Weierstrass approximation theorem

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• and really crap this page needs some formal logic saying what it shows, ie:
  • $\forall f\in C([0,1],\mathbb{R})\forall\epsilon>0\exists n\in\mathbb{N}[\Vert f-B_N(f)\Vert_\infty\le\epsilon]$ or something

This page is a dire page and is in desperate need of an update.

Statement

Let $C([a,b],\mathbb{R})$ denote the vector space of continuous functions from the closed interval $[a,b]:\eq\{x\in\mathbb{R}\ \vert\ a\le x\le b\}\subset\mathbb{R}$ to the real line, $\mathbb{R}$. We consider this space with the sup-norm on continuous real functions:

• $\Vert\cdot\Vert_\infty:C([a,b],\mathbb{R})\rightarrow\mathbb{R}$ given by $\Vert\cdot\Vert_\infty:f\mapsto\mathop{\text{Sup} }_{x\in [a,b]}(\vert f(x)\vert)$ where $\vert\cdot\vert:\mathbb{R}\rightarrow\mathbb{R}$ is, as usual, the absolute value.

Then we claim for $f\in C([a,b],\mathbb{R})$ and $\epsilon>0$ given:

• there exists a polynomial, $p(x):\mathbb{R}\rightarrow\mathbb{R}$ such that
  • $\Vert f-p\Vert_\infty\le\epsilon$ (i.e. $d_\infty(f,p)\le\epsilon$ where $d_\infty$ is the metric induced by the norm $\Vert\cdot\Vert_\infty$)

Proof

Here we consider the interval $[a,b]$ to be just $[0,1]$ - the closed unit interval, and $f\in C([0,1],\mathbb{R})$. It is easy to take a $g:[a,b]\rightarrow\mathbb{R}$, first "contract it" so it is on $[0,1]$ then apply the reverse of that "contraction" to put the resulting polynomial on $[a,b]$.

• The contraction might be: $c:t\mapsto t(b-a)+a$, for $t\eq 0$ this is $a$ and for $t\eq 1$ it is $b$. So $g(c(x))$ is now defined on $[0,1]$

As $f$ is uniformly continuous we know:

• $\forall\epsilon'>0\exists\delta>0\forall x,y\in[a,b]\big[d(x,y)<\delta\implies d(f(x),f(y))<\epsilon'\big]$

Pick $\epsilon':\eq\frac{1}{2}\epsilon$, then:

• $\exists\delta>0\forall x,y\in[a,b]\big[\vert x-y\vert<\delta\implies\vert f(x)-f(y)\vert<\frac{\epsilon}{2}\big]$

Note that:

• $f(x)-$$\mathcal{B}_n(f;x)$$\eq f(x)-\sum^n_{i\eq 0}f\left(\tfrac{i}{n}\right){}^nC_ix^i(1-x)^{n-i}$ - where $\mathcal{B}_n(f;x)$ is the $n$^th Bernstein polynomial of $f$ evaluated at $x$

  $\eq f(x)\underbrace{\sum^n_{i\eq 0}{}^nC_ix^i(1-x)^{n-i} }_{\eq 1} - \sum^n_{i\eq 0}f\left(\tfrac{i}{n}\right){}^nC_ix^i(1-x)^{n-i}$

  $\eq \sum^n_{i\eq 0}f(x){}^nC_ix^i(1-x)^{n-i} - \sum^n_{i\eq 0}f\left(\tfrac{i}{n}\right){}^nC_ix^i(1-x)^{n-i}$

  $\eq \sum^n_{i\eq 0}\big(f(x)-f\left(\tfrac{i}{n}\right)\big){}^nC_ix^i(1-x)^{n-i}$

Next see that:

• $\vert f(x)-\mathcal{B}_n(f,x)\vert\eq\left\vert \sum^n_{i\eq 0}\big(f(x)-f\left(\tfrac{i}{n}\right)\big){}^nC_ix^i(1-x)^{n-i} \right\vert$

