Maximal functions are naturally introduced when someone wants to study the pointwise convergence of functions. In particular, nontangential maximal functions are naturally used when we want to analyze a nontangential behavior of a function.

When I wrote a research paper [KK19], it was hard to find a proof of the following theorem.

Theorem 1. Let $\Omega$ be a bounded Lipschitz domain in $\mathbb{R}^n$, $n\geq 2$. Suppose that $u$ is harmonic in $\Omega$ and $u^* \in L^2 (\partial\Omega)$. Then there exists a function $g$ in $L^2(\partial\Omega)$ such that $u\rightarrow g$ nontangentially a.e. on $\partial\Omega$.

According to some literature, this theorem was shown by Dahlberg. Since I am not familiar with geometric measure theory, it was hard for me to follow the essence of the proof. The purpose of this post is to prove this well-known theorem by using layer potential technique. I want to point out that the proof is not original since the proof is well-known to experts. However, I will write this post for those who are not familiar with the concept `nontangential convergence’.

The organization of this post is as follows. We first present the role of nontangential maximal function to study the Dirichlet problem for Poisson equation in half-space. Next, we recall definitions of Lipschitz domains and introduce geometric objects that are useful when we study analysis on Lipschitz domains. Later, we define single and layer potentials and their mapping properties. Finally, we present a proof of the main theorem.

