Kenneth Kuttler

112 CHAPTER 4. CONTINUITY AND LIMITS

By Theorems 3.6.2 and 4.7.2, f is uniformly continuous. Let δ go with ε/2 in the definitionof uniform continuity. At this point you break the sum into two pieces, those values of ksuch that |k/n− x|< δ and those values of k where |x− (k/n)| ≥ δ . Then from the Lemma4.10.1,

| f (x)− pn (x)| ≤ ∑|x−(k/n)|<δ

(nk

)∣∣∣∣ f ( kn

)− f (x)

∣∣∣∣xk (1− x)n−k

+ ∑|x−(k/n)|≥δ

(nk

)∣∣∣∣ f ( kn

)− f (x)

∣∣∣∣xk (1− x)n−k (4.5)

≤ ∑|x−(k/n)|<δ

(nk

)ε

2xk (1− x)n−k +2M ∑

|nx−k|≥nδ

(nk

)xk (1− x)n−k

≤ ε

∑k=0

(nk

)xk (1− x)n−k +2M ∑

|nx−k|≥nδ

(nk

) ≥1

(k−nx)2

n2δ2 xk (1− x)n−k

≤ ε

2+2M

n2δ2 =

nδ2

Therefore, whenever n is sufficiently large that 4Mnδ

2 < ε

2 , it follows that for all n this largeand x ∈ [0,1] ,

| f (x)− pn (x)|<ε

2= ε.

These polynomials are called the Bernstein polynomials.Now this theorem has been done, it is easy to extend to continuous functions defined

on [a,b]. This yields the celebrated Weierstrass approximation theorem.

Theorem 4.10.3 Suppose f is a continuous function defined on [a,b]. Then thereexists a sequence of polynomials, {pn} which converges uniformly to f on [a,b].

Proof: For t ∈ [0,1] , let h(t) = a+(b−a) t. Thus h maps [0,1] one to one and onto[a,b] . Thus f ◦h is a continuous function defined on [0,1] . It follows there exists a sequenceof polynomials {pn} defined on [0,1] which converges uniformly to f ◦ h on [0,1]. Thusfor every ε > 0 there exists Nε such that if n ≥ Nε , then for all t ∈ [0,1] ,

| f ◦h(t)− pn (t)|< ε.

However, h is onto and one to one and so for all x ∈ [a,b] ,∣∣ f (x)− pn

(h−1 (x)

)∣∣ < ε.Nownote that the function x→ pn

(h−1 (x)

)is a polynomial because h−1 (x)= x−a

b−a . More specif-ically, if pn (t) = ∑

mk=0 aktk it follows

pn(h−1 (x)

∑k=0

(x−ab−a

which is clearly another polynomial.Weierstrass did not prove this theorem in this way. He used integrals instead of sums

to do it, but integrals have not been discussed yet. I think the Bernstein polynomials usedhere give the easiest proof. This amazing theorem shows that every continuous functiondefined on a finite closed interval is the uniform limit of polynomials. The analog does nothold for continuous functions of complex variables but this is another topic entirely.

112 CHAPTER 4. CONTINUITY AND LIMITSBy Theorems 3.6.2 and 4.7.2, f is uniformly continuous. Let 6 go with €/2 in the definitionof uniform continuity. At this point you break the sum into two pieces, those values of ksuch that |k/n —x| < 6 and those values of k where |x — (k/n)| > 6. Then from the Lemma4.10.1,Ved-miol SE (7) r(§) ~ f(x)].* Pies (;) i" (;) ~ FE)< E(k oem y ({)sa-94|x—( |nx—k|>nd(1—x)"*(1—x)"* (4.5)>1esE()eewrtom gi) Sageto|nx—k|>nd1 1 e 1M2M ~+=—><3 5t 42522 2nd?Therefore, whenever n is sufficiently large that + ao < 5, it follows that for all n this largeand x € [0, 1],€ €— —~+-==€.IF (4) — Pal) <5 +5 |These polynomials are called the Bernstein polynomials.Now this theorem has been done, it is easy to extend to continuous functions definedon [a,b]. This yields the celebrated Weierstrass approximation theorem.Theorem 4.10.3 Suppose f is a continuous function defined on |a,b|. Then thereexists a sequence of polynomials, { p,} which converges uniformly to f on {a,b}.Proof: For ¢ € [0,1], let h(t) =a+(b—a)t. Thus h maps [0,1] one to one and onto[a,b]. Thus f oh is a continuous function defined on [0, 1] . It follows there exists a sequenceof polynomials {p,} defined on [0,1] which converges uniformly to f oh on [0,1]. Thusfor every € > 0 there exists Ne such that ifm > Ne, then for all ¢ € [0,1],[foh(t)—pa(t)|<é.However, h is onto and one to one and so for all x € [a,b], n (A! (x)) | < €.Nownote that the function x + py (h- | (x)) i is a polynomial heute 1) = 5_,: More specif-ically, if pn (t) = Lrg agt* it followspelt") =a (F=2)which is clearly another polynomial. §jWeierstrass did not prove this theorem in this way. He used integrals instead of sumsto do it, but integrals have not been discussed yet. I think the Bernstein polynomials usedhere give the easiest proof. This amazing theorem shows that every continuous functiondefined on a finite closed interval is the uniform limit of polynomials. The analog does nothold for continuous functions of complex variables but this is another topic entirely.