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Feb 4 at 22:02 answer added Jingeon An-Lacroix timeline score: 1
Feb 4 at 19:46 vote accept Jingeon An-Lacroix
Feb 4 at 18:26 answer added Carlo Beenakker timeline score: 1
Feb 4 at 16:15 comment added Christian Remling This can be made sense of if one first view $f(x)=1/|x|^3$ as a distribution on $\mathbb R\setminus \{ 0\}$. There are various methods to extend this to a distribution on $\mathbb R$, discussed at length in Section 3.2 of Hormander I. Formula (3.2.17') relates these to $\delta''$, which is exactly what you want as $\delta''*u=\delta*u''=u''$.
Feb 4 at 15:45 history edited Christian Remling CC BY-SA 4.0
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Feb 4 at 11:29 comment added Jingeon An-Lacroix I was missing the dependence on the constant $c_s$ in the limit of $s\rightarrow 1-$. I think it can be understood that $c_s/|x|^{1+2s}\rightarrow \delta_0''$ in distribution, in the limit of $s\rightarrow 1-$, up to some constant.
Feb 4 at 1:35 history asked Jingeon An-Lacroix CC BY-SA 4.0