Some Preliminaries on the Fractional Laplacian

Suppose we have an orthonormal basis given by $\{ \phi_k \}_{k \geq 1}$ of $L^2(\Omega)$. This implies that we can write for all $w \in L^2(\Omega)$ as $$ w = \sum_{k=1}^{\infty} w_k \phi_k \, \, , \,\, \text{where} \,\, w_k = (w_k, \phi_k)_{L^2(\Omega)} = \int_{\Omega} w \phi_k \, dx \,\, .$$ Now, we also have a natural choice for the norm, induced from the inner product (since this is a Hilbert space): $$ \begin{align} \| w |_{L^2(\Omega)}^2 = (w,w)_{L^2(\Omega)} &= \int_{\Omega} \left( \sum_{k \geq 1} w_k \phi_k \right) \left( \sum_{j\geq 1} w_j \phi_j \right)dx \\
&= \sum_{k\geq 1} \sum_{j\geq 1} \int_{\Omega} w_k w_j \, \phi_k \phi_j \, dx \\
&= \sum_{k\geq 1} | w_k |^2 \, \, , \end{align} $$ where we used the orthonormality of the bases $\{ \phi_k \}_{k \geq 1}$.

From Linear Analysis, we can think of the Laplacian, as an operator, to be like a symmetric, positive definite matrix $A$ which has countable size. Hence, $A$ has a set of eigenvalues $\{ \lambda_k \}_{k \geq 1}$, which are all positive, and a corresponding set of eigenfunctions $\{ \phi_k \}_{k \geq 1}$. So, for $w \in L^2(\Omega)$, $$ - \nabla w = -\nabla \left( \, \sum_{k \geq 1} w_k \phi_k \right) = - \sum_{k \geq 1} w_k \, \nabla \phi_k = - \sum_{k\geq 1} w_k \lambda_k \phi_k \, \, .$$ Now, by a similar calculation as above, we can see that $$ \| \nabla \|_{L^2(\Omega)} = \sum_{k \geq 1} |w_k|^2 \, \lambda_k^2 $$

Now, formally, let’s write, for some $i \in \mathbb{N}$, $\nabla^i w$. Then, using the previous resuls, we get (at least formally) $$ \| \nabla^ iw \|_{L^2(\Omega)} = \sum_{k \geq 1} |w_k|^2 \lambda_k^{2i} $$ Now, suppose that $\lambda_k \to \infty$ as $k \to \infty$, but $\| \nabla w \|_{L^2(\Omega)} < \infty$. Now, consider: $$ \| \grad w \|_{L^2(\Omega)}^2 = (\grad w, \grad w)_{L^2(\Omega)} = \int_{\Omega} \grad w \cdot \grad w \, dx = - \int_{\Omega} w \nabla w \, dx \, \, , $$ using integration by parts and assuming $w$ is 0 on the boundary.

Now, using