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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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BV extension and translation bounds

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. The zero-extension map is bounded from BV(Ω) to BV(Rn), with support in Ω‾, and there is a finite CΩ such that ∥u~∥L1(Rn)+TV(u~;Rn)≤CΩ(∥u∥L1(Ω)+TV(u;Ω)) for every u∈BV(Ω). For every v∈BV(Rn) and h∈Rn, ∥v(⋅+h)−v∥L1(Rn)≤∣h∣TV(v;Rn).

Facts & Assumptions

Given: Countable Choice, a bounded Lipschitz domain Ω, u∈BV(Ω), v∈BV(Rn), and h∈Rn.

[F1]

The trace result gives a zero extension in BV(Rn) with variation bounded by CΩ(∥u∥1+TV(u;Ω)). (Trace formula and bounded zero extension on Lipschitz domains)

[F2]

Under Countable Choice, full-space BV functions admit compactly supported smooth strict approximations. (Strict smooth approximation in BV)

[F4]

A differentiable function with continuous derivative satisfies the one-dimensional fundamental theorem of calculus along compact intervals. (Fundamental theorem of calculus for Banach-valued continuous curves)

[F5]

Nonnegative integrands may be integrated by iterated integration. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

[A1]

Countable Choice is assumed for full-space strict approximation. (The Axiom of Countable Choice (ACω))

Proof

Given: Countable Choice, u∈BV(Ω), v∈BV(Rn), and h∈Rn.

1.1F1given

The zero extension is in BV(Rn) by [F1]. Its L1 norm equals ∥u∥L1(Ω), and [F1] bounds its full-space variation by CΩ(∥u∥1+TV(u;Ω)); its support is contained in Ω‾ by definition.

1.2F3F4F5givenalgebra

Let first v∈Cc∞(Rn). The fundamental theorem [F4] along the segment from x to x+h gives v(x+h)−v(x)=∫01h⋅∇v(x+th) dt. Integrate the absolute value in x, apply Tonelli [F5], and use translation invariance [F3] to obtain ∥v(⋅+h)−v∥1≤∫01∣h∣∫Rn∣∇v(x+th)∣ dx dt=∣h∣∫Rn∣∇v∣ dx.

2.1A1F2F3step 1.2choose∎

For general v∈BV(Rn), choose vk∈Cc∞ with vk→v in L1 and ∫∣∇vk∣→TV(v) by [F2] under [A1]. Translation is an L1 isometry [F3], so vk(⋅+h)−vk→v(⋅+h)−v in L1. Passing to the limit in step 1.2 proves the translation inequality.

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Sources