Functional analysis;: An introduction (Pure and applied mathematics, v. 15)

(1 < p
are Banach spaces, then so is

(Vn-11-11n )

is

with the norm topology equivalent to

I

the product topology if

where

(V,I,.jj)

11 -112

II

II1

V

be any n-dimensional linear space

11-112

are norms such that

are normed linear spaces.

are equivalent.

(V, II . II1)

Prove that the norms

Seminormed and Normed Linear Spaces

1.

28

6.

Let

(Corollary 1.3.3)

normed linear space over (a)

Prove that

(b)

If

W

(V,11.11)

t.

is a Banach space.

is a linear subspace of

is a closed linear subspace of If

(c)

be a finite-dimensional

E c :V

V,

prove that

(W,11.11)

('V,j(.jj).

is a closed bounded set, prove that

E

is

compact. 7.

Prove that a finite-dimensional subspace of a normed

(a)

linear space is closed. If

(b)

is a normed linear space and

V

linean subspace, prove that the quotient map by

cp(x) = x + W If

(c)

8.

W + Y

:

W C V

is a closed

V -- V/W

defined

is continuous. V

linear.subspace, and prove that

cp

is a normed linear space, w c V is a closed Y c V

is a finite-dimensional linear subspace,

is closed in

V.

Prove that a Banach space is finite dimensional if and only

if every linear subspace.is closed. *9.

Give

an example of a Banach space and a decreasing sequence

of nonempty bounded closed sets whose intersection is empty.

(Hint:

Theorem 1.3.3 tells you where not to look for an example.) 10.

Does

11.

(a)

p(x + iy) _ jxj

IR2,

but not in the complex linear space

C.

Prove that every set that is balanced in the complex

linear space C (c)

C ?

Give an example of a set in the plane that is balanced

in the real linear space (b)

define a seminorm on

is convex.

Give an example in a complex linear space of a nonconvex

balanced set. 12.

(a)

Give an example of a set in the plane that is absorbing,

but not convex.

29

Problems

1.6

Some authors define a set

(b)

Give an example of a set

V = Un=1nA.

sorbing, then

A

is ab-

in a linear

A

that is absorbing by Definition 1.4.1 but for which

V

space

Show that with this definition, if

(tl re.

and

t E i

whenever

tx E A

such that

e > 0

there exists some

x E V

every

to be absorbing if for

A

Un1nA.

V f

Given a convex balanced absorbing set

13.

exists a convex balanced absorbing set

gauges

U

Let

*14.

p

and (a)

and

B,

prove that there A + A C B.

such that

A

be convex balanced absorbing lets with

V

respectively.

q,

Prove that

is a convex balanced absorbing set

U fl V

r(x) = max{p(x),q(x)).

with gauge

(b)

with gauge

Prove that

(c)

is a convex balanced absorbing set

U + V

s(x) = inf{p(v)

q(v - x)

v E V).

I

U U V

Give an example to show that

is not necessarily

a convex balanced absorbing set. Conclude that

and

r

s

define seminorms, but

t(x) = min{p(x),q(x)) is not necessarily a seminorm. Let

15.

A

numbers for which co.

If

xn

0

Ix n

I

Prove that

< 1/n.

for only finitely many values of

absorbing in *16.

An absorbing set x E V

A C V

the set

(closed) relative to the interval

balanced absorbing set and

rays, then and

A

such that

{x n)

prove that

n,

A (I E

is

E.

rays if for each

that, if

is not absorbing in

denotes the space of all sequences

E

of complex

{xn)

be the set of all sequences

p

A = B1,

B1 = {x

Ix = {a

I

A = B1,

where, as usual,

x E V, p(x) < 1).

a > 0, x E aAj

= (O,m).

I

its gauge.

is open in rays, then

A

is said to be open (closed) in

Let

A

Use Theorem 1.4.1 to prove and if

Bo = {x i

is open

be a.convex

I

A

is closed in

x E V, p(x) < 1)

1. Seminormed and Normed Linear Spaces

30

*1'.

A series

Ek=lxk said to be sununable to a sum

in a normed linear space x

if

x E V

and limnlix -

that is, the sequence of partial sums converges to is said to be absolutely summable if normed linear space

(V,11-0

-k=lllxkll < m.

x.

is

=lxk11 = 0,

The series

Prove that a

is complete if and only if every abso-

lutely summable series is summable.

CHAPTER 2

TOPOLOGICAL LINEAR SPACES

2.0. Introduction. space

(V,P)

We have seen that, given a seminormed linear

we can introduce a Hausdorff topology

Tp

into

that is intimately connected with the family of seminorms

V

Now we

P.

wish to study such spaces -- that is, linear spaces equipped with a Hausdorff topology -- in their own right.

In order to make these

topological linear spaces interesting objects of investigation we shall demand that the linear space operations and the topology "fit together" properly, namely, that the operations be continuous.

All

seminormed linear spaces are examples of these topological linear spaces, but the converse is not true.

After introducing the basic notions and establishing some elementary results in Section 2.1, we shall look at finite-dimensional topological linear spaces in Section 2.2.

There it will be seen that a

topological linear space is finite dimensional if and only if its topology is locally compact.

Those topological linear spaces that

are seminormed linear spaces will be described in Section 2.3.

They

will be seen to be precisely those topological linear spaces whose topology has, at the origin, a neighborhood base consisting of convex open sets.

The notion of the gauge of a convex balanced absorbing

set will be instrumental in establishing this description,

In

Section 2.4 some properties of gauges in topological linear spaces will be given, and these results will then be used to characterize those topological linear spaces that are normable.

Metrizable top-

ological linear spaces will be discussed in the last section.

2.1. Topological Linear paces.

We begin this section with the

definition of topological linear spaces.

31

2. Topological Linear Spaces

32

Definition 2.1.1.

A linear space

topological linear space over

T on

V

over

f

is said to be a

if there'exists a Hausdorff topology

I

such that the following mappings are continuous:

V

The mapping from

(i)

V,

defined by

V,

defined by

V x V,

(x,y) -. x + y, x,y E V.

The mappings from

(ii)

I x V,

with the product topology, to

(a,x) -. ax, a E I, x E V.

The mapping from

(iii)

with the product topology, to

V

V, defined by

to

x -. -x, x E V.

Thus, somewhat loosely speaking, a linear space Hausdorff topology

T

V

with a

is a topological linear space if the linear

space operations of addition, inversion, and scalar multiplication are continuous.

by the pair

We shall generally denote a topological linear space

(V,T).

It should be noted that not all definitions of topological linear spaces include the assumption that the topology is Hausdorff. assumption is not made, for example, in [El, KeNa, T, and WI].

This More-

over, most of the results we shall establish are also valid in this more general context.

Nevertheless, we choose to include the hypo-

thesis of Hausdorffness in our definition since all of the topological linear spaces we shall discuss in the sequel have natural Hausdorff topologies.

For expositions without the Hausdorff assumption

we refer the reader to 177.].

[EI, pp. 56-66; T, pp. 123-133; Wit pp. 167-

In addition, it should be remarked that rather weak additional

restrictions on the topology of a"topological linear space

(V,T)

place of the Hausdorff assumption are sufficient to imply that Hausdorff.

T

For example, if

distinct points in

V,

T

in is

is a To-topology (i.e., given any two

at least one of them has an open neighborhood

not containing the other), then

T

is a Hausdorff topology.

case one can even prove that, besides being Hausdorff,

T

In this

also has

the property that any closed set and point not in the set have disjoint open neighborhoods.

For some discussion of these results see

[T,,p. ?26; Wit pp. 175-176].

33

2.1. Topological Linear Spaces

Examples of topological linear spaces are easy to come by, as' shown by the next result. Let

Theorem 2.1.1.

Then

be a seminormed linear space over

(V,P)

is a topological linear space over

(V,Tp)

f.

Since from Proposition 1.5,1 we see that

Proof.

Hausdorff topology on

neighborhood

is a

TP

it remains only to show that the linear

V,

For example, to see that addition is

space operations are continuous. .continuous, let

6.

xo,yo E V,

and

c > 0,

U(xo +

and consider the open of

x0 + yo

in

yo'E'pl'p2'.. "pn)

If

TP.

0

denotes the open neighborhood U(x0;c/2;p1,P2,...$Pn) X U(ye;c/2;P1,p2,...,pn) in the product topology on

then clearly

V x V,

(x,y) E U

implies

that

pk[x + Y - (xo + Yo)] < Pk(x- xo) + Pk(Y - Yo) (k = 1,2,...,n),

< E

from which we conclude that addition is continuous.

Similar arguments,

whose details are omitted, establish the continuity of inversion and

Q

scalar multiplication.

All the spaces described in Section 1.2 are topological linear spaces.

Now let us state some further results concerning topological linear spaces.

The proofs are reasonably straightforward and are

left to the reader.

Theorem 2.1.2.

Then for each defined by

y E V

Let

be a topological linear space over

(V,T)

and each

a E I, a # 0,

py(x) = x + y, x E V,

Ta(x) = ax, x E V,

and

:

the mappings V -' V,

ya are surjective homeomorphisms.

f.

pY: V - V,

defined by

Moreover, the

2. Topological Linear Spaces

34

r or Va

image under p the image under

Va

of a convex set is again a convex set, and

of a linear subspace is again a linear subspace.

The fact that translation in a topological linear space

(V,T)

is a homeomorphism is very useful, as it often allows us to reduce a "global" question to a "local" one.

For example, suppose that

is some collection of open sets in translation; show that

U E U

that is,

(V,T). that is invariant under

implies that

is a base for the topology

U

U

T

To

U + x E U, x E V.

it is then sufficient

to demonstrate that

U

base at the origin.

Similarly we shall see in Chapter 3 that a

contains some subset that is a neighborhood

linear transformation from

(V,T)

to

(V,T)

is continuous on

V

if and only if it is continuous at the origin. It is also worth remarking that the continuity of scalar multiplication in a topological linear space implies that every open neighborhood of the origin in Theorem 2.1.3. (i)

Let

W C V

If

V

is an absorbing set.

be a topological linear space over #.

(V,T)

is a linear subspace, then the closure of

W

is

a linear subspace.

W C V

If

(ii)

topology on

W

is a linear subspace and

induced by

then

T,

(W,T')

T'

is the relative

is a topological linear

space. (iii)

If

W C V

is a closed linear subspace, then

a topological linear space if

T'

open sets are sets of the form

is the topology on

{x + W I x E U), U E T,

(V/W,T')

V/W

is

whose.

that is,

T',

is the usual quotient topology. Theorem 2.1.4. over

#

a E A,

and let

V

Let

I.

be topological linear spaces

denote the product of the topological spaces

with the product topology

defined componentwise. over

(Va,Ta), a E A,

Then

(V,T)

T

V V.

and with linear space operations is a topological linear space

35

2.2. Finite-Dimensional Topological Linear Spaces

From

2.2. Finite-Dimensional Topological Linear Spaces.

Theorem 2.1.1 we see that the spaces &, II 1 < p <

respectively.

C,

Ilp)

and

(Cn, II Ilp)

are n-dimensional topological linear spaces over

and

U.

As is perhaps not surprising, in view of the

results on finite-dimensional normed linear spaces obtained in Section 1.3., all n-dimensional topological linear spaces are essentially

either

(1F2n, II

Ilp)

or

This is an

for any p, 1 < p < m.

(Cn, II Ilp)

immediate consequence of Theorem 2.2.1, whose proof is left to the reader. Let

Definition 2.2.1.

linear spaces over

(V1,T1)

Then

I.

and

(V1,T1)

and

(V2,T2) (V2,T2)

be topological are said to be

topologically isomorphic if there exists a surjective isomorphism tp

:

V1 .. V2

that is a homeomorphism.

As in the case of normed linear spaces, it is apparent that the relation of topological isomorphism between topological linear spaces is reflexive, symmetric and transitive. Theorem 2.2.1. linear space over

to o a

Let IR(C).

be an n-dimensional topological

(V,T)

Then

is topologically isomorphic

(V,T)

ill))

Corollary 2.2.1.

Let

(V1,T1)

topological linear spaces over

t.

and

(V2,T2)

be n-dimensional

Then

(V1,T1)

and

(V2,T2)

are

topologically isomorphic. Corollary 2.2.2.

Let

(V,T)

be a topological linear space over

W be a finite-dimensional linear subspace of

f

and let

W

is a closed linear subspace of

V.

Then

(V,T).

A characterization of finite-dimensional topological linear spaces analogous to that given for finite-dimensional normed linear spaces in Theorem 1.3.3 is also valid.

On the surface, the form of

the characterization is somewhat different from that for finite-dimensional normed linear spaces since Theorem 1.3.3 is no longer meaningful in the context of topological linear spaces.

The characteriza-

tion should be compared with the remarks following Theorem 1.3.3.

2. Topological Linear Spaces

36

Theorem 2.2.2. f.

Let

(V,T)

be a topological linear space over

Then the following are equivalent:

(ii)

is finite dimensional.

(V,T)

(i)

There exists a compact set

with a nonempty interior

B C :V

such that the origin belongs to the interior of Suppose

.Proof.

(V,T)

n

2.2.1, there exists some

V

cal isomorphism from

where Cn,

B1

Then, by Theorem

is finite dimensional. such that

(II ,

morphic to either

B.

or

is topologically iso-

(V,T)

If

(C

to either

or a

lid

is this topologi-

(p

then

T-

B =

is the closed unit ball about the origin in either e or

satisfies condition (ii), which is thus implied by (i). Conversely, suppose part (ii) holds.

the interior of

Since

is compact and

B

is a nonempty set that contains the origin, we

B

deduce at once that there exist a finite number of points ..,xn

in

such that

B

subspace of

V

with the quotient topology.

homomorphism determined by tient topology on

V/W

is a compact subset of the origin in

V/W.

B C W+ (1/2)B,

and so

V/W

Let W.

*

denote the canonical

V -» V/W

:

Then from the definition of the quois continuous and that

$

Moreover,, by the definition of #(B) C: (112)i(B).

we have

W,

By induction we then conclude

However, the continuity of

.

with the fact that the nonempty interior of

$(B)

entails that each point in

2k$(B)

Now, if Vj W,

$(B)

with a nonempty interior that contains

2kf(B) C $(B), k = 1,2,3,...

V/W = $(B),

be the linear

being finite

W,

scalar multiplication in the topological linear space

Hence

xl,x2,..

is a topological linear space

we see that V/W

Then

x1,x2,...,xn.

W

Let

B C Uk=l(xk +(1/2)B).

spanned by

dimensional, is closed, and so

that

(B1),

V/W,

combined

contains the origin,

V/W

belongs to

and so

V/W

is compact. -e

then

V/W

would not be'the zero space and so lt

for some

k.

0

as is easily verified, it would contain a linear sub§pace isomorphic to the scalar field

V/W. Consequently

thereby contradicting the compactness of V = W,

and so

V

is finite dimensional.

Therefore part (ii) of the theorem implies part (i).

0

37

2.3 Locally Convex Topological Linear Spaces

is a

(V,T)

An equivalent phrasing of this result is that

T

finite-dimensional topological linear space if and only if

is

a locally compact topology.

2.3. Locally Convex Topological Linear Spaces.

linear space over over

then

f,

is a seminormed

(V,P)

section of this chapter we saw that, if

In the first

is a topological linear space

(V,Tp)

We now wish to address our attention to the converse

f.

if

question:

(V,T)

is a topological linear space, does there

exist a family of seminorms

on

P

TP = T?

normed linear space and

V

such that

(V,P)

The answer is, in general, negative.

