Finding all entire functions such that for all $|z| \geq 1$ we have $|f(z)| \leq \frac{1}{|z|}$

Question

I am studying for my complex analysis exam and this question popped up few years ago on the "mini quiz":

Find all $f \in H(\mathbb{C})$ such that for all $|z| \geq 1$ we have $|f(z)| \leq \frac{1}{|z|}$.

The hint was to use some theorem about entire functions.

Could anyone give me a hand? I have no idea how to tackle this and this seems like one or two liner.

Answer 1

Let $M$ be the maximum of the restriction of $f$ to $\overline{D_1(0)}$. Then, for each $z\in\Bbb C$, $|f(z)|\leqslant\max\{M,1\}$. Therefore, by Liouville's theorem, $f$ is constant. It now follows from the fact that $\lim_{z\to\infty}\frac1{|z|}=0$ that that constant must be $0$.