[proofplan] The inclusion $\overline{A}^{\mathrm{SOT}}\subseteq \overline{A}^{\mathrm{WOT}}$ follows because strong operator convergence implies weak operator convergence. For the reverse inclusion, we take $T$ in the weak operator closure and prove that every basic strong neighbourhood of $T$ meets $A$. The key finite-dimensional packaging is to put the vectors $T\xi_1,\dots,T\xi_n$ into $H^{\oplus n}$ and prove that this vector lies in the closed subspace generated by $(a\xi_1,\dots,a\xi_n)$ with $a\in A$. Orthogonality against the complement is detected by weak-operator continuous matrix coefficient functionals, so weak closure forces the needed orthogonality. [/proofplan]

[step:Use strong convergence to obtain weak convergence] By [citetheorem:9260], if a net $(S_i)_{i\in I}$ in $\mathcal{L}(H)$ converges strongly to $S\in\mathcal{L}(H)$, then it converges to $S$ in the weak operator topology. Therefore every strong operator [limit point](/page/Limit%20Point) of $A$ is a weak operator limit point of $A$, and hence \begin{align*} \overline{A}^{\mathrm{SOT}}\subseteq \overline{A}^{\mathrm{WOT}}. \end{align*} [/step]

[step:Reduce strong approximation to a closed subspace in a finite direct sum]Let $T\in \overline{A}^{\mathrm{WOT}}$. To prove $T\in\overline{A}^{\mathrm{SOT}}$, fix an integer $n\in\mathbb{N}$, vectors $\xi_1,\dots,\xi_n\in H$, and a tolerance $\varepsilon>0$. It is enough to find $a\in A$ such that \begin{align*} \|a\xi_r-T\xi_r\|_H<\varepsilon \end{align*} for every $r\in\{1,\dots,n\}$, because a basic strong operator neighbourhood of $T$ is specified by finitely many vectors together with finitely many tolerances, and replacing those tolerances by a common smaller $\varepsilon$ still gives a neighbourhood of the same form. Let $H^{\oplus n}$ denote the [Hilbert space](/page/Hilbert%20Space) direct sum of $n$ copies of $H$, with [inner product](/page/Inner%20Product) \begin{align*} ((\eta_1,\dots,\eta_n),(\zeta_1,\dots,\zeta_n))_{H^{\oplus n}}=\sum_{r=1}^{n}(\eta_r,\zeta_r)_H. \end{align*} Define the subset $E\subseteq H^{\oplus n}$ by \begin{align*} E=\{(a\xi_1,\dots,a\xi_n):a\in A\}. \end{align*} Since $A$ is a linear subspace of $\mathcal{L}(H)$, $E$ is a linear subspace of $H^{\oplus n}$. Let \begin{align*} K=\overline{E}^{\|\cdot\|_{H^{\oplus n}}}. \end{align*} We will prove that \begin{align*} (T\xi_1,\dots,T\xi_n)\in K. \end{align*}[/step]

[guided]We want strong approximation of $T$ by elements of $A$ on the finite set $\{\xi_1,\dots,\xi_n\}$. The strong operator topology is exactly the topology of pointwise norm convergence on $H$, so a basic strong neighbourhood of $T$ asks for an operator $a\in A$ with $a\xi_r$ close to $T\xi_r$ for finitely many prescribed vectors. To package all these inequalities at once, we work in the Hilbert space $H^{\oplus n}$. This space consists of tuples $(\eta_1,\dots,\eta_n)$ with $\eta_r\in H$, and its inner product is \begin{align*} ((\eta_1,\dots,\eta_n),(\zeta_1,\dots,\zeta_n))_{H^{\oplus n}}=\sum_{r=1}^{n}(\eta_r,\zeta_r)_H. \end{align*} For each $a\in A$, the tuple $(a\xi_1,\dots,a\xi_n)$ records how $a$ acts on the chosen finite set of vectors. Define \begin{align*} E=\{(a\xi_1,\dots,a\xi_n):a\in A\}\subseteq H^{\oplus n}. \end{align*} Because $A$ is closed under addition and scalar multiplication, $E$ is a linear subspace of $H^{\oplus n}$. Let \begin{align*} K=\overline{E}^{\|\cdot\|_{H^{\oplus n}}} \end{align*} be its norm closure. The target tuple is $(T\xi_1,\dots,T\xi_n)$. If we prove \begin{align*} (T\xi_1,\dots,T\xi_n)\in K, \end{align*} then by the definition of norm closure there exists $a\in A$ with \begin{align*} \|(a\xi_1,\dots,a\xi_n)-(T\xi_1,\dots,T\xi_n)\|_{H^{\oplus n}}<\varepsilon. \end{align*} Since the direct-sum norm dominates each coordinate norm, this gives \begin{align*} \|a\xi_r-T\xi_r\|_H<\varepsilon \end{align*} for every $r\in\{1,\dots,n\}$. Thus membership of the target tuple in $K$ is precisely the finite-vector strong approximation we need.[/guided]

