Notes - Complex Analysis MT23, Liouville's theorem

Suppose

• $f : \mathbb C \to \mathbb C$
• $f$ entire (holomorphic on $\mathbb C$)

Then

• $f$ is constant

Suppose $ \vert f(z) \vert \le M$ for all $z$. Let $\gamma _ R$ be the circular contour of radius $R$.

Proof 1: Then

\[\begin{aligned} |f(w) - f(0)| &= \left|\frac {1}{2\pi i} \int_{\gamma_R} f(z) \left(\frac{1}{z-w} - \frac{1}{z}\right) \text dz \right| \\\\ &= \frac{1}{2\pi} \left| \int_{\gamma_R} \frac{f(z)w}{z(z-w)} \right| \\\\ &\le \frac{2\pi R}{2 \pi} \left| \frac{Mw}{R(R-w)} \right| \\\\ &= \frac{M|w|}{R-|w|} \end{aligned}\]

Letting $R \to \infty$, we see $f(w) \to f(0)$.

Proof 2: (basically the same, but using the generalisation of Cauchy’s integral formula, slightly simpler inequalities).

Since $f$ is holomorphic on $\mathbb C$,

\[f(z) = \sum^\infty_{n = 0} a_n z^n\]

where

\[a_n = \frac{1}{2\pi i} \int_{\gamma_R} \frac{f(w)}{w^{n+1} } \text dz\]

Then:

\[\begin{aligned} |a_n| &= \left| \frac{1}{2\pi i} \int_{\gamma_R} \frac{f(w)}{w^{n+1} } \text d w \right| \\\\ &\le \frac{1}{2\pi} \cdot 2\pi R \cdot \sup_{z \in \gamma_R^\star} \left|\frac{f(w)}{w^{n+1} }\right| \\\\ &= \frac{R}{R^{n+1} } \sup_{z \in \gamma^\star_R} |f(w)| \\\\ &= R^{-n} M \end{aligned}\]

Hence letting $R \to \infty$ for $n \ge 1$, we see $a _ n = 0$ for all $n \ge 1$, so $f(z) = a _ 0$, so a constant.