The extension of smooth function under the restriction of its Laplacian

Question

$u$ is a smooth bounded function on $\Omega-\{0\}$ where $\Omega$ is an open neighborhood of $0$ in $\mathbb R^n$. If $\Delta u$ is a bounded function on $\Omega-\{0\}$, then can we extend $u$ to be a smooth function on $\Omega$? (For example, if $\Delta u=0$, then we can do this.)

Answer 1

Actually, even when $\Delta u=0$ you have to assume $n>1$. In one dimension $u(x)=|x|$ is a counterexample.

Let $f=\Delta u$. Convolving $f$ with the fundamental solution of $\Delta$ (a multiple of $|x|^{2-n}$ or $\log |x|$) gets us a function $v$ which solves $\Delta v=f$ and belongs to the Sobolev class $W^{2,p}(\Omega)$ for every $p<\infty$ (see here). The Morrey-Sobolev embedding yields $v\in C^{1,\alpha}(\Omega)$ for every $\alpha<1$.

Since $u-v$ is harmonic in $\Omega\setminus\{0\}$ and is bounded near $0$, it extends to a harmonic function on $\Omega$ (this step fails in one dimension). We conclude that $u$ extends to a function in $C^{1,\alpha}(\Omega)$ for every $\alpha<1$.

We don't get a $C^2$ extension, however. For example, in the plane we could have $f=\mathrm{Re}\, z/|z|$ which is bounded and smooth everywhere except at $0$. The logarithmic potential of $f$ is $C^\infty$-smooth outside of $0$, but since $\Delta u=f$ is discontinuous at $0$, $u$ does not extend to a $C^2$ function in a neighborhood of $0$.

There must be also an example with $u\notin C^{1,1}$ (because the Riesz transforms do not preserve $L^\infty$) but I don't have one right now.