  $\le \sum^n_{i\eq 0}\left\vert \big(f(x)-f\left(\tfrac{i}{n}\right)\big){}^nC_ix^i(1-x)^{n-i} \right\vert$ - by triangle inequality

  $\eq \sum^n_{i\eq 0}\left(\left\vert \big(f(x)-f\left(\tfrac{i}{n}\right)\big)\right\vert {}^nC_ix^i(1-x)^{n-i}\right)$ - as ${}^nC_ix^i(1-x)^{n-i}$ is clearly $\ge 0$

  $$\eq \sum^n_{i\eq 0} {}^nC_ix^i(1-x)^{n-i}\vert f(x)-f\left(\tfrac{i}{n}\right)\vert$$

Note that if $\vert x-\tfrac{i}{n}\vert<\delta$ then $\vert f(x)-f(\tfrac{i}{n})\vert<\frac{\epsilon}{2}$ - as such our summation splits into two parts:

• $$\vert f(x)-\mathcal{B}_n(f,x)\vert\le \underbrace{\sum_{\begin{array}{c}0\le i\le n\\ i\text{ such that }\vert x-\frac{i}{n}\vert<\delta\end{array} }\left({}^nC_ix^i(1-x)^{n-i}\vert f(x)-f\left(\tfrac{i}{n}\right)\vert\right)}_{:\eq S_1} + \underbrace{\sum_{\begin{array}{c}0\le i\le n\\ i\text{ such that }\vert x-\frac{i}{n}\vert\ge\delta\end{array} } \big({}^nC_ix^i(1-x)^{n-i}\vert f(x)-f\left(\tfrac{i}{n}\right)\vert\big)}_{:\eq S_2}$$
  • Which we write more simply as $$\vert f(x)-\mathcal{B}_n(f,x)\vert\le S_1+S_2$$

Looking carefully at

$$S_1:\eq\sum_{\begin{array}{c}0\le i\le n\\ i\text{ such that }\vert x-\frac{i}{n}\vert<\delta\end{array} }\left({}^nC_ix^i(1-x)^{n-i}\vert f(x)-f\left(\tfrac{i}{n}\right)\vert\right)$$ we see that $\vert x-{i}{n}\vert<\delta$ for the things in this summation, by the uniform continuity property though we see $\forall x,y\in[a,b]\big[\vert x-y\vert<\delta\implies\vert f(x)-f(y)\vert<\frac{\epsilon}{2}\big]$, so we see:

• $\vert f(x)-f\left(\tfrac{i}{n}\right)\vert<\frac{\epsilon}{2}$
  • Thus: $$S_1:\eq\sum_{\begin{array}{c}0\le i\le n\\ i\text{ such that }\vert x-\frac{i}{n}\vert<\delta\end{array} }\left({}^nC_ix^i(1-x)^{n-i}\vert f(x)-f\left(\tfrac{i}{n}\right)\vert\right)$$

    $$<\sum_{\begin{array}{c}0\le i\le n\\ i\text{ such that }\vert x-\frac{i}{n}\vert<\delta\end{array} }\left({}^nC_ix^i(1-x)^{n-i}\frac{\epsilon}{2}\right)$$

    $$\eq\frac{\epsilon}{2}\sum_{\begin{array}{c}0\le i\le n\\ i\text{ such that }\vert x-\frac{i}{n}\vert<\delta\end{array} }\left({}^nC_ix^i(1-x)^{n-i}\right)$$

  • Note that $\sum^n_{i\eq 0}{}^nC_ix^i(1-x)^{n-i} \eq 1$, so we see:
    • $$\sum_{\begin{array}{c}0\le i\le n\\ i\text{ such that }\vert x-\frac{i}{n}\vert<\delta\end{array} }\left({}^nC_ix^i(1-x)^{n-i}\right)\le 1$$ as a subset of the exact same terms

