TITLE: Is entire function constant when $ |f(z)|\le \log|z|,\ |z|>1$.
QUESTION [9 upvotes]: Let $ f : \mathbb{C} \to \mathbb{C} ,$ entire and $|f(z)|\le \log|z|,\ |z|>1. $
Show that $f$ is constant.
What first comes to mind is Louville's theorem, but log 's problems with analyticity confuse me.

REPLY [8 votes]: If you use the Cauchy Integral Formula, the proof is very straight forward. Note, for $\forall n\ge 1$,
$$ |f^{(n)}(0)|=\frac{n!}{2\pi}\bigg|\int_{|z|=r}\frac{f(z)}{z^{n+1}}dz\bigg|\le\frac{n!}{2\pi}\frac{\ln r}{r^{n+1}}2\pi r=\frac{n!\ln r}{2\pi r^n}\to0\text{ as }r\to\infty $$
and hene $f\equiv C$ is a constant. Note $|f(z)|\le \ln |z|$ for $|z|>1$, namely, 
$$ |C|\le \ln|z| \text{ for }\forall |z|>1. $$
Letting $|z|\to1$ gives $C=0$, namely, 
$$ f(z)\equiv0.$$
You can use the same way to show a more general result: If there is a constant $\alpha>0$ such that
$$ |f(z)|\le C|z|^\alpha, \forall z\in\mathbb{C}, $$
then $f(z)$ is a polynomial.