A Short Course on Banach Space Theory

N L Carothers

Department of Mathematics and Statistics

Bowling Green State University Summer ii

Contents

Classical Banach Spaces

The sequence spaces and c

Finite dimensional spaces

The L spaces

The C K spaces

Hilb ert space

Neoclassical spaces

The big questions

Notes and Remarks

Exercises

Preliminaries

Continuous linear op erators

Finite dimensional spaces

Continuous linear functionals

Adjoints

Pro jections

Quotients

A curious application

Notes and Remarks

Exercises

Bases in Banach Spaces

Schauders basis for C

The Haar system

Notes and Remarks

Exercises

Bases in Banach Spaces I I

Awealth of basic sequences iii

iv CONTENTS

Disjointly supp orted sequences in L and

p p

Equivalent bases

Notes and Remarks

Exercises

Bases in Banach Spaces I I I

Blo ck basic sequences

Subspaces of and c

Complemented subspaces of and c

Notes and Remarks

Exercises

Sp ecial Prop erties of c and

True stories ab out

The secret life of

Confessions of c

Notes and Remarks

Exercises

Bases and Duality

Notes and Remarks

Exercises

L Spaces

Basic inequalities

Convex functions and Jensens inequality

A test for disjoin tness

Conditional exp ectation

Notes and Remarks

Exercises

L Spaces I I

The Rademacherfunctions

Khinchines inequality

The KadecPelczynski theorem

Notes and Remarks

Exercises

CONTENTS v

L Spaces I I I

Unconditional convergence

Orliczs theorem

Notes and Remarks

Exercises

Convexity

Strict convexity

Nearest p oints

Smo othness

Uniform convexity

Clarksons inequalities

An elementary pro of that L L

Notes and Remarks

Exercises

C K Spaces

The Cantor set

Completely regular spaces

Notes and Remarks

Exercises

Weak Compactness in L

Notes and Remarks

Exercises

The DunfordPettis Prop erty

Notes and Remarks

Exercises

C K Spaces I I

The StoneCec h compactication

Return to C K

Notes and Remarks

Exercises

C K Spaces I I I

The StoneCech compactication of a discrete space

A few facts ab out N

Top ological measure theory

The dual of

vi CONTENTS

The Riesz representation theorem for C D

Notes and Remarks

Exercises

ATop ology Review

Separation

Lo cally compact Hausdor spaces

Weaktopologies

Pro duct spaces

Nets

Notes and Remarks

References

Index

Chapter

Classical Banach Spaces

To b egin recall that a Banach space is a complete normed linear space That

is a Banach space is a normed vector space X kkwhich is a complete

metric space under the induced metric dx y kx y k Unless otherwise

sp ecied well assume that all vector spaces are over R although from time

to time wewillhave o ccasion to consider vector spaces over C

What follows is a list of the classical Banach spaces Roughly translated

this means the spaces known to Banach Once wehave these examples out in

the op en well haveplenty of time to ll in any unexplained terminologyFor

now just let the words wash over you

The sequence spaces and c

Arguably the rst innite dimensional Banach spaces to b e studied were the

sequence spaces and c To consolidate notation we rst dene the vector

space s of all real sequences x x and then dene various subspaces of s

For each p we dene

kxk jx j

p n

and wetake to b e the collection of those x s for which kxk The

p p

inequalities of Holder and Minkowski show that is a normed space from

there its not hard to see that is actually a Banach space

The space is dened in exactly the same wayforp but in this

case kk denes a complete quasinorm That is the triangle inequalitynow

CHAPTER CLASSICAL BANACH SPACES

holds with an extra constant sp ecically kx y k kxk ky k Its

p p y

worth noting that dx y kx y k denes a complete translation invariant

metric on for p

For p we dene to b e the collection of all b ounded sequences

that is consists of those x s for which

kxk sup jx j

Its easy to see that convergence in is the same as uniform convergence

on N and hence that is complete There are twov ery natural closed

subspaces of The space c consisting of all convergent sequences and the

space c consisting of all sequences converging to Its not hard to see that

c and c are also Banach spaces

As subsets of s wehave

c c

p q

for anypq Moreover each of the inclusions is normone

kxk kxk kxk kxk

p q

Its of some interest here to p oint out that while s is not itself a normed space

it is at least a complete metric space under the socalled Frechet metric

jx y j

n n

dx y

jx y j

n n

Clearlyconvergence in s implies co ordinatewise convergence

Finite dimensional spaces

We will also have o ccasion to consider the nitedimensional versions of the

n n n

spaces We write to denote R under the norm That is is the space

p p

of all sequences x x x of length n and is supplied with the norm

kxk jx j

p i

for p and

kxk max jx j

for p

Recall that all norms on R are equivalent In particular given any norm

kk on R we can nd a p ositive nite constant C suchthat

C kxk kxk C kxk

n n

for all x x x inR Thus convergence in anynormon R is the same

as co ordinatewise convergence and hence every norm on R is complete

Since every nitedimensional normed space is just R in disguise it

follows that every nitedimensional normed space is complete

The L spaces

We rst dene the vector space L to b e the collection of all equivalence

classes under equality almost everywhere of Leb esgue measurable functions

f RFor our purp oses L will serv e as the measurable analogue

of the sequence space s

For p the Banach space L consists of those f L

for which

kf k jf xj dx

The space L consists of all essentially b ounded f L under

the essential supremum norm

kf k esssup jf xj inffB jf jB ae g

in practice though we often just write sup in place of esssup Again

t role here the inequalities of Holder and Minkowski play an imp ortan

As b efore the spaces L are also dened for p but kk

p p

denes a complete denes only a quasinorm Again df gkf g k

translation invariant metric on L for p The space L is given

the top ology of convergence lo cally in measure For Leb esgue measure on

this top ology is known to b e equivalenttothatgiven by the metric

jf x g xj

df g dx

jf x g xj

As subsets of L wehave the following inclusions

L L L L

p q

CHAPTER CLASSICAL BANACH SPACES

for anypq Moreover the inclusion maps are all normone

kf k kf k kf k kf k

p q

The spaces L R are dened in much the same way but satisfy no such

inclusion relations That is for any p q wehave L R L R Never

p q

theless you may nd it curious to learn that L R and L are linearly

p p

isometric

More generallygiven a measure space X we might consider the

space L consisting of all equivalence classes of measurable functions

f X R under the norm

kf k jf xj dx

with the obvious mo dication for p

It is convenient to consider at least one sp ecial case here Given anyset

we dene L where is counting measure on What this

p p

means is that we identify functions f R with sequences x x in

the usual way x f and we dene

p p

jf kxk jx j j d kf k

p p

for p Please note that if x then x for all but countably

many For p we set

kxk sup jx j supjf j kf k

We also dene c to b e the space of all those x for which the set

f jx j g is nite for any Again this forces an elementof c to

have countable supp ort Clearly N and c Nc

p p

A priori the Banach space characteristics of L will dep end on the un

derlying measure space X As it happ ens though Leb esgue measure

on and counting measure on N are essentially the only two cases wehave

to worry ab out It follows from a deep result in abstract measure theory Ma

harams theorem that every complete measure space can b e decomp osed

into nonatomic parts copies of and purely atomic parts counting

measure on some discrete space From a Banach space p oint of view this

means that every L space can b e written as a direct sum of copies of L

p p

and or

Forthemostpartwe will divide our eorts here into three avenues of

attack Those prop erties of L spaces that dont dep end on the underlying

measure space those that are p eculiar to L and those that are p eculiar

to the spaces

The C K spaces

Perhaps the earliest known example of a Banach space is the space C a b of

all continuous realvalued functions f a b R supplied with the uniform

norm

kf k max jf tj

atb

More generallyifK is any compact Hausdor space wewriteC K to denote

the Banach space of all continuous realvalued functions f K R under the

norm

kf k maxjf tj

For obvious reasons we sometimes write the norm in C K as kf k and refer

to it as the sup norm In any case convergence in C K is the same as

uniform convergence on K

In Banachs day p oint set top ology was still very much in its developmental

stages In his b o ok Banach considered C K spaces only in the case of

compact metric spaces K We on the other hand mayhave o ccasion to

venture further At the very least we will consider the case where K is a

compact Hausdor space since the theory is nearly identical in this case

And if we really get ambitious wemaydelveinto more esoteric settings For

the sake of future reference here is a brief summary of the situation

X to denote the algebra of If X is any top ological space wewriteC

all realvalued continuous functions f X R For general X though

C X may not b e metrizable If X is Hausdor and compact say X

K thenC X is a complete metric space under the top ology of uniform

convergence on compacta or the compactop en top ology This top ology

is generated by the socalled Frechet metric

kf g k

df g

kf g k

where kf k is the norm of f j in C K

n K n n

CHAPTER CLASSICAL BANACH SPACES

If we restrict our attention to the bounded functions in C X then wemay

at least apply the sup norm for this reason we consider instead the Banach

space C X of all b ounded continuous realvalued functions f X R

endowed with the sup norm

kf k supjf xj

Obviously C X is a closed subspace of X If X is at least completely

regular then C X contains as much information as C X itself in the sense

that the top ology on X is completely determined byknowing the b ounded

continuous realvalued functions on X

If X is noncompact then we might also consider the normed space C X

of all continuous f X R with compact support That is f C X iff is

continuous and if the support of f namely the set

supp f f x X f x g

is compact While wemay apply the sup norm to C X its not in general

complete The completion of C X is the space C X consisting of all those

continuous f X R whichvanish at innity Sp ecically f C X iff

is continuous and if for each the set fjf j g has compact closure

The space C X is a closed subspace of C X hence is a Banach space under

the sup norm

If X is compact then of course C X C X C X For general X

C b

howev er the b est we can sayis

C X C X C X C X

C b

At least one easy example might b e enlightening here Consider the case

X Nobviously N is lo cally compact and metrizable Now every function

f N R is continuous and any such function can quite plainly b e identied

with a sequence namely its range f n That is we can identify C N with

s byway of the corresp ondence f C N x s where x f n

Convince yourself that

Nc C N R c C N C

and that

C Nfx s x for all but nitely many ng

C n

While this is curious it do esnt quite tell the whole story Indeed b oth and

c are actually C K spaces To get a glimpse into why this is true consider

the space N N fg the one point compactication of N that is we

appendapoint at innity If we dene a neighborhood of to b e any

set with nite compact complement then N b ecomes a compact Hausdor

space Convince yourself that

c C N and c ff C N f g

Well havemoretosay ab out these ideas later

Hilb ert space

As youll no doubt recall the spaces and L are b oth Hilb ert spaces or

complete inner pro duct spaces Recall that a vector space H is called a Hilb ert

space if H is endowed with an inner pro duct h i with the prop ert ythatthe

induced norm dened by

kxk hx xi

is complete Most imp ortant here is to recognize that the norm in H is inti

mately related to an inner pro duct byway of This is a tall order for

the run of the mill norm From this p oint of view Hilb ert spaces are quite

rare among the teeming masses of Banach spaces

There is a critical distinction to b e made here p erhaps an example will

help to explain Let X denote the space supplied with the norm kxk

kxk kxk ThenX is isomorphic linearly homeomorphic to b ecause

our new norm satises kxk kxkkxk But X is not itself a Hilb ert

space The test is whether the paral lelogram law holds

kxk ky k kx y k kx y k

k that the parallelogram law fails if x and And its easy to chec

y for instance The moral here is that its not enough to have

awelldened inner pro duct nor is it enough to have a norm that is close

to a known Hilb ert space norm In a Hilb ert space the norm and the inner

pro duct are inextricably b ound together through equation

Hilb ert spaces exhibit another prop erty that is rare among the Banach

spaces In a Hilb ert space every closed subspace is the range of a continuous

pro jection This is far from the case in a general Banach space In fact it

is known that any space with this prop erty is already isomorphic to Hilb ert space

