Unbounded punctured harmonic singularity is not removable

For $r=\mathrm{\mid }x\mathrm{\mid }>0$, differentiating $f(r)$ gives $\mathrm{\Delta }f(r)=f\mathrm{\prime }\mathrm{\prime }(r)+(n-1)f\mathrm{\prime }(r)\mathrm{/}r$. For $n=2$, $f\mathrm{\prime }=1\mathrm{/}r$ and $f\mathrm{\prime }\mathrm{\prime }=-1\mathrm{/}{r}^2$ cancel. For $n\ge 3$, $f\mathrm{\prime }=(2-n)r^{1-n}$ and $f\mathrm{\prime }\mathrm{\prime }=(2-n)(1-n)r^{-n}$ also cancel. Thus both profiles are harmonic away from zero.

The logarithm tends to $-\mathrm{\infty }$ as $r\downarrow 0$, and the negative power tends to $+\mathrm{\infty }$. Continuity at zero would require a finite limit, so no harmonic extension exists. These profiles violate boundedness near the point. Their signed ratios to the critical profiles in the stronger little-o removability condition are $-1$ for $n=2$ and $1$ for $n\ge 3$, so in both cases the absolute ratio is one rather than tending to zero.