Is this statement true?  
 
A bounded positive function $v\in C^{2}(\mathbb R^N)$ decaying to zero which is $s$-harmonic function ($s\in (0, 1)$) outside a ball behaves like $|x|^{2s-N}$ near infinity. That is, if $N>2s, $ then $|x|^{N-2s} v(x) \to l$ whenever $|x|\to \infty$ for some $l>0.$ 



On $s$-harmonic functions
Yes, it is. Although there are more elementary ways to prove this fact, I suppose the following argument is the shortest.


Every positive $s$-harmonic function in an arbitrary open set $D$ can be represented in terms of the Poisson kernel $P\_D(x, y)$ (with $y \in \mathbb{R}^N \setminus D\_M$) and the Martin kernel $M\_D(x, z)$ (with $z \in D\_M \setminus D$). Here $D\_M \subseteq \overline{D}$ is the Martin compactification of $D$.


For an exterior domain, these kernels decay as $|x|^{2s-N}$ as $|x| \to \infty$, except the Martin kernel $M(x, \infty)$ with pole at infinity, which is asymptotically constant (at least if $2s < N$; this is true for any exterior domain, and for the complement of a ball we even have explicit expressions at hand). If your function decays at infinity, the contribution of $M(x, \infty)$ is necessarily zero, and hence it is asymptotically equal to $|x|^{2s-N}$.


For details, see my article with Krzysztof Bogdan and Tadeusz Kulczycki:



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> K. Bogdan, T. Kulczycki, M. Kwaśnicki, *Estimates and structure of -harmonic functions*, Prob. Theory Rel. Fields 140(3–4) (2008): 345–381, [DOI:10.1007/s00440-007-0067-0](https://doi.org/10.1007/s00440-007-0067-0)
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> 
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The article is unfortunately not very reader-friendly; if you have any questions, feel free to ask.