For example, as noted in Example 1.2.4, the spaces

where dt

0 < p < 1,

is a semi-

is Lebesgue measure on

[0,1],

Lp([0,1],dt),

are neither

normed nor seminormed linear spaces.

However, they are topological

linear spaces if we use the topology

T

U(f,a) = {g where

f E Lp([O,l],dt)

I

generated by the neighborhoods

g e Lp([0,l],dt), I[g and

e > 0

- f1l p

is arbifrary.

< e) (See, for example,

[K,. pp. 157-158]).

What additional restrictions are then necessary to ensure that a topological linear space is a seminormed linear space?

The clue

to the answer lies in examining the nature of open neighborhoods in seminormed linear spaces.

To be more precise, we have the follow-

ing proposition: Proposition 2.3.1. f.

Then

TP

Let

(V,P)

be a seminormed linear space over

contai-is a neighborhood base at each point in

V

that

consists of convex open sets. Proof.

An elementary argument shows that for each

open neighborhoods U(x;s;pl,p2,...,pn)

x E V

the

are convex.

We shall see that the existence, at each point of a topological .linear space, of a neighborhood base consisting of convex open sets

2. Topological Linear Spaces

38

is sufficient to ensure that the topology comes from a family of Since the existence of such a neighborhood base is of

seminorms.

considerable importance, Definition 2.3.1. Then

4.

we make the following definition: Let

(V,T)

be a topological linear space over x E V

is said to be locally convex if each point

(V,T)

has a neighborhood base consisting of convex open sets. Lp([0,1),dt), 0 < p < 1,

The spaces

are not locally convex

topological linear spaces, as will-be shown in Section 4.2.

On the

other hand, the content of Proposition 2.3.1 is precisely that, if (V,P)

is a seminormed linear space, then

(V,TP)

is locally convex.

Before we establish the converse assertion we need to prove some preliminary reselts, which are also of independent interest. Proposition 2.3.2.

Let

(1V,T)

be a topological linear space

I.

over

Let

(i)

be a (open) neighborhood of the origin in

U

Then there exists a (open) neighborhood

N

V.

of the origin such that

aNCU, a E , lal <1. (ii) Let

U

be a (open) neighborhood of the origin in

there exists a balanced (open) neighborhood that

0 (iii)

Let

(iv)

U C V

Let

U

of the origin such

be a convex set.

U

Then the closure of

U

and

are convex sets.

U C V

be a balanced set.

balanced, and the interior of

U

Then the closure of

U

is

is balanced provided the interior

contains the origin.

Proof.

Let

U

be a (open) neighborhood of the origin.

scalar multiplication is continuous, there exists some (open) neighborhood

whenever and

Then

V.

N c U.

the interior of

of

No

lal < c

Ul

and

a E 4, Ial < 1,

of the origin in x E U1.

we have

Set

ax E U

V

N = 0U1.

because

such that

and a

ax E U

Then for u E U1

Since

e > 0

and

x = cu E N

2.3. Locally Convex Topological Linear Spaces

Hence

lacl = la's < c.

aN c U

whenever

39

and

lal < 1

N

is a (open)

Thus part (i) is established.

neighborhood of the origin.

To prove part (ii) we set

No

UlaL
where

N

is the

=

neighborhood we have just constructed. neighborhood of the origin in

Clearly N

and

c :U.

No

is a (open)

Furthermore, if

0

bNo c No.

then

b E t, lbl < 1,

V

Parts (iii) and (iv) are left to the reader.

If

is a locally convex topological linear space, then Propo-

V

sition 2.3.2 (ii) can be used to construct a neighborhood base at the origin consisting of convex balanced open sets. Proposition 2.3.3. linear space over origin in

(V,T)

be a locally convex topological

Then there exists a neighborhood-base at the

t.

consisting-of convex balanced open sets.

V

Proof.

Let

Let

U

be a neighborhood base at the origin consisting

of convex open sets.

If

U E U,

then by Proposition 2.3.2(ii) there

exists a balanced open neighborhood

N

of the origin such that

NcU. Let

be the collection of all finite convex linear combina-

N1

tions of elements of of the form

N;

F =lakxk

k = 1,2,...,n,

that is, let

for some

and q=lak = 1.

n,

N1,

consist of all elements. xk E N, 0 < ak < 1,

Then clearly

balanced neighborhood of the origin and By Proposition 2.3.2(iii)

N1

where

N1 c U,

U

is convex.

N1 D N, is a convex balanced open

neighborhood of the origin that is contained in U E U

as

and (iv) we conclude that the interior of

which is nonempty because

Thus every

is a convex

N1

U.

contains a convex balanced open neighborhood

of the origin and hence there exists a neighborhood base at the origin consisting of such sets.

2. Topological Linear Spaces

40

Now let us state and prove the theorem indicated earlier. Theorem 2.3.1. 0.

Let

(V,T)

be a topological linear space over

Then the following are equivalent: is a locally convex topological linear space over §.

(V,T)

(i)

is a seminormed linear space over

(V,P)

0

on

P

There exists a family of seminorms

(ii)

and

V

such that

TP = T.

Proposition 2.3.1 shows that part (ii) implies part (i).

Proof.

Conversely, suppose that part (i) holds and let V

base at the origin in

U

be a neighborhood

consisting of convex balanced open sets.

U E U

The continuity of scalar multiplication implies that each For each

is absorbing.

U E U

define

pU(x) = inf(a that is, Set

we see that

(

P

U E U).

Moreover, since

I

U

PU(x) < 1) c`U c (x

I

such that

is a neighborhood base at the origin in

x = 0

because

It is then evident that TP = T.

V

PU(x) < 1).

follows at once that, if. pU(x) = 0, U E U, and hence

U.

Then from Proposition 1.4.2 and Theorem 1.4.1

is a family of seminorms on (x

U E U,

(x E V),

is the gauge of the convex balanced absorbing set

PU

P = (pU

a > 0, x E aU)

(V,P)

T

then

x E U

V,

it

for all

is a Hausdorff topology.

is a seminormed linear space and

Consequently part (i) of the theorem implies part (ii).

O

Thus we see that seminormed linear spaces and locally convex topological linear spaces are the same, and we have a means of moving from one type of space to the other, either through the construction of the topology

TP

associated with a family of seminorms or through

the use of the gauges of certain convex balanced absorbing sets.

At

times there are significant advantages in being able to choose which type of space one wishes to work with while investigating various questions.

This will be apparent in the sequel.

41

2.4. Seminorms and Convex Balanced Absorbing Sets

We have

2.4. Seminorms and Convex Balanced Absorbing Sets.

already seen (Theorem 1.4.1) that, if ing subset of a linear space p

on

is a convex balanced absorb-

B

then there exists a unique seminorm

V,

such that

V

= {x

Bo

i

The seminorm

p(x) < 1} C B C (x

I

I

p(x) < 1) = B1.

is, of course, the gauge of the set

p

B.

In the next

theorem we shall investigate further the relationship between gauges and convex balanced absorbing sets in the context of topological Some preliminary definitions are desirable.

linear spaces.

Definition 2.4.1. 0

and let

Let

Then the interior of

E C V.

and the closure of

int(E)

be a topological linear space over

(V,T)

E

by

cl(E).

bounded if for each open neighborhood exists some

a > 0

such that

U

will bq denoted by

E

The set

F

is said to be

of the origin in

there

V

E C aU.

It is easily seen that the definition of boundedness in a topolo-

gical linear space reduces in the case of seminormed linear spaces to that given in Definition 1.3.2. Theorem 2.4.1.

,

the gauge of

Let

(V,T)

be a topological linear space over

be a convex balanced absorbing set, and let

B C V

let

B.

q

be

Then

int(B) C Bo C B C B1 C cl(B).

(i)

i

B = B1

(ii)

(iii) (iv)

if

B

is open.

B = B

if

B

is closed.

If

V -SIR

q

is continuous, then

Bo = int(B)

and

B1 = cl(B). q

(v)

If

(vi)

Proof.

x E int(B). U

of

x

:

V - IR B

is continuous if and only if

is bounded, then

q

is a norm.

From Theorem 1.4.1 we know that Since

such that

int(B)

0 E int(B).

Bi C B C B1.

Let

is open there exists an open neighborhood

U C int(B).

Moreover, by the continuity of

2. Topological Linear Spaces

.42

la - 11 <

then

that is,

int(B) C Bo1.

Hence

x E B.

q(x) =,I.

ality, we can assume that n = 2,3,...

If

Therefore, without loss of gener-

x E Bi C cl(B).

then

q(x) < 1.

so that

x E BI

On the other hand, let q(x) < 1,

q(x) < 1/a < 1,

Thus

x E (1/a)B.

and so

ax E int(B) C B,

1 < e,

0 < a -

In particular, if

ax E U.

then

e,

such that, if

e > 0

scalar multiplication, there exists an

an =

But let

1

- 1/n,

Then

.

1)q(x)

q(anx)

=

(anx) C :B1 C B, n = 2,3,...

and we conclude that

(n = 2,3,...),

- n < I

1

However, since

.

scalar multiplication is continuous, we deduce that Therefore

limnanx = x E cl(B).

and part (i) is proved.

B1 C cl(B),

Parts (ii) and (iii) follow immediately from part (i). q

If

:

V -+IR

is continuous, then

is open, as it is the

Bo i

that

int(B) C Bi C B

We conclude at once from

inverse image of Bo = int(B).

BI = cl(B).

A similar argument shows that

i

If as

Conversely, suppose

q(0) = 0 < 1.

an open neighborhood of we have

0

q(x) < 1, x E U.

neighborhood of that is,

Bo = int(B) i

is continuous, then

q

q

0

such that

such that, if

is continuous at

0.

and let

0 E int(B),

U c int(B).

But then, for any

0 E int(B),

entails that

e > 0, eU

x = eu E eU,

be

U

int(B) C BI,

Since

is an open q(x) = eq(u) < c;

then

The inequality

lq(x) - q(y)l < q(x - y)

(x,y E V),

which is valid for any seminorm by Proposition 1.1.1, then shows that q

is continuous, indeed even uniformly continuous, on

V.

This proves

part (v) of the theorem. Finally, suppose

B

is bounded.

If

then

x # 0,

Hausdorff, there exists an open neighborhood

U

of

0

since

T

such that

is

43

2.4. Seminorms and Convex Balanced Absorbing Sets

By Proposition 2.3.2 we may assume that

x f U.

Because

is bounded, there exists some

B

and since

if

ab < 1,

then

ab(x/ab) = x

such that

b

E abU d U,

Hence

U.

as

and this holds for

b > I/a,

By the definition of

x E bB.

ab > 1,

is balanced,

U

as

B C all,

such that

b > 0

But this implies that

x/ab E U.

contradicting the choice of any

for which

a > 0

is absorbing, there also exists some

B

Clearly then

x E bB.

is also balanced.

U

it then follows

q

q(x) > 1/a > 0.

that

Therefore it

then

x f 0,

q(x) 7 0,

and so

is a norm.

q

This completes the proof of part (vi) and the theorem.

The last portion of Theorem 2.4.1 allows us to characterize those topological linear spaces that are normed linear spaces. Theorem 2.4.2. @.

be a topological linear space over

(V,T)

Then the following are equivalent: (i)

T

Let

There exists a norm on

V

such that the norm topology and

coincide. (ii)

T

Proof.

contains a bounded convex open neighborhood of the origin.

Clearly, if there exists a norm

the norm topology coincides with

r,

then

(x

bounded convex open neighborhood of the origin.

(jx+j

such that

V

on

< 11

is a

Hence part (i)

implies part (ii).

Conversely, suppose of the origin.

B E T is a bounded convex open neighborhood

By repeating the argument used in proving Proposition

2.3.3, we may assume without loss of generality that balanced. B

B

From Theorem 2.4.1 (vi) we conclude that the gauge

is a norm on

V.

T.

First, from Theorems 2.4.1 (iv) and (v) we note that I

q(x) < 1),

q

of

It remains to prove that the topology generated

by this norm coincides with

B = (x

is also

and hence for each aB = (x

I

a > 0

q(x) < a).

we have

2. Topological Linear Spaces

44

To show that the norm topology coincides with T {aB

in view of Theorem 2.1.2, that the family

of open

a > 0)

forms a neighborhood

To accomplish this it suffices to prove that, if

base at the origin.

is any open neighborhood of the origin, then there is some

U E T a > 0

for which

since

B

aB CU.

But if

is such a neighborhood, then,

U

is bounded, there exists some where

aB c U,

Thus

I

T

neighborhoods of the origin in the topology

it suffices to show,

b > 0

such that

B C bU.

a = 1/b > 0.

Therefore part (ii)

of the theorem implies part (i).

O

Theorem 2.4.2 clearly provides us with a means of proving that certain topological linear spaces are not normed linear spaces. example, consider the seminormed linear space

(Cm([0,2n]),P),

For

where

n

P = {pn

Pn(f) =

I

E IIf(k)IL

n

k=0 To show that this is not. a normed linear space it is sufficient to

prove that there exists no norm on TP

topology and the topology

Cm([0,2n])

coincide.

for which the norm

In view of Theorem 2.4.2,

this assertion can evidently be established by showing that rP contains no bounded open sets.

Thus suppose that is, for each

E C Cm([0,2n])

n = 0,1,2,...,

is bounded and open -- that

and

E E

h E E

be fixed.

Then by the definition of

c > 0

and

i,

Pk 'pk

.

.

.

in

, p

Mn > 0

there exists an

pn(f) < Mn, f E Ef n = 0,1,2,...,

P

TP.

TP

such that

Furthermore, let there exists some

such that

n

1

U(h;c;Pk .Pk ,...,py ) C

Set

r = supj=1,2,,nkj, r > 0.

that

and so

Clearly

n

2

1

and assume, without loss of generality,

Pk (f) < pr(f), f E C '([0,2n]), j = 1,2,...,n,

U(h,e,pr) C E.

Now for each 0 < t < 2n.

b > 1

Obviously

define

hb(t) = h(t) + (c sin bt)/2rbr,

hb E Cm([0,2n]), b > 1.

Indeed, elementary

45

2.4. Seminorms and Convex Balanced Absorbing Sets

computations reveal that the kth derivative of value of

k,

depending on the

hb,

is one of the following four functions: c sin bt

h

(k)

(t) + 2rbr -k ,

h(k)(t) + c cos bt 2rbr-k

Consequently,

r pr(h - hb) =

E 11h(k)

k=0 e

2r

r k=0

-

1

br -

k

c(r + 1) 2r

< c, as

b > 1.

In particular, we can then conclude that 1hbn)11m < pn(hb.1 < Mn, n = 0,1,2,...,

for all

clearly leads to a contradiction since, when and

lI(e sin bt)/2rbr-n11.

E

b > 1.

and so

But this

n > r,

11(e cos bt)/2rbr-n1j.

as one chooses by a suitable choice of Therefore

hb E E, b > 1,

can be made as large

b > 1.

cannot be both bounded and open, and so

Tp

is

not a norm topology.

It is, however, the case that the metric topology on C ([0,2n)) determined by the metric P(f,g) =

does coincide with

E

n

pn (f - g)

n=0 2 [1 + pn(f - g)]

Tp.

(f,g E C '([0,2n]))

The details are left to the reader.

