# WOT convergence in the unit ball of B(X)

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?

• 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. Commented Mar 15, 2015 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. Commented Oct 2, 2019 at 13:01

1. Yes if and only if $H$ is separable.
2. Yes if and only if $H$ is separable.
• 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. Commented Feb 21, 2019 at 20:10