CHAPTER CLASSICAL BANACH SPACES

Neoclassical spaces

Wehave more or less exhausted the list of spaces that were well known in

Banachs time But wehaveby no means even b egun to list the spaces that are

commonplace these days In fact it would take pages and pages of denitions

to do so For nowwell content ourselves with the understanding that all of

the known examples are in a sense generalizations of the spaces wehave seen

so far As the need and the opp ortunity arises wemayhave o ccasion to

sp eak of some of these other spaces

The big questions

Were typically interested in b oth the isometric as well as the isomorphic

character of a Banach space For our purp oses all isometries are linear

Also as a reminder an isomorphism is a linear homeomorphism Here are

just a few of the questions wemight consider

Are all the spaces listed ab ove isometrically distinct For example is it

at all p ossible that and are isometric What ab out and L Or

p p

L and L R

p p

When is a given Banac h space X isometric to a subspace of one of the

classical spaces When do es X contain an isometric copy of one of the

classical spaces In particular do es L emb ed isometrically into L

Do es emb ed isometrically into C

Wemight p ose all of the same questions ab ove replacing the word iso

metric with isomorphic

Characterize all of the subspaces of a given Banach space X if p ossi

ble b oth isometrically and isomorphically In particular identify those

subspaces which are the range of a continuous pro jection that is the

complemented subspaces of X Knowing all of the subspaces of a given

space would tell us something ab out the linear op erators into or on the

space and vice versa After all the kernel and range of a linear op erator

are subspaces

All of the spaces weve dened ab ove carry some additional structure

C a b is an algebra of functions for example and L is a lattice

What if anything do es this extra structure tell us from the p ointofview

of Banach space theory Is it an isometric invariant of these spaces An

isomorphic invariant Do es it imply the existence of interesting sub

spaces Or interesting op erators

Its probably fair to say that functional analysis concerns the study of

operators between spaces Insert the adjective linear wherever p ossi

ble and you will havea working denition of linear functional analysis

Where do es the study of Banach spaces t into the bigger picture of

linear functional analysis In other words do es a b etter understanding

of Banach spaces tell us anything ab out the op erators b etween these

spaces

Go o d mathematics do esnt exist in a vacuum We also wanttokeep an

eye out for applications of the theory of Banach spaces to mainstream

analysis Converselywewillwant to b e on the lo ok out for applications

of mainstream to ols to the theory of Banach spaces Among others we

will lo ok for connections with probability harmonic analysis top ology

op erator theory and plain ol calculus By way of an example wemight

consider the calculus of vectorvalued functions f X where

X isaBanach space It would make p erfect sense to ask whether f is of

bounded variation for instance Or whether kf xk dx Well

put these tantalizing questions aside until were b etter prepared to deal

with them

Notes and Remarks

The space C a b is arguably the oldest of the examples presented here but

it was Frechet who oered the rst systematic study of the space as a metric

space b eginning in The space was intro duced in by Erhard

Schmidt of the GramSchmidt pro cess The space that b ears his name

held little interest for Hilb ert bytheway Hilb ert preferred the concrete

setting of integral equations to the abstractions of Hilb ert space theory

Schmidts pap er is notable in that it is b elieved to contain the rst app ear

ance of the familiar doublebar notation for norms Both the notation

and the attribution Hilb ert space though are due to Frigyes Riesz In fact

Riesz intro duced the L spaces and he Frechet and Ernst Fischer noticed

their connections with the spaces Although many of these ideas were in

theairforanumber of years it was Banach who launched the rst compre

hensive study of normed spaces in his dissertation culminating in his

b o ok For more on the prehistory of functional analysis and in partic

ular the development of function spaces see the detailed articles byMichael

CHAPTER CLASSICAL BANACH SPACES

Bernkopf the writings of A F Monna and the excellent

chapter notes in Dunford and Schwartz

For much more on the classical and neoclassical Banach spaces see

the b o oks by Adams Bennett and Sharpley Dunford and Schwartz

Duren Lacey Lindenstrauss and Tzafriri and Woj

taszczyk For more on the history of op en questions and unresolved issues

in Banach space theory see Banachs b o ok its review by Diestel and

its English translation with notes by Bessaga and Pelczynski see also Day

Diestel Diestel and Uhl Megginson and the articles by

Casazza Mascioni and Rosenthal

Exercises

If X kkisany normed linear space show that the op erations x y

x y and x x are continuous on X X and R X resp ectively

It do esnt much matter what norms we use on X X and R X for

example kx y k kxk ky k works just ne Why If Y is a linear

subspace of X conclude that Y is again a subspace

Show that X kk is complete if and only if every absolutely summable

series is summable that is if and only if kx k always implies

that x converges in the norm of X

ShowthatC the space of functions f R having a

continuous rst derivative is complete under the norm kf k kf k

kf k

Show that s is complete under the Frechet metric

Show that is not separable

Given p showthatkf g k kf k kg k for f g L A

p p p p

b etter estimate with a slightly harder pro of yields the constant

in place of

Let p and let p q Show that for p ositive real

p q

numb ers a and b wehave ab a p b q with equality if and only if

a bFor p and q show that the inequalityreverses

Let p and let p q If f and g are nonnnegative

R R R R

q p p q q

functions with f L and g show that fg f g

Given p and nonnegative functions f g L show that

kf g k kf k kg k

p p p

Prove the string of inequalities for x

Prove the string of inequalities for f L

Given p q p q showthatL R L R

p q

Given a compact Hausdor space X show that C X is a closed sub

space of C X and that C X isdenseinC X

b C

Let H b e a separable Hilb ert space with orthonormal basis e and let

K b e a compact subset of H Given show there exists an N such

that k h x e i e k for every x K ThatisifK is compact

n n

then these tail series can b e made uniformly small over K

CHAPTER CLASSICAL BANACH SPACES

Chapter

Preliminaries

We b egin with a brief summary of imp ortant facts from functional analysis

some with pro ofs some without Throughout X Y etc are normed linear

spaces over R If there is no danger of confusion wewillusekk to denote

the norm in anygiven normed space if two or more spaces enter into the

discussion we will use kk etc to further identify the norm in question

Continuous linear op erators

Given a linear map T X Y recall that the following are equivalent

i T is continuous at X

ii T is continuous

iii T is uniformly continuous

iv T is Lipschitz that is there exists a constant C suchthat kTx

Tyk C kx y k for all x y X

Y X

v T is bounded that is there exists a constan t C suchthat kTxk

C kxk for all x X

If a linear map T X Y is b ounded then there is in fact a smal lest

constant C satisfying kTxk C kxk for all x X Indeed the constant

Y X

kTxk

sup kT k sup kTxk

kxk

X x

kxk

CHAPTER PRELIMINARIES

called the norm of T works that is it satises kTxk kT kkxk and its

Y X

the smallest constanttodosoFurther its not hard to see that actually

denes a norm on the space B X Y of all b ounded continuous linear maps

T X Y

A map T X Y is called an isometry into if kTx Tyk kx y k

Y X

for all x y X Clearlyevery isometry is continuous If in addition T is

linear then its only necessary to checkthat kTxk kxk for every x

Y X

X Throughout these notes unless otherwise sp ecied the word isometry

always means linear isometry Its a curious fact due to Banach and Mazur

that any isometry mapping to is actually linear Please note that

an onto isometry is inv ertible and that the inverse map is again an isometry

A linear map T from X onto Y is called an isomorphism if T is onetoone

and b oth T and T are continuous That is T is an isomorphism onto if T

is a linear homeomorphism It follows from our rst observation that T is an

isomorphism if and only if T is onto and there exist constants c C

suchthat ckxk kTxk C kxk for all x X Ifwe drop the requirement

X Y X

that T be onto then this pair of inequalities denes an into isomorphism that

is an isomorphism onto the range of T Note that if X is a Banach space

then every isomorph of X is necessarily also a Banach space It follows from

the Op en Mapping theorem that a onetoone onto linear map T dened on

aBanachspaceX is necessarily an isomorphism

Phrases suchasX is isometric to Y or Y contains an isomorphic copy

of X should b e selfexplanatory

Finite dimensional spaces

Every nite dimensional normed space is just R in disguise To see this we

rst check that all norms on R are equivalent To this end let kk be any

norm on R We will nd constants c C suchthat

c kxk kxkC kxk

for all x R

n n

a e R Let e e b e the usual basis for R Then given x

i i n

wehave

n n n

X X X

ja jke k max ke k ja j C kxk e kxk a

i i i i i i

i i i

where C max ke k The hard work comes in establishing the other

in i

inequality

Now the inequalitythatweve just established shows that the function kk

is continuous on R kk Indeed

kxk ky k kx y kC kx y k

Hence its restriction to S f x kxk g is likewise continuous And since

S is compact k k must attain a minimum value on S What this means is

that there exists some constant c suchthatkxkc whenever kxk Thus

for any x R wehave kxk c kxk by homogeneity Since wemay assume

that the value c is actually attained wemust have c andsowere done

Next supp ose that X is any nite dimensional normed space with basis

x x Then the basistobasis map x e extends to a linear isomor

n i i

n n

phism from X onto R Indeed if we dene a new norm on R by setting

n n

X X

a x a e

i i i i

i i

then kk must b e equivalent to the usual norm on R Hence

n n n

X X X

a x C a e a x c

i i i i i i

i i i

X X

for some constants c C

As an immediate corollarywe get that every nite dimensional normed

space is complete And in particular if X is a nite dimensional subspace of

a normed space Y thenX must b e closed in Y

Related to this is the fact that a normed space X is nite dimensional if

and only if B f x kxk g the closed unit ball in X iscompact The

forward implication is obvious For the backward implication we app eal to

Rieszs lemma For each closed subspace Y of X and each there

exists a norm one vector x x X such that kx y k for all y Y

Thus if X is innite dimensional we can inductively construct a sequence of

norm one vectors x inX such that kx x k for all n m

n n m

Perhaps not so immediate is that every linear map on a nite dimensional

spaceiscontinuous Indeed supp ose that X is nite dimensional and that

is a basis for X Then from and our previous discussion x x

n n

X X

a x ja jC

i i i

i i X

CHAPTER PRELIMINARIES

for some constant CThus if T X Y is linear wehave

n n n

X X X

T a x ja jkTx k C max kTx k a x

i i i i Y i Y i i

i i i

Y X

C max kTx k That is kTxk K kxk where K

in i Y Y X

Continuous linear functionals

A scalarvalued map f X R is called a functional From our earlier

observations a linear functional is continuous if and only if its b ounded that

is if and only if there is a constant C suchthat jf xjC kxk for every

x X Iff is b ounded then the smallest constant C whichwillwork in this

inequality is dened to b e the norm of f again written kf k Sp ecically

jf xj

kf k sup sup jf xj

kxk

In particular note that jf xjkf kkxk for any f X x X

The collection X of continuous linear functionals on X canbeidentied

with the collection of b ounded in the ordinary sense continuous linear in

a restricted sense realvalued functions on B the closed unit ball of X

Indeed its easy to check that under this identication X is a closed subspace

of C B In particular X is always a Banach space In language b orrowed

b X

from linear algebra X is called the dual space to X or sometimes the norm

dual to distinguish it from the algebraic dual the collection of all linear

functionals continuous or not

ypically not The word dual may not b e the b est choice here for it is t

true that the second dual space X X can b e identied with X If for

example X is not complete then there is little hop e of X b eing isometrically

identical to the complete space X Nevertheless X is always isometric to a

subspaceof X Thisisavery imp ortant observation and is worth discussing

in some detail

Each element x X induces a linear functional xb on X byway of p oint

evaluation xbf f x for f X Using the inner pro duct notation

f xh x f i to denote the action of a functional f onavector x note that

the functional xb satises

h f xb i h x f i

Its clear that xb is linear in f Continuity is almost as easy jxbf j

jf xjkxkkf k Thus xb X and k xb kkxk But from the Hahn

Banach theorem we actually have k xb k kxk Indeed given any x

X the HahnBanach theorem supplies a norming functional A normone

element f X suchthat jf xj kxk All that remains is to note that the

corresp ondence x xb is also linear Thus the map x xb is actually a linear

isometry from X into X

We will also have o ccasion to write the hat map in more traditional terms

using the op erator i X X dened byixf f x In terms of the

inner pro duct notation then ix satises

h f ix i h x f i

We will write X to denote the image of X in X under the canonical hat

emb edding As a closed subspace of a complete space its clear that X the

b b

X is a completion of X closure of X in X is again complete It follows that

This is an imp ortant observation For one wenowknow that every normed

space has a completion which is again a normedspace and for another since

X is always isometric to a subspace of a complete space theres rarely any

harm in simply assuming that X itself is complete Note for example that

X and X necessarily have the same dual space

If it should happ en that X X wesaythat X is reexive Its imp ortant

to note here that this requirementismuch more stringent than simply asking

that X and X b e isometric Indeed there is a famous example due to

R C James of a Banach space J with the prop erty that J and J are

isometrically isomorphic byway of an unnatural map and yet J is a proper

closed subspace of J

Adjoints

Eachcontinuous linear map T X Y induces a continuous linear map

T Y X called the adjoint of T Indeed given f Y the comp osition

f f T Note that comp osition f T denes an elementof X We dene T

with T is linear That T is continuous is easy Clearly kT f kkT kkf k

and hence kT kkT kTo see that kT k kT k actually holds rst cho ose

a norm one vector x X suchthatkTxk kT k and then cho ose a norm

on functional f Y such that T f xf TxkTxkThen

kT kkT f kT f xkTxk kT k

CHAPTER PRELIMINARIES

Equivalently T is dened by the formula

h x T f i h Txf i

for every x X and f Y In this notation its easy to see that T reduces

to the familiar conjugate transp ose of T in the case of matrix op erators

n m

between R and R

Of course it also makes sense to consider T X Y where T

T Convince yourself that T is an extension of T that is under the

usual convention that X X and Y Y wehave T j T That

k T k kT k follows from our previous remarks

We will make o ccasional use of the following result for a pro of see

Theorem Theorem or Theorem

Theorem Let T X Y beabounded linear map between Banach spaces

X and Y Then

i T is onto if and only if T is an isomorphism into

ii T is an isomorphism into if and only if T is onto

Pro jections

Given subspaces M and N of a vector space X we write X M N if its

the case that each x X can b e uniquely written as x y z with y M and

z N In this case wesaythatX is the direct sum of M and N or that M

and N are c omplements in X Given a single subspace M of X wesay that

M is complemented in X if we can nd a complementary subspace N that is

if we can write X M N for some N Its easy to see that complements

need not b e unique

If X M N then we can dene a map P X X by Px y where

x y z asabove Uniqueness of the splitting x y z shows that P is

welldened and linear Clearly the range of P is M infactP P hence

P j is the identityon M Equally clear is that N is the kernel of P In short

P is a linear projection an idemp otent P P on X with range M and

kernel N

Converselygiven a linear pro jection P X X its easy to see that we

can write X M N whereM is the range of P andN is the kernel of P

Indeed x Px I P xand P I P x Px P x for any x X

While were at it notice that Q I P is also a pro jection the range of Q

is the kernel of P thekernel of Q is the range of P As b efore though there

are typically many dierent pro jections on X with a given range or kernel

In summary given a subspace M of X nding a complementfor M in

X is equivalent to nding a linear pro jection P on X with range M But

were interested in normed vector spaces and continuous maps If X is a

normed linear space when is a subspace M the range of a continuous linear

pro jection Well a moments reection will convince you that the range of a

continuous pro jection must b e closed Indeed if Px y then bycontinuity

P x Px PyThus w emust have Py y that is y must b e in the

n n

range of P Of course if P is continuous then N ker P is also closed

ConverselyifM is closed and if we can write X M N for some

closed subspace N of X then the corresp onding pro jection P with kernel N

is necessarily continuous This follows from an easy application of the Closed

Graph theorem Supp ose that x x and Px y SincePx M and

n n n

since M is closed wemust have y M too Nowwe need to show that

y Px and for this it suces to showthatx y N But x Px

n n

I P x x y and I P x N for each N Thus since N is also

n n

closed wemust have x y N Please note that we needed both M and N

to b e closed for this argument

Henceforth given a closed subspace M of a normed linear space X we will

say that M is complemented in X if there is another closed subspace N such

that X M N Equivalently M is complemented in X if M is the range of

a continuous linear pro jection P on X It will take some time b efore we can

ll in the details but you may nd it enligh tening to hear that there do exist

closed uncomplemented subspaces of certain Banach spaces In fact outside

of the Hilb ert space setting nontrivial pro jections are often hard to come by

ExampleIfM is a one dimensional subspace of a normed space X thenM

is complemented in X

ProofGiven y M the HahnBanach theorem provides a continuous lin

ear functional f X with f y CheckthatPx f xy is a continuous

linear pro jection on X with range M

Quotients

Given a subspace M ofavector space X we next consider the quotient space

XM The quotient space consists of all cosets or equivalence classes x

x M wherex X Two cosets x M and y M are declared equal if

CHAPTER PRELIMINARIES

x y M that is x and y are pro claimed equivalentifx y M and so

we identify xandy in this case to o Addition and scalar multiplication

are dened in the obvious way x y x y and ax ax Its easy

to check that these op erations make XM avector space with zero vector

M We also dene the quotient map q X XM by q x x

Under the op erations weve dened on XM its clear that q is linear and

necessarily onto with kernel M

Again our interest lies in the case where X is a normed space In this case

wewanttokno w whether there is a natural way to dene a norm on XM

and whether this norm will make the quotientmapq continuous Wetake the

easy way out here We knowthatwewant q to b e continuous so we will force

the norm on XM to satisfy the inequality kq xk kxk for all x X

But q x q x y forany y M thus we actually need the norm on XM

to satisfy kq xk kx y k for all x X and all y M This leads us

to dene the quotient norm on XM by

kq xk inf kx y k

XM X

y M

That is kq xk dx M the distance from x to M Given this we

evidently have kq xk precisely when x M Thus there is little hop e

of the quotient norm dening anything more than a pseudonorm unless we

also insist that M be a closed subspace of X And thats just what well do

Now most of what we need to check follows easily from the fact that M is

a subspace For example since M we clearly have kq xk kxk

kaq xk kq axk

XM XM

inf kax y k

y M

inf kax y k jajkq xk

y M

y is not much harder Given x x X and cho ose The triangle inequalit

y y M with kx y k kq xk and kx y k kq x k

X X

XM XM

Then since y y M weget

kq x q x k kq x x k

XM XM

kx x y y k kx y x y k

X X

kq xk kq x k

XM XM

What weve actually done here is to takethequotient topology on XM

induced by q that is the smallest top ology on XM making q continuous

o o o

where B denotes the open unit ball in X Indeed check that q B B

X X

Thusa setisopeninXM if and only if it is the image under q of an op en

set in X In particular note that q is an open map

We should also address the question of when XM is complete If X is

complete and if M is closed hence complete then its not terribly hard to

check that XM is also complete Indeed supp ose that x is a sequence

in X for which kq x k For each ncho ose y M such

n n

that kx y k k q x k Then kx y k and

n n X n XM n n X

P P

q x y x y converges in X Thus bycontinuity hence

n n n n

n n

q x converges in XM

Since were using the quotient top ology itseasytocheck that a linear

map S XM Y is continuous if and only if Sq is continuous Said in

other words the continuous linear maps on XM come from continuous linear

maps on X which factor through XM That is if T X Y satises

ker T M then there exists a unique linear map S XM Y which

satises Sq T and kS k kT k In the sp ecial case where T maps a Banach

space X onto a Banach space Y and M ker T it follows that the map S is

onetoone thus Xker T is isomorphic to Y

Finally its a natural question to ask whether the quotient space XM

is actually isomorphic to a subspace of X Now in the vector space setting

its easy to see that if N is an algebraic complementof M in X then XM

can b e identied with N Thus it should come as no surprise that XM is

isomorphic to a subspace of X whenever M is complemented in X

If X is a Banach space with X M N andifwe write Q X X for the

pro jection with kernel M then our earlier observations show that Xker Q

XM is isomorphic to N Conversely if the quotientmapq X XM is an

isomorphism on some closed subspace N of X thenQ q j q considered

as a map from X to X denes a pro jection with range N

In the sp ecial case of linear functionals these observations tell us that

XM M the annihilator of M in X That is the dual of XM

can b e identied with those functionals in X whic hvanish on M Indeed

if f XM R is a continuous linear functional then g fq X R

is a continuous linear functional that vanishes on M and satises kg k kf k

Converselyifg X R is continuous linear and satises ker g M then in

fact ker g M since theyre b oth co dimension one and hence there exists

a unique continuous linear functional f XM R satisfying g fq and

kg k kf k In either case its not hard to see that the corresp ondence f g

is linear and hence an isometry b etween XM and M Alternatively

CHAPTER PRELIMINARIES

check that the adjoint of the quotientmapq X XM is an isometry from

XM into X with range M

A similar observation is that M X M In other words each continu

ous linear functionals on M can b e considered as a functional on X thanks to

HahnBanach those functionals in X which agree on M that is functionals

which dier by an elementof M are simply identied for the purp oses of

computing M Alternatively the adjoint of the inclusion i M X is a

quotient map from X onto M with kernel M in fact i is the restriction

map i f f j Thus M X M

LastlyifX is a Banach space with X M N thenX is isomorphic

to M N Sp ecicallyif P X X is the pro jection with range M and

X X is again a pro jection with kernel N its not hard to see that P

range N and kernel M thatisX N M Thus M X M is

isomorphic to N and likewise N is isomorphic to M

A curious application

We round out our discussion of preliminary topics by presenting a curious re

sult due to Dixmier which highligh ts many of the ideas from this chapter

Theorem If i X X and j X X denote the canonical

embeddings then ji X X is a projection with range isometric to X

and kernel isometric to iX the annihilator of iX in X In short X

is always complementedinX

Proof Note that i X X and hence i j X X Toshowthatji

is a pro jection it suces to show that i j is the identityon X for then we

would haveji ji j i j i ji that is ji ji So given x X

lets compute the action of i j x onatypical x X

i j x x h x i j x i

h ixjx i

h x ix i

h x x i

x x

Thus i j x x and hence i j is the identityon X

Its not hard to see that i is onto thus the range of ji is j X which

is plainly isometric to X Along similar lines since j is onetoone its easy

to see that kerji ker i Finally general principles tell us that ker i

range i iX the annihilator of iX inX

Notes and Remarks

There are many excellent b o oks on functional analysis that oer more com

plete details for the topics in this chapter See for example Bollobas

Conway DeVito Dunford and Schwartz Holmes Jameson

Jarvinen Megginson Royden Rudin or Yosida

Megginsons b o ok in particular has a great deal on adjoints pro jections and

quotients

CHAPTER PRELIMINARIES

Exercises

Given normed spaces X and Y we write B X Y to denote the space of

b ounded linear op erators T X Y endowed with the op erator norm

Given a nonzero vector x in a normed space X show that there exists a

norm one functional f X satisfying jf xj kxk On the other hand

give an example of a normed space X and a norm one linear functional

f X such that jf xj kxk for every x X

Let Y b e a subspace of a normed linear space X and let T B Y R

Prove that T extends to a map T B X R withkT k kT k Hint

HahnBanach

Provethatevery prop er subspace M of a normed space X has empty

interior If M is a nite dimensional subspace of an innite dimensional

normed space X conclude that M is nowhere dense in X

Provethat B X Y is complete whenever Y is complete

If Y is a dense linear subspace of a normed space X showthat Y X

isometrically

Prove Rieszs lemma Given a closed subspace Y of a normed space X

and an there is a norm one vector x X such that kx y k

for all y Y IfX is innite dimensional use Rieszs lemma to construct

a sequence of norm one vectors x inX satisfying k x x k for

n n m

all n m

Given linear functionals f and g on a vector space prove that

T P

n n

ker f ker g if and only if f a g for some a a R

i i i n

i i

Given T B X Y showthatker T range T the annihilator of

range T in X and that ker T range T the annihilator of range T

in Y

X Y showthatT is an extension of T to X in the Given T B

following sense If i X X and j Y Y denote the canonical

emb eddings provethatT ix j T x In short T xTx

Let S b e a dense linear subspace of a Banach space X and let T S Y

be a continuous linear map where Y is also a Banach space Show that

T extends uniquely to a continuous linear map T X Y dened on

T k kT k Moreover if T is an isometry show that all of X and that k

T is again an isometry

Let X kk b e a Banach space and supp ose that jjj jj jis another norm

on X satisfying jjj x jjj kxk for every x X IfX jjj jj j is complete

prove that there is a constant csuchthat c kxkjjj x jjj for every

x X

Let M fx x Rg R Show that there are uncountably many

subspaces N of R suchthat R M N

Let M b e a nite dimensional subspace of a normed linear space X

Show that there is a closed subspace N of X with X M N In

fact if M is nontrivial then there are innitely many distinct choices

for N Hint Given a basis x x for M ndf f X with

n n

f x

i j ij

Let M and N b e closed subspaces of a Banach space X with M N

fg Prove that M N is closed in X if and only if there is a constant

C such that kxk C kx y k for every x M y N

Let M b e a closed nite co dimensional subspace of a normed space X

Show that there is a closed subspace N of X with X M N

Let M and N b e closed subspaces of a normed space X eachhaving the

same nite co dimension Showthat M and N are isomorphic

Let P X X b e a continuous linear pro jection with range Y and

let Q X X b e continuous and linear If Q satises PQ Q and

QP P show that Q is also a pro jection with range Y

A b ounded linear map U X X is called an involution if U I If

U is an involution show that P U I is a pro jection Conversely

if P X X is a b ounded pro jection then U P I is an involution

In either case P and U x the same closed subspace Y fx Px

xg fx Ux xg

A pro jection P on a Hilb ert space H is said to b e an orthogonal projection

if range P is the orthogonal complementofker P that is if and only if

H ker P range P Provethat P is an orthogonal pro jection if

and only if P is self adjoint that is if and only if P P

Let M b e a closed subspace of a normed space X IfbothM and XM

tsay that completeness is a are Banach spaces then so is X We migh

three space prop erty if it holds in anytwo of the spaces X M or

XM then it also holds in the third Is separability for example a

three space prop erty

Let T X Y b e a b ounded linear map from a normed space X onto

a normed space Y IfM is a closed subspace of ker T then there is a

CHAPTER PRELIMINARIES

e e

unique b ounded linear map T XM Y suchthat T Tq where q

is the quotient map Moreover kT k kT k

If X and Y are Banach spaces and if T X Y is a b ounded linear

map onto all of Y then X ker T is isomorphic to Y

Let X and Y b e Banach spaces and let T B X Y Then the follow

ing are equivalent

i X ker T is isomorphic to range T

ii range T is closed in Y

g iii There is a constant C such that inf fkx y k y ker T

C kTxk for all x X

Let M b e a closed subspace of a Banach space X and let q X XM

o o o

b e the quotientmapProve that q B B where B is the op en

X X

unit ball in X and B is the op en unit ball in XM

o o o

Wesaythat T B X Y isaquotient map if T B B whereB

X Y X

denotes the op en unit ball in X resp ectively Y Provethat T is a

quotient map if and only if X ker T is isometric to Y

Given T B X Y prove that T is a quotient map if and only if T is

an isometry into

Let M b e a closed subspace of a Banach space X and let q X XM

denote the quotient map Prove that q is an isometry from XM

into X with range M Thus XM can b e identied with M

Let M b e a closed subspace of a Banach space X and let i M X

denote the inclusion map Prove that i is a quotientmapfromX onto

M with kernel M Conclude that M can b e identied with X M

Let M b e a closed subspace of a normed space X For an y f X show

that minfkf g k g M g supfjf xj x M kxkg Please note the use of min in place of inf

Chapter

Bases in Banach Spaces

Throughout let X b e a real normed space and let x b e a nonzero sequence

in X Wesay that x isaSchauder basis for X if for each x X thereis

an unique sequence of scalars a such that x a x where the series

n n n

converges in norm to x Obviously a basis for X is linearly indep endent

Moreover any basis has dense linear span That is the subspace

span fx i Ng a x a a Rn N

i i i n

consisting of all nite linear combinations is dense in X In fact its not

hard to check that the set

a x a a Qn N

i i n

is dense in X

Wesay that x isa basic sequence if x is a basis for its closed linear

n n

span x span a space we denote byx or

n n

Example pand c

Its nearly immediate that the sequence e where that

single nonzero entry is in the nth slot is a basis for p and for

c Given an element x x x e for example the very fact

n n n p

that jx j tells us that

X X

jx j x x e

i i i

n i i

CHAPTER BASES IN BANACH SPACES

as n A similar argument applies to c

Essentially the same argument shows that any subsequence e ofe is

n n

a basic sequence in or c

But please note that a space with a Schauder basis must b e separable

Thus cannot have a basis In any case its easy to see that the calculation

ab ove fails irreparably in

Now all of the familiar separable Banach spaces are known to have a basis

This was well known to Banach which led him to ask

Does every separable Banach space have a basis

The answer is No and was settled byPer Eno presently at Kent State

in As you might imagine a problem that to ok years to solve has a

lengthy solution A more tractable question from our p oint of view is

Does every innite dimensional Banach spacecontain a basic sequence

The answer this time is Yes and is due to Mazur in Well givea proof

of this fact shortly A related question is

Does every separable Banach space embed isometrical ly into a spacewitha

basis

Again the answer to this question is Yes and is due to Banach and Mazur

who showed that every separable Banach space emb eds isometrically into

C Well display a basis for C very shortly The juxtap osition

of these various questions prompts the following observation Even though

C has a basis it apparently also has a closed subspace without a basis

Needless to say none of the spaces involved in this discussion are Hilb ert

spaces

ASchauder basis is not to b e confused with a Hamel basis A Hamel

basis e for X is a set of linearly indep endentvectors in X which satisfy

span e X That is each x X is uniquely representable as a nite

linear combination of e s Its an easy consequence of the Baire category

theorem that a Hamel basis for an innite dimensional Banach space must

in fact b e uncountable Indeed supp ose that e is a Hamel basis for an

innite dimensional Banach space X Then each of the nite dimensional

X But spaces X spanfe i ng is closed and clearly X

n n i

if X is innite dimensional then any nite dimensional subspace of X has

emptyinterior thus each X is nowhere dense contradicting Baires result

Hence e must actually b e uncountable Hamel bases are of little value to

us Henceforth the word basis will always means Schauder basis

X R by Given a basis x we dene the coordinate functionals x

n n

x xa where x a x Its easy to see that each x is linear

n i i

n n

and satises x x The sequence of pairs x x issaidtobe

m mn n

n n

biorthogonal Although in truth we often just say that x is biorthogonal

to x We also dene a sequence of linear maps P onX by P x

n n n

P P P

n n

a x It follows x xx That is P x a x wherex

i i i n i i

i i i

that the P satisfy P P P In particular P is a projection onto

n n m n

minfmng

span f x i n g Also since x isaSchauder basis wehave that

i n

P x x in norm as n for each x X Obviouslythex are all

continuous precisely when the P are all continuous Why The amazing

fact that all of these op erations are continuous whenever X is complete is due

to Banach

Theorem If x is a basis for a Banach space X then every P and

n n

hence also every x is continuous Moreover K sup kP k

Proof Banachs ingenious idea is to dene a new norm on X by setting

jjj x jjj supkP xk

Since P x x its clear that jjj x jjj for any x X The rest of the details

required to showthatjjjj jj is a norm are more or less immediate

In order to show that the P are uniformly b ounded wewanttoshow that

jjj x jjj K kxk for some constant K and all x X Tothisendwe app eal

to a corollary of the Op en Mapping theorem Notice that the formal identity

i X jjj jj j X kk is continuous since kxk lim kP xkjjj x jjj

n n

What we need to show is that this map has a continuous inverse And for this

weonlyneedtoshow that X jjj jj jisaBanach space That is weneedto

checkthat X is complete under jjj jj j

So let y beajjj jj jCauchy sequence Then for each n the sequence

P y is kkCauchy In fact since

n k

kP y P y kjj jy y jjj

n i n j i j

for any n the sequences P y are kkCauchy uniformly in nItfollows

n k

that if z lim P y in X kk then kP y z k as k

n k n k n k n

uniformly in nAndnow from this it follows that z iskkCauchy

kz z kkz P y k kP y P y k kP y z k

n m n n k n k m k m k m

Wecho ose k so that the rst and third terms on the righthand side are small

uniformly in n and m and then the middle term can b e made small b ecause

P y y as n

n k k

CHAPTER BASES IN BANACH SPACES

Now let z lim z in X kk We next showthatz lim y in

n n k k

X jjj jj j For this its enough to notice that P z z The key is that P is

n n n

continuous on the nite dimensional space P X

P z P lim P y

n m n m k

lim P P y

n m k

lim P y z

minfmng k minfmng

What this means is that there is a single sequence of scalars a such that

P P

z a x and so z a x and P z z follow Finally

n i i i i n n

i i

jjj y z jjj supkP y z k as k

k n k n

Of course if sup kP xk K kxk for all x then sup kP kK Also note

n n

that jx xjkx k kP x P xk K kxkThus kx kkx kK

n n n n

n n

The number K sup kP k is called the basis constant of the basis x

n n

A basis with constant is sometimes called a monotone basis The canonical

basis for for example is a monotone basis It follows from the pro of of

our rst theorem that any Banach space with a basis can always b e given an

t b ecomes Indeed jjj x jjj equivalent norm under which the basis constan

sup kP xk do es the trick

We next formulate a test for basic sequences this to o is due to Banach

Theorem Asequence x of nonzerovectors is a basis for the Banach

space X if and only if i x has dense linear span in X and ii thereis

aconstant K such that

m n

X X

a x K a x

i i i i

i i

for al l scalars a and al l nm

Thus x is a basic sequence if and only if ii holds

Proof The forward implication is clear note that for nm wehave

n m m

X X X

a x P a x sup kP k a x

i i n i i j i i

i i i

Now supp ose that i and ii hold and let S span f x i g Condition

i tells us that S is dense in X From here condition ii do es most of the

work

First ii and induction tell us that the x are linearly indep endent

nm a x ja jkx kK

i i n n

P P

n m

a x nm are welldened linear a x Thus the maps P

i i i i n

i i

pro jections on S Moreover condition ii says that each P has norm at most

K on S Hence each P extends uniquely to a continuous linear map on all

of X Ifwe conserve notation and again call the extension P then P is still

n n

a pro jection and still satises kP k K Wenow only need to show that

P x x for each x X

Given x X and let s a x S with kx sk Then

i i

for n m

kx P xk kx sk ks P sk kP s P xk

n n n n

kP k K

We conclude this rst pass at bases in Banach spaces with two classical

examples b oth due to J Schauder

Schauders basis for C

Schauder b egan the formal theory of bases in Banach spaces in by oering

up a basis for C that now b ears his name Rather than try to givean

analytical denition of the sequence consider the following pictures

1 f 2

f 0 f 1

1 f 3 f 4 f 5

1/21/2 1/4

CHAPTER BASES IN BANACH SPACES

If weenumerate the dyadic rationals in the order t t t

t t and so on notice that wehave f t andf t for

n n k n

kn This easily implies that the f are linearly indep endent Moreover its

is the set of all continuous piecewise now easy to see that spanf f

linear or p olygonal functions with no des at the dyadic rationals k

k Indeed the set of all such p olygonal functions is clearly a

m m

vector space of dimension whichcontains the linearly indep endent

functions f k Thus the t wospacesmust coincide Since the

dyadic rationals are dense in its not hard to see that the f have dense

linear span Thus f is a viable candidate for a basis for C

a f then If we set p

k k n

kp k max jp t j

n n k

k n

b ecause p is a p olygonal function with no des at t t And if we set

n n

p a f for mnthenwehave p t p t fork n b ecause

m k k m k n k

f t for jn k Hence kp k kp k This implies that f isa

j k n m n

normalized basis for C with basis constant K

Consequentlyeach f C can b e uniquely written as a uniformly

P P

convergent series f a f But notice please that a f van

k k k k

k k n

ishes at eachofthenodes t t Thus P f a f must agree with

n n k k

P f is the interp olating p olygonal approximation to f f at t t that is

n n

with no des at t t Clearly kP f k kf k

n n

a 2

a 1

a 0

f a f a f a f

Its tempting to imagine that the linearly indep endent functions t n

might form a basis for C After all the Weierstrass theorem

tells us that the linear span of these functions is dense in C But a

moments reection will convince you that not every function in C has a

uniformly convergentpower series expansion your favorite function that is not

dierentiable at for example Nevertheless as well see in the next chapter

C do es admit a basis consisting entirely of p olynomials

The Haar system

The Haar system h on is dened by h andfork

k k k

and i by h k x for i x i

k k

h k xfori xi and h k x otherwise A

i i

picture might help

h0 h1 h2 h3

1/2 1 1/2

– 1

As well see the Haar system is an orthogonal basis for L Each h

n is meanzero and more generally h h is either or h for any

n m m

nm In particular the h are linearly indep endent

Note that the Schauder system is related to the Haar system by the formula

f x h t dt for n and f In fact it was Schauder

n n

who proved that the Haar system forms a monotone basis for L for any

p While we could give an elementary pro of very similar to the

one we used for Schauders basis see Exercise it mightbeentertaining to

give a slightly fancier pro of The pro of well give b orrows a small amountof

terminology from probability

k k k

For each k let A f i i i g

Claim The linear span of h h k is the set of all step functions based

on the intervals in A That is

span f h h k g span f I A g

I k

Why Well clearly each h spanf I A g for j and span f

j I k I

I A g has dimension Thus the two spaces must coincide But it

functions at a should b e p ointed out here that its essential that we take

time The claim wont b e true if we consider an arbitrary batch h h

CHAPTER BASES IN BANACH SPACES

in place of the Haar This allows us to use the much simpler functions

functions in certain arguments For example its nowvery easy to see that

the h have dense linear span in L for any p Also please note

n p

that the set f I A g is again orthogonal in L although not the

I k

full set of for all dyadic intervals I

In particular if P is the orthogonal pro jection onto span f h h g

and if Q is the orthogonal pro jection onto span f I A gthenwemust

k I k

have P f Q f The savings here is that Q f is very easy to compute

k k k

Indeed

imI h f mI P f Q f

I I k k

I A

Thus P f is the conditional expectation of f given the algebra generated

k k

by A That is P f is the unique measurable function satisfying f

k k k

P f for all A

k k

Now lets see why P isacontraction on every L First consider f

k p

Z Z

jP f j f mI

I A

Z Z

jf j jf j

I A

Now for p and f L we use Holders inequality

Z Z

f mI jP f j

I A

p pq p

mI mI jf j

I A

Z Z

p p

jf j jf j

I A

Thus kP L L k for any p

k p p

The argument in the general case follows easily from this sp ecial case

We leave the full details as an exercise but heres an outline Given n with

k k

n letP b e the orthogonal pro jection onto the span of h h

Given f L write f f f where I and J are disjoint sets

p I J

whose union is all of suchthat I and J and such that f

k k I

span f h h k g and f span f h k h gspan f h h k g

J n

Then

Pf P f P f P f P f

I J k I I k J J

It follows that

p p p

kPfk k P f k k P f k

k I I k J J

p p p

kf k kf k kf k

I J

p p p

Notes and Remarks

The two main examples from this chapter are due to J Schauder

from however our discussion of the Haar system owes much to the

presentation in Lindenstrauss and Tzafriri See also the Monthly

article by R C James which oers a very readable intro duction to basis

theory as do es Megginson For more sp ecialized topics see Diestel or

the books by Lindenstrauss and Tzafriri already cited

There is a wealth of literature on b ounded orthogonal bases esp ecially

bases consisting of continuous or analytic functions See for example Linden

strauss and Tzafriri and Wo jtaszczyk

If f is an orthogonal basis for L then it is also a monotone

Schauder basis for L Moreover a function biorthogonal to f is g

n n

f kf k and in this case the canonical basis pro jection P coincides with the

n n n

orthogonal pro jection onto span f f f gHowever the typical orthogonal

basis for L will not yield a basis nor even elements of L for

cos t sin t p Its known cf that the sequence cos t sin t

forms a basis for L for p but not for p

It is a rather curious fact that a normalized Schauder basis for a separable

Hilb ert space H must b e an orthonormal basis Here is an elementary pro of

due to T A Co ok Supp ose that x isaSchauder basis for H with

asso ciated biorthogonal functionals x satisfying kx k kx k and

n n

supp ose that the inner pro duct h x x i for some k j Then there is a

k j

unit vector e in the span of x and x that is orthogonal to x Consequently

k j k

Thenjh x eij jh x eij Now write x h x ei e h x x i x

k j j j j k j

jh x x ij and hence jh x eij Finally notice that jf x j

k j j j j

CHAPTER BASES IN BANACH SPACES

jh x eijjf ej which implies that jf ej jh x eij This contradicts

j j j j

the fact that kf k j

Exercises

Let X beaBanach space with basis x Weknow that the expression

jjj x jjj sup kP xk denes an norm on X which is equivalentto k k

Showthatunderjjj jj jeach P has norm one That is we can always

renorm X so that x has basis constant K

Let f denote Schauders basis for C and let t denotetheas

k k

so ciated enumeration of the dyadic rationals If f C is written

P P

as f a f provethata f t a f t

k k n n k k n

k k

Heres an outline of an elementary pro of that the Haar system forms a

monotone basis in every L p

P P

n n

a Since a h and a h dier only on the supp ort of h

i i i i n

i i

p p p

conclude that we need to prove the inequality jabj jabj jaj

for all scalars a b

b The function f xjxj satises f xf y f x y for

all x y Hint f hence f is increasing

If x is a basis for a Banachspace X under what circumstances is

x kx k also a basis In other words can wealways assume that the

n n

basis vectors are norm one

If x is a basis for a Banach space X under what circumstances can

we renormalize so as to have kx k kx k forall n

Let f b e a disjointly supp orted normone sequence in L Show

n p

P P

ja a f converges in L if and only if that j What

n n n p

n n

if anything is the analogue of this result when p How ab out if L

is replaced by C or c

In any of the spaces porc show that wehave e

p n

Is the same true for p p

Dene x e e in c Isx a basis for c Whatisx in

n n n

this case Is x a basis for

Prove that a normed space X is separable if and only if there is a sequence

x inX such that spanx isdenseinX

n n

Let X b e a separable normed linear space If E is any closed sub

space of X show that there is a sequence of normone functionals f

in X such that dx E sup jf xj for all x X Conclude that

ker f Hint Given x dense in X use the HahnBanach E

n n

theorem to cho ose f so that f on E and f x dx E

n n n n n

CHAPTER BASES IN BANACH SPACES

Chapter

Bases in Banach Spaces II

Well stick to the same notation throughout with just a few exceptions Unless

otherwise sp ecied all spaces are innite dimensional Banach spaces Given

a sequence x in a Banach space X well use the shorthand x to denote

n n

the closed linear span of x Lastlye denotes the usual basis for

n n p

porc

Awealth of basic sequences

We b egin with a construction due to Mazur cf eg or showing

that every innite dimensional Banach space contains a basic sequence The

pro of features an ingenious application of Banachs criterion for basic sequences

which is of some interest in its own right

Prop osition Let F be a nite dimensional subspace of an innite dimen

sional normedspace X Then given thereisanx X with kxk

such that

ky k ky xk

for al l y F and al l scalars

Proof Let Recall that since F is nite dimensional the set

S f y F ky k g is compact in X Thus we can cho ose a nite

net y y for S that is each y S is within ofsomey Now

k F F i

for each icho ose a norm one functional y X suchthat y y

i i

Wewanttonda vector x which is in a sense p erp endicular to F

The next b est thing for our purp oses is to cho ose any norm one x with

CHAPTER BASES IN BANACH SPACES I I

y x y x How is this p ossible Well ker y is a subspace

of X of nite codimension and so must contain a nonzero vector The claim

is that anysuch norm one x will do To see this cho ose y S any scalar

R and estimate

ky xk ky xkky y k

i i

ky xk for some i

y y x

Thus ky k ky xk for all whenever ky k Since the inequality

is homogeneous b eing arbitrary this is enough

Corollary Every innite dimensional Banach spacecontains a closed

subspacewithabasis

Proof Let X b e an innite dimensional Banach space Given cho ose a

sequence of p ositivenumb ers with

n n

We next construct a basic sequence inductivelyby rep eated application

of Mazurs lemma To b egin cho ose anynormonex X Now cho ose a

norm one vector x so that

ky k ky x k

for all y x and all scalars Next cho ose a vector x of norm one so that

ky k ky x k

for all y x x and all scalars Cho ose x so that Well you get the

picture

The sequence x so constructed is a basic sequence with basis constant

at most K

Disjointly supp orted sequences in L and

p p

In preparation for later lets consider an easy source for basic sequences

Disjointly supp orted sequences in L orc In case its not clear the

p p

supp ort of a function f L is the set ff gTwo functions f g L

p p

are thus disjointly supp orted if ff gfg g that is if f g In

or c this reads x x andy y are disjointly supp orted if x y

p n n n n

for all n

Lemma Let f beasequence of disjointly supported nonzerovectors in

L p Then f isabasic sequenceinL Moreover f is

p n p n

isometric to and is complementedinL by a norm one projection

p p

Proof The linear span of f is unaected if we replace each f by f kf k

n n n n p

Thus wemay assume that each f is norm one

Next since the f are disjointly supp orted wehave

p p

Z Z

m m m m

X X X X

p p p

ja j jf j d ja j a f d a f

k k k k k k k

k n k n k n k n

for any scalars a This tells us that a f converges in L ifand

k n n p

only if ja j Why Thus the map e f extends to a

n n n

linear isometry from onto f In particular it follows that f isabasic

p n n

sequence in L

Lastly the existence of a pro jection is easy A sequence g biorthogonal

to f is clearly given by g sgn f jf j The g are disjointly sup

n n n n n

p orted norm one functions in L where q is the conjugate exp onentto p

Moreover g has the same supp ort as f namely A fg g ff g

n n n n n

Now consider

X X

fg f Pf hf g if

n n n n

n n

Clearly P is the identityonf To showthatP is norm one we essentially

rep eat our rst calculation

Z Z

X X

p p p

kPfk fg jf j kf k

p p

A A

n n

the inequality coming from Holders inequality

Its also true that a disjointly supp orted sequence in c spans an isometric

copyof c The pro of is a simple mo dication of the one wevejustgiven

Rather than rep eat the pro of in this case lets settle for p ointing out two such

