space $$\begin{gathered}
\label{D01}
D(g_0,g_1)=s(x)^{g_0}s(\pi/2r-x)^{g_1}{{\cal D}},\qquad g_0,g_1>-1/2,\end{gathered}$$ where $$\begin{gathered}
\label{cD}
{{\cal D}}\equiv \left\{ f\in C^2([0,\pi/2r])\mid f'(0)=f'(\pi/2r)=0\right\}.\end{gathered}$$ This domain choice ensures that the ONB functions $e_n(g;x)$ belong to the domain. Indeed, from the paragraph containing (\[ana\]) we see that the function on the r.h.s. of (\[ana\]) belongs to ${{\cal D}}$, so that we have more generally $$\begin{gathered}
\label{Ran}
{\rm Ran}\, ({{\cal I}}(g))\subset D(g_0,g_1).\end{gathered}$$

From $e_n(g)\in D(g_0,g_1)$ it already follows that $D(g_0,g_1)$ is dense. But this feature can also be derived quite easily without invoking ${{\cal I}}(g)$. Indeed, ${{\cal D}}$ contains the functions $\cos(2nrx)$, $n\in{{\mathbb N}}$, which are well known to yield an orthogonal base of ${{\cal H}}$. Therefore, the assumption $f\perp D(g_0,g_1)$ entails $$\begin{gathered}
\int_0^{\pi/2r}dxs(x)^{g_0}s(\pi/2r-x)^{g_1}\cos(2nrx)f(x)=0,\qquad \forall \, n\in{{\mathbb N}}.\end{gathered}$$ From the orthogonal base property we now deduce $f(x)=0$, so that $D(g_0,g_1)$ is dense.

We are now prepared for our second lemma, whose proof is relegated to Appendix \[appendix-A\].

\[lemma2.2\] Assume $g$ belongs to $\tilde{\Pi}$ . Then $H(g)$ is well defined and symmetric on the domain $D(g_0,g_1)$ given by , .

The proof of this lemma does not involve the HS operator ${{\cal I}}(g)$. In the next lemma we do use ${{\cal I}}(g)$ to obtain a short proof of essential self-adjointness on $D(g_0,g_1)$.

