The sharp upper bound for the area of the nodal sets of Dirichlet Laplace eigenfunctions

Abstract

Let $\Omega$ be a bounded domain in ${\mathbb {R}}^n$ with $C^{1}$ boundary and let $u_\lambda$ be a Dirichlet Laplace eigenfunction in $\Omega$ with eigenvalue $\lambda$. We show that the $(n-1)$-dimensional Hausdorff measure of the zero set of $u_\lambda$ does not exceed $C(\Omega )\sqrt{\lambda }$. This result is new even for the case of domains with $C^\infty$-smooth boundary.

1 Introduction

Let $\Delta _M$ be the Laplace operator on an n-dimensional smooth compact Riemannian manifold and let $u_\lambda$ be an eigenfunction of $-\Delta _M$ with the eigenvalue $\lambda$, i.e., $\Delta _M u_\lambda +\lambda u_\lambda =0$. Denote by $Z(u_\lambda )=\{u_\lambda =0\}$ the zero set of $u_\lambda$. S. T. Yau [21] conjectured that the surface area of the zero set of $u_\lambda$ satisfies the following inequalities

$$\begin{aligned}c\sqrt{\lambda }\le \mathcal {H}^{n-1}(Z(u_\lambda ))\le C\sqrt{\lambda },\end{aligned}$$

where the constants c, C depend on M. This conjecture was proved by Donnelly and Fefferman in [6] under the assumption that the metric is real analytic. The lower bound and a polynomial in $\lambda$ upper bound were obtained recently by the first author in [16] and [15] respectively.

In this article we consider the case of eigenfunctions of the Euclidean Laplace operator on a bounded domain with sufficiently regular boundary and the Dirichlet boundary condition. One of our results is the following.

Theorem 1

Let $\Omega$ be a bounded domain in ${\mathbb {R}}^n$ with $C^{1}$ boundary and let $u_\lambda$ be an eigenfunction of the Laplace operator with the Dirichlet boundary condition, $\Delta u_\lambda +\lambda u_\lambda =0$ and $u_\lambda |_{\partial \Omega }=0$. Then

$$\begin{aligned} \mathcal {H}^{n-1}(Z(u_\lambda ))\le C\sqrt{\lambda }, \end{aligned}$$

(1)

where C depends only on $\Omega$.

The lower bound

$$\begin{aligned} \mathcal {H}^{n-1}(Z(u_\lambda ))\ge c\sqrt{\lambda }, \end{aligned}$$

for sufficiently large $\lambda$, follows from the results of Donnelly and Fefferman in [6] combined with Lemma 10 below. We remark that this bound also holds for any solution of the equation $\Delta u_\lambda + \lambda u_\lambda =0$ and the boundary condition plays no role. This follows from the fact that the zero set is $C\lambda ^{-1/2}$ dense and a non-trivial result of [16]. The inequality (1) was also proved by Donnelly and Fefferman in [7] for the case of real analytic boundary $\partial \Omega$. Their result was generalized to eigenfunctions of elliptic operators with real analytic coefficients by Kukavica [9]. Similar estimates were recently obtained by Lin and Zhu [14] for eigenfunctions of the bi-Laplace operator with various boundary conditions under the assumption that the boundary is real analytic. Also, the polynomial (in the eigenvalue) upper bounds for the area of the zero set of the Dirichlet, Neumann, and Robin eigenfunctions in smooth bounded domains in ${\mathbb {R}}^n$ were proved by Zhu in [22].

Our proof of Theorem 1 is based on the results of Donnelly and Fefferman and the ideas developed in [15,16,17]. In particular, we reduce the statement of the theorem to an estimate of the size of the nodal set of a harmonic function with controlled doubling index (the doubling index in defined in Section 3 below). The novelty of the current work is the treatment of domains with non-analytic boundaries. More precisely, we work with Lipschitz domains in the Euclidean space and assume that (locally) the Lipschitz constant is small enough; the precise definition and the formulation of the main result are given in the next section. This class of domains was recently considered by Tolsa [20] in a different problem.

The rest of the article is organized in the following way. In Section 3 we first discuss the doubling index of harmonic functions and its (weak) monotonicity properties near the boundary of Lipschitz domains with small Lipschitz constant, and then we formulate the main estimate for the size of the zero set of harmonic functions in terms of the doubling index, see Theorem 2 below. Two auxiliary results are contained in Section 4, where the low regularity of the boundary requires some careful considerations. We prove Theorem 2 for harmonic functions in Section 5, and explain how Theorem 1 follows from Theorem 2 in Section 6.

2 Preliminaries

2.1 Smoothness of the boundary.

Some of the tools used in the current paper should be compared to those in [20], where the following boundary uniqueness conjecture is studied.

Let h be a bounded harmonic function in a Lipschitz domain $\Omega$. Assume that h vanishes on a relatively open set $U\subset \partial \Omega$ and $\nabla h$ vanishes on a subset of U of positive surface measure. Then $h=0$.

Recently Xavier Tolsa verified the conjecture for Lipschitz domains with small Lipschitz constant, see [20]. We use the following definition.

Definition 1

Let $\Omega$ be a domain in ${\mathbb {R}}^d$, $\tau \in (0,1)$, and let $B=B(x,r)$ be a ball centred on $\partial \Omega$. We say that $\partial \Omega$ is $\tau$-Lipschitz in B if there is an isometry $T:{\mathbb {R}}^d\rightarrow {\mathbb {R}}^d$ and a function $f: B^{d-1}(0,r)\rightarrow {\mathbb {R}}$ such that $T(0)=x$, f is a Lipschitz function with the Lipschitz constant bounded by $\tau$, $f(0)=0$, and

$$\begin{aligned} \Omega \cap B=T\left( \{(y', y'')\in B^{d}(0,r)\subset {\mathbb {R}}^{d-1}\times {\mathbb {R}}: y''>f(y')\}\right) . \end{aligned}$$

In this case we write $\partial \Omega \cap B\in Lip(\tau )$.

Remark 1

Our considerations are mostly local. When considering the part of the boundary $\partial \Omega \cap B\in Lip(\tau )$, we choose local coordinates in $B=B(x,r)$, $x\in \partial \Omega$, so that the isometry T in the definition is the identity. We denote by $e_d$ the unit vector in the direction of the last coordinate, so that $x+\varepsilon e_d\in \Omega$ for $0<\varepsilon <r$.

Remark 2

Note that if $\partial \Omega \cap B\in Lip(\tau )$ and $B_1\subset B$ is a ball centred on $\partial \Omega$, then $\partial \Omega \cap B_1\in Lip(\tau )$. Also, rescaling does not change the Lipschitz constant. So if $\partial \Omega \cap B\in Lip(\tau )$ and $x\in \partial \Omega$ is the centre of B, then, denoting $\Omega _c=\{x+c(y-x): y\in \Omega \}$ and $B_c=cB=\{x+c(y-x): y\in B\}$ for some $c>0$, we have $\partial \Omega _c\cap B_c\in Lip(\tau )$.

Definition 2

We say that $\Omega$ is a Lipschitz domain with local Lipschitz constant $\tau$ if there exists $r>0$ such that $\partial \Omega \cap B(x,r)\in Lip(\tau )$ for any $x\in \partial \Omega$.

Clearly, any bounded $C^1$ domain is a domain with local Lipschitz constant $\tau$ for any positive $\tau$. So Theorem 1 follows from the next result.

Theorem 1$^\prime$. For each n, there exists $\tau _n>0$ such that the following statement holds. Let $\Omega$ be a bounded Lipschitz domain in ${\mathbb {R}}^n$ with local Lipschitz constant $\tau _n$ and let $u_\lambda$ be an eigenfunction of the Laplace operator in $\Omega$ with the Dirichlet boundary condition, $\Delta u_\lambda +\lambda u_\lambda =0$ and $u_\lambda |_{\partial \Omega }=0$. Then

$$\begin{aligned} \mathcal {H}^{n-1}(Z(u_\lambda ))\le C\sqrt{\lambda }, \end{aligned}$$

where C depends only on $\Omega$.

The constant C depends only on the parameter r for $\Omega$ in the definition of a Lipschitz domain with local Lipschitz constant $\tau _n$, on the diameter of $\Omega$, and on the dimension n. In what follows we assume that the dimension of the ambient Euclidean space is fixed so usually we will not emphasize the dependence of our constants on it.

The rest of the article is devoted to a proof of Theorem ${1'}$. We start with the following property of Lipschitz domains.

Lemma 1

Suppose that $\partial \Omega \cap B\in Lip(\tau )$, $\tau <1/4$, where $B=B(x,r)$ and $x\in \partial \Omega$. We choose coordinates as in Remark 1. Let $x_0\in \overline{\Omega }\cap \frac{1}{4}B$, and let $x_1=x_0+\tau r e_d$. Then $\Omega \cap B(x_1, r/2)$ is star-shaped with respect to $x_1$.

A version of this lemma can be found in [10]. We provide a proof for the convenience of the reader.

Proof

Let $x_1=(x_1',x_1'')$. Suppose that $x_2=(x_2',x_2'')\in \Omega \cap B(x_1,r/2)$. Let now $x_3=(x_3',x_3'')$ be a point on the interval $(x_1,x_2)$. Clearly $x_3\in B(x_1, r/2)$. We want to check that $x_3''>f(x_3')$.

Let $x_3=ax_1+(1-a)x_2$, $a\in (0,1)$. We have $x_1''\ge f(x_1')+\tau r$ and $x_2''>f(x_2')$. Therefore, we obtain

$$\begin{aligned} x_3''=ax_1''+(1-a)x_2''> af(x_1')+a\tau r+(1-a)f(x_2'). \end{aligned}$$

Then, since $f(x_1')\ge f(x_2')-\tau |x_1'-x_2'|>f(x_2')-\tau r/2$, we have

$$\begin{aligned} x_3''>f(x_2')+a\tau r-\frac{a\tau r}{2}=f(x_2')+\frac{a\tau r}{2}>f(x_3'), \end{aligned}$$

where the last inequality holds since $|x_3'-x_2'|=a|x_1'-x_2'|<ar/2$ and f is $\tau$-Lipschitz. $\square$

2.2 Some observations.

In this section we recall some results about harmonic functions.

Suppose that h is a harmonic function in $\Omega$, $h\in C(\overline{\Omega })$, and $h=0$ on $\partial \Omega \cap B$, where $B=B(x,r)$ and $x\in \overline{\Omega }$. We define the function v in B by $v=h^2$ in $\Omega \cap B$, and $v=0$ in $B{\setminus } \Omega$. Then v is subharmonic in B and the mean-value theorem implies that for any $y\in B(x,r/2)\cap \Omega$,

$$\begin{aligned} h^2(y)\le \frac{1}{|B(y,r/2)|}\int _{B(y,r/2)\cap \Omega }h^2\le \frac{1}{|B(y,r/2)|}\int _{B(x,r)\cap \Omega }h^2, \end{aligned}$$

(2)

where |E| is the d-dimensional Lebesgue measure of the set E.

Another known fact that we use is the following quantitative version of the Cauchy uniqueness theorem.

