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The Sturm-Liouville eigenfunction expansion of a suitable real function $f$ isComparing this identity with the sampling property of the Dirac delta distribution gives the completeness kernelThis is an identity of distributions, rather than a pointwise-convergent series of ordinary functions.

Assuming zero is not an eigenvalue, define the spectral Green function of a regular Sturm-Liouville problem byApplying ${\mathcal{L}}_x$ gives $w(x){\sum }_n{y}_n(x)y_n(\xi )=\delta (x-\xi )$: multiplying the preceding completeness kernel by $w(x)/w(\xi )$ leaves the Dirac delta distribution unchanged. Each term also obeys the same homogeneous boundary conditions. There is no extra weight factor in $G$ for the ordinary source measure $d\xi$.

If zero is an eigenvalue, the inverse Green's function does not exist on all forcing functions: the forcing must obey the solvability condition at a Sturm-Liouville eigenvalue. Omitting the zero mode instead constructs a reduced inverse, with its projection removed from the right-hand side. This exception occurs in part (c) at $\alpha =1$.

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.

Write the equation as $(L+\beta )y=-\cos x/x$ with the positive Sturm-Liouville operator from part (a). A Sturm-Liouville eigenfunction expansion therefore gives $a_n=-f_n/(4n^2+\beta )$, where weighted orthogonal projection givesHere the product-to-sum formulas reduce the integral to half the sum of the integrals of $\sin ((2n+1)x)$ and $\sin ((2n-1)x)$. ConsequentlyAt $x=0$, use the continuous limits of the terms. The coefficients are $O(n^{-3})$, so the resulting endpoint series converges. Completeness is the ordinary Fourier sine series completeness after $v=xy$. For an independent check, solving $v^{\prime \prime }-\beta v=\cos x$ with zero endpoint values givesIts numerator vanishes at zero, and it has the same orthogonal projection coefficients. Positivity of $L+\beta$ proves uniqueness.

For a positive weight $w$, the weighted inner product of functions is $⟨f,g{⟩}_w=\int \overline{f}gw$. It is the usual inner product in the L2 space for the measure $w(x)dx$. A Sturm-Liouville eigenfunction expansion uses this weight to make its eigenfunctions orthogonal.

The weight specifies how different locations contribute to a squared error: $\Vert f-g{\Vert }_w^2=\int |f-g{|}^{2}w$. More weight makes an error in that region more costly. The weighted Euclidean inner product makes the same choice for a finite list of coordinates.

For an explicit real-function example on $[0,1]$, let $w(x)=1+x$ and approximate $f(x)=x$ by a constant function. The orthogonal projection onto a finite-dimensional subspace spanned by the function $1$ givesThe best unweighted constant is $1/2$. Giving greater weight to larger $x$ moves the best constant upward, so a weighting choice changes the approximation even though the allowed family of functions is unchanged.

On $(0,L)$, the differential operator $\mathcal{L}={\partial }_x^2+\beta {\partial }_x=e^{-\beta x}{\partial }_x(e^{\beta x}{\partial }_x)$ with homogeneous Neumann boundary conditions is self-adjoint in the weighted inner product with weight $e^{\beta x}$. Its eigenfunctions and nonnegative decay rates areTheir squared norms are $N_0=(e^{\beta L}-1)/\beta$ and $N_k=L[1+{\beta }^2/(4q_k^2)]/2$. The Sturm-Liouville eigenfunction expansion givesThis kernel acts against the measure $e^{\beta y}dy$, not Lebesgue measure alone. It is symmetric in $x,y$; the transition density against Lebesgue measure is $e^{\beta y}{K}_{\beta }(x,y,t)$.