A continuous function for which Poisson's equation has no C^2 solutions

Question

I am trying to solve exercise 4.9 of Gilbarg and Trudinger, and in particular need to show that for the function

$f(x)=\sum_{k=0}^{\infty}\frac{1}{k}\Delta(\eta{P})(2^kx)$

the problem $\Delta{u}=f$ has no $C^2$ solution in any neighbourhood of the origin. Here $\eta\in{C_c}^{\infty}$ is $1$ on $B_1(0)$ and $0$ on $B_2(0)^c$ and $P$ is a homogeneous degree $2$ harmonic polynomial satisfying $D^{\alpha}P\neq{0}$ for some $|\alpha|=2$.

I have so far shown that $f$ is continuous (as required by the exercise) and that the function $v(x)=\sum_{k=0}^{\infty}\frac{1}{2^{2k}k}(\eta{P})(2^kx)$ satisfies $\Delta{v}=f$ but is not $C^2$ (as $D^{\alpha}v$ diverges at $0$).

I now have 2 problems (the second is less important and just something that I started getting confused about while thinking about the problem):

1. How can I show that there exists no $C^2$ solution to the problem $\Delta{u}=f$? I know of uniqueness results for Poisson's equation relying on maximum principles, however these all held for $C^2$ solutions only, so I don't know how to use the function $v$.

2. By Weyl's lemma, any continuous weak solution of Laplace's equation is in fact $C^{\infty}$, yet $v$ is not even $C^2$ and satisfies Poissons' equation. Is this because Weyl's lemma does not hold for Poisson's equation with $f\neq{0}$, or because $v$ not being $C^2$ means it is not in fact a weak solution (as integration by parts is not valid)?

Answer 1

To answer 1 : Note that if $u$ is a $C^2$ solution in $B_r$, then $u-v$ satisfies $\Delta (u - v) = 0$ in $B_r\setminus \{0\}$. Since, $u-v$ is bounded as well, by removable singularity theorem ($\displaystyle \lim\limits_{|x| \to 0} \frac{u-v}{\Phi}(x) = 0$, where, $\Phi$ is the fundamental solution) it is harmonic in $B_r$, i.e., $u = v + \text{harmonic}$ which leads to a contradiction.

Weyl's lemma holds only for $$\Delta u = 0$$ in weak or more generally distributionl sense, as the above counterexample indicates it is not true for distributional solutions of Poisson equation in general.