# Superharmonic extension 2

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 $$U$$?

• 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. Feb 20, 2021 at 13:44
• Thanks. I corrected the question. Feb 20, 2021 at 19:06

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.

• Thanks. But isn't $\overline{u}$ still subharmonic (and not superharmonic) at infinity? Feb 25, 2021 at 20:32
• Ah, right, sorry. I mixed signs again... I'll try to fix it in a few minutes. Feb 25, 2021 at 23:03
• 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? Feb 26, 2021 at 16:34
• 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 $\mathbb R^m \cup \{\infty\} \setminus \{y_0\}$. Feb 26, 2021 at 20:15
• Then — time permitting — I will add some details about the $m = 2$ case. Feb 26, 2021 at 22:53