Liouville's theorem

Liouville’s theorem states that for a function $f(z)$ that is entire and bounded (i.e., analytic $\forall z\in \mathbb{C}$ with some $M$ such that $|f(z)|\le M\forall z\in \mathbb{C}$), then $f$ must be constant.

Proof

Let $z_0$ be an arbitrary number in $\mathbb{C}$, and the radius of a circle $R>0$ for any positive real number. We know that $f(z)$ is analytic for $|z-z_0|\le R$ because it’s analytic everywhere and $|f(z)|\le M$ everywhere, in particular on the circle $|z-z_0|=R$.

By the Cauchy inequality for $n=1$:

$$|f^{\prime }(z_0)|\le \frac{M}{R}$$

By squeeze theorem, if $R\to \infty$, then $f^{\prime }(z)\to 0$. Then:

$$f^{\prime }(z)=0\forall z_0\in \mathbb{C}⟹f(z_0)=c,\text{for some}c\in \mathbb{C}$$

where $c$ is a constant.