Compactness and convex closure for normal-weight proofs

Classical proof exposition assembled and corrected by OpenAI Codex, GPT-6 Astra / Ultra. Self-checked by the writing AI.

Bounded slices are often easier to control than an entire convex set. This note proves the passage from those slices to weak-star closedness, together with the real/complex and product-topology facts needed by the normal-weight construction. No separability, boundedness of the whole convex set, balancedness or sequential weak-star test is assumed.

The full Hahn–Banach proof and point-separation argument are written in NP1–NP2. AB04–AB05 prove compact products and Banach–Alaoglu, including arbitrary normed preduals. The real and complex Hilbert representation arguments are in the real Hilbert proof and BK01. The norm topology, complete scalar fields and maximal principle are the common entry assumptions.

The Krein–Šmulian argument preserves the course's previously written finite-cover proof and its corrections. The human mathematical sources are Javier Falcó and Daniel Isert, G-strong subdifferentiability and applications to norm attaining subspaces, §3.4, an open-access article under CC BY 4.0, and Jacob Lurie, Math 261y, Lecture 12, Theorem 6, page 3. The latter gives a second finite-cover construction. The full coefficient and affine-slice arguments needed here are written below. No citation substitutes for a proof.

CV1 — Extension with a seminorm and convex separation

NP1 proves the real dominated-extension theorem for every finite sublinear function. In particular, let $p$ be a seminorm and let $h$ be linear on a subspace with $|h|\leq p$. In the real case, extension dominated by $p$, applied also to $-x$, has the same absolute bound. In the complex case, extend $\operatorname{Re}h$ as a real functional $f\leq p$, then set $H(x)=f(x)-if(ix).$ As in NP1, evaluation at $ix$ proves complex linearity and agreement with $h$. Choose $|\lambda|=1$ with $\lambda H(x)=|H(x)|$. Complex absolute homogeneity of the seminorm gives $|H(x)|=f(\lambda x)\leq p(\lambda x)=p(x).$ Thus continuity of $p$ implies continuity of $H$.

If a functional is continuous on a subspace for the topology induced by a locally convex space, some finite maximum $p_0$ of continuous seminorms has $|h(x)|<1$ whenever $p_0(x)<\varepsilon$ on that subspace, for an $\varepsilon>0$. When $p_0(x)>0$, apply the bound to $tx$ for $0<t<\varepsilon/p_0(x)$, then let $t$ increase to that endpoint. This gives $|h(x)|\leq p_0(x)/\varepsilon$. When $p_0(x)=0$, arbitrary positive $t$ gives $h(x)=0$. The preceding extension with the continuous seminorm $p_0/\varepsilon$ therefore proves continuous extension. No closedness or completeness of the subspace is required.

NP2 proves strict separation of a point from a nonempty closed convex set, by a continuous real functional with a positive margin. It also proves that equal continuous real duals give equal closures of convex sets. We will apply precisely this point-separation statement to the norm closure of a convex image in $c_0$. This avoids requiring an additional open-set separation theorem.

CV2 — Weak compactness of Hilbert balls

Claim. Every closed norm ball of a real or complex Hilbert space $H$ is compact for its weak topology. No dimension or separability restriction is imposed.

Proof. Use the course convention that the inner product is linear in its first variable. Riesz representation gives the surjective isometry

J:H\longrightarrow H^*,\qquad (Jy)(x)=\langle x,y\rangle.

Over the complex scalars $J$ is conjugate linear. For every fixed $x$, the function $y\mapsto (Jy)(x)=\overline{\langle y,x\rangle}$ is weakly continuous. Since weak* topology is the initial topology of all evaluations, $J$ is continuous from weak $H$ to weak* $H^*$. Conversely, every bounded linear functional on $H$ is $y\mapsto\langle y,x\rangle$ for some $x$; its composite with $J^{-1}$ is $f\mapsto\overline{f(x)}$, which is weak* continuous. Thus $J^{-1}$ is continuous into weak $H$. This proves that $J$ is a homeomorphism for these two topologies. Its isometry property carries a closed ball onto the corresponding closed ball in $H^*$, which is weak* compact by AB05. Pull back this compact set by $J^{-1}$. The real case is the same with conjugation removed. $\square$

CV3 — Krein–Šmulian at arbitrary Banach-space generality

Theorem. Let $X$ be any real or complex Banach space, and let $C\subseteq X^*$ be convex. Then $C$ is weak* closed if and only if $C\cap r\overline B_{X^*}$ is weak* closed for every $r>0$. Equivalently one may test all closed norm balls, or just the zero-centered balls of positive integer radius. No boundedness, balancedness, separability or prior closedness assumption on $C$ is imposed.

