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seen Sep 21 at 18:09

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