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 <
m,
ak = 0
for
1.3-1.4
7
Necessity is by Theorem 6 and the fact that
ak
=
k limA6 ,
Sufficiency is by Theorems 6 and 8. The conditions of Theorem 9 are called the (L.L.) Silverman( 0 . ) Toeplitz conditions.
Their discovery in 1911 marked the
- as opposed to the study of
beginning of the general theory specific matrices. an excellent
The original proof by elementary analysis is
- and difficult - exercise for graduate students.
The "soft" proof given here is due t o S. Banach in 1932.
It marks
the beginning of the functional analysis approach. \
10. EXAMPLE.
matrix
A
The Cesaro matrix denoted by
such that
symbols, ank = l/n
(Ax),
for
k
is the average of =
1,2,. . . , n, and for
Every baseball player kcows that
(C,1) is the
x1,x2, . . . ,xn. k
>
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 <_
CC
11 tll
All 11 xl
1 1 1
'
and
interchange of summation for an absolutely convergent series is allowed. 5.
COROLLARY.
The set
@
and the set of row-finite matrices
have associative multiplication. For
A
E
A
Q, the rows of
so Theorem 4 (ii) applies.
are in
EXAMPLE.
BOA
em
the columns in
The other part follows from the observa-
tion that (i) of Theorem 4 holds for all 6.
2,
f BA, indeed
BOA
x
if
A
is row-finite.
is not a matrix map.
The example is nothing but the ordinary telescopic series of our childhood. l(Ax)
=
Let
Z(Ax),
(Ax), =
lim x
whose first row is in particular
=
xn for
- xn-1 (Convention: xo x
E
c.
1, other rows
BOA # 0, while
BA
Now let
0. Then =
B
=
0).
Then
be the matrix
B(Ax)
0. To see that
=
(lim x,O,O,..) BoA:c
+
c
is
not given by a matrix it is sufficient to observe that it vanishes
1.4-1.5
on
9
t h e terms o f a m a t r i x are d e t e r m i n e d by how i t maps
@;
m
since
= (M6
nk
k
)n
f o r any m a t r i x
More is t r u e :
14.
w e r e a m a t r i x i t would h a v e t o b e t h e m a t r i x by a n e a s y c a l c u l a t i o n i . e .
= (BA)nk
since
BA
B O A = BA
matrices
A, B
x
and
[B(Ask)]
t(Ax)
#
(tA)x.
and c o n s e r v a t i v e
A
I n cases l i k e C o r o l l a r y 5 i n which
7.
BOA
@.
on
Examples o f t h i s t y p e a r e a v a i l a b l e whenever The example i n Remark 2 l e a d s t o a r e g u l a r
if
4
(BA)x = B(Ax)
i n t h e s p a c e i n v o l v e d , B O A = BA
B
for
by d e f i n i -
tion. I f m a t r i x m u l t i p l i c a t i o n i s a s s o c i a t i v e , as i n C o r o l l a r y 5 ,
8.
a m a t r i x h a s an i n v e r s e m a t r i x (which is a u t o m a t i c a l l y u n i q u e ) i f f it has a unique r i g h t ( o r l e f t ) i n v e r s e . unique r i g h t inverse
and
bnn = l / a n n .
add
t
is one t o one.
A
Also
B = A-'.
5; thus
But
t o any row of
has a
A
i t may b e d i r e c t l y computed by i n d u c t i o n
B:
a n d is u n i q u e s i n c e
A triangle
BA = I A
B
since
A(1-BA)
is a t r i a n g l e
B
Moreover
by C o r o l l a r y
= 0
may h a v e a n o t h e r l e f t i n v e r s e : i n t h e example o f Remark 2 .
just
See also
1.8.5. Examples 6 and 8 show t h a t i t i s d a n g e r o u s t o d e n o t e t h e
9.
map
x
+
Ax
by
I n 1 . 3 . 6 w e d e n o t e d i t by
A.
Then
F-l
A-'.
I n 5 . 4 . 8 w e s h a l l see a case i n which
F
f o r t h i s reason
c o u l d b e w r i t t e n w i t h o u t w o r r y i n g a b o u t t h e ambiguous A-l
e x i s t s b u t does
n o t g i v e t h e i n v e r s e map.
THE ALGEBRAS
1.5.
1.
by
r.
DEFINITION.
r
AND
A
The s e t of c o n s e r v a t i v e m a t r i c e s i s d e n o t e d
1.5-1.6
10
r
Thus 2.
LEMMA.
Let
Sk = { A
0'
E
3.
since
@:limna nk
E
Let
So =
{A
un(A) =
ank
Z
Let
0.
0:lim
6
i s a c l o s e d v e c t o r s u b s p a c e of
Let
exists}.
Then each
un(A) = ank, defining
5 IIAll. The result follows by 1.0.5.
lun(A)I
LEMMA.
0.
C
i s a c l o s e d v e c t o r s u b s p a c e of
Sk
un
By 1.3.6, r
(c:c).
=
ank
Ck
exists].
Then
So
0.
so that
5 IlAIl.
lun(A)l
The result
follows as in Lemma 2. 4.
r
THEOREM.
i s a c l o s e d s u b a l g e b r a of
hence i t i s a
0,
Banach a l g e b r a .
.k = 0,1,2, ...1 k'
By 1.3.6, r subspace of
0
(BA)x = B(Ax) B ; thus
of
= n{S
by Lemmas 2, 3.
E
c
BA
E
hence is a closed vector
A,B
If
using 1.4.4 (ii) with
r.
r , let x
E
t
c.
E
Then
taken to be any row
It is routine to prove that
0
is a Banach
algebra using 1.4.5. 5.
THEOREM.
The s e t
i s a c l o s e d s u b a l g e b r a of
With
unk(A)
=
A
r.
of triangular conservative matrices Hence i t i s a Banach a l g e b r a .
ank, unke r '
and
A =
n{uik:k
>
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 <
A matrix
n.
is complete.
9,
Then
Cklankl
<
$:$
Choose
m
lank/ + 1
The result follows by 1.3.7.
m.
i s c o e r c i v e i f f i t has c o n v e r g e n t
A
c o l u m n s and a n y o n e of t h e s e e q u i v a l e n t c o n d i t i o n s h o l d s : ( i l Cklankl
a
E
L,
and
i s uniformly convergent,
( i i l the
f i i i l Eklankl
Cklank-akl+ 0 ,
~ O W Sof
A
and
lakl, b o t h s e r i e s
+ C
being convergent.
The series is uniformly convergent, so
(i) implies (ii): lim and
may be interchanged.
C
/C/ankl-cjak/ I 5 zlank-akl.
(ii) implies (iii): (iii) implies (i):
so that
n
>
finitely many
N
implies n
Ci=m
Iankl <
>
0
a.
N
Choose
This excludes only
E.
m
lak/ <
Ci=m
we can ensure that this last
n.
A
(ii) implies that is
E
so by increasing
inequality holds for all
and
Let
is coercive:
uniformly convergent so
lim
and
If
x
is bounded, Ckankxk
may be interchanged.
C
This yields limA x Next assume that
A
=
ax
for bounded
is coercive.
the column limits exist (1.3.6) and (iii):
a
x
(1)
Then it is conservative so E
L
(1.3.7). We shall prove
1.7
16
CASE I.
r e a l , ak = 0
A
By deleting rows of that
A
lank( > 1
Ck
Choose
so that 1
k=l lan(l),kl Choose
n. Ilan(l),k
1
>
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 2 m(2);
for sgn a n(3) ,k
m(2) < k 5 m(3)
and so on.
m
Then
(Ax)n(l)
-> 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
<
and the result follows from Theorem 9 En-l 1. To see this: k=l (bnk(
Iankl -
=
for
cA
=
=
For
2Q
-
Cm k= 1 Iankl
IIAll
-
-
a <
the second condition holds but
A = (C,1)
(1.2.5), ann
<
=
1 but
IIA/1 =
l Xn-1' Corollary 10 applies. n - 2
C
lak
+ 2;
20
1.8
The next example is an important application 12.
M
=
M(t)
Z \ t k \ and
EXAMPLE (S. MAZUR'S MATRIX), b e the matrix with mk
=
tk
Note also that f o r
(Mx),
=
Let
t
xn +; : : Z
tkxk; IIMII = 1
so Corollary 10 implies that
x
E
c,
Zimm x
=
Zim x
f
and let
t
E
M
tx ( 1 . 3 . 8 ) .
+
is Mercerian.
21
CHAPTER 2 CLASSICAL MATRICES
2.0.
BACKGROUND This section is exceptional, as is the whole chapter, in that it
is classical.
1. tion on n , t,
(Helly's choice theorem). [0,1]
Then
2.
for each
n
Let
gn
and assume that
Then
With
I
fdg,
gn +
(G.H. Hardy's big
is convergent.
1
fdg
M
<
0
for each continuous
Tauberian theorem). {n(yn-yn-l)l
for each [74], 9.61
gn
+
g
f. [74], 9.611
Let
is bounded.
y
be
Then
y
[ g o ] , p . 225.
The reason for the name is the notation
2.1
Ign(t)[
as in Theorem 1 suppose that
(C,1) summable and such that
Ixn/tnl
be an increasing func-
{gn} has a pointwise convergent subsequence.
THEOREM.
pointwise. 3.
See 52.1.
xn = O(tn)
to mean
is bounded.
INTRODUCTION The reader who is interested mainly in functional analysis may
proceed immediately to Chapter 3.
The rest of the book makes few
references to the present chapter and these can easily be consulted as needed. The classical matrices were introduced in the 19th century for application to problems in analysis such as analytic continuation of power series and improvement of the rate of convergence of numerical series.
2.2-2.3
22
2.2 H~LDERMATRICES
H
Let
= (C,l).
H",
is simply
the
nth
Hm
H", B
=
H.
power of
follows from 1.7.5 that strict, by 1.75 (A
n, n
The Hglder matrix of order
=I
Hn
if
m
Hn+l(Hn)-'
n.
>
1,2,.. . , H
=
it
The inclusion is
and 1.7.14 (A
H)
=
Since
=
H)
=
since
H
sums
{(-l)n}; it is also obvious from 1.7.8. We
the divergent sequence
have, however, the following Tauberian theorem:
I. THEOREM, A L L
un,
This means that if
x
mable H2
summable and let
summable, k
y
y
E
1, let
>
=
Hm
is bounded and summable
Hx.
Then
c , (2.0.3), i.e.
x
y = Hk-'x.
y
summable as just proved and so
x
it is sum-
Suppose first that
n(yn-yn-l)
Then
em.
a r e e q u i v a l e n t on
Hn to the same value for all n.
is bounded hence
H
,...,
n = I,Z
is
E
is
- yn-l which
xn
=
x
Hk+l If x is 2 is H summable, hence cH.
Hk
summable.
The limits
are equal as just mentioned.
2.3.
HAUSDORFF MATRICES A l l i n d i c e s w i l l s t a r t from
u
...)
= (uo,pl,p2,
matrix with
i n t h i s section.
be a complex sequence.
-- un
mnn
0
Let
.
n = 0,1,2,. . .
for
M
Let
be the diagonal
D be the matrix
Let
(binomial coefficients.) Then D is a triangle dnk = (-l)k(n) k with dnn = (-l)n, dno = 1 etc. A little calculation will suggest that
D2
=
I.
We shall prove this shortly. Hp = DMD is called the H a u s d o r f f m a t r i x a s s o c i a t e d
The matrix w i t h t h e sequence
p.
it is a triangle iff 11
It is triangular since D pn
#
0
for all
n
since
M
and hnn
is fixed throughout a discussion we shall write
=
H
are; and
un.
(When
for
Hp.)
To facilitate computation in this section we introduce two matrices.
Let
S
( S X ) ~= x
be the matrix such that
n, so, for example, the first two rows of
S
are
~
+for~ all
(O,l,O,O, ...I
2.3
and
(O,O,l,O,O,
...).
(Of course
i
use the formula
23
x
Let
( S x ) =~
(xo,xl,x,, . . . ) . )
=
A =
I - S
We shall
so that
(Ax),
= x n - xn+l'
By the definition
hnk
if
0
=
h
k 2 n:
n; and f o r
>
The last step was arrived at by the substitution noting that
i (-l)i+2k = (-1) .
i = j-k,
Continuing,
There a r e no symbolic operations here; Si and
An-k
are
simply powers of rpatrices; the matrices are row finite so all multiplication is associative. 1.
r A p
=
EXAMPLE.
0. Thus
as promised, D2 2.
Let
hnn =
THEOREM.
=
1, hnk
THEOREM.
0
=
if
A
0
u
= p =
k # n
1, and, for
i.e.
H1 = I.
r
>
0,
Thus,
D1D = HI = I. H H
u v
=
For the left side is
4.
Then
1.
p =
H
UV
.
DMDDND
=
DMND.
A l l Hausdorff matrices commute with each o t h e r .
Hence all regular Hausdorff matrices are consistent.
This follows from Theorem 2 and 1.7.12. EXAMPLE. r A r u = (1-t) p 5.
so
Fix
t c K
and
hnk
=
and set un = tn. Then n n-k k (k)(l-t) t from (2).
ALI =
(1-t)p
24
2.3
6. EXAMPLE.
=
I
(i)
(1-t)
Let
l/(n+l).
pn =
n-k tkdt = l/(n+l).
Then
Thus
0 m a t r i x i s a Hausdorff m a t r i x .
pn =
Hp = (C,l).
1'
tndt
so
S o t h e Ces%ro
Note that to agree with the conven-
tion of this section, subscripts start from
0 and the Ces&
matrix is altered accordingly 7. EXAMPLE.
Q
The matrix
(1.2.5) is not a Hausdorff
matrix by Theorem 2 since it does not commute with 8.
Indeed
The Hglder m a t r i c e s a r e a l l H a u s d o r f f m a t r i c e s .
EXAMPLE. Hr =
HIJ
(C,l).
with
pn =
(n+l)-r.
This follows from Theorem 2
and Example 6. Let
9. THEOREM.
and
A
p
be a o n e - t o - o n e
a row-finite matrix.
A commutes w i t h
Hll
Then
=
DHAD.
Substituting DMD
have DADM = MDAD. commute with using
D2
=
M
so
p,
# pnl
i s a Hausdorff m a t r i x i f f
A
.
Half of this is Theorem 4 . DAHD
sequence l i . e .
If
A
commutes with
for H
and using
H we have
D2 = I we
It is obvious that only a diagonal matrix can
DAD
I we have A
is a diagonal matrix; call it =
N.
Then
DND, a Hausdorff matrix.
W.A. Hurwitz and L.L. Silverman (1917), wishing to consider only summability methods consistent with
(C,l), applied the com-
mutativity criterion, 1.7.12, and investigated a p r i o r ; those matrices commuting with
(C,l), a special case of Theorem 9. In
1921 F. Hausdorff introduced his matrices from the point of view
of an analyst interested in the moment problem, as described below.
25
2.3
10. LEMMA.
- -m i l - n i l hn+l,rn+l
Let
mind that
=
0
k > 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) <
m m
lim un p =
6
0
r
-f
m,
fixed),
as just proved, so
exists it must be =
m
( l , O , O ,. . . ) .
0.
Then
H
is the
matrix whose first column is 1 and all of whose other columns are 0.
26
2.3
13. DEFINITION.
if
Hu
A sequence
is called t o t a Z Z y d e c r e a s i n g
p
is a non-negative matrix.
By ( 2 ) the condition is it says
u 2
it says
po
condition.
0 ; for
-
(An)p
2
n = 1 it says
2p1 + u 2 1. 0 ,
vl -
for all
0
2 ul
po
2p2
+
p3
2
( 3 , 2 , 0 , 0 , 0 ,. . . )
The sequence
The sequences in Example 5
p2
n.
n = 0
For
L . . . , for
n
= 2
etc., a convexity
0,
is not totally decreasing.
( 0 < t < l), 6, 8
are totally decreas-
ing. 14. THEOREM. H
matrix
=
H
u
Equivalent conditions for a real Hausdorff
are ( i ) H
is c o n s e r v a t i v e ,
( i i ) IIHll <
(iii) H
m,
is t h e d i f f e r e n c e o f t w o n o n - n e g a t i v e H a u s d o r f f m a t r i c e s ,
(ivl p
is t h e d i f f e r e n c e of t w o t c t a Z Z y d e c r e a s i n g s e q u e n c e s .
It is sufficient to prove that (ii) implies (iv). will follow from the existence of a sequence
since with
1
a = z(v+u),
the construction of
D
matrix
D
given by
and simply define As
we have
UD + TD
D
tors
(UD)nk
=
=
p
=
=
such that
v
p = a
-
6.
We turn to
The idea is a simple one. Consider the dnk = (A nv ) ~ . We express p as a function of v.
w
as the same function of
in the proof of Lemma 10 (iv) we have
i.e.
(U+T)mD
1
f3 = ~ ( v - p )
Now (iv)
(U+T)D
where
dn+l,k, (TD)nk
=
(r)Um-rTrD, i.e.
U, T
dnk
IDI. =
are the (symbolic) opera-
Dn,k+l.mSo, for any dnk =
1
dn+l,k + dn,k+l
m, D
(r)dn+m-r,k+r.
=
It follows
r=O
r=O
that
Now, as predicted, we set
A =
ID] and use it to define
v.
2.3
27
m
Let
1 (r)an+m-r,k+r m = (U+T)mA. We shall prove r=O f(m,n,k) is an increasing bounded function of m
f(m,n,k)
in Lemma 15 that for each
n, k.
=
Let
g(n,k)= lim f(m,n,k) and define v k = g(0,k). m The Proof of (3) is: I (A nU)kl = ldnkl = 11 (r)dn+m-r ,k+r 1 n < f(m,n,k) 5 g(n,k) = (A v ) ~ . (The last step is proved in Lemma 16)
15. LEMMA. rn
f o r each
(U+T)A.
is a n i n c r e a s i n g b o u n d e d f u n c t i o n of
n , k.
Note that =
f(rn,n,kl
f = [(U+T)mA]nk;
The operator
U
A
\Dl = IUD+TDI 5 UA + TA
=
+ T is monotone, so
(U+T)A 5
2
(U+T) A, i.e. f(l,n,k) 2 f(2,n,k) and so on. Next m n k we observe that f(m,n,k) = [(U+T) U T 2 [(U+T)m(U+T)n+kA]oo (U+T)(U+T)A
=
Aloe
=
f (m+n+k,0 , O ) . The inequality holds because
=
n+k 1 r=O
(
n+k n+k r lar,n+k-r ' ( n lan,k -> ank
16. LEMMA.
g ( n , k ) = (An!))
By induction on first, f(m+l,n,k)
n
(check
+
[ (u+T)mAln,k+l = f(m,n+l,k)
=
g(n+l,k) + g(n,k+l).
g(n+l,k) = g(n,k)
n k
( U T A)oo.
Finally
k'
n
= 0
[(U+T)m(U+T)Alnk
=
=
[ ( U+T)n+kA]oo
against the definition):
=
[(U+T)mA]n+l,k
+ f(m,n,k+l) and so g(n,k)
Hence, using the induction hypothesis,
- g(n,k+l)
=
( A " v ) ~ - (Anv)k+l
= (8""~)~.
The next definition will not be used outside this chapter. 17. DEFINITION. tive, regular if
H
U
Call
p
c o n s e r v a t i v e if
Hu
is c o n s e r v a -
is regular.
Thus a real sequence is conservative iff it is the difference of two totally decreasing sequences. of the moment p r o b l e m .
We proceed to the solution
28
2.3
18. THEOREM.
A real sequence
exists a f u n c t i o n pk =
g
p
is c o n s e r v a t i v e iff t h e r e [0,1] s u c h t h a t
of bounded v a r i a t i o n o n
lltkdg. 0
Sufficiency: We may assume that
g
is increasing.
Then, as
in Example 5, = /'tk(l-t)"dg
(Anp)k
> 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).
-.
<
(1) Condition (ii) ak -
It remains to prove that (i) implies that
exists f o r each
k, Reasoning by induction we have
-
an,k+l - pn-k-l/pn = (Pn-k-l/Pn-l)(Pn_l/Pn) = (an-l,k)(Pn-l/Pn) ak(l-X)
as
n
This completes the proof that
m.
-+
ak+l = ak(l-X)
tive. We also have
A
by (1); so 7.
A
is conservak = X(1-X)
ak = ao(l-X)k
so
is regular if and only if
X =
-+
0.
By (l), condition (i) of Theorem 6 can be replaced by:
(It is
lim Pn-l/Pn exists.
of Theorem 6 imply that
X =
1-A).
Also by 1.3.7, the conditions
0 or
Il-XI
<
1. Finally, i f
X
# 0,
C ak = 1 = lim 1; hence a con-
the proof of Theorem 6 shows that
A
I#
s e r v a t i v e Norlund m a t r i x m u s t be c o n u l l o r r e g u l a r . 8.
COROLLARY.
Let
p,
s e r v a t i v e i f and o n l y i f pn/P n
-+
for all
0
n.
Then I N , p )
i s con-
e x i s t s ; r e g u l a r i f and o n l y i f
Zim(pn/PnI
0
9. COROLLARY. (N,p)
-
( i ) p , 1. 0,
be r e g u l a r :
( i i i )z l p n l
< *,
Each o f t h e f o l l o w i n g i s s u f f i c i e n t t h a t
c p, #
0;
bounded; ( i i )p ,
Ipnl
(iv) p
E
0, p,
-+
0;
6.
Sufficiency of (i) and (iii) follow from Corollary 8 and Theorem 6, respectively, while (ii) is a special case of (i).
The
last is just Theorem 5. We now relate properties of 10. THEOREM. o f convergence o f i f and o n l y i f
Let P(zl.
R = 1.
(N,pl
Then
(N,p) with those of b e c o n s e r v a t i v e and
R 5 1.
Further
p(z),
P(z).
R
the radius
(N,p)
i s regular
2.5-2.6
35
Using Theorem 6 and its proof we have Il-XI 5 1
by Remark 7.
P(z)
But
has
(l+z)-'
=
R
Also
R
=
lim lPn-l/PnI
if and only if
= 1
R = 1
and
(N,p)
X
=
=
0.
is not even
conservative. THEOREM.
11. (N,ql,
let
Given
p(z), q(z)
r(zl = p(zlq(zl.
Ngrlund m a t r i x
(N,rl
generating
(N,p)
and
r , t h e sequence generating t h e
Then
r
i s given by
=
(p,q),,
Rn = ( r , 1 J n =
(p,Qln = ( P , q ) , .
This is standard multiplication of power series.
POLYNOMIAL MATRICES
2.6.
1.
THEOREM.
matrices
p, q
Let
b e p o l y n o m i a l s g e n e r a t i n g N:rlund
A = ( N a p ) , B = f",q),
(N,rl = AB
+
such t h a t
(N,rl n, k = (ABlnk
c(N,
where
F
r(z) = p(z)q(z).
and l e t
Then
i s a f i n i t e matrix i e . there e x i s t s
F
for
PI * Of course, ( N , I - )=~ ~(AB)nk
n
>
for
In p a r t i c u l a r
d.
k
>
d
d
cA B =
(when the theorem
has been proved.) d = u + v
Let
-
because
summation, either since
n-m
>
u, v
where
m 2 v
n 1. u
are the degrees of
so
Pn - Pu, and, in the range of
Qm - 9,
so
p,q. For
o r else
m
<
v
so
pn-m = 0
u.
We show that the same is true for
By 2.5.11 it is
C = (N,r).
n sufficient to note that for
n
d , Rn = (p,Q),
=
1
=
1
piQn-i
i=O
U
piQn-i
=
QvPu
since
n-i > n-u
=
v.
i=O 2.
COROLLARY. A l l p o l y n o m i a l N'drlund m a t r i c e s a r e c o n s i s t e n t .
36
2.6
It follows from Theorem 1 that
AB-BA
is a finite matrix,
0. The result follows as in 1.7.12.
hence sums all sequences to
We now give the K u b o t a - P e t e r s e n theorem which characterizes cA
A
if
(lzl
=
1).
=
(N,p) and
p(z)
is a polynomial with no zeros on
The idea is that
p(z)
=
C.*(z-ai)
and Theorem 1
allows us to treat the factors separately. 3.
LEMMA.
Mercerian. 4.
that
p(z) = 1
Let
-
tz
It1
with
1.
<
Then
(N,p)
is
This is the second part of 1.8.6.
LEMMA.
A = (N,p)
Let
p ( a l = 0.
Then
x
be a poZynomiaZ m a t r i x and s u p p o s e
i s summable
=
A.
Note by 2.5.4 that a # 0 , l . Now (Ax), = (p,x),/Pn and -n (p,x), - 1 pixn-i = a p(a) = 0 for sufficiently large n. 5.
THEOREM.
Let
A
(N,p)
=
p
i s M c r c e r i a n i f and o n l y i f
b e a poZynomia2 m a t r i x .
h a s no z e r o s i n
Then
A
i J z J2 1 ) .
Sufficiency is by Theorem 1 and Lemma 3 together with the obvious fact that a finite product of Mercerian triangles is Mercerian. 6.
Necessity is by Lemma 4
LEMMA.
( ~ , p ) . Then
Let
c = A
c
plz)
a =
It1 > 1
with
and l e t
A =
A
is dropped and the resulting matrix is
(t-l)/t, the first matrix of 1.8.6 appears with
eD
LEMMA. =
First By
0,l).
llt. 7.
Then
- tz
#
o Itn).
If the first row of multiplied by
= 1
(a
cB 0 y
Let
B
where
c D 3 cB
b e a t r i a n g l e and
a s i n Lemma 6 , D = A B .
y = B-lf{tn}).
A
since
cA. Conversely if
A
x
6
is conservative. Also
cD
then
Bx
E
cA
y
6
cD
so by Lemma 6 ,
since
37
2.6
u
Bx = u + v , where x
=
B-lu
c
Q
and
v
It"}, so
is a multiple of
+ B-lv.
Now in order to handle
(N,p) we can factor
( N , p ) = ABC
factors and write
...
p(z)
into linear
modulo a finite matrix as in
Theorem 1 where each of A,B,C is a first degree polynomial matrix. We begin with quadratic
p(z)
and there are a few cases; p(z) =
(1-sz)(l-tz):
Is\,It\
CASE I. Both
<
1. Then
(N,p)
is Mercerian by
Theorem 5. CASE 11.
Is1 < 1 < Itl.
A
Let
Then by Theorem 5 and Lemma 7, cAB
=
(N,1-tz), B = (N,1-sz).
=
where
c0 y
y = B ' (
{t"}).
The easiest way to compute the last expression is to replace (1-s)B
by
and then replace
bll
by
1.
B
This results in a matrix
which is equivalent for our purposes and has ( B x ) ~= xn - sxn-1 n n+letn+l and (B-lx), =k=O 1 s " - ~ x ~ .So yn = B-l({tnl)n = s-t y
Q
c 0 Itn} CASE 1 1 1 .
cB 0 y
with
cAB
and
Both y = B
-1
=
c 0 tn
Is1,ltl > 1, s # t. {t"})
E
C B O It"}
As in Case 1 1 , so
cAB =
n {t }
CAB = C B O
=
c 0 Is"} + It"). CASE IV.
Both
= c 0 {sn}0 {ns"}.
This is exactly like the case of repeated n-1 occurs. roots in differential equations where a derivative ns This then is the K u b o t a - P e t e r s e n theorem of 1952 which gives cA
explicitly if
on
I z ( = 1.
has the form /a( <
A = (N,p) with
Namely {nmcr-"}
p(z)
a polynomial with no zeros
cA = c 0 a 0 b 0.. . where each of according as
1 of multiplicity m.
p
has a zero
a
a,b,c...
with
Kubota's Theorem 5 was given in 1917.
2.6
38
8. o n l y if
THEOREM.
p(z)
A poZynomiaZ m a t r i x
has no z e r o 6 i n
1zI
=
(N,p)
i s Tauberian if and
I.
This is by Lemma 4 and Kubota-Petersen. 9.
EXAMPLE.
The matrix
large convergence domain.
Q = (N,p) with
p(z)
=
1
+ z
has a
None of the above theory applies t o it.
Some more information is contained in [55], pp. 161-167. the article [48] is highly recommended.
Also
39
CHAPTER 3 TRIANGLES AND BANACH SPACE
FUNCTIONAL ANALYSIS
3.0.
1.
if
is a vector subspace and
S
then
x
then
E
-
If
S.
i
E
X
=
is a subspace of
S
0 on
Y
implies
E
f = 0 on
i.e. the span of
The most useful is: f(x) = 0,
implies
f = 0,
E is dense.
and
f
is a linear func-
X which is continuous in the relative topology of Y,
tional on
f
if
f
is a set and
is f u n d a m e n t a l
Also:
then
This takes many forms.
Hahn-Banach.
can be extended to a continuous linear functional on
Y.
[80], Corollary 7-2-12, Theorem 7-2-11.
2.
Two comparable complete metrizable topologies on a vector
space must be equal. 3.
The sum of a closed and a finite dimensional subspace must
be closed.
4. Let each
E
([80],Corollary 5-2-7.)
>
([80], Theorem 6-3-3.)
p, q
be non-equivalent norms with
0 there exists
x
with
p(x) = 1, q(x) <
equivalence is guaranteed, for example, if is not.
q
p
Mp. E.
Then for Non-
is complete and
q
( [ 8 0 ] ,##2-1-9, 2-2-1.)
5. An absolutely convergent series (in a complete space) is
convergent.
3.1.
([SO],
Theorem 5.2.2.)
HISTORICAL O u r subject began here.
In 1927 S. Mazur won the prize of the
University of Lwow for solving the consistency problem, Theorem 3.3.6. His matrices were regular triangles and his functional analysis
3.1-3.2
40
was the brand new Banach space theory. To escape from triangles, /
Mazur and W. Orlicz and, independently, K. Zeller used Frechet spaces; still later even the matrices were abandoned! We shall show this process, doing just a little of the special theory in order to motivate the more general. NOTE.
The results in 83.5 are for general conservative
matrices, not just triangles.
CONVERGENCE DOMAIN
3.2.
1. g(y) =
t
E
x
II.
The general continuous linear function on c has the form k lim Y + ty where tk - g(6 1 , x = x ( g ) = g(1) - c tk, (1.0.2).
onto so
is a triangle, A : cA
A
Now if cA
c
is one to one, linear,
becomes a Banach space equivalent to
I] xI
identification norm 2.
+
THEOREM.
triangle); i.e.
A
=
c
I] Ax\]
m.
T h e c o o r d i n a t e s a r e c o n t i n u o u s on
f o r each
with the
n , t h e map
cA
i s continuous.
Pn
a
(A
Here
P n ( x i = xn.
Let
B = Ad',
also a triangle.
Then
Ixn( = IL bnk(Ax)kl
n
3.
THEOREM (INCLUSION AND INVARIANCE). c B x cA.
zngles with
Then t h e t o p o l o g y o f
t h e r e l a t i v e t o p o l o g y of
cB.
If
A,
Let cA
A,
be t r i -
B
i s larger than
are equipotent the
B
t o p o l o g i e s a r e t h e same.
Only the first statement needs to be proved. Then for
x
B
c A , IIxlIB
by 1.7.5 and 1.3.6.
=
IIBxll,
=
IICAxll
5
Let
IICII .I1 xIIA.
C
Also
=
BA-'.
11 CII
<
m
3.2-3.3
4.
f(x) =
h a s t h e form
Here
t, x
t i v e l y , and that
Let
THEOREM.
g
lim x A
+
the result.
Then t h e genera2
5. Then
Let
g(y) = f(x)
Then by Remark 1, f ( x ) =
x
+
lim y
x
and 1 . 4 . 4 .
E
Let
u(f)
using
f
ci.
E
(1.3.8.1
+
where
yx
~ ( f =) p . x ( l i m A ) + x ( u )
~ ( u = ) 0
and t h a t
y
c
B
2
= t A , by Theorem 4
where
I t is
u(x) = yx.
X(limA) = x(A).
THEOREM. (INCLUSION AND INVARIANCE).
servative triangles with
cA.
If
A
Let
A,
b e con-
B
i s conull, s o i s
B.
a r e e q u i p o t e n t t h e y a r e b o t h c o r e g u l a r o r b o t h conu22.
B
A,
which g i v e s
b e a c o n s e r v a t i v e t r i a n g l e and
A
c , f ( x ) = p limAx
Thus
trivial that
I f
ty
so
y = Ax
(15.6.1).
xlf) = ~ ( f ) x ( A )
6.
for
I t is not worthwhile at t h i s
p(f) = x(g).
Note t h a t
THEOREM.
For
cA
E
are t h o u g h t o f a s a r o w a n d column v e c t o r , r e s p e c -
point (although q u i t e easy) t o w r i t e a formula f o r Remark 1.
f
t ( A x ) , t c il.
t(Ax) = C tn(Ax)n.
c'.
E
be a t r i a n g l e .
A
p
41
Only t h e f i r s t s t a t e m e n t n e e d s t o b e p r o v e d .
w e may a p p l y Theorem 5 t o
limBlcA
yielding
By Theorem 3 ,
x ( B ) = px(A)
from
which t h e r e s u l t is o b v i o u s . 7.
A c o n u l l t r i a n g l e m u s t sum d i v e r g e n t s e q u e n c e s .
COROLLARY.
A Mercerian t r i a n g l e must be c o r e g u l a r .
The t w o s t a t e m e n t s a r e t h e same.
The s e c o n d i s Theorem 6 w i t h
B = I.
3.3.
THE PERFECT PART For a c o n s e r v a t i v e t r i a n g l e
closure dense, A
c
in
A, c
is a s u b s p a c e of
c A i s c a l l e d t h e p e r f e c t p a r t of
is c a l l e d p e r f e c t .
amsidering those
f
The Hahn-Banach
which v a n i s h on
c.
cA.
cA. If
c
theorem s u g g e s t s
Its is
42
3.3
on
Then
c.
Let
THEOREM.
1.
f(x)
=
be a c o r e g u l a r t r i a n g l e , f
A
t(Ax)
conditions are w r i t t e n
t A = 0, t
with
Let
COROLLARY.
p a r t of
cA
f
E
ck, f = 0
whenever
x
Q
cA ll
E
f
0
=
(The l a s t two
II.
A
p(f) = 0
and so
be a coregular t r i a n g l e .
i n c l u d e s a l l t h e bounded s e q u e n c e s i n
Let
x
el,
tL.4).
First, ~ ( f )= 0, hence by 3.2.5, Also 0 = f(6 k ) = (tA)k ( = c trark). 2.
E
E
on
c.
f(x)
=
t(Ax).
The p e r f e c t
cA.
By Theorem 1 and 1.4.4.
f(x)
0
=
By the Hahn-Banach theorem (3.0.1) such
Em.
c. The conclusion of Corollary 2 may, but need not, hold for
conull triangles.
(13.5.1, 13.5.2).
The next definition is motivated by Theorem 1.
t
(Theorem 1 ) i m p l i e s
tiA
THEOREM.
4.
i t i s of t y p e If
A
t(Ax)
by 1.4.4.
if
= 0.
A c o r e g u l a r t r i a n g l e i s p e r f e c t i f and o n l y i f
is of type
for
i s s a i d t o be of Type M
A
M.
Banach theorem. =
A matrix
DEFINITION.
3.
x
M
it is perfect by Theorem 1 and the Hahn-
A
Conversely if E
Since
cA. c
Then
f
E
is perfect and
CA
by 3.2.4, also
is dense, f = 0
uniquely determined by
f
in
tLA, let
so
t
=
f(x)
f = 0 on
0 since
t
c
is
3.2.4.
5. Sufficiency is false for conull; indeed a multiplicative
triangle c a n n o t be perfect since limA x
=
1
be of type
if
x
limA
is the solution of
M, f o r example, take
ck, limA
Q
Ax
(Ax),
=
=
0 on
0
c , but
1. Such a matrix may
= X ~ - X ~ - ~ A .conull
angle may be perfect (4.3.11); such a matrix must be of type the proof of Theorem 4.
tri-
M
by
43
3.3
We now give Mazur's famous consistency theorems. THEOREM. L e t
6.
be a r e g u l a r t r i a n g l e .
