TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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Every normalized univalent disc map contains the quarter disc

Statement

If $f \in S$ , then

$D (0, 1 / 4) \subset f (D) .$

Facts & Assumptions

Given: A function $f \in S$ .

[L1]

Every normalized univalent function satisfies $∣ a_{2} ∣ \leq 2$ (The second coefficient of a normalized univalent function has modulus at most two).

Proof

technique · direct

1.1L1givenassume-contraalgebra

Assume toward a contradiction that $w \in C$ with $0 < ∣ w ∣ < 1 / 4$ is omitted by $f (D)$ . Define $g (z) : = \frac{w f (z)}{w - f (z)} .$ Then $g$ is holomorphic and univalent on $D$ , $g (0) = 0$ , and $g^{'} (0) = 1, g (z) = z + (a_{2} + \frac{1}{w}) z^{2} + \dots .$

2.1L1step 1.1discharge-contradiction

Since $g \in S$ , fact [L1] gives $∣ a_{2} + \frac{1}{w} ∣ \leq 2.$ Applying [L1] again to $f$ yields $∣ a_{2} ∣ \leq 2$ , so $\frac{1}{∣ w ∣} \leq ∣ a_{2} + \frac{1}{w} ∣ + ∣ a_{2} ∣ \leq 4.$ Thus $∣ w ∣ \geq 1 / 4$ , contradicting step 1.1.

3.1step 2.1∎

Therefore no value of modulus less than $1 / 4$ is omitted, so $D (0, 1 / 4) \subset f (D)$ .

Depends on

Used by

Dependency tree · two levels

6 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, Corollary 7.5.7 (standard reference, not scraped)
Walter Rudin, Real and Complex Analysis, Theorem 14.14 (standard reference, not scraped)