Suppose that $p(z) = a_nz^n+\cdots+a_0$ and it has maximum modulus $1$ on the boundary of the unit disk.

Question

Suppose that $p(z) = a_nz^n+\cdots+a_0$ and it has maximum modulus $1$ on the boundary of the unit disk, show that $|p(z)| \leqslant max\{1,|z|^{n}\}$. How to show that $|p(z)| \leqslant |z|^n$?

Answer 1

Note that $q(z) = z^n p(z^{-1})$ is a polynomial. What do you know about $\lvert q(z) \rvert$ when $\lvert z\rvert=1$? Using the maximum modulus principle, what does that say about $\lvert p(z^{-1}) \rvert$ if $0 < \lvert z \rvert\leq 1$?

Answer 2

Your function $p$ is a polynomial; such functions are holomorphic functions on the whole $\Bbb C$.

Considering $p$ on the unit disk, maximum modulus theorem says that (since $p$ is not constant) $|p|$ assumes its maximum value on the boundary which is $1$ by hypotesis, that is $|p(z)|\le1$ when $|z|\le1$.

Since $q(z):=z^np(z^{-1})$ is again a polynomial, MMT still holds, and $|q|$ assumes its maximum on $|z|=1$; being this maximum $=1$ by previous considerations, we have that $|q(z)|\le1$ when $|z|\le1$, i.e. $|p(z^{-1})|\le|z|^{-n}$.

Thus taking $w=1/z$ we have that $|w|\ge1$ and $|p(w)|\le|w|^n$.

Resuming all $|p(z)|\le1$ inside the unitary disk, $|p(z)|\le|z|^n$ outside it. Briefly $|p(z)|\le\max\{1,|z|^n\}$, as wanted.

Answer 3

Hint:

$$\max_{1\le |z|\le R} \left |\frac{p(z)}{z^n} \right | \le \max (1,|a_n| + O(1/R)).$$