Proof of Liouville's Theorem in Complex Analysis 📂Complex Anaylsis

Proof of Liouville's Theorem in Complex Analysis

Theorem ¹

If the function f $f : \mathbb{C} \to \mathbb{C}$ is an entire function and there exists a positive number $M$ for all $z \in \mathbb{C}$ such that $|f(z)| \le M$ is satisfied, then $f$ is a constant function.

Explanation

Saying that $f$ is an entire function means it is analytic throughout the entire complex plane. The contrapositive statement is that if it’s not a constant function, then its absolute value cannot be bounded. For example, $\sin$ , when its domain is the set of real numbers, is trivially bounded by $-1$ and $1$ , but in complex analysis, because $| \sin i | = | i \sinh 1 | = \sinh 1 > 1$ it may not be bounded.

Proof

Let’s consider $\mathscr{C}$ as circle $| z - \alpha | = r$ with the radius $r$ and center $\alpha$ . Since $f$ is an entire function, we can consider its derivative $f ' (\alpha)$ at every point $z=\alpha$ .

Cauchy’s integral formula: $f^{(n)} (\alpha) = {{n!} \over {2 \pi i }} \int_{\mathscr{C}} {{f(z)} \over { (z - \alpha)^{n+1} }} dz$

From the generalized Cauchy integral formula regarding differentiation, if $n=1$ , then $|f ' (\alpha)| = {{1} \over {2 \pi}} \left| \int_{\mathscr{C}} {{f(z)} \over { (z- \alpha)^{2} }} dz \right|$

ML lemma: For a positive number $M$ that satisfies $|f(z)| \le M$ and the length $L$ of $\mathscr{C}$ , $\left| \int_{\mathscr{C}} f(z) dz \right| \le ML$

Since $|f(z)| \le M$ , $\displaystyle \left| { {f(z)} \over { (z - \alpha)^2 } } \right| \le { {M} \over {r^2} }$ , and the circumference of circle $| z - \alpha | = r$ is $2 \pi r$ , using the ML lemma, $|f ' (\alpha)| = {{1} \over {2 \pi}} \left| \int_{\mathscr{C}} {{f(z)} \over { (z- \alpha)^{2} }} dz \right| \le {{1} \over {2 \pi}} \left( { {M} \over {r^2} } \right) 2 \pi r = { {M} \over {r} }$ This inequality holds for any $r>0$ ; hence, $|f ' (\alpha)| = 0$ , that is, $f ' (\alpha) = 0$ . Since $f ' (\alpha) = 0$ for every point $z=\alpha$ , $f$ is a constant function.

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