= 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>. Hence
$$
U((\cdot-t)^2)(t)=0.
$$
Apply the same positive-barrier argument as in part (a), now with no moment errors. It gives $|Uf(t)-f(t)|\le\varepsilon$ for every $t$ and every $\varepsilon>0$. Therefore
$$
\boxed{Uf=f\text{ for every }f\in C[0,1],\qquad U=I.}
$$
The 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.
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