LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Hadamard gaps bound the additive representations used in even moments

Statement

Let $(λ_{j})$ be Hadamard-lacunary with ratio $q > 1$ , as in Hadamard-lacunary sequences and lacunary trigonometric series. For each integer $m \geq 1$ , choose $r \geq 1$ with $q^{r} > m$ . In every residue class modulo $r$ , an equality

$λ_{i_{1}} + \dots + λ_{i_{m}} = λ_{j_{1}} + \dots + λ_{j_{m}}$

forces the two index multisets to be equal.

Facts & Assumptions

Given: $m, q, r$ and two $m$ -term sums as in the Statement.

Proof

technique · compare the largest unmatched frequency

1.1givenalgebra

Suppose the multisets differ, cancel their common entries, and let [given, algebra] $λ_{s}$ be the largest remaining frequency. It occurs on only one side. If the other side has a remaining term, its index is at most $s - r$ because all indices lie in one residue class; in particular $s - r \geq 1$ , and every such frequency is at most $λ_{s - r} < λ_{s} / m$ . If the other side has no remaining term, the two sums are already unequal.

2.1step 1.1algebra∎

That other side has at most $m$ remaining terms, so its sum is strictly [step 1.1, algebra] less than $m λ_{s} / m = λ_{s}$ , whereas its opposing side is at least $λ_{s}$ . This contradicts the equality. Hence nothing remains after cancellation, which is exactly equality of multisets.

Depends on

Hadamard-lacunary sequences and lacunary trigonometric series

Used by

Dependency tree · two levels

2 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

Loukas Grafakos, Classical Fourier Analysis, 3rd ed., proof of Theorem 3.6.4 (standard reference, not scraped)