Proof. Integrating by parts, with \(u(a) = u(b) = 0\), \(\int pu'^2 = -\int u(pu')' \), so the numerator is \(\langle u, Lu\rangle \). For \(u = \phi _n\) this is \(\lambda _n\langle \phi _n, \phi _n\rangle _w\), so \(R[\phi _n] = \lambda _n\). For general \(u\), normalise the eigenfunctions so that \(\langle \phi _m, \phi _n\rangle _w = \delta _{mn}\) and expand \(u = \sum c_n\phi _n\) by Theorem 9.3.3(iii); then (with a limiting argument that we omit) the numerator is \(\sum \lambda _nc_n^2\) and the denominator \(\sum c_n^2\), so \[ R[u] = \frac {\sum _n\lambda _nc_n^2}{\sum _nc_n^2} \geq \lambda _1 , \] with equality only if \(c_n = 0\) whenever \(\lambda _n > \lambda _1\), that is \(u\) is a multiple of \(\phi _1\). If \(u\) is orthogonal to \(\phi _1, \dots , \phi _{n-1}\), the first \(n - 1\) coefficients vanish and the same argument gives \(R[u] \geq \lambda _n\). □