mo dications First if y is a sequence of disjointly supp orted norm one

vectors in c then a y converges in c if and only if a and in this

n n n

case

supka y k supja j a y

n n n n n

n n

CHAPTER BASES IN BANACH SPACES I I

The natural sequence of biorthogonal functionals in this case is simply an

appropriately chosen subsequence of e up to sign Sp ecically given y

k n

c cho ose k so that y attains its norm in its k th co ordinate that is

n n n

i ij hy e ky k jhy e

n k n n k

n n

The key fact in our last lemma is that k k is additive across sums of

disjointly supp orted functions But if all were willing to settle for isomor

phic in place of isometric then all we need is almost additive in place of

additive or almost disjointly supp orted in place of disjointly supp orted

This suggests the following generalization

Lemma Let f beasequence of norm one functions in L Sup

n p

pose that there exists a sequenceofdisjointmeasurable sets A such that

p p

jf j d where fast enough Then f is a basic sequence

n n n

in L Moreover f is isomorphic to and complementedinL

p n p p

Proof From our previous lemma weknowhow to deal with the disjointly

supp orted sequence f f The idea here is that f is a small p ertur

n n A n

p p p

jf j thus bation of f By design kf f k

n n n n

n p

X X X X

ja jkf f k ja j f a f a

n n n p n n n n n n

n n n n

This gives us a hint as to what fast enough should mean Given

lets supp ose that Then

X X X

sup ja j j ja j a j

n n n n n

n n n

From this and our previous lemma it follows that

p p

X X X X

p p

ja j ja j a f a f

n n n n n n

n n n n

p p

If we can establish a similar lower estimate wewillhaveshown that f is

isomorphic to But

X X X

p p p p

a f ja j ja j jf j d

n n n n n

n n n p

p p p

and hence

p p

X X X X

p p

ja j a f a f ja j

n n n n n n

n n n n

p p

In order to nd a b ounded pro jection onto f wenow mimic this idea to

show that our b est guess is another small p erturbation of the pro jection

given in the previous lemma The map that should work is

p p

Pf kf k f sgn f jf j f

n n n n

Rather than give the rest of the details now well save our strength for a more

general result that well see later

Equivalent bases

Two basic sequences x andy in p ossibly dierent spaces are said to b e

n n

P P

a y converge or diverge together A straight a x and equivalent if

i i i i

i i

forward application of the Closed Graph theorem allows us to rephrase this

condition x andy are equivalent if there exists a constantC

n n

such that

X X X

a y C a y a x C

i i i i i i

i i i

Y Y X

for all scalars a That is x and y are equivalent if and only if the

i n n

basistobasis map x y extends to an isomorphism b etween x and y

i i n n

Thus wemay and in practice will take as the dening statement

Stated in these terms the condition takes on new signicance If we start

with a basic sequence x and if we nd that some sequence y satises

n n

then y must also b e a basic sequenceequivalenttox of course

n n

Indeed if x has basis constant K and if y satises then y has

n n n

basis constantatmostC K

n n m m

X X X X

a y C a x CK a x C K a y

i i i i i i i i

i i i i

CHAPTER BASES IN BANACH SPACES I I

As a particular example a sequence x is equivalent to the usual basis

for if and only if

p p

X X X

p p

C ja j a x C ja j

i i i i

i i i

for some constant C and all scalars a

If holds we sometimes say that x andy are C equivalentThus

n n

we mightsay that a disjointly supp orted sequence of norm one vectors in

L is equivalent to the usual basis of Or to paraphrase Lemma

p p

an almost disjoint sequence in L is equivalenttothe basis As

p p

another example note that any orthonormal basis in a separable Hilb ert space

is equivalent to the usual basis of

Next lets generalize the content of Lemma

Theorem The Principle of Small Perturbations Let x be a normal

izedbasic sequence in a Banach space X with basis constant K and suppose

kx y k that y is a sequenceinX with

n n n

i If K then y is a basic sequenceequivalent to x

n n

ii If x is complementedbyaboundedprojection P X X and if

K kP k then y is also complementedinX

ProofWe b egin with a microlemma For any sequence of scalars a and

any n rst recall that

ja j ka x k kP x P xk K kxk

n n n n n

where x a x That is the co ordinate functionals all have norm at most

n n

K In particular we always have

X X

ja j a x ja j sup

n n n n

n n

Now on with the pro of To b egin notice that

X X X X

a x a y ja jkx y k sup ja jK a x

n n n n n n n n n n

n n n n

Thus

X X X

K a x a y K a x

n n n n n n

n n n

and hence y is a basic sequence equivalenttox

n n

P P

In other words weveshown that the map T a x a y is an

n n n n

n n

isomorphism b etween x and y Note that gives kT k K

n n

K and kT k

Toprove ii we next note that anynontrivial pro jection P has kP k

and hence the condition KkP k implies at the very least that K

A bit of arithmetic will convince you that this gives us kxk ky k where

x a x and y Tx that is kT k In particular it follows from

n n

our microlemma that the co ordinate functionals for the y havenormat

most K that is ja jK ky kwhere y a y

n n n

Next weshowthatTP is an isomorphism on Y y Indeed if y

P P

a y and x a x then

n n n n

n n

kTPy y k kTPy xk TP a y x

n n n

kT kkP k sup ja j ky x k

n n n

K kP kky k ky k

It follows see the next lemma that S TPj is an invertible map on Y

Hence Q S TP is a pro jection from X onto Y

Now the pro of that weve just given supplies a hintastohowwemight

further improve the result The microlemma tells us that wewant kx k

for all n where x is the nth co ordinate functional which by HahnBanach

we can take to b e an elementof X Or b etter still wemight ask for

kkx y k This sum estimates the norm of the map S X X kx

n n

dened by

xx y x Sx

n n

x What is the map S doing here Well if were given x xx in

x then the basistobasis map should send x into

x x y Tx

CHAPTER BASES IN BANACH SPACES I I

Thus at least on x wehave S I T IfS is small enough that is

if T is close enough to I thenT should b e an isomorphism on x That

this is true is a useful fact in its own right and is well worth including

Lemma If a linear map S X X on a Banach space X has kS k

then I S has a bounded inverse and kI S k kS k

Proof The geometric series I S S S converges in op erator norm to

a b ounded op erator U with kU k kS k The fact that U I S

follows by simply checking that I S Ux x U I S x for any x X

Given this setup see if you can supply a pro of for the following new and

improved version of Theorem

Theorem The Principle of Small Perturbations Let x beabasic

sequence in a Banach space X with corresponding coordinate functionals x

kx kkx y k Suppose that y is a sequenceinX with

n n n

i If then y is a basic sequenceequivalent to x

n n

ii If x is the range of a projection P X X andif kP k then

y is complementedinX

Hint For ii show that the map A X X dened by

Ax x Px x Px y

satises kI Ak and Ax y The pro jection onto y is then given

n n n

by Q AP A

Notes and Remarks

Mazurs construction Prop osition and Corollary is part of the folklore

of Banach space theory and as far as I know was never actually published

by Mazur Instead we know ab out it through word of mouth supplied

by the pioneering early pap ers of Pelczynski Pelczynski claims that

Prop osition was rst proved by Bessaga in his thesis but credits Mazur

for the idea b ehind b oth Bessagas pro of and Pelczynskis presented here

In a later pap er Pelczynski refers also to a pap er by Bessaga and

Pelczynski

The material on disjointly supp orted sequences in L and is old as the

p p

hills and was well known to Banach The variations oered by the notion of

almost disjointness and the principle of small p erturbations however are

somewhat more mo dern and can b e traced to the early work of Bessaga and

Pelczynski As a simple application of the principle of small p erturbations

it follows that C has a basis consisting entirely of p olynomials The same

is true of L

As a nal comment regarding equivalent bases we should p oint out that

there is no such thing as the basis for a given space Granted there are

some spaces in which a natural basis suggests itself for example but in

general a basis even if one exists is far from unique That this is so is shown

rather dramatically by the following theorem due to Pelczynski and Singer

Theorem If X is an innite dimensional Banach space with a Schauder

basis then thereareuncountably many mutual ly nonequivalent normalized

bases in X

CHAPTER BASES IN BANACH SPACES I I

Exercises

Recall that a sequence x in a normed space X converges weakly to x X

xiff x f xforevery f X Its easy to see that written x

n n

w w

x x if and only if x x A sequence tending weakly to is said

n n

to b e weakly nul l

Let f b e a sequence of disjointly supp orted functions in L p

n p

P P

kf k f converges in L ifandonlyif Provethat

n n p

n n

Let x b e a disjointly supp orted norm one sequence in c Prove that

x is isometric to c and complemented in c by a norm one pro jection

Let f b e disjointly supp orted norm one sequence in C Prove

that f is isometric to c Isf complemented in C Will these

n n

arguments carry over to disjointly supp orted sequences in L

Let x beabasisforaBanach space X andlety b e a sequence

n n

in a Banach space Y Supp ose that a y converges in Y whenever

n n

a x converges in X where a is a sequence of scalars Use the

n n n

e that formula holds for some con Closed Graph Theorem to prov

stantC

Prove Theorem

Let T X X be a continuous linear map on a Banach space X If

T is invertible and if S X X is a linear map satisfying kT S k

kT k prove that S is also invertible Thus the set of invertible maps

on X is op en in B X

In each of the spaces porc the standard basis e is

p n

weakly null but not norm null In fact the set fe n g is norm

closed Similarlyinany Hilb ert space an orthonormal sequence x is

always weakly n ull while the set f x n g is always norm closed

Let f beadisjointly supp orted sequence of norm one vectors in L

n p

p Prove that f Is the same true for p p

If T X Y is a b ounded linear map and if x inX prove that

Tx inY

Supp ose that X and Y are isomorphic Banach spaces and that Z is a

complemented subspace of Y Provethat X contains a complemented

subspace W that is isomorphic to Z

Chapter

Bases in Banach Spaces III

Recall Banachs basis problem Do es every separable Banach space havea

basis Although the question was ultimately answered in the negative several

p ositive results were uncovered along the way For example wemight ask

instead Do es every separable Banach space embed in a space with a basis

Or do es every Banach space contain a subspace with a basis that is do es every

Banach space contain a basic sequence The answers to b oth of these amended

problems are Yes and b oth were known to Banach The rst follows from

the amazing fact due to Banach and Mazur that every separable Banach

space emb eds isometrical ly into C One of our goals in this short course

is to give a pro of of this universal prop ertyof C

The second question the existence of basic seqeunces was settled by

Mazur as wesaw in the last chapter Corollary In this chapter we

give a second solution due to Bessaga and Pelczynski which features a

useful selection principle

Our goal here is to mimic the simple case of disjointly supp orted sequences

in The only real dierence is the interpretation of the word disjoint In

a space with a basis x we could interpret disjointly supp orted to mean

having nonzero co ecients relative to the basis x o ccurring over disjoint

P P

subsets of N That is we could saythat x a x and y b x

n n n n

n n

are disjointly supp orted relative to the basis x if a b forall n

n n n

Blo ck basic sequences

Let x b e a basic sequence in a Banach space X Given increasing sequences

x be any nonzero of p ositiveintegers p q p q lety b

i i k

CHAPTER BASES IN BANACH SPACES I I I

vector in the span of x x Wesay that y isa block basic sequence

p q k

k k

with resp ect to x Its easy to see that y is indeed a basic sequence with

n k

the same basis constantasx

q q

m m n n

k k

X X X X X X

a y K a b x a b x K a y

k k i k i k i i k k

ip ip

k k k k

k k

If x isxedwell simply call y a block basic sequence or even just a

n k

block basis Byway of a simple example note that any subsequence x of

a basis x isablock basis

We next showhowwe to extract basic subsequences The metho d well

use is a standard bit of trickery known as a gliding hump argument You

may nd it helpful to draw some pictures to go along with the pro of

Lemma Let x beabasis for a Banach space X and let x be the

associatedcoecient functionals Suppose that z is a nonzerosequencein

X such that lim x z for each i Then given thereisa

n n k

subsequence z of z and a block basic sequence y relative to x such

n n k n

that kz y k for every k

n k k

x Proof Let n Since z z x converges in X we can nd

n n i

some q such that

x z x

n i

x z x yields kz y k Setting p and y

n i n

The idea nowistondavector z which is almost disjoint from z

n n

For this we use that fact that lim x z for each i In particular by

n n

we can nd an n n such that applying this fact to only nitely many i

x x z

i n

The span of nitely many x is just R in disguise Let p q and

cho ose q p suchthat

x x z

i n

x then yields kz Setting y z x y k

i n n

Wecontinue nding z almost disjoint from z and z And so on

n n n

The last pro of contains a minor mistake Did you sp ot it The y in the

ointment is that wehavenowayofknowing whether the y are nonzero This

is easy to x though We should insist that the z are b ounded away from

zero The principle of small p erturbations tells us what to do next

Lemma Using the same notation as in Lemma suppose that in ad

dition lim inf kz k Then z has a subsequence that is basic and

n n n

that is equivalent to some block basic sequenceofx

Proof By passing to a subsequence if necessarywemay supp ose that kz k

for all n Now by taking fast enough in Lemma the

principle of small p erturbations Theorem will apply

By mo difying our gliding hump argument just slightlywe arrive at

Corollary Let X be a Banach space with a basis x and let E bean

innite dimensional subspaceof X Then E contains a basic sequenceequiv

alent to some block basis of x

Proof It suces to show that E contains a sequence of norm one vectors z

such that lim x z for each i But in fact well prove something

n n

ClaimFor each m there exists a norm one vector z E such

mthatisE contains a norm one vector of that x z for all i

a x the form z

n n m

is a subspace of X of co dimension ker x How can this b e Well

at most m and so must intersect every innite dimensional subspace of X

nontrivially Alternatively consider the linear map z x z x z

from E into R Since E is innite dimensional this map must havenontrivial

kernel that is there must b e some norm one z E whichismappedto

Once wehaveshown that every separable Banach space emb eds isometri

cally into C an application of Corollary will yield that every Banach

space contains a basic sequence Indeed it would then followthatevery sep

arable Banach space contains a basic sequence equivalenttoablock basis of

the Schauder basis for C Since every Banachspaceobviously contains

a closed separable subspace this do es the trick

A similar approach can b e used to show what we mightcall the Bessaga

Pelczynski selection principle

CHAPTER BASES IN BANACH SPACES I I I

Corollary Let X be a Banach space and suppose that z is a sequence

in X such that z but kz k Then z has a basic subsequence

n n n

Subspaces of and c

Temp orarily X will denote one of the spaces porc Well

use e to denote the standard basis in X ande to denote the asso ciated

sequence of co ordinate functionals in X Note that e is really just e

again but considered as a sequence in the dual space Lets summarize what

weknow ab out the closed innite dimensional subspaces of X

Prop osition Let y be any disjointly supported nonzerosequencein

X Then y is isometric to X and is complementedinX by a projection of

norm one

This is immediate from Lemma

Corollary Any seminormalized block basis y of e is equivalent to

n n

e Moreover y is isometric to X and is complementedinX by a pro

n n

jection of norm one

A seminormalized sequence y satises inf ky k and sup ky k

n n n n

This assumption is needed to check the equivalence with e Finally

contains a further Corollary Every innite dimensional subspaceofX

subspace that is isomorphic to X and complementedinX

As a brief reminder the pro of of this last fact consists of rst showing that

every innite dimensional subspace contains an almost disjoint sequence of

norm one vectors Such a sequence is a small p erturbation of a blo ckbasic

sequence and so is almost isometric to X and is the range of a pro jection

of norm almost one In fact given we can nd a subspace that is

isomorphic to X and complemented in X

The fact that every subspace of X contains another copyof X is just what

we need to prove that each memb er of the family p and c is

isomorphically distinct In fact no space from this class is isomorphic to a

subspace of another memb er of the class To see this lets rst consider a sp ecial case

Theorem Let pr and let T beaboundedlinear

r p

map Then kTe k Inparticular T is not an isomorphism The same

n p

is true of any map T c

That is given any f where Proof First note that Te

q p n

p q the claim here is that f Te asn But f T is an

elementof where r s andf Te f T e e f T

s r n n

is just the nth co ordinate of f T Thus since s wemust have

f Te asn

Heres the same pro of in dierentwords

h Te f i h e T f i as n

n n

in b ecause e

r n

Now supp ose that kTe k that is supp ose lim inf kTe k

n p n n p

Then by Lemma some subsequence of Te is basic and is equivalentto

a blo ck basis of e in which by Corollary is in turn equivalenttoe

n p n

In particular after passing to a subsequence we can nd a constant C

such that

X X

a e C a Te

k k k n

k k

p p

Thus

r p

X X X

r p

C kT k ja j ja j C kT k a e

k k k n

k k

Weve arrived at a contradiction If this inequalitywere to hold for all scalars

p r

then in particular wed have n C kT k n for all n Since pr this is

imp ossible Consequently kTe k

n p

The pro of in case T c is virtually identical

With just a bit more work we could improve this result to read A b ounded

linear map T pror T c is compact That is T

r p p

maps b ounded sets into compact sets

This pro of of Theorem actually shows something more T fails to be

an isomorphism on any innite dimensional subspaceof Indeed each in

nite dimensional subspace of contains an equivalent copy of e and we

r n

could rep eat the pro of for this copy Or in still other words each innite

CHAPTER BASES IN BANACH SPACES I I I

dimensional subspace of contains an isomorphic copyof and the restric

r r

tion of T to this copyof cannot b e an isomorphism A b ounded linear map

T X Y which fails to b e an isomorphism on any subspace of X is said to

be strictly singular Any compact op erator for example is strictly singular

but not conversely Thus every map T where pris

r p

strictly singular Likewise for maps T c Buteven more is true

Corollary Let X and Y be two distinct members of the family of spaces

c and p Then every boundedlinear map T X Y is strictly

singular In particular X and Y are not isomorphic

ProofWe consider the case T where rp and leave the

r p

remaining cases as an exercise

Supp ose that T is an isomorphism from a subspace of onto a subspace W

of Then there is a further subspace Z of W and an isomorphism S Z

p p

But then T S is an isomorphism from into which is imp ossible

p r

Complemented subspaces of and c

In this section we presentPelczynskis characterization of the complemented

subspaces of and c His pro of is based on an elegant and mysterious

decomposition method Before we can describ e the metho d well need a few

preliminary facts

Given two Banach spaces X and Y wecanenvision their sum X Y as

the space of all pairs x y where x X and y Y Up to isomorphism it

do esnt much matter what norm wetakeon X Y For example if we write

p p

ky k X Y to denote X Y under the norm kx y k kxk then

Y X

X Y X Y X Y

where means is isomorphic to This is a simple consequence of the fact

that all norms on R are equivalent As a particular example note that under

any norm

p p p p p p

where means is isometric to Why

Given a sequence of Banach spaces X X we dene the sum of

X X to b e the space of all sequences x with x X for which

n n n

incaseporkx k sup kx k kx k kx k

n n X n n

n n

in case p and we use the shorthand X X to denote this new

space In brief for any p wehave

X X f x x X and kx k g

p n n n n p

The c sum of spaces is dened in an entirely analogous fashion In this case

we write

X X fx x X and kx k c g

n n n n

Please note that in each case wehave dened X X to be a proper

subspace of the formal sum X X In particular wewillnolongerbe

able to claim that X X and X X are isomorphic for

p q

p q Notice for example that R R

p p

It should also b e p ointed out that the order of the factors X X in an

sum do es not matter that is if N N is anypermutation then

X X X X

p p

where means is isometric to Why

While this may sound terribly complicated all that we need for no w is one

very simple observation Wealways have

and c c c

p p p p

for any p And why should this b e true The pro of in essence is one

sentence N can be written as the union of innitely many pairwise disjoint

innite subsetsHow do es this help

Given this notation the pro of of Pelczynskis theorem is just a few lines

Theorem Let X be one of the spaces porc Then every

innite dimensional complementedsubspaceofX is isomorphic to X

ProofFor simplicity of notation lets consider X for some p

X c is identical The pro of in case

If Y is an innite dimensional complemented subspace of then we can

write Y Z for some Banach space Z And from Corollary we can

also write Y X W where W is some Banach space and where X

In brief Y W Thus

Y W W W Y

p p p p p p

CHAPTER BASES IN BANACH SPACES I I I

since Now for some prestidigitation

p p p

Y Y

p p p p

Y Z Y Z Y

Z Z Y Y Y

p p

Z Z Y Y

p p

Y Z Y Z

p p

Hence Y Y

p p

Notes and Remarks

Essentially all of the results in this chapter can b e attributed to Pelczynski

who might fairly b e called the father of mo dern Banach space theory

After the devastation of the Polish scho ol during World War I I the study of

linear functional analysis was slow to recover Aleksander Pelczynski and

Joram Lindenstrauss resurrected the lost arts in the late s and early s and

went on to form new centersinPoland and in Israel resp ectively Along with

Rob ert James in America they founded a new scho ol of Banach space theory

Needless to say all three names will b e cited frequently in these notes

Theorem in the form of Exercise is due to H R Pitt See

Lindenstrauss and Tzafriri for more on strictly singular op erators and

more imp ortantlymuch more on the subspace structure of and c

Corollary and Theorem mightleadyou to b elieve that the spaces

have a rather simple subspace structure Once we drop the word comple

mented however the situation changes dramatically For p the space

contains innitely many mutual ly nonisomorphic subspaces cf eg

Pelczynskis decomp osition metho d Theorem has one obvious prac

tical disadvantage Its virtually imp ossible to write down an explicit isomor

phism From this p oint of view its at b est an existence pro of

We might paraphrase the conclusion of Pelczynskis theorem bysaying that

the spaces c and p are prime b ecause they havenonontrivial

factors Its also true that is prime but the pro of is substantially harder

The heart of Pelczynskis metho d is that an innite direct sum Y Y

is able to swallow up one more copyof Y The decomp osition metho d will

generalize under the right circumstances leading to a SchroderBernsteinlike

theorem for Banach spaces Please take note of the various ingredients used in the pro of

i Y emb eds complementably in X and X emb eds complementably in Y

ii X X X and

iii X X X whichmay fail for certain spaces but actually follows from

ii in this case

There are several go o d survey articles on complemented subspaces and

SchroderBernsteinlike theorems for Banach spaces Highly recommended are

the articles by Casazza and Mascioni

The SchroderBernstein theorem for Banach spaces reads If X is isomor

phic to a complementedsubspaceof Y and if Y is isomorphic to a comple

mentedsubspaceofX then X and Y are isomorphic The theorem is known

to hold under rather mild restrictions on X and Y however it has b een re

cently shown to fail in general Indeed Fields medalist W T Gowers

constructed an example of a Banach space X which is isomorphic to its

cub e X X X but not to its square X X

Gowers put several longstanding op en problems to rest in recentyears

Another such problem asked whether and c were the only prime Banach

spaces Gowers again solved this problem in the negativeby providing an

example of a Banach space which fails to b e isomorphic to any of its prop er

subspaces Well havemoretosay ab out Gowerss work later in the course

CHAPTER BASES IN BANACH SPACES I I I

Exercises

The sequence x e x e e x e e e isablock

basis of e Showby hand that x is not equivalenttoe in

n n n p

for p but that x isequivalenttoe inc What can you

n n

say ab out the blo ckbasis y x n

n n

Prove Corollary

If T c p is b ounded and linear show that kTe k

p n p

Show that every b ounded linear map T pror

r p

T c p is compact Hint T is completely continuous

Show that every b ounded linear map T c or T c

r r

r is strictly singular

Show that X Y X Y isometrically

Prove that N can b e written as the union of innitely many pairwise

disjoint innite subsets

R Find a natural copyof in L

p p p p

If X is a sequence of Banach spaces prove that X X is a

n p

Banach space for any p

The pro of of Theorem requires that Y Z Y Z

Z Z Y Y Verify this claim

p p

Prove Theorem in the case X c

Chapter

Sp ecial Prop erties of c and

The spaces c and playvery sp ecial roles in Banach space theory

Youre already familiar with the space and its unique p osition as the sole

Hilb ert space in the family of spaces Wewont havemuchtosay ab out

here And bynowyou will have noticed that the space do esnt quite

t the pattern that weve established for the other spacesfor one its not

separable and so do esnt have a basis Nevertheless we will b e able to saya

few meaningful things ab out The spaces c and on the other hand play

starring roles when it comes to questions involving bases in Banach spaces In

the whole isomorphic theory of Banach spaces for that matter Unfortunately

we cant hop e to even scratch the surface here But at least a few interesting

results are within our reach

Throughout e denotes the standard basis in c or ande denotes

the asso ciated sequence of co ecient functionals As usual e ande are

really the same we just consider them as elements of dierent spaces

True stories ab out

We b egin with a universal prop ertyof due to Banach and Mazur

Theorem Every separable Banach space is a quotient of

fx kxk g Proof Let X b e a separable Banach space and write B

to denote the op en unit ball in X Whatwe need to show is that there exists

CHAPTER SPECIAL PROPERTIES OF C AND

o o

a linear map Q X such that QB B Whatwell actually do is

QB B Just as in the pro of of construct a norm one map Q suchthat

o o

the Op en Mapping theorem this will then imply that QB B

To b egin since X is separable we can nd a sequence x inB which

n X

is dense in B We dene Q X by setting Qe x and extending

X n n

P P P

linearlyThus kQk since k a x k ja j k a e k Clearly

n n n n n

n n n

o o

then QB B and QB B Since x is dense in B wealsohave

X n X

QB B

Now given x B wehave kxk for some Thus

y x B tooWe will nish the pro of byshowing that y Qz for

some z with kz k That is wewillshowthat x

y Q B Here we go Given wehave

y B n such that ky x k

y x B fx j n g

n X j

n such that ky x x k

n n

f x j n g y x x B

j n n X

n such that ky x x x k

n n n

andsoon

X X

j j

y x Q e

n n

j j

and z e has norm in

We will apply Banach and Mazurs result to showthat contains an

uncomplemented subspace To accomplish this we rst need some prop erty

of thats not shared byevery Banach space One suchproperty and a rather

dramatic example at that is due to J Schur from

Theorem In weak sequential convergence implies norm convergence

Before weproveSchurs theorem lets note that a weakly convergent se

quence is necessarily norm b ounded Given a weakly convergent sequence x

in a normed space X consider the sequence of functionals x X acting

on X For each f X the sequence f x x f is b ounded since its

n n

convergent That is the sequence x is pointwise bounded on X By the

BanachSteinhaus theorem this sequence must actually b e uniformly b ounded

on B In short

sup sup jx f j sup sup jx f j supkx k sup kx k

n n n n

f B f B

X X

Now its easy to see that norm convergence always implies weak conver

gence and so Schurs result states that the two notions of convergence coincide

for sequences in But since the weak top ology on any innite dimensional

space is guaranteed to b e weaker than the norm top ology there are nets in

whichareweakly convergent but not norm convergent Well havemoreto

say ab out this later

Proof of Schurs theorem Supp ose that x is a b ounded sequence in

but kx k Well arriveatacontradiction by such that x

n n

constructing an f suchthat f x

Wehave in particular that x k e x as n for each

n n

k Where at the risk of some confusion weve written x k for the k

th co ordinate of x That is x tends co ordinatewise to zero By a

n n

standard gliding hump argument weknow that some subsequence of x is

almost disjoint That is after passing to a subsequence and relab eling we

may supp ose that

i kx k for all nand

ii for some increasing sequence of integers p q p q we

P P P

have jx ij and jx ij hence jx ij

k k k

ip iq ip

k k k

Nowwe dene f by

sgn x i if p i q for some k

k k k

f i

otherwise

Then kf k and for any k

X X X

jx ij jf x j jx ij jx ij

k k k k

iq ip ip

k k k

which is the contradiction we needed

Corollary Let Q c be a quotient map Then ker Q is not comple

mentedin

Proof In the presence of a quotientmapQ c wehave that ker Q

c Ifker Q were complemented in then c ker Q would b e isomorphic

to a subspace of That is we could nd an into isomorphism T c

w w

But then But e inc which easily implies that Te in

k k

CHAPTER SPECIAL PROPERTIES OF C AND

from Schurs theorem Theorem wewould have kTe k whichis

imp ossible if T is an isomorphism

The conclusion here is that there can b e no into isomorphism T c

a fact that we already know to b e true for other reasons All thats really new

here is the existence of an onto map Q c

We could easily apply the same reasoning to any separable space that

contains a weakly null normalized sequence In particular we could just as

easily have used any pinplaceof c Thus has uncountably

many isomorphically distinct uncomplemented subspaces A far cry from what

happ ens in a Hilb ert space

Curiously there is a measure of uniqueness in our last two results Sp ecif

icallyif X is a separable space not isomorphic to then ker Q is isomorphi

cally the same no matter what map Q from onto X we mighttake Its

an op en problem whether this prop erty is an isomorphic characterization of

and See Theorem f and Problem f

While were sp eaking of quotients lets turn the tables and consider maps

onto

Theorem Let X be a Banach space If thereisabounded linear onto

map T X then X contains a complemented subspace isomorphic to

Proof From the Op en Mapping theorem T B B for some

Consequentlywe can nd a b ounded sequence x inX suchthatTx e

n n n

We rst show that x is equivalenttoe Now

n n

X X X

a x kC ja jkx ja j

n n n n n

n n n

since x is b ounded On the other hand

X X X

ja j a e a x kT k

n n n n n

n n n

Thus x is equivalenttoe That is the map S X dened by

n n

Se x and extended linearly to all of is an isomorphism from onto

n n

The fact that x is complemented is noweasy Indeed notice that TS is

the identityon Thus ST is a pro jection onto x

The last result is a neat and tidy curiosity with little consequence wouldnt

you think But to Pelczynski its the starting p oint for a deep theorem that

well prove later in this chapter And to Lindenstrauss its but one among a

wealth of results ab out liftings and extensions of op erators along with a variety

