LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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An extremizer onto a proper subdomain of the disc can be enlarged

Statement

Let $Ω ⊊ C$ be homologically simply connected, let $z_{0} \in Ω$ , and let $f \in F (Ω, z_{0})$ attain the extremal derivative $M$ . Then $f (Ω) = D$ .

Facts & Assumptions

Given: A proper homologically simply connected complex domain $Ω ⊊ C$ , a point $z_{0} \in Ω$ , and an extremizer $f \in F (Ω, z_{0})$ with $f^{'} (z_{0}) = M$ .

[L1]

The map $f$ is univalent (The extremal limit is univalent).

[L2]

For each $c \in D$ , the Blaschke factor $φ_{c}$ is a disc automorphism (Blaschke factors are automorphisms of the disc).

[L3]

On a homologically simply connected complex domain, every holomorphic nowhere-zero function has a holomorphic square root (A nonvanishing holomorphic function on such a domain has holomorphic roots of every positive order).

Proof

technique · direct

1.1

Assume toward a contradiction that $f (Ω) ⊊ D$ . Choose $c \in D ∖ f (Ω)$ . Since $f (z_{0}) = 0$ , one has $c \neq 0$ . Put $β : = φ_{c} \circ f$ . Then [L2] makes $β$ holomorphic and injective into $D$ , with $β (Ω) \subset D ∖ {0}$ .

L1L2givenassume-contrachoose

2.1

By [L3], the nowhere-zero holomorphic function $β$ has a holomorphic square root $q$ on $Ω$ with $q^{2} = β$ . If $q (z_{1}) = q (z_{2})$ then $β (z_{1}) = β (z_{2})$ , so injectivity of $β$ gives $z_{1} = z_{2}$ . If $q (z_{1}) = - q (z_{2})$ then again $β (z_{1}) = β (z_{2})$ , so $z_{1} = z_{2}$ and then $q (z_{1}) = 0$ , impossible because $β$ never vanishes. Hence $q$ is injective.

L1L3step 1.1algebra

3.1

Let $a : = q (z_{0})$ , so $a^{2} = β (z_{0}) = φ_{c} (0) = c$ . Define $g : = φ_{a} \circ q$ . Then [L2] makes $g$ holomorphic and injective from $Ω$ into $D$ , with $g (z_{0}) = 0$ . Thus after multiplying by a unimodular constant if needed, $g$ is another competitor in $F (Ω, z_{0})$ .

L2step 2.1construct

4.1

Differentiate $q^{2} = β$ at $z_{0}$ to obtain $2 a q^{'} (z_{0}) = β^{'} (z_{0}) = φ_{c}^{'} (0) f^{'} (z_{0}) = - (1 - ∣ c ∣^{2}) M$ . Since $φ_{a}^{'} (a) = - 1 / (1 - ∣ a ∣^{2})$ and $∣ a ∣^{2} = ∣ c ∣$ , one gets $∣ g^{'} (z_{0}) ∣ = \frac{1 - ∣ c ∣^{2}}{2 ∣ a ∣ (1 - ∣ a ∣^{2})} M = \frac{1 + ∣ c ∣}{2 \sqrt{∣ c ∣}} M > M,$ because $(1 + ∣ c ∣) / 2 > \sqrt{∣ c ∣}$ for $0 < ∣ c ∣ < 1$ . This contradicts the extremal definition of $M$ .

L2step 3.1algebradischarge-contradiction

5.1

Therefore the assumption of step 1.1 is false, so $f (Ω) = D$ .

step 4.1∎

Depends on

Used by

Every proper homologically simply connected plane domain is conformally equivalent to the unit disc Theorem

Dependency tree · two levels

22 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)