Dirichlet problem for the Poisson equation in half-space

In this section, we consider the Dirichlet problem for the Poisson equation in $\mathbb{R}^{n+1}_+$, where $\mathbb{R}^{n+1}_+ =\{ (x,y) : x\in \mathbb{R}^n, y>0 \}$. For $x\in \mathbb{R}^n$ and $y>0$, we define the Poisson kernel by \begin{equation}\nonumber P_y(x) = \frac{c_n y}{(|x|^2+y^2)^{(n+1)/2}},\quad c_n = \frac{\Gamma((n+1)/2)}{\pi^{(n+1)/2}}. \end{equation} The Poisson kernel has several interesting properties. One of important properties is that $\int_{\mathbb{R}^n} P_y(x)dx=1$ for any $y>0$. Hence by Young's convolution theorem, we see that \begin{equation} \Vert P_y *f \Vert_{L^p}\leq \Vert f \Vert_{L^p} \end{equation} for any $1\leq p \leq \infty$. For a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ and $y>0$, we define $u(x,y)=(P_y*f)(x)$ and we call it the *Poisson integral* of $f$. We write \begin{equation} \nonumber \Delta u = \sum_{j=1}^n \frac{\partial^2 u}{\partial x_j^2}+ \frac{\partial^2 u}{\partial y^2}. \end{equation} Then it is easy to check that for $f\in L^p(\mathbb{R}^n)$, $1\leq p\leq \infty$, then the Poisson integral is harmonic in $\mathbb{R}^{n+1}_+$. Since $\int_{\mathbb{R}^n} P_y (x)dx=1$ and $P_y(x)>0$, one can easily see that $(P_y *f)(x)\rightarrow f(x)$ as $y\rightarrow 0+$ if $f$ is sufficiently smooth. However, it is unclear if $f\in L^p$ because of the definition of $L^p$. To discuss the behavior of $u$ as $y\rightarrow 0+$, nontangential maximal function estimate gives an useful information.
Theorem 2. Let $1\leq p \leq \infty$ and $f\in L^p (\mathbb{R}^n)$, and let $\psi$ be a nonnegative, radial, decreasing and integrable function with $\int_{\mathbb{R}^n} \psi dx=1$.
  1. $\sup_{\varepsilon>0} |\psi_\varepsilon *f(x)|\leq Mf(x)$, where $M$ is the Hardy-Littlewood maximal operator and $\psi_\varepsilon(x)=\varepsilon^{-n} \psi(x/\varepsilon)$.
  2. $\lim_{y\rightarrow 0+} (\psi_\varepsilon *f)(x)=f(x)$ for almost every $x$.
  3. If $1\leq p <\infty$, then $(\psi_\varepsilon *f)\rightarrow f$ in $L^p$ as $\varepsilon\rightarrow 0+$.
Proof. (1) By scaling and translation invariance, it suffices to show that $$ |(\psi * f)(0)|\leq \left(\int_{\mathbb{R}^n} \psi dx \right)Mf(0).$$ Set $$ \lambda(r)= \int_{\mathbb{S}^{n-1}} f(r\omega) d\sigma(\omega),\quad \Lambda(r)=\int_0^r \lambda(t)t^{n-1}dt.$$ Since $\psi$ is integrable and decreasing radial function, it follows that $$ \int_{r\leq |x|\leq 2r} \psi(x) dx \geq cr^n \psi(r) $$ for some constant $c>0$. Hence $$ r^n \psi(r)\rightarrow 0$$ as $r\rightarrow 0+$ or $r\rightarrow \infty$. Therefore a change of variable and integration by part give $$ (\psi*f)(0)=\int_0^\infty \Lambda'(r)\psi(r) dr=-\int_0^\infty \Lambda(r)\psi'(r)dr. $$ (of course, we should change improper integral into finite integral...) Since $$ \Lambda(r)=\int_{B_r} f(y)dy \leq |B_r| Mf(0)= \frac{r^n \sigma(\mathbb{S}^{n-1})}{n} Mf(0), $$ it follows that $$ -\int_0^\infty \Lambda(r)\psi'(r)dr \leq -Mf(0)\sigma(\mathbb{S}^{n-1}) \int_0^\infty \frac{r^n}{n} \psi'(r)dr$$ since $\psi'(r)\leq 0$. Integration by part and a change of variable give $$\begin{aligned} -\sigma(\mathbb{S}^{n-1})\int_0^\infty \frac{r^n}{n} \psi'(r)dr &=\sigma(\mathbb{S}^{n-1})\int_0^\infty r^{n-1}\psi(r)dr\\ & = \int_{\mathbb{R}^n} \psi(x)dx. \end{aligned}$$ This implies that $$ (\psi*f)(0)\leq \left(\int_{\mathbb{R}^n} \psi(x) dx\right) Mf(0).