Kantorovich polynomials

Remarks

Let $F(x) = \int_0^x f(t) \, dt$ and let $B_{n+1}$ be the Bernstein polynomial operator of degree $n + 1$. Then $K_n(f) = (B_{n+1} F)'$, by the formula for the derivative of a Bernstein polynomial. Each $K_n$ is a positive linear operator. For $f \in L^p[0, 1]$ with $1 \le p < \infty$, $K_n(f) \to f$ in the $L^p$ norm, and for continuous $f$ the convergence is uniform.

Example: for $f(t) = t$, the integral over $[k/(n+1), (k+1)/(n+1)]$ is $\frac{2k + 1}{2(n+1)^2}$. Hence $K_n(f)(x) = \sum_{k} \binom{n}{k} x^k (1 - x)^{n-k} \frac{2k + 1}{2(n + 1)} = \frac{nx + 1/2}{n + 1}$, which tends to $x$.

Non-example: the Bernstein polynomial $B_n f(x) = \sum_k \binom{n}{k} x^k (1 - x)^{n-k} f(k/n)$ is not a Kantorovich polynomial. It uses the point values $f(k/n)$, so it changes when $f$ is changed on the null set $\set{0, 1/n, \dots, 1}$, while $K_n(f)$ does not.