Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / P / 3 / d / Solution

As explained by admissible polynomials for a singular inverse-linear weight, there is a domain issue: the constant polynomial does not belong to the weighted space, because ${\int }_0^1{x}^{-1}dx=\infty$. Hence the full space ${\Pi }_{n-1}$ is not a subspace of $V$, and the preceding inner-product projection theorem cannot be applied to it as stated.

Nevertheless a meaningful finite-error best approximation does exist after restricting to admissible polynomials. The target $x^n$ lies in $V$ for every $n\ge 1$, since its squared integrand near zero is $x^{2n-1}$. A polynomial $p\in {\Pi }_{n-1}$ has finite error exactly when $p(0)=0$: otherwise the residual tends to a nonzero constant and its weighted squared integral diverges; if $p(0)=0$, it has at least a simple zero and is integrable in the required norm. Thus the admissible approximation space isFor $n=1$ it is $\{0\}$, giving best approximation zero. For $n\ge 2$ it is a finite-dimensional inner-product subspace with positive definite Gram matrix, so it has a unique orthogonal projection of $x^n$. Therefore the full polynomial-space formulation is not defined inside $V$, but the finite-error interpretation has a unique best approximation. Declaring that no best approximation can exist under any interpretation would miss this admissible subspace.