A
s i s t e n t w i t h every regular matrix of t y p e
such t h a t
B
f i i ) For e v e r y r e g u l a r m a t r i x
M.
consistent with
cA
on
B
n
E.
W
(il A c B 2 cA
with
B
is con-
c
.
B
i f f
cA, A
2
is
A
is
We shall prove instead a more general result:
7. THEOREM. L e t
be a c o r e g u l a r t r i a n g l e .
A
s i s t e n t with every stronger matrix c
i f f
that
i s of type
A
limg = E m A
on
B
such t h a t
(il A
limB = limA
f i i ) For e v e r y s t r o n g e r m a t r i x
M. c, A
i s consistent with
i s con-
on
B
cA
on
B
such
n
m
.
Part (ii) and sufficiency in Part (i) will be trivial from Corollary 2 and Theorem 4, respectively as soon as it is shown that limBlcA
E
ck.
This follows from the formula limgx = lirnn
m limm
bnkxk for x E cA applying 3.2.2 and the Banachk=1 Steinhaus theorem (twice). To prove necessity in Part (i), suppose
A
that
is not of type
1.8.12, and
(BxIn
B
=
MA.
M.
Then
tiA, t # 0. Let
Let
cB = cA by 1.7.16.
Also by 1.4.4,
=
limBx - lirnA x = t(Ax). Theorem 4 . But if x
For
x
E
c
this is
is chosen so that
limBx # limBx; for example if
tk
+
o
0 as in the proof of
t(Ax) # 0
we may solve
This result fails for conull triangles. Remark 5 shows this with 8.
and
M = M(t),
EXAMPLE.
(C,1)
B
AX = gk
for x.
The one given in
= 2A.
It is easy to check directly that
are of type
then
M , hence perfect.
criterion given in the next result.
Q
(1.2.5)
They also satisfy the
3.3
44
THEOREM.
9. A'
Let
w i t h bounded columnij.
If
and s u p p o s e
A E @
Then
i s of t y p e
A
t = t(AA') = (tA)A' = 0
t l A , then
has a r i g h t i n v e r s e
A
is t a k e n t o be a n a r b i t r a r y column o f
M.
x
by 1 . 4 . 4 i n w h i c h
A'.
A v e r y e l e g a n t a p p l i c a t i o n o f t h e s e i d e a s i s t o t h e improvement o f c e r t a i n growth theorems " f r e e o f c h a r g e " : THEOREM.
10.
x
for all
= O(h,)
all
x
c
E
Let
Let
be a p e r f e c t t r i a n g l e and s u p p o s e t h a t
A
x E cA
w h e r e X,
-+
Then
m.
xn = o(An)
for
A'
f n ( x ) = xn/Xn.
ci
sequence i n
Then
by 3 . 2 . 2 .
cA.
l i n e a r subspace of
i s a p o i n t w i s e bounded
{fnl
By 1 . 0 . 5 ,
{x:fn(x)
0)
is a c l o s e d
it includes a l l of
c
Since it includes
+
CA'
11.
EXAMPLE.
x E c
For a l l
we h a v e
11.2.5)
Q
xn = o ( n ) .
By 1 . 7 . 2 a n d Theorem 10, t a k i n g a c c o u n t o f Example 8 . same is t r u e f o r
The e x t r a l i t t l e r e s u l t of 1 . 7 . 2 becomes:
fC,lj.
l i m ( x - x ~ + ~ e)x i s t s i m p l i e s n 12.
Exactly the
(xn-x
)/n
n+l
EXAMPLE (J. DeFranza a n d D . J .
s e q u e n c e w i t h no z e r o terms and
~
~
0.
+
Fleming) /
-+ u 1. ~
Let
u
L e- t ~A
be a r e a l
be t h e m a t r i x ,
c l e a r l y a c o r e g u l a r t r i a n g l e ( i n d e e d &A i s r e g u l a r ) , d e f i n e d by (Ax),
= x
~
(The m a t r i x Clearly
+- ( Q,
tA = 0
{l,-ul,u2,-u3,
~~ ~ / u ~ - n~ =) 1 x,~ 2 , ,. . .
i f and o n l y i f
. . .I ,
suppose t h a t
hence Let
p(z) = 0
A
t
f o r some
t
E
p.
by 2 . 5 . 1 0 ; a l s o
with
un = 1.)
M
u
i f f
L.
be a r e g u l a r Norlund m a t r i x and z, 0 < m
Then
$A
is a m u l t i p l e of t h e sequence
i s of t y p e
A = (N,p)
xo = O,uo = 1.
Take
1 . 2 . 5 , i s t h e s p e c i a l case of
EXAMPLE.
13.
.
IzI
< 1. m
Let
( t A ) k = 1 tnank = 1 z n=k n=O
n-k
tn = Pnzn. pn = z
-k
p(z)
3.3-3.4
=
A
It follows that
0.
3.4.
M.
is not of type
not Mercerian (Theo~em4).
Source :
45
In particular
is
A
This extends half of 2 . 6 . 5 .
[86].
WHEN
c
IS CLOSED
In contrast with 3 . 3 . 4 , I do not know any internal test of a matrix
A
to see if
c
is closed in
cA; nevertheless the condi-
tion is important. 1.
THEOREM.
-
(i.e. cl
perfect part
A
particular
X
Let
A
Let
=
If
A.
E
2. Then
A
In
must contain u n b o u n d e d sequences.
sums u n b o u n d e d s e q u e n c e s .
c
Xc
and assume
Em.
Then using 3 . 2 . 2 , the sequence c = Ix
{Pnl of coordin'ates is pointwise bounded and so exists]
c A , the
is not closed in
c
is closed in
THEOREM.
X.
Let
A
=
-c.
and assume that
c
(1.0.5). E
A
II.IIAJ II.Ilm
is c o r e g u l a r a n d
Thus
c
E
X:lim xn
is c l o s e d i n
are equivalent on
cA.
c.
The two norms mentioned are complete and comparable, indeed
1 1 ~ I 15~ 1lxll;IlAll.
Hence they are equivalent ( 3 . 0 . 2 ) .
where the closure is in the closure taken in
II.Ilm;
\l-l\A
hence in
3.
=
1.
Then
COROLLARY.
c, 1 f
~ ( f )= 1
so
A
f
E
To
the same is true with
since the norms are equj.valent on
By the Hahn-Banach theorem there exists c0 ' f(1)
cA
In
cA
with
is coregular.
If a c o n s e r v a t i v e t r i a n g l e
f
= 0
c. on
(3.2.5).
sums a b o u n d e d
A
d i v e r g e n t s e q u e n c e , i t m u s t sum u n b o u n d e d s e q u e n c e s .
Suppose that and dense in
cA
In contrast
c A c II
m
.
Then
c
is closed in
by Theorem 2 and 3 . 3 . 2 .
A
Hence
cA
CA =
by Theorem 1, c.
may sum unbounded and no bounded divergent
3.4
46
sequences, (1.8.5)
THEOREM.
4.
Let
A E A.
Then
is c l o s e d i n
c
cA
iff
A
i s Tauberian.
If
c
is closed, A
is coregular by Theorem 2, and Tauberian is not closed. To construct
by 3.3.2. Conversely, suppose that
c
a bounded divergent sequence in
we use the classical method of
If
the g l i d i n g hump:
cA
has a hump in
xn
[un,vn] (i.e. IxEl is n un 5 k 2 vn, while xk is 1 somewhere in
very small except when
u
~ it seems clear that C xn (in n’ some sense) will be bounded and divergent. The details follow:
this interval), and if
Fix an integer u
1
> +v
cA by 3.0.3. Thus
(Z,ll-llA)
1, 1lxIIA <
=
Given v
u
E
>
E.
=
in
x2 -...= xu-1 c, hence is not closed in {x
E
co:xl
=
is not complete. But
complete, s o , by 3.0.4, for any IIxllm
Z
and set
Z has codimension u
Then
= 0).
>
~
(Note that
0 and integer u
>
E
0 there exists x
11 xIIA 2 >
(Z,II.Ilm)
IIAll
-11 ~ 1 1 ~ ) .
E
is
Z with
To summarize:
1, there exists x
and integer
such that
and
The existence of v 1 Choose x , v1
E
to satisfy (l), (2) with
2
co. E
= 118, u1 = 2 ,
ul; then x , v2 with E = 1/16,u2 = v1 + 2, v2 . ;xn ,vn with E = 2-n-2 , un = vn-1 + 2, vn > un.
v1
.
is guaranteed since x
>
Now let (Ixn(l*<
C
x
= C
xn
E
u2; . . .
cA. The series converges since
2-”-2 (3.0.5), also xk - En
continuous (3.2.2).
>
XI:
since coordinates are
47
3.4-3.5
We first prove that
x
i s bounded:
fix
k ; now
lxkl n
2-n-2
<
if vn < k , xkn = 0 if un > k and this covers all values of n or all but one at which I x i l ~IIx n 5 1. Hence lxk] 5 1 + L 2-n-2. Next we prove that
x
i s divergent:
choose
k
in one of
N xk
= 1. Then /xkl 2 1 - C t intervals [uN,vN] so that -n-2 = 314 and this is true for infinitely many -> 1 - C 2
if
k
=
vN + 1, k
is not in any
[Un?vnl
and so
IXkl 5
:k # NI
k.
But
' lXk1n 5
c 2-n-2 = 114 and this also is true for infinitely many k. COROLLARY.
5.
Let
b e a conull t r i a n g l e .
A
Then
sums
A
bounded d i v e r g e n t and unbounded s e q u e n c e s .
By Theorem 2, c Tauberian.
is not closed; by Theorem 4 , A
is not
The rest is by Corollary 3 .
3.5. BOUNDED SEQUENCES AND NON-TRIANGLES
The extension of the previous material to general conservative matrices and beyond matrices altogether requires using more general structures than Banach spaces.
However the part that pertains to
bounded sequences can be extended by a simple trick. 1. THEOREM. triangle
coregular, M
-
=
x(A).
cBnk
m
=
r
E m
then there e x i s t s a multiplicative
= cA
n
fim,
c a n b e made r e g u l a r ; i f
x(MI = xIA). A
Thus i f
i s conull, M
a
bnk - ank -
- ak
Also, f o r
x
so
0
=
m
fi
Q
%. B
Then
is
A
i s multi-
0.
First let 1.3.7; bk
A
cM n fi
with
M
plicative
Let
IIB11 5 IIAll + L lakl <
-
by
is multiplicative, x(B) = limBl
, ( B x ) ~- (Ax)n
akxk
= C
for all
n, so
Eo
= c A n ~ . m
Next, for each dnk = bnk
for
k
<
n
choose m(n)
m(n),
0
1 lbnkl < l/n. Let k=m( n ) Then for x E Em ,
so that
for k 1. m(n).
3.5
48
I(Bx),-(Dx),l
IIxllm/n
<
on this set; D
+
cD n II
0 so
0)
=
cB
n
and
im
limD
limB
=
is also row-finite.
Next form a triangular matrix
E
by writing the first rows of
the identity matrix followed by the first row of
D when enough
rows of the identity matrix have been written to keep the resulting matrix triangular.
If the next row of
D
is no longer, write it
immediately; otherwise repeat the first row of
D
ncecessary until the next row of resulting matrix triangular.
E with
matrix
cE = cD
D
as many times as
can be written to keep the
Keeping this up we obtain a triangular
and
limE = lim,
on this set; E
is tri-
angular, enn = 0 for any
Finally if
n
replace it by
l/n.
The
result is the required triangle M. COROLLARY.
2.
Let
be a conuZZ m a t r i x .
A
Then
A
sums
A Tauberian m a t r i x must be c o r e g u l a r .
bounded d i v e r g e n t s e q u e n c e s .
This is by Theorem 1 and 3.4.5. COROLLARY.
3.
Let
be conull m a t r i c e s .
A, B
Then
cA
n
eB
c o n t a i n s bounded d i v e r g e n t s e q u e n c e s .
By Theorem 1 we may assume that
A, B
are multiplicative-0.
D by writing the rowsof A, B alternately.
Form a matrix
Then
is multiplicative-0 and so sums a bounded divergent sequence Then
by Corollary 2 .
(Ax)n = ( D x ) ~ and ~
(Bx),
=
D
x
( D X ) ~ ~ both +~
tend to limits. We are now able to improve 3.2.6 in several directions: THEOREM
4.
with
cB
so i s conuZ2.
A.
3
cA
If
n
(INVARIANCE). II
m
cB
.
n
If
A
am = c B
Let
A, B
i s conuli! s o i s
n
IIm,
be c o n s e r v a t i v e m a t r i c e s B.
If
B
i s coregular
they are b o t h coregular o r both
The same h o l d s a f o r t i o r i i f
cB
2
eA, cB = cA r e s p e c t i v e z y
3.5
49
Only the first statement needs to be proved. x(B)I.
divergent Dx
D
Then x
E
D
= B
-
is conull and by Corollary 3 there exists bounded
cD
il
cA.
is convergent, so 5.
Let
COROLLARY.
By hypothesis, x
E
cB.
Thus
x ( B ) x = Bx
x(B) = 0. No c o r e g u l a r m a t r i x may sum a l l t h e bounded
sequences.
Apply Theorem 4 taking This is the same as 1.7.21.
A
to be any conull matrix, e.g.
0.
-
This Page Intentionally Left Blank
51
CHAPTER 4 FK SPACES
FUNCTIONAL ANALYSIS
4.0.
1. A locally convex metrizable space has its topology defined
by a sequence p(x)
+
0
p
of seminorms in the sense that
for each
p.
The space
2.
/
+
p
If
3.
norms
K
+
A linear map
p, as in 1, f
E
X'
[80], Theorem 7-2-5.
4.
If
is a subset of graph.
XxY
X, Y
If
theorem.
-+
E
w'
pn(x) = lxnl
is continuous iff
n.
[SO], Remark 5-5-7.
M
p , such that
f
/
f
and
If(x)l
is
and if it is closed, f f
p1,p2, . . . ,pm,
2 ME p,(x)
I(x,f(x)):x
E
for
X I ; it
is said to have c Z o s e d is linear and has
is continuous. This is the c l o s e d g r a p h
[ 8 0 ] , Theorem 5-3-1.
$,
+ w
Y, the graph of
space has closed graph 5.
f : X
are Frechet spaces and
closed graph, then
n; f
The topology is
of seminorms where
iff there exist
x.
f : X
for each
0
is a space with topology determined by semi-
selected from the seminorms all
-+
f(x) = ax.
is continuous for each (X,p)
xn
[80], Examples 5-5-2, 4-1-8,
(See the preceding paragraph.)
Pn0f : X
iff
0
such that
$
E
x
also defined by the sequence
Theorem 7-2-2.
iff
of all sequences is a locally convex Frechet
w
a
0
-+
[80], #7-2-6, Theorem 7-2-2, Example 4-1-8.
space with the property that iff there exists
x
Any continuous map to a Hausdorff
[80], Corollary 5-3-3. 4
the space of finite sequences, has no Frechet topology.
[80], #6-4-107, Example 13-3-18.
4.0-4.1
52
X
maps:
.+
are Banach s p a c e s , B(X,Y), t h e c o n t i n u o u s l i n e a r
X, Y
If
6.
Y , i s a Banach s p a c e .
vector topology Let
8.
x.
Then t h e r e e x i s t
g, h
X
For e a c h
norms which g e n e r a t e s
T
+
+
iff
0
iff
0
x
+
p(x) 0
and
C
b e a set of semi-
P(T)
Let
P = UIP(T):T e C l .
X
called
for a l l
of locally
C
let
T E C 1.
+ 0
sup C
and a l l
p e P(T)
T
B
t h e sum o f t h e s e norms. f f ( X , sup C ) '
gi
(X,Ti)',
(see 9)
i = 1,2,,
.. , n
Then
with t h e property
C.
I f each
h a s f i n i t e l y many members, s u p C
10. E
have a c o l l e c t i o n
(X,T) f o r e a c h
in
a norm t o p o l o g y and by a norm:
for
f = g + h , Ig(x)( 5 p(x),
with
as i n
g e n e r a t e s a t o p o l o g y on
x
+ q(x)
Lemma 7-2-15.
[SO],
convex t o p o l o g i e s .
that
and s u p p o s e
X
If(x)I 5 p(x)
satisfies
Let a vector space
9,
x
f
Theorem 6-3-2.
[SO],
be seminorms o n a v e c t o r s p a c e
p, q
h(x) 5 q(x).
h a s a u n i q u e Hausdorff
i s a Banach s p a c e .
T; (X,T)
that a linear functional
P
X
A f i n i t e dimensional v e c t o r space
7.
all
Example 3-1-7.
[80],
T e C. T e C
Thus is
is g i v e n
[80] #7-2-2.
i f f there exist such t h a t
T1,T2..,TnF
f = Z gi.
C
[SO],
Theorem 7-2-16. 11.
A c o n t i n u o u s l i n e a r map between t o p o l o g i c a l v e c t o r s p a c e s
r e m a i n s c o n t i n u o u s when e a c h s p a c e i s g i v e n i t s weak t o p o l o g y . 1.0.7
and [80], 12.
tinuous
4.1.
See
Example 11-1-4.
E v e r y s c a l a r homomorphism o n a Banach a l g e b r a i s con[79],
Theorem 1 4 . 2 . 1 ; p . 277, l i n e 9 .
INTRODUCTION. A l l t h e s p a c e s m e t i n summability ( a t least i n t h i s book) are
4.1
53
sequence spaces i.e. linear subspaces of (of real o r complex numbers.) This cannot continue.
A
For
the set of all sequences
W,
So far they have been Banach spaces. = 0, cA = w
topology, is not a normed space.
which, with its familiar
(See 4.2.12)
To clarify this
last sentence, the various sequence spaces, such as
cA, will be
/
given topologies which make them locally convex Frechet spaces with the additional property that c o o r d i n a t e s a r e continuous.
We have
seen this in 3.2.2. with applications in 3.3.6, 3.3.9, 3.4.1.
A
strong impetus for the development by the Polish mathematicians of this generalization of Banach space came directly from summability. There are some exceptions to the above remarks
- on occasion
one deals with spaces of matrices or of operators as in 81.5; also there are fruitful generalizations to function spaces which we do not cover, What is going to happen now is that spaces of the type just mentioned, called briefly.
FK
/
(Frechet-Koordinat) spaces, will be studied
It will then be shown that the summability spaces can be
given this structure.
This leads to a host of fascinating, even
audacious, questions.
Let
have.
Is
P
invariant?
P
be some property which a matrix may
This usually means:
A, B both or neither have property Now suppose of the space
cA
P
is invariant.
can
cA
Then
P
apply to
FK
P
effects:
it automatically proves that FK
cB
must
is really a property
A.
This leads to a
be defined in terms that
spaces and with no mention of a matrix?
allows us to discuss
=
P?
rather than of the matrix
problem in the naming p r o g r a m :
if
P
This has two
is invariant, and it
spaces with property
P
whether or not
they arose from matrices. We have seen one instance:
type M ; restricting ourselves
entirely to coregular triangles the naming problem was solved in
4.1-4.2
54
3.3.4.
Hence i t is known ( u s i n g 3 . 2 . 3 , i n v a r i a n c e of t o p o l o g y )
that type
M
is invariant.
speak of p e r f e c t
FK
The name i s p e r f e c t .
Now o n e c a n
spaces.
The n e x t b e a u t i f u l p o s s i b i l i t y ( t h e
F K P r o g r a m ) i s t h a t some
c l a s s i c a l t h e o r e m f o r m a t r i c e s h a s t h e form Having s o l v e d t h e naming p r o b l e m f o r
P
and
implies
"P
'2''.
( h e n c e knowing
Q
t h a t t h e y a r e i n v a r i a n t ) w e make t h e a u d a c i o u s c o n j e c t u r e t h a t e v e r y FK
space with property
P
also has property
consider t h e f a c t t h a t a type
M
cA = c . )
and Z e l l e r r e s u l t ( 6 . 1 . 1 ) ,
A s an e x a m p l e ,
c o r e g u l a r t r i a n g l e must sum
bounded d i v e r g e n t s e q u e n c e s by 3 . 3 . 4 , 3 . 4 . 4 . t h e t r i v i a l case
Q.
(We a r e e x c l u d i n g //
T h i s i s s u p e r s e d e d by t h e Meyer-Konig t h a t any
FK
s p a c e i n which
n o t c l o s e d must c o n t a i n bounded d i v e r g e n t s e q u e n c e s .
c
is
My f a v o r i t e
example ( p a r t l y b e c a u s e I had t h e honor of s u g g e s t i n g i t t o A . K . Snyder who d i d n o t r e a l i z e t h a t I was j o k i n g ) is t h e s u p p l a n t i n g of Agnew's p r o o f t h a t e v e r y m u l t i p l i c a t i v e - 0
class ( c a l l e d every conull
n) FK
m a t r i x sums a l a r g e
o f s e q u e n c e s by S n y d e r ' s p r o o f ( 6 . 4 . 3 ) t h a t space includes
a.
(He, and i n d e p e n d e n t l y ,
J u r i m a e , had f i r s t s o l v e d t h e naming p r o b l e m f o r c o n u l l , 4 . 6 . 1 ) .
4.2.
FK
SPACES
These s p a c e s c a n be i n t r o d u c e d w i t h j u s t a t r i f l e more g e n e r a l i t y than is needed.
T h i s a c t u a l l y e n h a n c e s u n d e r s t a n d i n g and i n
a very s h o r t t i m e we s h a l l s p e c i a l i z e t o sequence spaces.
The more
g e n e r a l s e t t i n g a p p l i e s t o f u n c t i o n s p a c e s which a r i s e insumnability b u t w i l l n o t be t r e a t e d h e r e ; i t a l s o h a s a p p l i c a t i o n s i n Banach a l g e b r a and e l s e w h e r e . Let
H
b e a t o p o l o g i c a l v e c t o r ( H a u s d o r f f ) s p a c e 1801,
4.2
55
Definition 4.1.1. ( A large part of the theory about to be presentH , namely that it is a Hausdorff
ed uses much weaker assumptions on space and a vector space.
This assumption covers this and the
succeeding section for example) An
X
space
FH
/
is a locally convex Frechet space
is a vector subspace of
the restriction to
X
map:
-+
H
X
and the topology H
of the
topology
i.e.
X
such that
is larger than the inclusion
is continuous.
H
The reader may take will be 14.1.8 in which The theory of
1.
H
X
H FH
=
throughout.
(rl
The only casualty
is the dual of a Banach space. spaces can be developed without the
assumption of local convexity.
This is done in this section.
How-
ever, since the spaces actually arising in this book are all locally convex we have chosen to incorporate this assumption into the definition. If the phrase "let is assumed that
space
FK
An
H
X , Y be
FH
is the same for
is an
FH
spaces" is encountered it
X, Y.
H
space with
=
W,
i.e. an
FK
space
/
is a locally convex Frechet space which is made up of sequences and has the property that coordinates are continuous as in 3 . 2 . 2 . part of the
A
BH
FK or
This
program is carried out by fiat. BK
space is the special case of the foregoing in
4
which the Frechet space is a Ranach space. m
The most prominent examples of in each of these spaces tinuous.
The space
triangle, cA
is a
o
BK
BK
spaces
are
co, c , L , L ;
( x n (2 lixll, so that coordinates are conis an
FK
space ( 4 . 0 . 2 ) .
space b y 3.2.2.
If
A
is a
56
4.2
2.
THEOREM.
Let
Linear.
I f
f : X + Y
/
X
be a F r e c h e t s p a c e , Y
f : X
i s continuous,
H
+
an
s p a c e and
FH
then
f
X
:
Y
+
i s continuous.
Let
It is sufficient, by 4.0.4
to show that
TH be the topology of
restricted to
Now the graph of
has closed graph by 4 . 0 . 4 . (X,TX)
(Y,TH), hence in
x
logy is larger
and
f
: X
+
f : X
then
linear.
Y
i.e.
has closed graph.
Y. f
Then
is closed in
(Y,Ty) since the latter topo-
x
has more closed sets. 4
X
be a Frechet space, Y Pnof:X
If
so
4.
an
space
FK
i s continuous f o r each
K
+
Pn : Y
+
K
is defined by
Pn(y) = yn
given condition is equivalent to the continuity of 4.0.2
f:X+(Y,TH)
n,
i s continuous.
+ Y
The map
(X,TX)
(TyX TH)
COROLLARY. L e t
3.
H
f
as usual.
The
: X + o by
f
the result follows by Theorem 2.
COROLLARY.
t h e t o p o l o g y of
X
Let
X,
Y
be
i s l a r g e r (on
T h e y a r e equaZ i f and o n l y i f
X
p a r t i c u Z a r t h e t o p o Z o g y o f an
FH
spaces w i t h
FH
X c Y.
Then
Xl t h a n t h e t o p o Z o g y of
i s a cZosed s u b s e t of space i s unique
a t m o s t one way t o make a v e c t o r s u b s p a c e o f
H
i.e.
X
+
In
Y.
there i s
into a
Simply apply Theorem 2 to the inclusion map:
Y.
space.
FH
Y.
This
gives all the result except the statement about equality of the topologies.
If
X
is closed in
Y
it becomes an
TylX; by uniqueness this is equal to X
of
X
= (1
2
3
a , R , co, c ,
is seen to be larger the smaller 2 3
\ x i \)
2
IIXllrn
corresponding to
R
X 2
TY IX
=
TX,
Y.
This result explains a natural phenomenon: t3,
space with
TX. Conversely if
is a complete, hence closed, subset of
member of such a list as
FH
Em,
let W.
X
be any
The topology
is; f o r example
c c.
I\xI12
(Our inclusion of
4.2
the non-locally convex space
k*
57
in the list is covered by
Remark 1.) Corollary 4 is part of the
FK
program in that it supersedes
be
FH
3.2.3.
THEOREM.
5.
X, Y,
Let
and s u p p o s e t h a t
2
is c l o s e d i n
X
By hypothesis, X
spaces with Then
Z.
X
C
Y c Z
i s closed i n
X
is closed in
(Y,TZIY), hence in
X
s p a c e and
Y.
(Y,Ty)
by Corollary 4. EXAMPLE.
6.
i s closed in
c
If
is an
since
X
c
FK
is closed in
is invoked in the equivalent statement: then
X
m
15
if
c
.
X
C
The
c
Ilm, t h e n
C
FK
program
is not closed in
X
must contain unbounded sequences; see 3.4.1.
7. THEOREM. b e a s u b s e t of
X, Y
Let
Then
X.
be
FH
spaces w i t h
c l YE = c l y c l X E .
X
c Y.
Let
E
In p a r t i c u l a r
cl E c c l y E .
X
clyE
3:
2
clXE
closure of each side in 8. THEOREM.
A : X
Let
+.
TX 3 TylX
since
Y.
(Corollary 4 ) .
The opposite inclusion is trivial.
Any m a t r i x map b e t w e e n
Y
i.e A
c
spaces i s continuous.
FK
(X:Y) (1.2.7).
Corollary 3 , to show that for each n, x m map: X K. Now (Ax), = lim 1 ankxk m+m k=l +.
Take the
It is sufficient, by
is a continuous m and each map x -+ ankxk
+.
(Ax)n
k= 1
is continuous since it is a finite linear combination of coordinates. The result follows b y the Banach-Steinhaus theorem (1.0.4). It seems that are not
BK
BK
spaces are smaller than
FK
spaces which
spaces. This is made precise in the next few results.
4.2
58
DEFINITION,
9.
for a set
S
t
A sequence
of sequences i f
xn
i s c a l l e d a growth sequence Oftn)
=
c a l l e d a growth sequence f o r a m a t r i x sequence f o r
for all
x E S.
It i s
i f i t i s a growth
A
cA*
It was shown in 1.7.1 that every triangle has a growth sequence. This is included in the following FK
10. THEOREM. With
Pn(x)
Since f o r each
Every
xn
=
x
Pn
we have
X, lxnl
E
space
BK
11. THEOREM. i n c l u d e d i n some
Let
and so
X'
E
be a s e t o f sequences.
E
E.
any
ti
We may assume that =
0 by
sequence.
Let
ti
(1.0.1)
m.
+
t
i s
E
be a growth sequence
for all
0
Then
h a s a g r o w t h sequence.
E
Necessity is by Theorem 10. Now let for
<
5 ~ ~ P n ~ for ~ ~all i ~ nx, ~ ~
s p a c e i f and o n l y i f
BK
IIPnll
X.
is a growth sequencc for
IIIP,ll)
has a growth s e q u e n c e .
X
IPn(x)l
=
Program result.
since replacing
i
ti = 1 preserves the property of being a growth
X
Ix:xn
=
=
O(tn))
with
ilx/tllm. Then
=
.tm
under the map
space. Also, f o r
x
X, lxnl 5 Itnl.I[xII so X
+
X
so it is a Banach
is isometric with
E
x
llxll x/t
is a
BK
space.
(1.0.1). 12.
THEOREM.
i s included i n a
BK
Suppose that space, and is a w
BK
E
E c X. space.
topology on
No i n f i n i t e d i m e n s i o n a l c l o s e d s u b s p a c e of
E
space.
In p a r t i c u l a r
i s not a
w
is a closed subspace of Then
E
is closed in
By Corollary 4 the and so for
x
E
X
W ,
that
BK
X
w
space.
is a
by Theorem 5 so
BK
E
E
norm is continuous i n the n E, IIxli 2 M 1 lxkl using 4.0.3; k=1
M, n
are independent of
x.
Hence if
=...= xn = 0 (x1,x2,. . . , xn) is an x1
=
x2
follows that x = 0; thus the map x + morphism of E into Rn and so E is at most
it
ison-dimensional.
4.2
59
Of c o u r s e t h e l a s t p a r t i s i m m e d i a t e f r o m T h e o r e m 10. The p r o p e r t y g i v e n n e x t p l a y s a k e y r o l e i n s u m m a b i l i t y DEFINITION.
13.
an
x
space, i f
AK
FK
An
+
3
{snl
anti
x, x ( ~ ' + x , w h e r e
each
X
space
is a b a s i s f o r
the
x"),
AK, o r b e
i s s a i d t o haue
nth
X, i . e . f o r
s e c t i o n of
i s
x
n
1 xk6 k ; o t h e r w i s e e x p r e s s e d , x = C x k s k f o r a l l x E X . T h e k=l s p a c e i s s a i d t o h a v e AD, o r b e an A D s p a c e i f 0 i s d e n s e i n X . T h e i n i t i a l s come f r o m A b s c h n i t t s - K o n v e r g e n z , v e r g e n c e , and A b s c h n i t t s - d i c h t , 14.
s e c t i o n dense.
I
The s p a c e s
EXAMPLE.
s e c t i o n a l con-
and
cs
have
mean t h e s p a c e o f c o n v e r g e n t s e r i e s , i . e I x : C x k
cs
By
AK.
we It
converges}.
n
is
c A , where
(Ax),
1
=
i s a t r i a n g l e so
xk; A
cA is a
BK
k=l space,
n IIxllcs = s u p l c x k l . k=1
(3.2.2), with
AK
follows from t h e
m
calculation
1lx-x(")llA = s u p m l l xkl k=n+l t i o n s h o w s t h a t 9. h a s A K . 15. i s an
x
=
nxn Let
T
AK.
Tn
be
suplTnlXn}.
=
If e a c h
space. has
i s an
Xn
Xn.
he topology on
X
i t is a
hence convergent; s a y H
x
s o all t h e +
t
in
Tn
tn
Tn x
+
T.
If
is a
x
Cauchy s e q u e n c e i n
tn
n , so
second statement, observe t h a t by 4 . 0 . 9 .
X
P l a c e on
E
Xn.
then
AK
t h e topology
x
+
x(k)
T Xn
Then f o r e a c h
tn = t .
are t h e s a m e ; s a y
f o r each
space w i t h
FK
spaces
FH
The c o o r d i n a t e s a r e c o n t i n u o u s i n e a c h
(4.0.9).
Tn, hence i n t h e l a r g e r t o p o l o g y sequence i n
An e v e n e a s i e r c a l c u l a -
0.
T h e i n t e r s e c t i o n of c o u n t a b l y many
THEOREM
FH
-+
t
in +
x
X
Cauchy f o r each
n,
tn
in
n, x
Clearly by 4 . 0 . 9 .
i n each
Xn
+
t
E
X
and
For t h e hence i n
X
4.2-4.3
60
16. If there are only finitely many spaces in Theorem 15 the metric on
X may be taken to be the sum of the metrics on the Xn.
(4.0.9).
In particular t h e i n t e r s e c t i o n o f f i n i t e Z y many
is a
space.
BH
spaces
BH
17. Any sequence space which is a Banach algebra is auto-
BK
matically a
space, since the coordinates, being scalar homo-
morphisms, are continuous.
(4.0.12).
4.3. CONSTRUCTION Every locally convex metrizable space X
has its topology de-
fined by a finite or infinite sequence of seminorms which we write p
or
{pn}
(4.0.1).
sequence in example
X , then
(w,lPI) is an
of coordinates and in
iff :x
w
in which p(x) case, and
What this means is that if {xkl k xk -+ 0 iff p,(x. ) + 0 for each
xk
For
space where P = fPn] is the sequence
FK =
n.
lPnlI , a sequence of seminorms; xk
=
IIxII,; there is only one seminorm, a norm in this 0 in
c
iff
(4.0.2).
IIxkll,
-+
Another example is
0
0 for each
-+
n
+
-+
c
0.
(X,p), then, refers to a vector space X
The notation sequence p
IPI
is a
and a
of seminorms - the metrizable topology resulting has
the convergent sequences just mentioned.
FK
The theory of gence domains.
We begin with a basic construction:
1. THEOREM.
Let
m a t r i x d e f i n e d on Then
Z
is an
X
FK
This means that q l o A , q20A,..
spaces turns out to apply to all conver-
.
fX,pl, i.e.
(Y,ql
be
X c uA.
Let
space w i t h
FK
s p a c e s and
2 = X
n
A
a
YA ={x E X:Ax
E
Yl.
p U (qoAl.
Z is given all the seminorms p l I p 2 , ...
Since this is a countable set which includes p
and
4.3
61
X, hence of
it yields a metrizable topology larger than that of It remains to show that it is complete.
Z. Then
in
f o r each
n.
say
b
Ax
+
x
Also
Ax
Y.
in
b
=
At.
(qnoA)(x-t)
and
Hence
=
qn(Ax-b)
t
Ax Ax
and 0
in
-+
x
t
be a Cauchy sequence
X
in
since
qn(Ax) = (qnoA)(x);
A : X
since
w
i.e. pn(x-t)-tO
in
w
since
Y
x
t
since
pn(x-t)
b
+
E
-+
At
-+
x
Y
is Cauchy in
Then
continuous by 4 . 2 . 8 , Thus
X, say
is Cauchy in
Let
-+
for each
W.
is
+ w
is an
FK
space. and
0
-+
n.
The preceding proof may just as well have been written with instead of
To take advantage of the special character of
w.
we might have written the last few steps as:
(Ax)k
-t
each
(At)k
=
k
k
and
(Ax)k
so that
b = At.
2. THEOREM. A[X]
3
-+
bk
for each
Hence
space w i t h
hence
for
(At)k bk
w
for each
t E Z , etc. i s o n e t o one on
If i n T h e o r e m I, A p
t h e n t h e seminorms
Y,
k
can be o m i t t e d i . e .
and
X
i s an
2
FK
qoA. /
Z is a Frechet space with
It is trivial that
it is linearly isometric with using Theorem 1, Z
qoA, indeed
A : X
under the map
Y
-+
Y.
Thus
/
has two comparable Frechet c.etrics; they must
be equal by 3.0.2. Let
THEOREM.
3.
i s an
FK
I f
i s a triangle,
A
A
be a r o w - f i n i t e m a t r i x .
p, =
space where
11 + / I A ,
seminorms of q =
]I.Ilrn
so
X qoA
are =
is a
(cA,p0)
We apply Theorem 1 with p,,
II-IIA.
n
=
pn = lxn I
X
= w,
(cA,p)
n = I, 2,.
