Suppose $M$ is von Neumann algebra, $A$ is $*$-algebra in $M$, further if $(A)_1$, the unit ball of $A$ is strong operator closed, does it implies $A$ is von Neumann algebra? I started proving this using Kaplansky density theorem but stuck with the fact I cannot use uniform bounded principle for nets. Please give some hint.
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2 Answers
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Yes, on the unit ball the strong operator topology is stronger than the ultraweak topology, hence $(A)_1$ is also closed in the ultraweak topology (which is the weak$^{\ast}$-topology) of $M$. It follows from Krein-Smulian theorem that $A$ is closed in the ultraweak topology, therefore it is a von Neumann algebra.
answered Jun 25, 2018 at 12:10
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If $A$ is a C*-algebra, you can use Kaplansky denisity theorem like this: Suppose $\newcommand{\cl}{\operatorname{cl_{SOT}}}$$a\in \cl A$,then $a/\|a\|\in (\cl A)_1=\cl (A_1)=A_1$, hence $a\in A$.
