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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 $n(r)\log2\leq\log M(2r)$.


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.

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up vote 2 down vote accepted

$$ n(r) \log 2 \le \sum_{\substack{k=0 \\|a_k|\le r}}^n \log\frac{2r}{|a_k|}\le \sum_{k=0}^n \log\frac{2r}{|a_k|}=\frac{1}{2\pi}\int_0^{2\pi} \log|f(2re^{i\theta})|d\theta \leq M(2r). $$

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Hah! Awesome. Thanks. – John Adamski May 12 '12 at 19:22

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