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CorollaryStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-06
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Holomorphic functional calculus in the Wiener algebra

Statement

Assume the Axiom of Choice. Let f∈A(T) and let Φ be holomorphic on an open neighbourhood of f(T). Then Φ∘f∈A(T).

Facts & Assumptions

Given: The Axiom of Choice, f∈A(T), and Φ holomorphic near the compact set f(T).

[L1]

A nowhere-zero member of A(T) has its reciprocal in A(T) (Wiener's lemma for absolutely convergent Fourier series).

[L2]

A(T) is complete and closed under multiplication (The Wiener algebra is a unital commutative Banach algebra).

[L3]

Cauchy's formula holds for null-homologous complex cycles (Cauchy's integral formula for a null-homologous cycle).

Proof

technique · direct
1.1L1givenchoosealgebra

Let Ω be an open set on which Φ is holomorphic and which contains f(T). The compact-neighbourhood lemma A compact subset of an open Euclidean set has a compact Jordan neighborhood inside that open set gives a finite union J of closed grid rectangles with f(T)⊆int⁡J⊆J⊆Ω. Orient the frontier edges of the constituent grid cells positively and cancel each internal edge against its reverse. The resulting polygonal chain Γ is a cycle in Ω∖f(T), with n(Γ,w)=1 for w∈f(T) and n(Γ,w)=0 for w∉Ω: summing the cell indices first proves this away from the grid lines, and local constancy of the cycle index extends it to every point off the frontier. Thus Γ is null-homologous in Ω. For z∈Γ∗, z−f has no zero on T, so (z−f)−1∈A(T) by [L1].

2.1L2step 1.1algebra

The map z↦Φ(z)(z−f)−1 is continuous into A(T) (the inverse identity follows from (z−f)−1−(w−f)−1=(w−z)(z−f)−1(w−f)−1). Hence its normalized chain integral F:=12πi∫ΓΦ(z)(z−f)−1 dz is an A(T) element, as the finite sum of norm-limits of edgewise Riemann sums by [L2].

3.1L3step 1.1step 2.1∎

Evaluation at x commutes with those norm-limits. Since step 1.1 gives n(Γ,f(x))=1 and makes Γ null-homologous in Ω, [L3] applied to Φ gives F(x)=Φ(f(x)). Thus F=Φ∘f and belongs to A(T).

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