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This question is a simplified version of the one in the MO post Superharmonic extension.

Suppose $K$ is a compact of $\mathbb{R}^m$ ($m\geq2$), and $U(x)=\log\frac{1}{|x|}$ if $m=2$, and $=|x|^{2-m}$ if $m>2$. We know that this function is harmonic everywhere outside the origin, and subharmonic at infinity (see the aforementioned post for the definition).

Question. Can we extend $U$ to some function $\overline{U}$ that is superharmonic everywhere outside some point $y_0\not=\infty$ (so including at infinity), and such that the restriction of $\overline{U}$ to $K$ coincides with $U$?

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    $\begingroup$ Your function $U$ is superharmonic at the origin as well, therefore there is no such an extension if $K$ contains the origin, since there is no function which is superharmonic everywhere. $\endgroup$ Feb 20, 2021 at 13:44
  • $\begingroup$ Thanks. I corrected the question. $\endgroup$
    – M. Rahmat
    Feb 20, 2021 at 19:06

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For $m = 2$ you can simply use Kelvin transform to exchange the roles of $y_0$ and $\infty$, as pointed out by Alexandre Eremenko. Thus, we assume that $m \geqslant 3$.

Let $u$ be a superharmonic function in a neighbourhood of a compact set $K$ such that the complement of $K$ is connected. In your case $u$ is equal in $K$ to the Newtonian potential kernel $U(x)$. Let $y_0$ be a fixed point in $\mathbb R^m \setminus K$. By changing $x$ to $y_0 + x$, with no loss of generality we may assume that $y_0 = 0$.

As you write in the other question, by Theorem 6.10.1 in [Armitage, Gardiner, Classical Potential Theory], there is a superharmonic function $v$ in all of $\mathbb R^m$ such that $u = v$ in $K$ and $$v = \alpha + \beta U \qquad \text{ in } \mathbb R^m \setminus B(0, R)$$ for some $R > 0$, $\alpha \in \mathbb R$ and $\beta \geqslant 0$.

Now let $w(x)$ be the Green function of $\mathbb R^m \setminus K$ with pole at $0$ (that is, $w = U - h$ for a harmonic function $h$ in $\mathbb R^m \setminus K$ such that $h = U$ on $\partial K$ — up to irregular boundary points, $h \ge 0$, and $h \to 0$ at infinity). Consider $$\overline u = v - \gamma w$$ for $\gamma$ large enough, so that $$\gamma w \geqslant \beta U \qquad \text{ on } \partial B(0, R), \text{ and hence on all of } \mathbb R^n \setminus B(0, R)$$ (the latter claim follows from the former by the maximum principle applied to $\gamma w - \beta U$). Clearly, $\overline u = v = u$ in $K$. Furthermore, both $v$ and $-w$ are superharmonic in $\mathbb R^m \setminus \{y_0\}$, and so $\overline u$ is superharmonic in $\mathbb R^m \setminus \{y_0\}$, too. Finally, $$\overline u = (\alpha + \beta U) - \gamma w \leqslant \alpha \qquad \text{ in } \mathbb R^m \setminus B(0, R).$$ Since $\overline u \to \alpha$ at infinity, the mean value of $\overline u$ over any (large enough) sphere does not exceed $\alpha$. Thus, $\overline u$ is the desired extension of $u$.


Edit: Here are some details on the $m = 2$ case.

Let $u$ be a superharmonic function in a neighbourhood of a compact set 𝐾 such that the complement of 𝐾 is connected, and let $y_0 \in \mathbb R^2 \setminus K$. Again, with no loss of generality we assume that $y_0 = 0$.

For $x \ne 0$ let $x^*$ be the image of $x$ under the inversion with respect to the unit sphere: $$x^* = |x|^{-2} x.$$ Denote $$K^* = \{x^* : x \in K\}$$ and $$u^*(x^*) = |x|^{2 - m} u(x)$$ denote the Kelvin transform of $u$. Then it is a classical result that $u^*$ is superharmonic in a neigbourhood of $K^*$.

Let $v^*$ be the superharmonic extension of $u^*$ to all of $\mathbb R^m$, as in the first part of the answer (note that the complement of $K^*$ is connected), and let $$v(x) = v^*(x^*)$$ be the Kelvin transform of $v^*$. Then $v^*$ is superharmonic in $\mathbb R^m \setminus \{0\}$. Furthermore, if we set $v(\infty) = v^*(0)$ and if the dimension is 2 ($m = 2$), then $$v(\infty) = v^*(0) \geqslant \frac{r}{2\pi} \int_{\partial B(0, 1/r)} v^* = \frac{1}{2\pi r} \int_{\partial B(0, r)} v \tag{$\star$}$$ for $r$ large enough, which (almost) shows that $v$ is superharmonic at infinity. Almost, because we only consider balls centred at the origin. For the general case, one has the same argument, but this time involving the Poisson kernel: $$v(\infty) = v^*(0) \geqslant \int_{\partial B(x_0', r')} v^*(y) P_{B(x_0', r')}(0, y) dy = \frac{1}{2\pi r} \int_{\partial B(x_0, r)} v , $$ where $\partial B(x_0', r')$ is the image of $\partial B(x_0, r)$ under the inversion. (The calculations get a bit messy here, but they are rather standard.)

Note: in dimensions $m \geqslant 3$, one can follow the same argument, but the scaling in ($\star$) is wrong.

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  • $\begingroup$ Thanks. But isn't $\overline{u}$ still subharmonic (and not superharmonic) at infinity? $\endgroup$
    – M. Rahmat
    Feb 25, 2021 at 20:32
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    $\begingroup$ Ah, right, sorry. I mixed signs again... I'll try to fix it in a few minutes. $\endgroup$ Feb 25, 2021 at 23:03
  • $\begingroup$ Thanks. Just a question for the case $m=2$. Kelvin transform doesn't keep $K$ unchanged, but sends it to $K^*$. How should we proceed in this case? $\endgroup$
    – M. Rahmat
    Feb 26, 2021 at 16:34
  • $\begingroup$ I am afraid that do not quite understand your comment: you do the Kelvin transform, find a suitable extension of the function $u^*$ from $K^*$ to all of $\mathbb R^m$, and then do another Kelvin transform in order to get an extension of $u$ from $K$ to all of $\mathbb R^m \cup \{\infty\} \setminus \{y_0\}$. $\endgroup$ Feb 26, 2021 at 20:15
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    $\begingroup$ Then — time permitting — I will add some details about the $m = 2$ case. $\endgroup$ Feb 26, 2021 at 22:53

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