Timeline for Are the logarithms of the integer polynomials discrete in $L^1$ of the unit circle?
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12 events
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| Jun 27, 2019 at 3:59 | comment | added | Vesselin Dimitrov | Just saying where my question came up: from the observation with $m(wQ(z) - P(z))$. Your bound is much better, as it applies to all $T$ - thanks! Whereas the 'further generalization' ...was just for the sake of pointing out one :-). A side note which, as I wrote, is less interesting than the above. | |
| Jun 27, 2019 at 3:51 | comment | added | fedja | @VesselinDimitrov OK, OK, one thing a time ;-) If you get stuck again on some lemma, feel free to post it and, perhaps, I'll think of it, but I'd rather abstain from diving too deep into all this stuff now. :lol: | |
| Jun 27, 2019 at 3:45 | comment | added | Vesselin Dimitrov | And one further comment, although less interesting. Your proof was based on the Poisson kernel, but it ultimately comes down to the elementary inequality of your first display. The generalization to multivariate Laurent polynomials is trivial from here, for $r$ commuting variables. That makes me curious about the hypothetical non-commutative generalization as in this paper of Deninger ams.org/journals/jams/2006-19-03/S0894-0347-06-00519-4/… (cf. section 3 on the functional calculus defining $\log(ff^*)$, for $f \in \mathbb{Z}[\Gamma] \setminus \{0\}$). | |
| Jun 27, 2019 at 3:36 | comment | added | Vesselin Dimitrov | Ah! But of course, your lower bound is really by $\exp(-\int \log{\max(p,q)})$ and not merely by the final $\exp(-\int (|\log{p}| + |\log{q}|))$ of your answer... This is great and it answers all my issues - thanks very much! Just to point out my initial observation for small enough Mahler measures: $\| \log{|P|} - \log{|Q|} \|_{L^1} + m(P) + m(Q)$ turns out equal to none other than twice the Mahler measure of the bivariate asymmetric polynomial $wQ(z) - P(z)$, which is bounded away from zero by Breusch-Smyth (if $P/Q$ does not preserve the unit circle). | |
| Jun 27, 2019 at 3:26 | comment | added | fedja | @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. | |
| Jun 27, 2019 at 2:32 | comment | added | Vesselin Dimitrov | 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. | |
| Jun 26, 2019 at 22:29 | comment | added | fedja | @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 :-) | |
| Jun 26, 2019 at 10:16 | comment | added | Vesselin Dimitrov | 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. | |
| Jun 26, 2019 at 10:11 | comment | added | Vesselin Dimitrov | 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). | |
| Jun 26, 2019 at 10:00 | history | edited | Vesselin Dimitrov | CC BY-SA 4.0 |
Fixed the initially chosen normalization with $\max(p,q)$, rather than $p+q$, inside the integral in the last display.
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| Jun 24, 2019 at 14:23 | vote | accept | Vesselin Dimitrov | ||
| Jun 24, 2019 at 13:53 | history | answered | fedja | CC BY-SA 4.0 |