# Perron solution on punctured domain equals Perron solution on the whole domain?

## Definitions

Let $D \subset \Bbb C$ be a bounded domain, and let $h$ be a continuous function on $\partial D$.

We remove one point from the domain to obtain $D^*=D \setminus \{z_0\}$, and extend $h$ to a function $h^*$ on $\partial D^*$ by choosing some value $h^*(z_0)$.

The Perron solution on $D$ is the harmonic function defined by $\tilde h = \sup \mathcal{F}_h$, where $$\mathcal{F}_h = \{ u \in \mathcal{SH}(D) : \limsup_{z \to \zeta} u(z) \leq h(\zeta) \},$$

and we similarly define the Perron solution for $h^*$.

## Question

In a past post, it is proved that $\tilde h = \tilde h_0$ when we remove the center of a disk. Does it also hold true when removing a point from an arbitrary domain? In other words,

I want to know whether the Perron solution of the Dirichlet problem on $D^*$ is always equal to the Perron solution on $D$, regardless of the value of $h(z_0)$.

The insight of the proof in that linked post was to provide a subharmonic function $\lambda$ that vanishes on $\partial D$ but is equal to $-\infty$ on $z_0$. I don't know if such a function always exists, when the domain is arbitrary. Is the result true nevertheless?

## Added: The proof of Daniel Fischer for a disk

Copied here for ease of reference and to show where it seems not to work.

Assume $D=\{ |z-z_0| < r \}$. Letting $\lambda(z) = \ln \frac{|z-z_0|}{r}$, we will need the following three properties:

1. $\lambda$ is subharmonic on $D$
2. $\lambda(z) \to 0$ on $\partial D$
3. $\lambda(z_0) = -\infty$

First direction. Let $c>0$. The function $\tilde h+c\lambda \in \mathcal F_{h^*}$ (subharmonic, agrees with $h$ on all limits on $\partial D$, is $\infty$ on $z_0$). Letting $c \to 0$ gives $\tilde h \leq \tilde h^*$.

Second direction.

Let $c>0$. By continuity of $\lambda$ at $z_0$ there is a neighborhood of $z_0$ where $\widetilde {h^*} \leq \tilde h-c\lambda$.

Since $D$ is a soluble domain for the Dirichlet problem, we have on $\partial D$ that $\limsup \widetilde {h^*} \leq h^*=h=\tilde h \leq \tilde h-c\lambda$, so $\tilde h-c\lambda$ is a superharmonic function that has boundary limits greater than the harmonic function $\widetilde{h^*}$ on all $\partial D^*$. By the minimum principle, we then have $\tilde h-c\lambda \geq \tilde h^*$. Tending again $c \to 0$ finishes the proof.

• You want to know if it is possible to adapt DanielFischer's solution for $D = \{ 0 < |z| < 1\}$ but you don't copy and explain this solution, not even the notations. – reuns Aug 29 '17 at 17:18
• I see, good idea. Edited this in. – Pachirisu Aug 29 '17 at 19:47

Note that property 2 is not necessary. Instead you can have $\limsup\lambda\leq0$ at the boundary. This generalizes the proof to arbitrary bounded domain.
• Hi, thanks for answering. However, I suspect the proof for the disk uses the solvability of the Dirichlet problem, otherwise the inequality $\tilde h \leq \tilde h -c\lambda$ says little about $\widetilde {h^*}$. I edited the presentation of the proof in the post to explain what I mean by that. – Pachirisu Sep 9 '17 at 22:51