Laplace-Beltrami operator is essentially self-adjoint on a bounded domain

Question

I came across this lecture note The Poincaré inequality on domains on a webpage.

In the first section, it claims that $L$ is essentially self-adjoint on $\mathcal{D}^\infty$. It is known that the Laplace-Beltrami $\Delta$ is essentially self-adjoint on $C^\infty_c(M)$ when $M$ is complete, but not essentially self-adjoint on a general bounded domain. Of course, I'm fully aware that the author takes a different function class. Here $\mathcal{D}^\infty$ is defined as $$ \left\{f\in C^\infty(\bar{\Omega}): \int_\Omega h\Delta f= -\int_\Omega g(\nabla h, \nabla f),\ \forall h\in C^\infty(\bar{\Omega})\right\}. $$ Therefore, I'm wondering how to actually show that $\Delta$ is essentially self-adjoint on $\mathcal{D}^\infty$. The author claims it is easy, but I haven't been able to do it so far.

What I already have: since $\Delta$ is clearly negative and symmetric on $\mathcal{D}^\infty$, it is suffice to show that $$ \int_\Omega (\Delta h - h ) u =0,\ \forall h\in \mathcal{D}^\infty \Rightarrow u=0. $$ Note that $u$ is a weak solution of a elliptic equation, so $u$ is smooth inside $\Omega$.

I don't know how to proceed next. Does anyone know? Is this claim even correct?

Answer 1

I came up with this proof several days ago. It's for $\Omega$ with smooth boundary $\partial\Omega$. I highly doubt if it is correct for arbitrary boundary.

The proof is based on a lemma. The proof of the lemma makes use of the partition of unity, which is straightforward but rather cumbersome. So I'll not include it here.

For any smooth function $h_\partial$ on $\partial\Omega$, $\exists$ smooth function $h$ on $\bar{\Omega}$ s.t. $h|_{\partial\Omega}=h_\partial$ and $g(N,\nabla h)=0$.

Here $N$ is the normal vector field on $\partial\Omega$.

The basic idea is fundamental: $\Delta$ is essentially self-adjoint iff ker$(I-\Delta^*)=0$.

Note that $h\in\mathcal{D}^\infty\iff g(N,\nabla h)=0 $. Take $(I-\Delta)^*u=0$. By regularity argument (see example.2 $\S$X.3 in reed and $\S$5.7 taylor) $u\in C^\infty(\bar{\Omega})$ and thus $\Delta u = u$ in strong sense. Therefore, $$ \langle h,(I-\Delta) u\rangle = 0=\langle h, (I-\Delta)^*u\rangle =\langle (I-\Delta)h,u\rangle=\langle h,u\rangle+\langle \nabla h,\nabla u\rangle, $$ for all $h\in \mathcal{D}^\infty$. Note that $$ \int_{\partial\Omega} g(N,\nabla u) f=\int_\Omega div(f\nabla u) =\int_\Omega \Delta u f + g(\nabla u, \nabla f)=0, $$ by divergence theorem, for all $f\in C^\infty(\bar{\Omega})$. But the left hand side only involves the value of $f$ on $\partial\Omega$. Here we apply the lemma, so $u\in\mathcal{D}^\infty$. Consequently, $\langle u,(I-\Delta)u\rangle =\langle u,u\rangle+\langle\nabla u,\nabla u\rangle=0\Rightarrow u=0 $.

Answer 2

I did not look much in this kind of subject but apparently, Chapter 4.1 and Chapter 4.2 of "Heat Kernel and Analysis on Manifolds" by Alexander Grigor'yan contain a gentle introduction to the solution you want.