(\[Rs\]) it follows that the weight function (\[w\]) can also be written $$\begin{gathered}
\label{w2}
  w(g;x)=p^{2g_0+2g_1}s(x)^{2g_0}s(\pi/2r-x)^{2g_1}R(x)^{2g_2}R(\pi/2r-x)^{2g_3}.\end{gathered}$$ It is positive on $(0,\pi/2r)$ for all $g\in{{\mathbb R}}^4$, but integrable on $[0,\pi/2r]$ only for $g_0,g_1>-1/2$.

It now readily follows that for $\Psi(g;x,y)$ to belong to the kernel Hilbert space $$\begin{gathered}
\label{cHK}
 {{\cal H}}_K\equiv L^2\left([0,\pi/2r]^2,dxdy\right),
 \end{gathered}$$ the necessary and sufficient condition is that $g$ belong to the parameter set $$\begin{gathered}
\label{Pi}
 \Pi \equiv \left\{ g\in{{\mathbb R}}^4\mid g_0,g_1,g_0',g_1'>-1/2\right\}.
 \end{gathered}$$ Hence it is clear that the integral operator $$\begin{gathered}
\label{cI}
 ({{\cal I}}(g) f)(x)\equiv {\int_{0}^{\pi/2r}}dy\Psi(g;x,y)f(y),\qquad f\in{{\cal H}},\qquad
g\in\Pi,\end{gathered}$$ is well defined and HS.

It is not obvious, but true that its kernel is a zero-eigenvalue eigenfunction of a difference of Heun Hamiltonians: $$\begin{gathered}
\label{eiB}
(H(g;x)-H(g';y))\Psi(g;x,y)=0,\qquad x,y\in(0,\pi/2r).\end{gathered}$$ (Indeed, this is a corollary of Proposition 3.4 in I, cf. I(3.115).) At face value, this eigenfunction identity seems to have no bearing on the Hilbert space eigenfunctions of the two (generically) distinct Hamiltonians. A principal result of this paper is, however, that for $g$ in the restricted parameter set $$\begin{gathered}
\label{Pir}
\Pi_r\equiv \{ g\in\Pi \mid s_g>0\},\end{gathered}$$ the Hilbert space eigenvector ONBs (orthonormal bases) of $H(g)$ and $H(g')$ are given by functions $e_n(g;x)$ and $e_n(g';x)$, resp., which yield the singular value decomposition 