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

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Little Picard theorem

Statement

A nonconstant entire function omits at most one finite complex value.

Facts & Assumptions

Given: An entire function $f : C \to C$ .

[L1]

Schottky's theorem bounds a holomorphic map omitting $0$ and $1$ on every fixed smaller disc by a constant depending only on the center bound (Schottky's theorem).

[L2]

A bounded entire function is constant (Liouville's theorem: every bounded entire function is constant).

Proof

technique · direct

1.1

Assume toward a contradiction that $f$ omits two finite values. After an affine change of target, we may suppose those values are $0$ and $1$ . For every $R > 0$ , apply [L1] with the fixed inner radius $1 / 2$ to the map $f_{R} (z) : = f (R z)$ on $D$ . Since $f_{R} (0) = f (0)$ , the resulting bound is independent of $R$ and gives $∣ f (w) ∣ \leq C (∣ f (0) ∣, 1 / 2)$ whenever $∣ w ∣ \leq R / 2$ .

L1givenassume-contraalgebra

2.1

Since $R$ in step 1.1 is arbitrary, those discs exhaust $C$ while the same constant bounds all of them. Thus $f$ is bounded on $C$ . Fact [L2] then makes $f$ constant, contradicting the assumption.

L2step 1.1discharge-contradiction

3.1

Therefore a nonconstant entire function omits at most one finite value.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

14 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

Aleksander Simonic, The Ahlfors lemma and Picard's theorems, §6.2 (standard reference, not scraped)