Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Fixed-collar extension of C1alpha conductivities

Statement

Let d≥3 and let γ∈Xd(R) for the domain, collar, ellipticity bounds and exponent 0<α<1 defining that class. Extend γ by 1 outside Ω, using its continuous representative on Ω‾. Then γe∈C1,α(Rd), γe−1 has support contained in Ω′‾, and min⁡{m,1}≤γe≤M,sup⁡Rd∣γe∣≤max⁡{R,1},sup⁡Rd∣∂jγe∣≤R,[∂jγe]C0,α(Rd)≤R. Here the Hölder seminorm is [f]C0,α:=sup⁡x≠y∣f(x)−f(y)∣/∣x−y∣α, and j=1,…,d.

Facts & Assumptions

Given: A member γ∈Xd(R) with d≥3.

[F1]

The class has γ∈C1,α(Ω‾), each first derivative has supremum and Hölder seminorm at most R, m≤γ≤M almost everywhere, M≥1, and γ=1 almost everywhere outside Ω′⋐Ω. (Calderón admissible conductivity classes)

Proof

Given: The data in the statement.

1.1F1givenalgebra

Since Ω′‾ is compact in Ω, every boundary point of Ω has a neighborhood disjoint from Ω′‾. On the open set Ω∖Ω′‾, continuity and the almost-everywhere background condition force γ=1 everywhere: otherwise a sufficiently small ball would have positive measure and ∣γ−1∣ bounded below by a positive number. Thus γe is identically 1 in a neighborhood of every boundary point and is C1 globally. Its derivatives vanish off Ω and near ∂Ω. The same continuity argument promotes the almost-everywhere ellipticity bounds to pointwise bounds in Ω, and the extension gives the claimed global value and derivative bounds. It also shows γe−1=0 outside Ω′‾.

2.1F1step 1.1algebra∎

For x,y∈Ω, the Hölder estimate for each derivative follows from [F1]. For x,y∉Ω, both derivatives are zero. In the remaining case take x∈Ω and y∉Ω. The segment from x to y has a first exit point z∈∂Ω, with ∣x−z∣≤∣x−y∣. By step 1.1, ∂jγe(z)=∂jγe(y)=0. Applying the original Hölder estimate on Ω‾ to x,z gives ∣∂jγe(x)−∂jγe(y)∣≤R∣x−z∣α≤R∣x−y∣α. These cases exhaust all pairs, establishing the global seminorm and hence C1,α regularity.

Depends on

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Sources