TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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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 $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves 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

1.1

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 $∣ f^{'} (a) ∣ \leq M / r$ .

givenL1

1.2

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.

givenL3

2.1

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

step 1.1choosealgebra

3.1

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.

step 2.1step 1.2L2∎

Depends on

Used by

Every nonconstant entire function has dense image in the complex plane Corollary
FALSE: an entire function bounded on the real axis is constant False statement
Fundamental theorem of algebra by Liouville's theorem Theorem

Dependency tree · two levels

27 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

Lars Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §2.3 (standard reference, not scraped)
E. Stein and R. Shakarchi, Complex Analysis, Ch. 2, Corollary 4.5 (standard reference, not scraped)
Matthias Weber, Complex Analysis, Theorem 2.3.2 (standard reference, not scraped)
Steven G. Krantz, A Guide to Complex Variables, §3.1.3 (standard reference, not scraped)