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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]

Fourier coefficients on $T$ are defined by $\hat{f} (k) = \int_{0}^{1} f (t) e^{- 2 π i k t} d t$ (Period-one Fourier coefficients, partial sums, and convolution on the torus).

Proof

technique · direct

1.1

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 $∣ \hat{1_{[x_{j - 1}, x_{j})}} (k) ∣ \leq \frac{1}{π ∣ k ∣} .$

L1algebra

2.1

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.

step 1.1algebra

3.1

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

step 2.1algebra∎

Depends on

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

Used by

Riemann-Lebesgue lemma for Fourier coefficients Theorem

Dependency tree · one level

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Sources

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