While studying for an upcoming complex analysis qualifying exam, I found the following problem in Conway's Functions of One Complex Variable (XI.1 exercise #2).
Let $f$ be an entire function, $M(r)=\sup\{|f(re^{i\theta})|:0\leq\theta\leq2\pi\}$, $n(r)=$ the number of zeros of $f$ in $B(0;r)$ counted according to multiplicity. Suppose that $f(0)=1$ and show that $n(r)\log2\leq\log M(2r)$.
Attempt:
Let $a_1, a_2, \ldots, a_n$ denote the zeros of $f$, repeated according to multiplicity. From Jensen's formula and the condition that $f(0)=1$, we have
$$\sum_{k=0}^n\log\frac{r}{|a_k|}=\frac{1}{2\pi}\int_0^{2\pi}\log|f(re^{i\theta}|\,d\theta$$
$$\sum_{k=0}^n\log\frac{r}{|a_k|}\leq\log M(r).$$
And here is where I'm stuck. I could take exponentials of both sides, but I do not see any further progress being made toward the desired result. How would $\log2$ and $M(2r)$ come into the picture?
Thanks in advance.
Best Answer
$$ n(r) \log 2 \le \sum_{\substack{k=0 \\|a_k|\le r}}^n \log\frac{2r}{|a_k|}\le \sum_{k=0}^n \log\frac{2r}{|a_k|}=\frac{1}{2\pi}\int_0^{2\pi} \log|f(2re^{i\theta})|d\theta \leq M(2r). $$