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

Interpret convergence in $C[0,1]$ as uniform convergence, and work first with real continuous functions. Set $e_{n,j}=U_n(p_j)-p_j$, for $j=0,1,2$, and put $M=\Vert f{\Vert }_{\infty }$. The given barriers have one $\gamma \ge 0$ independent of both $t$ and $n$. A positive linear operator on continuous functions preserves non-strict inequalities, so application of $U_n$ and evaluation at $t$ giveStrict pointwise inequalities need not remain strict under a positive operator; non-strict ones suffice. Linearity and the cancellation of the exact moments yieldSubtracting $f(t)$ rather than $f(t)U_{n}1(t)$ therefore gives the quadratic barrier estimate for positive approximation operators:For fixed $\epsilon$, all three moment errors tend uniformly to zero. Thus ${\mathrm{limsup}}_n\Vert U_{n}f-f{\Vert }_{\infty }\le \epsilon$. Since $\epsilon$ was arbitrary,This completes the Korovkin theorem proof. Complex-valued functions follow by applying the real result to real and imaginary parts. Uniform convergence of the three test functions is essential to this uniform conclusion; purely pointwise test convergence should not be silently substituted.

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.