Liouville's Theorem (Complex Analysis)

Theorem
Let $f: \C \to \C$ be a bounded entire function.

Then $f$ is constant.

Proof
By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le M$ for all $z \in \C$.

For any $R \in \R: R > 0$, consider the function:
 * $\map {f_R} z := \map f {R z}$

Using the Cauchy Integral Formula, we see that:
 * $\ds \cmod {\map {f_R'} z} = \frac 1 {2 \pi} \cmod {\int_{\map {C_1} z} \frac {\map f w} {\paren {w - z}^2} \rd w} \le \frac 1 {2 \pi} \int_{\map {C_1} z} M \rd w = M$

where $\map {C_1} z$ denotes the circle of radius $1$ around $z$.

Hence:
 * $\ds \cmod {\map {f'} z} = \cmod {\map {f_R'} z} / R \le M / R$

Since $R$ was arbitrary, it follows that $\cmod {\map {f'} z} = 0$ for all $z \in \C$.

Thus $f$ is constant.

Remark
In fact, the proof shows that, for a nonconstant entire function $f$, the maximum modulus $\ds \map M {r, f} := \max_{\cmod z \mathop = r} \cmod {\map f z}$ grows at least linearly in $r$.