What is the domain of the Hermite operator?

Question

Good evening, the question is the following: what is the "natural" subspace of $L^2(\mathbb{R}^d)$ that provides the domain of the operator $H=-\Delta+|x|^2$? I mean, $H$ is an unbounded operator on $L^2$ and it is surely defined on the dense subset of $L^2$ given by the Schwartz functions, $\mathcal{S}$. However, it may be difficult to prove that $H$ is self-adjoint on $\mathcal{S}$ (as in the definition of self-adjointness for unbounded operators). I guessed that the domain of $H$ should sound like this: $D(H)$ is the set of those functions $f\in L^2(\mathbb{R}^d)$ s.t.

1. the partial derivatives of $f$ exist a.e. and they are $L^2$ functions;
2. the same for the second derivatives;
3. some further conditions on the decay of $f$, $\nabla f$ and $\Delta f$ in order to integrate by parts and "cancel the boundary term".

This would prove easily at least that $H$ is symmetric. The self-adjointness may still be problematic.

Thank you in advance.

Answer 1

The Hermite operator is essentially self-adjoint on $D(H)=C_c^\infty(\mathbb{R}^d)$: If $\psi\in D(H^\ast)$, then $$ \langle \phi, H^\ast \psi\rangle=\langle H\phi,\psi\rangle=-\int (\Delta \phi)\psi+\int|x|^2\phi\psi $$ for $\phi\in C_c^\infty(\mathbb{R}^d)$.

Thus $\Delta \psi\in L^2_{\mathrm{loc}}$ (distributional Laplacian) and $H^\ast\psi=-\Delta \psi+|x|^2\psi$. By elliptic regularity, $\psi\in W^{2,2}_{\mathrm{loc}}$. It is not hard to see that in fact $D(H^\ast)=\{\psi\in L^2\cap W^{2,2}_{\mathrm{loc}}\mid -\Delta \psi+|x|^2\psi\in L^2\}$.

For essential self-adjointness it is sufficient to prove that $\ker(H^\ast+1)=\{0\}$. For that purpose let $\psi\in \ker(H^\ast+1)$ and $\phi\in C_c^\infty(\mathbb{R}^d)$ with $0\leq \phi\leq 1$, $\phi=1$ on $B_1(0)$ and $\operatorname{supp}\phi\subset B_2(0)$. Set $\phi_R(x)=\phi(x/R)$.

As $\psi\in W^{2,2}_{\mathrm{loc}}$ we have $\psi\phi_R\in W^{2,2}$. In particular, $$ \langle H^\ast(\psi\phi_R),\psi\phi_R\rangle=\int|\nabla(\phi_R\psi)|^2+|x|^2(\phi_R\psi)^2\geq 0. $$ A little computation with the Leibniz rule shows $$ \langle (H^\ast+1)(\psi\phi_R),\psi\phi_R\rangle=\mathrm{Re}\langle \phi_R(H^\ast+1)\psi,\phi_R\psi\rangle+\int |\nabla \phi_R|^2\psi^2=\int|\nabla \phi_R|^2\psi^2. $$ Thus $$ \int_{B_R(0)}\psi^2\leq \langle (H^\ast+1)(\phi_R \psi),\phi_R\psi\rangle=\int|\nabla \phi_R|^2\psi^2\leq \|\nabla \phi_R\|_\infty^2\int \psi^2. $$ Letting $R\to \infty$, we obtain $\psi=0$.

Note that this argument works not only for $-\Delta+|x|^2$, but more generally for $-\Delta+V$ with lower bounded $V\in L^2_{\mathrm{loc}}$.