$$ This completes the proof of (1). To show (2), suppose first that $1\leq p<\infty$. Define $$ \Omega f(x) = \limsup_{\eta\rightarrow 0+} (\psi_\eta * f)(x)-\liminf_{\eta\rightarrow 0+} (\psi_\eta * f)(x). $$ By (1), we have $\Omega(x)\leq 2Mf(x)$. If $p=1$, then for any $\eta>0$, it follows from the weak $L^1$-bound of maximal function that $$ |\{x : \Omega f(x)>\eta \}| \leq |\{ x : Mf(x)>\eta/2 \}|\leq C\eta^{-1}\Vert f \Vert_{L^1}. $$ If $1<p<\infty$, then for any $\eta>0$, it follows from the $L^p$-boundedness of maximal function and Chebyshev's inequality that \begin{align*} |\{ x : \Omega f(x) > \eta \}| &\leq |\{ x : Mf (x)>\eta/2 \}| \\ &\leq C \eta^{-p} \Vert Mf \Vert_{L^p}^p \\ &\leq C \eta^{-p} \Vert f \Vert_{L^p}^p. \end{align*} From these estimate, we prove (2). Let $\varepsilon>0$ be given. Then there exists $f_1 \in C_c^\infty$ such that $f=f_1+f_2$, where $\Vert f_2 \Vert_{L^p}<\varepsilon$. Since $f_1 \in C_c^\infty$, it follows that $f_1 *\varphi_\eta \rightarrow f_1$ as $\eta\rightarrow 0+$. Hence $\Omega f_1=0$. When $p=1$, there exists a constant $C>0$ such that $$|\{\Omega f >\alpha\}| \leq |\{\Omega f_2>\alpha\}| \leq \frac{C}{\alpha}\Vert{f_2}\Vert_{L^{1}}\leq \frac{C\eta}{\alpha} $$ for all $\alpha>0$. Similarly, when $1<p<\infty$, there exists a constant $C>0$ such that $$|\{\Omega f >\alpha\}| \leq |\{\Omega f_2>\alpha\}| \leq \frac{C}{\alpha^p}\Vert{f_2}\Vert_{L^{p}}^p\leq \frac{C\eta^p}{\alpha^p} $$ for all $\alpha>0$. Hence if we choose sufficiently small $\eta>0$, then we conclude that $$ |\{\Omega f >\alpha\}|=0 $$ for all $\alpha>0$, which proves that $|\{ \Omega f>0 \}|=0$. Hence $\Omega f=0$ a.e. on $\mathbb{R}^n$. This proves that $f*\varphi_\eta\rightarrow f$ a.e. as $\eta\rightarrow 0+$. If $p=\infty$, fix a ball $B$. We show that $f*\varphi_\eta\rightarrow f$ a.e. $x\in B$. Let $B_1$ be any other ball which strictly contains $B$, and let $\delta=\mathrm{dist}(B,B_1^c)$. Let $f_1=f\chi_{B_1}$ and $f_2=f-f_1$. Then $f_1 \in L^1$ and $\Omega f_1=0$ a.e. on $\mathbb{R}^d$ by the previous step. Note also that for $x\in B$, we have \begin{align*} |(f_2 *\varphi_\eta)(x)|&\leq \int_{|y|\geq \delta>0} | f_2(x-y)| |\varphi_\eta(y)|dy\\ &\leq \Vert f \Vert_{L^\infty} \int_{|y|\geq \delta/\eta} |\varphi(y)|dy. \end{align*} Since $\varphi \in L^1$, it follows from the absolute continuity that $|(f_2*\varphi_\eta)(x)|\rightarrow 0$ as $\eta\rightarrow 0$. Since $\Omega f\leq \Omega f_1+\Omega f_2$, it follows that $\Omega f=0$ a.e. on $x\in B$. This completes the proof of (2).</br> (3) follows from the standard mollifier technique which could be found in a standard real analysis textbook.

