If $A$ is a $*-$ Banach algebra then $\bar A^{wot} = \bar A^{weak^*}$?

Question

If $A$ is a $*-$ subalgebra of $B(H)$, then clearly $\bar A^{weak^*}\subset \bar A^{wot}$ (wot means weak operator topology). Also on every bounded subset of $A$, two topologies equal. Now my question is that could we show that $\bar A^{wot} = \bar A^{weak^*}$?

Thanks

Answer 1

Yes, the two closures are equal.

Let $x\in \bar A^{wot}$. We have $\bar A^{wot}=\bar A^{sot}$, so there is a net $\{x_j\}\subset A$ such that $x_j\to x$ sot. By Kaplansky, there is a bounded net $\{x_j'\}\subset A$ with $ x_j'\to x$ sot, and so wot. That is, $x$ belongs to the wot closure of a ball in $A$. But on balls, the wot and ultraweak topologies agree, which tells us that $x$ is the ultraweak limit of a net in $A$.