Timeline for Restricted isometry
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 4, 2014 at 18:03 | vote | accept | nombre | ||
| Jan 4, 2014 at 12:19 | comment | added | nombre | @Igor Belegradek: Thanks for the info. Unfortunately the maths in your arxiv link go far beyond my level, to such extent that I couldn't even decide wether this proves my result in that peculiar case or not. Anton Petrunin: Same here, I am unfamiliar with the notion of volume so I'm going to look it up. Like I said, surjectivity can also be proven by studying $f - f(x)$ on every $\overline{B(x,r-\frac{1}{n})}$ and applying the result: Let $(X,d)$ be a metric space and let $g: X \rightarrow X$ be a distance-preserving map; if $X$ is compact then $g$ is surjective. | |
| Jan 4, 2014 at 4:06 | comment | added | Anton Petrunin | By the way, about your comment #4: the distance preserving map has to be volume preserving. Therefore if $C=\mathbb R^n$ then $f$ has to be surjective. | |
| Jan 4, 2014 at 3:10 | comment | added | Igor Belegradek | In the special case when $C$ is bounded, and the norm is Euclidean, this is known as rigidity problem for convex hypersurfaces. If the boundary of $C$ is $C^1$ smooth, the answer is yes, and is apparently due to Shenkin, see the paper [C. Vilcu, On typical degenerate convex surfaces. Math. Ann., 2008]. I have not read Sen'kin's paper though, and it is rather short, so this makes me worried a bit. There is an alternative proof in the $C^2$ case, see arxiv.org/abs/1306.1581. EDIT: actually, your setting is a bit different. | |
| Jan 4, 2014 at 0:30 | answer | added | Anton Petrunin | timeline score: 3 | |
| Jan 3, 2014 at 23:32 | history | asked | nombre | CC BY-SA 3.0 |