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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28
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Locally uniform convergence preserves the total multiplicity near an isolated zero

Statement

Let Ω⊆C be open, let fn:Ω→C be holomorphic, and suppose fn→f locally uniformly on Ω. Let a∈Ω be an isolated zero of f of multiplicity m≥1. Then there is r>0 such that D(a,r)‾⊆Ω, the function f has no zero on ∣z−a∣=r, and for all sufficiently large n the function fn has exactly m zeros in D(a,r) counted with multiplicity.

Facts & Assumptions

Given: Holomorphic functions fn on an open set Ω converging locally uniformly to f, and an isolated zero a of f of multiplicity m.

[L1]

A strict boundary perturbation preserves the total zero multiplicity in the disc (Small perturbations preserve the total local zero multiplicity).

Proof

technique · direct
1.1given

By the isolated-zero hypothesis, choose r>0 with D(a,r)‾⊆Ω such that a is the only zero of f in D(a,r)‾. Then ∣f∣ is a positive continuous function on the circle ∣z−a∣=r, so it has a positive minimum there. Call that minimum η.

2.1step 1.1L1

Because fn→f locally uniformly and the circle ∣z−a∣=r is compact, there is N such that ∣fn(z)−f(z)∣<η≤∣f(z)∣(∣z−a∣=r, n≥N). Applying [L1] on that circle shows that for every n≥N, the function fn has the same total zero multiplicity in D(a,r) as f, namely m.

3.1step 2.1∎

Step 2.1 is exactly the claimed persistence of the local multiplicity.

Depends on

Used by

Dependency tree · two levels

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Sources