Normal positive functionals in a cyclic natural cone

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

Let $M\subset B(H)$ have a cyclic separating vector $\Omega$ in its natural cone $P$. Every normal positive functional on $M$ has a unique representing vector in $P$. The target functional may have a proper support. The Hilbert space need not be separable. We give the positive Fourier averaging, the support-corner domain check, the quantitative correction, and the final norm-density argument.

Inner products are linear in the second variable. Write $\omega_u(a)=\langle u,au\rangle$, and write $u\le v$ in the cone when $v-u\in P$. We use the support, corner, and norm arguments CG01–CG09a, the concrete Hilbert/predual constructions H00–H03, and the real Hahn–Banach proof F01. The scalar exponential and continuous integral proofs are the elementary calculus component, Section 5 and R03–R07. The modular construction used here is NC00, NC04, NC08; its precise output is recorded next.

CR00. The modular input, with its domains

For every cyclic separating $\gamma\in P$, including one on a simultaneous support corner, let

NC08 supplies these actual graph cores and

CG09 identifies $J_\gamma$ with the given conjugation and $P_\gamma$ with the given cone. In particular

For a corner these formulas are read on that corner Hilbert space and with its algebra and commutant. A quarter-power expression in (2) is an actual domain value: the preceding half-power membership implies it by the spectral inequality $t^{1/2}\le1+t$. The operator $\Delta_\gamma$ is injective, although its spectrum can accumulate at zero and infinity. Its real powers have the spectral domains

The bounded calculus, logarithm, spectral bands, and strong continuity of $\Delta_\gamma^{it}$ are the concrete NC00 spectral output. Equations (1)–(3) are the only modular analytic input to the construction below. No positive-functional realization theorem is included in that input.

CR01. A positive Fourier kernel of mass one

With the Fourier convention $\widehat k(s)=\int_{\mathbb R}e^{its}k(t)\,dt$, we claim

To fix the scalar constant without importing trigonometric identities, put $A=\int_0^1(1+t^2)^{-1}\,dt$ and $\pi=4A$. Define $\alpha(t)=\int_0^t(1+s^2)^{-1}\,ds$ for $t\ge0$. The scalar exponential constructed in the elementary calculus component satisfies

Indeed the derivative of the right side is $i/(1+t^2)$ times that side; multiplication by $e^{-i\alpha(t)}$ gives derivative zero and value one at zero. For $t>0$, differentiation shows $\alpha(t)+\alpha(1/t)=2A$, with the constant obtained at $t=1$. Hence $\alpha(t)\uparrow2A$ as $t\to\infty$, and continuity of the exponential gives $e^{i2A}=i$, $e^{i\pi}=-1$. The right-side formula covers the first quadrant as $\alpha$ increases continuously from zero to $2A$; multiplication by $i$ covers the second quadrant. Thus $\operatorname{Re}e^{iy}$ has exactly one zero on $[0,\pi]$, at $y=\pi/2$. Conjugation and the exponential addition law give $|e^{iy}|=1$. Writing $\cos y=\operatorname{Re}e^{iy}$ and $\sin y=\operatorname{Im}e^{iy}$ now supplies the identities used below. These arguments use R03–R04's integral and fundamental theorem and the already proved exponential derivative.

Here is the Fourier calculation. For $u\ge0$, integrate $G(z)=e^{iuz}/\cosh z$ around the positively oriented rectangle with vertices $-R,R,R+i\pi,-R+i\pi$, where $\cosh z=(e^z+e^{-z})/2$. Its only pole is $z_0=i\pi/2$: a zero of $\cosh z$ satisfies $e^{2z}=-1$, whose modulus first forces $\operatorname{Re}z=0$, and the preceding quadrant calculation then gives that unique zero in the strip. Since $\sinh(i\pi/2)=i$, its simple principal-part coefficient is $e^{-u\pi/2}/i$. The rectangle integral is therefore $2\pi e^{-u\pi/2}$. The elementary contour fact used in this calculation can be obtained without a general residue theorem: for a continuously differentiable holomorphic function on a rectangle, the four side integrals sum to

by R04–R06's one-variable fundamental theorem, continuous rectangular Fubini proof, and the Cauchy–Riemann identity. Remove a small square about a simple pole and subdivide the remaining rectangle into rectangles. Cancellation of common sides reduces the boundary integral to the small-square integral. The remainder is bounded near the pole, so its integral tends to zero with the small-square perimeter. R07 computes the principal-part integral directly along the four sides as $8iA=2\pi i$ times its coefficient, invariant under positive dilation and translation. This proves the contour fact just used. For the explicit exponential quotient here the bounded remainder follows from its second-order Taylor remainders, obtained directly from the uniformly convergent exponential series.

