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Let $V : \mathbb{R}^2 \to \mathbb{R}$ be compactly supported, bounded away from the origin, and obey $$ |V(x)| \lesssim r^{-\delta_0}, \qquad 0 < |x| \le 1, \qquad r : =|x|,$$ for some $0 < \delta_0 < 2$. Note that this encompasses potentials which are "Coulomb-like" near zero. In this situation, the quadratic form $$H^1(\mathbb{R}^n) \ni u \mapsto q_V(u,u) : = \int |u|^2 Vdx$$ is "relatively form bounded" with respect to the natural quadratic form associated to the free Laplacian: $$H^1(\mathbb{R}^2) \ni u \mapsto \| \nabla u \|^2_{L^2}.$$ More precisely, what this means that is that for each $a > 0$, there exists $b > 0$ so that $$ \| \sqrt{|V|} u\|^2_{L^2} \le a \| \nabla u \|^2_{L^2} + b \|u\|^2_{L^2}.$$ A proof of this estimate may be found in Faris, Pacific J. Math. 1967. A key tool in the proof is the Hausdorff-Young equality.

Combining the above bound with the KLMN Theorem implies that the Schrödinger operator $$H : = -\Delta + V : L^2(\mathbb{R}^2) \to L^2(\mathbb{R}^2)$$ is self-adjoint with respect to the domain $$ \mathcal{D} := \{u \in H^1(\mathbb{R}^2): Hu \in L^2(\mathbb{R}^2) \},$$ where the membership $Hu \in L^2(\mathbb{R}^2)$ is interpreted in the distributional sense (note $uV \in L^1_{\text{loc}}(\mathbb{R})$ for any $u \in H^1(\mathbb{R}^2)$, thanks to our above bounds on $|V(x)| \mathbf{1}_{0 < |x| \le 1}$ and $\| \sqrt{|V|} u\|_{L^2}$). See also Proposition 1.1, Chantelau, Lett. Math. Phys 1990.

I would like to know whether there exists some subspace $D_0 \subseteq C^\infty(\mathbb{R}^2 \setminus \{0\})$ which is dense in $\mathcal{D}$ with respect to the standard graph norm $u \mapsto \| \cdot \|_{\mathcal{D}} := (\|u\|^2_{L^2} + \|Hu\|^2_{L^2})^{1/2}$ on $\mathcal{D}$.

In general, to have the inclusion $D_0 \subseteq \mathcal{D}$, it seems some (at least implicit) condition has to be imposed on the behavior of $u \in D_0$ as $|x| \to 0$, in order to counteract the blow-up of $V$. I have tried the barehanded approach of simply mollifying an element $u \in \mathcal{D}$,

$$ u_\varepsilon(x) : = \varepsilon^{-2} \int \varphi(y/\varepsilon)u(x - y)dy \in C^\infty(\mathbb{R}^2), \qquad 0< \varepsilon \ll 1,$$

where $\varphi \in C^\infty_0(\mathbb{R}^2)$ is an approximate identity. However, I fail to see how $u_\varepsilon$ even belongs to $\mathcal{D}$ (multiplication by $V(x)$ doesn't "play nicely" with the mollification).

On the other hand, I doubt that $(H, \mathcal{D})$ is the closure of $(H, C_0^\infty(\mathbb{R}^2 \setminus \{0\})$, due to the usual issue with controlling derivatives of cutoffs near zero (the situation would not be as severe in higher dimensions).

If the result holds only for a smaller range of $\delta_0$, I would still find this interesting and would like to know what is the optimal range. In terms of more recent literature on this topic, I came across Oliveira-Verri, Ann. Physics 2009. This paper discusses self-adjoint extensions for Coulomb potentials in dimensions 1, 2, and 3, but doesn't seem to directly address my density question here.

Comments, hints, or pointers to the literature are greatly appreciated.

Edit: When $0 < \delta_0 < 1$, it then holds that $V \in L^2(\mathbb{R}^2)$, in which case $V$ is self-adjoint with respect to the standard Sobolev space $H^2(\mathbb{R}^2)$ (in which $C^\infty_0(\mathbb{R}^n)$ is dense). This follows from Theorem 8 from Nelson 1964, J. Mathematical Phys.

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  • $\begingroup$ Do you assume $V \geq 0$? $\endgroup$ Apr 13 at 16:12
  • $\begingroup$ @GiorgioMetafune Thanks for your comment. No, I do not wish to assume $V$ has a fixed sign. $\endgroup$
    – JZS
    Apr 13 at 20:26
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    $\begingroup$ If $\delta <1$ then $D(-\Delta +V)=H^2(R^2)$ by the following argument (hence a core is given by any dense subset in $H^2$). If $u \in H^2$ then $\nabla u \in L^p$ for any $p<\infty$ and $u \in C^\alpha$ for every $\alpha<1$, by Sobolev embedding. Then the embedding $H^2 \to C(B)$, $B$ any ball, is compact. If the potential has support in a ball and behaves like $r^{-\delta}$, then $u \to Vu$ is then compact from $H^2$ to $L^2$. Then the potential is a small perturbation of the Laplacian and its domain is the same as for $\Delta$ (I am using tools from semigroup theory). $\endgroup$ Apr 14 at 15:39
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    $\begingroup$ Do you know whether the insersection of the domains, namely, $H^2 \cap D(V)$ is a core? This shoud be known but I do not know. $\endgroup$ Apr 23 at 20:13
  • $\begingroup$ @GiorgioMetafune Thank you for your comment. I do not know the answer to your question about the intersection of domains. I think I found an alternative reference for the case $0 < \delta_0 < 1$, please see my reference above. I am still not sure how to proceed in the case $1 \le \delta_0 < 2$. I was trying to make use of the results in Stetkær-Hansen 1966, Mathematica Scandinavica, see this question. $\endgroup$
    – JZS
    Apr 25 at 11:01

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