CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-04

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Continuity alone does not satisfy a Dini modulus

Statement refuted

Assume the Axiom of Countable Choice.

Every continuous one-periodic function automatically satisfies the Dini integrability condition at each point.

Facts & Assumptions

Given: The Axiom of Countable Choice and the Dini criterion on the Fourier page (Dini pointwise convergence criterion for Fourier series).

[L1]

Assuming the Axiom of Countable Choice, if $\int_{0}^{δ} \frac{∣ f (x + t) + f (x - t) - 2 s ∣}{t} d t < \infty,$ then the Fourier partial sums converge to $s$ at $x$ (Dini pointwise convergence criterion for Fourier series).

Counterexample

technique · direct

1.1

Let $d (x, Z) : = \min_{m \in Z} ∣ x - m ∣$ and define the one-periodic function $f (x) : = {\begin{cases} 0, & d (x, Z) = 0, \\ \frac{1}{\log (e / d (x, Z))}, & d (x, Z) > 0. \end{cases}$ Because $d (x, Z) \to 0$ exactly when $x$ approaches an integer and $1 / \log (e / r) \to 0$ as $r ↓ 0$ , the function $f$ is continuous on $T$ .

givenalgebra

2.1

At $x = 0$ one has $f (0) = 0$ and, for $0 < t \leq 1 / 4$ , $d (t, Z) = d (- t, Z) = t$ . Hence $f (t) + f (- t) - 2 f (0) = \frac{2}{\log (e / t)} .$ Therefore $\int_{0}^{1 / 4} \frac{∣ f (t) + f (- t) - 2 f (0) ∣}{t} d t = 2 \int_{0}^{1 / 4} \frac{d t}{t \log (e / t)} = \infty,$ because the change of variables $u = \log (e / t)$ turns the integral into $\int_{\log (4 e)}^{\infty} d u / u$ . So the hypothesis in [L1] fails at $x = 0$ : $f$ is continuous but does not satisfy the Dini condition there.

L1step 1.1algebra∎

Depends on

Dini pointwise convergence criterion for Fourier series

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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)