Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generated
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.

Higher elliptic regularity cannot gain more than two derivatives

Statement refuted

Assume Countable Choice. For every integer k≥0, smooth constant coefficients and f∈Hlock(Ω) force every H1 local weak solution of −u′′=f into Hlock+3(Ω).

Facts & Assumptions

Given: Countable Choice; a fixed integer k≥0; the interval Ω=(−1,1); the datum f(x)=∣x∣k+1/2; and the function u(x)=−A∣x∣k+5/2 with A=((k+52)(k+32))−1>0.

[F1]

On each half-interval the classical derivative of order j of f(x)=∣x∣k+1/2 is a nonzero constant times ∣x∣k+1/2−j, with a possible sign change across 0. For j≤k these derivatives tend to 0 at 0 and are in L2; the order k+1 derivative is locally integrable with magnitude a positive constant times ∣x∣−1/2, but is not in L2 near 0. Since all derivatives through order k extend continuously across 0, the piecewise classical derivatives are the weak derivatives through order k+1, with no point-mass terms. Thus f∈Hk((−1,1)) but f∉Hlock+1((−1,1)) near 0. (Integer-order Sobolev spaces and their norms, Weak derivative of a locally integrable function)

[F2]

A class u∈H1((−1,1)) is a local weak solution of −u′′=f if ∫−11u′φ′‾ dx=∫−11fφ‾ dx for every φ∈Cc∞((−1,1)). (Local weak solutions of a divergence-form operator)

[F3]

Put p=k+52 and A=(p(p−1))−1. For u=−A∣x∣p, its derivatives through order k+2 are piecewise constant multiples of ∣x∣p−j, hence lie in L2((−1,1)); the (k+3)rd derivative has magnitude a positive constant times ∣x∣−1/2 and is not locally in L2 at 0. The derivatives through order k+2 extend continuously across 0, so these piecewise formulas are the weak derivatives and no delta mass occurs. (Integer-order Sobolev spaces and their norms, Weak derivative of a locally integrable function)

[F4]

Assume Countable Choice. For the operator L=−u′′ in dimension one, a11=1 and b=c=0, so it is uniformly elliptic with θ=Ma=1, Mb=Mc=0, and its constant coefficients lie in Wlock+1,∞ for every k. If f∈Hlock(Ω) and u∈H1(Ω) is a local weak solution of Lu=f, the interior theorem gives u∈Hlock+2(Ω). (Interior Hk+2 elliptic regularity, Uniformly elliptic divergence-form operators and their sesquilinear forms, The Axiom of Countable Choice (ACω))

Counterexample

1.1F1algebragiven

Data regularity. The piecewise derivative calculation in [F1] shows f∈Hk((−1,1)) but f∉Hlock+1 near 0.

1.2F3algebragiven

The solution's regularity. With u=−A∣x∣p from [F3], one has −u′′=Ap(p−1)∣x∣p−2=∣x∣k+1/2=f on both sides of 0. Since u′ is continuous at 0 and u′′=−f is locally integrable, this also holds distributionally across 0. The piecewise derivative calculation in [F3] gives u∈Hk+2((−1,1)) but u∉Hlock+3 near 0.

2.1F2step 1.2algebra

The weak equation. For every φ∈Cc∞((−1,1)), the distributional identity −u′′=f gives ∫−11u′φ′‾ dx=∫−11fφ‾ dx. Thus u is a local weak solution by [F2].

3.1F4step 1.1step 2.1algebra

The interior Hk+2 theorem agrees with the direct calculation. The Laplacian has smooth constant coefficients, f∈Hlock, and the weak solution satisfies the hypotheses of [F4], so that theorem gives u∈Hlock+2. The explicit formula in step 1.2 gives the stronger global Hk+2 membership.

4.1F1step 1.1step 3.1algebra∎

No third extra local derivative. If u∈Hk+3(I) on a neighborhood I of 0, then its third weak derivative lies in Hk(I). But u′′′=−f′ in distributions, so f′∈Hk(I) and hence f∈Hk+1(I), contradicting step 1.1. Therefore this interior weak solution belongs to Hlock+2 but not to Hlock+3 near 0, even with constant smooth coefficients.

Source notes

Hunter's Theorems 4.28 and 4.31 and Teschl's Corollary 10.19 (printed pp. 114-116 and p. 243) record the gain of exactly two derivatives. The interior cusp f=∣x∣k+1/2 shows the sharpness at an interior point, rather than only at the boundary; the exact Hk+2 and failure of Hlock+3 follow directly from the explicit formula.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

31 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