Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 202 / 3 / d / ii

No such function exists. Continuity on the compact closed unit disk makes $u$ bounded near the origin. The removable singularity for a bounded harmonic function extends $u$ harmonically across the origin. The extended function is continuous on the closed disk and vanishes on its boundary, so the maximum principle for harmonic functions, applied to both $u$ and $-u$, forces $u=0$ throughout the disk. This contradicts $u(0,0)=1$.