proof of Tietze extension theorem

proof of Tietze extension theorem

To prove the

Tietze Extension Theorem

, we first need a lemma.

Lemma 1

.

If

X

is a

normal topological space

and

A

is closed in

X

, then for any

continuous function

f

:

A

→

R

such that

|

f

⁢

(

x

)

|

≤

1

, there is a continuous function

g

:

X

→

R

such that

|

g

⁢

(

x

)

|

≤

1

3

for

x

∈

X

, and

|

f

⁢

(

x

)

-

g

⁢

(

x

)

|

≤

2

3

for

x

∈

A

.

Proof.

The sets

f

-

1

⁢

(

(

-

∞

,

-

1

3

]

)

and

f

-

1

⁢

(

[

1

3

,

∞

)

)

are

disjoint

and closed in

A

. Since

A

is closed, they are also closed in

X

. Since

X

is normal, then by Urysohn’s lemma and the fact that

[

0

,

1

]

is

homeomorphic

to

[

-

1

3

,

1

3

]

, there is a continuous function

g

:

X

→

[

-

1

3

,

1

3

]

such that

g

⁢

(

f

-

1

⁢

(

(

-

∞

,

-

1

3

⁢

c

]

)

)

=

-

1

3

and

g

⁢

(

f

-

1

⁢

(

[

1

3

,

∞

)

)

)

=

1

3

. Thus

|

g

⁢

(

x

)

|

≤

1

3

for

x

∈

X

. Now if

-

≤

f

(

x

)

≤

-

1

3

, then

g

⁢

(

x

)

=

-

1

3

and thus

|

f

⁢

(

x

)

-

g

⁢

(

x

)

|

≤

2

3

. Similarly if

1

3

≤

f

⁢

(

x

)

≤

1

, then

g

⁢

(

x

)

=

1

3

and thus

|

f

⁢

(

x

)

-

g

⁢

(

x

)

|

≤

2

3

. Finally, for

|

f

⁢

(

x

)

|

≤

1

3

we have that

|

g

⁢

(

x

)

|

≤

1

3

, and so

|

f

⁢

(

x

)

-

g

⁢

(

x

)

|

≤

2

3

. Hence

|

f

⁢

(

x

)

-

g

⁢

(

x

)

|

≤

2

3

holds for all

x

∈

A

.
∎

This puts us in a position to prove the main

theorem

.

Proof of the Tietze extension theorem.

First suppose that for any continuous function on a

closed subset

there is a

continuous

extension

. Let

C

and

D

be disjoint and closed in

X

. Define

f

:

C

∪

D

→

ℝ

by

f

⁢

(

x

)

=

0

for

x

∈

C

and

f

⁢

(

x

)

=

1

for

x

∈

D

. Now

f

is continuous and we can extend it to a continuous function

F

:

X

→

ℝ

. By Urysohn’s lemma,

X

is normal because

F

is a continuous function such that

F

⁢

(

x

)

=

0

for

x

∈

C

and

F

⁢

(

x

)

=

1

for

x

∈

D

.

Conversely

, let

X

be normal and

A

be closed in

X

. By the lemma, there is a continuous function

g

0

:

X

→

ℝ

such that

|

g

0

⁢

(

x

)

|

≤

1

3

for

x

∈

X

and

|

f

⁢

(

x

)

-

g

0

⁢

(

x

)

|

≤

2

3

for

x

∈

A

. Since

(

f

-

g

0

)

:

A

→

ℝ

is continuous, the lemma tells us there is a continuous function

g

1

:

X

→

ℝ

such that

|

g

1

⁢

(

x

)

|

≤

1

3

⁢

(

2

3

)

for

x

∈

X

and

|

f

⁢

(

x

)

-

g

0

⁢

(

x

)

-

g

1

⁢

(

x

)

|

≤

2

3

⁢

(

2

3

)

for

x

∈

A

. By repeated application of the lemma we can

construct

a

sequence

of continuous functions

g

0

,

g

1

,

g

2

,

…

such that

|

g

n

⁢

(

x

)

|

≤

1

3

⁢

(

2

3

)

n

for all

x

∈

X

, and

|

f

⁢

(

x

)

-

g

0

⁢

(

x

)

-

g

1

⁢

(

x

)

-

g

2

⁢

(

x

)

-

⋯

|

≤

(

2

3

)

n

for

x

∈

A

.

Define

F

⁢

(

x

)

=

∑

n

=

0

∞

g

n

⁢

(

x

)

. Since

|

g

n

⁢

(

x

)

|

≤

1

3

⁢

(

2

3

)

n

and

∑

n

=

0

∞

1

3

⁢

(

2

3

)

n

converges

as a

geometric series

, then

∑

n

=

0

∞

g

n

⁢

(

x

)

converges absolutely

and uniformly, so

F

is a continuous function defined everywhere. Moreover

∑

n

=

0

∞

1

3

⁢

(

2

3

)

n

=

1

implies that

|

F

⁢

(

x

)

|

≤

1

.

Now for

x

∈

A

, we have that

|

f

⁢

(

x

)

-

∑

n

=

0

k

g

n

⁢

(

x

)

|

≤

(

2

3

)

k

+

1

and as

k

goes to

infinity

, the right side goes to zero and so the sum goes to

F

⁢

(

x

)

. Thus

|

f

⁢

(

x

)

-

F

⁢

(

x

)

|

=

0

Therefore

F

extends

f

.∎

Remarks:

If

f

was a

function

satisfying

|

f

⁢

(

x

)

|

<

1

, then the theorem can be strengthened as follows. Find an extension

F

of

f

as above. The set

B

=

F

-

1

⁢

(

{

-

1

}

∪

{

1

}

)

is closed and disjoint from

A

because

|

F

⁢

(

x

)

|

=

|

f

⁢

(

x

)

|

<

1

for

x

∈

A

. By Urysohn’s lemma there is a continuous function

ϕ

such that

ϕ

⁢

(

A

)

=

{

1

}

and

ϕ

⁢

(

B

)

=

{

0

}

. Hence

F

⁢

(

x

)

⁢

ϕ

⁢

(

x

)

is a continuous extension of

f

⁢

(

x

)

, and has the property that

|

F

⁢

(

x

)

⁢

ϕ

⁢

(

x

)

|

<

1

.

If

f

is

unbounded

, then Tietze extension theorem holds as well. To see that consider

t

⁢

(

x

)

=

tan

-

1

⁡

(

x

)

/

(

π

/

2

)

. The function

t

∘

f

has the property that

(

t

∘

f

)

⁢

(

x

)

<

1

for

x

∈

A

, and so it can be extended to a continuous function

h

:

X

→

ℝ

which has the property

|

h

⁢

(

x

)

|

<

1

. Hence

t

-

1

∘

h

is a continuous extension of

f

.

Title

proof of Tietze extension theorem

Canonical name

ProofOfTietzeExtensionTheorem

Date of creation

2013-03-22 14:08:58

Last modified on

2013-03-22 14:08:58

Owner

bbukh (348)

Last modified by

bbukh (348)

Numerical id

10

Author

bbukh (348)

Entry type

Proof

Classification

msc 54C20

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