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Riemann-Lebesgue lemma for Fourier coefficients

Statement

Assume the Axiom of Countable Choice.

Let $f$ be integrable on one period. Then

$\hat{f} (k) \to 0 as ∣ k ∣ \to \infty .$

Facts & Assumptions

Given: The Axiom of Countable Choice, a one-period integrable function $f$ , and a real $ε > 0$ .

[L1]

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

[L2]

One-period step functions have Fourier coefficients tending to $0$ as $∣ k ∣ \to \infty$ (Step functions on one period have vanishing Fourier coefficients).

[L3]

Assuming the Axiom of Countable Choice, one-period step functions are dense in $L^{1} (T)$ (Step functions on one period are dense in L^1 on the torus).

Proof

technique · direct

1.1givenL3choose

By [L3], choose a one-period step function $s$ with $\int_{0}^{1} ∣ f (t) - s (t) ∣ d t < ε / 3.$

1.2L2choose

By [L2], choose $K \geq 1$ such that $∣ k ∣ \geq K$ implies $∣ \hat{s} (k) ∣ < ε / 3$ .

2.1L1step 1.1algebra

For every integer $k$ , [L1, algebra] $∣ \hat{f} (k) - \hat{s} (k) ∣ \leq \int_{0}^{1} ∣ f (t) - s (t) ∣ d t < ε / 3.$

3.1step 2.1step 1.2algebra

If $∣ k ∣ \geq K$ , then step 2.1 and step 1.2 give $∣ \hat{f} (k) ∣ \leq ∣ \hat{f} (k) - \hat{s} (k) ∣ + ∣ \hat{s} (k) ∣ < \frac{2 ε}{3} < ε .$

4.1step 3.1∎

Since $ε > 0$ was arbitrary, step 3.1 is exactly $\hat{f} (k) \to 0$ as $∣ k ∣ \to \infty$ .

Depends on

Used by

Dependency tree · two levels

11 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)