Weierstrass approximation theorem

Stub grade: A*

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

Proper stub

• and really crap this page needs some formal logic saying what it shows, ie:
  • [ilmath]\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][/ilmath] or something

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

Statement

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

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

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

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

Proof

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

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

As [ilmath]f[/ilmath] is uniformly continuous we know:

• [ilmath]\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][/ilmath]

Pick [ilmath]\epsilon':\eq\frac{1}{2}\epsilon[/ilmath], then:

• [ilmath]\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][/ilmath]

Note that:

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

  [ilmath]\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} [/ilmath]

  [ilmath]\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} [/ilmath]

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

Next see that:

• [ilmath]\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[/ilmath]

  [ilmath]\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[/ilmath] - by triangle inequality

  [ilmath]\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)[/ilmath] - as [ilmath]{}^nC_ix^i(1-x)^{n-i} [/ilmath] is clearly [ilmath]\ge 0[/ilmath]

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

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

• [math]\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} [/math]
  • Which we write more simply as [math]\vert f(x)-\mathcal{B}_n(f,x)\vert\le S_1+S_2[/math]

Looking carefully at [math]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)[/math] we see that [ilmath]\vert x-{i}{n}\vert<\delta[/ilmath] for the things in this summation, by the uniform continuity property though we see [ilmath]\forall x,y\in[a,b]\big[\vert x-y\vert<\delta\implies\vert f(x)-f(y)\vert<\frac{\epsilon}{2}\big][/ilmath], so we see:

• [ilmath]\vert f(x)-f\left(\tfrac{i}{n}\right)\vert<\frac{\epsilon}{2} [/ilmath]
  • Thus: [math]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)[/math]

    [math]<\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)[/math]

    [math]\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)[/math]

  • Note that [ilmath]\sum^n_{i\eq 0}{}^nC_ix^i(1-x)^{n-i} \eq 1[/ilmath], so we see:
    • [math]\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[/math] as a subset of the exact same terms

So [ilmath]S_1<\frac{\epsilon}{2} [/ilmath]

Grade: A*

This page requires some work to be carried out

Some aspect of this page is incomplete and work is required to finish it
The message provided is:

[ilmath]S_2[/ilmath] is more tricky

• Recall [math]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) [/math]

  [math]\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)[/math] [math]\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) [/math]

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

  [math]\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) [/math]

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

  [math]\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) [/math]

  • As [ilmath](i-nx)^2\ge \delta^2n^2[/ilmath] we see [ilmath]\frac{(i-nx)^2}{\delta^2n^2}\ge 1[/ilmath]

  [math]\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) [/math]

  [math]\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) [/math]

  • As the added terms are clearly [ilmath]+\text{ve} [/ilmath]

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

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

  [math]\eq\frac{\Vert f\Vert_\infty}{2n\delta^2} [/math]

Note that:

• [math]\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[/math] - so that limit exists and converges, thus we see:
  • [math]\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][/math] (by definition of a convergent sequence)
    • Pick [ilmath]\epsilon':\eq\frac{1}{2}\epsilon[/ilmath] and pick [ilmath]N\in\mathbb{N} [/ilmath] with [ilmath]N\ge N'[/ilmath] which we know to exist by convergence, then:
      • [math]\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[/math]

And so: [ilmath]S_1+S_2<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon\eq\epsilon[/ilmath] as required.