Lemma 2

Let $B_+$ be the half-ball,

$$\begin{aligned} B_+=\{(x',x'')\in {\mathbb {R}}^{d-1}\times {\mathbb {R}}: |x'|^2+(x'')^2< 1, x''> 0\}. \end{aligned}$$

There exist $\gamma \in (0,1)$ and $C>0$ such that if h is harmonic in $B_+$, $h\in C^1(\overline{B}_+)$ and satisfies the inequalities $|h|\le 1, |\nabla h|\le 1$ in $B_+$ and $|h|\le \varepsilon$, $|\partial _d h|\le \varepsilon$ on $\Gamma =\{(x',x'')\in \overline{B}_+, x''=0\}$, $\varepsilon \le 1$, then

$$\begin{aligned} |h(x)|\le C\varepsilon ^\gamma \quad {\text {when}}\quad x\in \frac{1}{3}B_+=\left\{ (x',x''):|x'|^2+(x'')^2< \frac{1}{9}, x''> 0\right\} . \end{aligned}$$

The reader can find a proof of a similar statement in [13] and a general result on second order elliptic PDEs in Lipschitz domains in [5]. A simple proof is also given in Section A.3 for the convenience of the reader.

3 The Doubling Index

3.1 The doubling index inside the domain.

Let $h\in C(\overline{\Omega })$ be a non-zero harmonic function in a domain $\Omega \subset {\mathbb {R}}^d$. For each $x\in \overline{\Omega }$ and $r>0$, we define

$$\begin{aligned} H_h(x,r)=\int _{B(x,r)\cap \Omega }h^2\quad {\text {and}}\quad N_h(x,r)=\log \frac{H_h(x,2r)}{H_h(x,r)}, \end{aligned}$$

(3)

and, with some abuse of language, we call $N_h(x,r)$ the doubling index of h in $B=B(x,r)$.

Assume first that $B(x,2R)\subset \Omega$, then

$$\begin{aligned} N_h(x,r)\le N_h(x,R),\quad {\text {when}}\quad r<R. \end{aligned}$$

(4)

An elementary proof can be obtained by decomposing h into spherical harmonics, see, e.g., [19]. This is a simple and useful result, its various versions go back to the works of Landis [11, 12], Agmon [1], and Almgren [2].

Suppose that $B(x,4r)\subset \Omega$. Then we rewrite the inequality $N_h(x,r)\le N_h(x,2r)$ as

$$\begin{aligned} \left( \int _{B(x,2r)}h^2\right) ^2\le \int _{B(x,r)}h^2\int _{B(x,4r)}h^2. \end{aligned}$$

(5)

Similarly to (2), for any $y\in B(x, 3r/2)$, we have

$$\begin{aligned} h^2(y)\le \frac{1}{|B(y,r/2)|}\int _{B(y,r/2)}h^2\le \frac{1}{|B(y,r/2)|}\int _{B(x,2r)}h^2. \end{aligned}$$

Finally, applying (5) and using the trivial bound of the $L^2$ norms by the $L^\infty$ norms, we obtain

$$\begin{aligned} \sup _{B(x,3r/2)}|h|\le 2^d(\sup _{B(x,r)}|h|)^{1/2}(\sup _{B(x,4r)}|h|)^{1/2}. \end{aligned}$$

(6)

3.2 The doubling index on the boundary.

We need a version of the monotonicity formula (4) and the three ball inequality (6) near a part of the boundary on which the harmonic function vanishes. First, we recall a lemma that is proven in [10].

Lemma 3

(Kukavica, Nyström). Let $\Omega$ be a domain in ${\mathbb {R}}^d$ and let $B_1$ be a ball centred on $\partial \Omega$ such that $\partial \Omega \cap B_1$ is $C^3$ smooth. Let also $x\in \Omega$ be such that $\Omega \cap B(x,R)$ is star-shaped with respect to x, $B(x,R)\subset B_1$. Suppose that $h\in C(\overline{\Omega })$ is a non-zero harmonic function in $\Omega$ and $h=0$ on $\partial \Omega \cap B_1$. Then

$$\begin{aligned} \log \frac{H_h(x, r_2)}{H_h(x,r_1)}\le \frac{\log (r_2/ r_1)}{\log (r_3/r_2)}\log \frac{H_h(x, r_3)}{H_h(x,r_2)}, \end{aligned}$$

(7)

when $0<r_1<r_2<r_3<R$.

The assumption that the boundary of $\Omega$ is $C^3$ smooth implies that $h\in C^2(\overline{\Omega }\cap B_1)$, so every integration by parts in [10] can be easily justified.

Now we prove the following almost monotonicity property of the doubling index in Lipschitz domains.

Lemma 4

Let $\Omega$ be a domain in ${\mathbb {R}}^d$. For any $\varepsilon >0$, there exists $\tau _\varepsilon >0$ such that if $\tau <\tau _\varepsilon$, $\partial \Omega \cap B\in Lip(\tau )$, where $B=B(x,R),\ x\in \partial \Omega$, and $h\in C(\overline{\Omega })$ is a non-zero harmonic function in $\Omega$, $h=0$ on $\partial \Omega \cap B$, then

$$\begin{aligned} N_h(x_0,r)\le (1+\varepsilon ) N_h(x_0,2r), \end{aligned}$$

(8)

for any $x_0\in \overline{\Omega }\cap \frac{1}{4}B$ and $r<R/16$.

We remark that a stronger result holds when the boundary of the domain is smooth. For example for the case of a $C^{1,Dini}$ domain the inequality (8) can be replaced by $N_h(x_0,r_1)\le (1+\varepsilon )N(x_0, r_2)$ when $r_1<r_2<R/8$, see [10]¹. Thus for $C^{1,Dini}$ domains, we know that the doubling index over balls centred on $\partial \Omega \cap B$ stays uniformly bounded. We do not know if this is still true for Lipschitz domains. For the case of the domains with small Lipschitz constant, we can conclude only that the doubling index $N_h(x_0,r)$ does not grow faster than $r^{-a}$ with some small positive a as $r\rightarrow 0$, which is sufficient for our purposes.

Proof

First we assume that $\partial \Omega \cap B$ is a graph of a $C^3$-smooth function. Let $e_d$ be as in Remark 1 and let $x_1=x_0+16\tau re_d$. We assume that $\tau <1/16$. Then by Lemma 1 we see that $B(x_1,8r)\cap \Omega$ is star-shaped with respect to $x_1$. We apply (7) and obtain

$$\begin{aligned} N_h(x_0,r)= & {} \log \frac{H_h(x_0,2r)}{H_h(x_0,r)}\le \log \frac{H_h(x_1, (2+16\tau )r)}{H_h(x_1, (1-16\tau )r)}\\\le & {} \frac{\log ((2+16\tau )/(1-16\tau ))}{\log ((4-16\tau )/(2+16\tau ))}\log \frac{H_h(x_1, (4-16\tau )r)}{H_h(x_1, (2+16\tau )r)}\\\le & {} (1+O(\tau ))\log \frac{H_h(x_0, 4r)}{H_h(x_0, 2r)}=(1+\varepsilon ) N_h(x_0, 2r), \end{aligned}$$

when $\tau$ is small enough.

We want to drop the assumption that $\partial \Omega \cap B$ is $C^3$ smooth. We fix the ball B and assume that $\partial \Omega \cap B$ is given by the graph of a Lipschitz function $f: B^{d-1}(0,R)\rightarrow {\mathbb {R}}$ with the Lipschitz constant bounded by $\tau$. In this coordinate system the ball B is identified with $B^d(0,R)$

Let $\varphi$ be a mollifier supported in the unit ball of ${\mathbb {R}}^{d-1}$ and let, as usual, $\varphi _\delta (x)=\delta ^{-(d-1)}\varphi \left( \frac{x}{\delta }\right)$. We define $f_n=f*\varphi _{R/n}+\tau R/n$. Then $\{f_n\}$ is a sequence of $C^3$ smooth functions such that

$$\begin{aligned} f_n:B^{d-1}(0,(1-1/n)R)\rightarrow {\mathbb {R}},\quad f(y')<f_n(y')<f(y')+2\tau R/n, \end{aligned}$$

and the Lipschitz constant of $f_n$ is also bounded by $\tau$. We also define

$$\begin{aligned} \Omega _n=\{y=(y',y'')\in B^d(0, (1-1/n)R)\subset {\mathbb {R}}^{d-1}\times {\mathbb {R}}: y''>f_n(y')\}. \end{aligned}$$

Clearly, $\Omega _n\subset B^d(0, (1-1/n)R)\cap \Omega$. Let

$$\begin{aligned} \Gamma _n=\{y=(y',y'')\in B^d(0,(1-1/n)R): y''=f_n(y')\}. \end{aligned}$$

First, we see that $\delta _n=\sup _{\Gamma _n}|h|$ converge to zero as $n\rightarrow \infty$ since h is uniformly continuous on $\overline{\Omega }\cap \overline{B}$, $h=0$ on $\partial \Omega$, and ${{\text {dist}}}(y,\partial \Omega )<2\tau R/n$ when $y\in \Gamma _n$.

Next, we consider the harmonic function $h_n$ in $\Omega _n$ such that on $\partial \Omega _n$

$$\begin{aligned} h_n(x)={\left\{ \begin{array}{ll}h(x)-\delta _n,\ {\text {if}}\ h(x)>\delta _n,\\ 0,\quad \quad \quad \ \ \ {\text {if}}\ |h(x)|\le \delta _n,\\ h(x)+\delta _n,\ {\text {if}}\ h(x)<-\delta _n. \end{array}\right. } \end{aligned}$$

Clearly we have $h_n\in C(\overline{\Omega }_n)$, $h_n=0$ on $\Gamma _n$, and, by the maximum principle, $|h-h_n|\le \delta _n$ in $\Omega _n$. Thus $h_n\rightarrow h$ uniformly on compact subsets of $B\cap \Omega$.