It is easy to see that $P_1 (x)$ satisfies the desired property of $\psi$ in the above theorem. Also, $P_y(x)=y^{-n}P_1(x/y)$. Hence we conclude that the Poisson integral of $f \in L^p$ satisfies the following Dirichlet problem in $\mathbb{R}^{n+1}_+$: $$ \Delta u=0\quad \text{in } \mathbb{R}^{n+1}_+,\quad u=f\quad \text{on } \mathbb{R}^{n}. $$ Here we interpret the boundary condition as $$ \lim_{y\rightarrow 0+} u(x,y)=f(x)\quad \text{a.e. on } x\in \mathbb{R}^n.$$

Lipschitz domains

In this section, we list several well-known facts regarding Lipschitz domains that will be used in the next section.

We say that a domain $\Omega$ is a Lipschitz domain if for each $q\in \partial\Omega$, there exist a rectangular coordinate system $(x,s)$, $x\in \mathbb{R}^{n-1}$, $s\in \mathbb{R}$, a neighborhood, $U_q \subset \mathbb{R}^n$ containing $q$ and a function $\varphi_q = \varphi : \mathbb{R}^{n-1}\rightarrow \mathbb{R}$ such that
  1. There exists a constant $C>0$ such that $|\varphi(x)-\varphi(y)|\leq C|x-y|$ for all $x,y \in \mathbb{R}^{n-1}$,
  2. $U\cap \Omega = \{ (x,s): x>\varphi(x)\}\cap U$.

The coordinate systems may always be taken to be a rotation and translation of the standard rectangular coordinates for $\mathbb{R}^n$.

