Download The Stone-Weierstrass Theorem If X is a compact metric space, C(X

The Stone-Weierstrass Theorem
If X is a compact metric space, C(X) will denote the set of continuous functions f :
X → R. We can define the uniform norm on C(X) by ||f ||∞ = supx∈X |f (x)|.
We say that A ⊂ C(X) is an algebra if A is a linear subspace of C(X) with the additional
property that f, g ∈ A implies that f g ∈ A. (Here f g is just the function obtained by
pointwise multiplication: (f g)(x) = f (x)g(x).)
Theorem (Stone-Weierstrass Theorem). Let X be a compact metric space. Let A ⊂
C(X) be an algebra such that
(1) A contains a non-zero constant function;
(2) A separates points (i.e., if x, x0 ∈ X, x 6= x0 , then there exists f ∈ A such that
f (x) 6= f (x0 ).
Then A is uniformly dense in C(X).
Definition. For functions f, g ∈ C(X), we write
(f ∧ g)(x) = min{f (x), g(x)} =
1
(f (x) + g(x) − |f (x) − g(x)|)
2
(f ∨ g)(x) = max{f (x), g(x)} =
1
(f (x) + g(x) + |f (x) − g(x)|).
2
and
We have f ∧ g, f ∨ g ∈ C(X).
Definition. A subset L ⊂ C(X) is called a lattice if f, g ∈ L =⇒ f ∧ g, f ∨ g ∈ L. We call
a lattice L a closed lattice if it a closed subset of C(X).
Lemma 1.6. Suppose that X is a compact metric space. Let L ⊂ C(X) be a closed lattice
such that whenever x, y ∈ X, with x 6= y, and a, b ∈ R then there exists f ∈ L with
f (x) = a and f (y) = b. Then L = C(X).
Remark. Since A is an algebra, if A contains a non-zero constant function then it contains
all non-zero constant functions.
Example 1. Let X = [0, 1] and take A to be the set of polynomials on X. Then A contains
the non-zero constant function 1 and the function p(x) = x ∈ A separates points, so the
hypotheses of Theorem 1.3 are satisfied. This shows that Theorem 1.1 is a special case of
the Stone-Weierstrass Theorem.
Example 2. Let X = [0, 1] and
(
)
N
M
X
X
A = a0 +
an cos(2πnx) +
bn sin(2πnx) : an , bn ∈ R, N, M ≥ 1 .
n=1
n=1
1
Then A satisfies hypothesis (1) (since 1 ∈ A) but not hypothesis (2) (since f (0) = f (1)
for all f ∈ A). In fact, it is easy to see that A is not uniformly dense in C[0, 1]: choose
g ∈ C[0, 1] with g(0) 6= g(1) and suppose that 2² < |g(0) − g(1)|. Then, for any f ∈ A,
either |g(0) − f (0)| ≥ ² or |g(1) − f (1)| ≥ ², so that ||g − f ||∞ ≥ ².
However, the following simple modification of X allows us to make A a dense set.
Example 3. Let X = {e2πix : 0 ≤ x ≤ 1} (the unit circle). One can think of this as [0, 1]
with 0 and 1 identified. As before, take
(
)
N
M
X
X
A = a0 +
an cos(2πnx) +
bn sin(2πnx) : an , bn ∈ R, N, M ≥ 1 .
n=1
n=1
For f ∈ A, the fact that f (0) = f (1) now means that f is well-defined as a function on
X. Furthermore, one can easily check that A now satisfies the hypotheses of Theorem 1.5.
Therefore A is uniformly dense in C(X, R).
Proof of the Stone-Weierstrass Theorem. Since L is closed, we only need to show
that L is uniformly dense in C(X). In other words, given g ∈ C(X) and ² > 0, we need to
find f ∈ L such that ||g − f ||∞ < ².
