LemmaStatement: AI-adaptedProof: AI-generatedverified 2026-09-23 (gpt-6-sol)

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Step functions on one period have vanishing Fourier coefficients

Statement

Let $s$ be a step function on $[0, 1)$ , and extend it one-periodically to $R$ . Then there is a constant $C \geq 0$ such that

$∣ \hat{s} (k) ∣ \leq \frac{C}{∣ k ∣} (k \neq 0) .$

In particular, $\hat{s} (k) \to 0$ as $∣ k ∣ \to \infty$ .

Facts & Assumptions

Given: A one-period step function $s$ on $[0, 1)$ .

[L1]

For a step function, the displayed coefficient formula of Period-one Fourier coefficients, partial sums, and convolution on the torus is the ordinary Riemann integral of a finite piecewise-continuous function. This calculation needs no choice premise and agrees with the torus Lebesgue coefficient whenever Countable Choice is assumed.

Proof

technique · direct

1.1L1algebra

Write $s = \sum_{j = 1}^{m} a_{j} 1_{[x_{j - 1}, x_{j})}$ for a partition $0 = x_{0} < \dots < x_{m} = 1$ . For $k \neq 0$ , [L1, algebra] $\hat{1_{[x_{j - 1}, x_{j})}} (k) = \int_{x_{j - 1}}^{x_{j}} e^{- 2 π i k t} d t = \frac{e^{- 2 π i k x_{j - 1}} - e^{- 2 π i k x_{j}}}{2 π i k} .$ Therefore $∣ \hat{1_{[x_{j - 1}, x_{j})}} (k) ∣ \leq \frac{1}{π ∣ k ∣} .$

2.1step 1.1algebra

By linearity, $\hat{s} (k) = \sum_{j = 1}^{m} a_{j} \hat{1_{[x_{j - 1}, x_{j})}} (k) .$ Step 1.1 then gives $∣ \hat{s} (k) ∣ \leq \frac{1}{π ∣ k ∣} \sum_{j = 1}^{m} ∣ a_{j} ∣ .$ Taking $C : = π^{- 1} \sum_{j = 1}^{m} ∣ a_{j} ∣$ proves the displayed bound.

3.1step 2.1algebra∎

Since $C / ∣ k ∣ \to 0$ as $∣ k ∣ \to \infty$ , step 2.1 yields $\hat{s} (k) \to 0$ .

Depends on

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

Used by

Riemann-Lebesgue lemma for Fourier coefficients Theorem

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

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