.. .
space.
BK
Y
for
Then
=
1,2,. . . ; Y
c , so that is a
BK
Z = cA.
The
space with
The last part is by Theorem 2 or the
more elementary considerations in 53.2
H
62
4.3
To extend this to arbitrary matrices we introduce the important notion of a t t a c h m e n t : 4. DEFINITION. Then
z-'*Y
Ix
=
E
Let
Y
be an
FK
W : X - Z E
Y1
where
x ' z = I x nz n 1 .
s p a c e and
Note that the definition makes sense even if e.g.
0-l.Y =
Contrast
W .
x.z
with
a sequence.
z
z
-1
does not;
xz which means
xnzn.
C
It
As a special case we obtain a K o t h e - T o e p l i t z dual: 5. DEFINITION.
Then
z
-1
i s an
.Y
h(xI = q ( z - x I .
FK
If
FK
then
AK
z
-1
hn(x-x(m))
=
qn[z.
(X-X'~))]
7. THEOREM. space w i t h
pk
( S e e Remark 1 0 .
has
AK
= 0
for
qn[z-x-(z*x
ZI.
=
and
lxnl
also.
If
z E $,
>
(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
may b e o m i t t e d .
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
may b e o m i t t e d .
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
may b e o m i t t e d .
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
may b e o m i t t e d .
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
-<
Iu(x1)-u(x2)
w e may e x t e n d
q'[A(xl-x2)]
Y
t o a l l of
g
= 0.
Since
by t h e Hahn-Banach
theorem. 2. i s an
YA
x
THEOREM.
E
If
FK
Y A , where A
Let
s p a c e and B a E uA
f
and
i s a triangle, a
s p a c e and
FK
b e an
Y
E
YtA
g
E
Y'
+
f l x l = ax
i f f
(see 4.3.5
may be t a k e n t o b e
a matrix.
A
gfAxl
for the
Then
for
B-dual).
0.
I n Theorem 1 t a k e X = w A , a n AK s p a c e by 4.3.8. Then F ( x ) k k = F(C xk6 ) = C F ( 6 ) x k = a x . S u f f i c i e n c y i s by t h e BanachSteinhaus Theorem (1.0.4) FK
space.
If
since
A
a
E
B c YA B , and t h e f a c t t h a t
wA
i s a t r i a n g l e t h e map
A : YA +Y
YA
is an
is a n e q u i v a -
lence.
3. = p
Zim x A
THEOREM.
+
t(Axc)
Let
+
ax
A
be a m a t r i x .
where
t E II,
a E
Then
f
E
cA
i f f
flxl
cB
A'
T h i s is a s p e c i a l case o f Theorem 2 u s i n g t h e r e p r e s e n t a t i o n
4.4
of
c'.
67
(1.0.2.) w AB .
B by cA
In this theorem we can replace
4.
Sufficiencyis
true a f o r t i o r i and necessity is exactly what was proved in Theorem There are strong reasons for giving the result as we did, bound
2.
To see
up in questions of uniquess and invariance (see 5.4.12). the difference between the requirements on is a triangle we may take if we insist that
a
E
then we must have A
However, consider the matrx
= $.
+
\
where
for all
0
k.
limA x
Then
a
t
p =
#
=
0, a
=
a #
= C
since w i = w B n (Ax), = 1 akxk k=l so we have a $
E
u
=
ak xk 1, t = a
In the latter case
$.
representations for
Theorem 3 ; i f we may t a k e
f
a
= 0 E
and another
cAB
a B B but
=
as appropriate at the time. i s r o w - f i n i t e we may t a k e
If A
THEOREM.
i s a t r i a n g l e we may t a k e
A a E
a = 0.
4
E
CY
If
in
A
The second part was covered in Remark 4.
part, as in Remark 4, take 6.
a E
B c cB aA
In Remark 4 it was shown that
determined by
f.
A matrix is called
tions of each
f
ck
if some one 7. f
E
Let
A
p
=
cB c cB =
9,
A
@
a
E
[ak}
II
9,
(1.
is not always uniquely
u - u n i q u e if all representa-
u.
be a c o n s e r v a t i v e matrix, f
where
B A
For the third
Clearly
ak
Then
xlf)
=
E
c;.
u.x(A).
limn ank by 1.3.7, also (given in the proof of 1.3.2) so E
w
then they all do.
b e r e p r e s e n t e d a s i n Theorem 3 .
a
=
as in Theorem 3 have the same
has unique
THEOREM.
First a E
f
E
(1.
For as given in the proof of Theorem 2 we may choose = wB = 6 .
A
In applications of Theorem 3 we shall use particular
= $.
5.
a
defined by
representation for this function with with
we know that if
in Theorem 3, (3.2.4), moreover
a = 0
w AB
a,
=
Let
4.4-4.5
68
= 1-1 limAl+t(A1)+Cak-Ck(pak+Zntnank+ak)
x(A) + t(A1) - (tA)l
= 1-1
and the result follows by 1.4.4 (ii).
A coregular matrix i s
8. COROLLARY.
p-unique.
At this point in 53.2 we deduced invariance of conull and coregular for triangles from the formula above (3.2.6).
There is
no need for this now since a much more general result was establish(Deduction of the more special result from Theorem 7
ed in 3.5.4.
is easier in that it does not use 3.4.4). 9. THEOREM.
Let
b e a c o n s e r v a t i v e m a t r i x and
A
Then t h e r e e x i s t s a sequence f o r a l l bounded
x
suck t h a t
y
f(x1
= p
10.
THEOREM.
X = z
-1
.Y.
+
g(z.21, a
f ( x l = ax
Let
4.5
ann
=
Let
Then
Y
i s an
X E
zn ’ ank
b e an
$,
new
cA’’
Z i mA x
+
yx
=
s p a c e and
FK
s p a c e and
FK
z
y =
tA +
ci
a sequence.
f E X’
i f f
g E Y’.
0 if
n
+
k
and apply Theorem 2;
COMPLEMENTS In preceding sections FK
with
E
c A’
E
This follows from Theorem 3 and 1.4.4 (ii) with
Let
f
w
and
c
spaces were built up by starting
and using the basic construction of 4.3.1 to build
FK spaces as in 4.3.3. Other construction methods were inter
section (4.2.15), and the use of multipliers (4.3.15). Another important method of building new spaces is by adding spaces.
It
69
4.5
will be sufficient for our purposes to add just two spaces which have zero intersection.
H
In the next result continuity of addition in
is used f o r
the first time.
intersection Then are
z e
Each
s
E
-+
0
-+
+ q(y-t)
=
in
X
is a
Let
x + y
Z
=
+
X
Y = Ix+y x e x , Y S Y I
If x,
2.
-+
i.e.
t
E
Y.
z
-+
u
then
y
for
-+
0
u
Let
x
E
(X,p), (Y,q), hence convergent;
=
s+t. Then
r(z-u) = p(x-s)
H
z
s p a c e suppose
FH
andso
+
0.
so x
-+
0 in
z = x + y
-+
0
X , r x) = P(X) + q(0)
the relative topology of
(Z,r)
(Z,r) proving that t h i s s p a c e i s
in
in
and we
z is a Cauchy sequence in
If
p(x) 2 p(x) + q(y) = r(z)
H ; similarly
y
space.
BH
Z can be written uniquely in the form x+y
X, y
i s closed:
spaces w i h zero
FH
are closed i n
Y
X,
To see that i t i s a n
complete.
z
s p a c e and
FH
be
are Cauchy sequences in
x, y x
(Y,ql
Y = to}.
r(z) = p ( x ) + qfy).
define
say
n
X
spaces, Z
BH
then
i.e.
i s an
Z
(X,pi,
Let
1. THEOREM.
=
X, hence H.
in
X
so
P x)
Let
Next,
has
The last
Z, hence is closed by 4 . 2 . 4 .
statement is trivial. 2.
Let
COROLLARY.
Z
b e an
i c a l l y complementary subspaces o f Suppose t h a t closed i n
X, Y
can be g i v e n
X
Z
= X
+
topologies.
FH
X, Y
can be given an
are closed.
Y,
Then
algebra-
x n X,
Y
=
Y
to}. are
COROLLARY.
Let
X,
has f i n i t e codimension i n
FH topology in which its
Ry the uniqueness theorem ( 4 . 2 . 4 )
must be the original topology on 3.
i.e.
2
X, Y
s p a c e and
2.
By Theorem 1, Z subsets
FH
be
2
2.
this
Z. FH
Then
spaces w i t h X
X
i s closed i n
c 2.
Z.
Suppose
4.5-4.6
70
Y
The hypothesis means that sion.
Y
Let
in Corollary 2 has finite dimen-
H.
have the relative topology of
theorem (4.0.7), this makes
Y
a
BH
By the uniqueness
space and the result follows
by Corollary 2 . 4.
EXAMPLE.
Any a l g e b r a i c c o m p l e m e n t of
in
il
infinite dimensional, indeed it c a n n o t b e made i n t o an by Corollary 2 , since for
co,
co
co.
space
FK
(The same is true
has no closed complement. We make no use
The result may be found in [ 8 0 ] , Example 14-7-8.)
of this. 5.
because
Ern
is not closed in
il
must be
c
EXAMPLE (IMPORTANT).
X
Let
FH
be an
Let
Y be the (one dimensional) span of
ten
my. By Theorem 1 it is an
y.
space and X + Y
Then
FH space and
X
y
H\X.
E
is writ-
is a closed sub-
space. 6. THEOREM. Suppose t h a t
c
Let
Z
be a c o n s e r v a t i v e
i s n o t closed i n
Then
Z.
space
FK
i.e. Z
every non-trivial
space by 4.2.15. By Corollary 3
4.6
X
ilm!
l i n e a r combination i s unbounded.
Suppose the conclusion is false.
closed in
c.
contains i n f i n i t e l y
2
many unbounded s e q u e n c e s w h i c h a r e l i n e a r l y i n d e p e n d e n t (mod i.e.
3
X
Let
X
=
Z
is closed in
( 4 . 2 . 6 ) , hence closed in
n
ilm, an
Z. Now
FK c
is
Z.
COREGULAR AND CONULL We have seen that coregular and conull are invariant properties
(3.5.4.) We now solve the naming problem for these properties.
(A.K. Snyder and E. Jurimae). 1.
THEOREM.
Let
conull i f and o n l y i f
A
be a c o n s e r v a t i v e m a t r i x .
l(n) + 1
weakly i n
c
A'
Then
A
is
4.6
The notation 1
f(sk) = f(1)
4.4.7.
~ ( f )= 0, for all
f
E
ck.
Necessity is by
Conversely, the condition implies in particular, taking
f = limA, that
2. i f
was given in 4.2.13. The condition is that
l(n)
i.e.
71
1ln)
x(A)
weakly
I
=
0.
A conseroative
DEFINITION. -f
X(limA)
=
x(f)
i.e.
=
o
FK
space f
for all
is c a l l e d conuZZ
X 6
XI, otherwise
coregu l a r .
So a conservative matrix conull.
A
is conull if and only if' cA
is
It goes (almost) without saying that conull and coregular
are invariant properties since the topology is invariant (4.2.4). This, and the next result, are parts of the 3.
is conull s o i s
X, Y
Let
THEOREM.
Y.
X
I f
be
FK
FK
program for 3.2.6
spaces w i t h
X c Y.
i s c o r e g u l a r and c l o s e d i n
Y
If
X
then
Y
i s coregular.
Since the inclusion map:
X, Y
remains continuous when
X
+.
Y
is continuous (4.2.4), it
have their weak topologies (4.0.11).
This implies the first assertion.
If
X
is closed in
Y
the map
is a homeomorphism into, (4.2.4) hence its inverse is weakly continuous (4.0.11). If FK
Next comes the
Y
is conull this makes
X
conull.
program for the first part of 3.4.2.
(For
the second part it is 4.2.4). 4.
For
COROLLARY. c
=
cI
If c
is c l o s e d i n
is coregular.
Y,
Y
In particular
i s coregular. Ilm
is coregular.
We saw (3.2.7, 3.5.2) that a conull matrix must sum divergent sequences. c
The
is coregular.
FK
program for this is simply the statement that
But we can collect a much better result, and, in
Theorem 6 an even better one.
4.6
72
5.
THEOREM.
A conull
FII
s p a c e m u s t c o n t a i n i n f i n i t e z y many m
unbounded s e q u e n c e s w h i c h a r e l i n e a r l y i n d e p e n d e n t (mod !2 ) .
This is from 4 . 5 . 6 and Corollary 4 . 6.
THEOREM.
The i n t e r s e c t i o n of c o u n t a b l y many c o n u l l s p a c e s
i s conu2.l. ...
Let
f
E
XI
where
X
=
nXn.
(4.2.15)
Then
f
=
1
where
gk
k=l
We saw ( 3 . 4 . 3 ) that if a conservative triangle sums bounded divergent sequences it also sums unbounded sequences. This is extended in Corollary 8 to an arbitrary conservative matrix.
The p e r f e c t p a r t of any conserva-
gin with the extension of 3 . 3 . 2 . tive
FK
-c,
space is
We be-
and a p e r f e c t s p a c e is one in which
c
is
dense. 7.
COROLLARY.
Let
A
be c o r e g u l a r .
i n c l u d e s a l l t h e bounded s e q u e n c e s i n
Let
f
c;1, f
E
0 on
=
c.
Then
The p e r f e c t p a r t of
cA. ~ ( f )= 0
by 4 . 4 . 7 and
cA
f(x)
for bounded x E cA by 4 . 4 . 9 . Taking x = g k yields y k k m = f(6 ) = 0 so f = 0 on cA n 2! . The Hahn-Banach theorem gives = yx
the result. The conclusion of Corollary 7 may or may not hold for conull matrices ( 1 3 . 5 . 1 , 1 3 . 5 . 2 ) . 8. COROLLARY.
If A
c o n s e r v a t i v e m a t r i x sums a bounded
d i v e r g e n t s e q u e n c e i t m u s t sum unbounded s e q u e n c e s .
Suppose that and dense in
c A c em.
Then
c
is closed in
cA
by 4 . 2 . 6
cA by Corollaries 4 and 7. (Other p r o o f s :
6.5.6,
13.4.7). 9.
EXAMPLE.
For example
m
9.
The
FK
p r o g r a m f a i l s f o r C o r o l l a r i e s 7 and 8 .
i s coregular and its perfect part is
c.
Also an
4.6-4.7
FK
X
space strictly between
= Coz,
z
s
by 4 . 5 . 5 .
Ilm
c
and
Ilmcan
73
be obtained by taking
The proofs of the results must use
matrices in an essential way.
4.7 THE
FK
PROGRAM
With a few exceptions all earlier matrix results have been given their
formulation or else it was shown that the formula-
FK
tion is invalid. FK
From this point on all results will be given f o r
spaces when possible
assume that the
FK
- when a matrix result is given one may
program fails or has unknown status.
The reader should be aware, however, that many of these results were discovered first for matrices.
Since this is a summability book,
our primary motivation is in matrices so we resist the temptation to develop
FK
theory per se.
Bennett and Kalton pointed out that the
FK
program can be
applied more widely than was suspected - namely they observed that convergence domains are separable and so it is permissible to investigate separable
FK
spaces.
This works well
given in Chapter 16, especially 16.2.6.
- details are
This Page Intentionally Left Blank
75
CHAPTER 5 REPLACEABILIT Y AND CONSISTENCY
5.0.
FUNCTIONAL ANALYSIS 1. Let
f
space, M = f*. or dense; f
2.
Let
be a linear functional on a topological vector Then
M
is a maximal subspace; it is either closed
is continuous iff E
M
is closed.
[80], Theorem 4-5-10.
be a convex set in a locally convex space
The weak closure of
is the same as its
E
t
(X,t)
[80],
closure.
Corollary 8-3-6. 3.
Every continuous linear map:
w -+ w
is range closed.
[80],
This result is used only in 5.5.1 for which an
Theorem 12-4-20.
elementary proof is also cited. 5.1. FUNCTION AS MATRIX
f
Throughout this chapter we use the fact that each
c
cA
has
a representation f(x)
= 11
limAx + t(Ax)+
ax, t
€
as given in 4.4.3. From (1) it follows that 1.
THEOREM.
there exists
Note that
B
f
E
B cA
c A C aB
.
has a r e p r e s e n t a t i o n ( 1 ) w i t h
If f
with
L, a
cB
-
cAJ
lim,
=
p
# 0
then
f.
may also have a representation with
p =
0.
(4.4.4)
We may assume that
1-1 =
1.
Let
Mazur’s Mercerian matrix, 1.8.12.
D
=
MA
Let blk -
where
M
=
M(t),
ak, bnk = ak + dn-l,k
5.1-5.2
76
n
for
>
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
<
-
f o r each
is a n e i g h b o r h o o d o f
p
6
P, then
0, U
3
is
Z xn
n{x:p(x)
c
E )
where t h e i n t e r s e c t i o n i s t a k e n o v e r a f i n i t e s u b s e t F o f P . m [SO], Theorem 4 . 1 . 1 2 . With ym = 1 x n w e h a v e f o r p 6 F , n= 1 m r n k p ( y -y ) 5 1 p ( x n ) < E f o r l a r g e k , m . T h u s , f o r l a r g e k , rn, k+l
6.0-6.1
90
ym-yk
6
6.1.
c
Gym)
and s o
U
i s a Cauchy s e q u e n c e , h e n c e c o n v e r g e n t .
NOT CLOSED
The t w o main r e s u l t s o f t h i s c h a p t e r a r e t h a t i f closed i n a conservative
FK
X , then X
space
d i v e r g e n t s e q u e n c e s ( 6 . 1 . 1 ) and t h a t i f
case
is n o t
c o n t a i n s bounded
i s c o n u l l ( i n which
X
is n o t closed by 4.6.3), t h e n X
c
c
must i n c l u d e a l a r q e
s p a c e o f a c e r t a i n s p e c i f i e d form (6.4.2).
The f i r s t r e s u l t w a s
p r o v e d e a r l i e r u n d e r r a t h e r r e s t r i c t i v e h y p o t h e s e s (3.4.4); t h e e x t e n s i o n i s a n item i n t h e
is n o t c l o s e d , X
conservative
FK
Recall a l s o t h a t i f
program.
c
h a s l o t s o f unbounded s e q u e n c e s t o o (4.5.6).
THEOREM.
1.
FK
(W. Meyer-Konig
and K . Z e l l e r ) .
space i n which
c
i s not closed.
Let
be a
X
Then X
con-
t a i n s bounded d i v e r g e n t s e q u e n c e s . W e may assume t h a t t h e t o p o l o g y q n+l 2 q n
o f seminorms s a t i s f y i n g BK
s p a c e w e may t a k e a l l
9,
by
q
is g i v e n by a s e q u e n c e
X
for a l l
n, for i f
X
q
is a
t o be t h e same, o t h e r w i s e r e p l a c e
n
n
1
qk'
k=l
Fix an i n t e g e r Then
u > 1 and s e t
h a s codimension
Z
Z
So on
3.0.3.
i n d u c e d by any o n e exists Given integer
x E
6
X
>
u
=...=
c , hence is n o t c l o s e d i n
in
x
u- 1 = 01. X
by
t o p o l o g y is s t r i c t l y l a r g e r t h a n t h e
i.e.
such t h a t
Z
0
(4.2.4) h e n c e s t r i c t l y l a r g e r t h a n t h a t
qi
> 0, integer
v
II-Ilrn
the
r e l a t i v e topology of
u
Z = Cx E c : x1
IlxII,
f o r any =
1, q i ( x )
u > 1, integer
such t h a t
E
> 0 < t.
and any
there
i
To summarize:
i, there exist
x
and
6.1
91
and x The existence of 1
x , v1
Choose
v
E
x,
qi(x) <
E
is guaranteed since
x
to satisfy (11, (2) with
v1 > ul, i = 1; then x',v2 with n v2 > 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)
<
11
=
in the following definition.
and
“p(x)
X
Let
1. DEFINITION.
X
{ x : p ( x l 5 11
i f
p(z(k)j
<
p(x)
whenever
admissible f o r
Dn
where
X
FK
is an 3.
{x
=
E
E
X:
en + x 2 E
P
as
a
p
i s closed i n
X
and, f o r a l l
=
kth
{x
k,
m.
be an
section of
E
x.
(4.2.13).
II-II,
space, then
FK
X: IIxII, 5 11. Then
is D = nDn
Dn is closed in
Each
Let
Let
X
be an
E C X.
Then
t h e r e e x i s t s a sequence X
and
X
since
space.
DEFINITION.
in
4
i s called admissible
X: lxn[ 5 11.
a d m i s s i b Z e norm. {x
D
Let
X.
3
p
p(x) <
x
2. EXAMPLE. L e t
space
FK
Then
X.
x (k) is the
Recall that
The
is defined” are synonymous.
b e an
norm d e f i n e d on a s u b s p a c e of for
For purpose of this chapter the
and
{p(en)}
space
FK
P
bounded}.
4
p
and
an
i s defined t o b e
2 E
{en}
3
o f p o i n t s of
For
p =
ll.Ilm
E
with we w r i t e
2-E.
The use of
2
is to indicate the presence of two topologies
2 E is called the two-norm closure of E (even though X P need not be a normed space.) It may be a good deal smaller than
Thus
the ordinary closure, for example take
X
= w
in Theorem 6. It
6.2
94
may even be empty! 4.
2 = {x c
LEMMA.
With
x:
<
p(xl
fl
(Take E X,p,E
Z in Lemma 4).
as i n Definition 3, 2 E
-1.
n 2 E , say en z with p(e ) 5 M. P since the disc (p 5 M) is closed in X. Let
z c
X
Let
admissible norm.
Then
definition
Z
=
{z'")).
c 2 4. (For W P Ix c X: p(x) < - 1 .
an
p
see 5 . 6 . 1 ) .
Let
W
z c
n
By
2.
lies in the closure of the
z
an + z in X where ty 2 0, Xitin = 1 for each n. Then n Each an clearly lies ti = p(z). Say
(5.0.2).
=
4.
in
6. THEOREM.
Wb
and
2 @
Z
Z{trz(i): 1 5 i 5 m(n)1, n n 5 p(z) C p(a ) 5 C ti ~(2'~)) an
space
FK
weakly, so
z
-+
n
W
Here, as before,
convex hull of
be a n
p(z) 5 M
Then
-+
LEMMA.
5.
Uhere
2
C
P
=
Wb
X
be a conservative
space.
FK
=
W
n em.
Let
z c 2,c0.
In view of Lemmas 4, 5
and Example 2 it is sufficient to prove that z 6 W. let f c X'. Say an + z , an E co, IIan[lm < M. Then f(z) since
a c co
implies a and
=
=
To this end
n k) limnCk akf(6
c ak gk
in
Clf(6k)[
<
-
(1)
co, a fortiori, in since
X z co
X.
NOW
(1.0.2,
Hence the series in (1) is uniformly convergent and so
4.2.4).
-
we may let n + k I: zkf(6 ) since that
lim f(a n )
=
5 Mlf(sk)l
la!f(6k)l
Then
.
2-4 = 2-c
Here
Let
z
6
inside the summation. This leads to an
-+
z
and
X
is an
FK
f(z)
=
space. This proves
W.
7. COROLLARY ( A . K . Snyder). conull if and onZy if
1 E
2-4.
A
conservative
FK
space is
6.2-6.3
95
This is by Theorem 6 and 5.6.2. 8. The condition
1
is weaker than that of Corollary 7
E
and does not imply conull
(5.2.1,5.2.5).
Applications and historical discussion of two-norm topologies are given in r121, 564-565.
OSCILLATION
6.3.
Let
DEFINITION.
1.
r
positive integers with when
b e a s t r i c t l y i n c r e a s i n g s e q u e n c e of
r 1
Or(Xl,
Then
1.
=
i s f i x e d , i s t h e s e q u e n c e whose
r
w h i c h we w r i t e
nth
O(XI
term
(x) = m a x { l x -x I : r < u < v 5 r I and u v n n+l n ( r ) = { x : O n ( x ) + 0 as n -+ m } ; n b ( r l = n t r l
o
Properties of representation.
In the sequel
such t h a t
cz =
Of course
derived from a matrix
Ix:Ax
=
.
c0 1
E
as in 1.2.2. triangle
0
A
ntr).
A
depends on
- x
(Ax), = xn
ci
(1
There e x i s t s a m u l t i p l i c a t i v e
THEOREM.
2.
are most ConvenientlY
Sa
m
n
where
AX)^
Let
r.
=
is chosen so that
i
x1
and, for
n
n 5 ri+l.
ri
i'
A
Each row of
IIAll
are zero and ri+l 2 n
If
has two non-zero members Further
= 2.
so each column of
x
E
n(r),
A =
ri
0
implies
<
~ / 2 . Now let
u
=
ri, Ixu-xvI
proves that
Oi(x) <
=
E
Choose
> 0.
I ( A X ) ~ ~<
E.
n
+
since
m
x
so
0
E
cA.
1
6
If
as
Ixn-xr,I 5 Oi(x) * 0
x
I(Ax),I
-
terminates in zeros.
Conversely, let
cA,
-L
21, thus the row sums
Hence
i
> j
j
such that
and
~ / 2 while if
x
E
n(r).
ri 5 u
<
u > ri,
n > rj v 5 I'1+1' .
> 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
<
n
5 211yI1,/2
= lyn+l-yn\/2
+
In Lemma 5, { z n 1
6.
that
n
uln+Zl
by
u(n)
by
u(n+l)
-
uln)
+
-.
Let
0 so
x
6
go in equal steps from
xi = x k'
zn+l; finally let <
n+l 2 , Ok(x)
n(r).
Also
Ixr -x k+l k'
=
x
The same proof works with
I
is bounded.
c a n b e r e p Z a c e d b y any s e q u e n c e
such
u
replaced
2"
- u(n). Suppose t h a t
r
i s g i v e n and
vk.
more t h a n o n e
f o r each
Then t h e r e i s a subsequence
y c trn t h e r e e x i s t s
i s an i n c r e a s i n g
u
[r,, r n + l ) c o n t a i n s
s e q u e n c e of i n t e g e r s s u c h t h a t n o i n t e r v a l
n
starting from the
except in its last two occurrences where it is replaced
7. LEMMA.
property:
x
x
E
w
nb(r)
with the
v
of
such t h a t
Yn.
=
) and w n = vi so that Say Vk rrm(k)"m(k)+l n rrm(in)*rm(in)+l) = [ru(n)jru(n)+1 ) ; w can be chosen so that
'
wn
u(n+l)
- u(n)
x c Q(r)
+ m.
such that
By Remark 6 there exists, for each xr u(n)
-- yn. The construction
y
m
6
L , an
of Lemma 5,
6.3-6.4
with
u(n)
instead of
97
2", shows that
xw(n) - Yn
also.
Now attachment takes a third form: 8.
DEFINITION.
sequence.
Let
be a s e t of s e q u e n c e s a n d
E
z.E = {z-x:x E E}.
Then
The n o t a t i o n
a
z
was intro-
2-3:
duced i n 4 . 3 . 4 .
9. THEOREM. 2
Let
z
E
Qm\coJ
m
i s a n o n s e p a r a b Z e s u b s p a c e of
with
Z P.
= z-fi
.
(r) f o r some
b
Then
r.
I n p a r t i c u l a r this a p p l i e s
z = 1
Let
S
be a non-countable set of sequences of
0's
and
1's.
> E > 0; v can be chosen to satisfy the hypotheses Let IZv(n) of Lemma 7. Then with w as in Lemma 7, there exists, for each
s'
s e S
an
s # t
we have
nb(r)
E
Isn-t
-
Izw(nll I~;(~)-t;(~)l
with
s' = s . w(n) n for some n.
I
=
=
Izw(n)l 2
1
E.
Thus
Now if Then 2
s,t E S
with
IIz.s'-z.t'Ilm contains a non-
countable set all of whose pairwise distances 1.
E.
10. Lemma 7 may be interpreted as saying that there is a
regular matrix an,wn - 1, ank 6.4.
which maps
A =
0
for
)'(bif
onto
Q".
Namely, let
k f wn.
CONULL SPACES
In this section we give the bigness theorem of A . K. Snyder, Theorem 2 , which gives a best possible inclusion theorem for conull
FK
spaces; namely that such a space must include some
is best possivle in that each
O(r)
Q ( r ) ;it
is itself conull by 6 . 3 . 3 .
Partial results, now superseded, were given in 4 . 6 . 5 . Theorem 2 differs from earlier bigness theorems in that it gives a sufficient condition for summability, namely, every sufficiently slowly oscillating sequence is summable. theorems are called d i r e c t t h e o r e m s .
Such
They are opposite in spirit
6.4
98
t o Mercerian and Tauberian theorems
1 E
Let
LEMMA.
1.
2,$
X , a"
an + 1 i n
so t h a t
b"
+
space
FK
Then t h e r e e x i s t s
(6.2.3).
Let
b e an
X
0
X; b i
in
+
as
0
bf: = 1 f o r s u f f i c i e n t l y l a r g e
such t h a t
r
n
+
+
X
c
3
fl(rl.
b" = 1
Let
- an n,
k ; f o r each
f o r each
m
such t h a t
(5.2.8)
IIanllm b o u n d e d .
$,
E
$I
3
k ; IIbnllm < M .
The s e c o n d c o n d i t i o n h o l d s b e c a u s e c o o r d i n a t e s a r e c o n t i n u o u s . Let
r1 = 1 a n d c h o o s e
suppose t h a t
rl,r2,...,rn-l
Choose
Pn > P n - l
k = 1,2,
...
9
Now l e t
y
and
so t h a t
E
Pl'P2'.
IIbpnll < 2-"
and choose
'n-1
IIb p1 11 <
so that
p1
rn
r
>
* .
5.
,Pn-l
so t h a t
a n d w e s h a l l show t h a t
n(r)
have been chosen.
Ib:nl
and
n-1
Inductively
< 2-"
= 1
bn : y
E
for
+
X
Lrn.
for k
co,
completing t h e proof. F i r s t let
z"
= b
Pn
, tn
-
= yr
yr
n T h i s series converges by 3.0.5 s i n c e
IIznll
<
2-",
h e n c e , by 6.0.3,
11 tnznII
n-1
ltnl < O,-,(Y) <
2-"
and t h e p r o o f is c o n c l u d e d by showing t h a t
= T1
+
T2
+
n (t1=y 1) a n d x = Z t n z . +
eventually.
y
-
x
E
I n t h i s i n e q u a l i t y we choose
Tg, s a y .
and
0
Thus
c0 '
m
so t h a t
m
n - 1 2 m+l
5
11 tll
so t h a t
2-m-1.
Thus
terms e a c h o f w h i c h 2. some
THEOREM.
rn-l > rm+l > k
[ yk-xk -+
0
I
as
hence
Tg 2 1 ltn12-" n=m+2
i s dom n a t e d b y t h e sum of t h r e e m
E v e r y conuZZ
+
-, hence a s
FK
1 E 2m$
+
space i n c l u d e s
r. By 6 . 2 . 7 ,
k
a n d s o Lemma 1 a p p l i e s .
-. ntr)
for
x
E
X
6.4-6.5
3. cA
3
COROLLARY
nb(r)
r.
for some
i f it includes
A matrix
(a).
nlr)
( b ) An
f o r some
99
i s conull if and o n l y if
A
s p a c e i s conuZZ if and o n l y
FK
r.
Part (a) follows from Theorem 3, 6 . 1 . 2 and 6 . 3 . 3 .
Part (b)
follows from Theorem 3 , 4 . 6 . 3 and 6 . 3 . 3 . EXAMPLE.
4.
Since
is coregular and includes every
Ern
nb(r),
Part (b) of Corollary 3 cannot be improved to resemble Part (a). Let
COROLLARY.
5.
i s a non-separable
b e a conull
X
s u b s p a c e of
FK
space.
X n
Then
Ilm
km.
This is by Theorem 2 and 6 . 3 . 9 .
6.5.
COREGULAR SPACES AND MATRICES. The conull theory may be extended by using attachment. 1.
Y
general
Y
FK
is an
c c X, so
representation of
f
k
f ( 6 ) = a k + g(z.6
Necessity.
and so
=
k
).
Let
f(1)
z c W.
+
If
1.2 = z c X
Also
u
E
so
Ilm,
space.
FK
g(z-y),
n
X
E
g
u. E
$,
co
then
1 c Y.
The E
X'.
The
is c o n u l l
z c wb(X).
Now by definition of
g(z)
0'
is a c o n s e r v a t i v e f ( y ) = cty
z
c
is given by 4 . 4 . 1 0 .
Sufficiency: Let
Then
II
W i t h t h e h y p o t h e s e s of T h e o r e m I, Y
THEOREM.
if and o n l y if
space
FK
space is 4 . 3 . 6 .
u c Y.
0
2.
Y
i s g i v e n by
f c Y'
That
be an
X
Then
14.3.41.
= 2-I.X
U-z E
Let
THEOREM.
= C
That
f c Y'.
W,
X'
c
f(6
z c
= C ak
k k g(z) = 1 z k g ( 6 ) = c g ( z - 6 ) . k
Hence g
By Theorem 1, f(1)
k
f(1)
=
f(6
and define
) = 1 g(z.6 Qm
C
k
)
f
and
by
Y
f(y)
) = C g ( zk s k ) =
+ g(z)
Meanwhile is conull. =
g(z.y).
c zk g ( 6
is part of the hypothesis.
k
)
6.5
100
Lemma 5 . 6 . 8 i s a s p e c i a l case o f T h e o r e m 2 . 3.
Let
LEMMA.
r
Then t h e r e e x i s t s Y = z
Let
-1
be a c o n s e r v a t i v e
X
such that
X
z
W
E
b'
z-R(rl.
3
By T h e o r e m 2 , Y
.X.
s p a c e and
FK
is c o n u l l and t h e r e s u l t is
immediate from 6 . 4 . 2 . 4.
COROLLARY.
n
Wb # co, X
If
i s a conservative
X
FK
i s a n o n - s e p a r a b l e s u b s p a c e of
km
space w i t h km.
From T h e o r e m 3 a n d 6 . 3 . 9 .
Now w e c a n o b t a i n a s t r o n g i m p r o v e m e n t of T h e o r e m 6 . 1 . 1 : 5. c
THEOREM.
i s not closed.
such t h a t
X
s u b s p a c e of
3
Let
X
be a c o n s e r v a t i v e
Then t h e r e e x i s t s
zqfitrl.
FK
z E km\co
In p a r t i c u l a r
X
n
space i n which
r
and a s e q u e n c e i s a non-separable
km
am.
I t i s s u f f i c i e n t b y Lemma 3 a n d C o r o l l a r y 4 t o show t h a t
W
c o n t a i n s a b o u n d e d d i v e r g e n t s e q u e n c e , or, w h a t i s t h e s a m e , t h a t 2,c0
c o n t a i n s one ( 6 . 2 . 6 ) .
R e f e r r i n g t o t h e p r o o f of 6 . 1 . 1 ,
b y t h e same p r o o f w h i c h s h o w e d t h a t
x
let
is bounded i n 3 . 4 . 4 .
The r e s u l t of T h e o r e m 5 is new o n l y f o r c o r e g u l a r s p a c e s ; f o r c o n u l l s p a c e s i t is c o n t a i n e d i n 6 . 4 . 2 . 6.
THEOREM.
Let
A
bounded d i v e r g e n t s e q u e n c e . sequence
r
such t h a t
a non-separable By 6 . 1 . 3 Theorem 5 .
c
b e a c o n s e r v a t i v e m a t r i x w h i c h sums a Then t h e r e e x i s t s
cA 2 zefiirl.
s u b s p a c e of
( z = 1.)
m
k
z E km\co
In particular
eA
n
and a km
is
.
is not c l o s e d i n
cA a n d t h e r e s u l t f o l l o w s by
6.5-6.6
The
6.6.
FK
program f a i l s f o r Theorem 6 a s p o i n t e d o u t i n 6 . 1 . 4 .
