Let $f$ be a non-constant entire function with infinitely many $z_n$ such that $f(z_n)=0$.
Define $M(r):= \max_{|z|=r}|f(z)|$.
Prove that for all $n \in \mathbb{N}$:
- $\limsup \frac{M(r)}{r^n} = \infty$.
- $\liminf \frac{M(r)}{r^n} = 0$. (EDIT: found by the answerer Yiorgos to be incorrect)
Can someone give a hint? The only idea I have is that $f$ is not a polynomial and therefore has no limit at $\infty$. So it has an essential singularity at $\infty$, and for every possible complex value, there exists a subsequence with the limit being that value. This way, the $\limsup$ is $0$ and the $\liminf $ is $\infty$.
For the first, since $f$ is non-constant, the roots tend to infinity, and it is not a polynomial. Thus $$ \limsup_{r\to\infty}\frac{M(r)}{r^n}=\infty, $$ for otherwise the above limit would be finite, for $n$ equal to the degree of the polynomial.
The second is NOT true. For example $$ f(z)=e^z-1, $$ then $$ M(r)\ge e^r-1, $$ and hence $$ \lim_{r\to\infty}\frac{M(r)}{r^n}=\infty=\liminf_{r\to\infty}\frac{M(r)}{r^n} $$