Let $H$ be a Hilbert space and $B(H)$ the algebra of bounded operators in $H$ . Recall that the commutant of a subset $\mathcal{F} \subset B(H)$ is the set of all bounded operators that commute with those of $\mathcal{F}$ , i.e.

Proof: It is clear that $\mathcal{F}'$ contains the identity operator, since it commutes with all operators in $B(H)$ and in particular with those of $\mathcal{F}$ .

Let us now see that $\mathcal{F}'$ is a subalgebra of $B(H)$ . Let $T_1, T_2 \in \mathcal{F}'$ and $\lambda \in \mathbb{C}$ . We have that, for all $S \in \mathcal{F}$ ,

It remains to see that $\mathcal{F}'$ is weak operator closed. Suppose $(T_i)$ is a net in $\mathcal{F}'$ that converges to $T$ in the weak operator topology. Then, for all $x, y \in H$ we have that $\langle T_ix, y \rangle \to \langle Tx,y\rangle$ . Thus, for all $S \in \mathcal{F}$ , we have

\begin{eqnarray*} \langle (TS - ST)x, y \rangle & = & \langle TSx, y \rangle - \langle Tx, S^*y \rangle \\ & = & \lim \big( \langle T_iSx, y \rangle - \langle T_ix, S^*y \rangle \big)\\ & = & \lim \, \langle (T_iS - ST_i)x, y \rangle\\ & = & \lim\, \langle (T_iS - T_iS)x, y \rangle\\ & = & 0 \end{eqnarray*} Hence, $TS-ST=0$ , so that $T \in \mathcal{F}'$ . We conclude that $\mathcal{F}'$ is closed in the weak operator topology. $\square$