From: "Carl C. K. Mikkelsen" <spock@imf.au.dk>
Subject: Re: Reference on Bernstein Polynomials
Date: Fri, 26 Nov 1999 12:34:17 +0100
Newsgroups: sci.math.num-analysis,sci.math
Keywords: Weierstrass Approximation theorem

Matthew T. Brenneman wrote:
> 
> Hi,
> 
> Could you please tell me of a good (introductory) exposition on Sergi
> Bernstein's constructive proof (via Bernstein polynomials) of the
> Weierstrass Approximation theorem?
> 
> Thank you very much,
> 
> Matt Brenneman
Dear Mr. Brenneman.

I do not know the proper reference but the proof is very simple, if you
are familiar with Markov's inequality.
I have included a sketch of the proof written in latex, .dvi and
postscript. It is from a note written by one of my teachers, Professor
Ebbe Thue Poulsen of the University of Aarhus.

Your sincerely

Carl Christian Kjelgaard Mikkelsen
A student at the University of Aarhus, Denmark.


\documentclass[a4paper]{article}
\usepackage{amsmath,amssymb}

\begin{document}

Given $\delta \in (0,1)$ we have the following estimate
\begin{multline*}
\left| \left( \sum_{k=0}^n \binom{n}{k} x^k(1-x)^{n-k}f\left(\frac{k}{n}\right)\right) -
  f(x) \right| = \left| \sum_{k=0}^n \binom{n}{k} x^k(1-x)^{n-k}\left(f\left(\frac{k}{n}\right)
  - f(x) \right) \right| \\ 
\leqq \left| \sum_{k : |x - \frac{k}{n}| < \delta}^n \binom{n}{k} x^k(1-x)^{n-k}\left(f\left(\frac{k}{n}\right)
  - f(x) \right) \right| \\ + \left| \sum_{k : |x - \frac{k}{n}| \geqq \delta}^n
  \binom{n}{k} x^k(1-x)^{n-k} \left(f\left(\frac{k}{n}\right)
  - f(x) \right) \right|.
\end{multline*}
Using the uniform continuity of $f$ we chose $\delta$ 
\begin{equation*}
|f(x) - f(y)| < \epsilon, \qquad |x-y| < \delta
\end{equation*}
and we find
\begin{equation*}
\left| \sum_{k : |x - \frac{k}{n}| < \delta}^n \binom{n}{k} x^k(1-x)^{n-k}\left(f\left(\frac{k}{n}\right)
  - f(x) \right) \right | \leqq \epsilon
\end{equation*}
Let M be an upper bound for $|f|$. Then
\begin{equation*}
\left| \sum_{k : |x - \frac{k}{n}| \geqq \delta}^n
  \binom{n}{k} x^k(1-x)^{n-k} \left(f\left(\frac{k}{n}\right)
  - f(x) \right) \right| \leqq 2 M \sum_{k : |x - \frac{k}{n}| \geqq \delta}^n
  \binom{n}{k} x^k(1-x)^{n-k}
\end{equation*}
And now the miracle occurs. The sum
\begin{equation*}
\sum_{k : |x - \frac{k}{n}| \geqq \delta}^n
  \binom{n}{k} x^k(1-x)^{n-k}
\end{equation*}
is interpreted as the probability $P(|nx - Y| \geqq n\delta)$, where $Y$
is a stocastic variable with a binomial density $B(n,x)$. Behold $nx$
is exactly the mean value of such a distribution and we may now use
Markov's inequality
\begin{equation*}
P(|\mu - Y| \geqq t) \leqq \frac{Var(Y)}{t^2}
\end{equation*}
Thus
\begin{equation*}
\sum_{k : |x - \frac{k}{n}| \geqq \delta}^n
  \binom{n}{k} x^k(1-x)^{n-k} \leqq \frac{nx(1-x)}{n^2\delta^2} \leqq \frac{1}{4n\delta^2}
\end{equation*}
Ofcourse I didn't discover this wonderfull proff by myself. I found it
in a lecture note written by one of my teachers : Professor Ebbe
Thue Poulsen of The University of Aarhus, Denmark.
\end{document}

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