Summability Through Functional Analysis (North-Holland Mathematics Studies, No 85)

S U M M A B l L l T Y THROUGH F U N C T I O N A L ANALYSIS This Page Intentionally Left Blank NORTH-HOLLAND MATHEMA...

Author: Albert Wilansky

194 downloads 518 Views 3MB Size Report

This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form

DOWNLOAD PDF

S U M M A B l L l T Y THROUGH F U N C T I O N A L ANALYSIS

This Page Intentionally Left Blank

NORTH-HOLLAND MATHEMATICS STUDIES

85

Notas de Matematica (91) Editor: Leopoldo Nachbin Centro Brasileiro de Pesquisas Fisicas and University of Rochester

Sumability through Functional Analysis ALBERT WILANSKY L ehigh University

1984

NORTH-HOLLAND - AMSTERDAM

NEW YORK

OXFORD

ElsevierScience Publishers B. V.. I984 All rights reserved. No par/ oft/zi.spuhlicatior~may be reproduced. stored in a refriwalsysrern, or transmitred. i n m y form o r by any means, electronic, mectirinical. pho~ocopying,recording or o/herwi.ve. withortt /lieprior pertnissiori of / h e copyrig/i/ owner.

ISBN: 0444868402

Pu hlivlierv: ELSEVIER SCIENCE PUBLISHERS B.V P.O. B O X 1991 1000 BZ AMSTERDAM T H E NETHERLANDS Sol~~distrihutors for [he U.S.A . and Canada: ELSEVIER SCIENCE PUBLISHING COMPANY. INC. 52VANDERBlLT AVENUE NEW YORK, N.Y. 10017

Llbrary of Congress Cataloglng In PubUeatlon Data

Wilansky, Albert. Summability through functional analysis. (North-Holland mathematics studies ; 85) (Notas de matemtitica ; 91) Bibliography: p. Includes index. 1. Functional analysis. 2 . Summability theory. I. Title. 11. Series. 111. Series: Notas de matem6tica (Amsterdam, Netherlands) ; 91. 515 - 7 83-25398 @320.w56 1984 ISBN 0-444-86840-2

PRINTED IN T H E NETHERLANDS

To Carole, Eleanor, Johnny, Kathy, Laura, Leslie, Michael Unique members of a unique family

Herrn Karl Zeller zum 60. Geburtstag am 28. Dezember 1984 gewidmet.

vi i

PREFACE

Summability is an extremely fruitful area for the application of functional analysis; this book could be used as a source for such applications. Those parts of summability which have only "hard" ( = classical) proofs are omitted; the theorems given all have "soft" ( = functional analytic) proofs. exceptions.)

(There are a few

Here is an incomplete list of topics n o t covered:

normal spaces, perfect spaces, ordered spaces, classical Tauberian theorems, summability of Fourier series, absolute and strong summability, summability factors, discussion of dozens of special methods. These are all interesting and valuable topics but do not fall within our scope.

The monographs

r551,

1881 are recommended

for material not in this book.

I have found summability through functional analysis to be a most entertaining subject - full of interesting things to work on with amusing and challenging problems at every stage. Further, it rewards the expert in functional analysis with delightful short When I was in graduate school it was customary to assixn the necessity of the Silverman-Toeplitz conditions (1.3.6) as a homework problem. The best and bravest of us found the right proofs.

construction - those who did can really appreciate the soft proof given here; some of us can even remember the sensation when it first appeared in Banach's book In connection with this I must take exception to a remark of Ivor Maddox in [ 4 4 A ] , p. 162.

rz].

He indicates that functional analysis smooths the path of proof but does not obtain the actual results to be proved - the methods giving existence but not construction. On the contrary, as in Chapter 8, the results themselves appear while the classical writers had to guess what they were and then prove them. SOURCES.

I have given references and attributions for some of the results. The reader will not go far wrong to assume that every unattributed theorem outside of Chapter 2 is due to Karl Zeller.

viii

PREFACE

NOT ICE Coregular a n d conull h a v e s p e c i a l meanings in C h a p t e r s 1 - 8 In those chapters these spaces are conservative. Irz Chapters 3-16 they are assumed to have the weaker property variational semiconservative, written vsc throughout.

and 1 ) - 1 9 .

ACKNOWLEDGMENTS A. K. Snyder and W. H. Ruckle helped me over dozens of sticky points. A group of Lehigh University graduate students read over the manuscript and made many suggestions and corrections. They also attended a seminar on this material led by Professor R. M. DeVos of Villanova University. The group consisted of Deb Frantz, Abdullah Hakawati and Matt Schaffer. They were joined f o r a time by Jeff Connor of Kent State University who also proof-read part of the typescript. I thank all these friends for their help. Judy Arroyo undertook the difficult task of typinq cameraready copy. Her enthusiasm and proprietary attitude were indispensable. As to the quality of her work: "Si monumentum requiris, circumspice."

ix

CONTENTS

1.

MATRICES

1

1.0. Functional analysis 1.1. Introduction 2 1.2. Notation 3

2.

3.

1

1.3. 1.4.

Conservative and regular Associativity 7

5

1.5. 1.6.

The algebras r and A 9 Coregular and conull matrices

1.7. 1.8.

Types of summability theorems Inverses 17

10 11

CLASSICAL MATRICES 2.0. Background 21

21

2.1. 2.2.

Introduction 21 *I Holder matrices 22

2.3. 2.4. 2.5.

Hausdorff matrices 22 Cesgro and Hglder 30 N’Arlund matrices 32

2.6.

Polynomial matrices

35 39

TRIANGLES AND BANACH SPACE 3.0. Functional analysis 39 3.1. Historical 39 3.2. Convergence domain 40

3.3. The perfect part 41 3.4. When c is closed 45 3.5. Bounded sequences and non-triangles 4. FK SPACES 4.0. Functional Analysis 4.1. Introduction 52

4 . 2 . FK spaces

51

54

4.3. Construction 60 4.4. Dual space 66 4.5. Complements 68 4.6. Coregular and conull

4.7. The FK program

73

70

47 51

CONTENTS

X

5.

6.

7.

8.

9.

10.

REPLACEABILITY AND CONSISTENCY 5.1. Functional analysis 75 5.2. Replaceability 76 5.3. Consistency 81 5.4. Reversible matrices 82 5.5. Row-finite and one to one 85 5.6. Bounded consistency 86 BIGNESS THEOREMS 6.0. Functional analysis 89 6.1. c not closed 90 6.2. Two-norm convergence and W 93 6.3. Oscillation 95 6.4. Conull spaces 97 6.5. Coregular spaces and matrices 99 6.6. Subspace o f Em 101 6.7. Bigness 1 0 2 SEQUENCE SPACES 7.0. Functional analysis 103 7.1. Monotone norms 104 7.2. Duals 105 7.3. Nine spaces 109 7.4. Determining sets 111 INCLUSION AND MAPPING 8.0. Functional analysis 115 TABLE 116 8.1. Introduction 117 8.2. Inclusion 118 8.3. Mapping theorems 120 8.4. Examples 126 8.5. Mapping concluded 132 8.6. Functional dual 138 SEMICONSERVATIVE SPACES AND MATRICES 9.1. Introduction 141 9.2. Semiconservative spaces 141 9.3. Coregular and conull 144 9.4. Matrix domains 147 9.5. Matrices 149 9.6. Coregular and conull matrices 153 DISTINGUISHED SUBSPACES OF FK SPACES 10.0. Functional analysis 157

75

89

103

115

141

157

CONTENTS

BK 10.2. Distinguished 1 0 . 3 . The subspace 10.4. The subspace 10.1. Duals as

xi

spaces 157 subspaces 160 B 165 F 173

10.5. The subspace W 176 10.6. Basis 177 11. EXTENSION 11.0. Functional analysis 179 11.1. Row functions 179 11.2. Matrices 183 12. DISTINGUISHED SUBSPACES OF MATRIX DOMAINS 12.0. Functional analysis 187 12.1. Matrix domains 187 12.2. Associativity 191 12.3. AK spaces 191 12.4. Weak and strong convergence 194 12.5. c-like spaces 197 13. DISTINGUISHED SUBSPACES OF cA 13.1. B and the perfect part 201 13.2. The inset and replaceability 202 13.3. The main theorem 211 13.4. Applications 213 13.5. Almost coregular and very conull 218 14. THE FUNCTIONAL u 14.0. Functional analysis 221 14.1. Parts of the dual 222 14.2. The functional 11 224 14.3. The strong topology 225 14.4. 1-1 spaces 228 14.5. AB and closure of X’ 229 15. THE SUBSPACE P 15.0. Functional analysis 233 15.1. T and the test functions 233 15.2. The subspace P 235 15.3. Factorization 239 15.4. Inclusion and invariance 241 15.5. Replaceabili t . y 245 15.6. Miscellany and questions 247 16. SEQUENTIAL COMPLETENESS AND SEPARABILITY 16.0. Functional analysis 251

179

187

201

221

233

251

xi i

CONTENTS

16.1. Sequential completeness 16.2. Separability 255 17.

252

16.3. Dense subspaces of ~ l -258 MAPS OF BANACH SPACES 17.0. Functional analysis 263 17.1. Matrix maps of c 164 17.2. Maps of

c

263

265

17.3. The dual map

266 17.4. w-matrices 269 1 7 . 5 . w-almost matrices 17.6. Tauberian maps

270

274

17.7. Multiplicative abstracted 18.

ALGEBRA 18.0. Functional analysis

281 281

18.1. Topological divisors of 0 18.2. The maximal group 184 18.3. The spectrum 19.

291 291

19.1. What can YA be? 291 19.2. Toeplitz basis 293 19.3. Miscellany 296 19.4. Applications 297 19.5. Questions 299 19.6. History 300

INDEX

309

303

282

287

MISCELLANY 19.0. Functional analysis

BIBLIOGRAPHY

277

1

CHAPTER 1 M A TRICES

FUNCTIONAL ANALYSIS

1.0.

Reference:

Banach space i s a c o m p l e t e normed v e c t o r s p a c e

A

1.

Chapter 3.

[SO],

X; X'

t h e d u a l Banach s p a c e of c o n t i n u o u s l i n e a r f u n c t i o n a l s ( i . e .

X

v a l u e d f u n c t i o n s ) on

co

2. ilm

(the null sequences), c

( t h e c o n v e r g e n t s e q u e n c e s ) and

Each i s a c l o s e d s u b s p a c e o f t h e n e x t . g e n t s e r i e s ) i s a Banach s p a c e w i t h

iff f

E

a

with

E

il,

f ( x ) = xlimx+ax iff

il'

a

E

il,

a

E

with

IIxIIm = s u p l x n l .

( t h e a b s o l u t e l y conver-

il

jlxlll = I: l x k l .

-

X(f) = f(1) .km,

f

E

a x ) , IIfll = jlalll;

(see 1.2.1 for

with

f ( x ) = ax

scalar

with

( t h e bounded s e q u e n c e s ) a r e Banach s p a c e s w i t h

f ( x ) = ax

C

c; f

iff E

c'

k

f ( 6 ) (1.3.5),

IIfll = ~ ~ a ~ [SO], ~ m Examples .

2 . 3 . 5 , 2 . 3 . 6 , #2-3-3. 3.

UNIFORM BOUNDEDNESS.

and suppose t h a t

Ifn)

Let

{fn}c X ' , X

all

n.

See

4.

BANACH-STEINHAUS CLOSURE THEOREM,

suppose t h a t [SO],

{fn)

l]fn\] < M

for

Theorem 3-3-6.

f(x) = l i m fn(x)

Theorem 3-3-13.

Frgchet space

\ f n ( x ) I < Mx

is u n i f o r m l y bounded i . e .

n,x.

[SO],

a Banach s p a c e ,

is p o i n t w i s e b o u n d e d , i . e .

for all

Then

is

for a l l

T h i s h o l d s also i f

[SO], Theorem 9-3-7,

x

With E

X

X.

as i n 3

{fn) Then

f

E

X'.

i s a locally c o n v e x

Bxample 9-3-2.

2

1.0

5.

in 3 ) .

Then

Ix:lim fn(x)

X.

vector subspaces of 6.

I fn} be uniformly bounded (as

Let

C O N V E R G E N C E LEMMA.

exists}

and

Ix:fn(x)

01

+

are closed

[80], #3-3-3.

A Banach a l g e b r a is a Banach space which is also an algebra IIxyll 5 I]xII.Ily\].

and satisfies

IIx-lll < 1

identity and

X

If

then

x

is a Banach algebra with

is invertible.

14.2 Fact ii.

[79],

7. A topological vector space is a vector space and a topological space such that the vector operations are continuous.

X

such space

has a weak topology

for all

f

-+

1.1.

INTRODUCTION

0

E

X'.

which is the smallest vector

A sequence

topology with the same dual. f(an)

w

Each

{

an}

+

0

in

w

iff

[SO], Example 4-1-9.

In the beginning, the idea was conceived that there should be a way to find "sums" for divergent series. One popular procedure was to set

x

1

=

in the identity

(l+x)-'

=

c (-x)"

leading to 1

1-1+1-1+ ... = 2' Since it is obviously possible to assign a sum, for example

the mystically satisfying "result" that

0 , to any divergent series, we abandon this quest and simply

for some type of function

K

sequences and numbers) or

C

L:S

-+

K

where

S

look

is some set of

- always R

(the real

(the complex numbers) in this book.

Note that

is the set of scalars

"sequences" has been substituted for "series"; this is a mild and unnecessary change which is adopted f o r convenience. The function L

will be required to have certain explicitly stated properties;

for example, we usually require

S

to be a vector space which

includes all the convergent sequences and that

L(x)

=

lim xn

whenever

contains a divergent sequence

x

=

(xn}

L

to be linear and such

is convergent.

x , the number

L(x)

Then if

will be a

S

1.1-1.2

3

"limit of a divergent sequence" in the eldritch sense described above. It turns out that such functions may be very practical.

See

519.4.

1.2.

NOTATION. There is an Index of Symbols in the book which refers to the

place where each symbol is defined. fxnl, n = 1 , 2 , , . . , of scalars will be written

1. A sequence The number

x.

A = ( a , )

y

=

n,k

Ax

xkyk

Z

w i l l b e d e n o t e d by

...

Given a matrix

of scalars and a sequence

1,2,

=

xy.

to mean that for each

n,yn

(Ax)n

=

x , we write ankxk; each of

=

these series being assumed convergent. w A = {x:Ax

Let If

is defined}; this is called the d o m a i n of

is r o w - f i n i t e (i.e. each row of

A

all finitely non-zero sequences) then

A

lies in

A.

the set of

@,

u A = w , the s e t of all

sequences; it is a simple exercise to check that the converse is also true. 2.

Let

cA

Ix:Ax

=

E

c}

where

c

This is called the c o n v e r g e n c e domain of

sequences.

SA

generally we define

=

{x:Ax

Sl

E

where

this is consistent with the meaning of

Ax

is any subset of

as just given.

wA

We

Ax

exists, i.e.

By a historical accident, sequences in

cA

are called

SA c w A

3.

S.

instead of the more reasonable

For

limA:cA -+ K . c

S

E

for every

A-sumrnable

3

S

More

A.

always implies that

emphasize that

cA

is the set of convergent

x

E

cA

let

limAx

Finally, a matrix

(i.e. Ax

E

c

whenever

=

lim(Ax)n,

thus defining

is called c o n s e r v a t i v e if

A

x

A-limitable.

E

c), m u l t i p l i c a t i v e

m

if

W ;

. 1.2

4

limA x

=

m-lim x

x

c , and r e g u l a r if it is multiplicative 1.

E

EXAMPLE. The identity matrix

4.

also

for

wI =

5.

CI =

W,

EXAMPLE.

I

is regular since

Ix

=

x;

c. 9

Let

be the matrix given by

( Q x ) ~= ( X ~ - ~ + X ~ )for /~ n

>

1. Then

(Qx),

= X1,

is regular and sums the

Q

divergent sequence I 6. EXAMPLE.

A

For

one can construct A

= 0, cA = w

with

cA

while at the opposite extreme by taking Ax = (xl,O,x1’x2’

= { O }

o,x1’x2,x3,0, . . . I . 7. DEFINITION.

Let

i s t h e s e t of m a t r i c e s

Thus

A

E

# 0, sgn G

all

=

I.

such t h a t

always; A

( w ~ : w )

8. DEFINITION. z

A

b e s e t s of s e q u e n c e s .

X, Y

E

Ax

(c:c)

iff

For a c o m p l e x number

Thus

lsgn

21

and

= I

for a l l

Y

E

Then

z,

A

x

E

(X:Y)

X.

is conservative.

sgn z = I z l / z

z sgn z

=

121

if

for

z.

9. Here is a version of A b e l ’ s i d e n t i t y that I find very

easy to remember.

(Take +

Suppose vo

=

Uk(Vk?Vk-l)

=

in both o r

Then

Let

IZ

ukek 1

ukek I 5 mad

vk =

=

IZ

5 max ]vkl.

k

CiZl

ei

Then, for example, if

u

~

++0,~

to the left side.

Abel’s Inequality.

I Zi=I

0, then

=

n ‘k=l (Ukkuk+l)Vk‘

- in both).

one merely adds +Un+lvn 10.

u n+l

Suppose

2

u1

2 u2

...

> u > 0. n -

e k I :I 5 r 5 n 1 .

I

and set

Uk(Vk-Vk-l)l

1

=

IZ

u

~ = +0.~ Then by Abel’s identity,

( uk-uk+l)vk I 5 max[vkl.z

(U~-U~+~)

1.3

.

5

CONSERVATIVE AND REGULAR

1.3.

The main purpose of this section is to characterize members of

(c:c)

(1.2.7).

A unified approach to this problem for many

pairs of spaces is given in Chapter 8. 1. We now refer to the spaces

c0 '

and their norms as

Qm

Q,

given in 1.0.2. For a m a t r i x

2. THEOREM.

( i ) wA

Q",

3

( i i )W A

3

co,

A,

these are equivalent:

( i i i )~

~

\ < am

~f o r ~ e a l ch

n.

That (iii) implies (i) and (i) implies (ii) are trivial. Assuming (ii) let defining u

um

a

be any row of A and um(x) = Zm k=l akxk' m with IIumll = Zk=l lakl (1.0.2). The sequence

cb

is pointwise convergent, hence pointwise bounded so norm

bounded

(1.0.3).

This yields (iii).

3.

THEOREM.

For a m a t r i x

A E

fQm:Qml,

(il

ll All

=

fiil A E

s u p n Ck lank I

If

(ii) implies (iii): IIunlI

= Z

(co:Qmi,

lankl

(1.0.2).

0.

Thus

is easy to check that co 5.

for each bk

un

C;

4

by

The sequence

un(x) = (Ax)n u

SO

is pointwise

(1.0.3) and the result follows.

The set of matrices of finite norm (see Theorem 3(iii))

is denoted by

from

is bounded, I ( A x ) ~ ~2 IIAll*IIxil,.

x

Define

bounded, hence norm bounded 4.

where

( i i i l I~AiI < =

-

(iii) implies (i):

that

A, t h e s e a r e e q u i v a l e n t :

or

.tm

to

0 = (9.":~")

=

(cO:Qm) by Theorem 3.

IjAl/ is the usual norm of the map

It Ax

x

+

=

limnank

Q".

NOTATION (very Important!).

For a matrix

k , provided these (column) limits exist.

is defined for a matrix

B.

For each

A , ak

Similarly

k = 1,2,. . . ; g k

is the

1.3

6

sequence whose only non-zero term is a

in the

1

kth

place; 1

is either the integer o r the sequence all of whose terms are (according to context). It is crucial that ak = limA6k . 6. THEOREM. ( 5 ; ) ak

A matrix

e x i s t s f o r each

k,

liii) 1

a r e n e c e s s a r y and s u f f i c i e n t t h a t

A

Let

F(x)

gk

Ax

=

Then

cA.

E

cA.

E

eA

f i ) ,f i i )

T h e conditions

cA2 c

and, for example, ak

exists

If, conversely, the three conditions hold, let

define

U = F-l[c]

F:Lm

.+

a continuous linear map, by (i).

Em,

a m since c

is a closed vector subspace of

Now (ii) and (iii) imply that each

gk

E

U

is the smallest closed vector subspace of these sequences, it follows that tive.

0,

.

c

2

E

be conservative; (i) is a special case of Theorem 3 ;

(ii) and (iii) hold because

iff

f i )A

i s conservative i f f

A

1

U

2

c

and Lm

1

U.

E

Since

c

which contains all

and so

is conserva-

A

The second part of the theorem has the same proof with all

considerations involving

It is not true that

1 omitted, and

Z ank

c

replaced by

co.

must be uniformly convergent, as

the identity matrix shows. 8. THEOREM. l i mA x =

x- l i m

x

+

Let ax

A

where

b e c o n s e r v a t i v e and

x

=

x ( A l = l i mA 1

-

x

E

c.

Then

C

a k , ax

i s as

i n 1.2.1.

The Banach-Steinhaus Theorem (1.0.4)shows that

limA

E

c'

and the formula follows from 1.0.2. 9.

each

is.

k,

THEOREM. A m a t r i x i s r e g u l a r i f f and

lirn 1 = 1 . A

IIAll

In

n, ank = 0 .

(c,1) is regular: for example,

averaging 2 hits in 5 at bats over a long season will result in a batting average of .400 no matter how badly or how well the season started.

Mere mathematicians observe that

(C,l) obeys the

conditions of Theorem 9. THEOREM.

11.

Let

are closed s u b s p a c e s of

Here un

E

:c

=

A c

@.

m

c A fl 9.

and

0

m

cA I 7 9.

.km.

{x:Ax c col.

( e m ) ' , indeed

Then

Let

I\unll 5 IIAll.

un(x)

=

(Ax)n

so that

The result follows by 1.0.5.

1.4. ASSOCIATIVITY 1. DEFINITION. for

A matrix

A

is called triangular if

ank = 0

k > n, it is called a triangZe if it is triangular and

a # 0 nn

2.

for a l l

n.

The expressions

summability.

Here

t, x

t(Axl

and

(tAjx

are sequences and

arise often i n

A

is a matrix;

8

1.4

t(Ax)

tn(Ax)n

= Z

They may be different even if

x

E

cA

(Ax)n

t

- xn, so that tA

2xn-l

E

k,

A

(tA)kxk

= 0.

easy induction for arbitrary

tnankxk'

tn

=

2-",

It is very easy to find

t(Ax) # 0; indeed one can solve

such that

' k ' n

=

is a regular triangle,

For example let

and both numbers exist.

=

z

ZnZk tnankxk, (tA)x =

=

y, e.g.

y

for

y = Ax

x

x

by an

1 6 .

=

Valuable results ensue whenever it is known that equality

3.

holds for these two expressions. is just the appeal to this.

A key step in several arguments

The entry "associativity" in the index

may be consulted. 4.

and

x

THEOREM. E

The equality

o r (iil t

uA

E

k,

A

t A x ) = (tA)x Q,

E

holds if (il t

$

E

The first part is

x E Em.

obvious, involving only the adding together of finitely many convergent series.

If (ii) holds,

tnankXk I

nl

is closed

It is obviously closed under multiplication.

1.6.

COREGULAR AND CONULL MATRICES. 1. DEFINITION.

A matrix

A

i s c a l l e d conull, r e s p e c t i v e l y ,

c o r e g u l a r i f i t i s c o n s e r v a t i v e and

2.

x

=

The number

x(Al

x was given in

= 0,

1.3.8.

respectively, # 0.

A regular matrix has

1. 3.

It is possible to define

x

for matrices which are not

conservative but Definition 1 explicitly excludes such matrices. Beginning in

g9.4

we shall relax the assumption that

A

is

11

1.6-1.7

conservative, but it will never be entirely dropped i.e. a matrix x(A) = 0 will not be called conull unless it has an extra

with

property, namely (until further notice) that it is conservative. The formula

4.

x(A)

=

follows: the right side is = limn C k ank - C ak = x(A).

x

so

limmlimn C i = m ank

may be proved as m m- 1 limm(limn Zk=l ank - k=l ak) It follows that Ix(A)I 5 IIAll and r(l.O.1).

is a continuous linear function on

1.7. TYPES OF SUMMABILITY THEOREMS. GROWTH THEOREMS

(a)

THEOREM. L e t

I.

Then e v e r y

x

E

eA

-1

be a t r i a n g l e , B = A

A

x

satisfies

=

Olunl

, un

-

~ i I b=n k I . ~

Ixn/unl

i.e.

is

bounded.

From

un * II Axil

x

=

B(Ax)

=

namely, if

(b) 3.

Applying this to

lim(xn-xn+2)

I

E

I C

bnk(Ax)kl

5

(C,1) or

Q

(1.2.5) yields

exists then

=

O(n).

The proof

c 4'

T A U B E R I A N THEOREMS

THEOREM.

i s convergent.

For

=

An amusing, perhaps unexpected, result can be deduced,

O(n).

is that

lxnl

*

2. EXAMPLE. xn

(1.4.4 i ) follows

xn

=

x

Let

Then 1

x

E

c

Q

(2.2.51

and s u p p o s e t h a t

i s convergent. 1

2 ( ~ ~ + x ~+- ~~ () x ~ - x ~ - ~ ) .

A T a u b e r i a n T h e o r e m is one in which Convergence of a sequence is deduced from convergence of some transform together with a side

condition, i n this case the assumption that

{xn-xn-1 1

is

12

1.7

convergent. The first such theorem was given by A. Tauber in 1897.

(c) INCLUSION T H E O R E M S 4. DEFINITION. If

A.

eB = c A , A , B

THEOREM.

5. than

I f

i f f

A

cB

cAJ B

3

is s a i d t o b e s t r o n g e r t h a n

are called equipotent.

Let

be t r i a n g l e s .