We fix $x_0\in \Omega \cap \frac{1}{4}B$ and $r\in (0,R/16)$. Then $x_0\in \Omega _n\cap B(0,(1-1/n)R/4)$ and $r<(1-1/n)R/16$ for n large enough. Also, $|h_n|\le \max _{\overline{\Omega }\cap \overline{B}}|h|$ and $|(\Omega \cap B(x,R)){\setminus } \Omega _n|\rightarrow 0$ as $n\rightarrow \infty$. Then we have

$$\begin{aligned} N_h(x_0,r)=\frac{\int _{B(x_0,2r)\cap \Omega }h^2}{\int _{B(x,r)\cap \Omega }h^2}=\lim _{n\rightarrow \infty }\frac{\int _{B(x_0,2r)\cap \Omega _n}h_n^2}{\int _{B(x_0,r)\cap \Omega _n}h_n^2}=\lim _{n\rightarrow \infty }N_{h_n}(x_0,r). \end{aligned}$$

The inequality (8) is now obtained as the limit of the corresponding inequalities for $h_n$. Finally the required inequality (8) for $x_0\in \partial \Omega \cap \frac{1}{4}B$ follows by taking the limit as $\varepsilon \rightarrow 0+$ of the corresponding inequalities for $x_0+\varepsilon e_d$. $\square$

Corollary 1

Let $\partial \Omega \cap B\in Lip(\tau )$, $\tau <\tau _\varepsilon$, $B=B(x,R)$, $x\in \partial \Omega$, and let $x_1,x_2\in \overline{\Omega }\cap \frac{1}{4}B$ with $|x_1-x_2|<r/4$ and $r<R/8$. If $h\in C(\overline{\Omega })$ is a non-zero harmonic function in $\Omega$ such that $h=0$ on $\partial \Omega \cap B$, then

$$\begin{aligned} N_h(x_1,r/2)\le 3(1+\varepsilon )^2 N_h(x_2, r). \end{aligned}$$

Proof

Note that $B(x_1,r)\subset B(x_2, 2r)$ and $B(x_1, r/2)\supset B(x_2, r/4)$. Thus we obtain

$$\begin{aligned} N_h(x_1, r/2)\le \log \frac{\int _{B(x_2,2r)\cap \Omega }h^2}{\int _{B(x_2, r/4)\cap \Omega }h^2}=N_h(x_2, r/4)+ N_h(x_2, r/2)+N_h(x_2, r). \end{aligned}$$

Now Lemma 4 implies the required estimate. $\square$

3.3 Three ball inequality.

We apply the monotonicity lemma a number of times. First, we claim that it implies a version of the three ball theorem for harmonic functions vanishing on some part of the boundary.

Lemma 5

Let $\Omega$ be a domain in ${\mathbb {R}}^d$ and let B be a ball centred on $\partial \Omega$. We assume that $\partial \Omega \cap B\in Lip(\tau )$, where $\tau$ is small enough. Then for any function $h\in C(\overline{\Omega })$ harmonic in $\Omega$ and vanishing on $\partial \Omega \cap B$, we have

$$\begin{aligned} \sup _{ \frac{3}{2}B_0\cap \Omega }|h|\le 3^d(\sup _{B_0\cap \Omega }|h|)^{1/3} (\sup _{4B_0\cap \Omega }|h|)^{2/3}, \end{aligned}$$

for any ball $B_0$ with the centre in $\overline{\Omega }\cap \frac{1}{4}B$ and such that $16B_0\subset B$.

Proof

Assume that $\tau <\tau _1$ given by Lemma 4 for the case $\varepsilon =1$. Let $B_0=B(x_0,r)$. We apply Lemma 4. Taking the exponentials, we obtain

$$\begin{aligned} \int _{2B_0\cap \Omega } h^2\le \left( \int _{B_0\cap \Omega }h^2\right) ^{1/3}\left( \int _{4B_0\cap \Omega }h^2\right) ^{2/3}. \end{aligned}$$

Then (2) and the trivial bound of the $L^2$-norm by the $L^\infty$-norm imply that for any $y\in \frac{3}{2}B_0\cap \Omega$,

$$\begin{aligned} h^2(y)\le \frac{1}{|B(y,r/2)|}\int _{2B_0\cap \Omega }h^2 \le 8^d \left( \sup _{B_0\cap \Omega }|h|\right) ^{2/3}\left( \sup _{4B_0\cap \Omega }|h|\right) ^{4/3}. \end{aligned}$$

$\square$

3.4 The maximal doubling index.

Let $\Omega$ be a domain in ${\mathbb {R}}^d$ and let $\partial \Omega \cap B\in Lip(\tau )$ where B is centred on $\partial \Omega$. We consider a closed cube $Q\subset \frac{1}{32}B$ such that $Q\cap \Omega \ne \varnothing$. Assume that a non-zero function $h\in C(\overline{\Omega })$ is harmonic in $\Omega$ and vanishes on $\partial \Omega \cap B$ and let $\ell ={{\text {diam}}}(Q)$. We define the maximal doubling index of h in Q by

$$\begin{aligned} N^*_h(Q)=\sup _{x\in Q\cap \overline{\Omega }, \frac{\ell }{2}\le r\le \ell } N_h(x,r). \end{aligned}$$

(9)

Clearly the function $(x,r)\mapsto N_h(x,r)$ is continuous on $(Q\cap \overline{\Omega })\times [\ell /2,\ell ]$. Therefore the supremum above is finite.

Lemma 4 on the monotonicity of the doubling index implies that if $\varepsilon >0$ and $\tau <\tau _\varepsilon$, then for any cube $Q_1\subset Q\subset \frac{1}{32}B$ and $Q_1\cap \Omega \ne q\varnothing$, we have

$$\begin{aligned} N_h^*(Q_1)\le \left( \frac{2s(Q)}{s(Q_1)}\right) ^{2\varepsilon } N_h^*(Q), \end{aligned}$$

where s(Q) is the side length of the cube Q; we have used the inequality $\log _2(1+\varepsilon )\le 2\varepsilon$.

3.5 A version of the main result for harmonic functions.

Let $\Omega$ be a domain in ${\mathbb {R}}^d$ and let $h\in C(\overline{\Omega })$ be a non-zero harmonic function in $\Omega$. We assume that $h=0$ on the part $\partial \Omega \cap B$ of the boundary, where B is a ball centred on $\partial \Omega$ and $\partial \Omega \cap B\in Lip(\tau )$. Our aim is to estimate the $(d-1)$-dimensional measure of the zero set of h using the doubling index of h. We define the zero set of h by

$$\begin{aligned} Z(h)=\{x\in \Omega : h(x)=0\}, \end{aligned}$$

so that the boundary points are not included into the zero set.

Theorem 2

Let $\Omega \subset {\mathbb {R}}^{d}$, let $x\in \partial \Omega$ and let $r>0$ be such that $\partial \Omega \cap B(x,128r)\in Lip(\tau ),$ where $\tau$ is small enough. Then there exists $C=C(d)$ such that

$$\begin{aligned} \mathcal {H}^{d-1}(Z(h)\cap B(x,r))\le C(N_h(x,4r)+1)r^{d-1}, \end{aligned}$$

for any non-zero function $h\in C(\overline{\Omega })$ that is harmonic in $\Omega$ and satisfies $h=0$ on $\partial \Omega \cap B(x,128 r)$.

Theorem 2 is proved in Section 5.2. We then deduce Theorem 1’ in Section 6.2, where we consider the harmonic extension of the eigenfunction and use Lemma 10 below to estimate the doubling index of the extension by a multiple of the square root of the eigenvalue.

Theorem 2 allows us to estimate the area of the zero set of a harmonic function near the part of the boundary where the function vanishes. We remark also that the estimate for the zero set inside the domain was proved by Donnelly and Fefferman in [6].

Lemma 6

(Donnelly, Fefferman). Let h be a non-zero harmonic function in $\Omega \subset {\mathbb {R}}^{d}$. There exists C such that

$$\begin{aligned} \mathcal {H}^{d-1}(Z(h)\cap B)\le C (N_h(x,4r)+1)r^{d-1}, \end{aligned}$$

for any ball $B=B(x,r)$ that satisfies $\overline{B}(x,8r) \subset \Omega$.

The proof follows from the argument in [6], some versions of this result can be also found in [13] and [8]. We outline some steps of the proof for the interested reader in the Appendix, see A.1.

4 Two Auxiliary Lemmas

4.1 A standard construction.

In this section we give two versions of the Hyperplane Lemma. We suggest that the reader compares the statements to the one of [15, Lemma 4.1]. Both statements refer to the following construction.

Assume that $\Omega \subset {\mathbb {R}}^{d}$ and $\partial \Omega \cap B\in Lip(\tau )$, where B is a ball centred on $\partial \Omega$ and $\tau \in (0, (16\sqrt{d})^{-1})$. We fix a coordinate system as in Remark 1. Let Q be a cube centred at $x_Q=(x_Q',x_Q'')\in \partial \Omega \cap B$ whose sides are parallel to the axes of this coordinate system and such that $Q\subset B$. As above, the side length of Q is denoted by s(Q). Our choice of $\tau$ implies that $\partial \Omega$ does not intersect the two faces of the cube Q which are orthogonal to $e_d$, moreover, $\partial \Omega \cap Q$ is contained in the middle part $\{(x',x'')\in Q: |x''-x_Q''|<s(Q)/4\}$ of Q.

Let $k\ge 3$. We partition the projection $\pi (Q)$ of Q to the hyperplane ${\mathbb {R}}^{d-1}\times \{0\}$ into $2^{k(d-1)}$ small equal cubes w with the side length $s(w)=2^{-k}s(Q)$ in the usual way so that any two distinct small cubes have no common inner points. For each small cube w, there is a uniquely defined d-dimensional cube q such that $\pi (q)=w$ and the centre of q lies on $\partial \Omega \cap Q$. Furthermore, we cover $(\pi ^{-1}(w)\cap (\Omega \cap Q)){\setminus }{q}$ by at most $2^k$ cubes p such that $p\subset Q$, p, $\pi (p)=w$, p has no common inner points with q, and $s(p)=s(q)=2^{-k}s(Q)$, cubes p may overlap. See Figure 1.

Fig. 1

The standard construction.

We denote the set of all boundary cubes q by $\mathcal {B}_k(Q)$ and the set of all inner cubes p by $\mathcal {I}_k(Q)$. Note that for each $p\in \mathcal {I}_k(Q)$, we have ${{\text {dist}}}(p,\partial \Omega )>c s(p)$ for some absolute constant c. We call the triple $(Q,\mathcal {B}_k(Q), \mathcal {I}_k(Q))$ the standard construction. After we fix a coordinate system, our standard construction depends on the choice of the cube Q and the parameter k, the family $\mathcal {B}_k(Q)$ of the boundary cubes is defined uniquely and we may fix some choice for the inner cubes $\mathcal {I}_k(Q)$.

4.2 The first hyperplane lemma.

In the first lemma we assume that the maximal doubling index $N_h^*(Q)$ is large enough.

Lemma 7

There exist constants $k_0\ge 3$ and $N_0\ge 1$ such that for any integer $k\ge k_0,$ there exists $\tau (k)>0$ for which the following statement holds. Suppose that $\Omega$ is a domain in ${\mathbb {R}}^d$, $\partial \Omega \cap B\in Lip(\tau )$, $\tau <\tau (k),$ and $Q\subset \frac{1}{64}B$ is a cube as above centred on $\partial \Omega$. Then for any function $h\in C(\overline{\Omega })$ harmonic in $\Omega$, with $h=0$ on $B\cap \partial \Omega$, and $N^*_h(Q)>N_0$, there exists a cube $q\in \mathcal {B}_k(Q)$ such that $N^*_h(q)\le N_h^*(Q)/2$.

Proof

Let $x_Q$ be the centre of the cube Q and let $B_1=B(x_Q,\ell )$, where $\ell ={{\text {diam}}}(Q)$. We have $B_1\subset B$ and define $M^2=\int _{B_1\cap \Omega }h^2$.

