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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See Also
Osborne (1999). Complex variables and their applications: p94. ↩︎