B

A,

Then

i s stronger

B

is c o n s e r v a t i v e .

BA-I

The proof will make several uses of associativity (1.4.4 i). Necessity:

let

Sufficiency: hence

x E c

x

let

c

c.

x

E

c

Then

A'

Q

-1

The row sums are all are all

Q

Ax

so

B

E

so

c

is s t r o n g e r t h a n

IC,lI

note that a typical row of

sums of

cA c c

C

BA-.lx E c.

Bx = (BA-')Ax

E

c

B'

6. EXAMPLE.

4th row.)

A -1 x

Then

First

is

(-1,2,-2,2).

1.

This is true because the row

(Q-ll = Q

1.

(1.2.5).

Q

-1

91

(This is the

It was for the sake

= 1).

of this convenient check on calculation that the first entry of Q

of

was made

1 instead of the more natural

are also

CQ-'

1/2.

The row sums

1 by a similar argument; all the entries are

non-negative and its column limits are

0.

Thus

is regular

CQ-'

( 1 . 3 . 9 ) and the result follows by Theorem 5.

(d)

BIGNESS THEOREMS.

7. THEOREM. l/ann

unbounded.)

With

=

x

B

c0

x

E

so

be a triang-le w i t h

A

Then

B = A-',

there exists A(Bx)

Let

lbnnl

with

c0 Bx

E

+

(or just

0

sums an u n b o u n d e d s e q u e n c e .

d

IlBIl

ann

Bx

cA.

=

Il/ann( so

unbounded.

IjBII =

(1.3.3).

m.

Hence

Now

1.7

If

8. EXAMPLE.

i s not stronger than

A

are triangles with

B

A,

For

B.

AB-'

9.

as in Theorem 7 and

Q

is not stronger

C O N S I S T E N C Y THEOREMS

Matrices

DEFINITION.

Z i mA x = l i mR x

ohenever

stronger than

A

3

B

n

cB.

The c o n d i t i o n s : are w r i t t e n

A

i s

B B

A.

3

If

are called equivalent.

A,B

A,

cA

3: E

are called consistent i f

A,

and c o n s i s t e n t u i t h

Let

10. THEOREM.

B

A,

be t r i a n g l e s .

Then

B

i f f

A

3

i s regular.

BA-I

Necessity: =

Q

(C,1) i.e. (Example 6) they are not equipotent. (e)

A 2 B

,k

unbounded,

ann/bnn

In particular

the result follows by Theorem 5. than

13

limAA-lx

Theorem 5.

Let

lim x.

=

Then

x

Using Theorem 5, lim BA- 1x = limBA-lx

c.

E

Sufficiency:

Let

limBx = limBA-lAx

=

x

E

cA

lim Ax

cB

=

since

cA

by

BA-'

is

regular. For example, (C,1) 2 Q

11. x

n

=

EXAMPLE.

o(n), (e.9. i f

by the argument of Example 6.

A Tauberian theorem. i s bounded) t h e n

x

I f

-

xn

xn-I

Let

L = 0.

+

L

and

(Ax),

= xn - xn-l (xo = 0), ( B x ) ~= xn/n. Then BA-' = (C,1) is regular, hence B 2 A (Theorem 10). The hypothesis is limAx = L , lim x = 0 so L = 0. (A weaker result but with a nice functional B analysis proof is given in [ 8 0 ] , #2-3-106.) 12. that

THEOREM.

AB = BA.

Let

Then

A,

b e r e g u l a r and r o w - f i n i t e and a s s u m e

A,

B

B

are consistent.

Using the associativity result 1.4.5,f o r have

lim x B

=

lim Bx

=

lim ABx

=

x

lim BAx = lim Ax

n cB we

C

cA

=

limAx.

14

1.7

(f) MERCERIAN THEOREMS A Mercerian m a t r i x

13. DEFINITION.

i s o n e such t h a t

A

cA = "

The name derives from Mercer's theorem (2.4.1). A conservative triangle

14. THEOREM. A

-1

i s conservative.

By Theorem 5 with

B

=

I.

The result fails for non-triangles. Triangles

15. COROLLARY. with

i s M e r c e r i a n iff

A

are equipotent i f f

B

A,

(1.8.6). B = MA

a Mercarian t r i a n g l e .

M

Sufficiency: BA-' Necessity:

let

M

M, AB-'

=

= BA-';

= M-';

apply Theorem 5.

apply Theorem 5 .

It is useful to extend half of this result: Let

16. THEOREM. Then

B = MA.

If (M-'B)x

x =

E

M

-1

A,

(Bx)

E

be a m a t r i x , M

a Mercerian t r i a n g l e ,

are equipotent.

B

cA, Bx = M(Ax)

by 1.4.4(i) with

c

A

arbitrary column of (g)

A

E

c , using 1.4.4(i).

using 1.4.4(i).

M

replaced by

A.

Hence

Ax

But

and E

x

If

M-lB

=

x

E

cB,

M-l(MA)

=

A

taken to be an

c.

MAPPING THEOREMS

These give characterizations for membership in Examples are

( k " : ~ ) ,

(c:~"), (1.3.3) and

(co:w), (1.3.2); ( k m : k m ) , (c:c), (co:c), (1.3.6).

(X:Y).

(co,L") hence A unified treatment

using functional analysis is the subject of Chapter 8. We give here a characterization of

( Lm:c),

the so-called c o e r c i v e m a t r i c e s .

This mapping theorem is exceptional in that no functional analysis

1.7

treatment is known.

On the contrary an important result (Corollary

2 0 ) derives from it.

LEMMA.

17. that

15

See also [SO], Remarks 14-4-8, 15-2-3.

Let

b e a m a t r i x w i t h c o n v e r g e n t columns s u c h

A

i s uniformly convergent.

Cklank

a

Then

E

(For

L.

a

see

1.3.5).

This is really just the proof that so that

Ck=m lank I < 1 Eo

m- 1 lakl + xk=l

+

IIAll

and so

1

THEOREM.

18.

for all

21 , m ( 1 )

so that

m

2’ ‘k=m(l)+l

>

Suppose (iii) is false.

by a constant we may assume

and multiplyin

for all

n(1)

k.

for aZi!

lan(l),k

n(2)

‘E=m( 1)+1 I an( 2),k Choose m(2) m

‘k=m( 2)+1 I an( 2),k choose n(i)

‘k=m i- 1 +I m

(

)

‘m( i ) k=m(i-l)+l Let x m(1)

k=l lan(l),kl - ‘k=m(l)+llan(l),k I

(in the second sum

lxkl

=

1 for all

k

112 - 118 > 114

was used);

I

m(2) ,m(U I > 112 -(AX)n(2) 1. ‘k=m(l)+l - k=l ‘k=m(2)+1 “n(2) ,k 118 - 118 = 114 and so on, so that Ax has the subsequence C(-l)nunl

where

CASE 11. A

un > 114. Thus real.

coercive (1.3.7) and

Let

bk

which says exactly that

=

A

x c

bnk = ank

0

Rm

yet

Ax

c.

- ak. Then B

so by case I , B

satisfies (ii).

-

is also

satisfies (iii)

This is the same thing,

as proved earlier. CASE 111.

Let

A

=

B

+

since it is the real part of the other conditions.

iC.

Ax.

Similarly

Then for real

x

6

E m , Bx

Hence by Case 1 1 , B C

c

satisfies

does and so finally, does

We shall see in 5.2.12 that uniform convergence of not sufficient.

E

Ck ank

A.

is

1.7-1.8

19. THEOREM. xklankl

A

i f f i t h a s n u l l columns and

(Lm:col

6

17

i s uniformly convergent i . e .

(tm:c

n

= (L":c)

(+:c0).

The other equivalent conditions of Theorem 18 apply here too. Necessity is trivial by Theorem 18 and the fact that each

AAk

E

co

for

Sufficiency is immediate from (1) in the proof of

k.

Theorem 18.

20. COROLLARY. convergent.

an

+

L' =

sequences.

Set

=

(a:).

A

Then

(1.0.2,1.0.7), and Theorem 18 (ii)

9.

in norm.

COROLLARY.

21.

be weakly convergent, A m

a

v e a k l y c o n v e r g e n t s e q u e n c e s a r e norm

L,

I an}

Let

is coercive since says that

In

A c o r e g u l a r m a t r i x c a n n o t sum a l l t h e b o u n d e d

A c o e r c i v e m a t r i x must be c o n u l l .

x

=

in

1

in the proof of Theorem 18.

(1)

1.8 INVERSES

Applications of the inverse of a triangle were given in the preceding section.

Some results of this type hold in more

generality. Let

1. LEMMA.

Let

3 :

x

=

notation means. exists

u

E

c

u + v

with

Then Ax

with

be o n t o .

A : c + c

Au

=

=

Au

u E

Ax.

E

c , Av

Then

v

=

c 0 A'.

0, which is what the

=

c. c: Let Setting

cA

=

x

E

x-u

cA.

Then there

gives

x

=

u+v

as

required. 2. COROLLARY. right inverse

Given

x

B, E

c

is c o n s e r v a t i v e and h a s a c o n s e r v a t i v e

If A

cA = c

let

o A'.

u

=

Bx.

Then

u

B

c

and

using 1.4.4(ii). The result follows from Lemma 1.

Au

=

(AB)x

=

x

We shall see in

17.5.11 that the hypothesis of Lemma 1 does not imply that of

1.8

18

Corollary 2. 3.

EXAMPLE.

= ~-x (Bx)~

Let

(Ax)n

+ 2 ~ , + ~ . Then

n+l

-k

I: 2

=

A,

a r e r e g u l a r and

B

I (-l)n)

sums t h e b o u n d e d d i v e r g e n t s e q u e n c e

sequences such as called

A If

4. THEOREM. l e f t inverse

AB = I ; A

as well as unbounded

such that

cA

n

2,”

=

c

is

n em,

x

5.

EXAMPLE. Let B

cA

E

is regular.

i s c o n s e r v a t i v e and h a s a c o z s e r v a t i u s

i s Tauberian.

A

B,

A

is Tauberian even if A

If

A,

A

- Xn+2’

T a u b e r i a n ; so it appears that the hypothesis of Corollary 2

does not imply that

Then

A matrix

In}.

-

x ~ + ~ ~ - ~ , - 2xn

then

x

=

(BA)x

=

AX)^

=

2xn-l

- xn,

are regular, BA

t h e d i v e r g e n t sequence

=

E

c

using 1.4.4(ii).

( B X ) ~=

c 2-kxk+n.

i s a t r i a n g l e ! , and

I, A

{Znl.

B(Ax)

sums

A

It appears that the hypothesis of

A

Theorem 4 does not imply that

is Mercerian even if

A

is a

In the next example is shown a non-Mercerian

regular triangle.

regular matrix with a two-sided regular inverse.

Let

6. EXAMPLE. Let (Ax), = xn - axn+l where (Bx), = C CIk-1x ~ + ~ - Then ~ . A,B E r , indeed

(1-a)B

are regular, and

AB

=

BA

=

I.

0 < l a 1 < 1.

(l-a)-’A

An application of Corollary

2 and Theorem 4 yields the advanced calculus exercise: {xn-ax 1 n+l

is c o n v e r g e n t ,

convergent.

( a / < 1 , and

Indeed Corollary 2, alone

and

x

suppose

i s bounded; t h e n

implies that every

is

x

x

such

that this transform is convergent is a convergent sequence plus a multiple of

A superficially similar assumption yields a

{cI-”}.

quite different result: la1 < I ;

then

x

i s convergent.

to the transpose of now triangles.

suppose

A.

Ixn-crz n - 1 1

i s convergent,

This is proved by applying 1.7.14

The resulting matrix and its inverse are

1.8

The m a t r i x

7. EXAMPLE. c

A

-2

for if its first row is removed and it is divided 1 it has the form of Example 6 with cx = 2

A d j a c e n t c o n v e r g e n c e d o m a i n s , c B = cA 0

8. EXAMPLE.

A

i n Example 5 s a t i s f i e s

A

c 0 {znl

=

by

19

E

be a triangle and

(Example 6, or 7), B Thus

Ax

z

with

u + mv

=

A

=

a row-finite matrix such that

If

EA.

=

with

u

x c cB, E(Ax)

c

E

and

x

=

Let

2.

cE = c 0 v

c

using 1.4.4(i). -1 A-lu + mA v c cA 0 z c

-1

v

A useful sufficient condition for the existence of an inverse is dominance of the principal (or main) diagonal.

see [85], Theorem 2.

extensively modified, 9.

n

THEOREM. L e t r

where

A

Then

< 1.

r , a nn

E

Let

10. COROLLARY.

IIA-III 5 r < 1

for

+

0;

Then

i2 .

lak\

cx > IIAll

Then

n- 1 k=m+llankl

ann

r. and the result

be a c o n s e r v a t i v e t r i a n g l e w i t h

A

Z lak( >

cB

when we show that

11.

f o r aZZ

is M e r c e r i a n .

A

- 2

and choose

v

so that

n

>

v

ank I > a. Let m = max(u,v). Now let bnn = 1, for all n , k such that n > m and k 5 m, bnk = ank

otherwise.

-

I , ~ { ( a , ~ l :#k nI < r

being a Banach algebra (1.5.4).

follows by 1.0.6,'I

= 0

=

has an i n v e r s e i n

A

The hypothesis implies that

bnk

The idea can be

IIB-III

EXAMPLE.

A

(Ax)n

=

x

n.

i s conservative.

= pn,

D

and

2

D

Here we =

I.

Considering (iv), keep in

We have from (2)

n (k); subtract

hn+l,k

and transpose:

- khn+l,k]/(n+l).

= [(k+l)hn+l,k+l

Every n o n - n e g a t i v e

11. THEOREM.

( i ) hnn

= D(uo6 0) = po(D6 0 ) = pol.

in (1) yields (iii).

hnk = 0 where

- hn+l,k

n 2 0

Then f o r a l l

is the first column of

1

Multiply this by hnk

.

DMDl = DM6'

=

used the facts that k

lJ

m 2 0.

for all

For (ii), H1

Setting

H = H

This yields (iv).

( r e a l ) Hausdorff m a t r i x

A l l o f i t s column l i m i t s a r e

0

H

except possibly

the f i r s t .

From Lemma 10 (ii) it follows that conditions (i) and (iii)

of Theorem 1.3.6 hold.

Induction on

the remaining condition. negative function of yields, for any

n

12.

-.

since

r , (write

Since

EXAMPLE.

Let

in Lemma 10 (iv) yields

1 hnk is a decreasing nonk=O hn+l,m+l 2 0.) Next, Lemma 10 (iv)

(Note that

un - hn+l,m+l, holding

The right hand side converges as C un/(n+l)
0.

(5)

6 Necessity: We may assume that y is totally decreasing. By m m+k ( 4 1 , uk = 1 (:)hm+k,k+r/(k+r); replacing m by m-k, we have, f o r r=O

=

k-1 TI (i-r)/(rn-r)

for k 5 i 2 m , with the convention u(m,i,O) m pk = . l u(m,i,k)hmi because u = O if O ( i < k .

r=O

We may write Then

pk =

I

k-1 where

v(rn,t,k) =

0

gm

g,(O>

0 5 t 5 1; thus

hmi

at

= r(1-t)/(m-r)

v(m,t,k)

o(1)

-+

tk

g,

+

The result is

for 0 < t 5 1. This 1 j m thmj t = i/m for i = O , l ,... .

=

m

"+

yk =

r < k 2 m,

0 as

m

+

m ,

through a sequence of values

(some increasing function).

g

since

Since also

denotes a quantity which -+

1,

h

2 (k-l)/(rn-k+l)

uniformly.

In this formula we may let m such that

(t-r/m)/(l-r/m),

is increasing. For each fixed

g,

l(t-r/rn)/(l-r/m)-tl

where

= 0, gm(t)

has jump

Note also that

+ m

II

r=O

V(m,t,Ot = 1, and

n

1.

1=0

1 v(m,t,k)dgm

is because

=

This is b y 2.0.1.

lltkdg. 0

19. THEOREM.

Let

y

'

be conseruative, p k =

tk dg.

The first

0

column limit o f

is

H J !

g(O+l-g(O).

Thus

H

is multiplicative

2.3

We may assume that

g

1

(51, hno

2

(1-t)"dg

=

0

Conversely n

of

m

-.

-t

=

0.

is increasing and that

g(0)

2 (I-E)~~(E) + g(O+)

(1-t)"dg

=

0. By

as

0.

E +

lo

jE+j (1-t)"dg

Thus

lim hno 5 g(~) for all

5 R(E)

-+

(l--E)n[g(l)-fZ(E)I

g(E)

-+

E

The evaluation

0.

E >

follows from Lemma 10 (ii). EXAMPLE.

20.

H

E!

i s continuous a t

hno -

O

as

g

m = g(l/-glO/l i f f

(with

29

I.

(ii). With

(i) Taking g(t) = t

g = 0 on

we obtain

[O,l),

g(1) = 1 yields

H = (C,l).

We now show Hausdorff's elegant derivation of Mercerian theorems.

First with

g(t)

=

ta, a

> 0, pk =

l1tkdg = a/(k+a). 0

Then

+

p

(I-a)p

that

is regular by Theorems 18, 19, so for any

is regular sicce

HV = a1 +

(1-ci)H

!J

CI

> 0, v =

C

I

~

by Example 1. Note

vk = (ak+a)/(k+a).

21.

THEOREM.

Let

uk

=

I b k + l / / ( c k + l / , b,c > 0 .

Then

H

!J

is

r e g u l a r and M e r c e r i a n .

The substitution b = a/a, c = l/a makes which, as we just saw, is regular.

uk

=

(ak+a),'(k+a)

Similarly, 111-1 is regular and

the result follows by Corollary 3 and 1.7.14. 22.

COROLLARY.

Let

p(a), = (ak+l)-'.

811(bl a r e e q u i v a l e n t f o r aZl a,b

Then

and

0.

By 1.7.10.

A special case of Hausdorff's theorem is Mercer's theorem which we give in the next section. Hausdorff matrices are discussed also in [ 5 5 ] , pp. 167-173

2.4

30

2.4.

CESIROAND

H~LDER

Mercer's theorem (1907) states that the straight line joining the identity matrix to

(C,1) is made up of Mercerian matrices.

We shall write

C

for

(C,1) and remind the reader that indices

1. THEOREM.

For

a > 0, t h e m a t r i x

start from

0.

For it is Hp with

pk = a

a1

+ (1-a)/(k+l)

+ (1-aIC i s Mercerian. =

(ak+l)/k+l.

The

result follows by 2.3.21. For

for

a = 0

0 A

a

the matrix, call it

A, in Theorem 1

is

(C,l);

is not Mercerian, but has a very small convergence

domain : THEOREM (G.H. Hardy, 1913).

2. X

Then

= cA.

X

= c 0

where

v

u

The notation means that every vergent sequence

Let

a < 0,

A = aI

+

(I-alC,

is a d i v e r g e n t sequence.

A-sumable sequence is a con-

+ a multiple of v and conversely. (To moti-

vate the choices about to be made, let q be the first column of 1 CA-l. Then AC- q = 6 O , hence, since A, C commute, Aq = C6O so q

E

X.

Computation of

(AC-l)-'

- will

q

- which

will be a little easier using

reveal the origin of the sequence

v

defined in

the following argument.) Let

u = x-L.l

x

E

X.

Since

satisfies

x

=

limA u

u+L.I

where

L = limAx and

0 we may assume that limAx = 0. n Let x = 1 - a-1 > 1; yn = (n+l)(Cx), = 1 xi; zn = i=O ynr(n+2-A)/r(n+2) for all n if X is not an integer, while if =

n this way for n 2 A-1; for smaller n, zn may be defined arbitrarily. For sufficiently large n , X

is an integer, define

z

31

2.4

Thus 1 (~!-z~-~)is convergent so that z = M - 1 (z.-z ) = M + o (n'-'). n 1 i-1 i=n+l + Xyn/(n+l)

=

+ Xznr(n+l)/r(n+2-X)

o(1)

o(1) + XMv, + hvn.o(n

is

X c c

0

v

see this: =

(AV),

1- x

) =

=

M

=

lim z

Hence

n exists and xn = a-1(Ax)n

XMv, + o ( 1 ) .

This proves that

and the reverse inclusion merely requires for large

n, (C-lv),

avn + (1-a)X-lvn

{X:XI + (1-X)A

containing 1

M(A)

is Mercerian}.

regular triangular A

- but

v

X.

E

(n+l)vn - nvn-1 = XV,,

To

so

= 0.

A , let

Given a matrix

=

This

o(1) + Xznvn, say.

(the Mercerian set) be

In [45] it is shown that there is a

such that

M(A)

is any preassigned open set

there is also such an

A

that

M(A)

is not

open.

for

\

DEFINITION.

3.

a > 0,

T h e C e s a r o m a t r i x of o r d e r

i s t h e Hausdorff matrix

in which

Hu

a, m i t t e n

caJ

Pk =

a I1tk(l-tla-'dt. 0

(The definition of of values of

is usually extended to a la,*ger class

Ca

By 2.3.18 with

a.)

g(t) = -(l-t)a,

each

Ca

is

regular. Setting H = Ca a straightforward calculation gives

CY1

)/(nia),

hkk

=

with the Hglder matrix (C,l).

B

=

A

Also if

diag(l/n)

uk

=

(kia)-1 .

Ha which has

It is natural to compare this uk

=

(k+l)-", H1

=

C1 -

n

is the matrix such that

we have, for integers

-

hnk -

(Ax)n =

a , H a = (BA)',

1

xk

and

k=O

Ca = BaAa. A

more intimate connection is the theorem of Schnee and Knopp ([34I, p. 264) which states that these matrices are equivalent. We give it for integer

a

only.

2.4-2.5

32

4. THEOREM.

"

= l,Z,

The m a t r i c e s

H",

Ca

are equivalent f o r

... .

-1

They are H u , HV with -1 n{(k+r)/(k+l):r pk/vk = ( a ! )

=

yk

=

product of matrices of the form

Now

(k+l)-", vk = ( k+") 2,3,...,u}, HA

with

X

thus =

is s

Ha(Ca)-'

By

(k+r)/(k+l).

2.3.21 each such matrix is Mercerian. Some discussion and references for complete monotonicity may be found in Problem E2845 of the American Mathematical Monthly, January 1982, p. 6 4 .

2.5. N~~RLUND MATRICES We introduce the notation from two sequences

(a,b) for the sequence formed n akbn-k. a,b by the formula (a,b), = 1 k=O

(This is the standard c o n v o l u t i o n . )

I n d i c e s s t a r t from

every-

0

where i n t h i s c h a p t e r .

Let

p

be a complex sequence with

p0

=

1.

Let

Pn

= ( ~ , l ) ~

n

1

pk. We shall assume that Pn # 0 for all n. The NorZund k= 0 matrix A = (N,p) is defined by the formula ank - pn-k/Pn for =

k 2 n , 0 for

k

n.

Thus

lk

(Ax)n = (~,x)~/(p,l)~, ank

=

(Alln

= 1.

Manipulations with Norlund matrices are conveniently handled m

via the forma P(2)

=

1Pn2

power series

p(z)

=

1

For obvious reasons

pnz" = (1-z)P(z) where n=O (N,p) is called a poZynomia2

m a t r i x if

= 0 for sufficiently large n. The functions Pn may be defined first and the Nb'rlund matrix defined from it.

is done in Example 1. p(z)

p(0)

=

po

This

It is important to remember that whenever

is given and generates a Norlund matrix

the data are

p(z)

=

1, Pn # 0.

(N,p), parts of

2.5

1. EXAMPLE. (b) Let

p(z)

=

l+z.

p(z) = (l-z)-l. 2.

(a) Let

Then

EXAMPLE.

(N,p)

= Ca.

=

(N,p)

=

l,(N,p)

Then

P(z)

=

p =

(N,l+z)

=

6

0

, (N,p)

Q(1.2.5),

I.

=

(c) Let

= (C,l).

For positive integer

)zn. Then and

p(z) = 1.

Then p

33

p(z)/(l-z)

a , let =

p(z)

(l-z)-"-l

=

=

(l-z)-'

1

(n:a)zn

The latter are the only matrices which are both

Norlund and Hausdorff ! 3. Then

THEOREM.

f o r some

A = C

(n/(n+a))un-l.

>

Since

Let

Let

a.

A

=

uo

=

a00 = 1

1.

Thus

=

(H,').

p(1)

polynomial matrix, p(1) = Pu where was explicitly assumed non-zero. THEOREM.

Then, setting

Then for each 0.

u

is the degree of

Note again that

p(0)

=

For a p. po

This =

1.

Every polynomial m a t r i x i s r e g u l a r .

Each column

each row adds up to 1. Finally, for sufm n, 1 lank( = 1 Ipk/Pml where pk = 0 for k k=O

0's;

ficiently large

6.

n, IIAll

It follows

is not defined.)

The matrix consists of finitely many diagonals. terminates in

-

un

Thus

it follows by induction that

Pn is bounded away from

p(1) # 0. (In Example l(c),

5.

(N,p)

A = (N,p) have finite norm. 1

(ann( = \Pi

that

b e a N z r l u n d and a H a u s d o r f f m a t r i x .

A

- aann = pl/Pn = an,n-l - n(un-l-un).

a = p1,

4.

Let

THEOREM.

The Ngrlund m a t r i x

A = IN,pl

i s conservative

n i f f

( i l l i m p,/Pn

e x i s t s (call i t

XI,

(iil

1

k=O each

n; r e g u l a r iff

( i ) , ( i i )hol-d w i t h

A = 0.

lpkl 5 MlP,l

for

2.5

34

Note that ano = Pn/Pn - 1 - pn-l/pn

A

If

is ]]All limnank

is conservative, (i) follows by (1).

-.

