Liouville's Theorem (Complex Analysis)

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This proof is about Liouville's Theorem in the context of Complex Analysis. For other uses, see Liouville's Theorem.


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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 $z$.


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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 $z \in \C$.

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

$\blacksquare$


Remark


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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.


Sources

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