Denote $N=N_h^*(Q)$ and suppose that the inequality $N^*_h(q)>N/2$ holds for each cube $q\in \mathcal {B}_k(Q)$. Then for each such q, there exist $y_q\in q\cap \overline{\Omega }$ and $r_q\in (2^{-k-1}\ell , 2^{-k}\ell )$ such that $N_h(y_q,r_q)>N/2$. Suppose that

$$\begin{aligned} \tau<\tau _\varepsilon ,\quad {\text {and}}\quad (1+\varepsilon )^k<2, \end{aligned}$$

(10)

where we use the notation of Lemma 4. Then the almost monotonicity of the doubling index, Lemma 4, implies $N_h(y_q, 2^mr_q)>N/4$ when $0\le m\le k$.

Assuming that $k\ge 20$, we apply the estimate of the doubling index $k-4$ times and use that $B(y_q, \ell /2)\subset B_1$ to obtain

$$\begin{aligned} \int _{B(y_q, 2^{-k+2}\ell )\cap \Omega } h^2\le \int _{B(y_q, 8r_q)\cap \Omega } h^2\le e^{-N(k-4)/4}\int _{B(y_q,2^{k-1}r_q)\cap \Omega }h^2\\ \le e^{-N(k-4)/4}\int _{B(y_q,\ell /2)\cap \Omega }h^2 \le e^{-Nk/5}M^2. \end{aligned}$$

Next, we note that the integral estimate above implies a pointwise estimate in a smaller ball by (2). We have

$$\begin{aligned} \sup _{B(y_q,2^{-k+1}\ell )\cap \Omega }h^2\le C2^{dk}\ell ^{-d}\int _{B(y_q,2^{-k+2}\ell )\cap \Omega }h^2\le C2^{dk}\ell ^{-d}e^{-Nk/5}M^2, \end{aligned}$$

(11)

where $C=C(d)$.

As above, we assume also that $\tau < (16\sqrt{d})^{-1}$. For each cube $q\in \mathcal {B}_k(Q)$, denote by $q^+$ its upper quarter, where "up" is in the direction of $e_d$. Then $q^+\subset \Omega$ and ${{\text {dist}}}(q^+, \partial \Omega )\ge 2^{-k}s(Q)/10$. For $y\in q^+$, the standard Cauchy estimate implies

$$\begin{aligned} |\nabla h(y)|\le C2^k\ell ^{-1}\sup _{B(y, 2^{-k}s(Q)/10)}|h|. \end{aligned}$$

We note that $B(y, 2^{-k}s(Q)/10)\subset B(y_q, 2^{-k+1}\ell )\cap \Omega$. Then combining the above inequality with (11), we obtain

$$\begin{aligned} \sup _{q^+}|\nabla h|\le C 2^k\ell ^{-1}\sup _{B(y_q, 2^{-k+1}\ell )\cap \Omega }|h|\le C2^{k(d+2)/2}\ell ^{-(d+2)/2}e^{-Nk/10}M. \end{aligned}$$

(12)

Let $B_0=B(x_Q+3\cdot 2^{-k-3}s(Q)e_d, s(Q)/2)$ and let

$$\begin{aligned} B_{0,+}=\{x=(x',x'')\in B_0: x''\ge x''_Q+3\cdot 2^{-k-3}s(Q)\} \end{aligned}$$

be the upper half of $B_0$. We denote by $\Gamma _{0}$ the flat part of the boundary of $B_{0,+}$. We note that $2B_0\subset B_1$. Assuming that $\tau <2^{-k-3}$, we have ${{\text {dist}}}(B_{0,+},\partial \Omega )\ge 2^{-k-2}s(Q)$. Then using (2) and the Cauchy estimate, we get

$$\begin{aligned} \sup _{B_0\cap \Omega }|h|\le C\ell ^{-d/2}M,\quad \sup _{B_{0,+}}|\nabla h|\le C2^k \ell ^{-d/2-1}M. \end{aligned}$$

Also, by (11) and (12), we have

$$\begin{aligned} \sup _{\Gamma _0}|h|\le C2^{kd/2}\ell ^{-d/2}e^{-Nk/10}M,\quad \sup _{\Gamma _{0}}|\nabla h|\le C2^{kd/2+k}\ell ^{-d/2-1}e^{-Nk/10}M, \end{aligned}$$

since $\Gamma _0\subset \bigcup _{q\in \mathcal {B}_k(Q)}q^+$.

Applying Lemma 2 to $B_{0,+}$, we get

$$\begin{aligned} \sup _{\frac{1}{3} B_{0,+}}|h|\le C 2^{\gamma kd/2+k}\ell ^{-d/2}e^{-\gamma Nk/10}M. \end{aligned}$$

Let $y_Q=x_Q+s(Q) e_d/12$ and let m be the least integer such that $2^m>16\sqrt{d}$. Then $B_2=B(y_Q, 2^{-m}\ell )\subset \frac{1}{3} B_{0,+}$ when k is large enough (we remark that $B_{0,+}$ depends on k). Integrating the last inequality over $B_2$ and using that $vol(B_2)\le C\ell ^d$, we obtain

$$\begin{aligned} \int _{B_2} h^2\le C2^{\gamma kd+2k}e^{-\gamma Nk/5}M^2. \end{aligned}$$

Finally, we compare the last integral to $\int _{B_1\cap \Omega } h^2=M^2$. Note that $B_1\subset B(y_Q, 2\ell )=2^{m+1}B_2$. By the almost monotonicity of the doubling index, recalling that $\tau<\tau _\varepsilon <\tau _1$, we have

$$\begin{aligned}&2^{m+1}N_h(y_Q, \ell )\ge \sum _{j=0}^m N_h(y_Q, 2^{-j}\ell )= \log \frac{\int _{B(y_Q,2\ell )\cap \Omega }h^2}{\int _{B_2}h^2}\\&\quad \ge \log \frac{\int _{B_1\cap \Omega }h^2}{\int _{B_2}h^2}\ge \gamma N k/5-\gamma kd-2k-C. \end{aligned}$$

Since $N_h(y_Q,\ell )\le N^*_h(Q)=N$, we get $2^{m+1}N\ge \gamma Nk/5-\gamma kd-2k-C$. Taking k large enough we may achieve $\gamma k/5>2^{m+2}$. Then the inequality above implies

$$\begin{aligned} N\le \frac{10(\gamma kd+2k+C)}{\gamma k}\le 10\left( d+(2+C)\gamma ^{-1}\right) . \end{aligned}$$

Taking $N_0=10\left( d+(2+C)\gamma ^{-1}\right)$, we obtain a contradiction for $N>N_0$. We also choose $\varepsilon =\varepsilon (k)$ such that $(1+\varepsilon )^k<2$ and finally choose $\tau (k)=\min \{\tau _\varepsilon ,2^{-k-3}, (16\sqrt{d})^{-1}\}$. $\square$

4.3 The second hyperplane lemma: cubes without zeros.

For cubes with the maximal doubling index bounded by $N_0$, we use the following version of the above statement. The reader may compare it to Corollary in Section 3.4 of [18].

Lemma 8

For any $N>0$ there exist $\tau (N)$ and k(N) such that the following statement holds. Suppose that $\Omega$ is a domain in ${\mathbb {R}}^d$, $\partial \Omega \cap B\in Lip(\tau )$, $\tau <\tau (N)$, and $Q\subset \frac{1}{64}B$ is a cube centred on $\partial \Omega$. Let also $h\in C(\overline{\Omega })$ be a non-zero function harmonic in $\Omega$, with $h=0$ on $B\cap \partial \Omega$ and $N^*_h(Q)\le N$. Then for any $k\ge k(N)$, there exists $q\in \mathcal {B}_{k}(Q)$ such that $Z(h)\cap q=\varnothing$.

We remark that in this version both $\tau$ and k depend on N. First, we prove the following version of the lemma for a half ball.

Lemma 9

Let B be the unit ball in ${\mathbb {R}}^d$ and let $B_+$ be the half ball,

$$\begin{aligned} B_+=\{y=(y',y'')\in {\mathbb {R}}^{d-1}\times {\mathbb {R}}: |y'|^2+y''^2<1, y''>0\}. \end{aligned}$$

Let g be a function harmonic in $B_+$, $g\in C(\overline{B}_+)$, $g=0$ on $\overline{B}_+\cap \{y''=0\}$, and

$$\begin{aligned} \sup _{\frac{1}{4} B_+}|g|=1. \end{aligned}$$

For any $N>0$, there exist $\rho =\rho (N)\in (0,1/16)$ and $c_0=c_0(N)>0$ such that if $N_g(0, 1/4)\le N$, then there is $x'\in {\mathbb {R}}^{d-1}$ with $|x'|<1/16$ such that

$$\begin{aligned} |g(y)|\ge c_0y'',\quad {\text {for any}}\quad y=(y', y'')\in B((x',0), \rho )\cap B_+. \end{aligned}$$

Proof

Let $B_-$ be the reflexion of the half-ball $B_+$ with respect to the hyperplane $y''=0$. Then g can be extended to a harmonic function in B by $g(y',y'')=-g(y',-y'')$ when $(y',y'')\in B_-$. We denote this extension by g as well. The normalization $\sup _{\frac{1}{4} B_+}|g|=1$ and the standard Cauchy estimate imply that every partial derivative of g is uniformly bounded in B(0, 1/8).

Let $\delta =\max _{x'\in {\mathbb {R}}^{d-1},|x'|\le 1/16}|\nabla g(x',0)|$. Lemma 2, applied to the half ball $\frac{1}{16}B_+$ implies that

$$\begin{aligned} \sup _{B(0,\frac{1}{64})}|g|\le C\delta ^\gamma . \end{aligned}$$

Then $\int _{B(0, \frac{1}{64})}g^2\le C\delta ^{2\gamma }$ and $\int _{B(0,\frac{1}{2})}g^2\ge c\sup _{B(0,\frac{1}{4})}g^2=c$. On the other hand,

$$\begin{aligned} \log \frac{\int _{B(0,\frac{1}{2})}g^2}{\int _{B(0,\frac{1}{64})} g^2} \le 5N_g(0,\frac{1}{4})\le 5N. \end{aligned}$$

We have used that the doubling index of g in $B_+$ and of the extension are the same for balls centred at the origin and that the doubling index inside the domain is monotone by (4). We conclude that $\delta \ge ce^{-3N\gamma ^{-1}}$.

