Original text: CC0 1.0. Prerequisite proofs and component terms.

Recovering the commutant from right-bounded vectors

The elementary multiplication construction produces a right algebra inside the Hilbert completion, but it does not yet show that this algebra is dense. The missing approximation must control a vector and its adjoint-involution image at the same time. This unit constructs such approximations from the spectral cutoffs of a closed right multiplier. It then recovers the full commutant and proves the product-core and ideal-intersection statements needed for left/right dualization.

The mathematical antecedents are Takesaki, Theory of Operator Algebras II, Chapter VI, Lemmas 1.12–1.15. The arguments use closed operator domains, a cutoff ideal, and explicit contractive approximants. No modular commutant theorem is a premise.

OA-MOD-RD-01 — The exact starting interface

Use the arbitrary left Hilbert algebra $\mathcal A\subseteq H$ of OA-MOD-HA. Inner products are linear in the first variable. The algebra need not have a unit, and $H$ need not be separable. Write $$M=L(\mathcal A)'',\qquad S=\overline{(a\mapsto a^\sharp)},\qquad F=S^*,$$ $$\mathcal B_r=\{\eta:a\mapsto L_a\eta\text{ is bounded for the Hilbert norm}\}, \quad \mathfrak n_r=\{R_\eta:\eta\in\mathcal B_r\}, \quad \mathcal A_r=\mathcal B_r\cap D(F). \tag{RD.1}$$ Here $R_\eta a=L_a\eta$ for $a\in\mathcal A$. OA-MOD-HA-05–08 prove that $R$ is linear and injective, that $\mathfrak n_r$ is a left ideal of $M'$, and that $$R_{x\eta}=xR_\eta,\qquad R_\eta^*\zeta\in\mathcal A_r,\qquad F(R_\eta^*\zeta)=R_\zeta^*\eta \tag{RD.2}$$ for $x\in M'$, $\eta,\zeta\in\mathcal B_r$. For $\eta\in\mathcal A_r$, $$F\eta\in\mathcal A_r,\qquad R_{F\eta}=R_\eta^*. \tag{RD.3}$$ The product $\eta\zeta=R_\zeta\eta$ makes $\mathcal A_r$ an involutive algebra with bounded right multiplication. Its density is the first target here.

An immediate operator form of (RD.2) will be useful: $$\mathfrak n_r^*\mathfrak n_r\subseteq R(\mathcal A_r). \tag{RD.4}$$ This means every product $b^*c$, $b,c\in\mathfrak n_r$, belongs to $R(\mathcal A_r)$, and therefore so does its finite linear span. Indeed, write $b=R_\eta,c=R_\zeta$; covariance gives $b^*c=R_{R_\eta^*\zeta}$, and (RD.2) gives the required vector.

For $\eta\in D(F)$, OA-MOD-HA-07 supplies a closed densely defined operator $$T_\eta=\overline{\,a\mapsto L_a\eta\,},\qquad \mathcal A\subseteq D(T_\eta),\quad T_\eta a=L_a\eta. \tag{RD.5}$$ It is affiliated with $M'$, and $$\mathcal A\subseteq D(T_\eta^*),\qquad T_\eta^*a=L_aF\eta. \tag{RD.6}$$ Only the restriction in (RD.6) is assumed; equality between $T_\eta^*$ and the closure of the map for $F\eta$ is not a premise.

The other dependencies are OA-MOD-TC-03 for ordinary and conjugate-linear graph adjoints, OA-MOD-QF-03–04 for closed nonnegative forms and unitary symmetry, and OA-MOD-BK-02–08 for bounded operators, topology, inverse order, support cutoffs, and unitary tests. The unbounded self-adjoint spectral calculus is the explicit RD-DEP-SPECTRAL contract: on arbitrary Hilbert spaces it includes spectral integral domains, positive square roots, unitary transport, and dominated convergence for each vector's finite spectral measure. Uniform polynomial approximation on a compact real interval is also used in OA-MOD-RD-06. These are also the analytic prerequisites of Recovering operators from energy forms and Closing an involution and recovering its modular data.

OA-MOD-RD-02 — Polar cutoffs for a closed linear operator

The right multiplier in (RD.5) need not be bounded or injective. We first establish the exact polar facts needed in that generality.

