II Linear Analysis - Normed vector spaces

1Normed vector spaces

II Linear Analysis

1.7 Hahn–Banach Theorem

Let

be a real normed vector space. What can we say about

∗

(

V, R

For instance, If V is non-trivial, must V

∗

be non-trivial?

The main goal of this section is to prove the Hahn–Banach theorem (surprise),

which allows us to produce a lot of elements in

∗

. Moreover, it doesn’t just tell

us that

∗

is non-empty (this is rather dull), but provides a tool to craft (or at

least prove existence of) elements of V

∗

that satisfy some property we want.

Proposition.

Let

be a real normed vector space, and

W ⊆ V

has co-

dimension 1. Assume we have the following two items:

– p : V → R (not necessarily linear), which is positive homogeneous, i.e.

p(λv) = λp(v)

for all v ∈ V, λ > 0, and subadditive, i.e.

p(v

+ v

) ≤ p(v

) + p(v

)

for all

, v

∈ V

. We can think of something like a norm, but more

general.

– f : W → R a linear map such that f(w) ≤ p(w) for all w ∈ W .

Then there exists an extension

V → R

which is linear such that

and

f(v) ≤ p(v) for all v ∈ V .

Why do we want this weird theorem? Our objective is to find something in

∗

. This theorem tells us that to find a bounded linear map in

, we just need

something in

bounded by a norm-like object, and then we can extend it to

Proof.

Let

∈ V \ W

. Since

has co-dimension 1, every element

v ∈ V

can be written uniquely as

, for some

w ∈ W, a ∈ R

. Therefore it

suffices to define

f(v

) and then extend linearly to V .

The condition we want to meet is

f(w + av

) ≤ p(w + av

) (∗)

for all

w ∈ W, a ∈ R

. If

= 0, then this is satisfied since

restricts to

If a > 0 then (∗) is equivalent to

f(w) + a

f(v

) ≤ p(w + av

We can divide by a to obtain

f(a

−1

w) +

f(v

) ≤ p(a

−1

w + v

We let w

= a

−1

w. So we can write this as

f(v

) ≤ p(w

+ v

) − f(w

for all w

∈ W .

If a < 0, then (∗) is equivalent to

f(w) + a

f(v

) ≤ p(w + av

We now divide by a and flip the sign of the equality. So we have

f(a

−1

w) +

f(v

) ≥ −(−a

−1

)p(w + av

In other words, we want

f(v

) ≥ −p(−a

−1

w − v

) − f(a

−1

w).

We let w

= −a

−1

w. Then we are left with

f(v

) ≥ −p(w

− v

) + f(w

for all w

∈ W .

Hence we are done if we can define a

(

) that satisfies these two conditions.

This is possible if and only if

−p(w

− v

) + f(w

) ≤ p(w

+ v

) − f(w

)

for all w

, w

. This holds since

f(w

) + f(w

) = f(w

+ w

)

≤ p(w

+ w

)

= p(w

− v

+ w

+ v

)

≤ p(w

− v

) + p(w

+ v

So the result follows.

The goal is to “iterate” this to get a similar result without the co-dimension

1 assumption. While we can do this directly for finitely many times, this isn’t

helpful (since we already know a lot about finite dimensional normed spaces). To

perform an “infinite iteration”, we need the mysterious result known as Zorn’s

lemma.

Digression on Zorn’s lemma

We first need a few definitions before we can come to Zorn’s lemma.

Definition

(Partial order)

A relation

≤

on a set

is a partial order if it

satisfies

(i) x ≤ x (reflexivity)

(ii) x ≤ y and y ≤ x implies x = y (antisymmetry)

(iii) x ≤ y and y ≤ z implies x ≤ z (transitivity)

Definition

(Total order)

Let (

S, ≤

) be a partial order.

T ⊆ S

is totally ordered

if for all x, y ∈ T , either x ≤ y or y ≤ x, i.e. every two things are related.

Definition

(Upper bound)

Let (

S, ≤

) be a partial order.

