Timeline for Restricted isometry
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 17, 2015 at 21:56 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Jan 4, 2014 at 18:03 | vote | accept | nombre | ||
| Jan 4, 2014 at 17:47 | comment | added | Anton Petrunin | In Jvaisala's proof you may think that $g$ is a bijection from $Q=B_{|x-y|}(x)\cup B_{|x-y|}(y)$ to itself. No further changes are required. | |
| Jan 4, 2014 at 11:23 | comment | added | nombre | post-edit: What I tried in this direction was this: I replaced $W$ with the set of applications $f$ whose domain contain a neighborhood of the segment $[xy]$ (and is a subset of $\mathbb{R}^n$), that are distance-preserving, and that admit $x$ and $y$ as fixed points. The first step of the demonstration would require $x+y -f(\frac{x+y}{2})$ to be in $f(V)$ for any $f \in E$ and for some neighborhood $V$ of $[xy]$. I feel your condition $B_{100 - \|x-y\|}(x) \subset \Omega$ has a purpose related to this requirement; but I fail to see how you accomodate with Jvaisala's proof to make this work. | |
| Jan 4, 2014 at 6:00 | comment | added | Anton Petrunin | @nombre Okey, I think I see now what is your problem. The answer is rewritten. | |
| Jan 4, 2014 at 5:59 | history | undeleted | Anton Petrunin | ||
| Jan 4, 2014 at 5:59 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Jan 4, 2014 at 2:01 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Jan 4, 2014 at 1:49 | history | deleted | Anton Petrunin | via Vote | |
| Jan 4, 2014 at 1:10 | comment | added | nombre | Sorry for the double post. The problem is that given two points $X$ and $Y$ in $C$, you cannot expect $f(X)$ and $f(Y)$ to be in such places that there exists $Z \in C$ and a radius $r$ satisfying both $X,Y \in B(Z,r)$, $f(X),f(Y) \in B(f(Z),r)$ and $[XY] \subset \overline{B(Z,r)}$ and $[f(X)f(Y)] \subset \overline{B(f(Z),r)}$. Then again I might have misunderstood your point. | |
| Jan 4, 2014 at 0:56 | comment | added | nombre | I like the idea, but I am not sure it applies to any norm. | |
| Jan 4, 2014 at 0:30 | history | answered | Anton Petrunin | CC BY-SA 3.0 |