If $P$ is a polynomial and $max_{|z|=1}P(z)\leq1$ then $P(z)\leq|z|^n$ when $|z|$>1

Question

I've seen many questions that regard a monic polynomial, where the main trick is to use Cauchy's integral formula to deduce about $P(z)$.
Well, this question is concerned about a general $P(z)$.
If $P(z)=a_nz^n+\ldots+a_0$ is a polynomial of degree n, and $max_{|z|=1}|P(z)|\leq1$, then $|P(z)|\leq|z|^n$ when $|z|>1$.

I've managed to achieve $|a_n|\leq1, |a_0|\leq1, |a_n+\ldots+a_0|\leq1$, but I need to have a bound on $|a_n|+\ldots+|a_0|$, no?

Answer 1

Let $w=1/z$ for $|z| \ge 1$ and note that $Q(w)=\frac{P(z)}{z^n}=w^n\sum_{k=0}^n a_k/w^k=\sum a_{k}w^{n-k}$ is a polynomial in the closed unit disc that satisfies $|Q(w)|=|P(z)| \le 1$ for $|w|=1$ since then $|z|=1$ also.

By the maximum modulus one has $|Q(w)| \le 1$ in the unit disc or $|P(z)| \le |z|^n$ outside it