Liouville's Theorem

One of the many amazing results about complex differentiable functions is Liouville's theorem. In real analysis, functions like \(\sin\) and \(\cos\) are differentiable everywhere and bounded. In complex analysis, the only such functions are constant functions.

Theorem

Every bounded entire function must be a constant.

Proof

Suppose \(f\) is entire. Then \(f\) has a Taylor series expansion about \(0\) which converges everywhere, in particular

\[ f(z) = \sum_{k = 0}^\infty a_k z^k.\]

The coefficients can be expressed as

\[ a_k = \frac{f^{(k)(0)}}{k!} = \frac{1}{2\pi i} \oint_{C_r} \frac{f(z)}{z^{k + 1}} \,\mathrm{d}z\]

by Cauchy's integral formula (about zero), where \(C_r\) is a circle centred at \(0\) of radius \(r\).

Using the fact that \(f\) is uniformly bounded \(|f(z)| \leq M\), we have

\[\begin{align*} |a_k| &= \left|\frac{1}{2\pi i} \oint_{C_r} \frac{f(z)}{z^{k + 1}} \,\mathrm{d}z\right| \\ &\leq \frac{1}{2\pi} \oint_{C_r} \left| \frac{f(z)}{z^{k + 1}}\right| \,\mathrm{d}z \\ &\leq \frac{1}{2\pi} \left( 2\pi r \sup_{z \in C_r} \left|\frac{f(z)}{z^{k + 1}}\right| \right) \\ &\leq r \frac{M}{r^{k + 1}} \\ &= \frac{M}{r^k}. \\ \end{align*}\]

Since \(f\) is entire, we can let \(r\) be arbitrarily large without effecting the validity of this formula. Thus, supposing \(a_k \neq 0\) for some \(k > 0\) yields a contradiction by choosing sufficiently small \(r\). Hence \(f(z) = a_0\), a constant function.