Written by GPT-6 Astra (OpenAI), Ultra.

Bounded vectors, affiliated multipliers and the complete dual Hilbert algebra

CC0-1.0 to the extent of rights in this exposition.

The Hilbert space is arbitrary. Inner products are linear in the first variable. The inputs are CI-1–3, BD-1–4, their exact spectral inputs, and only the closed-form representation proved in FF-4. In particular the Fourier and modular parts of those later applications are not inputs here. The earlier Hilbert facts are SF-0 and SB-0; the full spectral-domain proof is SB-1–SB-6. Scalar spectral convergence uses SC-04–05.

A free human development route is Boey's institutional thesis, Definitions 4.1/4.3 and Lemmas 4.6–4.10, printed pp.25–31. The bounded-vector construction has this mathematical antecedent. Here the graph-core proof uses an explicit pair of resolvents and an adjoint-product calculation; the compact-cutoff identities are then derived for the later modular application.

HA-R1. The initial algebra and its bounded right vectors

Let $\mathcal A$ be a dense complex algebra in $H$, with a conjugate-linear involution $a\mapsto a^\#$, such that $\mathcal A^2$ is dense, each left multiplication $\lambda_a b=ab$ is bounded in the Hilbert norm, $\langle ab,c\rangle=\langle b,a^\#c\rangle$, and the involution is closable. Here $\mathcal A^2$ means the linear span of products. Then $\lambda_{ab}=\lambda_a\lambda_b$ and $\lambda_{a^\#}=\lambda_a^*$. The representation is nondegenerate because the span of its ranges contains $\mathcal A^2$. It is faithful: if $\lambda_a=0$, then $b^\#a^\#=0$ for all $b$; nondegeneracy makes the common kernel zero, hence $a^\#=a=0$.

Let $M=\lambda(\mathcal A)''$, $N=M'$, $S=\overline{\#}$ and $F=S^*$. CI proves that both $S,F$ are closed densely defined conjugate-linear involutions. Call $\eta\in H$ right bounded when the map $a\mapsto\lambda_a\eta$ on $\mathcal A$ is bounded in $\|a\|_H$; denote its extension by $R_\eta$. Write $B_r$ for these vectors. Associativity on $\mathcal A$ gives $R_\eta\in N$. The map $\eta\mapsto R_\eta$ is injective, because $R_\eta=0$ implies $\lambda_a\eta=0$ for every $a$. For $x\in N$, $x\eta\in B_r,\qquad R_{x\eta}=xR_\eta. \tag{HR1}$ Consequently $I=\{R_\eta:\eta\in B_r\}$ is a linear left ideal of $N$.

HA-R2. Adjoint products and the candidate dual algebra

For $\eta,\zeta\in B_r$, one has $R_\eta^*\zeta\in D(F),\qquad F(R_\eta^*\zeta)=R_\zeta^*\eta. \tag{HR2}$ Indeed for $a\in\mathcal A$, $\langle R_\eta^*\zeta,Sa\rangle =\langle\zeta,\lambda_{a^\#}\eta\rangle =\langle\lambda_a\zeta,\eta\rangle =\langle a,R_\zeta^*\eta\rangle.$ Since $\mathcal A$ is a graph core for $S$, the identity extends to $D(S)$ and is exactly the conjugate-linear adjoint criterion. By (HR1), this vector is also right bounded, with multiplier $R_\eta^*R_\zeta$.

Put $\mathcal D=B_r\cap D(F)$. For $\eta\in\mathcal D$, the identity $\langle R_\eta a,b\rangle=\langle\eta,a^\#b\rangle =\langle b^\#a,F\eta\rangle=\langle a,\lambda_bF\eta\rangle$ shows that $F\eta\in B_r,\qquad R_{F\eta}=R_\eta^*. \tag{HR3}$ Thus $\mathcal D$ is invariant under the involution $F$. Define $\eta\zeta=R_\zeta\eta\quad(\eta,\zeta\in\mathcal D),\qquad R_{\eta\zeta}=R_\zeta R_\eta. \tag{HR4}$ Membership in $\mathcal D$ follows before any involution calculation: the vector is $R_{F\zeta}^*\eta$, to which (HR2) applies. That formula then gives $F(\eta\zeta)=R_\eta^*F\zeta=(F\zeta)(F\eta). \tag{HR5}$ Associativity follows either by applying the bounded multipliers or by their injective reversed representation. Right multiplication is bounded, and its adjoint is right multiplication by $F\zeta$, by (HR3). Also $I^*I\subseteq R(\mathcal D). \tag{HR6}$ Here and below products of linear operator spaces denote finite sums of products. Density and the core assertion will be proved, rather than included in the definition of $\mathcal D$.

