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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Controlled derivative oscillation forces injectivity on a fixed subdisc

Statement

Let $g$ be holomorphic on $D$ with $g (0) = 0$ , $g^{'} (0) = 1$ , and $∣ g^{'} (w) ∣ \leq 2$ on $D$ . Then $g$ is univalent on $D (0, 1 / 6)$ and

$D (0, 1 / 12) \subset g (D (0, 1 / 6)) .$

Facts & Assumptions

Given: A holomorphic map $g : D \to C$ with $g (0) = 0$ , $g^{'} (0) = 1$ , and $∣ g^{'} ∣ \leq 2$ on $D$ .

[L1]

If $h : D \to D$ is holomorphic and $h (0) = 0$ , then $∣ h (w) ∣ \leq ∣ w ∣$ (Schwarz lemma with the equality cases).

[L2]

Rouche's theorem preserves zero count under a strict boundary perturbation (Rouche's theorem in the classical strict-inequality form).

Proof

technique · direct

1.1L1givenalgebra

Define $h (w) : = (g^{'} (w) - 1) / 3$ . Then $h$ is holomorphic on $D$ , $h (0) = 0$ , and $∣ h (w) ∣ \leq (∣ g^{'} (w) ∣ + 1) / 3 \leq 1$ . Hence [L1] gives $∣ g^{'} (w) - 1 ∣ \leq 3 ∣ w ∣$ for every $w \in D$ .

2.1step 1.1algebra

If $∣ w ∣ \leq 1 / 6$ , step 1.1 gives $∣ g^{'} (w) - 1 ∣ \leq 1 / 2$ . For $w_{1}, w_{2} \in D (0, 1 / 6)$ , one has $g (w_{1}) - g (w_{2}) = \int_{[w_{2}, w_{1}]} g^{'} (ζ) d ζ$ , so $∣ g (w_{1}) - g (w_{2}) - (w_{1} - w_{2}) ∣ \leq ∣ w_{1} - w_{2} ∣ / 2$ . Therefore $∣ g (w_{1}) - g (w_{2}) ∣ \geq ∣ w_{1} - w_{2} ∣ / 2$ , which shows injectivity on $D (0, 1 / 6)$ .

2.2L2step 1.1algebra

On $∣ w ∣ = 1 / 6$ , one has $∣ g (w) - w ∣ \leq \int_{0}^{1} ∣ g^{'} (t w) - 1 ∣ ∣ w ∣ d t \leq 3 ∣ w ∣^{2} / 2 = 1 / 24 < 1 / 6 = ∣ w ∣$ . Fix $ξ$ with $∣ ξ ∣ < 1 / 12$ . Then on $∣ w ∣ = 1 / 6$ , $∣ (g (w) - ξ) - w ∣ \leq ∣ g (w) - w ∣ + ∣ ξ ∣ < 1 / 24 + 1 / 12 = 1 / 8 < 1 / 6 = ∣ w ∣$ . Rouche [L2] gives the same zero count for $g (w) - ξ$ and $w$ , so $g (w) = ξ$ has exactly one solution in $D (0, 1 / 6)$ .

3.1step 2.1step 2.2∎

Step 2.2 shows every $∣ ξ ∣ < 1 / 12$ lies in $g (D (0, 1 / 6))$ , and step 2.1 shows the restriction there is univalent.

Depends on

Used by

Bloch's theorem Theorem

Dependency tree · two levels

13 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, §7.4 (standard reference, not scraped)