---
tag: J56JP23J57
kind: definition
name: Kantorovich polynomials
msc: [41A36, 41A10, 41A35]
defines: 8BXFDDSSG7
uses: [0JT92VK307, SASP14MA1Z, 092GEF38YD]
sources:
  - url: https://en.wikipedia.org/wiki/Bernstein_polynomial
---

## Definition

Let $f \colon [0, 1] \to \R$ be Lebesgue integrable and let $n \ge 0$ be an integer.
The *Kantorovich polynomial* of degree $n$ of $f$ is
$$K_n(f)(x) = (n + 1) \sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k} \int_{k/(n+1)}^{(k+1)/(n+1)} f(t) \, dt, \qquad x \in [0, 1].$$

## Remarks

Let $F(x) = \int_0^x f(t) \, dt$ and let $B_{n+1}$ be the [[Bernstein polynomial|0JT92VK307]] 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|SASP14MA1Z]], 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.