Let $x'_*\in {\mathbb {R}}^{d-1}$, $|x'_*|\le 1/16$, be such that $|\nabla g(x'_*,0)|=\delta$. Clearly we have $|\nabla g(x'_*,0)|=|\partial _d g(x'_*,0)|$ and we may assume that $\partial _d g(x_*',0)=\delta$, otherwise we consider the function $-g$. Then $\partial _d g(x)>\delta /2$ when ${{\text {dist}}}(x,(x'_*,0))<\rho =\min \{c_0\delta , 1/16\}$, where $c_0$ depends on the constant upper bound for the second derivatives of g in B(0, 1/8). Therefore

$$\begin{aligned} g(y)\ge \delta y''/2\ge ce^{-3N\gamma ^{-1}}y'', \end{aligned}$$

when $y=(y',y'')\in B((x_*',0), \rho )$. $\square$

Proof of Lemma 8

Now we deduce Lemma 8 from Lemma 9. By rescaling, see Remark 2, we can achieve that $s(Q)=4$. We may also assume that

$$\begin{aligned} \sup _{B(x_Q, 3)\cap \Omega }|h|=1. \end{aligned}$$

(13)

Let $x_1=x_Q-3\tau e_d$, $B_1=B(x_1,1)$, and let $B_{1,+}$ be the upper half of $B_1$. Let also $B_2=2B_1$. First, we consider the harmonic function $g_0$ such that $g_0=1$ on the upper half of the sphere $\partial B_2$ and $g_0=-1$ on the lower half of $\partial B_2$. We denote as usual $x_1''=x_1\cdot e_d$. Clearly $g_0=0$ on

$$\begin{aligned} \Gamma _0=\{x=(x',x'')\in B_2: x''=x_1''\} \end{aligned}$$

and $g_0\ge 0$ on $B_{2,+}$. We note that $\Gamma _0$ does not intersect $\overline{\Omega }$. Then $|h|\le g_0$ on $\Omega \cap B_2\subset B_{2,+}$ by the maximum principle. We also have $g_0(x)\le C_1 (x''-x''_1)$ when $x=(x',x'')\in B_{1,+}$, since $g_0=0$ on $\Gamma _0$ and $g_0$ has bounded derivatives in $B_1$. Therefore $|h(x)|\le g_0(x)\le C_1(x''-x''_1)$ when $x=(x',x'')\in \Omega \cap B_1$.

Let now g be the harmonic function in $B_{1,+}$ such that $g=h$ on $\partial B_{1,+}\cap \Omega$ and $g=0$ on $\partial B_{1, +}{\setminus }\Omega$. We have $|g(x)|\le C_1(x''-x_1'')$ in $B_{1,+}$ by the above estimate on h and the maximum principle. We consider the difference $g-h$. We have $g=h$ on $\Omega \cap \partial B_1$ and $|g-h|=|g|\le 4C_1\tau$ on $\partial \Omega \cap B_1$. Then, by the maximum principle, $|g-h|\le 4C_1\tau \quad {\text {in}}\quad \Omega \cap B_1.$ We extend h by zero to $B_{1,+}{\setminus } \Omega$. Then $|g-h|\le 4C_1\tau$ in $B_{1,+}$.

Let m be the integer such that $2\sqrt{d}\le 2^m<4\sqrt{d}$, clearly $m\ge 1$. Then the estimate $N^*_h(Q)\le N$ implies $N_h(x_Q,2^m)\le N$. We choose $\varepsilon$ such that $(1+\varepsilon )^{m+3}\le 2$ and assume that $\tau <\tau _\varepsilon$ using the notation of Lemma 4. Then $N_h(x_Q,2^{j})\le 2N$ when $-3\le j\le m$. We use (13) and (2) to conclude that

$$\begin{aligned} \int _{ B(x_Q,\frac{1}{8})\cap \Omega }h^2 \ge e^{-10N}\int _{B(x_Q,4)\cap \Omega }h^2\ge ce^{-10N}. \end{aligned}$$

Suppose that $\tau <\frac{1}{24}$. Then $B(x_Q, \frac{1}{8})\cap \Omega \subset \frac{1}{4}B_{1,+}$ and we have

$$\begin{aligned} \left( \int _{ \frac{1}{4}B_{1,+}}g^2\right) ^{1/2}\ge \left( \int _{\frac{1}{4}B_{1,+}}h^2\right) ^{1/2}-C_2\tau \ge \left( \int _{B(x_Q,\frac{1}{8})\cap \Omega }h^2\right) ^{1/2}-C_2\tau . \end{aligned}$$

Assuming that $\tau (N)$ is small enough, we conclude that

$$\begin{aligned} \int _{\frac{1}{4}B_{1,+}} g^2\ge c_1e^{-10N}. \end{aligned}$$

(14)

We also have $\sup _{\frac{1}{2}B_{1,+}}|g|\le \sup _{B_1\cap \Omega }|h|\le 1$ by (13). Then

$$\begin{aligned} N_g(x_1, \frac{1}{4})=\log \frac{\int _{\frac{1}{2}B_{1,+}}g^2}{\int _{\frac{1}{4}B_{1,+}}g^2}\le C(N+1). \end{aligned}$$

We note that (14) implies $\sup _{\frac{1}{4}B_{1,+}}|g|\ge ce^{-5N}$. Then, by Lemma 9, there exist $x_*\in \Gamma _0\cap \frac{1}{16}B_1$, $c_2=c_2(N)>0$, and $\rho =\rho (C(N+1))$ such that

$$\begin{aligned} |g(x)|\ge c_2(x''-x_1'')\quad {\text {for}}\quad x=(x',x'')\in B(x_*,\rho )\cap B_{1,+}. \end{aligned}$$

We may assume that $g>0$ in $B(x_*,\rho )\cap B_{1,+}$, otherwise we consider $-h$ in place of h. Then we obtain

$$\begin{aligned} h(x)\ge g(x)-4 C_1\tau \ge c_2(x''-x_1'')-4 C_1\tau \quad {\text {in}}\quad B(x_*,\rho )\cap \Omega . \end{aligned}$$

We note that $\rho$ does not depend on $\tau$ and for $\tau$ small enough we have $B(x_*,\frac{\rho }{4})\cap \partial \Omega \ne \varnothing$. We also have $B(x_*,\frac{\rho }{2})\subset Q$.

Our goal is to show that $h>0$ on $B(x_*,\frac{\rho }{2})\cap \Omega$. Let $y_*=(y_*',y_*'')\in B(x_*,\frac{\rho }{2})\cap \partial \Omega$. We note that

$$\begin{aligned} h(x)\ge c_2(x''-y_*'')-c_3\tau \quad {\text {in}}\quad B(x_*,\rho )\cap \Omega , \end{aligned}$$

(15)

where $c_3=4C+4c_2$. We consider the harmonic function

$$\begin{aligned} h_*(x)=\frac{1}{d\rho }\left( (d-1)(x''-y''_*)^2-|x'-y'_*|^2\right) , \end{aligned}$$

where $x=(x',x'')$. We claim that $h(x)\ge c_2 h_*(x),$ when $x\in B\left( y_*,\frac{\rho }{2}\right) \cap \Omega$ and $\tau$ is small enough.

First, we note that $h_*(x)\le 0$ if $|x''-y_*''|\le (d-1)^{-1/2}|x'-y_*'|$ and therefore $h_*\le 0$ on $\partial \Omega \cap B(y_*,\frac{\rho }{2})$ when $\tau$ is small enough, while $h=0$ on $\partial \Omega \cap B(y_*,\frac{\rho }{2})$. On $\partial B(y_*, \frac{\rho }{2})\cap \Omega$ we have

$$\begin{aligned} h_*(x)=\frac{(x''-y_*'')^2}{\rho }-\frac{\rho }{4d}. \end{aligned}$$

Comparing (15) to the last identity and denoting $t=x''-y_*''$, we reduce the inequality $h\ge c_2h_*$ on $\partial B(y_*,\frac{\rho }{2})\cap \Omega$ to the following one:

$$\begin{aligned} c_2t-c_3\tau \ge c_2\left( \frac{t^2}{\rho }-\frac{\rho }{4d}\right) , \end{aligned}$$

when $t\in (-\tau \rho /2, \rho /2)$ and $\tau$ is small enough. It suffices to check the inequality for $t=-\tau \rho /2$ and $t=\rho /2$. For $t=-\tau \rho /2$ we obtain the inequality

$$\begin{aligned} \frac{c_2\rho }{4d}\ge \tau \left( \frac{c_2\rho }{2}+\frac{c_2\tau \rho }{4}+c_3\right) , \end{aligned}$$

which holds when $\tau$ is small enough. On the other hand, for $t=\rho /2$, the inequality is reduced to

$$\begin{aligned} c_2\rho \left( \frac{1}{4}+\frac{1}{4d}\right) \ge c_3\tau . \end{aligned}$$

This one is also satisfied for small $\tau$.

Thus, by the maximum principle, $h\ge c_2h_*$ in $B(y_*,\rho /2)\cap \Omega$. In particular, $h(y_*',y'')\ge c_2 h_*(y_*', y'')>0$ when $y_*''<y''<\rho /2$. Therefore $h>0$ on $B(x_*,\frac{\rho }{2})\cap \Omega$.

Finally, since $B(x_*,\rho /2)$ contains a ball of radius $\rho /4$ centred on $\partial \Omega$, if k is large enough, there is $q\in \mathcal {B}_k(Q)$ such that $q\subset B(x_*,\frac{\rho }{2})$ and then $Z(h)\cap q=\varnothing$. $\square$

5 Proof of Theorem 2

Let $N_0$ be as in Lemma 7 and let $\Omega$, $B=B(x,r)$, and h be as in the statement of Theorem 2. We remind that the maximal doubling index $N_h^*(Q)$ of h in a cube Q was defined by (9). For the rest of the proof we modify the maximal doubling index and write $N^{**}_h(Q)=\max \{N^*_h(Q),N_0/2\}$. Then Lemmas 7 and 8 imply that there is k such that for $\tau$ small enough, if $Q\subset 2B$ and $(Q,\mathcal {B}_k(Q), \mathcal {I}_k(Q))$ is a standard construction, then there is a cube $q_0\in \mathcal {B}_k(Q)$ such that

$$\begin{aligned} {\text {either}}\quad \mathrm{(i)}\ N_h^{**}(q_0)<N_h^{**}(Q)/2\quad {\text {or}}\quad \mathrm{(ii)}\ Z(h)\cap q_0=\varnothing . \end{aligned}$$

(16)

5.1 Reduction to one cube.

Let Q be a cube as above. We claim that

$$\begin{aligned} \mathcal {H}^{d-1}(Z(h)\cap Q)\le CN_h^{**}(Q)s(Q)^{d-1}. \end{aligned}$$

(17)

Assume first that (17) holds. We show that Theorem 2 follows. We need to switch from cubes to balls and from the maximal doubling index to the doubling index at a single point.

To this end, we cover the ball B(x, r) with cubes $Q_j\subset B(x,2r)$ such that ${{\text {diam}}}(Q_j)=r/10$ and either ${{\text {dist}}}(Q_j, \partial \Omega )>s(Q_j)/10$ (inner cubes) or $Q_j$ satisfies the assumptions in the main construction (boundary cubes). We may assume that there are not more than $C=C(d)$ of such cubes.

First, for each cube $Q=Q_j$ in this cover, we have $Q\cap B(x,r)\ne \varnothing$, and we compare $N_h^*(Q)$ to $N_h(x,4r)$. There exists $y\in Q\cap \overline{\Omega }$ and $r_y\in [r/20, r/10]$ such that $N_h^*(Q)=N_h(y,r_y)$. Assuming that $\tau <\tau _1$ in the notation of Lemma 4, we get $N_h(y,32r_y)\ge 2^{-5}N_h^*(Q)$. We have ${{\text {dist}}}(x, y)\le \frac{11}{10}r$ and

$$\begin{aligned} N_h(y, 32 r_y)=\log \frac{\int _{B(y, 64 r_y)\cap \Omega }h^2}{\int _{B(y, 32 r_y)\cap \Omega }h^2}\le \log \frac{\int _{B(x, 8r)\cap \Omega }h^2}{\int _{B(x,r/2)\cap \Omega }h^2}\le 16 N_h(x,4r) \end{aligned}$$

by Lemma 4. Hence, $N^*_h(Q)\le 2^9 N_h(x,4r)$ and $N^{**}_h(Q)\le C(N_h(x,4r)+1)$.

