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Projecting the Sturm-Liouville eigenfunction expansion onto $y_m$ using the weighted inner product givesFor the partial sum $f_N={\sum }_{n=0}^N{a}_n{y}_n$, orthonormality therefore givesFor $f$ of finite weighted norm, the given completeness means that $f_N\to f$ in that norm. The triangle inequality implies $|\Vert f_N{\Vert }_w-\Vert f{\Vert }_w|\le \Vert f_N-f{\Vert }_w\to 0$. Taking the limit of the finite identity proves Parseval's identity:This argument justifies the limiting step without assuming that a double infinite series can be integrated term by term. As usual, the expansion is understood for square-integrable functions with respect to the weighted measure; the boundary and smoothness assumptions of the regular Sturm-Liouville problem provide the usual stronger interpretations when needed.

Only two coefficients in the Sturm-Liouville eigenfunction expansion are nonzero: $a_0=2$ and $a_1=2-\alpha$. Since $\alpha \ne 2$, the result of part (i) simplifies toUsing Parseval's identity, the prescribed weighted norm becomesThus $1-\alpha =\pm 2$, so $\alpha =-1$ or $3$. Positivity selectsThe standing nonresonance assumption still applies: this answer requires that $3$ is not one of the unspecified higher eigenvalues. Their actual values are not supplied. If a higher eigenvalue were $3$, the stated exclusion would fail and the data would not have an admissible $\alpha$ under that exclusion.

For a nonzero admissible radial profile define the radial diffusion Rayleigh quotientMinimizing the numerator subject to ${\int }_0^{1}r{R}^{2}dr=1$ introduces a multiplier $\nu$. The Euler-Lagrange equation for $r\kappa R^{\prime 2}-\nu r{R}^2$ isThe allowed variations vanish at $r=1$, and the natural condition at the centre is zero radial flux. Thus the minimum, when attained, is the smallest decay eigenvalue ${\nu }_0$. Equivalently, expansion in weighted-orthogonal eigenfunctions makes $\mathcal{R}[R]$ a weighted average of their decay eigenvalues. The Rayleigh-Ritz variational principle therefore givesAdmissible trial functions must obey the outer boundary condition, be nonzero and have finite weighted norm and energy; they need not solve the differential equation. If the diffusion operator has no lowest eigenfunction, the same variational argument bounds the infimum of its spectrum. Existence of a fundamental diffusion mode is an additional assumption for strongly vanishing diffusivity, not a consequence of the identity alone.

For the proposed trial profile, $R^{\prime }=-q{r}^{q-1}$. Its weighted energy isThe squared weighted norm isBoth are finite for $q>0$ and $n\ge 0$, and $R(1)=0$. Therefore the Rayleigh-Ritz variational principle givesWhen a fundamental diffusion mode exists, its decay rate is the infimum and is bounded above by this expression.

The existence qualification matters for the full range of $n$ printed in the question. For $n>2$, trial functions concentrated within radius $\epsilon$ have quotient proportional to ${\epsilon }^{n-2}$, proving the loss of a spectral gap under strongly degenerate radial diffusion. The infimum is zero and is not attained, because zero energy together with $R(1)=0$ forces the zero function. Thus there is no positive lowest-mode decay rate in that case. Even at $n=2$, the mode equation is $r^2{R}^{\prime \prime }+3r{R}^{\prime }+\nu R=0$: its characteristic powers are $r^{-1\pm \sqrt{1-\nu }}$, and for $\nu >0$ every nonzero solution is unbounded at the centre. Hence the asserted bounded fundamental mode cannot be assumed for all $n\ge 0$. The trial quotient remains valid, and the next part's $n=0$ case has the usual fundamental mode.

For $n=0$, a constant solution and its normalization give $y_0=1$. For $n=1$, substitute $y=ax+b$: the equation gives $b=0$, and the normalization gives $a=1$. For $n=2$, put $y=a{x}^2+bx+c$. Comparing coefficients gives $b=0$ and $c=-a/2$; then $y(1)=1$ fixes $a=2$. ThusThese are the first three Chebyshev polynomials. For their squared weighted norms, use $x=\cos \theta$, which converts $dx/\sqrt{1-x^2}$ into $-d\theta$:More generally, $y_n(\cos \theta )=\cos (n\theta )$ makes their orthogonality the usual cosine orthogonality.

The coefficient formula from part (b) and the Leibniz rule yield Rodrigues' formula directly:The leading coefficient is $(-1{)}^n/n!$, so $p_n^{(n)}=(-1{)}^n$. Substitute this differentiated expression into one factor in the squared weighted norm and apply integration by parts $n$ times:Every endpoint term contains a derivative of $e^{-x}{x}^n$ of order less than $n$: it vanishes at zero, and its exponential decay beats every polynomial at infinity. This proves the normalization of the ordinary Laguerre polynomials for all $n$, including $n=0$ without any integrations by parts.

Vector normalization converts a nonzero vector in a normed vector space into a unit vector. Absolute homogeneity of a norm givesThe direction is unchanged by multiplying by a positive real scalar. For $c>0$, $\Vert cv\Vert =c\Vert v\Vert$ and $\widehat{cv}=\hat{v}$; for a negative real scalar, $\widehat{cv}=-\hat{v}$. Thus $(3,4)$ and $(6,8)$ normalize to the same unit vector, while $(-3,-4)$ normalizes to its negative. This decomposition is useful for specifying a direction independently of a force, velocity or displacement magnitude.

In a complex normed vector space, a nonzero complex scalar $c$ instead gives $\widehat{cv}=(c/|c|)\hat{v}$. The remaining complex argument is not removed by normalization. For a pure quantum state, vectors differing only by this overall global phase represent the same physical state; vector normalization alone does not identify them.

The zero vector cannot be normalized this way. The construction also applies in a function space: an eigenfunction of a Sturm-Liouville problem can be scaled to unit weighted norm byThe positive weight $w$ specifies the weighted inner product. Multiplication by this nonzero constant preserves the eigenvalue equation and its homogeneous boundary conditions.