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A maximizing sequence has a locally uniform limit with extremal derivative

Statement

Assume the Axiom of Choice. Let $Ω ⊊ C$ be homologically simply connected, let $z_{0} \in Ω$ , and let

$M : = \sup {f^{'} (z_{0}) : f \in F (Ω, z_{0})} .$

Then there is a holomorphic $f : Ω \to D$ with $f (z_{0}) = 0$ and $f^{'} (z_{0}) = M$ .

Facts & Assumptions

Given: The Axiom of Choice, a proper homologically simply connected complex domain $Ω ⊊ C$ , and a point $z_{0} \in Ω$ .

[A1]

The Axiom of Choice supplies the maximizing sequence and the successive subsequence choices used by Montel's theorem (The Axiom of Choice).

[L1]

The derivative set of the extremal family is nonempty, positive, and has a finite supremum (The extremal derivatives are positive and have a finite supremum).

[L2]

Under the Axiom of Choice, every locally bounded holomorphic family is normal (Montel's theorem: every locally bounded holomorphic family is normal).

[L3]

Derivatives depend continuously on locally uniform convergence (Every derivative operator is continuous for locally uniform convergence on holomorphic functions).

[L4]

A nonconstant holomorphic map on a complex domain is open (Open mapping theorem for holomorphic functions).

Proof

technique · direct

1.1A1L1L2givenchoose

By [A1] and [L1], choose a sequence $(f_{n})$ in $F (Ω, z_{0})$ with $f_{n}^{'} (z_{0}) \to M$ . Because every $f_{n}$ maps $Ω$ into $D$ , the family is locally bounded, so [L2] gives a locally uniformly convergent subsequence, still denoted $(f_{n})$ , with holomorphic limit $f$ on $Ω$ .

2.1L3step 1.1algebra

For every $n$ , one has $f_{n} (z_{0}) = 0$ , so the locally uniform convergence of step 1.1 gives $f (z_{0}) = 0$ . Fact [L3] gives $f_{n}^{'} (z_{0}) \to f^{'} (z_{0})$ , hence $f^{'} (z_{0}) = M$ .

3.1L1L4step 2.1assume-contradischarge-contradiction

Because $M > 0$ by [L1], step 2.1 makes $f$ nonconstant. Also $∣ f ∣ \leq 1$ on $Ω$ as a locally uniform limit of disc-valued maps. If $∣ f (a) ∣ = 1$ at some $a \in Ω$ , then $f (Ω)$ would be an open subset of the closed unit disc by [L4], impossible. Hence $f (Ω) \subset D$ .

4.1step 2.1step 3.1∎

The map $f$ therefore has the required normalization and extremal derivative.

Depends on

Used by

Dependency tree · two levels

24 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 5.2.6 (standard reference, not scraped)
Walter Rudin, Real and Complex Analysis, Theorem 14.9 (standard reference, not scraped)