Timeline for Is the product of two Sobolev functions in L^p?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2018 at 15:17 | comment | added | Wenguang Zhao | Thanks for the answer. I am wondering whether it is meaningful in the Bony paraproducts sense ? | |
| Mar 31, 2018 at 21:48 | comment | added | Deane Yang | Note that even if $g$ is smooth, rapidly decaying, and equal to $1$ on an open set, it's hard to imagine a reasonable condition for $fg$ to be in $L^p$, except that $f$ is already in $L^p$. | |
| Mar 31, 2018 at 21:29 | comment | added | Mateusz Kwaśnicki | Not sure how is $fg$ defined: by duality, $(fg,\phi):=(f,g\phi)$? Even if $g$ is infinitely smooth (say, $g(x)=\exp(-|x|^2)$), $fg$ need not be any more regular than $f$: $fg \in L^p$ would mean $|(fg,\phi)|\le C\|\phi\|_q$, that is, $|(f,g\phi)|\le C\|g^{-1}g\phi\|_q$, and so $f\in L^p(|g(x)|^pdx)$. | |
| Mar 31, 2018 at 20:12 | comment | added | Nate Eldredge | Yes, sorry, I noticed that afterwards and so deleted my comment. | |
| Mar 31, 2018 at 20:10 | comment | added | Wenguang Zhao | Thanks! The problem here is that the function $f$ is in the Sobolev space with negative index. So sobolev embedding theorems seems not to be helpful. | |
| Mar 31, 2018 at 19:56 | history | asked | Wenguang Zhao | CC BY-SA 3.0 |