Is the SO-closure of a C*-subalgebra of B(H) always a von Neumann algebra?

Question

Let $A$ be a C*-subalgebra of $B(H)$ for some Hilbert space $H$. Let's denote by $B$ the strong operator closure of $A$ in $B(H)$.

Question: Is $B$ a von Neumann algebra? [Since a von Neumann algebra is defined as a C*-subalgebra of $B(H)$ that is closed in the strong operator topology, this comes down to determining whether or not $B$ is a C*-algebra].

I know from the literature, that the answer is "yes", but how can we be certain, that $B$ is closed under operator multiplication, given that multiplication $B(H) \times B(H) \to B(H)$ is not strong operator continuous?

Answer 1

That's taken care by von Neumann's Double Commutant Theorem, which gives you $B=\overline {A}^{\rm sot}=A''$, which is a C $^*$-algebra.

The non-continuity of multiplication is not really an issue, because multiplication is continuous on bounded sets and you have Kaplansky's Density Theorem (the sot-closure of the unit ball of $A $ is the unit ball of $B $).

Answer 2

It follows from the von Neumann double commutant theorem that $A^{''}= B$. Moreover, a *subalgebra $M\subseteq B(H)$ is called a von Neumann algebra if one of the following three equivalent conditions are satisfied,

1. $M$ is weakly closed

2. $M$ is strongly closed.

3. $M^{''}=M$

So clearly, $B$ is a von Neumann Algebra since $B^{''}=B$