# Extension of subharmonic functions at infinity

Let $$W$$ be the complement of a compact set $$K$$ in $$\mathbb{R}^{n}$$, and $$u$$ a subharmonic function on $$W$$. Can we find, under some conditions, a function $$\tilde{u}$$ that is subharmonic on $$W\cup\{\infty\}$$ and coincides with $$u$$ on $$W$$?

(Too long for a comment.)

What exactly is your definition of a (sub)harmonic function on $$W \cup \{\infty\}$$?

You can always project your function to the unit $$n$$-sphere using the stereographic projection: $$\mathbb{R}^n \ni x \mapsto \phi(x) = p + 2 |x' - p|^{-2} (x'-p) \in \mathbb{S}^n ,$$ where $$x' = (x_1, \ldots, x_n, 0) \in \mathbb{R}^{n+1}$$, $$p = (0, \ldots, 0, 1) \in \mathbb{R}^{n+1}$$ and $$\mathbb{S}^n = \{x \in \mathbb{R}^{n+1} : |x| = 1\}$$. More precisely, the function $$v(x) = |x' - p|^{2 - n} u(\phi^{-1}(x))$$ is subharmonic on $$\phi(W) \subseteq \mathbb{S}^n$$. (This is a close relative of the Kelvin transformation.)

With this identification, you can treat $$\infty$$ just as every other point of $$\mathbb{R}^n$$ (or $$\mathbb{S}^n$$). The problem is that the value of (the extension of) $$v$$ at $$p$$ is not the same as the hypothetical value of $$u$$ at infinity; in fact, $$v(p) = \lim_{|x| \to \infty} |x|^{n-2} u(x)$$.

I do not see any other reasonable notion of (sub)harmonicity on $$W \cup \{\infty\}$$. One could naively try to require that $$u(\infty)$$ is equal to (or does not exceed) the average of $$u$$ over an arbitrary sphere which contains $$\mathbb{R}^n \setminus W$$ in its interior. But this definition is not consistent: for example, the only harmonic functions in $$W \cup \{\infty\}$$ would be constants.

Edited: After reading another question by M. Rahmat, I realised that the last paragraph of my answer was wrong. What I called a "naive" approach actually works in dimensions $$n \geqslant 2$$, and it was an idea developed by Brelot in 1940s; see [M. Brelot. Sur le rôle du point à l'infini dans la théorie des fonctions harmoniques. Ann. Sci. ́Ecole Norm. Sup., 61:301–332, 1944].

• The situation is special when $n=2$. Then $\infty$ can be treated as any other point, and removable singularity theorem is applicable. – Alexandre Eremenko Feb 1 at 12:25