2. Topological Linear Spaces

46

We begin this section with two definitions.

2.5. Frechet SSpaces.

Definition 2.5.1. §.

Then

V

a

lim x

aor

Definition 2.5.2. linear space over then

be a topological linear space over

is a Cauchy net (sequence) in

such that

x E V

(V,T)

is said to be complete (sequentially complete) if, when-

(x ) c V

ever

Let

Let

If T

6.

there exists some

T,

= x.

(V,T)

be a locally convex topological

is metrizable and

(V,T)

is complete,

is'called a Frechet space.

V

It is obvious that every Banach space isa Frechet space. also be apparent from the following discussion that where

It will

(C"0([a,b}),[pn)),

is a Frechet space, but,

pn(f) = EZOljf(k),jm, n

as we have seen in the preceding section, it is not a Banach space.

As we know from Theorem 2.3.1, if

(V,T)

is a locally convex

topological linear space, then there exists a family of seminorms on

V

such that

(V,P)

is a seminormed linear space and

P

T = TP.

We shall see that the Frechet spaces are precisely those complete locally convex topological linear spaces such that

is countable.

P

Before we establish this we wish to prove a lemma, which is of some independent interest. Lemma 2.5.1.

such that

Let

P = (p

(V,P)

be a seminormed linear space over

is countable.

family of seminorms

Q = (qm)

such that

(i)

(V,Q)

is a seminormed linear space over

(ii)

(V,P)

and

(iii)

Proof.

(V,Q)

m = 1,2,3,...,

qm+1(x) <..., x E V.

define

qm(x) = sup(p1(x),...,pm(x)) We claim that each

qm

I.

are topologically isomorphic.

qI(x) < q2(x) <...< qm(x) < For each

t

Then there exists a countable

is a seminorm on

V.

Clearly

(x E V).

qm(x) > 0

and

47

2.5. Frechet Spaces

where

gm(ax) = jajgm(x)

and

x E V

a E . Moreover, if

x,y E V,

then qm(x + Y)

= sup(pl(x +

y))

< sup(pi(x) + P1(y),...,pm(x) + Pm(Y)} < sup(plt

,...,pm(x)) + suP(P1(y),...,pm(y))

= qT.,(x) + qm(y) . It is evident that (i) and (iii) hold. for each

(x

f

Moreover, we see that

e > 0

p1(x) < e/2,.. ,p m(x; < e%') c- (x

c (x and so the identity mapping

V

-n

qm (x) < s)

pI(x) < c,...,pm(x) < e),

is obviously a topological isomor-

phism.

O

We can now characterize the metrizable locally convex topological linear spaces.

Theorem 2.5.1. linear space over (i)

(ii)

V

T

is metrizable.

(V,P)

If

T

at the origin.

Let

pn

T =

TP.

(Un)

that forms a neighborhood base at

Clearly we may assume, without loss of generality,

balanced absorbing set. 2.3.3.

of seminorms on

is metrizable, then there exists a countable

Un+l, n = 1,2,3,...,

Un

P = (pn)

is a seminormed linear space and

family of convex open sets

that

be a locally convex topological

Then the following are equivalent:

There exists a countable family

such that Proof.

(V,T)

Let f.

and that each

Un

is a convex

The latter is possible because of Proposition

be the gauge of

Un, n = 1,2,3,...

.

Then from the

argument used in the proof of Theorem 2.3.1 we see that seminormed linear space and

T =

TP,

Thus part (i) implies part (ii).

where

P = (pn).

(V,P)

is a

2. Topological Linear Spaces

48

Because of Lemma 2.5.1,

Conversely, suppose that part (ii) holds.

we may assume that

Evidently the sets

n = 1,2,3,..., x E V. n,k = 1,2,3,...,

pn(x) < pn+l(x),

is such that

P = (pn)

Un k

(x

+

pn(x) < 1/k),

form at the origin, a neighborhood base for

T

consisting of convex balanced absorbing sets.

Define Pn(x - Y) P(x,Y) =

E

(x,y E V).

2n [1+pn(x-Y))

n= 1

defines a metric on

that is trans-

It is easily verified that

p

lation invariant, that is,

p(x - y,0) = p(x,y), x,y E V.

V

the metric topology and T agree it suffices, since

p

To see that is translation

invariant, to examine what happens at the origin. But let

and suppose

n > 1

pn+l(x) < 1/2n+1

pl(x) < p2(x) <...< pn+1(x) < 1/2n+1

and

Then since

pk(x)/[1 + Pk(x)) < Pk(x),

17

k = 1,2,3,...,

we see that

P(x,0) =

Pk (x) E

--k

k=1 2 [1 + pk(x)] P, (x)

k=n+ 2 2[1 + pk(x)]

k= 1 2 [1 + pk(x)]

1

2

n+l 1

n+T kE1

E

+

k=n+2 2

2

(

<2n+1 lk=1 2 2n

1

+

k=l

49

2.6. Problems

Hence

U

n+1,1/2

p(x,O) < 112nj, n = 1,2,3,...

n+l a (x

.

1/2n+1

where

p(x,O) <

On the other hand, suppose

n > 1.

1/2m+k+l

n = m + k, m > 1.

Write

Then

implies that

p(x,O) <

PM(x)

1

2m[l +

2m+k+1

pm(x)]

and so Pm(x) 1

Solving for

x)

1

<

+T

(2 (x

I

2k+l

we obtain

pm(x)

pm (

Thus

1

<

+ pm(x)

1

l

1)

p(x,O) < 1/2m+k+1) C :U

m,1/2

k m+ k = 1,2,3j... T = TP

From these observations it is apparent that metric topology determined by

p

.

and the

coincide.

Therefore part (ii) of the theorem implies part (i).

0

This is not the most general theorem about the metrization of topological linear spaces. linear space

Indeed, one can show that a topological

is metrizable if and only if there exists a

(V,T)

countable neighborhood base at the origin for

T.

The metric can

also be constructed in this case so as to be translation invariant. For details the reader is referred to [K, pp. 162-164; KeNa, pp. 48-49].

2_6. Problems. 1.

over

L.

mappings

(Theorem 2.1.2)

Let

(V,T)

be a topological linear space

Prove that for each

y E V

and each

yy

:

V

V,

defined by

a E f, a # 0,

yy(x) = x + y, x E V,

and

the

2. Topological Linear Spaces

50

cpa

:

defined by

V - V,

qa(x) = ax, x E V, are surjective homeo-

Prove that the image under

morphisms.

cpy

or

pa

of a convex set

cpa

of a linear sub-

be a topological linear space.

Prove that every

is again a convex set, and that the image under space is again a linear subspace. Let

2.

(V,T)

neighborhood of the origin in Let

3.

4.

over

Let

a < b,

be a base for a

is not a topological linear space.

(]R,T)

(Theorem 2.1.3)

where

(V,T)

be a topological linear space

f.

(a)

W

[a,b),

Prove that

T.

is absorbing.

have the half-open interval topology; that is, let

II2

all intervals of the form topology

V

W c V

If

is a linear subspace, prove that the closure.of

is a linear subspace. (b)

W CV is a linear subspace and

If

topology on

W

induced by

T,

prove that

T'

(W,T')

is the relative is a topological

linear space. (c)

W c V

If

is a closed linear subspace, prove that

is a topological linear space, where whose open sets are of the form

T' is,the topology on

(x + W I x ( U), U E T,

(V/W,T')

V/W

that is

T'

is the usual quotient topology. 5.

(Theorem 2.1.4)

spaces over spaces

f

and let

VC1 , a E A,

Let V

(Va,

ar),

topological linear space over 6.

(Theorem 2.2.1)

phic to

Let

T and with linear

Prove that

(V,T)

is a

f.

(V,T)

Prove that

,'l)((,'l)) IR(C).

be topological linear

denote the product of the topological

with the product topology

space operations defined componentwise.

linear space over

a E A,

be an n-dimensional topological (V,T)

is topologically isomor-

51

2.6. Problems

(Corollary 2.2.1)

7.

sional topological linear spaces over

be n-dimen-

(V2,T2)

Prove that

#.

and

(V1,11)

are topologically isomorphic.

(V2,T2)

Let

(Corollary. 2.2.2)

8.

over

and

(V1,TI)

Let

and let

#

Prove that

W

A

letting

be a finite-dimensional linear subspace of

is a closed linear subspace of

W

Letting

9.

be a topological linear space

(V,T)

and

B

be any subsets of

(a)

cl(x + A) - x + cl(A),

(b)

cl(aA) = a cl(A),.for each

(c)

cl(A) + cl(B) c cl(A + B).

(d)

If

(c)

A + int(B) c int(A + B).

(f)

If

U

U

(V,T).

be a topological linear space over

(V,T)

is open, then

V.

and

#

prove each of the following:

V,

for each

x E V.

a E #, a # 0.

A + U

is open.

is a neighborhood base at the origin in

then

V,

cl(A) ; fl{A + U I U E U). Let

*10,

A

and

(V,T)

be a topological linear space over

be any closed subsets of

B

Give

(b)

Prove that, if

Conclude that, when

A

exists a neighborhood

U

(a)

need not be closed.

A + B

Let

is closed.

cl(A) + cl(B) = cl(A + B).

is compact and

A fl B = 0,

of the origin for which

(Proposition 2.3.2)

space over

A + B

is compact, then

is compact,

Prove that, if A

(c)

11.

A

and let

V.

an example to show that

(a)

#

(V,T)

then there

(A + U) fl (B + U)

be a topological linear

#.

If

U C V

is convex, prove that

If

U c V

is balanced, prove that

cl(U)

and

int(U)

are

convex.. (b)

If

int(U)

contains the origin, prove that

cl(U)

is balanced.

int(U)

is balanced.

2. Topological Linear Spaces

52

V

Letting

12.

be a linear space over

prove each of the

if,

following: If

(a)

I11,...,Un

arbitrary scalars in

t(

U

If

(b)

then

U,

4,

are convex subsets of then

a x

a1 x1

n n

Letting

'cttinc,

is a convex set.

when

and

V

are elements

x1,...,x

a1,...,an

are nonnegative

=1ak = 1.

(V,T)

be a topological linear space over

convex subset of

I-e

If

are

a1,...,ah

and

n n

is a convex subset of

reel numbers such that 13.

a U

aIU1

E I1

V

V

and

4

with a nonempty interior, prove

of the following:

c1(O) = cl[int(U)). int(Ilj = int[ci(U)].

(a) (h) 14.

{a)

ipt(F

that

(h

if

B

,, examp1F to show that, if

(,4t-e

real linear -pa_c 1S.

let

xr: E V,

let

the sc(UErCC 16.

let

i; balanced in the complex linear space

n =

;>:ists some

B

is balanced in the

need not be balanced.

be a locally convex topological linear space and

1,2,3,...

If

.

(xn)

converges to

converges to

(1j

((i,'n)

rv,i)

int(B)

then

i

(1',Ti

dcfine-l a subset

be a seminormed linear space over

17.

letting

Letting

We have

of

0

such that p(x) < Mp, x E E (Definition 1.3.2).

E C aU (V,T)

B,B1,...,Bn

following:

I.

p E P

E

to be bounded if for each

V

fcr every open neighborhood such that

prove that

0,

0.

crave that this is equivalent to the definition that

a > 0

prove

C,

s also haIanL.ed.

U

of the origin in

V

E

there

is bounded if

there exists some

(Definition 2.4.1).

be a topological linear space over

be bounded subsets of

V,

4

and

prove each of the

2.6. Problems

53

is bounded.

(a)

Uk_1Bk

(b)

BI

(c)

aB

(d)

x + B

is bounded for each

(e)

cl(B)

is bounded.

Let

18.

is bounded.

Bn

is bounded for each

a E 4.

x E V.

be a topological linear space over

(V,T)

is a compact subset of

V,

prove that

V

Let

19.

of

E

V

i

any sequence of elements of

E, the sequence

Let

20.

21.

that converges to

V

V

is normable -- that is, there exists a norm on topology and the product topology number of spaces

V

Letting

*22.

(xn)

is

converges to

0.

Prove that every

be a family of normable topological linear

Let

letting

and

(an)

is bounded.

Prove that the topological product

spaces.

0

(anxn)

be a topological linear space.

(V,T)

Prove

4.

is bounded if and only if, whenever

is a sequence of scalars in

Cauchy sequence in

is bounded.

be a topological linear space over

(V,T)

that a subset

B

In particular,

is bounded.

B

conclude that every convergent sequence in

4.. If

a

of the spaces

V

a

V

such that the norm

T coincide --

if and only if the

that are non-zero is finite.

(V,T)

be a topological linear space over

and

W CV be a closed linear subspace, prove each of the

following:

(a)

If

V

is locally convex, then

If the topology on

V

V/W

is locally convex.

is given by the family of seminorms

(p J,

a

then the quotient topology on (Pa),

is given by the family- of scrin-mr',

where 'pa(x + W) =-infyEWpa(x + y), x E V. (b)

If

Note that, if then

V/W

V/W

is a Frechet space, then

V V

V/i'

is a Freshet sr,_

is an arbitrary, complete, locally convex spa c,

need not be complete.

2. Topological Linear Spaces

54

23.

Let

(Va,T a)

be a family of locally convex metrizable

topological linear spaces over duct

V

of the spaces

V

a

countably many of the spaces

Prove that the topological pro-

f.

is metrizable if and only if at most V

are nonzero. of

24.

A topological linear space

(V,T)

is said to be locally

bounded if there exists a bounded open neighborhood of the origin in V.

Prove that every locally bounded topological linear space is

metrizable. *25.

A locally convex topological linear space

to be bornological if, whenever V

that absorbs every bounded set in

B C V

there exists some

a 0

E f

is said

(V,T)

is a balanced convex subset of

A

V

(that is, for every bounded

such that

B C aA

whenever

jal > ja01), the origin is contained in the interior of

A.

Prove

that every metrizable locally convex topological linear space is bornological.

CHAPTER 3

LINEAR TRANSFORMATIONS AND LINEAR FUNCTIONALS

3.0.

Linear transformations and linear functionals

Introduction.

play a central role in functional analysis, and after defining these concepts, we shall examine a number of concrete examples.

Then we

shall discuss some basic results concerning linear transformations, the most important being the equivalence between th^ notions of continuity and boundedness, and the fact that the space of all contin-

uous linear transformations from a normed linear space to a Banach space can be made into a Banach space in a natural manner.

The last

section of the chapter contains a few fundamental results pertaining to linear functionals.

A characterization of continuity of linear

functionals is established, and the question of the existence of sufficiently many nonzero continuous linear functionals on separate the points of

V

V

to

This will set the scene

is considered.

for the Hahn-Banach Theorem to be discussed in the next chapter. 3.1.

Linear Transformations.

Let us begin by defining linear

transformations and linear functionals. Definition 3.1.1. A mapping

T

:

VI

V2

Let

VI

and

V2

be linear spaces over

is said to be a linear transformation

T(ax + by) = aT(x) + bT(y)

L'(V1,V2). If

logical linear spaces, those be denoted by

(V1,TI)

T E L'(V1,V2)

L(V1,V2).

55

if

(x,y E V; a,b E $).

The collection of all linear transformations from will be denoted by

b.

and

(V2,T2)

V1

to

V2

are topo-

that are continuous will

3. Linear Transformations and Functionals

56

Some authcrs use the term "linear operator' instead of "linear transformation". L'(V1,V2)

It is readily seen that space over

$

can be made into a linear

by defining

(T + S)(x) = T(x) + S(x)

(aT) (x) = aT(x)

(T,S E L'(V1)V2); x E V1; a E $).

