0
$\begingroup$

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 $W$?

$\endgroup$
1
$\begingroup$

(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].

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.