510 CHAPTER 19. HAUSDORFF SPACES AND MEASURES

and let θ : C0 (X)→D be given by θ f = f̃ . Then θ is one to one and onto and also satisfies∥ f∥

∞= ∥θ f∥

∞. Now D is complete because it is a closed subspace of a complete space

and so C0 (X) with ∥·∥∞

is also complete. ■The above refers to functions which have values in C but the same proof works for

functions which have values in any complete normed linear space.In the case where the functions in C0 (X) all have real values, I will denote the resulting

space by C0 (X ;R) with similar meanings in other cases.With this lemma, the generalization of the Stone Weierstrass theorem to locally com-

pact sets is as follows.

Theorem 19.4.3 Let A be an algebra of functions in C0 (X ;R) where (X ,τ) is alocally compact Hausdorff space which separates the points and annihilates no point. ThenA is dense in C0 (X ;R).

Proof: Let(

X̃ , τ̃)

be the one point compactification as described in Lemma 19.1.27.

Let à denote all finite linear combinations,{

∑ni=1 ci f̃i + c0 : f ∈A , ci ∈ R

}where for

f ∈C0 (X ;R) ,

f̃ (x)≡{

f (x) if x ∈ X0 if x = ∞

.

Then à is obviously an algebra of functions in C(

X̃ ;R)

. It separates points because thisis true of A . Similarly, it annihilates no point because of the inclusion of c0 an arbitraryelement of R in the definition above. Therefore from Theorem 5.10.5, Ã is dense inC(

X̃ ;R). Letting f ∈C0 (X ;R) , it follows f̃ ∈C

(X̃ ;R

)so there exists a sequence {hn}⊆

à such that hn converges uniformly to f̃ . Now hn is of the form ∑ni=1 cn

i f̃ ni + cn

0 and sincef̃ (∞) = 0, you can take each cn

0 = 0 and so this has shown the existence of a sequence offunctions in A such that it converges uniformly to f . ■

19.4.2 The Case of Complex Valued FunctionsWhat about the general case where C0 (X) consists of complex valued functions and thefield of scalars is C rather than R? The following is the version of the Stone Weierstrasstheorem which applies to this case. You have to assume that for f ∈A it follows f̄ ∈A .

Lemma 19.4.4 Let z be a complex number. Then Re(z) = Im(i z̄) , Im(z) = Re(i z̄) .

Proof: The following computation comes from the definition of real and imaginaryparts.

Re(z) =z+ z̄

2=

iz+ i z̄2i

=i z̄− (i z̄)

2i= Im(i z̄)

Im(z) =z− z̄

2i=

i z̄− iz2

=i z̄+(i z̄)

2= Re(i z̄) ■

Theorem 19.4.5 Suppose A is an algebra of functions in C0 (X) for X a locallycompact Hausdorff space which separates the points of X and annihilates no point of X ,and has the property that if f ∈A , then f̄ ∈A . Then A is dense in C0 (X).