3.4. THE CASE OF COMPLEX VALUED FUNCTIONS 85

of the form ∑mni=1 cn

i f̃ ni + cn

0 and since f̃((⃗

0,2))

= 0, you can take each cn0 = 0 and so this

has shown that in particular, specializing to Sp \{(⃗

0,2)}

,

limn→∞

supz∈X

∣∣∣∣∣mn

∑i=1

cni f n

i(θ−1 (θ (z))

)− f

(θ−1 (θ (z))

)∣∣∣∣∣= 0

where f ni ∈A . Thus

limn→∞

supz∈X

∣∣∣∣∣mn

∑i=1

cni f n

i (z)− f (z)

∣∣∣∣∣= 0

However, the sum gives a sequence of elements of A which are converging uniformly to fon X . ■

3.4 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 3.4.1 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 3.4.2 Suppose A is an algebra of functions in C0 (X) ,which separates thepoints of X and annihilates no point of X , a closed subset of Rp and has the property thatif f ∈A , then f̄ ∈A . Then A is dense in C0 (X).

Proof: Let ReA ≡ {Re f : f ∈A }, ImA ≡{Im f : f ∈A }.Claim 1: ReA = ImAProof of claim: A typical element of ReA is Re f where f ∈ A , then from Lemma

3.4.1, Re( f ) = Im(i f̄)∈ ImA . Thus ReA ⊆ ImA . By assumption, i f̄ ∈A . The other

direction works the same. Just use the other formula in Lemma 3.4.1.Claim 2: Both ReA and ImA are real algebras.Proof of claim: It is obvious these are both real vector spaces. Since these are equal, it

suffices to consider ReA . It remains to show that ReA is closed with respect to products.

f + f̄2

g+ ḡ2

=14[

f g+ f ḡ+ f̄ g+ f g]=

14[2Re( f g)+2Re

(f̄ g)]

Now by assumption, f g ∈A and so Re( f g) ∈ ReA . Also Re(

f̄ g)∈ ReA because both

f̄ ,g are in A and it is an algebra. Thus, the above is in ReA because, as noted, this is areal vector space.