whenever $g_0,g_1> 3/2$. This is basically known, but for completeness we sketch a proof. Choosing at first $g\in{{\mathbb R}}^4$, consider the differential operator $H(g)$ restricted to ${{\cal C}}$. This is obviously a well-defined symmetric operator on ${{\cal H}}$, and it is easy to verify that the domain of its closure contains the space $$\begin{gathered}
{{\cal C}}'    \equiv     \big\{ \psi\in C^1([0,\omega_1])\mid \psi(0)=\psi(\omega_1)=\psi'(0)=\psi'(\omega_1)=0,
\nonumber \\
\phantom{{{\cal C}}'    \equiv     \big\{}{}  \psi'(x)\in C^1((0,\omega_1)), \psi''(x), x^{-2}\psi(x),(\omega_1-x)^{-2}\psi(x)\in{{\cal H}}\big\}.\end{gathered}$$ Now assume $\phi$ is orthogonal to $ (H(g)+i){{\cal C}}$. Then $\phi$ is a weak solution to the ODE $(H(g)-i)\phi=0$ on $(0,\omega_1)$, and by hypo-ellipticity it is a classical solution.

Choosing next $g_0,g_1> 3/2$, it follows from the Frobenius theory that near 0 we have $$\begin{gathered}
\phi(x)= x^{g_0}h_0(x),\end{gathered}$$ with $h_0(x)$ analytic at $x= 0$. (Indeed, $x^{1-g_0}$ is not in $L^2$ near 0.) Likewise, near $x=\omega_1$, $$\begin{gathered}
 \phi(x)= (\omega_1-x)^{g_1}h_1(x),
 \end{gathered}$$ with $h_1(x)$ analytic at $x=\omega_1$. But then we deduce that $\phi$ has all of the properties defining ${{\cal C}}'$, and since the closure of $H(g)$ is symmetric on ${{\cal C}}'$, it cannot have an eigenvector with nonreal eigenvalue. Thus $\phi=0$ and the core property follows.

After this digression we proceed to obtain our first main result, for which we again 