Clearly, when

We shall always assume that this has been done. and

are topological linear spaces,

V2

V1

is a linear sub-

L(V1,V2)

L'(V1,V2).

space of

defined on all of

V1.

is always

T E L'(V1,V2)

Also it should be expressly noted that

It is, however, possible to discuss linear but this

V1,

transformations that are only defined on subsets of

introduces additional difficulties that we prefer to avoid -- for example, the definition of linear space operations in

L'(V1,V2).

It

is evident though that the domain of a linear transformation must be a linear space, in any case. V1 = V2 = V,

When for

and

L'(V,V)

V2 = $,

we shall generally write

we give the elements of

Definition 3.1.2.

Let

linear transformations in V.

If

(V,T)

L'(V,$)

be a linear space over

V

absolute-value topology, then the elements of

element in

V'

by

x';

conjugate,

Then the

$

L(V,$)

has the usual are called

V.

similarly

or dual space of

$.

will be denoted by V and 3 generic

L'(V,$)

and its generic elements by

and

are called linear functionals on

L'(V,$)

continuous linear functionals on

L(V)

a special name.

is a topological linear space and

The linear space

V1 = V

respectively; and when

L(V,V),

and

L'(V)

x*; V.

L(V,$) V*

will be denoted by

V*

will also be referred to as the

57

3.1. Linear Transformations

Obviously, since linear functionals are special instances of linear transformations, afy results that we establish for the latter objects will automatically hold for the former. Let us look at a few examples of linear transformations and As with the examples in Chapter 1, the verifications

functionals.

t

of the following assertions are left to the reader. Let

Example 3.1.1.

m x n

V1 = UP

A = (aij).

real matrix

and

V2 = U

,

and consider the

Then

(x « (xl,x2,...,xn) E IK )

T(x) = A )Cn

defines a linear transformation in

If we consider On

L'(42n,JF ).

and fl with any topologies under which they are topological linear spaces (e.g., any of the norms If

m

then

1,

T

p

,

1 < p < W)

then T E L(1

(C(f0,1]),Ij IIm).

Let

defined for each f E C([0,1])

Let

(V,P) = (C (f0,l]),(pn)),

pn(f) = IIflIn = =ollf(k)IIW (V,P)

(0 < t < 1).

where = 0,1,2,...).

is a seminormed linear space, and the transformation

defined for each

f E C "Q0,11)

1(V).

T

by

(Tf)(t) = fl(t)

belongs to

Then the trans-

L(V).

Example 3.1.3.

Then

).

by

(Tf)(t) = fp f(s) ds

belongs to

,11

T E

Example 3.1.2. formation

11-11

(0 < t < 1)

58

Linear Transformations and Functionals

3.

The elements of

Cm([0,1])*

are called distributions.

Discus-

sion of this important class of continuous linear functionals can be found, for example, in [E1, pp. 297-418; E3, pp. 46-132; Sh,

Y,

We, however, shall content ourselves with this

pp. 28-30, 46-52].

passing mention. Example 3.1.4.

Let

gical space and let For each

t

0

be a locally compact Hausdorff topolo-

X

(V,11.11.)-

E X

define

(f E C0 (X)) .

x*(f) = f(t0) Then

x* E C(X)*. More generally, if

µ E M(X),

then

x*(f) = SX f(t) dµ(t) defines an element

x* E C(X)*.

(f E C0(X))

The converse cf this last assertion

is also true, but we shall not prove it in its full generality.

We

shall, however, establish it for a spacial case in the next chapter.

let

Example 3.1.5.

Let

(V'11-11) = (M(X) ,

11 11) ,

total variation norm.

X

be a locally compact Hausdorff space and

where as usual the norm in

For each

f E C(X)

x* E M(X)*.

elements of

M(X)*

is the

define

fX f(t) dµ (t) Then

M(X)

E M(X)).

However, it is not generally the case that all are obtained in this manner.

For example, if

X

is not compact, then the-preceding formula defines an element of M(X)*

for any

f E C(X).

Example 3.1.6.

Let

(V,1I.Il) _

) CO

and let (an) E Q1.

Then

x*((bn}) =

E b

n=1

a n n

((bn3 E c0)

S9

3.1. Linear Transformations

defines an element of

Moreover all of

co.

is obtained in this

co

This is actually just a special case of Example 3.1.4 when

way.

X = (n

since in this case

n = 1,2,3,...)

I

Example 3.1.7.

(L1([-n,n],dt/2n),II.111),

Let

dt

denotes Lebesgue measure on

(C

where

o

and let

For each integer

fnnf(t)e-int

f(n) = zn

_

belongs to

we define

n

(f E L1([-n,n],dt/2Tr)).

dt

and the transformation

f E L1([-n,n],dt/2n),

(V2,II.1I2)

where

denotes the locally compact space of the

Z

integers with the discrete topology.

Then f E C0(Z),

M(X) = QI.

T,

T(f) = f,

defined by

That is, the Fourier

L(V1,V2).

transformation is a continuous linear transformation from L1((-n,n],dt/2n) into

Co(l)

to

The fact that

Co(Z).

T

L1([-n,n],dt/2n)

maps

is nontrivial and is known as the Riemann-Lebesgue

Lemma [E2, p. 36]. Example 3.1.8.

Let

and let if

p =

q, 1

1 < q < cD,

and

q =

be such that if

1

p =

(I < P < m),

(Lptµ),II Ilp

(v,11-11) = (Lp(X,S,µ),il Jlp)

where

1/p + 1/q = 1,

Then, if

g E Lq(p),

x*(f) = fX f(t)g(t) defines an element

x* E Lp(&)*.

If

If

p = 1,

if

(X,S,µ)

(Xa) C S

Lq(µ).

the formula

(f E Lp(µ))

1 < p < cc,

assertion is also valid; that is, every element of mined by an element of

q = m

then the converse Lp(N)*

is deter-

We shall prove this in Section 8.3.

then the converse assertion need not be valid.

However,

is a measure space such that there exists a family of disjoint subsets of

a-finite when restricted to

Xa,

X

with the properties that

and if

E E S

and

µ(E) < m,

µ

is

then

60

Linear Transformations and Functionals

3.

for at most a countable number of

E fl xa # 0

uous linear functional on for some

then every contin-

a,

is given by the preceding formula

L1(µ)

In particular this is true if

g E L.(µ).

(see, for example, [DS1, pp. 289-290]).

topological space and

If

X

is a-finite

µ

is a locally compact

is a regular Borel measure, then it is

µ

always the case that every continuous linear functional on obtained from some When

L1(µ)

is

[E1, pp. 215-220,239-240].

g E L.(&)

then, except in trivial cases, there always exist

p = m,

continuous linear functionals on preceding formula for, some

that are not given by the

Lm(µ)

g E L1(µ) (see [DS1, p.,296]).

3.2. Some Basic Results Concerning Linear Transformations.

The

linearity property of linear trapsformations is of fundamental importance, and its consequences will be used repeatedly in the remainder of this volume.

The reader should pay special note to the role lin-

earity plays in the proofs of the results in this section, as they are typical of the importance of linearity.

Subsequently the role of

linearity may not always be as clearly displayed as it is here. Proposition 3.2.1. and let

Let

T E L'(V1,V2).

V1

and

V2

be linear spaces over

t

Then

(i) T(0) = 0. The range of

(ii)

R(T) = {y

+

is a linear subspace of T

(iii)

Proof.

T

sinee

and hence

whenever

x E V1),

V2.

is injective, then

Clearly, if

T

Conversely, suppose that T

for some

T(x) = 0

T-1

implies

exists and

T-l

x = 0.

E L'(R(T),V1).

Since the proofs are quite elementary, we give only that

for part (iii). x = 0.

y E V2, y = T(x)

is injective if and only if

If

(iv)

T,

is injective, then x,y E V1

is a linear mapping, we have x - q

T(x) = 0

0,

that is,

implies

x = y.

x = 0.

and

T(x) = 0

T(x) = T(y).

implies Then,

0 = T(x) - T(y) = T(x - y), Consequently

T

is injective

0

61

3.2. Basic Results on Linear Transformations

Proposition 3.2.2.

linear spaces over

4

(V1,TI)

Let

(V2,T2) be topological

and

Then the following

T E L'(V1,V2).

and let

are equivalent: (i)

at

x

0 (ii)

(iii) (iv)

There exists some

T

x

E VI

such that

is continuous at the origin in

T

is uniformly continuous on It is obvious that

Proof.

(iv) implies (iii) implies (ii)

then

But then

T(x) E W2.

neighborhood of the origin in y - z E'U1,

Consequently

U2 E T2

If

V1,

W1 E TI

of

and hence, if

T(x0).

such that,

x0

is an open

U1 = W1 - x0

T(y) - T(z) = T(y -

then

is

then from Theorem 2.1.2

V2,

Thus there exists some open neighborhood x E W11

x0.

is an open enighborhood of

W2 = U2 + T(x0)

we see that

V1.

is continuous at

T

Suppose that

an open neighborhood of the origin in

such that

V1.

T E L(V1)V2).

implies (i).

if

is continuous

T

y,z E V1

are

z) E W2 - T(x0) = U2.

T, is uniformly continuous.

Therefore (i) implies (iv), and the proof is complete.

0

One of the must useful results concerning the continuity of linear transformations between normed linear spaces is the relationship between 'boundedness' and continuity given in the next theorem. Definition 3.2.1. linear spaces over

i

and

Let

and let

be bounded if there exists some

T E L'(V1,V2).

M > 0

be normed Then

T E L'(V1,V2)

is bounded, then we define

IIT1I = inf(M

I

is said to

such that

IIT(x)112 < MIIx111 If

T

IIT (x)112 < MllxjI1, x E V1)

(x E V1).

62

Linear Transformations and Functionals

3.

Let

Theorem 3.2.1.

linear spaces over

(V2,I1-i12)

T E L'(Vl,V2).

and let

i

and

(V1,11.111)

be normed

Then the following

are equivalent: T E L(Vl,V2).

(i)

T

(ii)

Moreover, if

is bounded. T

is bounded, then

IIT(x)112 IITI1

xEV

1

1

x#0 IiT(x)112

sup xEV

1

Ilx111 <1

sup

xEVI

IIT(x)112

Ilxiil = Proof.

If

is bounded, then the estimate

T

11T(x)

immediately shows that continuous. V1,

Clearly

11zI11 = s12 < e,

also valid when T

T

s > 0

and so

such that

'

0,

liT(z)112 < 1.

IIT(y)112 < 2llylil/e.

y = 0,

T

is

is continuous at the origin in

But then for any y E V1, y

then reveal that

Conversely, suppose

is continuous.

Then, in particular,

and so there exists some

11T(x)112 < 1.

that

T

(x,y E V1)

- T(y) 112 < Mlix - ylil

1lxlll < s,

implies

set z = cy/21lylil. Simple computations

Obviously this estimate is

and therefore, taking

M = 2/c, we conclude

is bounded.

The remainder of the proof is left to the reader.

0

63

3.2. Basic Results on Linear Transformations

The notation

L(V1,V2).

a norm on

Theorem 3.2.2.

spaces over

t.

is not haphazard, as

IITII

actually defines

11.11

Indeed, we have the following result: Let

and

(V1,11.111)

be normed linear

(V2,II.112)

Then

is a normed linear'space over

(i)

(T E L(V1,V2)).

xEpV IIT(x)IIZ

IITII

where

t

1

IIXIII =1 is a Banach space over

(ii)

is a Banach space over

(V2,11.112)

(V2,11.112)

is a Banach space, let

Cauchy sequence; that is, given integer

I.

The proof of part (i) is routine and is omitted.

Proof.

ing that

such that

N

n,m > N

s > 0,

we see at once that

x E V1

Thus, since

to

V2

- T(x)112 = 0.

V2

implies that

a,b E §

IITn

- Tall < e.

Since

(x E V1),

IIx1l1

is a Cauchy sequence in

T(x),

in

V2

V2

for each

such that

We claim that the mapping

defines a linear transformation

To prove the linearity of and

Assumbe a

is a Banach space, there exists for each

some element, call it

limnlITn(x) V1

(Tn(x))

(T.) C L(V1,V2)

there exists some positive

IITn(x) - T(x)112 <_ IITn - Tall

x E V1.

provided

$

T

x -. T(x)

from

T E L(V1,V2).

we note that for each

x,y E V1

we have

IIT(ax + by) - aT(x) - bT(Y)112
+ IaIIITn(x) - T(x) 112 + IblIITn(Y) - T(y)112.

64

Linear Transformations and Functionals

3.

Since the right-hand side tends to zero as conclude that

T

is linear.

will be established, by Theorem 3.2.1,

T

The continuity of once we have shown that Cauchy sequence in

T

is bounded.

then

L(V1,V2),

nonnegative numbers, as for which

M > 0

tends to infinity, we

n

(Tn)

But clearly, if

is a

is a Cauchy sequence of

(IITnII)

Hence there-exists some

is a norm.

IITnII < M, n = 1,2,3,...

Thus for each

.

x E V1

we have

IITn(x)112 < M11x111

from which it follows that

if

(Td

But, as already noted, given

e > 0

n,m > N,

Therefore

IIT(x)1I2 < MIIxIIl.

It remains only to show that

(n = 1,2,3,...),

converges to

there exists

for

n

T

in

N

such that,

then

1f n(x) - Tm(x)112 <_ IITn - TmIIIIXIIl < IIxIII Holding

T E L(v1,V2).

fixed and letting

m

(x E V1).

tend to infinity, we deduce that

n > N,

IITn (x) from. which it follows by the definition of for

n > N.

That is,

and so

(T n)

converges to

that T

IITn -

TII < c,

in

is a Banach space.

Since the scalar field

f

0

can always be considered to be a

Banach space, we have the following corollary: Corollary 3.2.1. i.

Then

Let

be a normed linear space over

is a Banach space over

f.

b5

3.2. Basic Results on Linear Transformations

Note that the norm in

here is, of course, that in

V*

and not the norm in

V* = L(V,4),

despite the notation.

V,

On the

surface this may appear confusing, but it will not be so in practice, Moreover, the

since the context will make clear which norm is meant. method adopted helps to simplify notation.

It is perhaps worthwhile to write out explicitly the various ex-

pressions for x* E V*

indicated in Theorem 3.2.1.

IIx*II, x* E V*,

Thus for

we have

We close this section with two further general results about linear transformations and,normed linear spaces. Proposition 3.2.3.

linear spaces over

6

Let

(V1,I1.111)

and let

and

be normed

(V2,11.112)

T E L(V1,V2).

Then the following

E L(R(T),V1),

where

are equivalent: T-1

(i)

T-1

exists and

sidered to be a linear subspace of There exists some

(ii)

mIIxI Proof.

m > 0

l

such that

< IIT(x)I12

If part (ii) holds, then clearly

L'(R(T),V1).

(x E V1).

T(x) = 0

T-1

implies

T-1

x = 0,

exists and

Inequality (ii), however, says precisely that

IlT-1(y) II1 < (1/m) IIY112

that is,

is con-

(V2,11.112).

and from Proposition 3.2.1(iii) and (iv) we see that belongs to

R(T)

(y E R(T)),

E L(R(T),V1).

The converse implication

s apparent:

0

Linear Trar.sfor,!-atiorr

3.

