Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedverified 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.

Matching boundary jets without interior equality

Example

On the unit ball Ω=B(0,1)⊂Rn, define ψ(x)={exp⁡ ⁣(−11−4∣x∣2),∣x∣<1/2,0,∣x∣≥1/2. For any ϵ≠0 with ∣ϵ∣∥ψ∥∞<1, the positive conductivities γ1≡1 and γ2=1+ϵψ agree with all orders of derivatives at ∂Ω but are unequal in the interior. This example shows that equality of all boundary conductivity jets alone does not imply interior equality; it makes no assertion that the two full DN maps agree.

Facts & Assumptions

Given: the unit ball and the explicit function ψ in the statement.

[F1]

Scalar conductivity and Schrödinger Dirichlet-to-Neumann maps: an admissible scalar conductivity is smooth and strictly positive on the closure of the domain.

[F2]

Boundary determination of all scalar conductivity jets: in dimensions n≥2, under its Axiom of Choice and Countable Choice hypotheses, equality of full conductivity DN maps implies equality of all boundary-normal jets. This is context only: the bump construction below establishes matching jets directly in every positive dimension, including n=1, without using a DN-to-jets implication or asserting equality of DN maps.

Proof

Given: the data in the statement.

1.1giveninductionconstructalgebra

For every integer m≥0, differentiation gives dmdtme−1/t=Pm(1/t)e−1/t on t>0 for a polynomial Pm, by induction on m. For every polynomial P, P(s)e−s→0 as s→∞; for each power sN, this follows from es≥sN+1/(N+1)!. Thus every right derivative tends to zero at t=0, so extension by zero to t≤0 is smooth and flat there. Composition with t=1−4∣x∣2 proves ψ∈Cc∞(B(0,1)), with compact support in B(0,1/2)‾. It is nonzero at x=0, and ∥ψ∥∞<∞.

2.1F1step 1.1givenalgebra∎

The bound ∣ϵ∣∥ψ∥∞<1 gives γ2(x)≥1−∣ϵ∣∥ψ∥∞>0, so it satisfies the positivity condition in [F1]. Since ψ vanishes on a neighborhood of ∂Ω, γ1 and γ2 agree on that neighborhood and all their boundary derivatives agree. At the origin, γ2(0)−γ1(0)=ϵe−1≠0. Thus identical boundary jets do not force interior equality.

Depends on

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Sources