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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.

The Bernstein polynomial isIt has degree at most $n$, since cancellation can lower the degree. For $0\le x\le 1$, the Bernstein basis weights are nonnegative and their sum is $(x+(1-x){)}^n=1$ by the binomial theorem. ConsequentlyThus the Bernstein polynomial defines a positive linear operator on continuous functions satisfying $\Vert B_{n}f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$. Since $B_{n}1=1$, its operator norm in the supremum norm is exactly one.

The Bernstein polynomial is the positive weighted averageFor $0\le x\le 1$, the Bernstein basis weights are nonnegative and the binomial theorem gives ${\sum }_{k=0}^n{b}_{n,k}(x)=1$. Hence $|B_n(f,x)|\le {\sum }_k\Vert f{\Vert }_{\infty }{b}_{n,k}(x)=\Vert f{\Vert }_{\infty }$. In particular,This is the contraction property of the positive linear operator on continuous functions defined by the Bernstein polynomial.

Averaging the given Dirichlet kernels gives the normalized Fejér kernelThe geometric-series identity, or summing the sine terms, proves the second equality. At multiples of $2\pi$ use its removable value $n/(2\pi )$. Therefore $K_n\ge 0$, andis a positive linear operator on continuous functions. The Fourier partial sums, viewed as orthogonal projections, fix $1$, so ${\sigma }_{n}1=1$ and $\int K_n=1$. They also fix $\sin x$ and $\cos x$ for degrees at least one, whereas $s_0$ annihilates them. ConsequentlyAll three periodic Korovkin theorem tests converge uniformly. Hence Fejér sums converge uniformly for every continuous periodic function: