# ShlMlkzdh

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bio website pitt.edu/~som13 location Pittsburgh, PA age 26 member for 4 years, 1 month seen Mar 3 at 21:30 profile views 502

# 62 Actions

 Mar1 asked Finding the lift of a curve under some assumptions Feb26 awarded Notable Question Jan21 accepted Is the speed of a curve in $\ell^\infty$ zero a.e. if the derivative of each component is zero a.e.? Jan21 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. Jan20 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? Jan18 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 $\mathbb{R}$. Jan18 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 $\vert h \vert \leq r$? Jan18 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 Jan18 asked Is the speed of a curve in $\ell^\infty$ zero a.e. if the derivative of each component is zero a.e.? Jan7 awarded Notable Question Nov10 comment Hausdorff measure and projections Thank you, Pietro. Nov10 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. $\pi_V(x_1, x_2, \dots) = (x_1, \dots, x_m)$. Nov9 asked Hausdorff measure and projections Jul4 awarded Nice Question Jun25 awarded Promoter Jun18 awarded Notable Question May20 asked Diameters of the images of two balls under a function Feb26 asked Estimating L1 functions over the ball with radius 2r Aug7 awarded Nice Question Aug26 awarded Notable Question