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: $$ \int_{B_{x/2}(0)} \frac{u(x+y)+u(xy)2u(x)}{y^{n+2s}}\,dy\leq cx^{n2s}, $$ where: $c=c(u,n,s)>0$ is a constant. I think that i have to use something like: $$ u(x+y)+u(xy)2u(x)\leqD^2u(y)y^2,$$ but after i can't go on. Any help would be appreciated.
By Taylor's theorem, for $x\ge1$, $y\lex/2$, and real $k$, $$u(x+y)u(x)=u'(x)(y)+\int_0^1 ds\,(1s)u''(x+sy)(y,y) =u'(x)(y)+O(y^2/x^k),$$ $$u(xy)u(x)=u'(x)(y)+\int_0^1 ds\,(1s)u''(xsy)(y,y), =u'(x)(y)+O(y^2/x^k).$$ Adding these, we get $$u(x+y)+u(xy)2u(x)=O(y^2/x^k).$$ Also, $$u(x+y)+u(xy)2u(x)=O(1/x^k).$$ So, your integral is $$O\Big(x^{k}\,\int_{\mathbb R^n}\frac{dy\,\min(1,y^2)}{y^{n+2s}}\Big)= O\Big(x^{k}\,\int_0^\infty\frac{dr\,r^{n1}\min(1,r^2)}{r^{n+2s}}\Big)=O(x^{k}),$$ for any real $k$.

$\begingroup$ In your opinion there is a way to use the fact that $\sup_{z\in R^n}(1+z)^ND^2u(z)<\infty$, for all $N$, and don't use $O(y^2/x^k)$? $\endgroup$ – inoc Oct 15 at 17:17

1$\begingroup$ @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