# Iterated integrations by parts using the fractional Laplacian

Let $$u \in C^\infty_c(\mathbb{\Omega})$$ and $$\varphi$$ be an eigenfunction of the fractional Laplacian $$(-\Delta)^s$$ in $$\Omega$$ with eigenvalue $$\lambda$$. In what sense, if any, is it true that $$\langle u, \varphi \rangle = \frac{1}{\lambda} \langle u, (-\Delta)^s\varphi \rangle = \frac{1}{\lambda} \langle(-\Delta)^s u, \varphi \rangle = \frac{1}{\lambda^2} \langle (-\Delta)^{2s} u, \varphi \rangle = ........?$$ Here $$\langle \cdot, \cdot \rangle$$ denotes the scalar product in $$L^2(\Omega)$$. In other words, is it true that we can integrate by parts using $$(-\Delta)^su$$, $$(-\Delta)^{2s}u$$ and so on even though their support is not compact anymore?

Formally, $$\varphi$$ is the eigenfunction of the unbounded operator $$L_s$$ on $$L^2(\Omega)$$, defined initially by $$L_s u(x) = (-\Delta)^s u(x) \qquad \text{for } x \in \Omega ,$$ where $$u \in C_c^\infty(\Omega)$$ (and it is understood that $$u(x) = 0$$ for $$x \notin \Omega$$), and then extended to an appropriate domain (e.g. by means of Friedrichs extension).
Now the key observation is that $$L_s L_s$$ is not equal to $$L_{2s}$$, unless $$\Omega = \mathbb R^d$$. Indeed, in $$\Omega$$ we have $$L_s L_s u = (-\Delta)^s (\mathbb 1_{\Omega} \times (-\Delta)^s u) ,$$ while $$L_{2s} u = (-\Delta)^s (-\Delta)^s u ,$$ and due to non-locality of $$(-\Delta)^s$$, the two are not equal.
For the above reasons, we have $$\langle u, \varphi \rangle = \frac{1}{\lambda^2} \langle u, L_s L_s \varphi \rangle = \frac{1}{\lambda^2} \langle L_s L_s u, \varphi \rangle$$ (provided that $$L_s u$$ belongs to the domain of $$L_s$$, which is rarely the case!), and in general the right-hand side is not equal to $$\lambda^{-2} \langle (-\Delta)^{2s} u, \varphi \rangle$$.
• Ok, I see. Then my real question is the following one: without this trick, how can we prove that $$U(t,x) = \sum_{k=1}^\infty \varphi_k(x) \left( \langle u, \varphi_k \rangle \frac{\sin(\sqrt{\lambda_k} t)}{\sqrt{\lambda_k}} \right)$$ (which is the solution of the fractional wave equation in a regular open set $\Omega$ with initial velocity $u$ and zero initial displacement) is smooth in $x \in \Omega$ if $u$ is smooth?