The vertical sides tend to zero as $R\to\infty$. Indeed, for $0\le y\le\pi$,

so each side has integral of modulus at most $\pi/\sinh R$. The top side contributes $e^{-u\pi}$ times the bottom side, because $\cosh(x+i\pi)=-\cosh x$ and its orientation is reversed. Absolute convergence follows from $(\cosh x)^{-1}\le2e^{-|x|}$. Consequently

Changing $x$ to $-x$ proves the same formula for negative $u$. Substitute $x=2\pi t$, $u=s/(2\pi)$, to obtain (4). In particular the normalization follows by setting $s=0$. The bound $k(t)\le4e^{-2\pi|t|}$ controls every tail integral below.

CR02. Spectral averaging preserves the cone

Fix a cyclic separating $\gamma\in P$, abbreviate $\Delta=\Delta_\gamma$, and put $B=\log\Delta$, $U_t=e^{itB}$. Define

On a compact interval the integrand is norm continuous, so its vector integral is the limit of Riemann sums in the complete Hilbert space. The tails have norm at most $\|u\|\int_{|t|>T}k(t)\,dt$. This defines (5) on all of $H$, with $\|K\|\le1$. Every Riemann sum with nonnegative weights preserves the closed convex cone, since $U_tP=P$. Thus

The exact spectral identity is

To justify it without interchanging an unbounded operator and an integral, let $E_n=1_{[-n,n]}(B)$. On $E_nH$, $B$ is bounded and $U_tE_n$ is norm continuous. Operator Riemann sums for a finite interval are the bounded continuous calculus of the corresponding scalar Riemann sums. Their convergence is uniform for $s\in[-n,n]$. The discarded tails are bounded uniformly in $s$ by $\int_{|t|>T}k(t)\,dt$. The scalar formula (4) therefore proves $KE_n=f(B)E_n$. Finally $E_nu\to u$ and both operators have norm at most one, proving (7). This also proves every truncation limit involved in the Fourier average.

CR03. A linear positive correction in a cyclic corner

Suppose first that $\xi\in P$ is cyclic and separating and that $0\le\psi\le\omega_\xi$. The bounded commutant-form proof CG09a supplies a positive contraction $b'\in M'$ with

By (2), $b'\xi\in D(\Delta_\xi^{-1/2})$. Put

The second quarter-power description in (2) shows that both $\zeta$ and $\xi-\zeta=\Delta_\xi^{-1/4}(1-b')\xi$ belong to $P$. Thus $0\le\zeta\le\xi$. Apply CR02 to this cyclic modular operator and define

The second equality includes a domain check. The spectral definition gives $\zeta\in D(\Delta_\xi^{1/4})$ and $\Delta_\xi^{1/4}\zeta=b'\xi$. For $\lambda>0$,

Spectral bands followed by convergence in the stated power domain prove (9). No multiplication of undefined unbounded powers occurs. Equations (6) give $0\le\eta\le\xi$.

The multiplier $r(\lambda)=2/(1+\lambda^{1/2})$ satisfies $\lambda^{1/2}r(\lambda)\le2$, so $\eta\in D(\Delta_\xi^{1/2})$. Since $J\eta=\eta$ and $J\Delta_\xi^{1/2}J=\Delta_\xi^{-1/2}$, it also lies in $D(\Delta_\xi^{-1/2})$, and

For the conjugate-linear adjoint, $\langle Su,v\rangle=\langle S^*v,u\rangle$. Thus for $a\in M$,

Combining this with (8)–(10) proves

This is a linear correction formula, rather than a claim that the square root of a commutant derivative already belongs to the cone.

CR04. The correction for a vector with proper support

Formula (11) holds for every $\xi\in P$ and every normal positive $\psi\le\omega_\xi$. For $\xi=0$ take $\eta=0$. Otherwise let $p=s_\xi\in M$, $j(p)=JpJ$, and $q=pj(p)$. CG01–CG02 give $\xi=q\xi$, $s(\psi)\le p$, and $\psi(a)=\psi(pap)$. CG07 gives the faithful unital order isomorphism

and the self-dual cone $P_q=qP=P\cap qH$. Transport the positive functional by $\psi_q(\theta(x))=\psi(x)$. It satisfies $0\le\psi_q\le\omega_\xi$ on $A$.

