Let $p(z)= \sum_{k=0}^n a_k z^k$ , $a_n \neq 0$ , be a polynomial of degree $n$ such that $|p(z)| \leq M$ for $|z| \leq 1$.

Show that $|p(z)| \leq M|z|^n$ for $|z| \geq 1$

This was an exam question. Part (a) of this question asked the converse: if $|p(z)| \leq M|z|^n$ and $|z|$ is sufficiently large, then $p(z)$ is a polynomial. This question came straight out of Ahlfors' Complex Analysis, which I solved by showing, using the Cauchy Estimate,

$|p^n (z)| \leq \frac{n!M|z|^k}{R^n} = 0$ as $R \to \infty$.

I would like to work backwards with this.

May I do this, or similar:

$p(z) = \frac{1}{2\pi i} \int_{c} \frac{f(\zeta)d\zeta}{\zeta-z}$

$\Rightarrow |p(z)| \leq \frac{2\pi M|z|^n}{1(2\pi)}= M|z|^n$


1 Answer 1


Consider the function $f(z)=\sum_{k=0}^n a_k z^{n-k}=z^n p(1/z)$. We have for $|z|=1$,$|f(z)| = |p(1/z)| \leq M$.

By maximum modulus principle, we have $|f(z)| \leq M, \forall |z|\leq 1$.

Thus $|z^np(1/z)| \leq M, \forall z| \leq 1$, i.e. $$|p(1/z)|\leq M|\frac{1}{z}|^n, \forall 0< \left|z\right| \leq 1$$

So $|p(z)| \leq Mz^n, \forall |z|\ge 1$

  • $\begingroup$ Quite nice. Thank you! $\endgroup$
    – user5262
    Commented Dec 18, 2014 at 19:26

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