proof of Stone-Weierstrass theorem

proof of Stone-Weierstrass theorem

Let

𝒜

¯

denote the

closure

of

𝒜

in

C

0

⁢

(

X

,

ℝ

)

according to the uniform convergence

topology

. We want to show that, if conditions 1 and 2 are satisfied, then

𝒜

¯

=

C

0

⁢

(

X

,

ℝ

)

.

First, we shall show that, if

f

∈

𝒜

¯

, then

|

f

|

∈

𝒜

¯

. Since

f

is a

continuous function

on a

compact space

f

must be

bounded

– there exists

constants

a

and

b

such that

a

≤

f

≤

b

. By the

Weierstrass approximation theorem

, for every

ϵ

>

0

, there exists a

polynomial

such that

|

P

⁢

(

x

)

-

|

x

|

|

<

ϵ

when

x

∈

[

a

,

b

]

. (By the way, one does not need the full-blown Weierstrass approximation theorem to show that

P

exists – see the entry “

proof of Weierstrass approximation theorem

” for an

elementary

construction of

P

) Define

g

:

X

→

ℝ

by

g

⁢

(

x

)

=

P

⁢

(

f

⁢

(

x

)

)

. Since

𝒜

¯

is an algebra,

g

∈

𝒜

¯

. For all

x

∈

X

,

|

g

⁢

(

x

)

-

|

f

⁢

(

x

)

|

|

<

ϵ

. Since

𝒜

¯

is

closed under

the uniform convergence topology, this implies that

|

f

|

∈

𝒜

¯

.

A

corollary

of the fact just proven is that if

f

,

g

∈

𝒜

¯

, then

max

⁡

(

f

,

g

)

∈

𝒜

¯

and

min

⁡

(

f

,

g

)

∈

𝒜

¯

. The reason for this is that one can write

max

⁡

(

a

,

b

)

=

1

2

⁢

(

a

+

b

+

|

a

-

b

|

)

min

⁡

(

a

,

b

)

=

1

2

⁢

(

a

+

b

-

|

a

-

b

|

)

Second, we shall show that, for every

f

∈

C

0

⁢

(

X

,

ℝ

)

, every

x

∈

X

, and every

ϵ

>

0

, there exists

g

x

∈

𝒜

¯

such that

g

x

≤

f

+

ϵ

and

g

x

>

f

. By condition 1, if

y

≠

x

, there exists a

function

h

~

x

⁢

y

∈

𝒜

such that

h

~

x

⁢

y

⁢

(

x

)

≠

h

~

x

⁢

y

⁢

(

y

)

. Define

h

x

⁢

y

by

h

x

⁢

y

⁢

(

z

)

=

p

⁢

h

~

x

⁢

y

⁢

(

z

)

+

q

, where the constants

p

and

q

have been chosen so that

h

x

⁢

y

⁢

(

x

)

=

f

⁢

(

x

)

+

ϵ

2

h

x

⁢

y

⁢

(

y

)

=

f

⁢

(

y

)

-

ϵ

2

By condition 2,

h

x

⁢

y

∈

𝒜

. (Note: This is the only place in the proof where condition 2 is used, but it is crucial since, otherwise, it might not be possible to construct a function which takes arbitrary preassigned values at two distinct points of

X

. The

necessity

of condition 2 can be shown by a simple example: Suppose that

𝒜

is the algebra of all continuous functions on

f

which vanish at a point

O

∈

X

. It is easy to see that this algebra

satisfies

all the hypotheses of the

theorem

except condition 2 and the

conclusion

of the theorem
does not hold in this case.)

For every

y

≠

x

, define the set

U

x

⁢

y

as

U

x

⁢

y

=

{

z

∈

X

∣

h

x

⁢

y

⁢

(

z

)

<

f

⁢

(

z

)

+

ϵ

}

Since

f

and

h

x

⁢

y

are

continuous

,

U

x

⁢

y

is an

open set

. Because

x

∈

U

x

⁢

y

and

y

∈

U

x

⁢

y

,

{

U

x

⁢

y

∣

y

∈

X

∖

{

x

}

}

is an

open cover

of

X

. By the

definition

of a compact space, there must exist a finite subcover. In other words, there exists a finite subset

{

y

1

,

y

2

,

…

,

y

n

}

⊂

X

such that

X

=

⋃

m

=

0

n

U

x

⁢

y

m

. Define

g

x

=

min

⁡

(

h

x

⁢

y

1

,

h

x

⁢

y

2

,

…

,

h

x

⁢

y

n

)

.

By the corollary of the first part of the proof,

g

x

∈

𝒜

¯

. By construction,

g

x

⁢

(

x

)

=

f

⁢

(

x

)

+

ϵ

/

2

and

g

x

<

f

+

ϵ

.

Third, we shall show that, for every

f

∈

C

0

⁢

(

X

,

ℝ

)

and every

ϵ

>

0

, there exists a function

g

∈

𝒜

¯

such that

f

≤

g

<

f

+

ϵ

. This will

complete

the proof becauase it implies that

𝒜

¯

=

C

0

⁢

(

X

,

ℝ

)

. For every

x

∈

X

, define the set

V

x

as

V

x

=

{

z

∈

X

∣

g

x

⁢

(

z

)

>

f

⁢

(

x

)

}

where

g

x

is defined as before. Since

f

and

g

x

are continuous,

V

x

is an open set. Because

g

x

⁢

(

x

)

=

f

⁢

(

x

)

+

ϵ

/

2

>

f

⁢

(

x

)

,

x

∈

V

x

. Hence

{

V

x

∣

x

}

is an open cover of

X

. By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset

{

x

1

,

x

2

,

…

⁢

x

n

}

⊂

X

such that

X

=

⋃

m

=

0

n

V

x

n

. Define

g

as

g

⁢

(

z

)

=

max

⁡

{

g

x

1

⁢

(

z

)

,

g

x

2

⁢

(

z

)

,

…

,

g

x

n

⁢

(

z

)

}

By the corollary of the first part of the proof,

g

∈

𝒜

¯

. By construction,

g

>

f

. Since

g

x

<

f

+

ϵ

for every

x

∈

X

,

g

<

f

+

ϵ

.

Title

proof of Stone-Weierstrass theorem

Canonical

name

ProofOfStoneWeierstrassTheorem

Date of creation

2013-03-22 14:35:01

Last modified on

2013-03-22 14:35:01

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

18

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 46E15

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