66

it < I /In.

Clearly, we see from the preceding, that Proposition 3.2.4.

linear spaces over

Let

(Vilii

he nnrmed

(V,

is finite dimensiona,t, then

If

t.

and

{1)

any;

L'(V1,V2) = L(V1,V2).

Some Basic Results Concerning Linear Functionals.

3.3.

is a linear space over

then it is easily seen that

C,

be considered as a linear space over

R.

V

If

V

can also

Our first concern in this

section is to investigate the connection between linear functionals on

V

and linear functionals on

C

as a linear space over

linear space over

as a

V

This connection will be useful in discussing

R.

Second, we show that a linear functional on

the Hahn-Banach Theorem.

a topological linear space is continuous if and only if its kernel, that is,

(x

(

x E V, x'(x) = 0),

is a closed linear subspace of

V.

This is also a particularly useful result, as will be seen in Chapter 5.

Finally, we shall give an example of a topological linear space

that has no nonzero continuous linear functionals. Definition 3.3.1.

then

x'

V

If

xi = Re(x')

Let

and

V

x' = Im(x'),

are real linear functionals on

x', V.

that is,

C xi

Let

a,b E R and

x'

x,y E V.

R,

and let and

respectively, then

xi

x2

and

Moreover,

x'(x) = xi(x) - ix'(ix) = xZ(ix) + ix(x) Proof.

If

V.

be a linear space over

are the real and imaginary parts of x;

C.

considered as a linear space over

is said to be a real linear functional on

Proposition 3.3.1. x' E V'.

be a linear space over

V

Let

is a linear functional on

(x E V)

Then on the one hand,

x'(ax + by) = axi(x) + aix2(x) + bxi(Y) + bix2(y) = axi(x) + bx;(Y) + i[ax2(x) + bx2(Y)], while on the other hand, x'(ax + by) = xi(ax + by) + ix;(ax + by).

67

8.3.Basic Results on Linear Functionals

Equating real and imaginary parts, we conclude at once that are linear functionals over

x2

xi

and

IF_

For the second part of the proposition we note first that for each

x E V

we have x'(ix) = xi(ix) + ix2(ix).

But also, since

x' E V',

x'(ix) = ix'(x) = i[xi(x) + ixZ(x)] ix

- x'(x).

'(x)

Again equating real and imaginary parts, we deduce that and so

xi(ix) = -x2(x), x2(ix) = xi(x), x E V, x'(x) = xj(x)

ix '(X)

= XI (x) - ixi (ix) = x2(ix) + ix2(x).

U

The last portion of this proposition shows us how to express a linear functional in terms of either its real or imaginary part alone.

Conversely, the next proposition shows us how to define linear func-

tionals over C

by using real linear functionals.

The details are

straightforward and are left to the reader. Proposition 3.3.2. x'

Let

be a linear space over C

V

be a real linear functional on

for each

x

in

V,

then

x'

V.

is in

If

and let

x'(x) = xi(x) - ixi(ix)

V'.

The next theorem can be easily derived from the preceding two propositions, we omit the proof. Theorem 3.3.1.

C

and let

x' E V'.

Let

(V,T)

be a topological linear space over

Then the following are equivalent:

68

Linear Transformations and Functionals

3.

x' E V*.

(i)

(ii)

Re(x')

is a continuous real linear functional on

V.

(iii)

Im(x')

is a continuous real linear functional on

V.

First we

Now we turn to the second concern of this section. need a few definitions. Definition 3.3.2. W

V

Let

be a linear space over

be a proper linear subspace of

Then

V.

W

W e Wl,

maximal linear subspace if, whenever

and let

i

is said to be a either W = WI

or

W1 = V.

W c V

It is easily seen that a proper linear subspace

is

maximal if and only if it is of codimension one, that is, for any xD E V - W

the linear subspace spanned by

Equivalently, W

is maximal if and only if

Definition 3.3.3. x' E V'. of

N(x') _ (x

Then

Another common name for

N(x')

Proposition 3.3.3.

V

Proof.

N(;,')

E V - N(x'),

But if

Let

is of dimension one. and let

§

is called the kernel

N(x')

x E V,

is the null space of

be a linear space over

N(x')

then, since

x'(xo) # 0,

N(x') U (x0) we see that

[x'(x)/x'(xo)}xo) = 0.

Hence

x - [x'(x)/x'(xo)]xo E N(x'), and so

and let V.

is maximal it clearly suffices to show that, if

then the linear span of

x''(x -

I

is a proper linear subspace.

x - [x'(x)/x'(xo)]x0 E V is such that

x'.

is a maximal linear subspace of

It is evident that

To prove that 0

x E V, x'(x) = 0)

V.

x'.

x' E V', x' 4 0.- Then

x

I

V/W

is all of

be a linear space over

V

Let

[xo) U W

x E N(x') + [x'(x)/x'(x0)]xo.

Therefore the linear span of

N(x') U (xo)

is

V.

is all of

V.

69

3.3. Basic Results on Linear Functionals

be a linear space over

and let

Corollary 3.3.1.

Let

V

x',xi,x,,,...,xn be in

V'.

Then the following are equivalent:

(i)

is linearly dependent on

x'

4

xi,x...,x'.

(ii) N(x')' flk_1N(xk). The proof is left to the reader.

Next we establish the indicated characterization of continuous linear functionals. Theorem 3.3.2.

Let

x' E V'.

and let

Then the following are equivalent:

x' E V*.

(i)

is closed.

N(x')

(ii)

Proof.

Clearly (1) implies (ii), and if

is closed, and that

tion 3.2.2; c > 0

some

x'

x'

is not continuous.

W = (x

then obviously

x' + 0,

that

Then, by Proposi-

is not continuous at the origin, and so there exists

such that every open neighborhood

contains some point Let

x'.= 0,

On the other hand, suppose that

(ii)'implies (i). N(x')

be a topological linear space over

(V,T)

(

for which

x

U

of the origin

Ix'(x)I > c.

x E V, x'(x) = e/2j.

We claim that

W = N(x') + y 0

for some

yo E V.

Indeed, if

y E V - N(x'),

x'(x + [c/2x'(y)jy) = s/2, x E N(x'),

that

then we see immediately and so

N(x') + [s/2x'(y)jy C W.

Conversely, since there exist some and so that

N(x')

a E I,

is of codimension one, if

z E W,

and

z = x + ay,

x'(z) = ax'(y) = s/2. W = N(x') + yo,

where

x E N(x') Thus

such that

a = s/2x'(y),

then

and we conclude

yo = [s/2x'(y)jy.

Consequently, by Theorem 2.1.2,

W

is a closed subset of

as it is the translate of the closed subspace Thus there exists an open neighborhood

U0

of

N(x'), 0

and

such that

V,

0 f W.

70

3.

U0 n W = 0.

x0 E Uo

Now there exists some Then clearly

a = s/2x'(x0).

x'(axo) = x'([s/2x'(xo))xo) = s/2

Therefore

+al < 1,

such that

Ix'(x0)I > s.

contradict-

axo E W,

implies that

But

axo E Uo.

and so

U0 tl W = .

ing the fact that

If

U0

Moreover, by Proposition 2.3.2, we may assume that

is balanced. Let

Linear Transformations and Functionals

is continuous.

x'

x' E V'

and

x' # 0,

then Theorem 3.3.2 gives a simple and

useful characterization of the continuity of

However, there is

x'.

a more fundamental question concerning linear functionals that we have Do there exist any nonzero (continuous) linear func-

not yet faced:

.shortly that it is always the case, when

while it may very well be the case that is whether V'

and

x,y E V. x # y,

or x*(x) # x*(y)?

x'(x) # x1(y) V # (0),

does there exist an

then

V'

(0),

V

(0),

that

V*

(0).

A related question

separate the points of

V*

We shall see

V?

tionals on a given (topological) linear space

V;

T

that is, given x* E V*

or

x' E V'

V'

such that

Again we shall see that, when

separates the points of

V,

but this need not

be the case for V*.

In the next chapter we shall establish general conditions on

V

that ensure the existence of sufficiently many nonzero continuous linear functionals on

to separate the points of

V

shall see that this is always the case whenever

V

V.

Indeed, we

is a locally

convex topological linear space -- equivalently, a seminormed linear space.

Somewhat different proofs of these facts will also be given In both cases the Hahn-Banach Theorem, in either its

in Chapter 5.

analytic or geometric form, will be instrumental in establishing the desired result.

that V'

The remainder of this section is devoted to showing

always contains sufficiently many nonzero elements to

separate the points of linear space

V

V

for which

and to giving an example of a particular V*

fails to have this property.

71

3.3 Basic Results on Linear Functionals

For the sake of completeness we make the following Cefinition: Definition 3.3.4.

be a linear space over

V

Let

F C V is said to separate points if,

family of linear functionals whenever

x' E F

there exists some

x,y E V, x # y,

Then a

E..

such that

x'(x) } x'(y). V # (0)

Note, in particular, that, if points, then

only if

V'

and that

(0)

'

x'(x) = 0, x' E F,

Theorem 3.3.3.

x = 0.

be a linear space over

V

Let

Proof.

xo E V,

Let

W0 C V

linear subspace

xo

Then

x0

W.

ordering into W1,W2 E W.

We claim that there exists a

0,

'

of codimension one such that

Indeed, consider the family that

Clearly

V,

that is,

Ua

E AWa

E W

and

Moreover,

W0

V/W0

Thus for each such that

dent that

W

Consequently,

has a maximal element,

will be greater than one and so

x1 E V - W0

and for which

that properly contains

y E W0

is an upper bound for

Wa < a E AWa, a E A.

such that

Wo,

x E V

x = axo + y.

x' E V', N(x') = W01

xI

will be linearly

x0 - ax, f Wo, a E f.

easily verified that the linear span of

W0.

E AWa

is of codimension one, since if this is not the

there will exist some x0

and

Wo.

case, then the dimension of

independent of

such

is a linearly ordered subset

Ua

by Zorn's Lemma [DS1, p. 6], the family which we denote by

WI C W2,

whenever

V = (Wa)a E A

W C V

We introduce a partial

(0) E W.

as

WI < W2

Then, as is easily verified,

W.

x0 f W0.

W of linear subspaces

01

W by setting

Suppose that

of

of

{,V # (0).

separates points.

V'

W

separates

F c V'

separates points if and

F C V'

implies

and

W0 U (x1)

It is then

is an element of

thereby contradicting the maximality there exists a unique Define and

x'(x) = a.

x'(xo) = 1.

a E 4

and

It is then evi-

Linear Transformations and Functionals

3.

72

Consequently, if construction, if x' E V'

x'(y

0

0

xo = yo

such that

Therefore

E V, y

y0,z

)

0

we see that there exists some

z0 # 0,

x'(x0) =

x'(yo) - x'(z0) = 1.

that is,

1;

# x'(z

then, by the previous

# z0,

and

),

separates points.

V'

0

As we have indicated, if then it may be the case that

LJ

V

to)

V*

(0),

is a topological linear space, and so a fortiori

does

V*

A concrete example of this is provided by the

not separate points.

topological linear spaces

([0,1],dt), 0 < p < 1,

L

where

dt denotes

P

Lebesgue measure on Theorem 3.3.4.

(0,1].

L([0,1],(jt)* _ [0j, 0 < p < 1.

Suppose that

Proof.

x* E L([0,1],dt)*

f E L([0,1],dt) p

exists some s, 0 < s < 1,

set

such that

fs = k[0"] f,

and

Ix*(f)I

where

teristic function of the interval

X[0 s] and set [O,s],

x* # 0. 1.

denotes the characf` = f - fl. S

S

Clearly

fs,fs

E Lp([0,1],dt), 0 < s < 1,

and

p)p = f'l f (t) `p dt reveals that of

s

(j1fIIIp)p

(0 < s < 1)

is a continuous monotone increasing function

and that

((IfIIIp)p = 0 and

p )p

Consequently there exists some 1

('Ifs( 1i1,)p =

(1

Then there

For each

= (1Ifllp)P.

so, 0 < so < 1,

For this

so

such that

we also have

73

3.3. Basic Results on Linear Functionals

(TIES 1IP)P =

j'o .f(t) - fs (t)Ip

dt

0

0 =

If(t)Ip dt

fsl 0

If(t)Ip dt

= f1

JSo If(t)Ip dt

-

(IIfIIp)P - (Ilfs

II

p)P

0

- (Ilfllp)P

Ix*(f)I > 1,

Now since

x*

and the linearity of

we deduce via the triangle inequality

that either

Ix*(fI )I > 1/2 0

or Ix*(fs

0

Define

fl = 2f5

where

,

i =

1

or

is such that

2

0

Ix*(fs )I > 1/2. 0

It is then evident that

fl E Lp([O,1],dt), Ix*(f1)I > 1,

IIfIIIp =

2(1 - 1/P)

Repeating the argument with f2 E Lp([0,1],dt)

such that

IIf2I1 p

=

f1

Ix*(f2)I

and"

IIfIIp

in place of > 1

f,

we obtain some

and

2(1-1/p)I(flop = 22(1-1/0IIfIIP.

Continuing in this manner, we construct a sequence

(fn} CL ([O,I],dt) p

such that

Ix*(fn)I > 1 and Ilfn(Ip = 2n(1-1/P)IIf1Ip, n = 1,2,3,...

.

Linear Transformations and Functionals

3.

74

we have

0 < p < 1,

But since

1

thereby contradicting the contin-

Ix*(fn)I > 1, n = 1,2,3,...,

while

limnllfnll p = 0,

and so

- 1/p < 0

uity of- x*.

L([0,1],dt)* _ (0).

Therefore

3_4. Problems.

1.

i

(Proposition 3.2.1)

(a)

T(0) = 0.

(b)

The range of

(c)' If

T

T

be linear spaces over

V2

and

V1

Prove each of the following:

T E L'(V1,V2).

and let

Let

is a linear subspace of

is injective, then

T-1

V2.

exists and

T-I E L'(R(T),VI). 2.

(Theorem 3.2.1)

linear spaces over

IITII

3.

and let

space over

1,

Prove that

= sup{jjT(x)112

1

= sup(IIT(x)I12

1

I.

'

0)

x E V1, j1x1l1 < 1) x E V1, 11x111 = 13. and

Let

Prove that

(V2,II.112)

(L(V1,V2),II.II)

be normed

is a normed linear

where

IITII

= sup(IIT(x)112

(Proposition 3.2.4)

linear spaces over sional, then

T E L(V1,V2).

= sup(IIT(x)jj2/1{x111 I'x E V1, x

(Theorem 3.2.2)

linear spaces over

4.

4

be normed

and

Let

6.

1

Let

)'

E V1, 11x111 = 1).

(V1,11-111)

Prove that, if

L'(V1,V2) - L(V1,V2).

and

(V1,11.111)

be normed is finite dimen-

75

3.4 Problems

S.

Let

VI

and

be linear spaces over

V2

E CV

be a linear transformation, and let

let

#,

T

:

be any subset of

V2

V1 VI.

I

is symmetric.

(a)

If

E

is symmetric, prove that

(b)

If

E

is balanced, prove that

(c)

Tf

E

is convex, prove that

(d)

Give an example to show that a nonconvex set may have a

T(E)

is balanced.

T(E) T(E)

is convex.

convex image.

6.

Let

and let

T

maps bounded sets in 7.

V1

Let

T E 1'(V1,V2). that

T

y E V2.

