Timeline for Zeros of polynomials with real positive coefficients
Current License: CC BY-SA 4.0
16 events
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| Apr 14, 2021 at 3:25 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Sep 16, 2014 at 13:49 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Sep 10, 2014 at 17:44 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Aug 23, 2013 at 20:51 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| S Aug 23, 2013 at 15:21 | history | suggested | CommunityBot | CC BY-SA 3.0 |
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| Aug 23, 2013 at 14:50 | review | Suggested edits | |||
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| Jul 19, 2013 at 10:29 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Jul 15, 2013 at 17:56 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Jul 14, 2013 at 11:15 | comment | added | Alexandre Eremenko | A good general reference for convergence of potentials is Hormander, Notions of Convexity, Birkhauser, 1994, especially Thm 3.2.13. | |
| Jul 13, 2013 at 20:51 | comment | added | Alexandre Eremenko | George, have you already resolved your difficulty, or you need an explanation? The hint is that u(|z|) is a subharmonic function which depends only on |z|. When measures converge weakly such functions converge uniformly on every compact set. | |
| Jul 13, 2013 at 14:28 | comment | added | George Lowther | The condition is then $\int v_\phi(t)d\mu(t)\ge0$ for all such $\phi$. That is, $\int(\log\lvert 1-\lvert z\rvert/t\rvert-\log\lvert 1-z/t\rvert)d\mu(t)\ge0$ holds for $z\in\mathbb{C}^*$ in the sense of distributions. I'm not sure if this is all obvious, or is how it was intended to be understood, but had me confused about the validity of the necessary condition for a while. | |
| Jul 13, 2013 at 14:24 | comment | added | George Lowther | This is nice. I was confused about the "evidently necessary" condition at first. I mean, it is true when $\mu$ is given by a polynomial with positive coefficients and giving no weight to $z$. However, $\log\lvert 1-z/t\rvert$ is not a continuous bounded function of $t$, so passing to the weak closure doesn't seem obvious. However, if you write $v_z(t)=\log\lvert(1-\lvert z\rvert/t)/(1-z/t)\rvert$ then this is bounded as $t\to\infty$. If you write $v_\phi(t)=\int v_z(t)\phi(z)d^2z$ for smooth nonegative $\phi$ with compact support in $\mathbb{C}^*$ then this is continuous and bounded. | |
| Jul 12, 2013 at 20:37 | history | bounty ended | ofer zeitouni | ||
| Jul 12, 2013 at 9:14 | vote | accept | ofer zeitouni | ||
| Jul 8, 2013 at 16:28 | history | answered | Alexandre Eremenko | CC BY-SA 3.0 |