Suppose $P(z) = a_0+a_1z + \dots + a_nz^n$ is bounded by 1 for $|z|\leq 1$. Show that $|P(z)|\leq |z^n|$ for all $|z^n|\geq 1$

Question

As title, I'm wondering is there any way to show that $|P(z)|\leq |z^n|$ for $|z|\geq 1$. I attempt to divide $|z^n|$ on both side, and get $$\bigg|\frac{P(z)}{z^n}\bigg| =\bigg |\frac{a_0}{z^n}+\frac{a_1}{z^{n-1}} + \dots a_n\bigg|$$ It's sufficient to show that$$\bigg|\frac{P(z)}{z^n}\bigg|\leq 1$$ But I can't figure out how to prove it, any comment about this question is welcome.

Answer 1

One has to consider analytic function $f(z) = P(z)/z^n$ on annulus $A(R) = \{{z:|z| >=1 , |z| <= R}\}$ where $R$ is large enough. Also, note that since $|P(z)| <=1$ on $|z| = 1$ one has $\frac{1}{2\pi}\int_0^{2\pi} |P(z)|^2 = \sum |a_k|^2$ so $|a_n| <=1$. Now, analytic function takes maximum on boundary and on unit circle, by assumption we know it is not larger than 1. But if $R$ is large enough the values of $|f(z)|$ are about $|a_n|$ which is not larger than 1 as well.

Answer 2

Let $w=1/z$ and note that $P(z)=P(1/w)= a_0+a_1/w + \dots + a_n/w^n$ so $Q(w)=w^nP(1/w)$ is analytic in the unit disc including at $0$. But on the unit circle $|w|=1$ clearly $|Q(w)|=|P(1/w)|=|P(\bar w)| \le 1$ so by maximum modulus we have $|Q(w)| \le 1, |w| \le 1$ which means precisely $|P(z)|/|z|^n \le 1$ for $|z|=|1/w| \ge 1$ and we are done!

Clearly, we get equality at a point iff $Q$ is constant of modulus $1$ so $P(z)=\alpha z^n$ and more generally we can extend this proof to show that if $F$ is a polynomial of degree (at least) $n$ (the degree of $P$) with no zeroes outside the unit disc and $|P(z)| \le |F(z)|$ on the unit circle, then $|P(z)| \le |F(z)|, |z|>1$ with equality at a point iff $P=\alpha F, |\alpha|=1$ as the result above corresponds to $F(z)=z^n$

Answer 3

Suppose $\vert P(z')\vert \gt \vert z'\vert^n$ for some $z'$ outside $S^1$, the unit circle.
$\implies q(z')=0$ with $q(z):=\delta\cdot P(z) +\alpha \cdot z^n =0$ for some $\delta \in (0,1)$ and some $\alpha \in S^1$.

But for $z\in S^1$ we have $\vert \alpha \cdot z^n\vert = 1\gt \delta\cdot \vert P(z)\vert=\vert \delta \cdot P(z)\vert$
$\implies q$ has exactly $n$ roots inside $S^1$ by Rouche. And $q$ is a polynomial with degree at most $n$, with exactly $n$ roots inside the unit circle (hence degree $n$). Conclude $q(z')=0$ is a contradiction$\implies \vert P(z)\vert \leq \vert z^n\vert$ for all $\vert z\vert \gt 1$.