II Linear Analysis - The topology of C(K)

3The topology of C(K)

II Linear Analysis

3.4 Stone–Weierstrass theorem

Here we will prove the Stone–Weierstrass theorem, which is a generalization of

the classical Weierstrass approximation theorem.

Theorem

(Weierstrass approximation theorem)

The set of polynomials are

dense in C([0, 1]).

This tells us that

([0

1]) is not too “big” because it is has a dense subset

of “nice” functions.

We will try to generalize this for more general domains. Note that for this

section, real-valued and complex-valued continuous functions are somewhat

differentiable. So we will write these as C

(K) and C

(K).

To state this theorem, we need some definitions.

Definition

(Algebra)

A vector space (

+) is called an algebra if there is an

operation (called multiplication)

V → V

such that (

, ·

) is a rng (i.e. ring

not necessarily with multiplicative identity). Also,

(

v ·w

) = (

λv

)

·w

v ·

(

λw

)

for all λ ∈ F, v, w ∈ V .

If V is in addition a normed vector space and

kv · wk

≤ kvk

· kwk

for all v, w ∈ V , then we say V is a normed algebra.

If V complete normed algebra, we say V is a Banach algebra.

is an algebra that is commutative as a rng, then we say

is a commutative

algebra.

If V is an algebra with multiplicative identity, then V is a unital algebra.

Example. C(K) is a commutative, unital, Banach algebra.

Example. Recall from the example sheets that if T

, T

: V → V , then

◦ T

B(V,V )

≤ kT

B(V,V )

So B(V, V ) is a unital normed algebra.

We will need this language to generalize the Weierstrass approximation

theorem. The main problem in doing so is that we have to figure out what we

can generalize polynomials to. This is why we need these funny definitions.

Theorem

(Stone-Weierstrass theorem)

Let

be compact, and

A ⊆ C

(

)

be a subalgebra (i.e. it is a subset that is closed under the operations) with the

property that it separates points, i.e. for every

x, y ∈ K

distinct, there exists

some

f ∈ A

such that

(

)

(

). Then either

(

) or there is some

∈ K such that

A = {f ∈ C

R(K)

: f(x

) = 0}.

Note that it is not possible that

is always zero on 2 or more points, since

A separates points.

This indeed implies the Weierstrass approximation theorem, since polynomials

separates points (consider the polynomial

), and the polynomial 1 is never 0

for all x. In fact, this works also for polynomials on compact subsets of R

Note, however, that the second case of Stone-Weierstrass theorem can actually

happen. For example, consider

= [0

1] compact, and

be the algebra

generated by x. Then

A = {f ∈ C

(K) : f(0) = 0}.

We will prove this in using two lemmas:

Lemma.

Let

compact,

L ⊆ C

(

) be a subset which is closed under taking

maximum and minimum, i.e. if

f, g ∈ L

, then

max{f, g} ∈ L

and

min{f, g} ∈ L

(with

max{f, g}

defined as

max{f, g}

(

) =

max{f

(

)

, g

(

)

}

, and similarly for

minimum).

Given

g ∈ C

(

), assume further that for any

ε >

0 and

x, y ∈ K

, there

exists f

x,y

∈ L such that

x,y

(x) − g(x)| < ε, |f

x,y

(y) −g(y)| < ε.

Then there exists some f ∈ L such that

kf − gk

(K)

< ε,

i.e. g ∈

This means that if we are allowed to take maximums and minimums, to be

able to approximate a function, we just need to be able to approximate it at any

two points.

The idea is to next show that if

is a subalgebra, then

is closed under

taking maximum and minimum. Then use the ability to separate points to find

x,y

, and prove that we can approximate arbitrary functions.

Proof. Let g ∈ C

(K) and ε > 0 be given. So for every x, y ∈ K, there is some

x,y

∈ L such that

x,y

(x) − g(x)| < ε, |f

x,y

(y) −g(y)| < ε.

Claim.

For each

x ∈ K

, there exists

∈ L

such that

(

)

− g

(

)

| < ε

and

(z) < g(z) + ε for all z ∈ K.

