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.