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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Banach-valued Cauchy integral vanishes

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let A be a unital complex Banach algebra, let UC be open, and let F:UA be continuous and weakly holomorphic: for every bounded linear functional φ:AC (The dual space X^* of a normed space and its dual norm) the scalar function φF:UC is holomorphic. Let Γ be a complex chain which is a cycle, with trace in U (Complex chains, their traces, and cycles) and null-homologous in U (Null-homologous cycles and homologous cycles in an open set). Then

ΓF(z)dz=0,

the integral being that of Banach algebra valued contour integral over the chain Γ. The Axiom of Choice is used exactly once, in the separation step supplied by The dual space separates points of a normed space.

Facts & Assumptions

Given: An assumed Axiom of Choice, an open UC, a continuous weakly holomorphic F:UA, and a chain Γ=k<rmkγk which is a cycle with trace ΓU and is null-homologous in U.

[F1]

For pΓ one has Γdz/(zp)=2πin(Γ,p), and Γgdz=k<rmk0mkγkgdz for every g continuous on Γ; integrals over chains are additive (Banach algebra valued contour integral, Integration over a complex chain and the index of a chain).

[F2]

For a bounded linear φ:AC and a single contour γ, bounded linearity commutes with the contour integral (Contour integral commutes with bounded linear maps). Hence for the finite chain Γ=k<rmkγk and every f continuous on its trace, φ ⁣(Γfdz)=k<rmk0mkφ ⁣(γkfdz)=Γ(φf)dz, by the chain-integral definition in [F1]. Each retained contour has trace contained in Γ, so its integral is defined; zero-coefficient contours are omitted even if their traces lie outside the domain of f. For an empty retained list, both sides are zero by linearity.

[F3]

If ΩC is open, g:ΩC is holomorphic, and Γ is a complex chain which is a cycle with trace in Ω and null-homologous in Ω, then Γgdz=0 (Cauchy's theorem for a null-homologous cycle).

[F4]

If xy in a complex normed space V then there is a bounded linear functional φ on V with φ(x)φ(y) (The dual space separates points of a normed space).

[A1]

The standing hypothesis is the Axiom of Choice, used here through [F4] and nowhere else (The Axiom of Choice).

Proof

technique · direct
1.1

For every bounded linear functional φ:AC the composition φF is holomorphic on U by weak holomorphy, and it is continuous; moreover Γ is a cycle with trace in U that is null-homologous in U by hypothesis, so [F3] applies to g:=φF and gives Γφ(F(z))dz=0.

F1F3
2.1

For every bounded linear φ, φ(ΓFdz)=Γφ(F(z))dz=0: the first equality is [F2], and the second is [step 1.1].

step 1.1F2
3.1

Suppose ΓFdz0. Then [F4] applied to the distinct points x:=ΓFdz and 0 produces a bounded linear functional φ with φ(ΓFdz)0, contradicting [step 2.1]; hence ΓFdz=0.

step 2.1F4A1

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