I suspect there is a slight error in Murphy's C*-algebras and Operator Theory:
Murphy defines a von Neumann algebra on a Hilbert space $H$ as a $*$-subalgebra of $B(H)$ that is strongly closed. I mention that because others define von Neumann algebras as those that are equal to their double commutants (if $id_H\in A$, the definitions are equivalent by the double commutant theorem, but if not, they are not the same; we can reduce to this definition by considering the unit of $A$ though, which is a projection and compressing to that subspace).
Anyway, after a number of results on von Neumann algebras and the strong and weak operator topologies, Murphy says "If $A$ is a $*$-subalgebra of $B(H)$, then its weak closure is a von Neumann algebra." and he refers to this as a simple observation BEFORE moving on to Kaplansky's density theorem. He also says that this will be used in the proof of Kaplansky's theorem (but I cannot spot where he uses it).
I don't think this is trivial without Kaplansky's help. I mean, obviously, since convex sets have equal strong and weak closures, if $A$ is a $*$-subalgebra of $B(H)$ then $\overline{A}^{WOT}=\overline{A}^{SOT}$, so this is indeed strongly closed. By weak continuity of involution, it is a self-adjoint set. It is obviously a linear subspace. But why is this a subalgebra?
Using Kaplansky's theorem, I can see why this is true: If $u,v\in\overline{A}^{SOT}$, then we can find a norm-bounded (by Kaplansky) net $(u_\lambda)\subset A$ with $u_\lambda\xrightarrow{SOT}u$ and let $(v_\lambda)\subset A$ be a net with $v_\lambda\xrightarrow{SOT}v$. Then since multiplication restricted on $S\times B(H)\to B(H)$ where $S$ is a bounded subset of $B(H)$ is strongly continuous, we get that $uv\in\overline{A}^{SOT}$ and we are done.
Is there something that obvious that I am missing out?
Best Answer
You are right. But you already know that in the non-degenerate case the wot/sot closure agrees with the double commutant, which is an algebra.
Even if $A$ is degenerate, unital or not, you can do the following.
Let $$ p=\inf\{q\in A'':\ q\ \text{ is a projection such that }qa=a\ \text{ for all }a\in A\}. $$ This infimum exists: the definition above is equivalent to $p$ being the projection onto the subspace $\bigcap_q qH$.
$p\in A''$. Indeed, if $T\in A'$ is selfadjoint, then $TqH=qTH\subset qH$, and so $TpH\subset pH$. This implies that $pTp=Tp$; taking adjoints, $Tp=pT$. If $T\in A'$ it is a linear combination of selfadjoints, so $p\in A''$.
You have $pA=A$, and $(1-p)A=0$, $p\in A''$. Let $A_1=A+\mathbb C\,(1-p)$. It is easy to check that $A_1'=pA'+(1-p)B(H)(1-p)$, and similarly that $A_1''=pA''+ \mathbb C(1-p)$.
You can also check that $\overline{A_1}^{SOT}=\overline{A}^{SOT}+\mathbb C(1-p)$, and that $p\overline{A}^{SOT}=\overline{pA}^{SOT}=\overline{A}^{SOT}_{\vphantom{SOT}}$.
It follows that $$ \overline{A}^{SOT}=p\overline{A_1}^{SOT}=pA'', $$ which is an algebra (note that $p\in A'\cap A''$).
The above is most often not necessary, because one considers von Neumann algebras represented non-degenerately (that is, "multiplied by $p$"), and that's probably while it is often glossed over (I'm not even sure if I have seen it explicitly in any textbook).