I have to prove that for $s\in(0,1)$, $u\in\mathcal{S}(\mathbb{R}^n)$, (i.e. $u$ is a Schwartz function): $$ (\Delta)^su(x)\leq c_{n,s}x^{n2s},\quad\forall x\in\mathbb{R}^n\setminus B_1(0), $$ for some $c_{n,s}>0$, where $$(\Delta)^su(x):=\frac{C(n,s)}{2}\int_{\mathbb{R}^n}\frac{u(x+y)+u(xy)2u(x)} {y^{n+2s}}\,dy,\quad\forall x\in\mathbb{R}^n, $$ is the fractional Laplacian. I have no idea. Any help would be appreciated.

$\begingroup$ The constant $c_{n, s}$ should depend on $u$. Or $u$ should be not arbitrary Schwartz function. $\endgroup$– Fedor PetrovOct 15, 2020 at 8:51

$\begingroup$ And if $c_{n,s}$ is allowed to depend on $u$, then simply split the integral into two parts. The integral over $B(x, 1)$ is bounded by the sup norm of second derivatives of $u$ over $B(x, 1)$ times a constant (by Taylor's theorem). The integral over the complement of $B(x, 1)$ is bounded by a constant times the convolution of $(1 + x)^{n  2s}$ with itself, which is again bounded by a constant times $(1 + x)^{n  2s}$. $\endgroup$– Mateusz KwaśnickiOct 15, 2020 at 8:59

$\begingroup$ @Mateusz Kwaśnicki .Can you give me the details please? $\endgroup$– inocOct 15, 2020 at 9:31
1 Answer
Write $$\begin{aligned} (\Delta)^s u(x) & = \frac{C(n,s)}{2} \int_{B(x,1)} \frac{u(x + y) + u(x  y)  2 u(x)}{y^{n + 2s}} \, dy \\ & \qquad + C(n,s) \int_{\mathbb{R}^n \setminus B(0,1)} \frac{u(x  y)  u(x)}{y^{n + 2s}} \, dy . \end{aligned}$$ Using Taylor's theorem and the fact that $u''$ is Schwartz class, we find that $$u(x + y) + u(x  y)  2 u(x) \leqslant C_n y^2 \sup_{B(x, 1)} u'' \leqslant C_{u,n} (1 + x)^{n  2s} y^2 $$ (here and below $C_p$ denotes some constant that only depends on the parameter $p$; the value of $C_p$ can be different each time it appears). Thus, $$\begin{aligned} \biggl \int_{B(0,1)} \frac{u(x + y) + u(x  y)  2 u(x)}{y^{n + 2s}} \, dy \biggr & \leqslant \frac{C_{u,n}}{(1 + x)^{n + 2s}} \int_{B(0,1)} y^{2n2s} dy \\ & = \frac{C_{u,n,s}}{(1 + x)^{n+2s}} \end{aligned}$$ Furthermore, since $u$ is Schwartz class, we have $$u(x + y)  u(x) \leqslant u(x  y) + u(x) \leqslant C_u ((1 + x  y)^{n2s} + (1 + x)^{n2s}).$$ It is a nice exercise to show that $$ \int_{\mathbb{R}^n \setminus B(0,1)} \frac{1}{y^{n + 2s} (1 + x  y)^{n + 2 s}} \, dy \leqslant \frac{C_{n,s}}{(1 + x)^{n + 2s}} \, . $$ It follows that $$\begin{aligned} \biggl \int_{\mathbb{R}^n \setminus B(0,1)} \frac{u(x  y)  u(x)}{y^{n + 2s}} \, dy \biggr & \leqslant C_u \int_{\mathbb{R}^n \setminus B(0,1)} \frac{1}{y^{n + 2s} (1 + x  y)^{n + 2 s}} \, dy \\ & \qquad + \frac{C_u}{(1 + x)^{n + 2 s}} \int_{\mathbb{R}^n \setminus B(0,1)} \frac{1}{y^{n + 2s}} \, dy \\ & \leqslant \frac{C_{u,n,s}}{(1 + x)^{n + 2 s}} + \frac{C_{u,n,s}}{(1 + x)^{n + 2 s}} = \frac{C_{u,n,s}}{(1 + x)^{n + 2 s}} \, .\end{aligned} $$ The desired result follows: $$ (\Delta)^s u(x) \leqslant \frac{C_{u,n,s}}{(1 + x)^{n + 2 s}} \, .$$

$\begingroup$ in equation (1) i don't understand because $C_ny^2\sup D^2u\leq C_{u,n}(1+x)^{n2s}y^2$. $\endgroup$– inocOct 15, 2020 at 14:48

$\begingroup$ @inoc: All secondorder derivatives are Schwartz class, and hence $D^2 u(x) \leqslant C_u (2 + x)^{n2s}$. Furthermore, $\sup_{z \in B(x,1)} (2 + z)^{n2s} \leqslant (2 + (x  1))^{n2s} = (1 + x)^{n2s}$. $\endgroup$ Oct 15, 2020 at 18:53

$\begingroup$ Can you give a me suggestion for prove that: $$ \int_{\mathbb{R}^n\setminus B_1(0}\frac{1}{y^{n+2s}(1+xy)^{n+2s}}\,dy\leq\frac{C_{n,s}}{(1+x)^{n+2s}},$$ please. $\endgroup$– inocNov 11, 2020 at 18:36

$\begingroup$ @inoc: Split the integral into two halfspaces: $\{y : x \cdot y \le \tfrac{1}{2} x^2\}$ and the other one. In each integral, estimate one term under the integral by the supremum (which is $(1+x/2)^{n2s}$), and the other one has finite integral even over all of $\mathbb{R}^n$. $\endgroup$ Nov 11, 2020 at 21:10

1$\begingroup$ The function $(1+x)/(1+x+y)$ seems to be unbounded when, for example, $y = x$, $x \to \infty$. $\endgroup$ Nov 11, 2020 at 22:32