5.10. THE STONE WEIERSTRASS APPROXIMATION THEOREM 143
Proof: By Corollary 5.10.2 there exists a sequence of polynomials, { p̃n} such thatp̃n→ |·| uniformly. Then let pn (t)≡ p̃n (t)− p̃n (0) . ■
Definition 5.10.4 An algebra of functions, A defined on A, annihilates no point ofA if for all x ∈ A, there exists g ∈ A such that g(x) ̸= 0. The algebra separates points ifwhenever x1 ̸= x2, then there exists g ∈A such that g(x1) ̸= g(x2).
The following generalization is known as the Stone Weierstrass approximation theorem.
Theorem 5.10.5 Let A be a compact topological space and let A ⊆C (A;R) be analgebra of functions which separates points and annihilates no point. Then A is dense inC (A;R).
Proof: First here is a lemma.
Lemma 5.10.6 Let c1 and c2 be two real numbers and let x1 ̸= x2 be two points of A.Then there exists a function fx1x2 such that
fx1x2 (x1) = c1, fx1x2 (x2) = c2.
Proof of the lemma: Let g ∈ A satisfy g(x1) ̸= g(x2). Such a g exists because thealgebra separates points. Since the algebra annihilates no point, there exist functions h andk such that h(x1) ̸= 0, k (x2) ̸= 0. Then let u ≡ gh− g(x2)h, v ≡ gk− g(x1)k. It followsthat u(x1) ̸= 0 and u(x2) = 0 while v(x2) ̸= 0 and v(x1) = 0. Let fx1x2 ≡
c1uu(x1)
+ c2vv(x2)
. Thisproves the lemma. Now continue the proof of Theorem 5.10.5.
First note that A satisfies the same axioms as A but in addition to these axioms, A isclosed. The closure of A is taken with respect to the usual norm on C (A),
∥ f∥∞≡max{| f (x)| : x ∈ A} .
Suppose f ∈A and suppose M is large enough that ∥ f∥∞< M. Using Corollary 5.10.3, let
pn be a sequence of polynomials such that
∥pn−|·|∥∞→ 0, pn (0) = 0.
It follows that pn ◦ f ∈A and so | f | ∈A whenever f ∈A . Also note that
max( f ,g) =| f −g|+( f +g)
2
min( f ,g) =( f +g)−| f −g|
2.
Therefore, this shows that if f ,g ∈ A then max( f ,g) , min( f ,g) ∈ A . By induction, iffi, i = 1,2, · · · ,m are in A then
max( fi, i = 1,2, · · · ,m) , min( fi, i = 1,2, · · · ,m) ∈A .
Now let h ∈C (A;R) and let x ∈ A. Use Lemma 5.10.6 to obtain fxy, a function of Awhich agrees with h at x and y. Letting ε > 0, there exists an open set U (y) containing ysuch that
fxy (z)> h(z)− ε if z ∈U(y).