80 CHAPTER 3. STONE WEIERSTRASS APPROXIMATION THEOREM

Proof: Let ∥ f∥∞

denote the largest value of ∥ f (x)∥. By uniform continuity of f ,there exists a δ > 0 such that if |x− x′| < δ , then ∥ f (x)− f (x′)∥ < ε/2. By the binomialtheorem,

∥pm (x)− f (x)∥ ≤m

∑k=0

(mk

)xk (1− x)m−k

∥∥∥∥ f(

km

)− f (x)

∥∥∥∥≤ ∑| k

m−x|<δ

(mk

)xk (1− x)m−k

∥∥∥∥ f(

km

)− f (x)

∥∥∥∥+2∥ f∥

∞ ∑| k

m−x|≥δ

(mk

)xk (1− x)m−k

Therefore,

≤m

∑k=0

(mk

)xk (1− x)m−k ε

2+2∥ f∥

∞ ∑(k−mx)2≥m2δ

2

(mk

)xk (1− x)m−k

≤ ε

2+2∥ f∥

1

m2δ2

m

∑k=0

(mk

)(k−mx)2 xk (1− x)m−k

≤ ε

2+2∥ f∥

14

m1

δ2m2

< ε

provided m is large enough. Thus ∥pm− f∥∞< ε when m is large enough. ■

Note that we do not need to have X be complete in order for this to hold. It would havesufficed to have simply let X be a normed linear space.

Corollary 3.1.3 If f ∈C ([a,b] ;X) where X is a normed linear space, then there existsa sequence of polynomials which converge uniformly to f on [a,b]. The coefficients of thesepolynomials are in X.

Proof: Let l : [0,1] → [a,b] be one to one, linear and onto. Then f ◦ l is contin-uous on [0,1] and so if ε > 0 is given, there exists a polynomial p such that for allx∈ [0,1] ,∥p(x)− f ◦ l (x)∥< ε . Therefore, letting y= l (x) , it follows that for all y∈ [a,b] ,∥∥p

(l−1 (y)

)− f (y)

∥∥< ε. ■

The exact form of the polynomial is as follows.

p(x) =m

∑k=0

(mk

)xk (1− x)m−k f

(l(

km

))

p(l−1 (y)

)=

m

∑k=0

(mk

)(l−1 (y)

)k (1− l−1 (y)

)m−kf(

l(

km

))(3.1)

80 CHAPTER 3. STONE WEIERSTRASS APPROXIMATION THEOREMProof: Let ||/||,, denote the largest value of ||f(x)||. By uniform continuity of f,there exists a 6 > 0 such that if |x—2’| < 6, then || f (x) — f (x’)|| < €/2. By the binomialtheorem,lows) S (f Jara") 29< x 1 )#a-a""'| (2) -re9|+rifle ( i aay’Therefore,m<)k=0( k ) kama kS 42 y ( ” ) Aaa(k—mx)? >m2.6?é 1 — m 2 m—k< _ —___ —_— _—< 5+ 2fla as ¥ ( fal mx)? x* (1 —x)€ 1< =42 —_m——S 5+ I flln garg <Eprovided m is large enough. Thus ||p,,— f||,. < € when m is large enough.Note that we do not need to have X be complete in order for this to hold. It would havesufficed to have simply let X be a normed linear space.Corollary 3.1.3 If f € C ({a,b];X) where X is a normed linear space, then there existsa sequence of polynomials which converge uniformly to f on [a,b]. The coefficients of thesepolynomials are in X.Proof: Let / : [0,1] — [a,b] be one to one, linear and onto. Then fo/ is contin-uous on [0,1] and so if € > 0 is given, there exists a polynomial p such that for allx € 0, 1], ||p (x) — fol (x)|| < €. Therefore, letting y =/ (x) , it follows that for all y € [a,b] ,|P(' 0) -f0)|| <aThe exact form of the polynomial is as follows.no E(t )ers(()