KindiakMath

The Bernstein Approximants

Recall the binomial theorem, which states that

\displaystyle (a + x)^n = \sum_{\nu = 0}^n {n \choose \nu} a^{n-\nu} x^\nu.

In particular, setting a = 1-x,

\displaystyle 1 = ((1-x) + x)^n = \sum_{\nu = 0}^n {n \choose \nu} x^{\nu} (1-x)^{n-\nu}.

Definition 1. For each n, define the Bernstein basis polynomial of degree n and parameter \nu = 0,1,\dots,n by

\displaystyle p_{\nu, n}(x) := {n \choose \nu} x^{\nu}(1-x)^{n-\nu}.

By the binomial theorem, \sum_{\nu = 0}^n p_{\nu, n}(x) = 1. A Bernstein polynomial is a linear combination of Bernstein basis polynomials.

Problem 1. Evaluate \displaystyle \sum_{\nu = 0}^n \frac{\nu}{n} \cdot p_{\nu, n}(x) and \displaystyle \sum_{\nu = 0}^n \left(x - \frac{\nu}{n}\right)^2 p_{\nu, n}(x).

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Solution. Given the original binomial theorem, we differentiate twice to obtain

\displaystyle\begin{aligned} n(a + x)^{n-1} &= \sum_{\nu = 1}^n \nu {n \choose \nu} a^{n-\nu} x^{\nu-1}, \\ n(n-1)(a+x)^{n-2} &= \sum_{\nu = 2}^n \nu(\nu-1) {n \choose \nu} a^{n-\nu} x^{\nu-2}.\end{aligned}

Multiplying the i-th equation by x^i,

\displaystyle\begin{aligned} nx(a + x)^{n-1} &= \sum_{\nu = 0}^n \nu {n \choose \nu} a^{n-\nu} x^{\nu}, \\ n(n-1)x^2(a+x)^{n-2} &= \sum_{\nu = 0}^n \nu(\nu-1) {n \choose \nu} a^{n-\nu} x^{\nu} \\ &= \sum_{\nu = 0}^n \nu^2 {n \choose \nu} a^{n- \nu}x^{\nu} - nx(a + x)^{n-1}.\end{aligned}

Setting a = 1- x, the first equation reduces to

\displaystyle \sum_{\nu = 0}^n \frac{\nu}n \cdot p_{\nu,n}(x) = x,

and the second equation reduces to

\displaystyle \sum_{\nu=0}^n \frac{\nu^2}{n^2} \cdot p_{\nu,n}(x) = x^2+\frac{x(1-x)}n .

Therefore, by the binomial theorem,

\begin{aligned} \sum_{\nu = 0}^n \left(x - \frac{\nu}{n}\right)^2 p_{\nu, n}(x) &= x^2 \cdot \sum_{\nu=0}^n p_{\nu,n}(x) - 2x \cdot \sum_{\nu=0}^n \frac{\nu}{n} \cdot p_{\nu,n}(x) + \sum_{\nu=0}^n \frac{\nu^2}{n^2} \cdot p_{\nu,n}(x) \\ &= x^2 \cdot 1 - 2x \cdot x + x^2+\frac{x(1-x)}n = \frac{x(1-x)}n. \end{aligned}

Problem 2. Let f : [0, 1] \to \mathbb R be (uniformly) continuous. Construct a sequence f_n \in \mathbb R[x] of Bernstein polynomials such that f_n \to f uniformly.

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Solution. For each n, use Definition 1 to define the Bernstein polynomial

\displaystyle f_n(x) = \sum_{\nu = 0}^n f(\nu/n) p_{\nu, n}(x).

We claim that f_n \to f uniformly. Since f is continuous, denote M := \max_{x \in [0,1]}f(x). Fix \epsilon > 0. Since f is uniformly continuous, for any k > 0, there exists \delta > 0 such that

|r-s| < \delta \quad \Rightarrow \quad |f(r) - f(s)| < k \cdot \epsilon.

By the binomial theorem,

\begin{aligned} f_n(x) - f(x) = \sum_{\nu = 0}^n (f(\nu/n) - f(x)) p_{\nu, n}(x). \end{aligned}

By the triangle inequality and Problem 1,

\displaystyle \begin{aligned} |f_n(x) - f(x)| &\leq \sum_{|\nu/n - x| < \delta} |f(\nu/n) - f(x)| p_{\nu, n}(x) + \sum_{|\nu/n - x| \geq \delta} |f(\nu/n) - f(x)| p_{\nu, n}(x) \\ &\leq \sum_{|\nu/n - x| < \delta} (k \cdot \epsilon) \cdot p_{\nu, n}(x) + \sum_{|\nu/n - x| \geq \delta} 2M \cdot p_{\nu, n}(x) \\ &= (k \cdot \epsilon) \cdot \sum_{|\nu/n - x| < \delta}p_{\nu, n}(x) + 2M \cdot \delta^{-2} \cdot \sum_{|\nu/n - x| \geq \delta} \left( \frac{\nu}n - x \right)^2 p_{\nu, n}(x) \\ &\leq k \cdot \epsilon \cdot 1 + 2M \cdot \delta^{-2} \cdot \frac{x(1-x)}{n} \\ &= k \cdot \epsilon + \frac{M}{2\delta^2} \cdot \frac 1n. \end{aligned}

To establish uniform convergence, set k = 1/2 and choose N = \lceil M/(\epsilon\delta^2) \rceil.

Problems 1 and 2 establish the Stone-Weierstrass theorem in one dimension.

Theorem 1 (Stone-Weierstrass Theorem). For any continuous function f : [0, 1] \to \mathbb R, there exists a sequence \{f_n\} of polynomials such that f_n \to f uniformly.

—Joel Kindiak, 30 May 25, 1605H

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