Liouville theorem (complex analysis)

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

More precisely, if $f : C \to C$ is holomorphic on the whole complex plane (i.e. entire) and there exists a constant $M$ such that $| f (z) | \leq M$ for all $z \in C$ , then $f$ is constant.

Proof

By the Cauchy integral formula, an entire function satisfies Cauchy's inequality on circles:

| f^{(n)} (0) | \leq \frac{n!}{2 π} \oint_{| z | = R} \frac{| f (z) |}{| z |^{n + 1}} | d z | \leq \frac{n! M}{R^{n}}, \forall R > 0.

Sending $R \to \infty$ gives $f^{(n)} (0) = 0$ for every $n \geq 1$ . Since an entire function equals its global Taylor series at $0$ , we get $f (z) = f (0)$ for all $z$ , i.e. $f$ is constant. $◼$

Connections

It is a direct corollary of the Cauchy integral formula, via Cauchy's inequality.
It implies the Fundamental Theorem of Algebra: a non-constant polynomial $p$ must vanish somewhere, since otherwise $1 / p$ would be a bounded entire function.
It is a key ingredient for the Hartogs' phenomenon in several variables, where the analogy fails in a striking way.
Related classical results with the same flavor: Weierstrass theorem differs (real case, compact domain) and the residue theorem.

Remark on naming

This theorem is unrelated to Liouville's theorem of Hamiltonian mechanics (phase-space volume preservation) and to Liouville's formula for the Wronskian. Same name, distinct statements.