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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Robust nullspace property gives stable noisy recovery

Statement

Let m,n,s∈N with m,n≥1 and 1≤s≤n. Use one-based coordinate labels [n]={1,…,n}, with label j corresponding to library coordinate j−1, and interpret natural numbers in real scalar expressions as their canonical real images. Let A∈Rm×n and suppose that for some 0<ρ<1 and τ>0, ∥hS∥2≤ρs∥hSc∥1+τ∥Ah∥2 for every h∈Rn and every S⊆[n] with ∣S∣≤s, where for any vector v∈Rn, vS agrees with v on S and vanishes on Sc=[n]∖S. Let x†∈Rn, ϵ≥0, and e∈Rm satisfy y=Ax†+e and ∥e∥2≤ϵ. Define σs(x†)1:=min⁡{∥xTc†∥1:T⊆[n], ∣T∣≤s}. The indexing family is finite and nonempty, so this minimum is attained; it also equals the minimum of ∥x†−z∥1 over vectors z with at most s nonzero coordinates. Then the basis-pursuit denoising problem min⁡z∈Rn∥z∥1subject to∥Az−y∥2≤ϵ has a minimizer. Every minimizer x^ satisfies ∥x^−x†∥2≤C1σs(x†)1s+C2ϵ,∥x^−x†∥1≤C3σs(x†)1+C4s ϵ, where C1=2(1+2ρ)1−ρ,C2=6τ1−ρ,C3=2(1+ρ)1−ρ,C4=4τ1−ρ. If the certified finite basis-pursuit denoising procedure returns success, its returned vector is a minimizer and satisfies these same bounds. Only reconstruction error is bounded; the theorem asserts neither uniqueness nor support recovery.

Facts & Assumptions

Given: Integers m,n≥1, 1≤s≤n, a real matrix A satisfying the robust nullspace inequality, a truth x†, a radius ϵ≥0, a perturbation e with y=Ax†+e and ∥e∥2≤ϵ, and any minimizer x^ of the residual-constrained ℓ1 problem.

[F1]

The ℓ2 robust nullspace property: for every h and every S⊆[n] with ∣S∣≤s, ∥hS∥2≤ρs∥hSc∥1+τ∥Ah∥2.

[F2]

Best s-term ℓ1 approximation error and The coordinate-tail and sparse-vector formulations of best s-term error agree: σs(x†)1 is the finite coordinate-tail minimum and equals the minimum ℓ1 error over vectors with at most s nonzero coordinates.

[F3]

The largest coordinates attain the best s-term ℓ1 error: the first s largest coordinates attain this minimum and maximize the retained ℓ1 mass over all sets of size at most s.

[F4]

A block-tail estimate for the largest coordinates: if S is the first s coordinates of h ordered by decreasing absolute value with increasing-index tie breaks, then ∥hSc∥2≤∥hS∥2+∥hSc∥1/s.

[F5]

A feasible finite-dimensional BPDN problem attains its minimum: nonempty finite-dimensional residual balls admit a global ℓ1 minimizer.

[F8]

Laws of finite sums and finite products: finite sums preserve coordinatewise inequalities and split over disjoint sets.

[F9]

The canonical natural ι(n)=n⋅1F of a field: s means s in real scalar expressions.

[F10]

Canonical naturals are positive and strictly increasing: s>0, so all divisions by s are defined.

[F11]

Certified basis-pursuit denoising by primal-dual hybrid gradient: a returned success vector is accompanied by the exact feasibility and primal-dual certificate checks stated in the Algorithm.

[F12]

Primal and infeasibility certificates for basis pursuit denoising: those exact checks imply that the returned feasible vector is a global minimizer.

Proof

1.1givenF5F6construct

The truth x† is feasible because ∥Ax†−y∥2=∥e∥2≤ϵ. Thus the feasible set is nonempty, and [F5] supplies at least one minimizer. Fix any minimizer x^ and put h=x^−x†.

1.2givenF2F3F6F8constructalgebra

Order the coordinates of x† by decreasing magnitude, breaking ties by increasing index, and let T be the first s indices. By [F2,F3], ∥xTc†∥1=σs(x†)1=:σ. Minimality of x^ and the triangle inequality give ∥xT†+hT∥1+∥xTc†+hTc∥1≤∥xT†∥1+∥xTc†∥1. Applying ∥a+b∥1≥∥b∥1−∥a∥1 to the two disjoint restrictions yields ∥xT†∥1−∥hT∥1+∥hTc∥1−∥xTc†∥1≤∥xT†∥1+∥xTc†∥1. After cancellation this is ∥hTc∥1−∥hT∥1≤2σ, hence the cone estimate ∥hTc∥1≤∥hT∥1+2σ.

2.1givenF3F6step 1.2algebra

Independently order h by decreasing magnitude with the same deterministic tie break and let S be its first s indices. Since T has size s, [F3] gives ∥hT∥1≤∥hS∥1 and ∥hSc∥1≤∥hTc∥1. Step 1.2 therefore yields ∥hSc∥1≤∥hS∥1+2σ. The data relation and feasibility give Ah=(Ax^−y)+e, so the triangle inequality gives ∥Ah∥2≤2ϵ.

3.1F1F7F9F10step 2.1algebra

Set u=∥hS∥2≥0. Applying [F1] to this S, then using step 2.1 and Cauchy–Schwarz on its s coordinates, gives u≤ρs∥hSc∥1+τ∥Ah∥2≤ρu+2ρσs+2τϵ. Since 1−ρ>0, rearranging proves u≤2ρ1−ρσs+2τ1−ρϵ.

4.1F4F6F7step 2.1step 3.1algebra

Apply [F4] to h and its chosen largest-coordinate set S. Step 2.1 and [F7] show ∥hSc∥2≤u+∥hSc∥1s≤u+s u+2σs=2u+2σs. By the triangle inequality on h=hS+hSc, ∥h∥2≤3u+2σ/s. Substitute step 3.1 and simplify: ∥h∥2≤(6ρ1−ρ+2)σs+6τ1−ρϵ=2(1+2ρ)1−ρσs+6τ1−ρϵ.

4.2F7F8step 1.1step 2.1step 3.1algebra

Disjoint coordinates and step 2.1 give ∥h∥1≤∥hS∥1+∥hSc∥1≤2s u+2σ by [F7]. Substituting step 3.1 yields ∥h∥1≤(4ρ1−ρ+2)σ+4τ1−ρs ϵ=2(1+ρ)1−ρσ+4τ1−ρs ϵ. This proves both bounds for the arbitrary minimizer fixed at step 1.1.

5.1F11F12step 1.2step 2.1step 3.1step 4.1step 4.2∎

If the finite procedure returns success, [F11] gives the same exact feasibility and certificate conditions, and [F12] makes its vector a minimizer. Steps 1.2, 2.1, 3.1, 4.1, and 4.2 therefore apply to that vector as well. No step uses uniqueness or identifies a support.

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