Timeline for Isodiametric Inequality
Current License: CC BY-SA 2.5
3 events
| when toggle format | what | by | license | comment | |
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| Oct 23, 2009 at 20:50 | comment | added | Darsh Ranjan | I think it's pretty easy to see that the worst case is some simplex. It's almost definitely the regular simplex, but I haven't thought of a slick way to prove that. Whatever the worst case is, though, there is a hard bound diam(A) >= sqrt(2)*rad(A) for a subset A of a real inner product space (where rad(A), the "radius" of A, is the radius of the smallest ball that contains A). | |
| Oct 22, 2009 at 20:59 | comment | added | Kevin P. Costello | Is the simplex in some sense the worst case. In other words, is it true that for any A in R^n we have that (diameter of smallest ball containing A)/diam(A) is at most what it is for the simplex? | |
| Oct 22, 2009 at 4:08 | history | answered | Anton Geraschenko | CC BY-SA 2.5 |