Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (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.

The W1,p→Lq bound fails for q>p∗ by dilation

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice (ACω)). Let n≥2, 1≤p<n and q>p∗=npn−p. The assertion that there is a finite constant C with ∥u∥Lq(Rn)≤C∥Du∥Lp(Rn)for every u∈Cc∞(Rn) is false. The stronger full-norm assertion ∥u∥Lq≤C∥u∥W1,p also fails for this family, so the exponent p∗ is sharp for both estimates.

Facts & Assumptions

Given: Countable Choice; n≥2; 1≤p<n; a fixed nonzero φ∈Cc∞(B(0,1)); and 0<λ≤1.

[F1]

The Sobolev conjugate satisfies 1p∗=1p−1n, so for finite q>p∗ one has 1q<1p−1n; the case q=∞ is treated separately with 1/q=0 (The Sobolev conjugate exponent and the scaling identity).

[F2]

An invertible linear map scales Lebesgue measure by ∣det⁡∣, so for a>0 the substitution y=x/a gives ∫Rng(x/a) dx=an∫Rng(y) dy; an Lq class is determined by its values almost everywhere (A linear map T of Rn sends Lebesgue measurable sets to Lebesgue measurable sets, with λn(T[E])=∣det⁡T∣ λn(E) when T is invertible and T[E] Lebesgue null when it is not, The space Lp(μ) as the quotient by null functions).

[F3]

The closed unit ball is compact, so a continuous function on it is bounded and the support of φ is compact (For n≥1, every Euclidean closed ball and every Euclidean sphere of positive radius is compact).

[F4]

A nonzero smooth bump in B(0,1) exists (A Euclidean bump for a compact set inside an open set); its gradient scales by the chain rule (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)). Its gradient norm is positive: otherwise continuity gives zero gradient everywhere, the fundamental theorem along segments makes it constant, and compact support makes it zero (If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz).

Counterexample

technique · direct
1.1F2F3F4givenalgebra

The norm scalings. For 0<λ≤1 put φλ(x):=φ(x/λ), so φλ∈Cc∞(B(0,λ))∖{0}. If q<∞, substituting y=x/λ and using [F2] gives ∥φλ∥Lqq=λn∥φ∥Lqq, hence ∥φλ∥Lq=λn/q∥φ∥Lq>0. If q=∞, then for each t>0, [F2] gives λn({∣φλ∣>t})=λnλn({∣φ∣>t}), so ∥φλ∥L∞=∥φ∥L∞>0. In either case Dφλ(x)=λ−1(Dφ)(x/λ) and ∥Dφλ∥Lp=λn/p−1∥Dφ∥Lp>0; the support statement uses [F3].

2.1F1step 1.1givenalgebra∎

The ratio diverges above p∗. For finite q>p∗, dividing the two scalings of step 1.1 gives ∥φλ∥Lq/∥Dφλ∥Lp=λ 1+n/q−n/p∥φ∥Lq/∥Dφ∥Lp; the exponent is negative exactly when 1q<1p−1n=1p∗ by [F1], so this ratio tends to +∞. For q=∞, step 1.1 gives ∥φλ∥L∞/∥Dφλ∥Lp=λ1−n/p∥φ∥L∞/∥Dφ∥Lp→+∞ because p<n. In either case any finite C satisfying the proposed inequality for every compactly supported smooth u would have to dominate this unbounded ratio, which is impossible. Also ∥φλ∥W1,pp=λn∥φ∥pp+λn−p∑i∥Diφ∥pp≤λn−p(∥φ∥pp+∑i∥Diφ∥pp) for λ≤1. Thus the full-norm ratio has the same divergent lower bound cλ1+n/q−n/p, with 1/q=0 for q=∞. Both bounds fail for every q>p∗.

Source notes

The dilation computation is Kinnunen's, printed pp. 61-62, and Laugesen's sharpness discussion, printed pp. 65-66; the counterexample is the standard concentrated-bump family.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

84 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