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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-09-23 (gpt-6-sol)
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Mazur theorem: weak and norm closure agree for convex sets

Statement

Assume the Axiom of Choice. For a convex subset C of a normed space X, its norm closure equals its weak closure. Here the weak topology is generated by the basic neighbourhoods of x

{y:∣fj(y−x)∣<ε (1≤j≤n)}

for finitely many fj∈X∗ and some ε>0.

Facts & Assumptions

Given: The Axiom of Choice, a convex set C⊆X, and D=C‾∥⋅∥.

[F1]

Disjoint convex sets, one open, are strictly separated by a nonzero continuous functional (Separation of disjoint convex sets when one is open).

[A1]

The Axiom of Choice is used through the Hahn–Banach extension behind [F1].

Proof

technique · direct
1.1given

Every f∈X∗ is norm-continuous, so every displayed basic weak neighbourhood is norm-open. Thus every weak-open set is norm-open, and D⊆C‾weak.

1.2givenF1A1choose

If C=∅, both closures are empty. Otherwise D is nonempty; if x∉D, choose r>0 with x∉D+B(0,r). This latter set is open and convex, so under [A1], [F1] yields f≠0 with Re⁡f(d+b)<Re⁡f(x) for d∈D, ∥b∥<r.

2.1step 1.2algebra

Taking suprema over b∈B(0,r) gives sup⁡d∈DRe⁡f(d)+r∥f∥≤Re⁡f(x). The weak neighbourhood {y:∣f(y−x)∣<r∥f∥/2} therefore misses D, hence misses C.

3.1step 1.1step 2.1∎

Thus x is not in the weak closure whenever it is not in D, proving the reverse inclusion and equality.

Depends on

Used by

Dependency tree · two levels

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Sources