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Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle

Statement

Let $f$ be holomorphic on $D (a, R)$ and let $0 < r < R$ . Suppose $M \geq 0$ and

$∣ f (ζ) ∣ \leq M whenever ∣ ζ - a ∣ = r .$

Then, for every $n \in N$ ,

$∣ f^{(n)} (a) ∣ \leq \frac{n! M}{r^{n}} .$

Facts & Assumptions

Given: A function $f$ holomorphic on $D (a, R)$ , a radius $0 < r < R$ , a bound $M \geq 0$ on the radius- $r$ circle, and a natural number $n$ .

[L1]

The higher-derivative Cauchy formula gives $f^{(n)} (a) = n! (2 π i)^{- 1} \int_{γ} f (ζ) / (ζ - a)^{n + 1} d ζ$ on the positively oriented radius- $r$ circle (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle).

[L2]

The ML estimate bounds the modulus of a contour integral by a bound for the integrand times the contour length (ML estimate: a contour integral is bounded by a supremum bound times path length).

[L3]

The once-traversed circle of radius $r > 0$ has length $2 π r$ (Every circle has circumference 2 pi r and circumference-to-diameter ratio pi).

Proof

technique · direct

1.1

On the circle, $∣ f (ζ) / (ζ - a)^{n + 1} ∣ \leq M / r^{n + 1}$ , so [L1], [L2], and [L3] give $∣ f^{(n)} (a) ∣ \leq n! (2 π)^{- 1} (M / r^{n + 1}) (2 π r)$ .

givenL1L2L3

2.1

Since $r > 0$ , simplifying step 1.1 gives $∣ f^{(n)} (a) ∣ \leq n! M / r^{n}$ . For $n = 0$ this is $∣ f (a) ∣ \leq M$ , and for $M = 0$ or constant $f$ the same computation remains valid.

step 1.1algebra∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 108 results over 23 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 (standard reference, not scraped)