(z@,p)

z

muy b e o m i t t e d .

m

n , while

0.

Then be a sequence. rn = supmII z k x k ( , p n ( x ) = Ixn k=l Let

po(xI

z k # 0,

=

Pn

E

a sequence.

z

where .Y

z

i s defined

n # k , and apply Theorem 1 with

To prove the last part, pn(x-x (m)

W .

s p a c e and

p U h

space w i t h

has

Y

.cs, i s w r i t t e n z B ; Z B

b e an

ann = zn, ank = 0 if

Let

that

(U,qI

Let

-1

i s convergent f o r a l l

n n

6. THEOREM.

=

z

n I z B : z E Z I = I ~ : xC z

t o be

X

The s p a c e

p

.

i s an

For a n y

k

AK

such

may b e o m i t t e d .

I

The first sentence follows from Theorem 6, Definition 5 and 4.2.14. The form of

po

can also be gotten from Theorem 3 with

n

1 xkzk. If zk # 0, then with the matrix k=1 mentioned, lxkl = I (Ax)k-(Ax)k-ll / 'kl 2 2Po(x)/lzkl (Ax)n

=

is redundant.

If

8. THEOREM. AK

space w i t h

such t h a t t h e

z

E

$,

Let

A

ZB =

just and so

Pk

W .

T h e n I w A , p U hI i s an rn = sup11 a n k x k l . For any k m k=l

be a m a t r i x .

pnIxl = lxnl, knlxl kth

A

c o l u m n of

A

h a s a t l e a s t one n o n - z e r o

term,

4.3

pk


f i n i t e Z e n g t h , hn

Since

For a n y

63

such t h a t t h e

n


diate from Theorem 7, and 4.2.15.

row has finite length, hn

has

A

A}, the first part is imme-

ank # 0, lxkl 5 2hn(x)/lankl

If

exactly as in the proof of Theorem 7 so nth

row of

( S e e Remark 1 0 . )

is a row of

w A = n{r6:r

nth

pk

is redundant.

If the

does not appear by the last part of

Theorem 7.

Of course Theorem 8 could be expressed as a statement

9.

about

(Definition 5) where

Z6

is essential, f o r example

0 which is not an

wB =

Z is a countable set.

which does not have

L B = (Im

FK

space at all (4.0.5).

Countability AK, and

In Theorem 15

we examine certain non-countable intersections. 10.

Let

be the identity matrix.

A

Then apparently Theorem

8 says that a l l the seminorms are redundant for

Theorem 8 says is that each sent. Actually hm - p, just

p1,p2,...

pn

wA - w.

is redundant if the

What

hm

are pre-

and so the first list of seminorms is

.

EXAMPLE. A p e r f e c t c o n u l l t r i a n g l e . Let a E L with n for all k, and (Ax)n = 1 akxk. Then cA = a B has AK

11. ak f 0

k=l

by Theorem 7, hence

0

is dense.

12. THEOREM. L e t

(Y,q)

Then

Y A = {x:Ax

p, h

a r e a s i n Theorem 8 .

of any

E

Y }

is a n

be an FK

FK

space w i t h

For any

k

n

such t h a t t h e

be o m i t t e d .

nth

row o f

( S e e Remark 1 0 ) .

Apply Theorem 1 with

X

I f = wA,

A A

an

p LJ h LJ

such t h a t t h e

h a s a t Z e a s t one n o n - z e r o t e r m , p k

A

s p a c e and

a matria.

A

qoA

where

kth


h a s f i n i t e l e n g t h , hn

i s a triangZe use only

FK

cozumn For

may qoA.

space by Theorem 8.

64

4.3

Then

Z

YA

=

and the seminorms may be read off from Theorems 1 and

8. The remaining parts follow from Theorem 8 and the fact that if A

is a triangle the map 13.

THEOREM.

Let

u

where

p

space w i t h for

n = 1,2,

h

... .

A : YA + Y

such t h a t t h e omitted.

be a m a t r i x .

A

po -

For any

h a s a t l e a s t one n o n - z e r o nth

is an equivalence.

k

II-llA,

p,,

hn

c

such t h a t t h e

has f i n i t e l e n g t h

A

column of

kth

A

For a n y


term, p k

row o f

i s an F K A a r e a s i n Theorem 8

Then

hn

n

may b e

( S e e Remark 1 0 . )

This is a special case of Theorem 12. See also Theorem 3. 14.

THEOREM.

a c l o s e d s u b s p a c e of

Define

f : YA

XA = fV1[X]

i s a c l o s e d subspace o f

If X

then

Y,

is

XA

YA.

Y

-+

by

f(y) = Ay, a continuous map.

Then

is closed.

We now discuss a general form of dual space, called the m u l t i p l i e r s p a c e , which specializes to various Kothe-Toeplitz duals.

15. = MIX,YI

is a

THEOREM. =

BK

n{x-l-Y:x

Y

E

B

=

be

Y

XI = {z:x-z

B(X,Y)

spaces w i t h

BK

z = 0.1

to show that +

Z

of all continuous linear maps from

X

Y

: X

-f

Y , i(x)

=

Z in

The induced norm is

T(x)

Z

B.

x

and, since

Z yields a

E

i

[If

= 0, zn =

i(6")

IIzl] = supIllx.zll:IIxII < 11.

is a closed subspace of

for each

z

x.z, which is continuous

To see that coordinates are continuous, fix n v = I16nlly. Then IIzll I[(uGn).zll = u]lzn6 11

&(x)

x E XI.

for all

Let

$.

2

Z

This clearly embeds

by 4.2.8. so

X

Then

E

is a Banach space (4.0.6). Each member

diagonal matrix map

= 0

X,

space.

The space to

Let

Y

n. =

u

Let

uvlznl.

B. is an

Let

FK

h

+

=

1/116"1Ix,

It remains T

E

space,

B.

Then

4.3

[S(x)],

+

T(x)~ for each k , i.e.

this gives xkzk SO

xktk

+

T

zk

k

+

T(6 )k

-f

T(x)~. With

which we shall write as -t

tk.

x = 6k Thus

(Tx)~. Hence

Tx

=

x.t

t.

=

X

space if

BK

xkzk

and, as just proved, xkzk

16. EXAMPLE. a

65

M(X,cs) = X B is.

11 41

= =

so Theorem 15 shows that

XB

is

Further SUP {II u .XI1 cs :II XI1 I 11 n u x (:n = 1,2,...;\I sup!lI k=l k

XI\

2 1).

II

17. EXAMPLE.

Other Kothe-Toeplitz duals are also special

cases of Theorem 15.

They are the

y-dual, M(X,bs) = X y , where i.e. y

E

hs

so it is a

bs

a-dual, M(X,L)

=

X", and the

is t h e space of b o u n d e d s e r i e s

iff

BK

space with the norm just shown, by Theorem 12

Further

18. THEOREM.

Let

X

cZosed subspace o f

Xy.

For the norms are the same, as shown in

be a

BK

space.

Then

is a

XB

the preceding examples. As pointed out in Remark 9, Theorem 15 fails if

BK

space; Z need not even he an Information about the spaces

pp. 11, 12, 122, 133.

FK

X

is not a

space.

M(X,Y)

may be found in

[%I,

4.4

66

DUAL SPACE

4.4

THEOREM.

1.

m a t r i x d e f i n e d on Then F

E

i s an

2

g

X',

i.e.

X

FK

lY,ql

be

X c uA.

Let

(X,p),

Let

s p a c e and

f

f

F

and

g O A

p

By 4.0.3 a n d 4.0.8, f = F

Z'.

lu(x)l 5 q ' ( A x )

F, u

where

If

has t h e given

f

and

+ u

qoA.

respectively.

IF(x)l 2 p ' ( x ) ,

with

are l i n e a r f u n c t i o n a l s on

(3.0.1). ever

W e may assume

F

E

A[X]

n

Y

Def n e

y = Ax.

y = Axl = AxZ Ig(y>l 2 q ' ( y )

g

on

To see t h a t then

X'

Z

Conversely,

a r e p o s i t i v e l i n e a r c o m b i n a t i o n s o f f i n i t e l y many

q'

with

a r e c o n t i n u o u s on

goA

w i t h smaller t o p o l o g i e s , r e s p e c t i v e l y E

+

Y'.

E

form i t is c o n t i n u o u s s i n c e

f

a

A

Y A = ixcX:AxEYI.

= F

The f i r s t p a r t i s a r e p e t i t i o n o f 4.3.1.

let

n

Z = X

i f f

Z'

E

s p a c e s and

FK

Z, p ' ,

p n , qn

by t h e Hahn-Banach t h e o r e m g(y) = u(x)

by s e t t i n g

when-

is w e l l - d e f i n e d , s u p p o s e t h a t

g

-

1. For

since M

x

cA we have

E

is conservative. Thus

(Bx)n = 1 akXk Also if

x

cB

E

(Dx)~-~ = (Bx), -

C

then

akxk

x

lirnBx

and so

c

3

=I

If

u

by 1 . 4 . 4

x

E

cA c

( B X ) ~=

C

akxk.

Bx

cD

c.

E

and and

a'

c

CD

The proof is con-

=

=

Hence

We have now proved

limM Ax + ax = f(x). cA by 1 . 7 . 1 6 .

limDx + ax

=

cluded by noting that 2.

c

since

since

aB

E

CA c CB

Further

cB

E

E

(DXIn-l*

+

x

M(Ax) = Dx

= 0 , Theorem 1 h o l d s w i t h t h e weaker c o n c l u s i o n

n-1 Take

c A'

(Mx),

1

=

tkxk.

The proof of Theorem 1 then applies

k= 1

with the omission of the last step. 3.

If

is uniquely determined by (l), by

i s coregular,

A

4 . 4 . 8 , and t h e c o n v e r s e of T h e o r e m 1 h o l d s :

This is because

exists.

x(B)

=

i f

X(f) = ux(A)

no s u c h

p = 0

( 4 . 4 . 7 ) and

B

B

is

coregular b y 3 . 5 . 4 . In case

4.

f o r arbitrary

f

A

is not

p-unique the result of Theorem 1 holds

since a representation may be chosen with

u # 0.

The remaining case is covered in 1 5 . 5 . 3 .

5.2. REPLACEABILITY

A matrix

A

is called r e p l a c e a b l e if there exists a matrix

with null columns and plicative-0

cB = cA.

THEOREM.

Let

equivalent:

( i )A

( i i i )I

(closure i n

g'

A

is conull, B

of course ( 3 . 5 . 4 ) and a coregular matrix

able iff a regular matrix 1.

If

A

B

exists with

be c o r e g u l a r .

i s replaceable, c,),

( i i )I

=

B

must be multi-

A

is replace-

'A'

Then t h e f o l l o w i n g are g'

Co

(closure i n

( i v ) l i m i s c o n t i n u o u s on

c

cA),

as a

77

5.2

s u b s p a c e of CA

(vl x

cAJ

+

a s a s u b s p a c e of

c

(ii) = (iii)

by 4.2.7 with

(i) implies (iv): =

i s continuous on

ax

limB

on

With

Y

= cA,

cB = cA, B

X

= co,

E

= $.

regular, we have

lim

c , so continuity is guaranteed by invariance of topology

(4.2.4).

(iv) implies (i): Banach theorem.

Let

Then

1

=

f c ci, f

=

lim on

c

by the Hahn-

(4.4.7), so

X(f) = p(f)X(A)

p(f) # 0.

The result follows by 5.1.1. (iv) implies (ii):

Since

(ii) implies (iv):

In

lim = 0 on

c , with the

maximal subspace which is not dense. But

so

co = lim* (iv)

limA

=

lim

For

(v):

2.

If

A

and

lim 1

topology, c0

=

1.

is a

Hence it is closed (5.0.1).

is continuous (5.0.1).

x

is continuous and

cA

co

c , limAx = x(A)lim

x + ax (1.3.8). Since

x(A) # 0 the equivalence follows.

is conull, (ii), (iii), (iv) are always false and

(v) is always true, while A may o r may not be replaceable. To see this suppose follows that

f = 0 on

co.

Since

~ ( f )= p.x(A)

= 0

it

f(1) = 0, so the falsity of (ii) follows by the Hahn-

Banach theorem.

Equivalence of this with (iv) is contained in the

proof of Theorem 1.

A s to (v), for

Finally any multiplicative

x

c , ax = limA x by 1.3.8. 0 matrix is automatically replaceable; c

a non-replaceable conull matrix is given in 13.2.7. 3.

if

DEFINITION.

A conservative

FK

space i s caZled r e g u l a r

I e' go. By 5.0.2 a regular space is coregular.

coregular matrix

A

By Theorem 1, a

is replaceable if and only if

cA

is regular.

It is not too difficult to construct a coregular space which is not regular

- it takes more effort to construct one of the form cA,

5.2

78

This is done in

equivalently a coregular non-replaceable matrix. Example 5. Let

THEOREM.

4.

c.

are fundamental i n

Let A-'ak

ak

be the and so

= gk

b e a c o n s e r v a t i v e t r i a n g l e whose c o l u m n s

A

kth {6

cA

Then

k

has

AD.

column of

1

A

i.e.

is fundamental in

:a

ank.

=

cA

since

Then

A-':c

-+

cA

is a linear isometry onto.

A

This result fails if example let

be the identity matrix with an extra column of

A

placed on its left side.

5. EXAMPLE.

Then

with =

x

even f o r

AK

bk .f 0 -k

i

211-1 k=l

X = cA, A

for all

Let

k.

cA.

triangle. does not

AD

a c o r e g u l a r t r i a n g l e . Let b n- 1 = x 2n-1 +k=l 1 bkx2k3

The columns of

bkx2k.

65+66 ,...; bl(l-6

space of t h e form

BK

1's

cA = c.

A coregular p e r f e c t non-replaceable

A coreguZar non-regular

irnpZy

is not assumed to be a triangle; for

A

are

1) , b2(1-6 1-6 2-6 3) , . . .

.

9.

E

AX)^^

3 4 6 +6 ,

6'+A2,

The span of this set i.s

the same as that of

Now suppose that 0 = f(l-62n-1) (1.0.2).

f(1)

-

It follous that

f(6"+6"+') finally

=

c', f = 0 on the columns of A. Then f(6 2n-1 ) hence f(1) = 0 since f(An) + 0

f

= 0

f

columns of

=

A

for each

E

f(62n-1) n

=

and so

for each

0

f(6")

= 0

But also

for all

n.

Thus

0. This shows, by the Hahn-Banach theorem that the

are fundamental in

c.

By Theorem 4, cA

A

hence Theorem 1 (iii) is satisfied and so It is easy to check that perfect.

n.

Finally

cA

A

is coregular.

does not have

AK

has

AD

is not replaceable. Of course

since

1

E

AD

cA

implies

and

A

is

5.2

79

coregular. A non-y-unique

6. THEOREM.

m a t r i x 14.4.4) m u s t b e r e p l a c e a b l e .

There is a representation of the function by 5.1.1, there exists 7.

EXAMPLE.

B

A

With

with

cB

=

0 with

u # 0 so

c A , limB = 0.

as in 4.4.4, one sees directly that

0 = limA x - ax.

The preceding theorem may be cited o r it may simply

be checked that

(Bx), = 1 akxk does the job. (Note that B is k=n It is severely limited in this direction - see

m

not a triangle. 15.2.14).

We continue with a few remarks to illuminate the conditions of Theorem 1. T h e l i n e a r s p a n of

8. DEFINITION.

+

and

1

i s denated by

I t i s t h e s e t o f e v e n t u a l l y constant sequences.

4,.

Let

9. THEOREM. iff

Ck

ank

A

satisfy

cA

Then

3

in

1 = C gk

c

A

i s uniformly convergent.

First observe that the equality holds iff

lll-l(m)llA

-+

0

(4.3.13) because this condition is automatically satisfied by the OD

other seminorms of 10. Then i f

cA

COROLLARY. Ck

ank

by 4.3.8.

Let

A,

This expression is

supnlC ankI. k=m+l

be equipotent matrices with

B

i s uniformZy convergent, so i s

cA

3+l.

C bnk.

This is an invariance result and Theorem 9 “names“ the property.

(Naming program.) 11.

conull i f

DEFINITION, 1 = I: 6

k

i s strongly conull.

A conservative

, a matrix

A

FK

space i s c a l l e d s t r o n g l y

i s called strongly conull i f

cA

80

5.2

Theorem 9 characterizes the spaces among conservative convergence domains. 12. EXAMPLE.

A c o e r c i v e m a t r i x i s strongly c o n u l l

Theorem 9 ) , but n o t conversely e.g. let 13. THEOREM. FK

tive

X

if

cA, A

=

X

space

log(n+l)].

E a c h of t h e following p r o p e r t i e s of a conservai m p l i e s the n e x t but i s n o t i m p l i e d by it e v e n

a triangle.

null, ( i v ) I E

ank = ( - l ) k / [ k

(1.7.18,

(iil strongly conull, ( i i i ) co-

( i ) AK,

7.

(iii) implies (iv):

If

f = 0 on

0 , f(1)

the result follows by the Hahn-Banach theorem. imply (iii) is Example 5 with Theorem 1.

X(f) = 0 and

=

That (iv) does not

That (iii) does not imply

xn - xn-1 and applying Theorem 9. That (ii) does not imply (i) is shown by taking (Ax)n = xn/n. It (Ax)n

(ii) is shown by taking

=

obeys (ii) by Theorem 9; it is multiplicative 0 hence not perfect ( 3 . 3 . 5 ) and so by (iv) it does not even have

If coregular is omitted in Example 5 an with easier calculations. (Note, Xf 14. EXAMPLE. Xf # X'.

X

Let

f

E

=

(tA)k

=

f = 0 on

X I ,

so that

of Type M) and

A

Q , then

1

(Ax)n f(x)

=

example can be given

is defined in 7.2.3). with =

AD

xn - x

t(Ax)

AK, a n d

a n d not

~ -(xo~= 0).

(4.4.2)

This easily implies that

and t

=

If

0 = f(6 0

(A

k

)

is

follows by the Hahn-Banach theorem (3.0.1).

Now, by Theorem 9, 1 # that

X

space

cA where

tAA. AD

BK

0

AD.

C

gk

in

cA, hence in

X

by 4.3.14 - note

X. Thus X

does not have AK. It turns out (10.6.3) that this is enough to imply X f f X B ; with o u r present state of E

knowledge this requires a construction: let t

E

1

so that k

uk = f(6 by 1 . 2 . 9 .

).

lim t,log

Then

u E

n

yn

=

log(n+l), choose

does not exist and set f(x) = t(Ax), n n X f b B since 1 ukyk = 1 tk(Ay)k - t,+lyn k=1 k=l

5.3

81

CONSISTENCY

5.3.

T h i s s u b j e c t i s c o n t i n u e d from S e c t i o n 3 . 3 i n w h i c h i t was treated f o r coregular triangles.

W e f i r s t see t h a t Type M d r o p s

o u t of t h e p i c t u r e f o r g e n e r a l c o r e g u l a r matrices.

i s shown t r i v i a l l y by t a k i n g

c o r e g u l a r m a t r i x n e e d n o t be o f t y p e M t o be t h e i d e n t i t y m a t r i x e x c e p t t h a t

A

That a p e r f e c t

= 0.

all

The f a i l u r e o f

t h e c o n v e r s e i s shown i n t h e n e x t example. 1. for

EXAMPLE.

If

ann

k < n , and i f

However

is a s u p e r d i a g o n a l m a t r i x

A

# 0

for all

n e e d n o t be p e r f e c t .

A

Tauberian matrix given i n 1.8.6. closed.

Of c o u r s e

a

i.e.

= 0 nk i s t r i v i a l l y o f Type M.

n, A

F o r example l e t

A

be t h e

c

I t f o l l o w s from 4 . 5 . 3 t h a t

is

is c o r e g u l a r ; i n d e e d a m u l t i p l e o f a r e g u l a r

A

matrix. 2.

THEOREM.

valent

lil A

Let

on

B

potent matrix

c; l i i l A

such t h a t

( i i i ) implies ( i ) :

limg

and

on

a r r a n g e t h a t l.~

necessary. of

B

=

3.

+

f.

A

c ; ( i i i )A

on

i s perfect. limA

with

is n o t p e r f e c t , t h e Hahn-Banach g = 0

on

g

C,

1 i n some r e p r e s e n t a t i o n o f p-unique;

f = limA

Let

A

If

g E c;1

is n o t

-

g.

# 0. g.

otherwise multiply

#

Then

B

0

W e can also

T h i s is a u t o g

by

if

2

i n some r e p r e s e n t a t i o n

equiwotent with

A,

T h i s shows t h a t ( i i ) i s f a l s e . THEOREM.

stronger matrix with

lim, = limA

and so, b y 5 . 1 . 1 , t h e r e e x i s t s

f

lim

i s consistent with every equi-

CA'

theorem s u p p l i e s

A

such

B

T h i s f o l l o w s from t h e c o n t i n u i t y o f

( i i ) implies ( i i i ) :

matic i f

The f o l l o w i n g c o n d i t i o n s a r e e q u i -

i s consistent with every stronger matrix

1imB = limA

that

r.

A E

on

cA

Let B

n

A

be a coreguZar m a t r i x .

such t h a t gm.

limg = 1imA

on

c

Then e v e r y i s consistent

5.3-5.4

82

This is immediate from 4.6.7. Theorem 2 is more general than 3.3.7 in that the latter is

- but

restricted to coregular matrices gives an internal test (Type M).

less specific in that 3.3.7

A remedy for this is suggested in

P, Chapter 15.

the introduction of

REVERSIBLE MATRICES

5.4.

1. DEFINITION. y c c

A matrix

there i s a unique

x

A

i s eaZled reverisibZe i f f o r e a c h

such t h a t

Ax = y .

For example, a triangle is reversible.

A matrix may have a

(two-sided) inverse without being reversible, for example the matrix

A

in 1.8.6 is not one to one.

all

2.

EXAMPLE.

Let

u,v be sequences with

n.

(For a first reading take u

=

un

+

1, vn

v = 1.) Let

(Ax),

+

for

0 =

unxl

+

vkxk; A is one t o one. Also, if y c c, let x1 = lim y , k=n+l xn = C(un-u n-1 )/vnllim Y + (yn-l-yn)/vn for n > 1. (Convention: uo = 0). Then

3. THEOREM.

II.IIA

with

+

so

Ax = y Let

A

A

is reversible.

be r e v e r s i b l e .

and t h e g e n e r a 2 f o r m of

t(Ax), t

f

E

Then eA

is

eA

is a

space

BK

f f x ) = p l i mA x

a.

6

As in 53.2 it is seen that

cA

is a Banach space with the dual

representation shown. The fact that coordinates are continuous lies much deeper. 4.

If

f

determined. 4.4.3 with

if

A

It follows from 4.3.2. is represented as in Theorem 3,

However a

f

is uniquely

will always have representations as in

# 0 and sometimes with different values of

is a triangle!

(4.4.4)

p,

even

5.4

83

Reversible matrices first rose in the problem of solving infinite systems of linear equations.

The existence and form of the

solution are shown as follows: Let

THEOREM.

5.

unique r i g h t inverse

a sequence

b

be a r e v e r s i b l e m a t r i x .

A

The r o w s o f

B.

B

unique s o l u t i o n

+

x = b lim y

belong t o

y = Ax

such t h a t t h e equation

Then

T h e r e is

9..

has, f o r

y E c, the

By.

Applying Theorem 3 to the coordinates we have, for y

and

=

setting

Ax, x

n

+ tn(Ax)

lim x n A

= p

bn - p n , B

=

has a

A

pnlim y +

=

tnkyk.

C

x

c

E

A

Now

we have all the theorem except that

(tnk)

B

is a right inverse. We see this by taking y = g k ; then xn k k The equation y = Ax becomes 'n - (B6 )n = bnk (i.e. tnk). = c . a .x. = c . a . b i.e. AB = I. 1 ni 1 1 ni ik Example 2 shows that shown by M.S. MacPhail.)

b

may be unbounded.

The next result shows the futility of

trying to modify Example 2 so as to make 6.

p-unique

THEOREM. ie.g.

A

Let

(This was first

A

regular o r row finite

be a revers-LbZe m a t r i x which i s e i t h e r

or row-finite.

c o r e g u l a r by 4 . 4 . 8 1 ,

Then

b

=

0

in

T h e o r e m 5.

If

A

is

Theorem 5 are

A =

p-unique, the numbers since

0

is row-finite, let (Ab), + [A(Bl)],

and so 7.

Ab

Hence

= 0.

B,

y

b

Let

b A

has 1

=

=

p =

Ax, then from Theorem 5 , 1 = (Ax),

+ [(AB)l],

= (Ab),

THEOREM.

coZumns a n d

xn

=

occuring in the proof of n t = 0, a = 6 in 4.4.3. If pn

by 1.4.4. This is

0. be a r e v e r s i b l e m a t r i x w i t h convergent

a s i n T h e o r e m 5.

Then

dnk

=

x

From Theorem 5 , x = b lim Ax + B(Ax) gives the result. dk

=

(Ab)n+l

BA

=

I

-

where

D

b n a k , ak = l i m n a n k '

for each

x

6

cA.

Taking

84

5.4

8. It follows that columns.

If

B = A-l

b = 0 A-l

if either

A

=

A

A

has null columns.

Theorem 11, there is given a multiplicative (hence B

0 or

=

has null

gives the inverse transformation by

Theorem 5 , but this is not so if

-1

b

as above) with

b

+

0

In [83],

re-Jersible matrix

A-'

0 and so

does not give

the inverse transformation. 9.

B = A

If

-1

.

A

is coregular or row-finite with convergent columns,

These follow from Theorems 6 , 7. With perserverence one

can squeeze out a little more: columns, BA

is row-finite with bounded

A(BA) = (AB)A = A

exists and

10. EXAMPLE.

A

if

(1.4.4) so B = A-I.

