Example 3.1.318C85C3D68394

Example 3.1.31.8C85C3D68394 Let $d\geq 2$. The punctured ball $\Omega =B(0,1)\setminus \{ 0\} $ is a bounded domain with boundary $S^{d-1}\cup \{ 0\} $, and the origin is an irregular boundary point, that is, a boundary point that is not regular. The punctured ball is connected, because it is the image of $(0,1)\times S^{d-1}$ under the continuous map $(t,\omega )\mapsto t\omega $ and the sphere $S^{d-1}$ is path-connected when $d\geq 2$. Indeed, two points $\omega ,\omega '$ with $\omega '\neq -\omega $ are joined by the path $s\mapsto ((1-s)\omega +s\omega ')/|(1-s)\omega +s\omega '|$, $0\leq s\leq 1$, whose denominator does not vanish, and two antipodal points can be joined through a unit vector orthogonal to them, which exists because $d\geq 2$.

To see that the origin is irregular, let $f$ equal one at the origin and zero on $S^{d-1}$; it is continuous on $\partial \Omega $ because the origin is an isolated point of $\partial \Omega $. Let $g(x)=\Gamma _ d(x)-\Gamma _ d(e_1)$, that is, $g(x)=(2\pi )^{-1}\log (1/|x|)$ if $d=2$ and $g(x)=((d-2)s_{d-1})^{-1}(|x|^{2-d}-1)$ if $d\geq 3$. This function is harmonic on $\mathbb R^ d\setminus \{ 0\} $ by proposition 3.1.13, it vanishes on $S^{d-1}$, and it tends to $+\infty $ at the origin. Let $v\in \mathcal S_ f$ and $\varepsilon >0$; recall that $v\leq \max _{\partial \Omega }f=1$. For every sufficiently small $\delta \in (0,1)$ we have $\varepsilon g\geq 1$ on the sphere $|x|=\delta $. On the annulus $D=\{ \delta <|x|<1\} $, the function $v-\varepsilon g$ is a continuous subsolution; it is continuous and nonpositive on the inner sphere $|x|=\delta $, and its upper limit at every point of $S^{d-1}$ is at most $f=0$, because $g$ is continuous with value $0$ there. By lemma 3.1.24, $v\leq \varepsilon g$ on $D$. Letting $\delta \downarrow 0$ and then $\varepsilon \downarrow 0$ gives $v\leq 0$ on $\Omega $. Since $0\in \mathcal S_ f$, the Perron envelope $U_ f$ vanishes identically, so it does not tend to $f(0)=1$ at the origin.

In dimension one the situation is different: the bounded open set $(-1,0)\cup (0,1)$ has boundary $\{ -1,0,1\} $. If $f$ is a function on this boundary, the function which is affine on each of the two intervals with the boundary values given by $f$ is harmonic and extends continuously to $[-1,1]$ with boundary values $f$, so it equals $U_ f$ by theorem 3.1.25. Hence the origin is a regular boundary point of $(-1,0)\cup (0,1)$.