Show that if $|f(z)| \leq M |z|^n$ then $f$ is a polynomial max degree n

Question

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: $$|f(z)| = \left| \sum_{n=0}^{\infty} a_n (z-z_0)^n \right| \le M |z|^n .$$

Remark: there is a list of similar questions here – Calvin Khor Feb 08 '22 at 01:23 — Calvin Khor, Feb 08 '22 at 01:23

score 16 · Accepted Answer · answered Nov 29 '11 at 17:52

16

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 $n$.

answered Nov 29 '11 at 17:52

Davide Giraudo

172,925

@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
@DavideGiraudo Hi, I faced the same problem but I am restricted to use the coefficients of the Taylor series for $f$ to prove it. So any hint or guidlines? Thanks – Wang Kah Lun Oct 09 '15 at 14:05
@DavideGiraudo can we solve it using rouché's theorem? – Math1995 May 09 '16 at 10:25
Sir, can you explain the last step? I mean how you get $M(r+|z_0|)^n$? – General Mathematics Mar 01 '24 at 03:54
1

@GeneralMathematics By the triangle inequality, $|z|\leq |z-z_0|+|z_0|=r+|z_0|$. – Davide Giraudo Mar 01 '24 at 09:39
Thanks sir, for replying to such an old post. – General Mathematics Mar 01 '24 at 14:51

Show that if $|f(z)| \leq M |z|^n$ then $f$ is a polynomial max degree n

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