The simplest case, u = v = 1, of Example 2

shows a reversible matrix with no left inverse.

11. THEOREM.

Let

A

be c o r e g u l a r and r e v e r s i b l e .

Then

A

i s p e r f e c t i f f i t i s of t y p e M .

If

f = 0

on

c

we have

f(x)

=

t(Ax)

with

This follows from Theorem 3 and 4.4.7 along with

k.

all

S o if

A

is of Type M , f

Hahn-Banach theorem. =

t(Ax).

choose

Then

x

f = 0

such that

=

0

Conversely, let

and tA

=

A

t f(6

Q

k

tA

R,

=

0.

0 for

) =

is perfect, by the

0, t E R .

Let

c (1.4.4) hence f = 0. Fix Ax = 6 n . Then 0 = f(x) = tn i.e. on

f(x) n

and

t = 0.

With 5.3.2 this gives a consistency theorem identical with that proved for triangles.

12. REMARK

(M.S. MacPhail).

of significance of weak if

p

(3.3.7) We can now point out the lack

p-uniqueness.

is uniquely determined by

f

Call

A

weakly

in 4.4.3 with

r o w - f i n i t e r e v e r s i b l e m a t r i x need not be a triangle, (4.4.4), but must b e w e a k l y

CY E

p-unique u8

A'

A

p-unique even i f it is p-unique.

[ I f not-there

5.4-5.5

exist

t

L, a

E

with

@

E.

m

1

and

y

o

Then

c,

y

ax

xi

y = Ax, we have

x

i

=.I

=

1 yields a contradiction.]

=

0. Each

1 ukyk where uk aibik. Hence for k=l 1=1 = lim y + ty + uy. Setting y = g k gives t + u

bikyk E

=

m

m

k= 1

all

limAx + t(Ax) + ax

by Theorems 5, 6 so, setting

= [B(Ax)li =

85

fails without row-finiteness.)

limB

=

so

0

B

0.

(Example 2 shows that this

It follows that weak

i s n o t i n v a r i a n t by considering 5.2.7. There

cA = cB, and

=

is not weakly

A

u-uniqueness

is a triangle,

u-unique.

5.5. ROW-FINITE AND ONE TO ONE

1. THEOREM.

Let

a c Z o s e d s u b s e t of

First, A

be a r o w - f i n i t e m a t r i x .

A

Then

A[w]

i s

w.

is continuous by 4.2.8. The result is that given

in 5.0.3. An elementary proof is given in [79] #6.4.28.

If

2. COROLLARY. y = Ax

tion

i s r o w - f i n i t e and r e v e r s i b l e t h e equa-

can be s o l v e d f o r e v e r y

A[w]

For

A

c

3

which is dense in

Row-finite cannot be dropped. that

yn

x1

-+

y.

i.e.

y

W.

In 5.4.2 if

y

=

Ax

it follows

c.

E

Row-finite one to one matrices behave like reversible matrices in that they obey analogoues of 5.4.3 and 5.4.11. 3.

is a f(x)

space w i t h

BK =

Let

Let

THEOREM.

ZirnAx

Y

space with

=

c

+

II.IIA

be r o w - f i n i t e a n d one t o o n e . and e v e r y

f

G

cA

Then

cA

can b e w r i t t e n

t ( A x ) , t E 1.

n

A[w]

then by Theorem 1 and 4.2.15

(llallm,p) where

all the seminorms

/I .Ilm .

A

p

pn(x) = lxnl.

are redundant so

Y

Since is a

BK

Y

is an

FK

lxnl 2 IIxllm space with

This could also be seen by checking directly that

(Y,TW)

5.5-5.6

86

is c l o s e d i n

c , hence a f o r t i o r i

A-l[Y]

= cA = Z ) .

y E Y , y = Ax.

Extending

Take

f i r s t part. g(y) = f(x)

for

c.

is c l o s e d i n

W e now

i s o n e t o o n e a n d o n t o and 4.3.2 y i e l d s t h e

cA -+ Y

A

have t h a t

(Y,Tc)

X = w,

If

f E ci, let

t o a l l of

g

f(x) = g(y) = p l i m y

t h e Hahn-Banach t h e o r e m g i v e s

+ ty

c

by limAx

= p

+ t(Ax). 4.

A r o w - f i n i t e one t o one c o r e g u l a r m a t r i x i s p e r -

THEOREM.

f e c t i f i t i s of t y p e M .

The c o n v e r s e i s f a l s e .

T h i s i s p r o v e d i n t h e same way a s 5.4.11.

For t h e c o n v e r s e ,

modify t h e i d e n t i t y m a t r i x by a d d i n g a row o f z e r o s on t o p o r rep e a t i n g e a c h row or some s u c h . 5. A

Let

THEOREM ( H . SKERRY).

hue a g r o w t h s e q u e n c e i f f Necessity:

If

A'

A'

be a r o w - f i n i t e m a t r i x .

A

= {x:Ax = 0 )

Sufficiency:

The r e s u l t f o l l o w s s i n c e

W e may assume

A

f i n i t e l y many f i n i t e rows on t o p o f

cA is a

BK

A : w + w

i s con-

cA 3 A'.

is one t o one s i n c e p l a c i n g A

does not alter

cA.

Then

s p a c e by Theorem 3 a n d t h e r e s u l t f o l l o w s by 4.2.11.

I t is n o t known w h e t h e r s u f f i c i e n c y h o l d s i n g e n e r a l .

6.

Necessity f a i l s s i n c e , f o r example, i f

cA = cs

has f i n i t e dimension.

h a s i n f i n i t e d i m e n s i o n i t h a s no g r o w t h

s e q u e n c e b y 4.2.11, 4.2.12 and t h e f a c t t h a t t i n u o u s (4.2.8).

Then

which h a s

alk = 1, ank = 0

1 as a g r o w t h s e q u e n c e .

for

n > 1

Of c o u r s e a m a t r i x

must h a v e a g r o w t h s e q u e n c e i f i t h a s a row w i t h no zeros i n i t .

5.6.

BOUNDED CONSISTENCY

Various subspace of an

FK

space, called distinguished, w i l l

be s t u d i e d s y s t e m a t i c a l l y b e g i n n i n g i n C h a p t e r 1 0 .

t h e space

W

which it w i l l be u s e f u l t o d e f i n e now.

One o f them i s

5.6

DEFINITION.

1.

be an

X

Let

87

FK

space

3

Then

$.

W = W(X)

= { x E X : x = Z x 6 k , c o n v e r g e n c e i n t h e zJeak t o p o l o g y o f k m W = W n 1 I f A i s a matrix W I A ) = W(cA). b

XI.

Also

.

Equivalently: x

E

W

x(~)

iff

section of

x.

-+

W

x x

iff

f(x)

xkf(6 k )

= C

weakly, where

standing for schwach AK

(=

x

A conservative space is conull iff

3.

DEFINITION.

ak = l i m ank.

as

nth

has

x

SAK,

1

E

W.

be a m a t r i x w i t h c o n v e r g e n t columns,

A

-

A(x1 = l i m x A

Then

W

E

X';

E

weak AK).

2.

Let

f

is, as usual, the

x(~)

It is customary to write

for all

ax

x

for

E

aB

n

cA.

4. The definition has been given in more generality than is required now.

Its ramifications will be explored beginning in

§13.2. For now i t w i l l b e assumed t h a t i s d e f i n e d on

5.

cA

n

.E

m

; for

In particular

a

E

I

A

E

(1.3.7)

A(1) = x(A)

A

so

r

i n which case

and so

aB

3

k-

is conull iff

A

.

A(1)

=

0.

(With the assumption of Remark 4 . ) 6. LEMMA.

Let

A

E

r, x

E

cA

n

c3.

1

.

Then

x

E

W

iff

A(xj

= 0.

If

x

E

W , apply the definition of W

to

f = limA; thus

limAx = c xk limA gk = ax. Conversely if ~ ( x )= 0 , let f E ci. Then by 4 . 4 . 9 , f(x) = p limA x + yx = pax + yx by hypothesis. Also Z

xkf(6 k ) = C xk (uak+y k ) = f(x), hence

x

E

W.

Attachment was introduced in 4.3.4; for matrices it takes this form: 7. DEFINITION. matrix

A.z

i s

Let

(ankzk).

z

be a s e q u e n c e , A

a matrix.

The

5.6

88

Y A e Z = z -1 .YA for any space

Thus

LEMMA.

8.

i f

Let

r,

A E

z E cA

n

Y. Ilm. T h e n

i f and o n l y

z E W

i s conull.

A-z

This follows from the fact that

= A(z),

x(A.2)

with Lemma 6.

This result is given a more natural setting in 12.1.6. 9.

LEMMA.

If

z E Wb(A),

Let

A,B

A.z

E

r

with

m

n

II

c cB.

Then W b ( A ) c W ( B ) .

is conull by Lemma 8 and so

z o W(B)

null by 3.5.4. Hence

cA

B.z

is co-

by Lemma 8.

We can now give a proof of the famous M a z u r - O r Z i c z Bounded About 5 proofs of this theorem are in the

C o n s i s t e n c y Theorem.

The

literature. The simplest one, given here, is from [ 7 0 ] .

result strengthens 3.3.6 (ii), which has the same conclusion in that its hypothesis is weaker. 10. THEOREM. c cB.

Then

A,

Let

A,

be r e g u l a r m a t r i c e s w i t h

B

cA

n

m

Q

a r e c o n s i s t e n t f o r bounded s e q u e n c e s .

B

We prove instead a more general result which similarly improves 3.3.7 (ii). 11. THEOREM,

Let

A,

c c B and

l i mA x = l i mB x f o r bounded s e q u e n c e s .

limBz

=

bz

and

CASE 11: Remark 5). limB 1

=

A,(z)

Then

limAl

x E c.

Then

bk = limBsk = limA6 =

limA z Let

for

k

Note first that CASE I:

be coregular matrices w i t h

B

y

0. Then = = z

az.

=

B

ak.

n

Ilm

are consistent

Let

z E cA

n

m

II

.

by Lemmas 9 and 6. Thus B These are equal as just mentioned. A (z) = 0

- tl where t

hA(y) = 0

A,

cA

so

it follows that

=

hA(z)/hA(l).

lirnB y = lirnA y limBz = limA z .

(See

by Case I.

Since

The failure of this result for unbounded sequences is demonstrated in [26].

89

CHAPTER 6 BIGNESS THEOREMS

FUNCTIONAL ANALYSIS

6.0.

1. = x

then

n'

[SO],

space

X

must be s e p a r a b l e :

is t o t a l o v e r

X

hence fundamental i n

A reflexive

{Pnl

BK

hence t h e l a t t e r s p a c e is s e p a r a b l e .

Theorem 8-1-10;

t h e weak

*

and weak t o p o l o g i e s c o i n c i d e

h e n c e norm s e p a r a b l e

Let

[SO],

X

(X',W*) Since

is weakly s e p a r a b l e ,

X'

so

#8-3-103;

Pn(x)

is s e p a r a b l e ,

[80],

#9-5-4.

2. Then

Let

X

be a s e p a r a b l e

[SO], Theorem 9-1-12,

= IIx-yIl

space,

the unit disc i n

D

#9-5-1.

The metric f o r a n where

FK

space

IIx11 = Z 2-"pn(x)/

W e s h a l l prove t h a t

(X,p)

[l+pn(x)].

11txll 2 IIxI1

if

i s g i v e n by

M u l t i p l y i n g by

P

Let

2-"

X

of seminorms.

convergent:

d(x,y)

[SO], Theorem 2-1-2.

It1 5 1 :

if

p

p ( t x ) / 1 1 + ~ ( t ~ =) i P+U r ~ t i ~ ( ~ ) i ) - l

4.

X'.

i s m e t r i z a b l e and c o m p a c t , h e n c e s e q u e n t i a l l y c o m p a c t .

(D,W*)

3.

BK

i s a seminorm

= p(x)/ rl+P(X)i.

and summing g i v e s t h e r e s u l t .

/

b e a F r e c h e t s p a c e w i t h t o p o l o g y g i v e n by a s e t If

f o r if

Z p(x")

U

u2, i = 2;. . . ; x ,vn with E

c

0'

E =

1/8, u1 = 2,

= 1/16, u2 = v1

E

=

E

2-n-2, un

+ 2,

vn-1 + 2, vn > Un' i = n. Now let x = Z xn E X. The series converges since f o r each i , qi(x n 1 5 qn(x n ) < 2-n-2 as soon as n 1. i and so C qi(x n ) is n since X is an FK space. convergent. (6.0.4). Also xk = Cnxk =

We have arranged the notation so that the check that

x

is

bounded and divergent is simply the last part of the proof of 3.4.4, the line (1) being the same in both places, as well as the choice of xn ,vn; the extra condition (i=n) being irrelevant. We can now incorporate 3.5.4 into the A complete extension would be:

partially.

X

=I

Y

Y CI

= W,

X

then

Qm

X = Q

m

.

FK program but only if

Y

is conull and

is conull; but this is false as shown by

In spite of this a complete extension will be

given in 16.2.7.

2. CA

=I

X

n

Let

THEOREM. La.

Then

Let

b e conu'l'l and

X

A

a matrix with

i s conul'l.

A

B = A - x(A)I

so that

B

is conull.

Hence

cB fl X

is

conull (4.6.6)and so, by Theorem 1, contains a bounded divergent x.

x(A)

By hypothesis

x

E

cA, then

X(A)x

Ax

- Bx

and

SO

i s T a u b e r i a n iff

c

is

Conversely, suppose that

c

is

=

E

c

= 0.

3.

cZosed i n

THEOREM.

Let

A E

r.

4.6.7.

A

cA.

Necessity is by Theorem 1. closed.

Then

Then

A

is coregular (4.6.4) and the result follows by

92

6.1

4.

Theorem 3 was g i v e n f o r t r i a n g l e s i n 3 . 4 . 4 .

X = c 0 z

program f a i l s ; f o r e x a m p l e , l e t

i s bounded

z

where

FK

The

and d i v e r g e n t ( 4 . 5 . 5 ) . The n e x t r e s u l t i m p r o v e s 1 . 8 . 4 :

5.

COROLLARY (J. C o p p i n g ) .

which has a l e f t i n v e r s e If

For

X 6

5 I I B ] ~ . [ I A ~ ~ * I I x ~Thus ~,.

c

x

i s Tauberian.

A

cA

IIBll*IIxll~

( 4 . 2 . 4 ) and

cA p r o v i n g t h a t

E

5

A

is

i s r o w - f i n i t e t h e same argument

B

If

may b e a p p l i e d t o e v e r y

Then

= IIB(Ax)llm ( 1 * 4 . 4 )

is c l o s e d i n

T a u b e r i a n , by Theorem 3.

be a c o n s e r v a t i v e m a t r i x

i s Mercerian.

II(BA)xll,

IIXllm =

C,

A

of f i n i t e norm.

B

i s also row-finite, A

B

Let

is bounded.

x

The

r e s u l t f o l l o w s by 4 . 6 . 8 . T h i s r e s u l t is p l a c e d i n a n o t h e r c o n t e x t and g i v e n a c o n v e r s e

i n 18.1.7, 18.1.10, 6. B

with

IIBll

Let and

Suppose t h a t

COROLLARY.

x

-.

< 6

c.

B

E

r.

Bx

E

cA n L m

Then Then

A E

IIBxll, 5 IIBII -IIxllrn. So

Bx

B

c

r

has a two-sided i n v e r s e

since

A(Bx) = x E c

(1.4.4)

by C o r o l l a r y 5.

( I made a n i c e l i t t l e p r o b l e m o u t o f t h i s

-

it appears

as #6414 i n t h e December 1982 American M a t h e m a t i c a l M o n t h l y . ) 7.

Let

LEMMA.

A,B

E

r

with

B

conuZl.

Then

AB

i s

conu Z l . If

x

is b o u n d e d , (AB)x = A(Bx)

( 1 . 4 . 4 ) , so

cAB3 cB n L a

The r e s u l t f o l l o w s by Theorem 2 . 8.

Let

COROLLARY.

M = B

-

0 = X(AM) = X(AB)

For

X(B)I.

-

A,B

Then

X(B)X(A).

E

r, AM

XfABl = XfAIXfBl.

i s c o n u l l by Lemma 7 s o

6.1-6.2

Let

9. COROLLARY.

A,B

r

E

93

with

conull.

B

Then

and

AB

a r e conull.

BA

By Corollary 8. SOURCES :

[141, [48Al, r831.

W.

6.2. TWO-NORM CONVERGENCE AND

The concept of two-norm convergence will be introduced with more generality than necessary. reader may take

p

phrases

-I’

”p(x)

1,

96

6.3

3.

COROLLARY.

i s a conull c o n s e r v a t i v e 0

In Theorem 2 , cA closed subspace of

n(r)

Each

is conull, cA

cA

since it is

includes limi.

and

@

space.

BK

and is a

1

The result follows by

4.6.3. 4.

COROLLARY.

"(r)

Each

i s a c l o s e d s u b s p a c e of

Lm

By Theorem 2 and 1.3.11. Suppose t h a t

5. LEMMA.

x E nb(r)

there e x i s t s

i s given.

r

such t h a t

xr

=

Then f o r each

y

Lm

6

Yn.

Zn

It remains to complete the definition of part given in the statement.

for

rk 2 i

k

goes from

xr k 2" to

r k+l'

Then, for

2" < k

as

to Yr+1

Yr

P n - l

k = 1,2,

...

9

Now l e t

y

and

so t h a t

E

Pl'P2'.

IIbpnll < 2-"

and choose

'n-1

IIb p1 11

* .

5.

,Pn-l

so t h a t

a n d w e s h a l l show t h a t

n(r)

have been chosen.

Ib:nl

and

n-1

Inductively

< 2-"

= 1

bn : y

E

for

+

X

Lrn.

for k

co,

completing t h e proof. F i r s t let

z"

= b

Pn

, tn

-

= yr

yr

n T h i s series converges by 3.0.5 s i n c e

IIznll

1).

(C,l), (Bz4)n = n-' m

(BTz P Ik

0

See

"j01.

1

it

8.6

138

FUNCTIONAL DUAL

8.6.

There is an inclusion theorem of a much simpler and more easily applicable form than the ones in 88.2. for

It shows the condition, e.g.

co, in a more transparent form. Recall the functional dual

tion, when it is used, that

X

given in 7.2.3 and the assump-

Xf

3

$.

THEOREM (A.K. Snyder and A. Wilansky).

1.

Then f o r any

space.

space

FK

we have

Y

Y

Let

be a n

i f f Yf

X

3

X

AD

Xf .

C

Necessity is by 7.2.6. To prove sufficiency we apply 8.2.3. Let

B

be a subset of

is bounded in

Y

which is bounded in

$

To this end let

exists g

6

on

$,

B

it is sufficient to show it to be weakly bounded

(8.0.2).

X'

To show that

X.

f

Y'. The hypothesis says that there k k g ( 6 ) = f(6 ) for all k , hence g = f

such that

in particular on

B.

6

Thus

f b] = g[B]

which is, by

assumption, a bounded set of scalars. 2.

1

(f(sk)l

for a22

to as:

x

EXAMPLE. C

-

for all

iff

f c Y'.

The

f c Y'.

k {6 1

co

3

first

yf c c!

Thus

=

A

condition

a

(co:Y)

6

i . e . iff

(7.3.1)

is

is weakZy a b s o l u t e l y summable.

{f(a

i f f

k

I}

6

11

referred See [49] 81.2.

It is fairly easy to show directly the equivalence between this result and that of 8.2.7. An easy remark is that any w e a k l y s e q u e n t i a l Z y c o m p l e t e l e . g . r e f t e x i v e ) space Em

Y

3

co

must a l s o s a t i s f y

Y

3

am.

(Hence since

is not weakly sequentially complete, the inclusion must be

proper.)

For

x

6

am

implies that

1 xk6k

is weakly Cauchy in Y,

hence weakly convergent. 3.

EXAMPLE. Y

is weakly bounded.

2

a

iff

yf c af = a m , (7.3.1) i.e. iff

{&n~

Here "weakly" may be omitted (8.0.2) and so

8.6

139

the result of 8.2.5 is reached again. {g(a k ) I 6 11for all g 6 Y ' . EXAMPLE.

4.

condition is:

Y 2 bvo

for each

Yf c bvof

iff

f

Y'

6

Also

A

( k : Y ) iff

6

bs, (7.3.5). This

=

M

there exists

such that

M

>

1

f(sk)l = f(l(m)), i.e. Il(m)} is (weakly) bounded. As in k=1 Example 3 we have the result of 8.2.6 again. Another important formulation is

Y

lar to this is the result belong t o

A

C g f ak i )

and

Y)

{f(6")1

if

bvo

3

6

for all

bs

6

Simi-

Y'.

iff ( t h e c o l u m n s of

(buo:Y)

6

f

f o r each

bs

6.

g

This is from

Y'.

E

A

8.4.2. and the fact that weakly bounded i s the same as bounded. (8.0.2).

Y

EXAMPLE.

5.

This condition is: (Convention: ' 6

for all

f

(p-'+q-' E

Y f c csf = bv, (8.3.5).

if and only if

1 f(sk) -

,

f

6

i.e.

1

If(dk)(

3 kp

(p > 1)

= I)

in

Y'

Y.

f(sk-l)l < -. where dk = 6 k - &k-1

-

1 - 2~

x

with

IIxII

=

yields

m

THEOREM.

X

be a

l u x ) 5 IIuIl.IIxl1.

f c X'

AZSO

with

5 IIfll .IIxI]. Taking the

1I~(~)-xll~~ > 1 - 2~ which

and yields

IIU(~)~~ by

Let

1,

su~llu(~)II 2 1. Con-

m, llu(m)-~llcs 5 I1u.xIIcs; taking

and any

cs

For any

there exists

2 ~ hence ,

>

is to replace each 6.

Then

11 xI

IIuIIB = 1

m.

By definition this implies the existence of

E.

i=l implies that

versely for

This is an increasing function of

inf

bs. space

BK

f o r each

f(6

The only change for

5 11uII B .

k

) =

uk

x

6

3

$,

$,

x

6

$

and

u 6 X

IlxII = s u p ~ l J u z l : I ) ~ 5l 1 3 .

we have

lux) = If(x)(

over all such f yields the first f result by definition of the norm on X . (10.0.1). To prove the

f.

10.1-10.2

160

second formula, let

f

Hahn-Banach theorem.

X'

E

Let

IIuII

tient norm (10.0.1)

with

\If11 = 1, f(x) = IIxII, using the k uk = f(6 ) . By defin tion of the quo-

5 1; also

lux1

(f(x)

=

=

IIxII.

This,

together with the first result proves the asserted equality.

A useful inclusion relation for multipliers follows from this. Let

M(X)

=

M(X,X), the latter symbol having been defined in 4.3.15.

7. LEMMA.

Let Since

u

6

Let

M(X),

v

k

f(6 ) = ukvk

shown that

f

be a

X

Xf .

6

BK

space

For

x

+

6

u.v

xI

where

Xf

6

X

=

C

M(XfI.

1 ukvkxk.

when it is

by the Hahn-Banach

This continuity holds since

(Theorem 6 ) 5 IIvII *IIuII-11

f(x)

MfX)

with the relative topology of

QI

X, since f can be extended to all of theorem (3.0.1).

define

it will follow that

is continuous on

Then

$.

2

IIuII

If(x)l

is computed in

5 IIvII .IIu.xII

M(X)

as

in 4.3.15.

10.2.

DISTINGUISHED SUBSPACES. The subspaces given here have proved to be of value in dis-

cussing the fine structure of matrices.

FK

spaces and the properties of

Early in the 20th century various classical results

were associated with boundedness properties of matrices. 513.4).

When placed in abstract form the property was seen to be

equivalent to sectional boundedness, called beschrankte.

AB

for Abschnitts-

(See 58.0 for bounded sets.)

1. DEFINITION. i f

(See

An

FK

space

{z'~)} i s a b o u n d e d s e t i n

It turned out that, for properties such as

S

0 1 =

F

X

3

+

for e a c h

X

X

=

cA, AB

=

B

=

X

i s s a i d t o have

x

X.

is equivalent to other

where

S, F, B

the distinguished subspaces soon to be defined. referred to are given in 13.3.1.

6

AB

are some of

The results

A natural process, then, is to

10.2

consider spaces without

AB

161

and investigate the subspaces on their

own merits. These investigations show the role of the subspaces in situations in which they do not satisfy the equations just given. Examples to illustrate the possibilities will be given in later An example of a non-AB space occurs in 7.2.5.

sections. llu(m)ll

~~u(m)~~ since m

=

u ( ~ ) Q co.

We begin with the smallest.

There

This is unbounded.

S stands for strong

The letter

(convergence) :

x 1 = {z: z

= Ir: x ( ~ +)

be an

X

Let

DEFINITION.

2.

has

in

AK

space

FK

XI

4.

3

S = S(X)

1 xk 6k 1.

{x: x =

=

Then

I f

A

a matrix, S l A ) = S ( c A ) .

7:s

X

Thus km,

S

=

is an

.

c

0

AK

Of course

S

space (4.2.13) iff

S

2

4

{x

E

X: f(x)

always.

X.

=

Sc X

Also

or

X = c

For

X

since

is

complete.

W

The subspace

=

=

1 xkf(6 k )

for all

f

E XI)

has been introduced earlier (5.6.1), and characterized in terms of two-norm convergence for conservative and vsc spaces. (6.2.6 ,9.2.10). It was also used to prove the Bounded Consistency Theorem 5.6.10. The letter 3.

+ F (XI E

=

F(A)

=

stands for weak (convergence).

