Proposition 37.4.2.label Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^{*}$-subalgebra, then $A$ is a maximal commutative von Neumann algebra if and only if $A = A'$.
Proof. ($\Leftarrow$): Let $B \supset A$ be a commutative von Neumann algebra, then $B \subset A' = A$.
($\Rightarrow$): For each $T \in (A')_{sa}$, the von Neumann algebra generated by $A$ and $T$ is commutative. As such, $T \in A$. As this holds for all $T \in (A')_{sa}$, $A' = (A')_{sa}+ i(A')_{sa}\subset A$.$\square$
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