Timeline for Local minimum from directional derivatives in the space of convex bodies
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 26, 2012 at 21:06 | comment | added | Yoav Kallus | P.S. It seems that for my generalized width, the ratio is smaller for the regular tetrahedron than for the ball, so the ball is not a global minimum (as it is for standard width amongst c.s. bodies), but I still think it's a local minimum. | |
| Jan 26, 2012 at 20:04 | comment | added | Yoav Kallus | Yes, for a $\mu$ with the property I gave (projection vanishes only for $n=1$), balls are the only body with $f_\vartheta$ independent of $\vartheta$. This is also true for the standard width if we restrict to centrally symmetric bodies (i.e. balls are the only c.s. bodies of const. width). Among c.s. bodies, the ball does minimize $\sqrt[3]{vol}/w$ (like you note, because we can put a ball inside $K$). That's why I think in my case I should also have a minimum. However, I don't think I can inscribed a unit ball whenever $\min f_\vartheta \ge 1$ (the tetrahedron is probably an easy example). | |
| Jan 26, 2012 at 18:04 | comment | added | Anton Petrunin | Is it true that ball is the only body with $f_\theta(K)=\mathrm{Const}(K)$? Can it happen that if $f_2(K)\ge 1$ then there is a ball $B$ in $K$ with $f_2(B)=f_2(K)$? | |
| Jan 26, 2012 at 6:15 | comment | added | Yoav Kallus | Actually, for symmetric $\mu$ (i.e. $w=$mean width), $\sqrt[3]{vol}/w$ is (globally) maximized by balls (aka Urysohn's inequality). For the standard width, the ball is locally neither maximal (see e.g. ellipsoids) nor minimal (e.g. bodies of constant width). What makes my $\mu$ different is the condition that the projection to the space of spherical harmonics of degree $n$ vanishes only for $n=1$. This is certainly untrue for the mean width (vanishes for all $n>0$) or for the standard width (vanishes for odd $n$). If it helps, my $\mu$ is supported at $12$ points with equal weight at each. | |
| Jan 26, 2012 at 4:24 | comment | added | Anton Petrunin | generalized widths is your $f_2$, say if support of $\mu$ is formed by two opposite points of $S^2$ then $f_2$ is the standard width of $K$. $$ $$ Completely symmetric means invariant w.r.t. all rotations. (I.e. $\mu$ is proportional to Lebesgue measure.) | |
| Jan 26, 2012 at 1:13 | comment | added | Yoav Kallus | It is true for the minimum width, of course. Can you explain what you mean by "generalized widths" and also by "completely symmetric $\mu$"? The minimum width corresponds to the above $f_2$ with a measure supported at a pair of antipodal points. | |
| Jan 25, 2012 at 23:18 | comment | added | Anton Petrunin | Blaschke compactness theorem alone will not help, you need to look at more properties of $f$. Your function $f_2$ is a generalized width and you want the ratio $\sqrt[3]{vol}/f_2$ to have minimum for ball --- for the usual width it is not true and I guess that for most of "generalized widths" it should not be true. (My intuition says: "it should be true only for completely symmetric $\mu$".) | |
| Jan 25, 2012 at 21:54 | history | edited | Yoav Kallus | CC BY-SA 3.0 |
added 18 characters in body
|
| Jan 25, 2012 at 20:42 | history | asked | Yoav Kallus | CC BY-SA 3.0 |