# Domain of Friedrichs extension of $-\partial^2_r + mr^{-2} : L^2(0,\infty) \to L^2(0,\infty)$

Consider the second order differential operator $$A = -\partial^2_r + mr^{-2} : L^2(0, \infty) \to L^2(0,\infty), \qquad m \ge -\frac{1}{4},$$ equipped with domain $$C^\infty_0(0, \infty)$$. Since $$\|r^{-1} u\|_{L^2} \le 2 \| u'\|_{L^2}$$ for any $$u \in C^\infty_0(0,\infty)$$, the condition $$m \ge -1/4$$ ensures $$A$$ is nonnegative. Thus we can speak of the Friedrichs extension of $$A$$, whose domain is a certain subspace of the closure of $$C^\infty_0(0, \infty)$$ with respect the norm $$\|u \|^2_{L^2} + \langle u, Au \rangle_{L^2}$$.

The equation $$(-\partial^2_r + mr^{-2})u = 0$$ has linearly independent solutions $$u_+ = r^{t_+}, \qquad u_- = \begin{cases} r^{t_-} & m = -\frac{1}{4}, \\ r^{t_-} \log r & m > -\frac{1}{4}, \end{cases}$$ where $$t_\pm = \frac{1 \pm \sqrt{1 + 4m}}{2}.$$ (Note there is a double root when $$m = -1/4$$.)

Let $$\chi \in C^\infty((0,\infty);[0,1])$$ be identically one on $$(0,1]$$ and vanishing near infinity.

I would like to show there exists a sequence $$u_n \in C^\infty_0(0, \infty)$$ such that $$\| u_n - \chi r^{t_+} \|^2_{L^2} + \langle u_n - \chi r^{t_+}, A(u_n - \chi r^{t_+}) \rangle_{L^2} \to 0, \qquad \text{as n \to \infty.}$$ I know how to do this when $$m > -1/4$$, but I don't know what to do in the endpoint case.

Roughly speaking, this is one step toward showing that functions $$u$$ with $$Au \in L^2$$, and $$u = O (r^{t_+})$$ as $$r \to 0$$, lie in the domain of the Friedrichs extension of $$A$$ (sometimes functions which are $$\sim r^{t_-}$$ as $$r \to 0$$ will not be in the domain of the Friedrichs extension, depending on the value of $$m$$).

Here is my attempt so far. Let $$\psi \in C^\infty((0,\infty); [0,1])$$ be supported away from zero and identically one near $$[1, \infty]$$. Put $$\psi_n(r) = \psi(2^{n}r)$$. It's clear that $$\psi_n \chi r^{t_+} \to \chi r^{t_+}$$ in $$L^2$$. Using $$Ar^{t_+}= 0$$ and support properties of $$\chi$$ and the $$\psi_n$$, I can manage to show $$\langle u_n - \chi r^{t_+}, A(u_n - \chi r^{t_+}) \rangle_{L^2} =\int(\psi'_n r^{2t_+} )'dr \\- \int 2t_+ \psi_n \psi'_n r^{2t_+ - 1} + \psi_n \psi''_n r^{2t_+} dr \\ = -2^{n(1- 2t_+)} \int 2t_+ \psi \psi' r^{2t_+ - 1} + \psi\psi'' r^{2t_+} dr.$$ When $$t_+ > 1/2$$ (i.e., when $$m > -1/4$$), this clearly goes to zero as $$n \to \infty$$. But when $$t_+ = 1/2$$, $$-2^{n(1- 2t_+)} \int 2t_+ \psi \psi' r^{2t_+ - 1} + \psi\psi'' r^{2t_+} dr = \int (\psi')^2 r dr.$$ So it seems I need to cut off away from zero more cleverly in the endpoint case. And from above, another way of phrasing this problem is, can one make some (perhaps nonlinear) scaling $$\psi_n$$ of $$\psi$$ so that $$\int (\psi'_n)^2 r dr \to 0 \qquad \text{ as } n \to 0.$$

Hints or suggestions are greatly appreciated.

Let me answer to the second question, which is the motivation for the first. The precise description of the domain of Schroedinger operator with inverse square potential in the critical case is known and can be found in Proposition 4.3 of https://arxiv.org/abs/2103.10314. From this it follows that functions in the maximal domain for which $$|u(r)| \leq Cr^{t_+}$$ lye in the domain. In Examples 7/2, 7.3 in https://arxiv.org/abs/1405.5657 it is also proved that this domain is the Friedrichs extension of your operator.