---
tag: 4PK2T949SQ
kind: counterexample
name: Zaremba's punctured ball
msc: [31B20, 31B25, 31A25]
uses: [B7948M365K, R12KQGVGN6, X4V2S56YQQ, 0YE5868HZB, 8D56GQ4AG8]
sources:
  - url: https://en.wikipedia.org/wiki/Dirichlet_problem
---

## Counterexample

Let $n \ge 2$ and let $\Omega = \set{x \in \R^n : 0 < \abs{x} < 1}$ be the unit [[open ball|0YE5868HZB]] with its centre removed.
Its boundary is the unit sphere together with the point $0$.
Let $f = 0$ on the unit sphere and $f(0) = 1$; this $f$ is continuous on $\partial\Omega$, since $0$ is an isolated point of $\partial\Omega$.
The [[Perron–Wiener–Brelot solution|R12KQGVGN6]] of $f$ is $H_f = 0$, so $0$ is not a [[regular boundary point|B7948M365K]].
Check: the constant $0$ is in the lower class, so $H_f \ge 0$.
For $\varepsilon > 0$, let $u_\varepsilon(x) = \varepsilon (\abs{x}^{2-n} - 1)$ if $n \ge 3$ and $u_\varepsilon(x) = \varepsilon \log (1/\abs{x})$ if $n = 2$.
This function is harmonic, hence [[superharmonic|X4V2S56YQQ]], on $\Omega$.
It is $0$ on the unit sphere and tends to $+\infty$ at $0$, so it is in the upper class and $H_f \le u_\varepsilon$.
Letting $\varepsilon \to 0$ gives $H_f \le 0$, so $H_f = 0$, which does not tend to $f(0) = 1$.

## Remarks

So the [[Dirichlet problem|8D56GQ4AG8]] on the punctured ball has no solution for this continuous boundary function.
A classical solution $u$ would be bounded and harmonic near $0$, so $0$ would be a removable singularity.
The extended function on the ball would vanish by the maximum principle, which contradicts $u(0) = 1$.
The point $0$ is too small to be seen by harmonic functions: it has capacity zero.

## Explanations

### L1 human

Take a ball, keep the surface at temperature $0$, and try to hold the single centre point at temperature $1$.
A single point cannot hold heat in equilibrium, so the temperature in the ball is $0$ everywhere.
This shows that the Dirichlet problem can fail for a very simple open set.