Analysis 2-The-Weierstrass Theorem, guaranteed and verified 100% Pass

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The Weierstrass Theorem


Our next goal is to study 𝐶[𝑎, 𝑏], the metric space of continuous functions on a
closed, bounded interval [𝑎, 𝑏] with the metric:

𝑑 (𝑓, 𝑔) = sup |𝑓(𝑥 ) − 𝑔(𝑥 )|.
𝑎≤𝑥≤𝑏



We want to be able to show that given any continuous function 𝑓 ∈ 𝐶[𝑎, 𝑏] we
can find a sequence of polynomials {𝑝𝑛 (𝑥)} on [𝑎, 𝑏] such that 𝑝𝑛 (𝑥) → 𝑓(𝑥)
uniformly on [𝑎, 𝑏] (which is the same as saying that 𝑝𝑛 (𝑥) → 𝑓(𝑥) in the
metric space 𝐶[𝑎, 𝑏]). This is useful because it means given any continuous
function 𝑓 ∈ 𝐶[𝑎, 𝑏], which could be fairly “complicated”, we can approximate it
with a polynomial, which is a fairly “simple” function to work with. That is, given
any 𝑓 ∈ 𝐶[𝑎, 𝑏] and any 𝜖 > 0 there is a polynomial, 𝑝(𝑥) ∈ 𝐶[𝑎, 𝑏] with

sup |𝑓(𝑥 ) − 𝑝(𝑥 )| < 𝜖.
𝑎≤𝑥≤𝑏



This is a common theme in analysis. We often try to approximate complicated
functions with less complicated functions. We then prove theorems with the less
complicated functions and then show the theorem still holds when you take a
limit of less complicated functions that converge to the more complicated
function.

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Lemma: There is a linear map 𝑇 from 𝐶[0,1] onto 𝐶[𝑎, 𝑏] such that

𝑑 (𝑓, 𝑔) = 𝑑(𝑇(𝑓), 𝑇(𝑔)) (this is called an isometry) and polynomials get
mapped to polynomials.

𝑥−𝑎
Proof. Let 𝜎: [𝑎, 𝑏] → [0,1] by 𝜎 (𝑥 ) = ; for 𝑎 ≤ 𝑥 ≤ 𝑏.
𝑏−𝑎
𝜎 is a continuous linear function from [𝑎, 𝑏] onto [0,1].
We can define: 𝑇: 𝐶[0,1] → 𝐶[𝑎, 𝑏]
𝑇(𝑓) = 𝑓 ∘ 𝜎.


𝑥−2 𝑥−2
For example, 𝜎: [2,5] → [0,1] by 𝜎 (𝑥 ) =
5−2
= 3
; for 2 ≤ 𝑥 ≤ 5.

𝑥−2 𝑥−2
If 𝑓(𝑥 ) = 𝑥 2 ∈ 𝐶[0,1], then 𝑇 (𝑓) = 𝑓(𝜎 (𝑥 )) = 𝑓 ( ) = ( )2
3 3
Is a continuous function on [2,5].


𝑇 is linear because for 𝑓, 𝑔 ∈ 𝐶[0,1] and 𝑎, 𝑏 ∈ ℝ:
𝑇(𝑎𝑓 + 𝑏𝑔) = (𝑎𝑓 + 𝑏𝑔)(𝜎(𝑥 ))
= 𝑎𝑓(𝜎(𝑥 )) + 𝑏𝑔(𝜎(𝑥 ))
= 𝑎𝑇(𝑓) + 𝑏𝑇(𝑔).
Since 𝜎 −1 (𝑡 ) = 𝑎 + 𝑡(𝑏 − 𝑎 ), 0 ≤ 𝑡 ≤ 1, we can define an inverse for 𝑇
by

𝑇 −1 (ℎ) = ℎ(𝜎 −1 (𝑡 ))
So 𝑇 is one to one and onto.