Lemma. Let $T:D(T)\subseteq H\to H$ be closed and densely defined. There are nonnegative self-adjoint operators $h,k$ and a bounded partial isometry $u$ such that $$\begin{aligned} h&=(T^*T)^{1/2},&D(h)&=D(T),\\ T&=uh=ku,&D(ku)&=D(h),\\ T^*&=hu^*,& D(T^*)&=\{\zeta:u^*\zeta\in D(h)\}=D(k),\\ k&=(TT^*)^{1/2}.&& \end{aligned} \tag{RD.7}$$ The initial and final projections are $$p=u^*u=P_{\overline{\operatorname{ran}T^*}}, \qquad q=uu^*=P_{\overline{\operatorname{ran}T}}. \tag{RD.8}$$ If $T$ is affiliated with a von Neumann algebra $N$, then $u,p,q\in N$, and all spectral projections of $h,k$ belong to $N$.

For every bounded Borel $f:[0,\infty)\to\mathbb C$, $$f(k)=u f(h)u^*+f(0)(I-q),\qquad f(k)u=u f(h). \tag{RD.9}$$ In particular, for $f\in C_c(0,\infty)$, extended by zero at zero, $$f(k)T\subseteq kf(k)u,\qquad f(h)T^*\subseteq hf(h)u^*, \tag{RD.10}$$ where the operators on the right are bounded on $H$. Their norms are at most $\sup_{t>0}|t f(t)|$.

Proof. The form $$q_T(\xi,\zeta)=\langle T\xi,T\zeta\rangle,\qquad D(q_T)=D(T),$$ is closed because its form norm is the graph norm of $T$. Its representing operator in OA-MOD-QF-03 is $T^*T$: for $\xi\in D(T)$, existence of $y$ with $\langle T\xi,T\zeta\rangle=\langle y,\zeta\rangle$ for all $\zeta\in D(T)$ is precisely $T\xi\in D(T^*)$, $T^*T\xi=y$. Consequently $$D(h)=D(T),\qquad \langle h\xi,h\zeta\rangle=\langle T\xi,T\zeta\rangle. \tag{RD.11}$$ Thus $\ker h=\ker T$. The rule $h\xi\mapsto T\xi$ is well-defined and isometric. Extend it to a unitary $U:pH\to qH$, where initially $p$ projects onto $\overline{\operatorname{ran}h}$ and $q$ onto $\overline{\operatorname{ran}T}$; set $u=U$ on $pH$ and zero on $(I-p)H=\ker h$. This gives $T=uh$.

For $\zeta\in H$, the pairing $$\langle T\xi,\zeta\rangle=\langle h\xi,u^*\zeta\rangle \quad(\xi\in D(h))$$ is bounded in $\|\xi\|$ exactly when $u^*\zeta\in D(h^*)=D(h)$. This proves the asserted domain and action of $T^*$. Its range equals $\operatorname{ran}h$: for any $h\xi$, replacing $\xi$ by $p\xi$ changes neither membership in $D(h)$ nor its image, and $T^*(u p\xi)=h p\xi=h\xi$. Thus the projection $p$ also has the description in (RD.8).

The spectral projection $p$ reduces $h$, so its restriction $h_p$ is self-adjoint on $pH$. Define $$k=(Uh_pU^*)\oplus0 \quad\text{on }qH\oplus(I-q)H.$$ This is nonnegative self-adjoint with $$D(k)=\{\zeta:u^*\zeta\in D(h)\},\qquad k\zeta=uh u^*\zeta.$$ Its square has domain $\{\zeta:u^*\zeta\in D(h^2)\}$. The product $TT^*$ has exactly the same domain and action $uh^2u^*$, by the already-proved formulas for $T,T^*$. Thus $k^2=TT^*$, including domains. Also $$u\xi\in D(k)\ \Longleftrightarrow\ p\xi\in D(h) \ \Longleftrightarrow\ \xi\in D(h),$$ because $\ker h\subseteq D(h)$. On that domain $ku\xi=uh\xi=T\xi$. This proves all of (RD.7).

Spectral calculus on the displayed orthogonal decomposition of $k$ proves (RD.9). Multiplication by $tf(t)$, for the compactly supported $f$ in (RD.10), is a bounded spectral function. On $D(T)$, the first identity in (RD.10) follows from $T=ku$; on $D(T^*)$, the second follows from $T^*=hu^*$. They are inclusions because the bounded extensions on the right have domain all of $H$.

