CV1–4: original expression under CC0.

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 and CV-4 are CC0-1.0 to the extent of rights held. CV-4 follows the classical bounded-slice argument for the Krein–Šmulian theorem; Javier Falcó and Daniel Isert give a version of it in G-strong subdifferentiability and applications to norm attaining subspaces, §3.4 (version of record, DOI 10.1007/s13163-025-00536-6). 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 every bounded slice $C\cap r\overline B_{X^*}$, $r>0$, is 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.

Lemma (a null sequence that detects a convex set). Let $D\subset X^*$ be nonempty and convex, disjoint from the closed unit ball, with every bounded slice $D\cap r\overline B_{X^*}$ weak* closed. There is a sequence $x_k\to0$ in $X$ such that every $f\in D$ satisfies $|f(x_k)|>1$ for some $k$.

Proof. We choose finite sets $A_1,A_2,\ldots$, with $A_m\subset m^{-1}\overline B_X$, such that for every $m\geq0$ $f\in D,\ \|f\|\leq m+1\ \Longrightarrow\ |f(a)|>1\ \text{for some } a\in A_1\cup\cdots\cup A_m . \tag{CV5}$ For $m=0$ the union is empty and (CV5) holds because no $f\in D$ has $\|f\|\leq1$. Suppose $A_1,\ldots,A_{m-1}$ satisfy (CV5) for $m-1$, and put $K_m=\{f\in D:\ \|f\|\leq m+1,\ |f(a)|\leq1\ \text{for all } a\in A_1\cup\cdots\cup A_{m-1}\}.$ This set is weak* compact: it is the slice $D\cap(m+1)\overline B_{X^*}$, weak* closed by hypothesis and contained in a weak* compact ball by CV-3, cut down by finitely many weak* closed conditions. By (CV5) for $m-1$, every $f\in K_m$ has $\|f\|>m$, so there is $x$ with $\|x\|\leq1/m$ and $|f(x)|>1$. The weak* open sets $\{f:|f(x)|>1\}$, $\|x\|\leq1/m$, therefore cover $K_m$. Let $A_m$ consist of the vectors $x$ of a finite subcover, with $A_m=\{0\}$ if $K_m$ is empty. Then (CV5) holds for $m$: an $f\in D$ with $\|f\|\leq m+1$ either fails one of the conditions defining $K_m$, or lies in $K_m$ and is caught by $A_m$.

List the vectors of $A_1$, then those of $A_2$, and so on, as one sequence $(x_k)$. It tends to zero because $A_m\subset m^{-1}\overline B_X$; no rate such as $\|x_k\|\leq1/k$ is asserted. For $f\in D$, (CV5) with an integer $m\geq\|f\|$ gives the required $k$. ∎

From the sequence to one vector. Keep $D$ and $(x_k)$ as in the lemma. The map $T:X^*\longrightarrow c_0,\qquad Tf=(f(x_k))_{k\geq1},$ is linear and bounded, and $T(D)$ is a convex set disjoint from the open unit ball of $c_0$. By CV-1 there is a real continuous linear functional $L$ on $c_0$ with $\|L\|=1$ and $L(Tf)\geq1$ for every $f\in D$. Put $a_k=L(e_k)-iL(ie_k)$ in the complex case and $a_k=L(e_k)$ in the real case. On finitely supported sequences $L(z)=\operatorname{Re}\sum_ka_kz_k$ (without real parts in the real case). Choosing coordinates $z_k$ of modulus one with $a_kz_k=|a_k|$ on a finite set of indices shows $\sum_k|a_k|\leq\|L\|=1$; by continuity the formula holds on all of $c_0$, and then $\sum_k|a_k|=1$. Since $\sum_k|a_k|\,\|x_k\|<\infty$ and $X$ is complete, $u=\sum_ka_kx_k$ exists, and $\operatorname{Re}f(u)=L(Tf)\geq1\qquad(f\in D). \tag{CV6}$

The theorem. The set $C$ is norm closed: a norm-convergent sequence in $C$ stays in one bounded slice, converges weak* there, and that slice is weak* closed. Let $C$ be nonempty and $f_0\notin C$. Choose $\delta>0$ with $(f_0+\delta\overline B_{X^*})\cap C=\varnothing$ and set $D=\delta^{-1}(C-f_0)$, which is convex and misses the closed unit ball. Its bounded slices are weak* closed: for $r>0$ and $R\geq\|f_0\|+\delta r$, $D\cap r\overline B_{X^*} =\delta^{-1}\Bigl(\bigl((C\cap R\overline B_{X^*})\cap(f_0+\delta r\overline B_{X^*})\bigr)-f_0\Bigr), \tag{CV7}$ where the set in parentheses is weak* compact and $g\mapsto\delta^{-1}(g-f_0)$ is a weak* homeomorphism. By (CV6) there is $u\in X$ with $\operatorname{Re}(f-f_0)(u)\geq\delta$ for all $f\in C$. The weak* open set $\{f:\operatorname{Re}(f-f_0)(u)<\delta\}$ contains $f_0$ and misses $C$. Hence $C$ is weak* closed. Neither countability, balancedness nor boundedness of $C$ is assumed, and $X$ is an arbitrary real or complex Banach space.