Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28
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Cauchy estimates propagate from a compact set to its hull

Statement

Let Ω⊆Cm be a domain, let ∅≠K⋐Ω be compact, let 0<r<δΩ(K), and define

Kr:=⋃a∈KΔ‾r(a).

Then Kr⋐Ω. Moreover, for every holomorphic f on Ω and every multi-index α,

sup⁡w∈K^Ω∣∂αf(w)∣≤α! r−(α1+⋯+αm)sup⁡z∈Kr∣f(z)∣.

Facts & Assumptions

Given: A nonempty compact set K⋐Ω, a number 0<r<δΩ(K), and f∈O(Ω).

[L1]

The hull K^Ω is characterized by the inequalities against holomorphic functions on Ω (Holomorphic hulls and holomorphic convexity).

[L2]

If a holomorphic function is defined on a polydisc, then its mixed derivatives satisfy the several-variable Cauchy estimates there (Cauchy estimates for mixed derivatives on a polydisc).

[L3]

The boundary-radius inequality r<δΩ(K) means that Δr(a)⊆Ω for every a∈K (The equal-radius polydisc boundary function).

Proof

technique · direct
1.1L3given

By [L3], every closed polydisc Δ‾r(a) with a∈K lies in Ω. Since K is compact, the union Kr is bounded and closed in Cm, hence compact, and it lies in Ω. Thus Kr⋐Ω.

1.2L2given

Fix a∈K. Because Δ‾r(a)⊆Ω, the function f is holomorphic on Δr(a), so [L2] gives ∣∂αf(a)∣≤α! r−(α1+⋯+αm)sup⁡z∈Δ‾r(a)∣f(z)∣≤α! r−(α1+⋯+αm)sup⁡z∈Kr∣f(z)∣. Therefore the holomorphic function ∂αf is bounded on K by the displayed constant.

2.1L1step 1.2∎

The function ∂αf is holomorphic on Ω, so [L1] propagates the step-1.2 bound from K to K^Ω. This is exactly the stated estimate.

Depends on

Used by

Dependency tree · two levels

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Sources