= Solution
Two <integrations by parts> give the boundary form of the <Sturm-Liouville operator>:
$$
\langle f,Lg\rangle-\langle Lf,g\rangle=\left[p\left(f'^*g-f^*g'\right)\right]_a^b.
$$
The interior terms cancel because $p,q$ are real. Thus the required endpoint condition is
$$
\boxed{\left[p(f'^*g-f^*g')\right]_a^b=0\quad\text{for every }f,g\text{ in the operator domain}.}
$$
For a regular interval, a standard <self-adjoint> domain is specified by one real homogeneous <Robin boundary condition> at each endpoint:
$$
A_a y(a)+B_a p(a)y'(a)=0,\qquad A_b y(b)+B_b p(b)y'(b)=0,
$$
where neither pair of real constants is identically zero. At each end, these conditions force the two boundary data vectors of $f,g$ to be proportional, making the displayed form zero. <Dirichlet boundary conditions> and <Neumann boundary conditions> are the cases $B=0$ and $A=0$. Coupled conditions also work, for example $y(a)=y(b)$ and $p(a)y'(a)=p(b)y'(b)$, whose two endpoint contributions cancel. Self-adjointness uses the maximal common domain with these conditions, not an arbitrary smaller symmetric restriction.
The <weighted inner product> must be nondegenerate. The usual interpretation is $w>0$ almost everywhere, allowing isolated endpoint zeros as in part (b), with the operator defined on its appropriate weighted domain. Merely allowing a weight that is zero on an entire interval would not by itself define the stated operator or a positive-definite <inner product> on ordinary functions.
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