2
$\begingroup$

I am trying to gather all I can know about the weak topology on a separable (infinite dimensional) Hilbert space. As far as I know, this space is :

  • Hausdorff Locally convex
  • compact on weakly-closed bounded subsets
  • metrizable on bounded subsets
  • quasi-complete
  • separable

(And it is not : complete, first-countable, metrizable, quasibarrelled)

I want to know weather or not this topological vector space is :

  • Nuclear

  • Mackey

I'm sure this is common knowledge but it's hard to find references on those specific questions.

$\endgroup$
1
$\begingroup$

A locally convex space $X$ is nuclear if for each continuous seminorm $p$ there is a bigger one $q$ such that the canonical map $X_q \to X_p$ is a nuclear operator between Banach spaces. Here, $X_p$ is the completion of $(X/Kern(p),p)$ and the canonical map is the extension to the completions of $X/Kern(q)\to X/Kern(p)$, $x+Kern(q)\mapsto x+Kern(p)$.

For the weak topology, those local Banach spaces are finite dimensional (and the quotients are already complete) so that every linear map between them is nuclear. This shows that the weak topology is nuclear.

The weak topology of an infinite dimensional Hilbert space is not Mackey (a locally convex space is Mackey if its topology coincides with the topology of uniform convergence on all absolutely convex weak-* compact sets of the dual). The reason is, that the Hilbert space $X$ and $(X,\sigma)$ have the same dual (again $X$, up to the identification by the Riesz theorem) and the unit ball of $X$ is weak-$*$ compact.

$\endgroup$

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.