Weierstrass Approximation Theorem/Proof 1

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Theorem

Let $f$ be a real function which is continuous on the closed interval $\Bbb I = \closedint a b$.


Then $f$ can be uniformly approximated on $\Bbb I$ by a polynomial function to any given degree of accuracy.


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 1

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

$\Box$


Lemma 2

$\ds \sum_{k \mathop = 0}^n \paren {k - n t}^2 \map {p_{n, k} } t = n t \paren {1 - t}$

$\Box$


Now we construct the estimates.

\(\ds \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n \map {p_{n, k} } t\) \(\le\) \(\ds \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n \map {p_{n, k} } t \frac {\paren {k - n t}^2} {\delta^2 n^2}\) as $\dfrac {\paren {k - n t}^2} {\delta^2 n^2} \ge 1$
\(\ds \) \(\le\) \(\ds \frac 1 {n^2 \delta^2} \sum_{k \mathop = 0}^n \paren{k - n t}^2 \map {p_{n, k} } t\)
\(\ds \) \(=\) \(\ds \frac {t \paren {1 - t} } {n \delta^2}\) Lemma 2
\(\ds \) \(\le\) \(\ds \frac 1 {4 n \delta^2}\) $\forall t \in \closedint 0 1: \dfrac 1 4 \ge t \paren {1 - t}$

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:

\(\ds \size {\map {B_n x} t - \map x t}\) \(=\) \(\ds \size {\map {B_n x} t - \map x t \sum_{k \mathop = 0}^n \map {p_{n, k} } t}\)
\(\ds \) \(=\) \(\ds \size {\sum_{k \mathop = 0}^n \map x {\frac k n} \map {p_{n, k} } t - \map x t \sum_{k \mathop = 0}^n \map {p_{n, k} } t}\)
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = 0}^n \size {\map x {\frac k n} - \map x t} \map {p_{n, k} } t\) as $\map {p_{n, k} } t \ge 0$
\(\ds \) \(=\) \(\ds \sum_{k \mathop : \size {\frac k n \mathop - t} < \delta}^n \size {\map x {\frac k n} - \map x t} \map {p_{n, k} } t + \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n \size {\map x {\frac k n} - \map x t} \map {p_{n, k} } t\)
\(\ds \) \(\le\) \(\ds \map {\omega_\delta} x \sum_{k \mathop : \size {\frac k n \mathop - t} < \delta}^n \map {p_{n, k} } t + 2 \norm x_\infty \frac 1 {4 n \delta^2}\)
\(\ds \) \(\le\) \(\ds \map {\omega_\delta} x \cdot 1 + \frac {\norm x_\infty} {2 n \delta^2}\)

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:

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

$\blacksquare$


Source of Name

This entry was named for Karl Weierstrass.


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