The vector $\xi$ is cyclic and separating for $A$ on $qH$. Indeed CG02 gives $\overline{M\xi}=j(p)H$; applying $q$ gives $\overline{A\xi}=qH$, since $qx\xi=qxq\xi$. If $x\in pMp$ and $x\xi=0$, commutation with $M'$ makes $x$ vanish on $M'\xi$, whose closure is $pH$; hence $x=0$. Faithfulness of $\theta$ proves separatingness. The bounded form CG09a, applied on this cyclic space, also proves that $\psi_q$ is a vector functional there; hence it is normal by H03. No converse between order-normality and ultraweak continuity is needed for this transport.

CG09 identifies the cyclic natural cone on $qH$ with $P_q$. Apply CR03 there. It gives $\eta\in P_q$ with $\xi-\eta\in P_q$ and the pairing formula for $\psi_q$. For $a\in M$,

Thus (11) holds on the original algebra. Every reduction used a projection and a dense orbit; none used a countable ambient Hilbert basis. The target functional may have support strictly below $p$.

CR05. Realizing a dominated functional by a geometric correction

Let $\psi$ be normal and $0\le\psi\le\omega_{\xi_0}$, with $\xi_0\in P$. Inductively suppose $R_n=\omega_{\xi_n}-\psi\ge0$, and put $E_n=R_n(1)$. Every residual is normal by H03. The norm of a positive functional is its value at one. For completeness, the scalar positive-form Cauchy–Schwarz inequality gives $|R_n(a)|^2\le R_n(1)R_n(a^*a)\le R_n(1)^2\|a\|^2$, and equality of the norm is attained at one.

Apply CR04 to $R_n\le\omega_{\xi_n}$, obtaining $0\le\eta_n\le\xi_n$ and (11) for $R_n$. Set

The cone membership follows from $\xi_n-h_n=(\xi_n-\eta_n)+\eta_n/2$. Expansion of the vector functional gives the exact residual identity

Self-duality gives $\langle h_n,\xi_n-2h_n\rangle\ge0$. Since cone pairings are real, (11) at one gives $E_n=\langle\eta_n,\xi_n\rangle=2\langle h_n,\xi_n\rangle$. Therefore

In particular

The last bound follows by summing the preceding geometric tail. Hilbert completeness and closedness of $P$ give $\xi_\infty\in P$. For arbitrary vectors,

by adding and subtracting $\langle v,au\rangle$ and taking the supremum over $\|a\|\le1$. Thus (13)–(15) show $\omega_{\xi_\infty}=\psi$. At a zero residual the correction is zero, and the sequence remains constant. The successive vectors are allowed to lose support; CR04 covers every such step.

CR06. Dominated normal functionals are norm dense

The following real separation argument supplies the density step without a positive vector-series representation. Put $\varphi=\omega_\Omega$. This is faithful and normal: $\varphi(a^*a)=\|a\Omega\|^2$, and $\Omega$ is separating. The orbit $M'\Omega$ is dense. In fact its closed-span projection $e$ belongs to $(M')'=M$ by H01; $e\Omega=\Omega$, so separatingness forces $e=1$.

Let $X=(M_*)_{\mathrm{sa}}$ as a real Banach space. The involution $\chi^*(a)=\overline{\chi(a^*)}$ preserves $M_*$, as one sees by reversing the two vectors in H03's norm-summable coefficient representation; it is an isometric conjugate-linear involution. Every $\chi\in M_*$ uniquely decomposes as

with both displayed real components in $X$. A continuous real-linear $f:X\to\mathbb R$ extends to the continuous complex-linear functional

H03's predual duality makes $F$ evaluation at some $a\in M$. The identity $F(\chi^*)=\overline{F(\chi)}$, and separation of $M$ by its predual, give $a=a^*$. Hence the continuous real dual of $X$ is evaluation by $M_{\mathrm{sa}}$.

Consider the convex cone

For $b'\in M'$, its vector functional is normal and belongs to $D$, because for $a\ge0$

If $a=a^*\in M$ has $\chi(a)\ge0$ for every $\chi\in D$, (18) makes its quadratic form nonnegative on the dense space $M'\Omega$. Continuity makes it nonnegative on all of $H$, so $a\ge0$. Conversely every $a\ge0$ pairs nonnegatively with $D$.

