Weierstrass Approximation Theorem/Proof 1

Proof
Let $\map f t: \Bbb I = \closedint a b \to \R$ be a continuous function.

Introduce $\map x t$ with a rescaled domain:


 * $\map f t \mapsto \map x {a + t \paren {b - a} } : \closedint a b \to \closedint 0 1$

From now on we will work with $x: \closedint 0 1 \to \R$, which is also continuous.

Let $n \in \N$.

For $t \in \closedint 0 1$ consider the Bernstein polynomial:


 * $\ds \map {B_n x} t = \sum_{k \mathop = 0}^n \map x {\frac k n} \binom n k t^k \paren {1 - t}^{n - k}$

For $t \in \closedint 0 1$, $0 \le k \le n$, let:


 * $\map {p_{n, k} } t := \dbinom n k t^k \paren {1 - t}^{n - k}$

By the binomial theorem:


 * $\ds \sum_{k \mathop = 0}^n \map {p_{n, k} } t = 1$

Lemma 2
Now we construct the estimates.

Here $\ds \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n$ denotes the summation over those values of $k \in \N$, $k \le n$, which satisfy the inequality $\size {\dfrac k n - t} \ge \delta$.

For some $\delta > 0$ denote:


 * $\ds \map {\omega_\delta} x := \sup_{\size {t - s} < \delta} \size {\map x s - \map x t}$

Then:

where $\norm {\,\cdot \,}_\infty$ denotes the supremum norm.

Let $\epsilon > 0$.

By Continuous Function on Closed Real Interval is Uniformly Continuous, $\map x t$ is uniformly continuous.

We choose $\delta > 0$ such that $\map {\omega_\delta} x < \dfrac \epsilon 2$.

Choose $n > \dfrac {\norm x_\infty} {\epsilon \delta^2}$

Then:


 * $\norm {\map {B_n x} t - \map x t}_\infty < \epsilon$.