= Solution
Use the standard regular self-adjoint assumptions on the <Sturm-Liouville problem>, including positive weight $w$. The normalized <eigenfunctions> obey <orthogonality>
$$
\int_a^bw(x)y_m(x)^*y_n(x)dx=\delta_{mn}.
$$
Completeness means every function in the weighted $L^2$ space has expansion $f(x)=\sum_ny_n(x)\int_a^bw(\xi)y_n(\xi)^*f(\xi)d\xi$, with convergence in that space. Comparing the action of this kernel on test functions with the <Dirac delta> gives, in the distributional sense,
$$
\boxed{w(\xi)\sum_{n=1}^\infty y_n(\xi)^*y_n(x)=\delta(x-\xi).}
$$
If zero is not an <eigenvalue>, divide each term by $\lambda_n$. Since $Ly_n=\lambda_ny_n$, application of $L$ recovers the displayed delta expansion, proving
$$
\boxed{G(x;\xi)=w(\xi)\sum_{n=1}^\infty\frac{y_n(x)y_n(\xi)^*}{\lambda_n}.}
$$
The factor $w(\xi)$ is required because the source equation uses an ordinary, unweighted delta. It distinguishes the <normalized Sturm-Liouville spectral Green kernel> from a kernel integrated against weighted source measure. If there is a zero <eigenvalue>, an inverse for arbitrary forcing does not exist; the unqualified division by that <eigenvalue> is invalid.
For $L=-d^2/dx^2+\omega^2$ with real $\omega$ and zero endpoints on $[0,1]$, the weight is one, $y_n=\sqrt2\sin(n\pi x)$ and $\lambda_n=n^2\pi^2+\omega^2>0$. Substitution yields
$$
\boxed{G(x;\xi)=\sum_{n=1}^\infty\frac{2\sin(n\pi\xi)\sin(n\pi x)}{n^2\pi^2+\omega^2}.}
$$
The series is absolutely and uniformly convergent, although applying the differential operator is understood distributionally at its source point.
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