8.

and

(V1,11-111)

T E L'(VI,V2).

E

and Let

be linear spaces over

T

A,

that is,

and

#

V1

T(x) = y, x E A,

let

and suppose for some

is identically zero.

Let

and

(V2,11.112)

spaces over

I.

composition

TS E L(VI,V3)

If

V2.

be an absorbing subset of

A

is constant on Prove that

into bounded sets in

V1

V2

be nonmed linear spaces over (V2111'112 ) Prove that T E L(VI,V2) if and only if

S E L(VI,V2)

and

(V3,11.113)

be normed linear

T E L(V2,V3),

prove'that the

and that

IITSII < IITIIIISII

I.

9. Let

Let (V1,II'111) and (V2,11.II2) lje normed linear spaces over T E L(VI,V2) be an isomorphism onto V2 and suppose

T-1 E L(V2,V1).

(a)

Prove that

(b)

If

V2

IIT-'II > IITII-I

is a Banach space, prove that

V1

is also a Banach

space.

10. with

Let

(V1,II.111)

V1 } (0).

is complete.

and

(V2,11'II2)

Prove that, if

be normed linear spaces over

L(V1,V2)

is complete, then

V2

3. Linear Transformations and Functionals

76

over

11.

Let

§,

let

be normed linear spaces

and

T,Tn E L(V1,V2), n = 1,2,3,...,

n = 1,2,3,...

If

and let

converges to (L(V1,V2),I1-II) and [xn) converges to x in iTn (xn)J converges to T (x) in (V II II2)

x,xn E 4'1,

Let

12. i

and let

(V1,P)

and

positive number

Let

c

T

:

in

T

prove that

be seminormed linear spaces over T E L(V1,V2)

Prove that

if and only

for all

q[T(x)] < c p(x)

such that

and a

p E P

there exists a seminorm

q E Q

x E V1.

he a mapping that is additive (that is,

U,.- 1k

T(x - y) = T(x) + T(y),

where

Prove

and continuous.

x,y E IR)

is linear; that is, there exists some, c EIR such that

T

that

[T n)

(V2,Q)

T E L'(V1,V2).

if for every seminorm

13.

.

T(x) = cx.

(Corollary 3.3.1)

*14.

x',xi,...,x'

let

on

xi..... x'

x E V,

16.

over

Q:

prove that

(Theorem 3.3.1) and let

x' E V'.

is linearly dependent

x'

Let

V

be a linear space over C V.

If

x'(x) = xl'(x)

Let

(V,T)

be a topological linear space

Prove that the following are equivalent:

x' E V.

(b)

Re(x')

is a continuous real linear functional on

V.

(c)

Im(x')

is a continuous real linear functional on

V.

17.

Let

V

.

be a linear space over C

If

x' E V'

and

Ix'(x)I < 1

for all

x E E.

V.

and

- ix(ix)

x' 6 V'.

(a)

subset of that

if and only if

Prove that

and

i

N(x') D 11 =1N(xk).

be a real linear functional on

xi

where

V.

(Proposition 3.3.2)

15.

let

be in

be a linear space over

V

Let

and let

Re[x'(x)] < I

E

for all

be a balanced x E E,

prove

77

3.4 Problems

Letting

18.

C

be a linear space over

V

x',y' E V',

and

prove each of the following: (a)

If

Re[x'(x)] < Re[y'(x)]

(b)

If

IRe(x'(x)]I < IRe[y'(x)]I

x' = ry' for some

20.

x' E V',

for all

= y'.

then

x E V,

be a normed linear space over (lx*II

Let

be a topological linear space over

(V,T)

If

C.

prove that

x'

x'

r E IR.

Let

19.

x* E V*,

then

x E V,

for all

= IIRe(x*)II. and let

§

Prove that the following are equivalent:

T 0.

(a)

x' E V*.

(b)

N(x')

(c)

x'

is not dense in

V.

is bounded on some subset

U c V

such that

int(U)

contains the origin.

subset

is a proper subset of

x'(U)

(d)

for some nonempty open

f

U C V.

be a Banach space over

Let

21.

be closed linear subspaces of

V.

unique representation in the form

and let

Suppose that each x = y + z,

Prove that there exists a constant

z E N.

f

K

M

x E V

with

and

N

has a and

y E M,

such that

Ilyll < Kllxll, 'x E V. Let

22.

W C V is,

V

be a linear space over

is a linear subspace of x'(W)

is a bounded subset of Let

23.

V

x' £ V'

24. (V1.I1.II1)

Let

and

V.

If

such that

x E V.-. W,

W = N(x')

f

and let

that

W

be a maximal

prove that there exists a and

x'(x) = 1.

be a normed linear space over

be its completion (see Theorem 1.1.1).

isometrically isomorphic to V.

W,

If

W C N(x').

prove that

f,

x' E V'.

and let

is bounded on

x'

be a linear space over

linear subspace of unique

V

f

f

and'let

Prove that

V*

is

78

Linear Transformations and Functionals

3.

25.

over

I.

Let

(V,7)

be an n-dimensional topological linear space is also an n-dimensional topological

Prove that

linear space.

26.

Give an example to show that

Let

V* # Vt 27.

Let

(V,II'11) = (C([0,1]),l1'II )

(a)

Let

T E L(V)

be defined by (t E [0,1); f E C([0,l])).

T(f)(t) = tf(t)/(l + t2)

Find

I`TII. (b)

Let

be defined by

S E L(V)

(t E (0,1]),

S(f)(t) = fI K(t,s)f(s) ds where

Find

K(t,s)

[0,1) x [0,1)

is continuous on

f E C([0,1]).

and

llS,I. (c)

Let

x* E V*

be defined by x*(f) = f(to)

for fixed (d)

to E [0,1) Let

y* E V*

and all

y*(f)

for fixed

*28. of

V

y E C([0,l])

and all

f E C([0,1]).

consisting of those functions

derivative of

Find

and let

T E L'(W,V) f.

(lx*It.

fp f(t)y(t) dt

Let

and second derivatives on Let

Find

f E C([0,1)).

be defined by

be defined by Prove that

f

T(f) = f", T-1

be the subspace

W

that have continuous first

and are such that

[a,b)

IJy*11.

where

exists and that

f(a) - f(b) = 0. f"

is the second

R(T) = V.

79

3.4 Problems

29.

and let

Let

family of linear functionals F = (p 9t (f) = f(t), f E C([0,1]). *30.

c* ..f1

I

Prove that

In Example 3.1.6 we saw that

tE F

F C V* [0,1]], where

separates points.

c* = 11.

and exhibit the general form of an

be the

Prove that

x* E c*.

CHAPTER 4

THE HAHN-BANACH THEOREM: ANALYTIC FORM

4.0. Introduction.

The Hahn-Banach Theorem is, together with

the Uniform Boundedness and Open-Mapping Theorems, one of the most important theorems of functional analysis.

In its analytic form,

discussed in this chapter, the theorem assures us that a linear functional on a linear subspace of a linear space that is bounded

by a seminorm can always be extended to the entire space in such a way that the seminorm boundedness is retained.

The proof of the

theorem and a number of its consequences will be given in Sections 4.1 and 4.2.

Among these consequences is the fact that there exist

sufficiently many continuous linear functionals on a nontrivial The

locally convex topological linear space to separate points.

remaining sections of this chapter are devoted to a sampling of various applications of the Hahn-Banach Theorem and its consequences, and to a proof, in the last section, of Helly's Theorem.

Although

the proof of the latter theorem makes no direct use of the HahnBanach Theorem as developed in this chapter, it does depend on a simple case of the geometric form of the theorem, and hence the proof provides a motivation for the exposition of the succeeding chapter.

4.1. The Hahn-Banach Theorem: Analytic Form. linear space over

*

and

W c V

also that

p

is a seminorm on

tional on

W

such that

Suppose

is a linear subspace.

V

and that

y'

V

is a

Suppose

is a linear func-

It

Iy'(x)I < p(x) Then can

y'

be extended to a linear functional on all of

80

(x E W). V

that

81

4.1. Hahn-Banach Theorem: Analytic Form

is also bounded by

That is, does there exist some

p?

such

x' E V'

that

(i)

(ii)

x'(x) = Y, (x)

(x E W).

Ix'(x)I < p(x)

(x E V).

The Hahn-Banach Theorem tells us that such an extension is always possible, although it is, in general, not unique. First, we

To prove this result we proceed in several stages. show how to make such an extension for linear spaces over the codimension of

IR

when

is one and then apply Zorn's Lemma (DS1, p.'61

W

to obtain the general result for arbitrary linear spaces over

II:.

t = C will then be established with the help of

The theorem when

some of the results in Section 3.3.

one.

exists some (i)

(ii)

be a linear space over W C V

and let

V,

is such that

y' E W'

If

V

Let

Lemma 4.1.1'.

seminorm on

x' E V'

p

y'(x) < p(x), x E W,

(x E V).

x0 E V - W.

Let

V

is spanned by

as

Then, since W U {xoj,

x = ax0 + y

x'(x) = ay + y1(y)

Clearly such an

x'

agrees with

W

is of codimension one,

and thus every

for some

xl,x2 E W.

and

a E IR

is a linear functional on y'.

x E V

can

y E W.

for some suitable choice of

To show that

x'

we need to look more closely at how one chooses Let

then there

x'(x) < p(x)

x'

W

be a linear subspace of codimension

such that

be uniquely expressed as

tion to

be a

p

(x E W).

we see that

y.

let

x'(x) = y'(x)

Proof.

We define

IR,

V

whose restric-

is also bounded by y.

Then

Y'(xl) - Y'(x2) = Y'(xl - x2) < p(xi - x2)

=P(xl+x0-x2-xo) P(xl + x0) + P(-x2 - xo),

4. Hahn-Banach Theorem: Analytic Form

82

and so we have -P(-x2 - x0) - y'(x2) < p(x1 + xo) x1 E W

Thus for fixed

we deduce that

(-p(-x2 - x0) - y'(x2)

is bounded above

Clearly

bl < b2.

x2 E W) C !R

and so has a least upper bound, call it

(p(xl + xo) - y'(xl)

larly

+

Let

y

Note that by the choice of

I

xl E W)

be any real number such that y

we have

To complete the proof we need to show that the (

chosen here is such that

x = axo + Y.

b2.

bl < y < b2.

(x E W).

-P(-x - x0) - Y'(x) < y < P(x + xo) - Y, (x)

the

Simi-

b1.

has a greatest lower bound

x'

x'(x) < p(x), x E V.

defined for Suppose

We consider three cases.

Case 1.

a = 0.

Then

x'(x) = y1(y) < p(y) - p(x).

Case 2.

a > 0.

Then from

y < p(y/a + xo) - y'(y/a)

we deduce

that

ay + y1(y) < ap(y/a + xo) = P(ax0 + Y),

and so

x'(x) < p(x).

Case 3.

a < 0.

Then from

-p(-y/a - x0) - y'(y/a) < y

we deduce

that

ay + y'(y) < -ap(-y/a - xo)

= P (axo + Y), and again we have

x'(x) < p(x).

This completes the proof.

0

4.1. Hahn-Banach Theorem: Analytic Form

83

A few remarks are in order before proceeding.

First we note that

the result is not more general than the one alluded to in our introIndeed, since

ductory remarks. that

is a seminorm, it is easily seen

p

if and only if

x'(x) < p(x)

The lemma was

Ix'(x)I < p(x).

stated in this seemingly stronger form since it remains true, as given, and with the same proof, if we replace the assumption that

p

is a seminorm with a slightly weaker assumption; namely, we need only

assume that

p

:

is such that

V -- IR

(1)

p(x) > 0,

(2) (3)

p(x + y) < p(x) + p(y). p(ax) = ap(x)

(a E I42;

a > 0; x,y E V) .

These remarks apply also to the next theorem and will be used in proving a geometric form of the Hahn-Banach Theorem (Theorem 5.2.1) However, when

in the next chapter. that

4 = C, we need the hypothesis

is a seminorm.

p

The ' functional given by Lemma 4.1.1 is clearly not generally unique.

Indeed, it is apparent that it is unique for a fixed

xo E V - W

if and only if

inf (p(x + x

- y'(x)) =

°)

xEW

sup { -p(- x - x) - y'(x)?.

xEW

°

Theorem 4.1.1 (Real Hahn-Banach Theorem). Let space over

let

142,

linear subspace.

be a seminorm on

p

If

y' E W'

then there exists some

V,

is such that

x' E V'

V

and let

be a linear W C V

y'(x) < p(x), x E W,

such that

x'(x) = y'(x)

(i)

(x E W).

(ii) x' (x) < p (x) Proof.

[DS1, p. 6].

be a

(x E V).

The proof is a standard application of Zorn's Lemma Let us call a pair

(u',U)

an extension of

y' provided

(1)

U C V

(2) (3)

u' (x) = y' (x)

(x E W).

u'(x) < p(x)

(x E U).

is a linear subspace such that

U D W.

4. Hahn-Banach Theorem: Analytic Form

84

Let

be the family of all extensions of

U

(y',W) E U.

(ui,U1) < Z,"2)

if

U1 C U2

is a partial ordering on

ordered subset of

U.

a

imal element, call it We claim that

by

Let

U = Ua Ua

and

V0 = V.

Clearly

V0

and let

Since

V0 # V1,

(x',V0),

V0 = V,

V1

of codimen-

z' E Vi

such

that is,(z',V1) > (x',Vo).

this contradicts the maximality of

Therefore

has a max-

be the linear space spanned

V1

is a linear subspace of

is an extension of

(z',VI)

U

If this were not true, we would get the

sion one, and so, by Lemma 4.1.1, there exists some that

is an

(u',U)

(x',Vo).

x0 E V -. V0

V0 U (x0).

where

Then <

is a linearly

Consequently, by Zorn's Lemma,

u'(x) = u'(x), x E U U.

a

((u',U))

Suppose

Then it is easily verified that

((u,,U,))

upper bound for

following:

U.

ui(x) = u2(x), x E U1.

and

as

U # 4),

then we write

(uI,U1), (uZ,U2) E U,

If

Clearly

y'.

(x',Vo).

and the theorem is proved.

0

We can now combine these results with our knowledge of real linear functionals obtained in Section 3.3 to establish the Hahn-Banach

Theorem for linear spaces over arbitrary Theorem 4.1.2 over

f,

p

let

subspace.

there exists some

(i)

(ii)

(Hahn-Banach Theorem). Let

be a seminorm on

y' E W'

If

f.

x' E V

and let

V,

is such that

V

be a linear space

W CV be a linear

ly'(x)l < p(x), x E W,

then

such that

x'(x) = Y, (x)

(x E W).

fx'(x)j < p(x)

(x E V).

Proof.

If

f = 1.,

then the result is an immediate consequence

of the Real Hahn-Banach Theorem and the remark immediately following Lemma 4.1.1.

So assume that

f = C and let

From Proposition 3.3.1 we see that

yi

on' W, that is, a linear functional on

yi(x) = Refy'(x)), x E W.

is a real linear functional W

considered as a linear

85

4.2. Consequences of the Hahn-Banach Theorem

space over

1R.

Moreover, we see that (x E W).

yi(x) = Re[y'(x)] < IY'(x)i < p(x)

Hence we may apply the Real Hahn-Banach Theorem and deduce the existence of a real linear functional and

such that

V

on

xi

x'(x) = yi(x), x E W,

xi(x)
Then from Proposition 3.3.2 we see that 3.3.1 that

x'(x) = y'(x), x E W.

