CV1–3 are original CC0 expression by GPT-6 Astra (OpenAI), Ultra. CV4 adapts the freely licensed Falcó–Isert article under CC BY4.0; attribution and changes are retained in the text. Independently reviewed; human review and formal verification are not asserted.

Convex separation and the full dual-ball criterion

This provider supplies the convex tools for extended-valued weights on arbitrary von Neumann algebras. Its only programme inputs are CF Section 1, CF Section 4, CF Section 8, for real sublinear Hahn–Banach, norm series, scalar compactness, product compactness and Hilbert Riesz representation. Every additional separation or dual-ball assertion used below is proved here.

Sections CV-1–3 are fresh local proofs, GPT-6 Astra (OpenAI), Ultra, 2026-10-04, CC0-1.0 to the extent of rights held. CV-4 adopts the corrected trivial-group specialization, based on Javier Falcó and Daniel Isert, G-strong subdifferentiability and applications to norm attaining subspaces, §3.4, Lemmas 32–33 and Proposition 34/Theorem 35, printed pp.254–258, version of record, DOI 10.1007/s13163-025-00536-6. That article is CC BY 4.0; retain this attribution and license for CV-4. Changes include the trivial-group specialization, corrected finite-block enumeration, an explicit real/complex coefficient argument in the correct space, the fixed separating vector, checked translated slices, and replacement of all formerly imported separation/compactness premises by CV-1–3. No general theorem from a restricted source is used.

Actual earlier proof ranges: OA-FLOW.CF.1, OA-FLOW.CF.4, OA-FLOW.CF.8.

CV-1. Separation from sublinear Hahn–Banach

Let $V$ be a real locally convex space, and let $U\subset V$ be nonempty, open and convex, with $0\notin U$. Choose $u_0\in U$, and put $W=U-u_0$. This open convex set contains zero and is absorbing. Its gauge $p(v)=\inf\{t>0:v\in tW\}$ is finite and nonnegative. Convexity gives subadditivity and positive homogeneity: if $v\in sW,w\in tW$, then $v+w\in(s+t)W$, and then take infima. Also $W=\{p<1\}$. One inclusion uses convexity and $0\in W$; for the other, openness permits a small radial enlargement of a point of $W$. Since $-u_0\notin W$, $p(-u_0)\geq1$.

On the real line through $-u_0$, the functional $t(-u_0)\mapsto t$ is at most $p$: for $t\geq0$ use $p(-u_0)\geq1$, and for $t<0$ use $p\geq0$. CF Section 1 extends it to a real linear $L\leq p$ on $V$. The inequalities $L(v)\leq p(v)$ and $-L(v)\leq p(-v)$ show continuity: on $\epsilon(W\cap(-W))$, $|L|<\epsilon$. For $u\in U$, $L(u)-L(u_0)<1=L(-u_0),\qquad L(u)<0. \tag{CV1}$ In particular $L\ne0$.

If $A,B$ are disjoint nonempty convex sets and $A$ is open, apply this construction to $A-B$. It yields $L(a)<L(b)$ for every $a\in A,b\in B$, and hence $\sup_{a\in A}L(a)\leq\inf_{b\in B}L(b). \tag{CV2}$ Both bounds are finite: fix one point in each set. The supremum on the open set is not attained, because $L\ne0$.

If $C$ is closed convex and $z\notin C$, choose an open convex balanced neighbourhood $V_0$ of zero with $(z+V_0)\cap C=\varnothing$. Then $z\notin C+V_0$. Apply (CV1) to $C+V_0-z$. For the resulting nonzero continuous $L$, put $d=\sup_{v\in V_0}L(v)>0$. This supremum is finite by fixing $c\in C$ in $L(c+v)<L(z)$. Thus $\sup_{c\in C}L(c)\leq L(z)-d<L(z). \tag{CV3}$ The empty set needs no separation. Complex spaces are treated as real spaces; a continuous real linear $L$ is the real part of the continuous complex linear functional $F(v)=L(v)-iL(iv)$.

Consequently, if every continuous real linear functional for a locally convex topology $\tau$ is also continuous for $\rho$, every convex $\tau$-closed set is $\rho$-closed: (CV3) separates each point outside it by a $\rho$-open half-space. This one-way statement is the form needed below.

CV-2. The operator and Hilbert product topologies

Let $M\subset B(K)$ be a concrete von Neumann algebra and let $H$ be any Hilbert space. The adjoint-strong topology on $M$ is generated by $x\mapsto\|x^*v\|$, $v\in K$. If a real linear functional on $M\times H$ is continuous for adjoint-strong times Hilbert norm, it is bounded by $C\left(\sum_{j=1}^n\|x^*v_j\|^2+\|\xi\|^2\right)^{1/2}.$ Factor through the real linear map $(x,\xi)\mapsto(x^*v_1,\ldots,x^*v_n,\xi)$. Its kernel is annihilated by that bound. Real norm-preserving Hahn–Banach and real Hilbert Riesz representation give the expression $\operatorname{Re}\left(\sum_{j=1}^n\langle x^*v_j,w_j\rangle+\langle\xi,\eta\rangle\right) =\operatorname{Re}\left(\sum_{j=1}^n\langle xw_j,v_j\rangle+\langle\xi,\eta\rangle\right). \tag{CV4}$ The real Hilbert representation used here follows from CF Section 8 on the ambient complex Hilbert space: for a bounded real functional $L$, the complex linear functional $v\mapsto L(v)-iL(iv)$ has a representing vector; taking real parts gives the required real representation. All inner products are linear in the first variable. This functional is continuous for concrete ultraweak topology on $M$ times weak topology on $H$. CV-1 therefore proves: every convex set closed for adjoint-strong times norm is closed for ultraweak times weak. No boundedness or metrizability is part of this topological assertion; boundedness, when needed to prove closedness of a particular set, must be supplied separately.

