Let $A$ be an abelian von Neumann algebra (say,acting on the Hilbert space $H$). I want to show that $A$ is a maximally abelian von Neumann algebra on $H$ if and only if $A=A'$.
I managed to prove that if $A=A'$, then $A$ is maximally abelian. I'm stuck on showing the converse though.
So assume $A$ is maximally abelian. Then $A \subseteq A'$. Since $A'$ is a VNA, one possible approach is showing that $A'$ is again abelian and then the maximality will imply $A=A'$. But I'm not sure this is even true. If $x \in A'$, then maybe the VNA generated by the commuting elements $A \cup \{x\}$ is abelian (?) so then we will get $x \in A$ once again by maximality. But I'm not sure why either one of thse claims should be true.
Thanks in advance!
Of course $(A\cup\{x\})''$ is abelian if $x\in A'$ and $A$ is abelian. For simplicity, you can assume without loss of generality that $x^*=x$. Then you have $$ (A\cup\{x\})''=\{p(a,x):\ a\in A,\ p\in\mathbb C[s,t]\}'', $$ and the latter algebra is clearly abelian since $xa=ax$ for all $a\in A$.