Maximum value of a complex polynomial on the unit disk

Question

The polynomial is $p(z)=\sum^n_{k=0} a_kz^k$. And I want to prove the following inequality on the unit disk$$\max_{B_1(0)}|p(z)|\geq |a_n|+|a_0|$$

By the maximum modulus principle, the maximum must be on the unit circle and greater than $|a_0|$ by considering $p(0)$. However, I cannot make further conclusions from this, since any attempt of using the triangle inequality will result in the opposite direction of the wanted result.

I have also seen a similar problem, although I can conclude $\max_{|z|=1}|p(z)|$is greater than any of the two on RHS, but since there is no relation of $\max_{k\in\{0,\ldots,n\}}|a_k|\geq|a_0|+|a_k|$, a tighter bound is needed.

I also tried expanding it into trig functions, and consider the roots, but it didn't work as expected.

Answer 1

This was trickier than I first thought. Perhaps there is a simpler solution, but here is one approach: Let $\omega = \exp(2\pi i/n)$. Then $$ \sum_{j=0}^{n-1} {(\omega^k)}^j = \begin{cases} n, & k \in \{ 0, n \} \\ 0, & 1 \le k \le n-1. \end{cases} $$ (This is a well-known property of roots of unity, or follows from computing the geometric sum if you prefer.)

Hence \begin{align} \frac1n \sum_{j=0}^{n-1} f(\omega^j z) &= \frac1n \sum_{j=0}^{n-1} \sum_{k=0}^{n} a_k {(\omega^j)}^k z^k \\ &= \frac1n \sum_{k=0}^{n} a_k z^k \sum_{j=0}^{n-1} {(\omega^k)}^j = a_0 + a_nz^n. \end{align}

Consequently, we get \begin{align} |a_0| + |a_n| &= \max_{|z|=1} |a_0 + a_nz^n | \\ &= \max_{|z|=1} \Big| \frac1n \sum_{j=0}^{n-1} f(\omega^j z) \Big| \\ &\le \frac1n \sum_{j=0}^{n-1} \max_{|z|=1} \big| f(\omega^j z) \big| = \max_{|z|=1} |f(z)|. \end{align}

Answer 2

I assume WLOG that $p$ is monic and that $p(0)\neq 0$
[e.g. $P(z):=\frac{p(z)}{z}$ and calling on an induction hypothesis easily handles the latter case]
This inequality is equivalent to this

$\text{this inequality }\implies \text{the link}$
when $\vert p(0)\vert \geq 1$ then (via max modulus) we have $\max_{\vert z \vert= 1}\big\vert p(z)\big\vert \geq 1 + \vert p(0)\vert \geq 2$

$\text{the link }\implies \text{this inequality }$
$p(z)=1\cdot z^n + \dots + a_1 z + a_0$
$r(z) := \overline a_0 z^n + \dots + \overline a_{n-1}z + 1$ [reciprocal polynomial; note $\vert p(z)\vert=\vert r(z)\vert$ for $\vert z\vert =1$]
$\lambda := \frac{1+\vert a_0\vert}{2}$
$q(z) := \frac{1}{\lambda}\cdot\frac{1}{2}\Big(p(z) +\frac{a_0}{\vert a_0\vert} \cdot r(z)\Big)$
where $q$ is monic and observes $\big\vert q(0)\big\vert =1$. The link tells us
$2\leq \max_{\vert z\vert =1} \big\vert q(z)\big\vert= \big\vert q(w)\big\vert$ for some $w\in \partial B\big(0,1\big)$, and re-scaling by $\lambda$ yields
$ 1+\vert a_0\vert =2\cdot \lambda \leq \lambda\cdot \big\vert q(w)\big\vert= \frac{1}{2}\Big\vert p(w) +\frac{a_0}{\vert a_0\vert} \cdot r(w)\Big\vert$ $ \leq \frac{1}{ 2}\Big(\big\vert p(w)\big\vert +\big\vert r(w)\big\vert\Big)\leq\max_{\vert z\vert =1} \big\vert p(z)\big\vert$
by triangle inequality

Answer 3

One gets that $$\max_{B_1(0)}|p(z)|\geq \sqrt{|a_n|^2+|a_{n-1}|^2+ \cdots+|a_0|^2}$$ by calculating $$\mathbb{E} \left[ |p(e^{i \Theta})|^2\right],$$ where $\Theta$ is a uniform random variable on $[0,2\pi]$.

Even though it does not answer precisely the question, it could be of interest.