SUBSPACES OF Although
101
Qm.
i s t o o b i g t o be included i n
.Lm
cA f o r a n y r e g u -
l a r o r c o r e g u l a r m a t r i x by 3 . 5 . 5 , and t h e same i s t r u e f o r c e r t a i n subspaces of
ilm
t h e r e are o t h e r subspaces of
by 6 . 1 . 2 ,
which c a n b e i n c l u d e d i n s u c h
Let
THEOREM.
1.
cA:
there e x i s t s a regular triangle g i v e n any m a t r i x
B c @,
such t h a t
A
cA
A
.
Then
Also,
E.
2
t h e r e i s a row s u b m a t r i x
p.
of
B
such
eA 3 E.
that
A row s u b m a t r i x o f
of
m
be a s e p a r a b l e subspace o f
E
Rm
B.
If
B
i s a m a t r i x o b t a i n e d by d e l e t i n g r o w s
B
is chosen r e g u l a r , t h e n
w i l l be r e g u l a r a l s o .
A
The f i r s t p a r t of t h e t h e o r e m f o l l o w s from t h e s e c o n d by 3 . 5 . 1 .
Now l e t
B c @ ; w e may assume
IIBII = 1.
u n ( x ) = ( B x ) ~ ; l u n ( x ) l 5 I(BX),l unit disc in
5
sequence
{ u ~ ( ~ )w)i t h t h e p r o p e r t y t h a t
f o r each
x c E.
m a t r i x of 2.
B
Thus e a c h
llXllm.
I t f o l l o w s from 6 . 0 . 2 t h a t
El.
{unl
lim u
E c cA where
This says
un c E '
Define
A
k(n)
by
un c D , t h e h a s a sub(x)
exists
i s t h e row sub-
g o t t e n by d e l e t i n g a l l b u t t h e rows numbered
COROLLARY.
Every m a t r i x i n
Apply Theorem 1 t o
E
=
@
h a s a row submatrix
k(n).
in
r.
c.
The same r e a s o n i n g c a n b e a p p l i e d t o r e f l e x i v e
BK
s p a c e s of
n
bounded s e q u e n c e s , s u c h a s
.LA.
However, any r e f l e x i v e
is s e p a r a b l e ( 6 . 0 . 1 ) s o t h i s is n o t h i n g n e w .
BK
space
6.7
102
6.7.
BIGNESS
W e h a v e s e e n o n l y o n e " a b s o l u t e " m e a s u r e of b i g n e s s , namely n o t h a v i n g a growth s e q u e n c e ( 4 . 2 . 9 ) .
a
An
s p a c e is i n c l u d e d i n
FK
s p a c e i f and o n l y i f i t h a s a growth s e q u e n c e ( 4 . 2 . 1 1 ) .
BK
111
a n o t h e r c o n t e x t w e h a v e s e e n t h a t c o n u l l s p a c e s are " r e l a t i v e l y " b i g g e r than c o r e g u l a r o n e s ; every s p a c e is i n c l u d e d i n
, which
w
i s c o n u l l , and no c o n u l l s p a c e c a n b e i n c l u d e d i n a c o r e g u l a r o n e , However, t h e r e i s n o t h i n g a b s o l u t e h e r e .
(4.6.3). if
i s a c o n u l l t r i a n g l e , t h e map
A
A : cA + c
F o r example
i s a Banach s p a c e
e q u i v a l e n c e between a c o n u l l and a c o r e g u l a r s p a c e , z.X
then
and
=
x
~
+- x n~;
example l e t doxically in
and
(Ax)n = x Z n , (Bx), X
n
n (Ax)n = (-1) (xnWl-xn),
A s a t h i r d e x a m p l e , t w o co-
may b e so l a r g e t h a t
Y
=
X
x ~ ~ X += ~c A , Y
+ Y =
for
= w,
cB.
Para-
must be l a r g e i . e . t h e r e i s a l o t o f "waste"
Y
+ Y , f o r example i f
X
X + Y,
xn
cB = z . c A .
then
X
regular spaces
z = {(-l)n}
are s u r e l y t h e "same" s i z e , y e t o n e c a n b e co-
X
r e g u l a r and o n e c o n u l l , f o r example l e t (Bx),
If
( 4 . 5 . 1 ) , so
X
X
n
Y = {O}, X
would b e c l o s e d i n
+ Y would be c o r e g u l a r ( 4 . 6 . 3 ) .
(In fact
Y J c.) F i n a l l y t h e r e are c o n u l l s p a c e s which h a v e a growth s e q u e n c e ,
cA w i t h
e.g.
a t r i a n g l e ; and c o r e g u l a r s p a c e s w i t h no g r o w t h
A
sequence e . g . t h e s p a c e
X
given i n t h e preceding sentence. X
The example j u s t g i v e n , o f c o r e g u l a r
with
Y = z.X
conull,
shows t h a t c o r e g u l a r and c o n u l l a r e "shape" r a t h e r t h a n " s i z e " p r o p e r t i e s ; more s p e c i f i c a l l y , t h e y measure t h e r e l a t i o n s h i p between weakly z c
w.
1 Z
and
(J;
in
X , 1 i s n o t weakly
z n s n ; 1 d o e s n o t have weak
AK
but
Z
6"
z
but
z
is
does; 1 $ W,
103
CHAPTER 7 SEQUENCE SPACES
7.0. FUNCTIONAL ANALYSIS. Let
1.
q: X
-f
space and such that
Y ' q
X.
on
[Sol,
f
X.
is closed in
into a Banach space such that linear functional
X
be linear and onto where
on
q
Y
Then
Corollary 6-2-15,
can be made
is continuous and open, and a
is continuous iff
Y
is a Banach
foq
Theorems 6-2-8,
is continuous
6-2-5.
2. A pointwise bounded sequence (1.0.3) of continuous linear /
maps from a Frechet space to any locally convex space must be
[SO], Theorem 9-3-4, Example 9-3-2.
equicontinuous.
3. The convergence lemma (1.0.5) holds assuming that is equicontinuous on a topological vector space.
4.
#9-3-104.
A continuous linear map from a
(Open Mapping Theorem).
/
[SO],
Ifn)
I
Frechet space onto a Frechet space must be an open map.
[SO],
Theorem 5-2-4. 5. Let
X, Y be Banach spaces and
linear bijection. such that
f(x)
=
this formula, let IIT-lII -I1yll. Now (1.0.1)*
I
Then f o r each g(Tx) y
=
llT-lll
for
x
E
f
6
X.
T : X
Y
a continuous
there exists
X'
[Proof:
Defining
g
5 IIfll.11xll
Then
<
by the open mapping theorem so
=
If(x)l
g E Y'
Tx. m
Ig(y)l
-+
by
zIIf/l. g
E
Y'
7.1
104
7.1. MONOTONE NORMS. In this chapter we present only those results from the extensive theory of sequence spaces which will be needed here.
To
follow the subject further one may consult [31] and r55], where many of the results are given in greater generality.
t o h a v e mo n o t o n e norm if sup I l r ( m ) l l
X
Let
1. DEFINITION.
IIr
be a
( m ) 11
BK IIr
space.
I n ) 11
Then m > n
for
i s said
X
11~11=
and
.
m
1 [ x i ( ,II1(n)ll = n SO this norm satisfies the first condition k= 1 but not the second. It is not a monotone norm. 3.
EXAMPLE.
co ,c , ( I r n , cs ,bs
The spaces
The next result shows that an
AK
have monotone norms.
space may be assumed to
have a monotone norm.
x
4. THEOREM. L e t s u p 112 I n ) 11 f o r x c X . i s equivalent with 5 E
For example, IIx+yll
IIxII *
+ 11 yII *.
let
fn: X
-+
Now X
tinuous since
Ilx(n)ll
In particular
=
11 XI]
be defined by
X
is an
FK
~~x\= l*
AK,
(n)
11
jlc
5 s~p(llx(~)~l+lly(~)II) 5
by
AK
fn(x)
=
IlxII * 2 IIxll .
so x(~).
Each
fn
space; and the sequence
5 Mllxll
that the norms are equivalent.
and
5 ~llrll f o r u z z
and so
Conversely is con-
{fnl
By uniform boundedness (1.0.3),
= llfn(x)ll
x
a m o n o t o n e norm f o r
sup
+
pointwise convergent. Hence
*
space w i t h
BK
I ~ . I I *is
Then
jl-II.
x.
be a
IIx\l* 5 M.Ilxll.
I] frill
is <
M.
This proves
To prove monotonicity:
if
m
>
n,
~ ~ * each is the sup of an eventually constant ~ ~ x ( ~ ) ~ ~ *x ( ~ )since sequence agreeing in the first supmsup~llx(n)ll :n 5 mj
=
*. IIXII
n
terms.
Also
sup
~Ix(~)~~*
=
7.1-7.2
105
Theorem 4 could be proved without using continuity of
5.
11.II*
coordinates by showing that closed graph theorem.
is complete and applying the
Since coordinates are obviously continuous
in this norm the same is true in the original norm.
This is an
important result in functional analysis which has no significance See [2] pp. 110-111, [79], Theorem 11.4.1.
here.
Since
A
write
u(m,n>
Then
A = IC,l).
is a triangle, this is a
AX(^))^\
=
-1
In
u(n,n)
=
is decreasing in
7.2.
Let
EXAMPLE.
6.
T
BK
h a s m o n o t o n e norm.
cA
space (3.2.2).
x
Fix
and
xkl. Then IIxlj = sup u(n,n). A l s o k=l if n 5 m, u(m,n) if n 2 m. Since u(m,n)
IIx(~)/I
n, it follows that
=
sup{u(n,n):n 2 m l .
DUALS
In this section we discuss various duals and give some inclusion relations. 1.
These definitions make sense if example
X’,
Recall the definitions of
Xy
=
n{xy:x
B
XI
where
X
Xy
given in 4.3.5, 4.3.17.
is any set of sequences, for
xy = {y: C xiyi
is a bounded
series I .
2. THEOREM. iil E
C
Let
E B B , (iil E B B ’
b e s e t s of s e q u e n c e s .
E, El =
same r e s u l t s h o l d for t h e
(iiil I f
E’,
El
3
E
then
Then
E;
C
The
E’.
dual.
Y
68
Parts (i) and (iii) are trivial. versely, by (i) and (iii), E
5
3
5 (EBB) .
By (i), E’ c (E’) Theproofs f o r the
; con-
y-dual
are the same. 3.
I{fisnll:
DEFINITION.
f
6
X‘l.
Let
X
be an
FK
space
3
4.
Then
Xf
=
106
7.2
It is important to observe that whenever is assumed that
X
THEOREM. L e t X Let u c Xf , say un
T.
to
Then
un = g(6"). f
E
un
X
is a
9.
Then
Let
g
3
Xf
= (Tjf.
be the restriction of
Conversely let
u
E
(T)f,say to
g
f(6"). co 0 z with z unbounded ( 4 . 5 . 5 ) . f space, X = L! by Theorem 4 and 1.0.2; X ff =
Let
BK
(1.0.2) so
Ern
space
PK
f(6").
=
g(6 n ) .
=
=
EXAMPLE.
5.
be an
By the Hahn-Banach Theorem (3.0.1) extend
X'. Then
Then
un
is written it
$.
3
4.
f
XI
xq
X
=
Xff.
However we get the analogue of part of Theorem 2: 6. THEOREM. Y , then
f
Xf
then
3
If
Yf.
is c l o s e d i n
X
= YJ@.
XJ@
If
Xc Y
If
Y'
6
then
flX c X'
by 4 . 2 . 4 .
The second part is by
Theorem 4
xY
7. THEOREM.
Let
x f.
x
c
(ii) If
X
be an
FK
has
AK, 'X
=
space
3
T h e n lil X B c
$.
~ f ,(iii) IJ@
x
has
AD,
x B = xY . Let f n'
u c XB
and define
f(x) = ux
by the Banach-Steinhaus theorem
c X'
u c Xf, Thus
so
(ii): f(c xn6n )
Let = Z
x
(1.0.4). Also
E
X.
Then
f(6")
=
X B c Xf
u c X f , say
xnun
(1.2.1) for
un
=
f(6").
For
x
E
X, f(x)
=
u c XB. The opposite inclusion was just
so
proved. iii): Then
{fn}
Y
n
1 ukxk for x E X. k=1 is pointwise bounded, hence equicontinuous (7.0.2). Let
u c X
Since
lim fn(x)
X E X
(7.0.3) i.e. u
and define
exists for all E
XB.
x
E
fn(x)
$,
=
it must exist for all
The opposite inclusion is trivial.
7.2
8.
(T)'
Xy c
(i):
(S)'
=
107
(by (iii)) c
given in 5.2.14. That X B # X
Y
X
AD
has
is not a sequence space.
of
X'
Let
i n t h e sense t h a t each
u
and t h e map
X'
E
b e an
xB
u c
u
is less clear since
The plain fact is that the members
(La)'.
X
X'
For example, no sequence space can
are functions, not sequences.
9. THEOREM.
is harder; it is
is possible is shown in 7.3.5 .
The relationship of the duals with
have the cardinality of
by Theorem 4 .
Xf
XY # X f ; one such is
It is easy to give examples in which
Example 5. To give one in which
XI
(T)f =
More on this in Example 11. space
FK
Then
$.
3
can be used t o r e p r e s e n t a f u n c t i o n
i s an isomorphism i n t o .
+
X B c X'
If
has
X
AK
t h e map is o n t o .
Namely, u(x) Theorem (1.0.4).
=
ux.
fi
If
the map is one to one.
Then
u
E
0, then
=
X
If
has k
X'
un
by the Banach-Steinhaus =
AK
6(6") and
=
f
E
0
so
X'
u let
0
=
un
i.e.
f(Z xk 6 ) = ux. Thus u 0 XB and f The converse of the last statement is given in 10.5.1.
Then for all
x, f(x)
THEOREM. L e t
10.
a q u o t i e n t of
XI
q l f ) = Ifl6'll
by
comrnu t e s
=
x
be an
i n t h e s e n s e t h a t t h e map i s onto.
=
=
3 $.
Then
xf
q : X'
+ ~f
given
Moreover t h e diagram
'X
+
.
is
X'
Xf
The second means that for
u
6
Let
X
=
c.
Then
XB
=
X
f
= L.
The embedding
of Theorem 9 is not onto, even though there is a sense in which =
II
B X ,
u , using the notation of Theorem 9.
11. EXAMPLE.
c'
6.
L +
The first part is trivial. q(;)
space
FK
f(6")
=
(1.0.2); for
follow that
un
=
lim
lim 6"
E
=
X', but it 0
so
lim x
lim u
=
=
ux
it would
0, contradicting
7.2
108
lim 1 since
=
1. Also the quotient map in Theorem 10 is not one to one
q(1im)
0.
=
12. THEOREM.
Let
be an
X
space
FK
( t h e map of Theorem 1 0 i s one to o n e ) iff
Sufficiency: Nexessity:
If
If
X
q(f)
does not have
(3.0.1) supplies a function
q(f)
0 then
=
f
E
f
has
X
on
= 0
=
X'
AD.
hence
$
f
= 0.
A D , the Hahn-Banach Theorem
f # 0, f
X I ,
=
0 on
Then
$.
Aficionados of duality theory can do a little Hellinger-
Z
Toeplitz number on this theorem. Let be inclusion.
Theorem 10.
Then
i': X '
(Note that
Z'
-f
Z'
=
Zf
bc X
=
=
Xf
particular
q
again!) Thus
is one to one iff
Z
= $"
Ri = Nq'
=
=
q
b.
Ri
= Nq'.
-+
X
of
A
by Theorems 12 and 4 ) .
is onto
i
i: Z
and let
is the quotient map
standard result (e.g. [ 8 0 ] , Lemma 11-1-7) says
(In
- this is Theorem
12
This shows that the value
Z is forced. 14. THEOREM.
BK
Xs
= 0.
13.
of
Then
II $ .
Let
X
be a
BK
space
3
Then
$.
Xf
is a
space.
The quotient topology by the map of Theorem 10 (See 7.0.1) makes it a Banach space since is closed in
-+
is one and 'q
=
If :f = 0 on
Continuity of coordinates follows from 7.0.1
X I .
since pnoq: X I
X'
K
IP,[~()I
satisfies
=
lf(6")I
2 Ilfll.ll6"ll
hence is continuous. We saw in Extample 5 that
X
need not be included in
Xff .
However part of it is: 15. THEOREM. ( c l o s u r e in
X).
Let
X
Hence if
be a
X
has
BK
space AD,
X
C
3
$.
Xff .
Then
Xff
3
$1
7.2-7.3
109
The second part is equivalent to the first by Theorem 4 so we shall assume that show that
Xf
Pk
true because space, and f(u)
(1.0.1).
for
Lf
=
7.3.
Xff
Pk(x)
to be meaningful. xk
=
since
(Theorem 10.) Now let
u c X'.
Then q: X '
Xf
+
=
=
Qm
3
X
Let
X X
but
and
X'.
E
Let f
X"
E
In particular
z
(4.5.5) with
does n o t have
EXAMPLE.
L
The familiar
BK
E
Then
L".
AD.
A).
(Consider one row of
( L ~ ) = f,.
spaces,
That m
similarly. Hence
cB = L .
y c L B , f(x) = xy.
Then
y e L".]
Finally :c
To introduce
bv
If
2.
LEMMA.
3.
DEFINITION.
Also lynl
and
IX
C
+ c
4. bvo
2 IIflla116"/1= =
Q"
IIfll.
Hence
by 7.2.7.
we prove: x E c.
then
m
b v , t h e space o f s e q u e n c e s o f bounded v a r i a -
v e r g e s ; / / z j l b v = l 2 i m x1 =
is proved
= 9.
b : Trivial; C : Let
= 9,
lf(6")I
<
-CC
x
3:
'Q
:c
Qf = L y = Q B
bvo
t i o n , i s t h e s e t of a 2 2
bv: 2 i m
=
and
= L
bs, cs, co, c , L m
It was pointed out in 1.3.2 that
have been introduced.
B
E
x c X.
FK
NINE SPACES 1.
ix
is an
(7.2.12): this yields
c0 0 z
=
X
This is
5 IIu/I.IIx11 so
If(u)l
u(x) = g(qu) for all u g(qPk) = g(6 k ) so x c x ff .
16. EXAMPLE.
Xff
But first it is necessary to
with
Pk(X)
Xk =
AD.
where
Apply 7.0.5 to
(XI)'
g E
X'
Q
q(Pk) = gk
u(x)
=
has
in order f o r
$
3
X
s u c h t h a t t h e s e r i e s i n Lemma 2 con-
Ixk-xk+ll; b u o = bu
n
co =
01.
is equivalent with
,P
under the map
x
+
y
=
m
the inverse map being
y
+
x
with
x
=
1
k=n
yk.
Moreover
{ X ~ - X ~ + ~ ~ ,
110
7.3
IIxIIbv
IIyill
=
for
x
bvo.
E
bvo
Thus
i s a Banach s p a c e .
That
m it i s a
space is proved as follows:
BK
x
1
=
x
~
1~
+
+
k=n Hence
bv
bvo 0 1
=
= bv,
( v i ) bs'
(ii):
bvo
li)
b s , ( i i i ) bv'
=
is also a
has codimension
5. THEOREM. bv!
lxnl 5 IIxII
(X~-X~+~). Thus
bvo = limL
that
I
1
ukl
= lf(l(")>l
1
bv
=
by 7 . 2 . 7 .
bv
=
Y buo =
c s B = c s Y = csf
bv.
bvoB
Also
u
let
( i i )bv:
(v)
bs,
=
Note
(5.0.1).
space.
( i v ) bv Y = b v f
=cs,
bvt = bv:
=
in
n.
(4.5.5).
space
AK
=boo, I v i i ) b s Y = b s f
bv;
for each
BK
i s an
test f o r convergence. Conversely Then
( X ~ - X ~ ++~ )
k=n
m
E
f bv,,
3
bs
by Dirichlet's
k uk = f(6 1.
say
5 ~ ~ f ~ ~ * ~ ~= lIIfll, ( n ) so \~ u
E
bs.
k=l (iii):
E
bvB c l B
hence
cs c bvB.
=
i s by Abel's test for convergence. Y Y Y (iv): bv = (bvo@l) = bvo n '1
= bs
The last inclusion
The rest is
by (ii).
by 7 . 2 . 6 and (ii).
(v): let
u
E
csf = csy = cs' = bvBB by (iii) and 7 . 2 . 7 . Conversely, Y cs . Now if y E c and xn = y, - yn-l (convention:
yo = 0) then
for
k
c
<
u
bv.
E
for
m
(vi)
Ay
(1.3.3).
bs'
and so
cs
E
n, un
follows that
]IA]l
x
6
k
=
n,
u - x E bs.
0
for
By Abel's identity (1.2.9),
k > n.
Since
u - x E bs
it
and since this i s true for each y E c , n-1 Thus for each n, \]A]] 1 Iuk-uk+ll and so k=l
= bv:B
by (v); finally let
km
2
bvo
by (ii).
u = {(-l)n}.
Conversely, bs B c csB = bv
Then
u
E
bs
so
bs B c u B c co
111
7.3-7.4
(vii) bsy by (v).
bvYV
bv
I>
by (iv).
Conversely, bsY c csY
spaces for
kp
p
bv
1 play an insignificant role in this
>
We give a brief discussion with references. 1lP 6. DEFINITION. ] ] x \ ] p= ( Z ~ Z ~ / ~ )f o r p 1. I ; Q P
kc:I1xllp
Qp
+
l/q
let
exists
B
(P)=
=
-1.
<
7. EXAMPLE. p > I,
=
The other part is by 7.2.4 and (v).
The book.
=
I/p
y c Eq
is a
then
1
=
BK
s u c h that
space w i t h f E
AK
(eP,J’
p 2 1.
for
For
i f and o n l y i f t h e r e
llfl]
f 1 x ) = x y ; moreover
=
IIyllq; a l s o
eq.
See [79], Examples 4.1.3, 6.4.3. Completeness follows from reflexivity. 8. The inequality
Example 7, H:Zder’s
If(x)l
inequality:
5 llfll*\Ix\l (1.0.1) yields, by I Z
I I ~ 1 1 ~ l l ~ l lthe ~,
xiyil 5
left
side being absolutely convergent if the right side is finite. 9. THEOREM.
x c bv.
Let
E a c h o f t h e n i n e s p a c e s h a s m o n o t o n e norm.
Then
m- 1
Also for
m
>
n , Ixnl
I 1
=
( X ~ - X ~ ++~ )
k=n Substituting this in (1) and using (1) with 5
gives bv
Il~(~)ll.
hasmonotone norm.
Also from (l), IIxII
1
5
Ixk-xk+l/
n
replaced by
=
lim
This implies the same for
Il~(~)ll.
+
IXml’
m Thus
For the
bvo.
other seven spaces the result is obvious.
7.4. DETERMINING SETS. 1.
{x c
$:
DEFINITION, L e t X b e a B K s p a c e . (We do not assume X 2 $ ) .
IIx11 5 I}.
Then
D
= D(X)
=
112
7.4
This is not the unit sphere in
X , but rather its intersection
9.
with
DEFINITION.
2.
Let
be a
X
w i l l be c a l l e d a d e t e r m i n i n g s e t f o r
c o n v e x h u l l of
X
A subset
@
i s the absolutely
DtX)
i f
of
E
E.
EXAMPLE.
3.
space.
BK
{6n}
i s a determining s e t f o r
A
be
1
tk6
Let
I?.
m the absolutely convex hull of this set.
If
x
E
A, x
=
k
k= 1 k
x E 0 and IIxII 5 Cltkl.llb 11 = C [ t k l 5 1. Conversely i f m 1 x g k and 1 2 IIxI1 = Z ( x k [ ;thus x 6 A. k= 1 k
so
x
=
EXAMPLE. { I ( ~ ) iIs a d e t e r m i n i n g s e t f o r
4.
f o r the notation).
set.
If
x
E
A, x
A
Let m
1
=
(Convention: 1") = 0). m 1 (Xk-xk+l)l (k) since k=l so x E A. EXAMPLE.
5.
Let
Let m
x
=
1
A
tksk so
y
be
x
xm+l = 0. Also
IIxII 2 Z ~ t k ~ ~ ~ ~ l ( k ) ~ ~
' Z I X ~ - X ~ +5~ (IIxII 5 1
b e t h e s e t of a l l s e q u e n c e s i n
E
9
6
x c D, x
i.
sj E E
x
and
m
Then
t l .
E
and
llx/j 5
Z
4
each
i s a determining s e t f o r
k jtkj-//s jj =
C
If
Itk/ 5 1
x
E
A,
Con-
m
versely if
Each
@
E
be the absolutely convex hull of this set.
k= 1
f o r all
x
so
x s D, x =
of whose n o n - z e r o t s r m s i s co.
b v o (see 4.2.13
1 xksk = c xk(l (k)-l(k-l)) k=l By Abel's identity (1.2.9) x =
5 1. Conversely if
= Cltkl
D,
6
be the absolutely convex hull of this
tkl(k)
k= 1
x
Let and
E .
1
x
=
1
xkAk; suppose first that
k= 1 = sgn x , m =
1
(
j=l rearranged so that
lxil 2
x. 1+1
I
i 1j for j = 1,2,... ,m. i=l x . 1 - 1 x . 1)sJ s A. In general, let J J+1 .
SJ
=
as a
7.4
member of
A.
Since
E
113
is invariant under permutation of the
terms of its members, so is
A ; hence
x
E
A.
This Page Intentionally Left Blank
115
CHAPTER 8 INCLUSION AND MAPPING
8.0.
FUNCTIONAL ANALYSIS 1.
A set in a topological vector space is called bounde d if it
is absorbed by every neighborhood of
0. A continuous linear map
preserves bounded sets; hence a bounded set remains bounded in any smaller topology.
If
A, B
are bounded sets,
A + B
is bounded.
[80], §4-4. 2.
In a locally convex space the absolutely convex hull of a
bounded set is bounded. topology is bounded i.e. is bounded in 3.
X.
Every set which is bounded in the weak f[E]
bounded for all
f E X'
implies
See 4.0.11 and [SO] #7-1-1, Theorem 8-4-1.
If a linear map from a metrizable topological vector space
preserves bounded sets it must be continuous.
[SO], Theorem 4-4-9,
Example 4-4-7. 4.
E
A set
E
in a locally convex space is bounded iff
is bounded for each continuous seminorm.
p [El
[80] Theorem 7-2-6.
116
8.1
TABLE
7 ( 8 . 6 . 3 ) (8.6.6)
b"o
bv
bs
C
0
C
cs
a
tp9 P>
Ern
w
-
T 8.4.1C 8.5.19 8.4.7A
-cL 8.4.1C 8.5.10 8.4.78
8.1
117
INTRODUCTION
8.1.
In this chapter we give a general principle which will yield, for a given
FK
space, criteria for an
FK
space to include it.
From these criteria we obtain conditions on a matrix
A
to
6
(X:Y)
A
i.e.
maps
X
A
squivalent
Y, this is done by simply
into
applying the inclusion criteria to decide when
YA
X.
3
The
classical treatment of these mapping theorems was to guess the conditions e.g. Silverman-Toeplitz (1.3.6), showing them sufficient by some standard inequality, and then (usually the difficult part) showing them necessary by contradiction: satisfy the conditions, a sequence that
Ax # Y.
x
E
X
if
A
does not
is constructed such
The construction might be by a gliding hump,
possibly extremely complicated.
For a very inclusive list of
references see 1721. The soft approach, initiated in Banach's book [2] in 1932, is the one we showed in 1.3.3, 1.3.6, it is based on uniform boundedness.
The approach of this chapter depends ultimately on
the closed graph theorem. tend to be easier.
Whenever this is true, calculations
The approach was worked out in a series of
seminars at Lehigh University involving Wilansky, Snyder, Bennett, Kalton, Garling, Ruckle and others.
It overlaps with ideas of
many other writers. The success of the method may be judged in that it gives short proofs a n d d e r i v a t i o n s (the classical methods gave only proofs) of 81 conditions out of a possible 90 in the Table, p. 116.
Of
the remainder, 8 are (IIm:co) and its modifications; no soft treatment of these is known.
(aP:llr) II
2
Finally there is the famous pair
about which we say very little.
It is interesting that
, the best of spaces, is the least tractable in the present
context.
The reason, ultimately, seems to be that its unit disc
8.1-8.2
118
has too many extreme points!
8.2.
INCLUSION In order for Y
to include
X
it is necessary that the
Y be smaller than that of X
topology of
A converse
(4.2.4).
result can be obtained as follows: 1.
=I
4.
LEMMA.
Let
Suppose t h a t
Let Iun)
x
6
X.
X Ty14
be an
There exist
is a Cauchy sequence in
X, Y are FK spaces, un
Then
TX($.
C
-+
s p a c e and
AD
6
Y
and so
un
-+
x
y
in
both
space
FK
Y 2 X.
un
with
$
an
Y
and
un
-+
x
in
X.
in Y.
y
W,
Then
Since
hence
x = y
x c Y.
and so
For the concept "bounded" refer to 8.0.1.
2. 3
4.
LEMMA.
X
Let
be an
s p a c e and
AD
S u p p o s e t h a t e v e r y s u b s e t of
a l s o bounded i n
Y.
Then
is continuous (8.0.3).
an
space
FK
w h i c h i s bounded i n
$
X
is
Y 2 X.
($,TX)-+ Y
The inclusion map:
Y
preserves bounded sets hence
This is equivalent to the hypothesis of
Lemma 1. If space
X
-
$,
AD
is not
these arguments may be applied to the
Also the properties of
do not enter.
$
general statement of what was proved.
AD
Here is a
(This all holds for
FH
spaces too. ) : 3.
LEMMA.
X, Y
Let
These are e q u i v a l e n t w h i c h i s bounded i n (i) = (ii):
(il X
be
FK
s p a c e s and l e t
T y l E c TXIE, ( i i l
i s a l s o bounded i n
E
c
X
Y.
e v e r y s u b s e t of Y,
(iiil
Y
3
E
clXE.
(i) is true iff the inclusion map is continuous;
this in turn, is true iff this map preserves bounded sets (8.0.1, 8.0.3).
n
8.2
(iii) implies (i):
FK
By the inclusion theorem 4.2.4 applied to
and 8.
spaces Y
(i) implies (iii):
replaced by
space Y
3
E, X
and
3 $,
i f f
X
by
THEOREM.
4.
119
The proof of Lemma 1 applies with E. be a
X
Let
i s bounded i n
E
space w i t h
BK
a determining s e t f o r
E
D
Lemma 2, let B c mD
=
D(X)
B
EXAMPLE.
5.
4 and 7.4.3.
B
m. hence Y
$
Y
2
X
by 8.0.2.
Now, applying
which is bounded in
X.
Then
Y.
i s bounded i n
Y.
By Theorem
(Another treatment in 8.6.3.)
if and only if
E , is bounded in
Y.
In fact, Example 5 fails if {&"I
space in which
FK
ISn]
Y
is not
&'I2. W. H.
is bounded in
Ruckle has shown, [56], that n{ilP:p > 01
intersection of all
we
D(X), the absolutely convex hull of
assumed locally convex since
semiconvex FK
For general Y
Y.
Theorem 4 uses local convexity of have
FK
Then
(7.4.21.
is bounded in
Isn}
i f f
1
3
an
Y
To prove sufficiency, note
Y
is bounded, in
be a subset of
for some
X
AD,
Y.
Necessity is by 4.2.4 and 8.0.1. first that
$
-
is the smallest locally
is bounded, and that the
I&")
spaces in which
is bounded is
$,
hence there is no smallest. He has also shown, [57], Theorem 4.1 that for each balanced and bounded set
B
in
space which is the smallest (locally convex) B
FK
there is a
BK
space in which
is a bounded set. 6.
EXAMPLE.
Y
Theorem 4 and 7.4.4.
0's
w
3
bvo
iff
i s bounded i n
Y.
By
(Another treatment in 8.6.4).
7. EXAMPLE. Let E be the set of all finite sequences of and 1's. Thus a member of E is the characteristic function
a . 2-8.3
120
of a finite set of positive integers. bounded in
Eo,
member of
iff
Y 3 c
is
E
(Another treatment in 8.6.2.) This is by Theorem 4
Y.
and 7.4.5 except for one detail: call it
Then
is not the determining set,
E
given in 7.4.5. However
Eo
E c Eo
E
=
- E
is the difference of two members of Eo
bounded if and only if
(i.e. each
E)
E
so
is (8.0.1). The condition that
bounded is referred to in the literature as:
E
is be
1 has unconditional
section boundedness. 8. Since
is not an
km
AD
space none of the techniques
given here applies to it.
Conditions for spaces
particular
km
Y , to include
in the Table p. 116.
i.e.
A
E
0's
8.3.
and
then
1's
are referred to
A sufficient condition of some interest is
given in [SO], Example 15.2.2, namely if of
(8":Y)
YA, with
Y
3
k".
Y
includes all sequences
(Compare 19.3.4).
MAPPING THEOREMS
A mapping theorem is a characterization of A: X
matrices
ization of
+
Y.
(X:Y)
i.e. of
An example is the Silverman-Toeplitz character-
(c:c), 1.3.6.
A
For a matrix
if and only if
YA3 X.
let
YA = {x
E
wA: Ax
E
A
Y); then
E
(X:Y)
Hence inclusion theorems lead to mapping
theorems. The presentation of results will be as follows: in the top row of the Table p. 116 are (a) only if an
AK
BK
spaces.
Direct application of the principle
Y A 3 X. space.
(b) Remark 6.
All 9 spaces
A
6
(X:Y) if and
This is restricted to the cases in which
X
The end result is Theorem 4.
Varying
Y
when
(X:Y) is known and
X
has
AK,
is
8.3
(c)
Enlarging
X
(d)
Obtaining
( Y 8 : X B ) from
121
by one dimension; Remark 7. (X:Y), Theorem 10.
The display of applications in the next section will follow this order The inclusion Theorem 8 . 2 . 4 shows that we need criteria for boundedness of sets in
Y A , an
FK
space by 4 . 3 . 1 2 .
(This refer-
ence will be omitted from now on.) 1. LEMMA.
E
bounded i n
Since
A : YA
Also
sets.
space, A
FK
i s b o u n d e d in
E
a matrix.
A subset
and
is
uA
A [E]
Y.
Necessity: 8.0.1.
be an
Y
i s bounded i f f
YA
of
Let
YA c w A , E
Y
+
is bounded in
by 4 . 2 . 4 ,
uA
is continuous, hence preserves bounded
(8.0.1.)
Sufficiency: f o r all
f
E
It is sufficient to prove that
Yi.
Trivially
(4.4.2.)
F $1
and
goA[E]
is bounded
+ goA where F
f = F
Such
(8.0.2.)
frE]
= g
$11
TA
Y'.
g
E wi,
are bounded.
We
could also have used 8 . 0 . 4 and 4 . 3 . 1 2 to give a direct proof. 2.
s p a c e and A
Let
THEOREM. A
belong t o
X
a matrix. Y,
b e an Then
( t i ) . A [El
w h i c h i s bounded i n
X,
A
space w i t h
FK
(X:YI
6
( i i i ) .T h e r o w s of
c o n d i t i o n ( i i )muy b e r e p l a c e d by l i i )
!.
an
Y
FK
( i ) . T h e coZurnns of
i f f
is bounded i n
AD,
Y
f o r each
A
belong t o
A: ($,TX1
-+
Y
E
C
4 The
XB.
i s con-
tinuous.
(ii)
=
by 8 . 0 . 1 ,
(ii)'
Necessity:
8.0.3.
Condition (iii) is simply the statement
Condition (i) says
YA2 4
which is certainly necessary.
(ii)' is immediate, indeed
A: X
+
Y
Sufficiency: We have to prove criterion of 8 . 2 . 2 .
Let
wA
E
3
X.
Finally
is continuous ( 4 . 2 . 8 ) .
YA
=I
X ; we shall use the
be a subset of
4
which is bounded
122
in
8.3
X.