Proof of the separation mechanism. Suppose $D\subseteq X^*$ is nonempty and convex, all its bounded ball slices are weak* closed, and $D\cap\overline B_{X^*}=\varnothing$. We will construct $x\in X$ such that

\operatorname{Re}f(x)\geq1\qquad(f\in D).

For a set $A\subseteq X$, write $P(A)=\{f:|f(a)|\leq1\text{ for all }a\in A\}$. Set $F_0=\{0\}$. We construct finite sets $F_n\subseteq n^{-1}\overline B_X$, each containing zero, such that

D\cap n\overline B_{X^*}\cap P(F_0\cup\cdots\cup F_{n-1})=\varnothing \qquad(n\geq1). \tag{KS1}

The case $n=1$ is the assumed disjointness. Suppose $KS1$ holds at $n$, and put

Q_n=D\cap(n+1)\overline B_{X^*}\cap P(F_0\cup\cdots\cup F_{n-1}).

This is weak* compact: the first ball slice is closed inside a weak* compact ball, and every polar constraint is weak* closed. Each $f\in Q_n$ has $\|f\|>n$, by $KS1$. By the definition of operator norm there is $a\in n^{-1}\overline B_X$ with $|f(a)|>1$. Thus the weak* open sets $\{f:|f(a)|>1\}$, indexed by such $a$, cover $Q_n$. Choose finitely many covering vectors, and add zero to obtain $F_n$. If $Q_n$ is empty, take $F_n=\{0\}$. Intersecting $Q_n$ with $P(F_n)$ gives the empty set, which is $KS1$ at $n+1$.

List the finite blocks $F_0,F_1,F_2,\ldots$ in order, keeping repetitions, to get a sequence $(u_j)_{j\geq1}$. Each block is nonempty; hence the sequence is infinite. It tends to zero in norm, because only finitely many vectors occur before any fixed block and every vector in $F_n$ has norm at most $1/n$. For any $f\in D$, choose an integer $n\geq\|f\|$. Formula $KS1$ shows that some vector in the first $n$ blocks satisfies $|f(u_j)|>1$.

Define the bounded linear map

T:X^*\longrightarrow c_0(\mathbb F),\qquad Tf=(f(u_j))_{j\geq1}, \quad\mathbb F=\mathbb R\text{ or }\mathbb C.

Indeed, $|f(u_j)|\leq\|f\|\|u_j\|\to0$, and $\|Tf\|_\infty\leq\|f\|\sup_j\|u_j\|$. Every element of $T(D)$ has norm strictly greater than one. Continuity of the norm implies that its norm closure $C_0$ is contained in $\{z:\|z\|_\infty\geq1\}$; in particular $0\notin C_0$. The closure is convex by NP2. Apply that note's point-separation theorem to $0$ and $C_0$, in the real normed space underlying $c_0(\mathbb F)$. Reverse the sign and divide by the resulting positive separation margin. This gives a continuous real linear functional $L$ with $L(Tf)\geq1$ for every $f\in D$. Its norm is finite and positive; it need not equal one.

Here is the coefficient representation, including its needed norm bound. In the real case put $a_j=L(e_j)$; in the complex case put $a_j=L(e_j)-iL(ie_j)$, where $e_j$ is the $j$-th coordinate vector. For every finitely supported $z$, real linearity gives

L(z)=\operatorname{Re}\sum_j a_jz_j.

For every finite index set, choose the coordinates $z_j$ of modulus one so that $a_jz_j=|a_j|$, taking any modulus-one value if $a_j=0$, and put the remaining coordinates equal to zero. Since $\|z\|_\infty\leq1$, this yields $\sum_{j\in F}|a_j|\leq\|L\|$. Thus $(a_j)\in\ell^1$. Truncations of any $z\in c_0$ converge in the sup norm, so continuity gives

L(z)=\operatorname{Re}\sum_{j=1}^{\infty}a_jz_j, \qquad \sum_{j=1}^{\infty}|a_j|=\|L\|.

The last equality follows because the displayed representation also gives $\|L\|\leq\sum_j|a_j|$. In the real case omit real parts and use signs instead of phases. The norm of any tail of $\sum_j a_ju_j$ is at most $(\sup_j\|u_j\|)\sum_{j\text{ in the tail}}|a_j|$. The scalar tail tends to zero. The partial sums are therefore Cauchy, and completeness of $X$ gives $x=\sum_j a_ju_j\in X$. Each $f\in X^*$ is continuous, hence

\operatorname{Re}f(x) =\operatorname{Re}\sum_j a_jf(u_j)=L(Tf)\geq1 \qquad(f\in D). \tag{KS2}

Proof of the full theorem. Assume the bounded-slice condition on $C$. Then $C$ is norm closed. To see this, a norm-convergent sequence in $C$ is bounded and converges weak*, so its limit belongs to one of the weak* closed ball slices. For a point in the norm closure, choose a point of $C$ within $1/n$ for each $n$; the resulting sequence converges in norm. The preceding observation puts its limit in $C$, proving norm closedness. This step does not make any assertion that weak* closure can be detected by sequences.