of interesting characterizations of c and cf eg Section f

We will settle for just one such result that is b oth simple and timely

Corollary Theorem characterizes up to isomorphism among al l

separable innite dimensional Banach spaces

Proof Supp ose that Y is a separable innite dimensional Banach space with

the prop erty that whenever T X Y is onto then X contains a comple

mented subspace isomorphic to Y Then in particular taking X and

T a quotientmaponto Y wemust have Y isomorphic to a complemented

subspace of But from Pelczynskis Theorem it follows that Y must

b e isomorphic to

Our last sp ecial prop ertyof in this section concerns what are sometimes

called almost isometries or small isomorphisms Sp ecicallyany space

that contains an isomorphic copyof contains an almost isometric copy

of The same is true for c although well omit the similar pro of This

result is due to James

Theorem Let jjjj jj beanequivalent norm on For every thereis

a subspace Y of such that Yjjj jj j is isomorphic to

ProofWemay supp ose that jjj x jkjj xk jjj x jjj for all x where

Let and let P b e the natural pro jections asso ciated with the usual

basis e of For each n put

supfkxk jjj x jjj P x g

n n

Thus kxk jjj x jjj whenever P x Since ker P ker P the s

n n n n n

decrease hence for some

Fix n suchthat Since for all nwe can

n n

y construct a blo ck basic sequence y ofe suchthatjjj y jjj P

k k n k n

and ky k for all k Now for every choice of scalars a wehave

k k

P a y and hence

n k k

X X

a y a y

k k k k

k k

CHAPTER SPECIAL PROPERTIES OF C AND

ja jky k

k k

ja j

P P

ja j follows from the triangle inequality Conse a y jjj Of course jjj

k k k

k k

quentlyy is equivalentto e that is y is isomorphic

k n k

The secret life of

We b egin by recalling a useful corollary of the HahnBanach theorem For

every vector x in a normed space X wehave

kxk supfjf xj f X kf k g

Indeed for a given xwe can always nd a norm one functional f in X such

that jf xj kxkOfinterest here is the fact that there are enough functionals

in the unit sphere of X to recover the norm in X The question for the day

is whether any smaller subset of X will work

Our rst result oers a simple example of just such a reduction

Lemma If X is a separable normedspace then we can nd a sequence

such that f in X

kxk sup jf xj

for every x in X Inparticular a countable family in X separates points in

Proof Given x dense in X cho ose norm one functionals f inX such

n n

that f x kx k Its not hard to see that then holds

n n n

Our interest in Lemma is that the conclusion holds for some nonsep

arable spaces to o For example it easily holds for the dual of any separable

space if y is dense in the unit sphere of Y whereX Y checkthaty

n n

works In particular has this property Indeed if e x x has its usual

meaning then the sequence e tsthebill kxk sup je xj

n n

If some collection of functionals X satises

kxk supjf xj

wesaythatisanorming set for X orthatnorms X In this language

Lemma says that separable spaces have countable norming sets Its easy to

see that any space with this property is isometric to a subspaceof Indeed

if kxk sup jf xj then dene T X by Tx f x

n n

Taken together these observations pro duce an easy corollary

Corollary A normedspace X has a countable norming set if and only if

it is isometric to a subspaceof

There are a couple of reasons for having tak en this detour For one its now

pretty clear that every normed space X emb eds isometrically into for

some where dep ends on the size of a norming set in X For another these

observations lend some insightinto the nature of complemented subspaces of

Heres why

If T is continuous and linear then ker T kere T

In particular any complemented subspace of isacountable intersection of

kernels of continuous linear functionals Our next task is to showthatc fails

to b e so writeable that is c is not complemented in This result is due to

Phillips and Sob czyk The pro of well give is due to Whitley

Theorem c is not complementedin

Proof Well showthatc cant b e written as ker f for any sequence

of functionals f on Wewont need to knowvery muchabout

in order to do this What this amounts to indirectly is showing that c

has no countable norming set hence cant b e a subspace of The pro of

sidesteps consideration of the quotient space though We pro ceed in three

steps

There is an uncountable family N of innite subsets of N such that

N N is nite for

is innite If For each the function x is in n c since N

f with c ker f thenf f x g is countable

CHAPTER SPECIAL PROPERTIES OF C AND

T T

If c ker f then some x is in ker f too Thus c

n n

ker f

First lets see how follows from and b esides its easiest From

wewould have that the set f f x for some n g is countable Thus

some isnt in the set That is there is an for which f x forall n

and so follows

is due to Sierpinski Heres a clever pro of that Whitley attributes

to one Arthur Kruse Let r b e a xed enumeration of the rationals For

each irrational cho ose and x a subsequence r converging to andnow

dene N f n k gEach N is innite there are uncountably

many N and N N has to b e nite if Very clever no

Finally we prove Supp ose that f vanishes on c For each n

let A f jf x jn g Well showthat A is nite Supp ose that

n n

are distinct elements in A Ifwe dene

k n

x sgn f x x

i i

then f x kn But since f vanishes on c were allowed to change x in

nitely many co ordinates without eecting the value of f x In particular if

we dene M N n N thenM diers from N by a nite set and

i i

i j i

j i

so setting

sgn f x y

i i

gives f y f x But nowthatweve made the underlying sets disjoint we

also have ky k and hence

k nf xnf y nkf k

Thus A is nite

The functionals on whichvanish on c are precisely the elements of

c In fact our pro of of actually computes the norm of x in the

quotient space c

Corollary c is not isomorphic to a dual space

Proof Supp ose to the contrary that there is an isomorphism T c X

from c onto the dual of a Banach space X ThenT X is again

an isomorphism and further T j T Now from Dixmiers theorem

there exists a pro jection P X X mapping X onto X But

then T PT is a pro jection from onto c contradicting

Theorem

As it happ ens a closed innite dimensional subspace of is complemented

precisely when its isomorphic to We can prove half of this claim by

showing that is injective This means that has a certain Hahn

Banach prop erty This to o is due to Phillips

Theorem Let Y be any subspaceofanormedspace X and suppose that

T Y is linear and continuous Then T can be extendedtoacontinuous

linear map S X with kS k kT k

Proof Clearly

Ty e Ty y y

n n

y e T Y Ifweletx X b e a HahnBanach extension of y where

n n n n

do es the trick then Sx x x

kSxk sup jx xj

sup kx k kxk

k kxk kT kkxk sup ky

Thus kS kkT k That kS k kT k is obvious

As an immediate corollary notice that if is a closed subspace of some

Banach space X then is necessarily the range of a norm one pro jection on

X Indeed the identity I considered as a map into X extends

to a norm one map P X also considered as a map into X Clearly

P is a pro jection In short is norm one complementedinanysuperspace

This fact is actually equivalent to the extension prop erty of the previous

theorem

Its not hard to see that anyspace has this same HahnBanach

extension prop erty

CHAPTER SPECIAL PROPERTIES OF C AND

Confessions of c

Although c is obviously not complemented in every sup erspace it do es share

s extension prop erty to a certain extent In particular c is separably

injective This observation is due to Sob czyk the short pro of of the

following result is due to Veech If you dont know the BanachAlaoglu

theorem dont panic Well review all of the necessary details in a later chap

ter Time p ermitting well even present a second pro of of Sob czyks theorem

that sidesteps certain of these issues

Theorem Let Y be a subspace ofaseparable normed linear space X If

T Y c is linear and continuous then T extends to a continuous linear

map S X c with kS k kT k

Proof As b efore let y e T Y and let x X b e a HahnBanach

n n n

extension of y with kx k ky kkT k Wewould like to dene Sx

n n n

x x as b efore but we need to know that Sx c That is wewantto

replace x by a sequence of functionals which tend p ointwise to on X but

we dont want to tamp er with their values on Y What to do

Since X is separable weknowthatB kT k B is b oth compact and

Let d b e a metric on B whichgives this metrizable in the weak top ology

top ologyNowletK B Y and notice that anyweak limit p ointofx

must lie in K since x y y y for any y Y Thatisdx K

n n n

This means that we can p erturb each x by an elementof K without eecting

its value on Y

Sp ecicallywecho ose a sequence z inK suchthat dx z and

n n n

x z Then Sx c Sy x x y Ty for we dene Sx x

n n n

y Y and kS kkT k

As an immediate corollary we get that c is complementedbyaprojection

of norm at most in any separable space that contains it Our nal result

brings us full circle This is the promised and deep result of Bessaga and

Pelczynski mentioned earlier

Theorem Let Y be a Banach space If Y contains a subspace isomor

phic to c then Y contains a complementedsubspace isomorphic to Thus

Y contains a subspace isomorphic to al l of In particular no separable

dual spacecan contain an isomorphic copy of c and of course c itself is not

isomorphic to any dual space

Proof In order to prove the rst assertion well construct a map from Y onto

The second assertion will then follow easily

To b egin let T c Y b e an isomorphism into Then T Y is

onto by Theorem But wewantamaponY so lets consider S T j

Now h e Sy i h y T e i for all y Y and all n and so wemust have

n n

Sy h y T e i h y T e i Te y Te y

where e is the usual basis for c

Next we pull back a copyofe the basis for SinceT is onto there

exists a constant K and a sequence y inY suchthatky kK and

n n

T y e for all nWewould prefer a sequence in Y with this prop erty

n n

for then wed b e nished Well settle for the next b est thing Since K B

we can nd a sequence y inY with ky k K is weak dense in K B

n n Y

such that

jTe y jTe y j n j n and

n n i n

Weve used the functionals Te Te to sp ecify a certain weak neighbor

ho o d of y for each n Of course y Te T y e e e

k k k nk

n n n n

Thus Sy has a large nth co ordinate and very small entries in its rst n

co ordinates What this means is that Sy has a subsequence equivalent

to the usual basis in ie an almost disjoint subsequence whose span

In particular after passing to a is complementedin by a pro jection P

subsequence we can nd a constant M so that

X X X X

a y K ja j KM a Sy KM kS k a y

n n n n n n n

n n n n

That is S is invertible on y and hence Q S PS is a pro jection from

Y onto y This completes the pro of of the rst assertion In other words

wenowhav e Y X for some X Y It follows that Y X

which nishes the pro of

Notes and Remarks

The pro of of Theorem is closely akin to Banachs pro of of the Op en Map

ping theorem Its quite clear that Banach understo o d completeness com

pletely Schurs Theorem is an older gem which also app ears in Banachs

CHAPTER SPECIAL PROPERTIES OF C AND

b o ok It provides a classic example of a gliding hump argument Our pro ofs

of Theorem and its corollary are largely b orrowed from Diestel who

attributes these results to Lindenstrauss Theorem is sometimes called

Jamess nondistortion theorem In brief it says that and c are not dis

tortable a notion that wewont pursue further here Emb eddings into

spaces along the lines of Corollary are semiclassical and can b e traced

backtoFrechet see Theorem and its companion Theorem

were among the rst results in a frenzy of research during the late s and

early s concerning injective spaces and HahnBanachlike extension prop

erties Wewillhavemoretosay ab out such topics in a later chapter For

more on the spaces c and see Day Diestel Jameson and

Lindenstrauss and Tzafriri

Exercises

Find a weakly null normalized sequence in L and conclude that L and

are not isomorphic

Complete the pro of of Lemma

Show that the conclusion of Lemma also holds in case X Y where

Y is separable

If X is separable showthatX emb eds isometrically in

Let X and Y b e Banach spaces and let A b e a b ounded linear map from

X onto Y Then for every b ounded linear map T Y there exists

a lifting T X suchthat AT T Hint Use the Op en Mapping

theorem to cho ose a b ounded sequence x inX such that Ax Te

n n n

The op erator dened by Te x then do es the trick

n n

If H is any Hilb ert space prov ethatevery b ounded linear op erator from

H into is compact

Prove Theorem for c

Prove Theorem for maps into That is prove that is

injective

Prove that the following are equivalent for a normed space X

i If X is contained isometrically in a normed space Y then X is

complemented by a norm one pro jection on Y

ii If E is a subspace of a normed space F thenevery b ounded linear

map T E X extends to a b ounded linear map S F X with

kS k kT k

Hint Weve already seen that ii implies i Toprove that i implies

ii rst embed X isometrically in some space and extend T as

a map into using Theorem

CHAPTER SPECIAL PROPERTIES OF C AND

Chapter

Bases and Duality

If x is a basis for X you mayhavewondered whether the sequence of

co ordinate functionals x forms a basis for X Ifwe consider the pair e

then its easy to see that the answer is sometimes Yes in case and e

X c and X for example and sometimes No in case X

isalways a basic and X for instance As it happ ens though x

sequence that is its always a basis for x Lets see why

Given a a and cho ose x c x with kxk such that

n i i

n n n

X X X

a x a x a c x

i i i i

i i

i i i

NowifK is the basis constantofx then

K K kxk c x

i i

Thus for any m nwehave

m n m

X X X

a x c x a x K

i i i i

i i

i i i

n m

X X

c x a x

i i i

i i

a c

i i

a x

CHAPTER BASES AND DUALITY

That is since was arbitraryx is a basic sequence with the same basis

constant K

Its also easy to see that if P is the canonical pro jection onto x x

n n

then P is the canonical pro jection onto x x Indeed given m kn

n n

wehave

m k m k

X X X X

a x b x P a x P b x

i j j i n j j

i n i

i j i j

n m

X X

b x a x

j j i

j i

b a

i i

n k

X X

a x b x

i j j

i j

P P

m n

That is P a x a x Whats more since P x x for any x

i i n

n i i

i i

isweak its easy to checkthat P x x for any x andhencespanx

n n

dense in X IfX is reexive then this already implies that x is a basis

for X since the weak weak closure of a subspace coincides with its norm

closure

In any case x is a basis for X if and only if x X Our next

n n

result supplies a test for this condition

Theorem Let x beabasis for X with coordinate functionals x

Then x is a basis for X if and only if lim kx k for each x X

n n

where kx k is the norm of x restricted to the tail space span f x ing

n i

Proof The forward implication is easyIfx is a basis for X and if x

a x then

a x kx k

i n

Now supp ose that lim kx k foreach x X Given x X

n n

rst notice that the functional x hx x i x vanishes on the span of

x x Thus given x c x wehave

n i i

X X X

hx x i x x x c x x kx k c x

i i i n i i

i in in

But

X X

c x kxk c x K kxk

i i i i

in i

where K is the basis constantforx Thus

K kx k x hx x i x

n i

and hence x is a basis for X

Wesay that a basis x isshrinking if lim kx k foreach x X

n n n

as in our last result That is x is shrinking if x is a basis for X The

natural basis e is shrinking in c and p but not in

n p

If X has a shrinking basis then X can b e represented in terms of the

basis to o

Theorem If x is a shrinking basis for a Banach space X then X

can be identied with the spaceofallsequences of scalars a for which

a x k The c sup k orrespondence is given by

i i

x x x x x a a

and the norm of x is equivalent to and in case the basis constant is

actual ly equal to sup k x x x k

n i

Proof By renorming X if necessarywemay assume that the basis constant

of x is This will simplify our arithmetic

Mimicking a few of our previous calculations and using the fact that x

x x is a basis with constant its not hard to see that P x x

n i

P x x and kP x kkP x kkx kThus wehave

n n n

kx k lim kP x k supkP x k

n n

Converselyifa is a sequence of scalars such that sup k a x k

n i i

x satises a then anyweak limit p oint x of the b ounded sequence

i i

x x a for all iThus since x isabasisfor X the functional x is

i i

uniquely determined and it follows that x weak lim a x

n i i

CHAPTER BASES AND DUALITY

Notethatifx is a shrinking basis for X then the canonical image of X

in X corresp onds to those sequences a for which a x is not only

n i i

b ounded but actually converges in norm Indeed from wehave

x xx x x x xx xa a

for any x a x in X

n n

Related to our rst question is this If X has a basis do es X havea

basis The answer turns out to b e Yes and in fact X can b e shown to have

a shrinking basis but this is hard to see A somewhat easier question is If

Y has a basis y what prop ertyofy will tell us whether Y is actually

n n

the dual of some other space with a basis X Our next result supplies the

answer

A basis x for a Banach space X is called boundedly complete if a x

n n n

a x k Note that the usual basis e for converges whenever sup k

i i n

p clearly has this prop erty On the other hand e fails to b e

p n

a b oundedly complete basis for c

Theorem If a basis x for a Banach space X is shrinking then x is

aboundedly complete basis for X Ifabasis x for a Banach space X is

boundedly complete then X is isomorphic to the dual of a Banach space with

a shrinking basis

Proof Supp ose that x is a shrinking basis for a Banach space X We already

know that x isabasisforX we need to showthatx is b oundedly com

n n

kM a x plete So supp ose that a is a sequence of scalars suchthatk

i n

for all n

Toshowthat a x converges in X its enough byBanachSteinhaus

to show that the series a hx x i converges for each x X For this we

check that the sum is Cauchy for each x c x But

n n

m m

X X

a hx x i x a x

n n

in in

n m

X X

a x c x

n i i

i in

m n

X X

c x x a

i i n

in i

M c x as m n

i i

Now supp ose that x is a b oundedly complete basis for X Well prove

that X is isomorphic to the dual of Y x X To b egin let J X Y

b e dened byJxy y x That is Jx xj the functionalx considered

as a functional on Y Note that kJxk kxk for free and so J is at least

continuous The claim here is that J is an onto isomorphism

ToshowthatJ is an isomorphism its enough to consider the action of J

on span x that is on vectors of the form x a x Given suchan x

n i i

cho ose x X with kx k and x xkxk Then

kxk x x x P xP x x JxP x

n n

b ecause P x Y And since kP x kK where K is the basis constantof

n n

x wehave that kxk K kJxkThatisJ is b ounded b elowandthus is an

isomorphism

Toshowthat J is onto observe that the sequence Jx Y is biorthogo

nal to x Thus Jx is a basic sequence in Y with the same basis constant

K as x In particular given y Y wehave k y x Jx k K ky k

n i

y x x for all n Thus since x is b oundedly complete the series

n n

converges to an element x X Its not hard to see that Jx y

The notions of shrinking and b oundedly complete taken together charac

terize reexivity for spaces p ossessing a basis

Theorem Let X be a Banach space with a basis x Then X is reexive

if and only if x is both shrinking and boundedly complete

ProofIfx is shrinking then from Theorem X corresp onds to the

a x k and X corre collection of all sequences a for whichsup k

i i n

sp onds to the collection of all sequences a for which a x converges

n i i

in X Ifx is also b oundedly complete its clear that these two collections

are identical Comparing and wemust have X X

w supp ose that X is reexive As weve seen this already implies that No

x is a basis for X thatisx is shrinking It remains to showthatx

n n

is b oundedly complete

Now since x is a basis for X it has a corresp onding sequence of co ef

x that is inX But its easy to see that x cient functionals x

n n

x is just x considered as an elementof X Why Since X is reexive

its clear then that x is a basis for X X From Theorem this

is a shrinking basis for X which in turn means that x is means that x

a b oundedly complete basis for X Of course all of this means that x itself

n is b oundedly complete

CHAPTER BASES AND DUALITY

Notes and Remarks

All of the results in this chapter are due to R C James in fact our

presentation is largely based on his Monthly article Time p ermitting

we will give alternate versions of the main results of this chapter in a slightly

dierent setting In particular Theorem will b e transformed to read A

Banach space X with an unconditional basis a notion that we will encounter

later is reexive if and only if X do es not contain an isomorphic copy of either

c or This to o is due to James

Exercises

Let e denote the usual basis of Whatise in

If x is a basis for X show that x X is a sequence biorthogonal

n n

to the basic sequence x In particular x is a basic sequence with

the same basis constantasx

If X has a basis x with canonical pro jections P then P x x

n n

for every x X In other words x isa weak basis for X

x in X showthatk If x x klim inf kx kIfkx k

n n n

kx k for all n conclude that kx k lim kx k

Check directly that e is not a shrinking basis for and not a b ound

edly complete basis for c

Provethat C is not separable and conclude that C do es not

have a shrinking basis Hint Consider the collection f t g

where is the p oint mass at t that is the Borel measure dened by

A if t A and A otherwise

t t

Provethat C is not isomorphic to a dual space and conclude that

C do es not have a b oundedly complete basis Hint Theorem

and Exercise on page

If x is a b oundedly complete basis for X show that anyblock basis

y ofx is also b oundedly complete

k n

CHAPTER BASES AND DUALITY

Chapter

L Spaces

Throughout L L L B m where B is the Borel algebra

p p p

and m is Leb esgue measure As a collection of equivalence classes under equal

ity ae it makes little dierence whether we use the Borel algebra or the

larger algebra of Leb esgue measurable sets We will almost surely b e glib

ab out the use of almost everywhere and its relatives Muchofwhatwell

havetosay holds in any space L butasweve already p ointed out the

spaces L and are the most imp ortant for our purp oses If and when we

p p

have recourse to use other measure spaces we will b e careful to say so Given

anyinterval on the line though Leb esgue measure is always understo o d

Basic inequalities

We b egin with a survey of imp ortant inequalities most of these you already

know

First on the list has to b e Holders inequality Given pletq

denote the conjugate exp onen t dened byp q If f L and

g L then fg L and kfgk kf k kg k Equality can only o ccur if jf j

q p q

and jg j are prop ortional that is jf j jg j for some constants not

b oth zero

In the case p and q the inequality is nearly obvious Equalityin

this case can only o ccur if jg j kg k almost everywhere on the supp ort of f

It follows that kf k kf k whenever p q and f L More

p q q

generallyifX p q andf L then kf k

q p

kf k Equality can only o ccur if jf j c that is jf j must b e X

ae constant Well give another pro of of this fact shortly

CHAPTER L SPACES

Twovariants of Holders inequality are also useful Eachisproved byan

appropriate application of the regular version of the inequality The rst is

simply a fancier formulation that well call the generalizedHolder inequality

Let p q r satisfy r p q If f L and g L

p q

then fg L and kfgk kf k kg k The second is called Liapounovs

r r p q

inequality Let p q and If r p q then

r p

kf k kf k kf k for any f L L What this means is that the

p q

r p

function log kf k is a convex function of r Well have more to say ab out this

shortly

Most authors use Holders inequality to deduce Minkowskis inequality al

though its p ossible to do this just the other way around Given p

and f g L wehave kf g k kf k kg k Equality can only o ccur if

p p p p

f g for some nonnegative constants not b oth zero

Again the inequality is nearly ob vious in case p orp although the

case for equalitychanges For p equality can only o ccur if jf g j jf j jg j

that is only if f and g have the same sign or in brief only if fg For

p the case for equality is harder to state more or less equality o ccurs

only if f and g have the same sign on some set of p ositive measure where b oth

functions attain their norms

Its of interest to us that Minkowskis inequality reverses for p

That is if f and g are nonnegative functions in L for p then

kf g k kf k kg k Well sketch a pro of of this later

p p p

Everything weve had to sayth us far with one obvious exception has

little if anything to do with the underlying measure space Nevertheless its

worthwhile summarizing the dierences in at least one imp ortant case Recall

that we can identify with L where is counting measure on f ng

Given pq and x x x R wehave kxk kxk

n q p

n kxk Equality in the rst inequality forces x ce for some k that is

q k

x can have at most one nonzero term Equality in the second inequality forces

jxj c that is jxj must b e constant The rst inequality is easy

to pro ve using induction and elementary Calculus the second is just Holders

inequality in this new setting

Convex functions and Jensens inequality

The fact that the various inequalities weve just discussed do not dep end on

the underlying measure space is easiest to explain bysaying that the function

x jxj is convex Recall that a function f I R dened on some

interval I is said to b e convex if

f x y f x f y

for all x y I and all Geometricallythissays that the chord

joining x f x and y f y lies ab ove the graph of f Equivalently f is

convex if and only if its epigraph

epi f f x y f x y g

is a convex subset of I R

Nowiff is convex and if a c b in I then

b c c a

c a b

b a b a

and hence

b c c a

f c f a f b

b a b a

By juggling letters we can rewrite this inequalityas

f b f a f b f c f c f a

c a b a b c

whenever a cbIfwe apply this same reasoning to any x y c d

a b weget

f y f x f b f d f c f a

c a y x b d

It follows that f is lo cally Lipschitz and hence absolutely continuous on

each closed subinterval of I Thus f exists ae and from f is increas

ing If f is twice dierentiable for example wewould have f

The last conclusion is the one were really after If f is twice dierentiable

everywhere in I then f is convex if and only if f The backward

implication is easy to ll in If f then f is increasing Thus from the

mean value theorem

f x y f x f x y y x

f y f x y

which after a bit of symb ol manipulation yields

CHAPTER L SPACES

Given this simple criterion there are plentyofconvex functions around

Any concave up function from elementary Calculus will do In particular e

and jxj where p are convex A function f for which f is convex

is called concave thus jxj where p and log x are concave

As a quick application of convexity lets prove Minkowskis inequality

Given f g L pwithf and g not b oth zero wehave

jf g j jf xj jg xj jf j jg j

kf k kg k kf k kg k kf k kg k

p p p p p p

where kf k kf k kg k and kg k kf k kg k Thus since

p p p p p p

jxj is convex

p p p

jf g j jg j jf j

p p

kf k kg k kf k kg k

p p

Integrating b oth sides leads to kf g k kf k kg k which is the same

p p

as Minkowskis inequality Note that for pand f g the inequality

reverses since jxj is concave

Returning to our geometric interpretation of convex functions note that if

f is convex on I andif x I then f dominates any of its tangent lines at

x That is if m lies b etween the left and righthand derivatives of f at x

then

x f x f x mx

In particular if f is dierentiable at x thenf x f x f x x x

For dierentiable functions this last inequality implies that f is increasing

and so characterizes convexity using the same mean value theorem pro of we

gave a moment ago Royden calls y f x mx x asupporting line

for the graph of f at x

Finally lets return to our discussion of L spaces byproving a classical

inequality due to Jensen in

Theorem Jensens Inequality Let beaconvex function on R and let

f L Iff L then

Z Z

f t dt f t dt

Proof Let f and let y mx b e a supp orting line at

Then

f t mf t

Integrating b oth sides of the inequality do es the trick

f f

for example Closer to home if It follows that if f L then e e

R R

f L p then jf j jf j This is the alternate pro of that we

alluded to earlier Well see another application of Jensens inequality shortly

A test for disjointness

We next discuss a few inequalities that are intimately related to the lattice

structure of L For example its clear that if f and g are disjointly supp orted

p p p p p

in L then jf g j jf g j jf j jg j ae and hence kf g k

p p p

kf g k kf k kg k More interesting though would b e to rephrase this

p p p

fact as a test for disjointness of two functions in L For this we need another

elementary inequality

Lemma Given p and a b R we have

p p p p

j a b j j a b j jaj jbj

For p we get equality for any a bFor p the inequality reverses

For p equality can only occur if ab

Proof We apply our basic inequalities in the tw odimensional L space

Sp ecically

p p p p

j a b j j a b j j a b j j a b j byHolder

p p p

jaj jbj

p p

jbj since p jaj

Equality in the rst inequality forces j a b j j a b jorab The same

conclusion holds if wehave equality in the second inequality The pro of in

case p is essentially identical

Theorem Let f g L p p Then f and g are

disjointly supported if and only if

p p p p

k f g k k f g k kf k kg k

p p p p

CHAPTER L SPACES

Proof The function

p p p p

h j f g j j f g j jf j jg j

is of constant sign and integrates to give

p p p p

hd k f g k k f g k kf k kg k

p p p p

Thus f and g are disjointly supp orted if and only if fg if and only if

h if and only if hd

Corollary If T L L is a linear isometry p p

p p

then T maps disjoint functions to disjoint functions

Proof If T is an isometrythen kTfk kf k kTgk kg k and kTf

p p p p

Tgk kf g k for any f and g in L Consequently f and g are disjoint

p p p

if and only if Tf and Tg are disjoint

This last result tells us something ab out the subspace structure of L

We already know that L contains an isometric copyof spanned byany

p p

sequence of disjointly supp orted nonzero functions and nowweknow that

this is the only waytogetan isometric copyof inside L

p p

L contains other natural sublattices For example if weidentify the space

L with the collection of all functions f L which are supp orted

p p

in for example then L is a complemented sublattice of L

p p

which is isometric to all of L The pro jection is obvious

Pf f

t and the isometry is nearly obvious If we set Tft f t for

and f L then Tf L and

p p

Z Z

p p p p

jTftj dt jf tj dt kf k kTfk

p p

More generally L a b is isometric to L and from this its a short leap to

p p

the conclusion that L and L R are isometric to L Indeed if we

p p p

partition into countably many disjoint nontrivial intervals I then

L L I L I

p p p

L L L

p p p

Of course L is also isometric to a sublattice of L

p p

Conditional exp ectation

Less obvious examples of L subspaces of L are generated by sub algebras

p p

of the Borel algebra B Given a sub algebra B of B we dene the space

L B L B m to b e the collection of all B measurable functions

p p

in L Thus L B is a closed subspace and even a sublattice of L In fact

p p p

L B is complemented in L by a norm one pro jection Whats more any

p p

subspace of L which is the range of a norm one pro jection has to b e of this

form Well prove the rst claim in some detail the second claim isnt terribly

hard to prove but well leave the details for another time

By way of a simple example supp ose that we partition into nitely

many disjoint measurable sets A A withmA for each k and let

n k

B b e the algebra generated by the sets A TheB measurable functions are

a and the space L B precisely the simple functions of the form

k A p

is then isometric to

It might also help matters if we wrote out the pro jection onto L B In

this case its given by

Pf f

Recall that kPfk kf k review the computation we used to show that the

p p

Haar system is a basis for L Notice that Pf is constant on each A of

p k

R R

course and that Pf f for every B measurable set A Indeed

A A

Z Z Z Z

f f Pf

A A A A

j j j j

This simple example is not far from the general situation In some sense

f is averaged over all B measurable sets to arriveatits B measurable

counterpart

In order to build a norm one pro jection from L onto L B we need

p p

some waytomapB measurable functions into B measurable functions with

out increasing their norms The secret here is what is known as a conditional

exp ectation op erator Given an integrable B measurable function f we de

ne the conditional expectation of f given B to b e the ae unique integrable

B measurable function g with the prop ertythat

Z Z

f g for every A B

A A

CHAPTER L SPACES

and we write g E f jB The existence and uniqueness of g follow from the

RadonNikodym theorem That is if wedeneA f for A B and

if wethinkofm as b eing dened only on B then is absolutely continuous

with resp ect to m and g is the RadonNikodym derivative ddm

Its often simpler to think of as the dening statement though after

all g is completely determined as an elementof L bythevalues g as A

runs through B For example its easy to see how forces E f jB to

b e b oth linear and p ositive Consequentlywemust also have jE f jB j

E jf jjB and now its clear that conditional exp ectation is a contraction

on b oth L and L Indeed if f L then

Z Z Z

jE f jB j E jf jjB jf j kf k kE f jB k

And if f L then

jE f jB jE jf jjB E kf k jB kf k

since constant functions are B measurable Thus kE f jB k kf k

As it happ ens a p ositive linear map which is a simultaneous contraction

on L and L is also a contraction on every L p That this is

so follows from an application of an ingenious interp olation scheme due to

Marcinkiewicz in

Theorem Suppose that T L L is a linear map satisfying

i Tf whenever f ie T is positive

ii kTfk kf k for al l f L and

kf k iii Tf L whenever f L and kTfk

Then Tf L whenever f L for any p andkTfk kf k

p p p p

Proof Since jTfjT jf j its enough to consider the case where f So

let p and let f L Now for each xed y we can write

and where y is f f R wheref f y and R f f f y

y y y y y

also used to denote the constant function y L

Since f y wehave that T f T y y from

y y

iii and so f

T f T R y T R

Tf y

y y y

f y

But T R b ecause R hence

y y

Tf y T R T f y

Next integration gives

Z Z Z

Tfy T f y kT f y k kf y k f y

The rest of the pro of consists in upgrading this inequality to an estimate

involving the L norms of Tf and f

We dene auxiliary functions and by

y y

if f x y if Tfx y

x and x

y y

otherwise otherwise

Then f y f y and Tf y Tf y In this notation

y y

our last estimate says that

Z Z

Tf y x x dx f y x x dx

y y

for every y Consequently

Z Z

x dx dy y Tfx y

Z Z

f x y x dx dy y

Since all the functions involved are nonnegative Fubinis theorem yields

Z Z Z Z

p p

f x y x dx dy y f x y x dy dx y

y y

Z Z

f x

y f x y dy dx

f x dx

p p

CHAPTER L SPACES

and similarly

Z Z Z

p p

y Tfx dx Tfx y x dx dy

p p

Thus Tf L and kTfk kf k

p p p

Corollary Conditional expectation is a simultaneous contraction on every

Some authors prove Corollary by rst proving a version of Jensens in

equalityvalid for conditional exp ectations in particular it can b e shown that

jE f jB j E jf j jB One pro of of this fact rst reduces to the case

of nonnegative simple functions where the inequalityismoreorlessimmedi

ate and then app eals to a monotone convergence theorem for conditional

exp ectations recall that conditional exp ectation is monotone In anyevent

integration then yields kE f jB k kf k

p p

There is an another approach to conditional exp ectation that warrants at

least a brief discussion In this approach one b egins by dening conditional

exp ectation for L functions Sp ecicallywe dene E jB to b e the or

thogonal pro jection from L onto L B Since L is dense in L we can

extend this denition to L functions by taking monotone limits Since each

f L is the limit in L norm of an increasing sequence of b ounded

hence L simple functions we can take E f jB lim E jB

n n

The savings in using Marcinkiewicz interp olation is clear Wehave b een

spared the tedium of the typical limit of simple functions calculation In

stead we cut to the heart of the matter Each f L can b e written as the sum

of an L function and an L function namely f f R Marcinkiewicz

y y

then estimates the L norm of E f jB andtheL norm of E R jB

y y

Finallywe should at least state the converse due to Douglas for

p and Ando for p

Theorem Let p and let P L L beapositive linear

p p

contractive projection with P Then there exists a sub algebra B

such that Pf E f jB for every f L

Notes and Remarks

An excellent source for inequalities b oth big and small including a discussion

of convex functions and Jensens inequality Theorem is the classic b o ok

Inequalitiesby Hardy Littlewood and Polya The results on disjointly

supp orted functions in L and their preservation under isometries were essen

tially known to Banach and form the basis for a more general result due to

Lamp erti Lo osely stated here is Lamp ertis result for isometries on L

Theorem Let p p and let T be a linear isometry on

L Then thereisaBorel measurable map from onto almost

al l of and a norm one h L such that

Tf h f

The function h T is uniquely determined ae and is uniquely deter

mined ae on the support of h Moreover for any Bor el set E we have

mE jhj

Lamp ertis full result handles isometries from L into L details can b e

p p

found in Royden as can much of the material in the rst three sections

Conditional exp ectation op erators on L are bynow part of the folklore

of Banach space theory but can b e traced to a rash of pap ers from the s

and s almost all of which app eared in the Pacic Journal of Mathemat

ics probably b eginning with Moy and certainly culminating in Ando

Marcinkiewiczs theorem Theorem is a classic and can b e found in any

numb er of b o oks see for example The pro of given here is taken from

my notes for a course oered by D J H Garling at The Ohio State University in the late s