By a cylinder $Z_r(x)$, we mean an open, right circular, doubly truncated cylinder centered at $x\in \mathbb{R}^n$ with radius equal to $r$. A coordinate cylinder $Z=Z_r(x)$, $x\partial\Omega$, will be defined by the following properties
  1. The bases of $Z$ have positive distance from $\partial\Omega$.
  2. There is a rectangular coordinate system for $\mathbb{R}^n$, $(x,s)$, $x\in \mathbb{R}^{n-1}$, $s\in \mathbb{R}$, with $s$-axis containing the axis of $Z$.
  3. There is an associated function $\varphi=\varphi_Z:\mathbb{R}^{n-1}\rightarrow \mathbb{R}$ that is Lipschitz, i.e., $|\varphi(x)-\varphi(y)|\leq C |x-y|$, $C=C_Z<\infty$ for all $x, y \in \mathbb{R}^{n-1}$.
  4. $Z\cap \Omega = Z\cap {(x,s) : s >\varphi(x)}$
  5. $Q=(0,\varphi(0))$
The pair $(Z,\varphi)$ is called a coordinate pair. For any positive number, $\nu$, $\nu Z_r(q)$ will denote the cylinder $\{x\in \mathbb{R}^n : q+(x-q)/\nu \in Z \}$, i.e., the dilation of $Z$ about $q$ by a factor $\nu$.

Since $\Omega$ is a bounded Lipschitz domain, it follows from the compactness of $\partial\Omega$ that we can cover the boundary by finitely many coordinate cylinders, say $Z_1$, $Z_2$, ..., $Z_N$. We may choose such cylinders by enlarging these for a later purpose. We also choose our coordinate functions $\varphi_j$ to have compact support in $\mathbb{R}^{n-1}$. Also, there exists a number $M>0$ such that $\max_{1\leq j\leq N} \Vert \nabla \varphi_j \Vert_{L^\infty} \leq M$. The smallest such number is called the Lipschitz constant for $\Omega$.

To discuss the nontangential behavior of functions, it is useful to consider cones associated to the domains since we are working on Lipschitz domains.

By a cone, it is an open, circular, doubly truncated cone with two non-empty, convex components. If $q\in \partial\Omega$, $\Gamma(q)$ will denote a cone with vertex at $q$ and one component in $\Omega$ and the other in $\mathbb{R}^n\setminus \overline{\Omega}$. The component interior to $\Omega$ will be denoted by $\Gamma_i(q)$ and the component exterior to $\overline{\Omega}$ is denoted by $\Gamma_e(q)$.

We say that a family of cones $\{\Gamma(q) : q\in \partial\Omega \}$ is regular if there is a finite covering of $\partial\Omega$ by coordinate cylinders as described above, such that for each
$(Z_r,\varphi)$, there are three cones $\alpha,\beta$, and $\gamma$, each with vertex at the origin and axis along the axis of $Z$ such that $$ \alpha \subset \overline{\beta}\setminus \{0\}\subset \gamma$$ and for all $(x,\varphi(x))=q\in (4/5)Z^*\cap \partial\Omega$, $$ \alpha + q \subset \Gamma(q)\subset \overline{\Gamma(q)}\setminus \{q\}\subset \beta +q, $$ $$(\gamma+q)_i \subset \Omega \cap Z^*,\quad \text{and}\quad (\gamma+q)_e \subset Z^*\cap \overline{\Omega}$$

Given a function $u$, we define $u^*$, $u^*_i$, $u^*_e$ by \begin{align*} u^*(z) &=N(u,\Gamma)(z)=\sup_{x\in \Gamma(z)}|u(x)|,\\ u^*_i(z) &=N(u,\Gamma_i)(z)=\sup_{x\in \Gamma_i(z)}|u(x)|,\\ u^*_e(z) &=N(u,\Gamma_e)(z)=\sup_{x\in \Gamma_e(z)}|u(x)| \end{align*} for $z\in \partial\Omega$. We call these functions as nontangential maximal functions.

