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

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  • $\begingroup$ The constant $c_{n, s}$ should depend on $u$. Or $u$ should be not arbitrary Schwartz function. $\endgroup$ Oct 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 $(1 + |x|)^{-n - 2s}$. $\endgroup$ Oct 15, 2020 at 8:59
  • $\begingroup$ @Mateusz Kwaśnicki .Can you give me the details please? $\endgroup$
    – inoc
    Oct 15, 2020 at 9:31

1 Answer 1

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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, $$\begin{aligned} \biggl| \int_{B(0,1)} \frac{u(x + y) + u(x - y) - 2 u(x)}{|y|^{n + 2s}} \, dy \biggr| & \leqslant \frac{C_{u,n}}{(1 + |x|)^{n + 2s}} \int_{B(0,1)} |y|^{2-n-2s} dy \\ & = \frac{C_{u,n,s}}{(1 + |x|)^{n+2s}} \end{aligned}$$ Furthermore, since $u$ is Schwartz class, we have $$|u(x + y) - u(x)| \leqslant |u(x - y)| + |u(x)| \leqslant C_u ((1 + |x - y|)^{-n-2s} + (1 + |x|)^{-n-2s}).$$ 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: $$ |(-\Delta)^s u(x)| \leqslant \frac{C_{u,n,s}}{(1 + |x|)^{n + 2 s}} \, .$$

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  • $\begingroup$ in equation (1) i don't understand because $C_n|y|^2\sup |D^2u|\leq C_{u,n}(1+|x|)^{-n-2s}|y|^2$. $\endgroup$
    – inoc
    Oct 15, 2020 at 14:48
  • $\begingroup$ @inoc: All second-order derivatives are Schwartz class, and hence $|D^2 u(x)| \leqslant C_u (2 + |x|)^{-n-2s}$. Furthermore, $\sup_{z \in B(x,1)} (2 + |z|)^{-n-2s} \leqslant (2 + (|x| - 1))^{-n-2s} = (1 + |x|)^{-n-2s}$. $\endgroup$ Oct 15, 2020 at 18:53
  • $\begingroup$ Can you give a me suggestion for prove that: $$ \int_{\mathbb{R}^n\setminus B_1(0}\frac{1}{|y|^{n+2s}(1+|x-y|)^{n+2s}}\,dy\leq\frac{C_{n,s}}{(1+|x|)^{n+2s}},$$ please. $\endgroup$
    – inoc
    Nov 11, 2020 at 18:36
  • $\begingroup$ @inoc: Split the integral into two half-spaces: $\{y : x \cdot y \le \tfrac{1}{2} |x|^2\}$ and the other one. In each integral, estimate one term under the integral by the supremum (which is $(1+|x|/2)^{-n-2s}$), and the other one has finite integral even over all of $\mathbb{R}^n$. $\endgroup$ Nov 11, 2020 at 21:10
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    $\begingroup$ 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

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