DEFINITION. L e t X b e a n F K s p a c e ( n 1 is w e a k l y Cauchy in X ) { z E w: { z

for

cs

W

irZZ

f

6

X 1.

AZso

F c X.

If

X

=

co,

functional (convergence) since is convergent for all z

F

X.

1

F \F.

If

4.

=

12 6

A

Then

F+ =

o:{znftsn)l

is a m a t r i x

F(cA).

Thus

as

+

F =

2

has

FAK

f

E

X'.

6

z

6

+ +

F

F

The letter if and only i f

It is customary to write

i.e. functional

AK.

(Compare AK

and

stands for {f(z("))}

z

E

F

+

SAK.)

10.2

162

DEFINITION. Let

4.

+ B IX)

{zln')

= { z 6 w:

= IZ c w:

{znfisn)l

6

be an

X

FK

is b o u n d e d in

bs

space

Then

B+ =

XI

f c X'I.

for a l l

4.

2

A Z S ~

B = B

+ n x.

If

is a matrix, B l A l = B l c A l .

A

B c X.

Thus

The two definitions are equivalent since

m

1

znf(6") = f(z(m)),

bounded and weakly bounded being the same

n=l (8.0.2. )

5. EXAMPLE. = c

X

For

c , F+

+

-

m

B

=

=

a , F = B = c , + = S = W

0'

By Definition 1, X

is an

AB

space iff

B+ = a , a proper subspace of

7.1.2 has =

=

B = X.

X; 1 $ B

The space of

since 111'")1]

n. 6. THEOREM.

wc

F C

B C

x

and

b e an

X

Let

FK

space

2

4.

4 c

Then

S

c

wcT.

gc s c

Wc

The only non-trivial part is g.

A glance at the definition of W

on

W.

T.

Let

f

XI, f

6

=

0 on

just given shows that

f

=

0

The Hahn-Banach theorem (3.0.1) gives the result.

Example 5 shows that

F

6

need not be included in

These ideas are intimately bound with the ideas of Chapter 9: inclusions of

bvo, bv, sc and vsc

7. THEOREM. L e t (i)

x

3

6

B+

z

6

B

iff

z

-1

.X

2

FK

variational semiconservative): space

3

+,

then

z c w,

b v o , in particular

I c B+

i f f

boo,

(iil (iii)

X

z

be an

X

(=

is s c ,

z 6 F+

iff iff

2-I.X z

-1

2

.X

b v , in particular

I

is s c , in particular

6

B

iff X 3 bv,

I c F+

i f f

163

10.2

(iv)

-1

z

6

F

iff

z

6

GI

iff z-'*x

S

iff

z

is

.X

in particuZar

VSC,

1

iff

F

6

X

is use, (v)

X

I

is conulz, in particuzar

w

6

iff

is conuZ2. (vi) z

1

6

6

iff X

S

Let

f

(z-'*X)'.

6

+ g(Z n 6 ) = an + Zng(6 z

6

is s t r o n g Z y conuZZ; in particuZar

.X

is strongly conuZZ.

n

iff

-1

2

B+, F+

By 4.4.10, f(6'")) n

), a

$,

6

respectively.

g

6

X'.

+ g(z-6n )

= an

Thus

{f(Sn)}

6

= a

n

bs, cs

Parts (i), (iii) follow from 8.6.4

and 9.2.1. If z

-1

z

6

bv.

(i), z

6

B

z

then

so

X

6

Conversely

1

6

bv

and so by (i), -1 z .X i.e. z 6 X.

1 6 2-l.X

so

1

C

By

B.

Part (iv) follows from (iii) in the same way. Each condition in Part (v) implies that f

6

(z-1.X)', f(l-l(n))

proof of (i) (iii).

g(z.l-z.l("))

=

z

X

6

= g(z-z("))

as in the

The result is immediate from 5.6.1, 9.3.1.

For (vi) use 4.3.6 to obtain the seminorms of ~ ~ ( 1 - 1 ' ~ )=) (l-l(m))n = g(z-z

(m)).

so if

=

0 for

m > n

while

2-l.X.

First

g[~.(l-l(~))]

The result is clear from this.

The first three conditions of 5.2.13 are

S

=

X, 1 6 S, 1 6 W ;

6.5.2 is a special case of Theorem 7 (v). 8. Theorem 7 suggests definitions of "distinguished subspaces" -1 at will by choosing the set of z such that z X includes c

.

or

co

or is conservative and conull etc.

will not be used later) let From 8.6.2 it follows that

C+ = { z :

C+ = Xfa

A s one example (which

z-l-X3 co), C

=

C

+n

X.

(4.3.17) and one can improve

6.5.1 (replacing c by co for convenience) as follows. Let X 2 co, then z -1 .X 3 co iff z 6 Xfa.

10.2

164

In view of Remark 8, Theorem 7 (iv) is seen to be the analogue of 6.5.1. The subspaces mentioned in Remark 8 could well merit serious study

- however

the subspaces S , W , F, B

arose independently of

the results of Theorem 7 at an early stage of the development of summability through functional analysis.

Semiconservative spaces

also arose independently as a natural home for some of the conservative theory.

Only later, when both studies were established, was

the simple link given in Theorem 7 observed. THEOREM.

9.

if

X c Y

then

also for

T h e d i s t i n g u i s h e d s u b s p a c e s a r e m o n o t o n e i.e.

E(XI

ElY)

C

The inclusion map x ( ~ + ) x

W

S. For

X

in

E = S,W, F , F

+, B , B + .

This holds

c ely+.

i.e.

E =

where

i: X

+

Y

is continuous (4.2.4), so

Y. This is the assertion f o r

implies the same in

it follows from the fact that

i

is weakly continuous

(4.0.11).

Now f

6

z 6 F+, B+

iff

XI, hence for all B+

follows for ,'F (The results f o r

g

{znf(6")}

6

Y'

6

since

cs, bs glX

respectively for all

X'

6

(4.2.4).

The result

F, B. The last part is by 4.2.7.

and so f o r

are also immediate from 10.3.4, 10.4.2.)

F+, B+

This section is concluded with an unusual theorem which will not be referred to again.

Y.z we mean

By

{y.z: y

6

Y, z

6

Z}.

It need not be a vector space. 10.

THEOREM.

S u p p o s e t h a t an

Let Z

c F+

=

z

6

XfB

Let

FK space

Z. Then

b e a s c s p a c e and

Y

z

X

-1

=I

.X

3

Then

Y. Z.

Y

so

z

X 6

(obvious from Definition 3).

=I

F+

2

an

AD s p a c e .

Z.

by Theorem 7. Hence

Thus

10.2-10.3

Zf 2 ZB

2

XfBB

and the result follows by 8.6.1.

(If

3

165

Xf 1 6 Y, the result is trivial.)

Theorem 10 has obvious connections with the construction of See [80], Theorem 15-2-7.

barrelled subspaces.

Source 1861.

B.

10.3. THE SUBSPACE Although

X

may have various pathological properties we shall

see that its subspace it has

AB

is relatively well-behaved, for example

B

(Corollary 14) includes the other distinguished sub-

spaces as closed subsets and includes (Theorem 19.) But see Remark 27.

as a basic set

ISn)

Of course an

AB

space will

have these properties too. We begin with a growth theorem and an application.

z

6

Let

THEOREM.

1.

b e an

X

FK

space

and e a c h c o n t i n u o u s s e m i n o r m ' p

Bf

2

on

4.

Then f o r each

we h a v e

X

zn =

O(p(6")-I).

lzn]p(6 n )

For

G +

F

there.

gnk = mk

is a triangular matrix. c

is not

k 5 n,

for

It follows from (M-G)l $ co.

X-continuous since

There is something paradoxical about Example 2; namely A: Q Q

+

Y

is

X-continuous.

than that of

If

YA

induces a smaller topology on

X , it follows that

A

that the continuity of

YA 2

8.

The fact is

(8.2.1).

does not imply this.

More on this in the

following optional remark. REMARK

5.

on

YA, written

that

A: YA

-+

Y

is continuous [80], Example 4-1-9. For

is the indiscrete topology

of

YA -

I $ , whereas

The map

(8.2.1).

YA If

7

2

A: Q

iff

E

W.

is

FK

let

Then

is

E

is b o u n d e d in

E

is

wA YA.

topology of

wA

Ec

@

YA

T, the

and

bounded

Since

far from the

FK

C

TX

FK

=

0,

topology

on

on

@.

topology of

and s u p p o s e t h a t

(i.e.

A

w A C TX

X-continuous iff

is row-finite, wA

A

Conversely, if

- thus

Y

+

iff the

YA, share this property: e d in

A

wA, is defined to be the smallest topology such

wA

o.

The weak t o p o Z o g y b y

(This may be omitted).

E

is b o u n d

ArE] is bounded in

Y)

w A c T, half of this is trivial.

bounded it is

T

bounded by 8.3.1.

This can be used to prove the row-finite case in Theorem 4 ; if E

is bounded in

X

it is bounded in

w

and

A[E]

is bounded in

186

Y.

11.2

Thus by 8.3.1, E

b y 8.2.3.

Source :

[86].

is bounded i n

YA

and t h e r e s u l t follows

187

CHAPTER 12 DISTINGUISHED SUBSPACES OF MATRIX DOMAINS

FUNCTIONAL ANALYSIS

12.0.

I f a Banach s p a c e h a s t h e p r o p e r t y t h a t w e a k l y c o n v e r g e n t

1.

s e q u e n c e s are norm c o n v e r g e n t t h e same is t r u e f o r Cauchy s e q u e n c e s . [SO],

#8-1-10. I f a Banach s p a c e

2.

h a s a c o s e d max m a 1 s u b s p a c e w i t h

Y

t h e p r o p e r t y g i v e n i n 1; t h e n

Y

Say

extend

g

tng(u)

Let

yn = x

n

+ tnu

X, f ( u ) = 1 (5.0.1).

on

f = 0

-

X 0 u.

=

also has t h i s property.

Y

t o a l l of +

0

so

xn

Then

(Hahn-Banach,

Y +

-+

i n norm.

0

weakly.

0

Let

tn = f ( y n ) 3.0.1).

g

6

X'

g(xn) = g(yn)

Hence a l s o d o e s

yn.

c A . More g e n e r a l

s p a c e s were c o v e r e d i n C h a p t e r 1 0 ; i n t h i s c h a p t e r w e d i s c u s s I t s p r o p e r t i e s depend on t h e c h o i c e o f

A ; our procedure w i l l be t o f i x

of

Y',

For

0.

Then

The o r i g i n a l g r o u n d s p a c e of s u m m a b i l i t y i s

YA.

6

MATRIX DOMAINS.

12.1.

FK

-+

f

Proof:

YA

depend on t h o s e o f

A.

Y

Y

and t h e c h o i c e of

and d i s c u s s how t h e p r o p e r t i e s

T h i s d i s c u s s i o n w i l l d e p e n d on which

Y

is chosen.

Y

w i l l be s p e c i a l i z e d i n some way.

I n e a c h of t h e s u b s e q u e n t s e c t i o n s of t h i s c h a p t e r The h i s t o r i c case

Y = c

will

be t a k e n up i n 15 and C h a p t e r 13.

W e p a u s e t o n o t e t h a t m a t r i x maps n e e d n o t p r e s e r v e d i s t i n g u i s h ed subspaces.

For e x a m p l e , l e t

anl = 1, ank = 0 = 1 , iB ( Y ) s i n c e here. )

for

k

> 1.

Ill(n)ll = n .

X

Then

be any

A: X

FK -f

space

Y, 6 l

(In particular

x

6

2 $,

B(x) c a n be

Y = II 0 1, 1 but. A 6

yA = w

188

12.1

To characterize distinguished subspaces of trivial, for example

z

6

+ B

iff

{p(~(~))}

YA

is quite

is bounded for each

seminorm defining the topology of

YA. These may be read from It turns out however that a simpler criterion is available,

4.3.12.

namely that

{AZ(~)}

is bounded in

Y

simpler because

Y.

(Theorem 3).

has fewer seminorms than

This is

(4.3.12) and

YA

fewer functions in its dual (4.4.2); and more useful because properties of

Y

will force similar properties on

instance of this is shown in 12.4.3, 12.4.7. the simplification is that the seminorms of those of

uA

Y.

and those inherited from

YA. A specific

The basic reason f o r YA

are made up of

Since

has

wA

AK

(4.3.8), it offers no obstacle to boundedness o r convergence,

leaving the issue to be decided by the seminorms of

Y.

This proof

i s used in Theorem 6 but in Theorems 3, 4, 5 other techniques are

more natural at this stage. 1.

In this section,

i s a matrix such t h a t

A

Y. The subspaces

to yA

REMARK.

YA 2

S,

W,

@

F, B

z

w,

6

Y

i s an

FK

i.e. the columns of

s p a c e , and

A

belong

are caZcuZated i n t h e

space

FK

*

W i t h t h e n o t a t i o n of Remark I, A z (ml -

LEMMA.

2.

rn

1

zka

k

k=l

ak

where

3.

i s the

THEOREM.

kth

coZumn of

With

z, Y, A

A.

a s i n Remark 1 , t h e s e a r e e q u i -

valent: (il

z 6 B

+,

( i i ) I A z ' ~ ) } i s bounded i n (iiil (iv)

Y,

Y A S z 2 bvo,

k {zkg(a

6

bs

f o r each

g

6

Y',

where

ak

i s the

kth

12.1

c o l u m n of

A.

Also these are equivalent: ( i v l and

189

z

z

6

B,

YA.z

bu,

3

by 1 0 . 2 . 7 ( i )

since

z -1 .YA -

(ii) = (iv):

Y'.

YAmZ. ( i i i )

of

by t h e l a s t p a r t o f 8 . 6 . 4 s i n c e t h e k t h column

6

z

6

Y

A.z

= (ii)

zkak .

is

( i i ) i s t r u e i f f { g ( A ~ ( ~ ) ) i3s bounded f o r e a c h By Lemma 2 , t h i s i s t h e same a s ( i v ) .

(8.0.2).

The s e c o n d s e t o f e q u i v a l e n c e s i s c l e a r s i n c e

z c YA

iff

'A.Z.

I t i s p o s s i b l e t o make t h e s e r e s u l t s l o o k l i k e t h o s e o f

4. 510.3.

Suppose, w i t h

a l l sequences

fixed, we define

A

for

Ig(ak)l

B+ = Ygy, v e r y l i k e 1 0 . 3 . 4 . has

AB

if

5.

THEOREM.

Yg c

Yi.

g

6

Y'.

t o be t h e set of

Yg

Then c o n d i t i o n ( i v ) s a y s

The a n a l o g u e o f 1 0 . 3 . 8 would b e :

S i m i l a r remarks hold f o r

With

z , Y,

F

YA

i n Theorem 5 .

a s in Remark 1 t h e s e a r e e q u i v a -

A

lent: li)

z

6

F

+,

( i i ) I A Z ( ~ ) } is w e a k l y Cauchy i n f o r each

g

liiii

6

f o r each

YA,

{ ~ [ A Z ' ~ ) ]6 } L'

g c Y'

where

ak

is t h e

kth

A.

A l s o these are equivalent: 6

i.e.

Y',

i s sc, k { z k g ( a I } 6 cs

c o l u m n of

z

Y

YA.z

(iv)

AJ

YA.

6

(i) = ( i i i l :

g

( i i ) and

( i v ) and

z

6

z 6 F,

YA.z

i s use,

l i i ) and

YA.

( i ) = ( i i i ) by 1 0 . 2 . 7 ( i i i ) ; ( i i i ) = ( i i ) by 9 . 4 . 1 s i n c e t h e k k t h column of A-z i s z k a ; ( i i ) = ( i v ) by Lemma 2 . The l a s t p a r t i s c l e a r a s i n Theorem 3 . A weak form of p a r t of t h e n e x t r e s u l t was g i v e n i n 5 . 6 . 8 .

190

12.1

With

6. THEOREM.

Y,

z,

a s i n Remark 1 t h e s e a r e e q u i v a -

A

lent: li)

z

6

W,

(ii)

~

3

'

liii)

weakly i n

~z )

-+ ~

Y,

Y A m z i s conull,

1 z k g ( ak I

livl

glAzl

=

f o r each

Y'.

g

(i) z (iii) by 10.2.7 (v); (iii) = (ii) by 9.4.9 since the kth column of A-z is zkak ; (ii) = (iv) by Lemma 2. With

7. THEOREM.

Y, A

z,

a s i n Remark 1 t h e s e a r e e q u i v a -

lent: lil

z c S,

(ii)

~

fiiil

2

1 zkak

column of

(i)

= Ax,

(iii) by 10.2.7 (vi).

=

=

z =

1 zksk

1 zkak .

hence u(~-z(~)) and 4.3.12.

Y , where

ak

i s the kth

(ii) = (iv) by Lemma 2.

A : YA

and

(ii) implies (i):

-+

Thus

0 for any z

typical seminorm of in

convergence i n

A.

implies (ii):

1 zkAGk

Y,

i s strongly conull,

YASz

livl

in

+ ~ ~z )

'

6

S

+

where

q[A(z-~(~))]

Az

is continuous so

First

z c wA

if

Y

(i)

+

wA

AK

has

u = p

or

0 where

q

Y by 4.3.12. But this is simply

=

by 4.3.8 h

in 4.3.8 is a

AZ(~)

+

Az

Y. 8.

A host of invariance theorems can be read from these

results. The subspaces S, W , F, I3 for matrices

A, M

in terms of the

FK

if

YA = ZM; indeed they are named i.e. defined

space.

Thus the other properties are invariant;

let us give just one example. AZ(~)

+

Az

weakly in

are invariant i.e. the same

Suppose YA = ZM, z

Y ; then

This follows from Theorem 6 (ii).

MZ(~)

-+

Mz

6

w

weakly in

and

Z also.

12.2

191

ASSOCIATIVITY

12.2.

There are two subspaces of

which play an important role

YA

in summability. 1.

REMARK.

In this section

m a t r i x such t h a t ed i n

YA

3

i s an

Y

FK

s p a c e and

i s a

A

The d i s t i n g u i s h e d s u b s p a c e s a r e c a l c u t a t -

$.

YA.

DEFINITION.

2. (tAlz

e x i s t s f o r aZZ La = { x

After t h i s chapter

W i t h t h e n o t a t i o n of Remark I , L L =

t c Y 01 , L e

+ L~ n

=

6

YA:(tA)x = tlAx)

Y

w i l Z always be

6

{ Z

W:

yA,

for a l l

t

6

YO}.

c.

The letter L was used because in summability one takes Y’

II.

=

that

The

t(Ax)

e

stands for existence, a

always exists since

Ax

= c,

for associativity. Note

Y, t

6

Y

Y O ; also

6

$

c La

Le.

C

Another fact which may be of interest some day is that the assumption

YAz

is not always needed.

$

We have seen these formulas before: bv

and

Y

has

1

(e.g.

Y

Y

A

m

=

co,c,II

)

Y =

then

La - uA

W,

La

2

YA

and if

A

6

B

and

m

II , the latter played

the same role for 4.4.7. Also 4.4.9 is an associativity result and was applied to prove 4.6.7. 3.

+

Le

c

wA

since taking

Theorem 4 implies then that definition of

F+

using

t

= 6

+ F c

uA.

f(x)

some

x

=

t(Ax)

and be

then f(x)

1 f(6

k

)xk

for some

n

yields

(tA)z = (Az),.

This a l s o follows from the

AX)^.

f(x) =

The key motivation is that if by

~

AD, then L a x bv, this was a key step in the proof

of 9.6.8.; 1.4.4 shows that if Y0 =

9.6.7 says that if

=

x.

t

6

Y0

(tA)x

and

f

6

Y;1

is defined

and this will exist for

In the latter case it gives

192

12.2

good information, for example if such

THEOREM.

z c

W i t h t h e n o t a t i o n of Remark I, F

W

+

F , t

definition of 6

f(x) = 0 for

t

t

c Le, F c Le,

La.

If

x

then

x. 4.

wc

4

f = 0 on

F+,

YB

6

=

f(x)

1 f(6k)zk

the equality

t(Ax)

let

f(x)

1

=

=

t(Ax)

define

f

6

Yi.

By

converges and this is (tA)z. If k f(6 )xk with the same f says

t(Ax).

In case

Y B has

AD

this result can be sharpened. Note that

this includes the all-important case

Y

= c.

The "improvement of

mapping" plays a significant role. 5. that

THEOREM.

Y8

That

has

Let

A

Then

AD.

Le - La

b e a m a t r i x and Le = L a ,

+ B n

uA

a

Y

c

+ LeJ

c

B

Z = Y8 , x

is simply 9.6.5 with

space such

BK

Le =

6

YA.

La'

Condition

(ii) is automatic as pointed out just after Definition 2. Next let z

6

B

+n

(YB:bs).

uA.

By 12.1.3, A-z c (bvo:Y) and so by 8.3.8, ( A - z ) ~6 z

Since

6

uA, each row of

(A.z)~ belongs to cs. Hence

A*z i.e. each column of

(Y8:cs) by the second part T of 8.3.8 which improves the map. So, for t 6 Y B , (Aaz) t 6 cs yielding the convergence of 1 [(A-z)T tlk = 1 1 ankZktn = ( t A ) z . k n + Thus z c Le. ( A . z ) ~6

We can deduce 9.6.7 thus:

La - Le

=

B(YA)>

(10.2.9)

B(bv)

= bv.

6.

Take

A

=

I.

Y B B . By choosing Y

F, W, B

Then

YA

=

Y.

Clearly

Le

-- La

=

L

spaces.

The case

shows that s u b s e q u e n t r e s u l t s o f t h i s c h a p t e r p r o v e d f o r Y

=

suitably one sees that, (pace Theorems 4 , 5),

are fairly independent of the

a l s o hoZd f o r

+

Y, Le

itself.

YA

A

=

will

I

12.2-12.3

7. EXAMPLE.

Let

Remark 3 .

Y

Here

q! L:,

B + q! w A .

1.

Then

z

BK

space; use 12.1.3, AZ(~)

I n T h e o r e m 5 , E'

alk = 1, ank = 0

for

n

>

can be any

A

Recall that since

193

Let

B +\ u A

6

AT

6

z $

so

(YA:Y) it follows that

6

z

bs\cs. Le +

by

=

(YB :YA) f

6

The next result shows when a sharper inclusion is true.

(8.3.8).

(Not always - 13.2.1, 13.4.10). 8.

THEOREM.

W i t h t h e n o t a t i o n o f Remark I , A T

lYB:Yi)

6

i f f

Le = Y A .

Clearly

Le

precisely that

YA

=

AT

iff

tA

YAB

6

f o r all

t

Y B and this says

6

(YB : Y BA ) .

6

An important aspect of associativity is its role in classifying various members of the dual space. 9. DEFINITION.

G = {f

Y ' : f l x ) = tlAxl

6

Recall also the embedding: G(x)

ux.

=

XB

+

X'

t c YBI .

with

A

u

in which

with

+

(7.2.9). B W i t h t h e n o t a t i o n o f Remark I, G c Y A

10. THEOREM.

i f f

La = Y A .

Necessity: Let By hypothesis

(tA)k f

6

= ak

G, say

AK

12.3.

t(Ax)

and so f(x)

=

t c YB =

ax

t(Ax) t(Ax).

and define

for all =

(tA)x

x.

f

6

With

for all

G

by

x

= gk

x.

f(x) = t(Ax) this yields

Sufficiency:

By hypothesis this is

(tA)x

let B x c YA.

so

SPACES

This section and the next one are of secondary importance. 1.

REMARK.

In this section

a m a t r i x whose c o l u m n s a r e i n W,

F,

B

are calculated in

'A

Y '

Y

i .e .

i s an YA

3

AK

s p a c e and

0.

The s u b s p a c e s

A

i s S,

194

12.3-12.4

La =

a s i n Remark I, L:

Y, A

F',

=

Le = F ,

w.

+

z c Le,

Let

Y

With

THEOREM.

2.

has

equal to

+

we have

t n = g(6")

With

t c YB

A p p l y i n g 12.1.2 and t h e f a c t t h a t

AK (7.2.7).

z c F

Thus

g c Y'.

by 12.1.5 ( i i ) .

so

t ( A z ) = g(Az)

If

z c W

z c La

Y

since

has

AK,

the l a s t e x p r e s s i o n is

by 12.1.6 ( i i ) .

The G p p o s i t e

i n c l u s i o n s were g i v e n i n 12.2.4.

I t follows t h a t t h e three

3.

L

Y, 2

s p e l l e d o u t i n 12.1.8 p r o v i d e d t h a t

4. +

+

EXAMPLE.

are

AK

spaces.

Y = c s , (Ax), = xn - x ~ - ~ Then .

Let

m

Le = F

subspaces are i n v a r i a n t as

- R , L e = F = c , L

YA = c

so

a = W = c0 '

1131, [861, r871.

Sources:

WEAK AND STRONG CONVERGENCE

12.4.

T h i s s e c t i o n is of s e c o n d a r y i m p o r t a n c e .

1.

REMARK.

In t h i s section

Y

i s an

space such t h a t

FK

weakZy c o n v e r g e n t s e q u e n c e s a r e c o n v e r g e n t i n t h e

i s a m a t r i x such t h a t calculated i n 2.

bvo

bv = bvo 0 1 3. = F+

the

If

S,

W, F,

B

are

R and

h a s t h e p r o p e r t y m e n t i o n e d i n Remark l(1.7.20) bv

because

bvo

is e q u i v a l e n t w i t h

R

and

(7.3.4, 12.0.2). Y

since i f FK

The s u b s p a c e s

topology, A

YA.

EXAMPLE.

So a l s o d o

Y A 2 $.

FK

i s a s i n Remark 1, t h e n f o r x c F',

topology of

Ix(")I Y

Y

itself, S = W = F

i s weakly Cauchy, h e n c e Cauchy i n

(12.0.1), s o c o n v e r g e n t , s a y

x ( ~ )+ y .

