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I can't prove this statement, can anybody show me how to prove it?

$$f:\mathbb{C}\rightarrow \mathbb{C} \in \mathcal{O}(\mathbb{C}), \exists n\in \mathbb{N}, R >0 , M>0 : |f(z)| \le M|z|^{n} \ \ \forall |z|>R \Rightarrow \deg(f)\le n $$

To show is that if there exists such an $M$, that then $f$ is a polynomial of max degree $n$. I started like this:

$$f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n$$ So if I put this into the inequality: $$|f(z)| = \left| \sum_{n=0}^{\infty} a_n (z-z_0)^n \right| \le M |z|^n .$$

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1 Answer 1

up vote 7 down vote accepted

We have for $k\geq 1$, $z_0\in\mathbb C$ fixed and $r$ such that $\{z,|z-z_0|=r\}\subset\{z\in\mathbb C,|z|\geq R\}$, we have thanks to the Cauchy integral formula $$f^{(n+k)}(z_0) =\frac{(n+k+1)!}{2i\pi}\int_{C(z_0,r)}\frac{f(z)}{(z-z_0)^{n+k+1}}dz,$$ hence $$|f^{(n+k)}(z_0)|\leq \frac{(n+k+1)!}{2\pi}\int_{C(z_0,r)}M\frac{|z|^n}{r^{n+k+1}}\leq \frac{(n+k+1)!}{r^{n+k+1}}M(r+|z_0|)^n.$$ Since it's true for $r$ large enough, we get that $f^{(n+k)}(z_0)=0$ for each $z_0$ and $k\geq 1$, which show that $f$ is a polynomial of max degree $n$.

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Thank you. (╯°□°)╯︵ ┻━┻ –  user20318 Nov 29 '11 at 17:57
You're welcome. –  Davide Giraudo Nov 29 '11 at 19:15
@DavideGiraudo I got a bit confused, why exactly do we deduce that n is the maximum degree? –  tmac_balla Nov 28 '14 at 21:45
The degree cannot be larger than $n$ because $|f(z)|\leqslant |z|^n$ (consider this inequality for $z$ with large modulus). –  Davide Giraudo Nov 28 '14 at 21:51

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