HA-R3. The closed multiplier associated to any adjoint-domain vector

For $\eta\in D(F)$, define $T_\eta^0a=\lambda_a\eta$ on $\mathcal A$. The same adjoint pairing as above gives $\mathcal A\subseteq D((T_\eta^0)^*),\qquad (T_\eta^0)^*a=\lambda_aF\eta. \tag{HR7}$ Thus $T_\eta^0$ is closable. Write $T_\eta=\overline{T_\eta^0}$. Then $\mathcal A\subseteq D(T_\eta)\cap D(T_\eta^*),\qquad T_\eta a=\lambda_a\eta,\quad T_\eta^*a=\lambda_aF\eta. \tag{HR8}$ The second formula uses equality of the adjoints of an operator and its closure. It does not assert that $\mathcal A$ is a core for $T_\eta^*$, or that $T_{F\eta}=T_\eta^*$.

The graph of $T_\eta^0$, and then its closure, is invariant under every $\lambda_b\oplus\lambda_b$, by associativity and boundedness. It is invariant under their adjoints as well. Hence its orthogonal projection has all four entries in $N$. It therefore commutes with $x\oplus x$ for $x\in M$. For every unitary $x\in M$, $xD(T_\eta)=D(T_\eta),\qquad T_\eta x=xT_\eta. \tag{HR9}$ This is affiliation with $N$. Taking adjoints gives the corresponding statement for $T_\eta^*$.

BD-1 supplies a net $\lambda_{a_i}\to I$ strongly; no norm bound is needed here. Testing only the fixed vectors $\eta,F\eta$ in (HR8) gives $T_\eta a_i\longrightarrow\eta,\qquad T_\eta^*a_i\longrightarrow F\eta. \tag{HR10}$ Thus $\eta\in\overline{\operatorname{Ran}T_\eta}$ and $F\eta\in\overline{\operatorname{Ran}T_\eta^*}$. Actual range membership is neither proved nor used.

HA-R4. Closed linear polar decomposition with its domains

We supply the precise unbounded linear polar result needed next. Let $T$ be any closed densely defined linear operator on $H$. Its closed form $q(v,w)=\langle Tv,Tw\rangle$ on $D(T)$ has, by FF-4, a positive self-adjoint representative $A$ with $D(A^{1/2})=D(T),\qquad \langle Tv,Tw\rangle=\langle A^{1/2}v,A^{1/2}w\rangle. \tag{HR11}$ The representative satisfies $v\in D(A)$ exactly when $v\in D(A^{1/2})$ and there is $z\in H$ such that $q(v,w)=\langle z,w\rangle$ for every $w\in D(A^{1/2})$, and then $Av=z$. To check the nontrivial implication directly, test on each spectral subspace $E_A([0,n])H$. One obtains $E_A([0,n])z=A E_A([0,n])v$. These norms are bounded by $\|z\|$; spectral monotone convergence gives $v\in D(A)$ and $Av=z$. Applied to $q$, this criterion is exactly $Tv\in D(T^*)$, with $z=T^*Tv$. Hence $A=T^*T$ on the full product domain.