Finally suppose $T$ is affiliated with $N$. For each unitary $v\in N'$, the form $q_T$ has invariant domain and satisfies $q_T[v\xi]=q_T[\xi]$. Uniqueness and unitary symmetry of closed-form representation imply that $v$ commutes with $h$ and its spectral projections. On $\operatorname{ran}h$, $$uvh\xi=uhv\xi=Tv\xi=vT\xi=vuh\xi.$$ On $\ker h$ the two sides also agree, since that kernel is invariant under $v$. Hence $u$ commutes with every unitary of $N'$, so $u\in N$ by the unitary test and bicommutant theorem. This gives $p,q\in N$, and spectral transport in (RD.9) gives the spectral projections of $k$ in $N$. No range or kernel has been assumed trivial. $\square$

OA-MOD-RD-03 — Bounded cutoffs land in the operator ideal

Fix $\eta\in D(F)$. Apply OA-MOD-RD-02 to $T=T_\eta$, and retain $h,k,u,p,q$. All their bounded spectral functions and $u$ lie in $M'$.

Cutoff theorem. For $f\in C_c(0,\infty)$, extended by zero at zero, $$\begin{aligned} f(k)\eta&\in\mathcal B_r, &R_{f(k)\eta}&=kf(k)u,\\ f(h)F\eta&\in\mathcal B_r, &R_{f(h)F\eta}&=hf(h)u^*. \end{aligned} \tag{RD.12}$$ Moreover, $$f(h),f(k)\in R(\mathcal A_r), \tag{RD.13}$$ and both cutoff vectors in (RD.12) belong to $\mathcal A_r^2$. Their adjoint-involution relation is $$F(f(k)\eta)=\overline f(h)F\eta. \tag{RD.14}$$ Here $\overline f(t)=\overline{f(t)}$, and $\mathcal A_r^2$ denotes the linear span of products.

Proof of the bounded formulas. For $a\in\mathcal A$, the cutoff operators commute with $L_a$. Equations (RD.5–6) and (RD.10) give $$\begin{aligned} L_a f(k)\eta&=f(k)T a=kf(k)u a,\\ L_a f(h)F\eta&=f(h)T^*a=hf(h)u^*a. \end{aligned}$$ The final operators are bounded on $H$, proving (RD.12) directly from the definition of right boundedness.

Proof of the ideal assertion. By covariance (RD.2), $$u^*R_{f(k)\eta}=h f(h)\in\mathfrak n_r, \qquad uR_{f(h)F\eta}=k f(k)\in\mathfrak n_r. \tag{RD.15}$$ The polar transport identities justify these equalities also on the kernel subspaces, since $f(0)=0$. Now $g(t)=f(t)/t$ is again in $C_c(0,\infty)$. Apply (RD.15) to $g$ to conclude $f(h),f(k)\in\mathfrak n_r$.

Choose a real $\chi\in C_c(0,\infty)$ that equals one on $\operatorname{supp}f$. Such a function can be chosen piecewise linear on an interval with positive endpoints containing that compact support. Both $\chi(h)$ and $f(h)$ lie in $\mathfrak n_r$, and $$f(h)=\chi(h)^*f(h)\in\mathfrak n_r^*\mathfrak n_r \subseteq R(\mathcal A_r).$$ The same argument applies to $k$, proving (RD.13).

Proof of the vector and domain assertions. Set $v=f(k)\eta$, already known to lie in $\mathcal B_r$. As $\chi(k)$ is self-adjoint and belongs to $R(\mathcal A_r)$, it can be written $R_\alpha^*$ for some $\alpha\in\mathcal A_r$, by (RD.3). Since $\chi(k)v=v$, equation (RD.2) gives $v\in\mathcal A_r$. Write also $\chi(k)=R_\beta$, $\beta\in\mathcal A_r$. Then $$v=\chi(k)v=R_\beta v=v\beta\in\mathcal A_r^2.$$ The same reasoning using $h$ proves $f(h)F\eta\in\mathcal A_r^2$.

