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Let $u_n : \mathbb{R}^n \to \mathbb{R}$ be a sequence of harmonic functions which converge uniformly on compact subsets. The limit function $u$ (which we assume to be not identically $0$) is clearly harmonic (via mean value property). Suppose we denote by $Z_{f}$ the set of zeros of a function $f$.

My question is, is there a convergence of $Z_{u_n}$ to $Z_u$ in any reasonable sense (maybe when restricted to compact regions)? This is probably well-known, and in that case, this is mainly a reference request. Thanks in advance!

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  • $\begingroup$ I guess you maybe want to impose $u \neq 0$ as well? $\endgroup$
    – Leo Moos
    Oct 7, 2022 at 14:12
  • $\begingroup$ @LeoMoos Thanks, edited. $\endgroup$
    – user492517
    Oct 7, 2022 at 14:17
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    $\begingroup$ 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. $\endgroup$ Oct 7, 2022 at 14:57
  • $\begingroup$ 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 $Z(u)\cap U$? $\endgroup$ Oct 24, 2022 at 15:36

1 Answer 1

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Yes.

With an appropriate topology, the function mapping a non-zero harmonic function to its zero set is continuous. The result actually applies to open mappings from a locally compact locally connected space $X$ to $\mathbb{R}$, and harmonic functions are just a particular example of such open mappings.

If $X$ is a topological space, then let $H(X)$ denote the collection of all closed subsets of $X$. The set $H(X)$

If $X$ is a compact Hausdorff space and $U,V_1,\dots,V_n$ are non-empty open subsets of $X$, then let $[U;V_1,\dots,V_n]$ be the collection of all closed subsets $C\subseteq X$ where $C\subseteq U$ and where $C\cap V_1\neq\emptyset,\dots,C\cap V_n\neq\emptyset$. Then the Vietoris topology on $H(X)$ is the topology with basis consisting of all sets of the form $[U;V_1,\dots,V_n]$.

Proposition: Let $X$ be a locally compact locally connected Hausdorff space. Let $H$ be a collection of continuous open mappings $f:X\rightarrow\mathbb{R}$. Give $H$ the topology of uniform convergence on compact sets. Define a mapping $T:H\rightarrow X\cup\{\infty\}$ by letting $T(h)=Z(h)\cup\{\infty\}$. Then the function $T$ is continuous.

Proof: Suppose that $h\in H$. Let $z\in Z(h)$. Let $U$ be a neighborhood of $z$. Then there is some connected open set $V$ with $z\in V\subseteq\overline{V}\subseteq U$ and where $V$ is compact. In this case, we have $h(r)>0,h(s)<0$ for some $r,s\in V$. Let $\delta=\min(h(r),-h(s))$. Therefore, if $g\in H$ and $\|g|_U-h|_U\|<\delta$, then $g(r)>0,g(s)<0$. Since $V$ is connected and $r,s\in V,g(r)>0,g(s)<0$, there is some $t\in V$ with $g(t)=0$ (this is the reasoning that Giorgio Metafune has made in the comments).

We conclude that if $Z(h)\cap U\neq\emptyset$, then there is some $\delta>0$ where if $\|(g-h)|_U\|<\delta$, then $Z(g)\cap U\neq\emptyset$ as well.

Suppose now that $h\in H$, and $T(h)\in[U;V_1,\dots,V_n]$ for open subsets $U,V_1,\dots,V_n\subseteq X\cup\{\infty\}$. Then $Z(h)\cup\{\infty\}\subseteq U$ and $Z(h)\cup\{\infty\}\cap V_j\neq\emptyset$ for $1\leq j\leq n$. Without loss of generality, assume that $1\leq m\leq n$ and $\infty\not\in V_j$ for $1\leq j\leq m$ where $\infty\in V_j$ for $m<j\leq n$. Then $V_j$ is an open subset of $X$ with $Z(h)\cap V_j\neq\emptyset$ for $1\leq j\leq m$.

Therefore, for $1\leq j\leq m$ there is a relatively compact open $W_j\subseteq V_j$ and some $\delta_j$ where if $\|(g-h)|_{W_j}\|<\delta_j$, then $Z(g)\cap V_j\neq\emptyset$. Now, set $W=W_1\cup\dots\cup W_m$ and $\delta=\min(\delta_1,\dots,\delta_m)$. Then whenever $\|(g-h)|_W\|<\delta$, we have $Z(g)\cap V_j\neq\emptyset$ for $1\leq j\leq m$.

Now let $C=X\setminus U$. Then $C$ is a compact subset of $X$ with $0\not\in g[C]$. Therefore, there is some $\epsilon>0$ where $|g(c)|\geq\epsilon$ for each $c\in C$. Therefore, if $\|(h-g)|_C\|<\epsilon$, then $g(c)\neq 0$ whenever $c\in C$, so $Z(g)\subseteq U$.

Therefore if $g\in H$ and $\|(h-g)|_{C\cup W}\|<\min(\delta,\epsilon)$, then $g\in[U;V_1,\dots,V_n]$ as well. We therefore conclude that the function $T$ is continuous at the point $h$.

Q.E.D.

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    $\begingroup$ Very nice answer. I was totally unaware of this topology! The one which I had in my mind was the Hausdorff metric, but it seems it is known that they induce the same topology (is that correct?). After some googling, I found the book "Topologies on closed and closed convex sets" by Gerald Beer. However the book is written in a very abstract language, and I cannot locate the region of the book (I suspect it is Chapter 6) related to your answer, and I would appreciate if you point out where to read. $\endgroup$
    – user492517
    Oct 8, 2022 at 6:09
  • $\begingroup$ Yes. The Vietoris topology is always induced by the Hausdorff metric from a compact metric space. More generally, every compact Hausdorff space has a unique uniformity. And every uniform space $X$ induces a uniformity on $H(X)$ called the hyperspace uniformity; the hyperspace uniformity is a generalization of the Hausdorff metric. The hyperspace uniformity on a compact Hausdorff space always induces the Vietoris topology. $\endgroup$ Oct 8, 2022 at 15:29
  • $\begingroup$ Is there a book or a handy reference? I would very much like to look up some of the details. Thanks! $\endgroup$
    – user492517
    Oct 9, 2022 at 3:41
  • $\begingroup$ The book on uniform spaces by John Isbell is a standard reference for uniform spaces and a good reference for the hyperspace uniformity, but I do not know of any reference for the result about open mappings that I have proven. $\endgroup$ Oct 11, 2022 at 19:27
  • $\begingroup$ If $f:U\rightarrow\mathbb{R}$ and the gradient of $f$ vanishes nowhere on $Z(f)$, then $Z(f)$ is an analytic submanifold of $U$. The Vietoris topology does not take into consideration the differentiable structure of $Z(f)$. I will need to think about this question some more. $\endgroup$ Oct 17, 2022 at 14:56

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