Timeline for Could the convex hull of $\operatorname{Lip}_1(\mathbb R)$ be dense in $\operatorname{Lip}_1(\mathbb R^d)$?
Current License: CC BY-SA 4.0
6 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Nov 7, 2019 at 3:25 | history | edited | user128095 | CC BY-SA 4.0 |
deleted 89 characters in body
|
Nov 7, 2019 at 3:19 | comment | added | user128095 | @SamZbarsky Thanks for pointing out that. | |
Nov 7, 2019 at 3:12 | comment | added | Sam Zbarsky | It seems unlikely, barring stupid answers like the trivial topology. For any function $g=T(f,v)$ in the image of $T$, we have that $|g(1,0,\ldots,0)-g(0,\ldots,0)|+|g(0,1,0,\ldots,0)-g(0,\ldots,0)|\le |v_1|+|v_2|\le\sqrt{2}$. After taking convex combinations, we get $|F(1,0,\ldots,0)-F(0,\ldots,0)|+|F(0,1,0,\ldots,0)-F(0,\ldots,0)|\le\sqrt{2}$. However, for the function $h(x)=|x|$, this quantity is 2. It's possible you can fix this by only trying approximating, e.g. $Lip_{\alpha(d)}$ functions for some $\alpha(d)$. I'm not sure what the right dependence on $d$ is. | |
Nov 7, 2019 at 3:10 | history | edited | user128095 | CC BY-SA 4.0 |
added 1 character in body
|
Nov 7, 2019 at 3:04 | history | edited | LSpice | CC BY-SA 4.0 |
TeX fixes
|
Nov 7, 2019 at 2:57 | history | asked | user128095 | CC BY-SA 4.0 |