The punctured disc has an irregular boundary point and a continuous boundary datum with no harmonic solution

Suppose $u$ were a continuous harmonic solution of this boundary-value problem on $\mathrm{\Omega }$. Then $u$ is bounded on every punctured neighbourhood of $0$ because it extends continuously to the puncture with value $1$. By [L1], $u$ extends to a harmonic function $U$ on the full unit disc.

The extension $U$ still has boundary value $0$ on the unit circle, so [L2] forces $U\equiv 0$ on the closed unit disc. But then $U(0)=0$, contradicting the prescribed puncture value $1$. Therefore no such harmonic solution exists.

Since one continuous boundary datum is not solvable on $\mathrm{\Omega }$, [L3] shows that $\mathrm{\Omega }$ cannot have all boundary points regular. In particular, the puncture is an irregular boundary point.