Theorem 37.2.6 (Von Neumann’s Bicommutant Theorem).label Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a unital, self-adjoint subalgebra, then $A''$ is the strong-operator closure of $A$.
Proof, [Section 18.3, Zhu93]. ($\overline{A}^{\text{\small SOT}}\subset A''$): By separate continuity of composition, $A''$ is a strong-operator closed subset that contains $A$. Hence $A''$ contains the strong-operator closure of $A$.
($\overline{A}^{\text{\small SOT}}\supset A''$): Let $T \in A''$ and $x = \seqf{x_j}\in H^{n}$. For each $S \in B(H)$, denote $S^{(n)}= (S, \cdots, S)$ ($n$-copies), then
- (i)
$A^{(n)}= \bracsn{S^{(n)}|S \in A}$ is a unital, self-adjoint subalgebra of $B(H^{n})$.
- (ii)
$T^{(n)}\in (A^{(n)})''$.
By Lemma 37.2.5,
so $T \in \ol{A}^{\text{\small SOT}}$.$\square$
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