Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 61 / 1 / b / Solution

The reproduction hypothesis gives $U1=1$, $U(\cdot )(t)=t$ and $U((\cdot {)}^2)(t)=t^2$. If “quadratic” were read as degree exactly two, the same identities follow by subtracting pairs of such reproduced polynomials. HenceApply the same positive-barrier argument as in part (a), now with no moment errors. It gives $|Uf(t)-f(t)|\le \epsilon$ for every $t$ and every $\epsilon >0$. ThereforeThe identity is itself a positive linear operator on continuous functions reproducing all quadratics, so existence as well as uniqueness is established. In particular, reproducing three small test functions is much stronger for a positive operator than merely fixing one finite-dimensional polynomial space without positivity.