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Tautologically, the integer polynomials form a discrete set in $L^1$ of the unit circle. On the other hand, a set of logarithms ordered by norm becomes generally rather denser than the original set.

Is the set $$ \big\{ \log{|P|} \, : \, P \in \mathbb{Z}[X] \setminus \{0\} \big\} \subset L^1(\mathbb{T}) $$ of functions on the complex unit circle $\mathbb{T} = \{ z \mid |z| = 1 \}$ discrete in $L^1$, or does it have an accumulation point?

I am equally happy with the $L^2$ norm, if it makes a difference.

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  • $\begingroup$ I don't quite understand your first remark - could you elaborate? Probably I am just being slow and missing something $\endgroup$
    – Yemon Choi
    Commented Jun 22, 2019 at 15:11
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    $\begingroup$ @Yemonchoi if $P(z)=\sum c_n z^n$, then $\frac1{2\pi}\int_0^{2\pi} P(e^{it})e^{-ikt} dt=c_k$. Thus $\|P\|_{L^1}\geqslant \max_k |c_k|$ and the distance between two distinct integer polynomials is not less than $1$. $\endgroup$
    – Fedor Petrov
    Commented Jun 22, 2019 at 16:29
  • $\begingroup$ Thanks @FedorPetrov - indeed I was just being slow $\endgroup$
    – Yemon Choi
    Commented Jun 22, 2019 at 16:34

1 Answer 1

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It is still discrete though not uniformly. Since $\log|P|=\frac 12\log|p|$ where $p=P\bar P$ is a real non-negative trigonometric polynomial with integer coefficients, it is enough to work with $p$ instead of $P$. We have $$ |\log p-\log q|\ge \frac{|p-q|}{\max(p,q)}. $$ Now consider the outer function $f$ with $|f|=\frac 1{\max(p,q)}$, so $$ |f(0)| = \exp\Big(-\int_{\mathbb T}\log{\max(p,q)}\Big) \ge \exp\Big(-\int_{\mathbb T}[|\log p|+|\log q|]\Big). $$ Then, denoting by $r$ the difference $p-q$ multiplied by an appropriate power of $z$ so that $r(0)\in\mathbb Z\setminus\{0\}$ and $r$ is analytic, we get $$ \int_{\mathbb T}|\log p-\log q|\ge \int_{\mathbb T}\frac{|p-q|}{\max(p,q)} =\int_{\mathbb T}|r||f|\ge |r(0)||f(0)| \\ \ge \exp\Big(-\int_{\mathbb T}[|\log p|+|\log q|]\Big)\,, $$ finishing the story.

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  • $\begingroup$ Really nice! Your inequality doesn't need that $P,Q$ have integer coefficients, but only that $\big| \|P\|_{L^2}^2 - \| Q\|_{L^2}^2 \big| \geq 1$ if it is not zero. Here is another consequence of applying this remark to $Q = 1$. For every complex polynomial $P = \sum_{i=0}^d a_iz^i$ of degree $d \geq 2$ and with non-zero free term, there is a positive constant lower bound on the Mignotte height $\widetilde{h}(P / \sqrt{|a_0a_d|}) := \int_{\mathrm{T}} \frac{\log^+ |P|}{\sqrt{|a_0a_d|}} > 0.1$ (that appears in the Amoroso-Mignotte refinement of the Erdos-Turan equidistribution theorem). $\endgroup$
    – Vesselin Dimitrov
    Commented Jun 26, 2019 at 10:11
  • $\begingroup$ One outstanding question here is whether or not the discreteness should be uniform if we restrict to the subset of the integer polynomials $P \in \mathbb{Z}[X]$ with a bound $M(P) \leq T$ on their Mahler measure. $\endgroup$
    – Vesselin Dimitrov
    Commented Jun 26, 2019 at 10:16
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    $\begingroup$ @VesselinDimitrov 1) I'm not so sure: I need the geometric mean in the story (the condition you wrote is way off what you really meant, so I'm trying to answer the implied question) 2) Yeah, one has to think a bit more in this case. I have no time now, but you've seen all my tricks by this time :-) $\endgroup$
    – fedja
    Commented Jun 26, 2019 at 22:29
  • $\begingroup$ 1) Oh, but of course $p(z)−q(z)$ is not analytic at $z=0$, and it was absolutely crucial that $r(z)=z^m(p(z)−q(z))$ was a polynomial in $z$ and not in $z,z−1$... I see. Ah well, but that only makes this proof more interesting! 2) I don't think this is nearly so easy: a positive constant lower bound by a positive function of $m(P)+m(Q)$ instead. But I do know it at least for small enough Mahler measures. $\endgroup$
    – Vesselin Dimitrov
    Commented Jun 27, 2019 at 2:32
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    $\begingroup$ @VesselinDimitrov 2) It is not so hard either. Note that either $\int|\log p-\log q|\ge 1$, or $\int \log\max(p,q)\le 1+\int\log p=1+2M(P)$, after which the same bounds work. You can optimize this dichotomy a bit, of course. $\endgroup$
    – fedja
    Commented Jun 27, 2019 at 3:26

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