Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 2 / 16D / Solution

Multiply the Gegenbauer differential equation byIt becomes the Sturm-Liouville formFor two polynomial solutions of degrees $m\ne n$, multiply their equations crosswise, subtract, and integrate over $(-1,1)$. The boundary term vanishes because $\alpha >0$ and $(1-x^2{)}^{\alpha +1/2}\to 0$. Since the eigenvalues $n(n+2\alpha )$ are distinct,Thus $a=-1$, $b=1$, and the displayed function is the weight.

Every interior root is simple: a solution and its derivative cannot both vanish at an ordinary point of a second-order linear differential equation unless the solution is identically zero. Let $x_1,\dots ,x_k$ be all interior roots and put $q(x)={\prod }_i(x-x_i)$. If $k<n$, then $q$ has degree below $n$, so orthogonality givesBut $C_n^{\alpha }/q$ has no zero and constant sign on $(-1,1)$, whence the integrand $q(x{)}^2(C_n^{\alpha }(x)/q(x))w(x)$ has one strict sign except at finitely many points. Its integral cannot vanish, a contradiction. Therefore $k=n$: all $n$ roots are real, simple, and lie in $(-1,1)$.