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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?

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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 $).

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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 $B^{''}=B$

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  • $\begingroup$ Thank's for this answer too. I see now, that the answer to my question follows from the double commutant theorem. But in the book I am reading, the claim in my question is just stated without justification two chapters before the double commutant theorem is even mentioned. So I thought, that I was missing something. It's nice to see an explanation. $\endgroup$ – Kimarokko Mar 24 '17 at 18:10
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    $\begingroup$ One should mention that the subalgebra has to contain the identity on $H$ since it is always contained in the double commutant. Otherwise you have to restrict the algebra to a smaller Hilbert space. $\endgroup$ – Sebastian Bechtel Mar 31 '17 at 17:08

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