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?


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.