Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 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.

Primal and infeasibility certificates for basis pursuit denoising

Statement

Let m,n∈N with m,n≥1, let A∈Rm×n act by matrix multiplication in the standard finite coordinate bases, with row and column labels 1 through their dimensions corresponding to library coordinates 0 through one less, let y∈Rm, and let ϵ≥0. Define g:Rm→{0,+∞} by g(u)=0 when ∥u−y∥2≤ϵ and g(u)=+∞ otherwise. For p∈Rm, define g∗(p):=sup⁡u∈Rm{⟨p,u⟩−g(u)}. Then g∗(p)=⟨p,y⟩+ϵ∥p∥2(p∈Rm). Let A∗:Rm→Rn be the Euclidean adjoint, characterized by ⟨A∗p,z⟩=⟨p,Az⟩ for all p,z. If x∈Rn satisfies ∥Ax−y∥2≤ϵ and a vector p∈Rm satisfies ∥A∗p∥∞≤1 and ∥x∥1=−⟨p,y⟩−ϵ∥p∥2, then x minimizes ∥z∥1 over that residual ball. If instead p≠0, A∗p=0, and ∣⟨p,y⟩∣>ϵ∥p∥2, then the residual ball is disjoint from the range of A.

Facts & Assumptions

Given: Integers m,n≥1, a real linear map A:Rn→Rm, its Euclidean adjoint A∗, a vector y∈Rm, a radius ϵ≥0, and the residual ball {u:∥u−y∥2≤ϵ}.

[F1]

Fenchel conjugate of an extended-real function: g∗(p)=sup⁡u∈Rm{⟨p,u⟩−g(u)}.

[F2]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: ⟨p,u⟩ is the finite Euclidean coordinate pairing and ∥p∥22=⟨p,p⟩.

[F3]

The p-norms ∥x∥p for rational p≥1, and ∥x∥∞: ∥z∥1=∑j=1n∣zj∣, ∥u∥2=(∑i∣ui∣2)1/2, and ∥v∥∞=max⁡j∣vj∣ in positive dimensions.

[F5]
[F7]

Laws of finite sums and finite products: finite sums of nonnegative terms are nonnegative.

Proof

1.1givenF1F2F3F6F8choosealgebra

In the supremum defining g∗(p), only u in the residual ball contributes. Write u=y+v with ∥v∥2≤ϵ. For p∈Rm, define g∗(p):=sup⁡u∈Rm{⟨p,u⟩−g(u)}. Then g∗(p)=⟨p,y⟩+sup⁡∥v∥2≤ϵ⟨p,v⟩. By [F6], the last supremum is at most ϵ∥p∥2. If p=0 or ϵ=0, v=0 attains this value. Otherwise v=ϵp/∥p∥2 is feasible and gives ⟨p,v⟩=ϵ∥p∥2. This proves the conjugate formula, including the zero cases.

1.2givenF3F4F5F6F7algebra

Let z be any feasible vector and put q=A∗p. The condition ∥q∥∞≤1 means ∣qj∣≤1 for every coordinate. Hence each term ∣zj∣+qjzj is nonnegative, and [F7] gives ∥z∥1+⟨q,z⟩=∑j(∣zj∣+qjzj)≥0. By [F4], ∥z∥1≥−⟨p,Az⟩. Feasibility and [F6] give ∣⟨p,Az−y⟩∣≤ϵ∥p∥2, so ∥z∥1≥−⟨p,Az⟩≥−⟨p,y⟩−ϵ∥p∥2. If x is feasible and attains equality with the last quantity, it has objective no larger than any feasible z, proving the optimality certificate.

2.1givenF4F6F8algebra∎

If p≠0, A∗p=0, and ∣⟨p,y⟩∣>ϵ∥p∥2, then for every z∈Rn, [F4] gives ∣⟨p,Az−y⟩∣=∣⟨A∗p,z⟩−⟨p,y⟩∣=∣⟨p,y⟩∣>ϵ∥p∥2. By [F6], this forces ∥Az−y∥2>ϵ. Thus no z is feasible, proving the infeasibility certificate.

Depends on

Used by

Dependency tree · two levels

55 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