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Fourier partial sums are Dirichlet convolutions

Statement

Let $f$ be a one-period integrable function on $R$ . Then for every $N \geq 0$ and every $x \in R$ ,

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

Facts & Assumptions

Given: A one-period integrable function $f$ , an integer $N \geq 0$ , 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 $D_{N} (t) = \sum_{∣ k ∣ \leq N} e_{k} (t)$ , where $e_{k} (t) = e^{2 π i k t}$ (Dirichlet and Fejer kernels).

Proof

technique · direct

1.1

Expanding the convolution against the finite sum [L1, L2, algebra] gives $\int_{0}^{1} f (x - t) D_{N} (t) d t = \sum_{∣ k ∣ \leq N} \int_{0}^{1} f (x - t) e^{2 π i k t} d t .$

L1L2algebra

2.1

For each $k$ , substitute $u = x - t$ . Then $\int_{0}^{1} f (x - t) e^{2 π i k t} d t = e^{2 π i k x} \int_{x - 1}^{x} f (u) e^{- 2 π i k u} d u .$ The integrand $u \mapsto f (u) e^{- 2 π i k u}$ is one-periodic, so its integral over $[x - 1, x]$ equals its integral over $[0, 1]$ , namely $\hat{f} (k)$ . Therefore $\int_{0}^{1} f (x - t) e^{2 π i k t} d t = \hat{f} (k) e_{k} (x) .$

L1step 1.1algebra

3.1

Summing step 2.1 over $∣ k ∣ \leq N$ yields $\int_{0}^{1} f (x - t) D_{N} (t) d t = \sum_{∣ k ∣ \leq N} \hat{f} (k) e_{k} (x) = S_{N} f (x) .$ By [L1], the integral is also $(f * D_{N}) (x)$ .

L1step 2.1algebra∎

Depends on

Used by

Symmetric difference formula for Fourier partial sums Lemma

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)