E is bounded in YA. Lemma
It is required to prove that
1 applies: E
X c uA
X hence in
is bounded in
by (iii), and
since
(8.0.1)
wA
is bounded in Y by (ii).
A[E]
X
Under a more stringent assumption on the number of conditions in Theorem 2 .
or
A
we can reduce
The result is Theorem 4 .
In Example 5 it is shown that the extra assumptions cannot be dropped. We begin by reducing the conditions of Lemma 1 for a particular E, Y.
choice of
LEMMA. L e t
3. Then in
D
DlXl
=
be a
X
space w i t h
BK
i s bounded i n
(7.4.1)
wA
a matrix.
i s bounded
ArD]
are given in 4.3.8.
is bounded since X
is an
I
x c D we have
exists K
where
w
argument.)
X
i f f
ankxk
AX(^)).
=
(7.1.4), so
X
Y
c
n
i.e.
BK
b e an
hn[D]
is bounded.
E
s p a c e and
A
( i l . The columns of
A
belong t o
i s a bounded s u b s e t of
AK
<
C
This
a determining
i s row-finite.
has
a
wA.
A
X
A[K.D],
(which is fixed in this
s p a c e , and FK
m
There
AX(^))
It follows that
is bonnded in
be a
Ax(m)
or
Suppose t h a t e i t h e r A c (X:Yl
2 K
D
Let
(7.4.2).
X so each pn[D]
k=l
From this, hn(x) 5 C
Let
(m)l
1
ankxkl = k= 1
of
Next, fixing n, for any
m
completes the proof that
set for
space.
by hypothesis.
THEOREM.
The seminorms pn, hn
is bounded in
is a constant depending on
C
4.
D
FK
IIX(~)~~
such that
bounded set in
First
is continuous. To prove
+ w
D c 0 c wA.
sufficiency, note first that
and
i f f
A
w.
Necessity is trivial since A: wA
wA
and
AK
a matrix. Then
Y, and (ii). A [El
Y.
Necessity is immediate exactly as in Theorem 2.
As to
sufficiency: condition (i) of Theorem 2 is the same as here; condition (ii) of Theorem 2 holds, for if
E1c 4
is bounded in
X,
8.3
then
El
C
for some K , hence
K.D
by (ii) since A[D]
123
$,] c
A
X have AK; A b ]
Lemma 3 , E
b]
is bounded in
is the absolutely convex hull of
tion (iii) of Theorem 2 is trivial if let
K.A
is bounded in
is bounded in
is row-finite.
Y
by (ii), hence in
By 8.2.4, uA
uA.
X
2
Condi-
ArE].
A
Y
Finally By
W.
which is just
condition (iii) of Theorem 2.
5.
EXAMPLE.
Theorem 4 f a i l s i f
i s n o t r o w - f i n i t e even if
with
>
Let
(5.2.14).
1. Then
if Y
is an
a conservative triangle.
A
X B # Xf n
X
X
uA
is any
FK
so
space
The first is trivial. A : ($,TX) + Y
space. Let f(x>s'
so
u
6
Let
X
or
A
cA
be such a space with alk - uk, ank = 0 for
Let
A $ (X:Y)
AK
space of the form
AD
Xf\XB.
no matter what
Y
is.
However,
does satisfy (i) (ii) of Theorem 4.
A
2 $,
d o e s n o t have
X
The second will follow when we show that
is continuous since E is bounded in the domain f E X' with f(6 k ) = uk. Then, for x c +, Ax = is continuous.
A
6. The following technique called improvement of mapping will allow knowledge of space of Yl
Y.
Let
(X:Y) to yield
a cZosed s u b s p a c e of
and t h e columns of $
b e an
X
c X c Y1)A.
A [ $ ] c Y1.
A
space, Y
Then
beZong t o
A
Yl.
Sufficiency: A : X + Y
Since
$
7. Suppose that A c (X1:Y)
Y.
AD
(X:Y1) with
iff
A
6
is dense in X
and
Yl
(X:YII
6
Y1
i f f
a closed subspaces w i t h
FK A
6
IX:Yl
Necessity is trivial since is continuous (4.2.8) and
X, A[X] c Y1
is an
FK
(X:Y) and
A1
space and 6
XI = X 0 1 .
Then
Y.
Next is an important set of results connecting properties of A
with those of
AT, the transposed matrix.
but different from the dual (or adjoint) map
This is related to T':Y'
+
X' (14.0.5).
124
8.3
To illustrate Theorem 8, let
x n-Xn-1‘
X =
A
E
(X:Y )
n T since for x = (-1) , A x
=
2x.
(AX),
=
Then
1 (4.5.5), Y = 1 , T A $ (YB:XB)= (Qm:cs)
9.0
but
Also it is shown in 14.5.7 that
AT
8. THEOREM.
s p a c e and
Y
(Y’:x~).
If
Let
sequences.
I f
A
s p a c e s and
YB
has
taken i n
be an
X
FK
T A
then
IX:YI
then
AD,
AT E
6
need not be in
IY 8 .. X I
any s e t o f Y
X,
are
BK
where t h e c l o s u r e i s
( I m p r o v e m e n t of m a p p i n g ) ,
Xf.
All spaces mentioned in the second part are Let
z(Ax) =
This converges by hypothesis and
zn(Ax),.
z E YB
and define
f
BK
4.3.16 and 7.2.14. C
(YB:Xy):
E
spaces by
X’ by
f(x)
f
=
is
continuous by the Banach-Steinhaus Theorem (1.0.4). Then f(6 k ) T C znank = (A z ) ~ SO ATz E X f . To prove the second part note that
uA
ZJ
columns of
AT
belong to
X’; these are the
and the result follows by Remark 6.
9. THEOREM. Then
A
so the rows of
X
=
Let
and
IX:Y) = (XBB:Y)
Note first that 4.3.16 and 7.2.7.
Z
X,
X8
Let
(Z:XB). Conversely if
A
be
A
BK
(X:Y).
E
AT
(X:Y)
6
is a
E
spaces w i t h
BK
i f f
A
T
E
Y = ZB
AK,
.
(Z:XB).
space and is equal to By Theorem 8 , AT
by
Xf
(ZBB:XB)c
E
(Z:XB) then by Theorem 8, A
E
(XBB:ZB)
A
6
This proves the second part. To prove the first T (X:Y) then A E (Z:X8) as just proved. By
Theorem 8, A
E
(XBB:Y).
c (X:ZB) = (X:Y). part:
if
10. THEOREM.
The r e s u l t of T h e o r e m 9 h o l d s i f
i s any one
X
o f t h e n i n e s p a c e s l i s t e d i n t h e t o p row o f t h e T a b l e , p .
(Recall that
Y = Z8 , w h e r e
t h e n i n e spaces, Y to
2 = cs, b v o ,
may b e kq(l/p+l/q
Z
i s any
BK
space w i t h
b v , b s , k P ( p 2 I), =
I),
co,
and
k.)
and
km
116.
AK.
Of
corresponding
8.3
125
The proof given covers the cases when m
A
Next let
T
(Z:e)
E
A
E
(YB :bv), hence
uA
2
bs
i.e.
(co:Y)
so Theorem 9 holds for
X
=
a ,
A
(Z:bv)
E
A
0
since
lies in
m
if
C.
Z c ZBB= YB. Now bsB
=
bvo, hence each
=
By Remark 6, A T
(bs:ZB )
E
(c :Y)
=
By Theorem 8 this implies that
each row of
this implies that
space.
E
A T lies in bvo.
column of
(Z:bvo).
E
Conversely
(bs:Y) by Theorem 8. Since
the other part of Theorem 9 is automatically true.
bsBB = b s
A
Finally let bs, bv, bvo
11.
AT
AK
A
(c:Y)
=
(bs:Y).
6
AT
is an
and
Also by Theorem 9 , (II :Y) and only if
X
E
(bv:Y).
replaced by
The preceding argument applies with
bv, bs, cs
respectively.
The reader may have wondered why the last part of Theorem
9 was not written as
AT
(Z:XB)
(YB:XB) since this is equal to
E
The reason is that in this form Theorem 10
by the first part.
would not be true (8.4.9).
that
YA
given
It is trivial that:
12. 3
X
i f f
uA
X, t h e r e e x i s t s
and i f f t h e r o w s of
X
3
A,
suck
Y
belong t o
A
XB.
This gives a set of minimal conditions of which the following is typical: such t h a t
i f t h e rows of YA
are not a l l i n
A
6
of
rB
(X:W)
3
= (XBB:w)
belong t o
A
X B B 2 X.
Y
i s conservative.
13. Theorems 9, 10 hold also for A
t h e r e e x i s t s no
II,
x’.
i f f
If
Conversely
($:X
AT E
r rB
B
)
Y =
X
X,
and i f and o n l y i f t h e r o w s
A, r
is any row of 3
indeed, f o r any
W;
implies
r
E
E
xB
implies
r B Bc X B .
Thus
(X:W) = (XBB:0). Also AT c ($:XB) if and only if the columns of A T ( = rows of A) belong to XB. Taking X = w we have A 6 ( $ : + ) i f f
A
T
finite.
6
(W:W)
i f f
i s r o w - f i n i t e i . e . iff
A
i s column-
8.4
126
EXAMPLES.
8.4.
We now proceed to list spaces
(X:Y) following the order of
presentation given at the beginning of 88.3 1. EXAMPLE. t h e columns o f
E
k 16 1
=
Let
Y be an
A.
From 8.3.13, e a c h roL) o f A
6
i f f
We apply 8.3.4 with
Y.
A
(Q:Yl
6
is the kth
must be i n
and
Qm
(Q:wl.
s u p lankl < m . n, k ( Q : c l add t o t h i s t h e r e q u i r e m e n t t h a t t h e columns of
1A. EXAMPLE.
A
A
The result follows since A6k
t h i s i s t h e c o n d i t i o n for
For
space. Then
f o r m a bounded s e t i n
A
using 7.4.3.
column of
FK
(k:col,
From Example 1, A
6
(Q:Q"l
i f f
a r e n u l l , c o n v e r g e n t , r e s p e c t i v e l y , as in 8.3.6.
We have
The argument of 6.2.6 Droves
actually used this result already! that the matrix given there lies in
(Q:c). m
1B. EXAMPLE. <
-.
( 1 : ~ s ) add:
For
i f f sup
I 1
m,k is convergent f o r each
n=I
From Example 1, A Cn
ank
6
(Q:bsl
k
ankl
as i n
8.3.6.
1C. EXAMPLE. Fran Example 1, A <
(Convention: aok
m.
(Q:bvol
= 0.) F o r
A
is in bv
by the other condition.)
1D. EXAMPLE.
1
sup k
lanklP <
n
E
=
{l(m)}
by 7.4.4.
the kth colum of series i n A
6
6
Thus, for example To consider A
EXAMPLE.
2.
From Example 1, A
-.
i f f sup lank-an-l,k k n
add, (8.3.6), e a c h
(This implies that it is in bvo
i s null.
column of
Now
A , so
Al(m)
6
=
i f f
A
6
(C,l) 6 (Q:QP) iff p > 1.
(bvo:Y) apply 8.3.4 with m 1 = 1 ak where ak k-1
A E (bvo:Y) i f f
(bvo:Yl
since it
( Q : Q P ) p, 2 I, i f f
C ak
i s a bounded
Y, (bv:Y)
I
1
fQ:bvl
6
and
A1 6 Y
(8.3.7),
is
127
8.4
A
i f f t h e rows of
(bv:w)
(bvo:w),
6
lie in
A
bs, cs
r e s p e c t i v e l y (8.3.13).
2A. sup,
11
EXAMPLE. k
a
(Important).
From Example 2, A m
1,
<
I 1
sup m,n
i.e.
m
k=l
ankl <
k=l
-.
6
iff
(bvo:tm)
Interchanging m, n
i f f t h e s e t of r o w s of
i s a bounded
we see that
A
set i n
Another derivation of this is to observe the equiva-
bs.
lence of A
A
T
6
6
(bvo:Qml
(Q:bs) by 8.3.10 and apply Example 1.
(buo:co), ( b v o : c )
6
A
i f f
A
a n d t h e columns of
(bvo:Qm)
6
are n u l l , convergent r e s p e c t i v e l y .
A
A
(bv:Qm)
6
i f f t h e s e t of rows of
A
i s a bounded s e t i n
cs.
The easiest way to see this is to apply 8.3.10 and then apply Example 1 to A T E (Q:cs). Finally, (8.3.7), A 6 I b v : c o ) i f f A
(bvo:co)
6
A1
6
and
c
6
A
0'
(bv:c)
6
i f f
A
6
and
(bvo:c)
C.
n
28. EXAMPLE. <
A1
m.
For
From Example 2 , A
1 ank
(bvo:cs) a d d :
6
1
i f fs u p (
(bvo:bs)
m
c
Qikl
m,n k = I i = l
i s convergent f o r each
k
(8.3.6);
n for
(bv:bsl,
cS),
A1
6
2C.
add (8.3.7) A 1
6
bs
From Example 2, A
6
(bvo:bv)
(bv:csl
(or
A
has rows in
CS.
EXAMPLE.
iff
m
(bvo:bvo) a d d (8.3.6) t h a t e a c h
For
column of
A
i s n u l l (i.e. in
( b v : b u o ) add (8.3.7) A 1 A1
6
bv
(or
(bu:bv),
has convergent rows),
A
bVo.
2D. EXAMPLE.
From Example 2, A
-.
1I anklP < n k=l has convergent rows. ) sup m
6
as in Example 1C); f o r
bvo
For
6
(bvo:tp), p 2 1
( b v : e P ) a d d (8.3.7)
A3
6
Qp
i f f
(or A
8.4
128
3.
EXAMPLE.
{A(KI: K
and onZy i f bounded i n
1
{ank:k
By
Y.
tion of
K.
i f f
(co:YI
A
6
be an
FK
y
=
Ax
where
1 I k 61K
n
A
E
and
A1
by 8.3.7.
Y
6
ank-an-l,k
EXAMPLE.
1I 1
4.
i s
where
6
Also
A
(c:YI
6
As in Example 2 the last
the rows of
For
EXAMPLE.
y = {ann}.
(co:bvl
A
are convergent.
i f f m.
y B 3 X, i.e.
y
6
E
the conditions are
bvo. 6
( c o : I I p I , p 1. I , i f f
a f i n i t e s e t of p o s i t i u e i n t e g e r s } < =.
We can take advantage of the preceding work to 6-duals.
Then
(c:bvo)
From Example 3, A
anklP:x
kcK
calculate a few
Taking
i f
a f i n i t e s e t of p o s i t i v e i n t e g e r s } <
(:K
(co:bvo) and (8.3.7) A1
n
y
as in 8.2.7.
From Example 3 , A
as in Example 1C).
3B. sup{
(co:Y)
6
(co:buol add (8.3.6) t h a t e a c h coZumn i s n u l l (i.e. in
To get
bvo
A
Yn = is the characteristic func-
x
E
We apply 8.3.4 with
EXAMPLE.
Then
we mean the sequence
condition can often be replaced by: 3A.
space.
i s a f i n i t e s e t of p o s i t i v e i n t e g e r s }
A(K)
K}; thus
6
Y
Let
csA = y B X6 .
X = R , we have
Thus y
A
Let
6
be used to show :c
= R
X , csA
and so, for any space A
(X:csl
6
.t6
bu: = b s .
I I B = II
m
.
A
E
Taking
3
X
(L:cs). X
iff
y E XB
i f and onZy i f
if and only if
Example 1B this gives immediately using Example 2B gives
be a diagonal matrix and set
=
.
From
bvo
and
An extension of Example 3 could
- we chose to postpone (co:cs) to
Example 6 A for convenience. Use of this would be circular. We now come to the fourth stage in the order of presentation given at the beginning of 18.3:
this is the application of 8.3.10.
Several pairs of spaces will fall into more than one category; in such cases we choose the simpler treatment if possible one at random. For example (co:%") but far more easily by Example 5A.
- or choose
could be treated by Example 3 ,
8.4
5.
EXAMPLE.
Take
our nine spaces, ( X : L m l EXAMPLE.
5A.
in 8.3.10. For
= 8”
= IXBB:8”)
and A
X
i f f AT E
fX:L”l
E
From Example 5, (co:8”)
any one of
= (c:a
m
I
(8
=
The last equality is by Example 1D applied to
(1.3.4).
(co:cl, (co:col
add (8.3.6) t h e c o n d i t i o n :
convergent, n u l l .
co.
Y
129
To g e t
and
(c:c)
(8:x’).
m
m
:a.
= Q,
)
AT.
For
t h e c o l u m n s of
are
A
Ie:col add (8.3.7) A 1 c c,
This is a complete proof of the Silverman-Toeplitz Theorem
1.3.6.
Note that
( 8
m
:co) and
(!Lm:c) are not covered by these
techniques. From 8.3.13, ( c o : w l s p a c e s i f f t h e rows of
5B.
EXAMPLE.
Ic:wl
=
A
m
=
belong t o
(cs:cl
and
A
A
belongs t o these
A
8.
AT
(cs:ilmJ i f f
s u p 1 lank-an,k-lI < n k
From 8.3.13 A
belong to
bv
c
(8:bv).
c
For
m.
are null,
A
(cs:Y) for some
Y
and t h i s i s t h e c o n d i t i o n f o r
(cs:~).
6
EXAMPLE.
5C.
From Example 5 , A
This may be read from Example 1C: and e a c h row of
bvo
to
E
6upn
EXAMPLE.
(8:8‘),
1
A
i s null.
E
lank]‘
<
m.
For
i f f
A
From 8.3.13, t h e r o w s o f
From Example 5, A
l / p + l/q = 1.
Ibs:kml
T
(8:bvJ.
c
t h e c o n d i t i o n of E x a m p l e 5 B ,
and t h i s i s t h e c o n d i t i o n f o r
5D.
AT
and
add (8.3.6) t h a t t h e c o l u m n s of
convergent, respectively.
iff the r o w s of
:wl
From Example 5, A c
This may be read from Example 1C: (cs:cJ
(8
A c c
A
belong
(bs:wl.
(LP:Lml,
p
>
I, iff
This may be read from Example 1D:
(8P:eol
and
( Q p : c ) add (8.3.6) t h a t
k
t h e c o l u m n s of
A
t h e rows of
belong t o
A
6
(LP:u).
A
are n u l l , convergent respectively. Qq
From 8.3.13,
and t h i s i s t h e c o n d i t i o n f o r
130
8.4
EXAMPLE.
6.
Take
nine spaces, (X:bsl
( X B B : b s l and
=
A c
(X:bsl
From Example 6 , (co:bsl T b e l o n g s t o these spaces iff A c (bvo:a).
1I
sup,
Example 2D:
k
t h e column-sums of
A
1 1 ank
(8.3.7) t h a t
n
-.
For
=
if
AT
(c:bsl
(bvo:XBl.
E m
( a :bsl
=
and
This may be read from add (8.3.6) t h a t
(co:csl
are convergent; f o r
any one of our
add t o t h i s
(c:csl
i s convergent.
k
The fact that result.
ank( < n=I
X
in 8.3.10. For
EXAMPLE.
6A.
A
Y = bs
( c o : c s ) ~@
is not at all obvious from this
However it is immediate from the inclusion
(co:cs) c
(co:co). EXAMPLE.
6B.
From Example 6, A
6
(cs:bsl m
i f f
AT
sup 1 I 1 ank-anJ k - I m k n=1 i s convergent f o r each k .
This may be read from Example 2C:
1 ank n the historic series-series map. add (8.3.6)
(cs:csl
1 (Ax)n
i.e.
1 xn
=
are
A
To see this note that
cs
1 (Ax),
=
we have
f(x)
1
6
6C.
EXAMPLE.
x = 6
k
(cs:csl
is an AK f(6 k )xk =
6D.
AT sup m
c
A
k
+
ankIq <
n=l
gent f o r each
FOP
This is
1 ank = 1 f o r each n space and s o , with f(x) = and
1 1 ankxk,
k.
giving sufficiency.
n
From Example 5, A c ( b s : b s l
EXAMPLE.
I
m.
i f f
A
T
6
(bvo:bvo).
t h e c o n d i t i o n of Example 6 B and
i s a n u l l sequence.
(bv:tq), l/p
1
<
.
This may be read from Example 2C: e a c h row of
I
The c o n d i t i o n s f o r r e g u l a r i t y
k For necessity take
(bvo:bv).
6
From Example 6, A c l ! Z p : b s l , p l / q = 1. m.
For
> I ,
This may be read from Example 2D: ( E P : c s l add (8.3.6)
1 ank n
k.
i f f
i s conver-
131
8.4
7. EXAMPLE.
Take
ninespades, ( X : b v l 7A.
=
bv
in 8.3.10. For
( X B B : b v ) and
=
EXAMPLE.
Y
A
(X:bv)
6
From Example 7, ( c o : b v l
any one of o u r
X i f f
AT
(cs:XB1 .
6
(c:bvl = ( a m : b v ) .
=
We shall show that
The condition is given in Example 3A.
(co:bvol
= (c:bvol = ( Q m : b v o l
as well, and this condition can also be
found in Example 3A.
The argument is this:
= (em:bv) c
(Em:c) from Example 7; also m
(co:bvo)c ( a :c) n (co:co)
= (E
(co:bvo) c (am:bv) n (Rm:c0)
=
m
(co:bvo)c (co:bv)
(co:bvo) c (co:co)
so
:co) by 1.7.19. From this
(am:bvo).
The opposite inclusion is
trivial. 7B. have
A
From Example 7 with A, AT interchanged, we T i f f A 6 (co:bv). This may be read from
EXAMPLE. Its:!,)
6
Example 3A:
1I 1
sup{
k
n6K
ank-an,k-l
a f i n i t e s e t of p o s i t i v e i n t e g e r s }
I:K
-. 8. EXAMPLE.
Take
Y
=
ap, p
one of o u r nine spaces, ( X : ! L p ) AT
(tq:XB),
6
8A. p
1.
z
+
l/p
EXAMPLE.
EXAMPLE.
changed, we have
9.
A
1I 1
k
6
A
6
X
any
(X:EPI
i f f
l / q = 1.
From Example 8, (co:Lp) = (c:!LP) = (Em:%'),
From Example 8 with 6
(ap:,), p > I ,
iff
A, AT AT
and p, q
inter-
(co:kq), l/p + l/q
6
ank14:,
EXAMPLE.
(co:XBI.
a f i n i t e s e t of p o s i t i v e i n t e g e r s ) <
-.
n6N
nine spaces, ( X : & l A
and
= (XB8:Qp)
For
This may be read from Example 3B: SUP{
T
1, in 8.3.10.
The condition is given in Example 3B.
8B.
= 1.
>
Take
Y
= II
= (XBB:kl
in 8.3.10. and
A
6
(X:al
For
X
any one of o u r
i f and o n l y i f
(Remark 8.3.11 cautions against using
(&":XB) here.)
132
8.4-8.5
9A.
EXAMPLE.
From Example 9, (co:Ql = (c:Ql
= (Q
m
The
:Q).
condition is given in Example 3B. Note that 8.3.8 implies that if A 6 (Q":Q) then A T 6 (Q":k) also. The condition given in Example 3B does not obviously have this symmetry property. 9B. AT
6
EXAMPLE.
(co:bvo).
From Example 9, A
6
the
The condition may be read from Example 3A:
c o n d i t i o n of Example 7 B and e a c h r o w of
8.5.
if and o n l y if
(bs:kl
i s null.
A
MAPPING CONCLUDED
For this section only we introduce a new space. include
@.
1. DEFINITION.
cs:
6
0
Let
LEMMA.
{x
cso =
cs = cs 0 6 1
2. Clearly 3.
It does not
.
1 xi
Hence
A
be a m a t r i x and
vk
=.I
= 03.
csB = cs' = bv. 0
bnk
a nk
=
- 'n,ktl'
Let
k
x
be a s e q u e n c e and
x
z=l
i'
(i
I f t h e rows of
A
a r e n u l l s e q u e n c e s and
(ii
I f t h e rows of
A
a r e bounded s e q u e n c e s and
then
Ax
=
Bv
and
x
6
if
wA
v
6
wB.
m
+ -;
x
bs, or
6
x
6
cs
0'
By Abel's identity (1.29) we have
In case (i) v vm
+
0
and
is bounded and {anm]
anm
-+
0 as
is bounded for each
n.
in case (ii)
Letting
m
+
m
yields
the result. 4.
LEMMA.
Let
Y
be an
i t has c o n v e r g e n t r o u s and Lemma 3 .
B
FK 6
space.
(co:YI
Then
where
B
A
6
(cso:Yl
i s d e f i n e d in
iff
8.5
Necessity: uA
2
let
cs xn
Since
uA
A
and so the rows of vn
=
it follows from Remark 2 that
cso
2
- vn-1' Then x
133
are in
so
cso
E
csB
bv c c.
=
Ax
v c co
If
is defined.
By Lemma
Bv = Ax c Y.
3(ii)
Sufficiency:
so
v c co
If
v
define
as in Lemma 3.
Then
is defined and by Lemma 3(ii), Ax = Bv c Y.
Bv
Y
Let
EXAMPLE.
5.
x a cso
b e an
space.
FK
Then
A
(cs:YI
6
i f f i t has c o n v e r g e n t r o w s , i t s f i r s t c o l u m n
c Y
where
The first column of
A
i s d e f i n e d i n Lemma 3 .
B
is
A6'
6
Y.
Necessity:
and
The rest is by Lemma 4. Sufficiency:
B
(co:Y/
6
By Lemma 4
and Remark 2.
5A.
EXAMPLE.
A
(cs
6
/ P : K a finite,
Necessity: replaced by contain
From Example 5 and 8.4.3B we get
k+l.
Thus
(*)
By Example 5 the first column tion to allow
K
to contain
Sufficiency: column of that bk
=
A
c
Taking
K
is correct if
since we can substitute
1
of p o s i t f v e i n t e g e r s )
pc?t
c tp
1
m
=
k+l
(*)
with
is forbidden to and change the signs.
and this extends the condi-
again.
K = C11
in
(*)
shows that the first
Condition ( * ) implies the existence of
Lp.
)Ip
I 1
k-1
(ank-an,k-l < M f o r all K , n. kc K ank - an and take the pth r o o t . Then
Fix
n , let
I 1
bkl
<
M
C.
such
Fix
ka K
an arbitrary positive integer K2
=
2C.
{k:k
-< m , bk
Hence the
<
Then
01.
m m
1
bkl
k=l nth
row of
by Example 5 and 8.4.3.
A
K1 = {k:k 2 m, bk 1.
and let
E
=
1 bk:k
bv c c.
a K1} -
1
{bk:k
c
K2
The result now follows
134
8.5
5B.
EXAMPLE.
(cs:bvI
A c
Imitating Example 5A, using 8.4.3A, we have
i f f
(cs:bvol add (8.3.6) t h a t t h e columns o f
Fo r
6.
EXAMPLE.
Let
i t h a s n u l l rows and
v c am
If
Sufficiency: If Lo
6
so
6A.
Bv
3
let
bs
xn
where
Then
A
are in
- v ~ - ~ Then , x
v
/bs:Y)
i f f
i s d e f i n e d i n Lemma 3 .
B
the rows of =
A c
6
bsB
=
so
bs
Ax
Bv = Ax c Y.
By &ma 3(i)
is defined.
v
uA
space.
FK
B c (Lm:YI
Necessity: Since b v o C co.
be an
Y
are n u l l .
A
x
6
bs
define
v
as in Lemma 3.
Then
is defined and by Lemma 3(i), Ax = Bv c Y.
EXAMPLE.
A
6
(bs:bv) iff it has null rows and
This is by Example 6 and 8.4.78. From 8.4.38,
A 6
B e (co:bv)
(bs:bvl
i f f it
h a s nu22 r o w s and
I 1
s'p'c
n kcK integers) <
6B.
that
I:K
a f i n i t e s e t of positive
-.
EXAMPLE.
B e (co:bvo).
, k f an - I , k t l
a n k - a n J k+l-'n-1
A c (bs:bvo)
iff it has null rows and
This is by Example 6 and 8.4.7A. We shall show
A c (bs:bvo)
iff
A c
(bs:bv)
( E x a m p l e 6 A I and
A
has
n u l l columns.
Sufficiency: Clearly B Example 6A. A
6
Thus
has null columns and
B c (co:bv)
by
B c (co:bvo) by 8.4.3A. So, as j u s t remarked,
(bs:bvo).
Necessity: We have only to show that A has null columns. Now, taking x = 6 m in Lemma 3(i), A6m = Bv where v is a certain bounded sequence. Since follows that
Bv
B
m
6
(co:bvo) = ( E :bvo) by 8.4.7A it
is a null sequence, hence so is
This is
8.5
the
B
f
mth
column of
6C.
EXAMPLE.
A. A
( b s : ! L P ) , p 1. 1
6
i f f it has null rows and
( c ~ : % ~ ) . This by Example 6 and 8.4.8A. Thus by 8.4.3B the
condition is:
s ' p f l n I kcK 1
h a s n u l l rows and
A
ank-an,k+l I P : K a f i n i t e s e t of p o s i t i v e i n t e g e r s } <
6D. EXAMPLE.
B
rows and
6
columns and
From Example 6 , A
(~":c) i.e.
i f f
and
A
(bs:c)
6
has n u l l rows, convergent
A
Sufficiency
B.
B
Conversely if
(!L":c)
6
has null rows we obtain the fact that the columns of x = 6m
are convergent by taking
in Lemma 3(i);
is a certain bounded sequence so 6E.
EXAMPLE.
columns,
A
6
(bs:co)
m.
iff it has null
is u n i f o r m l y c o n v e r g e n t .
ilank-an,k+ll
is obvious by applying 1.7.18 to
v
135
A6m
iff
6
A 6 m = Bv
A
where
c.
A e (bs:c) and
A
The conditions may be read from Example 6D.
has null
The proof
is the same as the preceding, using 1.7.19, n 7. LEMMA.
Let
be a m a t r i x , , b n k
A
1
=
'ik'
Then
wB = wA
i=I
For x
6
w
A
X
6
wB, (BX)n
=
1 (bnk-bn-l,k)xk
x
8. EXAMPLE.
6
w
A
6
hence
=
k
and the stated identity holds.
is trivial that
B
- (BX)n-l
Conversely if
x
wA
6
it
and again the identity holds.
B
(!Lm:csl i f and o n l y i f
i s g i v e n i n Lemma 7 .
B
where
(km:c)
6
T h i s i s c l e a r f r o m Lemma 7 .
Hence by
m
1.7.78(ii),
A c
( ! L ~ : C S J
i f f
1 1.1
k m
Because
(co:Y)
= (!L
aikl
+
0.
z=n
:Y) for several spaces Y, it is worth-
while to point out that this is false f o r The identity matrix lies in
Y
=
c0
(c,:c~)\(~?~:c~) while
and
Y
=
cs.
136
A
8.5
(co:cs)\(Lm:cs)
6
where
1I
A are null and
k 10. EXAMPLE.
AT
6
For
=
xn - xn+l
From Examples 8 and 6, A c (bs:cs) iff the rows
9. EXAMPLE.
of
(Ax)n
i=n
aik-ai,k+lI
0.
-+
From Example 8.4.7,A
6
fLP:bvl, p > 1, i f f
(cs:Lq), l/p + l/q = 1. This may be read from Example 5A:
(aP:bvo) add (8.3.6) that each column of
A
is null (as in
8.4.1C). 11. EXAMPLE.
AT
6
(cs:bvo).
From 8.4.7,A
6
(bs:bv) if and only if
This may be read from Example 5B.
The remaining case is tions are known.
( Lp: L r )
, p
1, r > 1. No tidy condi-
>
Elegant theorems of Crone and Pitt are given in
[55] p . 114, Theorem 5.4 and p. 140, Theorem 9.2, respectively.
In 1381 an explicit norm is given for a Hausdorff matrix; in this special case the results are easier to apply than those of Crone. Other conditions are given in [72]. A very nice discussion of the 2 2 Hilbert matrix, which lies in (P. :p. ) , is given in [22B], p. 208, Solution V. 12. EXAMPLE. Bl
and
Let
A
b e a m a t r i x , B = IAl and a s s u m e t h a t
a r e b o t h bounded.
BTl
This is the special case 13. sequence < f3z:
EXAMPLE, z
Let
l/p
+
Then z = 1
l/q = 1
A
6
n,k;
a,@
for
p > 1.
of:
and s u p p o s e t h e r e e x i s t s a
o f p o s i t i v e numbers such t h a t
for a l l
fLP:LpI
l B z q I n 5 a z qn'
c o n s t a n t , B = ( A ] . Then
A
6
(BTzPIk fI1p:I1p).
137
8.5
Let
x
E
with
$
xk 2 0
for all
k.
Fix
m
and set
Applying HGlder's inequality (7.3.8) in which we take the
pth
power of each side:
u 5
This shows that
( + : Q p ) since the argument could have been
written with it shows that that
A
E
A
1x1
E
instead of
x ; also, giving
IlAll 5 al/qgl/q.
Since
.tp
the
$
has
norm
LP
it follows
AK
(QP:Qp).
14. EXAMPLE.
(e:e)n
(Qm:Qm)
c
i n c l u s i o n i s proper i n t h e strong sense t h a t right side for a l l
f o r a22
(aP:kP)
The
beZongs t o t h e
IC,li
p > 1 , but not t o the Zeft.
p > 1.
The first part is
immediate from Example 12 with 8.4.1D; 8.4.5A. That
(C,1) f
(Q:e)
was pointed out in 8.4.1D. Finally we shall prove that the criterion of Example 13 applies.
zn = n--E
Let
with
n
B
with
=
For
k
8 = 2
= =
r-1
c
E
< l/q.
:1
1 n-lZ--EP = 1 n=k 1 this is a certain (finite) number, while f o r
k
>
/(r-1).
Some classical spaces are unions of
FK
spaces.
Mapping
theorems f o r these may be discussed in terms of maps of the constituent spaces.
Then,
t "qdt =z;/(l--Eq); 1 k-Eq < n-' k=l n-r (where r = 1+Ep so that r > 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
-
<
k I6 1
iff
i s weakly absolutely
This,means that
Also
A c (iP:Y) iff {g(a
Theorem 1 f a i l s for
EXAMPLE.
k
with 8.
Y qt X, Y B
=
%-duals.
AX
f-duals:
Of course
s p a c e s w i t h t h e same
cf
Two
AD
f co
= k
=
X
= is,
for all
g
6
such that
X B B = k m ; indeed if
co X
Y'.
In 10.3.23,
X
and an
AK
space
spaces w i t h the
but
@-dual are equal.
An amusing corollary of Theorem 1 is that space
m
In contrast with Example 7, Theorem 1 implies this dual
f-dual are equal.
Hence t w o
<
XB.
characterization theorem for same
(kp)f
) } 6 iq
10.3.24 it is shown that there exists a space
Y
Ilf(sk)lq
This follows from Theorem 1 and
Y'.
Example 7.3.7. 7.
Y
cs
for each
= 0.)
6. EXAMPLE. q-summable
3
c # co. (7.2.7).
is the largest is a n
AD
AD
s p a c e and
140
Y
8.6
an
X B c Xf
YB
space w i t h
AK
fCS
k
I # 0
For such a
Z3
u # $ , let
Y $.
if
Y
4 C Y #
f o r i n f i n i t e l y many
by Theorem 1, Y
space
then
3
X
since
Yf = Y B
=
by 7.2.7.
A few more remarks:
with
= XB
= w.
w
t h e r e must e x i s t
k
for if not, Y
Finally, t h e r e e x i s t s n o
would, by Theorem 1, satisfy (Here
Yo = cky$.)
u - v # co, then
Yo
Y C
f
f c Y'
= $
and so
such t h a t
Yf
Z for every
This implies Yo = 4
=
W.
FK
[if
u # v-l-co1 which is impossible by
4.0.5.
Conditions of the type given in 8.4.3B are intimately connected with unconditional convergence. absolute convergence.)
