LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Cauchy estimates on a smaller concentric disc

Statement

Let $0 \leq r < R < S$ , let $f$ be holomorphic on $D (a, S)$ , and suppose $M \geq 0$ satisfies $∣ f (ζ) ∣ \leq M$ whenever $∣ ζ - a ∣ = R$ . If $n \in N$ and $∣ z - a ∣ \leq r$ , then

$∣ f^{(n)} (z) ∣ \leq \frac{n! R M}{(R - r)^{n + 1}} .$

Facts & Assumptions

Given: Reals $0 \leq r < R < S$ , a function $f$ holomorphic on $D (a, S)$ , a bound $M \geq 0$ on the radius- $R$ circle, a natural $n$ , and a point $z$ with $∣ z - a ∣ \leq r$ .

[L1]

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

[L2]

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).

[L3]

If an integrand has modulus at most $B$ on a rectifiable contour $γ$ , then the modulus of its integral is at most $B L (γ)$ (ML estimate: a contour integral is bounded by a supremum bound times path length).

[L4]

A 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

Formula [L1] applies because $∣ z - a ∣ \leq r < R$ , and for $∣ ζ - a ∣ = R$ the triangle inequality in [L2] gives $∣ ζ - z ∣ \geq ∣ ζ - a ∣ - ∣ z - a ∣ \geq R - r > 0$ , so the integrand has modulus at most $M / (R - r)^{n + 1}$ .

L1L2

2.1

Applying [L3] to step 1.1 and using the circle length from [L4] gives $∣ f^{(n)} (z) ∣ \leq (n! / (2 π)) (M / (R - r)^{n + 1}) (2 π R) = n! R M / (R - r)^{n + 1}$ .

step 1.1L3L4algebra

3.1

The bound in step 2.1 is independent of $z$ on the closed radius- $r$ disc and includes derivative order $n = 0$ , inner radius $r = 0$ , and bound $M = 0$ ; the strict inequality $r < R$ keeps every denominator positive.

step 2.1∎

Depends on

Used by

Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly Theorem

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Matthias Weber, Complex Analysis, Theorem 2.4.2 (standard reference, not scraped)
E. Stein and R. Shakarchi, Complex Analysis, Ch. 2, Theorem 5.3 (standard reference, not scraped)