Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Dominated extension conditional on the relative principle

Statement

Assume HB. Let X be a real vector space, M≤X a real linear subspace, p:X→R sublinear, and g:M→R real linear with g(m)≤p(m) for every m∈M. There is a real linear extension F:X→R satisfying −p(−x)≤F(x)≤p(x)(x∈X). No topology, closedness, or completeness is required.

Facts & Assumptions

[F1]

HB asserts a dominated real linear extension for every such quadruple (X,M,p,g) (The real dominated-extension principle as an additional hypothesis over ZF).

Proof

Given: HB and X,M,p,g as in the statement.

1.1

All four objects have the types required by HB, and the given inequality holds for every m∈M. Applying HB to this quadruple yields real linear F with F∣M=g and F(x)≤p(x) for every x∈X.

givenF1
2.1

For each x∈X, also −x∈X, so F(−x)≤p(−x). Since F(−x)=−F(x), multiplication by −1 gives −p(−x)≤F(x). Together with the upper bound this proves the claim. At zero, p(0)=F(0)=0, so both inequalities are equalities.

step 1.1algebra∎

Source notes

Brezis Theorem 1.1, p.1; Teschl Theorem 4.13 and following lower-bound observation, pp.112–113.

Depends on

Used by

Dependency tree · two levels

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Sources