View source for Weierstrass approximation theorem

{{Stub page|grade=A*|msg=Proper stub
* '''and really crap''' this page needs some formal logic saying what it shows, ie:
** {{M|\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}}
{{Dire page}}
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==Statement==
Let {{M|C([a,b],\mathbb{R})}} denote the [[vector space of continuous functions from a closed interval to the reals|vector space of continuous functions]] from the [[closed interval]] {{M|[a,b]:\eq\{x\in\mathbb{R}\ \vert\ a\le x\le b\}\subset\mathbb{R} }} to the [[real line]], {{M|\mathbb{R} }}. We consider this space with the [[sup-norm on continuous real functions]]:
* {{M|\Vert\cdot\Vert_\infty:C([a,b],\mathbb{R})\rightarrow\mathbb{R} }} given by {{M|\Vert\cdot\Vert_\infty:f\mapsto\mathop{\text{Sup} }_{x\in [a,b]}(\vert f(x)\vert)}} where {{M|\vert\cdot\vert:\mathbb{R}\rightarrow\mathbb{R} }} is, as usual, the [[absolute value]].
Then we claim for {{M|f\in C([a,b],\mathbb{R})}} and {{M|\epsilon>0}} given:
* there exists a [[polynomial]], {{M|p(x):\mathbb{R}\rightarrow\mathbb{R} }} such that
** {{M|\Vert f-p\Vert_\infty\le\epsilon}} (i.e. {{M|d_\infty(f,p)\le\epsilon}} where {{M|d_\infty}} is the [[metric induced by a norm|metric induced by the norm]] {{M|\Vert\cdot\Vert_\infty}})
==Proof==
Here we consider the interval {{M|[a,b]}} to be just {{M|[0,1]}} - the [[closed unit interval]], and {{M|f\in C([0,1],\mathbb{R})}}. It is easy to take a {{M|g:[a,b]\rightarrow\mathbb{R} }}, first "contract it" so it is on {{M|[0,1]}} then apply the reverse of that "contraction" to put the resulting polynomial on {{M|[a,b]}}.
* The contraction might be: {{M|c:t\mapsto t(b-a)+a}}, for {{M|t\eq 0}} this is {{M|a}} and for {{M|t\eq 1}} it is {{M|b}}. So {{M|g(c(x))}} is now defined on {{M|[0,1]}}

As {{M|f}} is [[uniformly continuous]] we know:
* {{M|\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 {{M|\epsilon':\eq\frac{1}{2}\epsilon}}, then:
* {{M|\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:
* {{M|f(x)-}}[[Bernstein polynomial|{{M|\mathcal{B}_n(f;x)}}]]{{M|\eq f(x)-\sum^n_{i\eq 0}f\left(\tfrac{i}{n}\right){}^nC_ix^i(1-x)^{n-i} }} - where {{M|\mathcal{B}_n(f;x)}} is the {{M|n}}<sup>th</sup> [[Bernstein polynomial]] of {{M|f}} evaluated at {{M|x}}
*: {{M|\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} }}
*: {{M|\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} }}
*: {{M|\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:
* {{M|\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}}
*: {{M|\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]]
*: {{M|\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 {{M|{}^nC_ix^i(1-x)^{n-i} }} is clearly {{M|\ge 0}}
*: {{MM|\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 {{M|\vert x-\tfrac{i}{n}\vert<\delta}} then {{M|\vert f(x)-f(\tfrac{i}{n})\vert<\frac{\epsilon}{2} }} - as such our summation splits into two parts:
* {{MM|\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 {{MM|\vert f(x)-\mathcal{B}_n(f,x)\vert\le S_1+S_2}}
Looking carefully at {{MM|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 {{M|\vert x-{i}{n}\vert<\delta}} for the things in this summation, by the uniform continuity property though we see {{M|\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:
* {{M|\vert f(x)-f\left(\tfrac{i}{n}\right)\vert<\frac{\epsilon}{2} }}
** Thus: {{MM|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)}}
**: {{MM|<\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)}}
**: {{MM|\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 {{M|\sum^n_{i\eq 0}{}^nC_ix^i(1-x)^{n-i} \eq 1}}, so we see:
*** {{MM|\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 {{M|S_1<\frac{\epsilon}{2} }}
{{Requires work|grade=A*|msg={{M|S_2}} is more tricky}}
* Recall {{MM|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) }}
*: {{MM|\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)}} {{MM|\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: {{M|1=\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.
*: {{MM|\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 {{M|\vert x-\frac{i}{n}\vert\ge\delta}} -see annotations on lecture notes if stuck, shouldn't be stuck though
*: {{MM|\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 {{M|(i-nx)^2\ge \delta^2n^2}} we see {{M|\frac{(i-nx)^2}{\delta^2n^2}\ge 1}}
*: {{MM|\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) }}
*: {{MM|\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 {{M|+\text{ve} }}
*: {{MM|\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 {{M|\eq}} this)
*: {{MM|\leq\frac{1}{4}\frac{2\Vert f\Vert_\infty}{n\delta^2} }} as {{M|x(1-x)}} attains its maximum at {{M|x\eq\frac{1}{2} }} and that maximum is {{M|\frac{1}{4} }}
*: {{MM|\eq\frac{\Vert f\Vert_\infty}{2n\delta^2} }}

Note that:
* {{MM|\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:
** {{MM|1=\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 {{M|\epsilon':\eq\frac{1}{2}\epsilon}} and pick {{M|N\in\mathbb{N} }} with {{M|N\ge N'}} which we know to exist by convergence, then:
**** {{MM|\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: {{M|S_1+S_2<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon\eq\epsilon}} as required.
==References==
<references/>
{{Theorem Of|Analysis|Real Analysis|Functional Analysis}}