Let $s\in(0,1)$, $u\in\mathcal{S}({\mathbb{R}^n})$, $x\in\mathbb{R^n}$ with: $|x|\geq1$, i have to prove that:
$$ \int\_{B\_{|x|/2}(0)} \frac{|u(x+y)+u(x-y)-2u(x)|}{|y|^{n+2s}}\,dy\leq c|x|^{-n-2s}, $$
where: $c=c(u,n,s)>0$ is a constant. I think that i have to use something like:
$$ |u(x+y)+u(x-y)-2u(x)|\leq|D^2u(y)||y|^2,$$
but after i can't go on. Any help would be appreciated.



Integral inequality for Schwartz function
By Taylor's theorem, for $|x|\ge1$, $|y|\le|x|/2$, and real $k$,
$$u(x+y)-u(x)=u'(x)(y)+\int\_0^1 ds\,(1-s)u''(x+sy)(y,y)
=u'(x)(y)+O(|y|^2/|x|^k),$$
$$u(x-y)-u(x)=-u'(x)(y)+\int\_0^1 ds\,(1-s)u''(x-sy)(y,y),
=-u'(x)(y)+O(|y|^2/|x|^k).$$
Adding these, we get
$$u(x+y)+u(x-y)-2u(x)=O(|y|^2/|x|^k).$$
Also,
$$u(x+y)+u(x-y)-2u(x)=O(1/|x|^k).$$
So, your integral is
$$O\Big(|x|^{-k}\,\int\_{\mathbb R^n}\frac{dy\,\min(1,|y|^2)}{|y|^{n+2s}}\Big)=
O\Big(|x|^{-k}\,\int\_0^\infty\frac{dr\,r^{n-1}\min(1,r^2)}{r^{n+2s}}\Big)=O(|x|^{-k}),$$
for any real $k$.


