Removable singularities for harmonic functions

Remarks

Proof idea for $n \ge 3$ and $a = 0$: take a closed ball $\overline{B}(0, r) \subseteq U$. Let $h$ be harmonic on $B(0, r)$ and continuous on its closure, with $h = u$ on the sphere, given by the Poisson integral formula. For $\varepsilon > 0$, the functions $u - h \mp \varepsilon(\abs{x}^{2-n} - r^{2-n})$ are harmonic on the punctured ball and have the right sign near $0$ and on the outer sphere. The maximum principle on annuli gives $\abs{u - h} \le \varepsilon(\abs{x}^{2-n} - r^{2-n})$, and letting $\varepsilon \to 0$ gives $u = h$ off $0$. In the plane the same argument works with $\log(r/\abs{x})$. The growth condition is sharp: the fundamental solution of the Laplacian, $\abs{x}^{2-n}$ for $n \ge 3$ or $\log(1/\abs{x})$ for $n = 2$, is harmonic off $0$ and has no harmonic extension. Example: $u(x, y) = \frac{x}{x^2 + y^2}$ is harmonic on $\R^2 \setminus \set{0}$, and on the positive $x$-axis it equals $1/x$, which grows faster than $\log(1/x)$. So the hypothesis fails, and indeed $u$ is unbounded near $0$ and has no harmonic extension.

Going deeper

For a student

This is the harmonic counterpart of Riemann's theorem on removable singularities of holomorphic functions. The precise threshold is the growth of the fundamental solution, which is the only kind of singularity a positive harmonic function can have at an isolated point, by Bôcher's theorem. For larger exceptional sets, bounded harmonic functions extend across closed sets of zero capacity, which shows the role of capacity in describing negligible sets.