Each inner cube $Q\subset \Omega$ can be covered by at most C balls b with centres in Q and with radii s(Q)/100. Then $8\overline{b}\subset \Omega$. Moreover, if $b=B(y, s(Q)/100)$, we have $N_h(y,s(Q)/25)\le CN_h^{*}(Q)$ by Lemma 4 again. Then we use Lemma 6 to estimate the area of the zero set of h in each of the balls b and obtain

$$\begin{aligned}&\mathcal {H}^{d-1}(Z(h)\cap b)\le C(N_h(y, s(Q)/25)+1)r^{d-1}\\&\quad \le C'(N_h^*(Q)+1)r^{d-1}\le C''(N_h(x,4r)+1)r^{d-1}. \end{aligned}$$

For the boundary cubes, we use the inequality (17). Thus for every $Q_j$, we obtain

$$\begin{aligned} \mathcal {H}^{d-1}(Z(h)\cap Q_j)\le C(N_h(x,4r)+1)s(Q_j)^{d-1}. \end{aligned}$$

Summing these inequalities over all cubes, we obtain the required estimate. It remains to prove (17).

5.2 Proof of (17).

We fix a compact set $K\subset \Omega$ and prove that

$$\begin{aligned} \mathcal {H}^{d-1}(Z(h)\cap Q\cap K)\le C_0N_h^{**}(Q)s(Q)^{d-1}, \end{aligned}$$

(18)

where $Q\subset 2B$ is a cube as in the standard construction and $C_0$ is independent of K. Then (17) follows.

First, note that (18) holds for all cubes Q small enough, since $Q\cap K=\varnothing$ for such cubes. We prove (18) by induction on the size of Q, going from small cubes to larger ones. Assume that it holds for cubes with $s(Q)<s$, we want to prove it for cubes with $s(Q)<2^ks$, where k is as in (16).

We consider the standard construction $(Q, \mathcal {B}_k(Q), \mathcal {I}_k(Q))$. Each inner cube $q\in \mathcal {I}_k(Q)$ can be covered by balls b centred in q with radii s(q)/100 and such that $8\overline{b}\subset \Omega$, so that the number of balls is bounded by a dimensional constant. For each such ball $b=B(y, s(q)/100)$, applying Lemma 4, we get $N_h(y,s(q)/25)\le C(k)N^*_h(Q)$ when $\tau$ is small enough. Then by Lemma 6, we have

$$\begin{aligned} \sum _{q\in \mathcal {I}_k(Q)}\mathcal {H}^{d-1}(Z(h)\cap q)\le C (N_h^*(Q)+1)s(Q)^{d-1}\le C_1N_h^{**}(Q)s(Q)^{d-1}, \end{aligned}$$

(19)

where C and $C_1$ depend on k.

For all other boundary cubes q, we have $N^{**}_h(q)\le (1+\varepsilon )^kN^{**}_h(Q)$. Also (16) implies that there is a cube $q_0\in \mathcal {B}_k(Q)$ such that either $N^{**}_h(q_0)\le N_h^{**}(Q)/2$ or $Z(h)\cap q_0=\varnothing$. We apply the induction assumption to each boundary cube and obtain

$$\begin{aligned}&\mathcal {H}^{d-1}(Z(h)\cap K\cap (\cup _{q\in \mathcal {B}_k(Q)}q))\\&\quad \le \sum _{q\in \mathcal {B}_k(Q),q\ne q_0}\mathcal {H}^{d-1}(Z(h)\cap K\cap q)+\mathcal {H}^{d-1}(Z(h)\cap K\cap q_0)\\&\quad \le \sum _{q\in \mathcal {B}_k(Q),q\ne q_0}C_0 N_h^{**}(q)s(q)^{d-1}+\frac{C_0}{2} N_h^{**}(Q)s(q_0)^{d-1}\\&\quad \le \left( \frac{2^{k(d-1)}-1}{2^{k(d-1)}}(1+\varepsilon )^k+\frac{1}{2}\cdot \frac{1}{2^{k(d-1)}}\right) C_0N_h^{**}(Q)s(Q)^{d-1}. \end{aligned}$$

Finally, we choose $\varepsilon$ small and $C_0$ large enough so that

$$\begin{aligned} C_1+\left( \frac{2^{k(d-1)}-1}{2^{k(d-1)}}(1+\varepsilon )^k+\frac{1}{2}\cdot \frac{1}{2^{k(d-1)}}\right) C_0<C_0. \end{aligned}$$

Note that $C_0$ does not depend on K. Then, assuming that $\tau$ is small enough and taking into account (19), we obtain

$$\begin{aligned} \mathcal {H}^{d-1}(Z(h)\cap K\cap Q)\le C_0N_h^{**}(Q)s(Q)^{d-1}. \end{aligned}$$

This concludes the induction step and the proof of (17).

6 Dirichlet Laplace Eigenfunctions

6.1 Harmonic extension and an estimate of the doubling index.

Let $\Omega _0\subset {\mathbb {R}}^n$ be a bounded Lipschitz domain. Let $u_\lambda$ be an eigenfunction of the Dirichlet Laplace operator, $u_\lambda \in W_0^{1,2}(\Omega _0),$ $\Delta u_\lambda +\lambda u_\lambda =0$. Then $u_\lambda \in C(\overline{\Omega }_0)$. This fact is well-known, we provide a proof in the Appendix below, see Section A.2.

We consider the harmonic extension of $u_\lambda$ to the domain $\Omega =\Omega _0\times {\mathbb {R}}\subset {\mathbb {R}}^{n+1}$, given by

$$\begin{aligned} h(x,t)=u_\lambda (x)e^{\sqrt{\lambda }t}. \end{aligned}$$

Then $h\in C(\overline{\Omega })$ and, clearly, $Z(h)=Z(u_\lambda )\times {\mathbb {R}}$, where the zero sets are sets inside the domains $\Omega$ and $\Omega _0$ respectively. We need the following estimate of the doubling index of this harmonic extension.

Lemma 10

Let $\Omega _0$ be a bounded domain in ${\mathbb {R}}^n$ with a sufficiently small local Lipschitz constant $\tau$. Let $r_0>0$ be such that $\partial \Omega _0\cap B(x,r_0)\in Lip(\tau )$ for any $x\in \partial \Omega _0$. Then for any $r\in (0, r_0/16)$, there exists $C=C(r,\Omega _0)>0$ such that for any Dirichlet Laplace eigenfunction $u_\lambda$, the corresponding harmonic extension $h(x,t)=u_\lambda (x)e^{\sqrt{\lambda }t}$ satisfies $N_h(y,r)\le C\sqrt{\lambda }$ when $y=(x,t)\in \overline{\Omega }$.

This result is similar to the results of Donnelly and Fefferman, [6, 7], who considered eigenfunctions on compact manifolds and on domains with $C^\infty$-smooth boundaries and obtained the above estimate of the doubling index for eigenfunctions. However, in contrast to the previous results, the doubling index is allowed to blow up as $r\rightarrow 0$ in the above lemma. The statement of the lemma follows by application of Lemma 5 and inequality (6) to a chain of balls, the argument is similar to the one in [18, Section 2.4]. For the convenience of the reader, we provide the details below.

Proof

We consider any $y=(x,t)\in \overline{\Omega }$ and let $y_0=(x,0)$. Since $h(x,t+s)=e^{\sqrt{\lambda }t}h(x,s)$, we have $N_h(y,r)=N_h(y_0,r)$. So it is enough to estimate the doubling index of h in the balls centred on $\overline{\Omega }_0\times \{0\}$.

We fix $r\in (0, r_0/16)$ and let $\mathcal {S}\in \overline{\Omega }_0$ be a finite r/8-net for $\overline{\Omega }_0$, i.e., $\overline{\Omega }_0\subset \bigcup _{p\in \mathcal {S}} B(p,r/8)$. Let $B_*=B(y,r)$ be a ball of radius r centred at $y=(y_*,0)\in \overline{\Omega }_0\times \{0\}$. Assume that $\max _{\Omega _0}|u_\lambda |=|u_\lambda (x_0)|=1$. We consider a path $\gamma :[0,1]\rightarrow \overline{\Omega }_0$ from $y_*$ to $x_0$ such that $\gamma ((0,1))\subset \Omega _0$. Now we construct a chain of balls $\{B_j\}_{j=0}^J$. Let $B_0=B(y_*,r/2)$. Assuming that $B_j=B(y_j,r/2)$ is constructed, we define

$$\begin{aligned} s_j=\sup \{s\in [0,1]: |\gamma (s)-y_j|\le r/8\}. \end{aligned}$$

If $s_j<1$, we have $|\gamma (s_j)-y_j|=r/8$ and we choose $y_{j+1}\in \mathcal {S}$ such that $|y_{j+1}-\gamma (s_j)|<r/8$. If $s_j=1$, we define $y_{j+1}=y_J=x_0$ and stop the chain. We have $|y_j-y_{j+1}|< r/4$ and define $B_{j+1}=B(y_{j+1}, r/2)$. We note that $s_{j+1}>s_j$ when $0\le j<J-1$ and that $y_{j+1}\in \mathcal {S}{\setminus }\{y_0,\ldots ,y_j\}$ when $0\le j< J-1$. We also have $B_{j+1}\subset \frac{3}{2}B_j$. The resulting chain is finite, moreover, the number of balls in the chain is bounded by the number of elements in $\mathcal {S}$ plus two.

