Timeline for Hausdorff dimension of the graph of a BV function
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 15, 2019 at 12:31 | vote | accept | Riku | ||
| Apr 10, 2019 at 17:57 | comment | added | Riku | I've asked the question in my last comment in a different post: mathoverflow.net/questions/327698/… | |
| Apr 9, 2019 at 13:15 | comment | added | Riku | Do you have a simpler proof (that does not rely on Theorem 1) in the case $N=M=1$ that the graph of a BV function has Hausdorff dimension equal to 1? | |
| Apr 7, 2019 at 11:36 | comment | added | Riku | I see. Thank you. | |
| Apr 6, 2019 at 22:10 | comment | added | Skeeve | @Riku because $\Gamma_{D} = \bigcup_{\lambda \in \mathbb N} \Gamma_{D_\lambda}$, and Hausdorff dimension is stable under taking countable unions. And thank you for reposting a more general version of my question in a separate thread. I do not claim that the $N-$dimensional statement can be deduced from the one-dimesional. But I think that the simpler version is easier to address, and maybe its solution could be adopted for the general case. | |
| Apr 6, 2019 at 17:09 | comment | added | Riku | Also, why is it sufficient to show that the Hausdorff dimension of the graph $\Gamma_{D_\lambda}:=\{(x,u(x)) : x\in D_\lambda\}$ is $N$ for each $\lambda\in \mathbb N$? That is, why can we "pass to the limit" in that expression? | |
| Apr 6, 2019 at 17:08 | comment | added | Riku | Thank you. I've asked that last question as a separate post (mathoverflow.net/questions/327331/…). Why do you pose it only in $1$ dimension? That is, how can you deduce the $N$-dimensional statement from that? | |
| Apr 6, 2019 at 10:22 | history | answered | Skeeve | CC BY-SA 4.0 |