Timeline for Isoperimetric-like inequality for non-connected sets
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 13, 2019 at 14:02 | history | edited | Guillaume Aubrun | CC BY-SA 4.0 |
more accurate title
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| Oct 15, 2011 at 19:53 | vote | accept | Guillaume Aubrun | ||
| Oct 11, 2011 at 13:59 | answer | added | Jean-Marc Schlenker | timeline score: 3 | |
| Oct 11, 2011 at 9:40 | comment | added | André Schlichting | I'm not sure whether considering open sets changes the following argument substantially: Take two closed balls with volume one half and radii $r$. Put them with canters apart by a distance of $L$ in the plane. Then for $L\to\infty$ the mean shadow will be $4\pi r$. Further the mean shadow is monotone in $L$ and decreases for decreasing $L$ as the overlap in the projection becomes larger if the balls are nearer. Eventually, for $L=2r$ the balls will touch and you have a connected set, where the mean shadow conincides with the perimeter of the convexification. | |
| Oct 11, 2011 at 3:40 | answer | added | Anton Petrunin | timeline score: 5 | |
| Oct 10, 2011 at 12:40 | history | edited | Guillaume Aubrun | CC BY-SA 3.0 |
deleted 2 characters in body
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| Oct 10, 2011 at 10:36 | history | edited | Willie Wong | CC BY-SA 3.0 |
fixed math display
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| Oct 10, 2011 at 10:09 | comment | added | Guillaume Aubrun | For connected sets, taking the convex hull does not change the mean shadow, but for non-connected sets it increases ... | |
| Oct 10, 2011 at 10:03 | comment | added | Jean-Marc Schlenker | I don't quite understand the question. You can always compare an open set to its convex hull, it will have the same mean shadow, but a smaller area. So the isoperimetric inequality for open subsets should follow from the convex case? | |
| Oct 10, 2011 at 9:58 | history | asked | Guillaume Aubrun | CC BY-SA 3.0 |