CHAPTER L SPACES

Exercises

Let p q r satisfy r p q Iff L andg L

p q

showthatfg L andkfgk kf k kg k

r r p q

Let p q and If r p q show that

r p

kf k kf k kf k for an y f L L

p q

r p

Given p and f g L use Holders inequalitytoprove that

kf g k kf k kg k Further show that equality can only o ccur if

p p p

f g for some nonnegative constants not b oth zero

For p show that equality in Minkowskis inequalityfor L can

only o ccur if jf g j jf j jg j that is only if f and g have the same

sign Find a necessary and sucient condition for equality when p

Given pq and x R prove that kxk kxk

q p

n kxk Further show that equality in the rst inequality forces

x ce for some k while equality in the second inequality forces jxj to

b e constant

Let f I R b e convex and let x I

a Showthatf x f x where f resp f is the lefthand

l r l r

resp righthand derivativeof f at x

b If m lies b etween the left and righthand derivativeof f at x

prove that f x f x mx x for all x I

Prove Lemma in the case p

If B is a sub algebra of B provethatE jB is b oth linear and

p ositive and conclude that jE f jB j E jf jjB for every f L

If B is a sub algebra of B provethat E jB is a pro jection on every

L Hint Use to compute E E f jB jB

Given pa sub algebra B of B and a nonnegative simple

function f L showthatjE f jB j E jf j jB

Let B be a sub algebra of B ProvethatPf E f jB is the

orthogonal pro jection from L onto L B Hint Use to show

measurable simple function g that g f E f jB for every B

and from this conclude that E f jB is orthogonal to f E f jB

Chapter

L Spaces II

Weve seen that L contains copies of and L but do es L contain or

p p p p q

L for p q In this chapter well settle the question completely in the

case pWell consider the remaining cases in the next chapter

We b egin though with a classical result showing that every L contains a

natural copyof

The Rademacher functions

We b egin by describing an orthonormal system of functions on of great

imp ortance in b oth classical and mo dern analysis The Rademacher functions

r are dened by r t sgn sin t

n n

r 1 r 2 r 3

h The Rademacher functions are related to the Haar system by r

k n

and from this it follows that r is an orthonormal sequence in L Alterna

tively notice that if nm then r is constantoneach of the p erio ds of r

n m

CHAPTER L SPACES I I

and hence r r Since each r is mean zero though its clear that r

n m n n

isnt complete that is the linear span of r isnt dense in L Along the

same lines notice that if n then r r is constantoneach of the p erio ds

of r and hence r r r Thus r r is orthogonal to the closed linear

n n

span of r inL

If wewere to work just a bit harder we could show that r r r

n n n

for any n n n The Rademacher functions along with the

functions of the form r r r and the constant function sometimes

n n n

lab eled r form the collection of Walsh functions a complete orthonormal

basis for L

The Rademacher functions are imp ortant from a combinatorial p ointof

view to o Notice that as t ranges across the vector r tr t

represents all p ossible choices of signs where Moreover

n i

each particular choice is equally likely o ccurring on a set of mea

P P

n n

a for some random a r t sure Wemightsay that

k k k

k k

choices of signs In particular

n n

X X X

a r t dt a

k k k k

k k

all

Average of a over all choices

An imp ortant classical inequality due to Khinchine in b elow tells us

how to estimate this average Note also that

n n

X X

a r t ja j max

k k k

k k

The RieszFischer theorem tells us that a r converges in L if and

n n

P P

a r converges in L if and a As it happ ens though only if

n n

P P

a Thus for example a r converges ae if and only if only if

n n

the series converges for almost all choices of signs

In probabilistic language the Rademacher functions form a sequence of

independent Bernoul li random variables That is the r are indep endent

identically distributed mean zero random variables with P f r g

a r then P f f A g P ff A g P f r g Also if f

k k n

that is f is a symmetric random variable

There are a couple of go o d ways to see that the Rademacher functions are

indep endent Probably easiest is this Each t has a binary decimal

expansion t t where t The are clearly indep en

n n n

dent and its easy to checkthat r t t Alternatively but along

n n

the same lines we could identify with the pro duct space f g by

means of the map t r t in whichcase r is the nth co ordinate pro jec

n n

tion t r t If weendow f g with the pro duct mea

k n n

sure induced by taking counting measure in each co ordinate then f g is

measuretheoretically equivalentto The p oint to this approach is that

distinct co ordinate pro jections in a pro duct space are the canonical example

of indep endent functions No matter howwe decide to say it what wewould

need to checkis

P fr r g P fr gP fr g and

n m n m

P fr r g P fr gP fr g

n m n m

Induction takes care of the rest

Because the Rademacher functions are indep endent it can b e shown that

Z Z Z

f f r t dt f r t dt r t f r t dt

n n n n

for any n and any continuous functions f f Some authors take this

formula as the dening prop erty of indep endence and we could do the same

its the only consequence of indep endence that well need

Khinchines inequality

Its a fact from classical Fourier analysis that any lacunary trig sequence such

as sin t satises

n n

X X

a sin t ja j

k k

for every p Its not to o surprising that the Rademacher sequence

shares this prop erty

Theorem Khinchines Inequality Given p there exist con

stants A B such that

p p

n n n

X X X

B ja j A ja j a r

p k p k k k

k k k p

CHAPTER L SPACES I I

for al l n and al l scalars a

Proof It should b e p ointed out that Khinchines inequalityworks equally well

for complex scalars although our pro of will use real scalars but that it do es

not hold for p recall equation

Note that when p we can take A B That is we can write

n n n

X X X

A a r a r B a r

p k k k k p k k

k k k

Thus when pwe need only nd A and wetake B B and

p p

when p we need only nd B and wetake A A

p p

As well see shortly its enough to establish the righthand side of for

all p In fact its enough to consider the case where p is an even integer

p m m In this case we use the binomial theorem to compute

a r t dt

k k

k k k k

k k

n n

r r r a a a

n n

k k k

k k m

where wehave used the multinomial coecients

m m

k k k k k

n n

and where the sum is over all nonnegativeintegers k k k whichsumto

mNow the integral in this formula is unless every k is even in which case

its Thus

X X

k k

dt a a r t a a

k k

k k k

k k m k

If we could replace the multinomi al co ecientinthisformula by

k k k

a on the righthand side So lets compare these then wed have

two co ecients

k k m

k k k

k k m

k k k

mm m

k k k k

n n

m m

k k

Thus

X X

k k

k m

a a a r t a dt m

k k

k k k

k k k m

a m

P P

n n

or k a r k m a That is B m will work In

k k m

k k

the general case we can take B p since p p

p That is

n n n

X X X

a r a r p a

k k k k

k k k

p p

Now when p we use Liap ounovs inequality Cho ose

such that p Then f a r satises

k k

p p p

kf k B kf k kf k kf k kf k kf k kf k

p p p

Thus kf k kf k b ecause p and since B

hence

p p

kf k kf k kf k

b ecause p That is we can take A

Just for fun heres another pro of of the hard part of Khinchines in

equality As b efore the hard work is establishing the righthand side of

for p Also just as b efore wemay assume that p is an integer but

p jf j

not necessarily even In this case jf j p e and so its enough to show

P P

n n

jf j

since is that e C whenever f a r and a

k k

homogeneous Here go es

Z Z

jf j f

e e b ecause f is symmetric

CHAPTER L SPACES I I

Z Z

n n

X Y

a r

k k

exp a r e

k k

a r

k k

e by indep endence

cosh a

n x

But cosh x x n e Thus

n n

X Y Y

a e exp cosh a e

k k

And hence

p jf j p

kf k p e p e

p p

That is B p e in case p is an integer and so B p e in

p p

general Obviously this value of B isnt quite as go o d as p but

then this pro of is shorter

Khinchines inequality yields an immediate corollary

Corollary For any p the Rademacher sequence r spans an

then r is even complemented isomorphic copy of in L If p

n p

in L

Proof The rst assertion is clear In any L p the Rademacher

sequence is equivalent to the usual basis of Thus the closed linear span of

r inL is isomorphic to

n p

The second assertion is also easy For p wehave the orthogonal

pro jection dened by

Pf h f r i r

n n

Of course

jh f r ij kf k kPfk

for any f L The claim here is that this same pro jection is b ounded in every

L for pFor p thisisimmediate from Khinchines inequality

B kPfk B kf k B kf k kPfk B jh f r ij

p p p p p p n

For pwe use duality Since r is orthonormal the pro jection P

is selfadjoint that is

h f r ihg r i h Pgf i h g P f i

n n

Thus P P is a pro jection onto r with norm at most B whereq

n q

pp is the conjugate exp onent and q here Since were new to these

sorts of computations it couldnt hurt to say this another way Notice that if

f L is xed and g L is arbitrarywehave

p q

jh g P f ij jh Pgf ij kPgk kf k B kf k k g k

q p q p q

Thus since g is arbitrarywehave Pf L and kPfk B kf k

p p q p

Its known that the closed linear span of the Rademacher functions is not

complemented in L Indeed as well see no complemented innite dimen

sional subspace of L can b e reexive In L the Rademacher functions span

an isometric copyof recall equation whichisknown to b e uncom

plemented in L Nevertheless L do es contain an isometric copyof

Indeed has a countable norming set and hence is isometric to a subspace

of This subspace is likewise uncomplemented

Khinchines inequality tells us even more

Corollary For p p the spaces L and are not isomor

p p

phic

Indeed as weve seen p p cannot contain an isomorphic

copyof Curiously though just as with and L the spaces and L

are even isometric but this is hard

The KadecPelczynski theorem

Were now in a p osition to make a big dent in the problem raised at the

b eginning of this chapter In particular well completely settle the question of

which L spaces embed into L for p The key is to generalize one of the

q p

prop erties of the Rademacher sequence We will consider subspaces of L on

which all of the various L norms for qp are equivalent The results in

this section are due to Kadec and Pelzynski from While this is one

CHAPTER L SPACES I I

of the newer results that well see it requires no machinery that was unknown

to Banach in this sense it qualies as a classical result

For andp consider the following subset of L

M p f f L mf x jf xjkf k g g

p p

M p L Notice that if thenM p M p Also

since for any nonzero f L wehave mfjf j gmf f g as

In fact any nite subset of L is contained in an M p for some

Finallynotethatiff M p then so is f for any scalar in

particular M p

The elements of a given M p have to b e relatively at Notice for

example that every elementof M p is constant ae Indeed

mfjf jkf k gjf j kf k ae

p p

Along similar lines notice that M p do esnt contain the spike f

f f g for any In this case kf k while m

The following Lemma puts this observation to go o d use

Lemma For a subset A of L the fol lowing areequivalent

i A M p for some

ii For each q p there exists a constant C such that kf k

q q

kf k C kf k for al l f A

p q q

iii For some q p there exists a constant C such that kf k

q q

kf k C kf k for al l f A

p q q

Proof i ii If f A M p then

Z Z Z

q q q

jf j jf j jf j

fjf jkf k g fjf jkf k g

p p

q q

mfjf jkf k g kf k

q q

kf k

q q

That is kf k kf k

q p

ii iii is clear so next is iii i Supp ose that f A but that

f M p for some Wewillshow that iii then implies for

some and hence that A M p forall

Let S fjf jkf k g Then of course mS b ecause f M p

Next write q p r for some r Thus qp qr and so pq

and rq are conjugate indices Now lets estimate

Z Z

q q q

kf k jf j jf j

S S

p qr q q

jf j mS kf k

qr q q

kf k

q q q q qr q q

Thus from iii kf k C kf k C kf k Since f this means

p q q q p

that must be bounded away from that is there exists some such that

If an entire subspace X of L is contained in some M p then the L

p p

and L top ologies evidently coincide on X for every q p In fact its

even true that the L and L top ologies agree on X whereL carries the

top ology of convergence in measure Most imp ortant for our purp oses is this

If p and if X is a closed subspace of L whichiscontained in some

M p then the L and L top ologies coincide on X That is X must b e

isomorphic to a Hilb ert space How so Well if kf k kf k C kf k for all

f X then the inclusion map from X into L is an isomorphism Since every

closed subspace of L is isometric to a Hilb ert space X must b e isomorphic

to a Hilb ert space

Nowiff L kf k and f M p then we can write f g h

p p

where g is supp orted on a set of measure less than and where khk

Indeed

and h f g f g f

fjf j g fjf j g

do the trick Note that the function g is a spike since kg k while

msupp g

Next supp ose that a subspace X of L fails to b e entirely contained in

M p forany Then in particular the set S of all norm one vectors in

X isnt contained in any M p Thus given a sequence of p ositivenumbers

we can nd a sequence of norm one vectors f S such that

n n X

f M p Each f can b e written f g h where g is supp orted

n n n n n n n

on a set of measure less than and where kh k That is f isa

n n p n n

small p erturbation of the sequence of spikes g The claim here is that if

fast enough then the seqence g is almost disjoint If so then g

n n n

and hence also f will b e equivalent to the usual basis in This claim is

n p

worth isolating as a separate result

CHAPTER L SPACES I I

Lemma Let f beasequence of norm one functions in L p

n p

If msupp f then some subsequenceoff is equivalent to a disjointly

n n

supportedsequenceinL

Proof The key observation here is that if f L is xed then the measure

A jf j is absolutely continuous with resp ect to m In particular for

each there is a suchthat A whenever mA

Let A supp f Then mA By induction we can cho ose a

n n n

subsequence of f whichwe again lab el f such that

n n

p np

jf tj dt

for all n Now let B A n A and let g f

n n B i n n

Obviously the sets B are pairwise disjoint and hence the sequence g is

n n

disjointly supp orted in L Finally

Z Z

X X

ip np p p p

g k jf tj dt jf tj dt kf

n n n n

A A nB

n n

in in

and so kg k Since g is a monotone basic sequence

n p n

in L the functionals g biorthogonal to g satisfy kg k

p n q

n n

where p q Thus kg k kf g k An

q n n p

app eal to the principle of small p erturbations Theorem completes the

pro of

Its easy to mo dify our last result to work for seminormalized sequences

In our particular application this means that if we can nd a sequence of

norm one vectors f M p where fast enough then f isa

n n n n

small p erturbation of a sequence of spikes g with kg k and

n n p n

msupp g The seminormalized sequence g is in turn a small p ertur

n n n

bation of a disjointly supp orted sequence in L Thus f is isomorphic to

p n

and complemented in L Its high time we summarized these observations

p p

Theorem Let p and let X be an innite dimensional closed

subspaceofL Then either

i X is containedinM p for some in which case X is isomorphic

to and the L and L topologies agreeonX or

ii X contains a subspace that is isomorphic to and complementedinL

p p

Corollary For p and r r p nosubspaceofL

is isomorphic to L or to

r r

Proof Since L contains an isometric copyof it suces to check that

r r r

is not isomorphic to a subspace of L But for r pweknowthat

p r

is neither isomorphic to nor do es it contain a subspace isomorphic to

Recall the discussion surrounding Theorem Thus neither alternativeof

Theorem is available

As it happ ens every copyof in L p is necessarily complemented

Curiously this fails in general for p

Corollary Let X be a subspaceofL p IfX is isomorphic to

thenX is complementedinL

Proof Since X cannot contain an isomorphic copyof wemust have X

M p for some Thus there is a constant C suchthatkf k kf k

C kf k for all f X Asweve seen this means that X can also b e considered

as a subspace of L As such w e can nd an orthornormal sequence in

L such that X is the closed linear span of where the closure can b e

computed in either L or L since the two top ologies agree on X Now let P

b e the orthogonal pro jection onto X sp ecicallyput

Pf h f i

n n

Just as wesaw with the Rademacher functions the pro jection P is also

boundedasamaponL Indeed since Pf X wehave

kPfk C kPfk C kf k C kf k

p p

Corollary Let p and let X be an innite dimensional closed

subspaceofL Then either X is isomorphic to and complementedinL or

p p

X contains a subpsace that is isomorphic to and complementedinL

p p

By considering only complemented subspaces we can use duality to transfer

a mo died version of these results to L for p The principle at work

is very general and well worth isolating

Lemma Let Y be a closedsubspace of a Banach space X If Y is

is isomorphic to a complemented subspaceof complementedinX then Y

CHAPTER L SPACES I I

Proof First recall that Y can b e identied with a quotientof X Sp ecically

Y X Y isometrically Recall that if i Y X is inclusion then

i X Y is restriction and has kernel Y Sincei is an isometry into i

is a quotient map

NowifP X X is a pro jection with range Y then P X X is a

pro jection with kernel Y Asweve seen P is indeed a pro jection To see

that ker P Y consider

P x h x P x i h Pxx i for all x X

h y x i for all y Y

x Y

Thus the range of P can b e identied isomorphically with X Y whichwe

knowtobeY isometricallyThatisP is a pro jection onto an isomorphic

copyofY

We will also need the following observation

Lemma A closed subspaceofareexive space is again reexive

Proof Let Y b e a closed subspace of a reexive Banachspace X Let i

Y X denote inclusion and let j Y Y and j X X denote

Y X

the canonical emb eddings Then i Y X is an isometry into which

makes the following diagram commute

X X

i i

Y Y

Thus all four maps are isometries into and j is onto b ecause X is reexive

Next we compare ranges for this we will need to compute annihilators but

only Our calculations will b e slightly less cumbersome if we in X and X

o ccasionally ignore the formal inclusion i

Now the range of i is Y b ecause the kernel of i is Y Thus Y is

reexive if and only if j is onto if and only if i j Y j iY Y

Y Y X

But for any subset A of X its easy to checkthat j AA since j

X X

is onto thus j Y Y In other words Y is reexive if and only if

Y Y iY Y

Corollary Let X be an innite dimensional complemented subspaceof

L p Then either X is isomorphic to or X contains a subspace

that is isomorphic to and complementedinL

p p

Proof Of course we already know this result when p So supp ose

that p and supp ose that X is a complemented subspace of L By

Lemma we knowthat X is reexiveandby Lemma weknow that

X is isomorphic to a complemented subspace of L where p q and

q Thus either X is isomorphic to orelseX contains a complemented

subspace isomorphic to NowifX is isomorphic to then surely X X

is to o FinallyifX contains a complemented subspace which is isomorphic

to thenX X contains a complemented subspace isomorphic to

q p

Corollary L is not isomorphic to L for any p q

p q

Now that weknow L pcontains complemented subspaces

isomorphic to and its p ossible to construct other less apparent com

plemented subspaces For example its not hard to see that and

Z are isomorphic to complemented subspaces of L For

p p p

p the spaces L and Z are isomorphically distinct although

p p p p

thats not easy to see just now In particular L do es contain complemented

subspaces other than and L InfactL contains innitely man y iso

p p p

morphically distinct complemented subspaces

Notes and Remarks

An excellent source for information on the Rademacher functions is the

pap er byREACPaley see also Zygmund The fancy pro of

of Khinchines inequality whichimmediately follows the classical pro of is a

minor mo dication of a pro of presented in a course oered by Ben Garling at

The Ohio State University in the late s Stephen Dilworth tells me that this

clever pro of was shown to Garling by Simon Bernau The constants A and

B arising from our pro ofs of Khinchines inequality are not b est p ossible

for example its known that A and that B m

for m are b est See Szarek and Haagerup Most of our

applications of Khinchines inequality to subspaces of L and indeed nearly

all of the resutls from this chapter are due to Kadec and Pelczynski but

much of the avor of our presentation is b orrowed from Garlings masterful

interpretation For more on the subspace structure of L see and p

CHAPTER L SPACES I I

Lemma is due to B J Pettis in but is bynow standard fare in

most textb o oks on functional analysis see Megginson for much more on

reexive subspaces

Exercises

If showthatM p M p

Provethat M p L

Prove Theorem

Showthat c is not isomorphic to a subspace of L for p

However c is isometric to a subspace of L

For p p provethat and Z are

p p p

isomorphic to complemented subspaces of L

Let Y b e a closed subspace of a Banach space X and let i Y X denote

inclusion Showthati X Y is restriction dened by i f f j

and that ker i Y Further showthati is an isometry into with

range Y

Let X b e reexive and let j X X denote the canonical emb ed

ding Showthatj AA for any subset A of X Is this true

without the assumption that X is reexive

Fill in the details in the pro of of Corollary Sp ecically supp ose

that X is a complemented subspace of L p and that X is

not isomorphic to Prove that X contains a complemented subspace

isomorphic to

Prove Corollary

CHAPTER L SPACES I I P

Chapter

L Spaces III

As p ointed out earlier the spaces L and exhaust the isomorphic typ es

p p

of L spaces Thus to b etter understand the isomorphic structure of L

p p

spaces we might ask as Banachdid

For p r can or L be isomorphical ly embeddedinto L

r r p

We know quite a bit ab out this problem Weknow that the answer is always

Yes for r and in case p the KadecPelczynski theorem

Corollary tells us that r is the only p ossibility Inthischapter well

prove

If p and r live on opposite sides of therecan be no isomorphism from

L or into L

r r p

This leaves op en only the cases rpand pr The rst

case can also b e eliminated as well see but not the second

Unconditional convergence

We next intro duce the notion of unconditional convergence of series What

follows are several plausible denitions

x in a BanachspaceX is Wesay that a series

a unorderedconvergent if x converges for every p ermutation one

toone onto map N N

CHAPTER L SPACES I I I

b subseries convergent if x converges for every subsequence x of

n n

k k

x converges for anychoice of signs c random signs convergent if

n n

d bounded multiplier convergent if a x converges whenever ja j

n n n

As well see in a complete space all four notions are equivalent Henceforth

we will say that x is unconditional ly convergent if any one of these four

conditions holds

In R all four of these conditions are equivalentto kx k that

is all are equivalent to the absolute summabilityof x In an innite

dimensional space however this is no longer the case It is a deep result

due to Dvoretzky and Rogers in that every innite dimensional

space contains an unconditionally convergent series which fails to b e absolutely

summable

We will briey sketch the pro of that a through c are equivalentand

leave condition d as an exercise but rst lets lo ok at an example or two If

P P

we x an element a e in pthen a e is unconditionally

n n p n n

n n

convergent but not absolutely convergent in general unless p Why

P P

Because a e converges in if and only if ja j which clearly

n n p n

n n

allows us to change the signs of the a For this reason we sometimes say that

has an unconditional basis Similarlyife isany orthonormal sequence

p n

P P

in a Hilb ert space H then a e converges if and only if ja j if

n n n

n n

and only if a e is unconditionally convergent

n n

Theorem Given a series x in a Banach space X the fol lowing are

equivalent

i The series x converges for every permutation of N

ii The series x converges for every subsequence n n n

iii The series x converges for every choice of signs

n n n

iv For every there is exists an N such that k x k for every

nite subset A of N satisfying min A N

Proof That ii and iii are equivalent is easy The fact that iv implies

i and ii is pretty clear since iv implies that each of the series in i and

ii is Cauchy The hard work comes in establishing i ii iv To this

end supp ose that iv fails Then there exist an and nite subsets A

of N satisfying max A min A and k x k for all n But then

n n i

P S

x do esnt converge A denes a subsequence of N for which A

i n

iA n

Thus ii fails Lastlywe can nd a p ermutation of N such that maps

the interval min A max A onto itself in sucha waythat A B is

n n n n

P P P

an interval Thus k x k But then x do esnt x k k

n i

i iA B

n n

converge and so i also fails

If x converges unconditionally to x then condition iv implies that

every rearrangement x likewises converges to x The same of course is

no longer true of the sums x On the other hand condition iv tells

n n

us that the set of all vectors of the form x is a compact subset of X

n n

Indeed from iv the map f f g X dened by f x is

n n n

continuous In particular if x is unconditionally convergent then

sup sup x

i i

It also follows from iv that if x is unconditionally convergent then

v a x converges for every boundedsequenceofscalars a

n n n

and the map T X dened by T a a x is continuous more

n n n

over the restriction of T to c is even compact Since v clearly implies

iii this means that condition v is another equivalentformulation of un

conditional convergence Since wehavenoimmediate need for this particular

condition wewillleave the details as an exercise

Orliczs theorem

In terms of the Rademacher functions or had you forgotten already we can

write another way If x is unconditionally convergent then

r tx sup sup

i i

t n

Armed with this observation wecanmake short work of an imp ortant theorem

due to Orlicz in We b egin with a useful calculation

CHAPTER L SPACES I I I

Prop osition For any f f L p we have

n p

n n n

X X X

A r sf ds B jf j j f j

p i i p i i

i i i

p p

where A B are Khinchine constants

p p

Proof This is a simple matter of applying Fubinis theorem First

p p

Z Z Z

n n

X X

r sf ds r sf t dt ds

i i i i

i i

Z Z

ds dt r sf t

i i

And now from Khinchines inequality

p p

Z Z Z

n n

X X

A dt r sf t ds dt jf tj

p i i i

i i

A jf j

p i

The upp er estimate is similar

We could paraphrase Prop osition by writing

n n

X X

r sf t jf tj

i i i

i i

The idea now is to either compare the lefthand side to say k f k as

we might for an unconditionally convergent series or to compare the right

P P

n p n

aswemight for disjointly kf k or to kf k hand side to

i i

i p i p

supp orted sequences As a rst application of these ideas we present a classical

theorem due to Orlicz

Theorem Orliczs Theorem If f is unconditional ly convergent

in L p then kf k converges

p n

Proof Since f is unconditionally convergent there is some constant K

suchthat k r sf k K for all n and all sThus from Prop osition

i i

there is some constant C suchthat jf j C for all n All that

remains is to compare this expression to kf k

But for pwehave

n n

X X

jf j jf tj dt

i i

n p

jf tj dt

kf k

where the inequality follows from the fact that p and the triangle

inequalityin L is reversed

Orliczs theorem reduces Banachs problem by eliminating one case

Corollary If p r orif r p then there

can be no isomorphism from L or into L

r r p

Proof Its clearly enough to showthat do esnt embed in L and weve

r p

already settled this question in case r p

Now supp ose that p r and that T L is an

r p

P P

isomorphism Then given a e in the series a Te is unconditionally

n n r n n

convergentin L Hence by Orliczs theorem wehave

X X

ja j kTe k kT k ja j

n n n

P P

That is ja j converges whenever ja j converges This is clearly imp os

n n

sible for r as the series n plainly demonstrates

For p the conclusion of Orliczs theorem changes

Theorem If f is unconditional ly convergent in L p

n p

then kf k converges

CHAPTER L SPACES I I I

ProofFor pwe use a dierent trick

n n

X X

jf j jf tj dt

i i

jf tj dt

kf k

where here weve used the fact that kk kk for p

To this p oint wehave completely settled Banachs question in all but the

cases rpand pr As wementioned at the b eginning of

this chapter the rst case can b e eliminated The argumentinthiscaseisvery

similar to that used in the pro of of Orliczs theorem this time we compute

k upp er estimates for r sf k ds

i i

Theorem For any f f L

n p

n n

X X

r sf ds kf k for p

i i i

i i

and

n n

X X

r sf ds B for p kf k

i i p i

i i

ProofAsweve already seen

Z Z

n n

X X

r sf ds r sf ds

i i i i

i i

p p

Z Z

ds dt r sf t

i i

dt jf tj B

i p

and B for p All that remains is to estimate this last expression from

ab ove But if p then p and so

Z Z

n n n

X X X

p p

dt jf tj dt kf k jf tj

i i i

While if pthen p andso

p p

Z Z

n n n

X X X

jf tj dt jf tj dt kf k

i i i

by the triangle inequalityin L

The choice of the expression k r sf k ds as opp osed to the ex

i i

in our last result is of little consequence pression k r sf k ds

i i

r sf k ds for any r All such We could have in fact used k

i i

jf j expressions are equivalent and hence all are equivalentto

see Theorem e or I I IA

Finally were ready to deal with the last case that can b e handled by

elementary inequalities

Corollary If rp therecan be no isomorphism from L or

into L

r p

Proof As b efore we only need to consider So supp ose that T L is

r r p

an isomorphism Then

Z Z

n n

X X

r se n r sT e ds kT k ds

i i i i

i i

r p

kT k kT e k

kT kkT k n

r p

That is n Cn for all nwhich is clearly imp ossible since r p

Banachs approach to the case r pwas somewhat dierent from

ours Instead he app ealed to the BanachSaks theorem

CHAPTER L SPACES I I I

Theorem Every weakly nul l sequence f in L p has a

n p

subsequence f with f O k

n n

i i

Thus if r p and if T L is an isomorphism then T e

r p i

O k T e would have a subsequence satisfying But just

as ab ove the fact that T is an isomorphism implies that T e

k which is a contradiction This argument do esnt apply in e

the case r since e isnt weakly null in But since isnt reexive

it cant p ossibly b e isomorphic to a subspace of the reexive space L for any

The remaining case pr is substantially harder than the

rest but theres a payo For p and r in this range L actually contains a

closed subspace isometric to all of L The pro of requires several to ols from

probability theory that taken one at a time are not terribly dicult but

taken all at once would require more time than wehav e

Notes and Remarks

For more on unconditional convergence see Day or Diestel For more

on unconditional bases see also Lindenstrauss and Tzafriri and Meggin

son Prop osition and its relatives come to us partly through folklore

but largely through the workofsuchgiants as J L Krivine and B Mau

reySuch squarefunction inequalities are bynow commonplace in harmonic

analysis and probabilitytheoryaswell as in Banach space theory See also

Zygmund For more on the unresolved case pr as well

as more on the work of Krivine Maurey and others see Lindenstrauss and

Tzafriri and the Memoir by Johnson MaureySchechtman and Tzafriri

Exercises

If x converges unconditionally to xprovethatevery rearrangement

x likewise converges to x

x is unconditionally convergentinX show that the map f If

f g X dened by f x is continuous where

n n n

N n

f g is supplied with the metric d j j

n n n n

Supp ose that the series a x converges in X for every b ounded se

n n

quence of scalars a Use the Closed Graph theorem to prove that the

map T X dened by T a a x is continuous

n n n

Let r and let f x y b e a nonnegative function in L

Provethat

Z Z Z Z

r r

f x y dx dy f x y dy dx

Let r p q and let f f L Show that

n p

q q

n n

X X

q q

kf k jf j

i i

r r

n n

X X

r r

jf j kf k

i i

vided that weusemax kf k The inequality also holds for q pro

in i p

and k max jf jk as the rst two terms

in i p

CHAPTER L SPACES I I I P

Chapter

Convexity

Several of the inequalities that weve generated in L spaces are generalizations

of the parallelogram law To see this lets rewrite the usual parallelogram law

for

ky k kxk kx y k kx y k kx y k kx y k

That is the average value of kx y k over the four choices of signs is

kxk ky k In other words

t x r t y k dt kxk ky k k r

Now the parallelogram law tells us something ab out the degree of round

ness of the unit ball in Indeed if kxk ky k andifkx y k

then

kx y k kxk ky k kx y k

Thus kx y k foranytwo distinct p oints on the unit sphere in That

is the midp ointx y has norm strictly less than and so lies strictly

inside the unit ball In fact we can even determine just how far inside the

ball

x y

Its not hard to see that every p ointonthechord joining x and y has norm

strictly less than thus there can b e no line segments on the sphere itself

In this chapter well investigate various analogues of the parallelogram law

and their consequences in a general normed linear space X

CHAPTER CONVEXITY

x x

δ ε (x+y)/2 (x+y)/2

y y

SX S

0 0 X

Strictly Convex Uniformly Convex

Strict convexity

Its often convenient to know whether the triangle inequality is strict for non

collinear p oints in a given normed space Wesay that a normed space X is

strictly convex if

kx y k kxk ky k

whenever x and y are not parallel That is not on the same line through

hence not multiples of one another The triangle inequalityisalways strict if

y is a negativemultiple of x while its always an equalityif y is a p ositive

multiple of x Since strict convexity is more accurately attributed to the norm

in X we sometimessay instead that X has a strictly convex norm Like the

parallelogram law itself this is very much an isometric prop ertyaswell see

shortly

Lets b egin with several easy examples

a A review of the pro of of the triangle inequality for Hilb ert space and

the case for equality in the CauchySchwarz inequality shows that an y

Hilb ert space is strictly convex Also for any pitfollows

from the case for equalityinMinkowskis inequalitythat L is strictly

convex If youre still skeptical well give several dierent pro ofs of these

facts b efore were done

b Consider that is R under the norm kx y k maxfjxj jy jg

Its easy to see that is not strictly convex For example the vectors

and are not parallel and yet k k

k k k k Whats more the entire line segment joining

and lies on the unit sphere since for any we

have k k k k

For later reference lets rephrase this observation in two dierentways

First the p oint is not in the unit ball of a compact convex

subset of It follows that must have a nearest p oint on the

sphere In fact it has many every p oint on the segment joining

and is nearest k yk k y k for

(1,1)

0 (2,0)

(1,–1)

