International Mathematics Competition
for University Students
2021

IMC2021: Day 2, Problem 7

Problem 7. Let \(\displaystyle D \subseteq \mathbb{C}\) be an open set containing the closed unit disk \(\displaystyle \{z\ :\ |z|\le 1\}\). Let \(\displaystyle f:D \to \mathbb{C}\) be a holomorphic function, and let \(\displaystyle p(z)\) be a monic polynomial. Prove that

\(\displaystyle \big|f(0)\big| \le \max_{|z|=1} \big|f(z)p(z)\big|. \)

Lars Hörmander

Solution.

Let \(\displaystyle q(z)=z^n \cdot \overline{p\left(1/\overline{z}\right)}\), or more explicitly, if

\(\displaystyle p(z) = z^n + a_{n-1}z^{n-1} + \dots + a_0,\)

let

\(\displaystyle q(z) = 1 + \overline{a_{n-1}} z + \dots + \overline{a_0} z^n. \)

Note that for \(\displaystyle \vert z\vert=1\) we have \(\displaystyle 1/\overline{z}=z\) and hence \(\displaystyle \vert q(z)\vert=\vert p(z)\vert\). Hence by the maximum principle or the Cauchy formula for the product of \(\displaystyle f\) and \(\displaystyle q\), it follows that

\(\displaystyle |f(0)| = |f(0)q(0)| \le \max_{|z|=1} |f(z)q(z)| = \max_{|z|=1} |f(z)p(z)|. \)