(x E V) .

- ixi(ix)

and from Proposition

x' E V'

Finally, given

we see that

x E V,

Ix'(x)l = exp[-i arg x'(x)]x'(x) = x'(exp[-i arg x'(x)]x) = xi(exp[-i arg x'(x)]x)

< p(exp[-t arg x'(x)]x)

= p (x) as

x'(exp[-i arg x'(x)]x) = lx'(x)( EIR,

where exp

denotes

the usual exponential function.

0

Some Consequences of the Hahn-Banach Theorem.

4.2.

We now

consider many of the most useful and iomediate consequences of the Hahn-Banach Theorem.

the points of

In particular, we shall see that

and so

V,

V* # (01,

whenever

V

V*

separates

is a nontrivial

seminormed linear space. Theorem 4.2.1.

and let some

W C V

x* E V* Proof.

Let

(V,P)

be a seminormed linear space over

be a linear subspace. such that

Since

y*

P1'P2 " " 'pn

P

y* E W*,

is continuous on

such that

(W,TP),

g > 0

jy*(x)l < 1

E

then there exists

x*(x) = y*(x), x E W.

at the origin, and so there exists an in,

If

it is continuous

and seminorms whenever

4. Hahn-Banach Theorem: Analytic Form

86

Define

pk(x) < c, k = 1,2,...,n, x E W.

sup{pl(x),...,pn(x))

P(x) =

(x E V) .

c

Then, by essentially the same argument as that used in the proof of Lemma 2.5.1, we see that

Moreover, given we deduce that that.is,

x E W,

and so

p(x/a) < 41

ly*(x/a)l < 1;

However, since this holds for each

ly*(x)l < p(x), x E W.

x' E V'

x'(x) = y*(X), x E W,

and

lx'(x)l < p(x), x E V.

6 > 0

and

x E V

Furthermore, if k = 1,2,...,n,

a > p(x),

Consequently by the

there exists some

Hahn-Banach Theorem (Theorem 4.1.2) that

V.

Then from

a > p(x).

let

pk(x/a) < c, k = 1,2,...,n,

ly*(x)l < a.

we conclude that

is a seminorm on

p

is such that

sucl.

pk(x) < 6c,

then

sup(p1(x),...,Pn(x))

p(x) = and so

lx'(x)l < 6.

by Proposition 3.2.2,

< 6,

c

Thus

x'

is continuous at the origin, and so,

x' = x* E V*. Cl

Corollary 4.2.1. and let -Proof.

W

(V,P)

Let

V # [0).i Then Suppose

V*

be a seminormed linear space over

separates points.

x,y E V, x # y.

Then

xo = x - y # 0.

be the one-dimensional linear subspace of

define

y*(axo) = a, ax0 E W.

to verify that x* E V*

y* E W*.

points.

spanned by

x0

and

Using Theorem 2.2.1 it is not difficult

Hence, by Theorem 4.2.1, there exists some

such that x*(x) = y*(x),

In particular,

V

Let

x E W.

x*(x0) = x*(x - y) -

I

# 0,

and

V* separates

0

4.2. Consequences of the Hahn-Banach Theorem

over

V

and let

f

'

[0).

f.

T

(0).

be a seminormed linear space over

(V,P)

x* E V*

then there exists some

x E V, x # 0,

If

V*

Then

Let

Corollary 4.2;3.

be a seminormed linear space

(V,P)

Let

Corollary 4.2.2.

such that

x*(x) = 1.

Since, by Theorem 2.3.1, the concepts of seminormed linear spaces and locally convex topological linear spaces are equivalent, we conclude at once that Theorem 4.2.1 as well as Corollaries 4.2.1, 4.2.2, and 4.2.3 are all valid when

such a space and

V # (0).

separates points whenever

V*

In particular,

linear space.

is a locally convex topological

(V,T)

V

is

Thus we see that the topological linear where

is Lebesgue measure on

spaces

Lp([O,1],dt), 0 < p < 1,

[0,1],

are not locally convex since, by Theorem 3.3.4,

dt

Lp([0,1],dt)* = (0), 0 < p < 1.

These results are, of course, also valid in the special case in which

V

Moreover, in this case several

is a nonmed linear space.

of these results can be sharpened. Theorem 4.2.2. and let some

W C V

x* E V*

be a normed linear space over

Let

be a linear subspace.

If

y* E W*,

such that

(i) x*(x) = Y*(x)

(ii)

t

then there exists

(x E W) .

Ux*II = IIY*II

Proof.

Let

p(x) = IIY*II IIxII, x E V, IIY*II

II

p

where as usual

IY*(x)I.

X11 =1 xEW

Then

p

is evidently a seminorm on

V

and

Iy*(x)I < p(x), x E W.

Hence by the Hahn-Banach Theorem (Theorem 4.1.2) there exists some x' E V'

such that

x'(x) = y*(x), x E W,

and

I x' (x) 1 < p (x) = IIY*III1xII

(x E V) .

4. Hahn-Banach Theorem: Analytic Form

88

This, by Theorem 3.2.1, x' = x* E V* and

I1x*11 < Ily*II

Moreover, from

sup

IIY*II

114 -1

IY*(x)I °

II X11p= 1

Ix*(x)I

sup

Ix*(x)I

xEW

xEW

<

IIxII = 1

xEV

= IIx*li,

we conclude that

IIx*II = 11Y*11-

0

A substantial improvement of Corollary 4.2.3 is also available in normed linear spaces. Theorem 4.2.3.

W C V

let

Let

(V,II.11)

be a normed linear space over

be a linear subspace, and let

d = infy E WIIxo - ytl > 0,

x0 E V - W.

then there exists some

If

such that

x* E V*

x*(x) = 0

(i)

f,

(x E W).

x*(xo) = d.

(ii)

(iii)

llx*11 = 1.

Proof. Let

W0 CV be the linear subspace spanned by

W U (x 0)

and define

y* (axo + y) = ad Clearly

y*

is a linear functional on

y*(y) - 0, y E W.

W0, y*(xo) = d,

Moreover, we claim that

Indeed, note-'first that if

IIxII -

I1ax0

(a E f ; y E W).

IlY*ll

and

I.

x = axo + y, a f 0,.y E W,

then

+ YII - Ila (xo - (-Y/a) l II > laid.

Hence

Iy*(x)I = laid < Ilxil;

by the definition of

d,

that is,

given

s > 0

iiy*II < 1.' On the other hand,

there exists some

y E W

such

4.2.

Consequences of the Hahn-Banach Theorem

that

llx0 -

Set

It < d + a.

z E WO,llzll = 1.

89

z = (x0 - y)/11x0 - yll.

Then

and

Y11

d > -(-d--+--CT Since

c > 0

is arbitrary, it follows from the definition of

that lly*ll > 1,

ilr*lI

and thus llr*ll = I.

An application of Theorem 4.2.2 completes the proof.

0

Note, in particular, that Theorem 4.2.3 applies whenever a closed linear subspace and

W

is

x0 E V.- W.

A number of corollaries can be obtainel from the preceding results. We reave the proofs to the reader. Corollary 4.2.4. let

W C V

Let

d - infy E Wllxo - yll > 0, (i)

x0 E V - W.

x*(x0) * 1.

(iii)

llx*ll = 1/d.

x0 E V, x0 + 0,

(i) x*(x0)

If

then there exists some x* E V* such that (x E W) .

(ii)

(ii)

be a normed linear space over

x*(x) = 0

Corollary 4.2.S. If

(V,It.l1)

be a linear subspace, and let

be a norued linear space over

Let

then there exists some

x* E V*

1lx0II.

Ilx*II - 1.

Moreover,

sup l[x Oil - llx*ll

_l x* E V*

l x* (X ) I 0

.

such that

I.

4.

90

Corollary 4.2.6. and let

x0 E V.

Let

be a normed linear space over

(V,1j.I1)

x*(xo) = 0, x* E V*,

If

Corollary 4.2.7.

Let

let

W C V

and

x*(x) = 0, x E W,

then

W C V

imply that Let

x*(x0) = 0,

(V,1+.!)

be a linear subspace.

4,

x0 = 0.

be a normed linear space over

(V,11-11)

be a closed linear subspace, and let

Corollary 4.2.8. and let

Hahn-Banach Theorem: Analytic Form

x0 E V. then

If

4,

x* E V*

x0 E W.

be a normed linear space over Then the following are equiva-

lent:

(i)

(ii)

cl(W) - V. If

x* E V*

is such that

then

x*(x) = 0, x E W,

x* = 0.

As the reader may suspect, a number of these results for normed linear spaces have valid analogs in the context of seminormed linear spaces, (i.e., locally convex topological linear spaces).

We shall

return to this in Section 5.3 after we obtain a geometric version of the Hahn-Banach Theorem.

In the remaining sections of this chapter we shall examine various applications of the Hahn-Banach Theorem and its consequences. Other uses of this important theorem will occur in subsequent chapters. 4.3. The Hahn-Banach Theorem and Abelian Semigroups of Transformations.

We wish to establish an extension of the Hahn-Banach

Theorem asserting that linear functionals on subspaces bounded by a' seminorm can be extended to the entire space in such a way as to

preserve not only this boundedness but also the action of the linear functionals with regard to certain families of linear transformations. The statement of the next theorem will make this rather vague assertion precise.

First, however, we need a definition.

Definition 4.3.1. that

G C LT(V)

Let

V

be a linear space over

i.

Suppose

is a family of linear transformations such that

(i)

T,S E G

(ii)

TS = ST

implies

TS E G. (S,T E 'v).

4.3. Abelian Semigroups of Transformations

Then on

G

91

is said to be an Abelian semigroup of linear transformations

V.

Theorem 4.3.1. Let seminorm on

formations on Suppose

let

f,

p

be a

G be an Abelian semigroup of linear transp[T(x)] < p(x), x E V

such that

V

W C V

be a linear space over

V

and let

V,

y' E W'

is a linear subspace and

and

T E G.

is such that (x E W).

(a)

ly'(x)l < p(x)

(b)

T(x) E W

(x E W; T E G).

(c)

y'[T(x)J = y'(x)

(x E W; T E G).

Then there exists some (i)

(ii)

(iii)

x' E V'

such that

x'(x) = y1(x)

(x E W).

IX'(x)) < p(x)

(x E V).

x'[T(x)] = x'(x)

Proof.

(x E V; T E G).

As might be expected, the idea of the proof is to arrange

things so that the Hahn-Banach Theorem (Theorem 4.1.2) can be applied.

To this end we begin by defining a new seminorm on x E V

for each

V:

we define

P[TI(x)+...+Tn(x)j po(x) - inf(

)

n

where the infimua is taken over all possible finite subsets (T1,T2,...,Tn) c G.

Since

we see at once that

0 < po(x) < p(x), x E V.

verify that

p

is a seminorm and

p0(ax) - ja`p0(x), x E f,

Furthermore, let

definition of

po

x,y E V

there exist

as

and suppose

p

has this property. s > 0.

T1,...,Tn; S1,...,S®

p[T1(x)+...+Tn(x)J < po(x) + n P[S1(Y)+...+s (y)] m

p[T(x)] <_ p(x), T E G,

It is elementary to

2 s

< p (Y) +

°

2

Then by the in

G

for which

4.

92

Hahn-Banach Theorem: Analytic Form

Hence m

n

II

E T S.(x + y) pD(x+Y)< 1k-1 j=1kJ p

E

nm

n

< pLk E

m

1

j

E nm

n

m

p

E S.( ET

+

p k

m 11

1

j

+

E 1TkSj (Y)J nm

1 k(x)/n)l PEn

=1 ]k=]

_

n E

(x)1

m

T ki E S.(y)/m)

k=1

m

E

<j=1

pISj(

E

k=1

j=1 n

m

E p (Tkj E S. (y)/mJI J +k=l LL j=1

Tk(x)/njl

n

m

E pI E T (x)/n k

< j=1 k=1

E P

E S.(Y)/ml

J+ k=1 11j=1

JJ

n

m

p[n

E 1Tk

(x)I J

+

pI

E S (Y)J

`j=1

n

m

< Po (x) + Po (Y) + c

.

Note that the commutativity and linearity of the elements of as well as the assumption that

p[T(x)] < p(x), x E V, T E G,

all used in establishing the preceding estimates. arbitrary, we conclude that Thus

po

Since

G,

are

a > 0

is

Po(x + Y) < Po(x),+ P0(Y), x,y E V.

is a seminorm on

V

such that

p(x), x E V.

93

4.3. Abelian Semigroups of Transformations

Moreover, if

x E W,

T1,T2,...,Tn

then for any

G

in

we have

jy[T1(x)]+. ..+y'[Tn(x)]I

ty`()I =

n

n =

Iy'[ E T (x) k=1k n

<

.p[ E Tk(x k=1 n

T1,T2,...,Tn

Because the

y'[T(x)] = y'(x), T E G.

since

in

G

!y'(x)j < po(x), x E W.

are arbitrary, we can conclude that

Consequently, appealing to the Hahn-Banach Theorem (Theorem 4.1.2) with regard to the seminorm x' E V'

for which

the following grounds:

p[x - T(x)] < o

we see that there exists some

x'(x) = y'(x), x E W,

Furthermore we claim that

positive integer

p0.

and

1x'(e)I < po(x), x E V.

x'[T(x)] = x'(x), x E V

Suppose

x E V

and

and

T E G,

T E G,

on

then for each

n

pjT[x - T(x)] + T2(x

T(x)] +...+ Tn[x - T(x)]) -

pL(x) - Tn+l (x) ] n

P[(T(x)] +P[Tn+l(x)] n

from which we deduce at once that Ix' [x - T(x)]l < po[x - T(x)]

p0[x - T(x)]

we have

0.

However, since

x'[T(x)] = x'(x).'

94

Hahn-Banach Theorem: Analytic Form

4.

An examination of the proof shows that Theorem 4.3.1 also holds for linear spaces

x,y E V.

over

1R

if one assumes only the following: and

p(ax) = ap(x), a > 0

and

The reader should compare the remarks following Lemma 4.1.1. G

Clearly, if V,

V

p(x + y) < p(x) + p(y),

p(x) > 0,

consists only of the identity transformation on

then Theorem 4.3.1 reduces to the Hahn-Banach Theorem. Let us look at an application of Theorem 4.3.1 to the existence

of so called Banach limits.

Consider the faiAily

sequences of complex numbers.

c

of all convergent

Then from well-known properties of

convergent sequenceg we know that to every such sequence corresponds a unique complex number

limkak

there

(akj

that has the following

properties: If

(I)

ck - aak + bbk, k = 1,2,3,..., then

[ak),(bk) E c, If

(ii)

ek = 1, k

For each

(iii)

where

(ck) E c

(ak) E c,

and

1,2,3,...,

n = 1,2,3,...,

then

(ck) E c

where

a,b E C

and

limkck = a limkak + b limkbk. then if

and

limkek = 1.

ck = ak+n, k = 1,2,3,..., limkck = limkak.

Is it possible to define a notion of "limit" for all bounded sequences of complex numbers that will satisfy all of these properties and will reduce to the ordinary limit if the sequence under consideration is convergent?

A rephrasing of this question in func-

tional analytic terms will indicate how we might proceed to answer it.

Note first that the collection of all bounded sequences of complex numbers is a Banach space

le with the norm

((akj E

11(a011- - sup (ak+ k

and that

c

is a closed linear subspace of

in Example 1.2.4. that seen,

limkak

f .

