Consider the following problem, which was extracted from page 13 of these lecture notes. Consider $w$ to be radially symmetric ($w(x) = w(|x|)$), compactly supported and nonnegative. Let $f$ be radially symmetric solution of: $$(-\Delta + \frac{1}{2}w)f = 0$$ with $\lim_{|x| \to \infty}f(x) = 1$. When $x$ is really large, we have $f(x) = C-\frac{a_{0}}{|x|}$, and since $\lim_{|x|\to \infty}f(x) = 1$, this implies $f(x) = 1 - \frac{a_{0}}{|x|}$. So far, so good.
However, the problem now is to prove that $a_{0}$ is also given by: $$8\pi a_{0} = \int_{\mathbb{R}^{3}}dx w(x)f(x).$$ My reasoning is the following. Because $f$ solves the above equation, we have (passing to spherical coordinates): $$\int_{\mathbb{R}^{3}}dx w(x)f(x) = 2\int_{\mathbb{R}^{3}}dx \Delta f = 8\pi \int_{0}^{\infty}dr r^{2}\frac{1}{r^{2}}\frac{d}{dr}\bigg{(}r^{2}\frac{df}{dr}\bigg{)} = r^{2}\frac{df}{dr}\bigg{|}_{0}^{\infty}$$ I know that $\lim_{r\to \infty}r^{2}\frac{df}{dr} = a_{0}$, but how to ensure that $\lim_{r\to 0}r^{2}\frac{df}{dr} = 0$? I have no further hypothesis on $f$ and it seems not possible to conclude this. Any help is welcomed!
It is because $w$ is compactly supported, thus far away the solution looks like that ($\Delta f = 0$ for $|x|>R$). Near the origin $\frac{df}{dr}$ should be bounded (because $f\in C^2$ near the origin), and thus that zero limit when $r\to 0^+$.
Edit based on the feedback in comments: if $f\in H^1(\mathbb{R}^3)$, which implies $\displaystyle \int_D|\nabla f|^2$ must be finite for a domain including the origin, the most singular function to satisfy the integrability is $f\sim 1/|x|^{1/2-\epsilon} = 1/|r|^{1/2-\epsilon}$ near the origin for an arbitrary small $\epsilon>0$. This will make the limit zero as well.