Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
How statement and proof provenance work

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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Full-space BV Sobolev inequality for L² distributional TV

Statement

Assume Countable Choice. Every u∈L2(R2;R) with TV2(u;R2)<∞ satisfies ∥u∥L2(R2)≤TV2(u;R2).

Facts & Assumptions

Given: Countable Choice and u∈L2(R2;R) with finite TV2(u;R2).

[A1]

Countable Choice supplies the finite-Radon representation used in the cutoff approximation. (The Axiom of Countable Choice (ACω))

[F1]

Compactly supported BV cutoffs for finite-variation L² functions: there are vk∈BV(R2)∩L1(R2) with vk→u in L2 and lim sup⁡kTV(vk;R2)≤TV2(u;R2).

[F2]

BV Sobolev inequality in full space: every globally integrable v∈BV(R2) satisfies ∥v∥2≤TV(v;R2).

Proof

1.1A1F1F2givenchoose

By [F1], choose vk∈BV(R2)∩L1(R2) with vk→u in L2 and the stated variation upper bound. Apply [F2] to each vk to obtain ∥vk∥2≤TV(vk;R2).

2.1step 1.1algebra∎

Norm continuity gives ∥vk∥2→∥u∥2, so ∥u∥2=lim⁡k∥vk∥2≤lim sup⁡kTV(vk;R2)≤TV2(u;R2). This is the required inequality.

Depends on

Used by

Dependency tree · two levels

18 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