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

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Symmetric difference formula for Fourier partial sums

Statement

Assume the Axiom of Countable Choice. Let $f$ be a one-period integrable function, let $x, s \in R$ , and let $N \geq 0$ . Then

$S_{N} f (x) - s = \int_{0}^{1 / 2} (f (x + t) + f (x - t) - 2 s) D_{N} (t) d t .$

Equivalently,

$S_{N} f (x) - s = \int_{0}^{1 / 2} (f (x + t) + f (x - t) - 2 s) \frac{\sin ((2 N + 1) π t)}{\sin (π t)} d t .$

Facts & Assumptions

Given: Countable Choice, a one-period integrable function $f$ , reals $x, s$ , and an integer $N \geq 0$ .

[L1]

Fourier partial sums are Dirichlet convolutions: $S_{N} f (x) = \int_{0}^{1} f (x - t) D_{N} (t) d t$ (Fourier partial sums are Dirichlet convolutions).

[L2]

The Dirichlet kernel is even and $\int_{0}^{1} D_{N} (t) d t = 1$ (Dirichlet and Fejer kernels).

[L3]

For $t \notin Z$ , $D_{N} (t) = \sin ((2 N + 1) π t) / \sin (π t)$ (Closed form and size bounds for the Dirichlet kernel).

Proof

technique · direct

1.1L1L2algebra

By [L1], $S_{N} f (x) - s = \int_{0}^{1} (f (x - t) - s) D_{N} (t) d t,$ because [L2] gives $\int_{0}^{1} D_{N} (t) d t = 1$ .

2.1step 1.1L2algebra

Split the integral in step 1.1 at $1 / 2$ and substitute $u = 1 - t$ on $[1 / 2, 1]$ . Since $f$ is one-periodic and $D_{N} (1 - u) = D_{N} (- u) = D_{N} (u)$ by [L2], this yields $S_{N} f (x) - s = \int_{0}^{1 / 2} (f (x - t) + f (x + t) - 2 s) D_{N} (t) d t .$

3.1step 2.1L3algebra∎

The first displayed formula is step 2.1 with the two summands reordered. Replacing $D_{N} (t)$ by the closed form from [L3] gives the second displayed formula.

Depends on

Used by

Dini pointwise convergence criterion for Fourier series Theorem
Dirichlet-Jordan pointwise convergence Theorem
Riemann localisation principle for Fourier series Theorem

Dependency tree · two levels

4 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)