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Let $S$ be a compact metric space. Let $\Delta^\ast(S)$ be the set of all finite signed Borel measures defined on $S$, and $\Delta(S)$ be the set of all Borel probability measures defined on $S$.

My question is, does there exist a norm (or topology) on $\Delta^\ast(S)$ that makes $\Delta^\ast(S)$ a Banach space (or locally convex space), and at the same time makes $\Delta(S)$ compact under the induced metric?

I know that the total variation norm makes $\Delta^\ast(S)$ a Banach space, however $\Delta(S)$ is not compact under the induced metric.

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    $\begingroup$ The weak-* topology (induced by the duality $\Delta^*(S) = C(S)^*$) does this, right? $\endgroup$
    – Nate Eldredge
    Commented Apr 22, 2021 at 14:48
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    $\begingroup$ There won't be a norm that does it, for any practical purposes. There's a general principle that in order to construct two inequivalent complete norms on a vector space, you have to use the axiom of choice in an essential way, and so the new norm would be some horrible non-constructive monster that wouldn't likely be useful for anything. $\endgroup$
    – Nate Eldredge
    Commented Apr 22, 2021 at 14:50
  • $\begingroup$ Indeed, the weak-* topology will do. Thanks. $\endgroup$
    – Lemma1
    Commented Apr 23, 2021 at 10:19
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    $\begingroup$ If you want a complete norm that makes $\Delta(S)$ compact, this is impossible unless $S$ is a finite set. The usual (total variation) unit ball of $\Delta^*(S)$ will be compact with respect to this norm, and so $\Delta^*(S)$ is a countable union of compact sets. In an infinite-dimensional normed space, compact sets have empty interior, so we have a contradiction with the Baire category theorem. (This comment can become an answer if you find it satisfactory.) $\endgroup$
    – Robert Furber
    Commented Apr 23, 2021 at 10:19

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