Norming, convex separation and bounded functional supports

Original mathematical exposition by GPT-6 Astra / Ultra. Self-checked by the writing AI.

These proofs supply the elementary functional-analysis inputs used in the concrete predual construction. We assume the complete real and complex scalar fields, the inner-product axioms and the set-theoretic maximal principle. Hilbert spaces have arbitrary dimension. Locally convex spaces below need not be Hausdorff or complete, and operator limits may be arbitrary directed nets.

For the norm-extension formulation, the free human source is Yury Kudryashov and Heather Macbeth's mathlib Hahn–Banach module, exists_extension_norm_eq and exists_dual_vector. NP1 writes the real extension argument and its complex conversion in full. No theorem is supplied by that citation in place of a proof. The bounded operator inputs are the written BK01, BK03, BK04 and BK06; their free-source reconstruction is separate from this note.

NP1 — Norm-preserving extension and the norming identity

Let $p$ be a finite sublinear function on a real vector space $X$, and let $f:D\to\mathbb R$ be linear on a subspace with $f\le p$. For $x\notin D$, a linear extension assigning value $a$ to $x$ is dominated by $p$ exactly when $\sup_{d\in D}\{f(d)-p(d-x)\}\le a \le\inf_{e\in D}\{p(e+x)-f(e)\}. \tag{NP.1}$ Each expression on the left is at most each expression on the right, because $f(d+e)\le p(d+e)\le p(d-x)+p(e+x)$. Taking $d=0$ and $e=0$ shows that both endpoints of this interval are finite. Real completeness supplies a choice of $a$. The positive and negative coefficients of $x$, divided by their absolute values, give precisely the two inequalities in (NP.1); hence the extension is dominated on all of $D+\mathbb Rx$. The union of a chain of dominated extensions is a dominated extension. The maximal principle and the one-dimensional step therefore give an extension to $X$.

For a bounded real functional on a subspace of a normed space, apply this with $p(x)=C\|x\|$, where $C$ is its norm. Domination at $x$ and $-x$ gives an extension bounded by $C$, and restriction gives the reverse norm inequality.

For a complex-linear functional $h$ on a complex subspace, extend $\operatorname{Re}h$ as a real functional $f$, dominated by $\|h\|\|x\|$, and set $L(x)=f(x)-if(ix). \tag{NP.2}$ Real linearity and $L(ix)=iL(x)$ prove complex linearity. On the original subspace, $f(ix)=\operatorname{Re}(ih(x))=-\operatorname{Im}h(x)$, so $L=h$. For each $x$, rotate by a scalar $\lambda$ of modulus one making $\lambda L(x)$ nonnegative real. Then $|L(x)|=f(\lambda x)\le\|h\|\|x\|$. Restriction again gives equality of norms. This includes the zero functional and does not require the subspace to be closed.

If $z\ne0$, the functional $\lambda z\mapsto\lambda\|z\|$ on its real or complex span has norm one. Extend it by the corresponding result. Testing $z$, together with the defining bound for a functional, proves $\|z\|=\sup\{|L(z)|:L\in X^*,\ \|L\|\le1\}. \tag{NP.3}$ For $z=0$ both sides vanish. This is the exact norming identity needed for the quotient predual norm.

Two elementary finite-dimensional steps in that construction can also be made explicit. Given a finite spanning list of Hilbert vectors, successively subtract its projections on the orthonormal vectors already chosen. Skip a zero remainder and normalize a nonzero remainder. Inner-product expansion proves orthogonality at each step, and each discarded or normalized vector remains in the span accumulated so far. After finitely many steps this gives an orthonormal basis of the original span. For a subspace $W\subset\mathbb C^n$ and $v\notin W$, successively adding vectors outside the current span constructs a basis of $W$, then a basis of $\mathbb C^n$ starting with that basis and $v$. Assigning values zero on the former basis, one on $v$, and zero on the remaining basis vectors gives a linear functional annihilating $W$ but not $v$. These are the finite coefficient and annihilator tests in CP02 and CP05.

