Timeline for Regularity of Newtonian potential along smooth boundary
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 12, 2023 at 19:53 | comment | added | Terry Tao | Ah, nice, that seems to work. Interestingly it seems that the Littlewood-Paley components are not nearly as well behaved now - each component can have rather bad normal derivative behavior, but somehow when one sums there is some cancellation and one recovers regularity up to the boundary again. | |
| May 12, 2023 at 18:34 | comment | added | Giorgio Metafune | Sorry for the late replay. The argument I have in mind is the following. Take $g \in C^\infty (\bar \Omega)$ and let $V$ be its newtonian potential. Then $V \in C^1(R^n) \cap C^2(\Omega)$ and $\Delta V=g$ in $\Omega$. Your argument gives $V \in C^\infty (\partial \Omega)$. Next, solve the problem $\Delta u=g$ in $\Omega$, $u=V$ at the boundary. Since $g$ is smooth up to the boundary and $V$ is smooth at $\partial \Omega$, by elliptic regularity $ u \in C^\infty (\bar{\Omega})$. Since $u-V$ is harmonic in $\Omega$, continuous up to the boundary where it vanishes, we have $u=V$. | |
| May 10, 2023 at 0:48 | comment | added | Terry Tao | @GiorgioMetafune As mentioned by the OP, $V$ is not going to lie in $C^\infty(\overline{\Omega})$ in general; note that lying in $C^\infty(\Omega)$ and in $C^\infty(\partial \Omega)$ is insuffiicent to guarantee $C^\infty(\overline{\Omega})$ regularity (in particular, the normal derivatives of $V$ could well blow up as one approaches the boundary; the argument here only controls tangential derivatives). | |
| May 9, 2023 at 21:37 | comment | added | Iosif Pinelis | @TerryTao : Thank you for your explanation! | |
| May 9, 2023 at 17:30 | comment | added | Giorgio Metafune | Thank you again. I have some comments. First of all, the proof seems to work if instead of $1$ there is a function $g \in C^\infty (\bar \Omega)$. Then $V$ is actually in $C^\infty (\bar \Omega)$ since $\Delta V=g$ and $V$ is smooth at the boundary. I cannot prove this, however, by the same method and I have to use the solvability of an elliptic problem. Is there a direct way? Finally, I think that the same proof applies for $n=2$ after differentiaitng first. | |
| May 9, 2023 at 14:28 | comment | added | Terry Tao | $\Omega$ is bounded by hypothesis, so $y-z$ is bounded, hence any summand with $\varepsilon$ sufficiently large vanishes. | |
| May 9, 2023 at 9:11 | comment | added | Giorgio Metafune | To check if I have understood: you need $\epsilon$ small to sum $\sum_\epsilon \epsilon^2$ (over small diadic numbers) and for $\epsilon$ large you sum $\sum_\epsilon \epsilon^{2-n-k}$ where $k$ denotes a number of derivatives. I do not see a restricion on $\epsilon$ when localizing. Is this correct? Thank you | |
| May 9, 2023 at 7:13 | history | edited | Terry Tao | CC BY-SA 4.0 |
added 159 characters in body
|
| May 9, 2023 at 7:07 | comment | added | Terry Tao | Starting with a standard Littlewood-Paley decomposition $1 = \sum_j \phi(2^j \xi)$ (with $\phi$ smooth and supported on an annulus $\{ \xi: |\xi| \sim 1\}$) we have $\frac{1}{|z-y|^{n-2}} = \sum_\varepsilon \varepsilon^2 \varepsilon^{-n} \varphi(\frac{y-z}{\varepsilon})$, where $\varepsilon$ ranges over powers of two and $\varphi(x) := \phi(x)/|x|^{n-2}$. Because of the boundedness of $\Omega$ we can discard the contribution of all sufficiently large $\varepsilon$. | |
| May 9, 2023 at 4:09 | comment | added | Iosif Pinelis | Could you please say more about the smooth dyadic decomposition in the first sentence: what it is and why "it suffices to show that [...]"? | |
| May 9, 2023 at 0:45 | comment | added | student | Professor Tao: This makes perfect sense. Thank you so much! | |
| May 9, 2023 at 0:44 | vote | accept | student | ||
| May 8, 2023 at 20:52 | history | answered | Terry Tao | CC BY-SA 4.0 |