(They are obviously implied by
An interesting discussion of this point
may be found in 14781, pp. 11-13.
141
CHAPTER 9 SEMICONSER VATIVE SPACES AND MATRICES
INTRODUCTION.
9.1
Because of the historical roots of summability in convergence, conservative spaces and matrices play a special role in its theory. However, the results seem mainly to depend on a weaker assumption, that the spaces be semiconservative.
This is a significant general-
ization of the theory, but more than that natural home for it.
- it seems to be a more
Many of the results still to come were first In addition to these we also extend
given for conservative spaces.
results given earlier so that the reader may compare the settings. (See for example 9.2.10,
9.3.6,
9.3.7,
9.6.8,
9.6.9).
Even if none of the previous remarks is sufficient to motivate the concept, semiconservativity must arise in the discussion of distinguished spaces in view of 1 0 . 2 . 7 .
SEMICONSERVATIVE SPACES
9.2.
1. (=
DEFINITION.
sc f o r s h o r t ) i f
This means that f
E
An
FK
space
i s called semiconservative
X
X f c cs.
X
2
+
and
1 f(sk)
is convergent for each
X'.
EXAMPLE. E v e r y c o n s e r v a t i v e s p a c e i s sc, for X conservaf tive implies X c cf = n. c cs by 7 . 2 . 6 and 7 . 3 . 1 . The same proof, 2.
with
c
replaced by
co, shows that
if
X
3
co
then
X
i s sc.
The introduction of sc is motivated by the observation that in the
142
9.2
conservative theory, the convergence of
-
sufficient
i s s c i f and o n l y i f
zB
AK
(4.3.7), we have
if
z B Oc cs.
z
6
c8.
and
zBf= z B B by 7.2.7 so z B is sc if and only z c cs,
z B Oc bvB = cs, using 7.3.5 and 7.2.2.
c8 = l Bi s n o t
special case
z B has
Since
This implies z c z B Oc cs; conversely if
z B 3 csO = bv
As a
80.
gives an example, z0 , of a sc space which
z c cs\E
Taking
was usually
absolute convergence being more than was needed.
EXAMPLE.
3.
1 f(6 k )
does not include co.
f i )Every s c space
THEOREM.
4.
of a l l s c s p a c e s i s
bvo,
l i i i ) bvo
3
l i i ) the intersection
buo,
i s not
l i v ) t h e r e i s no
SO,
s m a l l e s t sc space.
Xf c cs c bs
(i)
Example 3 and 7.3.5.
I c co since co
Also
=
bv
using
is sc by Example 2,
n
I c bv
co = bvo. The opposite inclusion is by (i). f = bs c$ cs by 7.3.5. bvo
(iii). (iv).
By (ii), (iii).
5. THEOREM. 4
subspaoe
Y
where
is s o ,
X
3
(i) A
The f o l l o t l i n g a r e s c s p a o e s :
of a s o space
f i i i ) .n X n
(i) is true since
c cs.
by 8.6.4.
I c n i z B : z c cs} = csO
The intersection
(ii)
so
X I bvo
so
To prove (iii):
X,
( i i ) .Any
where each
Yf = Xf
(7.2.6).
Xn,
space
FK
n
= 1,2
closed Y
,...,
3
X
is
SC.
(ii) holds since Yf c X f
First, the intersection, X, is an
FK
space
m by 4.2.15. gk c XA
Every
for some
the sum of
m
f c X' n.
can be written
f =
1
gk
k=l This is by 4.0.3, 4.0.8. Thus
sequences each of which is in
cs.
where each Ef(6"))
is
9.2
6.
EXAMPLE.
spaces bvo
and
and
bv
cs
7. X B c cs
143
a r e n o t s c since their closed sub-
bs
are not.
(Theorem 4, Example 3).
is not sufficient for X
to be sc since bvB = cs.
This is hardly surprising since this condition holds for every space containing 1. 8.
For purposes of contrast we write three lists, each list
consisting of a set of equivalent conditions. The first is: f X 2 bvo, X c bs, 11(")1 is (weakly) bounded in X (8.2.6) f second is: X is sc, X c cs, l(n) is weakZy Cauchy i.e. {f(l(n))}
is convergent for each The third is: X
is convergent. each
f c X'.
3
f c X', equivalently co, Xf C a , 1 If(S k ) I
1f <
-
The
Sk)
for
(8.6.2).
The extension of two-norm closure (16.2) to sc spaces is made by using the
bv
9. LEMMA. admissible f o r
Let
I X'")~
-+
m- 1
xn
=
1
norm: Let X.
X
be an
space
2 $.
Then
Il*llbv
because
IIx'~)~~
( X ~ - X ~ ++ ~xm; )
IIxII. First, for x 6 bv n-1 = 1 Ixk-xk+ll + lxnl. k=1
we have Also
combining these
k=n
the result just proved shows that since X
is an
FK
space, so
D
D
=
nDn. Each Dn is closed
is closed. The other part of
the definition was part of the above proof.
-
is
(6.2.1).
IIxI1bv be written IIx11
FK
2bv
10. THEOREM. bvo.
Let
X
be a s c space.
Then
W
n
bv = ZbV$
9.2-9.3
144
n
By 6.2.4, 6.2.5, W z
6
abvbvo.
f
6
X'.
It is sufficient to prove z E W. To this end let n Say an + z , a 6 bv,, IIanllbv < M. Then f(z)
A
where
Finally let
bv c 2bv$ c 2bvbvo c bv.
=
n k = limAu 1 akf(6 k n i.e. an is the ak
lim f(an) = limn
is the matrix with ank = and uk = f(6 k ) . This is true since
of
A,
a =
1 ak6k
in
bvo
(7.3.5) a fortiori, in
a
E
bvo
(1)
nth
row
implies
X. Now A
E
(cs:~)
n ak
A
has convergent columns, indeed + zk since m in1 n n n + 1 lak-ak+ll x is an FK space, also 1 lai-aE-ll = lall + k=l k=l lzll + IIanllbv as m + m since an E bvo. The last term is bounded from 8.4.5B for
by assumption and so the criteria of 8.4.5B are satisfied. Since
A
6
(cs:~), limA
1.0.4, or because
X
1 f(6
=
)limnaE =
cs'
limA = limoA
is sc, and since cs k
E
1 f(6
k
has )zk.
(either by Ranach-Steinhaus
with 4.2.8), also
u E cs since k AK, limAu = limA (1 uk 6 ) = 1 u k limA6k With (1) this shows that z E W.
11. Naturally Theorem 10 holds if
X
=I
co.
In this case W
n
Em =
2-0
X
is conservative or
as well, by 6.2.6. In r69]
p . 601, it is shown that the latter result fails if
assumed sc, even if of
X = I bv
is only
In the same place an example is given
such that the conclusion of Theorem 10 fails.
also shown that conull.
X = cA.
X
1 6 2m$
in
cA, A sc does not imply that
It is is
A
Some further examples and comments may be found in [68],
especially 4.12.
9.3.
COREGULAR AND CONULL. 1. DEFINITION.
weakly.
A sc s p a c e is called c o n u l l
if
+
1
9.3
145
l(n) = " 1 6k . The definition is the same as that of k= 1 conull for conservative spaces, 4 . 6 . 2 . A sc space need not contain Here
1, but must, by definition, if it is conull. 2. DEFINITION.
A sc s p a c e c o n t a i n i n g
1
i s called coregular
i f i t i s n o t conull.
Thus
co
is sc but is not classified as coregular.
an arbitrary convention. a closed subspace of regular.
Any sc space
X @ 1
(4.5.5)
X
This is
not containing
is
1
and the latter space is co-
(X is convex closed hence weakly closed ( 5 . 0 . 2 ) . )
Thus
it would do no harm to define a coregular space as a sc space which is not conull. 3.
Henceforth a coregular o r conull space will be assumed to
be sc and to contain
FK
1. Thus by 9.2.4 it must include bv.
spaces which include
bv
are called v a r i a t i o n a l by J. J .
Sember, who studied their properties. bv
is not sc. 4.
They need not be sc e.g.
(9.2.6.)
DEFINITION.
An
FK
s p a c e i s called v a r i a t i o n a l s e m i c o n -
v s c f o r s h o r t ) i f i t i s s c and i n c l u d e s f Thus X 3 bv iff X c b s and 1 E X, ( 8 . 6 . 4 ) , f vsc iff X c cs and 1 6 X , also iff X is sc and servative
(=
bv.
while 1
X
x.
Q
(9.2.4(i). 5.
A conull space is automatically vsc while a coregular
space is defined to be a vsc space which is not conull. c o r e g u l a r o r conull s p a c e
II
bv
Every
by Definition 4.
Some of the conservative theory extends very easily:
is
146
9.3
Let
6. THEOREM.
conull
is, Y .
80
If
be
X, Y
spaces w i t h
FK
X
i s c o r e g u l a r and c l o s e d i n
X
c Y.
If
then
Y
is
X
Y
is
coregular.
The proof 4.6.3 applies, taking account of 9.2.5. 7.
THEOREM.
The i n t e r s e c t i o n of c o u n t a b l y many conull spaces
i s conull.
By the proof of 4.6.6 taking account of 9.2.5. Since
X
1 c W, we have
is conull iff
X
1 c 2bv4 by 9.2.10. If
X
is conull iff
is conservative this is equivalent to
the formally weaker condition
1 c 2,bv
by 6.2.10. The first
example cited in 9.2.11 shows that this equivalence fails for sc
X in general. 8. THEOREM.
If b v o
i s c l o s e d i n a v s c space
then
X,
X
i s coregutar.
Compare 4.6.4. Since
and
bvo
is weakly closed
l(n) $i 1.
(5.0.2) it follows that 9.
l(n) c bvo
It is equivalent to assume in Theorem 8 that bv
is closed.
This is by 4.5.3. Recall the functional X defined on X' by X ( f ) = f(1) k 1 f ( 6 ) where X is an FK space containing the sequences mentioned e.g. a vsc space. 10. THEOREM. i f f
X = 0
on
Let
X
be a v s c Space.
Then
X
i s conull
X'.
This is precisely the content of Definition 1. (See Remark 5). 11. APPLICATION. of
CS.
1 ankxk
Suppose g i v e n a sequence
{a,)
Then t h e r e e x i s t s an unbounded sequence 58 c o n v e r g e n t f o r each
n.
Let
X
x
of members such t h a t
n' Then X is
= naB
9.3-9.4
conull by 9.2.3 and Theorem 7.
AK
(4.3.7) and contains
147
ah
(Each
X @ am
Thus
1.)
is conull since it has by Theorem 6.
(The
latter space is coregular by 4.6.4.)
MATRIX DOMAINS
9.4.
The problem of characterizing matrices conservative (for a given
such that YA
A
Y) was solved in 8.4.3.
is
In this section
we solve the same problem for sc. If
A
is a triangle, f
since A : YA ak for
is the g
6
+
Y
Let
be an
Y
i s sc i f f t h e columns of g c Y ’ , where
each
1 g(ak)
iff
converges
This turns out to be the correct condition in general.
1. THEOREM. YA
f = goA, g 6 Y ’ , f(6 k ) = g(a k ) where
if and only if
is an equivalence. Thus f c cs column of A , and YA
kth
Y‘.
Y;1
6
ak
are i n
A
i s the
kth
Necessity: The columns of definition of s c .
Given
f c Y;1
Then
by 4.4.2. since YAf c cs.
let k f(6 )
space and
FK
f(x)
g,
=
Y
=
g(Ax)
g(ak’)
6
Then for
08
k
A , an = a n k ’
column of
are in
A
k {gfa 1 )
and
Y
a matrix.
A
since YA for x
3
41
by
YA, so
6
and the result follows
Sufficiency: We first note that each row of since in the hypothesis we may take g = Pn k this yields Ig(a ) } = {ank}. Hence u A 3 bv. cs
A
belongs to
where
Pn(x) = xn;
Then by 4.4.2, f(x) = ax + g(Ax) with k k g 6 Y‘, a 6 uAB C b v B = cs. Thus f(6 ) = ak + g(a ) ; by the hypok thesis and the fact that a 6 cs we have If(6 ) } 6 cs. Thus YAf c cs and YA is sc. Now let
2.
f
Given
t h e rows o f
A
6
A,
YH.
i f t h e r e i s any
l i e in
cs.
For
Y uA
Alternatively, one can argue that if
3
such t h a t
bv r
YA
i s sc, t h e n
as proved in Theorem 1. is a row of
A,
148
r8
=I
9.4
uA =I YA
hence
I n d e e d , j u s t as i n 8 . 3 . 1 2 , w e h a v e g i v e n
9.2.3.
such t h a t
is s c if and o n l y if t h e rows of
YA
was j u s t p r o v e d .
i s sc i f f
uA
Y =
r A,
A
6
then
W;
cs
6
by
there exists
Y
Necessity
cs.
uA = n{r 8 : r
is
Another f o r m u l a t i o n :
bv.
13
is s c i f
YA
Theorem 1 s a y s
3.
Y.
For s u f f i c i e n c y t a k e
i s sc by 9 . 2 . 3 and 9 . 2 . 5 ( i i i ) .
A1
a row o f uA
i s s c by 9 . 2 . 5 ( i i ) a n d s o
r8
{I ak 3
Also 8 . 4 . 2 s a y s t h a t
(Compare 9 . 2 . 8 . )
i s w e a k l y Cauchy i n
YA
3
bvo
{I ak 1
iff
i s ( w e a k l y ) bounded i n 'Y. The two c o n d i t i o n s c a n a l s o be w r i t t e n : k { g ( a ) ) 6 cs, b s , r e s p e c t i v e l y , f o r a l l g 6 Y' Remark 2 s a y s t h a t t h e columns o f
l i e i n cs.
AT
The n e x t
r e s u l t improves t h i s : THEOREM.
4.
First
A
If
(bvo:Y)
6
is sc t h e n
YA
by 9 . 2 . 4 .
AT
(Y8 :cs).
6
Hence
A
T
(YB:bs)
6
(8.3.8).
Under s t r i n g e n t a s s u m p t i o n s w e c o u l d now a p p l y t h e improvement o f mapping i n 8.3.6 s i n c e t h e rows of result i n general, let
z
6
Y0
A
are i n
and d e f i n e
g
u s i n g t h e Banach-Steinhaus theorem ( 1 . 0 . 4 ) . that =
f
6
Yk
1 znank =
Then
A
Hence
~ so 2 ) A T~
~
COROLLARY.
5.
i s sc.
by 4 . 4 . 2 . (
6
Let
(bV:Y
Y
k
If(& ) I
6
cs.
cs.
To p r o v e t h e
Y'
6
Let
by
g ( y ) = zy
f ( x ) = g(Ax)
But
so
f ( s k ) = g(Ask)
cs.
be a
BK
s p a c e and s u p p o s e t h a t
YA
Bf 1 .
By Theorem 4 and 8 . 3 . 8 . 6.
but
EXAMPLE.
YA = bv
i s faZse.
Let
Y = bv, A = I.
is n o t sc ( 9 . 2 . 6 ) .
Also
Then
Thus t h e c o n v e r s e of C o r o l l a r y 5
AT = I c ( c s : c s ) = ( Y B : c s )
Theorem 4 is f a l s e .
A c ( b v : b v ) = (bv:bvBf)
s o t h e c o n v e r s e of
I n t h i s e x a m p l e t h e c o l u m n s of
A
6
Y,
9.4-9.5
149
We can obtain a converse for Theorem 4 in the unimportant case in which 7.
Y
has
AK:
THEOREM.
Let
sc iff t h e c o l u m n s of
Sufficiency:
be an
Y
space w i t h
FK
belong t o
A
and
Y
Apply Theorem 1.
Let
AT
g
C
by 7.2.7,so A Tz c cs. But -- g(1 anks") = g(a k ) since Y has AK. z
6
AK.
YA
is
(Y*:csI.
6
Y' , zn = g( 6").
(AT~)k
Yf = Y8
Then
Then
1 znank
=
8. One can now apply Theorems 1 and 7 to give a list of conditions equivalent to the semiconservativity of
YA for Y
through the spaces in the Table p. 116. This was done for in Remark 2.
For
Y = c
9. THEOREM. that all
g
6
Let
where
Y'
ak
Y
= w
it is 9.5.6. be an
Y
Then
is vsc.
YA
running
s p a c e and
FK
1 g ( ak I
i s c o n u l l iff
YA
i s the
kth
a matrix such
A
c o l u m n of
= g(A1)
for
I f n o t conuli!
A.
it i s c o r e g u l a r . Let
Necessity: g(A1) = f(1)
=
Let
First, f(x)
f(l-l(m))
=
1
ak
g(Ax)
so
f
= lim g(Al(am))
f
C
Y;1.
YA
Y;2 by 4.4.2. Then
=
1g
k
a 1.
By 4.4.2 there are two cases to
A c bvB = cs [YA 2 bv by 9.2.43. 0. Second, f(x) = g(Ax) for which the
+
m+ 1 calculation given in the first part shows If
C
= ax, a 6 a '
m
So
=
1 m f(l(m))
Sufficiency consider.
f(x)
f(l(m))
+
f(1).
is not conull we are allowed to call it coregular by
9.3.3.
9.5. MATRICES.
The historical order has now been reversed. vative matrices, those for which to
FK
cA
2
c.
First came conser-
When attention widened
spaces it was very natural to define one to be conservative
if it includes
c.
But now we have semiconservative (sc)
FK
150
9.5
spaces and are contemplating what class of matrices might deserve the title.
(This trend continues in 99.6, in which coregular and
conull matrices are introduced after the corresponding
FK
spaces
in 99.3.)
A
By analogy with the fact that it seems reasonable to look for an A
semiconservative iff
6
(X:X).
A c (c:c),
is conservative iff
FK
space X
such that
XA
is
In this section we give a proof
(due to A.K. Snyder) that no such
X
exists.
The space X, to be
sought in vain, must satisfy two conditions and we discuss them separately in Lemmas 1 and 2.
1. LEMMA. ( i ) .I f
The f o l l o w i n g a r e e q u i v a l e n t f o r an then
A c (X:X)
(i) implies (ii):
XA
Take
i s sc
A
=
I.
(iil. X
FK
space
X.
i s sc.
(ii) implies (i):
If
A c (X:X) then XA 3 X, hence XA is sc by 9.2.5. 2.
Suppose t h a t an
LEMMA.
(ill. I f
A c (X:X).
is sc then
XA
FK
space Then
If the conclusion is false, X 8 9 cs.
=
cs8 = bv
let
by 7.2.3 and 7.3.5.1
ank = vnuk.
Then
Let
u
6
has t h e p r o p e r t y
X X
c bv.
[X8
= J
cs
implies XcX”
cs\X8 , 0 # v c X, and
XA = uA = u B which is sc by 9.2.3; but
A # (X,X) since uA $ X. 3. i s sc i f f
COROLLARY. A c
There e x i s t s no
FK
space
X
such t h a t
XA
(X:Xl.
By Lemmas 1, 2 and 9.2.5, bv would be sc, contradicting 9.2.6. However J.J. Sember [60] has proved that
A
c
(bv:bv) and
A:bv
Theorems 8, 9, 20.
+
bv
is a compact map.
bvA
is sc iff
See also [lo],
9.5
151
It is amusing and-alittle unexpected that (Lemma 2).
Y
This follows from 9.4.5 with
largest space satisfying (i)'. (i)' by 9.2.4(i).
satisfies (i)'
= bv.
Thus bv
X c bvo
Also any
is the
must satisfy
This leads to the conjecture that the converse
of Lemma 2 is true.
I have not been able to settle this.
4. DEFINITION. A m a t r i x conservative
bv
i s c a l l e d ( v a r i a t i o n a l ! semi-
A
f=(v)sc for short) i f
i s ( v a r i a t i o n a l ) semi-
cA
conservative.
The reason for this definition is that sumability theory deals with spaces of the form
cA
and with
generalize those of such spaces. conservative spaces to sc spaces 5. EXAWLE.
A
SO
much the better.
may b e s c and
is sc by 9.2.3 but
A2
spaces whose properties
If we can extend theorems about
v c cs\u B , alk = v k , anl = un, ank cA = vB
FK
A2
= 0
not.
if
n
Let
u
1
or
>
co\bv,
c
k
Indeed
does not exist.
>
1. Then A(A6')
does not exist so no natural modification of Corollary 3 will change its conclusion; e.g. there is no
X
such that
A
is sc iff
A c (X:X).
6. THEOREM. A
(i) (iil
A
h a s c o n v e r g e n t columns i . e .
a c cs T
(iii) A
i s sc i f f
where
cA
;3
Q,
a = { a k } , ak = l i m a n k ,
c ( ~ : c s ) ,e q u i v a l e n t l y
A 6
( b v : ~ ~ ) .
The equivalence of the conditions in (iii) is by 8.3.10. Necessity: with
g
=
(i) is by definition; to prove (ii) apply 9.4.1
lim; (iii) is by 9.4.4. (The shorter proof mentioned
there is available.) Sufficiency: Apply 9.4.1. g(y)
= X
lim y + ty, t c L ,
g c c' so that (1.0.2) k g(a ) = X ak + (tA)k and the
Let
Then
9.5
152
result follows from (ii), (iii). Theorem 6 provides a partial converse to 9.4.5,a modification of 9.4.7 for a space, c , without 7. THEOREM.
AK.
i s s c i f and o n l y i f C o n d i t i o n s ( i ) , ( i i )of
A
Theorem 6 h o l d and
(iv)
t h e r o u s of
A
belong t o
This is by Theorem 6 and 8.4.1B.
cs
fi.e.
uA
2
bv).
The equivalences in paren-
theses are by 8.4.28, 8.4.2,respectively. 8. COROLLARY.
i s sc i f and o n l y i f
A
cA
3
bvo
and
C o n d i t i o n s ( i v ) ( T h e o r e m 7 ) and ( i i ) T he or e m 6 h o l d .
This is by Theorem 7 and 8.4.28. n
1 (X2k-1-x2k). k=l either directly or by 8.4.2A. But A i s n o t s c 9. EXAMPLE.
(-l)k-l
Let
(Ax),
=
Then
cA
because
2
ak
bv =
so (ii) fails in Theorem 6. The example can be modified
to give a triangle by repeating each row and filling in the diagonal with a null sequence. We add a few remarks in the spirit of Lemmas 1, 2.
FK
10. Certain
include
spaces X
(all of which will be assumed to
in this paragraph) have the property
C$
(*I
i f
xA
3
bv
For example e v e r y
then AX
XA
i s sc.
space s a t i s f i e s
(8.3.8) and 9.4.7 gives the result. However (*)
Y
by Example 9, hence satisfies
satisfies
( * I and
(*).
For
X
Em
(*)
c
A
E
(XB:cs)
does not s a t i s f y
d o e s n o t by the following fact:
i s a c l o s e d subspace of
YA 3 XA
for
2
bv
implies YA
Y
then
i f
X
is sc; XA is a
9.5-9.6
153
If X 2 b v
closed subspace (4.3.14), hence is sc by 9.2.5. X
satisfies
(*I
then X
(*I.
not s a t i s f y
- take A
i s sc
I
=
-
bv
thus
and does
I do not know any internal characterization of
(*).
11.
The converse property
(**I
If
i s sc t h e n
XA
BK
is satisfied by every example
9.6.
X
= L,bv,bs.
XA
2
space with
bv, i . e .
XBf
X A i s vsc
X
=
(9.4.5).
For
Another condition is given in 12.4.11.
COREGULAR AND CONULL MATRICES To characterize vsc matrices we use again the notation
ak
=
lim ank, a 1.
=
{a,].
THEOREM.
A matrix
cA
i s vsc i f f
A
bv
3
and
a
6
cs.
This is clear from 9.5.6. 2.
Theorem 1 says that
i s vsc i f f
A
cA
bv
2
and
XlA)
exists.
3. columns,
COROLLARY.
f i i )a
bounded s e t i n
6
cs,
( i i i )A 1
A 6
i s vsc i f f
has convergent
(il A
c, ( i v ) The roWs o f
A
form a
cs.
Necessity: imp ied by
A matrix
(i), (ii) are the same as in 9.5.6.
(iii) is
c A 2 bv, (iv) is equivalent to (iii), (iv) in 9.5.7.
Sufficiency:
(i), (iii) and (iv) imply
cA
2
bv
by 8.4.2A.
The rest is by Theorem 1. 4.
tive iff
Compare Corollary 3 with the fact that A
A
is conserva-
has convergent columns, A 1 c c , and the rows of
form a bounded set in
II.
The condition
a
6
L
A
corresponding to
(ii) is true but need not be listed since it follows from the others
154
9.6
(1.3.7).
Before proceeding we need a result about associativity (partly due to R. DeVos) 5.
.
LEMMA.
Let
( i l . ItAlx
sequence such t h a t
t c Z,
and
be a m a t r i x , Z
A
( i i l . t(Axl
(=
1 1)
(=
an
AD
s p a c e , and
x
a
1 1 t n a n k x k ) converges f o r each k n c o n v e r g e s f o r e a c h t 6 Z. Then
n k t h e two sums a r e e q u a l f o r e a c h
t c 2.
If
i s an
Z
AK
space,
c o n d i t i o n l i i ) and t h e c o n c l u s i o n a r e i m p l i e d by ( i l .
Example 13.4.16 will show that (even a greatly strengthened form of) (i) does not imply (ii) without the extra assumption.
AD
Example 12.3.4 will show that f(t) = (tA)x, g(t) = t(Ax).
cannot be omitted.
Then
f,g c Z '
Let
by the Banach-Steirhaus
Further f = g on $I (1.4.3) so f = g on Z has AK, f(t) = 1 ti f(6 i ) = 1 ti 1 aikxk = t(Ax). i k Note that (i) says AT c (Z:x*). By 8.3.8, A c (xBB:Zf)
theorem (1.0.4).
If
which is a stronger statement than (ii) when 6. In writing
(tA)x
can also be written
(ATt)x
7. LEMMA. L e t Suppose t h a t
Clearly
exists.
t
t
Then f o r
6
This
is a column vector.
s p a c e such t h a t
BK
Y*, x c b v
Y*
has
we h a v e
AD.
ItAlx
exists so it is sufficient by Lemma 5 to show
(tA)x.
Now
A
c
by the second part of 8.3.8 hence (ATt)x
as a row vector.
(7.2.7).
t(Ax).
t(Ax)
the existence of
t
in which
be a
Y
YA 2 bv.
e x i s t s and e q u a l s
we treat
Z has AK.
Z.
This is
In Lemma 7, bv
(tA)x.
(bv:Y) and so AT 6 (YB:cs) A Tt c cs. Since x 6 bv = csB , (Remark 6 ) .
may be replaced by
bvo.
9.6
155
The next result is the extension of 4.4.7 to vsc matrices. Recall the representation f(x) = u limAx + t(Ax) + ax, t c L , a 6 cA B for f 6 ci given in 4.4.3; X(f) = f(1) - 1 f(sk) and X(A)
=
X(limA) = lim 1 -
A
1 ak.
Note that
X(A)
A
exists if
is
vsc by Remark 2. 8. THEOREM.
Let
sentation j u s t given.
First
a = {a,)
A
be a vsc m a t r i x , f
Then 6
Y
c
=
cs by Theorem 1, a by
II
cA
c o r e g u l a r and c o n u l l a c c o r d i n g a s
cA 8 c bv'
6
cs
=
cs
so
the
and using Lemma 7
are divided into
A
is c o r e g u l a r or conull.
Moreover the membership can be decided by the
(9.3.1, 9.3.2).
X
with the repre-
for the last step.
A s with conservative, vsc matrices
value of
cA
= p X(A).
X(f)
proof of 4.4.7 applies, replacing with
6
which was a matter of definition for conservative The result is 4.6.1 turned around and
matrices, (1.6.1). generalized.
The proof now given for Theorem 9 short-cuts 9.4.9 by using only
lim
g =
rather than all
g
6
cisely the condition of 9.4.9 with THEOREM.
9.
X(A)
Let that
X
X(1im A )
A vac matrix
X(A)
=
0
X(A) = 0
is pre-
g = lim.
i s c o n u l l i f f XIA) = 0 .
Then by Theorem 8, X
= 0.
is conull by 9.3.10. =
A
c', for
=
0 on
Conversely if
X
XI.
This implies
is conull
by 9.3.10.
It should be emphasized that the generalization from conservative to vsc is as far as one can go with these ideas. if
A
nxn - nxn-1' there is no use in conull just because X(A) = 0. Nothing follows. For
is defined by
calling A
For example
(Ax),
=
156
9.6
example it is false that implies that
A
l(n)
-t
1
in
cA
since this condition
is sc by definition, but this
A
does not satisfy
Condition (iii) of 9.5.7. To attempt to discuss variational matrices is equally futile since for such a matrix not even exist unless 10 COROLLARY.
A
is sc.
does
(Remark 2).
A coregular matrix
is
p-unique.
By Theorems
This generalizes 4 . 4 . 8 .
8, 9.
There are bigness theorems for namely
cA
3
V
where
V
since it is conservative). FK
A , X(A)
A
conull and sc
is a certain oscillation space, (not
0
However R. DeVos has scotched the
program with an example.
see [22].
cA with
For details and extensive discussion
157
CHAPTER 10 DISTINGUISHED SUBSPACES OF FK SPACES
10.0. FUNCTIONAL ANALYSIS Let
1.
on
Y
is given by
Y
closed iff 2.
I1yII
q-'[E]
=
inflllxl]:y
6
U
3.
U.
p
Examples of
a".
BK
spaces
Em,
as in (a), (b), (d), X
is a
shown to be a
such that
U
p(x)
p(x) 5 1; p
BK
X
hence a BK
in a vector implies
1
<
is called the
c
=
space.
0
If
BK
co
i.e.
X
+ E
E
c
0'
U
(b) Let
n
where
X
is
be a subspace
a",
(d) E
=
,'U
is a
[SO], #15-3-301.
SPACES
Before launching the main topic in the next section we give some topological properties of duals of general are not
FK
BK
spaces, for example,
spaces.
wB = @
E
co = { 0 3
[SO], Example 15-4-7, E
is dense in
X
with
(a) The easiest
(c) E = bs, 1111, p. 29, Theorem 3 .
quasi-complement for
DUALS AS
BK
X
space in [80], Example 13-4-5.
which is not norming over
[SO] #15-4-31.
co c X c em
with
space by 4.5.1; if not
example is given in [82], p. 12, 06.
10.1.
[SO], Theorem 6-2-11.
These are all of the form
is a closed subspace of
a
is
[SO], Theorem 2-2-1.
not closed in
of
A set E c Y
q(x)).
Given an absolutely convex absorbing set
and the latter implies that
g a u g e of
=
X.
is closed in
space, there exists a seminorm x
as in 7.0.1. Then the norm
X
be a quotient of
Duals in
is not (4.0.5)
For reference a few earlier results are collected:
10.1
158
1. THEOREM.
x
(assuming
3
4)
Let
X
be a
~f
are
BK
space t h e n
BK
X’,
Xy
and
spaces.
The first two are covered by 4.3.16, 4.3.17, the third by 7.2.14.
When any of these spaces is recognizable as a
BK
space, e.g.
cf = 1, the norm is known and need not be calculated from Theorem 1
The calculation must yield an equivalent result by 4.2.4.
X B c Xy c Xf
We know that
ordered in the opposite way.
(7.2.7) and so the topologies are
Although it is not impor-
(4.2.4).
tant, we take a moment to prove the stronger result 1. ~ ~ u l l f o~r
u c
x’.
Let
f(x) = ux
for
x c X.
n
have
If(x)l
IIfll 5
IIuIIB.
IIull, = IIuII If
II~II 5 1
we
lux1 5 sup1 1 uixil = IIu-xIIcs 5 11ull,. Hence i =1 n Since f ( 6 ) = un the definition of the quotient norm
=
2 IIullf. The next example shows that the norms
(10.0.1) yields
need not be equivalent. EXAMPLE.
2.
not closed i n
X
= co
Q
c XB c Xf
0z
X’
XJ.
not closed i n
(4.3.18).
(4.5.5).
Then
= a
X’
and
X’
Let =
Q
Xf; t h i s i m p l i e s t h a t z
n
is dense.
Xy
is
be an unbounded sequence and f zB, X = Q (7.2.4) so More instructive considera-
tions of this type are given in 14.5.4. XB to be closed in
Criteria f o r
Xf
are given in 10.3.9,
14.5.3. 3.
in 7.2.9.
EXAMPLE. With
X
X’
not closed i n
X‘.
as in Example 2 , X‘
=
The embedding is given 10 f
f o r some
f
and
the result is clear. Criteria f o r 14.5.3.
X’
to be closed in
X‘
are given in 10.3.11,
10.1
The inclusion maps from
XB
159
into
Xy
and
Xf
(7.2.7(i)) are
The same is true for
automatically continuous by 4 . 2 . 4 .
XI.
The
embedding is shown in 7.2.9: THEOREM.
4.
in
Let
space.
BK
The embedding of
XB
i s continuous.
X'
IIxII
If
5
1,
m
1 UkXkl = Il'*xIlcs 5 ll'll~. k=l The result is also immediate from the diagram
l)x(;(
IIfiI] 5 IIu11 B.
Thus
be a
X
luxl
=
5l P ' s
of 7.2.10. BK
A
(7.1.4).
space with
For some of the duals the norm is already monotone:
THEOREM.
5.
may be assumed to have amnotone norm
AK
Let
be a
X
have monotone norms.
space.
BK
Then
and
XB
Xy
17.2.1).
A direct check of these norms from 4.3.16, 4.3.17 shows this.
For example
SUPIII u-xllcs:l
I]uII =
IIu(m).
Now
xi1 5 11.
xII
cs -
n
sup{[ 1 ukxkl:n 5 m}. k=l Further supposing that l]u.xllcs
>
1 -
I
such that
uixil
Il~(~)ll
SUP over 'X
x c X I 1
-
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 <
M
p(zn6 n )
=
=
+ p(z (n-1)
p(~(~)-z(~-~))5 p(z("))
since a continuous seminorm is bounded on bounded sets (8.0.4). This result does not necessarily give any information, for
example if what
p
x
= w,
space and
p
U
116"11
tinuous seminorm
X
Let
some n e i g h b o r h o o d
take
for sufficiently large
n, no matter
is.
2. EXAMPLE.
BK
o
~(6") =
p
of
2
E
b e an 0
> 0;
FK
space
excludes a l l then
Bf c
Lm.
2
6n
$
and s u p p o s e t h a t
e.g. if
X
is a
There exists a con-
such that
~(6") 2 1
for all
U
(10.0.2).
Theorem 1 gives the result.
to be the gauge of
n, for example
10.3
166
EXAMPLE.
3.
An
AB
space with the property of Example 2 must
contain only bounded sequences.
B
We now discuss ,'B tions of
in some detail and give characteriza-
BK
spaces. Often the discussion is restricted to
AB
spaces for convenience. Remark 26 discusses restoring generality. THEOREM.
4.
Let
By 10.2.4, z
B+
6
be an
X
iff
space
FK
z-u
bs
6
4.
3
for each
u
Then f 6
X
B+ = X f y
.
This is
precisely the assertion. This makes it easy to compute
B+
and
The next result
B.
makes it even easier although there is a little less here than meets the eye
Y may be AB
- namely and
THEOREM.
5.
same f o r a l l implies
First
Y
not, even if
Let
be an
X
spaces
FK
B'(Yl
X
will be different for different Y
B
= B+(X).
Y
FK
between
is closed in space
$
and
4
(The c l o s u r e o f
B+(T) c B+(Y) c B+(X)
2
4. X;
X
Then i.e.
(7.1.2).
is t h e
B'
7C
Y C X
i s c a l c u l a t e d in
by 10.2.9.
e.g.
Xl.
The first and last
are equal by Theorem 4 and 7.2.4.
Y
In particular this holds if
Y
the same whether calculated in 6. EXAMPLE.
= 1'
Then
X. = Em.
x (n) c co. 7. in
am.
X
(assuming
Under the conditions of Theorem 5 the closure of
2 $).
space
is closed in
(10.2.9).
co
All this is actually quite trivial since for each The same r e s u l t h o l d s i f
EXAMPLE.
