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The following is stated as an exercise in the "Classical Potential Theory" of Armitage and Gardiner (pg 195). Let $K$ be a compact of $\mathbb{R}^m$ ($m\geq2$) and $\Omega$ an open set containing $K$. Suppose each bounded component of $\mathbb{R}^m\setminus K$ contains a point of $\mathbb{R}^m\setminus \Omega$. If $u$ is superharmonic on a neighborhood of $K$, then $u$ can be extended to a superharmonic function $\overline{u}$ whose restriction to $K$ equals $u$. I have two questions:

  1. Do you have any idea of the proof?

  2. Suppose $K$ is the boundary of a bounded open set $V$ and let $z\in V$. Let $\Omega$ be a neighborhood of the closure of $V$ from which we remove the point $z$. By the above result, if $u$ is a superharmonic function on a neighborhood of $K$, then $\overline{u}$ is superharmonic on $\Omega$. My question is: can we arrange that $$-\infty<\liminf \overline{u}(x),$$ as $x\to z$ from inside $\Omega$?

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  • $\begingroup$ It seems that the answer to the second question is no. In fact, we can take a nonconstant superharmonic function $u$ on a neighborhood of $\overline{R}^m\setminus V$ (one point compactification). Then we extend it to $\overline{u}$, superharmonic everywhere in $\overline{R}^m$ outside $z$. Finally, if this later function is bounded below at a neighborhood of $z$, it can be extended to a superharmonic function on the whole of $\overline{R}^m$ without being constant, by removable singularity theorem. Impossible. $\endgroup$
    – M. Rahmat
    Commented Jun 27, 2021 at 17:07
  • $\begingroup$ Still I have no answers to 1). Any hint would be appreciated. $\endgroup$
    – M. Rahmat
    Commented Jun 27, 2021 at 17:08

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