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

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

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:

\[\begin{aligned} \vert a _ n \vert &= \left \vert \frac{1}{2\pi i} \int _ {\gamma _ R} \frac{f(w)}{w^{n+1} } \text d w \right \vert \\\\ &\le \frac{1}{2\pi} \cdot 2\pi R \cdot \sup _ {z \in \gamma _ R^\star} \left \vert \frac{f(w)}{w^{n+1} }\right \vert \\\\ &= \frac{R}{R^{n+1} } \sup _ {z \in \gamma^\star _ R} \vert f(w) \vert \\\\ &= R^{-n} M \end{aligned}\]

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.