Liouville's theorem

Liouville’s theorem. A bounded entire function is constant.

That is: if $f$ is analytic on all of $\mathbb{C}$ and $|f(z)|\le M$ for all $z$, then $f$ is constant.

A striking statement with no real-variable analog — $\sin x$ is real-analytic on all of $\mathbb{R}$ and bounded, but certainly not constant.

Proof

By Cauchy’s estimate with $n=1$, for any $z_0\in \mathbb{C}$ and any $r>0$,

$|f^{\prime }(z_0)|\le \frac{M}{r}.$

Letting $r\to \infty$ (which is allowed because $f$ is analytic on all of $\mathbb{C}$): $f^{\prime }(z_0)=0$. Since $z_0$ was arbitrary, $f^{\prime }\equiv 0$. Connected domain, zero derivative, so $f$ is constant. ∎

Why entire-and-bounded is a strong constraint

The proof shows that analyticity-everywhere forces derivatives to be controlled by the global bound on $f$ — and stretching the bound to apply on arbitrarily large disks shrinks the bound on derivatives to zero.

In real analysis, the analogous statement fails: $\sin x$ is bounded by $1$, infinitely differentiable, but has ${\sin }^{\prime }(x)=\cos (x)$, also bounded. No rigidity. The complex version is rigid because the Cauchy integral formula gives integral representations of derivatives in terms of boundary values, and ML-bounding those integrals propagates the boundedness of $f$ into a boundedness of every derivative — which becomes vacuous (zero) when the boundary can be pushed to infinity.

$\sin z$ and $\cos z$ must be unbounded

Liouville’s theorem implies: any non-constant entire function is unbounded. Since $\sin z$ and $\cos z$ are entire and non-constant, they must be unbounded somewhere in $\mathbb{C}$. Where? Along the imaginary axis, where they grow like $\cosh y$, $|\sinh y|$ — exponentially. See Complex sine and cosine.

This is one of the cleanest illustrations of why complex analysis behaves differently from real analysis.

Consequences

Fundamental theorem of algebra. Every non-constant polynomial has a complex root. Proof: if not, $1/p(z)$ is entire; $|p|\to \infty$ as $|z|\to \infty$ means $|1/p|\to 0$, so $1/p$ is bounded near infinity; continuous bounded on the compact complement gives global boundedness; Liouville says $1/p$ is constant, hence $p$ constant — contradiction.

Generalizations. Picard’s little theorem: a non-constant entire function takes every value in $\mathbb{C}$ with at most one exception. Liouville is the much weaker “doesn’t omit all of $\{|z|>M\}$.” Picard is dramatically stronger but requires more machinery.

In context

Liouville’s theorem is one of the central “magical” consequences of analyticity. The full chain from Cauchy integral formula:

CIF → derivatives via generalized CIF → Cauchy’s estimate → Liouville → FTA → … → maximum modulus principle → … → residue theorem.

All of complex analysis cascades from CIF, and Liouville is the first dramatic payoff.