A
BK
is
i s a c l o s e d s u b s p a c e o f an fY From Theorem 5 , B+ (X) = B+ (co) = co
Suppose t h a t
Bf = am.
or in Y.
X
4
space
X
c
FK
x,
is a c l o s e d subspace o f
with
This turns out to be quite hard.
c c X c
am,
X
X.
not closed
Since it rests mainly on
functional analysis prerequisites we shall merely refer to a list
10.3
of constructions in 10.0.3.
167
The easiest one to say is
c + bs,
although it is far from the easiest to check! 8. THEOREM. L e t Xf c XY
i f f
AB
i.e.
X
be an
xf
=
space
FK
Then
$.
3
3
X
.
Sufficiency: B+ = X f Y
xY c xf
always.
then
i s closed i n
XB
3
Xyy
X.
3
XfYy
3
The last part is because
be a
X
space
BK
X B in
3
If
$.
has
X
AB
Xf.
By Theorem 8, since X B is closed in Closure of
Xy
(7.2.7).
COROLLARY. L e t
9.
has
xy.
Necessity: Using Theorem 4, X c B+ = XfY. Hence f
X
X'
Xy.
(4.3.18.)
can be evoked from Corollary 9 by
means of a property of the quotient map.
The embedding is given
in 7.2.9. 10. Let i f f
XB
+
4'
be a
BK
is c l o s e d i n
i s closed i n 7.2.10.
X
For
X'.
XB
space 3 9 .
=
4.5.2 with
H = X',
+
is c l o s e d i n
$';
Xf XB
q-' rXB], using the notation of
The result follows from 10.0.1. B Z = X
X0
This condition implies that
X'.
+
Then
XB
The second part is by
is a
BH
space by 10.1.4,
10.1.1.
11. COROLLARY (W.L.C. Sargent). with
AB.
Then
X0
i s closed i n
Let
X
be a
BK
space
X'.
This is by Corollary 9 and Remark 10. Corollaries 9, 11 and their converses hold in a very strong form for every
cA.
(14.5.3).
Next we give a short discussion of monotone norms, (7.1.1), culminating in Corollary 15. automatically a
BK space
3
Any space with a monotone norm is $.
We have seen that any
AK
space,
168
10.3
B-dual or
Y-dual has (or may be assumed to have) a monotone norm.
(7.1.4, 10.1.5.)
The first result is trivial: THEOREM, A n y s p a c e w i t h a monotone norm has
12.
AB.
We shall see in Example 16 that the converse is false. THEOREM.
13. is a
Let
f X , l ~ b~e ~a ~ BK ~
\lzllB+
s p a c e w i t h m o n o t o n e norm
BK
IIrllB = llxllB+
space w i t h
+
If
11x11.
They are
BK
=
has
X
llrfm’ll 5 ~ l l x l l f o r ~ Z Z m
such t h a t
space
+.
2
supllz‘n’ll; AB,
Then
Bf
is a
B
BK
there e x i s t s
M
and ~ Z Zz c X.
spaces by two applications of 10.1.1 (using
Theorem 4) and the intersection Theorem 4.2.15, the latter giving the norm on a
B.
The norm on
B+
is monotone since this space is
(10.1.5); this will also be obvious when the asserted
Y-dual.
identity is proved:
<:
Let
z c
B+ , u c X f .
1
Then
ukzkl 2 IIuII .IIz(”)II
by
k=l 10.1.6.
IIu.zllbs 5
Hence
IIuII *su~llz(~)ll and by definition of the
Y-norm, 4.3.17, (using B+ = Xfy, Theorem 4), 11 zllB+ 5 supl[~(~)ll. m f -> : Let z 6 B+ , For each m , ~ ~ z ( m= )sup{ ~ ~ 1 ukzk:u E X , k=1
I~uII
5 11 (10.1.6) 5 s ~ ~ i ( ( ~ . z l l ~ ~ :5l 11 l ~ l=l / I ~ l l ~ + .
If of
M.
X
has
AB
there are two easy ways to see the existence First, apply uniform boundedness to the maps x x (n)., -+
second, one can argue that the valent since
X
is a
BK
B
are
AB
COROLLARY.
Let
X-norm are equi-
space with each ( 4 . 2 . 4 ) ,
follows from the form of the 14.
B-norm and the
X
and the result
B-norm. be a
BK
space
3
$.
Then
’B
and
spaces.
The first is by Theorems 12, 13 (or 4 and 10.1.5).
As to the
10.3
second, for 15.
x
B,
6
+
=
COROLLARY.
X
Let
be a
by Theorem 13.
i f f
B+
Necessity is by Theorem 13:
I]X(~)~]
space
BK
a n d ) i s a c l o s e d s u b s p a c e of
AB
169
X
3
0.
Then
h a s a m o n o t o n e norm.
has the same topology a s
X
by 4.2.4 so its norm is equivalent to that of .'B the norm is monotone, for Theorem 13, so
X
x
Conversely if
X , /IxII = supllx(")/l
=
IIxIIB+
by
AK
spaces (7.1.4) and
B
and
Y
(10.1.5).
duals.
16. EXAMPLE. Thus
closed i n
B+ =
B+
is closed in .'B
This result applies to
norm.
6
(has
X
X
Bf.
An
space
AB
w i t h no e q u i v a l e n t m o n o t o n e
X
i s a n o n - c l o s e d s u b s p a c e of
Bf
and
B
i s not
This is the space of Example 7. By Theorem 5,
Ilm.
It appears that there are two kinds of with monotone norms (X
is closed in )'B
AB
spaces
X, those
and the other kind as
shown in Example 16. An amusing but perhaps not significant remark is that =
B+
=
XfY.
for
(3:
by Corollary 14; c : by Theorem 13).
Many such formulas could be found.
W
This says
B+(B') XfYfY
The corresponding result
is false (12.5.7).
To prove the next theorem we need a result about multipliers. Let
M(X) = M(X,X); the latter symbol having been defined in 4.3.15. 17.
( i )X
has
(ii) norm for =
AB, =
be a
(ii) M ( X )
(iii):
M(X)
X
Let
LEMMA.
2
bv,
BK
space
(iii)1
6
2
B
0.
These are e q u i v a l e n t :
[Mtx)].
by 10.2.7 (ii); (i) implies (iii):
given in 4.3.15, IIl(m)ll
s~p{llx(~)ll :I1xll 5 1)
<
M
=
by Theorem 13.
sup{llx.l(m)ll
Using the :IIXII
2 1)
10.3
170
(ii) implies (i):
Fix
x
6
X.
The map
T: bv
-f
Tu = u - x is a matrix map) hence continuous (4.2.8).
given by
Since 1 has
is bounded in X
bounded sections in bv, x(~) = T1'")
I do not know what conditions on M(X)
X
X
(8.0.1).
would correspond to
being sc, coregular etc. Let
18. THEOREM.
Xf
then
X c X ff
.
has
be a
X
space
BK
=I
(10.1.7) and
6
(Lemma 17) c B[M(Xf)]
B[M(X)]
B
If
has
X
AB
The c o n v e r s e i s f a l s e i n g e n e r a l b u t i s t r u e i f
AB.
I n p a r t i c u l a r the converse holds i f
First, 1
6.
has
X
AD.
since M ( X )
C
M(Xf)
is monotone (10.2.9). The result follows by
Lemma 17. The falsity of the converse is shown in 7.2.5. sentence is by 7.2.15
The last
-
it is strictly a particular case by 7.2.16. ff Finally, assume that Xf ha AB and X c X Let x 6 X.
.
Then :IIullf 2 11 (10.1.6). Also,
I
ukxkl
= ~U(~)X( 5
(10.1.6).
k=1
By Theorem 13, ~ ~ u (< ~M ) ~ so ~ {x'~)] A corollary is that if
X c X ff .
Xf
is bounded.
has
AB,
then
X
has
Sufficiency is by Theorem 18; necessity is because
Xff = XfY = Xyy by Theorems 8 and 18. Another is that AB
i f f
5
has
AB
has
Xf
by 7.2.4 and Theorem 18.
A useful smoothness property of contain
iff
AB
AB
spaces is that they
I 6 n ~nicely :
19. THEOREM.
Let
be an
X
41
( S e e Corollary 2 1 ) .
Then
i n particular i f
i s closed e.g.
B
has
FK
space such t h a t
AK
and
i f
X
S = W =
has
AB.
T.
B
=I
T.
This holds
A special
10.3
case i s
implies
AD-AB
AK.
Suppose first that f (x) = x - x(~).
Then
tinuous (7.0.2).
Since
n
171
X
has
{fnl fn
AB.
fn: X + X
Define
is pointwise bounded, hence equiconon
0
-+
then also
Q
(7.0.3). This is the desired conclusion. If
result just proved to the Since S
5.
FK
is the set of
by
space
x
B
fn
3
AK
AB
-
0 on
B, apply
which has
with
-+
$
the
by Theorem -
it follows that
$
c S.
The opposite inclusion is 10.2.6. LEMMA.
20. B 3
E
(closure i n
for all
x
6
8, IIx[iB+
be a
X
space
BK
and
2
c
i f f there exists
X )
by 4.2.4.
Sufficiency: Let
x
E
6
c X.
Then
such t h a t
(E,TX) is an FK spacec B
CllxI1
E
IE.
T ~ + ] E CT X
i.e.
E
6
Necessity: x
Let
+.
Hence for all
The result follows by Theorem 13.
IIx11 2 1.
with
Theorem 13. By 8.2.3, B + =
so
BZI
be a
BK
E.
Then
(That
IIxIIB+ 5 C by
E c X ll B+
is
immediate from the hypothesis.) COROLLARY. L e t
21. B
AB,
i s closed i n
tively, i n
X
B
Let
X
#
Xf,
be a AD
BK
space with
B
Xy
c Xff
has
X
x, r e s p e c -
AD
but not
+,
i s not s u f f i c i e n t that
X
the remarks following Theorem 18, and 7.2.15.)
AK.
# X f , a fortiori
i s not closed i n
(Theorem 19), i t i s d e n s e since it includes X
Then
$.
space w i t h
by Theorems 8 and 19. Also
also that
+.
3
i f f ( 1 ) hoZds f o r a l l
3
(5.2.5, 5.2.14) this is an XB
space
lor a dense s u b s e t ) , B,
EXAMPLE.
22.
X,
X
X
This example shows have
AB.
(See
172
10.3
The ideas of this section lead to a useful smoothing theorem: THEOREM.
23. exists a
space
BK
Y
Such closure of
X
Let
be a
with
Y
Also since Y c X B 8
10.1.5.
Y
in X B B . Then
$
YB
X
2
AD.
Y% = x 8
such t h a t
AK
Y
is unique and
space w i t h
BK
(8.6.1, 7.2.7).
has
AK
Then t h e r e
. Let
Y
be the
by Theorems 12, 19 and
we have
2
XBBB
XB
3
YB
2
using 7.2.2. 24.
There exist non-trivial examples as in Example 22.
25.
EXAMPLE.
a
XB8;
space w i t h
BK
Examples 22 and 7. it has
AK.
For an
space w i t h
BK
A
AB
X
and t h e same p r o p e r t y .
Clearly an AB
and
AD
AD
space
X
not closed i n
These are
is closed in
X B B iff
space the condition is that it have a mono-
tone norm. 26.
s:
w
-P
XN
To extend the results to given by
s(z) =
sequences of members of meaning.
Thus
B',
B
X. are
Iz(~)}. Then FK
FK
spaces we may consider
Here
XN
is the space of
B+ = s - ' [ Q m ( X ) ]
with obvious
spaces. Another possibility is to
apply techniques like those of Theorem 13 to each defining seminorm as was done in Theorem 1 and Example 2.
Any form of Theorem 18
will have to depart from our present ideas, for example if then Xf = $ which is not an FK space (4.0.5). 27. on
B
REMARK.
It should be emphasized that t h e
BK
X
= w
topologies
and t h e o t h e r s p a c e s a r e r a r e l y u s e d ; m o s t l y t h e s e s p a c e s
a r e t r e a t e d a s s u b s p a c e s of
X
with the r e l a t i v e topology.
This
is why the preceding Remark was made so casually. The first use of the topology of
Bf
was in Theorem 13. This is unambiguous
10.3-10.4
since usually
X, thus inherits no
is not a subspace of
B+
topology from it.
In Corollary 14 the
X, B
a subspace of
173
is not an
FK
These are the only places where the
B+
norm is being used; as
space unless it is closed in
BK
topologies of
B
B+,
X.
are
used.
References:
10.4.
1331, D91, 1611.
THE SUBSPACE F.
1. DEFINITION. (functional AKI i f
2.
An
F+
THEOREM.
x
3
i.e.
X
Let
space
FK
i s s a i d t o have
X
FAK
F = X.
b e an
space
FK
3
Then
$.
This is proved in the same way as was 10.3.4, with of
cs
instead
bs. 3.
COROLLARY.
t h e same f o r a l l
c
= Xf'.
F'
X
implies
Let
X
b e an
spaces
FK
-
between
Y
$
Then
$.
3
and
X
4
( T h e c l o s u r e of
= F'(XI.
Ft(Y)
space
FK
Ff
-$ C
i.e.
is
Y
i s calculated i n
Xl.
The proof is similar to that of 10.3.5. 4. FAK
COROLLARY.
i f f
xf
c 'X
Let
i.e.
X ~f
be an =
space
FK
3
$.
Then
has
X
x'.
The proof is similar to that of 10.3.8. An easy corollary is that
for
AD
spaces, FAK
and
AB
are
e q u i v a l e n t (7.2.7, 10.3.8).
Since
F c B, FAK
5, EXAMPLE. =
bvfB = ' : v b 0
=
bv
bvo.
implies AB does n o t have
Thus
F(bv)
=
but not conversely: FAK
bv,.
for F+(bv)
=
+
F (bvo)
(We used Corollary 3 and
174
10.4
the fact that
bvo
sc (9.2.6) thus have =
+ B
AB;
F(bs) = 6.
B
=
CS,
1
d F+
=
bv
An instructive proof is: bv 1 6 bv\F+.
so
(10.2.7)
by a similar argument. so
B+ bs) = bs
EXAMPLE
F+ = Xfg
AK.)
has
is
X
bs
sc i f f
it is clear that
has
F+
bv c Fi
Similarly
does
F+(bs)
FAX.
Since bv = lBB and
bv.
2
bv
However
but not
AB
is not
iff
1
am.
Necessity:
6
F+.
The result
follows by 10.2.7. 7.
EXAMPLE.
X
i f f
2 co
Ff
2
F+(X)
by the monotonicity Theorem 10.2.9 and the latter is a =
Sufficiency: Also
F+(cl = F f ( a m ) = 8.
9.
THEOREM.
(closure i n F
F+ ( c o )
X
X f g B 2 Xf
= am
so
X
2
co
am.
=
by 8.6.2.
by Corollary 3.
Let
closed i n
X
iff F
X)
=
F+(co)
Monotone n o r m d o e s n o t i m p l y F A K (Example 5)
EXAMPLE.
does n o t i m p l y
FAK
' ) F (
2
c:~
3
b e an
XBB
by 10.3.7; compare 10.3.25.
F K space
i s closed i n
X.
3
0.
Thus
Then B
B 2
F
closed implies
closed.
Sufficiency is trivial.
B
2
T.
Fix
f
Q
XI
g(x) = {f(x("))]. Then Pnog: F -+ K n k given by Pn h(x)l = f(x(")) = 1 f(6 )xk is continuous, 7 k= 1 having the relative topology of X. Thus g: F + am is continuous and define
g:
P
Now suppose that
(4.2.3), hence
F
=
nrg-l[cl:f
+ Pm
g-'[c] 6
by
is closed in
-F, and
s o , consequently, is
XI).
A criterion for Theorem 9 is given in 10.3.20. 10.
If
X
is a
i s a c l o s e d subspace o f
BK
space Theorem 9 can be improved to: Bf
the norms given in 10.3.13. Thus
Ff
is a
BK
and
F
i s a c l o s e d s u b s p a c e of
B
Ff
in
This is by 4.3.17, Theorem 2 and 10.3.4.
s p a c e w i t h monotone norm.
(For FK
spaces see
10.4
175
Theorem 9 t h e n f o l l o w s s i n c e
1 0 . 3 . 2 6 , and f o r a c a v e a t , 1 0 . 3 . 2 7 . )
Fc
F c
and s o
B
is c l o s e d i n
F
An e x a c t d e s c r i p t i o n of X =
Tu
(T)BB.
from
AK
has
Q
F
by 4 . 2 . 5 .
FAK
s p a c e s c a n b e g i v e n ; namely
and
E
i s made up o f p o i n t s s e l e c t e d
A similar d i s c u s s i o n could be given
(Theorem 1 2 ) .
AB.
for
11.
Then
FAK
-
where
E
-
LEMMA.
Let
be an
X
For
F+ = X f B
(Theorem 2 ) = ( T ) f B ( 7 . 2 . 4 ) =
12.
THEOREM.
Let
iff
7
has
Necessity:
AK, bv
has
X
X c
and AB
FK
1 $ bv:’
contains
T
space
FC B
X
FAK
= bvo.
has
If
X
i s maximal
applies t o such t h a t
(F)” Then
Q.
-
so
I$
has
(7.2.7).
has
X
AK
by
because, although Similarly
a
@
-
4 = bvo
has
1, co 0 z ( z $ a ” )
FAK. AK
has FAK
s u b s p a c e of any
X
AK.
S u f f i c i e n c y i s by Lemma 11.
w e can see how much C o r o l l a r y 4 f a i l s by u s i n g X 6 = Xf
t h e f o r m u l a (whose e a s y proof w e o m i t ) 13.
=I
(TIBB.
since
f a i l s t o have
bv
f a i l t o have If
AK
be an
X
The rest i s by Lemma 11.
Thus
has
(TJBB.
Ff =
10.3.19.
-Q
space i n which
FK
FAK
and i s a
AK
6
i n t h e sense t h a t space.
a , bvo, cs.
X
(i.e.
X = X
BB
), then
is n o t a p r o p e r c l o s e d
c0
is a c l o s e d s u b s p a c e of any
( 4 . 2 . 5 ) and e a c h s u c h
Theorem 1 2 o r b e c a u s e i t h a s
(X\T)B.
T h i s i s immediate from Theorem 1 2 and
However
co c X c am
space
n
X
has
FAK
by
AB.
The c o n d i t i o n s o f Theorem 1 2 imply t h a t
S =
T.
But making
t h i s s u b s t i t u t i o n a l l o w s u s t o prove t h e seemingly s t r o n g e r r e s u l t
176
10.4-10.5
X
that
has
omit t h e c o n d i t i o n space without 14.
X c SBB
iff -
FAK
has
$
C e r t a i n l y w e cannot
(Theorem 1 4 ) .
from Theorem 1 2 , w i t n e s s any
AK
AD
( 5 . 2 . 5 or 5 . 2 . 1 4 . )
AK.
Let
THEOREM.
valent:
(i)
x
~ v xB ) =
s',
( u i ) 'X
b e an
X
has F A K ,
(ii)
x
space
FK
4.
3
These are equi-
x c w**,
c s B * , (iii)
(iv)
x
c F~~~
= F ~ .
That ( i i ) i m p l i e s ( i i i ) and ( i i i ) i m p l i e s ( i v ) a r e t r i v i a l
S c W c F.
since
I f ( i v ) is t r u e , t h e n
( i ) is t r u e by C o r o l l a r y 4 . that
S =
T
XB
3
F0
XfBB
=
3
Xf
so
I f ( i ) i s t r u e , Theorem 1 2 i m p l i e s
and t h a t ( i i ) i s t r u e .
The e q u i v a l e n c e of ( v ) , ( v i ) ,
( v i i ) w i t h t h e o t h e r s is clear.
W = {x: f ( x ) =
Recall t h a t W C
Members of
X.
s a i d t o have
1.
valent:
(i)
x
for a l l
a r e s a i d t o have
S
i s f o r Schwach = Weak.
f
S A K , and i f
6
XI].
Thus
W = X, X
is
I t t u r n s out t h a t
space is redundant:
SAK
Let
THEOREM.
1 f ( 6 k )xk
W
SAK.
t h e c o n c e p t of
[61].
W.
THE SUBSPACE
10.5.
1591,
[33],
References:
has
X
SAK,
b e an
(ii)
space
FK
x
has
These are equi-
3 $.
(iii) X* = X I .
AK,
The meaning of c o n d i t i o n ( i i i ) w i l l be e x p l a i n e d d u r i n g t h e proof.
Clearly ( i i ) implies ( i ) .
must have
W c B.
Thus
W c
for
AD X
has
AK
T
Conversely i f
by 1 0 . 2 . 6 . by 1 0 . 3 . 1 9 ,
X
I t a l s o has
ux i.e.
for
X
x c X. has
f
6
X ' ; there exists
Since
SAK.
f ( 6 " ) = un
AB
SAK
it
since
That ( i i ) i m p l i e s ( i i i )
i s 7 . 2 . 9 where a l s o t h e meaning of ( i i i ) i s g i v e n . ( i i i ) holds, let
has
u
6
XB
Conversely i f
such t h a t
i t follows t h a t each
f(x) = x E W
10.5-10.6
The discussion of
c
177
in 7.2.11 is pertinent here.
Very sharp conditions can be given for W THEOREM.
2. valent:
Let
(il W
is c l o s e d i n
b,
S =
(vl
p,
(ivl W
=
in
a l l these hold.
X
b e an
X
(ii) implies (v):
( i i )B
X,
(vil S
space
FK
3
3
b,
has
4.
X.
X:
These are equi-
fiii). F
is c l o s e d i n
By 10.3.19,
to be closed in
If B
8,
2
is c l o s e d 4 c S.
AK, i.e.
The
opposite inclusion is 10.2.6. That (v) implies (iv), (iv) implies
SC W c
(iii) and (iii) implies (ii) are because
(i) implies (iv) and (vi) implies (v) since
$
T, W
c F c B;
c S c W c
T.
Finally
(iv) implies (i) and (v) implies (vi). Example 13.2.24 shows that for
cA with
A
W
is not sufficient here even
S
=
conservative.
Condition (ii) was discussed in 10.3.21.
W
X
If
3.
space parts of this result are improved by:
BK
i s a c l o s e d s u b s p a c e of
10.3.13. since
on
is a
F
h(x)
by
has
AK
AB.
with the norms given in
=
where, for any f 6 X I , h = hf is defined f(x) - 1 f ( 6 k )x,, and is continuous by the Banach-
If
W
has
(See 10.4.10.)
X may have
Sources:
1331,
AD
1611,
and
1863,
Hence
it must have
AD
with the (smaller) topology of
see that
10.6.
B
W = nh'
Steinhaus theorem (1.0.4). s p a c e oitlz
and
F
W
X
AK
and so
W
is a
BK
by 10.3.19, hence W = S.
We shall
not. (12.5.6).
r871.
BASIS
There is a fairly difficult Functional Analysis theorem which /
says that a weak basis for a k.c. Frechet space is a Schauder basis
178
10.6
i.e. if
x
=
1 tkbk
uniquely, the series converging weakly, then 0
x
the equality holds in the Frechet topology, and the maps
+
tk
are continuous. See CEO] #5.4.302. The difficult part of the proof is the second assertion. Since we are dealing with
FK
spaces, continuity of coordinates is assumed and the whole theorem
(A similar point was made in 7.1.5):
becomes easy. 1. f o r each
THEOREM. x,
Let
1 xksk
X
be an
space
FK
converges weakly t o
3
0.
Suppose t h a t
Then
X.
has
X
AK.
This is 10.5.1. 2. series
THEOREM.
1 rksk
Let
X
be an
space and suppose t h a t t h e
AD
i s bounded f o r e a c h
x
Then
X.
6
has
x
x
This is 10.3.19 since the hypothesis is that each
AK.
has
bounded sections. For instance, in Theorem 2 it would be sufficient to assume that the series is weakly Cauchy. special case AD.
FAK implies AK.
COROLLARY. L e t X f i n particuzar i f X8 = Xf i 3.
be an
This is 10.3.19 since X 4.
This just gives the
(8.0.2).
COROLLARY.
Let
X
then
FK
space
$ has
3
0.
Xy = X f
If
AK.
has
AB
(10.3.8).
be an
AD
space.
Then
X
has
AK
179
CHAPTER 11 EXTENSION
FUNCTIONAL ANALYSIS
11.0.
1.
Let
E
be a dense subspace of a metrizable vector space
0
X, Y a Frechet space and
T: E
Y
-+
a continuous linear map.
T has an extension to a continuous linear map: X
+
Y.
Then
1801,
#2-1-11.
11.1.
ROW FUNCTIONS
Extension of functions is an important topic.
The Hahn-Banach
theorem (3.0.1), which extends functionals, does not apply to maps of Banach spaces, for example the identity map: extended to
cu
11
.
( 1 8 0 1 , Example 14.4.9).
co
-+
c0
cannot be
The problem of which
spaces allow extensions is a well-studied one which will not concern us.
The best result which holds in all situations is:
E c X, is continuous and linear, then T: B -+ Y.
if
T: E
-+
T has an extension to
(11.0.1.) The problem we shall study is to inquire when
such an extension of a matrix is a matrix. tion about the rows of 1. DEFINITION.
A
so we begin by isolating this aspect.
calZed 4 row f u n c t i o n if h = B u c E . EquivaZentZy, h l x l =
u
on
1 h(6
h = u
on a v e c t o r space
h
A functional
Once a row function
8, i.e.
k
)xk
x, and
h
for
B
E
2
0
is
h ( x l = ux, f o r s o m e
x
C
E.
has been defined on
immediately has a natural domain, namely defined for such
This is really a ques-
u ; i.e.
@
it
h(x) = ux
is
has a unique extension as a row function
Y,
11.1
180
to this domain.
Thus several questions arise if
function defined on
is the natural domain of
h
how much of the domain is tinuous on
X
and
$
h
FK space
is an
continuous?
X?
Third:
Let
THEOREM.
Then
$.
be an
X
is
h
space
FK
Necessity:
for
on
h
is con-
X?
Then
Extend
h
to
un = f(6").
f
6
X )
X'
i f f
a rou
hl$
is
u 6 ~ f .
by the Hahn-Banach theorem
h = f
so
(i.e.
$
c
h =
and
Sufficiency: Say
x c $ , h(x) = ux = f(x)
This result
$
2
X - c o n t i n u o u s on
c o n t i n u o u s in t h e r e l a t i v e t o p o l o g y of
(3.0.1).
Second:
what is its Hahn-Banach extension to all of
b
f u n c t i o n on
how
if
We begin with a special case of continuity. justifies taking u 6 Xf . 2.
First:
$.
3
located relative to h
is a row
on
un = f(6"). i.e.
$
h
Then
is
continuous. Of course
h
6
X'
iff
u
6
X'.
Sufficiency is by the Ranach-
Steinhaus theorem (1.0.4). The situation with respect to continuity is bizarre, namely a row function can be continuous on a proper subset of its natural
domain. 3.
tinuous
Let
THEOREM. POW
f u n c t i o n on
n a t u r a l domain of
on of
F.
h
However
h h
be an
X $
(so
includes
space
FK
h = Ff
u
with
but
h
m u s t b e c o n t i n u o u s on
need n o t i n c l u d e
3
and
$
u 6
an
X-con-
Then t h e
need n o t be c o n t i n u o u s
W.
The n a t u r a l domain
-
$.
Viewed as an extension theorem this says that tinuous extension as a row function to row function to
h
X f) .
W
any extension as a row function to
$.
has a con-
and an extension as a
F , not necessarily continuous.
-
h
Of course
It need not have h
has a
11.1
continuous extension t o a l l of
more d i r e c t l y , s i n c e
let
u
f o r some
h = f1$
by Theorem 2 .
Xf
6
f c X'
i.e.
= h(x)
h = f
on
For
u'
h x
theorem o r ,
f , by h y p o t h e s i s .
is
h
Second
f ( 6n ) = u n .
with
by t h e Hahn-Banach
X
F i r s t t h e n a t u r a l domain o f since
181
3
= F+
Xf'
i s c o n t i n u o u s on
W, f(x) =
6
X
for
W
1 f ( 6 k )xk
=
ux
W.
The r e s t o f Theorem 3 w i l l be shown i n e x a m p l e s .
cases
(10.4.2)
cA w i t h
may b e c h o s e n t o b e
A
In both
a c o n s e r v a t i v e coregu-
lar triangle. 4. -
A c o n t i n u o u s row f u n c t i o n w i t h no e x t e n s i o n t o
EXAMPLE.
( a s a row f u n c t i o n . )
$
(10.3.22).
Say
tion since
u
un = f ( 6 " ) ,
# XB, b u t
f u n c t i o n a l s on 5. Let
f
fl$
Then
X'.
6
AD
s p a c e and
u
6
X f \Xf 3
is n o t a r o w func-
f
i s a row f u n c t i o n , a s a r e a l l l i n e a r
$.
A row f u n c t i o n c o n t i n u o u s on
EXAMPLE. b e an
X
be a n
X
Let
space such t h a t
AD
W # F.
b u t n o t on
W
F.
Any c o r e g u l a r s p a c e
h a s t h e l a t t e r p r o p e r t y by 1 0 . 2 . 7 ; t o make i t
i s more
AD
d i f f i c u l t , a n e x a m p l e is t h e c o n v e r g e n c e domain o f t h e c o n s e r v a t i v e coregular non-replaceable t r i a n g l e given i n 5.2.5. x
6
There e x i s t s
F\W.
f
6
un = f ( 6 n ) , by d e f i n i t i o n of take
x = 1, f = l i m A , u = a .
i s c o n t i n u o u s on
but not at
x.
$
Since
such t h a t
X'
has
# ux
where
( I n t h e case of 5 . 2 . 5 w e may f A See 5 . 2 . 1 ( v ) . ) S i n c e u 6 X , u
W.
(Theorem 2 ) , a s i s X
f(x)
Let
AD
f.
They a r e e q u a l o n
u
t h i s shows t h a t
is n o t
continuous. If
uB
n
X
is c l o s e d i n
domain by t h e B a n a c h - S t e i n h a u s
X, then
u
is c o n t i n u o u s on its
theorem 1 . 0 . 4 (so
c o n v e r s e i s f a l s e by t h e n e x t t w o e x a m p l e s .
u
6
f X ).
The
T h e s e e x a m p l e s show
$
11.1
182
an extension of
u
to all of
EXAMPLE.
6.
Let
X
(Ax),
of a very special form.
xn + xnml so that
=
for one row) the regular triangle Q
that
is
ui
with
vk
Now
since
x
6
Let
X.
yk
=.I
uixi.
+
f(x) =
limAx + t(Ax)
p
(4.4.5) it follows
and s o , by Abel's identity (1.2.9),
ir,
say.
ti(xi+xi-l)
t(Ax)
hence
+
tk+lYk = Vk
i s
v
+
Wk'
summable t o
(C,l)
t(Axl
is
w
(C,1) is regular, and we shall show that
(C,II
n summable t o
0.
To see this, observe that
(l/n)
1
wk
=
(Et),
k= 1
enk = xk/n
where
=
o(n)
k 5 n, t = (t2,t3,t4,. . . ) .
for
has null columns and xn
sup[enk(
by 3.3.11. Thus
7. EXAMPLE.
E
since
sup{ (xk/n(:k 5 n)
6
(e:co) by 8.4.1A.
<
for all
-
E
since
x c uB
(C,1) sum of the series is
n
X.
t(Ax)
x, hence if the series is convergent, this is its sum
t(Ax)
is continuous in
domain n e e d n o t be c Z o s e d for
since A
is of type
a p r o p e r s u b s e t of Remark): lim tk+lxk
i s c o n t i n u o u s on i t s
is regular. This shows that
(C,1)
domain since
With
The matrix
=
In Example 6, ux = t ( A x l
For it was proved that the
for all
.
1=1
ti + ti+l k Yk -
=
u 6 Xf
Let
(1.2.5), X = cA.
summable f o r a l l
(C,l/
uk = f(6 k )
Now if
is (except k
c ukxk
Then
*A
let
M
u*
2
c
x
on
since
X.
Finally t h i s
u
a, c
6
is dense
hence perfect (3.3.4), and t h e domain i s
if u is chosen in Xf\XB (See the next k xk = (-1) log k and choose t 6 II so that X
does not exist; then
uk = tk + tk+l, f(x)
=
y
t(Ax),
(Example 6 ) is divergent. f(6 k ) = uk and so u has the
required property. 8. The existence of
u
in the latter part of Example 7 can
be deduced from general theorems: X
does not have
FAK
by
11.1-11.2
Xf # X B
13.4.10, 13.3.1, so
183
A very similar situation
by 10.4.4.
may be observed in 5.2.14 where a construction much like that just given could be avoided by an appeal to a general theorem.
A row function composed with a matrix need not be a row
9.
function (1.4.6).
11.2.
MATRICES
Suppose that
X, Y
are
FK
0
into
Y
is a matrix mapping
Y.
A
That
belong to
be
Such
A
A
[a c Y.
valent to
co, Y = I I , A
=
T:
T
g.)
on
+
Y
=
A
A
W, F
A
i.e.
YA
3
Let
b e an
X
A: 0
a m a t r i x such t h a t
WI, a n d
is
A
extends to
( i n particular
nth row of
A A
-f
is
Y
W, F
as a matrix.
A
[a # -Y I .
to
T: g
and
P,oT
+
Y
FI.
W
3
Also
g,
an
Y
uA
3
i s dense i n
Finally
X.
are subspaces of
uA
space
FK
X-continuous.
W.
even i f
Y
X - c o n t i n u o u s on
Extend
space
FK
X - c o n t i n u o u s on
need n o t be a s u b s e t o f need n o t be
A
are equi-
Then
as a matrix, (equivalent statements are
W
Thus the
This is equivalent to asking whether
on
THEOREM.
extends t o
It is
I.
=
Similarly the conditions A [Wl c Y, A [F] c Y
T
A
is given by a matrix (which must be
-
and
belong to
(ll.O.l), and the
In this section the subspaces g , F, W 1.
A
is a stronger assumption.
question arises whether
A
A
and that
is continuous, using 11.1.2.
can be extended to
since T = A
g
3
i.e. the columns of
not sufficient, however, e.g. Cake X continuity of
X
X-continuous it is necessary that each row of since PnoA
Xf
spaces with
A
A [W] c Y ,
F
but
F
(thus
A [F] A
need n o t i n c l u d e
(11.0.1). Applying 11.1.3 to the
for each
n
shows that
T
=
A
on
W,
0
184
11.2
A [W] c Y , and
hence
uA
=J
F
s i n c e e a c h row o f
A E FB
The rest o f Theorem 1 w i l l b e shown i n e x a m p l e s .
cM w i t h
may b e c h o s e n t o b e
Y
by 11.1.3.
I n e a c h case
X
a conservative coregular t r i a n g l e ,
M
= co.
2.
EXAMPLE.
3.
EXAMPLE
uA
9
=J
T h i s i s t h e same a s 8 . 3 . 5 .
i n Theorem 1 .
A[F] d Y
Consider f i r s t t h e
i n Theorem 1 .
s i m p l e e x a m p l e of t h e i d e n t i t y m a t r i x which maps d o e s n o t map
c
into
example w i t h
W
dense i n
where
x
For
M: X + c
mk = l i m mnk, m = {mk}.
Let
given i n 5.2.5.
Now
E
(Ax)n = (Mx),
$,
is c o n t i n u o u s , also
-
Let
X = c
mk = l i m M 6 k
and
limM
X-continuous,
hence s o is
A: $
+
X(M) # 0
A 1 f co
so
but
F
1
so t h a t
mx
ank - mnk
Let
-
Ax = Mx
E
c 0'
0
c
is
Thus
A:
But
(Al),
= (M1)N -
Also
(10.2.7).
1 mk
+
AD
so
Then
A
has
X
mk9
by
L$
X'.