NP2 — Point separation and equality of convex closures

Let $X$ be a real locally convex topological vector space, $C\subset X$ a nonempty closed convex set, and $x_0\notin C$. The definition of a locally convex topology gives a symmetric open convex neighborhood $V$ of zero such that $(x_0+V)\cap C=\varnothing$. Symmetry implies $x_0\notin C+V$. Choose $c_0\in C$ and put $A=C+V-c_0,\qquad y=x_0-c_0.$ The set $A$ is open and convex, contains zero, and does not contain $y$. It is absorbing: continuity of $t\mapsto tx$ at zero places a sufficiently small positive multiple of every $x$ in $A$. Its gauge is consequently finite: $p_A(x)=\inf\{t>0:x\in tA\}. \tag{NP.4}$ It is nonnegative and positively homogeneous. If $s>p_A(x)$ and $t>p_A(z)$, then $x\in sA$ and $z\in tA$. To see the first assertion, choose a smaller admissible dilation and use convexity with zero to enlarge it to $sA$; the second is identical. Convexity now gives $x+z\in(s+t)A$. Letting $s,t$ decrease to their infima proves subadditivity.

Moreover $A=\{x:p_A(x)<1\}$. An admissible dilation smaller than one lies in $A$ by convexity. Conversely, if $x\in A$, openness and continuity of $t\mapsto tx$ allow a scalar $t>1$ with $tx\in A$; then $p_A(x)\le1/t<1$. Thus $p_A(y)\ge1$.

On $\mathbb Ry$, define $f(ty)=t p_A(y)$. For positive $t$ this is $p_A(ty)$; for negative $t$, domination follows from $-p_A(y)\le p_A(-y)$, a consequence of subadditivity at zero. NP1 extends $f$ to a real linear $\ell\le p_A$. On the open neighborhood $A\cap(-A)$ we have $|\ell|<1$, so $\ell$ is continuous: scaling this neighborhood by any positive $\varepsilon$ makes $|\ell|<\varepsilon$.

We have $\ell(y)=p_A(y)\ge1$. A sufficiently small positive multiple $v_0$ of $y$ belongs to $V$; set $\delta=\ell(v_0)>0$. For every $c\in C$, the vector $c+v_0-c_0$ belongs to $A$. Therefore $\ell(c)<\ell(c_0)+1-\delta \le\ell(x_0)-\delta. \tag{NP.5}$ This proves strict point separation with a common positive margin. No Hausdorff assumption entered the proof.

For completeness, the closure of a convex set is convex. If $a,b$ lie in that closure and $0\le t\le1$, continuity of $(u,v)\mapsto tu+(1-t)v$ lets points of the original set approximate $ta+(1-t)b$ in every neighborhood; each approximating convex combination belongs to the set. Applying (NP.5) to its closure shows that the closure of a nonempty convex set is the intersection of all closed half-spaces, defined by continuous real linear functionals, that contain it. The empty set is closed separately. Thus two locally convex topologies with the same continuous real dual have exactly the same closures of convex sets. There is no boundedness, completeness or sequential-closure restriction. CP09–CP10 establish the requisite dual equality for the operator topologies before using this conclusion.

NP3 — Supports of positive ultraweakly continuous functionals

Let $M\subset B(H)$ be a weak operator closed unital *-algebra and let $\omega:M\to\mathbb C$ be positive and linear. The bounded calculus in BK01 writes a self-adjoint element as a difference of positive elements, so $\omega$ is real on self-adjoint elements and preserves adjoints. Positivity of $\omega((x+\lambda y)^*(x+\lambda y))$, followed by minimizing in the complex scalar $\lambda$, gives $|\omega(y^*x)|^2\le\omega(x^*x)\omega(y^*y). \tag{NP.6}$ If $\omega(y^*y)=0$, varying the magnitude and argument of $\lambda$ instead forces the mixed coefficient to be zero, proving the same inequality. Taking $y=1$ and using $x^*x\le\|x\|^2 1$ gives $|\omega(x)|\le\omega(1)\|x\|$; testing the unit proves $\|\omega\|=\omega(1)$. For the zero algebra all assertions are immediate.

Assume now that $\omega$ is ultraweakly continuous. This is the hypothesis called normal here; no order-normal-to-ultraweak theorem is used. BK04 shows that a bounded increasing positive net converges strongly and ultraweakly to its supremum. Applying $\omega$ proves preservation of such suprema.

