Timeline for Limit of zero sets of harmonic functions

11 events

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Oct 24, 2022 at 15:36	comment	added	Joseph Van Name		If $H$ is the set of harmonic functions, and $Z$ is the collection of all zero sets of non-zero harmonic functions, then we can give $Z$ the topology where $U\subseteq Z$ is open precisely when $Z^{-1}[U]$ is an open subset of $H$. I wonder if this topology is well-behaved. I wonder how closely related a harmonic function is to its zero set. If $u:U\rightarrow\mathbb{R}$ is harmonic function and $Z(u)\cap U$ is a smooth manifold, then is the gradient of $u$ necessarily non-zero on $Z(u)\cap U$?
Oct 9, 2022 at 13:36	vote	accept	user492517
Oct 7, 2022 at 22:26	answer	added	Joseph Van Name		timeline score: 5
Oct 7, 2022 at 14:57	comment	added	Giorgio Metafune		Just a remark. If $u(x_0)=0$ and $u$ is not identically zero, then $u$ assumes positive and negative values in any ball centered at $x_0$, by the mean value property. The same then holds for $u_n$ for large $n$ and then $u_n$ has a zero.
Oct 7, 2022 at 14:17	history	edited	user492517	CC BY-SA 4.0	added 44 characters in body
Oct 7, 2022 at 14:17	comment	added	user492517		@LeoMoos Thanks, edited.
Oct 7, 2022 at 14:12	comment	added	Leo Moos		I guess you maybe want to impose $u \neq 0$ as well?
Oct 7, 2022 at 13:56	history	edited	user492517	CC BY-SA 4.0	added 18 characters in body
Oct 7, 2022 at 13:56	history	edited	user492517	CC BY-SA 4.0	added 4 characters in body
S Oct 7, 2022 at 13:56	review	First questions
Oct 7, 2022 at 15:17
S Oct 7, 2022 at 13:56	history	asked	user492517	CC BY-SA 4.0