In what function space is the fractional Laplacian $(-\Delta)^su$ essentially bounded?

Question

Can it be said that if $u(x)=O(|x|^\beta)$, with $\beta\geq N+2s$, for $x\in\Omega\subset\mathbb{R}^N$ a bounded domain, then $(-\Delta)^su\in L^{\infty}(\Omega)$?. Is there a generalized result corresponding to this if true?. In other words what will be the space in which $u$ belongs to if $(-\Delta)^su\in L^{\infty}(\Omega)$?.

Answer 1

Whatever size restriction one has on $u$ do not imply its smoothness and smoothness is reflected in the size of the Laplacian. So the answer is negative regardless of $\beta$.

The inverse of $(-\Delta)^s$ is convolution with Riesz kernel $|x|^{-2s}$. So the space of functions such that $(-\Delta)^su\in L^{\infty}(\Omega)$ is precisely the Riesz potential space (look it up) $I_{2s}(L^\infty)$. On a bounded domain, the Riesz potential is comparable to Bessel potential, which is more widely used.

Bessel potential inverts $(I-\Delta)^s$ but the distinction matters mostly for infinity. Unfortunately $p=\infty$ is a hairy case, and I believe the space of Bessel potentials of $L^\infty$ functions will be slightly larger than the fractional Sobolev space $W^{2s,\infty}$, but it should be "close" to it.

A comprehensive reference is On spaces of potentials connected with $L^p$ classes by Aronszajn, Mulla, and Szeptycki.