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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.

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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 $k$.

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  • $\begingroup$ In your opinion there is a way to use the fact that $\sup_{z\in R^n}(1+|z|)^N|D^2u(z)|<\infty$, for all $N$, and don't use $O(|y|^2/|x|^k)$? $\endgroup$ – inoc Oct 15 at 17:17
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    $\begingroup$ @inoc : I did use the same fact, in the equivalent form $\|u''(x)\|=O(|x|^{-k})$ for $|x|\ge1$ and any real $k$. So, for the quadratic form $u''(x)(y,y)$ corresponding to the linear operator $u''(x)$, we have $|u''(x)(y,y)|\le\|u''(x)\|\,|y|^2=O(|x|^{-k}|y|^2)$. (Cf. en.wikipedia.org/wiki/… .) Here I identify, as usual, a bounded linear operator $A$ with a bounded quadratic form $y\mapsto A(y,y):=\langle Ay,y\rangle$ or, equivalently, with a bounded bilinear form $(y,z)\mapsto A(y,z):=\langle Ay,z\rangle$. $\endgroup$ – Iosif Pinelis Oct 15 at 18:02

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