Limitation of the fractional laplacian operator

Question

Let $\Omega \subset \mathbb{R}^N$ be a bounded smooth domain. In which space $X \subset L^2(\Omega)$ the fractional Laplacian operator $(-\Delta)^s : X \subset L^2(\Omega) \to L^2(\Omega)$, for some $s \in (0,1)$, is bounded (continuous)? Recall that $$(-\Delta)^s u(x) = c_{n,s}\lim_{\varepsilon \to 0^+}\int_{\mathbb{R}^N \backslash B_\varepsilon (x)} \frac{u(x) - u(y)}{|x - y|^{N + 2s}}dy.$$ I've been looking a lot in articles and books, without success. Someone told me here it would be in the $H^{2s}(\Omega)$ space, but I couldn't prove it, nor did I find this.

Answer 1

Indeed, the fractional Laplacian operator $(-\Delta)^s$ is bounded from $H^{2s}(\Omega)$ to $L^2(\Omega)$, for $s \in (0,1)$.

We can first define the fractional Sobolev space $H^s(\Omega)$ as the set of functions $u \in L^2(\Omega)$ whose fractional Sobolev norm $||u||_{H^s(\Omega)}$ is finite, where

$$||u||_{H^s(\Omega)}^2 = ||u||_{L^2(\Omega)}^2 + \int_{\Omega}\int_{\Omega} \frac{|u(x) - u(y)|^2}{|x-y|^{N+2s}} dx dy.$$

We can then define the space $H^{2s}(\Omega)$ as the set of functions $u \in L^2(\Omega)$ that have two fractional derivatives in $L^2(\Omega)$, i.e., $u \in H^{2s}(\Omega)$ if and only if $u$ and its two fractional derivatives are in $L^2(\Omega)$.

Using the fractional Sobolev inequality, which relates the fractional Sobolev norms and the norm in $L^q(\Omega)$, we can show that the fractional Laplacian operator $(-\Delta)^s$ is bounded from $H^{2s}(\Omega)$ to $L^2(\Omega)$.

In summary, the fractional Laplacian operator is bounded from $H^{2s}(\Omega)$ to $L^2(\Omega)$, where $H^{2s}(\Omega)$ is the Sobolev space of order $2s$ in $\Omega$.

Answer 2

There's a bit of a technicality with the fact that the fractional Laplacian is nonlocal and you are working on a bounded domain, but this just comes down to extending by 0, so I think really the space $X$ should be $X=H^{2s}_0(\Omega)$ and maybe this is why you were unable to find it. In the case of $\Omega=\mathbb{R}^N$, consider the representation of the fractional Laplacian as pseudo-differential operator with symbol $\psi(\xi)=|\xi|^{2s}$, i.e. $$(-\Delta)^su(x)=\mathcal{F}^{-1}(|\xi|^{2s}\mathcal{F}f)(x).$$ Furthermore, we have the representation of the Sobolev space $$H^{2s}(\mathbb{R}^N)=\{f\in L^2(\mathbb{R}^N):\|f\|_{H^{2s}}=\Big(\int_{\mathbb{R}^N}(1+|\xi|^2)^{2s}(\mathcal{F}f(\xi))^2d\xi\Big)^{\frac{1}{2}}<\infty\}$$ Then by Plancherel's Theorem we have $$\|(-\Delta)^su\|_{L^2}=\||\cdot|^{2s}\mathcal{F}u(\cdot)\|_{L^2}=\Big(\int_{\mathbb{R}^N}|\xi|^{4s}\mathcal{F}u(\xi)^2d\xi\Big)^\frac{1}{2}\leq\|u\|_{H^{2s}}$$