proof of Hahn-Banach theorem

proof of Hahn-Banach theorem

Consider the family of all possible

extensions

of

f

, i.e. the set

ℱ

of all

pairings

(

F

,

H

)

where

H

is a

vector subspace

of

X

containing

U

and

F

is a

linear map

F

:

H

→

K

such that

F

⁢

(

u

)

=

f

⁢

(

u

)

for all

u

∈

U

and

|

F

⁢

(

u

)

|

≤

p

⁢

(

u

)

for all

u

∈

H

.

ℱ

is naturally endowed with an

partial order

relation

: given

(

F

1

,

H

1

)

,

(

F

2

,

H

2

)

∈

ℱ

we say
that

(

F

1

,

H

1

)

≤

(

F

2

,

H

2

)

iff

F

2

is an extension of

F

1

that is

H

1

⊂

H

2

and

F

2

⁢

(

u

)

=

F

1

⁢

(

u

)

for all

u

∈

H

1

.
We want to apply Zorn’s Lemma to

ℱ

so we are going to prove that every chain in

ℱ

has an

upper bound

.

Let

(

F

i

,

H

i

)

be the elements of a chain in

ℱ

. Define

H

=

⋃

i

H

i

. Clearly

H

is a vector subspace of

V

and contains

U

. Define

F

:

H

→

K

by “merging” all

F

i

’s as follows. Given

u

∈

H

there exists

i

such that

u

∈

H

i

: define

F

⁢

(

u

)

=

F

i

⁢

(

u

)

. This is a good definition since if both

H

i

and

H

j

contain

u

then

F

i

⁢

(

u

)

=

F

j

⁢

(

u

)

in fact either

(

F

i

,

H

i

)

≤

(

F

j

,

H

j

)

or

(

F

j

,

H

j

)

≤

(

F

i

,

H

i

)

.
Notice that the map

F

is linear, in fact given any two vectors

u

,

v

∈

H

there exists

i

such that

u

,

v

∈

H

i

and hence

F

⁢

(

α

⁢

u

+

β

⁢

v

)

=

F

i

⁢

(

α

⁢

u

+

β

⁢

v

)

=

α

⁢

F

i

⁢

(

u

)

+

β

⁢

F

i

⁢

(

v

)

=

α

⁢

F

⁢

(

u

)

+

β

⁢

F

⁢

(

v

)

.
The so constructed pair

(

F

,

H

)

is hence an upper bound for the chain

(

F

i

,

H

i

)

because

F

is an extension of every

F

i

.

Zorn’s Lemma then assures that there exists a

maximal element

(

F

,

H

)

∈

ℱ

. To

complete

the proof we will only need to prove that

H

=

V

.

Suppose by

contradiction

that there exists

v

∈

V

∖

H

. Then consider the

vector space

H

′

=

H

+

K

⁢

v

=

{

u

+

t

⁢

v

:

u

∈

H

,

t

∈

K

}

(

H

′

is the vector space generated by

H

and

v

).
Choose

λ

=

sup

x

∈

H

⁡

{

F

⁢

(

x

)

-

p

⁢

(

x

-

v

)

}

.

We notice that given any

x

,

y

∈

H

it holds

F

⁢

(

x

)

-

F

⁢

(

y

)

=

F

⁢

(

x

-

y

)

≤

p

⁢

(

x

-

y

)

=

p

⁢

(

x

-

v

+

v

-

y

)

≤

p

⁢

(

x

-

v

)

+

p

⁢

(

y

-

v

)

i.e.

F

⁢

(

x

)

-

p

⁢

(

x

-

v

)

≤

F

⁢

(

y

)

+

p

⁢

(

y

-

v

)

;

in particular we find that

λ

<

+

∞

and for all

y

∈

H

it holds

F

⁢

(

y

)

-

p

⁢

(

y

-

v

)

≤

λ

≤

F

⁢

(

y

)

+

p

⁢

(

y

-

v

)

.

Define

F

′

:

H

′

→

K

as follows:

F

′

⁢

(

u

+

t

⁢

v

)

=

F

⁢

(

u

)

+

t

⁢

λ

.

Clearly

F

′

is a

linear functional

.
We have

|

F

′

⁢

(

u

+

t

⁢

v

)

|

=

|

F

⁢

(

u

)

+

t

⁢

λ

|

=

|

t

|

⁢

|

F

⁢

(

u

/

t

)

+

λ

|

and by letting

y

=

-

u

/

t

by the previous estimates on

λ

we obtain

F

⁢

(

u

/

t

)

+

λ

≤

F

⁢

(

u

/

t

)

+

F

⁢

(

-

u

/

t

)

+

p

⁢

(

-

u

/

t

-

v

)

=

p

⁢

(

u

/

t

+

v

)

and

F

⁢

(

u

/

t

)

+

λ

≥

F

⁢

(

u

/

t

)

+

F

⁢

(

-

u

/

t

)

-

p

⁢

(

-

u

/

t

-

v

)

=

-

p

⁢

(

u

/

t

+

v

)

which together give

|

F

⁢

(

u

/

t

)

+

λ

|

≤

p

⁢

(

u

/

t

+

v

)

and hence

|

F

′

⁢

(

u

+

t

⁢

v

)

|

≤

|

t

|

⁢

p

⁢

(

u

/

t

+

v

)

=

p

⁢

(

u

+

t

⁢

v

)

.

So we have proved that

(

F

′

,

H

′

)

∈

ℱ

and

(

F

′

,

H

′

)

>

(

F

,

H

)

which is a contradiction.

Title

proof of Hahn-Banach theorem

Canonical name

ProofOfHahnBanachTheorem

Date of creation

2013-03-22 13:31:58

Last modified on

2013-03-22 13:31:58

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

8

Author

paolini (1187)

Entry type

Proof

Classification

msc 46B20

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