N}}}$, of eigenvectors that are also eigenvectors of $H(g)$. The functions $e_n(g;x)$ are real-valued and belong to ${D(g_0,g_1)}$. They are of the form , where $d_n(g;x)$ extends to a function that is analytic in $|\Im z|<\alpha/2$, even and $\pi/r$-periodic. Moreover, we have $$\begin{gathered}
\label{dnnz}
d_n(g;0)\ne 0,\qquad d_n(g;\pi/2r)\ne 0,\end{gathered}$$ and the spectrum of $H(g)$ is nondegenerate.

We have just proved the first assertion. The second and third one have been shown earlier too, cf. the paragraphs containing (\[ed\]) and (\[cD\]). To prove the last one, we first note that it follows from $e_n(g)\in{D(g_0,g_1)}$ that $e_n(g;x)$ is a classical solution to the ODE $H(g)e_n=\lambda_ne_n$ for a certain $\lambda_n\in{{\mathbb R}}$. From the Frobenius method it then follows that $d_n(g;x)$ satisfies (\[dnnz\]). Moreover, any solution to the same ODE that is linearly independent of $e_n(g;x)$ does not have the features encoded in (\[ed\]), so the eigenvalue $\lambda_n$ is simple.

Quite likely, for $g\in\tilde{\Pi}\setminus \Pi_r$ the $H(g)$-eigenvectors still belong to ${D(g_0,g_1)}$ and have the same features as for $g\in\Pi_r$. (If so, it follows as before that $\sigma(H(g))$ is simple.) However, we cannot prove this via the above arguments, since they hinge on exploiting the HS operator ${{\cal I}}(g)$ with $g$ restricted to $\Pi_r$.

It is 