= Solution
The <Von Neumann double commutant theorem> states that for a unital star-subalgebra $A\subseteq B(H)$,
$$
\boxed{\overline A^{\mathrm{SOT}}=\overline A^{\mathrm{WOT}}=A''.}
$$
The <commutant of an operator algebra> consists of operators commuting with every element. Each commutation equation is closed in the <weak operator topology>, so $A''$ is weakly closed and contains both indicated closures.
Let $T\in A''$ and fix $\xi_1,\ldots,\xi_m\in H$. On $H^m$ let $A$ act diagonally and put $\xi=(\xi_1,\ldots,\xi_m)$. The subspace $K=\overline{\{(a\xi_1,\ldots,a\xi_m):a\in A\}}$ reduces the diagonal star-algebra. Its <orthogonal projection> $P$ therefore commutes with every diagonal $a$. Writing $P$ as an $m\times m$ operator matrix shows that each entry belongs to $A'$. Since $T\in A''$, the diagonal $T$ commutes with $P$. Unitality gives $\xi\in K$, so $P\xi=\xi$ and $P(T\xi)=T\xi$. Thus $T\xi\in K$. By definition of $K$, for any $\varepsilon>0$ a single $a\in A$ satisfies $\sum_j\|(T-a)\xi_j\|^2<\varepsilon^2$. These are precisely the finite-vector neighbourhoods of the <strong operator topology>. Hence $T$ belongs to the strong closure, completing the proof. Unitality, or the corresponding nondegeneracy hypothesis, is essential to the theorem's stated form.
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