Proof: Weierstrass Approximation via Bernstein Polynomials

This proof of the Weierstrass Approximation Theorem uses Bernstein polynomials, which provide an explicit, constructive approximation. The proof combines probability (law of large numbers) with analysis (uniform continuity).

Statement

Theorem7.1Weierstrass Approximation Theorem

If $f : [0, 1] \to \mathbb{R}$ is continuous, then for every $\epsilon > 0$ , there exists a polynomial $p$ such that $|f(x) - p(x)| < \epsilon$ for all $x \in [0, 1]$ .

Proof via Bernstein Polynomials

Proof

Define the Bernstein polynomial of degree $n$ :

$B_n(f)(x) = \sum_{k=0}^n f(k/n) \binom{n}{k} x^k (1-x)^{n-k}.$

We show $B_n(f) \to f$ uniformly on $[0, 1]$ .

Step 1: Since $f$ is continuous on the compact set $[0, 1]$ , $f$ is uniformly continuous. Given $\epsilon > 0$ , choose $\delta > 0$ such that $|f(x) - f(y)| < \epsilon/2$ whenever $|x - y| < \delta$ .

Step 2: For fixed $x \in [0, 1]$ , the Bernstein polynomial can be interpreted probabilistically: if $X \sim \text{Binomial}(n, x)$ , then

$B_n(f)(x) = \mathbb{E}[f(X/n)].$

By the law of large numbers, $X/n \to x$ in probability as $n \to \infty$ . For large $n$ , $|X/n - x| < \delta$ with high probability, so $|f(X/n) - f(x)| < \epsilon/2$ with high probability.

Step 3 (Rigorous): Let $M = \sup_{[0,1]} |f|$ . Split the sum:

$|B_n(f)(x) - f(x)| = \left|\sum_{k=0}^n \left(f(k/n) - f(x)\right) \binom{n}{k} x^k (1-x)^{n-k}\right|.$

$|B_n(f)(x) - f(x)| < \epsilon.$

The convergence is uniform in $x$ (the same $n$ works for all $x$ ).

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Summary

Bernstein polynomials provide a constructive proof of Weierstrass:

Explicit formula: $B_n(f)(x) = \sum_{k=0}^n f(k/n) \binom{n}{k} x^k (1-x)^{n-k}$ .
Convergence follows from uniform continuity and probabilistic estimates.
Gives a practical method for polynomial approximation.

See Weierstrass Theorem for applications.