We may now apply (RD.3) to $v$. Taking adjoints in its bounded formula in (RD.12) gives $$R_{Fv}=R_v^*=h\overline f(h)u^* =R_{\overline f(h)F\eta}.$$ Both vectors are right bounded, so injectivity of $R$ proves (RD.14). The argument established membership in $D(F)$ before computing this value. $\square$

OA-MOD-RD-04 — Graph density of the right algebra

The graph norm of $F$ is $$\|\eta\|_F^2=\|\eta\|^2+\|F\eta\|^2. \tag{RD.16}$$

Theorem. Both $\mathcal A_r$ and $\mathcal A_r^2$ are graph cores for $F$. In particular, both are dense in $H$, $\mathcal A_r$ is a right Hilbert algebra, and $$\overline{F|_{\mathcal A_r}}=F,\qquad (F|_{\mathcal A_r})^*=F^*=S. \tag{RD.17}$$

Proof. Fix $\eta\in D(F)$ and its polar data from OA-MOD-RD-03. First we need the support identities $$q\eta=\eta,\qquad pF\eta=F\eta. \tag{RD.18}$$ For $a\in\mathcal A$, the vector $L_a\eta=T a$ lies in $qH$. Since $q\in M'$, $$L_a(I-q)\eta=(I-q)L_a\eta=0.$$ The common-kernel criterion OA-MOD-HA-02 gives $(I-q)\eta=0$. Similarly $L_aF\eta=T^*a\in pH$, and the same criterion proves $pF\eta=F\eta$. Thus the support projections need not be the identity, but they fix the two particular vectors we approximate.

For $n\geq1$, define on $(0,\infty)$ $$f_n(t)=\max\!\left(0,\min\!\left(1,\,2-\frac{|\log t|}{n}\right)\right). \tag{RD.19}$$ Each function is continuous, supported in $[e^{-2n},e^{2n}]$, takes values in $[0,1]$, and increases pointwise to one on $(0,\infty)$. Extend it by zero at zero. The spectral calculus and scalar dominated convergence give $$f_n(k)\longrightarrow q,\qquad f_n(h)\longrightarrow p \quad\text{strongly}. \tag{RD.20}$$ OA-MOD-RD-03 supplies $\eta_n=f_n(k)\eta\in\mathcal A_r^2$ and $F\eta_n=f_n(h)F\eta$. Equations (RD.18–20) therefore imply $$\|\eta_n-\eta\|^2+\|F\eta_n-F\eta\|^2\longrightarrow0. \tag{RD.21}$$ This proves that $\mathcal A_r^2$ is a graph core. Since $\mathcal A_r^2\subseteq\mathcal A_r\subseteq D(F)$, the same is true of $\mathcal A_r$. The domain $D(F)$ is dense in $H$, so both subspaces are Hilbert-norm dense too.

OA-MOD-HA-08 established the first three right Hilbert algebra properties; density of $\mathcal A_r^2$ supplies the fourth. Graph density proves the first equality in (RD.17), and taking adjoints gives the second by $F=S^*$, $S^{**}=S$. $\square$

The approximating sequence depends on $\eta$, through its multiplier $T_\eta$. It is not a countable dense family for $H$, nor an assertion that a global net of algebra operators has a sequential replacement.

OA-MOD-RD-05 — Contractive approximation of the entire commutant

Put $\mathcal D=R(\mathcal A_r)\subseteq M'$. It is a *-algebra, by the anti-representation and adjoint formulas in OA-MOD-HA-08. It is nondegenerate because its action on $\mathcal A_r$ contains the dense product span $\mathcal A_r^2$.

Theorem. There is a net of positive contractions $e_i\in\mathcal D$ with $e_i\to I$ strongly. For every $x\in M'$, $$e_i x e_i\in\mathcal D,\qquad \|e_i x e_i\|\leq\|x\|,\qquad e_i x e_i\longrightarrow x \quad\text{strongly*}. \tag{RD.22}$$ Consequently $$R(\mathcal A_r)''=M'. \tag{RD.23}$$ Strong* convergence means strong convergence of both the operators and their adjoints.

Proof. Index a net by pairs $(E,\varepsilon)$, where $E$ is a finite subset of $\mathcal D$ and $\varepsilon>0$; enlarge $E$ and decrease $\varepsilon$. Put $$b_E=\sum_{r\in E}r^*r,\qquad c_{E,\varepsilon}=b_E(b_E+\varepsilon I)^{-1}. \tag{RD.24}$$ These are positive contractions $c_{E,\varepsilon}\in M'$. By inverse order, they increase when $E$ enlarges and $\varepsilon$ decreases. For fixed $E$, letting $\varepsilon\downarrow0$ gives the support projection $s(b_E)$, by OA-MOD-BK-06.

