Timeline for Generalisation of a multivariable calc problem
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 13, 2010 at 13:45 | vote | accept | Thierry Zell | ||
| Sep 13, 2010 at 8:58 | comment | added | Pietro Majer | An application of these unit-gradient functions is a nice proof for the affirmative answer to the Chebyshev problem for convex sets in finite dimension. If a subset $C\subset\mathbb{R}^n$ has a unique closest point to any given point (a Chebyshev set), then it is a closed convex set. (One considers the distance function from C, etc) | |
| Sep 13, 2010 at 4:02 | comment | added | fedja | That's true, but note that they are all (signed) distances to some sets (any level set would work) and they are isometries on each gradient line. Now, if you look at pairs of functions that have unit gradients orthogonal to each other, you'll realize that they have "gradient planes" on which they are isometries and free $n-2$-dimensional joint level sets that parametrize those planes in the obvious way. By the moment you reach dimension $n$, you'll get full rigidity. | |
| Sep 12, 2010 at 22:28 | comment | added | Sergei Ivanov | By the way, this is a global fact (unlike the one in the question). Locally there are plenty of distance-like functions. | |
| Sep 12, 2010 at 22:08 | comment | added | Deane Yang | Very nice and much more interesting than my answer. | |
| Sep 12, 2010 at 21:31 | comment | added | Victor Protsak | Never mind, I got it: otherwise there exists a point $r$ on the gradient line through $p$ such that $pr>qr,$ so that $|f(p)-f(r)|=pr>qr\geq |f(q)-f(r)|,$ contradiction. | |
| Sep 12, 2010 at 20:56 | comment | added | Thierry Zell | I like this version: it does go in a different direction than what I expected, and I was not aware of this result. Thanks! | |
| Sep 12, 2010 at 18:41 | comment | added | Victor Protsak | "if you have any point p with f(p)=y, then any other point q with f(q)=y is on the plane perpendicular to the gradient line through p (otherwise we can go far enough on that gradient line and then take a shortcut to q from there)": I cannot follow the parenthetical statement, can you, please, elaborate on it? | |
| Sep 12, 2010 at 17:29 | history | answered | fedja | CC BY-SA 2.5 |