# Limitation of the fractional laplacian operator

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.

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 $$\Omega$$.

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 $$\|(-\Delta)^su\|_{L^2}=\||\cdot|^{2s}\mathcal{F}u(\cdot)\|_{L^2}=\Big(\int_{\mathbb{R}^N}|\xi|^{4s}\mathcal{F}u(\xi)^2d\xi\Big)^\frac{1}{2}\leq\|u\|_{H^{2s}}$$