TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Lebesgue constants grow logarithmically

Statement

There are absolute constants $c, C > 0$ such that, for every $N \geq 1$ ,

$c \log (N + 1) \leq \int_{0}^{1} ∣ D_{N} (t) ∣ d t \leq C \log (N + 1) .$

Facts & Assumptions

Given: An integer $N \geq 1$ .

[L1]

For $t \notin Z$ , $D_{N} (t) = \frac{\sin ((2 N + 1) π t)}{\sin (π t)},$ and in particular $D_{N} (1 - t) = D_{N} (t)$ for $0 < t < 1$ (Closed form and size bounds for the Dirichlet kernel).

[L2]

For $0 < u \leq 2$ , $\sin u \geq u / 3 > 0$ (Sine is positive and cosine is strictly decreasing on (0,2), with cos 2 at most -1/3).

Proof

technique · direct

1.1L1L2algebra

For $0 < t < 1$ , [L1] gives $D_{N} (1 - t) = D_{N} (t)$ , so $\int_{0}^{1} ∣ D_{N} (t) ∣ d t = 2 \int_{0}^{1 / 2} ∣ D_{N} (t) ∣ d t .$ Split the last integral at $1 / (2 N + 1)$ . On $(0, 1 / (2 N + 1)]$ , [L1] and the bound $∣ D_{N} (t) ∣ \leq 2 N + 1$ give a contribution at most $1$ . On $[1 / (2 N + 1), 1 / 2]$ , one has $0 < π t \leq π / 2 < 2$ , so [L2] implies $\sin (π t) \geq π t / 3$ . Using [L1], $∣ D_{N} (t) ∣ \leq \frac{1}{\sin (π t)} \leq \frac{3}{π t} .$ Therefore $\int_{0}^{1} ∣ D_{N} (t) ∣ d t \leq 2 + \frac{6}{π} \int_{1 / (2 N + 1)}^{1 / 2} \frac{d t}{t} \leq C \log (N + 1)$ for a universal $C$ .

1.2L1algebra

For $m = 0, \dots, N - 1$ , let $J_{m} : = [\frac{m + 1 / 6}{2 N + 1}, \frac{m + 5 / 6}{2 N + 1}] .$ These intervals lie in $(0, 1 / 2)$ and are disjoint. If $t \in J_{m}$ , then $(2 N + 1) π t \in [(m + 1 / 6) π, (m + 5 / 6) π]$ , so $∣ \sin ((2 N + 1) π t) ∣ \geq 1 / 2$ . Also $\sin (π t) \leq π t$ . Hence [L1] gives $∣ D_{N} (t) ∣ \geq \frac{1}{2 π t} (t \in J_{m}) .$

2.1step 1.1step 1.2algebra

Integrating the lower bound from step 1.2 over each $J_{m}$ and summing yields $\int_{0}^{1 / 2} ∣ D_{N} (t) ∣ d t \geq \frac{1}{2 π} \sum_{m = 0}^{N - 1} \int_{J_{m}} \frac{d t}{t} = \frac{1}{2 π} \sum_{m = 0}^{N - 1} \log ⁣ (\frac{m + 5 / 6}{m + 1 / 6}) .$ Since $\log ⁣ (\frac{m + 5 / 6}{m + 1 / 6}) \geq \frac{2 / 3}{m + 5 / 6} \geq \frac{2}{3 (m + 1)},$ one gets $\int_{0}^{1 / 2} ∣ D_{N} (t) ∣ d t \geq \frac{1}{3 π} \sum_{m = 0}^{N - 1} \frac{1}{m + 1} \geq \frac{1}{3 π} \int_{1}^{N + 1} \frac{d t}{t} = \frac{1}{3 π} \log (N + 1) .$ Using step 1.1 once more, $\int_{0}^{1} ∣ D_{N} (t) ∣ d t \geq \frac{2}{3 π} \log (N + 1) .$

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 give the two-sided logarithmic bound.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (standard reference, not scraped)
Richard S. Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)