If $C$ is empty there is nothing to prove. Given $f_0\notin C$, norm closedness supplies $\delta>0$ with $(f_0+\delta\overline B_{X^*})\cap C=\varnothing$. Put $D=\delta^{-1}(C-f_0)$. It is nonempty, convex, and disjoint from the closed unit ball. Its slices have the required weak* closedness: for $r>0$, choose $R\geq\|f_0\|+\delta r$. Then

D\cap r\overline B_{X^*} =\delta^{-1}\left(((C\cap R\overline B_{X^*}) \cap(f_0+\delta r\overline B_{X^*}))-f_0\right).

The set inside the parentheses is weak* compact, using the hypothesis and Banach–Alaoglu. Translation and multiplication by a nonzero scalar are weak* homeomorphisms, so the displayed slice is compact and therefore closed in the Hausdorff weak* topology. Apply $KS2$ to $D$. There is a fixed $x\in X$ with $\operatorname{Re}(f-f_0)(x)\geq\delta$ for every $f\in C$. The set

\{g\in X^*: \operatorname{Re}(g-f_0)(x)<\delta\}

is a weak* open neighborhood of $f_0$ disjoint from $C$. Every point outside $C$ therefore lies outside its weak* closure. This proves weak* closedness. The converse follows by intersecting two weak* closed sets; the alternate ball formulations follow from CV4. $\square$

CV4 — Products, real parts and arbitrary ball centers

Claim. If $\tau_j$ and $\rho_j$ are locally convex topologies on $V_j$, for $j=1,2$, with the same continuous scalar dual on each factor, then the product topologies $\tau_1\times\tau_2$ and $\rho_1\times\rho_2$ have the same convex closures.

Proof. A linear functional on $V_1\times V_2$ has the unique expression $L(x,y)=L_1(x)+L_2(y)$, with $L_1(x)=L(x,0)$ and $L_2(y)=L(0,y)$. If $L$ is continuous, each restriction is continuous because the coordinate inclusions are continuous. If both restrictions are continuous, their sum after the coordinate projections is continuous. Equality of the duals on each factor therefore gives equality of the product duals. Apply NP2. $\square$

If $V$ is a complex topological vector space and $u:V\to\mathbb R$ is continuous and real linear, put $F(x)=u(x)-i u(ix)$. Then $F(ix)=iF(x)$, and additivity and real homogeneity show that $F$ is complex linear. It is continuous because $u$ and multiplication by $i$ are continuous, and $\operatorname{Re}F=u$. Conversely the real part of a continuous complex linear functional is continuous and real linear. This proves the real-part identification needed when comparing complex scalar duals for real-convex separation. It does not identify the continuous dual of $M_{\mathrm{sa}}$ with a particular predual; that remains the separate predual contract.

For a subset $C\subseteq X^*$, suppose $C\cap n\overline B_{X^*}$ is weak* closed for every positive integer $n$. Every closed ball $\overline B(f,r)$, with $r\geq0$, lies in $n\overline B_{X^*}$ for an integer $n\geq\|f\|+r$. Hence

C\cap\overline B(f,r) =(C\cap n\overline B_{X^*})\cap\overline B(f,r)

is weak* closed, because the ball is weak* closed by AB05. The converse is immediate by choosing the centered integer balls. Thus the two formulations of the bounded-slice hypothesis are equivalent. CV3 supplies the further passage from these slices to the whole convex set.

CV5 — Exact use for weight sublevel sets

The convex-analysis contract in NW01 requires extension, separation, weak-star compactness, weak compactness of Hilbert balls, and the full convex Krein–Šmulian theorem. CV1 together with NP1–NP2 supplies extension and separation; AB05 supplies dual-ball compactness; CV2 supplies Hilbert-ball compactness; CV3 proves the arbitrary real or complex Banach-space theorem; CV4 supplies the product and scalar-field conversions and arbitrary-center ball tests.

These facts alone do not identify any concrete operator predual or its topology. CP06 supplies the onto predual duality, CP08–CP10 identify the continuous duals, and CP11 handles corners. Nor does this note prove that an order-normal weight has closed sublevel sets: those weight-theoretic arguments remain NW02–NW11. It supplies their convex-analysis input without assuming normality of a GNS representation, so it does not introduce the WG007/NP02 circularity that must be avoided.