Let $\Omega \subseteq \mathbb{R}^n$ be open. Suppose $u : \Omega \to \mathbb{R}$ is a $C^2$ function satisfying $-\Delta u = \lambda u$ for some $\lambda \ge 0$. I would like to show that $|u|$ also belongs to $C^2(\Omega)$ and that $-\Delta |u| = \lambda |u|$.
It is clear that the desired smoothness and equality holds in a neighborhood of any point $x$ at which $u(x) > 0$ or $u(x) < 0$. The problem is what to do near a point $x$ where $u(x) = 0$.
Some thoughts: integrating $|u|$ against $\partial_j \phi$, where $\phi \in C^\infty_0(\Omega)$, and using $|u| = \lim_{\varepsilon \to 0}\sqrt{u^2 + \varepsilon}$ and the dominated convergence theorem, we do get that the distributional gradient of $|u|$ is $\textrm{sgn}(u)\nabla u$. This same sort of calculation does not seem to work for tackling the distributional Laplacian of $u$ because of buildup of negative powers of $\varepsilon$ after taking two derivatives.
Hints or solutions are appreciated.