# Liouville’s theorem

We can use the Cauchy estimates from last time to prove a very beautiful theorem:

Theorem 1 If ${f\in{\mathcal H}({\mathbb C})}$ is bounded (i.e., ${|f| for some ${M\in{\mathbb R}}$), then ${f}$ is constant.

Proof:Expand ${f}$ as a power series at 0. We know that this representation is valid everywhere in ${{\mathbb C}}$. Write ${f(z)=a_0+a_1z+a_2z^2+\cdots}$. Recall that

$\displaystyle a_n=\frac{f^{(n)}(0)}{n!}.$

The Cauchy estimates tell us that

$\displaystyle |a_n|=\frac{f^{(n)}(0)}{n!}\le\frac{M}{r^n}$

for every positive ${r}$ less than the radius of convergence of ${f}$. Of course, the radius of convergence is infinite, so this estimate is valid for all ${r>0}$. Thus, for ${n>0}$, ${|a_n|}$ can be bounded by arbitrarily small positive numbers. That is, ${a_n=0}$ for all ${n>0}$. In other words, ${f(z)=a_0}$, a constant. $\Box$

It’s worth mentioning that trigonometric functions such as ${\sin}$ and ${\cos}$ are holomorphic on all of ${{\mathbb C}}$ (often called entire) functions. Hopefully this doesn’t bother you, even in light of Liouville’s theorem. If it does, you’re remembering that ${\sin x}$ is always between ${1}$ and ${-1}$. You are correct for ${x\in{\mathbb R}}$, but Liouville’s theorem doesn’t say anything about functions on ${{\mathbb R}}$. Indeed, if you plug in complex numbers, you’ll see that ${\sin z}$ can be arbitrarily large. In fact, look at ${\sin(ix)}$ for ${x\in{\mathbb R}}$. You can use Wolfram Alpha to see that on the imaginary axis, ${\sin z}$ takes on arbitrarily large values.