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Let $P(z) = \prod_{i = 1}^n (z - z_i) \in \mathbb{C}[z]$ be a monic polynomial having all roots $z_1, \dots, z_n$ on the unit circle $\mathbb{T} := \{z \in \mathbb{C} : |z| = 1\}$.

What is known about upper bounds for $$M(P) :=\max_{|z| = 1} |P(z)|$$ in terms of the distribution of $z_1, \dots, z_n$ in $\mathbb{T}$ ?

On the one hand, if $z_1, \dots, z_n$ are equally spaced on $\mathbb{T}$ then we have that $|P(z)| = |z^n - 1|$ and so $M(P) \leq 2$. In other words, $M(P)$ is small.

On the other hand, if $z_1, \dots, z_n$ all belong to one half of $\mathbb{T}$, say $\Im(z) < 0$, then it follows easily that $M(P) \geq P(1) \geq (\sqrt{2})^n$, and so $M(P)$ is large.

My guess is that knowing that $z_1, \dots, z_n$ are "well distributed", in some precise quantitative way, would provide a nontrivial upper bound for $M(P)$.

I searched in the literature but I could not find anything like that. Thanks for any suggestion.

Note. This is a somehow similar question but regarding a lower bound for $M(P)$ and for "random" $P$.

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    $\begingroup$ Why on earth would anyone vote to close this? $\endgroup$
    – Lucia
    Commented Oct 4, 2021 at 14:07
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    $\begingroup$ @Lucia My sentiments exactly, I've just seen this in the "close" queue and am baffled why $\endgroup$
    – Yemon Choi
    Commented Oct 5, 2021 at 11:24

1 Answer 1

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Let me first start with the other side: does the maximum being small guarantee that the roots are equidistributed? This is indeed the case, and is a beautiful theorem of Erdos and Turan. For a recent exposition see Equidistribution of zeros of polynomials (published paper in the Amer. Math. Monthly). This shows that $$ {\mathcal D}(P) \le C \sqrt{N \log M(P)}, $$ with $C= 8/\pi$, and ${\mathcal D}(P)$ denotes the discrepancy of the angles of the roots. That is, ${\mathcal D}(P)$ is the maximum over all intervals $(\alpha, \beta)$ of $(0, 2\pi]$ of the absolute value of the difference between $N(\beta-\alpha)/(2\pi)$ and the number of roots with argument between $\alpha$ and $\beta$. In fact the result is a bit more precise, and recently there's been progress in improving the constant $C$ by Shu and Wang (see also Carneiro et al).

Your question is about the other direction: to bound $M(P)$ in terms of the discrepancy ${\mathcal D}(P)$. This can also be done, using a result of Carneiro and Vaaler for example. Their Theorem 8.1 (with small changes of notation) shows that for any $1\le K\le N$ $$ \log M(P) \le \frac{N\log 2}{K+1} + \sum_{k=1}^{K} \frac 1k \Big| \sum_{j=1}^{N} z_j^k\Big|. $$ If the roots are equidistributed then the power sums $\sum_{j=1}^{N} z_j^k$ are small, and the desired bound would follow. In fact, with ${\mathcal D}(P)$ being the discrepancy, partial summation shows that $$ \Big| \sum_{j=1}^{N} z_j^k \Big| \le 2\pi k {\mathcal D}(P), $$ so that from Carneiro--Vaaler we would get $$ \log M(P) \le \frac{N\log 2}{K+1} + 2\pi K {\mathcal D}(P). $$ Choosing $K$ to minimize this, we obtain the complementary bound $$ \log M(P) \le C \sqrt{N {\mathcal D}(P)}, $$ for a suitable constant $C$.

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  • $\begingroup$ Thanks! It seems to me that in some inequalities $M(P)$ should be $\log M(P)$. Am I right? $\endgroup$
    – Erik4
    Commented Oct 4, 2021 at 7:43
  • $\begingroup$ Yes, $\log M(P)$ throughout. Will fix. $\endgroup$
    – Lucia
    Commented Oct 4, 2021 at 14:05
  • $\begingroup$ All right. Thanks again for this very complete answer! $\endgroup$
    – Erik4
    Commented Oct 4, 2021 at 15:03

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