Generators of von Neumann algebras Let us suppose that M is a von Neumann algebra on some hilbert space $H$ such that $M = A''$ for some $C^*$-algebra $A\subseteq B(H)$. 
I am wondering when $A$ will be dense in $M$ and by which topology. I also want to know that if I have two functional $S,T \in M^*$ such that $S|_A=T|_A$, can I say that $T=S$?
Thank you very much.
 A: Let $K\subseteq H$ be the closed subspace generated by $AH$.  Then von Neumann's double commutant theorem says that if $K=H$, then $M$ is the closure of $A$ in the weak operator, strong operator, ultraweak, or ultrastrong topologies.  If $K$ is a proper subspace of $H$, then $A=A|_K\times \{0\}|_{K^\perp}$, $A''=(A|_K)''\times \mathbb C I_{K^\perp}$, and $A$ is weakly dense in $(A|_K)''\times \{0\}$ but not $M$.  The condition $K=H$ is also known as $A$ acting nondegenerately on $H$.  In particular, it holds if $A$ contains the identity operator on $H$.
So assuming that $A$ acts nondegenerately, $A''=M$ is equivalent to density of $A$ in $M$ in various topologies that are coarser than the norm topology.  If $M^*$ is the set of norm-continuous linear functionals on $M$, then it is possible in general for distinct functionals in $M^*$ to agree on $A$, if at least one of them is not ultraweakly continuous.  $A$ is a norm-closed subspace of $M$, so if $A\neq M$ then there are nonzero continuous functionals on the Banach space $M/A$, hence by composing with the quotient map, there are nonzero continuous functionals on $M$ that vanish on $A$.
