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

Proof of Liouville's Theorem in Complex Analysis

Theorem ¹

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

Explanation

Saying that $f$ is entire means that it is analytic over the entire complex plane. Stated as a contrapositive, it means that if it is not a constant function, then its absolute value cannot be bounded. For example, $\sin$ is trivially bounded between $-1$ and $1$ when its domain is the set of real numbers, but in complex analysis $$ | \sin i | = | i \sinh 1 | = \sinh 1 > 1 $$ so we can see that it may not be bounded.

Proof

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

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

Setting $n=1$ in the Cauchy integral formula generalized to derivatives gives $$ |f ' (\alpha)| = {{1} \over {2 \pi}} \left| \int_{\mathscr{C}} {{f(z)} \over { (z- \alpha)^{2} }} dz \right| $$

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

Since $|f(z)| \le M$, we have $\displaystyle \left| { {f(z)} \over { (z - \alpha)^2 } } \right| \le { {M} \over {r^2} }$, and since the circumference of the circle $ | z - \alpha | = r $ is $2 \pi r$, applying the ML lemma gives $$ |f ' (\alpha)| = {{1} \over {2 \pi}} \left| \int_{\mathscr{C}} {{f(z)} \over { (z- \alpha)^{2} }} dz \right| \le {{1} \over {2 \pi}} \left( { {M} \over {r^2} } \right) 2 \pi r = { {M} \over {r} } $$ The inequality above holds for any $r>0$, so $|f ' (\alpha)| = 0$, that is, $f ' (\alpha) = 0$. Since $f ' (\alpha) = 0$ at every point $z=\alpha$, $f$ is a constant function.

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