As a second restatement consider the element the predual

of Notice that every functional of the form y y

norms h y i k k kyk

It follows that any space that contains an isometric copyof is likewise

not strictly convex Thus c L andC are not strictly convex

and In L for example the functions span an isometric

copyof

c Its nearly obvious that and L are not strictly convex since the tri

angle inequalityisanequalityonanypairofvectors of the same sign

co ordinatewise or p ointwise Again for later reference lets state this

fact in a couple of dierentways In L consider the p ositive face of

the unit sphere

K f L f f

Clearly K is a closed convex subset of the unit sphere of L its the

intersection of two closed convex sets and K contains lots of line seg

ments Why Also note that every pointinK is distance away from

K Finally notice that the functional f f attains its norm at

K every pointof

As a second example consider

K x x

CHAPTER CONVEXITY

Once more K is a closed convex set in its a hyp erplane and K

This time however there is no p ointinK nearest that is no element

of smallest norm in K Indeed its easy to check that distK

while for every x K wehave

X X

jx j kxk x

n n

Said still another way this same calculation shows that the functional

do esnt attain its norm on the sphere of f

d Strict convexityisvery much an isometric prop erty and isnt typically

preserved by isomorphisms or equivalent renormings For example con

sider the norm

jjj x y jjj maxf jxj kx y k g

on R Clearly jjjj jj is equivalent to the Euclidean norm kk Butjjjj jj

isnt strictly convex for if wetake R then jjj jjj

jjj jjj while jjj jjj jjj jjj

Its ab out time wegave a formal pro of or two First lets givetwo equiv

alentcharacterizations of strict convexity

Theorem X is strictly convex if and only if either of the fol lowing holds

x y

i For x y in X with kxk ky k we have

p p

kxk ky k x y

ii For p and x y in X we have

Proof Condition i is surely implied by strict convexity if x y are norm

one vectors then either x and y are nonparallel or else y x and in either

case i follows Supp ose on the other hand that i holds but that we can

nd nonparallel vectors x and y in X with kx y k kxk ky k Wemay

clearly assume that kxk ky kThus

x y y y y x

kxk ky k kxk kxk kxk ky k

kxk ky k

ky k

kxk kxk ky k

whichcontradicts i

Next supp ose that X is strictly convex and let pGiven non

parallel vectors x and y in X wehave

p p

x y kxk ky k kxk ky k

since the function jtj is convex Similarlyif y tx t then

p p

p p p

kxk ky k jtj x y t

p p

kxk kxk

since the function jtj is strictly convex Thus ii holds That ii implies

strict convexitynow follows easily from i

The fact that L is strictly convex for p follows from ii and the

fact that jtj is strictly convex Well give another pro of of this fact later

Its clear from our pictures that if X is strictly convex then S the unit

sphere in X contains no nontrivial line segments What this means is that

each p ointofS sticks out from its neighb ors A fancier waytosay this is

to saythateach p ointof S is an extreme point of B the closed unit ball of

X X

X A p oint x in a convex set K is said to b e an extreme p ointifx cannot b e

written as a nontrivial convex combination of distinct p oints in K Thus x is

an extreme p ointofK if and only if

x y z yz K y z x

Now a convex set need not have extreme p oints as the closed unit ball of

c will attest but if it do es they havetolive on the b oundary of the set In

particular any extreme p ointof B must livein S Indeed if kxk then

X X

kxk for some Thus k xk andwewould then

x x Hence x is not extreme NowifX is strictly have x

convex and if kxk then x must b e an extreme p ointfor B by condition

i of our last result In a strictly convex space then not only do es B have

extreme p oints but every p ointinS is extreme

But we can sayeven more If X is strictly convex then eachpointofS

is even an exposedpoint of B That is for each x S there exists a norm

X X

one functional f X which exp oses xthatis f x while f y

x x x

Now an exp osed p ointisalsoanextremepoint for if for all other y B

x y z then f x f y f z and hence at least one of

x x x

f y orf z holds

x x

CHAPTER CONVEXITY

Theorem X is strictly convex if and only if any one of the fol lowing

holds

iii S containsnolinesegment

iv Every point of S is an extreme point of B

X X

v Every point of S is an exposedpoint of B

X X

Proof Its nearly obvious that strict convexity is equivalenttoeach of iii

and iv so lets concentrate on the equivalence of strict convexity with v

As weve already p ointed out v implies iv so we just need to show that

something on the list implies v Lets go for the obvious Given x S the

HahnBanach theorem supplies a norm one functional f X with f x

What happ ens if f y for some y x in B Well for one wed have

to have ky k since f y ky k But then f x y likewise

forces kx y k in violation of the strict convexityof X Consequently

wemust have f y forevery y x in B

Nearest p oints

We next turn our attention all to o briey to the issue of nearest p oints Given

a nonemptyset K in a normed linear space X and a p oint x X wesay

that the p oint x K is nearest x if kx x k inf ky x kObviously

y K

if x K then x is its own unique nearest p ointinK If K is closed

then this inmum exists and is p ositive for any x K but it need not b e

attained in general Moreover even if nearest p oints exist they need not b e

unique as our earlier examples p oint out The questions that arise then are

Under what conditions on K will nearest p oints always exist When are they

unique If each x has a unique nearest p oint px K what can wesay

ab out the typically nonlinear nearest p oint map p

Now its an easy exercise to showthatifK is compacttheneach p oint

x K has a nearest p ointinK In fact its not much harder to show

that the same holds for a weakly compact set K Once webringtheweak

top ology into the picture though its typical to add the requirementthat K

b e convex since weak and normclosures coincide for convex sets This is a

consequence of the HahnBanach theorem For these reasons nearest p oint

problems are often stated in terms of closed convex sets Well adopt this

convention wholeheartedlyFor now lets settle for a few simple observations

Theorem Let K be a nonempty compact convex subset of a normed

linear space X

a Each x X has a nearest point in K

b If X is strictly convex then there is a unique point px K nearest to

x Moreover the map x px is continuous

Proof A carefully constructed pro of will give all three conclusions at once

Here go es Let x K and let d inf ky x k For each n consider

y K

K x K d K x d B kx x kd

n X

n n

Each K is a nonempty closed convex subset of K and K K for all n

n n n

T T

Thus K Clearlyany p oint y K is a nearest p ointto x

n n

in K

Nowifx y K are each nearest to x then so is their midp oint z

x y since

d kz x k x x y x d

Thus if X is strictly convex then wemust have x x y x z x

hence x y z That is there can b e at most one nearest p oint Whats

more this tells us that the diameters of the sets K must tend to whenX

is strictly convex for otherwise we could nd twopoints nearest to x

Finally supp ose that X is strictly convex and let x b e a sequence in X

converging to x Theres no great harm in supp osing that kx x kn

k this will simplify the pro of It then follows that d inf ky x

n n y K

d n and hence that kx px kkx px k kx x kd n

n n n n n

d n where px is the unique p ointinK nearest x Consequently

n n

wemust have px K Since the diameter of K tends to we get

n n n

px px

Smo othness

Related to the notion of strict convexity is the notion of smo othness Wesay

that a normed space Y is smooth if for each y Y there exists a unique

norm one functional f Y suchthat f y ky k Of course the Hahn

Banach theorem insures the existence of at least one such functional f whats

at issue here is uniqueness In any case its clearly enough to check only norm

CHAPTER CONVEXITY

one vectors y when testing for smo othness As it happ ens a normed space

is smo oth if and only if its norm has directional derivatives in each direction

Hence the name of the prop ertyWewont pursue this further our interest in

smo othness is as a dual prop erty to strict convexity

Its immediate that a Hilb ert space H is smo oth Indeed eachvector norms

itself in the sense that hx xi kxk for all x H and uniqueness follows from

the converse to the CauchySchwarz inequality Similarlyitfollows from the

converse to Holders inequalitythat L is smo oth for pInthiscase

each f L is normed by the function jf j sgn f L

p q

Our examples at the b eginning of this chapter should convince you that

c L L and C all fail to b e smo oth In c for instance the

vector e e is normed by e e and e e Similarlythevector

e is normed by e e e and e e e

Theorem If X is strictly convex then X is smooth If X is smooth

then X is strictly convex

ProofIfX is not smo oth then we can nd a norm one vector x X which

is normed bytwo distinct norm one functionals f g X But then

kf g kandf g x hence kf g k This denies the strict

convexityof X

If X is not strictly convex then wecananormonevector x X whichis

not an exp osed p oint Thus if wecho ose a norm one f X with f x

then wemust have f y for some other norm one y x But thenx

y X b oth norm f X denying the smo othness of X

Smo othness and strict convexity arent quite dual prop erties There are

examples of strictly convex spaces whose duals fail to b e smo oth But there is

at least one easy case where full duality has to hold

Corollary If X is reexive then X is strictly convex resp smooth if

and only if X is smooth resp strictly convex

Notice that the spaces L p andany Hilb ert space H necessarily

enjoy both prop erties

Uniform convexity

A normed linear space X is said to b e uniformly convex if for each

there is a suchthat

x y

kxk ky k kx y k

Obviouslyany uniformly convex space is also strictly convex but there are

strictly convex spaces that arent uniformly convex An easy compactness

argument will convince you that if X is nite dimensional then X is strictly

convex if and only if X is uniformly convex Just as with strict convexity

uniform convexity is actually a prop erty of the norm on X so it mightbe

more appropriate to saythat X has a uniformly convex norm whenever

holds

Similar to the strictly convex case we could reformulate our denition in

terms of pairs of vectors x y with kxk ky k and kx y k Likewise

we could use some p ower of the norm say kk for p The details in this

case are rather tedious and not particularly imp ortant to our discussion so

we will omit them

As wesaw at the b eginning of this chapter the parallelogram law implies

that every Hilb ert space is uniformly convex In fact weeven computed

We will see that uniformly convex spaces share

many of the geometric niceties of Hilb ert space Our primary aim is to prove

the famous result due to Clarkson in that L is uniformly convex

whenever p Using this observation we will give a geometric pro of

essentially devoid of measure theory that L L In fact wewillshow that

every uniformly convex Banach space is necessarily reexive

Note that since L and L fail to b e even strictly convex they cant b e

uniformly convex The same is true of c andC Also note

ery much an isometric prop erty the equivalent that uniform convexityisv

renorming of R wegave earlier that fails to b e strictly convex obviously also

fails to b e uniformly convex Along similar lines its easy to check that the

expression jjj x jjj kxk kxk denes an equivalent strictly convex norm on

but that supplied with this norm cant b e uniformly convex since its

not reexive

We b egin with a simple but very useful observation

CHAPTER CONVEXITY

Lemma X is uniformly convex if and only if for al l pairs of sequences

x y with kx k ky k we have

n n n n

x y

n n

kx y k

n n

Proof One the one hand if X is uniformly convex and if kx y k

n n

then wemust have kx y k for otherwise kx y k implies

n n n n

kx y k

n n

On the other hand if X is not uniformly convex then wecanndan

and sequences x andy inB such that kx y k while

n n X n n

x y k k

n n

Corollary Let X be uniformly convex If x in X satises kx k

n n

and kx x k as m n then x is Cauchy

n m n

Note that the condition kx x k also forces kx kby the

n m n

triangle inequality kx x kkx k kx k Thus if X is

n m n m

complete x converges to some norm one elementof X

Our last two results can b e used to prove several classical convergence

theorems but well settle for just one such result In the case X L its

sometimes called the RadonRiesz theorem

Theorem Let X be a uniformly convex Banach space and suppose that

x in X satises x x and kx kkxk Then kx xk

n n n n

ProofIfkxk theres nothing to show so wemay supp ose that x

Now since kx k kxk its easy to see that we can normalize our sequence

w w

x x implies that y y If weset y xkxk and y x kx kthen

n n n n n

also ky y k will imply that kx xk

n n

Next cho ose a norm one functional f X with f y Then

y y y y

n m n m

But f y f y as n and so wemust have ky y kas

n n m

m n From Corollary this means that y is Cauchy and hence

y converges in norm to some p ointinX But since norm convergence

implies weak convergence and since weak limits are unique wemust have

y y in norm n

Using a very similar argument wecangive a short pro of of an interesting

fact due to Milman in and Pettis in If youre not familiar

with nets just pretend theyre sequences you wont b e far from the truth

Theorem A uniformly convex Banach spaceisreexive

Proof Let X b e uniformly convex and let x X with kx k We

need to showthat x x for some x X Now since B is weak dense in

x But B we can nd a net x inX with kx ksuchthat x

since kx kkx k it follows that kx kkx k A slightvariation on

Theorem shows that x is Cauchyin X and hence must converge to

some x X It follows that x x

Clarksons inequalities

To round out our discussion of uniform convexitywe next prove Clarksons

Theorem

Theorem L is uniformly convex for p

As it happ ens the pro of of Clarksons theorem is quite easy for p

and not quite so easy for p For this reason dozens of pro ofs have

b een given Wehave the luxury of selecting bits and pieces from several of

these pro ofs Now Clarkson proved several inequalities for the L norm which

mimic the parallelogram law We will do the same but without using his

original pro ofs It should b e p ointed out that all of the results well givein

this section hold for every L space after all this has more to do with the

L norm than with measure theory

We rst prove Clarksons theorem for the case p this particular

pro of is due to Boas in

Lemma Given p and real numbers a and b we have

p p p p p

j a b j j a b j jaj jbj

ProofWeve seen this inequality b efore more or less See if you can ll in

the reasons b ehind the following arithmetic

p p p

j a b j j a b j j a b j j a b j

jaj jbj

p p p p

aj jbj j

p p p p

jaj jbj

CHAPTER CONVEXITY

Theorem For p and any f g L we have

p p p p p

kg k kf k kf g k kf g k

p p p p

Corollary L is uniformly convex for p

ProofIff g L with kf k kg k and kf g k then

p p p p

kf g k

and this is known to b e exact That is

Unfortunately inequality reverses for p so weneeda

dierent pro of in this case the one well give is due to Friedrichs from

As with Boass pro of we b egin with a p ointwise inequality

Lemma If p q pp and x then

q q p q

x x x

Proof Consider the function

q q q q

f x x x x x

for and x Then f x is the lefthand side of and

f x x is the righthand side of b ecause p q Thus

since x wewanttoshowthatf But

q q q

q x x x

q q q

and since and x x since

For p the inequality in holds with the roles of p and q exchanged

Our pro of shows that the inequality in as written reverses for p

For p the inequality in easily implies that

q q p p q

j a b j j a b j jaj jbj

for all real numb ers a and b But this time we cant simply integrate b oth

sides to arriveatan L result Instead well need Friedrichs clever extension

of this inequalityto L p

Theorem For p q pp and any f g L we have

p p q q

kg k kf k kf g k kf g k

p p p p

Proof First notice that

Z Z

p p

q p q p q

kf k jf j jf j kj f j k

and p Now since the triangle inequalityreverses in L

q q q q

k f g k k f g k kjf g j k kjf g j k

p p

q q

kjf g j j f g j k

p p

k jf j jg j k

p p

k f k kg k

p p

where the last equalityfollows from the fact that q p

Corollary L is uniformly convex for p

Notice that if f g L p with kf k kg k and

p p p

q q

kf g k then just as b efore wed get

where q pp This however is to o small And is to o big Its

known that p is asymptotically the correct value for L

p Notice that this is essentially the value of for Hilb ert space

the p oint here is that no space can b e more convex than Hilb ert space

For the sake of completeness we list all of Clarksons inequalities well

half of them anyway Let p and q pp Then for any f

g L wehave

p p p p p p p

kf kg k kf k kg k kf k k f g k g k

p p p p p p

q q p p

kg k kf k k f g k k f g k

p p p p

p q p q

kf k kf g k kg k k f g k

p p p p

for p In particular all reduce to the These inequalities all reverse

parallelogram identity when p

CHAPTER CONVEXITY

An elementary pro of that L L

In our discussion of the dualitybetween strict convexity and smo othness we

made the claim that smo othness had something to do with dierentiability

of the norm It should come as no surprise then that uniform convexityis

dual to an ever stronger dierentiability prop erty of the norm Rather than

formalize the condition well settle for a few simple observations Most of

these results are due to McShane from

We b egin with a General Principle If a linear functional attains its norm

atapoint of dierentiability of the norm in X then its value is actually given

by an appropriate derivative This is the content of

Lemma McShanes Lemma Let X beanormedlinear space and let

T X Suppose that f g X satisfy

i kg k and T g kT k

p p

kg tf k kg k

ii lim exists for some p

p p

kg tf k kg k

Then T f kT k lim

It should b e p ointed out that the pth p ower in ii is purely cosmeticthe

case p is enoughbut it will make life easier when we apply the Lemma

p p

to the L norm The number lim kg tf k kg k pt is the directional

p t

derivativeof kk at g in the direction of f

Proof First we use lHopitals rule to compute T f as a derivative

p p p p

T g tf T g T g tTf T g

lim lim

t t

pt pt

lim T g tTf T f

T g T f

kT k T f

Now since kT kkg k T g and kT kkg tf kT g tf weget

p p

T g tf T g kg tf k kg k

lim lim kT k

t t

pt pt

kT k T f

p p

T g tf T g

lim

p p

kg tf k kg k

lim kT k

pt t

McShanes Lemma gives us a schedule First we show that each T L

p actually attains its norm and then we compute the limit given in

ii Curiously the fact that eachelementof L attains its norm is equivalentto

the fact that L is reexivewhich do esnt actually require knowing anything

ab out the dual space L

First a general fact

Lemma If X is a uniformly convex Banach space then each T X

attains its norm at a unique g X kg k

ProofChooseg inX kg k suchthat T g kT kThen

n n n

kg g kkT k j T g g j

n m n m

as m n Thus g isCauchy and hence converges to some g X

Clearly T g kT k and kg k Uniform or even strict convexitytellsus

that g must b e unique

Next lets do the Calculus step

Lemma Let f g L p Then

p p

kg tf k kg k

f jg j sgn g lim

Proof For any a b R the function tj a bt j is convex and has

t pj a bt j b sgn a bt The limit nowfollows from the Dominated

or even Monotone Convergence theorem

p p p p

k g tf k kg k j g xtf x j jg xj

lim lim dx

t t

pt pt

f jg j sgn g

CHAPTER CONVEXITY

Alternativelywe could use the standard approach Let

Z Z

p p

j g xtf x j dx x t dx F t k g tf k

Then F is dierentiable and F t x t dx provided that exists and

t t

is integrable But for say t wehave

j x tj pfx jg xtf xj sgn g x tf x

p p p

p jf xjjg xj jf xj

whichisintegrable b ecause jg j L

Now given T L take g L with kg k and T g kT k Then for

p p

every f L

p p

kg tf k kg k

T f lim kT k

kT k f jg j sgn g

That is T is represented byintegration against kT kjg j sgn g L andof

p p

course kjg j k kg k Thisproves that L L

q q

p p

Notes and Remarks

Our presentation in this chapter b orrows from a great numb er of sources for

example the b o oks by Beauzamy Day Diestel Hewitt and

Stromb erg Holmes Kothe and Rayaswell as the several original articles cited in the text

Exercises

Let f I R b e a continuous function dened on an interval I Iff

satises f x y f xf y for all x y I prove that f is

convex

A convex function f I R is said to b e strictly convex if f x

y fx f y whenever andx y Show that

f xjxj is strictly convex for p

Give a direct pro of that L is strictly convex

Prove that a normed space X is strictly convex if and only if kk is a

strictly convex function

Show that every subspace of a strictly convex space is strictly convex Is

the same true for uniformly convex spaces For smo oth spaces

Provethatjjj x jjj kxk kxk denes an equivalent strictly convex norm

Supp ose that T X Y is continuous and onetoone and that Y is

strictly convex Show that jjj x jjj kxk kTxk denes an equivalent

strictly conv ex norm on X

Give a direct pro of that p is strictly convex Generalize

your pro of to conclude that if X is strictly convex for each n then

X X is strictly convex for anyp

Let p and dene g L by g t for t

and g t for t If we set f show that

p p

kf k kg k kf g k andkf g k

p p p p

p p

Conclude that is b est p ossible for L

Cho ose a sequence p with p suchthat p orp

n n n n

and dene X ShowthatX X is strictly convex and

even reexive but not uniformly convex

kx kGive an example If x x in X show that kxklim inf

n n n

where this inequality is strict If kx kkxk for all n conclude that

we actually have kxk lim kx kShow that the same holds for a

n n

x in X sequence x

Apply McShanes Lemma to show that each linear map T R R is

given by inner pro duct against a vector x R with kxk kT k

Let x with kxk Showthatx is an exp osed p oint of the closed

for some k unit ball of if and only if x e

CHAPTER CONVEXITY

Let f f beinL p Iff f ae and kf k kf k show

n p n n p p

p p p p

that kf f k Hint Use the fact that jf j jf j jf f j is

n p n n

p p

nonnegative and tends p ointwise ae to jf j Show that the result

also holds if we assume instead that f f in measure

a Show that a p oint x in the closed unit ball of is an extreme

point if and only if x where fori n

n i

can b e written as a Provethateachpoint in the unit ball of

convex combination of extreme p oints

b Show that the set of extreme p oints of the unit ball of consists

of all p oints of the form where for all i

n i

Show that the set of extreme p oints of the unit ball of consists

of the p oints e k For p every norm one

vector in is an extreme p oint of the unit ball

c In sharp contrast to the previous cases show that the closed unit

ball in c has no extreme p oints

Chapter

C K Spaces

The Cantor set

In this chapter we will catalog several imp ortant prop erties of the Cantor set

Our goal in this endeavor is to uncover the universal nature of C For

starters well prove that is the biggest compact metric space byshowing

that every compact metric space K is the continuous image of Now the

presence of a continuous onto map K tells us something ab out C K

Comp osition with ie the map f f denes a linear isometry and

an algebra isomorphism from C K into C Since is itself a compact

metric space this means that C is biggest among the spaces C K for

K compact metric

To b egin recall that the eachelement of the Cantor set has a ternary base

a wherea or and that decimal expansion of the form

n n

P P

n n

the Cantor function a a denes a continuous map

n n

from onto This proves

Lemma The interval is the continuous image of

As an immediate corollarywehavethat C is isometric to a closed

subspace and subalgebra of C Do es this sound backwards If so have

patience

For our purp oses a convenient representation of the Cantor set is as the

countable pro duct of twop oint discrete spaces This representation is easy

to derive from the ternary decimal representation of That is since the

This represen elements of are sequences of s and s wehavef g

tation also yields a natural metric on Setting dx y ja b j

n n

CHAPTER C K SPACES

where a b are the ternary decimal digits for x y resp ectively

n n

denes a metric equivalent to the usual metric on That is is homeo

morphic to the pro duct space f g supplied with the metric d Moreover

this particular metric has the additional prop ertythat dx y dx z now

implies that y z Whenever we sp eak of the metric on this will b e the

one wehave in mind

Many authors take f g since this choice makes a group under

co ordinatewise multiplication

Now since N can b e partitioned into countably many countably innite sub

sets the pro duct space representation of yields an immediate improvement

on our rst Lemma

Corollary The cube is the c ontinuous image of

Again this means that C is isometric to a closed subspace of C

Completely regular spaces

Recall that a top ological space X is said to b e completely regular if X is

Hausdor and if given a p oint x X outside a closed set F X there is

a continuous function f X such that f x while f on

F In other words X is completely regular if C X separates points

from closed sets in X Since singletons are closed in X itsimmediate that

C X also separates points in X

From Urysohns lemma every normal space is completely regular Thus

metric spaces and compact Hausdor spaces are completely regular Lo cally

compact Hausdor spaces are completely regular to o In fact nearly every

top ological space encountered in analysis is completely regular Even top olog

ical groups and top ological vector spaces F rom our p oint of view theres no

harm in simply assuming that every top ological space is completely regular

As it happ ens the class of completely regular spaces is precisely the class

of spaces whichemb ed in some cub e To see this we start with

Lemma Emb edding Lemma Let X becompletely regular Then X

C X

is homeomorphic to a subset of the cube

ProofFor simplicity lets write C C X Let e X b e the

evaluation map dened by exf f xforx X and f C That is x

is made to corresp ond to the tuple exf x Thate is onetoone

f C

is obvious b ecause C separates p oints in X Thate is continuous is easytoo

since each of its co ordinates e f f C iscontinuous

Now let U b e op en in X WewanttoshowthateU isopenineX

Heres where we need complete regularity Given x U cho ose an f C

c c

such that f x while f on U and let V fg Clearly V is

op en in the pro duct and ex V b ecause exf Finally

ey V f y y U and hence ex V eX eU

Our pro of of the Emb edding Lemma shows that if FC X and

if e X is dened by exf f x then e is continuous If

in addition F separates p oints then e is onetoone FinallyifF separates

points from closed sets then e is a homeomorphism into

The Emb edding Lemma also tells us that X carries the weak top ology

induced by C X In terms of nets

x x in X ex exin

ex ex for every f C X

f f

f x f x for every f C X

x x in the weak top ology induced by C X

Lets consolidate several of these observations

Theorem For a Hausdor topological space X the fol lowing areequiv

alent

a X is completely regular

b X embeds in a cube

c X embeds in a compact Hausdor space

d X has the weak topology inducedby C X

has the weak topology inducedby C X e X

Proof The Emb edding Lemma shows that a implies b implies d Of

course b and Tychono s theorem imply c Since compact Hausdor

spaces are completely regular and since every subspace of a completely regular space is again completely regular c implies a

