= Solution
For the real <Sturm-Liouville problem>, multiply the equations for $y_n$ and $y_m$ by $y_m$ and $y_n$, respectively, and subtract. <Integration by parts> gives
$$
(\lambda_n-\lambda_m)\int_0^1wy_ny_mdx
=\left[p(y_ny_m'-y_my_n')\right]_0^1=0.
$$
The boundary term vanishes because both functions are zero at both endpoints. Distinct <eigenvalues> therefore imply weighted orthogonality. For equal indices the integral is the squared weighted norm, so
$$
\boxed{\int_0^1wy_ny_mdx=\delta_{nm}\int_0^1wy_n^2dx.}
$$
Let $c_n=\int_0^1wy_n^2dx>0$. For $y=\sum a_ny_n$ in the operator domain, with convergence sufficient to evaluate its quadratic energy, weighted orthogonality gives
$$
\int_0^1wy^2dx=\sum a_n^2c_n,\qquad
\int_0^1y\mathcal Ly\,dx=\sum\lambda_na_n^2c_n.
$$
For finite sums these identities follow immediately; the corresponding convergent energy expansion follows by approximation in the operator or quadratic-form domain. For nonzero $y$, the <Rayleigh quotient> is a weighted average of the <eigenvalues>, so
$$
\boxed{\frac{\int_0^1y\mathcal Ly\,dx}{\int_0^1wy^2dx}\ge\lambda_1.}
$$
Equality occurs precisely when only a lowest-eigenvalue component is present.
Assume completeness as permitted and take a sufficiently regular solution of the diffusion equation. It belongs to the weighted expansion space at each positive time; its modal coefficients satisfy $\dot a_n=-\lambda_na_n$. Alternatively differentiate its energy directly and use the preceding inequality:
$$
\boxed{\frac12\frac d{dt}\int_0^1wy^2dx
=\int_0^1wy\,y_tdx=-\int_0^1y\mathcal Ly\,dx
\le-\lambda_1\int_0^1wy^2dx.}
$$
Thus the squared weighted norm is at most its initial value times $e^{-2\lambda_1t}$. This describes decay when $\lambda_1>0$; positivity of $p,w$ alone does not require $\lambda_1>0$ if the potential $q$ can be negative. The displayed inequality remains valid regardless of that sign.
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