I am reading Spectral Theory (Chapter 6.3) by David Borthwick and having a question about how to argue the smoothness of eigenfunctions.
Consider the Dirichlet problem on a bounded open set $\Omega \subset \mathbb{R}^N$. The domain of the Dirichlet Laplcian is $$D(-\Delta) = \{ u\in H^1_0(\Omega): -\Delta u \in L^2(\Omega) \}$$ The eigenvectors form an orthonormal basis of $L^2(\Omega)$.
Now, for a compactly supported smooth function $\xi \in C_0^\infty (\Omega)$ and an eigenfunction $u$, we can view $\xi u$ as a function in $H^1(\mathbb{R}^N)$ by zero extension. Then the eigenvalue equation gives $$ -\Delta (\xi u) = \lambda\xi u - [\Delta,\xi] u $$ where $\lambda>0$ is the eigenvalue of $u$ and $[\Delta,\xi] = \Delta\xi - \xi \Delta$ is the commutator.
Because $[\Delta,\xi]$ is a first-order differential operator and $u \in H_0^1(\Omega)$, the RHS of the eigenvalue equation is in $L^2(\Omega) \subset L^2(\mathbb{R}^N)$. By taking Fourier transform, it can be shown that $\xi u \in H^2(\mathbb{R}^N)$.
Then, the book says that $\xi u \in H^2(\mathbb{R}^N)$ implies the RHS is indeed a function in $H^1(\mathbb{R}^N)$ so $\xi u \in H^3(\mathbb{R}^N)$ by again taking Fourier transform. Iterating this process shows that $\xi u \in H^m(\mathbb{R}^N)$ for all $m\in\mathbb{N}$ so $\xi u$ is smooth. This is true for all $\xi$, so $u$ is smooth.
I don't see why we have the implication in the bold font. To have $\lambda\xi u - [\Delta,\xi] u \in H^1(\mathbb{R}^N)$, we need $u\in H^2(\mathbb{R}^N)$, but we merely have $\xi u\in H^2(\mathbb{R}^N)$. Did I miss anything?
I think, you have an error in the right-hand side of the equation (or I do not understand the meaning of $[\xi,\Delta]$). It should read $$ -\Delta(\xi u) = \lambda \xi u -(\Delta \xi)u -2\nabla \xi \cdot \nabla u. $$ Now elliptic theory tells us, that $\xi u\in H^2$ for all $\xi$. That is, the restriction of $u$ to compact subsets of $\Omega$ is $H^2$, or $u \in H^2_{loc}(\Omega)$.
This implies that the right-hand side of the equation above is in $H^1$, as $\nabla \xi \cdot \nabla u$ is in $H^1$ because $\xi$ has compact support.