= Solution
The appropriate <weighted inner product> and <Sturm-Liouville operator> are
$$
\boxed{\langle u,v\rangle_w=\int_1^R x^2\overline{u(x)}v(x)\,dx,\qquad
Lv=\frac1{x^2}\left[-(x(x-1)v')'+l(l+1)v\right].}
$$
For two functions obeying the endpoint conditions, <integration by parts> gives the <Lagrange identity for a Sturm-Liouville operator>
$$
\langle u,Lv\rangle_w-\langle Lu,v\rangle_w
=\left[x(x-1)(\overline{u'}v-\overline u v')\right]_1^R=0.
$$
At $R$, both $u$ and $v$ vanish. At $1$, the factor $x(x-1)$ tends to zero while the functions and their first <derivatives> remain bounded. Thus the boundary form vanishes at both ends.
This gives the regular <self-adjoint differential operator> realization: its singular-end condition is $\lim_{x\downarrow1}x(x-1)v'=0$, together with $v(R)=0$. The bounded classical <eigenfunctions> satisfy exactly this no-log choice. More precisely the operator domain is the maximal differential-operator domain in the weighted space with these separated endpoint conditions; the preceding calculation verifies its self-adjoint boundary form, rather than merely the differential expression's formal symmetry.
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