12.4

since

x ( ~ + ) x

we can conclude t h a t

AK

(7.2.7) = F

YBO = Yfg

y = x , so

it follows t h a t

w

is also

Y

If

in

195

+

= F c Y.

co

Thus

x

S.

6

since

YBB = Y

does not have t h e

p r o p e r t y o f Remark 1. THEOREM.

4.

row f i n i t e .

With

Then

a s i n Remark 1 s u p p o s e t h a t

A

Y,

i s

A

has t h e p r o p e r t y t h a t weakly convergent

YA

sequences are convergent.

Let

xn

-+ 0

hence Axn+ 0

in

weakly.

Axn +. 0

Then

YA

C o n s u l t i n g t h e seminorms of

Y.

4.3.12 w e see t h a t e a c h hn n n proved, and pk(x ) = lxkl

i s r e d u n d a n t , qoA(x") -+

f o r each

0

Y

weakly i n

k

(4.0.11)

given i n

as j u s t

0

-+

is an

YA

since

FK

space. 5.

EXAMPLE.

Ax = z . x

Let

be a s e q u e n c e .

z

z'

Then

= LA

where

Hence t h i s s p a c e a l s o h a s t h e p r o p e r t y of

(4.3.17).

Remark 1. The r e s u l t o f Remark 3 h o l d s f o r

if

YA

i s r o w f i n i t e , by

A

Theorem 4.

alk

6.

EXAMPLE.

= 1,

ank = 0

Let

Row f i n i t e c a n n o t be o m i t t e d i n Theorem 4 .

for

Then

IIunII = 1, y e t

= bv

and

n > 1. un

f ( u n ) = f(6")

-

gA =

cs.

weakly, f o r i f

0

-+

Then

f(6"")

+ 0

since

Let f

6

u n = 6"

-

6"".

cs', If(&"))

bv c c

csf

6

(7.3.2).

I n s p i t e o f Example 6 t h e r e s u l t o f Remark 3 d o e s h o l d f o r 7. If

THEOREM. z

6

+ F ,

With

(Ax(m))

Y, A

a s i n Remark I, S = W = F = F

i s w e a k l y Cauchy i n

Cauchy (12.0.2), h e n c e c o n v e r g e n t in

uA

s i n c e t h i s is an

AK

say

Y

YA:

.

by 12.1.5, h e n c e

A Z ( ~ -+ ) y.

s p a c e (4.3.8) h e n c e

4-

Now

z ( ~ + ) z

A Z ( ~ )+ A z

in

12.4

196

W.

But

so

z

6

AZ(~)

+

y

in

since Y

w

FK

is an

y = Az

space hence

by 12.1.7.

S

8. EXAMPLE.

which has

B

=

Let

Y

bv, S = W

e , (Ax),

= =

F

+ F

=

=

xn - xn-1' bvo. =

Then

LA = bv

From the results we can obtain i n e q u i v a l e n c e t h e o r e m s : 9. EXAMPLE.

A

T h e r e exists n o m a t r i x

This is,immediate from Theorem 7 and the fact that for

W

=

c0'

Compare Example 8. Also c0 ' e m , F+

since for

10. EXAMPLE.

=

em,

w

Suppose that

with columns in

c0

or

ern

X

=

c, F

=

c,

is not possible

co.

=

Y

sequences are convergent, e.g. A

kA =

RA = c .

such that

Y

has

AK

bv0 .

= .k,

and weakly convergent Then for any matrix (Theorem 7,

+ Le - Le - La = S = W = F = F + -

Y:

12.3.2).

Recall the condition

11.

(i.e. YA YBf = Y

is sufficient.

F+

holds.

=

is sc then

YA

3

bv

Another sufficient condition is that are convergen

for this implies

(Theorem 7) and so 10.2.7 (iii), ( v) show that ( * * )

F

Thus bvo

12.

YA

There it was pointed out that

is vsc), given in 9.5.11.

weakly convergent sequences in Y that

if

(**)

also satisfies ( * * ) .

COROLLARY.

With

Y, A

as i n R e m a r k 1

if

is sc it

YA

m u s t b e conull.

For

1

13.

EXAMPLE.

conull.

cA

6

F+

by 10.2.7, hence If

1

6

b v A , b u )A , o r

W !LA

For each of these is included in

is conull ( 9 . 3 . 6 ) .

by Theorem 7. is s c then

cA

A

m u s t be

and is conull, hence

It is proved in [42] Theorem 3 that if

is a regular Hausdorff matrix then see that it cannot be sc.

bvA

is variational.

A

Here we

12.4-12.5

A space Y

with the property of Remark 1 is in particular

So also is any reflexive space.

weakly sequentially complete.

weakly sequentially complete space Suppose that

197

X

co

2

bigness theorem.

as well.

X

Then

obviously has X

=

F+

=I F+ (co)

The inclusion must be proper since

1861,

[13],

=

= F =

Any

W.

= Em, a

is not

Rm

m

II , W = c

0’

[87].

C-LIKE SPACES

12.5.

and

F

2

weakly sequentially complete (because F

Sources:

F+

A

c - l i k e s p a c e is a

X

is closed in

BK

XBf.

space

X

For example

such that

X6

co, c,

bv, bs

Em,

has

AD

are

Their study brings us close to summability.

c-like spaces.

1.

If

X

Also if

X6

has

is

c-like then

AK

and

is closed in

X

is closed in

X

XB6

(4.2.5).

X B 5 then

X

is

c-

like by 7.2.7 (ii). 2.

REMARK.

In this section

m a t r i x such t h a t ed f o r

3.

c - l i k e s p a c e and

a

A

The d i s t i n g u i s h e d s u b s p a c e s a r e c a l c u l a t -

$.

(J.J. Sember and G. Bennett.)

THEOREM =

+ B n

z c Le.

For

f

Let

Hence

The columns of columns of 12.1.3.

2

is a

YA.

of Remark 2 , L:

(YB:cs).

YA

Y

A.z A.z

L e = La = B .

wAJ

t

6

T Y B , (tA)z = ( A - z ) t

c (bv:YBf) by 8.3.8, and

are in

A , so A.z

6

With the notation

( A - z ) 6~

so

so

f3f

A - z 6 (bvo:Y

)

Y , for they are multiples of the

(bvo:Y) by 8.3.6.

Hence

z

6

B

+

by

The rest of the result is given in 12.2.3 and 12.2.5.

Example 12.2.7 shows that Theorem 3 is sharp. 4.

EXAMPLE.

Suppose

Y

is

c-like and has

AK

e.g. Y = co.

12.5

198

A

Then for any matrix

B

=

F = W.

with columns in

(Example 6 shows that

Q l r ) has

5. EXAMPLE.

n(r)

If tk

A

0

for

ri

0

1 ankxk

BY 5.2.9,

and choose

ko

is

such that

m

I 1

ankxkl

ko.

Then

lankxkl =

k=r m

m

I 1 - 1 i=k

i=k+l

a .x.I

Taking

sup,

gives the

n1

result. 3.

EXAMPLE.

With the notation of Remark 1, i f

s e r v a t i v e and each column has a

in Theorem 2 )

then

S =

c

.

For

1

A

i s con-

i n i t l o r more g e n e r a l l y , u k >

S

;3

S(c)

=

co

(10.2.9), and

conversely by Theorem 2. APPLICATION I.

A b e s t p o s s i b l e growth theorem. (That it is -1 6 cA.): best possible will be clear since the columns of A

E

214

13.4

4.

suppose t h a t

For obvious. 5.

Let

COROLLARY. ann

x c

be a c o r e g u l a r m a t r i x w i t h

Then e a c h

0.

+

A

x c cA

xn

satisfies

this follows from Theorem 2 .

S

For

AB

and

= oll/annl.

x = 1 it is

But this is all there is by 13.3.1, (vi)'. EXAMPLE.

Since

by 7.1.6, and has

AB

First

{l/annl

co

S =

subspace of

X

=

A

(it has a monotone norm

be a c o n s e r v a t i v e m a t r i x w i t h

i s bounded.

by Example 3.

cA

AB

by 10.3.12) this yields the result of 3.3.11. Let

6. COROLLARY. suppose t h a t

(C,l) has

(13.3.1).

Then

Now

Since

i s Mercerian.

A

S

and

AB

is at least a maximal

1 c X\S

it follows that

X = S O l = c . A p r o o f of

APPLICATION 11.

4 . 6 . 8 for triangles.

(The proof

can easily be adapted to apply in general.): Let

7. COROLLARY.

A

o n l y bounded s e q u e n c e s .

For

A

has

u - z c E";

i s Mercerian.

A

+ =I B+ ( c )

is bounded by 1.7.7.

satisfies

Then

since B

AB

8. THEOREM.

be a c o n s e r v a t i v e t r i a n g l e w h i c h sums

=

Em

(10.2.9)

and

{l/ann3

The result follows by Corollary 6.

W i t h t h e n o t a t i o n o f Theorem 2 , e a c h i n particular

{annzn}

t

i s bounded.

This is the special case of 10.3.1 in which Thus, f o r example, i f A

z c B

p(x)

h a s no z e r o c o l u m n s t h e n

=

IIxIIA. Bt

has

a growth sequence. 9.

then

Bf

EXAMPLE. =

a",

s a t i s f i e s t h e c o n d i t i o n s of Example 3

If A

B = cA

n

II

m

.

For

B'2

B+(c)

= 1"

(10.2.9) and

conversely by Theorem 8. 10. EXAMPLE.

The matrix

Q

(1.2.5) is a perfect (Type M)

13.4

215

(Example 9 ) hence B is regular triangle (3.3.8); B = c n L" 9 dense in X = cQ; B is a proper subset of X, i.e. Q does not have

AB, by Corollary 6 and the fact that

{(-l)nl

also immediate from Theorem 14.) Thus finally, B

6

X.

(This is

i s not closed.

In order to apply these results one needs methods to recognize AB

matrices.

A sufficient condition will now be given.

Others

may be found in [84]. 11. EXAMPLE.

w i t h convergent columns, dnk = an+l,k/ank that

0

5 dnk 5

a f u n c t i o n of

k

1

Let

(Bosanquet's criterion).

and t h a t f o r each

n.

We shall show that for all

This is trivial for

dnk

m,

n

be a t r i a n g l e

k 5 n.

for

Suppose

i s decreasing ( n o t s t r i c t l y ) as Then

I x(~)[I

n = 1 since

prove it by induction on

A

A

has

AB.

Fix

x

6

5 IIxII, i.e. for every

I(AX("))~~

n

=

AX (m

we consider

X = cA'

(AXIn+1

I

n + 1 > m the express-on is r 1 ankxkl: 1 2 r 2 ml (Abel's I k= 1 an+l,kxkI = 11 dnkankXk -< maxIl k=l inequality 1.2.10) 5 by the induction hypothesis that (1) holds

5 IIxII

if

n + 15 m

while

f

i

IIxII

for

n. The

(C,1)

matrix has

dnk = n/(n+l).

criterion of Example 11, hence is an

AB

This satisfies the

matrix.

a little more as pointed out in Example 5.

We already knew

The methods of Example

5 can be applied to any matrix which satisfies Bosanquet's criterion. APPLICATION 111. assume

0 < ti

M a z u r ' s m a t r i x (1.8.12).

For simplicity

2 1. The matrix is Mercerian by Example 11 and

13.4

2 16

Corollary 6. APPLICATION IV. that for

A k

+ CYI is Mercerian for

= (C,1)
0. Here

n.

=

dnk

n/(n+l)

=

The result follows by

Example 11 and Corollary 6. We now give some conditions for

triangles.

AB

The classical

theory dealt mainly with triangles and these results were extensively used. 12. THEOREM. X = cA.

Then

for all

m , n, x .

Let has

A

be a t r i a n g l e w i t h c o n v e r g e n t columns,

A

AB

i f f there exists

(Note

i s not required.)

X

Necessity: Define

such t h a t

M

by

um(x)

Ax(m).

=

is pointwise, hence uniformly, bounded (1.0.3). Then for all

m, n

Let

{urn)

Then M

supllumll.

=

5 Ilu(x(m))ll

and any sequence x,

-< Mll~(~)ll which gives the result. Sufficiency: Trivial since for

[ (Ax(m) )n [

and so

By taking

[IX(~)(I

x

X

6

the first term is

is bounded.

y = Ax, z = A-’Y(~)

(m

fixed) so that

is easy to calculate that Theorem 12 holds with

MilxllA

z

6

for

X

x

it 6

X

on the right side of the inequality. 13. COROLLARY. exists

M

Taking

such t h a t

x

= 6m

Let

A

be a t r i a n g l e w i t h

l a n k ] 5 Mlakkl

f o r a12

in Theorem 12 yields

AB.

Then t h e r e

n,k.

lanml 5 Mlaml.

This result shows that the substitution of

akk

for

uk

Theorems 2 and 8 and elsewhere is not a genuine improvement.

in

13.4

2 17

Corollary 13 of course supplies easy examples of non-AB matrices.

An improvement now given will make this even easier as For one thing there is no need to

the subsequent examples show.

involve Mercerian considerations. THEOREM.

14.

-

Let

be a r e g u l a r t r i a n g l e w i t h

A

Then

AB.

If the conclusion is false, fix m. Then for n > m , I(Al),I m un + Vn where Un = k= 1 lank + 0 as n + m and m

2.M

c

which can be made arbitrarily small.

((C,1)

has

AB,

AB

limAl = 0.

Thus

As an application we get that t h e need n o t have

(Corollary 13)

lakk/

k=m 1

square

of an

A = (C,1)2 = H2,

indeed if

1

triangle

AB

1 l/n2 .

(ann( =

as pointed out after Example 11.)

Indeed all the Cesaro and Holder matrices of order greater than 1 fail to have

15.

EXAMPLE.

AB

by the same reasoning.

n(r)

Some more information abou 03

is available.

( c ~ ) ~ .So Theorem 8 and

Clearly Theorem 8 applies to

IIA

Example 3 with 12.5.6

B = F = W = n b (r , s = c .

16. EXAMPLE. a matrix such that let

Absence of

some

Z = cM

has

AD

but not

Let

M

be

(For example

AB.

is guaranteed by 13.2.20, 13.3.1 (v).).

AB

exists f o r all t c a, x

setting t(Ax)

( i )d o e s n o t i m p l y l i i ) i n 9 . 6 . 5 .

be the coregular non-replaceable triangle of 5.2.5.

M

t(Mx)

yield

and

A = MT

6

Z

t c I?, x

6

Z, but

by 13.3.1 (iii).

gives:

(tA)x

fails t o e x i s t f o r some

6

fails to exist f o r

(tM)x

Interchanging t , x

exists for all

z

Then

II,

t

6

2.

x

6

II,

t

6

and

Z

but

13.4-13.5

218

r391, 1411,

History and other applications may be found in [84],

[58],

and the references given there.

13.5. ALMOST COREGULAR AND VERY CONULL Conull spaces show a wider variety of behavior than do coregular spaces. We have already seen strongly conull spaces, those for which

(5.2.11,5.2.13, 10.2.7).

16 S

Here we discuss a classification which is suggested by two examples. 1. EXAMPLE. A c o n u l l m a t r i x w i t h W # F. n (-1) ( X ~ - X ~ - ~(Convention: ) xo = 0). Then A 0 triangle with

tive

13.2.6). part

2

Also

limA

0 on

=

Moreover

B.

F = B

{(-l)n}

6

=

X n II

c

but

(D

Let

limA(-l)"

=

is a multiplica-

X

where

(Ax)n

=

cA.

so t h e p e r f e c t

2

=

(13.4.9,

by 13.2.14. T h i s m a t r i x i s

F\W

p - u n i q u e by 13.2.25.

2. EXAMPLE.

A

conull m a t r i x w i t h

W = F.

Let

(This is a special case of the matrix given in 6.3.2

As in Example 1, F f = 0 on

Since

t

6

4

=

B = X

we have

0

=

n

am. k

f(6

it follows that

f(x) =

Let

part is

2

B.

That

W

=

F

rn = n.)

limAx + t(Ax).

If

uak + (tA)k = (tA)k = tk-t k+l' t = 0 so f = p limA. By the

x c X

$I

=

(c,),.

Now

(1.7.11) so the perfect

follows from 13.2.14.

(That t h i s m a t r i x

p - u n i q u e is shown in 15.2.9.)

The matrix of 4 . 4 . 4 also has

S

- take

) =

Hahn-Banach theorem (3.0.1) it follows that limAx = 0 for every bounded

p

(Ax), =xn-xnV1.

=

cA).

However it is not

3. DEFINITION. W # F, v e r y c o n u l l i f

W

=

F

(indeed

cA = a B

so

p-unique.

A v s c space is c a l l e d a l m o s t c o r e g u l a r i f

W = F.

The a d j e c t i v e s a r e a l s o a p p l i e d t o

13.5

a matrix

219

according t o which p r o p e r t y a p p l i e s t o

A

eA.

The p r e c e d i n g e x a m p l e s show t h a t c o n u l l matrices o f e a c h t y p e

exist.

A c o r e g u l a r m a t r i x i s almost c o r e g u l a r

(1 6 F\W)

and

almost c o r e g u l a r matrices h a v e many o f t h e p r o p e r t i e s o f c o r e g u l a r ones. 4.

THEOREM.

(necessarily i n

i s almost coregular i f f there e x i s t s

A

FI

such t h a t

A-x

Y

= c.

By 1 2 . 1 . 5 , 1 2 . 1 . 6 w i t h 5.

THEOREM.

An a l m o s t e o r e g u l a r m a t r i x i s

If i t i s r e v e r s i b l e

13.2.21).

If

i s coregular.

p-unique

I f i t has c l o s e d i n s e t i t must be r e p l a c e a b l e

9.6.15).

cA = S 0 u

then

has

A

x

b = 5

i n 5.4.5

(compare

(compare

(compare 5 . 4 . 6 ) .

(compare 1 3 . 3 . 3 ) .

AB

The f i r s t p a r t i s b y 1 3 . 2 . 2 5 .

(The c o n v e r s e f a i l s by Example

2) * has closed inset

A

i s c o n t i n u o u s by t h e Banach-

S t e i n h a u s theorem ( 1 . 0 . 4 ) so

W

is

If

A

(13.2.14).

But

implies t h a t That

A

b = 0

F = B

X-closed i n

(13.2.21) so

W

is

F

X = cA

where

X-closed i n

B.

This

i s r e p l a c e a b l e by 1 3 . 2 . 1 8 ( i i i ) . f o l l o w s from t h e

u-uniqueness and 5 . 2 . 6 .

The l a s t p a r t i s i m m e d i a t e from 1 3 . 3 . 4 . The r e a d e r i s i n v i t e d t o s c a n t h e t e x t f o r t h e o r e m s w i t h coT h i s i s a r i c h s o u r c e of p r o b l e m s .

regular i n t h e i r hypothesis.

Here are two n e g a t i v e r e s u l t s : 6.

3.5.5).

EXAMPLE. Let

(Mx),

(Compare

A coercive almost coregular t r i a n g l e . =

by 1 2 . 1 . 5 , 1 2 . 1 . 6 w i t h

xn/n, x = {n}.

A

=

M, Y = c.

The

Max = I

so

x

6

F\W

220

13.5

EXAMPLE.

7.

coregular

B

with

(1.7.9)

M 2 A

(compare 3.5.4).

Let

v e r y conull, M

A

be the matrices called A ,

A, M

in 1.7.11 and apply Examples 2, 6. Note also that

almost coregular w i t h t r i a n g Z e namely

C

=

C

almost

a r e g u l a r t r i a n g l e and

(C,l) in which case

may be

CA

a v e r y conull

A

CA = M.

Two other classifications of conservative conull matrices were introduced by E . Jurimae for the purpose of extending the MazurA c o n s e r v a t i v e matrix

Orlicz bounded consistency theorem 5.6.11. A

is called a

stands for on

c

cA

matrix

J

n

t m ) , an

if 0

-c 2

b

(where, in this section b

matrix

if

implies the latter equality on

cM

b.

both of these properties (4.6.7, 5.6.11). facts may be found in [ 9 ] , 2

b.

r i x since

W

i f f

A'

[22A]:

b

and

limM = limA

A coregular matrix has Proof

of the following

A conull m a t r i x i s an

0 matrix

(Hence a r e p l a c e a b l e v e r y c o n u l l m a t r i x i s an =

F

n

Example 6.) E v e r y

and

A'

F

=

B

0 matrix i s a

3

J m a t r i x and t h e c o n v e r s e h o l d s

J matrix.

Sources:

[71,

[91,

D61,

0 mat-

b , but not conversely by

f o r r e p Z a c e a b l e m a t r i c e s b u t n o t i n general.. matrix i s a

3

P5A1,

r821.

Every non-replaceable

221

CHAPTER 14 THE FUNCTIONAL p

FUNCTIONAL ANALYSIS

14.0. 1.

The sum of two

space 17761, pp. 39, 62. 2.

Then

'f

Let

X

FH

(or

BH)

f

a linear functional.

can be given a complete norm smaller than the restriction

Define

'g

q: ' f

new norm of

+

x.

f(u) = 1.

q ( x ) = x - g(x)u

by (If

Let

q(x) = 0 ,

o

Let

g

and take

= f k-g(x)u]

X ' , g(u)

6

IIq(x)ll

=-g(x)

so

=

1.

as the x

= 0.)

A linear functional which is bounded on some neighborhood

3.

0 must be continuous.

4.

X

Let

The set of all

[80] Theorem 4-5-9.

be a locally convex space.

Eo

with

E

For

a bounded set in

local base of neighborhoods of

X

FH

spaces is an

be a Banach space and

X. Proof:

If

BH)

[SO], Example 13-4-5.

of the norm of

of

(or

0'

X

E c X,

(8.0.1) is a

f o r the s t r o n g t o p o Z o g y on

X'.

is a normed space, the strong topology is given by the norm

(1.0.1) [80], p. 119.

5.

Let

X, Y

be locally convex spaces and

the dual m a p . topologies.

T

=

T': Y '

X'

by

T'(g)

It is continuous when

Y'

and

and continuous. Define

+

A special case is X c Y

inclusion.

Then

T'(g)

=

glX

=

T: X

+

goT.

Y

linear

This is

X' have their strong

(with the same topology) and

[80], Example 11-2-3.

222

14.0-14.1

6. Let q: X

+

Y

X

be a locally convex space, Y

a linear map such that

is closed iff

q-l [El

is closed.

q'

locally convex topology such that

Y

has a EcY

X. This quotient topology is [80],

continuous.

q

Then

is continuous and a set

q

is closed in

the largest which makes

a vector space and

16-2.

14.1 PARTS OF THE DUAL One of the most intriguing problems in our subject is to decide

u , (14.2.2), is continuous. Apart from the

whether the function

intrinsic interest in a problem which is so easy to state is the importance of the function tinguished subspace P

in connection with the fifth dis-

p

(Chapter 15).

It also appears in results

in which its role is unexpected, e.g. 14.5.3.

X

1. With

in which 2.

recall

f(x) = t(Ax),

with the form

X'

cA

=

u

If

consisting of the set of X8

t c k ; also the embedding of

u

corresponds to

EXAMPLE.

X' = II 0 lim

G c X'

X = c

with

6(x) = ux

(i.e. A = I), then

in the sense that each

f c X'

is

f in

(7.2.9).

G

= XB =

f(x) =

p

Q,

lim x

+ tx (1.0.2). 3.

Let

EXAMPLE.

A

be

r o w - f i n i t e and one t o o n e .

X B = G. I

p - u n i q u e and e i t h e r r e v e r s i b l e , o r

Then

X8 c G

(Hence

Note that every coregular matrix is

G(x) =

1.1

limAx + t(Ax)

this is true for

G

Let

(5.4.3, 5.5.3), and

by definition.

AB

i f f

p-unique (9.6.10),

indeed every almost coregular matrix is (13.5.5). Then

has

A

Thus ii c G

p =

.

u c X'. 0 since

(The last

part is by 12.5.8.) 4. REMARK. assumed t o b e a

In this section A BK

space.

i s a m a t r i x and

X = cA

is

14.1

THEOREM.

5.

includes

X0

223

W i t h t h e n o t a t i o n of Remark 4 , t h e c l o s u r e of

G.

The t o p o l o g y i n v o l v e d i s t h e norm t o p o l o g y o f

Let

X I .

f(x)

m

= t(Ax),

Choose

> 0.

E

m

1

so t h a t

1

Uk =

tnank'

Obviously

Itn!

E

Then

1 f(x(n))-lim

Let

(13.2.5, 13.2.14).

If not, W = cA =

1

p-con-

B.

by Theorem 5 .

f c limA + CEO.

7. COROLLARY (J. Boos).

limA x

limA c X8

a bounded set in

The last term

x c W.

Thus

W # -

Indeed it i s s u f f i c i e n t t h a t

If it is not x

An a l m o s t c o r e g u l a r m a t r i x i s

THEOREM (J. Boos).

(4.4.4).

is AB

p-unique; this matrix need not

228

14.4

14.4.

SPACES

p

The

program is applicable to some of the results in the

FK

last section. DEFINITION.

1.

'X

An

FK

By 14.3.5, a m a t r i x

is as

space i f

p

is

A

u - u n i q u e and

u-continuous i f f

c

A

If the main conjecture is true (that every matrix

space.

p

i s caZZed a

X

x'.

i s n o t dense i n

is a

space

p-continuous) then this condition characterizes p-uniqueness it does in any case where THEOREM.

2.

A

is known to be

p-continuous.

u

Every coreguZar space i s a

space

(compare

14. 3. 4 ) .

If

G.)

u c XB

THEOREM.

is a

I.I

Let

Let

b e an

X

(See 14.1.1 for

0.

h c

h -

h = 0

X I ,

$ Eo

X

For

7.