Finally, we end this section by introducing well-known approximation scheme of Lipschitz domains.
Theorem 3. Let $\Omega$ be a bounded Lipschitz domain in $\mathbb{R}^n$, $n\geq 2$. Then the following hold:
  1. There is a regular family of cones $\{\Gamma\}$ for $\Omega$ as described in the above.
  2. There is a sequence of $C^\infty$ domains, $\Omega_j\subset \Omega$ and homeomorphisms, $\Lambda_j:\partial\Omega\rightarrow \partial\Omega_j$ such that $\sup_{z\in \partial\Omega} | z-\Lambda_j(z)|\rightarrow 0$ as $j\rightarrow \infty$ and for all $j$ and all $z\in \partial\Omega$, $\Lambda_j(z)\in \Gamma_i(z)$,
  3. There is a covering of $\partial\Omega$ by coordinate cylinders, $Z$, so that given a coordinate pair $(Z,\varphi)$, then $Z^*\cap\partial\Omega_j$ is given for each $j$ as the graph of a $C^\infty$ function $\varphi_j$ such that $\varphi_j\rightarrow \varphi$ uniformly, $\Vert \nabla \varphi_j \Vert_{L^\infty} \leq \Vert \nabla \varphi \Vert_{L^\infty}$, and $\nabla \varphi_j\rightarrow \nabla \varphi$ pointwise a.e. and in every $L^q(Z^*\cap \mathbb{R}^{n-1})$, $1\leq q<\infty$.
  4. There are positive functions $\omega_j : \partial\Omega \rightarrow \mathbb{R}_+$ bounded away from zero and infinity uniformly in $j$ such that for any measurable set $E\subset \partial\Omega$, $\int_E \omega_j d\sigma=\int_{\Lambda_j(E)} d\sigma_j$, and so that $\omega_j\rightarrow 1$ pointwise a.e. and in every $L^q(\partial\Omega)$, $1\leq q<\infty$.
  5. The normal vectors to $\Omega_j$, $N(\Lambda_j(z))$ converges pointwise a.e. and in every $L^q(\partial\Omega)$, $1\leq q<\infty$, to $N(z)$. An analogous statement holds for locally defined tangent vectors.
  6. There exist $C^\infty$ vector fields $\mathbf{h}$, in $\mathbb{R}^n$ such that for all $j$ and $z\in \partial\Omega$, $\left<\mathbf{h}(\Lambda_j(z)),N(\Lambda_j(z)) \right>\geq C>0$, where $C$ depends only on $\mathbf{h}$ and the Lipschitz constant for $\Omega$.
Proof of the above approximation scheme can be found in the Ph.D. dissertation of Verchota.
# Single and double layer potentials
In this section, we define single and double layer potentials that are crucial to discuss $L^p$-solvability for Dirichlet and Neumann problems for the Poisson equation on Lipschitz domains. Let $\Gamma$ be the fundamental solution of the Laplacian defined by \begin{equation} \Gamma(x)=\begin{cases} -\frac{1}{2\pi} \log |x|,&\quad (n=2),\\ \frac{1}{n(n-2)\omega_n} \frac{1}{|x|^{n-2}},&\quad (n\geq 3), \end{cases} \end{equation} where $\omega_n$ is the Euclidean volume of the unit ball in $\mathbb{R}^n$. Recall the Green formula that $$ \int_\Omega u\Delta v-v\Delta u dy =\int_{\partial\Omega} u \frac{\partial v}{\partial \nu}-v\frac{\partial u}{\partial \nu} d\sigma,$$ where $\nu$ is the unit normal to the boundary $\partial\Omega$, $d\sigma$ is the surface measure on $\partial\Omega$ If we assume that $u$ is harmonic in $\Omega$ and $v(y)=\Gamma(x-y)$, then by the Green formula on $\Omega\setminus B_\varepsilon(x)$, we have $$ \int_{\Omega\setminus B_\varepsilon(x)} u(y)\Delta \Gamma(x-y)dy=\int_{\partial(\Omega\setminus B_{\varepsilon(x)})} u(y) \frac{\partial}{\partial \nu_y}\Gamma(x-y) - \Gamma(x-y)\frac{\partial u}{\partial \nu_y}(y)d\sigma.$$ Letting $\varepsilon \rightarrow 0$, we get \begin{equation}\label{eq:green-identity-potential} -u(x)=\int_{\partial\Omega} u(y)\frac{\partial }{\partial \nu_y}\Gamma(x-y)-\Gamma(x-y) \frac{\partial u}{\partial \nu_y}(y)d\sigma(y). \end{equation} This identity motivates us to define the following operators $$ \mathcal{S}f(x)=\int_{\partial\Omega} \Gamma(x-y)f(y)d\sigma(y),\quad x\in \Omega$$ and $$ \mathcal{D}f(x)=\int_{\partial\Omega} \frac{\partial}{\partial \nu_y}\Gamma(x-y)f(y)d\sigma(y),\quad x\in \Omega.$$ We call $\mathcal{S}f$ the single layer potential of $f$ and $\mathcal{D}f$ the double layer potential of $f$. Observe that $$ \frac{\partial}{\partial \nu_y}\Gamma(x-y) =\frac{(x-y)\cdot \nu(y)}{n\omega_n |x-y|^n}. $$ Then the double layer potential operator is given by $$ \mathcal{D}f(x)=\frac{1}{n\omega_n}\int_{\partial\Omega} \frac{(x-y)\cdot \nu(y)}{|x-y|^n} f(y)d\sigma(y). $$ To investigate the boundary behavior of the double layer potential operators, for each $\varepsilon>0$, we define $$ K_\varepsilon f(x) =\frac{1}{n\omega_n} \int_{y\in \partial\Omega, |x-y|>\varepsilon} \frac{(x-y)\cdot \nu(y)}{|x-y|^n} f(y)d\sigma(y) $$ for $x\in \partial\Omega$.