The family of these supports spans $H$. Indeed, $$\bigcap_E\ker b_E=\bigcap_{r\in\mathcal D}\ker r=\{0\}. \tag{RD.25}$$ The first equality follows by expanding $\langle b_E\xi,\xi\rangle=\sum_{r\in E}\|r\xi\|^2$. For the second, a common-kernel vector is orthogonal to every $r^*\zeta$; *-closure and nondegeneracy of $\mathcal D$ make these vectors span a dense subspace.

The increasing contraction net $c_{E,\varepsilon}$ has a strong limit $c$ by OA-MOD-BK-04. This limit satisfies $s(b_E)\leq c\leq I$ for every $E$. A positive contraction dominating a projection acts as the identity on its range, by OA-MOD-BK-06. Equations (RD.25) thus imply $c=I$.

We next make approximants that lie in $\mathcal D$ itself. Since $b_E\in\mathcal D\subseteq\mathfrak n_r$, the left-ideal property gives $c_{E,\varepsilon}\in\mathfrak n_r$. Define $$e_{E,\varepsilon}=c_{E,\varepsilon}^*c_{E,\varepsilon} =c_{E,\varepsilon}^{\,2}. \tag{RD.26}$$ Equation (RD.4) puts $e_{E,\varepsilon}$ in $\mathcal D$. It is a positive contraction, and $$\|(e_{E,\varepsilon}-I)\xi\| \leq 2\|(c_{E,\varepsilon}-I)\xi\|\longrightarrow0.$$ These self-adjoint operators converge strongly*. Their squares are not asserted to form an increasing net.

Finally $e_i\in\mathfrak n_r$ and $xe_i\in\mathfrak n_r$, so $$e_i x e_i=e_i^*(xe_i)\in\mathfrak n_r^*\mathfrak n_r \subseteq\mathcal D.$$ Their norms are at most $\|x\|$, and $$\|(e_i x e_i-x)\xi\| \leq\|x\|\|(e_i-I)\xi\|+\|(e_i-I)x\xi\|\longrightarrow0.$$ Replace $x$ by $x^*$ to obtain the adjoint convergence. Since $\mathcal D\subseteq M'$, (RD.22) proves that its strong closure is $M'$, and hence its bicommutant is $M'$, using the nondegenerate density lemma OA-MOD-HA-03. $\square$

This explicit net proves the bounded approximation needed here without assuming a density theorem for unit balls. On a bounded set, its strong convergence is also ultraweak convergence by OA-MOD-BK-03.

OA-MOD-RD-06 — Products form a core for the original involution

Theorem. The product span $\mathcal A^2$ is dense in $D(S)$ for the graph norm of $S$. Equivalently, for $v,w\in H$, $$\left[ \langle a^\sharp b,v\rangle=\langle w,b^\sharp a\rangle \quad\text{for every }a,b\in\mathcal A \right] \quad\Longleftrightarrow\quad v\in D(F),\ Fv=w. \tag{RD.27}$$

Proof of graph density. It suffices to approximate each $a\in\mathcal A$, because $\mathcal A$ is already a graph core for $S$. Write $A=L_a$. The net in OA-MOD-RD-05 has $e_i=R_{\beta_i}$, $\beta_i\in\mathcal A_r$. Therefore $$e_i a=L_a\beta_i=A\beta_i\longrightarrow a.$$ This proves $a\in\overline{\operatorname{ran}A}$. Applying the same argument to $a^\sharp$ gives $a^\sharp\in\overline{\operatorname{ran}A^*}$.

For $\delta>0$, let $g_\delta(t)=t/(t+\delta)$ on $[0,\infty)$. The support-cutoff theorem gives $$g_\delta(AA^*)a\longrightarrow a,\qquad g_\delta(A^*A)a^\sharp\longrightarrow a^\sharp \quad(\delta\downarrow0). \tag{RD.28}$$ These cutoffs need not themselves be left multipliers of vectors in $\mathcal A$, so we approximate their scalar functions by polynomials.

