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LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-23 (gpt-6-sol)
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Fourier partial sums are Dirichlet convolutions

Statement

Assume the Axiom of Countable Choice. Let f be a one-period integrable function on R. Then for every N≥0 and every x∈R,

SNf(x)=∫01f(x−t)DN(t) dt=(f∗DN)(x).

Facts & Assumptions

Given: Countable Choice, a one-period integrable function f, an integer N≥0, and a real x.

[L1]

Fourier coefficients, Fourier partial sums, and torus convolution are defined exactly as in Period-one Fourier coefficients, partial sums, and convolution on the torus.

[L2]

The Dirichlet kernel is DN(t)=∑∣k∣≤Nek(t), where ek(t)=e2πikt (Dirichlet and Fejer kernels).

Proof

technique · direct
1.1L1L2algebra

Expanding the convolution against the finite sum [L1, L2, algebra] gives ∫01f(x−t)DN(t) dt=∑∣k∣≤N∫01f(x−t)e2πikt dt.

2.1L1step 1.1algebra

For each k, substitute u=x−t. Then ∫01f(x−t)e2πikt dt=e2πikx∫x−1xf(u)e−2πiku du. The integrand u↦f(u)e−2πiku is one-periodic, so its integral over [x−1,x] equals its integral over [0,1], namely f^(k). Therefore ∫01f(x−t)e2πikt dt=f^(k)ek(x).

3.1L1step 2.1algebra∎

Summing step 2.1 over ∣k∣≤N yields ∫01f(x−t)DN(t) dt=∑∣k∣≤Nf^(k)ek(x)=SNf(x). By [L1], the integral is also (f∗DN)(x).

Depends on

Used by

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Sources