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Riemann localisation principle for Fourier series

Statement

Assume the Axiom of Countable Choice.

Let $f$ and $g$ be one-period integrable functions, and let $x \in R$ . Assume there is $δ \in (0, 1 / 2)$ such that $f (y) = g (y)$ for almost every $y \in (x - δ, x + δ)$ . Then

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

Facts & Assumptions

Given: The Axiom of Countable Choice, one-period integrable functions $f, g$ , a real $x$ , and a real $δ$ with $0 < δ < 1 / 2$ such that $f (y) = g (y)$ for almost every $y \in (x - δ, x + δ)$ .

[L1]

The Dirichlet kernel satisfies $D_{N} (t) = \sin ((2 N + 1) π t) / \sin (π t)$ away from the integers (Closed form and size bounds for the Dirichlet kernel).

[L2]

Assuming the Axiom of Countable Choice, Fourier coefficients of an $L^{1} (T)$ function tend to $0$ at infinity (Riemann-Lebesgue lemma for Fourier coefficients).

[L3]

For every real $s$ , $S_{N} h (x) - s = \int_{0}^{1 / 2} (h (x + t) + h (x - t) - 2 s) D_{N} (t) d t$ (Symmetric difference formula for Fourier partial sums).

Proof

technique · direct

1.1

Apply [L3] to $h : = f - g$ with $s = 0$ . Since $h (y) = 0$ for almost every $y \in (x - δ, x + δ)$ , the integrand vanishes for almost every $t \in (0, δ)$ , so $S_{N} h (x) = \int_{δ}^{1 / 2} (h (x + t) + h (x - t)) D_{N} (t) d t .$

givenL3algebra

2.1

Define a one-period function $ψ$ on $[0, 1)$ by $ψ (t) : = 1_{[δ, 1 / 2]} (t) \frac{(h (x + t) + h (x - t)) e^{i π t}}{\sin (π t)} .$ Because $\sin (π t)$ is bounded away from $0$ on $[δ, 1 / 2]$ and $h \in L^{1} (T)$ , one has $ψ \in L^{1} (T)$ . Using [L1], step 1.1 becomes $S_{N} h (x) = Im \int_{0}^{1} ψ (t) e^{2 π i N t} d t = Im \hat{ψ} (- N) .$

L1step 1.1algebra

3.1

By [L2], $\hat{ψ} (- N) \to 0$ as $N \to \infty$ . Step 2.1 therefore gives $S_{N} h (x) \to 0$ . Since $h = f - g$ , this is exactly $S_{N} f (x) - S_{N} g (x) \to 0.$

L2step 2.1algebra∎

Depends on

Used by

Localisation for functions equal on an arc Example

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)