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

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One ell-2 weak derivative implies an absolutely convergent Fourier series

Statement

Assume the Axiom of Countable Choice. If $f \in L^{2} (T)$ has periodic weak derivative $g \in L^{2} (T)$ , then $(\hat{f} (k)) \in ℓ^{1} (Z)$ .

Facts & Assumptions

Given: Countable Choice and $f, g \in L^{2} (T)$ with $g = f^{'}$ periodically.

[L1]

$\hat{g} (k) = 2 π i k \hat{f} (k)$ (Fourier coefficients of a periodic weak derivative).

[L2]

A finite weighted $ℓ^{2}$ sum at exponent $1$ implies $ℓ^{1}$ (Weighted ell-2 decay implies absolute convergence).

Proof

technique · direct

1.1givenalgebra

For a finite $E \subset Z$ , orthogonality of the characters gives $\sum_{k \in E} ∣ \hat{g} (k) ∣^{2} \leq ∥ g ∥_{2}^{2}$ .

2.1L1step 1.1algebra

By [L1], $\sum_{k \in E} (1 + ∣ k ∣)^{2} ∣ \hat{f} (k) ∣^{2} \leq ∣ \hat{f} (0) ∣^{2} + {(1 + \frac{1}{2 π})}^{2} ∥ g ∥_{2}^{2} .$ Taking increasing finite $E$ gives the weighted $ℓ^{2}$ hypothesis.

3.1L2step 2.1∎

Apply [L2] with $s = 1$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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, Exercise 3.3.5 (standard reference, not scraped)