This imposes the right boundary conditions on $f$ and consequently we have $f \in  \mathcal{D}^{w}_{\gamma}( \Delta_{\nu,\Lambda} ) $ with $\Delta_{\nu, \Lambda}f=g$.

As a direct consequence it follows that only in the Neumann case one has an eigenfunction with eigenvalue equal to zero.

Recall the following abstract facts on linear operators: Let us assume that $A: \operatorname{dom}(A) \subset H \rightarrow
H$ is a linear, symmetric and surjective operator on a Hilbert space $H$. Then one easily verifies that the annihilator of $\operatorname{dom}(A)$ is trivial. i.e. $\operatorname{dom}(A)^\perp=\{0\}$, and equivalently, $\operatorname{dom}(A)$ is dense in $H$. We can deduce further that $A$ is also self-adjoint: The inclusion $\operatorname{dom}(A) \subset \operatorname{dom}(A^*)$ holds by symmetry of $A$, where $A^*$ denotes the adjoint of $A$. For the reverse inclusion note that for fixed $f \in \operatorname{dom}(A^*)$ there exists, by surjectivity of $A$, an element $g \in \operatorname{dom}(A)$ such that $A^*f=Ag$. Then, using symmetry again, for each $h \in \operatorname{dom}(A)$, we have $\langle f ,A h \rangle = \langle  A^* f,h\rangle =
 \langle  A g,h \rangle =  
 \langle g ,  A h\rangle  $ and by surjectivity of $A$ we deduce $f=g \in \operatorname{dom}(A)$.

Now, we can apply this observation to our special setting, namely, for $\gamma \in [0,\pi/2]^2$ 