Operator density from finite vector tests

Written by GPT-6.1 Sol (OpenAI), Ultra, October 2026. Original text and embedded diagram: public domain (CC0).

A strong operator neighbourhood asks for good approximation on finitely many vectors. Commutants turn those finite tests into one invariant-subspace problem. A rational function then converts self-adjoint approximants into contractions. These two constructions prove the positive matrix density needed to reconstruct completely positive maps with values in a C*-algebra.

We use the complete Hilbert-space, projection and series-vector proofs H00–H03, their arbitrary-Hilbert-space identification P00, C*-unitization, calculus and approximate identities F02–F09, and real convex separation SEP. The C*-norm and order on finite matrices, including entrywise representations, are proved in Completely positive maps, Section 2. Inner products are linear in their second variable. No separability assumption is imposed.

For a representation $\pi:A\to B(H)$, put

The notation $\overline{D H}$ means the closed linear span. We work on $K$, and extend operators on $K$ by zero on $K^\perp$ whenever the original space $H$ is used. Write

where these commutants are taken in $B(K)$. The unit of this algebra on $H$ is $P$. When $K=0$, all supported operators and all the approximations below are zero.

ODF01. Bounded strong limits and ultraweak tests

Strong convergence means $T_i\xi\to T\xi$ in norm for every vector $\xi$. Weak operator convergence means convergence of every scalar $\langle\eta,T_i\xi\rangle$. Strong* convergence means that both $T_i\to T$ and $T_i^*\to T^*$ strongly. H03 describes the ultraweak topology by the functionals

P00 makes that description available on every Hilbert space, including spaces with uncountable orthonormal bases.

Bounded product lemma. If $S_i\to S$ and $T_i\to T$ strongly and $\sup_i\|S_i\|\le C<\infty$, then $S_iT_i\to ST$ strongly.

Proof. For each $\xi$,

The last vector $T\xi$ is fixed. No bound on the norms of the $T_i$ is needed for this assertion. If both nets and their adjoints converge strongly and both nets have uniform norm bounds, applying the same argument to $T_i^*S_i^*$ also proves strong* convergence of the products. $\square$

Bounded topology lemma. If $T_i\to T$ in the weak operator topology and $\sup_i\|T_i\|\le C$, then $T_i\to T$ ultraweakly. In particular every uniformly bounded strongly convergent net converges ultraweakly.

Proof. For unit vectors $\eta,\xi$, the scalar limit gives $|\langle\eta,T\xi\rangle|\le C$, and the Hilbert norming formula gives $\|T\|\le C$. Given (ODF1), its tail after $N$, evaluated at $T_i-T$, has modulus at most

Choose $N$ to make this small. The initial finite sum tends to zero by weak operator convergence, so the whole series tends to zero. If $C=0$, every operator involved is zero. This is a net argument: convergence of the finite sum requires only finitely many eventual conditions. $\square$

The uniform bounds in these lemmas are hypotheses. Strong convergence of an arbitrary net supplies no uniform operator-norm bound by itself.

ODF02. The essential representation and its support

Every $d\in D$ preserves $K$. Since $D=D^*$, it also preserves $K^\perp$. A vector $\zeta\in K^\perp$ satisfies $\langle d^*\eta,\zeta\rangle=0$ for every $d\in D$, $\eta\in H$, so $d\zeta=0$. Thus $d=PdP$.

Let $(e_\lambda)$ be the positive contractive approximate identity in $A$ supplied by F09. On a vector $\pi(a)\eta$,

F05 gives $\|\pi(e_\lambda)\|\le1$. Finite sums of these vectors are dense in $K$; the uniform bound extends convergence to all of $K$. Every $\pi(e_\lambda)$ vanishes on $K^\perp$. Consequently

On $K$, this is a strongly convergent approximate identity with limit $1_K$. In particular the restricted representation is nondegenerate. These statements include $A=0$ and $\pi=0$.

The supported algebra $M$, rather than the full bicommutant in a degenerate ambient space, is the closure to be approximated. For example, if $D=0$ on a nonzero $H$, its strong closure is zero although its full bicommutant contains $1_H$.

ODF03. The bicommutant theorem on finitely many vectors

Theorem. On $K$, $D$ is strongly dense in $M=D''$. More precisely, for $T\in M$, $\xi_1,\ldots,\xi_r\in K$, and $\varepsilon>0$, there is $a\in A$ such that

Proof. The zero essential space and the assertion with no test vectors are immediate: take $a=0$. For $r\ge1$, write $\boldsymbol\xi=(\xi_1,\ldots,\xi_r)\in K^r$, and consider the closed linear subspace

It is invariant under every diagonal operator $\Delta(d)=\operatorname{diag}(d,\ldots,d)$, by multiplication in $D$. It is reducing, since $d^*\in D$. Its orthogonal projection $Q$ therefore commutes with each $\Delta(d)$, by H01.

