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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30
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A nonconstant locally uniform limit of univalent functions is univalent

Statement

Let Ω be a complex domain, let fn:ΩC be univalent for every n, and suppose fnf locally uniformly on Ω. If f is nonconstant, then f is univalent.

Facts & Assumptions

Given: A complex domain Ω, univalent maps fn:ΩC, and locally uniform convergence fnf to a nonconstant holomorphic limit.

[L1]

A univalent map is injective (Univalent holomorphic functions).

[L2]

A locally uniform limit of nowhere-zero holomorphic functions is either identically zero or nowhere zero (Hurwitz's zero-free limit theorem).

Proof

technique · direct
1.1

Fix aΩ. For each n, define hn(z):=fn(z)fn(a)za(zΩ{a}). Since each fn is injective by [L1], the function hn has no zeros on Ω{a}. The removable singularity at a is filled by hn(a):=fn(a), so each hn is holomorphic and nowhere zero on Ω.

L1givenalgebra
2.1

The functions hn converge locally uniformly to h(z):={f(z)f(a)za,za,f(a),z=a, because fnf locally uniformly and derivatives converge locally uniformly as well. Fact [L2] therefore makes h either identically zero or nowhere zero.

L2step 1.1algebra
3.1

Since f is nonconstant, the function h is not identically zero. Hence step 2.1 makes h nowhere zero. If f(z)=f(a), then h(z)=0 unless z=a, so necessarily z=a. As a was arbitrary, f is injective and therefore univalent by [L1].

L1step 2.1discharge-construct

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