consists of complex valued functions and the field of scalars is ℂ
rather than ℝ? The following is the version of the Stone Weierstrass theorem which applies to this case.
You have to assume that for f ∈A it follows f∈A.

Theorem 2.8.12Suppose A is an algebra of functions in C_{0}

(X )

, where X is a closed subset ofℝ^{n}which separates the points, annihilates no point, and has the property that if f ∈A, thenf∈A.Then A is dense in C_{0}

(X )

.

Proof: Let ReA≡

{Ref : f ∈ A }

, ImA≡

{Im f : f ∈ A }

. First I will show that
A =ReA + iImA =ImA + iReA. Let f ∈A. Then

1( ¯) 1 ( ¯)
f = 2 f + f + 2 f − f = Ref + iIm f ∈ Re A + iIm A

and so A⊆ReA + iImA. Also

( ---)
f = 1-(if + i¯f)−-i if + (if) = Im (if)+ iRe(if) ∈ Im A + iReA
2i 2

This proves one half of the desired equality. Now suppose h ∈ReA + iImA. Then h = Reg_{1} + iImg_{2}
where g_{i}∈A. Then since Reg_{1} =

1
2

(g1 + g1)

, it follows Reg_{1}∈A. Similarly Img_{2}∈A. Therefore, h ∈A.
The case where h ∈ImA + iReA is similar. This establishes the desired equality.

Now ReA and ImA are both real algebras. I will show this now. First consider ImA. It is obvious this
is a real vector space. It only remains to verify that the product of two functions in ImA is in ImA. Note
that from the first part, ReA,ImA are both subsets of A because, for example, if u ∈ImA then
u + 0 ∈ImA + iReA = A. Therefore, if v,w ∈ImA, both iv and w are in A and so Im

(ivw)

= vw and
ivw ∈A. Similarly, ReA is an algebra.

Both ReA and ImA must separate the points. Here is why: If x_{1}≠x_{2}, then there exists
f ∈A such that f

(x1)

≠f

(x2)

. If Imf

(x1)

≠Imf

(x2)

, this shows there is a function in ImA,
Imf which separates these two points. If Imf fails to separate the two points, then Ref
must separate the points and so you could consider Im

(if)

to get a function in ImA which
separates these points. This shows ImA separates the points. Similarly ReA separates the
points.

Neither ReA nor ImA annihilate any point. This is easy to see because if x is a point there exists
f ∈A such that f

(x)

≠0. Thus either Ref

(x)

≠0 or Imf

(x)

≠0. If Imf

(x)

≠0, this shows this point is not
annihilated by ImA. If Imf

(x)

= 0, consider Im

(if)

(x )

= Ref

(x)

≠0. Similarly, ReA does not
annihilate any point.

It follows from Theorem 2.8.11 that ReA and ImA are dense in the real valued functions of
C_{0}

(X)

. Let f ∈ C_{0}

(X)

. Then there exists

{hn}

⊆ReA and

{gn}

⊆ImA such that h_{n}→Ref
uniformly and g_{n}→Imf uniformly. Therefore, h_{n} + ig_{n}∈A and it converges to f uniformly.
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