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Difference between revisions of "Liouville's Theorem (complex analysis)"

(statement and proof)
 
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<cmath> f(B) - f(A) = \int_{A}^B f'(z)dz  = 0 , </cmath>
 
<cmath> f(B) - f(A) = \int_{A}^B f'(z)dz  = 0 , </cmath>
 
so <math>f</math> is constant, as desired.  <math>\blacksquare</math>
 
so <math>f</math> is constant, as desired.  <math>\blacksquare</math>
 +
 +
== Extensions ==
 +
 +
It follows from Liouville's theorem if <math>f</math> is a non-constant [[entire]]
 +
function, then the [[image]] of <math>f</math> is [[dense]] in
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<math>\mathbb{C}</math>; that is, for every <math>z_0 \in \mathbb{C}</math>, there exists
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some <math>z \in \text{Im}\,f</math> that is arbitrarily close to <math>z_0</math>.
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 +
=== Proof ===
 +
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Suppose on the other hand that there is some <math>z_0</math> not in the
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image of <math>f</math>, and that there is a positive real <math>\epsilon</math> such
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that <math>\text{Im}\,f</math> has no point within <math>\epsilon</math> of <math>z_0</math>.  Then
 +
the function
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<cmath> g(z) = \frac{1}{f(z) - z_0} </cmath>
 +
is holomorphic on the entire complex plane, and it is bounded by
 +
<math>1/\epsilon</math>.  It is therefore constant.  Therefore <math>f</math> is constant.
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<math>\blacksquare</math>
 +
 +
[[Picard's Little Theorem]] offers the stronger result that if
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<math>f</math> avoids two points in the plane, then it is constant.  It is
 +
possible for an entire function to avoid a single point, as
 +
<math>\exp(z)</math> avoids 0.
 +
  
 
== See also ==
 
== See also ==

Revision as of 23:18, 6 April 2009

In complex analysis, Liouville's Theorem states that a bounded holomorphic function on the entire complex plane must be constant. It is named after Joseph Liouville. Picard's Little Theorem is a stronger result.

Statement

Let $f : \mathbb{C} \to \mathbb{C}$ be a holomorphic function. Suppose there exists some real number $M \ge 0$ such that $\lvert f(z) \rvert \le M$ for all $z \in \mathbb{C}$. Then $f$ is a constant function.

Proof

We use Cauchy's Integral Formula.

Pick some $z_0 \in \mathbb{C}$; let $C_R$ denote the simple counterclockwise circle of radius $R$ centered at $z_0$. Then \[\lvert f'(z_0) \rvert = \biggl\lvert \frac{1}{2\pi i} \int_C -\frac{f(z)}{(z-z_0)^2}dz \biggr\rvert \le \frac{M}{R^2} .\] Since $f$ is holomorphic on the entire complex plane, $R$ can be arbitrarily large. It follows that $f'(z) = 0$, for every point $z \in \mathbb{C}$. Now for any two complex numbers $A$ and $B$, \[f(B) - f(A) = \int_{A}^B f'(z)dz = 0 ,\] so $f$ is constant, as desired. $\blacksquare$

Extensions

It follows from Liouville's theorem if $f$ is a non-constant entire function, then the image of $f$ is dense in $\mathbb{C}$; that is, for every $z_0 \in \mathbb{C}$, there exists some $z \in \text{Im}\,f$ that is arbitrarily close to $z_0$.

Proof

Suppose on the other hand that there is some $z_0$ not in the image of $f$, and that there is a positive real $\epsilon$ such that $\text{Im}\,f$ has no point within $\epsilon$ of $z_0$. Then the function \[g(z) = \frac{1}{f(z) - z_0}\] is holomorphic on the entire complex plane, and it is bounded by $1/\epsilon$. It is therefore constant. Therefore $f$ is constant. $\blacksquare$

Picard's Little Theorem offers the stronger result that if $f$ avoids two points in the plane, then it is constant. It is possible for an entire function to avoid a single point, as $\exp(z)$ avoids 0.


See also

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