bio | website | |
---|---|---|
location | ||
age | 27 | |
visits | member for | 4 years, 11 months |
seen | Sep 21 at 18:09 | |
stats | profile views | 520 |
Sep 9 |
awarded | Nice Answer |
Jul 2 |
awarded | Curious |
Mar 1 |
asked | Finding the lift of a curve under some assumptions |
Feb 26 |
awarded | Notable Question |
Jan 21 |
accepted | Is the speed of a curve in $ \ell^\infty $ zero a.e. if the derivative of each component is zero a.e.? |
Jan 21 |
comment |
Is the speed of a curve in $ \ell^\infty $ zero a.e. if the derivative of each component is zero a.e.?
Thanks again. Yeah, you're absolutely right. |
Jan 20 |
comment |
Is the speed of a curve in $ \ell^\infty $ zero a.e. if the derivative of each component is zero a.e.?
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 18 |
comment |
Is the speed of a curve in $ \ell^\infty $ zero a.e. if the derivative of each component is zero a.e.?
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 |
comment |
Is the speed of a curve in $ \ell^\infty $ zero a.e. if the derivative of each component is zero a.e.?
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 |
revised |
Is the speed of a curve in $ \ell^\infty $ zero a.e. if the derivative of each component is zero a.e.?
added 174 characters in body; edited title |
Jan 18 |
asked | Is the speed of a curve in $ \ell^\infty $ zero a.e. if the derivative of each component is zero a.e.? |
Jan 7 |
awarded | Notable Question |
Nov 10 |
comment |
Hausdorff measure and projections
Thank you, Pietro. |
Nov 10 |
comment |
Hausdorff measure and projections
The standard metric on $ \ell^\infty $ which is the supremum metric. And you're right. I should've been more careful about the projections. I meant to say projection onto the first m components. $ \pi_V(x_1, x_2, \dots) = (x_1, \dots, x_m) $. |
Nov 9 |
asked | Hausdorff measure and projections |
Jul 4 |
awarded | Nice Question |
Jun 25 |
awarded | Promoter |
Jun 18 |
awarded | Notable Question |
May 20 |
asked | Diameters of the images of two balls under a function |
Feb 26 |
asked | Estimating L1 functions over the ball with radius 2r |