WeierstrassApproximationTheorem

The Weierstrass approximation theorem states that every continuous function defined on a closed interval [a, b] can be uniformly approximated as closely as desired by a polynomial function.

We can illustrate the proof using Bernstein polynomials, which are a sequence of polynomials that can uniformly approximate any continuous function on a given interval.

Theorem:

If $f: [0,1] \to \mathbb{R}$ is continuous, then $B_n(f)(x) \to f(x)$ uniformly as $n \to \infty$.

Proof:

Fix $x \in [0,1]$ and $\epsilon > 0$. Since $f$ is continuous on a closed interval, it is uniformly continuous. So, there exists a $\delta > 0$ such that $|f(u) - f(v)| < \epsilon / 3$ whenever $|u - v| < \delta$.

Now let's consider the summation that defines $B_n(f)(x)$. It has a finite number of terms, let's say $M$ terms, for which $|\frac{k}{n} - x| < \delta$. We bound the remaining terms in the summation.

Since $f$ is continuous on a closed interval, it is bounded. Let $M = \max_{u \in [0,1]} |f(u)|$. Then for any $k$ such that $|\frac{k}{n} - x| \geq \delta$, we have $|f(\frac{k}{n}) B_{n,k}(x)| \leq M B_{n,k}(x)$.

By the properties of Bernstein polynomials, we have

$$ \sum_{|\frac{k}{n} - x| \geq \delta} B_{n,k}(x) \leq \sum_{j=0}^{\delta n} B_{n,j}(x) + \sum_{j=(1-\delta)n}^n B_{n,j}(x) \leq 2 \delta, $$

which tends to $0$ as $n \to \infty$.

Therefore, for sufficiently large $n$, $|B_n(f)(x) - f(x)| < \epsilon$ for every $x \in [0,1]$, and so $B_n(f)(x) \to f(x)$ uniformly as $n \to \infty$.

This completes the proof.