Countable amplification and all three operator closures

This supplemental proof supplies the full mathematical contract imported in BA-01. This is a complete local proof for the cumulative course.

The exact inputs are BK-01 for Hilbert projection, bounded operators and adjoints; BK-02 for the finite-vector bicommutant proof; and CP-03–06 for the concrete predual and its vector-series description. No tensor-product theorem, separability assumption or uniform bound on a strongly convergent net is required.

The countable Hilbert sum

For any complex Hilbert space $H$, let $L$ consist of sequences $\xi=(\xi_n)$ with $\sum_n\|\xi_n\|^2<\infty$. Its inner product, linear in the first variable, is $\langle\xi,\eta\rangle_L =\sum_n\langle\xi_n,\eta_n\rangle_H.$ The series is absolutely convergent by finite Cauchy–Schwarz and passage to the increasing real partial sums. Truncating a sequence proves density of the finite-support sequences.

For completeness, a Cauchy sequence $\xi^{(j)}$ in $L$ is Cauchy in every coordinate, giving limits $\xi_n\in H$. The original sequence has a common norm bound $C$. Taking coordinate limits in each finite sum gives $\sum_{n\leq N}\|\xi_n\|^2\leq C^2$; hence $\xi\in L$. Given $\varepsilon>0$, choose $j_0$ with $\|\xi^{(j)}-\xi^{(k)}\|_L\leq\varepsilon$ for $j,k\geq j_0$. Fix $j\geq j_0$, take $k\to\infty$ in every finite partial sum of squared differences, and then take the supremum over the number of coordinates. This gives $\|\xi^{(j)}-\xi\|_L\leq\varepsilon$. Thus $L$ is a Hilbert space.

Let $J_n:H\to L$ insert a vector in coordinate $n$, and let $P_n:L\to H$ extract that coordinate. They have norm at most one and satisfy $J_n^*=P_n$. If $H\ne0$, both norms are one. For $a\in B(H)$, the diagonal operator $D(a)$ on $L$ satisfies $\|D(a)\|=\|a\|,\qquad D(a)^*=D(a^*).$ The upper norm bound follows by summing the coordinate estimates; a vector in a single coordinate gives the lower bound. The adjoint identity follows first on finite-support vectors, then on all vectors by continuity. These facts also give multiplicativity and linearity of $D$.

Approximation of a square-summable family

Let $A\subset B(H)$ be a unital complex *-subalgebra, and let $T\in A''$. For any $\xi\in L$ and $\varepsilon>0$, we prove that there is $a\in A$ with $\|D(T-a)\xi\|_L<\varepsilon. \tag{CA.1}$ On the zero Hilbert space the statement is immediate. Otherwise let $Q$ be the orthogonal projection onto $V=\overline{\{D(a)\xi:a\in A\}}\subset L.$ The set being closed is a linear space because $A$ is linear. Multiplicativity and the *-property show that $V$ is invariant under both $D(a)$ and $D(a)^*$. Its orthogonal complement is therefore invariant too, and $Q$ commutes with every $D(a)$.

Each bounded matrix entry $Q_{mn}=P_m QJ_n$ lies in $A'$: multiply $QD(a)=D(a)Q$ on the left by $P_m$ and on the right by $J_n$. Consequently $TQ_{mn}=Q_{mn}T$. For a finite-support vector $\eta$, the $m$-th coordinates of $D(T)Q\eta$ and $QD(T)\eta$ are respectively $T\sum_n Q_{mn}\eta_n,\qquad \sum_n Q_{mn}T\eta_n.$ The sums are finite, so they agree. Agreement of all coordinates gives agreement of the two vectors. Since the two operators are bounded and finite-support vectors are dense, $D(T)Q=QD(T)$ on $L$.

The unit of $A$ puts $\xi$ in $V$. Thus $D(T)\xi=D(T)Q\xi=QD(T)\xi$ also lies in $V$. The defining norm closure of $V$ supplies (CA.1). No bound on the norm of the chosen $a$ has been asserted.

Concatenating finitely many square-summable families into one sequence proves simultaneous approximation for every finite family of sigma-strong seminorms. Choose the single tolerance smaller than all the prescribed positive tolerances. Every separate squared norm is a partial sum of the concatenated squared norm and is therefore no larger.

Strong, weak and ultraweak closures

The preceding approximation includes the finite-vector strong tests, so $A''$ lies in the strong closure of $A$. Conversely $A''$ is weak-operator closed: for each $b\in A'$, the equations $xb=bx$ are closed under all matrix coefficients, as seen by testing the fixed vectors $b\xi$ and $b^*\eta$. It follows that $\begin{gathered} \overline A^{\,\mathrm{strong}}\\ =\overline A^{\,\mathrm{weak}}=A''. \end{gathered} \tag{CA.2}$ This also follows from BK-02, but the direct argument records the full imported conclusion.

For the ultraweak statement, take a finite collection of normal functionals $f_k$ on $B(H)$. The exact CP vector-series representation gives $\begin{gathered} f_k(x)=\sum_n\langle x u_{k,n},v_{k,n}\rangle,\\ \sum_n\|u_{k,n}\|\,\|v_{k,n}\|<\infty. \end{gathered}$ Omit zero terms. Rescale each pair by reciprocal positive numbers so that both norms equal $\sqrt{\|u_{k,n}\|\|v_{k,n}\|}$. This preserves every coefficient and makes both sequences square summable.

Put $U_k=(u_{k,n})_n$ and $V_k=(v_{k,n})_n$ in the countable Hilbert sum. For each $k$, Cauchy–Schwarz gives $\begin{gathered} |f_k(T-a)|\\ \leq E_k\|V_k\|_L,\\ E_k\\ =\|D(T-a)U_k\|_L. \end{gathered} \tag{CA.3}$ Concatenate all the $u$-families and apply (CA.1), with tolerance small enough for the finitely many desired bounds in (CA.3). Thus every ultraweak neighborhood of $T$ meets $A$. The reverse inclusion follows because every vector coefficient is normal, so the weakly closed set $A''$ is ultraweakly closed as well. We have proved $\overline A^{\,\mathrm{ultraweak}}=A''. \tag{CA.4}$ In a representation with identity projection $p$, apply these statements on $pH$. Compression and zero extension preserve the vector-series coefficients after replacing both vectors by their projections under $p$. This identifies the inherited ultraweak topology with the corner topology. It covers precisely the supported application in BA-01.