# Limit of zero sets of harmonic functions

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!

• I guess you maybe want to impose $u \neq 0$ as well? Oct 7, 2022 at 14:12
• @LeoMoos Thanks, edited. Oct 7, 2022 at 14:17
• 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 14:57
• 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 24, 2022 at 15:36

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 $$m. Then $$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.

• 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. Oct 8, 2022 at 6:09
• 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. Oct 8, 2022 at 15:29
• Is there a book or a handy reference? I would very much like to look up some of the details. Thanks! Oct 9, 2022 at 3:41
• 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. Oct 11, 2022 at 19:27
• 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. Oct 17, 2022 at 14:56