Upper bound of max product

Let $z_1, z_2 , \dots, z_n \in \mathbb{C}$ be the roots of the polynomial

$f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \mathbb{C}[x]$

Prove that:

$\prod_{k = 1}^{n} \max (1,|z_k|) \leq \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2}$

Solution

Suppose the roots of polynomial $f$ are $z_1, z_2, \dots, z_n$ where

$|z_1| \geq |z_2| \geq \cdots \geq |z_m| > 1 \geq |z_{m+1}| \geq \cdots \geq |z_n|$

Let $g(z)=z^nf \left(\frac{1}{z}\right) = 1 + a_{n-1}z + \cdots + a_0z^n$ . Then, the $\{1/z_k\}_{1\leq k \leq m}$ are the zeros of $g$ in the disk $|z| \le r = 1-\epsilon < 1$ where $\epsilon$ is chosen such that $g(re^{i\theta}) \neq 0$ for $\theta \in [0, 2\pi]$ .

Jensen’s inequality implies that

$\begin{align*} \log|r^m z_1 \cdots z_m| = \frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d} \theta & \Leftrightarrow \\ |r^m z_1\cdots z_m| = \exp \left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d}\theta\right) &\Leftrightarrow \\ \exp \left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d}\theta\right) \le \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,\mathrm{d}\theta \end{align*}$

Applying Cauchy – Schwartz yields,

$\begin{align*} \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,\mathrm{d}\theta &\leq \frac{1}{2\pi}\left(\int_0^{2\pi}\,\mathrm{d}\theta\int_0^{2\pi} |g(re^{i\theta})|^2\,\mathrm{d}\theta\right)^{1/2} \\ &= \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2} \end{align*}$

Therefore,

$\left|r^m z_1\cdots z_m\right| \leq \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2}$

Letting $r\rightarrow 1$ and $\epsilon \rightarrow 0$ we get the result.

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One thought on “Upper bound of max product”

One may also use induction… For example:

The 2-norm of a polynomial $f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0$ is defined as

$\lVert f \rVert_2 := \sqrt{|a_n|^2+|a_{n-1}|^2 + \cdots + |a_{0}|^2}$

and we have

$\begin{align*} \lVert (x-z)f(x) \rVert_2^2 &= \sum\limits_{j=0}^{n+1} |a_{j-1} - za_j|^2 \\ &= \sum\limits_{j=0}^{n+1} (a_{j-1} - za_j)\overline{(a_{j-1} - za_j)} \\ &= (1+|z|^2)\lVert f(x) \rVert_2^2 - \sum\limits_{j=0}^{n+1} (a_{j-1}\overline{za_j} + za_j\overline{a_{j-1}}) \\ &= (1+|z|^2)\lVert f(x) \rVert_2^2 - \sum\limits_{j=0}^{n+1} (z\overline{a_{j-1}}a_{j} + \overline{z}a_{j-1}\overline{a_{j}}) \\ &= \sum\limits_{j=0}^{n+1}(\overline{z}a_{j-1} - a_j)(z\overline{a_{j-1}} - \overline{a_j}) \\ &= \sum\limits_{j=0}^{n+1}|\overline{z}a_{j-1} - a_j|^2 = \lVert (\overline{z}x-1)f(x) \rVert_2^2 \end{align*}$

As before let us write the roots of the polynomial in the order presented above. Consider the polynomial

$g(x) = \prod\limits_{j=1}^m (\overline{z_j}x-1)\prod\limits_{j=m+1}^n (x - z_j) = \sum\limits_{j=0}^{n}b_jx^j$

where $|b_n| = |z_1z_2\cdots z_m|$ . Now,

$\begin{align*} |z_1z_2\cdots z_m| = |b_n| &\le \sqrt{|b_n|^2+\cdots+|b_0|^2} \\ & = \lVert g(x) \rVert_2 \\ &= \lVert \frac{g(x)}{\overline{z_1}x - 1}(x-z_1) \rVert_2 \\ &= \left \| \frac{g(x)}{\prod\limits_{j=1}^m (\overline{z_j}x-1)}\prod\limits_{j=1}^m(x-z_1) \right\|_2 \\ &= \lVert f \rVert_2 \end{align*}$

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One thought on “Upper bound of max product”

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