Is there a smooth cut off function with compact support such that $\phi: \Omega \subset \mathbb R^N \to \mathbb R$, $\mathrm{supp} \phi \subset B_R(0) \subset \Omega$ and $$(-\Delta)^s \phi \le C R^{-2s} \text{ and } D^k(-\Delta)^s \phi \le C R^{-2s - k}?$$

Here (-\Delta)^s denotes the spectral fractional Laplacian

  • $\begingroup$ Yes, any function $\phi_R(x) = \phi(x/R)$ will do the job. This is because if $0 \in \Omega$, then the kernel of the "spectral" fractional Laplacian $(-\Delta_\Omega)^s$ is locally very similar to that of $(-\Delta)^s$. $\endgroup$ – Mateusz Kwaśnicki Jun 28 at 23:04
  • $\begingroup$ @MateuszKwaśnicki If the center of the ball is not $0$ and $0\notin \Omega$,is this true anyway? $\endgroup$ – user139845 Jun 29 at 7:57

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