[step:Annihilate the orthogonal complement using weak operator continuity] Let $\zeta=(\zeta_1,\dots,\zeta_n)\in K^\perp$. Define a linear functional \begin{align*} \Phi:\mathcal{L}(H)\to\mathbb{C} \end{align*} by \begin{align*} \Phi(S)=\sum_{r=1}^{n}(S\xi_r,\zeta_r)_H. \end{align*} Each map $S\mapsto (S\xi_r,\zeta_r)_H$ is one of the defining matrix coefficient maps for the weak operator topology, so $\Phi$ is weak-operator continuous as a finite linear combination of weak-operator continuous maps. For every $a\in A$, the vector $(a\xi_1,\dots,a\xi_n)$ belongs to $E\subseteq K$. Since $\zeta\in K^\perp$, we have \begin{align*} 0=((a\xi_1,\dots,a\xi_n),\zeta)_{H^{\oplus n}}=\sum_{r=1}^{n}(a\xi_r,\zeta_r)_H=\Phi(a). \end{align*} Thus $\Phi$ vanishes on $A$. Since $T\in\overline{A}^{\mathrm{WOT}}$ and $\Phi$ is weak-operator continuous, $\Phi$ also vanishes at $T$: \begin{align*} \Phi(T)=0. \end{align*} Equivalently, \begin{align*} ((T\xi_1,\dots,T\xi_n),\zeta)_{H^{\oplus n}}=0. \end{align*} Therefore $(T\xi_1,\dots,T\xi_n)$ is orthogonal to every vector in $K^\perp$. [/step]

[step:Convert orthogonality into membership in the closed subspace] Since $K$ is a closed linear subspace of the Hilbert space $H^{\oplus n}$, the Hilbert space orthogonal complement identity gives \begin{align*} (K^\perp)^\perp=K. \end{align*} The previous step proved that $(T\xi_1,\dots,T\xi_n)\in (K^\perp)^\perp$. Hence \begin{align*} (T\xi_1,\dots,T\xi_n)\in K. \end{align*} By the definition of $K$ as the norm closure of $E$, there exists $a\in A$ such that \begin{align*} \|(a\xi_1,\dots,a\xi_n)-(T\xi_1,\dots,T\xi_n)\|_{H^{\oplus n}}<\varepsilon. \end{align*} Consequently, for each $r\in\{1,\dots,n\}$, \begin{align*} \|a\xi_r-T\xi_r\|_H<\varepsilon. \end{align*} [/step]

[step:Conclude equality of the strong and weak operator closures] The preceding step shows that every basic strong operator neighbourhood of $T$ intersects $A$. Therefore $T\in\overline{A}^{\mathrm{SOT}}$. Since $T\in\overline{A}^{\mathrm{WOT}}$ was arbitrary, we have \begin{align*} \overline{A}^{\mathrm{WOT}}\subseteq \overline{A}^{\mathrm{SOT}}. \end{align*} Together with the reverse inclusion proved at the beginning, this gives \begin{align*} \overline{A}^{\mathrm{SOT}}=\overline{A}^{\mathrm{WOT}}. \end{align*} [/step]