3

My questions is (probably) related to:

On separable Hilbert space $H$, weak operator topology is metrizable on bounded parts of $B(H)$

  1. Does the theorem quoted in the above question, together with the fact that the unit ball of $B(H)$ (or of any $B(X)$ where $X$ is reflexive) is WOT compact, imply that any sequnce in the unit ball of $B(H)$ has a convergent subsequence? In other words, is the unit ball sequentially-WOT compact?

  2. Since WOT and SOT coincide on convex sets, does this mean that any sequence in the unit ball of $B(H)$ has a SOT-convergent subsequence?

  3. Do the above (if indeed true) hold when $X$ is a separable reflexive Banach space? Does the proof about metrizability of WOT on the unit ball of $B(H)$ hold in $B(X)$ as well?

Community
  • 1
Markus
  • 2,439
  • 12
  • 17
  • 1
    Regarding #1, if $H$ is separable then the norm-closed unit ball in $B(H)$ is sequentially WOT-compact. I'm not sure if it is true when $H$ is nonseparable, but I suspect not. – Ben W Mar 15 '15 at 23:59
  • I think your assertion in 2 is wrong. The closures of every convex sets coincides in both topologies, but the topologies are DIFFERENT in general. As a counterexample, $\mathcal{B}(H)$ is a convex set and WOT and SOT do not coincide unless $H$ is finite-dimensional. – javi1996 Oct 02 '19 at 13:01

1 Answers1

1
  1. Yes if and only if $H$ is separable.
  2. Yes if and only if $H$ is separable.
  3. If I remember correctly yes. Please check this survey paper.
Tomasz Kania
  • 16,361
  • Thank you again :). The survey paper is something that I should really read. – Markus Aug 13 '15 at 00:15
  • 1
    The assertion made about 2. is incorrect. Consider the unilateral shift $S.$ The sequence $S^n$ goes to $0$ in WOT but not SOT, but all are indeed contractions. More generally, when $H$ is countably infinite dimensional, the set of isometries is SOT-closed, however their WOT closure is the entire unit ball, if I recall correctly. – J. E. Pascoe Feb 21 '19 at 20:10