We include the separation step. For a closed convex cone $C$ in a real Banach space and $z\notin C$, choose $r>0$ such that $z\notin U=C+B(0,r)$. The open convex set $U$ contains $B(0,r)$, and its gauge

is finite, nonnegative, positively homogeneous, and at most $\|w\|/r$. Convexity gives subadditivity: $w\in sU,v\in tU$ imply $w+v\in(s+t)U$, after which one takes infima. Openness and convexity give $U=\{w:p(w)<1\}$: a multiple $tU$ with $t<1$ is contained in $U$, whereas for $w\in U$ a slightly expanded multiple of $w$ still lies in $U$. Thus $p(z)\ge1$. On $\mathbb Rz$, define $g(tz)=tp(z)$; it is dominated by $p$ for $t\ge0$ by homogeneity, and for $t<0$ since then $g(tz)\le0\le p(tz)$. F01's real Hahn–Banach proof extends $g$ with $g\le p$. Applying that bound to $w$ and $-w$ gives $|g(w)|\le\|w\|/r$, hence continuity. Every positive multiple $\lambda c$, $c\in C$, lies in $U$, so $\lambda g(c)<1$ for every $\lambda>0$; therefore $g(c)\le0$. The functional $-g$ is nonnegative on $C$ and strictly negative at $z$.

Apply this to $C=\overline D$. Positivity is preserved under norm limits, so $C\subset M_*^+$. If a positive normal $\nu$ lay outside $C$, separation and the real dual identification would give a self-adjoint $a$ nonnegative on $D$ with $\nu(a)<0$. The preceding quadratic-form test forces $a\ge0$, a contradiction. We have proved

CR07. Cyclic realization and uniqueness

Let $\phi\in M_*^+$. For $\phi=0$ take zero. Otherwise use CR06 to choose $\psi_n\in D$ with $\|\psi_n-\phi\|\le4^{-n}$. Each $\psi_n\le c_n\omega_\Omega$ is represented by CR05, starting with $\xi_0=\sqrt{c_n}\Omega$; if $c_n=0$, use zero. Let its representative be $u_n\in P$. CG05 gives

Hence $u_n$ converges in Hilbert norm to $u\in P$. Equation (16) proves $\omega_u=\phi$. If $v\in P$ also represents $\phi$, the same CG05 bound gives $\|u-v\|^2\le0$. Thus

is well defined and unique. The estimates include $\|u_\phi\|^2=\phi(1)$ and $\|u_\phi-u_\psi\|^2\le\|\phi-\psi\|$. The target functional can be nonfaithful, and the proof places no countability restriction on the ambient Hilbert space. The zero algebra on the zero Hilbert space has only the zero functional and the same conclusion. CG10's support-corner reduction then applies (20) to arbitrary algebras in the constructed standard form.

An exact two-point iteration

In this figure the target functional is $\phi=(1,0)$, represented by $u=(1,0)$. Starting at $\xi_0=(2,1)$, the modular operator is one and the correction (11) becomes $\eta=(x-1/x,y)$ for $\xi=(x,y)$, $x\ge1,y\ge0$. Thus (12) gives $\xi_1=(5/4,1/2)$ and $\xi_2=(41/40,1/4)$. The functional residual norm is $E=x^2-1+y^2$, giving the exact displayed values $4,13/16,181/1600$. The target has the proper support $\operatorname{diag}(1,0)$. This illustrates the support loss allowed by CR04 and the quantitative correction in CR05; the general cone and its modular operator need not be finite dimensional. The iterative method is Hiai's Lemma 3.18, printed page 30; the coordinates are explicit substitutions into (11)–(14).

Sources and method attribution

The positive modular Fourier correction and its quarter-error iteration are the methods in Fumio Hiai, Concise lectures on selected topics of von Neumann algebras, arXiv:2004.02383v1, Lemmas 3.17–3.18, printed/PDF page 30, in the proof of Theorem 3.12. CR01 spells out the scalar Fourier formula with the mass-one normalization for the convention used here. CR02–CR04 spell out the spectral bands, vector power domains, and corner transport. CR06 supplies a real Hahn–Banach proof of the density conclusion in Hiai's Lemma 3.19, printed page 30 and proof page 31, so that its positive vector-series representation is not an imported theorem here.

Huzihiro Araki's Some properties of modular conjugation operator of von Neumann algebras and a non-commutative Radon–Nikodym theorem with a chain rule, Theorem 6, printed pages 335–339, is an earlier primary proof of the cyclic realization conclusion. Its longer affiliated-operator and strip-estimate route is historical attribution, rather than a prerequisite of this proof. The preceding natural-cone construction credits its own free operator and cone antecedents. Credit for these mathematical methods is separate from the original CC0 expression above.