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!



Limit of zero sets of harmonic functions
**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.