The following theorem was first shown by Calderon and later by Coifman-Meyer-McIntosh.
Theorem 4. Suppose that $\varphi:\mathbb{R}^{n-1}\rightarrow \mathbb{R}$ is Lipschitz with compact support. For $x,y\in \mathbb{R}^{n-1}$, $x\neq y$, put $$ k_j(x,y)= \frac{x_j-y_j}{(|x-y|^2+|\varphi(x)-\varphi(y)|^2)^{n/2}},\quad j=1,2,\dots,n-1 $$ and $$ k_n(x,y)=\frac{\varphi(x)-\varphi(y)}{(|x-y|^2 +|\varphi(x)-\varphi(y)|^2)^{n/2}}.$$ Letting $k(x,y)$ stand for any $k_j$, $1\leq j,\leq n$, define $$T_\varepsilon f(x)=\int_{|x-y|>\varepsilon} k(x,y)f(y)dy,\quad \varepsilon>0.$$ Then $$ T_*f(x)=\sup_{\varepsilon>0}|T_\varepsilon f(x)|$$ is a bounded operator on $L^p(\mathbb{R}^{n-1})$, $1<p<\infty$. Moreover, we have $$ \Vert T_* f \Vert_{L^p} \leq C \Vert f \Vert_{L^p}, $$ where the constant $C$ depends only on $n$, $p$, and $\Vert \nabla \varphi \Vert$.
As an application of the above theorem, we present the $L^p$-boundedness of truncated boundary double layer potential operator.
Theorem 5. Let $1<p<\infty$. Then there exists a constant $C>0$ depending only on $n$, $p$, and the Lipschitz character of $\Omega$ such that $$ \left\Vert {\sup_{\varepsilon>0} |K_\varepsilon f|} \right\Vert_{L^p(\partial\Omega)}\leq C \Vert f \Vert_{L^p(\partial\Omega)}$$ for all $f\in C(\partial\Omega)$.
Proof. Let $(Z_i,\varphi_i)$ be a coordinate pair associated to the Lipschitz domain $\Omega$ in $\mathbb{R}^n$, $n\geq 2$. Choose a partition of unity $\{\zeta_i\}_{i=1}^m$ so that $0\leq \zeta_i\leq 1$, smooth, finite partition of unity subordinate to the $Z_i$'s. Clearly, \[ K_\varepsilon f(x)=\frac{1}{n\omega_n}\sum_{i=1}^m \int_{y \in \partial\Omega,|x-y|>\varepsilon} \frac{(x-y)\cdot \nu(y)}{|x-y|^n}\zeta_i(y)f(y)d\sigma(y). \] Since $(Z_i,\varphi_i)$ is a coordinate pair, it follows that \begin{align*} &K_{\varepsilon}f(x)\\ &=\frac{1}{n\omega_n} \sum_{i=1}^m \int_{|x'-y'|^2+|\varphi_i(x')-\varphi_i(y')|^2 >\varepsilon^2} \frac{\varphi_i(x')-\varphi_i(y')-\nabla \varphi_i(y')\cdot (x'-y')}{|x-y|^n} \zeta_i(y',\varphi_i(y'))f(y',\varphi_i(y'))\sqrt{1+|\nabla \varphi_i(y')|^2}dy'. \end{align*} Hence if we define $$ K_i(x',y')= \frac{\varphi_i(x')-\varphi_i(y')-\nabla \varphi_i(y')\cdot (x'-y')}{(|x'-y'|^2+|\varphi_i(x')-\varphi_i(y')|^2)^{n/2}},$$ then $$ K_\varepsilon f(x)=\frac{1}{n\omega_n} \sum_{i=1}^n \int_{U_\varepsilon^i(x')} K_i(x',y')g(y')dy',$$ where $$ g_i(y')=\zeta_i(y',\varphi_i(y'))f(y',\varphi_i(y'))\sqrt{1+|\nabla \varphi_i(y')|^2} $$ and $$ U_{\varepsilon}^i(x')=\{y' : |x'-y'|^2+|\varphi_i(x')-\varphi_i(y')|^2>\varepsilon^2\}. $$ Note that $y'\in U_{\varepsilon}^i(x)$ implies $|x'-y'|>\varepsilon /\sqrt{1+\Vert{\nabla \varphi_i}\Vert_{L^\infty}^2}$. Furthermore, we define $$ V_\varepsilon^i(x')=\{ y' : |x'-y'|>\varepsilon/\sqrt{1+\Vert \nabla \varphi_i\Vert_{L^\infty}^2}.$$ Then it is easy to check that $U_\varepsilon^i(x')\subset V_{\varepsilon}^i(x')$. Then for $\varepsilon>0$, we have \begin{align*} &\left|\int_{U_{\varepsilon}^i(x')}K_i(x',y') g(y')dy'\right|\\ & \leq \left|\int_{V_{\varepsilon}^i(x')} K_i(x',y')g(y')dy' \right| + \left|\int_{V_{\varepsilon}^i(x')\setminus U_\varepsilon^i(x')} K_i(x',y')g(y')dy' \right| . \end{align*} It is easy to check that there exists a constant $C$ depending only on $n$ and the Lipschitz character of $\Omega$ such that $$ \left|\int_{V_{\varepsilon}^i(x')\setminus U_\varepsilon^i(x')} K_i(x',y')g(y')dy' \right|\leq C Mg(x') $$ for all $\varepsilon>0$. Hence it follows from the Coifman-Meyer-McIntosh theorem that $$ \Vert \sup_{\varepsilon>0}|K_\varepsilon f| \Vert_{L^p(\partial\Omega)} \leq \sum_{i=1}^m \left(\Vert{T_* g_i}\Vert_{L^p}+C\Vert Mg_i \Vert_{L^p}\right) \leq C \sum_{i=1}^m \Vert g_i \Vert_{L^p}\leq C \Vert f \Vert_{L^p(\partial\Omega)}. $$ This completes the proof.

