In a book on Operator Theory there is the following statement:
If $\mathscr A$ is a commutative $C^*$-subalgebra of $\mathscr B(\mathcal H)$, where $\mathcal H$ is a Hilbert space, then the weak operator closure of $\mathscr A$ is also commutative.
I can not prove this. Please help me. Thanks.
Apparently, $A_n\to A$ weakly iff $$ (u,A_nv)\to(u,Av), \quad \text{for all $u,v\in H$}. $$ Assume that $A,B$ belong to the weak closure of the commutative sub-algebra $\mathscr A$, and $A_n\to A$, $B_n\to B$, weakly, with $\{A_n\}_{n\in\mathbb N},\{B_n\}_{n\in\mathbb N}\in\mathscr A$. Then $A_nB_m=B_nA_m$, for all $m,n\in\mathbb N$. Fix know $u,v\in H$. Then, as $n\to \infty$, we have that $$ (u,A_n B_m v)\to (u,A B_m v) $$ and $$ (u,A_n B_m v)= (u,B_mA_nv)=(B_m^*u,A_nv)\to (B_m^*u,Av)=(u,B_mAv) $$
and hence $(u,A B_m v)=(u, B_mA v)$, for all $u,v\in H$, and hence $AB_m=B_mA$, for all $m\in\mathbb N$. Repeating this for $(u,A B_m v)=(u, B_mA v)$, and letting $m\to\infty$, we obtain that $(u,A B v)=(u, BA v)$, and hence $AB=BA$.