CHAPTER C K SPACES

Next lets check that d implies a Supp ose that X has the weak top ol

ogy induced by C X and let x X b e a p oint outside the closed set

F X Then for some f f C X and some the basic

op en set U N x f f iscontained in F Thatis

U N x f f f y X jf y f xj i ngF

n i i

Now each of the functions g jf f xj is in C X as is the function

i i i

g maxfg g g Moreover g x whileg y for all y F U

Hence h minf g gC X satises hx and hy for

all y F U Thus X is completely regular

Finally the fact that d is equivalent to e is easySinceC X

C X its clear that eachC X op en set is also C X op en On

b b

the other hand given x X f C X and put M kf k and

notice that g f M M C X satises

f y X jf x f y j g f y X jg x g y j M g

Thus eachC X op en set is also C X op en

If X is completely regular and if some countable family FC X

separates p oints from closed sets in X thenX emb eds in the cub e and

hence is metrizable Compare this with Urysohns metrization theorem A

normal second countable space is metrizable This result has a converse of

sorts to o If for example X is a separable metric space then some count

able family FC X will separate p oints from closed sets in X In

particular X will embed in More to the p oint for us is

Lemma Every compact metric space is homeomorphic to a closed subset

ProofLetK b e a compact metric space and let x b e dense in K Wemay

assume that the metric d on K satises dx y Given this dene

by xndx x Clearly is onetoone and continuous K

each co ordinate ndx iscontinuous Since K is compact and

is Hausdor is a homeomorphism into and the result follows

Lest you b e fo oled into thinking that only countable pro ducts of intervals

are separable here is an easy counterexample

Example is separable but not metrizable

Proof Consider the collection D of all functions of the form q where

i J

q q are rationals in and where J J are disjoint closed in

n n

tervals in with rational endp oints Then D is a countable subset of

Nowthetypical basic op en set in is given by

N f x x f g jg x f x j ing

n i i

where x x and f Given a basic op en

set N f x x cho ose rationals q q in with jq f x j

n n i i

for each i and cho ose disjoint rational intervals J with x J Then g

i i i

DN f x x Thus D is dense and so is q

n i J

separable

To see that is not metrizable note that its not sequentially

compact Y our favorite sequence of functions f with no

pointwise convergent subsequence will do the trick The sequence r

where r is the nth Rademacher function comes to mind

Were now ready to establish our claim that the Cantor set is the biggest

compact metric space

Theorem Every compact metric space K is the continuous image of

ProofWe know that K is homeomorphic to a closed subset of and

hence that K is the continuous image of some closed subset of To nish the

pro of then it suces to show that each closed subset of is the continuous

image of

ja b j So let F b e a closed subset of and let dx y

n n

b e the metric on Given x notice that the distance dx F

inf dx y is attained at a unique p oint y F and in particular y x

y F

whenever x F Dene f xy for this unique y Tocheckthat f is

continuous let x x in Now since F is compact wemay assume that

y f x converges to some p oint z But then dx y dx z and

n n n n

dx y dx F dx F dx y so wemust have y z thatis

n n n

y y f x

Corollary If K is a compact metric space then C K is isometric to a

closed subspace even a subalgebra of C

CHAPTER C K SPACES

This completes the rst circle of ideas in this chapter C is universal

for the class of spaces C K K compact metric This may seem o dd in view

of the small role that typically plays in a rst course in analysis Indeed

C is usually given much more emphasis The reader who is uncomfortable

with this turn of fate can take heart in the fact that C emb eds isometrically

into C That is C is universal to o Well need a bit more machinery

before we can give a pro ofnote that we cant hop e to prove this claim by

nding a continuous map from onto

Now each f C extends to a continuous function on This

already would follow from Tietzes extension theorem for example but there

may b e many such extensions Wewanttocho ose a metho d for extending f

that will lead to a linear map from C into C And the most natural

extension do es the job

The complementof in is the countable union of disjointopen

intervals The endp oints of these op en intervals are of course elements of

Simply connect the dots in the graph of f across the endp oints of each

of these op en intervals to dene an extension f for f This is quite like the

pro cedure used to extend the Cantor function to all of Note that if

x then f xistheaverage of twovalues of f on Clearly then

sup jf xj sup jf xj Whats more its not hard to see that if

x x

were given f g C then f g f g Well take this as pro of of

from C into C is a Lemma The extension map E f f

linear isometry

Theorem C is isometric to a complementedsubspaceofC

ProofTo complete the pro of we need only note that restriction to denes a

linear map R C C with the prop ertythatP ER is the identity

on E C That is E C is an isometric copyof C and is the range

of a b ounded pro jection P C C

Corollary If K is a compact metric space then C K is isometric to

a closed subspaceofC Inparticular C K is separable

Our last Corollary should b e viewed as a rough analogue of the Weierstrass

theorem valid in C K for a compact metric space K Whats more the

converse is also true If C K is separable for a given compact Hausdor

space K then K is metrizable Well forego the details just now but this

issue will come up again

The universalityof C or C reaches b eyond the class of C K

spaces in fact every separable normed space is isometric to a subspace of

C Even more is true Every separable metric space is isometric to a

subset of C By our last Corollarywe only need to checkthateach

separable normed linear space emb eds in some C K K compact metric At

least part of this claim is easy to checkandisvalid for any normed linear space

As a consequence of the HahnBanach theorem a normed linear space X is

under the sup norm via p ointevaluation isometric to a subspace of C B

b X

x x j wherex x x x That is the emb edding of X into C B is

B b X

nothing other than the canonical emb edding of X into X with the additional

restriction that each elementof X is to b e considered as a function just on

What remains to b e seen is whether B can b e given a suitable top ol

ogy that will turn it into a compact metric space thus making C B

b X

C B For an innitedimensional space X the norm top ology on B

X X

wont do its never compact Well need something weaker The top ology

that ts the bill is the weak top ology on X restricted to B

The weak top ology is the top ology that X inherits as a subset of the pro d

uct space R That is each x X is identied with its range x x

xx considered as an elementof R Under this identicationx is the

x X

pro jection onto the xth co ordinate of X Thus the weak top ology on X

is the smallest top ology on X making everyx continuous or in still other

words the weak top ology is the weak top ology on X induced by X X

A neighb orho o d base for the weak top ology is generated by the sets

N x x x f y X jy x x j for i k g

k i

X and In terms of nets where x X x x

x x for each x X x in X x x x

x x x x for each x X

x x in R

Of particular merit here is that X is still isometric to a subspace of C B

b X

is supplied with the weak top ology since eachx is weak even when B

continuous That this observation has brought us one step closer to our goal

is given as

CHAPTER C K SPACES

Theorem BanachAlaoglu If X is a normedlinear space then B

is compact in the weak topology on X

ProofAsweve already seen X under the weak top ology is homeomorphic

thenwe can do a bit b etter Again to a subset of R Ifwecutdown to B

with its range x x but now considered we identify each x B

xX X

as an element of the pro duct kxk kxk This identication is still

a homeomorphism into by virtue of the denition of the weak top ology

kxk kxk is compact we will b e done once weshow that the Since

is closed in the pro duct top ology But what do es this really image of B

of linear mean We need to check that the p ointwise limit of a net x B

functions on X is again linear and has norm at most onewhich is clear

Corollary Every normed linear space is isometric to a subspaceof

C K for some compact Hausdor space K

Wewanttoemb ed a separable normed space X into C K where K is a

compact metric space Our next result shows the way

Theorem If X is a separable normed linear space then the weak topol

ogy on B is both compact and metrizable

Proof That B is compact in the weak top ology is immediate Now let

jx y x bedenseinS and dene dx y x j for x

n X n

Next notice that Its easy to see that d is a metric on B y B

X X

n n

dx y max jx y x j

max jx y x j

Heres what this do es Given x y B B b ecause kx y kforx

X X

and if M is chosen so that then

N x x x B fy B dx y g

N X X

discontinuous But weak toB That is the formal identityfromB

X X

d is Hausdor the two spaces must weak is compact and B since B

X X actually b e homeomorphic

Corollary BanachMazur Every separable normedlinear spaceis

isometric to a subspaceofC K for some compact metric space K Thus

every separable normed linear space is isometric to a subspaceofC

If follows from the BanachMazur theorem that C isnt reexive since

it contains an isometric copy of the nonreexive space In fact this same

observation tells us that C must b e nonseparable

Corollary Every separable metric space is isometric to a subset of

Proof Supp ose that M d is a separable metric space and let x b e dense

in M d Fix a p oint x M and dene a map from M into by

x x dx x dx x

n n

By the triangle inequalityinM d wehavex and kx k dx x

Essentially the same observation shows that the map is an isometry

kx yk supjdx x dy x j dx y

n n

Indeed its clear from the triangle inequalityinM d that kx yk

gives kx yk dx y On the other hand taking x with dy x

n n

jdx x dy x jdx y Thus M is isometric to a subset M of

n n

In particular the sequence x in is dense in M It follows that

the closed linear span of M is a separable subspace of An app eal to the

BanachMazur theorem nishes the pro of

Here is our longawaited application of the BanachMazur theorem

Corollary Every normed linear spacecontains an innite dimensional

closed subspacewitha basis

Finally were ready to complete a second circle of ideas from this chapter

The separabilityof C K is equivalent to the metrizabilityof K We consider

the apparently more general case of a completely regular space T to save

wear on tear on the letter X

Theorem If T is a completely regular topological space then T is

weak homeomorphic to a subset of B

C T b

CHAPTER C K SPACES

Proof Each p oint t T induces an element C T byway of what else

t b

pointevaluation f f t The functional is the point mass or Dirac

t t

Moreover it measure at t Note that is norm one hence B

t t C T

follows easily that the map t is actually a homeomorphism from T into

B weak The key is that the completely regular space T carries the

C T

weak top ology induced by C T In terms of nets

t t in T f t f t for all f C T

f f for all f C T

t t b

t t

Corollary Let T becompletely regular If C T is separable then T

is metrizable

weak is metrizable By Proof Since C T is separable we get that B

b C T

our previous result T is then metrizable to o

Corollary Let K bea compact Hausdor space Then C K is sepa

rable if and only if K is metrizable

Before we leave these topics wewould b e wise to say a few words ab out

the weak top ology on X The weak top ology on X is the top ology X

that is the smallest top ology on X making inherits as a subspace of R

each elementof X continuous In terms of nets

x x x x x x for all x X

A neighb orho o d base for the weak top ology is given by the sets

N x x x f y X jy x x x j ing

n i i

where x X x x X and

As a rst observation notice that the weak top ology on X when re

stricted to X is nothing but the weak top ology on X In terms of nets

x X x in X x x x x for all x

x x x x for all x X

x x in X

Thus X weak is homeomorphic to X weak or in other words the map

x x is a weaktoweak homeomorphism from X into X

Next from the BanachAlaoglu theorem we knowthatB is compact

in the weak top ology on X Hence if X is reexive then B is weakly

compact in X The converse is also true If B is weakly compact in X then

is weak compact X is reexive Indeed in this case wewould havethatB

and weak dense in B from Goldstines theorem Thus B B and

X b X

it follows that X X W etake this as pro of of

Prop osition X is reexive if and only if B is weakly compact

Compare this result to the fact that X is nite dimensional if and only if

B is norm compact

Notes and Remarks

Our presentation in this chapter b orrows heavily from Lacey but see also

Folland Chapter Kelley and Willard Theorem is semi

classical Kelley attributes the result to Alexandro and Urysohn from

I b elieve but gives an ambiguous reference Banach refers to the edi

tion of Hausdor s b o ok The BanachAlaoglu theorem Theorem

was rst proved byBanach for separable spaces and later improved to

its present state by Alaoglu Alaoglus pap er also contains some interest

ing applications including some contributions to the problem of emb edding

normed spaces into C K spaces along the lines of Corollary The

BanachMazur theorem Theorem app ears in Banachs b o ok

CHAPTER C K SPACES

Exercises

Let X and Y b e Hausdor top ological spaces with X compact If f

X Y is onetoone and continuous provethat f is a homeomorphism

into

Let d b e the metric on dened by dx y ja b j where

n n

a b are the ternary decimal digits for x y

n n

a If dx y show that the rst n ternary digits of x and y

agree

b Showthatd is equivalent to the usual metric on Hint The

identity map from dto usual is continuous

c Show that dx y dx z implies y z

Prove Corollary

If X is a separable metric space prove that some countable family F

C X will separate p oints from closed sets in X

Prove Lemma llingin any missing details ab out the extension

op erator E

Show that every subspace of a completely regular space is completely

regular

a If A is a subset of X show that A is a closed subspace of X

b If A is a subset of X show A is a weak closed subspace of X

b b

c If A is a subset of X showthat A A where A is the canon

ical image of A in X

d If Y is a subspace of X show that Y is the weak closure of

Y in X IfY is a subspace of X conclude that Y is the weak

closure of Y in X

e Goldstines theorem junior grade Provethat X is weak dense

in X

Goldstines theorem utility grade ProvethatB is weak dense in

B Hint If F B is outside the weak closure of B then there

X X b

is a weak continuous linear functional separating F from B

b X

Chapter

Weak Compactness in L

In this chapter we examine the structure of weakly compact subsets of L

L Since the closed unit ball of any reexive space is weakly compact

this undertaking will lead to a b etter understanding of the reexive subspaces

of L It should b e p ointed out that many of our results will hold equally well

in L where is a nite measure

We b egin with a denition WesaythatasubsetF of L is uniformly

integrable or as some authors say equiintegrable if

sup jf tj dt as a

f F

fjf ja g

What this means is that all of the elements of F can b e truncated at height a

with uniform error in the L norm A few examples mighthelp

Examples

Given f L its an easy consequence of Chebyshevs inequality that

mfjf j ag asa Thus since A jf j is absolutely

continuous with resp ect to mweget

jf j as a

fjf ja g

In other words any singleton ff g is a uniformly integrable set in L In

fact any nite subset of L is uniformly integrable

If there exists an element g L suchthat jf j jg j for all f F then

F is uniformly integrable Indeed given f F and a Rwewould

CHAPTER WEAK COMPACTNESS IN L

have fjf j agfjg j ag and hence

Z Z

sup jf tj dt jg tj dt

f F

fjf ja g fjg ja g

What this means is that the order interval

g g f f g f g g

is uniformly integrable Its not hard to see that any order interval g h

in L is likewise uniformly integrable

The sequence f n is not uniformly integrable Why

n n

Note that a uniformly in tegrable subset F of L is necessarily norm

jf j for all b ounded Indeed if wecho ose a R suchthat

fjf ja g

f F then kf k jf ja for all f F

As our rst example might suggest F is uniform integrable whenever the

jf j f F is uniformly absolutely continuous with family of measures

resp ect to m or as some authors say equiabsolutely continuous or simply

equicontinuous This is the content of our rst result

Prop osition A subset F of L is uniformly integrable if and only if the

family of measures F jf j f F is uniformly absolutely continuous

with respect to m that is if and only if for every there exists a

jf j for al l f F and for al l Borel sets A with such that

Proof First supp ose that F is uniformly integrable Given cho ose a

suchthat jf j forallf F and nowcho ose such that

fjf ja g

a Then for f F and A wehave

Z Z Z

jf j jf j jf j

A Af jf j a g Af jf ja g

a mA

whenever mA Thus F is uniformly absolutely continuous

Now supp ose that F is uniformly absolutely continuous Then F is norm

bounded Indeed cho ose such that jf j whenever f F and A

mA and partition into M subintervals each of length

at most It follows that kf k M for each f F

Given cho ose suchthat jf j whenever f F and

mA Next from Chebyshevs inequality notice that for f F wehave

kf k M

mfjf j ag

a a

Thus if a is chosen so that Ma then jf j Thus F is

fjf ja g

uniformly integrable

To place our discussion in the prop er context it will help to recall the

measure algebra asso ciated with B m To b egin it is an easy exercise

to check that dA B mAB denes a pseduometric on B Thus if

we dene an equivalence relation on B by declaring A B if and only if

mAB and if wewriteB to denote the set of equivalence classes under

e e

this relation then d denes a metric on B In fact d is a complete metric B

This is easiest to see if we rst make an observation For A and B in B we

have

mAB dA B k k

A B

Thus if we agree to equate sets that are ae equal then B inherits a complete

metric from L The complete metric space B d is called the measure algebra

asso ciated with B m

In this setting a Borel measure is absolutely continuous with resp ect to

m if and only if is continuous at in B d Measures b ehave rather like

linear functions on B In particular we only need to worry ab out continuity

at zero And a subset F of L is uniformly integrable if and only if the

corresp onding set of measures F is equicontinuous at when considered as a

collection of functions on B d

A sp ecial case is worth isolating

Lemma Given f L dene BR by A f Then is

uniformly continuous on B d

Proof Note that is welldened This is a simple computation

Z Z Z Z Z

f f jf j f f

AnB B nA AB A B

Thus jA B j can b e made small provided that dA B mAB is

suciently small where small dep ends on f but not on A or B

CHAPTER WEAK COMPACTNESS IN L

f j is closed Corollary Given f L and the set A j

in B d

Finally were ready to make some connections with weak compactness or

in this case weak convergence in L

f exists Prop osition Let f beasequenceinL such that lim

n n n

in R for every Borel subset A of Then

a f is uniformly integrable and

R R

b f converges weakly to some f in L inparticular f f for

n n

A A

every Borel subset A of

ProofFor andN dene

F A B f f for all m n N

N m n

e e

f Since is Cauchyforeach A B wehave B F But each F

n N N

must have is closed and B is complete By Baires theorem then some F

nonemptyinterior Thus there exists some A B and some r such that

mAA r implies A F

Supp ose nowthatmB r Then for m n N wehave

Z Z Z

f f f f f f

m n m n m n

B A B A nB

and also

mA B A r and mA n B A r

f f Applying the same argument to the sets Hence

m n

B ff f g and B ff f g

m n m n

then yields jf f j whenever mB r and m n N

m n

Now the set F f f j N g is uniformly integrable and so there

exists an sr suchthat jf j whenever j N and mB s

Finallyif mB s and jN wehave

Z Z Z

jf j jf f j jf j

j j N N

B B B

Thus f is uniformly integrable whichproves a

Next for each A B putA lim f Then is countably

n n

additive this follows from the fact that f is uniformly integrable and ab

solutely continuous with resp ect to mThus by the RadonNikodym theorem

R R R

f for ev f f That is there exists an f L such that A

A A A

R R

ery A B It follows from linearityoftheintegral that f g fg for

every simple function g Since the simple functions are dense in L we get

R R

f in L whichproves b Thatisf f g fg for every g L

n n

Corollary L is weakly sequential ly complete

Proof Supp ose that f isweakly CauchyinL Then in particular f

n n

converges for every A B Hence f converges weakly to some f L by

Prop osition

Finally were ready to characterize the weakly compact subsets of L

Theorem A subset F of L is relatively weakly compact if and only if it

is uniformly integrable

Proof Supp ose that F is not uniformly integrable Then there exists an

such that for each n we can nd an f F with jf jIn

n n

fjf jn g

particular f has no uniformly integrable subsequence It then follows from

Prop osition that f hasnoweakly convergent subsequence Thus F

cannot b e relatively weakly compact This follows from the Eb erleinSmulian

theorem A subset of a normed space is weakly compact if and only if its

weakly sequentially compact

Now supp ose that F is uniformly integrable We will show that the weak

and hence that closure of F is contained in L considered as a subset of L

F is weakly compact To this end supp ose that G is in the weak closure

of F Given cho ose suchthat jf j whenever f F and

mA Since G is a weak cluster p ointof F it follows that jG j

is absolutely continuous whenever mA Thus the measure A G

with resp ect to m By the RadonNikodym theorem there is a g L such

that G g and it now follows from the linearity and continuit yof G

that Gh hg for every h L ThatisG g L

The same pro of applies to any L where is a p ositive nite measure

Thus if is p ositive and nite the weakly compact subsets of L are

precisely the uniformly integrable subsets This leads us to our nal but

most useful corollary

CHAPTER WEAK COMPACTNESS IN L

Corollary VitaliHahnSaks Theorem Let beasequence of signed

measures on a algebra such that A lim A exists in R for each

n n

A Then is a signedmeasureon

ProofLet j j k k where kk jjX the total variation

n n

of applied to the underlying measure space X Then each is absolutely

continuous with resp ect to thatis d f d for some f L Thus

n n n

f d A A for each A and so b y Prop osition there

n n

exists an f L such that

Z Z

fd lim f d A

A A

for every A That is d fd and it follows that is a signed measure

ie is countably additive on

Notes and Remarks

Our presentation in this chapter b orrows heavily from some unpublished notes

for a course on Banach space theory given by Stephen Dilworth at the Uni

versity of South Carolina The Eb erleinSmulian theorem can b e found in any

numberofbooksbutseealsoand

Exercises

Prove that the sequence n is not uniformly integrable in L

If f converges to f in L give a direct pro of that f is uniformly

n n

integrable

Let f b e a sequence in L Iff f ae show that the following are

n n

equivalent

a f is uniformly integrable

b kf f k asn

c kf k kf k as n

If f converges weakly to f in L showthatf is uniformly inte

n n

grable

Provethat dA B mAB denes a complete pseudometric on B

Given a measure BRprove that the following are equivalent

a is absolutely continuous with resp ect to m

b is continuous at in B d

c is uniformly continuous on B d

Prove that a subset F of L is uniformly integrable if and only if the

corresp onding set of measures F f fdm f F g is equicontinuous

at in B d

CHAPTER WEAK COMPACTNESS IN L

Chapter

The DunfordPettis Prop erty

A Banach space X is said to have the DunfordPettis property if whenever

w w

x inX and f inX wehavethat f x Our main result

n n n n

in this chapter will show that the DunfordPettis prop ertyisintimately related

to the b ehavior of weakly compact op erators on X Recall that a b ounded

linear op erator T X Y is said to b e weakly compact if T maps b ounded

sets in X to relatively weakly compact sets in Y Thus T is weakly compact

if and only if T B isweakly compact in Y Since weakly compact sets are

norm b ounded it follows that a weakly compact op erator is b ounded

Our rst result provides several equivalentcharacterizations of weak com

pactness for op erators

Theorem Let T X Y bebounded and linear Then T is weakly

compact if and only if any one of the fol lowing hold

a T X Y

b T Y X is weak toweak continuous

c T is weakly compact

Proof Supp ose that T is weakly compact Then T B is relatively weakly

compact in Y Regarding T B as a subset of Y wehave

Y weak Yweak

T B T B

X X

b ecause weakly compact sets in Y are weak compact in Y SinceT B is

convex this simplies to read

weak

T B T B Y

X X

CHAPTER THE DUNFORDPETTIS PROPERTY

But B is weak dense in B by Goldstines theorem and T is weak to

X X

weak continuous so

weak

T B Y T B

X X

Thus T X Y

Converselyif T X Y thenT B isaweakly compact subset of

Y by the BanachAlaoglu theorem and the weak toweak continuityof T

and again the observation that the weak top ology on Y when considered as

a subset of Y reduces to the weak top ology on Y Thus T B is relatively

This proves that T is weakly compact in Y b eing a subset of T B

weakly compact if and only if a holds

Now from a

T is weakly compact T X Y

T x Y for each x X

T x is weak continuous for each x X

T x y T x y for each x X

whenever y y in Y

T y for each x X x T y x

y in Y whenever y

w w

T y T y whenever y y in Y

T is weak toweak continuous

This proves that T is weakly compact if and only if b holds

Finallyfrombif T is weakly compact then T is weak toweak con

tinuous But B is weak compact in Y sowehave that T B isweakly

Y Y

compact in X Thus T is weakly compact ConverselyifT is weakly

compact then T B isweakly compact in Y Thus T B is relatively

X X

weakly compact in Y

Armed with these to ols we can nowprovide our main result in this chapter

Theorem A Banach space X has the DunfordPettis property if and

only if every weakly compact operator from X into a Banach space Y maps

weakly compact sets in X to norm compact sets in Y ie if and only if every

weakly compact operator is completely continuous

Proof Supp ose rst that X has the DunfordPettis prop erty and let T X

Y be weakly compact To b egin we consider the action of T onaweakly

null sequence x inX For each ncho ose a norm one functional g Y

n n

such that g Tx kTx k That is T g x kTx kNow T is weakly

n n n n n n

compact so there is an f X and a subsequence g ofg such that

n n

f But then T g

g Tx T g x f x T g f x

n n n n n n n

k k k k k k k

Indeed f x b ecause x isweakly null and T g f x

n n n n

k k k

b ecause X has the DunfordPettis prop ertyThus

g Tx kTx k

n n n

k k k

That is Tx has a norm null subsequence whenever x isweakly null By

n n

linearity it follows that Tx has a norm convergent subsequence whenever

x isweakly convergent Consequently T K is norm compact whenever K

is weakly compact

Now supp ose that every weakly compact op erator on X maps weakly com

w w

pact sets to norm compact sets Given x inX and f inX

n n

consider the map T X c dened by Tx f x Its easy to see that

T X satises T e f In particular given x X wehave

n n

T e x f T x e x

n n n

since f isweakly null That is T x c In other words T is weakly

compact But then T maps weakly compact sets in X to norm compact sets

in c and it follows that wemust have kTx k Why Consequently

f x

n n

Corollary Suppose that X has the DunfordPettis property If T X

X is weakly compact then T X X is compact

Proof Since T is weakly compact T B is relatively weakly compact And

since X has the DunfordPettis prop erty T B is then relatively norm com

pact

Corollary Suppose that X has the DunfordPettis property If Y is a

complementedreexive subspaceof X then Y is nite dimensional

CHAPTER THE DUNFORDPETTIS PROPERTY

ProofIfP X X is any pro jection onto Y thenP is weakly compact

Indeed P B B is weakly compact since Y is reexive But then P P

X Y

is compact Consequently B is norm compact and it follows that Y must b e

nite dimensional

Corollary Innite dimensional reexive Banach spaces do not have the

DunfordPettis property

Our task in the remainder of this chapter will b e to show that C K and

L have the DunfordPettis prop ertyGiven this it will followthat C K and

L have no innite dimensional reexive complemented subspaces

Theorem Let K beacompact Hausdor space Then C K has the

DunfordPettis property

w w

Proof Supp ose that f inC K and that inC K Let

n n

j j

k k

where k k j jK Then is uniformly absolutely continuous with

n n n

resp ect to by Prop ositions and Thus given there exists

a such that j jA for all nprovided that A

Now f p ointwise on K Thus by Egorovs theorem f uniformly

n n

on some set K n A where A Next

Z Z Z

f f d f d f d

n n n n n n n n

K K nA A

and f d asn b ecause f uniformly on K n A Finally

n n n

K nA

kf k j jA sup kf k f d

n n k n n

b ecause A Thus f

n n

Corollary If X is a separable innite dimensional reexive Banach

space then X is isometric to an uncomplemented subspaceofC

We next attack the DunfordPettis prop ertyinL To this end we will

need some additional information ab out weak convergence in L

in L Then given there Prop osition Suppose that g

exists a Borel set A with mA such that g uniformly

on A

Proof By rep eated application of Lusins theorem we can nd a Borel subset

B of with mB and a sequence of functionsg each contin

uous on B suchthatg g ae on B Moreover by the Leb esgue density

n n

theorem wema y assume that eachpoint x B has density that is

mx rx r B

lim

for each x B thus in particular B has no isolated p oints It follows that

n n n

X X X

a g x esssup a g a g

k k k k k k

k k k

for all scalars a and all x B Hence for each x B themapg g x

k n n

extends to a b ounded linear functional on g But then since g isweakly

n n

null wemust haveg x asn for each x B Hence g

n n

ae on B FinallybyEgorovs theorem there exists a Borel set A B with

mA such that g uniformly on A

Corollary L has the DunfordPettis property

w w

inL Then f is uniformly inL and let g ProofLetf

n n n

integrable in L Thus given there is a suchthat jf j

for all nprovided that mB By Prop osition there is a Borel set A

with mA such that g uniformly on AThus

Z Z Z

jf g f g jf g j j

n n n n n n

A A

kg k supkg k sup kf k sup

n n A n

n n n

which tends to as n

Corollary Every complemented innite dimensional subspaceofL is

nonreexive

CHAPTER THE DUNFORDPETTIS PROPERTY

Notes and Remarks

Our presentation in this chapter b orrows heavily from some unpublished notes

foracourseonBanach space theory given by Stephen Dilworth at the Univer

sity of South Carolina Theorem is often called Gantmachers theorem

after Vera Gantmacher who proved the theorem for separable spaces The

general case was later settled by Nakamura Theorem in dierent

language is due to Dunford and Pettis The DunfordPettis prop ertywas

sonamed by Grothendieck who gave us Theorem Theorem

andawealth of other results For more on the DunfordPettis prop erty as

well as its history see Diestel and Uhl

Exercises

If T X Y is b ounded and linear prove that T X Y is

weak toweak continuous

If K is weakly compact in X provethatK is weak closed as a subset

of X

If K is a subset of X such that the weak closure of K is again contained

in X when considered as a subset of X prove that K is relatively

weakly compact

If K X is weak compact as a subset of X prove that K is weakly

compact in X

Provethe has the DunfordPettis prop erty

Let A b e a Borel subset of with Leb esgue density and let g

A R b e continuous Provethat jg xj esssup jg j for every x A

Prove that if X has the DunfordPettis prop ertythenX do es to o

Thus c has the DunfordPettis prop erty

Prove that a Banach space X has the DunfordPettis prop erty if and only

if every weakly compact op erator T X c maps weakly compact sets

in X to norm compact sets in c In other words we need only consider

Y c in Theorem

CHAPTER THE DUNFORDPETTIS PROPERTY

Chapter

C K Spaces II

By now even a skeptical reader should b e thoroughly sold on the utilityof

emb eddings into cub es But the sales pitch is far from over We next pursue

the consequences of a result stated earlier If X is completely regular then

C X completely determines the top ology on X In brief to know C X

b b

is to know X Just how far can this idea b e pushed If C X andC Y

b b

are isomorphic as Banach spaces as lattices or as rings must X and Y

b e homeomorphic Which top ological prop erties of X can b e attributed to

structural prop erties of C X and conversely

These questions were the starting place for Marshall Stones landmark

pap er Applications of the theory of Boolean rings to general topology Its

in this pap er that Stone gave his account of the StoneWeierstrass theorem the

BanachStone theorem and the StoneCech compactication These few are

actually tough to nd among the dozens of results in this mammoth page

work Paraphrasing a passage from his intro duction We obtain a reasonably

complete algebraic insightinto the structure of C X and its correlation with

the structure of the underlying top ological space Stones work proved to b e a

goldminethe digging continued for yearsits inuence on algebra analysis

and top ology alike can b e seen in virtually every mo dern textb o ok

Indep endentlybutlaterthatsameyear Eduard Cechgave

another pro of of the existence of the compactication but strangely credits

a pap er of Tychono for the result see Shields for more on this

Cechs approach which leans story To a large extent we will b e faithful to

more toward top ology than algebra

CHAPTER C K SPACES I I

The StoneCech compactication

Given a Hausdor top ological space X any compact Hausdor space Y which

contains a dense subspace homeomorphic to X is called a Hausdor compact

ication for X What this means in practice is that welookforany compact

Hausdor space Y which admits a homeomorphic emb edding f X Y from

X into Y the closure of f X inY then denes a compactication of X

There is a hierarchy of compactications the full details of which arent

necessary just now It shouldnt come as a surprise that for lo cally compact

spaces the onep oint compactication is the smallest compactication in this

hierarchyWhatwere after is the largest compactication

Given a completely regular space X we dene X the StoneCech com

C X

pactication of X tobetheclosure of eX in From the

Emb edding Lemma Lemma X is then homeomorphic to a dense subset

of the compact Hausdor space X Note that if X is compact then e is a

homeomorphism from X onto X

Strictly sp eaking the compactication is dened to b e the pair e X

but we will have little need for such formality In fact we often just think of

X as already living inside X and simply ignore the emb edding e

X is characterized by the following extension theorem

Theorem Extension Theorem Let X bea completely regular space

and let e X X be the canonical embedding

a Every bounded continuous function f X R extends to a continuous

function F X R in the sense that F e f

e then each continuous function f b If Y is a compact Hausdor spac

X Y extends to a continuous function F X Y in the sense that

F e f If Y is a compactication of X thenF is onto In particular

every Hausdor compactication of X is the continuous image of X

Proof a Supp ose that f X R is continuous and b ounded By comp os

ing with a suitable homeomorphism of Rwemay assume that f X

C X

But then f is one of the co ordinates in the pro duct space

We claim that the co ordinate pro jection when restricted to X is the

when extension wewant Indeed X is continuous and

f f

restricted to eX is just f since ex exf f x Note that

F j is unique since eX is dense in X

f X

b Now supp ose that Y is a compact Hausdor space and that f X

Y is continuous Let C C X and C C Y and let

X Y

C C

X Y

e X and e Y b e the canonical emb eddings which

X Y

wemay consider as maps into X and Y resp ectively Note that since Y

is compact e is a homeomorphism from Y onto Y

In order to extend f to X we rst lift f to a mapping from to

To b etter understand this lifting recall that f x Y corresp onds

under the evaluation map e Also to the tuple g f x in

Y g C

note that g f C whenever g C

X Y

C C

X Y

id id

Y X

e e e

Y X

C C

X Y

Wenow dene by sp ecifying the co ordinates of its

images p p for each g C and each p The map

g g f Y

iscontinuous since its co ordinates are and e e f

X Y

e x e x g f x e f x

g X g f X g Y

x e Y wehave that X e Y Next since e x e f

Y Y X Y

e Y Consequently F e j e extends f Again uniqueness of F

Y X X

follows from the fact that eX is dense in X

Finallyif f X Y is a homeomorphism from X onto a dense subspace

of Y thenF X Y maps X onto a dense compact subset of Y That

is F is onto

The conclusion of part b of the Extension Theorem is that the Stone

Cech compactication is the largest Hausdor compactication of X in a

categorical sense

The Extension Theorem tells us something new ab out the space C X

note that the map F F e denes a linear isometry from C X onto

CHAPTER C K SPACES I I

C X That is each f C X is of the form F e for some F C X In

b b

particular wenow know that C X isa C K space for some rather sp ecic

compact Hausdor space K

Corollary Let X becompletely regular

i C X is isometrical ly isomorphic to C X

ii If Y is completely regular then each continuous function f X Y

lifts to a continuous function F X Y satisfying F e e f

X Y

iii If Y is any Hausdor compactication of X enjoying the property of X

describedinpart a of the Extension Theorem then Y is homeomorphic

to X

Proof Only iii requires a pro of Let Y b e a compact Hausdor space and

supp ose that f X Y is a homeomorphism from X onto a dense subspace

of Y Supp ose further that each h C X extends to a continuous function

g C Y with g f h Then in particular each h C X isofthe

form g f for some g C Y This implies that the lifting of f

constructed in the pro of of part b of the Extension Theorem is onetoone

Thus the extension F X Y of f is b oth onetoone and onto hence a

homeomorphism

For later reference we next presenttwo simple metho ds for computing the

StoneCech compactication

Lemma Let X becompletely regular

a Let T X X If each bounded continuous real valued function

on T extends continuously to X then T cl T the closureof T in

b If X T X then T X

Proof a By design each b ounded continuous real valued function on T

extends all the waytoX hence also to cl T Since cl T is a compacti

X X

cation of T itmust then b e T by Corollary iii

b First observethat T is dense in X and so X is a compactication

of T Next each b ounded continuous real valued function on T extends

continuously to X since its restriction to X do es The fact that X is dense

in T takes care of any uniqueness problems caused by extending the restriction

or restricting the extension Thus by Corollary iii X T

Were wayoverdue for a few concrete examples

Examples

Why Because sinx has no continuous extension

to But sinx do es of course extend continuously to

whatever that is As well see in a moment is much larger than

If D is any discrete space then D C D C D isometrically

In particular C N Since isnt separable wenowhavea

pro of that N isnt metrizable In fact as well see N is in no way

sequentially compact

card N Heres a clev er pro of Recall that Y is

separable Hence there is a continuous map from N onto a dense subset

of Y This map extends to a continuous map from N onto all of Y

Consequently cardY card N while from the construction of

N it follows that card N card

card card R card N The rst equalityisobvious

since is homeomorphic to R Next as ab ove R is the continuous

image of N b ecause R is separable Thus card R card N

To nish the pro of well nd a copyof N living inside R Heres

how Each b ounded continuous realvalued function on N extends to

a b ounded continuous function on all of R No deep theorems needed

here Just connect the dots Thus by Lemma a cl N N

Hence card R card N

Banach limits invariant means We will use the fact that C N

to extend the notion of limit to include all b ounded sequences First

given x lets agree to writex for its unique extension to an element

of C N Now given any xed p oint t N n Nwe dene Lim x

x t This generalized limit often called a Banach limit satises

Lim x lim x if x actually converges

lim inf x Lim x lim sup x

Lim ax by a Lim x b Lim y

Lim xy Lim xLimy

CHAPTER C K SPACES I I

Just for fun lets check the rst claim The key here is that t is in the

closure of fn n mg for each m Thus if L limx exists and if

then x L L for all n m for some m Hence

x t L L to o That isx t L

What weve actually found is a particularly convenient HahnBanach

extension of the functional lim on the subspace c of What makes

this example interesting as well see later is that no p oint t N n N

can b e the limit of a sequence in N

From our discussion of completely regular spaces in the last chapter we

can give an alternate denition of the StoneCech compactication Each

completely regular space T lurks within C T under the guise of the

point masses P f t T g It follows that we can dene T to b e

the weak closure of P in C T Why Well since P weak cl P is a

compactication of T we only need to show that eachelement f C T

extends to an element f C P and this is easier than it might sound

Just dene f ptobepf In other words the canonical emb edding of

C T into C T supplies an emb edding of C T into C T

b b b

Finally heres a curious pro of of a sp ecial case of Tychono s theorem

based on the Extension Theorem Let X X where each X is

a nonempty compact Hausdor space Then X is completely regular

Hence the pro jection maps X X have continuous extensions

X X But this means that the map p p from X

uous Thus X is compact onto X is contin

Return to C K

N is a most curious space and will play a ma jor role in the next chapter More

generally as our examples might suggest the StoneCech compactication of

a discrete space is a p otentially useful to ol This is further highlighted by the

