Timeline for Density of restrictions of harmonic functions inside a ball
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 16, 2022 at 8:42 | comment | added | Mateusz Kwaśnicki | @TerryTao: I believe du Plessis proved this in 1969, see doi.org/10.1112/jlms/s2-1.1.404 | |
| Feb 16, 2022 at 8:17 | comment | added | username | I read too quickly, I thought it was in dimension 2, @RBega2. | |
| Feb 16, 2022 at 2:33 | history | edited | LSpice | CC BY-SA 4.0 |
Greens' -> Green's
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| Feb 16, 2022 at 1:21 | comment | added | Terry Tao | In 2D I think Runge's theorem (or Mergelyan's theorem) gives the required density in the simply connected case. For higher dimensions I think an analogous theorem should hold for harmonic functions on a contractible domain (with a connected exterior), though I don't have a reference at hand for this. | |
| Feb 16, 2022 at 1:12 | comment | added | RBega2 | @username Since the ambient space is $\mathbb{R}^3$, $U$ is already simply connected. Even in $\mathbb{R}^2$ what I think you are proposing would make the situation even worse, e.g., one could have something like the real part of $\sqrt{z}$ being harmonic. | |
| Feb 16, 2022 at 0:40 | history | edited | Terry Tao | CC BY-SA 4.0 |
edited body
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| Feb 15, 2022 at 23:23 | comment | added | Giorgio Metafune | If $u$ can be approximated then it admits an harmonic extension to $B$. In fact one can replace the $L^2$ norm of $u-v$ by the sup-norm in a smaller annulus $U'$. If $v_n$ correspond to $\epsilon=1/n$, then $(v_n)$ is Cauchy in $U'$ in the sup-norm and, by the maximum principle, in a ball $B'$. | |
| Feb 15, 2022 at 22:45 | comment | added | Ali | Is there any hope of characterizing the subset of $S_U$ that can be approximated or even say something roughly about the size of such a subset? For instance the example by @Terry Tao shows that one needs to add the assumption that for elements in $S_U$ one additionally imposes that the neumann derivative of $u$ on the boundary of the smaller ball integrates to zero. | |
| Feb 15, 2022 at 21:58 | comment | added | username | But if you remove a thin slice of $U$ so that it becomes simply connected, you are back in business. | |
| Feb 15, 2022 at 21:23 | vote | accept | Ali | ||
| Feb 15, 2022 at 21:12 | history | answered | Terry Tao | CC BY-SA 4.0 |