Timeline for Is the speed of a curve in $ \ell^\infty $ zero a.e. if the derivative of each component is zero a.e.?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2014 at 18:23 | comment | added | Nik Weaver | No problem. You are welcome. | |
| Jan 21, 2014 at 13:46 | vote | accept | Axiom | ||
| Jan 21, 2014 at 2:40 | comment | added | Nik Weaver | I think I understand your concern, but you see this is not a problem because every Lipschitz function on ${\bf R}$ is differentiable almost everywhere. So the worst that can happen is that there is a measure zero subset of $A$ on which $f$ is differentiable but $F$ is not. Everywhere they are both differentiable, since $F$ extends $f$ it is clear their derivatives must agree. | |
| Jan 20, 2014 at 23:22 | comment | added | Axiom | Ok, so I think there might be a small problem with your argument. The problem is when you extend $ f $ to the entire set of $ \mathbb{R} $, let's call this extension $ F $, then $ F(t+h) $ is not necessarily equal to $ f(t+h) $ because $ t+h $ is not in $ A $ necessarily. Hence, the estimate $ \frac{\vert f(t+h)-f(t)\vert}{\vert h \vert} \leq 2L\epsilon $ doesn't hold. Am I missing something here possibly? | |
| Jan 19, 2014 at 4:05 | history | edited | Nik Weaver | CC BY-SA 3.0 |
added 1 characters in body
|
| Jan 18, 2014 at 23:11 | comment | added | Nik Weaver | (If $f$ is complex-valued, then you may have to increase its Lipschitz number by at most a factor of $\sqrt{2}$ when extending it.) | |
| Jan 18, 2014 at 23:10 | comment | added | Nik Weaver | Extend $f$ to all of ${\bf R}$ without increasing its Lipschitz number, then it is the integral of its derivative which has $L^\infty$ norm at most $L$, and you get the inequality by integrating its derivative. | |
| Jan 18, 2014 at 23:10 | comment | added | Axiom | And yes, $ \mathcal{H}^1 $ is just the 1-dimensional Hausdorff measure on $ \mathbb{R} $ which coincides with the Lebesuge measure on $ \mathbb{R} $. | |
| Jan 18, 2014 at 23:07 | comment | added | Axiom | Thanks Nik. I just have a quick question. How did you conclude that $ \frac{\vert f(t+h) - f(t) \vert}{\vert h \vert} \leq L\epsilon $ for all $ \vert h \vert \leq r $? | |
| Jan 18, 2014 at 22:59 | history | answered | Nik Weaver | CC BY-SA 3.0 |