Set $h=A^{1/2}$. The rule $u(hv)=Tv$ defines an isometry from $\operatorname{Ran}h$ onto $\operatorname{Ran}T$, extends to their closures, and is set to zero on $\ker h$. The resulting partial isometry has initial space $(\ker T)^\perp$ and final space $\overline{\operatorname{Ran}T}$. The adjoint pairing gives $T=uh,\qquad T^*=hu^*,\qquad D(T^*)=\{w:u^*w\in D(h)\}. \tag{HR12}$ Write $p=u^*u,q=uu^*$. Transport $h|_{pH}$ by the unitary $u:pH\to qH$, and set it equal to zero on $q^\perp H$. This defines a positive self-adjoint $k$. Direct substitution in (HR12) shows $k^2=TT^*$, with the full product domain, so $k=(TT^*)^{1/2}$. It also proves $T=ku,\qquad u f(h)=f(k)u \tag{HR13}$ for Borel functions with $f(0)=0$, including equality of the transported domains. The first equality has domain $\{v:uv\in D(k)\}=D(h)=D(T)$; the kernel summands cause no loss.

If $T$ is affiliated with $N$, every unitary of $N'$ commutes with $T,T^*$, their positive products and their spectral projections. Uniqueness of the isometry defined on $\operatorname{Ran}h$ makes it commute with $u$ as well. The commutant-unitary test proved in SF-0 then gives $u,p,q\in N$, and $h,k$ are affiliated with $N$.

HA-R5. Resolvent pairs give the graph core

Fix $\eta\in D(F)$, and use HA-R4 for $T=T_\eta=uh=ku$. We first record a bounded multiplier calculation. If $\phi$ is a bounded Borel function on $[0,\infty)$ and $t\phi(t)$ is bounded, then $R_{\phi(k)\eta}=k\phi(k)u,\qquad R_{\phi(h)F\eta}=h\phi(h)u^*. \tag{HR14}$ In particular both vectors are right bounded. To prove this, commute the bounded operator $\phi(k)$ past $\lambda_a$, for $a\in\mathcal A$, and use (HR8): $\lambda_a\phi(k)\eta=\phi(k)T a=k\phi(k)u a$. The other formula uses $T^*a=\lambda_a F\eta$, only on its already proved test domain. This calculation makes no assertion about $F\phi(k)\eta$; that domain statement comes next.

Two resolvents and one adjoint pairing. For $\epsilon,\delta>0$, put $r(t)=\frac1{1+\epsilon t^2},\qquad s(t)=\frac1{\delta+t^2},\qquad \chi(t)=t^2r(t)s(t). \tag{HR-R1}$ The functions $r,s,ts$, as well as their products with $t$, are bounded. Define three right-bounded vectors $a=u r(h)F\eta,\qquad b=u s(h)F\eta,\qquad c=k s(k)\eta .$ Equations (HR1), (HR13) and (HR14) give $R_a=k r(k),\qquad R_b=k s(k),\qquad R_c=k^2s(k)u.$ All operators displayed here are bounded; in transporting $h$ and $k$, the scalar functions being transported vanish at zero. Now apply (HR2), rather than assume that $F$ commutes with a spectral multiplier: $R_a^*c=\chi(k)\eta\in\mathcal D,\qquad F(\chi(k)\eta)=R_c^*a=\chi(h)F\eta . \tag{HR-R2}$ For the last equality, substitute the three bounded formulas to obtain $u^*k^2s(k)u r(h)F\eta=h^2s(h)r(h)F\eta$. Also $R_a^*R_b=\chi(k)\in I^*I\subseteq R(\mathcal D)$. The vectors $u^*r(k)\eta,u^*s(k)\eta$ have multipliers $hr(h),hs(h)$, respectively. Their adjoint product similarly proves $\chi(h)\in R(\mathcal D)$.

Thus $\chi(k)^2\eta$ is a product of two members of $\mathcal D$: one application of $\chi(k)$ puts the vector in $\mathcal D$, and the second is multiplication by a represented member of $\mathcal D$. The same holds for $\chi(h)^2F\eta$. Their bounded right multipliers, computed by (HR14), are adjoints: $\big(k\chi(k)^2u\big)^*=h\chi(h)^2u^* .$ Using (HR3) and injectivity of $R$, we obtain the full-domain identity $\chi(k)^2\eta\in\mathcal D^2,\qquad F(\chi(k)^2\eta)=\chi(h)^2F\eta . \tag{HR-R3}$

