= Solution
On its specified domain of admissible functions, a <self-adjoint operator> equals its adjoint, including the adjoint domain. Its integration identity on the interval is
$$
\boxed{\langle u,Lv\rangle=\langle Lu,v\rangle,\qquad\langle u,v\rangle=\int_a^b u^*(x)v(x)\,dx.}
$$
For a differential operator, the boundary conditions must make the boundary form from <integration by parts> vanish. Formal symmetry of the differential expression alone does not specify a self-adjoint boundary-value problem.
Apply this identity to two <eigenfunctions>. Since the weight is real,
$$
\lambda_n\int_a^b wy_m^*y_n\,dx=\int_a^b y_m^*Ly_n\,dx=\int_a^b(Ly_m)^*y_n\,dx=\lambda_m^*\int_a^b wy_m^*y_n\,dx.
$$
First take $m=n$: positivity of $\int_a^b w|y_n|^2dx$ shows $\lambda_n=\lambda_n^*$, so the <eigenvalues> are real. For distinct eigenvalues, subtraction therefore gives
$$
\boxed{\int_a^b w(x)y_m^*(x)y_n(x)\,dx=0\quad(\lambda_m\ne\lambda_n).}
$$
This is <orthogonality of Sturm-Liouville eigenfunctions> in the <weighted inner product>, rather than necessarily in the unweighted inner product.
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