give here a quick overview and reduce all properties essentially on two key observations, namely the symmetry as a consequence of the Gauss-Green formula and surjectivity as stated in the following lemma.

\[Surjec\] For $\gamma=(\alpha, \beta)  \in [0,{\pi/2}] ^{2}$ and $* \in \{s,w\}$ we have that the map $\Delta_{\nu,\Lambda}: \mathcal{D}^{*}_{\gamma}( \Delta_{\nu,\Lambda} ) \rightarrow \S_{\gamma}^* $ is surjective, where $\S_{(\pi/2,\pi/2)}^*\coloneqq\left\{ g \in \S^* : \int g {\mathrm{d}}\nu=0 \right\}$ and $\S_{\gamma}^* \coloneqq \S^
*$ for $\gamma\not=(\pi/2,\pi/2)$.

For $\gamma=(\alpha, \beta)  \in [0,{\pi/2}]^{2}\setminus\{(\pi/2,\pi/2)\}$ we have that $\Delta_{\nu,\Lambda}$ is also injective and for its inverse $\Delta_{\nu,\Lambda}^{-1}: \S_{\gamma}^* \rightarrow   \mathcal{D}^{*}_{\gamma}( \Delta_{\nu,\Lambda} )$ we have the following kernel representation $$\begin{aligned}
 \Delta_{\nu,\Lambda}^{-1}g:x\mapsto \int K_{\alpha,\beta}(x,y) g(y) \,{\mathrm{d}}\nu(y)\end{aligned}$$ with continuous kernel given by $$\begin{aligned}
 K_{\alpha,\beta}(x,y)\coloneqq 
 \begin{cases}
 A_{\alpha,\beta}(1+\tan(\beta)-y)( \tan(\alpha)+x)+{\mathbbm{1}}_{[0,x]}(y)(x-y) &\mbox{for $\alpha,\beta\in [0,\pi/2)$},\\
 (y-1)(x+\tan(\alpha) )+{\mathbbm{1}}_{[0,x]}(y)(x-y)   &\mbox{for $\beta=\pi/2$ and $\alpha \in [0,\pi/2)$},\\
 (y-1-\tan(\beta))+{\mathbbm{1}}_{[0,x]}(y)(x-y)   &\mbox{for  $\alpha= \pi/2$ and $\beta \in [0,\pi/2)$}.
 \end{cases}\end{aligned}$$ with $A_{\alpha,\beta}:= {-1}/(1+\tan(\alpha)+\tan(\beta))$, for $\alpha,\beta\in (0,\pi/2)$. For the Neumann case $\alpha=\beta=\pi/2$ the operator $\Delta_{\nu,\Lambda}$ is not injective with kernel $ \Delta_{\nu,\Lambda}^{-1}(\{0\})=\R {\mathbbm{1}}$.

We only consider the case $\alpha,\beta \in [0,\pi/2)$ the other cases can be proved along the same lines. For fixed $ g \in \S^{*}$ and $x \in [0,1]$ we set $$f(x):=b\tan(\alpha)+bx+\int_{[0,x]} (x-y)g(y) {\mathrm{d}}\nu(y),$$ with $$b\coloneqq\dfrac{-1}{(1+\tan(\alpha)+\tan(\beta))} \left(\tan(\beta)\int_{[0,1]}g(y) {\mathrm{d}}\nu(y)+
\int_{[0,1]} (1-y)g(y) {\mathrm{d}}\nu(y)\right).$$ 