Choose $\epsilon=\delta=1/n$, and denote the resulting function by $\chi_n(t)=\frac{t^2}{(t^2+1/n)(1+t^2/n)}.$ For $t>0$, both factors $t^2/(t^2+1/n)$ and $1/(1+t^2/n)$ increase to one; hence $0\leq\chi_n(t)\uparrow1$. At zero, $\chi_n(0)=0$. By (HR10), $\eta$ is in the support of $k$ and $F\eta$ is in the support of $h$. Spectral dominated convergence therefore gives $\big(\chi_n(k)^2\eta,\ F(\chi_n(k)^2\eta)\big) \longrightarrow(\eta,F\eta). \tag{HR-R4}$ This proves that $\mathcal D^2$, and consequently $\mathcal D$, are graph cores for $F$. In particular they are dense in $H$. The sequence depends on the fixed vector $\eta$; it imposes no separability assumption on $H$.

Compact cutoffs for the later modular argument. We also prove the precise compact-cutoff statements used downstream. Let $f\in C_c((0,\infty))$, extended by zero. Set $v(t)=\frac{(1+t^2)f(t)}{t^2},\qquad w(t)=\frac{(1+t^2)f(t)}t,$ with value zero at zero. Both functions are bounded and supported away from zero, and so are $tv$ and $tw$. The right-bounded vectors $a=u(1+h^2)^{-1}F\eta,\qquad b=u v(h)F\eta,\qquad c=w(k)\eta$ have multipliers $k(1+k^2)^{-1},kv(k),kw(k)u$. Consequently $R_a^*R_b=f(k)\in R(\mathcal D),\qquad R_a^*c=f(k)\eta\in\mathcal D,\qquad F(f(k)\eta)=R_c^*a=\overline f(h)F\eta .$ The symmetric operator calculation, using $u^*(1+k^2)^{-1}\eta$ and $u^*v(k)\eta$, gives $f(h)\in R(\mathcal D)$. Choose a real compactly supported $f_1$ equal to one on the support of $f$. Then $f(k)\eta=f_1(k)[f(k)\eta]\in\mathcal D^2$. The same reasoning gives $\overline f(h)F\eta\in\mathcal D^2$. It also gives $f(h),f(k)\in R(\mathcal D^2)$, by factoring $f=f_1 f$ within the represented algebra. In particular, for real $f$, $F(f(k)\eta)=f(h)F\eta . \tag{HR15}$ Every vector in this equality has already been placed in its domain.

For an explicit compact family, define $g(t)=1$ on $0<t\leq1$, $g(t)=2-t$ on $1\leq t\leq2$, and $g(t)=0$ on $t\geq2$. Set $f_n(t)=g(t/n)g(1/(nt))$. These functions have support in $[1/(2n),2n]$, satisfy $0\leq f_n\uparrow1$ on $(0,\infty)$, and give $f_n(k)\eta\in\mathcal D^2,\qquad f_n(k)\eta\longrightarrow\eta,\qquad F(f_n(k)\eta)=f_n(h)F\eta\longrightarrow F\eta . \tag{HR16}$ This supplies the compact family as a consequence of the explicit adjoint-pairing calculation; the graph-core proof above already follows from the resolvents.

HA-R6. The represented dual generates the whole commutant

The algebra $\mathcal D$, with its involution $F$, is now a right Hilbert algebra: all its algebra/adjoint axioms were proved in HA-R2, its products are dense by HA-R5, and its involution has closure $F$. Its right representation is nondegenerate. BD-3 gives positive contractions $c_i\in R(\mathcal D)$ converging strongly to $I$. For $x\in N_+$, (HR1) gives $x^{1/2}c_i\in I$, so $c_i x c_i=(x^{1/2}c_i)^*(x^{1/2}c_i)\in I^*I\subseteq R(\mathcal D). \tag{HR17}$ The common bound on $c_i$ makes $c_i x c_i\to x$ strongly. Every element of $N$ is a complex linear combination of four positive elements, by the positive/negative spectral parts of its real and imaginary parts. The reverse inclusion was established in HA-R1. Hence $R(\mathcal D)''=N=M'. \tag{HR18}$

