Equivalence between two fractional Sobolev spaces

Question

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?

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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?

Answer 1

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.