⊆ S

subset. We

say b ∈ S is an upper bound of this subset if x ≤ b for all x ∈ S

Definition

(Maximal element)

Let (

S, ≤

) be a partial order. Then

m ∈ S

is a

maximal element if x ≥ m implies x = m.

The glorious Zorn’s lemma tells us that:

Lemma

(Zorn’s lemma)

Let (

S, ≤

) be a non-empty partially ordered set such

that every totally-ordered subset

has an upper bound in

. Then

has a

maximal element.

We will not give a proof of this lemma here, but can explain why it should

be true.

We start by picking one element

. If it is maximal, then done.

Otherwise, there is some

> x

. If this is not maximal, then pick

> x

. We

do this to infinity “and beyond” — after picking infinitely many

, if we have

not yet reached a maximal element, we take an upper bound of this set, and call

it x

. If this is not maximal, we can continue picking a larger element.

We can do this forever, but if this process never stops, even after infinite

time, we would have picked out more elements than there are in

, which is

clearly nonsense. Of course, this is hardly a formal proof. The proper proof can

be found in the IID Logic and Set Theory course.

Back to vector spaces

The Hahn–Banach theorem is just our previous proposition without the constraint

that W has co-dimension 1.

Theorem

(Hahn–Banach theorem*)

Let

be a real normed vector space, and

W ⊆ V a subspace. Assume we have the following two items:

– p

V → R

(not necessarily linear), which is positive homogeneous and

subadditive;

– f : W → R a linear map such that f(w) ≤ p(w) for all w ∈ W .

Then there exists an extension

V → R

which is linear such that

and

f(v) ≤ p(v) for all v ∈ V .

Proof. Let S be the set of all pairs (

V ,

f) such that

(i) W ⊆

V ⊆ V

(ii)

f :

V → R is linear

(iii)

= f

(iv)

v) ≤ p(

v) for all

v ∈ V

We introduce a partial order

≤

by (

)

≤

(

) if

⊆

and

. It is easy to see that this is indeed a partial order.

We now check that this satisfies the assumptions of Zorn’s lemma. Let

{(

)}

α∈A

⊆ S be a totally ordered set. Define (

V ,

f) by

V =

[

α∈A

f(x) =

(x) for x ∈

This is well-defined because

{

(

V ,

)

}

α∈A

is totally ordered. So if

x ∈

and

x ∈

, wlog assume (

)

≤

(

). So

. So

(x) =

(x).

It should be clear that (

V ,

)

∈ S

and (

V ,

) is indeed an upper bound of

{(

)}

α∈A

. So the conditions of Zorn’s lemma are satisfied.

Hence by Zorn’s lemma, there is an maximal element (

W ,

)

∈ S

. Then by

definition,

is linear, restricts to

, and bounded by

. We now show

that

W = V .

Suppose not. Then there is some

∈ V \

. Define

span{

W , v

}

Now

is a co-dimensional 1 subspace of

. By our previous result, we know

that there is some

V → R

linear such that

and

(

)

≤ p

(

) for all

v ∈

V .

Hence we have (

W ,

)

∈ S

but (

W ,

)

(

V ,

). This contradicts the

maximality of (

W ,

f).

There is a particularly important special case of this, which is also known as

Hahn-Banach theorem sometimes.

Corollary

(Hahn-Banach theorem 2.0)

Let

W ⊆ V

be real normed vector

spaces. Given

f ∈ W

∗

, there exists a

f ∈ V

∗

such that

and

∗

kfk

∗

Proof.

Use the Hahn-Banach theorem with

(

) =

kfk

∗

kxk

for all

x ∈ V

Positive homogeneity and subadditivity follow directly from the axioms of the

norm. Then by definition

(

)

≤ p

(

) for all

w ∈ W

. So Hahn-Banach theorem

says that there is

V → R

linear such that

and

(

)

≤ p

(

) =

kfk

∗

kvk

Now notice that

f(v) ≤ kf k

∗

kvk

, −

f(v) =

f(−v) ≤ kf k

∗

kvk

implies that |

f(v)| ≤ kfk

∗

kvk

for all v ∈ V .

