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.

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 $\frac{1}{3}B_+$.