Let $D$ be a bounded domain and let $h$ be a continuous function on $\partial D$.
We consider the Laplace equation on $D$ with boundary values given by $h$. Let $\tilde h$ be the Perron solution $\tilde h = \sup \mathcal F_h$, where $\mathcal F_h$ is the family of subharmonic functions $f$ on $D$ with $\limsup f(\zeta) \leq h(z)$ as $\zeta \to z$.

Fix $z_0 \in D$. For every $\varepsilon > 0$, define $D_{\varepsilon} = D \backslash \{ z \in \mathbb{C} : |z - z_0| \leq \varepsilon \}$. Define $h_{\varepsilon}$ on $\partial D_{\varepsilon}$ by $h_{\varepsilon} = h$ on $\partial D$ and $h_{\varepsilon} = 0$ on $\{ z \in \mathbb{C} : |z - z_0| = \varepsilon \}$.


Show that $\tilde h_{\varepsilon} \to \tilde h$ uniformly as $\varepsilon \to 0$ on any subset of $D$ that has a positive distance from $z_0$.


Say $X$ is a subset of $D$ that has positive distance from $z_0$. Then there is some $\varepsilon$ such that $X \cap \{ z \in \mathbb{C} : |z - z_0| \leq \varepsilon \} = \varnothing$. Thus one may equivalently prove that there is a uniform convergence on every subset of the form $D_\frac 1n$.

Let $\mathcal{F}_h$ be the Perron family of subsolutions of $h$ and $\mathcal{F}_{h_{\varepsilon}}$ be the Perron family of subsolutions of $h_{\varepsilon}$. On $X$, we have that $\mathcal{F}_{h_{\varepsilon}} \subseteq \mathcal{F}_h$, but this may not yet say $\tilde{h_{\varepsilon}} \leq \tilde{h}$, because $\tilde h = \sup \mathcal F^{D}_h \geq \sup \mathcal F^{X}_h$ and it is not clear that the inequality is an equality.

This is where I've stopped. I'm unsure if this method is correct, or if there is some theorem I should be using. Any hints or suggestions are welcomed!

User Daniel Fischer noted (please correct me if I got it wrong) that for every $\eta>0$ there a set $D_{\frac 1n}$ and a function $\lambda$ defined on $D\setminus {z_0}$ such that $\tilde h+\lambda \in \mathcal F_{h_\epsilon}$ and $\lambda > -\eta$ on $D_{\frac 1n}$, so $\tilde h_\varepsilon \geq \tilde h-\eta$ on this $D_{\frac 1n}$.

  • $\begingroup$ I'm sorry, the $\tilde{h}_{\varepsilon} \leqslant \tilde{h}$ part of my answer is wrong. I'll delete my answer later today unless I have an idea how to fix it before. $\endgroup$ Commented Aug 15, 2017 at 13:12


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