The

either i s a

not have

3

0.

If

B @

7

then

space.

a bounded set in

W C

=

space

FK

on

$I, h(z) = 2

the Hahn-Banach theorem (3.0.1).

SO

X(;)

The result follows from 14.3.3 and 9.3.10. 3.

X

it is immediate that

X.

and

A

FK p

Then for any

XB

yn

= z

6

B; using

- z(n) and E

=

{yn},

u c X8 ,

is not dense.

this folllows from 14.3.6 and 13.2.25 since

c

=

Let

for some z

program for 14.3.7 is: An space o r has

AK.

AD; otherwise by 10.3.19.

FK

space w i t h

AB

This is by Theorem 3 if .X does For

cA

the two cases are

p-unique and non-u-unique. 4.

EXAMPLE.

An

FK

space o f t h e f o r m

X

=

ci

cannot be a

14.4-14.5

l~

space.

(In this assertion

space.)

By 4.4.2each

i.e.

=

X'

G + XB

f c X'

X

=

0

cA.

THEOREM.

5.

u

w h i c h is a

Thus

If a n

has the form

+

u c

w AB

XB

is dense in

FK

space, then H : Y'

Define

can be replaced by any f(x)

t(Ax)

=

AK

+ ax

in the language of 14.1.2. Examination of the

proof of 14.3.1 shows that where, now,

co

229

X'

Y

by

so, in particular, u c X B ; X'.

space

Y

has a c l o s e d subspace

is a

FC

space.

H(f) = flX.

H

Then

X

is onto by the

YB

Hahn-Banach theorem (3,0.1), continuous (14.0.5), and takes into

XB.

Such a map preserves dense sets.

6. This is as far as we can go.

Example 4 shows how to get

space which is a closed, even maximal, subspace of a

a non-U space.

multiplicative

0. A

u-unique multiplicative

shown in 13.5.1, also 13.5.2. the form

cD

n

is even, 0 if

triangle is

0

The smaller space can also be given

by the simple device of defining

if

(Ax)n/2

n

is odd.

D

Then

by cD

(Dx), =

other direction, every space is included in the non-u

14.5.

A

The smaller space can be made conservative by making

AND CLOSURE OF

AB

0

cA.

=

In the

space

w.

XB

We can now give a satisfactory discussion of some results which were given in piecemeal fashion and restricted to The improvement of mapping and continuity of role.

BK

spaces.

play an important

LI

We need a preliminary result which will allow improvement

of mapping as in 8.3.8 to be applied when the range is not an space e.g. 1.

A

=

LEMMA.

is c o n t i n u o u s .

0

in the next result.

Let

A

(4.0.5).

b e a m a t r i x , X = c A'

Then

AT:

1 -+

Xf

FK

230

14.5

A c (X:c)

Since

in the case of

BK

(14.0.6) by

where

q

AT c (k:Xf )

it follows that

spaces, Xf

is given the quotient topology q(f) = If(& k ) I for f c X ' , the latter

space having the strong topology (14.0.4). the dual map (14.0.5, A' Note that

c'

9. C

As

by 8.3.8.

A': c'

Let

X'

+

be

is continuous with the strong topologies.)

(properly) with the embedding

t

with

+

i(y) = ty, (14.1.1, 14.1.2), and the strong topology agrees with

the norm topology on Then

H

Now let

(14.0.4).

9.

H = qoA':

9. +

Xf .

is continuous as the composition of two continuous func-

tions. The proof is concluded by showing that H = A T . Let t c k . Then H(t)i = q[A'(i)Ik = A'($)(6 k ) = $(A6 k ) = (tA)k = (ATt)k.

2.

I c o n j e c t u r e t h a t e v e r y m a t r i x map:

REMARK.

continuous i f

i s an

X

space (14.2.8, 7.2.14). more generally if

Xf

space.

FK

This is true if

It is also true if

X

XI, hence

Ptak's closed graph theorem applies.

is

X

is a

BK

is reflexive or

has the quotient topology by

Mackey topology, since then

Xf

9. -+

XI

with its

Xf, is fully complete and

See [80], Example 12-3-8,

Theorems 12-4-5, 12-5-7. 3.

THEOREM.

These are e q u i v a l e n t f o r a matrix

convergent columns, X = cA.

X I , ( i i i )'X closed i n

(i) A

is c l o s e d i n

x', ( v i )

A'

(i) implies (ii) (12.5.8) hence

c

~

f

has

AB,

(, i v ) X B =

( i i )X B

xf,

( v ) 'X

A

with

i s closed i n

+ '6

is

(9.:~').

(Given in 10.3.11 for

XB = X8 + G =

p'

(14.2.3).

BK

spaces):

X'x

G

This is closed by

14.3.7.

(ii) implies (i). implies AB

XB

2

G

by 14.3.1 and the hypothesis.

(12.5.8).

(i), (iv), (vi) are equivalent. (iii)

=

(v)

(10.4.4, 12.5.8, 13.3.1).

as in 10.3.10, using 14.0.6.

This

14.5

(iii) implies (vi): are in

and so AT

X’

231

A , hence the columns of

The rows of

[$]

c XB.

Since

AD, Lemma 1 implies

has

k

AT,

the result. (iv) implies (iii). One application of Theorem 3 is to give easy examples in

4.

which

is not closed e.g. 13.4.10 or any non-perfect regular

X’

Ad hoc constructions were given in 10.1.2,

triangle (13.1.7).

X’ # Xf.

These are also easy examples for

10.1.3.

The

Compare 7.2.8.

FK

program fails for Theorem 3, for example bv’ is f (7.3.5). This shows again that a proper closed subspace of bv 5.

.

as in 13.3.2.

‘A # bv

All the conditions of Theorem 3 are meaningless, if

6.

fails to have convergent columns, except (ii).

It is interesting

to ask what conditions equivalent to (ii) could be found. example, suppose that for each

x(~) 6 X

X

=

cA

A

For

x c X

has the property that

implies

n; would (i) and (ii) be equivalent in this

case? 7. EXAMPLE.

It was proved in 8.3.8 that

T A c (Y’:Xf).

But

X

space.

=

cA

a

BK

AT

angle (e.g. 13.4.10).

AT

6

(!L:X’)

the rows of

B

Then

A

be any non-AB

A

may apply 8.3.6.

X’; X’

A

(X:Y)

even f o r

c

has

implies Y = c,

(a:XY) then

the columns of

is closed in

It follows that

6

conservative tri-

AT

(X:c); if

6

by the following argument:

A, are in

Y

( Y :X I

n e e d n o t b e in

Let

A

AB

Xy

AT, being

(4.3.18) so we

(Theorem 3 ) and

this is a contradiction. 8. Let

X

A useful extension of the ideas mentioned in Example 7 is: be a

BK

space

2 $,

then

A

6

(X:Y)

implies

14.5

232

A c (F+(X):YBy) = (Xfe:YBy).

space this is immediate since YBy = YBf

any set of sequences) let

x c X.

I hn(x) I

boundedness (1.0.3) this inequality to

r

since

-+ m

u c Y B , hn(x)

By hypothesis this converges as

x

6

x(~).

uA

5 KllxII.

Since

x

is

a

BK

( 1 0 . 3 . 8 , 10.3.12, 10.1.5)

and two applications of 8 . 3 . 8 give the result. (Y

Y

We note first that if

n

-+

=

m

Now fix

In the general case n 1 uj(Ax) j f o r j=l so by uniform

+

x c F

and apply

has bounded sections,

by 1 2 . 2 . 3 ; this says

C 1. (hn(x)\

for all

n

and so Ax c Y”. 9.

EXAMPLE.

A special case of Remark 8 is

(co:Y) c (Lm:YBY).

233

CHAPTER 15 THE SUBSPACE P

FUNCTIONAL ANALYSIS

15.0.

1. The sum of a closed and a finite dimensional subspace is

closed.

-

E0 u

Thus if

E

is a subspace the closure of

Y#

let

the weak

6

Y

*

EL

Let

subspace of

-s

HI

{g

Z: g(y) = 0

6

g

6

Y'#

X.

6

Y: h(y)

for all

Let

5.

implies

T

=

0 for all

f

H

* 6

y

El.

6

h

6

or

E

Then, in the weak

f, g

*

Then

HA*

continuous is given by Y'

[80] Theorem 8-1-7.

[81],

f

S

a

X", each point in

Lemma 3.3.

be linear functionals such that

f(x) = 0. Then

is

[80] #8-3-110.

topology of

S

i.e.

HI; for

be a Banach space with separable dual and

is a sequential limit point of

15.1.

-

Z be a to-cal subspace of

for all

which is weak

g(f) = f(y)

X

{y

=

linear closure of

= o(2,Y)

i.e.

4.

let =

Each

3.

y

H c 2

For

Y

denotes the a l g e b r a i c d u a l of a vector space

the set of all linear functionals. Let

Ec Y

is

[80] Theorem 6-3-3.

2.

Y#.

E 0 u

is a multiple of

g.

g(x)

=

0

1801 Theorem 1-5-1.

AND THE TEST FUNCTIONS

The subspace

P

of

cA

(15.2.1)

was introduced by H. R.

Coomes and V. F. Cowling to generalize the Type non-triangles.

(See 15.2.2.)

M

condition to

The question of its invariance

(15.4.12) was settled affirmatively in [6]. (15.2.10) is available in the presence of

far as we know, covers all cases.

A simpler proof u-continuity which, as

234

15.1

We begin by associating with a matrix a certain subset of 1. DEFINITION. L e t

be a m a t r i x .

A

T = TA = I t c 1: ( t A l x e x i s t s f o r a l l x

Clearly

T = .F

Le = X

i f f

I t c II: t A c X').

A

and t h i s (if

(12.2.21,

convergent columns) i s e q u i v a l e n t t o Type M

Then X = cA 1 =

6

II:

One thinks of

A B (13.3.1).

in connection with Definition 1 since if

has

t A A , certainly

t c T.

DEFINITION.

2.

a t e s t function for

X

understood that

EXAMPLE.

3.

A function

=I

If

EXAMPLE. W # B.

satisfy function.

f(x)

Let

0

Taking x

Ax

=

Yx

= gk

This implies W

B

(It is

A

f

p-unique t h e n

In the first

In the second case it is by definition.

be a l m o s t c o r e g u Z a r o r , more g e n e r a l l y ,

= 0

on

B, p(f)

(12.5.9) and so

gives =

0.

on

i s a t e s t function.

Then e v e r y f u n c t i o n v a n i s h i n g on

If not, let

= lim

= 0

i s not replaceable or not

A

~ ( f )= 0 by 13.2.18. 4.

f

X = cA, i s c a l l e d

0).

e v e r y f u n c t i o n v a n i s h i n g on

case

and

plfl = 0

if

A

where

f c X',

Yk =

limAx

-% and

= 1.

For x c B,

-Yx

for x c B.

=

limAx

so

is a test

B

=

ax

for x c B.

by 13.2.5, 13.2.14.

The connection between the two definitions follows: 5. THEOREM. Then

f

Let

A

be a m a t r i x w i t h convergent columns.

i s a test function for

A

i f f f ( x ) = tlAxl

-

(tAlx

t c T.

f o r some

Sufficiency is trivial. Now suppose that f is a test function. Then f(x) = t(Ax) + ax (4.4.3). Also 0 = f ( 6 k )

(tA)k +

ak

and

so

tA c X B

the representation of

f

since

follows.

a

does.

Hence t

c

=

T and

15.1-15.2

235

We shall see that the set of test functions is invariant. (15.4.11).

The proof is complicated and it is worthwhile to prove

a (possibly) special case: 6. THEOREM.

Let

b e a m a t r i x w i t h c o n v e r g e n t columns s u c h

A

t h a t a l l equipotent matrices are t e s t functions i s invariant for

By 14.3.2,

p-continuous. A.

is invariant; trivially

p'

T h e n t h e s e t of

is invariant.

$

Paradoxically, in view of Theorems 5 and 6, T even for equipotent triangles in which case

is not invariant,

p-continuity is assur-

ed by 14.2.4. ( [46], p. 240.) Many sufficient conditions for

p-continuity are given in

914.3.

THE SUBSPACE P.

15.2.

1.

DEFINITION.

{ x c cA: ( t A ) x 2.

=

Let

t(Ax)

be a m a t r i x .

A

for a l l

t

6

TI

Then

P = P

=

(15.1.11.

P

The original motivation for the definition and name of

was Theorem 6 which is the analogue of the Type M general matrices. 3.

A

THEOREM.

condition for

It was originally given for conservative matrices. Let

X

=

cA

3

bu

(e.g. i f

i s c l o s e d and i n c l u d e s t h e p e r f e c t p a r t of

X

i s uscl.

A

Then

P

and a l l t h e o t h e r

d i s t i n g u i s h e d subspaces (hence t h e i r c l o s u r e s a l s o ) .

For each

t c T

set

ft(x) = (tA)x - (tA)x.

This is con-

tinuous by the Banach-Steinhaus theorem (1.0.4) and Next

B

=

Le

(12.5.3) c P

trivially. Also

dl

C

B

P = n{

1 and set

17.1-17.2

-

z = { x E c : x 1 = x2

< u

in

X

with

Z, Y

=

c , for each

=

IIzllm = 1, IIAxllm

To summarize:

E.

265

A[Z]

Arc]

has codimension

(3.0.3). By 17.0.5

there exists

z

Given

and integer

> 0

E

E

with

2

1, there exists

>

z

z1

Choose 1

lznl

>

c0 such that zk = 0 for 1 5 k

E

to satisfy (1) with

1

3 , otherwise x1 n

possible since

z

1

E

=

u

=

u , IIzllm = 1, IIAzll,
= <xn ,T'g> = + 0 and so

xn

-+

f

F

X

T"

Thus

T' [Y'] implies

0 weakly by 1 7 . 0 . 4 .

The classical theorem given next is due in part and in full with various proofs to I.D. Berg, J.P. Crawford, R.J. Whitley and The proof given here is that of 1323.

to S. Mazur and W. Orlicz.

Other references are given there.

An easy proof for row-finite

matrices is given in 1821, Remark 4.3. that i f f o r maps:

The more general result

h a s no r e f l e x i v e s u b s p a c e , T a u b e r i a n = s e m i - F r e d h o l m

X

x

-+

Y

10. THEOREM.

is given in [PO],

4.3.

A conservative matrix i s Tauberian i f f it i s a

s e m i - F r e d h o l m map f r o m

to

c

c.

This follows from Example 3, Theorem 5 and 17.1.4 along with the fact that

c

has no infinite dimensional reflexive subspace

( 17.0.3).

11. COROLLARY.

Let

A

be a c o n s e r v a t i v e t r i a n g l e .

Then

A

i s Tauberian i f f it i s range c l o s e d , Mercerian i f f i t i s o n t o .

These refer to the map A

is one to one.

A: c

-+

c.

The conditions hold since

17.6-17.7

A Tauberian matrix

12.

c

(17.1.4).

A

maps

277

c

onto a closed subspace of

It follows from Theorem 10 that

closed linear subspaces.

[80],

Reference:

(See [82],

preserves a 2 2

A

Theorem 2.3.)

pp. 175-177, 14-1-304, 305.

MULTIPLICATIVE ABSTRACTED

17.7.

T".

In the preceding sections we considered eigenvalues of An obvious modification is to do the same for

TI.

Actually,

more than idle curiosity suggests this; it is motivated by applying to multiplicative matrices the process of generalization which led

from matrices to

w-matrices.

We shall show the abstract formula-

tion first. DEFINITION.

1. h # 0.

For

Let

M = M(X,h/

write

T E M

LEMMA.

2.

For 3.

b e a Banach s p a c e a n d fix

X

Let

= IT

E

For e a c h

T

E

LEMMA.

That m

TI).

h(Txl. = mh(xl

M,

f o r a12

x; m = m ( T ) .

= <x,T'h> = <x,mh> = m<x,h>. T E M

T [ h * ] c h'.

iff

<x,T'h> = = 0 and so

that

i s an e i g e n v e c t o r of

T ' h = m(Tlh.

Necessity is by Lemma 2. =

BIX): h

h E X',

M

Sufficiency: T'h

If

h(x) = 0, (T'h)(x)

is a multiple of

h

is a c osed suba gebra, with identity, of

(15.0.5).

B(X)

and

is a scalar homomorphism are proved like 17.5.4.

To discuss multiplicative m a t r i c e s in the general setting we shall assume t h a t and

Mra 4.

refer to THEOREM.

w(h) = 1

M Let

n

and

I'

T

where

E

Mr,.

M

w

is as in 17.4.2.

n ra

respectively.

Then

mlTl = p ( T )

+

Let

h(x)

Mr

where

278

17.7

+

Trrw= pw

x.

Thus

they agree on

Mra

has two distinct scalar homomorphisms;

on ldhich they are equaZ t o

Mr

w(h) = 1 we have

Since

Note also that

x

Then

g.

In [ 1 5 ] it is proved that of

g = A'h

However if

ra

=

B(X)

X"

and

X =

p

This Page Intentionally Left Blank

281

CHAPTER 18 ALGEBRA

FUNCTIONAL ANALYSIS

18.0.

Let

1.

with

co.

#9-6-110.

S

Then

be a subspace of

S

(Note.

co

which is linearly homeomorphic

is complemented in

[80], Corollary 9-6-5,

co

It is proved in [89] that no other Banach space

has this property.) 2.

(a)

G

The set

r/9],

with identity is open.

E

>

0 there exist

p9],

14.2 Fact iii.

(18.2.1) is a topological divisor of

aG

of

of invertible members of a Banach algebra

x, y

(c)

#14.2.39.

G

with

IIx11

=

IIyll

(b) 0

Every member

i.e. for each

1, IIxzll

=

z

IIzYll

ktz

( l e f t t o p o l o g i c a l d i v i s o r of 0 1 i f f o r

there e r i s t s

0

b e a member of a Banach a l g e b r a .

z

\ly\l

with

y

= I ,

~ ~ z y5 ~E.l

The next result is an adaptation of a theorem of B. Yood to

r

(17.4.2):

THEOREM.

3. T

i s an l l t z i f f

Necessity: subspace of

W i t h t h e n o t a t i o n of 1 7 . 4 . 2 , T

from 0 i.e. T

T: X

Say

U

=

11 Ull .I1TVll

0.

T

Choose

E

X

with

IIaII = 1,

is one to one, otherwise we can make

IIgll = 1, w(g) E

a

X, IIVyll

=

=

0

Ig(y)l

and set so

IIVII

V

=

a8 g

=

1, IITVyll

=

18.7

Ig(y)l.IITall

5 cIIy/I so

V”w = w(g)a

=

IITVll 5

283

Finally

E.

holds for any subalgebra, such as Let

THEOREM (I.D. Berg).

5.

since (17.5.12)

B(X),

the same result

6

0.

T

Since this proof works for any

4.

r

V

w h i c h i s one t o one on

Then

c.

E

ra

(17.5.1), which includes

A

be a c o n s e r v a t i v e m a t r i x

r.

i s T a u b e r i a n i f f i t i s n o t an

A

r

&tz i n

From Theorem 3 and 17.6.10. 6. THEOREM. Tauberian i f f

A

Let

be a c o n s e r v a t i v e t r i a n g l e .

i s n o t an k t z i n

A

is

A

(1.5.51.

A

Necessity is as in Theorem 5.

Then

Conversely if

A

is not Tauber-

ian it is not ranged closed by 17.6.10;the proof of Theorem 3 provides a matrix

V = a

vnk

g(x) = bx.

=

=

anbk

x(g).)

(1 BII

=

1

where Let

and

g

@

IIVll = 1, IIAVll 5

with

bnk

=

vnk

for

11 ABll

=

11 AVll

5

(g

By 17.2.7,

E.

has this form since

k 5 n, 0 for

k

n.

>

0 = w(g)

Clearly

A

if

it is not an Ltz, so is Tauberian.

@

E

A,

E.

7. A proof of Copping’s theorem (6.1.5) follows: has a left inverse in

B

the result of 1.8.4 can be given in full generality:

r

e

Also

T e B(Xl

i f

h a s a l e f t i n v e r s e it is not an k t z , hence is a linear homeomorphism

by Remark 4 applied to 8.

LEMMA.

Let

T

l i n e a r homeomorphism i f f OP

By 17.6.5 i t is T a u b e r i a n .

B(X). E

Blcl T

r

lor

r,).

or

has a l e f t i n v e r s e i n

is a

T

B(cl

lor

r

r,). Sufficiency is obvious.

Necessity:

S =

UOP. Then for

x e c , STx

=

U

Let

By 18.0.1, there is a continuous projection Let

Then

U [PTx]

=

P

of

=

UTx

-1 T : TIC] c =

onto

x.

The

+ C.

T[c].

284

18.1-18.2

r , ra

results for For

9.

T

6

follow from 17.5.7.

r

in Lemma 8, the left inverse map

S

is also a

ST = I, by 1.4.4. This is the same as

left inverse matrix i.e. Remark 1. 10. THEOREM.

W i t h t h e a s s u m p t i o n s o f Theorem 5 , A

is

r.

T a u b e r i a n i f f i t has a l e f t i n v e r s e i n

Sufficiency: Remark 7. Necessity: Theorems 3 , 5 and Lemma 8. For general exists

D

B

E

r

A

r

E

we have that

is Tauberian iff there

A

BA = I + D, D a f i n i t e member of

with

has only a finite set of non-zero columns.

A

by adding rows to the top of

r

i.e.

This is accomplished

to make it one to one on

c;

possible by 17.6.10. 11. J. Copping, [ 2 4 ] Lemma 1, has an interesting extension: Suppose t h a t exists

D

E

i s coregular, B

A

r

with

BA

and

0

BA E

I , in which case

=

r.

A

is automatically

coregular since it is Tauberian (Remark 6, 3 . 5 . 2 ) . that i f

A

i s of type

M

Then t h e r e

We have been concerned up till now

DA = B A .

with the special case

E

then

B E

r

since

It is trivial

(D-B)A = 0. This

is a l e f t i n v e r s e c l o s u r e result.

Sources:

1141,

[23],

[77]

18.2. THE MAXIMAL GROUP

1. DEFINITION. L e t The maximal g r o u p Z;

aC

Z

G = G(Zl

i s t h e boundary o f

b e a Banach a l g e b r a w i t h i d e n t i t y . i s t h e s e t of i n v e r t i b l e members o f

G

i.e.

C\G.

18.2

285

This agrees with the usual definition of boundary: since

G

b

with

E

Z

2.

is open (18.0.2a). ab

=

ba

=

a

G

E

iff there exists

1.

A matrix

THEOREM.

Note that

E\Gi,

A

( 1 . 5 . 5 1 i f f i t i s a Mercer-

G(d)

E

ian triangle. A

Necessity:

is a triangle.

is triangular and has an inverse matrix, so it

It is Mercerian by 1.7.14.

Sufficiency is by

1.7.14. 3.

If A

THEOREM.

E

Girl

By 1.8.3 (or 18.1.7).

i t must be T a u b e r i a n .

It was pointed out in 1.8.6 that

A

need not be Mercerian. 4.

prises.

So far the discussion has been easy and lacking in surThe interest lies in looking at

The first hint that

aG.

it contains objects of interest came with the discovery of Mercer’s theorem 2.4.1. for

a > 0.

In view of Theorem 2 it says that

Letting

a

+

C

0 we see that

this matrix is Tauberian (2.4.2).

E

aG.

a1

For

+ (1-a)C

E

G

a < 0

Similarly 1.8.6 shows that if

-

axn-1’ A is Mercerian for la1 < 1, Tauberian for la1 > 1, (replace a by l / a in the first matrix given in 1.8.6,)

(AX),

=

xn

and not Tauberian for gent unless

a = 1

la

=

1.

in wh ch case

(It sums

A

{an}

which is diver-

is conull and 3.5.2 applies.)

T h i s s u g g e s t s t h e c o n j e c t u r e t h a t t h e non-Tauberian m a t r i c e s are precisely those i n

aG.

The conjecture turns out to be false but

is close enough to the truth to make it interesting. To avoid trivialities we add the assumption that

A

is one to one on

c;

for example the identity matrix with one diagonal member replaced by

0

is in

aG

but is Mercerian.

Half of the conjecture is true

18.2

286

5.

THEOREM.

Let

A

r

E

be one t o one on

c.

If A

E

aG(r)

i t i s non-Tauberian.

By 18.0.2(b) and 18.1.5. The extreme non-Tauberian matrices give evidence for the converse: 6. EXAMPLE, E v e r y c o e r c i v e m a t r i x

A E

aG(r).

is weakly

A

compact (17.5.9), hence compact (18.0.4). Hence there exists a sequence un

A

Since

f

G

-

unI

G

E

for each

is T a u b e r i a n i f f

Let A E

A

be a t y p e

cA

(18.0.4).

Gtr).

Sufficiency:

triangle (3.3.3).

M

A

(This is the same as Indeed

A

(6.1.1)it follows that

A

E

Then

G(A)

A

in

Necessity: A

is Mercerian.

is coregular (3.5.2), hence perfect (3.3.4).

in

n

(Theorem 3), the result follows.

7. EXAMPLE.

this case.)

A

0 such that

+

Since

c

is closed

is Mercerian.

8. Example 7 and Remark 4 suggest the easy looking question: find a type

M

triangle not i n

-

GlAl.

A good deal of effort has

gone into this but the question is still open. 9. EXAMPLE.

The c o n j e c t u r e i n Remark 4 i s f a l s e .

J. Copping

(Ax)n = xn - 99~,-~ - ~ O O X ~then - ~ the (essentially regular) triangle A is an interior point of the

[23], p. 193, shows that if

set of non-Tauberian matrices in with Remark 8, A

is not of Type M

A.