Let now $\widetilde{B}_j=B((y_j,0),r/2)$ be the corresponding ball in ${\mathbb {R}}^{n+1}$. Then $\sup _{4\widetilde{B}_j\cap \Omega }|h|\le e^{2\sqrt{\lambda } r}$. If $4\tilde{B}_j\subset \Omega$, then (6) gives

$$\begin{aligned} \sup _{\frac{3}{2}\widetilde{B}_j}|h|\le & {} 2^{n+1}(\sup _{\widetilde{B}_j}|h|)^{1/2}(\sup _{4\widetilde{B}_j}|h|)^{1/2}\\\le & {} 3^{n+1}(\sup _{\widetilde{B}_j}|h|)^{1/3}(\sup _{4\widetilde{B}_j}|h|)^{2/3}\le 3^{n+1}e^{4\sqrt{\lambda }r/3}(\sup _{\tilde{B}_j}|h|)^{1/3}. \end{aligned}$$

Otherwise we have ${{\text {dist}}}(y_j,\partial \Omega _0)<2r<r_0/8$. In this case, there is a ball $\tilde{B}$ of radius $r_0$ centred on $\partial \Omega _0\times \{0\}$ such that $(y_j,0)\in \overline{\Omega }\cap \frac{1}{4}\tilde{B}$ and $16 \tilde{B}_j\subset \tilde{B}$. Then Lemma 5, applied to the ball $\tilde{B}_j$, implies that

$$\begin{aligned} \sup _{\frac{3}{2} \tilde{B}_j\cap \Omega }|h|\le 3^{n+1}(\sup _{\tilde{B}_j\cap \Omega }|h|)^{1/3}(\sup _{4\tilde{B}_j\cap \Omega }|h|)^{2/3}\le 3^{n+1}e^{4\sqrt{\lambda }r/3}(\sup _{\tilde{B}_j\cap \Omega }|h|)^{1/3}. \end{aligned}$$

Therefore, we obtain for each j,

$$\begin{aligned} \sup _{\widetilde{B}_{j}\cap \Omega }|h|\ge 3^{-3(n+1)}(\sup _{\frac{3}{2}\widetilde{B}_{j}\cap \Omega }|h|)^3 e^{-4\sqrt{\lambda } r}\ge 3^{-3(n+1)}(\sup _{\widetilde{B}_{j+1}\cap \Omega }|h|)^{3}e^{-4\sqrt{\lambda } r}. \end{aligned}$$

We also have $\sup _{\widetilde{B}_J\cap \Omega }|h|=e^{\sqrt{\lambda } r/2}$. Combining the above inequalities, we get

$$\begin{aligned} \sup _{\widetilde{B}_0\cap \Omega }|h|\ge c_1e^{-C_2\sqrt{\lambda }}, \end{aligned}$$

where $c_1$ and $C_2$ depend on r and J but not on $\lambda$. We can choose the r/8-net $\mathcal {S}$ so that the number of points in $\mathcal {S}$ depends only on ${{\text {diam}}}(\Omega _0)$, r, and the dimension. Thus we conclude that the constants in the last inequality depend only on r, the diameter of $\Omega _0$, and n.

Finally, applying (2), we obtain

$$\begin{aligned} N_h(y,r)= & {} \log \frac{\int _{4\widetilde{B}_0\cap \Omega }h^2}{\int _{2\widetilde{B}_0\cap \Omega }h^2}\\\le & {} \log \frac{\sup _{4\widetilde{B}_0\cap \Omega }|h|^2}{\sup _{\widetilde{B}_0\cap \Omega }|h|^2}+C\\\le & {} (4r+2C_2)\sqrt{\lambda }+C\le C\sqrt{\lambda }, \end{aligned}$$

where $C=C(\Omega _0,r)$. We remark that $\lambda \ge \lambda _1(\Omega _0)>0$, where $\lambda _1(\Omega _0)$ is the first Dirichlet Laplace eigenvalue in $\Omega _0$. Moreover, if $B^*$ is a ball of radius ${{\text {diam}}}(\Omega _0)$ then $\lambda _1(\Omega _0)\ge \lambda _1(B^*)$. Thus the constant C in the conclusion of this Lemma depends only on r, ${{\text {diam}}}(\Omega _0)$, and n. $\square$

6.2 Proof of Theorem 1 $^\prime$.

Let $\Omega _0\subset {\mathbb {R}}^n$ be a bounded domain with a sufficiently small local Lipschitz constant $\tau$. Let also $r_0>0$ be such that $\partial \Omega _0\cap B(x,r_0)\in Lip(\tau )$ for every $x\in \partial \Omega _0$. We consider the domain $\Omega =\Omega _0\times {\mathbb {R}}\subset {\mathbb {R}}^{n+1}$ and let $\Omega _1=\Omega _0\times [-1,1]$. For each $x\in \partial \Omega \times [-1,1]$ we consider a ball centred at x of radius $2^{-9}r_0$. These balls cover the closed $2^{-10}r_0$-neighborhood of the set $\partial \Omega \times [-1,1]$. We can choose a disjoint collection of these balls $b_j$ such that the balls $B_j=4b_j$ cover the same closed neighborhood of $\partial \Omega \times [-1,1]$. Then for each point of $\Omega _1{\setminus }\cup _j B_j$, we choose a ball b centred at the point of radius $2^{-15}r_0$, so that $32b\subset \Omega$. Once again, we find a finite sub-collection of disjoint balls $b'_k$ such that $B_k'=4b_k'$ cover $\Omega _1{\setminus }\cup _j B_j$. We note that $8B_k'\subset \Omega$. We fix this covering of $\Omega _1$ and remark that radii of all balls depend only on $r_0$ and the number of balls depends on $r_0$, the diameter of $\Omega _0$, and n.

Let now $u_\lambda$ be a Dirichlet Laplace eigenfunction in $\Omega _0$: $\Delta u_\lambda +\lambda u_\lambda =0$ in $\Omega _0$ and $u_\lambda =0$ on $\partial \Omega _0$. We consider its harmonic extension $h(x,t)=e^{\sqrt{\lambda }t}u_\lambda (x)$. Then $h\in C(\overline{\Omega })$ is non-zero, and $h=0$ on $\partial \Omega$. Let $C_0=\max \{C(2^{-5}r_0,\Omega _0), C(2^{-11}r_0,\Omega _0)\}$, where $C(r,\Omega _0)$ is as in Lemma 10. Then for $B(x,r)\in \{B_j\}\cup \{B_k'\}$, we have $N_h(x,4r)\le C_0\sqrt{\lambda }$. Finally, we apply Theorem 2 to each of the balls $B_j$ and Lemma 6 to each of the inner balls $B_k'$. We conclude that

$$\begin{aligned}&\mathcal {H}^{n}(Z(h)\cap \Omega _1)\le \sum _j\mathcal {H}^n(Z(h)\cap B_j)+\sum _k\mathcal {H}^n(Z(h)\cap B'_k)\\&\quad \le C(C_0\sqrt{\lambda }+1)\left( \sum _j r(B_j)^n+\sum _kr(B'_k)^n\right) \le C_1\sqrt{\lambda }. \end{aligned}$$

Then $\mathcal {H}^{n-1}(Z(u_\lambda )\cap \Omega _0)\le C_1\sqrt{\lambda }$, which finishes the proof of Theorem 1$^\prime$.

Appendix: Proofs of Some Auxiliary Results

1.1 Estimates for the zero set of harmonic functions inside the domain.

We outline some steps of the proof of Lemma 6. First the harmonic function h is extended to a holomorphic function H on a domain in $\mathbb {C}^{d}$, see Lemma 7.2 in [6]. Our situation is particularly simple, since we only consider the standard Laplace operator on Euclidean domains. For this case the holomorphic extension is given by the complexification of the Poisson kernel. The Poisson kernel in a ball $B(x,r)\subset {\mathbb {R}}^d$ is given by

$$\begin{aligned} P_r(z,y)=c_d\frac{r^2-|z-x|^2}{r|z-y|^{d}},\quad |z-x|<r,\ |y-x|=r. \end{aligned}$$

For any $y\in \partial B(x,r)$, the function $z=(z_1,\ldots ,z_d)\mapsto \sum _j(z_j-y_j)^2$ maps the complex ball $B_\mathbb {C}(x,r/\sqrt{2})\subset \mathbb {C}^d$ of radius $r/\sqrt{2}$ centred at $x\in {\mathbb {R}}^d\subset \mathbb {C}^d$ to the half-plane $\mathfrak {R}\xi >0$. Then the Poisson kernel has the holomorphic extension to $B_\mathbb {C}(x,r/\sqrt{2})$. Moreover, for any $a<1/\sqrt{2}$,

$$\begin{aligned} |P_r(z,y)|\le C(a)r^{-(d-1)},\quad z\in B_\mathbb {C}(x,r_0),\ r_0\le ar. \end{aligned}$$

We consider a ball $B=B(x,8r)$ such that $\overline{B}\subset \Omega$. Then there exists a holomorphic extension H(z) of h defined on a ball $B_{\mathbb {C}}(x, 3r)$,

$$\begin{aligned} H(z)=\int _{\partial B(x,6r)}P_{6r}(z,y)h(y)d\sigma (y), \end{aligned}$$

such that $|H(z)|\le C\max _{\overline{B}(x, 6r)}|h|$. Then

$$\begin{aligned} \sup _{B_{\mathbb {C}}(x,3r)}|H(z)|\le C'r^{-d/2}\left( \int _{B(x,8r)}h^2\right) ^{1/2}. \end{aligned}$$

Now we can cover the set $Z(h)\cap B(x,r)$ by a finite number of balls with centrs in B(x, r) of radii r/20 so that the number of the balls is bounded by a constant depending on the dimension only. Let B(y, r/20) be one of such balls. By a version of Corollary 1 for the doubling index inside the domain, we have $N_h(y, 2r)\le 3N$, where $N=N_h(x,4r),$ and, therefore, $N_h(y, r_1)\le 3N$ when $r_1<2r$. Thus

$$\begin{aligned} \sup _{B(y,\frac{r}{16})}h^2\ge cr^{-d}\int _{B(y,\frac{r}{16})}h^2 \ge cr^{-d}e^{-15N}\int _{B(y,2r)}h^2\ge cr^{-d}e^{-15N}\int _{B(x,r)}h^2. \end{aligned}$$

Therefore,

$$\begin{aligned} \sup _{B(y,\frac{r}{10})}|H|\ge \sup _{B(y,\frac{r}{16} )}|h|\ge cr^{-d/2}e^{-7.5N}\left( \int _{B(x,r)}h^2\right) ^{1/2}. \end{aligned}$$

Combining the inequalities above, we obtain

$$\begin{aligned} \frac{\sup _{B_{\mathbb {C}}(y,2r)}|H|}{\sup _{B(y,r/10)}|H|}\le \frac{\sup _{B_{\mathbb {C}}(x,3r)}|H|}{\sup _{B(y,r/10)}|H|}\le Ce^{7.5N}\left( \frac{\int _{B(x,8r)}h^2}{\int _{B(x,r)}h^2}\right) ^{1/2}\le Ce^{9N}. \end{aligned}$$

Finally an estimate for the size of the zero set of a holomorphic function, Proposition 6.7 in [6], implies that

$$\begin{aligned} \mathcal {H}^{d-1}(Z(h)\cap B(y, r/20))\le C(N_h(x,4r)+1). \end{aligned}$$

We sum these inequalities over all balls B(y, r/20) to obtain the required estimate for $\mathcal {H}^{d-1}(Z(h)\cap B(x,r))$.

1.2 Continuity of eigenfunctions in Lipschitz domains.

First we prove the following regularity result.

Lemma 11

Let $\Omega$ be a domain in ${\mathbb {R}}^d$ and let h be a harmonic function in $\Omega$. Suppose that B is a ball centred on $\partial \Omega$ and that there exists a sequence of functions $\{h_n\}$, $h_n\in C^\infty _0({\mathbb {R}}^d)$ with the support of $h_n$ contained in $\Omega$, such that $h_n\rightarrow h$ and $\nabla h_n\rightarrow \nabla h$ in $L^2(B\cap \Omega )$. Assume also that $\partial \Omega \cap B\in Lip(\tau )$ and define $h=0$ on $\partial \Omega \cap B$. Then $h\in C(\overline{\Omega }\cap \frac{1}{2}B)$.

