prove that $\lim\sup_{r \to \infty} \frac{M_{f}(r)}{r^n} = \infty$

I need some help with this question.

Let $f(z)$ an entire function, with infinite number of zeros. I want to prove that $\lim\sup_{r \to \infty} \frac{M_{f}(r)}{r^n} = \infty$

The definition of $M_{f}(r)$ is $\sup_{|z|=r}|f(z)|$, $r\in(0,R)$ when $f$ analytic in ${|z|<R}$

Assume that $f(z) \not\equiv 0$.

Thanks.

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What about $f(z) \equiv 0$? – JavaMan Nov 29 '11 at 13:42
Basically you're right but I think that the intention was to exclude this case. I've edited the question accordingly. – bond Nov 29 '11 at 14:09

Suppose that $\limsup_{r \rightarrow \infty}\frac{M_f(r)}{r^n}<\infty$.

This implies that there exists a constant $M$ such that for all $r$ sufficiently large, $$|f(re^{i\theta})| \leq M r^n$$ for every $\theta$.

Write $f(z)= \sum_{k=0}^{\infty}a_kz^k$. Fix $r$ large. By Taylor's theorem, $$a_k=\frac{1}{2\pi i}\int_{\{|z|=r\}} \frac{f(z)}{z^{k+1}}dz$$ and thus $$|a_k| \leq \frac{2\pi r Mr^n}{2\pi r^{k+1}} = Mr^{n-k}.$$ This holds for all $r$ sufficiently large, so if $k>n$, we find $a_k=0$, by letting $r \rightarrow \infty$. Thus, $f$ is a polynomial of degree at most $n$. Since $f$ is not identically zero, then $f$ has a finite number of zeros.

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This proof is ok but it's a proof by contrapositive, not by contradiction: if the lim sup is not infinity then the function has a finite number of zeros. – lhf Nov 29 '11 at 15:57
hmm you're right, I edited the proof accordingly. Thanks for the comment. – Kalim Nov 29 '11 at 16:11

The Riemann sphere is compact, so the zeroes have to accumulate somewhere. Since the zeros in the plane are a discrete set (otherwise $f(z) = 0$ everywhere), they must accumulate at $\infty$. Then $f$ has an essential singularity at $\infty$ and the desired result follows.

(More generally, any non-polynomial entire function has an essential singularity at $\infty$.)

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