\neq t$ and recall that $S$ is uncountable, so $M(\Omega)$ is not separable. Thus in many cases $M(\Omega)$ is not separable. $\endgroup$ – Norbert Aug 21 '17 at 7:02
  • $\begingroup$ @Norbert Then are Dirac measures dense in $\mathcal{M}(\Omega)$? I don't know how to prove/disprove it. $\endgroup$ – Arun Aug 21 '17 at 7:04
  • $\begingroup$ Nevermind, I missed that you are talking about weak* topology. $\endgroup$ – Norbert Aug 21 '17 at 7:06
  • $\begingroup$ Sorry if I miss anything but as far as I understand you need separately of the space of continuous function on $\Omega $, not separately of the space of Borel measures $\endgroup$ – Bananach Aug 21 '17 at 7:21

The dual of an infinite dimensional normed space is NEVER metrizable in its weak$^*$ topology (otherwise it would be a Frechet space and the closed graph theorem then implies that the weak$^*$ topology coincides with the topology of the dual norm which onyl holds in finite dimension).

  • $\begingroup$ but there is a result: If X is separable then its dual $X^*$ is metrizable $\endgroup$ – Arun Aug 21 '17 at 14:03
  • $\begingroup$ No! If $X$ is separable then the unit ball of $X^*$ is metrizable in the weak$^*$ topology. $\endgroup$ – Jochen Aug 22 '17 at 6:22
  • $\begingroup$ please give me some examples!! $\endgroup$ – Arun Aug 22 '17 at 8:15
  • $\begingroup$ I don't understand what kind of example you are looking for. Perhaps: If $\Omega$ is a compact metric space then ${\scr M}(\Omega)$ is weak$^*$ metrizable if and only if $\Omega$ is finite. $\endgroup$ – Jochen Aug 22 '17 at 8:25

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