So $S_1<\frac{\epsilon}{2}$

Grade: A*

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$S_2$ is more tricky

• Recall $$S_2:\eq\sum_{\begin{array}{c}0\le i\le n\\ i\text{ such that }\vert x-\frac{i}{n}\vert\ge\delta\end{array} } \big({}^nC_ix^i(1-x)^{n-i}\vert f(x)-f\left(\tfrac{i}{n}\right)\vert\big)$$

  $$\le\sum_{\begin{array}{c}0\le i\le n\\ i\text{ such that }\vert x-\frac{i}{n}\vert\ge\delta\end{array} } \big({}^nC_i x^i(1-x)^{n-i}\ 2\Vert f\Vert_\infty\big)$$ $$\eq 2\Vert f\Vert_\infty\sum_{\begin{array}{c}0\le i\le n\\ i\text{ such that }\vert x-\frac{i}{n}\vert\ge\delta\end{array} }\big({}^nC_i x^i(1-x)^{n-i}\big)$$

  • By the lemma: $\forall x,y\in[0,1]\subset\mathbb{R}[\vert f(x)-f(y)\vert\le 2\Vert f\Vert_\infty]$ - which is easily proved.

  $$\eq 2\Vert f\Vert_\infty\sum_{\begin{array}{c}0\le i\le n\\ i:\ (i-nx)^2\ge \delta^2n^2\end{array} }\big({}^nC_i x^i(1-x)^{n-i}\big)$$

  • By simple re-writing of $\vert x-\frac{i}{n}\vert\ge\delta$ -see annotations on lecture notes if stuck, shouldn't be stuck though

  $$\eq 2\Vert f\Vert_\infty\sum_{\begin{array}{c}0\le i\le n\\ i:\ (i-nx)^2\ge \delta^2n^2\end{array} }\left(\frac{(i-nx)^2}{\delta^2n^2}{}^nC_i x^i(1-x)^{n-i}\right)$$

  • As $(i-nx)^2\ge \delta^2n^2$ we see $\frac{(i-nx)^2}{\delta^2n^2}\ge 1$

  $$\eq\frac{ 2\Vert f\Vert_\infty}{\delta^2n^2}\sum_{\begin{array}{c}0\le i\le n\\ i:\ (i-nx)^2\ge \delta^2n^2\end{array} }\left((i-nx)^2\ {}^nC_i x^i(1-x)^{n-i}\right)$$

  $$\leq\frac{ 2\Vert f\Vert_\infty}{\delta^2n^2}\sum^n_{k\eq 0}\left((i-nx)^2\ {}^nC_i x^i(1-x)^{n-i}\right)$$

  • As the added terms are clearly $+\text{ve}$

  $$\eq\frac{2\Vert f\Vert_\infty}{\delta^2n^2} nx(1-x)$$ - by the third part of the lemma seemingly missing from this page (the summation $\eq$ this)

  $$\leq\frac{1}{4}\frac{2\Vert f\Vert_\infty}{n\delta^2}$$ as $x(1-x)$ attains its maximum at $x\eq\frac{1}{2}$ and that maximum is $\frac{1}{4}$

  $$\eq\frac{\Vert f\Vert_\infty}{2n\delta^2}$$

Note that:

• $$\lim_{n\rightarrow\infty}\left(\frac{\Vert f\Vert_\infty}{2n\delta^2}\right)\eq\frac{\Vert f\Vert_\infty}{2\delta^2}\cdot\lim_{n\rightarrow\infty}\left(\frac{1}{n}\right)\eq 0$$ - so that limit exists and converges, thus we see:
  • $$\forall\epsilon'>0\exists N'\in\mathbb{N}\forall n\in\mathbb{N}\left[n\ge N'\implies \left\vert\frac{\Vert f\Vert_\infty}{2n\delta^2}\right\vert<\epsilon'\right]$$ (by definition of a convergent sequence)
    • Pick $\epsilon':\eq\frac{1}{2}\epsilon$ and pick $N\in\mathbb{N}$ with $N\ge N'$ which we know to exist by convergence, then:
      • $$\frac{\Vert f\Vert_\infty}{2N\delta^2}\eq\left\vert\frac{\Vert f\Vert_\infty}{2N\delta^2}\right\vert<\epsilon':\eq\frac{1}{2}\epsilon$$

And so: $S_1+S_2<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon\eq\epsilon$ as required.