A f aG(A).

So

since

In connection

L A.

Copping's

example is amplified in 1141. 10. This form of the question in Remark 4 is still open: must a l l n o n - T a u b e r i a n m a t r i c e s l i e i n

to one on

c.)

aG(rl?

(Again, assume one

287

18.2-18.3

11.

EXAMPLE.

perfect triangle i n

(Actually BA each

bn, n

Am

I f we define

Sue me!)

1 + l/m

m, by

~ +- bnxn ~ with bn

(Ax)n = x

Let

aGfA).

is regular. >

A r e g u l a r non-

(J. DeFranza and D.J. Fleming).

Am

we shall have

F

-+

1

by replacing as in

G(A)

Remark 4. Also Am + A. To make A not perfect we simply choose -2 in 3.3.12, taking account of 3.3.4. Earlier examples un = n [40A],

[62] had

12.

conull (and not of type M).

A

B.E. Rhoades [ 5 3 ] has located many of the classical

matrices (Chapter 2) in

see also [21],

aG(A);

p. 65.

Naturally

some N2rlund matrices are Tauberian (2.6.8,also 1141, Lemma 4 )

so they cannot be there, by Theorem 5.

1621 and, of course, Example 11.

ed in [ 5 3 ] are answered in [ 4 0 A ] , 13.

Membership in

G(A)

Some of the questions rais-

is not invariant, for example if

(Ax)n = xn-1' A is equipotent with I but A f G. Even more simply, replace a diagonal element in I by 0; this yields a matrix in

which shows that this set is also not invariant.

aG

However membership i n t h e s e s e t s i s i n v a r i a n t f o r t r i a n g l e s :

G

is trivial for B

=

MA

where

Theorem 2.

M

by Theorem 2.

=

Since

follows that

B

E

BA-'

-G

Next let

A

E

aG,

cB = cA.

G

is Mercerian (1.7.5) hence in

this Then

by

is closed under multiplication (18.0.2~) it

E ; finally B

f G

by Theorem 2.

(This argument

is due to B. E. Rhoades).

Source:

18.3.

F71

THE SPECTRUM. 1. Any operator T

(18.0.3) as a member of

algebra 2

on a Banach space B(X).

If

T

X

has a spectrum

belongs to some closed sub-

with identity its spectrum as a member of

2

might

18.3

288

a p r i o r i be l a r g e r . here:

r

If

T

or i n

r

E

T h i s i s by 1 7 . 5 . 7 which a l s o y i e l d s a s i m i l a r

ra,

statement f o r

2.

i t s s p e c t r u m i s t h e same computed i n

(77.4.3),

B(XI.

( 1 . 5 . 1 ) and

T h i s c a n n o t happen i n t h e c a s e s o f i n t e r e s t

The same is t r u e f o r

Any q u e s t i o n a b o u t membership i n a s p e c t r u m can b e r e s o l v -

t

E

u(z)

is i n a s p e c t r u m .

0

For example, i f

is i n v e r t i b l e i f f i t is a M e r c e r i a n t r i a n g l e n

Fix

LEMMA.

This is

Thus a l l comes t o d e c i d i n g

0 c o(z-tl).

iff

which o b j e c t s are i n v e r t i b l e .

3.

r

vis a vis

B(c).

e d i f it is known how t o t e l l when because

(1.5.5)

A

and d e f i n e

A

A

E

(1.5.1)

A

A.

Then

(1.7.14).

f(nl = a

nn

for

A

E

is a non-zero s c a l a r homomorphism.

f

4.

For

COROLLARY.

each

A E A,

ann

E

o(Al.

By Lemma 3 and 1 8 . 0 . 5 . 5.

[85A], Theorem 2 , t h a t i f

I t is proved i n

EXAMPLE.

A c A

is c o e r c i v e , o ( A ) = {Ol U { a n n ] . 6.

t

+

Let

EXAMPLE.

1 and l e t

A

Then

= C-tI.

a = t/(t-1).

with

(C,l)

be t h e

C

is a Mercerian t r i a n g l e i f f

is c l o s e r t o

> 1 . Now

It

-

21

value

>

1

t-3'2 5.

+

(1-a)C

a > 0.

[34],

Ra > 0

p . 1 0 6 , Theorem 5 2 .

i.e.

1 than i t is t o

1 a+l - a 1

The proof of 2 . 4 . 2

and givdn t h e same r e s u l t f o r

a

The d e t a i l s may be s e e n i n

a

a1

Theorems 2 . 4 . 1 , 2 . 4 . 2 show t h a t t h e l a t t e r

a c t u a l l y works f o r complex

iff

Fix

(1.3.10).

is a m u l t i p l e of

A

m a t r i x i s a Mercerian t r i a n g l e i f f

0.

matrix

a(C)

Thus

A

( i n t h e complex p l a n e )

-1 which h o l d s i f f

is i n v e r t i b l e i f f

so, finally, A

We have p r o v e d t h a t

Ra >

=

{ t : It

-

TI1

5 71.

The

t = 1 w a s e x c l u d e d , b u t i s i n t h e s p e c t r u m by C o r o l l a r y 4 ,

18.3

or because in that case 7.

EXAMPLE.

A

is conull

N.K. Sharma [62A] shows that for

only possible isolated points of In particular

289

o(A)

A

E

the

A

are its diagonal elements.

cannot be an isolated point of the spectrum of

0

a triangle. These examples show that it is fairly difficult to compute the spectrum of a particular matrix.

The spectra of Hausdorff

and weighted mean matrices are considered in [21], The spectrum of the Hilbert matrix in [22B], Note, p. 307.

1621 and [62A].

is considered in

B(L2)

In [27A], Theorem 3 , the spectrum of a cey-

6)

tain Mercerian Norlund matrix with

ann

is shown to contain

1

=

negative numbers.

A

It is curious that the behavior of

outside

consulted to learn about its spectrum e.g. if divergent sequence into 8.

c, A

A

r

tional on

X

is a Banach space and

X ' , not necessarily in

w

can be

maps some

A

is not Mercerian hence

We now turn to a generalization of

17.5.1) in which

E

c

and

0

ra

a(A).

E

(17.4.3,

is a linear func-

(Suggested by D. Franekic.)

X".

See also 17.5.16.

It is unknown whether

algebras of

in this case; however they are inverse closed

B(X)

these are closed sub-

(like 17.5.7) and this is sufficient to imply that every scalar homomorphism is continuous.

The continuity of

p

will be proved

directly, however (Corollary 13). 9.

DEFINITION.

m a t e e i g e n v a Z u e of jixjj

=

I,

11 ~

T

Let

T

E

B(XI; t h e n

i f f o r every

E

>

X

0

i s caZZed a n a p p r o x i there exists

x

with

~ - ~ <x jE .l

For example any eigenvalue is an approximate eigenvalue and each approximate eigenvalue lies in the spectrum since

T-XI

is

290

18.3

not range closed (17.0.5). 10.

EXAMPLE.

T be a linear homeomorphism of

Let

T

a proper subspace. Then

is not invertible so

0

X

onto

o(T), but

E

is not an approximate eigenvalue.

0

11.

THEOREM.

Let

T

a p p r o x i m a t e e i g e n v a l u e of

Let

m.

(Remark 8).

E

Then

i s an

p(T)

T.

U = T - p(T)I.

Then

U"w

=

x

E

X.

We shall show that

x

E

=

U"w(g) = w(U'g) = 0. The Hahn-Banach theorem implies the asser-

tion.

If

Let

U

XI, g = 0 on

g E

Then

U'g = 0

and so

g(x)

is not range closed the conclusion of the Theorem is

If

true by 17.0.5. Then

RU.

U"w = U"z

U

is range closed, x

and so

onto (17.0.1) and

U

U"

E

RU, say

is not one to one.

x = Uz.

Hence

U'

is not

is not a linear homeomorphism (17.0.1) so

the conclusion follows again. Let

A

be a conull triangle.

is not an eigenvalue of

A

since A

12.

13.

EXAMPLE.

COROLLARY.

p

BY Theorem 11, p(T)

Then

p(A)

=

0

is one to one.

i s continuous. E

a(T), hence

(P(T)( 5 \IT\\ (18.0-3)-

291

CHAPTER 19 MISCELLANY

FUNCTIONAL ANALYSIS

19.0.

1.

Every s e p a r a b l e q u o t i e n t of

2.

Let

X = nXn. then

{Xnl

is not a

X

be a d e c r e a s i n g s e q u e n c e o f

X

Suppose t h a t BK

[SO] #15-3-1.

is r e f l e x i v e

Ilm

FK

s p a c e s and

is a dense proper s u b s e t of each

space.

is an

(X

FK

Xn;

s p a c e by 4.2.15.)

[79] #11.3.27.

19.1.

WHAT CAN

1.

BE?

YA

cA is s e p a r a b l e and t h i s l e a d s

W e h a v e s e e n (16.2.1) t h a t

t o i n e q u i v a l e n c e r e s u l t s such as t h a t w i t h i t s s t r o n g topoZogy is s e p a r a b Z e ,

e x a m p l e , Il

am

181

cannot be

v a l e n c e r e s u l t s a r e g i v e n i n 13.3.2.

X

X

space such t h a t c a n n o t be

n

Some o t h e r i n e q u i -

X

Also i f

is a conservative

is separable but not equal t o

km

c A ; t h i s f o l l o w s from 6.5.6.

FK

space

i f t h e codimension of be

XB

in

X'

cA.

FK

program e . g .

F i n a l l y we mention t h a t

is more t h a n o n e , t h e n

X

cannot

c A ; t h i s f o l l o w s from 14.2.3 and 14.3.1.

2.

EXAMPLE.

e v e r y row o f take

cannot be

3 Em

c, then

In g e n e r a l an inequi-

valence theorem r e s u l t s from e v e r y f a i l u r e o f t h e

a coregular

cA. The

cA or e v e n a c l o s e d s u b s p a c e o f

c a n n o t be

A =

Any be

0,

z.

e.g.

zB

Also =

c xn.

B

cs = 1 , c a n be and

co

cA

Theorem 5, and s o , f o r

f i r s t was p r o v e d i n a d i f f e r e n t way i n 12.4.9.

FK

cA. Also

cA. J u s t l e t

cA. F o r

c a n be

Finally every

FK

c0,

s p a c e of t h e

19.1

292

.

n c 0 [x1,x2,.. ,x ]

form

can be

(mod k")

cA.

where the

xi

are linearly independent

This is proved in 1851, Theorem 3 , with a

correction on p. 386. cA

A different sort of inequivalence theorem is that

3.

not b e a c l o s e d s u b s p a c e of

For

in 1.2.6).

ci

can-

II, unless it is finite dimensional (as

would then be a quotient of

(17.0.1) and

Qm

separable, as mentioned in Remark 1. This makes it reflexive by 18.0.1 so

cA

is a reflexive subspace of

R , hence finite dimen-

sional (17.0.3).

have the form

cA

I;

=

RP.

We shall show this for

result is similar.

A

A

11

=

IIi

E

p

=

28'

has

I

(txk)

fix any

5 IIxII

SO

1

1. The more general

as the next eight and so on.

21

or 0.

x

For

E

II,

IIx[IA 5 IIxI1 1. Conversely given

m

m ; then

such

A

lxkl

=

1

(txk)

=

(Ax)n 5

IIxI~~

SO

k=l

11

1 5

I1

A matrix

as its first two rows,

is row-finite and all its terms are

I (AX), I x

The matrix

as the next four, +61+622+63

So

spaces

seems extremely difficult.

5. EXAMPLE (G. Bennett and G. Meyers). that

FK

The general problem of characterizing which

4.

A'

6. With

A

as in Example 5, cA

is a closed subspace of

L

(4.3.14), hence finite dimensional as pointed out in Remark 3.

cA = {O).

Actually it is clear that

A : 9.

The map

II

+

m

cannot

come close to being onto of course; it does n o t even cover just mentioned, o r even xn = y

that x

E

for

II.

y

2"-1 Since II E ACE],

potent since

$.

- y2"-l -1

=I

but

It is one to one and = (By),

say.

this shows that

$

AB

ABAB = AIB

BA

is nothing like =

AB.)

Thus =

I.

x

y =

=

Ax

c

as

implies

By = B(Ax)

I. Also

A(By) = y

(Yet it is idem-

for

19.1-19.2

7.

We now turn to

R

c B B f co

B

It is a

AK

(4.3.8) so can-

dual, namely

A , so it cannot be

is the set of rows of

dual since

This space has

uA.

c , bv, em, bs etc.

not be

293

R',

where

is not a

B

where

i s

co; co

(7.2.2).

0

8. THEOREM.

i s a

uA

space i f f it i s

BK

a f i n i t e s e t of s e q u e n c e s n o t a l l i n

m

b e t h e first

Necessity:

r o w s of

En be the first n

Let

is their intersection.

n

EXAMPLE.

Let

otherwise.

1, 0

=

It is dense in each

wA

ank = 1 Then

such that

A, Em

$

:E

A.

Then

4.2.15)

E;

and

{EiI

uA

since it includes

for all sufficiently

2

:E

rn

for all

n , k.

is a

wA

n , where

rows.

(Also an

bnk = 1

P. = E B

with

E

an E

space there must be is the

nth

row of

EiB).

a finite set, so

which is impossible.

for

Theorem 8 shows that

BK

for all

Let

It seems fairly obvious from Theorem 8 that

Otherwise

+

: a

is the first 10,

bv

=

uA = uB = cs.

this situation is typical; if m

rows of

Sufficiency: By 4 . 3 . 7 .

m.

9.

may b e t a k e n t o

E

spaces ( 4 . 3 . 7 ,

AK

It follows from 1 9 . 0 . 2 that

large

E

A.

is a decreasing sequence of

$.

Here

$.

EB

wA

# L.

ern = E B B = (Span E).

(The last mentioned identity is not

too hard to prove.)

19.2. TOEPLITZ BASIS A

1.

b i o r t h o g o n a l s y s t e m is a sequence of points

of continuous linear functionals i.e. 1 is

if

=

k , 0 otherwise.

{Pn}. It is a basis

{An},

1 fk(x)b

n

k

; thus an

FK

{fn}

The most familiar such system

for

space has

such that

{bn} and k fn(b ) = 6n k

X

AK

if every iff

x c X

is

({Snl,IPnl) is a basis.

19.2

294

2. Henceforth we shall refer to

Ibn)

X

is a basis, o r

saying, for example, {b"}

(with has

Ifn} AK

understood)

iff

ISn}

is a basis. We say t h a t

3. DEFINITION.

i f there e x i s t s a regular matrix

T-summable t o

x

f o r every

Ibn)

i s a ToepZitz basis f o r

such t h a t

T

1

fk(xlbk

X

is

x E X.

k

1 fi(x)b i , then 1 tnkyk + x i=l A b a s i s i s a T o e p Z i t z b a s i s since we may take T = I in

This means that if we set in X.

yk =

that case. 4. EXAMPLE.

The set of trigonometric functions (and Fourier

coefficients) forms a Toeplitz basis, with basis, f o r the (periodic 2s)

T = (C,l), and not a

continuous functions on

[-s,s]

with

the topology of uniform convergence P 4 A ] 13.33, p. 441 #15, p 9 ] 7.6 Application 3, and for

L[-a,s]

with almost everywhere conver-

gence [ 7 4 ] 13.34. 5. EXAMPLE.

Let

(Ax)n = xn + x ~ - ~ Then .

0

ToepZitz basis f o r

c A , hence

Isn}

is a

Il,Sn) is a Toeplitz basis for

cA.

(But not a basis:

see 13.3.1 (iv)', 13.4.10) The (C,l) matrix n works: Let hn(x) = x - (l/n) 1 x(~). Consulting 11.1.6, we see k=1 exists and so for each x, Ihn(x)l that for f E c;1, lim f[hn(x)] is weakly bounded, hence bounded (8.0.2). By the uniform boundedness principle (7.0.2) Ihnl this

case.)

(7.0.3).

Now

But

hn(x)

-Q =

0

cA

+

0

is equicontinuous ( = norm bounded in for

(3.3.8).

x

E

Q , hence also for

AB

E

5

This argument is similar to that

used in 10.3.19; the latter is not available here since an

x

A

is not

matrix.

Since the whole gamut

of regular matrices is available it

19.2

295

seems at first glance difficult to find an example in which

Ibn}

is not a Toeplitz basis, given the trivially necessary assumption

that it is fundamental ( 3 . 0 . 1 ) .

However an example is easily

deduced from : LEMMA.

6.

If

i s a Toeplitz basis,

IIbn)Ifnl)

i s

{fn}

total.

x must be

This means that

if

0

fn(x) = 0 for all

n.

The proof is trivial. 7. EXAMPLE.

Let

bn

=

1 + An,

This is a biorthogonal system. let

f

E

for all t

E

L, t

=

1 ti

X +

Then 0, so

+ tn

X = 0, f = 0.

Banach theorem 3.0.1. However

=

=

-

xn

tn

for

x

E

c.

i s fundamental,

and

(1.0.2),

0 so

lim x

{bn}

To see that

f(x) = X lim x + tx

c', say n.

fn(x)

f(b")

= 0

is constant.

Since

The result follows by the Hahnfn(l) = 0

so

{bn}

i s not a

T o e p Z i t z b a s i s by Lemma 6 .

Finally we consider the problem of giving examples in which Ibn}

is fundamental and

AD

({~"I,{P~})in an 8.

LEMMA.

Let Then

Here

W

10.2.3,

F

and

5.6.1.

Let

by definition of

F.

space

X

Toeplitz basis.

{fn}

is total.

In particular, consider

X:

be an

FK

space i n which

i s a

{An}

F = W.

are the distinguished subspaces defined in k x E F, f E X I . Then 1 xkf(6 ) is convergent I t s sum must be

f(x)

(forcing

x

E

concluding the proof) since there exists a regular matrix that

lim x ( ~ ) = x T 9.

with basis.

EXAMPLE.

{bn}

implying that

limTf(x(k))

=

W

and

T such

f(x).

A Banach s p a c e w i t h b i o r t h o g o n a l s y s t e m {bn}{fn},

f u n d a m e n t a l and

{fn}

t o t a l , which i s n o t a ToepZitz

By Lemma 8 it is sufficient to take

A

to be a coregular

296

19.2-19.3

triangle so that

1

E

F\W

by 10.2.7 and such that

One is shown in 5.2.5. Of course

bn

cA

A.

(Example 9) by the map

system out:

the

bn

the rows of

A-l

i.e.

(By),

=

where

For

c

is

We can write the

are simply the columns of fn(y)

AD.

has a fundamental-total

c

biorthogonal system which is not a Toeplitz basis. equivalent to

has

g n , fn - Pn.

=

The Banach space

10. EXAMPLE.

cA

A , the

fn

are

B = A-'.

19.3. MISCELLANY. 1. There are of course many ways to define generalized limits

other than by means of matrices.

For example, methods associated

with Abel and Bore1 are well known; others of a very general nature have been introduced by L . Wlodarski and by A. Persson.

We refer

to [88]. 2.

One can also extend

lim

from

c

to

Qm

by the Hahn-

Banach theorem and obtain a generalized limit for bounded sequences. This cannot be given by a matrix since such a matrix would be regular and coercive contradicting 3.5.5. limit with the additional property

A special case is the Banach

LIx,}

= L I X ~ + ~ } . [SO]

#2-3-106. 3.

M. Henriksen 1351 introduced another sort of limit on

]Im

which cannot be a Banach limit ([80] #9-6-103); namely let

t

E

BN\N

positive of

where

BN

is the Stone-Cech compactification of the

integers and define

L(x)

to be the unique value at

x, extended as a continuous function on

a fruitful theory; we refer to [lA], 4.

gN.

[35], 1641,

t

This idea led to [651,

r661.

In a recent conversation Jeff Connor told me an elegant

proof of the fact that no regular matrix can sum every member of

297

19.3-19.4

2N, the set of all sequences of

is an immediate consequence.)

The space

topology (the relative topology of If it is included in

cA

with

[44A],

regular, limA

A

Ch 2, 927, Th6or;me

1 + x, x

1.

But every point is the limit the other of points of

$,

This proof shows that the generalized

$.

E

is a function of

hence is continuous a,?some

of two sequences, one of members of the form

2N, with the product

is a complete metric space.

W)

Baire class 1 since it is lim(Ax)n, point

1's. (Of course 3.5.5

and

0's

limits mentioned above are not of Baire class 1

- and probably

can be adapted to show that they are not of any Baire class by Baire's continuity theorem [ 4 4 A ] , 5.

Let

A

Ch 2, 128.

be a regular matrix and

It is easy to prove that if

space.

xn

+

X

a locally convex

0

then

1 ankxk

+

0.

k However, let

X

Q1I2,

=

xn

&"In.

=

Then

ll(l/n) " 1 xk k=l

-

convex.

6.

Various sufficient conditions for a space to be conull

are given in 6.4.3, 9.3.6, 16.2.6, 16.2.7.

A quotient, or even an

isomorphic image of a conull space need not be conull, for example

A

if

is a triangle

vsc matrix w i t h

X(A) x

E

=

c,\c

19.4.

1. Let

c

A

M

=

(9.3.6).

cA

and

c

are equivalent.

If A

is a

= uA, t h e n

A

must be c o n u l l .

A-I.

Then

M

is conull so there exists

Then

Ax

=

Mx + x f c

so

x

E

If not, say

uA\cA.

APPLICATIONS

The first applications were possibly mystical in nature attach a "limit" to a divergent series as mentioned in 11.1. the last century these were used for analytic continuation

-

to

In

- for

298

19.4

example

1 zn

diverges for

IzI > 1

but is 1/ (1-2) for z = -1

one can declare this to be true for

1

and get

IzI

1;

143

A' Conservative 128 Conull 149,196 Dual 66 FK 6 3 Possibilities 291-293 sc 147,150 Subspaces 64,188-199

0

17,30,70

@

266

NAMES Agnew, R . P . 54 Aissen, M. 299 SYMBOLS Banach, S . 3 0 1 (These are listed alphabetically Beekmann, W. 2 4 2 , 2 4 5 in the above index where possible) Bennett, G. 117,197,206,209,252, A - x 87 259,292 x.E 97 Berg, I.D. 276,282,283 x.y 6 2 Boos, J. 227,245 XY 3 Bosanquet, L . S . 215 x - l E 62,68 Connor, J . 296 x(n) = section Coomes, H . R . 2.73 T' = dual map Copping, J . 9 2 , 2 8 3 , 2 8 4 , 2 8 6 XI = dual of X Cowling, V.F. 233 X# 233 Crawford, J . P . 2 6 6 , 2 7 6 XB,Xy = B-,Y-dual DeFranza, J. 4 4 , 2 8 7 Kf 105 DeVos, R . 1 5 4 , 1 5 6 , 2 3 7 , 2 3 8 Eo 221,225 Fleming, D . J . 44,287 Franekic, D. 289 = transpose of A Garling, D.J.H. 1 1 7 Ax,(AxIn 3 Gelfand, I. 3 0 1 akjbk 5 Hardy, G.H. 2 1 , 3 0 t(Ax),(tA)x 7 Hausdorff, F. 2 4 , 2 9 B 3 A 13 Helly, E . 2 1 (a,b) 32 Henriksen, M. 296 AoB = composition H u m i t z , W . A . 24 (X:Y) 4 S ,X (See Y A in alabetical Jurimae, E . 5 4 , 7 0 , 2 2 0 A ihdex) Kalton, N.J. 1 1 7 , 2 5 1 , 2 5 2 , 2 5 9 II'IIA 40 Knopp, K. 3 1 Zeros

75,221,233

318

INDEX

Kothe, G.

62,64,65

Rhoades, B . E .

287,298

Kuan, S-Y.

245

Ruckle, W.H.

Kubota, T .

37

S a r g e n t , W.L.C.

Mackey, G .

230

Sember, J . J .

145,150,197

Sharma, N . K .

289

MacPhail, M.S. Mazur, S.

20,39,43,88,276,301

Mercer, J.

30

#4

Meyer-Konig, W . Meyers, G.

O r l i c z , W.

S k e r r y , H.

86

Tauber, A .

40,88,276 296 136 230

36,37

7,24

54,70,94,97,117, Snyder, A . K . 138,150,252,258,298 T o e p l i t z , 0.

P e t e r s e n , G.M. P t a k , V.

54

167

Silverman, L.L.

292

P e r s s o n , A. P i t t , H.R.

83,84

236,248,249

Madden, C .

117,119,225

12 7,62,64,65,301

Whitley, R . J .

266,276

W l o d a r s k i , L.

296

Yood, B.

Zeller, K.

282 See p . v i i

Summability Through Functional Analysis (North-Holland Mathematics Studies, No 85)

Summability through functional analysis

Recent Progress in Functional Analysis (North-Holland Mathematics Studies, 189)

Functional Analysis: An Introduction (Graduate Studies in Mathematics)

Principles of functional analysis (Graduate Studies in Mathematics)

Principles of Functional Analysis, Second Edition (Graduate Studies in Mathematics)

Functional Analysis (Pure and Applied Mathematics)

Elementary Functional Analysis (Graduate Texts in Mathematics)

Functional Analysis in Modern Applied Mathematics

Elementary Functional Analysis (Graduate Texts in Mathematics)

Linear Functional Analysis (Springer Undergraduate Mathematics Series)

Functional Analysis

Functional analysis

Functional Analysis

Functional analysis

Functional analysis

Functional Analysis

Functional Analysis

Functional analysis

Functional analysis

Functional Analysis

Functional Analysis

Functional Analysis

Functional analysis

Functional analysis

Functional analysis

Functional Analysis

Functional analysis

Functional analysis

Functional Analysis

Functional Analysis

Summability Through Functional Analysis (North-Holland Mathematics Studies, No 85)