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The higher-derivative form of the global Cauchy formula

Statement

Let Ω⊆C be open, let f:Ω→C be holomorphic, and let Γ be a complex chain which is a cycle, with trace in Ω and null-homologous in Ω. Then for every natural number m and every z∈Ω∖Γ∗

n(Γ,z) f(m)(z)=m!2πi∫Γf(ζ)(ζ−z)m+1 dζ,

with f(0)=f. The case m=0 is the integral formula already proved.

Facts & Assumptions

Given: An open Ω, a holomorphic f:Ω→C, and a cycle Γ with Γ∗⊆Ω which is null-homologous in Ω.

[L1]

Under these hypotheses, n(Γ,z)f(z)=(2πi)−1∫Γf(ζ)(ζ−z)−1 dζ for every z∈Ω∖Γ∗ (Cauchy's integral formula for a null-homologous cycle).

[L2]

For a chain Γ and φ continuous on Γ∗, the functions Fj(z)=(2πi)−1∫Γφ(ζ)(ζ−z)−j dζ are holomorphic on C∖Γ∗ for every natural j≥1 and satisfy Fj′=jFj+1 (The Cauchy transform of a cycle is holomorphic off its trace, with the expected derivatives).

[L3]

For a cycle Γ the trace is compact, the index is constant on every connected component of C∖Γ∗, and each such component is open (The index of a cycle is locally constant off its trace and vanishes far from it).

[L4]

A holomorphic function is smooth in the real coordinates (Holomorphic functions are real analytic and smooth in their two real coordinates) and has complex derivatives of every natural order (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle); a complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L5]

If a property holds at 0 and passes from j to j+1, it holds for every natural number (The principle of mathematical induction).

[L7]

A constant multiple of a function complex differentiable at a point is complex differentiable there with the corresponding derivative (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L8]

Negative integer powers are defined exactly for nonzero complex bases (Integer powers in the complex field).

[L9]

n(Γ,z)=(2πi)−1∫Γdζ/(ζ−z) for z∉Γ∗ (Integration over a complex chain and the index of a chain), and null-homology in Ω means the index vanishes at every point outside Ω (Null-homologous cycles and homologous cycles in an open set).

[L10]

The connected component of a point is the union of all connected subsets containing it (Connected components, quasicomponents, and totally disconnected spaces), and a set is closed exactly when its complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Proof

technique · direct
1.1givenL2L3L4L8L10

The trace Γ∗ is compact by [L3], hence closed, so C∖Γ∗ is open by [L10] and Ω∖Γ∗ is open. The restriction of f to Γ∗ is continuous by [L4], so the functions Fj(z)=(2πi)−1∫Γf(ζ)(ζ−z)−j dζ of [L2] are defined and holomorphic on C∖Γ∗ with Fj′=jFj+1, the powers being legitimate by [L8].

1.2givenL4

By [L4] the function f has complex derivatives f(m) of every natural order on Ω.

2.1step 1.1L5L6L7

An induction on j ([L5]) using F1′=F2, the relation Fj′=jFj+1 of step 1.1, [L6] and [L7] gives F1(j)=j! Fj+1 on C∖Γ∗ for every natural j, the case j=0 reading F1=0! F1.

2.2step 1.1L3L9L10

Fix z0∈Ω∖Γ∗ and let C be the connected component of z0 in C∖Γ∗. By [L3] the set C is open and n(Γ,⋅) is a constant k on it, so W=C∩Ω is an open subset of Ω∖Γ∗ containing z0 on which the index has the constant value k.

3.1step 1.2step 2.2L1L7

By [L1] the identity k f=F1 holds on W; both sides are holomorphic there by steps 1.1 and 1.2, and complex differentiation is a local operation, so differentiating m times on W and using [L7] gives k f(m)=F1(m) on W.

4.1step 2.1step 3.1L1L6∎

Combining step 3.1 with step 2.1 at the point z0 gives n(Γ,z0)f(m)(z0)=k f(m)(z0)=m! Fm+1(z0), which is the displayed formula; since z0∈Ω∖Γ∗ was arbitrary and m was an arbitrary natural number, the formula holds throughout, and at m=0 it is [L1] again by [L6].

Depends on

Used by

Dependency tree · two levels

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Sources