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Let $H$ be a Hilbert space and $A$ be a $*$-subalgebra of $B(H)$ and $\mathbf 1\in A.$ Let $U$ denotes the closed unit ball of $A$. Given that $U$ is weakly closed. Then I want to show that $A$ is weakly closed.
I know that, weak topology and ultraweak topology coincides on unit ball and then I stuck here and do not know how to proceed. Please help me to solve this. Thank you for your help and time.

  • Doesn't this follow from the fact that a weakly convergent net is norm bounded and thus its limit lies in a closed ball of $A$? – Evangelopoulos Foivos Jan 07 '23 at 18:22
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    @EvangelopoulosPhoevos A weakly convergent sequence is norm bounded, but a weakly convergent net is not necessarily so. See e.g. Nate Eldredge's answer here – Robert Furber Jan 08 '23 at 02:11
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    @DenOfZero Do you know the Krein-Smulian theorem or the Banach-Dieudonné theorem? A linear subspace $E$ of a dual Banach space $V^$ is weak- closed iff the intersection $E \cap \mathrm{Ball}(V^)$ is weak- closed. Are you trying to prove the Kaplansky density theorem? – Robert Furber Jan 08 '23 at 02:16
  • @RobertFurber Sorry, I didn't know the Krein-Smulian theorem. If I follow this theorem then, $U$ is weakly closed implies $U$ is weak-$$ closed, as ultra weak topology and weak-$$ topology both coincides on $B(H)$. So from the theorem, $A$ is weak-$$ closed. But I have two questions, first is how can I assume $A$ as $V^$, that is dual of a Banach space and also from the theorem we have $A$ is weak-$*$ closed but I want to show $A$ is weakly closed. –  Jan 08 '23 at 09:34
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    @DenOfZero For Krein-Smulian you don't need that $A$ is $V^*$, only that $B(H)$ is (and it's the dual of the trace-class operators). To get from weak-* closed to weakly closed, use the fact that a weak-*-closed set is ultrastrongly closed, and an ultrastrongly closed *-algebra is equal to its bicommutant, and therefore weakly closed by the bicommutant theorem. (You have to use the *-algebra structure somehow because there are weak-*-closed hyperplanes that are not weakly closed.) – Robert Furber Jan 09 '23 at 12:21
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    But maybe, if you are not trying to reprove it, what you want is the Kaplansky density theorem. – Robert Furber Jan 09 '23 at 12:26

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