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A locally uniform limit of injective holomorphic functions is injective or constant

Statement

Let $Ω \subseteq C$ be a complex domain, let each $f_{n} : Ω \to C$ be holomorphic and injective, and suppose $f_{n} \to f$ locally uniformly on $Ω$ . Then $f$ is injective or constant.

Facts & Assumptions

Given: A complex domain $Ω$ , injective holomorphic functions $f_{n}$ on $Ω$ , and locally uniform convergence $f_{n} \to f$ .

[L1]

The limit $f$ is holomorphic (Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly).

[L2]

A nonzero holomorphic function on a complex domain has isolated zeros (Zeros of a nonzero holomorphic function are isolated).

[L3]

Near an isolated zero of the limit, sufficiently late approximants preserve the total zero multiplicity (Locally uniform convergence preserves the total multiplicity near an isolated zero).

Proof

technique · contradiction

1.1

By [L1], the limit $f$ is holomorphic. Assume it is not constant.

givenL1assume-contra

2.1

Suppose toward a contradiction that $f$ is not injective. Then there are distinct points $a, b \in Ω$ with $f (a) = f (b) = : w$ . Because $f$ is nonconstant, the function $f - w$ is not identically zero, so [L2] makes both $a$ and $b$ isolated zeros of $f - w$ . Choose disjoint closed discs $\overline{D (a, r_{a})}, \overline{D (b, r_{b})} \subseteq Ω$ containing no other zeros of $f - w$ .

step 1.1L2choose

3.1

Apply [L3] to the sequence $f_{n} - w$ on each of those discs. For all sufficiently large $n$ , the function $f_{n} - w$ has at least one zero in $D (a, r_{a})$ and at least one zero in $D (b, r_{b})$ . Since the discs are disjoint, those are two distinct preimages of $w$ , contradicting injectivity of $f_{n}$ .

step 2.1L3discharge-contradiction

4.1

The contradiction in step 3.1 shows that a nonconstant limit must be injective. Therefore every limit is injective or constant.

step 1.1step 3.1discharge-contradiction∎

Depends on

Used by

The sequence z/(n+1) shows why the injective-limit theorem needs the constant escape clause Counterexample
FALSE: a locally uniform limit of injective holomorphic functions is always injective False statement

Dependency tree · two levels

20 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

J. Lebl, Guide to Cultivating Complex Analysis, §5.4, Corollary 5.4.9 (standard reference, not scraped)
R. W. Howell and J. H. Mathews, Complex Analysis, §8.7 (standard reference, not scraped)