Timeline for Diameter-area ratio for affine tranformations.
Current License: CC BY-SA 3.0
13 events
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| Sep 29, 2013 at 10:25 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
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| May 14, 2013 at 16:52 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
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| May 14, 2013 at 16:20 | history | edited | Nikita Kalinin | CC BY-SA 3.0 |
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| May 14, 2013 at 15:10 | answer | added | Nikita Kalinin | timeline score: 1 | |
| May 14, 2013 at 12:24 | comment | added | alvarezpaiva | @Misha: Actually, I think you may be right and that the trick is to use the Loewner-John ellipsoid (ellipse in this case), but one needs to use the sharp bound for quotient between the area of the ellipse and the area of the body (instead of using John's theorem). | |
| May 14, 2013 at 9:17 | answer | added | Dietrich Burde | timeline score: 3 | |
| May 13, 2013 at 22:54 | comment | added | Misha | Yes, the Jones ellipse method gives $diam^2 \le \frac{8}{\pi} Area \approx 2.5 Area$ which is a bit worse than $diam^2 \le \frac{4}{\sqrt{3}} Area \approx 2.3 Area$. | |
| May 13, 2013 at 21:02 | comment | added | Cristos A. Ruiz | I guess $F$ must have positive area. This is trivialy false if $F$ is a line segment. | |
| May 13, 2013 at 20:31 | history | edited | Nikita Kalinin |
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| May 13, 2013 at 20:15 | comment | added | alvarezpaiva | @Misha: this will give a similar bound, but I don't think it will be sharp. Behrend's bound is sharp with equality if and only if $F$ is a triangle. But of course one should check ... | |
| May 13, 2013 at 19:52 | comment | added | Misha | Did you try to take the largest-area (Jones-Loewner) ellipsoid E in F and then apply affine transformation which sends E to the unit circle? | |
| May 13, 2013 at 19:24 | comment | added | alvarezpaiva | Funny, I'm also looking at this reference and I can't even find the statement of this result. It is supposed to be in pages 745 and 746. | |
| May 13, 2013 at 18:02 | history | asked | Nikita Kalinin | CC BY-SA 3.0 |