LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-23 (gpt-6-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.

A dyadic Fourier-coefficient square-sum bound for Hölder functions

Statement

Let $0 < α \leq 1$ and let $f \in C^{α} (T)$ : $∣ f (x) - f (y) ∣ \leq C d_{T} (x, y)^{α}$ . Then there is $C_{α}$ such that, for every integer $N \geq 1$ , $\sum_{N \leq ∣ k ∣ < 2 N} ∣ \hat{f} (k) ∣^{2} \leq C_{α} C^{2} N^{- 2 α} .$

Facts & Assumptions

Given: $f, C, α, N$ as in the statement and the Fourier convention of Period-one Fourier coefficients, partial sums, and convolution on the torus. Since $f$ is continuous, all coefficient and norm integrals in this proof are ordinary Riemann integrals; this argument is choice-free and agrees with the torus Lebesgue convention under Countable Choice.

Proof

technique · direct

1.1givenalgebra

Put $h = (4 N)^{- 1}$ . For $N \leq ∣ k ∣ < 2 N$ , $∣ e^{- 2 π i k h} - 1 ∣ \geq \sqrt{2}$ , since $2 π ∣ k ∣ h \in [π / 2, π)$ .

1.2givenalgebra

For $u_{h} (x) = f (x - h) - f (x)$ , translation in the coefficient integral gives ${\hat{u}}_{h} (k) = (e^{- 2 π i k h} - 1) \hat{f} (k)$ .

2.1step 1.1step 1.2algebra

Let $B = {k : N \leq ∣ k ∣ < 2 N}$ and $p = \sum_{k \in B} {\hat{u}}_{h} (k) e_{k}$ . Finite character orthogonality and expansion of the nonnegative Riemann integral give $0 \leq \int_{0}^{1} ∣ u_{h} - p ∣^{2} = \int_{0}^{1} ∣ u_{h} ∣^{2} - \sum_{k \in B} ∣ {\hat{u}}_{h} (k) ∣^{2} .$ By steps 1.1--1.2, each summand on the right is at least $2 ∣ \hat{f} (k) ∣^{2}$ . Hence $2 \sum_{N \leq ∣ k ∣ < 2 N} ∣ \hat{f} (k) ∣^{2} \leq \int_{0}^{1} ∣ u_{h} (x) ∣^{2} d x .$

3.1step 2.1givenalgebra∎

The Hölder bound gives $∣ u_{h} (x) ∣ \leq C (4 N)^{- α}$ , so step 2.1 proves the assertion.

Depends on

Period-one Fourier coefficients, partial sums, and convolution on the torus

Used by

Hölder Fourier coefficients have subcritical weighted ell-2 decay Lemma

Dependency tree · two levels

8 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, proof of Theorem 3.3.16 (standard reference, not scraped)