Take $\delta_n=1/n$. If $A=0$, faithfulness gives $a=0$, and there is nothing to prove. Otherwise choose real polynomials $p_n$ with $p_n(0)=0$ such that $$\sup_{0\leq t\leq\|A\|^2} |p_n(t)-g_{\delta_n}(t)|\leq n^{-1}. \tag{RD.29}$$ Uniform polynomial approximation supplies these: approximate the real function with error $1/(2n)$, then subtract the value of the approximating polynomial at zero. Since $g_{\delta_n}(0)=0$, the final error is at most $1/n$.

Define the algebra vector $$a_n=p_n(aa^\sharp)a. \tag{RD.30}$$ Because the polynomial has zero constant term, this is a finite linear combination of products of at least three algebra vectors and hence belongs to $\mathcal A^2$. Algebraic associativity and the real coefficients give $$a_n=p_n(AA^*)a,\qquad a_n^\sharp=p_n(A^*A)a^\sharp. \tag{RD.31}$$ For the second identity, each term satisfies $\bigl((aa^\sharp)^m a\bigr)^\sharp =a^\sharp(aa^\sharp)^m=(a^\sharp a)^m a^\sharp$. Uniform functional calculus and (RD.28–29) now prove $a_n\to a$ and $a_n^\sharp\to a^\sharp$. Thus $\mathcal A^2$ is graph dense in $\mathcal A$, and therefore in $D(S)$. Explicitly, approximate a vector of $D(S)$ by algebra vectors with graph error $1/n$, and for each of those choose a product-span vector with a further graph error $1/n$.

Proof of the pairing characterization. The forward implication of the right side of (RD.27) to the left follows from the adjoint identity, applied to $b^\sharp a$, whose image under $S$ is $a^\sharp b$. Conversely, the assumed pairings extend by linearity to $$\langle S\xi,v\rangle=\langle w,\xi\rangle \quad(\xi\in\mathcal A^2).$$ Graph density extends this equality to all $\xi\in D(S)$. It is exactly the adjoint-domain test $v\in D(S^*)$, $S^*v=w$. Since $F=S^*$, this proves (RD.27). $\square$

In the modular notation of OA-MOD-TC, the equality of graph norms $\|\xi\|^2+\|S\xi\|^2=\|\xi\|^2+\|\Delta^{1/2}\xi\|^2$ also makes $\mathcal A^2$ a form core for $\Delta$. This does not assert that every product vector belongs to $D(\Delta)$, or that $\mathcal A^2$ is an operator core for $\Delta$.

OA-MOD-RD-07 — The adjoint intersection of the right ideal

Theorem. With $\mathfrak n_r^*=\{b^*:b\in\mathfrak n_r\}$, $$R(\mathcal A_r)=\mathfrak n_r\cap\mathfrak n_r^*. \tag{RD.32}$$ More precisely, if $\eta,\zeta\in\mathcal B_r$, then $$R_\eta^*=R_\zeta \quad\Longleftrightarrow\quad \eta\in D(F)\text{ and }F\eta=\zeta. \tag{RD.33}$$

Proof. If $\eta\in\mathcal A_r$, (RD.3) gives $R_\eta^*=R_{F\eta}$, proving one inclusion in (RD.32) and one implication in (RD.33).

Conversely suppose $\eta,\zeta\in\mathcal B_r$ and $R_\eta^*=R_\zeta$. For $a,b\in\mathcal A$, $$\begin{aligned} \langle a^\sharp b,\eta\rangle &=\langle b,L_a\eta\rangle =\langle b,R_\eta a\rangle\\ &=\langle R_\eta^*b,a\rangle =\langle L_b\zeta,a\rangle =\langle\zeta,b^\sharp a\rangle. \end{aligned}$$ The graph-core characterization (RD.27) proves $\eta\in D(F)$, $F\eta=\zeta$. Finally every operator in $\mathfrak n_r\cap\mathfrak n_r^*$ has such a pair of representing right-bounded vectors, by the definitions, proving the remaining inclusion. $\square$

The theorem applies to any left Hilbert algebra satisfying the original four axioms. In particular, after OA-MOD-RD-04, it can be applied to the opposite algebra of the right Hilbert algebra $\mathcal A_r$. That supplies the corresponding left-ideal statement once the left-bounded-vector spaces for that application have been defined. No identification of a full left completion is implicit in this observation.

