Section 1.4 Interlude: Liouville's Theorem
In the calculus of real functions, it isn't at all unusual to deal with power series that converge everywhere in \(\R\) that represent bounded functions. (A whole class of these in the real world are signals that can be expressed as Fourier series). For example, \(f(x) = \sin x\) has a power series representation valid for all real \(x\) and it bounded in absolute value by 1. The extension theorem tells us that this same series should have infinite radius of convergence on \(\C\) as well, but we should note what happens to the size of the values of the function. Let \(z = x + iy\text{.}\) Then
\begin{equation*}
f(z) = \sin z = \frac{1}{2i} (e^{iz} - e^{-iz}),
\end{equation*}
and
\begin{equation*}
e^{iz} = e^{ix - y} = e^{ix} e^{-y}
\end{equation*}
which will be unbounded as we let \(y \to -\infty\text{.}\) So even though \(\sin z\) remains a function with a convergent power series representation on \(\C\) (a so-called entire function), it is an unbounded function. Liouville's theorem asserts that this is the general situation - an entire function that is bounded must necessarily be a constant function.
The proof of Liouville's theorem follows from a useful set of estimates about the behavior of derivatives of bounded analytic functions called the Cauchy inequalities. First, recall that Cauchy's integral theorem for derivatives gives
\begin{equation*}
f^{n}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} \, dz
\end{equation*}
where \(C\) is a simple closed curve about \(z_0\) inside the domain of \(f\text{.}\)
Theorem 1.4.1. Cauchy inequalities.
Let \(f\) be an analytic function on a domain \(A\) with bound \(\abs{f(z)} \lt M\) on \(A\text{,}\) and let \(\gamma\) be a circle of radius \(R\) centered at a point \(z_0 \in A\) with \(\gamma \subset A\text{.}\) Then
\begin{equation*}
\abs{f^{n}(z_0)} \leq \frac{n!}{R^n} M.
\end{equation*}
Proof.
Suppose that \(\gamma(t) = z_0 + R e^{it}\) parametrizes \(\gamma\text{.}\) Note that restricted to \(\gamma\text{,}\) we have \(z - z_0 = R\text{.}\) By the integral theorem for derivatives, we have
\begin{align*}
\abs{f^{n}(z_0)} \amp= | \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}} \, dz |\\
\amp= \frac{n!}{2\pi} |\oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}} \, dz|\\
\amp\leq \frac{n!}{2\pi} \frac{M}{R^{n+1}} 2 \pi R,\\
\amp= \frac{n!}{R^n}M
\end{align*}
where the inequality follows from the ML-inequality.
We're now ready for the main result of this section
Theorem 1.4.2. Liouville's Theorem.
A bounded entire function is necessarily constant.
Proof.
Let \(z_0\) be a complex number and let \(\gamma\) be a circle of radius \(R\) centered at \(z_0\text{.}\) Suppose that \(f\) is bounded in modulus on \(\C\) by a constant \(M\text{.}\) Now apply the Cauchy inequality with \(n = 1\) to \(f\) at \(z_0\text{.}\) This gives
\begin{equation*}
|f\dd(z_0)| \leq \frac{M}{R}.
\end{equation*}
This inequality holds for all radii \(R\text{,}\) because every circle centered at \(z_0\) is contained in the domain of \(f\text{.}\) In particular, this means that as we let \(R \to \infty\text{,}\) we have
\begin{equation*}
|f\dd(z_0)| = 0,
\end{equation*}
which means that \(f\dd(z_0) = 0\text{.}\) Since this holds for every \(z_0 \in \C\text{,}\) we have \(f\dd(z) = 0\) and so \(f(z) = c\) for some constant \(c \in \C\text{.}\)
