ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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.

Localisation for functions equal on an arc

Example

Assume the Axiom of Countable Choice.

Let $f \equiv 0$ and let $g = 1_{[1 / 4, 3 / 4]}$ , both extended one-periodically. Then $f$ and $g$ agree on the arc $(- 1 / 8, 1 / 8)$ modulo $1$ , so

$S_{N} g (0) - S_{N} f (0) \to 0.$

Since $S_{N} f (0) = 0$ for every $N$ , this gives

$S_{N} g (0) \to 0.$

Facts & Assumptions

Given: The Axiom of Countable Choice and the one-periodic functions $f \equiv 0$ and $g = 1_{[1 / 4, 3 / 4]}$ .

[L1]

Assuming the Axiom of Countable Choice, if two one-period integrable functions agree almost everywhere on a neighborhood of $x$ , then their Fourier partial sums at $x$ differ by a term tending to $0$ (Riemann localisation principle for Fourier series).

Verification

technique · direct

1.1

On the interval $(- 1 / 8, 1 / 8)$ modulo $1$ , the indicator $g$ vanishes, so $f = g = 0$ there. Thus the hypothesis of [L1] holds at $x = 0$ .

L1given

2.1

Applying [L1] at $x = 0$ gives $S_{N} g (0) - S_{N} f (0) \to 0$ . But every Fourier coefficient of the zero function is $0$ , so $S_{N} f (0) = 0$ for all $N$ . Therefore $S_{N} g (0) \to 0$ .

L1step 1.1algebra∎

Depends on

Riemann localisation principle for Fourier series

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

Richard S. Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)
Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (standard reference, not scraped)