9

Suppose $X$ is a topological vector space, $X^*$ is its topological dual space. Let the topology of $X^*$ is weak*-topology, Is $X^*$ complete?

Suppose $f_s$ is a Cauchy net in $X^*$, it is easy to see that $f=\lim f_s$ exists. We can prove that $f$ is linear, but I couldn't see if it is continuous.

89085731
  • 7,932
  • 5
  • 37
  • 80
  • A reference for a proof of this statement is Proposition 2.6.13 of An Introduction to Banach Space Theory by Megginson. – Dean Miller Apr 29 '25 at 12:04

1 Answers1

9

No. For $X$ Hausdorff locally convex, the completion of $(X^*,\sigma^*)$ consists of all linear functionals on $X$ (the reason is that the semi-norms of $\sigma^*$ are determined by finite subsets of $X$ and on a finite dimensional subspace each linear functional is continuous and can be extended to a an element of $X^*$ by Hahn-Banach).

For example, if $X$ is an infinite dimensional Banach space the weak$^*$ dual is incomplete (this depends on the axiom of choice).

Jochen
  • 13,235
  • But why is this not contradicting the Banach-Alaoglu theorem. Is there a reference of the proof of incompleteness? – Liding Yao Jan 21 '25 at 02:20
  • @LidingYao Can you explain why you believe this would be a contradiction to the Banach-Alaoglu Theorem? – Smiley1000 Apr 16 '25 at 07:59
  • @Smiley1000 My confusion comes from the sequentially completeness vs the true completeness, or alternatively, how to understand a Cauchy but non-convergemt net. – Liding Yao Apr 17 '25 at 12:52
  • @LidingYao Sorry, I don't understand what you're trying to say. What was your argument for why this should be a contradiction? – Smiley1000 Apr 17 '25 at 13:01