 Solution

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.