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.