Let $R_j:K\to K^r$ be the coordinate inclusion. The entries $Q_{ij}=R_i^*QR_j$ are bounded operators, and the equation $Q\Delta(d)=\Delta(d)Q$ says $Q_{ij}d=dQ_{ij}$. Hence every $Q_{ij}$ lies in $D'$. Since $T\in D''$, it commutes with every $Q_{ij}$, and therefore $\Delta(T)Q=Q\Delta(T)$. Equation (ODF2) puts $\boldsymbol\xi$ in $L$. It follows that

The definition of $L$ now provides one $d=\pi(a)$ satisfying (ODF3).

Conversely, $D''$ is strongly closed and contains $D$. Indeed, for $b\in D'$, the equation $S_ib=bS_i$ passes to a strong limit, since both evaluations use fixed vectors. Thus every strong limit from $D$ belongs to $D''$. The two inclusions prove the theorem. $\square$

The indices consisting of a finite vector set and a positive error, ordered by enlarging the set and decreasing the error, give a directed set. Choosing one approximant for each such index turns (ODF3) into a strong-convergent net. No countable cofinal family is required.

Figure. The maps $\Delta(d),\Delta(T):K^r\to K^r$ and $Q:K^r\to L\subset K^r$ are the actual operators in the ODF03 proof. The diagram depicts their invariant-subspace mechanism; it asserts no finite-dimensionality of $K$ or $L$.

ODF04. Self-adjoint strong density by real separation

Lemma. Every $y=y^*\in M$ is a strong limit of operators $\pi(a_i)$ with $a_i=a_i^*\in A$. Their norms are not asserted bounded.

Proof. ODF03 supplies a net $d_i\in D$ converging strongly to $y$. Strong convergence implies weak operator convergence. The adjoints converge weakly to $y$, because

Thus $h_i=(d_i+d_i^*)/2$ converges weakly to $y$. Each $h_i$ has a self-adjoint lift $(a_i+a_i^*)/2$ in $A$.

Fix $\xi_1,\ldots,\xi_r$. The real linear map

takes the real vector space $D_h$ into the real Hilbert space underlying $K^r$. Every continuous real linear functional on that Hilbert space has the form $\operatorname{Re}\sum_j\langle\eta_j,v_j\rangle$. To see this from H00, if $\ell$ is real linear, then $F(v)=\ell(v)-i\ell(iv)$ is bounded complex linear and has a Hilbert representative; taking its real part recovers $\ell$.

The weak operator convergence just proved therefore puts $\Phi(y)$ in the Banach weak closure of $\Phi(D_h)$. This image is a real linear subspace, hence convex. The real separation proof SEP says that its weak and norm closures agree. Choose $h\in D_h$ with $\sum_j\|(h-y)\xi_j\|^2<\varepsilon^2$. Directing these choices by finite sets and errors proves strong density. $\square$

ODF05. The rational contraction cutoff

Resolvent lemma. If $b_i=b_i^*\to y=y^*$ strongly, then

strongly, even when the $\|b_i\|$ have no uniform bound.

Proof. For a self-adjoint bounded operator $b$,

The operator has closed range and zero kernel. Its adjoint $b\mp i1_K$ has zero kernel too, so that range is dense and therefore all of $K$. Its inverse has norm at most one. The elementary inverse identity gives, for $z=i$ or $z=-i$,

For a fixed $\xi$, the norm of the right side applied to $\xi$ is at most $\|(y-b_i)(y-z)^{-1}\xi\|\to0$. The rightmost vector is fixed; the leftmost inverse has its own uniform bound. This proves the lemma without estimating $\|b_i\|$. $\square$

Put

It is real, $f(0)=0$, and $|f(t)|\le1$, since $2|t|\le1+t^2$. F06 gives $f(a)\in A_h$ with $\|f(a)\|\le1$ for every $a=a^*\in A$. This assertion uses the forced unitization when necessary, and the vanishing value at zero puts the result back in $A$. The resolvent lemma and (ODF4) give $f(b_i)\to f(y)$ strongly.

Theorem (self-adjoint contraction density). Every $h=h^*\in M$ with $\|h\|\le1$ is a strong* limit of $\pi(a_i)$, where $a_i=a_i^*\in A$ and $\|a_i\|\le1$.

Proof. If $K=0$, take the constant zero net. Otherwise, in $B(K)$, put

These operators belong to $M$, by its C*-calculus and inverse calculus. They commute, and $y=y^*$. Indeed $s$ is obtained by continuous functions of $h^2$, and $1_K+s\ge1_K$ is invertible. The identity $h^2=1_K-s^2$ gives

Consequently $f(y)=2y(1_K+y^2)^{-1}=h$. This includes $\|h\|=1$.