Fix x ∈ X. Then, by the hypothesis on L, for each y ∈ X, we have fy ∈ L such that
fy (x) = g(x) and fy (y) = g(y).
Let Vy = {z ∈ X : fy (z) < g(z) + ²}; then Vy is open (why?) and x, y ∈ Vy .
The collection of sets {Vy }y∈X is an open cover of X and therefore, since X is compact,
we can choose a finite subcover {Vy1 , . . . , Vyn }.
Consider the corresponding functions fy1 , . . . , fyn and let Fx = fy1 ∧ · · · ∧ fyn . Then,
since L is a lattice, Fx ∈ L.
Observe that:
(1) since fyi (x) = g(x), i = 1, . . . , n, Fx (x) = g(x); and
(2) for z ∈ Vyi , we have Fx (z) ≤ fyi (z) < g(z) + ², so that Fx (z) < g(z) + ², for all
z ∈ X.
Let Wx = {z ∈ X : Fx (z) > g(z) − ²}; then Wx is a open set containing x. Thus,
{Wx }x∈X is an open cover for X. Again by compactness, it has a finite open subcover
{Wx1 , . . . , Wxm }. This gives us functions Fx1 , . . . , Fxm ∈ L. Let G = Fx1 ∨ · · · ∨ Fxm ∈ L
(since L is a lattice).
For z ∈ Wxi , G(z) ≥ Fxi (z) > g(z) − ², so that G(z) > g(z) − ², for all z ∈ X.
Combining inequalities, we have that g(z) − ² < G(z) < g(z) + ², for all z ∈ X, i.e.,
||g − G||∞ < ². So the lemma is proved by taking f = G.
Lemma 1.7. Suppose that X is a compact metric space. Let B ⊂ C(X) be a closed
algebra. Then B is a closed lattice.
Proof. First we show that if f ∈ B then |f | ∈ B. Consider the interval I = [−||f ||∞ , ||f ||∞ ]
and the (continuous) modulus function | · | : I → R : t 7→ |t|.
2
Given ² > 0, we can approximate | · | by a polynomial p(t) on I, i.e.,
|p(t) − |t|| < ²,
for all t ∈ I. Furthermore, we may replace p(t) by p0 (t) = p(t) − p(0), which a polynomial
with no constant term, and still have
|p0 (t) − |t|| < 2².
If p0 (t) =
Pn
k=1
ak tk then p(f (x)) =
Pn
k=1
ak (f (x))k , f (x) ∈ I, and
|p0 (f (x)) − |f (x)|| < 2²,
for all x ∈ X. Since B is an algebra, the function (p ◦ f )(x) = p(f (x)) ∈ B. Since B is
closed and ² > 0 is arbitrary, this implies that |f | ∈ B.
To complete the proof, we recall that f ∧g = 12 (f +g−|f −g|) and f ∨g = 12 (f +g+|f −g|),
so f, g ∈ B implies f ∧ g, f ∨ g ∈ B, i.e., B is a (closed) lattice.
Now we prove the Stone-Weierstrass Theorem.
The statement of the Theorem is equivalent to showing that C(X) is the closure of A in
the uniform norm. Let B denote the closure of A in C(X). It is easy to see that B is
also an algebra so, by Lemma 1.7, B is a closed lattice. Furthermore, B contains non-zero
constant functions because A does. Thus, we only need to show that if x, y ∈ X, with
x 6= y, and a, b ∈ R then there exists f ∈ B with f (x) = a and f (y) = b, since then Lemma
1.6 will tell us that B = C(X).
Choose g ∈ B such that g(x) 6= g(y) (we may do this since B ⊃ A). If we define f ∈ B
by
·
¸
·
¸
g(z) − g(y)
g(z) − g(x)
f (z) = a
+b
g(x) − g(y)
g(x) − g(y)
then f has the desired property. (Note that we have used the property that A (and hence
B) contains non-zero constant functions to ensure that this f lies in B.)
The theorem is proved.
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