Liouville's Theorem (Complex Analysis)

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

This article has been identified as a candidate for Featured Proof status. If you do not believe that this proof is worthy of being a Featured Proof, please state your reasons on the talk page. To discuss this page in more detail, feel free to use the talk page.

Theorem

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

Then $f$ is constant.

Corollary 1

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.

Corollary 2

Let $f: \C \to \C$ be a non-constant entire function.

The maximum modulus:

$\ds \map M {r, f} := \max_{\cmod z \mathop = r} \cmod {\map f z}$

grows at least linearly in $r$.

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:

where:

$\map {g_z} w := \map f {w + z}$

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

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

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

$\blacksquare$

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): entire function
• 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): entire function (integral function)
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Liouville's theorem