Let $\psi(x) \in C^1(\mathbb{R}^n)$ with $\psi \in C^\infty (\{|x| \le 1\})$ and $\psi \in C^\infty(\{ |x| \ge 1\})$ (the point being that the higher order derivatives of $\psi$ are not necessarily continuous on the unit sphere).
I would like to understand how it is possible to arrange a sequence $\psi_k \in C^\infty(\mathbb{R}^n)$ such that $\psi_k \to \psi$ in $L^2(\{ |x| \le 2\})$ and also that the $\Delta \psi_k$ converge weakly to $\Delta \psi$ (the latter function defined Lebesgue a.e.)
So far my straightforward approach has not worked. Fix a cutoff $\chi \in C^\infty([0, \infty); [0,1])$ that is identically zero near $[0, 1/4]$ and identically one near $[1/2, \infty)$. Then the function $\chi(k(||x| - 1|)$ is smooth and vanishes in neighborhood of the unit sphere, hence $\psi_k : = \chi(k(||x| - 1|)\psi(x) \in C^\infty(\mathbb{R})$. It's easy to see that $\psi_k \to \psi$ in $L^2(\{ |x| \le 2\})$. However, I am unable to conclude $\Delta \psi_k \to \Delta \psi$ weakly. The issue is that while $\Delta \chi(k(||x| - 1|) = O_{L^\infty}(k^2)$, I am only able to bound the support of $\Delta \chi(k(||x| - 1|)$ by the $n$-dimensional annulus $\{ 1 - \frac{1}{k} \le |x| \le 1 + \frac{1}{k} \}$ which has measure $O(k^{-1})$. So it seems a better method of cutting off away from the unit sphere is needed.