Liouville's Theorem

Recall from the Cauchy's Derivative Inequalities page that if $A \subseteq \mathbb{C}$ is open, $f : A \to \mathbb{C}$ is analytic, $z_0 \in A$, $r > 0$ is such that $D(z_0, r) \subseteq A$, and there exists an $M > 0$ such that $\mid f(z) \mid \leq M$ for all $z$ on the circle $\mid z - z_0 \mid = r$ then for all $k = 0, 1, 2, ...$ we have that:

(1)

\begin{align} \quad \mid f^{(k)}(z_0) \mid \leq \frac{k! M}{r^k} \end{align}

We proved this using Cauchy's Integral Formula for Derivatives. We will now use Cauchy's derivative inequalities to prove a simple yet very important result known as Liouville's theorem.

Theorem 1 (Liouville's Theorem): If $f : \mathbb{C} \to \mathbb{C}$ is analytic on all of $\mathbb{C}$ and if there exists an $M > 0$ such that $\mid f(z) \mid \leq M$ for all $z \in \mathbb{C}$ then $f$ is a constant.

• Proof: Cauchy's derivative inequality holds for all $z_0 \in \mathbb{C}$ and so for $k = 1$ we have that:

(2)

\begin{align} \quad \mid f'(z_0) \mid \leq \frac{M}{r} \end{align}

• But this holds for all $r > 0$ by taking larger and larger disks centered at $z_0$ with radius $r$. So taking the limit as $r \to \infty$ gives us:

(3)

\begin{align} \quad \mid f'(z_0) \mid \leq \lim_{r \to \infty} \frac{M}{r} = 0 \end{align}

• Hence $f'(z_0) = 0$ for all $z_0 \in \mathbb{C}$. But then this means that $f$ must be a constant on all of $\mathbb{C}$. $\blacksquare$

So Liouville's theorem states that if a function $f : \mathbb{C} \to \mathbb{C}$ is analytic on all of $\mathbb{C}$ is such that the absolute value of $f$ is bounded by some $M > 0$ on $\mathbb{C}$ then $f$ must be constant on $\mathbb{C}$. So if $f$ is analytic on all of $\mathbb{C}$ and not a constant function that the absolute value of $f$ cannot be bounded on $\mathbb{C}$.