For a positive $a\in M$, BK06 constructs the range-support projection $s(a)$ and proves $a(a+\varepsilon I)^{-1}\uparrow s(a) \quad(\varepsilon\downarrow0),\qquad 0\le a(a+\varepsilon I)^{-1}\le\varepsilon^{-1}a. \tag{NP.7}$ Consequently $\omega(a)=0$ implies $\omega(s(a))=0$. For two projections $e,f$, the identity $\langle(e+f)\xi,\xi\rangle=\|e\xi\|^2+\|f\xi\|^2$ gives $\ker(e+f)=\ker e\cap\ker f$. Thus $s(e+f)$ projects onto the closed span of their ranges: it is their least upper projection, denoted $e\vee f$. If both projections are $\omega$-null, (NP.7) proves that their join is null as well. No commutation assumption is needed.

Let $\mathcal N$ be the set of all $\omega$-null projections. For each finite subset $F\subset\mathcal N$, let $q_F$ be its join, with $q_\varnothing=0$. The preceding argument proves that every $q_F$ is null. These projections form an increasing net bounded by the identity. By BK04 its supremum $q\in M$ is its strong and ultraweak limit. It is a projection: for every $\xi$, $\|(q_F^2-q^2)\xi\| \le\|(q_F-q)\xi\|+\|(q_F-q)q\xi\|\longrightarrow0,$ because $\|q_F\|\le1$. Since $q_F^2=q_F\to q$ strongly, we obtain $q^2=q$; self-adjointness follows from the positive limit. Ultraweak continuity gives $\omega(q)=0$. It contains every member of $\mathcal N$, so it is the largest null projection.

Set $p=1-q$. Using (NP.6) with the pair $q,x^*$ gives $\omega(xq)=0$; adjoint preservation gives $\omega(qx)=0$. Expanding with $1=p+q$ proves $\omega(x)=\omega(pxp)\qquad(x\in M). \tag{NP.8}$ If a projection $e$ has this compression property, then $\omega(1-e)=0$. Maximality of $q$ implies $1-e\le q$, hence $p\le e$. Thus $p$ is the least compression projection.

The restriction to $pMp$ is faithful. If $a\in pMp$ is positive and $\omega(a)=0$, (NP.7) makes $s(a)$ a null projection, so $s(a)\le q$. Since the range of $a$ lies in $pH$, also $s(a)\le p$. Hence $s(a)=0$, and $a=0$. The zero functional has $q=1$ and $p=0$. Otherwise $\omega(p)=\omega(1)>0$, so division by this value gives a state on the support corner. None of these projections is required to be central.

NP4 — A noncommuting null-projection example

Take $M=B(\mathbb C^3)$ and $\omega(x)=\langle xe_3,e_3\rangle$. Let $q_1$ project onto $\mathbb Ce_1$ and $q_2$ onto $\mathbb C(e_1+e_2)$. Their matrices give $a=q_1+q_2= \begin{pmatrix}3/2&1/2&0\\1/2&1/2&0\\0&0&0\end{pmatrix}, \qquad q=q_1\vee q_2=\operatorname{diag}(1,1,0), \qquad p=\operatorname{diag}(0,0,1). \tag{NP.9}$ The upper two-by-two block has characteristic polynomial $t^2-2t+1/2$, so the two positive eigenvalues are $\lambda_\pm=1\pm1/\sqrt2$. The third eigenvalue is zero. The product matrices $q_1q_2$ and $q_2q_1$ differ, although both projections are null for $\omega$. The calculus in (NP.7) has eigenvalues $\lambda_\pm/(\lambda_\pm+\varepsilon)$ and zero; their limits are exactly the eigenvalues of $q$.

Two noncommuting null projections span the null plane; the resolvent cutoff converges to its projection.

Figure. Left: the real slice of the two complex range lines inside $\operatorname{span}_{\mathbb C}\{e_1,e_2\}$. Their join is the entire complex plane indicated; the orthogonal support line $\mathbb Ce_3$ is described by $p$ and is outside this slice. Right: the exact two nonzero eigenvalues of $a(a+\varepsilon I)^{-1}$, with the horizontal axis traversed toward $\varepsilon=0$, and its identically zero third eigenvalue. This finite-dimensional example illustrates the arbitrary-dimensional net proof in NP3; it does not replace it. Reproducible figure source.

NP5 — Prerequisite boundary

NP1–NP2 prove norm extension, norming and locally convex separation. NP3 uses only the previously written bounded Hilbert, calculus, topology, monotone-net and range-support proofs in BK01/BK03/BK04/BK06. It does not use a predual, generic weight supports, an approximate-unit theorem, the universal bidual or the characterization of order-normal functionals. The conditional converse paragraph in CP12 remains a later corollary requiring the separately completed normal-weight theorem.