LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06

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Absolutely summable Fourier coefficients give uniform convergence

Statement

Assume the Axiom of Countable Choice. If $(a_{k})_{k \in Z} \in ℓ^{1} (Z)$ , then $\sum_{k} a_{k} e_{k}$ converges absolutely and uniformly to a continuous function $g$ , and $\hat{g} (m) = a_{m}$ . Consequently every $f \in A (T)$ has $f = g$ almost everywhere for $a_{k} = \hat{f} (k)$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and an $ℓ^{1} (Z)$ sequence $(a_{k})$ .

[L1]

Abel means of an $L^{1} (T)$ function converge to that function in $L^{1}$ as $r ↑ 1$ (Abel means converge in L^p, uniformly, and at Lebesgue points).

Proof

technique · direct

1.1

Since $∣ a_{k} e_{k} (x) ∣ = ∣ a_{k} ∣$ and $\sum_{k} ∣ a_{k} ∣ < \infty$ , the Weierstrass M-test gives absolute uniform convergence to a continuous $g$ .

givenalgebra

2.1

Uniform convergence permits integration term by term against $e_{- m}$ ; character orthogonality gives $\hat{g} (m) = a_{m}$ .

step 1.1algebra

3.1

For $a_{k} = \hat{f} (k)$ , the Abel means are $\sum_{k} r^{∣ k ∣} a_{k} e_{k}$ . They converge uniformly to $g$ by dominated tail control, while [L1] says they converge to $f$ in $L^{1}$ ; hence $f = g$ almost everywhere.

L1step 1.1algebra∎

Depends on

Used by

Wiener inversion needs nonvanishing Counterexample
The Wiener algebra is a unital commutative Banach algebra Theorem
Wiener's lemma for absolutely convergent Fourier series Theorem

Dependency tree · two levels

8 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, Chapter 4 (standard reference, not scraped)