86 CHAPTER 3. STONE WEIERSTRASS APPROXIMATION THEOREM

Claim 3: A = ReA + i ImAProof of claim: If f ∈A , then

f =f + f̄

2+ i

f − f̄2i∈ ReA + i ImA

so A ⊆ReA + i ImA . Now for f ,g ∈A

Re( f )+ i Im(g)≡ f + f̄2

+ i(

g− ḡ2i

)=

f +g2

+f̄ − ḡ

2∈A

because A is closed with respect to conjugates. Thus ReA + i ImA ⊆A .Both ReA and ImA must separate the points. Here is why: If x1 ̸= x2, then there exists

f ∈A such that f (x1) ̸= f (x2) . If Im f (x1) ̸= Im f (x2) , this shows there is a function inImA , Im f which separates these two points. If Im f fails to separate the two points, thenRe f must separate the points and so, by Lemma 3.4.1,

Re f (x1) = Im(i f̄ (x1)

)̸= Re f (x2) = Im

(i f̄ (x2)

)Thus ImA separages the points. Similarly ReA separates the points using a similar argu-ment or because it is equal to ImA .

Neither ReA nor ImA annihilate any point. This is easy to see because if x is apoint, there exists f ∈ A such that f (x) ̸= 0. Thus either Re f (x) ̸= 0 or Im f (x) ̸= 0. IfIm f (x) ̸= 0, this shows this point is not annihilated by ImA . Since they are equal, ReAdoes not annihilate this point either.

It follows from Theorem 3.3.3 that ReA and ImA are dense in the real valued func-tions of C0 (X). Let f ∈C0 (X) . Then there exists {hn} ⊆ReA and {gn} ⊆ ImA such thathn→ Re f uniformly and gn→ Im f uniformly. Therefore, hn + ign ∈A and it convergesto f uniformly. ■

3.5 Exercises1. Let φ n (x) =

(1− x2

)n for |x| ≤ 1. For f a continuous function defined on [−1,1] ,extend it to have f (x) = f (1) for x > 1 and f (x) = f (−1) for x < −1. Considerpn (x) ≡

∫ x+1x−1 φ n (x− y) f (y)dy =

∫ 1−1 φ n (y) f (x− y)dy. This involves elementary

calculus and change of variables. Show that pn (x) is a polynomial and that pn con-verges uniformly to f on [−1,1]. This is the way Weierstrass originally proved thefamous approximation theorem.

2. In fact the Bernstein polynomials apply for f having values in a normed linear spaceand a similar result will hold. Give such a generalization.

3. Consider a continuous function f defined on the box ∏pk=1 [0,1]≡ [0,1]p . Then con-

sider f (t1)≡ f (t1, · · ·) as a continuous function having values in C([0,1]p−1

). Then

the Bernstein polynomials are of the form ∑mk=0(m

k

)f( k

m , · · ·)

tk1 (1− t1)

m−k . Now re-peat the process on these coefficients f

( km , · · ·

)which can be considered functions

in C([0,1]p−2

), t2→ f

( km , t2, · · ·

). Continuing this way, show there is a polynomial

∑k1,··· ,km

ak1,··· ,kntk11 · · · t

km