Weierstrass approximation theorem
- and really crap this page needs some formal logic saying what it shows, ie:
- ∀f∈C([0,1],R)∀ϵ>0∃n∈N[∥f−BN(f)∥∞≤ϵ] or something
Contents
[hide]Statement
Let C([a,b],R) denote the vector space of continuous functions from the closed interval [a,b]:={x∈R | a≤x≤b}⊂R to the real line, R. We consider this space with the sup-norm on continuous real functions:
- ∥⋅∥∞:C([a,b],R)→R given by ∥⋅∥∞:f↦Supx∈[a,b](|f(x)|) where |⋅|:R→R is, as usual, the absolute value.
Then we claim for f∈C([a,b],R) and ϵ>0 given:
- there exists a polynomial, p(x):R→R such that
- ∥f−p∥∞≤ϵ (i.e. d∞(f,p)≤ϵ where d∞ is the metric induced by the norm ∥⋅∥∞)
Proof
Here we consider the interval [a,b] to be just [0,1] - the closed unit interval, and f∈C([0,1],R). It is easy to take a g:[a,b]→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↦t(b−a)+a, for t=0 this is a and for t=1 it is b. So g(c(x)) is now defined on [0,1]
As f is uniformly continuous we know:
- ∀ϵ′>0∃δ>0∀x,y∈[a,b][d(x,y)<δ⟹d(f(x),f(y))<ϵ′]
Pick ϵ′:=12ϵ, then:
- ∃δ>0∀x,y∈[a,b][|x−y|<δ⟹|f(x)−f(y)|<ϵ2]
Note that:
- f(x)−Bn(f;x)=f(x)−∑ni=0f(in)nCixi(1−x)n−i - where Bn(f;x) is the nth Bernstein polynomial of f evaluated at x
- =f(x)∑ni=0nCixi(1−x)n−i⏟=1−∑ni=0f(in)nCixi(1−x)n−i
- =∑ni=0f(x)nCixi(1−x)n−i−∑ni=0f(in)nCixi(1−x)n−i
- =∑ni=0(f(x)−f(in))nCixi(1−x)n−i
Next see that:
- |f(x)−Bn(f,x)|=|∑ni=0(f(x)−f(in))nCixi(1−x)n−i|
- ≤∑ni=0|(f(x)−f(in))nCixi(1−x)n−i| - by triangle inequality
- =∑ni=0(|(f(x)−f(in))|nCixi(1−x)n−i) - as nCixi(1−x)n−i is clearly ≥0
- =n∑i=0nCixi(1−x)n−i|f(x)−f(in)|
Note that if |x−in|<δ then |f(x)−f(in)|<ϵ2 - as such our summation splits into two parts:
- |f(x)−Bn(f,x)|≤∑0≤i≤ni such that |x−in|<δ(nCixi(1−x)n−i|f(x)−f(in)|)⏟:=S1+∑0≤i≤ni such that |x−in|≥δ(nCixi(1−x)n−i|f(x)−f(in)|)⏟:=S2
- Which we write more simply as |f(x)−Bn(f,x)|≤S1+S2
Looking carefully at S1:=∑0≤i≤ni such that |x−in|<δ(nCixi(1−x)n−i|f(x)−f(in)|) we see that |x−in|<δ for the things in this summation, by the uniform continuity property though we see ∀x,y∈[a,b][|x−y|<δ⟹|f(x)−f(y)|<ϵ2], so we see:
- |f(x)−f(in)|<ϵ2
- Thus: S1:=∑0≤i≤ni such that |x−in|<δ(nCixi(1−x)n−i|f(x)−f(in)|)
- <∑0≤i≤ni such that |x−in|<δ(nCixi(1−x)n−iϵ2)
- =ϵ2∑0≤i≤ni such that |x−in|<δ(nCixi(1−x)n−i)
- Note that ∑ni=0nCixi(1−x)n−i=1, so we see:
- ∑0≤i≤ni such that |x−in|<δ(nCixi(1−x)n−i)≤1 as a subset of the exact same terms
- Thus: S1:=∑0≤i≤ni such that |x−in|<δ(nCixi(1−x)n−i|f(x)−f(in)|)
So S1<ϵ2
The message provided is:
- Recall S2:=∑0≤i≤ni such that |x−in|≥δ(nCixi(1−x)n−i|f(x)−f(in)|)
- ≤∑0≤i≤ni such that |x−in|≥δ(nCixi(1−x)n−i 2∥f∥∞) =2∥f∥∞∑0≤i≤ni such that |x−in|≥δ(nCixi(1−x)n−i)
- By the lemma: ∀x,y∈[0,1]⊂R[|f(x)−f(y)|≤2∥f∥∞] - which is easily proved.
- =2∥f∥∞∑0≤i≤ni: (i−nx)2≥δ2n2(nCixi(1−x)n−i)
- By simple re-writing of |x−in|≥δ -see annotations on lecture notes if stuck, shouldn't be stuck though
- =2∥f∥∞∑0≤i≤ni: (i−nx)2≥δ2n2((i−nx)2δ2n2nCixi(1−x)n−i)
- As (i−nx)2≥δ2n2 we see (i−nx)2δ2n2≥1
- =2∥f∥∞δ2n2∑0≤i≤ni: (i−nx)2≥δ2n2((i−nx)2 nCixi(1−x)n−i)
- ≤2∥f∥∞δ2n2n∑k=0((i−nx)2 nCixi(1−x)n−i)
- As the added terms are clearly +ve
- =2∥f∥∞δ2n2nx(1−x) - by the third part of the lemma seemingly missing from this page (the summation = this)
- ≤142∥f∥∞nδ2 as x(1−x) attains its maximum at x=12 and that maximum is 14
- =∥f∥∞2nδ2
- ≤∑0≤i≤ni such that |x−in|≥δ(nCixi(1−x)n−i 2∥f∥∞) =2∥f∥∞∑0≤i≤ni such that |x−in|≥δ(nCixi(1−x)n−i)
Note that:
- lim - 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
- 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:
- \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)
And so: S_1+S_2<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon\eq\epsilon as required.
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