The same argument with $xv_j$ proves that a strongly closed convex subset of $M$ is ultraweakly closed. On the real space $M_{\rm sa}$ use real restrictions of these functionals. Hilbert norm and weak topology have the same scalar continuous dual, directly by Riesz representation and the definition of weak topology.

CV-3. Compact dual balls

For a real or complex normed space $X$, map the radius-$r$ closed ball of $X^*$ into $\prod_{x\in X}\{z:|z|\leq r\|x\|\}$ by evaluation. The product is compact by the actual CF Section 4 proof. Additivity and scalar homogeneity are closed coordinate equations. They cut out exactly the image of the dual ball, since the displayed bounds give a bounded functional of norm at most $r$. The coordinate topology is exactly weak*. Thus every centered closed dual ball is weak* compact and closed; translating proves the same for all closed balls. No completeness or separability of $X$ was used. By Riesz representation the analogous assertion holds for weak Hilbert balls, including arbitrary Hilbert dimension; in the complex case the conjugate-linear Riesz isometry and its inverse are still homeomorphisms for these topologies.

CV-4. Full Krein–Smulian, with affine slices checked

Let $X$ be a real or complex Banach space, and $C\subset X^*$ convex. We prove that $C$ is weak* closed if all $C\cap r\overline B_{X^*}$, $r>0$, are weak* closed. The converse is immediate. Testing positive integer radii suffices, since every other closed ball is contained in one such ball and is itself weak* closed by CV-3.

First suppose $D\subset X^*$ is nonempty convex, has closed bounded slices, and misses the closed unit ball. For $F\subset X$, write $P(F)=\{f:|f(v)|\leq1\ (v\in F)\}$. Starting with $F_0=\{0\}$, construct finite $F_n\subset n^{-1}\overline B_X$, containing zero, so that $D\cap n\overline B_{X^*}\cap P(F_0\cup\cdots\cup F_{n-1})=\varnothing. \tag{CV5}$ The case $n=1$ is given. At stage $n$, the set $Q_n=D\cap(n+1)\overline B_{X^*}\cap P(F_0\cup\cdots\cup F_{n-1})$ is weak* compact by CV-3 and the slice hypothesis. Every $f\in Q_n$ has norm greater than $n$, so some $v\in n^{-1}\overline B_X$ satisfies $|f(v)|>1$. These open conditions cover $Q_n$; a finite subcover, together with zero, gives $F_n$ and proves (CV5) at $n+1$. If $Q_n$ is empty, take $F_n=\{0\}$.

Enumerate the nonempty finite blocks $F_0,F_1,\ldots$, preserving repetitions, as a sequence $u_j\to0$ in norm. There is no claimed bound $\|u_j\|\leq1/j$. Every $f\in D$ satisfies $|f(u_j)|>1$ for some $j$, by choosing an integer $n\geq\|f\|$ in (CV5). Therefore $T:X^*\longrightarrow c_0,\qquad Tf=(f(u_j))_j$ is bounded linear and $T(D)$ misses the open unit ball of $c_0$. CV-1 separates these convex sets. After normalization it gives a real continuous functional $L$ on $c_0$ with norm one and $L(Tf)\geq1$ for $f\in D$.

In the complex case set $a_j=L(e_j)-iL(ie_j)$; in the real case set $a_j=L(e_j)$. On finite sequences, $L(z)=\operatorname{Re}\sum_j a_jz_j.$ Finite coordinate phase choices with $a_jz_j=|a_j|$ show $\sum_{j\in F}|a_j|\leq1$ for every finite $F$. Truncation of $z\in c_0$ then gives the same expression for all $z$ and $\sum_j|a_j|=\|L\|=1$. In the real case omit real parts and use signs. Completeness of $X$ gives $u=\sum_j a_ju_j\in X$, and $\operatorname{Re}f(u)=L(Tf)\geq1\qquad(f\in D). \tag{CV6}$

Return to $C$. It is norm closed: a norm-convergent sequence in $C$ lies in a fixed bounded slice and converges weak*, so its limit lies in that slice. Norm topology is metrizable; no weak* sequential-closure claim is made. For $f_0\notin C\ne\varnothing$, choose $\delta>0$ with $(f_0+\delta\overline B)\cap C=\varnothing$, and put $D=\delta^{-1}(C-f_0)$. Its slices are weak* closed, since for $R\geq\|f_0\|+\delta r$, $D\cap r\overline B =\delta^{-1}\left(((C\cap R\overline B)\cap(f_0+\delta r\overline B))-f_0\right). \tag{CV7}$ The set in parentheses is compact, and the affine map is a weak* homeomorphism. Apply (CV6): some fixed $u\in X$ satisfies $\operatorname{Re}(f-f_0)(u)\geq\delta$ for all $f\in C$. The strict reverse inequality defines a weak* neighbourhood of $f_0$ missing $C$. This proves the theorem for arbitrary real or complex Banach $X$, with no countability, balancedness or boundedness restriction on $C$.