Since f

x,y

is continuous, there is some U

x,y

containing y open such that

x,y

(z) − g(z)| < ε

for all z ∈ U

x,y

. Since

[

y∈K

x,y

⊇ K,

by compactness of K, there exists a some y

, ··· , y

such that

[

i=1

x,y

⊇ K.

g + ε

We then let

(z) = min{f

x,y

(z), ··· , f

x,y

(z)}

for every

z ∈ K

. We then see that this works. Indeed, by assumption,

∈ L

If z ∈ K is some arbitrary point, then z ∈ U

x,y

for some i. Then

x,y

(z) < g(z) + ε.

Hence, since f

is the minimum of all such f

x,y

, for any z, we have

(z) < g(z) + ε.

The property at x is also clear.

Claim. There exists f ∈ L such that |f(z) − g(z)| < ε for all z ∈ K.

We are going to play the same game with this. By continuity of

, there is

containing x open such that

(w) − g(w)| < ε

for all w ∈ V

. Since

[

x∈K

⊇ K,

by compactness of K, there is some {x

, ··· , x

} such that

[

j=1

⊇ K.

Define

f(z) = max{f

(z), ··· , f

(z)}.

Again, by assumption, f ∈ L. Then we know that

f(z) > g(z) − ε.

We still have our first bound

f(z) < g(z) + ε.

Therefore we have

kf − gk

(K)

< ε.

Lemma.

Let

A ⊆ C

(

) be a subalgebra that is a closed subset in the topology

of C

(K). Then A is closed under taking maximum and minimum.

Proof. First note that

max{f(x), g(x)} =

(f(x) + g(x)) +

|f(x) − g(x)|,

min{f(x), g(x)} =

(f(x) + g(x)) −

|f(x) − g(x)|.

Since

is an algebra, it suffices to show that

f ∈ A

implies

|f| ∈ A

for every

such that kfk

(K)

≤ 1.

The key observation is the following: consider the function

(

) =

√

x + ε

Then

(

) approximates

|x|

. This has the property that the Taylor expansion

(

) centered at

is uniformly convergent for

x ∈

1]. Therefore there

exists a polynomial S(x) such that

|S(x) −

x + ε

| < ε.

Now note that

(

)

− S

(0) is a polynomial with no constant term. Therefore,

since A is an algebra, if f ∈ A, then S(f

) − S(0) ∈ A by closure.

Now look at

k|f|−(S(f

) −S(0))k

(K)

≤ k|f|−

+ ε

k+k

+ ε

−S(f

)k+kS(0)k.

We will make each individual term small. For the first term, note that

sup

x∈[0,1]

|x −

+ ε

| = sup

x∈[0,1]

|x +

√

+ ε

= ε.

So the first term is at most

. The second term is also easy, since

is chosen

such that

(

)

−

√

x + ε

| <

1 for

x ∈

1], and

(

)

| ≤

1 for all

x ∈

1].

So it is again bounded by ε.

By the same formula, |S(0) −

√

0 + ε

| < ε. So |S(0)| < 2ε. So

k|f| − (S(f

) − S(0))k

(K)

< 4ε.

Since

ε >

0 and

is closed in the topology of

(

f ∈ A

and

kfk

(K)

≤

implies that |f| ∈ A.

Note that if we have already proven the classical Weierstrass approximation

theorem, we can just use it to get a polynomial approximation for |f| directly.

We will now combine both lemmas and prove the Stone-Weierstrass theorem.

Theorem

(Stone-Weierstrass theorem)

Let

be compact, and

A ⊆ C

(

)

be a subalgebra (i.e. it is a subset that is closed under the operations) with the

property that it separates points, i.e. for every

x, y ∈ K

distinct, there exists

some

f ∈ A

such that

(

)

(

). Then either

(

) or there is some

∈ K such that

A = {f ∈ C

R(K)

: f(x

) = 0}.

Proof.

Note that there are two possible outcomes. We will first look at the first

possibility.

