When you start studying certain free boundary problems you soon appreciate the importance of understanding the boundary behavior of harmonic functions, especially harmonic functions which vanish on a portion of the boundary. Motivated by questions in the Hele-Shaw problem and related free boundary problems, I am going to review several important theorems proved by Dahlberg in the late 1970’s.

Let $D\subset \mathbb{R}^n$ be a Lipschitz domain. For such a domain it is well known that for any continuous function $f:\partial D\to\mathbb{R}$ there is a unique function $U_f$ which is continuous in $\overline D$ and twice differentiable in $D$ solving

\[\left \{ \begin{array}{rl}\Delta U_f = 0 & \text{ in } D \\ U_f = f & \text{ on } \partial D\end{array} \right.\]This defines an operator $f \mapsto U_f$ which is linear, it is also monotone in the sense that if $f\leq g$ in $\partial D$ then $U_f \leq U_g$ in $D$. In particular, for every $X \in D$ the linear functional

\[f \mapsto U_f(X)\]is positive. By the representation theorem for each $X \in D$ there is a Borel measure $\omega_X$ in $\partial D$ such that

\[U_f(X) = \int_{\partial D}f(y)d\omega_X(y).\]The measure $\omega_X$ is called the harmonic measure associated with the domain $D$ at the point $X$. When $D$ is a Lipshitz domain, the measure $\omega_X$ is absolutely continuous with respect to the surface measure of $\partial D$, so

\[d\omega_X(y) = k(X,y)\;d\sigma(y)\]The function $K$ is called the Poisson kernel of the domain $D$. The best known instances of the Poisson kernel are those where $D$ is a ball or a half space. For the unit ball in $\mathbb{R}^n$ the kernel takes the form

\[k(X,y) = \frac{1}{|\partial B_1|}\frac{1-|X|^2}{|X-y|^n}\]Dahlberg showed that for a general Lipschitz domain the Poisson kernel for each fixed $X$ yields a function on the boundary which satisfies a reverse Harnack inequality.

** Theorem. **

There is a constant $C$ such that given $X\in D$, $y_0 \in \partial D$, and $r\in(0,1)$ we have
$$ \left ( \frac{1}{\sigma(\Delta)}\int_{\Delta} k(X,y)^2\;d\sigma(y) \right )^{\frac{1}{2}} \leq \frac{C}{\sigma(\Delta)}\int_\Delta k(X,y)\;d\sigma(y),\;\;\Delta := B_r(y_0) \cap \partial D.$$

There is the Boundary Harnack Principle. It says that if $U$ is a positive harmonic function in $D$ which vanishes on a portion of $\partial D$, then the value of $U$ at a point properly place away from said portion will control the values of $U$ everywhere.

Before we state the principle, let us state two preliminary facts about harmonic functions.

$\bullet$ (De Giorgi’s oscillation lemma) Fix $\delta \in(0,1)$. Let $U$ be a subharmonic function $U$ in some ball $B_{2r}(X_0)$, and suppose that $U \leq 1$ in $B_{2r}(X_0)$ and

\[|\{ U\leq 0 \} \cap B_{2r}(X_0)| \geq \delta |B_r|\]then \(U\leq \mu\) in $B_r(X_0)$, where $\mu = \mu(\delta) \in (0,1)$.

$\bullet$ (Harnack inequality) If $P \in D$ is such that $d(P,\partial D) = \rho$, then

\[U(P) \leq C \left ( \frac{r}{\rho}\right )^\alpha U(X_1)\]
** Theorem. **

Let $U$ be a positive harmonic function in $B_r(X_0) \cap D$ such that $U \equiv 0$ on $B_r(X_0) \cap \partial D$ and let $X_1$ be a point in $B_{r/2}$ such that $B_{r/4}(X_1) \subset B_R(X_0) \cap D$. Then
$$U(X) \leq CU(X_1) \text{ for any } X \in B_{r/2}(X_0) \cap D$$
where the constant $C$ is determined by $D$.

**Proof.**

For every $X$, we shall denote by \(X^*\) a point on \(\partial D\) such that \(d(X,\partial D) = d(X^*,\partial D)\).

The proof hinges on a conflict between two effects:

- The positivity and harmonicity of $U$ force it to grow, meaning if $U$ reaches a certain value $h$ in a ball $B_r$ then in $B_{2r}$ it must reach at least a value above $\kappa h$, where $\kappa>1$ is a dimensional constant.
- The Harnack inequality prevents $U(X)$ from being much larger than $U(X_1)$ if the distance from $X$ to $X_1$ is comparable to the distance from $X$ to $\partial D$.

Let $P_0$ be the point where $U$ is the largest in $B_{r/2} \cap D$. Using the two effects above we are going to show that if the value $U(P_0)$ is much larger than $U(X_1)$, then there is a sequence of points $P_k$ all lying in $B_r \cap D$ with the property $U$ takes arbitrarily large values along this sequence.

The sequence \(\{P_k\}_k\) is built recursively, starting from $P_0$. Having determined the sequence up to some $P_k$, we determine the next element $P_{k+1}$. Let us write

\[\rho_k := d(P_k,\partial D)\]The oscillation lemma says that

\[\sup \limits_{B_{\rho_k}(P_k)} U \leq \theta \sup \limits_{B_{2\rho_k}(P_k)} U\]Applying the oscillation lemma in $B_{2\rho_k}(P_k^*)$ we obtain a new point $P_{k+1} \in D$ such that

\[|P_{k+1}-P_k| \leq 2\rho_k,\;\; U(P_{k+1}) \geq \theta^{-1} U(P_k)\]This can be done as long as … so the sequence is either finite and ends at some $k=k_0$ or it goes on indefinitely.

In either case, the second inequality applied recursively means that for every $P_k$ in the sequence we have

\[U(P_k) \geq \theta^{-k} U(P_0)\]We now use the Harnack inequality. From the H

Note there is a ball of radius $r+\rho_k$ ….containing $X_1$ and $P_k$… and which is contained in $D$, then, thanks to Harnack’s inequality

\[U(P_k) \leq C \left (\frac{r}{\rho_k} \right )^\alpha U(X_1)\]In other words,

\[\rho_k \leq C r\left (\frac{U(X_1)}{U(P_k)}\right )^{1/\alpha}\] \[\rho_k \leq C r \left (\frac{U(X_1)}{U(P_0)}\right )^{1/\alpha}(\theta^{1/\alpha})^k\]In summary, the sequence ${P_k}_k$ has the following properties

\[U(P_k) \geq \theta^{-k}U(P_0)\] \[|P_k-P_{k+1}| \leq Cr \left (\frac{U(X_1)}{U(P_0)} \right )^{1/\alpha} (\theta^{1/\alpha} )^{k}\]Observe that

\[\sum \limits_{k=0}^\infty |P_k-P_{k+1}| \leq Cr\left (\frac{U(X_1)}{U(p_0)} \right )^{1/\alpha} \frac{1}{1-\theta^{1/\alpha}}\]∎

Let $\phi:\mathbb{R}^n \to \mathbb{R}$ be a bounded, Lipschitz function,

\[\|\phi\|_\infty \leq C,\; \|\nabla \phi\|_\infty \leq \delta\]and consider the domain

\[D(\phi) = \{ X = (x,x_{n+1})\in \mathbb{R}^{n+1} \mid x_{n+1} > \phi(x) \}\]**Proof.**

Let $V$ be superharmonic and such that $CV(y) \geq k(P,y)^q$ for $y\in \partial D$. Then

\[CV(P) \geq \int_{\partial D}k(P,y)^{1+q}\;d\sigma(y)\]