130 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES

Theorem 5.9.5 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 5.9.6 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 5.9.5.

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 5.9.3, 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 5.9.6 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).

Since A is compact, let U (y1) , · · · ,U (yl) cover A. Let

fx ≡max(

fxy1 , fxy2 , · · · , fxyl

).

Then fx ∈A and fx (z)> h(z)−ε for all z∈ A and fx (x) = h(x). This implies that for eachx ∈ A there exists an open set V (x) containing x such that for z ∈ V (x), fx (z) < h(z)+ ε.Let V (x1) , · · · ,V (xm) cover A and let f ≡ min( fx1 , · · · , fxm). Therefore, f (z) < h(z)+ ε

for all z ∈ A and since fx (z)> h(z)−ε for all z ∈ A, it follows f (z)> h(z)−ε also and so| f (z)−h(z)|< ε for all z. Since ε is arbitrary, this shows h ∈A and proves A =C (A;R).■