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.

3.4. THE CASE OF COMPLEX VALUED FUNCTIONS 85of the form Yc fr +l and since f f((6. 2)) = 0, you can take each cj = 0 and so thishas shown that in particular, specializing to S? \ { (0, 2) \ ;MnDela (@ (012) £6" (0) ~0i=jim n sup° 2EXwhere f/’ € &. Thuslim sup =0Nr 2EXLets — f(z)i=However, the sum gives a sequence of elements of < which are converging uniformly to fon X. &3.4 The Case of Complex Valued FunctionsWhat about the general case where Cp (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 € © it follows f € &.Lemma 3.4.1 Let z be a complex number. ThenRe(z) =Im(iz), Im(z) = Re (iz)Proof: The following computation comes from the definition of real and imaginaryparts.z+Z iztiz iz—(izZ)Re(s) =)2-2 igi 18+ .Img) = 28 = 8S ES) _ peinTheorem 3.4.2 Suppose & is an algebra of functions in Co (X ) which separates thepoints of X and annihilates no point of X, a closed subset of R? and has the property thatif f € &, then f € &. Then of is dense in Cy (X).Proof: Let Re” = {Ref: fé€ #}, Im# ={Imf: fea}.Claim 1: Rev = Im #Proof of claim: A typical element of Re. is Ref where f € &, then from Lemma3.4.1, Re(f) =Im(if) € Im. Thus Rex C Ima#/. By assumption, if € &. The otherdirection works the same. Just use the other formula in Lemma 3.4.1.Claim 2: Both Re. and Im & are real algebras.Proof of claim: It is obvious these are both real vector spaces. Since these are equal, itsuffices to consider Re. It remains to show that Re. is closed with respect to products.+fg+g_ | cif | FPALS 8 _ 7 [fet fat fe Fal = 7 2Re( fe) +2Re (Fs)]Now by assumption, fg € & and so Re(fg) € Rex”. Also Re ( fg) € Rex because bothf,g are in & and it is an algebra. Thus, the above is in Re. because, as noted, this is areal vector space.