Timeline for Sobolev functions on $\mathbb{R}^N$ cannot be discontinuous on a $(N-1)$-dimensional submanifold
Current License: CC BY-SA 4.0
15 events
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| Apr 12, 2019 at 17:55 | comment | added | Riku | Finally, your second approach inspired a related question: mathoverflow.net/questions/327913/… | |
| Apr 12, 2019 at 10:55 | comment | added | Riku | Also, what does this say about the traces of a Sobolev function on "both sides" of the submanifold? | |
| Apr 12, 2019 at 10:51 | comment | added | Riku | Thank you again. I have an additional question: could you clarify what you mean precisely by saying that $u$ behaves nicely on $N-1$ dimensional manifolds? Does it somewhat imply that Sobolev functions cannot have jumps? Can something similar be said for BV functions? | |
| Apr 11, 2019 at 15:21 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Apr 11, 2019 at 14:32 | comment | added | Piotr Hajlasz | @Skeeve As I said, $f$ restricted to $M$ may be discontinuous, because traces are not necessarily continuous. Traces will be continuous on almost all $(N-1)$-dimensional manifolds if $p>N-1$. | |
| Apr 11, 2019 at 14:30 | comment | added | Skeeve | sure, I agree that $\chi_{\mathbb Q}$ is a bad representative of $0$. But my comment was just to show that even if some function $g$ is continuous along lines then I don't see how to conclude that $g$ cannot be discontinuous on $M$. I mean, how do you proceed after constructing good representative $g$? | |
| Apr 11, 2019 at 14:23 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Apr 11, 2019 at 14:16 | comment | added | Piotr Hajlasz | @Skeeve First of all I do not claim that $f$ restricted to $M$ is continuous only that it is well defined a.e. on $M$. Secondly $\chi_{\mathbb{Q}}$ is in fact continuous since it equals zero a.e. so you can find a continuous representative of $\chi_{\mathbb{Q}}$ in the class of functions equal a.e. | |
| Apr 11, 2019 at 8:24 | comment | added | Skeeve | Approach 1 is very nice to have an insight. But can it be turned into a rigorous proof? I don't quite see how to do this because e.g. $f(x,y) = \chi_{\mathbb Q}(x)$ is discontinuous everywhere, but continuous along all vertical lines. | |
| Apr 10, 2019 at 21:19 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Apr 10, 2019 at 19:21 | vote | accept | Riku | ||
| Apr 10, 2019 at 19:21 | history | bounty ended | Riku | ||
| Apr 10, 2019 at 19:04 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Apr 10, 2019 at 18:50 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Apr 10, 2019 at 18:43 | history | answered | Piotr Hajlasz | CC BY-SA 4.0 |