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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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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≥0 and

∣f(ζ)∣≤Mwhenever ∣ζ−a∣=r.

Then, for every n∈N,

∣f(n)(a)∣≤n!Mrn.

Facts & Assumptions

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

[L1]

The higher-derivative Cauchy formula gives f(n)(a)=n!(2πi)−1∫γf(ζ)/(ζ−a)n+1 dζ 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 2πr (Every circle has circumference 2 pi r and circumference-to-diameter ratio pi).

Proof

technique · direct
1.1givenL1L2L3

On the circle, ∣f(ζ)/(ζ−a)n+1∣≤M/rn+1, so [L1], [L2], and [L3] give ∣f(n)(a)∣≤n!(2π)−1(M/rn+1)(2πr).

2.1step 1.1algebra∎

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

Depends on

Used by

Dependency tree · two levels

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Sources