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Let $f:\Bbb C \to \Bbb C$ be an entire function. I wanna show that if there is an $M >0$ and an $n\in \Bbb N$ s.t. $$ |f(z)|\leq M \cdot |z|^n$$ forall $z\in \Bbb C$ with $|z|\geq 1$. Then $f$ is a polynomial of degree at most $n$.

Suppose we have a $M$ and a $n$ s.t. that $ |f(z)|\leq M \cdot |z|^n$. Now because $f$ is an entire function, we can represent $f$ as a power serie $$f(z)=\sum^{\infty}_{i=0}a_iz^i $$ We have that the coefficients are bound by the cauchy bound $$|a_i| \leq r^{-i} \max\{|f(u)|:u \in B(0,r)\}, \; \forall r >0$$ but with our $M$ and $n$ we have that $|a_i| \leq r^{-i} \cdot M \cdot |z|^n$, $\forall r>0$.Now if $r \to \infty$ t follows that $a_i=0$ $\forall i >n$. Why is it true? I have some problems to see this statement.

José Carlos Santos
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Frederick Manfred
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1 Answers1

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The inequality$$|a_i|\leqslant r^{-i}\max\{|f(z)|\mid z\in B(0,r)\},$$together with $|f(z)|\leqslant M|z|^n$ allows you to deduce that$$|a_i|\leqslant r^{-i}Mr^n=Mr^{n-i}.$$So,$$i>n\implies|a_i|\leqslant\lim_{n\to\infty}Mr^{n-i}=0.$$

José Carlos Santos
  • 427,504