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

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A quarter-disc inclusion at every point of a univalent disc map

Statement

Let $f : D \to C$ be holomorphic and univalent, and let $a \in D$ . Then

$D ⁣ (f (a), \frac{(1 - ∣ a ∣^{2}) ∣ f^{'} (a) ∣}{4}) \subset f (D) .$

Facts & Assumptions

Given: A holomorphic univalent map $f : D \to C$ and a point $a \in D$ .

[L1]

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

[L2]

Every normalized univalent disc map contains the quarter disc (Every normalized univalent disc map contains the quarter disc).

Proof

technique · direct

1.1

Define $g (ζ) : = \frac{f (φ_{a} (ζ)) - f (a)}{- (1 - ∣ a ∣^{2}) f^{'} (a)} .$ By [L1], the map $φ_{a}$ is an automorphism of $D$ with $φ_{a} (0) = a$ and $φ_{a}^{'} (0) = - (1 - ∣ a ∣^{2})$ , so $g$ is holomorphic and univalent on $D$ , with $g (0) = 0$ and $g^{'} (0) = 1$ .

L1givenalgebra

2.1

Thus $g \in S$ , and [L2] gives $D (0, 1 / 4) \subset g (D)$ . Multiplying by the affine factor from step 1.1 and translating back yields $D ⁣ (f (a), \frac{(1 - ∣ a ∣^{2}) ∣ f^{'} (a) ∣}{4}) \subset f (D) .$

L2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Walter Rudin, Real and Complex Analysis, Theorem 14.15 (standard reference, not scraped)
Matthias Weber, Complex Analysis, Theorem 7.5.8 (standard reference, not scraped)