On the other hand, we have (again taking supremum over non-zero v)

∗

= sup

v∈V

f(v)|

kvk

≥ sup

w∈W

|f(w)|

kwk

= kfk

∗

So indeed we have k

∗

= kfk

∗

We’ll have some quick corollaries of these theorems.

Proposition.

Let

be a real normed vector space. For every

v ∈ V \ {

}

there is some f

∈ V

∗

such that f

(v) = kvk

and kf

∗

= 1.

Proof.

Apply Hahn-Banach theorem (2.0) with

span{v}

(

) =

kvk

Corollary.

Let

be a real normed vector space. Then

if and only if

f(v) = 0 for all f ∈ V

∗

Corollary.

Let

be a non-trivial real normed vector space,

v, w ∈ V

with

v 6= w. Then there is some f ∈ V

∗

such that f (v) 6= f (w).

Corollary.

is a non-trivial real normed vector space, then

∗

is non-trivial.

We now want to restrict the discussion to double duals. We define

V →

∗∗

as before by φ(v)(f ) = f(v) for v ∈ V, f ∈ V

∗

Proposition. The map φ : V → V

∗∗

is an isometry, i.e. kφ(v)k

∗∗

= kvk

Proof. We have previously shown that

kφk

B(V,V

∗∗

)

≤ 1.

It thus suffices to show that the norm is greater than 1, or that

kφ(v)k

∗∗

≥ kvk

We can assume v 6= 0, for which the inequality is trivial. We have

kφ(v)k

∗∗

= sup

f∈V

∗

|φ(v)(f)|

kfk

∗

≥

|φ(v)(f

∗

= |f

(v)| = kvk

where

is the function such that

(

) =

kvk

, kf

∗

= 1 as we have

previously defined.

So done.

In particular,

is injective and one can view

as an isometric embedding of

V into V

∗∗

Definition (Reflexive). We say V is reflexive if φ(V ) = V

∗∗

Note that any reflexive space is Banach, since V

∗∗

, being the dual of V

∗

, is

reflexive.

You might have heard that for any infinite dimensional vector space V , the

dual of

is always strictly larger than

. This does not prevent an infinite

dimensional vector space from being reflexive. When we said the dual of

always strictly larger than

, we are referring to the algebraic dual, i.e. the set of

all linear maps from

. In the definition of reflexive (and everywhere else

where we mention “dual” in this course), we mean the continuous dual, where

we look at the set of all bounded linear maps from

. It is indeed possible

for the continuous dual to be isomorphic to the original space, even in infinite

dimensional spaces, as we will see later.

Example.

Finite-dimensional normed vector spaces are reflexive. Also

reflexive for p ∈ (1, ∞).

Recall that given T ∈ B(V, W ), we defined T

∗

∈ B(W

∗

, V

∗

) by

∗

(f)(v) = f(Tv)

for v ∈ V, f ∈ W

∗

We have previously shown that

∗

B(W

∗

)

≤ kT k

B(V,W )

We will now show that in fact equality holds.

Proposition.

∗

B(W

∗

)

= kT k

B(V,W )

Proof. We have already shown that

∗

B(W

∗

)

≤ kT k

B(V,W )

For the other inequality, first let ε > 0. Since

kT k

B(V,W )

= sup

v∈V

kT vk

kvk

by definition, there is some

v ∈ V

such that

kT vk

≥ kT k

B(V,W )

kvk

− ε

wlog, assume kvk

= 1. So

kT vk

≥ kT k

B(V,W )

− ε.

Therefore, we get that

∗

B(W

∗

)

= sup

f∈W

∗

(f)k

∗

kfk

∗

≥ kT

∗

T v

∗

≥ |T

∗

T v

)(v)|

= |f

T v

(T v)|

= kT vk

≥ kT k

B(V,W )

− ε,

where we used the fact that

T v

∗

and

kvk

are both 1. Since

is arbitrary,

we are done.