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?


1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$ Commented Jan 19, 2023 at 17:54
  • $\begingroup$ 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 $\nabla \xi \cdot \nabla u\in H^1$. $\endgroup$
    – daw
    Commented Jan 20, 2023 at 6:53
  • $\begingroup$ 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 $\xi$). $\endgroup$
    – daw
    Commented Jan 20, 2023 at 6:54
  • $\begingroup$ 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 $\phi$. $\endgroup$ Commented Jan 20, 2023 at 7:28
  • 1
    $\begingroup$ yes, that was my idea. $\endgroup$
    – daw
    Commented Jan 20, 2023 at 11:10

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