# Why are the Dirichlet eigenfunctions smooth

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.

• Yes, I arrived at the same equation for $-\Delta(\xi u)$. But $\xi u \in H^2$ means $\nabla (\xi u) \in H^1$. Why does it imply $\nabla \xi \cdot \nabla u \in H^1$? My thought is that if $K$ is an open set containing the support of $\xi$, then $\nabla \xi \cdot \nabla u = \nabla \xi \cdot \nabla (\phi u )$ where $\phi$ is a compactly supported smooth function that is $1$ on $K$. Thus, $\nabla \xi \cdot \nabla u \in H^1$. Or is there an easier way to see this? Also, since the local Sobolev spaces are not introduced in the book, could you also provide a link or reference? Thank you! Commented Jan 19, 2023 at 17:54
• My idea was to use $\psi \approx \chi_K$ with $K$ compact. Then $\xi u\in H^2$ for all $\xi$ implies $u\in H^2_{loc}$, the latter implies that the rhs is in $H^1$. I do not see how $\xi u\in H^2$ for some particular $\xi$ implies $\nabla \xi \cdot \nabla u\in H^1$.
– daw
Commented Jan 20, 2023 at 6:53
• So the logic: prove some claim that is valid for all $\xi$. Get some property of $u$ out of that. Use this to prove another claim (which happens to involve the same letter $\xi$).
– daw
Commented Jan 20, 2023 at 6:54
• Yeah, so first $\xi u \in H^2$ for all compactly supported smooth $\xi$. Then $\nabla \xi \cdot \nabla u = \nabla \xi \cdot \nabla(\phi u) \in H^1$ for each $\xi$ since $\nabla (\phi u)$ is in $H^1$ for again compactly supported smooth $\phi$. Commented Jan 20, 2023 at 7:28
• yes, that was my idea.
– daw
Commented Jan 20, 2023 at 11:10