OA-MOD-RD-08 — A weighted discrete model of the approximations

Let $I$ be any set and let $\mu_i>0$. On $\mathcal A=c_{00}(I)$, use pointwise products, complex conjugation, and $$\langle a,b\rangle=\sum_{i\in I}\mu_i a_i\overline{b_i}, \qquad H=\ell^2(I,\mu). \tag{RD.34}$$ Every nonnegative sum is the supremum of its finite subsums. Left multiplication by $a$ has norm $\sup_i|a_i|$, as follows by the coordinatewise upper bound and testing one coordinate. Pointwise products give the adjoint compatibility. Conjugation extends to an antiunitary involution on $H$, so is closable. Each finite-support vector is its product with the indicator of its support, giving $\mathcal A^2=\mathcal A$. These verify the four left Hilbert algebra axioms.

Both $S,F$ are conjugation on all of $H$. Right boundedness is exactly $$\mathcal B_r=\mathcal A_r=\ell^2(I,\mu)\cap\ell^\infty(I), \qquad R_\eta=\operatorname{diag}(\eta_i),\quad \|R_\eta\|=\sup_i|\eta_i|. \tag{RD.35}$$ Testing coordinate vectors proves necessity of the bound; summing the coordinate inequalities proves sufficiency.

For an arbitrary $\eta\in H=D(F)$, the closed operator $T_\eta$ is multiplication by $\eta_i$, with exact domain $$D(T_\eta)= \left\{\xi\in H:\sum_i\mu_i|\eta_i\xi_i|^2<\infty\right\}. \tag{RD.36}$$ This operator is closed by coordinatewise limits, and finite-coordinate truncations approximate both $\xi$ and its displayed image, so $c_{00}(I)$ is a graph core. Its polar data are $$h=k=\operatorname{diag}(|\eta_i|),\qquad u_i=\begin{cases}\eta_i/|\eta_i|,&\eta_i\neq0,\\0,&\eta_i=0.\end{cases} \tag{RD.37}$$ The cutoff vector has entries $f_n(|\eta_i|)\eta_i$; it is bounded because $t f_n(t)$ has compact support. The graph convergence for $F$ is simply $$2\sum_i\mu_i|f_n(|\eta_i|)-1|^2|\eta_i|^2\longrightarrow0. \tag{RD.38}$$ The zero coordinates contribute zero, and scalar dominated convergence applies to the summable family.

The product span has the exact description $$\mathcal A_r^2=\ell^1(I,\mu)\cap\ell^\infty(I). \tag{RD.39}$$ A product of two $\ell^2(I,\mu)$ vectors belongs to $\ell^1(I,\mu)$ by Cauchy–Schwarz, and a product of bounded vectors is bounded. Conversely, for $v\in\ell^1(I,\mu)\cap\ell^\infty(I)$, let $b_i=|v_i|^{1/2}$ and let $a_i=v_i/|v_i|^{1/2}$ when $v_i\neq0$, with $a_i=0$ otherwise. Both factors are bounded and square summable, and $v=ab$. Also $v\in H$, since $\sum_i\mu_i|v_i|^2\leq\|v\|_\infty\sum_i\mu_i|v_i|$. This proves (RD.39).

For counting weights on $\mathbb N$, the vector $v_n=1/n$ belongs to $\mathcal A_r$ but not to $\mathcal A_r^2$. Nevertheless its finite truncations converge in the graph norm of $F$, which is $\sqrt2$ times the Hilbert norm. Thus a product core can be a proper subspace.

Here $M=M'$ is the algebra of all bounded diagonal multipliers. To check the commutant statement directly, an operator commuting with every one-coordinate projection preserves each coordinate line and hence is diagonal; boundedness bounds its diagonal coefficients. Conversely every bounded diagonal multiplier commutes with $\mathcal A$. The finite-support indicators give positive contractions in $R(\mathcal A_r)$ converging strongly to the identity. When $I$ is uncountable, no sequence of finite-support indicators can do so, since a coordinate outside the union of its supports is annihilated by the whole sequence. Thus the global net in OA-MOD-RD-05 and the vectorwise sequence in OA-MOD-RD-04 serve different purposes.

OA-MOD-RD-09 — Problems and worked solutions

Problem 1: a genuine unbounded multiplier with a nontrivial kernel. In the weighted discrete model take $I=\{0\}\cup\mathbb N$, $\mu_0=1$, $\mu_n=n^{-4}$, and $\eta_0=0,\eta_n=n$. Find its right multiplier and polar supports. Explain why the spectral cutoffs converge strongly to a proper projection but still approximate $\eta$ in the graph norm of $F$.

