# Liouville's Theorem (Complex Analysis)

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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 $z$.

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: $\int_{\map {C_1} z} M \cmod {\d w}$ must be defined.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

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

This page or section has statements made on it that ought to be extracted and proved in a Theorem page.In particular: Remove this section and add this as a corollaryYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by creating any appropriate Theorem pages that may be needed.To discuss this page in more detail, feel free to use the talk page. |

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

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.29$: Liouville ($\text {1809}$ – $\text {1882}$) - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next):**Liouville's theorem** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next):**Liouville's theorem**