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The Cauchy transform of a cycle is holomorphic off its trace, with the expected derivatives

Statement

Let $Γ = \sum_{k < r} m_{k} γ_{k}$ be a complex chain with trace $Γ^{*}$ and let $φ$ be continuous on $Γ^{*}$ . Put $V = C ∖ Γ^{*}$ , which is open, and for every natural $n \geq 1$ define the Cauchy transform

$F_{n} (z) = \frac{1}{2 π i} \int_{Γ} \frac{φ (ζ)}{(ζ - z)^{n}} d ζ (z \in V) .$

Then each $F_{n}$ is holomorphic on $V$ and

$F_{n}^{'} (z) = n F_{n + 1} (z) (z \in V) .$

Facts & Assumptions

Given: A complex chain $Γ = \sum_{k < r} m_{k} γ_{k}$ and a continuous $φ$ on its trace.

[L1]

Let $γ : [α, β] \to C$ be a rectifiable contour, let $φ$ be continuous on its trace and let $W \subseteq C$ be open and disjoint from that trace. For every natural $n \geq 1$ the function $z \mapsto (2 π i)^{- 1} \int_{γ} φ (ζ) (ζ - z)^{- n} d ζ$ is holomorphic on $W$ and its derivative is $n$ times the corresponding function with exponent $n + 1$ (Cauchy-kernel contour integrals may be differentiated by a direct difference-quotient estimate).

[L2]

$\int_{Γ} f d z = \sum_{k < r, m_{k} \neq 0} m_{k} \int_{γ_{k}} f d z$ (Integration over a complex chain and the index of a chain).

[L3]

A complex chain is a finite list of pairs $(m_{k}, γ_{k})$ of integers and complex contours, and its trace is the union of the $γ_{k}^{*}$ with $m_{k} \neq 0$ (Complex chains, their traces, and cycles).

[L4]

Finite linear combinations and products of complex-differentiable functions are complex differentiable, as are reciprocals and quotients wherever their denominators do not vanish; constants and the identity are complex differentiable (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L5]

A compact subset is closed and bounded (A compact subset of a metric space is closed and bounded); the continuous image of a compact subset is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset); a finite union of compact subsets is compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact); a closed bounded interval is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L6]

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

[L7]

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).

[L8]

A sum over a finite index set in the additive commutative monoid of $C$ is well posed and additive, with empty sum $0$ ; complex-field distributivity permits scaling term by term (A finite sum in a commutative monoid indexed by an arbitrary finite set, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[L9]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

Proof

technique · direct

1.1

Each $γ_{k}^{*}$ is the continuous image of a compact interval, hence compact by [L5], so the trace $Γ^{*}$ of [L3] is a finite union of compact sets, compact by [L5] and closed by [L5]; therefore $V = C ∖ Γ^{*}$ is open by [L7].

givenL3L5L7

1.2

For $z \in V$ and $ζ \in Γ^{*}$ one has $ζ - z \neq 0$ , so the powers $(ζ - z)^{- n}$ are defined by [L6]. For fixed $z$ , the map $ζ \mapsto (ζ - z)^{- n}$ is holomorphic on $C ∖ {z}$ by repeated products and nonvanishing quotients, using [L4], hence continuous by [L9]; multiplying by the continuous function $φ$ makes the integrand of each $F_{n}$ continuous on $Γ^{*}$ , so [L2] defines $F_{n} (z)$ .

givenL2L4L6L9

2.1

Fix $k < r$ with $m_{k} \neq 0$ . Then $γ_{k}^{*} \subseteq Γ^{*}$ by [L3], so the open set $V$ of step 1.1 is disjoint from $γ_{k}^{*}$ , and $φ$ is continuous on $γ_{k}^{*}$ ; hence [L1] makes $F_{n}^{(k)} (z) = (2 π i)^{- 1} \int_{γ_{k}} φ (ζ) (ζ - z)^{- n} d ζ$ holomorphic on $V$ with $(F_{n}^{(k)})^{'} = n F_{n + 1}^{(k)}$ .

step 1.1step 1.2L1L3

3.1

By [L2] and [L8], $F_{n} = \sum_{k < r, m_{k} \neq 0} m_{k} F_{n}^{(k)}$ on $V$ , a finite linear combination with constant coefficients of the functions of step 2.1; so [L4] makes $F_{n}$ holomorphic on $V$ with $F_{n}^{'} = \sum_{k} m_{k} (F_{n}^{(k)})^{'} = n \sum_{k} m_{k} F_{n + 1}^{(k)} = n F_{n + 1}$ . The empty chain, and a chain with all coefficients zero, give $F_{n} \equiv 0$ and the identity holds trivially.

step 2.1L2L4L8∎

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Sources

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §4.4 (standard reference, not scraped)
J. Lebl, Complex Analysis, Ch. 4 §4.2 (standard reference, not scraped)