following observation

Lemma Every compact Hausdor space K is the continuous image of

D for some discrete space D Consequently C K is isometric to a subspace

of C D D

Proof Let D be any dense set in K and let D be D with the discrete

top ology Then the formal identity from D into K extends to a continuous

map from D onto K The comp osition map f f denes a linear

isometry from C K into C D

As an immediate Corollarywe get a result that weve essentially seen

before

Corollary If K is a compact metric space then C K embeds isometri

cal ly into Hence every separable normed linear spaceembeds isometrical ly

into

In the next chapter we will compute the dual of C K by instead com

puting the dual of D That is wewillprove the Riesz representation

theorem for spaces and then transfer our work to the C K spaces This

b eing the case wemight b e wise to quickly summarize a few features of the

fancy versions of the Riesz representation theorem

First the spaces C K and its relatives C X C T and so on are

C b

probably b est viewed as vector lattices Under the usual p oint wise ordering of

functions C K is an ordered vector space

f g C K f g f h g h for all h C K and

f C K a Rf a af

In addition C K is a lattice

f g C K f g maxff gg and f g minff gg are in C K

This last observation allows us to dene p ositive and negative parts f f

and f f Thus each f C K can b e written as f f f

Moreover jf j f f

Now C K enjoys the additional prop erty that its usual norm is compatible

with the order structure in the sense that

jf jjg j kf k kg k

Since C K is also complete under this norm wesaythat C K isa Banach

lattice

As it happ ens the dual of a Banach lattice can b e given an order structure

by dening S T to mean that S f T f to o We dene an order on C K

for all f In particular a linear functional T on C K ispositive if

CHAPTER C K SPACES I I

T f whenever f Its easy to see that every p ositive linear functional

is b ounded indeed if T is p ositive then

jT f jT jf j T kf k kf k T

where denotes the constant function Hence kT k T

Whats a little harder to see is that the dual space will again b e a Banach

lattice under this order We rst need to check that each b ounded linear

functional can b e written as the dierence of p ositive functionals Given T

C K we dene

f supf T g g f g for f T

Its tedious but not dicult to check that T is additive on p ositive ele

ments and that T af aT f fora Now for arbitrary f we dene

T f T f T f It follows that T is p ositive linear and satises

T f T f forevery f Thus T T T is likewise p ositiveand

linear That is weve written T as the dierence of p ositive linear functionals

Consequently a linear functional on C K is bounded if and only if it can be

written as the dierenceofpositive linear functionals

Finally lets compute the norm of T in terms of T and T Clearly

kT k kT k kT k T T

On the other hand given f wehave jf j and hence kT k

T f T f T By taking the supremum over all f we

the get

kT k T T T T

Hence kT k T T If we dene j T j T T asonewould

exp ect then wehave kT k kjT jk jT j It follows from this denition of

jT j that C K is itself a Banach lattice

In terms of the Riesz representation theorem all of this tells us that we

only need to represent the p ositive linear functionals on C K As you no

doubt already know each p ositive linear functional on C K willturnoutto

be integration against a p ositive measure The generic linear functional will

then b e given byintegration against a signed measure In terms of measures

is the Jordan decomp osition of while jj is the total

variation of Not surprisinglywe dene kk jjK d jj kjjk K

Notes and Remarks

The StoneCech compactication is discussed in anynumber of books see for

example Folland Chapter Gillman and Jerison Wilansky

or Willard For more on Banach limits and their relationship to N

see Nakamura and Kakutani Banach lattices are treated in a number

of b o oks see for example Aliprantis and Burkinshaw Lacey Meyer

Nieb erg or Schaefer

CHAPTER C K SPACES I I

Exercises

Prove Corollary ii

Let X b e completely regular ShowthatX is lo cally compact if and

only if X is op en in X

Complete the pro of of the claims made in Example concerning the

Banach limit Lim x on

Chapter

C K Spaces III

In this chapter we present Garlings pro of of the Riesz representation

theorem for the dual of C K K compact Hausdor This theorem go es by

avariety of names The RieszMarkov theorem the RieszKakutani theorem

and others The version that well prove states

Theorem Let K beacompact Hausdor space and let T beapositive

linear functional on C K Then there exists a unique positive Bairemeasure

on K such that T f fd for every f C K

As we p ointed out in the last chapter our approach will b e to rst prove

the theorem for spaces Tothisendwe will need to know a bit more ab out

the StoneCech compactication of a discrete space and a bit more measure

theory First the top ology

The StoneCech compactication of a discrete space

A top ological space is said to b e extremal ly disconnected orStonean if the

closure of every op en set is again op en Obviously discrete spaces are ex

tremally disconnected Less mundane examples can b e manufactured from

this starting p oint

Lemma If D is a discrete space then D is extremal ly disconnected

Proof Let U b e op en in D and let A U D Then A is dense in U

since U is op en and so cl A cl U Nowwejustcheck that cl A is

D D D

CHAPTER C K SPACES I I I

D f g a continuous function also op en The characteristic function

on D extends continuously to some f D f gThus bycontinuity

cl A f fgisopen

By mo difying this pro of its not hard to show that a completely regular

space X is extremally disconnected if and only if X is extremally discon

nected Since wewont need anything quite this general well forego the

details

Notice that if A and B are disjoint op en sets in a discrete space D

then cl A and cl B are disjointinD Indeed just as in the pro of of

D D

Lemma the function extends continuously to a function f D

f g which satises cl A f fg and cl B f fg In particular

D D

any set of the form cl A where A D isclopen that is simultaneously

op en and closed In fact every clop en subset of D is of this same form

Lemma Let D be a discrete space Then the clopen subsets of D ar e

of the form cl A where A is open in D Further the clopen sets form a base

for the topology of D

ProofIfC is a clop en subset of D then just as in Lemma

C cl C cl C D

D D

Now let U b e an op en set in D and let x U Since D is regular we can

nd a neighborhood V of x suchthat x V cl V U Since cl V is

D D

clop en this nishes the pro of

A few facts ab out N

We can now shed a bit more lighton N Note for example that N is open in

N Indeed given n N the set cl fng is op en in N But fng is compact

hence fng cl fng That is fng is op en in Ntoo In particular each

n N is an isolated p ointin N

It follows that a sequence in N converges in N if and only if it is eventual ly

constant that is if and only if it already converges in N Supp ose to the

contrary that x is a sequence in N whichisnoteventually constant and

supp ose that x converges to a p oin t t N n N Then the range of x

n n

must b e innite for otherwise x would have a subsequence converging in

NThus by induction we can cho ose a subsequence x of distinct integers

n k

x x if i j Butnow the sets A fx g and B fx g are disjoint

n n n n

i j

k k

in N while t is in the closure of eachin Nacontradiction In particular

weveshown that no p oint t N n N can b e the limit of a sequence in NAs

a consequence N isnt sequentially compact and thus isnt metrizable

A similar argumentshows that for any t N n N the compact set ftg isnt

a G in N Indeed if ftg were a G thenwe could nd a sequence of clop en

sets of the form B cl A whereA N such that ftg B The

n N n n n

sets A have the nite intersection prop ertysowe can cho ose a sequence of

distinct p oints x A A Putting A fx n Ngwewould then

n n n

havecl A n N B ftg But since A is an innite subset of Nit

N n

follows that cl A is homeomorphic to N and in particular has cardinality

a contradiction What weveshown of course is that every nonempty G

subset of N n N has cardinality This observation will b e of interest in our

discussion of measures on N

If A is an innite subset of Nthencl A is homeomorphic to N Since N

can b e partitioned into innitely many disjoint innite subsets A it follows

that N contains innitely many pairwise disjoint clop en sets B cl A

n N n

B is not all of N each homeomorphic to N Note however that T

since a compact space cant b e written as a disjoint union of innitely many

disjoint op en sets Since N T Nwedohavethat T is dense in N

moreover T N

Using this observ ation we can build a copyof N inside the closed set

N n NTo see this let t B n N cl A n N and set D ft n g

n n N n n

Obviously D is a discrete subspace of N and as such is homeomorphic to N

Thus D is homeomorphic to NNow given a b ounded realvalued function

f on D we can easily extend f to a b ounded continuous function on T by

setting f xf t for every x B SinceD T N T itfollows

n n

that D cl D But since D is a subset of the closed set N n NsoisD

In short weve just found a copyof N in N n N

As a very clever argument demonstrates there are in fact c disjoint copies

of N living inside N n N Indeed recall that we can nd c subsets E

of N suchthateach E is innite and anytwo E have at most a nite

intersection For each the set F cl E n N is then homeomorphic to

N n N and so contains a copyof N Finally notice that the F are pairwise

thesetcl E n E diers from cl E in only disjoint since for each

N N

a nite subset of N

CHAPTER C K SPACES I I I

Top ological measure theory

Now for some measure theory Our job rememb er is to compute the dual

of C D where D is discrete Weknow that there are enough clop en sets

in D to completely determine its top ology and so enough clop en sets to

completely determine C D It should come as no surprise then that there

are also enough clop en sets to determine C D The clop en sets in D form

an algebra of sets whichwe will denote by Athe algebra generated by A

will b e denoted by Twomore algebras will enter the picture B the Borel

algebra on Dand B the Baire algebra on D The Baire algebra is

the smallest algebra B such that each f C DisB measurable Its

not hard to see that B and are sub algebras of B The next lemma sho ws

that we also have B

Lemma Each f C D is measurable Moreover the simple func

tions basedonclopen sets in A are uniformly dense in C D

Proof Let f C D and let RThen

f f g f f n g f f n g

n n

f f n g bycontinuity

f f g

Thus wehave equality throughout It then follows from Lemma that the

set f f g is the countable intersection of clop en sets and as such is in

Hence f is measurable

The second assertion follows from the fact that the nitelymanyvalued

functions are dense in D

Since each f C D is measurable wemust have B On the

other hand since each clop en subset of D can b e realized as a zero set

for some f C D wealsohave AB and hence B Thus the

algebra of Baire sets on D coincides with the algebra generated by the

clop en sets in D

Please note that our pro of also shows that f f g is a compact G in D

The Baire algebra on any reasonable space turns out to b e the algebra

generated by the compact G sets In contrast note that the Borel algebra

on any compact Hausdor space could b e dened as the algebra generated

by the compact sets We briey describ e a few such cases b elow

For a lo cally compact space X the Baire algebra B is dened to b e the

smallest algebra on X such that eachelementof C X is measurable where

C X is the space of continuous realvalued functions on X with compact

supp ort

Lemma Let X bea local ly compact Hausdor space

a If f C X is nonnegative then f f g is a compact G for every

b If K is a compact G in X then thereisanf C X with f

such that K f fg

c The Baire algebrain X is the algebra generated by the compact G

sets in X

Proof a For the set f f g is a closed subset of the supp ort of f

f f n g is also a hence is compact And as b efore f f g

b Supp ose that K U where U is op en Apply Urysohns

n n

lemma to nd an f C X with f on K and f o U

n C n n n

f is in C X and f precisely on K Then f

n C

c Let G b e the algebra generated by the compact G sets in X From

a each f C X isG measurable hence B GFrom b each compact

G is a Baire set and so GB

If D is discrete then the Baire sets in D are typically a prop er sub

algebra of the Borel sets in DIfD is innite then a cardinality argument

similar to the one weusedfor Nwould show that for t D n D the

compact set ftg is not a G in D

Our next Lemma explains whywe never seemed to need the Baire sets

before On R or on a compact metric space the Baire sets coincide with the

Borel sets

Lemma Let X beasecond countable local ly compact Hausdor space

Then

i Every open set in X is a countable union of compact sets

CHAPTER C K SPACES I I I

ii Every compact set in X is a G

iii The Baire and Borel algebras on X coincide

Proof Since X is lo cally compact it has a base of compact neighb orho o ds

Since X is second countable we can nd a base consisting of only countably

many such compact neighborhoods Thus i follows And ii clearly fol

lows from i by taking complements Finally iii follows from i ii and

Lemma c

For go o d measure heres another example

Example

On an uncountable discrete space D the Baire sets are a prop er sub

algebra of the Borel sets

Proof Since D is discrete the only compact subsets of D are nite It follows

that the Baire algebra on D is the algebra generated by the singletons

B f E E or E is countable g

Since we can write D as the union of two disjoint subsets eachhaving the

same cardinalityas D we can obviously nd an op en subset of D whichis

not a Baire set

The dual of

We are nowwellprepared to compute the dual of D As you might imag

ine the continuous linear functionals on D should lo ok likeintegration

against some measure on D As a rst step in this direction weintro duce

the space ba the collection of all nitely additive signed measures of nite

variation on supplied with the norm kk jjD where jj is the total

variation of The name ba stands for b ounded variation and nitely

additive

Recall that the total variation of is dened by

n n

E jE j E E disjoint E jjE sup

i i n

i i

ell to nitely additive measures which applies equally w

But what is meantbyintegration against such measures Well if

ba then fd is well dened and linear for simple nitelymany

valued functions f Now given a simple function f write f a

i E

where E E are disjoint and partition D Then

n n

X X

fd ja E jkf k jE jkf k kk

i i i

i i

fd denes a b ounded linear functional on the subspace of simple Thus

functions in D Since the simple functions are dense in D this means

fd extends unambiguously to all f D This unique linear that

fd With this understanding extension to D iswhatwe mean by

our work is half done

Theorem D ba isometrical ly

ProofAsweve just seen each ba denes a functional x D

by setting x f fdforf simple and extending to all of D And

as the calculation ab ove shows kx k kk

Next the hard direction Let x D and dene E x

for E D Clearly is nitely additive we just need to checkthat is of

bounded variation

Given disjoint subsets E E of D wehave

n n

X X

jE j jx j

i E

i i

x for some

i E i

by linearity x

i E

kx k since

i E

Thus kk jjD kx k Also by linearitywehavethat x f fd

for any simple function f Since b oth functionals are continuous and agree

f fd for all on a dense subspace of D we necessarily have x

f D

CHAPTER C K SPACES I I I

Combining the two halfs of our pro of we arrive at the conclusion that the

corresp ondence x is a linear isometry b etween the spaces D and

Please note that our pro of actually shows something more The positive

linear functionals in D correspond to positive measures in ba

The Riesz representation theorem for C D

While knowing the dual of D should b e reward enough we could hop e for

more from our result It falls just short of the full glory of the Riesz represen

tation theorem for C D Optimisticallywed like to represent the elements

of C D as regular countably additive measures on DAsyou can imag

ine we mightwant to explore the p ossibility of applying say Caratheo dorys

extension theorem to the elements of ba But notice please that the

natural algebra asso ciated to integration on C D is the Baire algebra

in this case The approachwell take only supplies a Baire measure Its a

fact however that every Baire measure on a compact Hausdor space is reg

ular moreov er there is a standard technique for extending a Baire measure

to a unique regular Borel measure see Chapter for example

Now while the elements of ba are not typically countably additive

they do satisfy a somewhat weaker prop erty Given a sequence of disjoint sets

A inD wehave

X X

A jA jjj kk A

i i i

Hence if is nonnegative then

A A

i i

tably additive then what could the sum A miss that If is not coun

A picks up The answer comes from D

The fact that disjoint sets in D have disjoint closures in D allows us to

A cl A dene a twin of on DWeidentify each subset A of D with

in D and we dene A A The set function is a nitely additive

to explain measure on A the algebra of clop en subsets of DWe can use

how might fall short of b eing countably additive

If A is a sequence of disjoint subsets of D then in general

A A

i i

Why Because the union on the left is open while the union on the rightis

compact equality can only o ccur if all but nitely manyofthe A s are empty

P S

Thus in general A A thatis fails to account for the

i i

A closure of

But this same observation shows us that is actually countably additive

A is clopen then its actually on A the algebra of clop en sets Indeed if

a nite union and so must equal A In the terminology of is a

premeasure on AThus we can invoke Caratheo dorys theorem to extend

to a regular countably additive measure on the algebra generated by

A The extension will still satisfy A A whenever A D ofcourse

In particular if f a is a simple function based on disjoint subsets

i A

a of D thenf is a simple function based on disjointsetsinA

and wehave

Z Z

n n

X X

A fd a a A fd

i i i i

D D

i i

What this means is that we can represent the elements of C D as in

tegration against regular countably additive measures on If T D

C D is the canonical isometry notice that T maps the characteristic func

tion of a set A in D to the characteristic function of cl A in DThus T

maps simple functions based on sets in to simple functions based on clop en

sets in ANow a functional x C D induces a functional x T on D

Hence there is a measure ba such that x Tf fd for every

f D If f is a simple function then Tf f in the notation we used

ab ove and so

Z Z

x f x Tf fd fd

D D

Since the simple functions based on clop en sets are uniformly dense in C D

wemust have x g gd for every g C D That is weve arrived

at the Riesz representation theorem for C D In symbols

C D D ba rca

where rca denotes the space of regular countably additive measures on

CHAPTER C K SPACES I I I

Theorem Given a continuous linear functional x C D there ex

ists a unique signedmeasure on the Baire sets in D such that

x f fd for al l f C D

Moreover kx k jjD

Although wehave not addressed uniqueness in this representation it fol

lows the usual lines The fact that wehave equality of norms for the repre

senting measure can again b e attributed to the fact that the simple functions

are dense in C D

Nowwere ready to apply this result to the problem of representing the

elements of C K as integration against Baire measures on K To b egin

let K b e a compact Hausdor space Next wecho ose a discrete space D

and a continuous onto map D K Then as youll recall the map

f f denes a linear isometry from C K into C D in other words

each continuous f K R lifts to D bywayof f D RThus each

x C K extends to a functional y C D satisfying x f y f

k That is y is a HahnBanach extension of for all f C K and ky k kx

the functional g x g dened on the image of C K inC D And

its not hard to check that a p ositive functional has a p ositive extension

From the Riesz representation theorem for C D we can nd a measure

dened on the Baire sets in D such that

y g gd for all g C D

Hence

Z Z

x f y f f d fd for all f C K

D K

and A A denes a Baire measure on K

The uniqueness of follows from the regularityof and Urysohns lemma

it requires checking that fd for all f C K forces The

details are left as an exercise

Notes and Remarks

Our presentation in this chapter b orrows heavily from Garlings pap er but see also Diestel Gillman and Jerison Hartig Holmes

Kelley and Yosida and Hewitt An approach to Rieszs theorem

that would have pleased Riesz can b e found in Dudleys b o ok

CHAPTER C K SPACES I I I

Exercises

Show that X is extremally disconnected if and only if disjoint op en sets

in X have disjoint closures

Let X b e completely regular ShowthatX is extremally disconnected

if and only if X is extremally disconnected

If D is an innite discrete space prove that D is not sequentially com

pact and hence not metrizable

Let K b e a compact Hausdor space and let B and B denote the Borel

and Baire algebras on K resp ectivelyProvethatB B

App endix A

Top ology Review

We denote a top ological space byX T where X is a set and T is a topology

on X That is T is a collection of subsets of X called open sets satisfying

i X T ii U V T U V T and iii A T AT

The closed sets in X are the complements of the op en sets that is a subset

E of X is closed if E is op en As shorthand reference to the top ology T is

often only implicit as in the phrase Let X b e a top ological space

Every set X supp orts at least two top ologies Indeed its easy to check

that fXg is a top ology on X called the indiscrete topology and that P X

the p ower set of X is a top ology on X called the discrete topology Wesay

that X is a discrete space if X is endowed with its discrete top ology Please

note that every subset of a discrete space is b oth op en and closed

Once wehave the notion of an op en set we can consider continuous func

tions b etween top ological spaces A function f X Y from a top ological

space X to a top ological space Y is continuous if f U isopeninX when

ever U is op en in Y The collection of all continuous functions from X into Y

is denoted by C X Y In case Y Rwe shorten C X Rto C X Various

subsets of C X suchas C X C X etc have the same meaning as in

b C

the intro ductory chapter

We can also consider compact sets A subset K of a top ological space

X is said to b e compact if every covering of K by op en sets admits a nite

sub cover that is K is compact if given any collection of op en sets U satisfying

f V V UgK we can always reduce to nitely many sets V V U

with V V K Its easy to see that compact sets are necessarily closed

APPENDIX A TOPOLOGY REVIEW

Separation

Recall that a top ological space X T is said to b e Hausdor if distinct p oints

in X can always b e separated by disjoint op en sets that is given x y X

we can nd disjoint sets U V T suchthat x U and y V Please

note that in a Hausdor space each singleton fxg is a closed set More

generally each compact subset of a Hausdor space is closed Just as with

metric spaces closed will mean closed under limits well make this precise

shortly while compact will mean limits exist in abundance Wewould

prefer those existential limits to land back in our compact set so its helpful to

know that a compact set is closed For this reason among others its easier

to do analysis in a Hausdor space In fact its quite rare for an analyst to

encounter or even consider a top ological space that fails to b e Hausdor

Henceforth we will assume that ALL top ological spaces are Hausdor Be

forewarned though that this blanket assumption may also mean that a few

of our denitions are fated to b e nonstandard

Metric spaces and compact Hausdor spaces enjoyaneven stronger sepa

ration prop erty in either case disjoint closed sets can always b e separated by

disjoint op en sets A Hausdor top ological space is said to b e normal if it has

this prop erty Given disjoint closed sets E F in X there are disjointopen

sets U V in T such that E U and F V

Normalityhastwo other characterizations each imp ortant in its own right

The rst is given by Urysohns lemma In a normal top ological space disjoint

closed sets can b e completely separated That is if E and F are disjoint

closed sets in a normal space X then there is a continuous function f

C X such that f on E while f on F The second is Tietzes

extension theorem If E is a closed subset of a normal space X then each

continuous function f C E extends to a continuous function f

C X on all of X Urysohns lemma and Tietzes theorem are each

equiv alent to normalityTheinterval can b e replaced in either statement

by an arbitrary interval a b Further it follows from Tietzes theorem that

if E is a closed subset of a normal space X thenevery f C E extends to

an elementof C X simply by comp osing f with a suitable homeomorphism

from R into

Lo cally compact Hausdor spaces

If X has enough compact neighb orho o ds then C X will have enough

functions to take the place of C X in certain situations In this context

enough means that X should b e local ly compact A lo cally compact Haus

dor space is one in whicheachpoint has a compact neighb orho o d ie given

x X there is an op en set U containing x suchthat U is compact Its an

easy exercise to show that a lo cally compact Hausdor space has a wealth of

compact neighb orho o ds in the following sense Given K U X whereK

is compact and U is op en there is an op en set V with compact closure such

V U This observation along with a bit of hard work leads that K V

to lo cally compact versions of b oth Urysohns lemma and Tietzes extension

theorem Heres howtheynow read please note that C X is used in place

of C X ineach case Let X b e a lo cally compact Hausdor space and let

K b e a compact subset of X Urysohn If F is a closed set disjoint from K

then there is an f C X such that f on K while f on F

Tietze Each elementof C K extends to an elementof C X

An alternate approach is to consider the one point compactication of X

The one p oint compactication X of a top ological space X is dened to b e

the space X X fgwhere is a distinguished p oint app ended to X

and where we dene a top ology on X by taking the neigh borhoods of to

b e sets of the form fg U where U is compact in X Its easy to see

that X is a compact space that contains X as an op en dense subset this is

what it means to b e a compactication of X It is likewise easy to see that

X is Hausdor precisely when X is lo cally compact and Hausdor Thus

if X is a lo cally compact Hausdor space then X is a dense op en subset of

the compact Hausdor hence normal space X Consequentlywe can now

takeadvantage of such niceties as Urysohns lemma and Tietzes extension

theorem in X and then simply translate these prop erties to X In particular

the lo cally compact versions of Urysohns lemma and Tietzes theorem stated

ab ove are direct consequences of considering the full versions in X and

then cutting back to X

If X is lo cally compact then the completion of C X under the sup norm

is the space C X the functions in C X thatvanish at innity that is

those f C X for whic h the set fjf j g is compact for every

Clearly C X is a Banach space and a Banach algebra under the sup norm

The phrase vanish at innity b ecomes esp ecially meaningful if we consider

X the one p oint compactication of X In this setting the space C X is

isometrically the collection of functions in C X that are zero at

It may come as a surprise to learn that the discrete spaces are a very

imp ortant class of lo cally compact spaces In the case of a discrete space D

the various spaces of continuous functions are often given dierentnamesFor

example since every function f D R is continuous the space C D is b

APPENDIX A TOPOLOGY REVIEW

simply the collection of al l b ounded functions on D and this is often written

as D in analogy with the sequence space N The space C D is

the collection of functions with nite supp ort in D and the space C D

is often written as c D again in keeping with the sequence space notation

c C N As a last curiosity notice that the space c of all convergent

sequences is just a renaming of the space C N fg

Weak top ologies

A familiar game is to describ e the continuous functions on X after weve b een

handed a top ology on X But the inverse pro cedure is just as common and

p erhaps even more useful In other words given a collection of functions F

from a set X to some xed top ological space Y can we construct a top ology

on X under whicheach elementof F will b e continuous If X is given the

discrete top ology then every function from X to Y is continuous while if X

is given the trivial or indiscrete top ology then only constant functions are

continuous Wetypically want something in b etween In fact wed liketo

know if there is a smal lest or weakest top ology that makes each elementof

F continuous As well see the answer is Yes and follows easily from an

imp ortant bit of machinery that provides for the construction of top ologies

having certain predetermined op en sets

Lemma A Subbasis Lemma Suppose that X is a set and that S is a col

lection of subsets of X Then there is a smal lest topology T on X containing

S Moreover S fXgS forms a subbase for T In other words the sets

of the form S S wher e S S for i nareabase for T

n i

Proof Let T denote the intersection of all top ologies on X containing S or

S Its easy to see that T is itself a top ology on X containing S as well

as S and clearly T is the smallest such top ology Consequently T also

contains the collection T whichwe dene to b e the set of all p ossible unions

of sets of the form S S wheren and where S S for i n

n i

All that remains is to show that T T But since T contains S it suces

X In fact the only detail that we need to to showthatT is a top ology on

checkisthat T is closed under nite intersections since its obviously closed

under arbitrary unions Here go es Let U V T and let x U V

Take A A and B B in S suchthat x A A U and

n m n

x B B V Then x A A B B U V

m n m

That is U V T

And how do es Lemma A help Well given a set of functions F from X

into a top ological space Y take the smallest top ology on X containing the

sets

S f f U f FUis op en in Y g

Since each elementof S will b e op en in the new top ologyeach elementof

F will b e continuous This top ology is usually referred to as the the weak

topology inducedby F

We dont really need the inverse image of every op en set in Y we could

easily get by with just the inverse images of a collection of basic op en sets or

even subbasic op en sets In particular if Y R then the collection of sets

N x f f f y X jf x f y j i ng

n i i

where x X f f Fand is a neighb orho o d base for the weak

top ology generated by F

If X carries the weak top ology induced by a collection of functions F from

X into Y then its easy to describ e the continuous functions into X note that

f Z X is continuous if and only if g f Z Y is continuous for every

g F

Finallyitsworth p ointing out that our construction of weak top ologies

in no way requires a xed range space Y In particular given a collection of

functions f where f maps X into a top ological space Y we can easily

apply the subbasis lemma to nd the smallest top ology on X under which

each f is continuous In this setting we consider the top ology generated by

the collection

S f f U U is op en in Y for A g

Pro duct spaces

The subbasis lemma readily adapts to more elab orate applications The pro d

uct or Tychono top ology provides an excellent example of such an adapta

tion First recall that the Cartesian pro duct of a collection of nonempty

sets Y is dened to b e the set of all functions f A Y satisfying

A A

Y Ifweidentify an element f f Y the pro duct space is written

of the pro duct space with its range f thenwerecover the familiar no

tion that the pro duct space consists of tuples where the th co ordinate

of each tuple is to b e an elementof Y Wealsohave the familiar co ordinate

pro jections Y Y dened by the formula f f or

APPENDIX A TOPOLOGY REVIEW

f f If each Y is the same set Y we usually write the

pro duct space as Y the set of all functions from A into Y

In case each Y is a top ological space we top ologize the pro duct space

Y by giving it the weak top ology induced bythe s That is we

dene the product topology to b e the smallest top ology under which all of the

co ordinate pro jections are continuous In terms of the subbasis lemma this

means that each of the sets U where U is op en in Y and Aisa

subbasic op en set in the pro duct The basic op en sets in the pro duct are of

course nite intersections of these

This particular choice for a top ology on the pro duct space results in several

useful consequences For example a function X Y from a

top ological space X into a pro duct space is continuous if and only if eachof

its co ordinates X Y is continuous Well see other b enets of

the pro duct top ology shortly

Nets

Now this b eing analysis or had you forgotten we need a valid notion of

limit or convergence in a general top ological space An easy choice from

our p oint of view is to consider nets The reader who is unfamiliar with nets

would b e well served by thinking of a net as a generalized sequence We

start with a directedset D thatisD is equipp ed with a binary relation

satisfying i for all D iiif and then and

iii given any D there is some D with and Several

standard examples come to mind N with its usual order is a directed set

The set of all nite subsets of a xed set is directed by inclusion ie A B

if A B The set of all neighborhoods of a xed pointinany top ological

space is directed by reverse inclusion ie A B if A B And so on As

usual we also write to mean

Now a net in a set X is any function into X whose domain is a directed

set A sequence recall is a function with domain N and so is also a net Just

as with sequences though wetypically identify a net with its range In other

words wewould denote a net in X by simply writing x whereD is a

directed set and where each x X

In dening convergence for nets we just tailor the terminology we already

use for sequences For example wesay that a net x is eventual ly in the

if for some D wehave fx gA And x is frequently set A

in A if given any D there is some D with suchthat x A

Finally a net x in a top ological space X converges to the p oint x X

if x is eventually in each neighborhood of x As with sequences we use

the shorthand x x in this case

Many top ological prop erties can b e characterized in terms of convergent

nets Indeed sequential characterizations used in metric spaces can typically

b e directly translated into this new language of nets Heres an easy example

AsetE in a top ological space X is closed if and only if eachnetx in

E that converges in X must actually converge to a p ointofE On the

one hand supp ose that E is closed and let x beanetinE converging to

x X Ifx E an op en set then wewould havetohavex eventually

in E an imp ossiblility On the other hand supp ose that eachconvergent

net from E converges to a p ointinE Nowletx E and supp ose that

for every neighborhood U of x there is some p oint x U E Ifwe direct

the neighborhoods of x byreverse inclusion then weve just constructed a net

x inE that converges to a p oint x not in E yielding a contradiction Thus

c c

some neighborhood U of x is completely contained in E that is E is op en

Another example thats easy to check A function f X Y between

top ological spaces is continuous if and only if the net f x conv erges to

f x Y whenever the net x converges to x X Supp ose rst that f is

continuous Let x x in X and let U b e a neighborhood of f xinY Then

f U is a neighborhood of x in X and hence x iseventually in f U

Consequentlyf x is eventually in U That is f x converges to f x

Next supp ose that f x f x whenever x xLet E b e a closed set in

Y and let x b e a net in f E that converges to x X Thenf x is a

net in E that converges to f x Y Hence f x E or x f E Thus

f E is closed and so f is continuous

Nets will prove esp ecially useful in arguments involving weak top ologies

If X carries the weak top ology induced by a family of functions F itfollows

that a net x converges to x X if and only if f x converges to f x for

each f F Why

Now since the pro duct top ology is nothing more than a weak top ology

our latest observation takesonavery simple guise in a pro duct space A net

Y converges to f if and only if f converges f in a pro duct space

co ordinatewise that is if and only if f converges to f for each

AFor this reason the pro duct top ology is sometimes called the topology

of pointwise convergence Beginning to sound like analysis

The only p otential hardship with nets is that the notion of a subnet is a

bit more complicated than that of a subsequence But were in luck We will

have no need for subnets and so we can blissfully ignore their intricacies

APPENDIX A TOPOLOGY REVIEW

Notes and Remarks

There are many excellent b o oks on top ology or on top ology for analysts that

will provide more detail than wehavegiven here See for example Folland

Chapter Jameson Kelley Kelley and NamiokaKothe Simmons or Willard

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Index

A Z

p p

B X Y X

B N

C K

C a b

C X

sum

C X

c sum

supp f

rca

adjoint

M p algebra of sets

M algebraic complements

s algebraic dual

almost disjoint sequence

almost isometry

XM Baire algebra

Banach lattice

X Banach limit

INDEX

BanachAlaoglu theorem direct sum

directed set

BanachMazur theorem

discrete space

BanachSaks theorem

discrete top ology

basic sequence

disjointly supp orted functions

basis

basis constant

disjointly supp orted sequence

basis problem

BessagaPelczynski selection principle

disjointly supp orted sequences

Dixmiers theorem

biorthogonal

dual norm

biorthogonal sequence

dual of a quotient

blo ck basic sequence

dual of a subspace

blo ck basis

dual space

b ounded linear map

DunfordPettis prop erty

b ounded multiplier convergent

Eb erleinSmulian theorem

b oundedly complete basis

emb edding lemma

Clarksons inequalities

epigraph

Clarksons theorem

equicontinuous

clop en set

equivalent bases

closed sets

exp osed p oint

co ecient functionals

extension of an op erator

compact op erator

compact set

Extension Theorem

compactication

extremally disconnected

complemented subspace

extreme p oint

complemented subspaces of c

nite co dimensional subspace

complemented subspaces of L

nitely additive measures

Frechet metric

complemented subspaces of

functional

complemented subspaces of

completely continuous op erator

Gantmachers theorem

generalized Holder inequality

completely regular

gliding hump argument

completely separated

Goldstines theorem

concave function

Grothendiecks theorem

conditional exp ectation

Haar system

convex function

convex sets HahnBanach extension prop erty

co ordinate functionals

co ordinate pro jection Hamel basis

INDEX

one p oint compactication Hausdor compactication

op en sets Hausdor top ological space

Hilb ert space

op erator norm

Holders inequality

order interval

Orliczs theorem

indep endent random variables

orthogonal pro jection

indiscrete top ology

orthonormal basis

injective spaces

involution

parallelogram law

isometry

Pelczynskis decomp osition metho d

isomorphism

isomorphism into

Phillips lemma

isomorphism theorem

Pitts theorem

p oint mass

Jamess nondistortion theorem

p olygonal functions

p ositive linear functional

Jensens inequality

prime spaces

Jordan decomp osition

principle of small p erturbations

KadecPelczynski theorem

pro duct top ology

Khinchines inequality

pro jection

Lamp ertis theorem

quotient map

Liap ounovs inequality

quotient norm

lo cally compact top ological space

quotient space

quotient top ology

Marcinkiewiczs theorem

Rademacher functions

Mazurs lemma

RadonRiesz theorem

Mazurs theorem

random signs convergent

McShanes lemma

reexive space

measure algebra

Minkowskis inequality

Riesz representation theorem

monotone basis

multinomial co ecient

Rieszs lemma

nearest p oints

RieszKakutani theorem

net

RieszMarkov theorem

norm dual

Schauder basis

normal top ological space

Schauders basis for C

normalized basis

norming functional SchroderBernstein theorem for

Banach spaces

norming set

nowhere dense Schurs theorem

INDEX

self adjoint uniformly integrable

seminormalized sequence

unordered convergen t

separably injective spaces

Urysohns lemma

separates p oints

separates p oints from closed sets

vector lattice

shrinking basis

VitaliHahnSaks theorem

small isomorphism

smo oth space

Walsh functions

smo othness

weak convergence

Sob czyks theorem

StoneCech compactication

weak top ology

Stonean top ological space

weak basis

strictly convex function

weak compact sets

strictly convex norm

weak convergence

strictly convex space

weak top ology

weakly compact op erator

strictly singular op erator

subbasis lemma

weakly compact sets

sublattice of L

subseries convergent

weakly null sequence

subspaces of c

weakly sequentially complete

subspaces of L

Weierstrass theorem

subspaces of

supp ort

supp orting line

three space prop erty

Tietzes extension theorem

top ological space

top ology

top ology of p ointwise convergence

total variation measure

uncomplemented subspace

unconditional basis

unconditionally convergent

uniform convexity

uniformly absolutely continuous

uniformly convex norm

uniformly convex space