Equivalence between two fractional Sobolev spaces

For $$s \in (0,1)$$, we consider the spectral fractional Laplacian \begin{align} (-\Delta)^{-s}u = \sum_{k=1}^{\infty}\lambda_k^{-s}(\phi_k,u)_{L^2}\phi_k \end{align} where \begin{align*} \begin{cases} -\Delta\phi_k = \lambda_k\phi_k &\mbox{in }\Omega,\\ \partial_\nu \phi_k = 0 &\mbox{on }\partial\Omega,\qquad k\geq 1. \end{cases} \end{align*} Then, for $$\alpha \in (0,1)$$, we define the space \begin{align*} H^\alpha(\Omega) = \{ u\in L^2(\Omega):~ \nabla (-\Delta)^{(\alpha-1)/2}u\in L^2(\Omega) \}, \end{align*} and the space \begin{align*} \tilde H^\alpha(\Omega) = \{ u\in L^2(\Omega)~:~ \sum_{j=1}^\infty \lambda_j^\alpha|(\phi_j,u)_{L^2(\Omega)}|^2<\infty\} \end{align*}

Questions:

1. Are these two spaces equivalent?

2. Do Sobolev embeddings, Poincaré inequalities, etc hold in them? If yes, where can I find a reference?

UPDATE. After the discussion in the comments, let me reformulate the question in a way that avoids (hopefully) some issues:

Let $$(\phi_k)\subset L^2(\Omega)$$ be a complete orthonormal system for $$\mathcal L^2(\Omega)= \{ u\in L^2(\Omega):\int_\Omega u d x =0 \}$$ composed of eigenfunctions of $$-\Delta$$ with homogeneous Neumann boundary conditions with eigenvalues $$(\lambda_k)\subset (0,\infty)$$.

Q1: Is \begin{align} (-\Delta)^{-s}u = \sum_{k=1}^{\infty}\lambda_k^{-s}(\phi_k,u)_{L^2}\phi_k \end{align} well-defined?

Q2: Define \begin{align*} \mathcal H^\alpha(\Omega) = \{ u\in \mathcal L^2(\Omega):~ \nabla (-\Delta)^{(\alpha-1)/2}u\in L^2(\Omega) \}, \end{align*} and the space \begin{align*} \tilde{\mathcal H}^\alpha(\Omega) = \{ u\in \mathcal L^2(\Omega)~:~ \sum_{j=1}^\infty \lambda_j^\alpha|(\phi_j,u)_{L^2(\Omega)}|^2<\infty\} \end{align*} Are these spaces well-defined and are they equivalent?

• Not sure what $(-\Delta)^{-s}$ does to the constant function $\phi_1$ (which corresponds to $\lambda_1 = 0$). If $\phi_1$ is not in the domain of $(-\Delta)^{-s}$, then it belongs to $\tilde{H}^\alpha$, but not to $H^\alpha$, I guess. Commented Sep 16, 2022 at 14:06
• @MateuszKwaśnicki I'm not quite sure what you mean. What is the domain of $(-\Delta)^{-s}$? Why are constants in $\tilde H^\alpha$ but not in $H^\alpha$?
– Zac
Commented Sep 16, 2022 at 14:42
• Just asking for clarification. Currently, as far as I understand, we have $\lambda_1 = 0$ and the term $k = 1$ in the definition of $(-\Delta)^{-s}$ involves $\lambda_1^{-s} = 0^{-s}$. Commented Sep 16, 2022 at 19:32
• @MateuszKwaśnicki is right: the Neumann laplacian is not invertible because its leading eigenvalue is zero. Commented Sep 17, 2022 at 10:49
• Here are two possibilities for correcting this bug with the leading eigenvalue to avoid division by zero: (1) you can change the boundary conditions to Dirichlet or (2) you can shift the operator as in, e.g., Section 3.2 of link.springer.com/article/10.1007/s40072-020-00175-6 Commented Sep 17, 2022 at 14:22

Yes, they are. Since $$(\phi_j,(-\Delta)^\eta u) = \lambda_j^\eta (\phi_j,u)$$, it suffices to consider the case $$\alpha = 1$$. Since, for any closed operator $$A$$, the domain of $$A$$ coincides with that of $$(A^*A)^{1/2}$$ (their graph norms are the same), it remains to note that, for $$\nabla$$ as in the question, $$\nabla^*\nabla$$ coincides with the Neumann Laplacian, which is the case if $$\Omega$$ is regular enough.
• Hi Martin, How are you? It seems $\nabla$ is a derivative operator and not endowed with any boundary conditions. So, it's unclear why "$\nabla^*\nabla$ coincides with the Neumann Laplacian." That's why the answer I provided discussed the boundary terms that arise from integrating by parts. Not sure why those boundary terms are necessarily bounded for functions in $H^{\alpha}(\Omega)$ ... Commented Sep 16, 2022 at 10:18
• @NawafBou-Rabee Take the case of $\Omega = [0,1]$. By definition, the domain of $\nabla^*$ consists of those $f \in L^2$ such that there exists $g \in L^2$ with $(g,u) = (f,\nabla u)$ for all $u$ in the domain of $\nabla$. For $f$ smooth, this forces it to vanish at the boundary since otherwise a boundary term appears which cannot be generated by $g$. As a consequence, functions in the domain of $\nabla^*\nabla$ must have vanishing gradient at the boundary. Commented Sep 16, 2022 at 13:21