Let $H$ be a Hilbert space and $C \subseteq B(H)$ be a convex set. Is it well-known that the closure of $C$ with respect to the strong operator topology agrees with the closure of $C$ with respect to the weak operator topology.
Is it true that the closure of $C$ with respect to the weak operator topology also coincides with the closure of $C$ with respect to the $\sigma$-weak (= the weak $^*$-topology on $B(H)$, or equivalently, the topology generated by the normal functionals) topology?
If the answer is positive, a reference suffices as an answer.
Best Answer
Two locally convex topologies on a vector space share closed convex sets iff they have the same dual.
The dual of $B(H)$ with the weak topology is formed by the functionals $$ \phi_S: T\in B(H)\mapsto \text{tr}(TS)\in\mathbb C, $$ where $S$ is a finite rank operator, while the dual with the $\sigma$-weak topology is as above, but for the larger class of trace class operators $S$.
Therefore the answer to the question posed by the OP is no.
For a concrete example, given a trace class operator $S$ of infinite rank, the null space of $\phi_S$ is $\sigma$-weakly closed, but not weakly closed.