# Liouville's Theorem (Complex Analysis)

## Theorem

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

Then $f$ is constant.

### Corollary

Let $f: \C \to \C$ be an entire function such that:

$\size {\map f z} \ge M$

for all $z \in \C$ for some real constant $M > 0$.

Then $f$ is constant.

### Banach Space

Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\C$.

Let $f : \C \to X$ be an analytic function that is bounded.

Then $f$ is constant.

## Proof 1

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:

$\ds \cmod {\map {f_R'} z} = \frac 1 {2 \pi} \cmod {\int_{\map {C_1} z} \frac {\map {f_R} w} {\paren {w - z}^2} \rd w} \le \frac 1 {2 \pi} \int_{\map {C_1} z} M \cmod {\d w} = M$

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.

$\blacksquare$

## Proof 2

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

Let $r > 0$.

consider:

$D_r = \set {z \in \C: \cmod z \le r}$

Then, consider the parameterization $\gamma_r : \closedint 0 {2 \pi} \to \partial D_r$ given by:

$\map {\gamma_r} t := r e^{2 \pi i t}$

For all $z \in \C$, we have:

 $\ds \map {f'} z$ $=$ $\ds \map {g_z'} 0$ $\ds$ $=$ $\ds \frac 1 {2 \pi} \oint_{\partial D_r} \frac {\map {g_z} w} {w^2} \rd w$ Cauchy Integral Formula $\ds$ $=$ $\ds \frac 1 {2 \pi} \int_0^{2 \pi} \frac {\map {g_z} {\map {\gamma_r} t} } { {\map {\gamma_r} t}^2} {\map {\gamma_r '} t} \rd t$ Definition of Complex Contour Integral $\ds$ $=$ $\ds \frac 1 {2 \pi} \int_0^{2 \pi} \frac {\map {g_z} {\map {\gamma_r} t} } { {\paren {r e^{2 \pi i t} } }^2} {2 \pi i r e^{2 \pi i t} } \rd t$ $\ds$ $=$ $\ds \frac i r \int_0^{2 \pi} \frac {\map {g_z} {\map {\gamma_r} t} } { e^{2 \pi i t} } \rd t$ $\ds$ $=$ $\ds \frac i r \int_0^{2 \pi} \frac {\map f { {\map {\gamma_r} t} + z} } { e^{2 \pi i t} } \rd t$

where:

$\map {g_z} w := \map f {w + z}$

Hence, for all $z \in \C$:

 $\ds \cmod {\map {f'} z}$ $=$ $\ds \cmod {\frac i r \int_0^{2 \pi} \frac {\map f { {\map {\gamma_r} t} + z} } { e^{2 \pi i t} } \rd t}$ $\ds$ $\le$ $\ds \frac 1 r \int_0^{2 \pi} \cmod {\frac {\map f {\map {\gamma_r} t + z} } { e^{2 \pi i t} } } \rd t$ Modulus of Complex Integral $\ds$ $\le$ $\ds \frac 1 r \int_0^{2 \pi} M \rd t$ $\ds$ $=$ $\ds \dfrac {2 \pi M} r$ $\ds$ $\to$ $\ds 0$ as $r \to \infty$

Thus it follows that $\map {f'} z = 0$ for all $z \in \C$.

$\blacksquare$

## 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 $r$.

## Source of Name

This entry was named for Joseph Liouville.