Solution. The vector belongs to $H$, since $\sum_{n\geq1}n^{-4}n^2=\sum_n n^{-2}<\infty$, but it is unbounded and hence not right bounded. Its closed right multiplier has $$(T_\eta\xi)_0=0,\quad (T_\eta\xi)_n=n\xi_n,\qquad D(T_\eta)=\left\{\xi\in H:\sum_{n\geq1}n^{-2}|\xi_n|^2<\infty\right\}.$$ This is a nonnegative self-adjoint diagonal operator, with $h=k=T_\eta$. Its partial isometry is $u=p=q=I-P_0$, where $P_0$ projects onto coordinate zero. Thus $f_n(k)\to I-P_0$, not to $I$. Since $\eta_0=0$, this projection fixes both $\eta$ and $F\eta=\eta$. Formula (RD.38) proves graph convergence. Each cutoff vector has finite support in the positive coordinates, because $f_n(m)=0$ for integers $m>e^{2n}$, so in this example the approximants even lie in $c_{00}(I)$.

Problem 2: Hilbert-norm approximation alone does not control the involution. In OA-MOD-HA-09 take $I=\mathbb N$, $D_n=\operatorname{diag}(1,n^4)$. Let $v^{(n)}$ have only one nonzero block, its $n$-th block equal to $n^{-3}e_{12}$. Show that $v^{(n)}\to0$ in $H$ but not in the graph norm of $F$.

Solution. The weighted squared norm of $e_{12}$ in block $n$ is $n^4$, so $\|v^{(n)}\|=n^{-1}\to0$. The exact formula in OA-MOD-HA-09 gives $$Fv^{(n)}=n e_{21}$$ in that block and zero elsewhere. The weighted norm of $e_{21}$ is one, so $\|Fv^{(n)}\|=n$. All these vectors have finite support and lie in $\mathcal A_r^2$, but their graph norms diverge. This does not conflict with graph density: a core theorem provides appropriate approximating vectors, and does not make every norm-convergent sequence converge in graph norm.

Problem 3: squaring a monotone contraction family needs care. Set $$c=\frac14\begin{pmatrix}2&0\\0&0\end{pmatrix}, \qquad d=\frac14\begin{pmatrix}3&1\\1&1\end{pmatrix}.$$ Verify $0\leq c\leq d\leq I$, but $c^2\not\leq d^2$. Explain which conclusion about the squared approximants in (RD.26) is valid.

Solution. The difference $d-c=\frac14\begin{pmatrix}1&1\\1&1\end{pmatrix}$ is positive. The matrices $d$ and $I-d=\frac14\begin{pmatrix}1&-1\\-1&3\end{pmatrix}$ have nonnegative diagonal entries and positive determinants, so both are positive. However, $$d^2-c^2=\frac1{16}\begin{pmatrix}6&4\\4&2\end{pmatrix}, \qquad \left\langle(d^2-c^2)\binom1{-2},\binom1{-2}\right\rangle=-\frac18.$$ Thus the square map does not preserve this order. In (RD.26), positivity and contractivity of each square remain true. Strong convergence $c_i\to I$ and the uniform contraction bound give $c_i^2\to I$ strongly, by the displayed estimate there; self-adjointness gives strong* convergence. Monotonicity of the squares is unnecessary.

OA-MOD-RD-10 — Exported conclusions and the remaining modular boundary

This unit closes the general cutoff, right-algebra density, commutant-generation, original product-core, and ideal-intersection statements:

All proofs retain arbitrary Hilbert spaces, nonunital algebras, nontrivial operator kernels, and possibly unbounded right multipliers. The graph-density sequence and the global operator-approximation net have their separate topologies stated. The analytic prerequisites remain the explicit contracts in OA-MOD-RD-01.

Further work must define the left-bounded-vector completion, prove its repeated-dualization and fullness identities, establish the modular resolvent estimates and the fundamental modular theorem, and construct the analytic Tomita algebra. The equality $R(\mathcal A_r)''=M'$ proved here does not identify $R(\mathcal A_r)$ with $J M J$. No assertion $JMJ=M'$ or $\Delta^{it}M\Delta^{-it}=M$ has entered the proof. The general weight-to-Hilbert-algebra and weight-reconstruction theorems also retain their separate obligations.

Editable source · Proof dependencies and component terms