This was mentioned

The preceding discussion says, among other things,

defines a linear functional on

lli,mka.j < II(akIll., (ak) E c.

c,

and, as is easily

Thus the question we have posed is

4.3. Abelian Semigroups of Transformations

95

equivalent to asking whether there exists a (continuous) linear functional on

f

tional on

c

that satisfies the three properties of the limit funcindicated above and agrees with the ordinary limit

functional on

c.

Obviously a straightforward application of the'}lahn-Banach

Theorem (Theorem 4.1.2) with the seminorm

p(ta k)) = U(ak)l1.

(or of Theorem 4.2.2) yields an extension to e m of the limit functional on c that satisfies properties (i) and (ii). In order to

ensure that property (iii) will also hold for the extension obtained we appeal to Theorem 4.3.1.

To this end we define ck = ak+l, k Clearly Moreover,

G

T E L(Q

by

m)

T((ak)) = {ckwhere

(ak) E em,. and set G - (Tn In = 1,2,3,...). is an Abelian semigroup of linear transformations on

if

p((ak)) = jh(ax)JIm, (ak) E em,

pCT((ak))

I

t

m.

then

= JJT((uk))11m

sup Iak

k>2

sup iak

k>1 - 11(a OIL = p((ak)),

and, from property (iii) of the limit functional on then

limkTn((ak)) = limkak, n = 1,2,3,...

.

c,

if

(ak) E c,

Thus all the hypotheses

of Theorem 4.3.1 are fulfilled, and an application of the theorem immediately gives the next result. Theorem 4.3.2. (i)

(ii)

If

(ak) E c,

Given

k = 1,2,3,...,

There exists some thin

n = 1,2,3,..., then

x* E * such that

x*((ak1) = limkak. if

(ak) E Im and

x*((ck)) = x*((ak)).

ck = ak+n,

96

Hahn-Banach Theorem: Analytic Form

4.

The continuous linear functional

obtained in Theorem

t

on

x*

m

We shall take a look at a

4.3.2 is generally called a Banach limit.

different method of obtaining such functionals in Section 9.6. 4.4. Adjoint Transformations.

Suppose

are nontrivial seminormed linear spaces over 4.2.2 we know that

(V2,P2)

Then from Corollary

4.

There exists a simple, and

Vk # (0), k = 1,2.

quite useful, relationship between transformations certain elements in ' L'(V2,Vi).

and

(V1,P1)

T E L(V1,V2)

and

To be precise we make the following

definition:

Definition 4.4.1.

linear spaces over

4,

Let

(V1,P1)

and

let

then we define a mapping

and

Vk f (0), k = 1,2.

T* E L'(V*,V*)

T E L(V1,1'2),

(x1 E V1; X2 E V*),

2

2

being called the adjoint of

T.

It is easy to verify that the definition of

an element of

If

by setting

T*(x* )(xl) = x2[T(xl)]

T*

be seminormed

(V2,P2)

T*

indeed defines

L'(V*,V-).

With the aid of the Hahn-Banach Theorem we can easily establish several results about the adjoint when the spaces

V1

and

V2

are

normed linear spaces. Theorem 4.4.1. spaces over

Let

and suppose

4

(i) T* E L(V*,V*) (ii)

T*

and

T E L(V1,V2).

and

(V2,11.112)

Then

IIT*I1 = IIT11

is injective if and only if

R(T)

(V2.II.112). Proof.

Let

x1 E V1

and

be normed linear

x*2 E V.

Then

IT*(x2)(x1)I = Ix*2[T(xl)II

IIx*llliTUIIxllll,

is dense in

97

4.4. Adjoint Transformations

Thus from Theorem 3.2.1 we see that T* E I,(VZ,V,*) and IIT*II < IITII. Moreover, suppose c > 0 and let xl E VI be such that IIx1II = 1 and IIT(x1)II2 > IITII - a, which is possible by the definition of IITII. Then by Corollary 4.2.5 there exists some x2 E V2* such that 11x211 = 1 and x2[T(x1)] = IIT(xl)II2' But then we have and 30

IIT* (x2) II < IITII IIx2I1.

IT*(x2)(x1)I = Ix2[T(x1)]I

(IT(x1)II

-a

> IITII from which we conclude at once that since

IIx2II = 1,

IIT*(x2)II > IITII

this entails also that

s > 0 is arbitrary, we obtain

IIT*II > IITII

IIT*II > IITII.

However,

- a.

- a,

and since

Thus Theorem 4.4.1(i)

is proved.

In order to establish the equivalence in Theorem 4.4.1(ii) us first suppose that T* it suffices, by Corollary that

we see that

6ince

T*(x2) = 0.

be such that

xl E V1,

shows that

Therefore

T*

is dense in

T*(x2) = 0.

Then

vanishes on

x2

via the continuity of

However, given such an

then x2 = 0.

R(T)

and so

(V 2'11-112 )

x2 = 0..

and let

x*[T(xl)] = T*(x2)(xl) = 0,

R(T),

from which we deduce,

and the denseness of

x

is such

x2 E V*

is injective, it follows that

T*

x2 E VZ

is dense

R(T)

T*(x2)(x1 ) = x2[T(xl)] = 0, xl E V1,

Conversely, suppose

x2 = 0.

4.2.8, to prove that, if

0, x2 E R(T),

x2(x2)

x2 E V*,

To show that

is injective.

let

R(T),

that

is injective, and this completes the proof.

Theorem 4.4.1 easily yields the following corollary.

The

details are left to the reader. Corollary 4.4.1.

Let

a normed linear space over that

(V1,II'II1)

I,

IIT(x)112 > =11x111, x E V1,

are equivalent: (i) 'T (ii)

T*.

is surjective. is injective.

be a Banach space and

and suppose for some

T E L(V1,V2)

m > 0.

is such

Then the following

4.

98

Hahn-Banach Theorem: Analytic Form

We shall return to the notion of the adjoint in subsequent secA

tions, particularly when we discuss Hilbert spaces in Chapter 13.

similar idea will also appear in the investigation of reflexivity of normed linear spaces in Section 8.1. 4.5. Separability of bility of

V*

V

at least in the case that

is a

We recall the following definition:

Definition 4.5.1.

Then

V,

implies that of

normed linear space.

i.

Here we wish to show that the repara-

V*.

(V,T)

Let

be a topological linear space over

is said to be separable if it contains a countable dense

V

subset.

Theorem 4.5.1. If

is separable, then

V*

Proof.

(xn) C V and

be a normed linear space over

Let

Let

is separable.

V

be a countable dense subset of

(x*J

4.

be so chosen, using the definition of

Ixn(xn)I ? IIxnII/2, n = 1,2,3,...

D C V

Let

.

IIxnII,

V*

and let

that

1

IIxnII

denote the family

of all finite linear combinations with (complex) rational coefficients of (xn).

of

V

Clearly

is countable.

D

spanned by

(xn).

Thus it is apparent that dense in

Obviously

is dense in

D

W denote the linear subspace

D C W

particular,

x* E V*

D

is dense in

if and only if

is such that

x*(xn) = 0, n = 1,2,3,...

.

W

W.

is

- x*II.= 0.

limkllxn

x*(x) a 0, x E W.

Since

there exists some sequence from the set

such that

and

V

V.

Suppose then that

V*,

Let

(xn)

(xn),

However,

In

is dense in call it

(x* k

k

Ilxnk - x*II

=

(xnk - x*) (x) { IIxII p

1 I

I (x*nk - x*) (xnk ) I

= Ixn (xnJI k

> Ilxn II/2 k

k

(k - 1,2,3,...),

99

4.6 Annihilators

from which it follows that

= 0.

linkllxn k

implies that

limkllxn

11

=

But

k

and so llx*ll = 0,

lIx*II,

Consequently from Corollary 4.2.8 we conclude that is dense in

Therefore

V.

- X*" = 0

limkllxn

11

that is, x* = 0. W,

and hence

D,

is separable.

V

It should be noted that the converse of this result need not he For example,

valid.

li = .f

is not.

CD

is a separable Banach space, whereas

fI

In this section we wish to introduce the

4.6. Annihilators.

notion of annihilator and to use it for describing the dual spaces of subspaces and quotient spaces of normed linear spaces.

It will be

apparent when we discuss Hilbert spaces in Chapter 13 that the concept of the annihilator is a natural generalization to normed linear spaces of the notion of an orthogonal complement in Euclidean spaces and in inner-product spaces in general. Let

Eppfinition 4.6.1.

over

f.

If

E C V,

E1

then

(x*

x* E V*, x*(x) = 0, x E E)

I

is called the annihilator of (E*)j W (x

I

on

V

EL C V*

E.

If

E* C V*,

then

x E V, x*(x) - 0, x* E E*)

is called the annihilator of Thus

be a topological linear space

(V,T)

E*.

is the set of all continuous linear functionals

that vanish identically on

E,

and

(E*)1 C V

common zeros of the continuous linear functionals on. V to

E*.

It is evident that

E'

and

(E*)1

is the set of that belong

may be rather trivia],

for example, if "V* _ (0).

The proof of the next proposition is straightforward and is left to the reader.

Hahn-Banach Theorem: Analytic Form

4.

100

Proposition 4.6.1. over

f,

and

E C V,

El C V*

is a normed linear space, then

V

If

E* c :V*.

Then

is a linear subspace.

F.l C V*

(i)

(ii)

be a topological linear space

(V,T)

Let

is a closed

linear subspace. (iii)

is a closed linear subspace.

c V

(E*) A.

(iv) (v)

E C (E)t.

1

V T (0),

is a normed linear space,

V

If

E CV

and

El # (0).

is a proper linear subspace, then

Among other things, the precedirg proposition says that E C (El)l

whenever

The next result gives us a condition

E c :V.

under which the containment is equality. Theorem 4.6.1.

W C V

If

be a normed linear space over

Let

is a linear subspace, then Since

Proof.

c,t!W` C (W,)l.

W C (WL)t

But if

and

cl(W) = (WL)1 is closed, we see that

(Wl)1

X0 E (W')1 - cl(W),

then,

since

cl(W)

i- a closed lirear subspace, we see from Theorem 4.2.3 that there x' E V*

exists some x t cl(W;. that

such that

In particular,

x*(xo) = 0, T'aerefore

x*(xo) # 0

x* E W.

x*(x) - 0,

and

However,

x

0

E (Wl)1

thereby contradicting the choice of

implies

x*.

cl(W) _ (W). C)

Corollir n

and let

42.6.1.

V

Let

(V,11.11)

be a norm-ad linear space over

be a linear subspace.

Then the following are

equivalent: (i)

(ii)

W

is a closed linear subspace.

W - (W1)1

Next we shall use the concept of an annihilator to describe the spaces of continuous linear functionals on the subspaces and quotient spaces of normed linear spaces.

101

4.6. Annihilators

First we make a general definition. Definition 4.6.2. linear spaces over isometry if

Let

(Vl,ll-ji

y

A mapping

1.

be normed

and

)

V1 - V2

:

is said to be an

Such a mapping is also said to

IIcP(x)ji2 = Wxljl, x E V1.

be isometric.

Theorem 4.6.2. and let

W C V

be a normed linear space over

Let

be a closed linear subspace.

jective isometric isomorphisms between (V/W)*

and

Proof.

If

cp

cp

V*/Wl - W*,

Moreover, suppose

is an isomorphism.

x*(x)

and between

y*(x), x E W.

defined by It is easily seen

is well defined.

Theorem 4.2.2, there exists some and

W*,

defines a continuous linear functional :

cp(x* + WL)(x) = x*(x), x E W, that

and

then clearly the equation

x* + Wl E V*/WL,

Thus the mapping

W.

V*/Wi

Wi.

(x* + Wl)(x) = x*(x), x E W, on

I

Then there exist sur-

x* E V*

y* E W*.

such that

IIx*ll = IIY*II


Clearly,

Then, by

and so

tf

is surjective.

Furthermore, the preceding argument, combined with the definition of the norm in cp(x* + WL) = y*,

V*/Wl

(Theorem 1.1.1), reveals that, if

where

x* E V*

y* E W*

is that chosen as in the preceding

paragraph, then

IlY*II = Ilx*II ? IIIx* +

that is,

Il,(x* + Wl) II' > Ill x* + W"III

Y* = (P(x* + W1).

On the other hand, if

.

then IIY*11 =

and

sun IY*(x)I

xEW

llxll= 1 sup

(x* + Z*)(x)

xEW IixII = 1

<_ IIx* + Z*ll

(Z* E W1)

4.

102

Hahn-Banach Theorem: Analytic Form

IIW(x' + 't)II <

and hence Therefore

+

IIIx'`

III

is an isometry, and the first assertion of the

cp

theorem is proved.

The second assertion is even easier to prove and is left to the reader.

0

Analogous results are also valid for locally convex topological linear spaces (see, for example, [K, pp. 276-279]). 4.7.

Ideals in

L1 R,dt).

As noted in Example 1.2.4, the space

is a Banach space, where as usual dt denotes

(L

Lebesgue measure on

!R.

It is also possible to introduce an operation

L1(Rdt)

of multiplication into

so that

becomes an algebra.

L1(1R,dt)

The main purpose of this section is to use the Hahn-Banach Theorem to give a description of the closed ideals in this algebra. First, however, we must see what the multiplication in is to be.

L1(IR,dt)

The following definition and proposition are more general

than needed for our present purposes, but we shall have use for these more general results in subsequent sections. Definition 4.7.1.

tions defined on

IR.

Let

f

and

h

be Lebesgue-measurable func-

Then the convolution

defined formally for almost all

f * h

of

f

and

h

t EIR by the expression

f + h(t) = J'JRf(t - s)h(s) ds. Various properties of convolution are collected in the next proposition.

Proposition 4.7.1. 1 < p < w.

Let

f,g E L1(IR,dt)

Then

(i) f hEL (iR,dt).

(iii

Ilf * hllp

Uflil IIhIIp.

and

h E Lp(lR,dt),

is

4.7.

Ideals in LI (IR,dt).

103

(iii) f * (g * h) _ (f * g) * h. (iv)

(f+g) *h =f *h+g*h.

a(f * h) _ (af) * h = f * (ah) (vi) f * h = h * f. (v)

(a E C).

Moreover, if f E L ("tit) and h E L OR,dt), 1/p + 1/q = I, P I < p < m, then f * g E C(IR,dt) and q IIf * hL <_ IIflip Ilhllq We do not give the proof of this result, but instead refer the reader to (HS, pp. 396-3991.

The proof of parts (i) and (ii) of the

proposition is nontrivial and is obtained essentially as a consequence of Fubini's Theorem [Ry, p. 269].

Parts (i) and (iii) through (vi)

of the proposition say precisely that

is a commutative

L1(IR,dt)

algebra over C with convolution as multiplication. incidentally, that the algebra

LI(IR,dt)

Part (ii) says,

is a Banach algebra, but

this does not concern us here.

A few more preliminary notions are required. Definition 4.7.2. For each then we define

T

f

s E IIR,

if

f

is a function on

IR,

by the equation

s

(TSf) (t) = f (t

- s)

(t E IR).

If one makes the usual agreement about identifying Lebesgue measurable functions that are equal almost everywhere, then it is easily seen that

TS E L(Lp(IR, dt)), s E IR, 1 < p < w,

TS (f * g) = (TS f) * g = f * (TSg) 1