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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 $n \in \mathbb{N}$:

  1. $\limsup \frac{M(r)}{r^n} = \infty$.
  2. $\liminf \frac{M(r)}{r^n} = 0$. (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 $\infty$.

co.sine
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1 Answers1

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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 $$ \lim_{r\to\infty}\frac{M(r)}{r^n}=\infty=\liminf_{r\to\infty}\frac{M(r)}{r^n} $$

Yiorgos S. Smyrlis
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    Thank you very much. Also I think this question is worth mentioning here: https://math.stackexchange.com/questions/143468/entire-function-bounded-by-a-polynomial-is-a-polynomial – co.sine Nov 19 '17 at 19:04
  • Why does $\limsup_{r\to\infty}\frac{M(r)}{r^n}$ not being $\infty$ implies $f$ must be a polynomial? I guess one would have to prove that this implies $|f(z)|< k|z|^n$ for large $|z|^n$ as it was suggested in the above comment, but I don't see clearly how would the proof go. Any suggestions? – user392559 Jul 08 '22 at 16:18
  • I've thought of using Maximum Modulus Principle, but I'm not sure how a proof would go since $M(r)\to \infty$ but $\frac{1}{r^n} \to 0$ as $r\to \infty$. – user392559 Jul 08 '22 at 16:55