Liouville's Theorem (Complex Analysis)/Proof 2

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 $z \in \C$.

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 $z \in \C$.

By Zero Derivative implies Constant Complex Function, $f$ is constant.

$\blacksquare$

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Source of Name

This entry was named for Joseph Liouville.