Therefore, we have $$\begin{aligned}
    \Delta_{\nu \circ F_{\mu}^{-1}, \Lambda}(f   \circ \check{F}_{\mu}^{-1} ) \circ F_{\mu}
    = \Delta_{\nu,\mu} ( f ) \quad \text{and} \quad f\in  \mathcal{D}^{*}(\Delta_{\nu, \mu}).
    \end{aligned}$$ It remains to verify that the boundary conditions are preserved. Combining the above with gives $$\begin{aligned}
 \nabla_{\mu}f(1)= \nabla_{\mu}f(0)+\int_{[0,1]} \Delta_{\nu,\mu}f(y)\;{\mathrm{d}}\nu(y) 
 &= \nabla_{\Lambda}f(0)+\int_{[0,1]} \Delta_{\nu,\mu}f ( \check{F}_{\mu}^{-1}(y) )\;{\mathrm{d}}\nu \circ F_{\mu}^{-1}(y) \\
 &= \nabla_{\Lambda}f(0)+\int_{[0,1]} 
 \Delta_{\nu \circ {F}_{\mu}^{-1}, \, \Lambda} (f \circ \check{F}_{\mu}^{-1})
 \;{\mathrm{d}}\nu \circ F_{\mu}^{-1}(y)
 = \nabla_{\Lambda} (f \circ \check{F}_{\mu}^{-1} )(1).  \qedhere\end{aligned}$$

For $\gamma=(\alpha,\beta )\in [0,{\pi/2}]^{2} $, set $$\begin{aligned}
    C_{\nu,\mu}^{\gamma} 
    \coloneqq \begin{cases}
    \left\{ f \in C_{\nu,\mu}  \colon f(0)=0, \ f(1)=0\right\}& \text{ if } \alpha=\beta=0, \\
    \left\{ f \in C_{\nu,\mu}  \colon f(0)=0\right\}& \text{ if }  \alpha=0, \ \beta \in (0,\pi/2],\\
    \left\{ f \in C_{\nu,\mu}  \colon f(1)=0\right\} &\text{ if } \alpha \in (0,\pi/2], \ \beta=0, \\
     C_{\nu,\mu
     }& \text{ if } \alpha,\beta \in (0,\pi/2].
    \end{cases}
    \end{aligned}$$ The set $\mathcal{D}^s_{\gamma}(\Delta_{\nu, \, \mu})$ is dense in $(C_{\nu,\mu}^{\gamma}, \lVert \cdot \rVert_{\infty})$.

This follows from , Lemma \[densecm\] and Theorem \[prop:laplace\_backward\].

For each $\gamma \in [0,{\pi/2}]^{2}$, the operator $\Delta_{\nu, \, \mu}$ with domain ${\mathcal{D}}_{\gamma}^w(\Delta_{\nu,\mu})$ is densely defined and self-adjoint.

The denseness follows from , Lemma \[adjoint\] and Theorem \[prop:laplace\_backward\]. Furhermore, for $f \in \operatorname{dom}(\Delta_{\nu,\mu}^*)$ by Theorem \[prop:laplace\_backward\] and Lemma \[identity\], $$\begin{aligned}
g\mapsto
\left\langle  f   ,\Delta_{\nu,\mu} 