By the above theorem, we can conclude that $ \lim_{\varepsilon \rightarrow 0+ }K_\varepsilon f$ exists a.e. and in $L^p(\partial\Omega)$. We define $$ Kf(x)=\lim_{\varepsilon\rightarrow 0+} K_\varepsilon f(x).$$ Moreover, we can show the following result.
Theorem 6. Let $1<p < \infty$. Then $$ \lim_{x\rightarrow x_0; x\in \Gamma_i(x_0)} \mathcal{D}f(x) = \left(\frac{1}{2}I+K \right)f(x_0), $$ where $\Gamma_i(x_0)$ denotes the inner regular cone with vertex at $x_0 \in \partial\Omega$.

Nontangential behavior of harmonic function

Now we are ready to present the main theorem of this post.
Theorem 7. Suppose that $u$ is harmonic in $\Omega$ and $u^*\in L^2(\partial\Omega)$. Then there exists $f\in L^2(\partial\Omega)$ such that $$ \lim_{z\in \gamma(x), z\rightarrow x} u(x)=f(x)\quad \text{a.e. } x\in \partial\Omega.$$
Choose a sequence of smooth domains $\{\Omega_j\}$ as in Theorem 3 and let $f_j= u|_{\partial\Omega_j}$. Since $u^*\in L^2(\partial\Omega)$, it follows that $\Vert f_j \Vert_{L^2(\partial\Omega)}\leq C<\infty$. Note also that the identity $$ u= \mathcal{D}_j\left(\left(\frac{1}{2}I+K_j\right)^{-1}f_j \right)\quad \text{on } \Omega_j $$ holds, where $\mathcal{D}_j$ and $K_j$ are the double layer potential and the boundary integral operator associated to $\Omega_j$, respectively. Moreover, we have $$ \left\Vert \left(\frac{1}{2}I+K_j\right)^{-1}f_j \right\Vert_{L^2(\partial\Omega_j)}\leq C \Vert f_j \Vert_{L^2(\partial\Omega_j)}, $$ where the constant $C$ does not depend on $j$. If we define $$ g_j = \left(\frac{1}{2}I+K_j\right)^{-1}f_j\circ \Lambda_j,$$ then $\{g_j\}$ is bounded in $L^2(\partial\Omega)$. Hence by the weak compactness result, there exists a function $g\in L^2(\partial\Omega)$ such that $g_j\rightarrow g$ weakly in $L^2(\partial\Omega)$. Moreover, it follows that $\mathcal{D}g_j\rightarrow u$ in $\Omega$. Therefore, $$ \mathcal{D}g=u\quad \text{on } \Omega.$$ Hence, we conclude that $$\lim_{x\rightarrow x_0;x\in \gamma(x_0)} u(x)=\lim_{x\rightarrow x_0;x\in \gamma(x_0)} \mathcal{D} g(x)=\left(\frac{1}{2}I+K \right)g(x),$$ which proves the desired result since $\left(\frac{1}{2}I+K \right)g \in L^2(\partial\Omega)$.

Remark. There is another way to prove this theorem without using layer potential technique. Let $\sigma$ be the surface measure on $\partial\Omega$. By the Perron method, given $f\in C(\partial\Omega)$, we can construct a unique harmonic function $u$ in $\Omega$ satisfying $u=f$ on $\partial\Omega$. They by the maximum principle and the Riesz representation theorem, for each $x\in \Omega$, there exists a harmonic measure $\omega^x$ on $\partial\Omega$ satisfying $$ u(x)=\int_{\partial\Omega} f(y) d\omega^x(y).$$ It was shown by Dahlberg that the surface measure and harmonic measure are mutually absolutely continuous. Moreover, if $d\omega^x =h(x)d\sigma$, then $h$ satisfies $$ \left(\frac{1}{\sigma(B_\delta(x))} \int_{B_\delta(x)} h^2 d\sigma \right)^{1/2}\leq \frac{C}{\sigma(B_\delta(x))} \int_{B_\delta(x)} h d\sigma.$$ From this, we conclude that $\Vert u^*\Vert_{L^2(\partial\Omega)} \leq C \Vert f \Vert_{L^2(\partial\Omega)}$. Since it requires a delicate geometric measure theoratic argument, we omit the details.


  1. B. Dahlberg, Estimates of harmonic measure, Arch. Rational Mech. Anal. 65 (1977), 275-288.
  2. M. Mitrea and M. Taylor, Boundary layer methods for Lipschitz domains in Riemannian manifolds, J. Funct. Anal. 163 (1999), 181-251.
  3. E. M. Stein, Singular integrals and differentiability properties of functions, Princeton University Press, 1970.
  4. G. Verchota, Layer potentials and Regularity for the Dirichlet problem for Laplace's equation in Lipschitz domains, J. Funct. Anal. 59 (1984), 572-611.
  5. </ol>