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First some references:

  • [M] Megginson - An Introduction to Banach Space Theory
  • [D] Denkowski, Migórski, Papageorgiou, Socrates - An Introduction to Nonlinear Analysis
  • [B] Brezis - Functional Analysis, Sobolev Spaces and Partial Differential Equations
  • [DS] Dunford and Schwartz - Linear Operators Part I: General Theory

The following theorem appears in [M, p.231], [D, p.305], [B, p.74], [DS, p.426]. I'm sure it appears elsewhere too.

Theorem. Let $X$ be a Banach space. Then $X$ is separable iff the closed unit ball of $X^*$ is metrizable in the weak* topology (inherited from X*).

However, [M] only assumes $X$ is a normed space, not a Banach space. I've been through the proofs, and I can't see where completeness is being used.

Question 1. Is completeness of $X$ really necessary?

There is also a closely related theorem that appears in [D, p.305], [B, p.74], [DS, p.426]:

Theorem. Let $X$ be a Banach space. Then $X^{*}$ is separable iff the closed unit ball of $X$ is metrizable in the weak topology (inherited from X).

Question 2. Is completeness of $X$ really necessary?

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No, in both caes. The dual does not "see" if $X$ is complete or not; the dual of a space and of its completion is the same. And separability and metrizability pass to subsets, so $X$ is separable/metrizable if and only if $\overline X$ is.

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  • $\begingroup$ I forgot that the dual of a dense subspace is isometrically isomorphic to the dual of the original space. $\endgroup$ – MichaelGaudreau Jun 29 '18 at 3:01
  • $\begingroup$ I would say it is equal, more than isometrically isomorphic. It's just that by taking the completion you cannot introduce any new functionals. $\endgroup$ – Martin Argerami Jun 29 '18 at 3:04
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    $\begingroup$ But the weak$^*$ topology sees the completion: $\sigma(X^*,X)$ is strictly coarser than $\sigma(X^*,\tilde X)$. $\endgroup$ – Jochen Jun 29 '18 at 7:09
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    $\begingroup$ But on the dual unter ball the weak topologies coincide: The ball is $\sigma(X^*,\tilde X) $-compact and hence the other weak topology can't be strictly coarser. $\endgroup$ – Jochen Jun 29 '18 at 19:55
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    $\begingroup$ You need not touch nets or filters or neighbourhoods: on a compact Hausdorff space there is no strictly coarser Hausdorff topology. $\endgroup$ – Jochen Jun 30 '18 at 8:08

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