HA-R7. The original product core and the full adjoint intersection

We still need the graph core of the original products; using only their Hilbert density would be insufficient. Fix $a\in\mathcal A$, rescaling so that $\|\lambda_a\|\leq1$. Let $p\in M$ be the projection onto $\overline{\operatorname{Ran}\lambda_a}$. For $d\in\mathcal D$, $R_da=\lambda_a d\in pH$. Since $p$ commutes with $R_d$, this implies $R_d(pa-a)=0$ for every $d$. Nondegeneracy of $R(\mathcal D)$ makes its common kernel zero, so $pa=a$. Applying the same argument to $a^\#$ gives the corresponding support statement for $\lambda_a^*$.

Put $p_n(t)=1-(1-t)^n$ on $[0,1]$. These polynomials vanish at zero and increase to one on $(0,1]$. Algebraic expansion gives $a_n=p_n(\lambda_a\lambda_a^*)a\in\mathcal A^2,\qquad a_n^\#=p_n(\lambda_a^*\lambda_a)a^\#. \tag{HR19}$ For the second identity, a monomial on the left is $(aa^\#)^j a$; its involution is $a^\#(aa^\#)^j=(a^\#a)^j a^\#$. The support statements and spectral dominated convergence show $a_n\to a$, $a_n^\#\to a^\#$. As $\mathcal A$ itself is a core for $S$, a diagonal choice of these approximants proves that $\mathcal A^2$ is a graph core for $S$.

Suppose $\eta,\zeta\in B_r$ and $R_\eta^*=R_\zeta$. For $b,c\in\mathcal A$, $\langle\eta,b^\#c\rangle =\langle R_\eta b,c\rangle =\langle b,R_\zeta c\rangle =\langle c^\#b,\zeta\rangle. \tag{HR20}$ Set $a=c^\#b$; linearity extends this to the span $\mathcal A^2$. The preceding graph core extends it to all $a\in D(S)$, proving $\eta\in D(F)$ and $F\eta=\zeta$. Together with (HR3), this gives the exact operator-space equality $R(\mathcal D)=I\cap I^*. \tag{HR21}$

HA-R8. Exact modular uses and the arbitrary-weight boundary

HA-R1–2 and HA-R7 supply the bounded-vector ideal, adjoint-product and adjoint-intersection inputs previously taken from HA-05, HA-08 and RD-07. HA-R3–6 supply the affiliated multiplier, its test-domain adjoint formulas, both polar decompositions, cutoff involution identity, graph-core density and commutant-generation inputs previously taken from RD-02–06. The opposite-algebra version has the same proof with left and right exchanged; it uses the same arbitrary Hilbert space and no countability assumption.

In the later application, the WH-04 proof supplies the second-dual fullness statement and equality of its closed involution with $S$. Applied to a full left Hilbert algebra, HA-R3–5 give precisely the closed left-multiplier and cutoff identities needed in MF-06's resolvent argument; CI supplies its polar and imaginary-power facts. Thus the replacement preserves the general Hilbert-algebra scope of those analytic inputs. It does not replace the later MF-06 modular averaging certificate or its whole bounded-vector conclusion.

These constructions do not assert that an arbitrary n.s.f. weight's finite-star GNS algebra satisfies the initial Hilbert-algebra axioms, do not reconstruct a weight from a full left Hilbert algebra, and do not prove the order/ultraweak characterization of arbitrary normal weights. The later WF construction and domain transport, all-weight normality and faithful finite-star closability–reverse/opposite finite-cone correspondence supply these separate branches. They are not premises of this pure Hilbert-algebra provider.

Exact graph spectrum and cutoff functions

Original CC0 illustration. The left panel shows the reciprocal scalar spectrum in CI-2–3; the upper-right panel shows the exact compact cutoff functions of HA-R5, HR14–HR16; the middle row shows the squared rational regularizers and adjoint-pairing mechanism of HR-R1–HR-R4. The bottom row retains both Hilbert spaces and the full graph-domain identities. The sampled curves illustrate the proved formulas. Reproduction source and editable SVG.