 Solution

The Lagrange identity for a Sturm-Liouville operator givesIn the intended class of functions whose values and first derivatives stay finite at the endpoints, the endpoint factor $a$ makes this entire boundary term vanish. Thus no separately prescribed endpoint values are needed to eliminate the boundary form. The same integration by parts argument then proves that the eigenvalues are real.

The admissible endpoint behaviour is part of this conclusion: it is the vanishing endpoint coefficient and the Sturm-Liouville boundary form, together with endpoint regularity, that replaces explicit boundary conditions. If singular functions are allowed, a vanishing coefficient alone need not suffice: $a(x)=x(1-x)$ and $v(x)=\log x$ have $a(x)v^{\prime }(x)\to 1$ as $x\downarrow 0$. A self-adjoint differential operator therefore still needs a specified domain of admissible functions.