Timeline for Higher integrability for Sobolev functions
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 8, 2023 at 17:17 | comment | added | Willie Wong | @GiorgioMetafune: I edited my own answer to give a complete construction, taking care to prove the uniform bound. | |
| Feb 7, 2023 at 22:21 | comment | added | Giorgio Metafune | @WillieWong How do you define f, now? Why not edit the answer to give the full solution? | |
| Feb 7, 2023 at 21:45 | comment | added | Willie Wong | In particular, divide the interval $[2^{-n}, 2^{1-n})$ into $2^{n^2}$ pieces of $2^{-n-n^2}$ width, in each piece taking the leading $2^{-n-2n^2}$ length interval. | |
| Feb 7, 2023 at 19:25 | comment | added | Christian Remling | @GiorgioMetafune: I think we can fix it though by just spreading out the interval over many separated smaller intervals, still of total length $2^{-{n^2}-n}$. | |
| Feb 7, 2023 at 15:49 | comment | added | Christian Remling | @GiorgioMetafune: Yes, I read the question too quickly and didn't notice that the OP imposed a uniform constant. | |
| Feb 7, 2023 at 9:34 | comment | added | Giorgio Metafune | I have doubts on this example. If you take $a=2^{-n}, b=a+2^{-n-n^2}$ the integral of $f$ on the interval is $2^{-n}$ and cannot be bounded by a positive power of $b-a$, independently of $n$. | |
| Feb 6, 2023 at 22:16 | comment | added | Christian Remling | @WillieWong: This looks good. However, since the exponent doesn't matter here (that's the meaning of my $\alpha$, some positive exponent), my more confused and weaker version of essentially the same reasoning also suffices. | |
| Feb 6, 2023 at 22:09 | comment | added | Christian Remling | @IosifPinelis: $f=|\nabla u|^2$, except that we also need to modify to go from one to two dimesions. | |
| Feb 6, 2023 at 21:22 | vote | accept | Adi | ||
| Feb 8, 2023 at 15:48 | |||||
| Feb 6, 2023 at 21:22 | vote | accept | Adi | ||
| Feb 6, 2023 at 21:22 | |||||
| Feb 6, 2023 at 20:22 | comment | added | Willie Wong | I must be dense, but where did $r^\alpha$ come from? For $\int_{-r}^r f$, you sum over dyadic intervals with $2^k < r$, on each interval the integral contribute $2^k$, so the integral should be of size $r^1$. Is there supposed to be an $\alpha$ in the definition of $f$ somewhere? | |
| Feb 6, 2023 at 20:16 | vote | accept | Adi | ||
| Feb 6, 2023 at 20:36 | |||||
| Feb 6, 2023 at 20:06 | comment | added | Iosif Pinelis | How is your $f$ related to the $u$ in the OP? | |
| Feb 6, 2023 at 19:46 | history | answered | Christian Remling | CC BY-SA 4.0 |