is also bounded.

\[Spectral\] Let $(\alpha,\beta) \in [0,\pi/2]^2$ , then the operator $\Delta_{\nu,\mu}$ with domain ${\mathcal{D}}_{\gamma}^{w}(\Delta_{\nu,\mu} )$ gives rise to an orthonormal (possibly finite) basis of eigenfunctions with eigenvalues $\lambda_{n}\leq 0$. If $L^{2}(\nu)$ is not finite dimensional then we have a countable number of eigenvalues with $\lim_{n \rightarrow \infty} -\lambda_n=\infty$, in particular $\Delta_{\nu,\Lambda}$ is an unbounded operator, otherwise there are only finitely many eigenfunctions and $\Delta_{\nu,\Lambda}$ is bounded.

At first we consider the case $\gamma=(\alpha,\beta) \in [0,\pi/2]^2\setminus\{ (\pi/2, \pi/2)\}$. Note that if $f \in {\mathcal{D}}^w_{\gamma}\left(\Delta_{\nu, \Lambda} \right)$ is an eigenfunction of $\Delta_{\nu, \Lambda}$ with eigenvalue $\lambda<0$, then applying Lemma \[Surjec\] gives $$\Delta_{\nu, \Lambda}f=\lambda f \Longleftrightarrow  \lambda^{-1} f = \Delta_{\nu, \Lambda}^{-1}f.$$ Then the statement follows directly from the spectral theorem for linear, compact and self-adjoint operators applied to $ \Delta_{\nu, \Lambda}^{-1}$.

For the case $\gamma=(\alpha,\beta)=(\pi/2,\pi/2)$ we have to consider the resolvent operator $R^{\lambda}_{\nu,\Lambda}\coloneqq(\lambda I -\Delta_{\nu,\Lambda})^{-1}$ with $\lambda>0$ , from the integral representation of the resolvent operator $R^{\lambda}_{\nu,\Lambda}\coloneqq(\lambda I -\Delta_{\nu,\Lambda})^{-1}$ of $\Delta_{\nu,\Lambda}$ with domain ${\mathcal{D}}^w_{\gamma}\left(\Delta_{\nu, \Lambda} \right)$ given in [@MR0314125 1.2], one may conclude that $R^{\lambda}_{\nu,\Lambda}$ is compact and self-adjoint, see also [@MR574035 Theorem 1, p. 251]. Again applying the spectral theorem proves the statement.

Clearly, $L^{2}(\nu)$ is finite dimensional if and 