TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Liouville's theorem: every bounded entire function is constant

Statement

Every bounded entire function is constant.

More explicitly, if $f : C \to C$ is holomorphic and there is a real $M \geq 0$ such that $∣ f (z) ∣ \leq M$ for every $z \in C$ , then $f$ is constant.

Facts & Assumptions

Given: An entire function $f : C \to C$ and a real $M \geq 0$ with $∣ f (z) ∣ \leq M$ for every $z$ ; the Euclidean identification of The complex coordinate isometry transports topology and affine segments and the definition of complex domain in A complex domain is a nonempty connected open subset of $C$ .

[L1]

If $f$ is holomorphic on $D (a, R)$ , $0 < r < R$ , and $∣ f ∣ \leq M$ on the radius- $r$ circle, then $∣ f^{(n)} (a) ∣ \leq n! M / r^{n}$ for every natural $n$ (Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle).

[L2]

A holomorphic function on a complex domain whose derivative vanishes everywhere is constant (A holomorphic function with zero derivative on a domain is constant).

[L3]

The Euclidean plane $R^{2}$ is polygonally connected and connected ( $R^{n}$ is polygonally connected, connected, locally path-connected and locally connected).

Proof

technique · direct

1.1givenL1

Fix $a \in C$ and $r > 0$ . Since $f$ is holomorphic on $D (a, r + 1)$ and its modulus is at most $M$ on the radius- $r$ circle, [L1] with derivative order one gives $∣ f^{'} (a) ∣ \leq M / r$ .

1.2givenL3

Under the identification in the given data, [L3] makes $C$ connected; it is also nonempty and open in itself, so it is a complex domain.

2.1step 1.1choosealgebra

If $∣ f^{'} (a) ∣ > 0$ , choose $r = M / ∣ f^{'} (a) ∣ + 1$ ; then $M / r < ∣ f^{'} (a) ∣$ , contradicting step 1.1, so $f^{'} (a) = 0$ .

3.1step 2.1step 1.2L2∎

Since $a$ was arbitrary, step 2.1 gives $f^{'} = 0$ throughout the domain of step 1.2, and [L2] makes $f$ constant; this also covers $M = 0$ and every constant entire function.

Liouville's theorem: every bounded entire function is constant

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources