FIVE MAJOR RESULTS IN ANALYSIS AND TOPOLOGY Aaron ...

3. THE STONE-WEIERSTRASS THEOREM 19
Lemma (Part 1). If f ∈ B, then |f| ∈ B, where |f|(x) = |f(x)| for every x ∈ K.
Proof. Let a = sup |f(x)| and ǫ > 0. By the Weierstrass approximation theorem,
x
choose a sequence of polynomials {P ∗ n(x)} such that P ∗ n(x) → |x| uniformly on [−a,a]. Define
Pn(x) = P ∗ n(x)−P ∗ n(0). Then one can find real numbers c1,...,cn such that
ǫ for every −a ≤ y ≤ a. Since B is an algebra, g =
n
|y|
− ciy
i=1
i
<
n
cif i is in B, so that |g(x) − |f(x)|| < ǫ
for each x ∈ K. Since B is uniformly closed, |f| ∈ B.
Before proceeding, we must define the maximum and minimum of two functions.
Definition. If f and g are real functions defined on a space X, then the maximum of f
and g is a real function max(f,g) defined on X such that max(f,g)(x) = max (f(x),g(x))
for each x ∈ X. The function min(f,g) is defined analogously. If f1,...fn are real functions
defined on a space X, then we define max(f1,...,fn) recursively by
min(f1,...,fn) is defined analogously.
i=1
max (...max (max(f1,f2),f3),...,fn).
Lemma (Part 2). If f1,...,fn ∈ B, then max(f1,...,fn),min(f1,...,fn) ∈ B.
Proof. If f,g ∈ B, then since
max(f,g) =
min(f,g) =
f + g
2
f + g
2
|f − g|
+ and
2
|f − g|
− ,
2
the fact that max(f,g) ∈ B and min(f,g) ∈ B is a consequence of Part 1. The conclusion
immediately follows via induction and the preceding definition.
Lemma (Part 3). If f ∈ C(K, R) and ǫ > 0, then for each point x of K there is a function
gx in B such that gx(x) = f(x), and gx(t) > f(t) − ǫ for every t ∈ K.
Proof. Choose x ∈ K. Since A ⊂ B, A separates points of K and vanishes nowhere on
K, then Lemma 1 guarantees, for each y ∈ K, the existence of a function hy ∈ B such that
hy(x) = f(x) and hy(y) = f(y). The continuity of hy allows us to choose an open set Jy of