1. Stone-Weierstrass Theorem
Let X be a compact metric space and C(X) be the space of continuous real-valued functions on X. On C(X), we define
kf k∞= sup
x∈X
|f (x)|.
Then the normed space (C(X), k · k∞) is a Banach space over R.
A subalgebra of C(X) is a vector subspace A such that for any f, g ∈ A, f · g ∈ A. We say that a subalgebra A separates points if for any x, y ∈ X, there exists f ∈ A so that f (x) 6= f (y). We say that A is unital if A contains 1.
Theorem 1.1. (Stone-Weierstrass Theorem) Suppose A is a unital subalgebra of C(X) such that A separates points of X. Then A is dense in C(X), i.e. the closure A of A is the whole space C(X).
Proof. The proof will be omitted here.
Corollary 1.1. (Weierstrass second approximation theorem) Let A be the vector subspace of C([0, 2π]) spanned by {1, cos nx, sin nx : n ≥ 1}. Then A is a unital subalgebra of A which separates points of [0, 2π] and hence A is dense in C(X).
Proof. Elements of A are are of the form
t(x) = a0
2 +
N
X
k=1
{akcos kθ + bksin kθ},
where a0, a1, · · · , aN and b1, · · · , bN are real numbers. To show that A is a subalgebra of C([0, 2π]), we need to show that t(x)s(x) ∈ A for all t(x), s(x) ∈ A. To do this, we use the following formulas:
(1) 2 cos kθ cos jθ = cos(k + j)θ + cos(k − j)θ, (2) 2 sin kθ sin jθ = cos(k − j)θ − cos(k + j)θ, (3) 2 sin kθ cos jθ = sin(k + j)θ + sin(k − j)θ.
Hence t(x)s(x) is also a linear combination of {1, cos nx, sin nx : n ≥ 1}. It is not hard to see that A separates points. By Stone-Weierstrass theorem, A is dense in C([0, 2π]). Lemma 1.1. Let {fn} be a sequence of real-valued continuous functions on [a, b] which converges uniformly to a continuous function f on [a, b]. Then
n→∞lim Z b
a
fn(x)dx = Z b
a
f (x)dx.
Proof. Since {fn} converges to f uniformly on [a, b], for any > 0, there exists N > 0 so that for any n ≥ N, and any x ∈ [a, b]
|fn(x) − f (x)| ≤ kfn− f k∞< .
1
2
Hence if n ≥ N,
Z b a
fn(x)dx − Z b
a
f (x)dx
=
Z b a
(fn(x) − f (x))dx
≤ Z b
a
|fn(x) − f (x)|dx
≤ Z b
a
kfn− f kdx
< (b − a).
By definition, we prove our assertion.
Corollary 1.2. Let {fn} be a sequence of continuous functions on [a, b] so that the infinite series
∞
X
n=1
fn(x) converges uniformly on [a, b]. Then Z b
a
∞
X
n=1
fn(x)
! dx =
∞
X
n=1
Z b
a
fn(x)dx.
Proof. Let us consider the n-th partial sum sn(x) =
n
X
k=1
fk(x). Then {sn} converges to a
function s =
∞
X
n=1
fn in C([a, b]). By the previous lemma,
n→∞lim Z b
a
sn(x)dx = Z b
a
s(x)dx.
Suppose ψ ∈ C([0, 2π]) periodic of period 2π such that the following series converges uniformly on [0, 2π] :
ψ(θ) = a0 2 +
∞
X
n=1
{ancos nx + bnsin nx}.
By the uniform convergence, we can interchange summation and integral Z 2π
0
ψ(θ)dθ = Z 2π
0
ψ(θ)dθ +
∞
X
n=1
{an Z 2π
0
cos nθdθ + bn
Z 2π 0
ψ(θ) sin nθdθ}.
Hence we obtain
a0 = 1 π
Z 2π 0
ψ(θ)dθ.
Multiplying ψ by cos kθ, we obtain ψ(θ) cos kθ = a0
2 cos kθ +
∞
X
n=1
{ancos nθ cos kθ + bnsin nθ cos kθ}.
Integrating this infinite series, by the uniform convergence of the infinite series, we have Z 2π
0
ψ(θ) cos kθdθ = Z 2π
0
a0
2 cos kθdθ +
∞
X
n=1
Z 2π 0
{ancos nθ cos kθ + bnsin nθ cos kθ}dθ.
3
Similarly, multiplying ψ by sin kθ and integrating it, we obtain Z 2π
0
ψ(θ) sin kθdθ = Z 2π
0
a0
2 sin kθdθ +
∞
X
n=1
Z 2π 0
{ancos nθ sin kθ + bnsin nθ sin kθ}dθ.
Using the trigonometric identity, one can show that for any n, k, (1)
Z 2π 0
cos nθ cos kθdθ = πδnk, (2)
Z 2π 0
sin nθ sin kθdθ = πδnk, (3)
Z 2π 0
cos nθ sin kθdθ = 0.
These imply that for n ≥ 1, an= 1
π Z 2π
0
ψ(θ) cos nθdθ, bn= 1 π
Z 2π 0
ψ(θ) sin nθdθ.
