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Cauchy-kernel contour integrals may be differentiated by a direct difference-quotient estimate

Statement

Let $γ : [α, β] \to C$ be a rectifiable contour, let $φ$ be continuous on its trace, and let $V \subseteq C$ be open and disjoint from that trace. For every natural number $n \geq 1$ , with powers understood as in Integer powers in the complex field, define the following The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral:

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

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

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

Facts & Assumptions

Given: A rectifiable contour $γ$ , continuous boundary data $φ$ , an open set $V$ disjoint from the trace, and a natural $n \geq 1$ .

[L1]

A continuous integrand has a complex line integral along every rectifiable contour (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L2]

Complex line integrals are linear, and the ML estimate bounds an integral by a uniform integrand bound times the contour length (Complex line integrals are linear in the integrand, ML estimate: a contour integral is bounded by a supremum bound times path length).

[L3]

A closed bounded interval is compact, continuous images of compact metric spaces are compact, and compact subsets of metric spaces are bounded (Heine-Borel by bisection: every closed bounded interval $[a, b]$ 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 compact subset of a metric space is closed and bounded).

[L4]

The complex modulus is multiplicative and satisfies the triangle inequality (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1

For every $z \in V$ , the function $ζ \mapsto φ (ζ) / (ζ - z)^{n}$ is continuous on the trace, so $F_{n} (z)$ exists by [L1]. Moreover, [L3] applied to $φ \circ γ$ gives a finite $M \geq 0$ with $∣ φ (ζ) ∣ \leq M$ on the trace.

givenL1L3

1.2

Fix $z_{0} \in V$ . Choose $ρ > 0$ with $B (z_{0}, ρ) \subseteq V$ ; because the trace is disjoint from $V$ , $w = ζ - z_{0}$ satisfies $∣ w ∣ \geq ρ$ , and if $0 < ∣ h ∣ < ρ / 2$ then $∣ w - h ∣ \geq ρ / 2$ .

givenL4choose

2.1

The finite power identity gives $((w - h)^{- n} - w^{- n}) / h = \sum_{j = 0}^{n - 1} 1 / (w^{j + 1} (w - h)^{n - j})$ , and after subtracting $n / w^{n + 1}$ the remainder is $h \sum_{j = 0}^{n - 1} \sum_{k = 0}^{n - j - 1} 1 / (w^{j + k + 2} (w - h)^{n - j - k})$ ; by step 1.2 and [L4], its modulus is at most $∣ h ∣ n (n + 1) (2 / ρ)^{n + 2} / 2$ , uniformly on the trace.

step 1.2L4algebra

3.1

By [L2], the difference between $(F_{n} (z_{0} + h) - F_{n} (z_{0})) / h$ and $n F_{n + 1} (z_{0})$ has modulus at most a fixed finite constant times $M L (γ) ∣ h ∣$ , which tends to zero. Hence $F_{n}^{'} (z_{0}) = n F_{n + 1} (z_{0})$ ; since $z_{0}$ was arbitrary, $F_{n}$ is holomorphic on $V$ . The estimate also covers $n = 1$ and a constant contour, while $h = 0$ is only the excluded difference-quotient value.

step 1.1step 2.1L2∎

Depends on

Used by

All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 135 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Lars Ahlfors, Complex Analysis, third edition, Ch. 4, Section 2.3, Lemma 3 (standard reference, not scraped)