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.



On topology in von Neumann algebras
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.