Consider the case where for all

x ∈ K

, there is some

f ∈ A

such that

(

)

= 0.

Let

g ∈ C

(

) be given. By our previous lemmas, to approximate

, we

just need to show that we can approximate

at two points. So given any

ε >

x, y ∈ K, we want to find f

x,y

∈ A such that

x,y

(x) − g(x)| < ε, |f

x,y

(y) −g(y)| < ε. (∗)

For every

x, y ∈ K

x 6

, we first show that there exists

x,y

∈ A

such that

x,y

(

)

= 0, and

x,y

(

)

x,y

(

). This is easy to see. By our assumptions, we

can find the following functions:

(i) There exists h

(1)

x,y

such that h

(1)

x,y

6= h

(1)

x,y

(y)

(ii) There exists h

(2)

x,y

such that h

(2)

x,y

(x) 6= 0.

(iii) There exists h

(3)

x,y

such that h

(3)

x,y

(y) 6= 0.

Then it is an easy exercise to show that some linear combination of

(1)

x,y

and

(2)

x,y

and h

(3)

x,y

works, say h

x,y

We will want to find our

x,y

that satisfies (

∗

). But we will do better. We

will make it equal

and

. The idea is to take linear combinations of

x,y

and

x,y

. Instead of doing the messy algebra to show that we can find a working

linear combination, just notice that (

x,y

(

)

, h

x,y

(

)) and (

x,y

(

)

, h

x,y

(

)

are linearly independent vectors in

. Therefore there exists

α, β ∈ R

such that

α(h

x,y

(x), h

x,y

(y)) + β(h

x,y

(x)

, h

x,y

(y)

) = (g(x), g(y)).

So done.

In the other case, given

, suppose there is

∈ K

such that

(

) = 0 for

all f ∈ A. Consider the algebra

= A + λ1 = {f + λ1 : f ∈ A, λ ∈ R}

Since

separates points, and for any

x ∈ K

, there is some

f ∈ A

such that

f(x) 6= 0 (e.g. f = 1), by the previous part, we know that

= C

(K).

Now note that

A ⊆ {f ∈ C

(K) : f(x

) = 0} = B.

So we suffices to show that we have equality, i.e. for any

g ∈ B

and

ε >

0, there

is some f ∈ A such that

kf − gk

(K)

< ε.

Since

(

), given such

and

, there is some

f ∈ A

and

∈ R

such

that

kg −(f + λ1)k

(K)

< ε.

But

(

) =

(

) = 0, which implies that

|λ| < ε

. Therefore

kg −f k

(K)

So done.

What happens for C

(K)?

Example.

Let

B(0, 1) ⊆ C

, and let

be the set of polynomials on

B(0, 1)

We will show that

A 6= C

(B(0, 1)).

Consider

(

) =

¯z

. This is not in the closure of

, since this is not holomor-

phic, but the uniform limit of any sequence of holomorphic is holomorphic (by

Morera’s theorem, i.e.

is holomorphic iff

(

) d

= 0 for all closed piecewise

smooth curve γ).

Hence we need an additional condition on this. It turns out that this rather

simple example is the only way in which things can break. We have the following:

Theorem

(Complex version of Stone-Weierstrass theorem)

Let

be compact

and

A ⊆ C

(

) be a subalgebra over

which separates points and is closed

under complex conjugation (i.e. if

f ∈ A

, then

). Then either

(

)

or these is an x

such that

A = {f ∈ C

(K) : f(x

) = 0}.

Proof.

It suffices to show that either

A ⊇ C

(

) or there exists a point

such that

A ⊇ {f ∈ C

(

) :

(

) = 0

}

, since we can always break a complex

function up into its real and imaginary parts.

Now consider



f +

: f ∈ A



∪



f −

: f ∈ A



Now note that by closure of

, we know that

is a subset of

and is

a subalgebra of

(

) over

, which separates points. Hence by the real

version of Stone-Weierstrass, either

(

) or there is some

such that

= {f ∈ C

(K) : f(x

) = 0}. So done.