TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-09-24 (gpt-6-sol)

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Geometric Hahn--Banach theorem for subspaces

Statement

Assume the Axiom of Choice. For a linear subspace $M \subseteq X$ and $x \notin \overline{M}$ , there is $f \in M^{⊥}$ with $f (x) = 1$ .

Facts & Assumptions

Given: The Axiom of Choice, A subspace $M \subseteq X$ and $x \notin \overline{M}$ .

[A1]

AC is used through norm-preserving extension [F2] in step 3.1.

[F1]

$x \notin \overline{M}$ exactly when $δ = dist (x, M)$ is positive (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

[F2]

A bounded functional on any subspace extends to the ambient normed space without increasing its norm (A bounded linear functional on an arbitrary subspace extends with the same norm, without assuming the subspace is closed).

Proof

technique · direct

1.1F1givenconstruct

By [F1], $δ = dist (x, M) > 0$ . On $M + K x$ define $g (m + λ x) = λ$ ; the representation is unique because $x \notin M$ .

2.1step 1.1F1algebra

For $λ \neq 0$ , $∥ m + λ x ∥ = ∣ λ ∣ ∥ x + m / λ ∥ \geq ∣ λ ∣ δ$ , while the case $λ = 0$ is immediate. Thus $∥ g ∥ \leq 1 / δ$ .

3.1step 2.1F2∎

Extend $g$ by [F2] to $f \in X^{*}$ . Then $f ∣_{M} = 0$ and $f (x) = 1$ , so $f \in M^{⊥}$ as required.

Depends on

Used by

The annihilator detects the closure of a subspace Corollary

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Buehler and Dietmar Salamon, Functional Analysis, Theorem 2.53 (standard reference, not scraped)