Proof

We define the function

$$\begin{aligned} v={\left\{ \begin{array}{ll} h^2\ {\text {in}}\ \Omega \cap B,\\ 0\ {\text {in}}\ B{\setminus } \Omega .\end{array}\right. } \end{aligned}$$

Then $v\in L^1(B)$. Let $\varphi \in C_0^\infty (B)$. We have

$$\begin{aligned} \int _B v\Delta \varphi= & {} \lim _{n\rightarrow \infty } \int _B h_n^2\Delta \varphi \nonumber \\= & {} -2\lim _{n\rightarrow \infty } \int _B h_n\nabla h_n\cdot \nabla \varphi =-2\int _{B\cap \Omega } h\nabla h\cdot \nabla \varphi . \end{aligned}$$

(20)

On the other hand, since h is harmonic in $\Omega$, we obtain

$$\begin{aligned} 0=\int _{\Omega } \nabla h\cdot \nabla (h_n\varphi )=\int _\Omega h_n\nabla h\cdot \nabla \varphi +\int _\Omega \varphi \nabla h\cdot \nabla h_n. \end{aligned}$$

Taking the limit as $n\rightarrow \infty$, we get

$$\begin{aligned} \int _\Omega h\nabla h\cdot \nabla \varphi =-\int _{\Omega } \varphi |\nabla h|^2. \end{aligned}$$

Combining the last identity and (20) gives

$$\begin{aligned} \int _B v\Delta \varphi =2\int _{B\cap \Omega }|\nabla h|^2\varphi . \end{aligned}$$

In particular, v is subharmonic in B in the weak sense: If $\varphi \ge 0$, $\varphi \in C_0^\infty (B)$, then $\int _B v\Delta \varphi \ge 0$. If $\alpha$ is a standard mollifier, $\alpha _\delta (x)=\delta ^{-d}\alpha (\delta ^{-1}x)$, and $v_\varepsilon =v*\alpha _{\varepsilon r}$, where r is the radius of B. Then $v_\varepsilon$ is subharmonic in $(1-\varepsilon )B$ and $v_\varepsilon \rightarrow v$ in $L^1(B)$ and almost everywhere. In particular, v satisfies the mean value inequality at each of its Lebesgue points. Clearly any $y\in \Omega \cap B$ is a Lebesgue point of v as $v=h^2$ in $\Omega \cap B$ and $h\in C(\Omega )$. So for any $y\in \Omega \cap B$ and any ball $B_1\subset B$ centred at y we have

$$\begin{aligned} v(y)\le \frac{1}{|B_1|}\int _{B_1}v. \end{aligned}$$

In particular,

$$\begin{aligned} \sup _{\frac{2}{3}B\cap \Omega } h^2\le \frac{3^d}{|B|}\int _{B\cap \Omega }h^2<\infty . \end{aligned}$$

Suppose that $x_1\in \partial \Omega \cap \frac{1}{2} B$. There exists a cone $\mathcal {C}$ with the vertex at $x_1$ such that ${\mathcal C}\cap (\Omega \cap B)=\varnothing$ and the aperture of $\mathcal {C}$ does not depend on $x_1$ (it depends on $\tau$ only). We use the following simple fact. If $y_1\in {\mathbb {R}}^d$ and $\rho >2\,{{\text {dist}}}(x_1,y_1)$, then

$$\begin{aligned} | B(y_1, \rho )\cap \mathcal {C}|\ge \alpha | B(y_1,\rho )|, \end{aligned}$$

for some $\alpha =\alpha (\tau )\in (0,1)$.

Let $m_k=\sup _{B(x_1, 3^{-k}r)\cap \Omega }|h|$ for $k\ge 2$. We know that $m_k<\infty$. Let $y\in B(x_1, 3^{-k}r)\cap \Omega$, $k\ge 3$. By the mean value inequality applied to v, we obtain

$$\begin{aligned} v(y)\le \frac{1}{|B(y, 2\cdot 3^{-k}r)|}\int _{B(y, 2\cdot 3^{-k}r)} v\le (1-\alpha )m_{k-1}^2. \end{aligned}$$

Thus $\sup _{B(x_1, 3^{-k}r)\cap \Omega }|h|\le (1-\alpha )^{(k-2)/2}\sup _{\frac{2}{3}B\cap \Omega }|h|$. We conclude that

$$\begin{aligned} \lim _{y\rightarrow x_1, y\in \Omega } h(y)=0. \square \end{aligned}$$

We remark that the argument above implies that h is Hölder continuous in $\overline{\Omega }\cap B$ and there exist $C>0$ and $\beta \in (0,1)$ such that

$$\begin{aligned} |h(y)|\le C{{\text {dist}}}(y,\partial \Omega )^\beta r^{-\beta }\sup _{\Omega \cap \frac{2}{3}B}|h|, \quad y\in \Omega \cap \frac{1}{2}B. \end{aligned}$$

Corollary 2

Let $\Omega _0\subset {\mathbb {R}}^n$ be a bounded Lipschitz domain. Let $u_\lambda$ be Laplace Dirichlet eigenfunction in $\Omega _0$. Then $u_\lambda$ extended by zero to $\partial \Omega _0$ is continuous on $\overline{\Omega }_0$.

Proof

We have $u_\lambda \in W^{1,2}_0(\Omega _0)\cap C^\infty (\Omega _0)$ and $\Delta u_\lambda +\lambda u_\lambda =0$ in $\Omega _0$. We consider the harmonic function $h(x,t)=e^{\sqrt{\lambda }t}u_\lambda (x)$ in $\Omega =\Omega _0\times {\mathbb {R}}$. We note that for any B centred on $\partial \Omega$, h satisfies the assumptions of Lemma 11. Then h is continuous in $\overline{\Omega }$ and vanishes on $\partial \Omega$. This implies that $u_\lambda \in C(\overline{\Omega }_0)$ and $u_\lambda =0$ on $\partial \Omega _0$. $\square$

1.3 Quantitative Cauchy uniqueness.

We give an elementary proof of Lemma 2 in this section for the convenience of the reader.

Let $G(x,y)=-c_d|x-y|^{2-d}$ be the fundamental solution of the Laplace equation in ${\mathbb {R}}^d$ when $d\ge 3$ (similar computations can be done with $G(x,y)=c_2\log |x-y|$ for $d=2$). We write $\partial B_+=\Gamma \cup \Sigma$, where $\Gamma$ is the flat part of the boundary and $\Sigma =\partial B_+{\setminus }\Gamma$. We denote by n the outer normal to $\partial B_+$. Then for $x\in B_+$, the Green formula implies

$$\begin{aligned} h(x)= & {} \int _{\partial B_+} \left[ \frac{\partial G}{\partial n}(x,y)h(y)-G(x,y)\frac{\partial h}{\partial n}(y)\right] dy\\= & {} \int _\Gamma \left[ \frac{\partial G}{\partial n}(x,y)h(y) -G(x,y)\frac{\partial h}{\partial n}(y)\right] dy\\&+\int _{\Sigma } \left[ \frac{\partial G}{\partial n}(x,y)h(y)-G(x,y)\frac{\partial h}{\partial n}(y)\right] dy\\= & {} h_1(x)+h_2(x). \end{aligned}$$

The functions $h_1$ and $h_2$ are defined in the complements of $\Gamma$ and $\Sigma$ respectively and are harmonic in the corresponding domains. Moreover, for $x\not \in \overline{B}_+$, applying the Green formula to the functions h and $G(x,\cdot )$ in $B_+$, we obtain $h_1(x)+h_2(x)=0$.

First, we estimate the value of $h_1$ at some point $x=(x',x'')\in B{\setminus }\Gamma \subset {\mathbb {R}}^{d-1}\times {\mathbb {R}}$. We divide the integral into two

$$\begin{aligned} h_1(x)=\int _{\Gamma } \frac{\partial G}{\partial n}(x,y)h(y)dy-\int _{\Gamma }G(x,y)\frac{\partial h}{\partial n}(y)dy=I_1(x)+I_2(x). \end{aligned}$$

Since $|\partial h/\partial n|<\varepsilon$ on $\Gamma$, the second integral is bounded by

$$\begin{aligned} |I_2(x)|\le c_d\varepsilon \int _{B^{d-1}(x',2)}|x'-y'|^{2-d}dy'\le C\varepsilon . \end{aligned}$$

To estimate the first term, we note that for $y\in \Gamma$,

$$\begin{aligned} \frac{\partial G}{\partial n}(x,y)=c_d'x''|x-y|^{-d}, \end{aligned}$$

and thereby

$$\begin{aligned} \int _{\Gamma } \left| \frac{\partial G}{\partial n}(x,y) \right| dy\le c_d'\int _{{\mathbb {R}}^{d-1}}\frac{|x''|}{(x''^2+|x'-y'|^2)^{d/2}}dy'=c''_d. \end{aligned}$$

Using that $|h(y)|<\varepsilon$ on $\Gamma$, we conclude that $|I_1(x)|<C\varepsilon$ in $B{\setminus }\Gamma$. Therefore $|h_1(x)|\le C\varepsilon$ in $B{\setminus }\Gamma$. Since $h_1(x)+h_2(x)=0$ when $x\in {\mathbb {R}}^d{\setminus } \overline{B}_+$, and $|h_1+h_2|=|h| \le 1$ in $B_+$, we obtain that $h_2(x)$ satisfies

$$\begin{aligned} |h_2(x)|<C\varepsilon \ \ {\text {in}}\ B_-=B{\setminus } \overline{B}_+\quad {\text {and}}\quad |h_2(x)|\le 1+C\varepsilon \ \ \text {in}\ B_+. \end{aligned}$$

Now we apply the three sphere inequality (6). We note that $h_2$ is harmonic in B. First we take $x=(0,-1/5)$ and $r=1/5$ and obtain

$$\begin{aligned} \sup _{B(0,1/10)}|h_2|\le \sup _{B(x,3/10)}|h_2|\le 2^d(\sup _{B(x,1/5)}|h_2|)^{1/2}(\sup _{B(x, 4/5)}|h_2|)^{1/2}\le C\varepsilon ^{1/2}. \end{aligned}$$

Next, we apply inequality (6) to the ball centred at the origin with $r=1/10$. We obtain

$$\begin{aligned} \sup _{B(0,3/20)}|h_2|\le C\varepsilon ^{1/4}. \end{aligned}$$

Iterating two more times, by applying the same inequality to the balls centred at the origin and $r=3/20$ and, finally, $r=9/40$, and noticing that $27/80>1/3$, while $9/10<1$, we conclude that

$$\begin{aligned} \sup _{\frac{1}{3}B}|h_2|\le C\varepsilon ^{1/16}. \end{aligned}$$

Finally, combining the last inequality with the bound $|h_1|\le C\varepsilon$ in $B_+$, we get the required estimate $|h|\le C\varepsilon ^\gamma$ in $\frac{1}{3}B_+$.