ODF04 supplies self-adjoint lifts $c_i\in A_h$ with $\pi(c_i)\to y$ strongly. Take $a_i=f(c_i)$. The scalar bound gives $\|a_i\|\le1$ in $A$, not just in its represented image. Extending $\pi$ to the forced unitization on $K$ sends the new unit to $1_K$. The extension is a unital *-homomorphism, so it preserves inverses and the rational identity (ODF4); hence

All these operators are self-adjoint, so this is strong* convergence. $\square$

ODF06. Positive matrices with the same norm bound

Theorem (positive matrix Kaplansky density). For every $n\ge1$ and every $X\in M_n(M)_+$, there is a net $C_i\in M_n(A)_+$ such that

The same convergences hold after extension by zero to $H^n$.

Proof. First identify the strongly generated algebra of $\pi^{(n)}(M_n(A))$. Its elements have entries in $D$. The algebra $M_n(M)$ is strongly closed: a strong limit has entries extracted by the coordinate inclusions and projections, and each limiting entry belongs to $M$. It contains $M_n(D)$.

Conversely, let $Z=[z_{pq}]\in M_n(M)$, and fix finitely many vectors in $K^n$. For each entry $z_{pq}$, ODF03 supplies an element of $D$ approximating that entry on all the finitely many $q$-th vector components. If each entry error on each component is less than $\varepsilon/(n\sqrt n)$, each output row error is less than $\varepsilon/\sqrt n$, and the full output vector error is less than $\varepsilon$. Thus $M_n(D)$ is strongly dense in $M_n(M)$. Its essential space is $K^n$: diagonal copies of $\pi(e_\lambda)$ converge strongly to $1_{K^n}$. Applying ODF03 to the representation $\pi^{(n)}$ identifies its bicommutant with $M_n(M)$.

If $X=0$, take the constant zero net. Otherwise put $c=\|X\|>0$, and let $Z=(X/c)^{1/2}\in M_n(M)$. It is a self-adjoint contraction. Apply ODF05 to the C*-algebra $M_n(A)$ and its representation $\pi^{(n)}$. We obtain self-adjoint $B_i\in M_n(A)$, $\|B_i\|\le1$, with $\pi^{(n)}(B_i)\to Z$ strongly. Set

These matrices are positive and satisfy $\|C_i\|\le c$. ODF01's bounded product estimate gives $\pi^{(n)}(C_i)\to cZ^2=X$ strongly. The matrices and the target are self-adjoint, so convergence is strong*. Their represented norms are bounded by $c$, and ODF01 then gives ultraweak convergence.

For extension by zero, a vector of $H^n$ is tested only through its orthogonal projection onto $K^n$. This proves strong and strong* convergence on $H^n$; the same norm bound and ODF01 give ultraweak convergence there. $\square$

In particular the positive part of the unit ball of a faithfully represented C*-algebra is strongly and ultraweakly dense in the positive unit ball of its generated von Neumann algebra. The statement for Choi matrices uses (ODF5) with that same finite matrix size.

ODF07. The entire contraction ball and the closure algebra

Every contraction $T\in M$ is a strong* limit of $\pi(a_i)$ with $\|a_i\|\le1$. To prove this, the self-adjoint matrix

has norm $\|T\|$. Its square is $\operatorname{diag}(TT^*,T^*T)$, and the C*-identity and the norm of a block diagonal operator prove this equality. Apply ODF05 at matrix size two to obtain self-adjoint contractions

The matrix norm inequality $\|a_i\|\le\|B_i\|\le1$ is part of the matrix construction linked above. The two off-diagonal corners give $\pi(a_i)\to T$ and $\pi(a_i)^*\to T^*$ strongly. This proves the assertion. The uniformly bounded approximants also converge ultraweakly, by ODF01.

It follows that the strong, strong* and ultraweak closures of $D$, extended by zero on $K^\perp$, all equal the supported algebra $M$. For the reverse inclusion in the ultraweak closure, $D''$ on $K$ is ultraweakly closed: for each $b\in D'$, the equations $\langle\eta,(Tb-bT)\xi\rangle=0$ are differences of one-term functionals (ODF1). On $H$, the additional equations $T=PTP$ are also ultraweakly closed by the same one-term tests. The strong* and strong reverse inclusions follow from their fixed-vector versions. On a nondegenerate representation $K=H$, so each closure is precisely $\pi(A)''$.

References

The bicommutant theorem is due to John von Neumann. The finite-vector projection argument and the rational cutoff above give its density consequence and Kaplansky's theorem at the hypotheses stated here.

Irving Kaplansky, A theorem on rings of operators, Pacific Journal of Mathematics 1 (1951), 227–232. Theorem 1, printed pp.227–231, is the bounded density theorem; Lemmas 1–5, printed pp.228–230, develop the bounded-calculus and Cayley-resolvent method. The real separation, finite amplification and nonunital support arguments needed in this lesson are proved above from the linked programme foundations.