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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Compactness and lower semicontinuity for bounded-domain BV

Statement

Assume Countable Choice. Let Ω⊆Rn be a bounded Lipschitz domain, meaning a bounded open set whose boundary has a finite cover by Lipschitz graph neighborhoods with Ω on one side of each graph. Total variation is lower semicontinuous under L1(Ω) convergence of BV functions. Every sequence (uk)⊂BV(Ω) satisfying sup⁡k(∥uk∥L1(Ω)+TV(uk;Ω))<∞ has a subsequence converging in L1(Ω) to a function in BV(Ω).

Facts & Assumptions

Given: Countable Choice, a bounded Lipschitz domain Ω, and a sequence bounded in the BV norm.

[F1]

Zero extensions have common bounded support and uniformly bounded BV norms, and translations obey the linear BV estimate. The extension theorem includes the endpoint case in dimension one. (BV extension and translation bounds)

[F2]

Under Countable Choice, an L1-bounded sequence with common bounded support and uniform translation continuity has an L1-convergent subsequence. This is proved by finite cell averages and a deterministic diagonal selection, without the metric compactness/sequential-compactness equivalence. (Sequence compactness in L1 by finite cell averages)

[F3]

Under Countable Choice, an L1 limit with bounded liminf variation belongs to BV and satisfies lower semicontinuity; if the limit is already BV, the inequality uses no choice axiom. (Test-field formula and L1 lower semicontinuity of spatial TV)

Proof

Given: The hypotheses in the Statement; let M bound the displayed BV norms.

1.1F3givenalgebra

For uk→u in L1(Ω) with all functions in BV(Ω), each compactly supported smooth unit test field has ∫udiv⁡ϕ=lim⁡k∫ukdiv⁡ϕ≤lim inf⁡kTV(uk;Ω). Taking the supremum using [F3] proves the lower-semicontinuity assertion. If only u∈L1 is initially known and that liminf is finite, the Countable Choice characterization in [F3] also proves u∈BV.

2.1F1F2F3step 1.1construct∎

Extend every uk by zero. By [F1], the functions vk=u~k have support in the bounded set Ω‾, satisfy sup⁡k∥vk∥1≤M, and have TV(vk;Rn)≤CΩM. Their translation differences are at most ∣h∣CΩM, uniformly in k. Hence [F2] gives a subsequence vkj→v in L1(Rn). Restriction yields ukj→u=v∣Ω in L1(Ω). The variations in Ω are bounded by M, so step 1.1 and [F3] prove u∈BV(Ω) and TV(u;Ω)≤lim inf⁡jTV(ukj;Ω). This proves the full compactness conclusion.

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