Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The coordinate-tail and sparse-vector formulations of best s-term error agree

Statement

Let n,s∈N with n≥1 and 1≤s≤n, and let x∈Rn use the one-based labels [n]={1,…,n}, with label j corresponding to library coordinate j−1. For T⊆[n], xTc agrees with x off T and is zero on T. Define σs(x)1:=min⁡{∥xTc∥1:T⊆[n], ∣T∣≤s}. Define supp⁡(z):={j∈[n]:zj≠0} using the one-based labels [n]={1,…,n}, with label j corresponding to library coordinate j−1. Then σs(x)1=min⁡{∥x−z∥1:z∈Rn, ∣supp⁡(z)∣≤s}, and this sparse-vector minimum is attained.

Facts & Assumptions

Given: A vector x∈Rn, positive integers n,s with 1≤s≤n, and the coordinate-tail quantity σs(x)1.

[F1]

Best s-term ℓ1 approximation error: σs(x)1=min⁡{∥xTc∥1:T⊆[n], ∣T∣≤s}, and the finite nonempty indexing family has a minimizing member.

[F2]

The p-norms ∥x∥p for rational p≥1, and ∥x∥∞: ∥x−z∥1=∑j=1n∣xj−zj∣.

[F3]

Laws of finite sums and finite products, clause 4: finite sums preserve termwise inequalities, and a finite sum of nonnegative terms is nonnegative.

Proof

1.1givenF1F2F3algebra

Let z be any vector with ∣supp⁡(z)∣≤s, and put T=supp⁡(z). For every j∈Tc, zj=0, so ∣xj−zj∣=∣(xTc)j∣. For every j∈T, ∣(xTc)j∣=0≤∣xj−zj∣. Thus [F2,F3], applied in the original coordinate order, give ∥x−z∥1=∑j=1n∣xj−zj∣≥∑j=1n∣(xTc)j∣=∥xTc∥1≥σs(x)1, where the final inequality follows from [F1] because ∣T∣≤s. Thus every admissible sparse vector has error at least σs(x)1.

2.1givenF1F2step 1.1choosealgebra∎

By [F1], choose a minimizing set T0⊆[n] with ∣T0∣≤s. Define z0=(xT0), equal to x on T0 and zero on T0c. Define z0 coordinatewise by (z0)j=xj for j∈T0 and (z0)j=0 otherwise. Then supp⁡(z0)⊆T0, so it is admissible, and direct coordinate summation gives ∥x−z0∥1=∥xT0c∥1=σs(x)1. Together with step 1.1 this proves equality and attainment. The proof chooses one witness from a finite nonempty set and uses no choice axiom.

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