 Solution

Write the endpoint condition in its linear Robin boundary condition form $y^{\prime }(1)=y(1)/\alpha$. For a positive eigenvalue $\lambda =q^2$, the left Dirichlet boundary condition gives $y=C\sin (qx)$, and the right condition givesFor each positive root $q_n$, the squared norm isConsequently the sine coefficients in any complete expansion areWhen $\alpha >1$, these are all modes and the Sturm-Liouville eigenfunction expansion gives exactly the stated sine series, with the usual convergence interpretation. Positivity of all eigenvalues follows from integration by parts and the Cauchy-Schwarz inequality:for a nonzero admissible $y$, since $y(1)={\int }_0^1{y}^{\prime }dx$.

However, the original PDF states only $\alpha >0$, and with that range its sine-only completeness claim is false. The missing modes are as follows. At $\alpha =1$, $\lambda =0$ admits $y=x$, so the correct expansion isFor $0<\alpha <1$, a negative eigenvalue $\lambda =-{\kappa }^2$ admits $y=\sinh (\kappa x)$, withThere is exactly one positive root: $\tanh \kappa /\kappa$ decreases strictly from $1$ to zero. For example, its derivative has the sign of $\kappa {\mathrm{sech}}^2\kappa -\tanh \kappa <0$, because $\sinh \kappa \cosh \kappa >\kappa$. Thus the corrected expansion isThese are the nonpositive modes in a Dirichlet-Robin interval expansion. As an explicit counterexample to the printed claim, at $\alpha =1$ take $f(x)=x$: it satisfies both boundary conditions, and each positive root obeysbecause $\tan q_n=q_n$. Thus all $a_n$ vanish despite $f\ne 0$. Likewise $f=\sinh (\kappa x)$ is a counterexample for $\alpha <1$, by orthogonality of Sturm-Liouville eigenfunctions with distinct eigenvalues. The formulas above give the full solution without silently assuming away the omitted modes.