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Banach-valued Cauchy integral vanishes
Statement
Assume the Axiom of Choice (The Axiom of Choice). Let be a unital complex Banach algebra, let be open, and let be continuous and weakly holomorphic: for every bounded linear functional (The dual space X^* of a normed space and its dual norm) the scalar function is holomorphic. Let be a complex chain which is a cycle, with trace in (Complex chains, their traces, and cycles) and null-homologous in (Null-homologous cycles and homologous cycles in an open set). Then
the integral being that of Banach algebra valued contour integral over the chain . 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 , a continuous weakly holomorphic , and a chain which is a cycle with trace and is null-homologous in .
For one has , and for every continuous on ; integrals over chains are additive (Banach algebra valued contour integral, Integration over a complex chain and the index of a chain).
For a bounded linear and a single contour , bounded linearity commutes with the contour integral (Contour integral commutes with bounded linear maps). Hence for the finite chain and every continuous on its trace, by the chain-integral definition in [F1]. Each retained contour has trace contained in , so its integral is defined; zero-coefficient contours are omitted even if their traces lie outside the domain of . For an empty retained list, both sides are zero by linearity.
If is open, is holomorphic, and is a complex chain which is a cycle with trace in and null-homologous in , then (Cauchy's theorem for a null-homologous cycle).
If in a complex normed space then there is a bounded linear functional on with (The dual space separates points of a normed space).
The standing hypothesis is the Axiom of Choice, used here through [F4] and nowhere else (The Axiom of Choice).
Proof
For every bounded linear functional the composition is holomorphic on by weak holomorphy, and it is continuous; moreover is a cycle with trace in that is null-homologous in by hypothesis, so [F3] applies to and gives .
For every bounded linear , : the first equality is [F2], and the second is [step 1.1].
Suppose . Then [F4] applied to the distinct points and produces a bounded linear functional with , contradicting [step 2.1]; hence .
Depends on
- Contour integral commutes with bounded linear maps
- Cauchy's theorem for a null-homologous cycle
- The dual space separates points of a normed space
- The Axiom of Choice
- Banach algebra valued contour integral
- Complex chains, their traces, and cycles
- Null-homologous cycles and homologous cycles in an open set
- The dual space X^* of a normed space and its dual norm
- Integration over a complex chain and the index of a chain
Used by
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Sources
- Theo Bühler and Dietmar A. Salamon, Functional Analysis — Lemma 5.11 and Theorem 5.25(i), printed pp. 217–219 and 228 (standard reference, not scraped)
- Vahid Shirbisheh, Lectures on C-star Algebras, v2 — §2.5, printed pp. 43–47 (standard reference, not scraped)