Proof of Liouville's Theorem

We present a detailed proof that every bounded entire function is constant, along with several important generalizations.

Statement

Theorem4.12Liouville's theorem (restated)

If $f: \mathbb{C} \to \mathbb{C}$ is entire and bounded (i.e., $|f(z)| \leq M$ for all $z \in \mathbb{C}$ and some constant $M$ ), then $f$ is a constant function.

First Proof (via Cauchy's inequality)

Proof

Fix any two points $z_1, z_2 \in \mathbb{C}$ . We will show $f(z_1) = f(z_2)$ .

Let $R > 2\max(|z_1|, |z_2|)$ and let $\gamma_R$ be the circle $|z| = R$ . By the Cauchy integral formula:

$f(z_1) - f(z_2) = \frac{1}{2\pi i}\oint_{\gamma_R} f(z)\left(\frac{1}{z-z_1} - \frac{1}{z-z_2}\right)dz = \frac{z_1-z_2}{2\pi i}\oint_{\gamma_R}\frac{f(z)}{(z-z_1)(z-z_2)}\,dz.$

For $|z| = R$ : $|z - z_j| \geq R - |z_j| \geq R/2$ for $j = 1, 2$ . By the ML inequality:

$|f(z_1) - f(z_2)| \leq \frac{|z_1 - z_2|}{2\pi} \cdot \frac{M}{(R/2)^2} \cdot 2\pi R = \frac{4M|z_1-z_2|}{R}.$

Letting $R \to \infty$ : $f(z_1) = f(z_2)$ . Since $z_1, z_2$ were arbitrary, $f$ is constant. $\blacksquare$

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Second Proof (via derivative estimate)

Proof

For any $z_0 \in \mathbb{C}$ and $R > 0$ , Cauchy's inequality on $|z-z_0| = R$ gives:

$|f'(z_0)| \leq \frac{1! \cdot M}{R^1} = \frac{M}{R}.$

Since this holds for all $R > 0$ and $M$ is fixed, letting $R \to \infty$ gives $f'(z_0) = 0$ . Since $z_0$ was arbitrary, $f' \equiv 0$ on $\mathbb{C}$ , so $f$ is constant (a holomorphic function with zero derivative on a connected domain is constant). $\blacksquare$

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Generalizations

Theorem4.13Extended Liouville theorem

If $f$ is entire and $|f(z)| \leq C(1 + |z|)^n$ for some constant $C$ and non-negative integer $n$ , then $f$ is a polynomial of degree at most $n$ .

Proof

Apply Cauchy's inequality for the $(n+1)$ -th derivative on $|z-z_0| = R$ :

$|f^{(n+1)}(z_0)| \leq \frac{(n+1)! \cdot C(1+R+|z_0|)^n}{R^{n+1}}.$

The right side behaves like $C' R^{-1} \to 0$ as $R \to \infty$ . Thus $f^{(n+1)} \equiv 0$ , so $f$ is a polynomial of degree at most $n$ . $\blacksquare$

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ExampleEntire function with bounded real part

If $f$ is entire with $\text{Re}(f) \leq M$ , then $g(z) = e^{f(z)}$ is entire with $|g(z)| = e^{\text{Re}(f(z))} \leq e^M$ . By Liouville, $g$ is constant, hence $e^{f(z)} = e^{f(0)}$ for all $z$ . This gives $f(z) = f(0) + 2\pi i k(z)$ for integer-valued continuous $k$ , forcing $k$ to be constant. So $f$ is constant.

This generalizes: an entire function that omits a half-plane of values is constant.

RemarkPicard's theorem

Liouville's theorem says: an entire function that omits a disk of values (i.e., is bounded) must be constant. Picard's little theorem dramatically strengthens this: an entire function that omits two values must be constant. For instance, $e^z$ omits only $0$ , but if $f$ omits both $0$ and $1$ , then $f$ is constant. The proof requires the theory of modular functions or the elliptic modular lambda function.