-+
-
(mx)l.
h i s X - c o n t i n u o u s on
11.1.2 s i n c e
W
To g i v e a n
r e q u i r e s more e f f o r t .
F
but
M is a c o n s e r v a t i v e c o r e g u l a r non-replaceable t r i a n g l e as
M
Y = co.
W = c0 ' F = c .
Here
co.
co
into
$
is d e n s e . With more a s s u m p t i o n s t h e e x t e n s i o n c a n be made: 4.
THEOREM.
-L$
extends t o continuous.
Let
x
I n Theorem 1 assume a Z s o t h a t
as a matrix i . e .
uA.
in
w
w
E
p; s a y
x
m -+
Y , and
since Y
c o n t i n u i t y of
A: uA
is a n A
-+
FK
w
x, x
m
E
$.
A
Then
is
X
r e p l a c e d by
b.)
I t follows t h a t
f o l l o w s from 4 . 2 . 8 .
q.
p
+
Y
POW
{Axm]
is c o n t i n u o u s ( 4 . 2 . 8 ) . space.
=J
A:
L e t i t s l i m i t be c a l l e d
Y.
(Apply 4 . 2 . 4 w i t h
since
51 c
I n p a r t i c u Z a r t h i s hoZds i f
hence convergent, i n in
A
uA
i s
X-
finite.
i s Cauchy, y.
Hence Also Ax = y
Now
xm
Axm Axm E
Y.
Ax
+
+
x
+
y
in The
11.2
185
A IF] c Y
It does not follow that
even if
A
row finite,
i s
as is shown by the first part of Example 3 . Lemma 8.2.1 is the special case of Theorem 4 in which
A
is
the identity matrix. One might try to improve Theorem 4 by concluding that in Theorem 1 the extension can be made to
uA
is shown by Example 3 since this subspace
n T . That this is false
3
With the notation of Example 3 define n.
Then
Theorem 4 that
G: Q
0
for
k
>
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
<
=
1
t ri ri+l f o r each
k
=
ci
La -
(6.3.2)
given in 6.3.2 is of type
t A A , a calculation yields =
F+, Le
It is sufficient, just as in
AD.
5.2.14, to show that the matrix
= B+ =
may be different.)
S
We shall apply Example 4 to
Y, :L
{tk:ri
M.
k < ri+13 = 0,
<
i = 1,2,...
.
Thus
t
=
ri
tri+l and so
t = 0.
atr)
6. EXAMPLE. B = F = W
BK
#
i s a conull c o n s e r v a t i v e
I t has n e i t h e r
S.
t o p o l o g y i s n o t an
AD
is from Examples 4, 5. by 5.2.9. X
Finally if
space.
W
and
AB;
AD
W , w i t h i t s own
Call the space AK
X.
Some of this 1 # 1 6k
is because
1 !f S; 1 6 W
since this would imply
AB
had
nor
The absence of
This shows also that
does not have
AK
space w i t h
AD
it would follow that
since
X
is conull,
AK
by 10.3.19.
S =
W
by 10.5.3. A
more exact description is given in 13.4.15.
7. An amusing consequence is that since W topology) does not have W(W) # W.
yA
it does not have
We shall see 13.2.26 that
has
W
SAK by 10.5.1
is not even conull.
A B , A*
6
(Y~:Y;), G c
xB .
This is immediate from Theorem 3, 12.2.8, 12.2.10. 8.3.8 says that
AT
6
Em.
(Note that
f (YB :YA) always.)
The best possible extension of 4.4.9 is now given. extension since A
i.e.
W i t h t h e n o t a t i o n of Remark 2 t h e s e a r e e q u i v a -
8. COROLLARY. lent:
AK
(with its BK
conservative implies BC
It is best possible by Theorem 3.
3
B+(c)
It is an
(10.2.9)
=
199
12.5
9. f
6
+
Yx
ci.
COROLLARY.
Let
A
be a m a t r i x w i t h convergent columns,
Then t h e r e e x i s t s a sequence
f o r aZZ
x
6
Y
such t h a t
ffx)
=
lim A x
B.
This follows from 4.4.3 with Theorem 3, Y
=
tA +
a.
The reader should note that most of the applications of Theorem 3 will be through this corollary.
This Page Intentionally Left Blank
201
CHAPTER 13 DISTINGUISHED SUBSPACES OF CA
B
13.1.
AND THE PERFECT PART
1. Many results of earlier chapters assume a more definitive
and instructive form when phrased in terms of distinguished subspaces. Recall the set
2.
X
Its closure in
sequences.
X
(5.2.8) of eventually constant
$1
is called the perfect p a r t of
is called perfect if this set is dense.
vious definitions (53.3) since if
is conservative, c
-
4.2.7 which also implies that the perfect part is e.g.
if
X
3.
Let
FK
immediate from 10.2.7:
+ B
2
conull, S
=
La
by X 2 bv,
if
1
is sc, F iff
@1
2
X
space $1
2
2
2
X
iff iff
$1
The following facts are
$.,
X
2
bvo, B
$1
2
is vsc, W
3
$1
X
iff iff
2
X
bv, is
is strongly conull.
Numerous references will be made to 12.5.3 which says that
4.
B
X
iff
$1
bv
-
= $
is vsc (9.6.1).
X be an
+ F
and
This agrees with pre-
-
X
X
Le.
=
Let
THEOREM.
5. part of
X
i s equal t o
be c a r e g u l a r , X = c A .
A
Then t h e perfect
B.
-
Since Let
f
B
X', f
6
by 9.6.8.
For
so
on
f
=
0
(3.0.1).
$1
3
on
= 0
x
B
(Remark 3) it is sufficient to show that
6
In particular
$l.
B, f(x)
=
Yx
(12.5.9).
~ ( f )= 0 so Now
Y k = f(6
li =
k
$ 1 2B.
0
) = 0
and t h e result follows b y the Hahn-Ranach theorem
13.1-13-2
202
The p e r f e c t p a r t may be l a r g e r t h a n closed.
i.e.
B
(10.3.22; but 13.4.10 is easier.)
The
n e e d n o t be
B
situation for
conull matrices is discussed in 13.5.1, 13.5.2. Theorem 5 extends 4.6.7 (which says that the perfect. part includes the bounded sequences) in two directions: first !Lm
A
if
is conservative since
Second, A
B
is allowed to be sc.
+ 3 B+ (c)
DEFINITION, A m a t r i x
c a l l e d an
AB
m a t r i x if
cA
By Theorem 5 .
n
(10.2.9).
by
$1, cA fl !L
i s s a i d t o have
A
has
m
B.
by
or is
AB
AB.
A coregular
7. COROLLARY.
cA
3
There is an obvious extension of
the consistency theorem 5.3.3: replace c 6.
= Lm
B
AB
matrix i s perfect.
The converse is false.
(13.4.10).
Theorem 5.4.11 is now extended to sc matrices. 8. THEOREM.
Let
b e c o r e g u l a r and r e v e r s i b l e .
A
i s p e r f e c t i f f i t i s of T y p e
Sufficiency:
if
M.
Also
$1, f(x) = t(Ax) by 5.4.3 0 = f(6 k ) = (tA)k so t = 0. Thus
f = 0 and the Hahn-Banach theorem yields the result.
0,
t A A , f ( x ) = t(Ax).
(Remark 3) so
Then
0 = f ( x ) - tn
A
f = 0 on
since u = 0 by 9.6.8.
let
Then
Then
f
f = 0. Fix so
=
n
0 on
B
Necessity:
(Remark 4), hence on
and choose x
such that Ax
= 6”.
t = 0.
THE INSET AND REPLACEABILITY
13.2.
A
If
is a matrix with convergent columns i.e.
c A 3 9 , we
have had numerous encounters with the sequence a = {ak), ak = If A c o
lim ank.
then
a c
!L
(1.3.7); if
A
is sc then
a c cs
(9.5.6). It is slightly paradoxical and very important
that x
6
cA
does not imply
x c aB :
13.2
203
2n
1. EXAMPLE. B
I f - l l n l c cA\a
.
1 xk. Then a = 1 and k=1 The example can easily be modified to give a conLet
(Ax),
=
servative triangle by repeating each row, replacing the a , and replacing diagonal 0 ' s
members of a sequence in small numbers.
1's by
with
There will be plenty of coregular examples also.
(Example 10.) 2.
REMARK.
coZumns, a
=
Iak 1
where
of
A,
The d i s t i n g u i s h e d sub-
a k = l i m an k .
spaces are c a l c u l a t e d i n
3.
i s a matrix with convergent
In this section A
X
=
cA '
DEFINITION. W i t h t h e n o t a t i o n of Remark 2 , In, t h e i n s e t
is
aB
n
X.
If
In
=
X, A
i s s a i d t o h a v e maximal i n s e t .
Thus a multiplicative matrix, or more generally, any matrix with null columns, has maximal inset.
The matrix of Example 1
does not. The crucial relations among
W , F, B
are shown by means of
the inset. W i t h t h e n o t a t i o n of Remark 2 , Ff c a',
LEMMA.
4.
Let
f = limA, z
6
+ k F . Then 1 f(6 )zk +
Y
A similar argument shows B c a
.
=
az
F
In.
C
converges.
This seems less signifi-
cant. 5. a B , =~
B
THEOREM.
n
Let {zkg(a
= B
f
n
uA
n
In.
B+
z c
k
W i t h t h e n o t a t i o n of Remark 2 , F+
)}
6
n
wA
n
a B . By 12.1.5 (iv) we have to show that
cs for each
g 6
c' where ak is the kth column of A (1)
The first case is
g
=
lim
and (1) says
z - a c cs which is
13.2
204
z
true since
aB .
6
The second case is
g(y) = ty, t
tion of 12.1.3 (iv) will yield only =
z
6
bs
A simple minded applicak in (l), but 1 zkg(a )
1.
1 n1 tnankzk = (tA)z and this converges, verifying (I), since + n W A = L+~ . (12.5.3). Thus B+ n 6 B n a ’ c F+ . But also k
WA
F+ c B+, F+ c
WA
(12.2.31, F+ c a’.
6. COROLLARY.
Let
If
cA = cM where
By Theorem 5, F = B
for
(Lemma 4 ) .
is r e p l a c e a b l e , F
A
M
= B.
has null columns so that
cM, hence f o r
cA
In(M)
=
cM
since it is the same
space. In [ 7 1 ] ,
it is pointed o u t that the converse of Corollary 6 is
false, even for a coregular conservative matrix, also that the matrix of 5.2.5 has
7. EXAMPLE.
F # B.
A non-replaceable
conull conservative matrix.
The example can easily be modified to be a triangle by repeating rows and replacing zeros on the diagonal with a null sequence. n Let (AX), = 1 (1/2)k-1(x2k-1+x2k). If x = (1,-1,2,-2,3,-3,. .. ) k=1 1 1 then Ax = 0, x ff In since a = ( 1 , 1 , ~ , ~ , 1 / 3 , 1... / 3), so that ax is divergent.
m so
is odd,
Also
AX'^))^
IIAX(~)~~_= 0 o r
12.1.3.
x c B
for if
= 0
for
n
1
for
m
<
m
is even, Ax(m)
(m-1)/2, 1
for
even o r odd.
It has now been proved that
=
0, while if
n 1. (m-1)/2;
Thus
x c B
by
Bq! In, hence
B # F
by
Lemma 4 and the result follows by Corollary 6. (This m a t r i x is u - u n i q u e by Theorem 2 5 . )
There is an intimate relationship between (Theorem 9 ) .
F
and the inset.
We first introduce some natural invariants:
13.2
8.
s p a c e of t h e f o r m
ve,
eAl,
Let
DEFINITION.
b e any s e t a s s o c i a t e d w i t h a
= U(A)
Then
cA.
the external
U
205
Ui, t h e i n t e r n a l
U, i s
u{u(M):eM
=
U, i s
nIUIMI:eM =
CAI.
For example, the i n t e r n a Z i n s e t , Ini = nIIn(M):cM =
mk
n { m B n cM:cM
=
A
cA n(n{m B :cM
cAl, where
=
m
=
{mkl,
The next result solves the naming problem for the
lim mnk.
=
c I
=
= cA)
internal inset. W i t h t h e n o t a t i o n of Remark 2 , F
THEOREM.
9.
c: By Lemma
4 and the invariance of
It is sufficient to show that Let
(12.5.3).
D
and set
=
t
= limMak = lim ak
Since
x
6
dB
n
cD
+ (tA)k
aB
6
B
=
cA
M
=
+ (tA)k.
it follows that
(tA)x
.
x
i.e.
Ini .
6
x
F
Le
6
(1.8.12),
dk = limD6 Thus
tA
converges i.e.
But there may be no minimum inset i.e.
191 p.
I ni
Let
M(t)
by 1.6.4. Then
(1.8.12) = ak
inset of any equipotent matrix (
3:
(Theorem 5)
L , form Mazur's matrix
6
so that
MA
x
F.
=
k
=
d-a.
x
6
Le.
need not be the
229, Ex 2).
Example 13
shows the opposite extreme. 10. EXAMPLE.
In
i s not invariant.
matrix which does not have
AB.
Let
AB, a fortiori
F # cA.
matrix with smaller inset. this, B
n In
cA
has
It
However it does not
By Theorem 9 there is an equipotent
It is interesting that in spite of
F.
i s i n v a r i a n t since it is
Of course if
be a regular
(One is given in 13.4.10.)
has maximal inset since its columns are null. have
A
FAK
then
A
(Theorem 5).
has
PMI
(Propagation
of Maximal Inset) i.e. every equipotent matrix has maximal inset,
by Theorem 9.
The ingenious example given next shows a matrix
which does not have PMI but has invariant inset i.e. In(M) = In(A) whenever
cM
=
cA.
A few results are given in preparation:
206
13.2
LEMMA.
11.
z c bu\bvo
If
Since lim z # 0
then
x
6
then
z0
l/z c bv
x = (x.z).(l/z)
Let
LEMMA.
12.
and assume t h a t
Let bv.
b0
3
z = b/a.
6
cs
by Abel's test.
(That
=
properly. z0
bvo
C
Izn-zn+ll / lznzn+ll< Mlzn- z ~ + 1~ \ . be s e q u e n c e s w i t h ai # 0 f o r a 1 1 i
a, b aB
Then
By Lemma 11, z
csebv
Izn+l-zn -1 -1I
is because
since
i
doesn't affect anything. Let
z
6
08.
zi 9 0 for all
we can assume
changing finitely many terms of
c
zB
3
so
Then
1 ' = cs z0
3
bB
a'.
3
properly.
bs = ' 1
z c z B Oc
Thus
from which the result
f01 lows. A matrix with invariant i n s e t
EXAMPLE (G. Bennett).
13.
w h i c h i s n o t maximal.
shown then that
A
Let
A
be the matrix of Example 7; it was
does not have maximal inset.
+
S
(AZ'~))~
since
1
=
akzk
for suitable
B
First,
s , so
F
= a'
+
=
Y
a
by Theorem
k=l
F
5, hence
aB since obviously
=
In(M) # In(A) In(A) so m 6 In(M)
3
B.
a 0 c cA.
Now if
cM = cA
we must have, using Lemma 4, m B
3
In(M)
By Lemma 12, m B
3
aY = Bf
3
a'
properly.
By Theorem 5 this implies F = B
with
F = a'
3
=
so
contradicting the
result of Example 7. A(x) = limAx
The inset is the natural domain of the function
-
ax which was defined for a special domain in 5.6.3. If
vsc then
A(1)
coregular
If
A(x)
+
Yx
or
W = F
is
x(A).
=
W i t h t h e n o t a t i o n of Remark 2 , W
14. THEOREM. either
A
W
i s a maximaZ s u b s p a c e of
=
If
F.
F Il A' A
so
is
F = W 0 1. x
= 0.
6
W
the definition of
W
n
A',
Conversely if
(12.5.9) = uax
+
x c F
Yx = gx, say.
applied to
limA
f(x) = p limAx k f(6 ) so x c W.
f c ci, then Then
Bk =
shows that
13.2
207
Xf
since
is possible.
Thus
15. With the notation of Remark 2, a (We saw in Example 1 that continuous on
#
a
XB
(11.1.2) but actually
W
6
0
A =
on
W
ak = limA6k . is
A
X-
by Theorem
14. W i t h t h e n o t a t i o n of Remark 2 s u p p o s e t h a t
16. THEOREM.
i s coregular able,
(ivl A
on
$,
X - c o n t i n u o u s on
is
l i ) A i s replace-
X l , ( i i i )l i m i s
(closure i n
l i i ) 1 p'
$,
Then t h e s e are equivaZent
(VSC).
(v) x
$,
lvil In ( i v ) , I v ) replace
by
$,
X - c o n t i n u o u s on
ax
-f
A
is
X-continuous
F.
This is the extension of 5.2.1 to sc matrices. (iv) = (v): A(x)
= lirn x A
- ax and limA
E
The two parts
X I .
of (vi) are equivalent f o r the same reason.
(i) implies (iii):
Let
cM
limM = 0 on
1. Then
=
cA, M
having null columns,
limM 1 = 1 so limM = lirn on $1 and continuity is guaranteed by invariance of topology (4.2.4). X(M)
=
(iii) implies (i):
f
Let
Banach theorem (3.0.1).
$,
X', f = lim on
6
Then
1 = X(f)
=
u X(A)
$1
by the Hahn-
so
(9.6.8)
l.~
# 0.
The result follows by 5.1.1. (iii) implies (ii):
Since lim = 0 on
(ii) implies (iii):
In
b1
with the
X
maximal subspace which is not dense -hence Since it is (iii)
=
X(A)
Thus
lim 1
topology, $
is a
it is closed (5.0.1).
l i d , lim is continuous (5.0.1).
For x
(iv):
particular f o r =
and lim 1 = 1.
$
x
6
$1
and f o r
A = X-lim on
$1
x
since
a
c $,
A(x)
6
cs (9.5.6). =
ax
- ax
=
Thus
In
F
2
$1
A(1)
0 = X(A)
and the result follows since
(vi) implies (iv) since (ii) implies (vi):
- ax. This holds in
In, h(x) = limA x
6
= X(A)
lim x.
X # 0.
(13.1.3).
F with the X
topology, W
mal subspace (Theorem 14) which is not dense since
W C
7
is a maxi(10.2.6)
208
13.2
1 c F
and
(10.2.7). Hence
(Theorem 14) and so If
17.
is closed
$1, F , W, I n
are all non-closed subspaces of
tinuous o n any closed subspace of its domain.
F with the X
it is closed in
makes
with its
FK
W
If
is con-
is closed,
topology and as proved above this
continuous. Of course
A
= A'
is continuous (5.0.1).
A
for, by the Banach-Steinhaus theorem (1.0.4), A
= cA,
W
(5.0.1). But
is coregular and non-replaceable it follows from
A
Theorem 16 that
X
W
F
is always continuous on
A
topology (10.4.10, 10.3.26). 1
18. THEOREM.
W i t h t h e n o t a t i o n of Remark 2 , s u p p o s e t h a t
i s non-repZaceable,
f
4,
on
= 0
i s dense i n
With
A
x c B.
p-unique,
f
=
Then
M
f, ( i i i l
f
has a representation as
4.
By (ii) and 12.5.9, f(x) f
=
0 on
M
f(6 k ) = 0, so
This proves (ii), and (i) follows.
0 on
$
X1.
has null columns since
Yk = f(6 k ) = 0 so
But
f c X',
v # 0. By 5.1.1 there is an equipotent matrix f.
=
liil i f
i n e v e r y r e p r e s e n t a t i o n of
as in (ii) suppose that
is replaceable.
(iii) let
= 0
is
(with the topology of
B
f
limM
u
then
in 4.4.3 with with
Then l i l A
A
=
Yx
To prove for
B , and the result
follows from the Hahn-Banach theorem (3.0.1). 19. either
-
B
Since
OF
F
-
W i t h t h e n o t a t i o n of Remark 2 , F = B .
COROLLARY.
is c l o s e d t h e n
If
F = B.
4 c F C B , the first part follows for a non-replaceable
matrix from Theorem 18 (iii). (Corollary 6).
If
B
If
A
is replaceable, F = B.
is closed, so also is
F
(10.4.9) and the
last result follows. The next result shows that the only role of the assumption of null columns is to ensure maximal inset; for replaceablity it is
13.2
209
enough t o h a v e a n e q u i p o t e n t m a t r i x w i t h maximal i n s e t . c o r o l l a r y is t h a t i f
i s g i v e n a n e x t r a row, namely
A
resulting matrix, call it able.
20.
THEOREM.
For
A
=
In].
c A , h e n c e c o n t i n u o u s by t h e Banach-
is d e f i n e d on
If If
The r e s u l t f o l l o w s by 5 . 1 . 1 .
has c l o s e d i n s e t , F = B .
cA A
is c o r e g u l a r .
A
Any m a t r i x w i t h m a x i m a l i n s e t i s r e p l a c e a b l e .
S t e i n h a u s theorem 1.0.4. 21.
a, then t h e
M , h a s maximal i n s e t , h e n c e is r e p l a c e -
cM = cA n a B i s r e g u l a r i f
Also
An a m u s i n g
[B c
B
=
m
=
c
i s a l s o c o r e g u l a r i t i s r e p l a c e a b l e by Remark 1 7 .
I d o n o t know w h e t h e r
A
must b e r e p l a c e a b l e i n g e n e r a l .
See
13.5.5. 22.
EXAMPLE.
A
continuous but
regular replaceable triangle.
u
6
Xf\XB
(11.1.7,
(11.1.7); let
Q)
Q
need t o c o n t i n u e s i n c e
In
(11.1.7).
X
D
for
x
So
However
1.7.15).
A
replaced
A
In(A) = u
B
n
Then
A(x) = l i m A x
k ak = f ( 6 )
which is n o t
X
-
ux
EXAMPLE
(G. B e n n e t t ) .
In
but
The m a t r i x o f Example 7 h a s F # B 2m I f x 6 I n , AX)^ = akxk by Remark 2 1 . k= 1
so
In
I\ = 0
closed.
Note a l s o t h a t F # B In
X
is a t r i a n g l e and t h e r e is no
on
for
n
is continuous
(11.1.7).
23.
since
uB
6
There e x i s t s a t r i a n g l e
(3.2.4,
a = 0
as i n 1 1 . 1 . 7 .
= uk
closed i n on
Q’ l e t
ux = t ( Q x )
f ( x ) = l i m x + t(Qx).
i n t h e proof of which
= (tQ),
Now
a co-
as i n 1 . 2 . 5 , X = c
Q
cA = X , l i mA = f ; t h i s i s by 5 . 1 . 1 ( w i t h
such t h a t by
11.1.8).
With
not closed, A
In
In(M)
is an i n v a r i a n t c o n d i t i o n .
A
so
ax
not
is n o t c l o s e d A = 0
on
is n o t c l o s e d f o r any e q u i p o t e n t m a t r i x
t o be closed in
example if
+
In
cA
and n o t i n
is p e r f e c t , r e g u l a r and n o t
In. M
In general i t i s possible cM
AB
with
cA = cM:
for
( 1 3 . 4 . 1 0 ) , I n = cA
13.2
2 10
but there exists M
In
c
2
cM
=
cA, In
+
cM
(Theorem 9); thus
so it is dense.
EXAMPLE.
24.
(Example 13) and and 14).
The matrix of Example 7 has invariant inset A
=
0 (Example 23), so
W
=
F = In
Any easy calculation shows S = In.
from [87] 8.2: impty that
not
W
if
A*
=
W
i s ctosed.
Then
p-unique.
W
then
=
(This also follows
S.) Thus
(Example 23.)
(Theorems 9
W
does n o t
= S
Compare Theorem 10.5.2.
W i t h t h e n o t a t i o n of Remark 2 s u p p o s e t h a t
THEOREM.
25. i 8
with
A
i s repZaceabZe a n d
W = F = B
A
=T.
This is just a summary of Corollary 6, Theorems 14 and 18. 26.
EXAMPLE.
n
with
B
Then
c c W c Ilm
BK
=
cA
topology 27.
II
m
Let
.
A
be any conull conservative triangle
(There are plenty of these so
W
-
see 13.4.9.)
is coregular and conservative in its own
(10.5.5, 4.6.3).
There is a subtle commentary on the use of abstract
methods available here.
First note that
W(AI =
nIAk:
cM = c A l ,
This is proved in the same way as Theorem 9. Agnew and Zeller showed that if such that n(r) c
A cA
W
A*.
A
is conull and conservative, there exists r
n(r) c cA.
(Generalized in 6.4.3.)
This implies that the sequence r
and cannot be defined invariantly from the as in 6.4.3; for
n(r)$
W
if
A
They showed more: chosen depends on
FK
structure of
is as in Example 26 since
is coregular (6.4.4). and so the formula for W
ago shows that there exists M
with
Sources and further results:
given a moment
cM = cA, n ( r ) $ A;.
81,
1161,
1443.
13.3
THE MAIN THEOREM
13.3.
1.
X
=
211
cA.
Let
THEOREM.
b e a m a t r i x w i t h c o n v e r g e n t columns,
A
These are e q u i v a l e n t : (il
X
has
AB,
(ii) X
has
FAK,
(iii) Le
= X
(Definition 12.2.2 with
B = X, F = X,
(iv)
A
i s associative i.e.
(v)
A
has
PMI
(ail
X
has
AK
(viii) If
A
(vil' (vii)'
(Text following 13.2.101, or
or
W = X
AT c
La = X ,
with
X = S0 u
(E:X
= W 0 u
X
with
B
=
i.e.
i s c o r e g u l a r r e p l a c e ( v i l , ( v i i ) by n X = S@ 1 i.e. X has 1 1 , s 3 a s b a s i s ,
X
W@
=
1.
X, F
B = X
=
by 13.2.19.
=
(v) by 13.2.9; (i) = (ii):
The converse is trivial since
W c B; (ii) implies (vi):
(vii) implies (i) since
F
is defined on
(13.2.4) hence
W
theorem (1.0.4)and so
Fn
=
X
since
W
S =
W
since
6 c Sc W c
If
A
has
u c B,
S C
W;
Note that
A
F c B ; (vi) implies (vii) and (vi)' implies (vii)' since
W
X
B).
(i) = (iii) = (iv) by 12.5.3; (ii) If
u c B
as basis,
{u,sn)
(vii)
Y = c),
=
A'
c X'
A
by the Banach-Steinhaus
is either a closed maximal subspace or (13.2.14).
0
By 10.5.2, S
is closed so
(10.2.6); (iii) = (viii) by 12.2.8.
is coregular, 1 c fiW
by 10.2.7 so the last part is
a special case. Other conditions equivalent to the important AB
conditions
are given in 14.5.3. 2.
Some
easy inequivalence theorems follow, namely, there
exists no matrix
A
such that
cA = cs 0 u , u c bs, for the latter
13.3
212
cA # bv
space satisfies (vi) but not (ii) by 10.4.12. Similarly The last result can be phrased:
(10.4.5). (e.g. i f bv
-
i s sc),
A
then
cA
i f
3
bvo
properly
m u s t sum a s e q u e n c e w h i c h i s n o t i n
A
a bigness theorem. In (vi), (vii), u c B
u
may be replaced by
However, it may not be replaced by
u
6
In
c
F
even if
obviously.
A
is a multi-
plicative triangle and all the distinguished subspaces are closed (The last condition is bravado - see Remark 4 . ) :
and equal.
EXAMPLE.
3.
but
S O u
A
A
A m u l t i p l i c a t i v e matrix
does n o t have
such t h a t
A
cA =
It is sufficient to construct
AB.
with these properties but with "multiplicative" replaced by
"conservative and replaceable". Let A = DE where (Dx), = xn n 2 ~ , - ~ ,(Ex)n = 1 ekyk with e c L , ek # 0. Then A is a conk=l -1 servative triangle and cA = cE O u where u = E I2-") (1.8.8). Now
cE
=
e B has
AK
(4.3.7) and is a closed subspace of
(4.5.5) hence is equal to
with AB
tn = 2-" we have
by 13.1.8, 13.1.7.
A
S. Also
trD
hence
is not of type
tAA, thus
A
M
cA since
does not have
Finally we have to show that
A
is replace-
m
able. cDM
Let
1 ekyk. Then cM = cE = e' and so cA -k=n+l is a triangle. Clearly DM has null columns. (By
( M x ) ~=
since D
the next Remark, B
=
F
=
W = S.)
4. If 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 of e q u a l and
X = S 0 u
t i o n holds i f
X
A
then
A
has
F = S
or
F = X
the same is true.
by 13.2.14.
If
F
=
S
=
cA
are not a l l
In particular t h e implica-
AB.
Since
i s coregular.
X
S
then
is closed, S = W
B # F
so
B
=
so
X; if
13.4
213
APPLICATIONS
13.4.
Obviously Theorem 13.3.1 is very powerful. AB
It shows that an
matrix has a convergence domain which is either an
(like
or an
co)
either case A
cA
the basis is
AK
AK
space
space with one more sequence (like c).
has a basis of very special form. For coregular 1 6 k 1 u { 1 } , exactly the same as that of c. In
order to show applications we shall give growth theorems for and
and a criterion f o r
S,
In
AB
B
which will identify matrices
falling within the scope of the applications. REMARK.
1.
columns, X in
cA
=
In this section
A
i s a matrix with convergent
and 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 c a l c u l a t e d
X.
THEOREM.
2.
.
= IIakll
W i t h t h e n o t a t i o n of Remark I , l e t
Then each
x c S
satisfies
ukxk
+
u k = supnla*l
0; i n particular
m
akkxk
0.
+
BY 12.1.7, A - X is strongly conull. uniformly convergent.
Let
E
~ / 2for all
n
>
0
1 ankxk
BY 5.2.9,
and choose
ko
is
such that
m
I 1
ankxkl
<
if
r > ko.
Then
lankxkl =
k=r m
m
I 1 - 1 i=k
i=k+l
a .x.I <
E
for
k
ko.
>
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) <
M e r c e r ‘ s t h e o r e m (2.4.1). A form of this is
n, n/(n+l)(na+l)
for
k
a > 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!
<
n=m+l For x c X
m
u c XB.
and l e t
E
with
IIxII 5 1 w e
n=l have m
If(x)-u(x)l
11 t n ( A x ) n
= <
1 UkXkl = I 1
n=m+l
tn(Ax)nl 5
W i t h t h e n o t a t i o n of Remark 4 , G
a Banach s p a c e whose t o p o l o g y i s l a r g e r t h a n t h e
Let II
onto
q : II
+
b e d e f i n e d by
XI
nCE(x):x c X I
where
closed since
Ax c c
f u n c t i o n a l on
q : II
Ilq(t)ll 5 IItll
-+
and e a c h t(Ax)
E(x)
is
is a c o n t i n u o u s
I q ( t ) ( x ) l = It(Ax>l 5
NOW
G
T h a t t h i s norm makes
G
a Banach s p a c e is t r u e , w i t h t h e A.
e x p e c t t o o much, f o r example i f
then
LEMMA.
IItll.IIXIIA
as required.
same p r o o f , w i t h o u t a n y r e s t r i c t i o n on
8.
maps
To show t h a t t h e q u o t i e n t t o p o l o g y is l a r g e r
is c o n t i n u o u s .
7.
t
q
i s c l o s e d , f o r i t is
q
E ( x ) = I t : t ( A x ) = 01
so t h a t t h e map
Then
G.
X I - t o p o l o g y i t is s u f f i c i e n t ( 1 4 . 0 . 6 ) t o show t h a t
+
hence
X I - t o p o l o g y on
Thus t h e q u o t i e n t t o p o l o g y ( 7 . 0 . 1 ) makes
II.
i n t o a Banach s p a c e . than t h e
c a n b e made i n t o
q ( t ) ( x ) = t(Ax).
F u r t h e r t h e null-space of
G.
1 It,!
IIAxll,
E
LEMMA.
6.
-
A = 0
However o n e must n o t G = {Ol.
W i t h t h e n o t a t i o n of Remark 4 , G
+
X0
c a n b e made
i n t o a Banach s p a c e whose t o p o l o g y i s l a r g e r t h a n t h e r e l a t i v e t o p o l o g y of
X'.
14.1-14.2
224
Here we are identifying X B
H
Remark 1. With
=
G
X I ,
and
with its image in XB
BH
spaces by Lemma 6 and
The result follows by 14.0.1.
10.1.4.
LEMMA.
9.
The same a s Lemma 8 w i t h l a r g e r r e p l a c e d by s m a l l e r .
First, G + X 5 trivial.
A
if
X'
=
is not
(Smaller allows equal.)
question is a maximal subspace of for which 10. in
are
shown in
X'
p =
p-unique so the result is
If
A
X'
since it is the set of
is
p-unique the set in
0. The result follows from 14.0.2.
THEOREM.
W i t h t h e n o t a t i o n o f Remark 4 , G
+
'X
i s closed
X'.
By 3.0.2 the norms given in Lemmas 8, 9 are equivalent. norm of
X'
is between so they are all equivalent.
on this proof is as follows: 4.5.2 applies with and
f
X
replaced by
G
+
X5, a
BK
Y
The
A variation one-dimensional
space by Lemma 8. This avoids
using Lemma 9. 11. COROLLARY. (in
W i t h t h e n o t a t i o n o f Remark 4 , G
+
X B = c15X B
f
c X'
X').
By Theorems 5 and 10.
14.2. THE FUNCTIONAL
1-1.
For reference we rewrite Theorem 4.4.3. 1. f f s ) =
If
THEOREM. Z i mA x
A
is
+
Let tlAx)
be a m a t r i x , X = cA.
A
+
ax, a
6
Then
X5.
u-unique it is reasonable to write
X'. Otherwise
p(f)
defining a linear functional
)1
sentations with arbitrary
in particular 0. The set
defined in 14.1.1.
p,
i f f
on
f
thus has repreG
was
14.2-14.3
DEFINITION.
2. t i o n a l on
3. and i f f
for a l l
f
LEMMA. G
+
i s
A
X’
If
i s not
A
p-unique,
+ x’;
G
A
p-unique i f f l i m A
is
G
+
X’
+ XI.
This is clear from Definition 2 and Theorem 1. is
take
X.
6
=
pL
i s the linear func-
u-unique
d e f i n e d i n Theorem 1 .
X’
p(fi = 0
If
225
u-unique iff
V(lim
A)
Note that
A
1.
=
As stated at the beginning of this chapter, the continuity of p
in general is an open question.
In this and the next section
several sufficient conditions will be given.
i s a
(W.H. Ruckle).
THEOREM
4.
space i s
BK
Every m a t r i x
u-continuous i . e .
u*
By Lemma 3 and 14.1.10
A
such t h a t
i s c o n t i n u o u s on
p
X =c X’.
is closed, and the result follows
by 5.0.1. 5. space.
Let
THEOREM. Then
pL
and
M
is
is a
cA
p-uniqueness ape i n v a r i a n t f o r
This means that if and
be a matrix such t h a t
A
cM
=
A
p-unique iff
cA then
is.
pM(f)
=
0
BK
A.
iff
pA(f)
The expression for
in Lemma
3 is given an invariant formulation (i.e. named) in 14.1.11.
A
is
14.3.
u-unique iff
pL
+
ci
= 0,
Also
as in Lemma 3.
THE STRONG TOPOLOGY In order to extend the results of the preceding section to
more general matrices we need to topologize FK
space; then
X’
ci.
Let
X
be any
w i l l be assumed t o have t h e s t r o n g t o p o l o g y
(14.0.4). This is the norm topology if
X
is a
p - c o n t i n u i t y w i l l be i n t e r p r e t e d a c c o r d i n g l y .
BK
space).
A
226
14.3
The proof of the invariance of
P
(15.4.12) does not use the
results of this section. Let
1. THEOREM.
x6
of
Compare 14.1.5. c E
Let
Then t h e cZosure
cA.
Itn! < 1/M
E
X. Then
be a bounded set in
f(x) = t(Ax), choose m uk = tnank* Obviously n=l
IIAxll,
(8.0.4). Let
m
and let
u c X'.
m
1
=
c.
includes
< M for x
be a m a t r i x , X
A
n=m+l
that
so
For
m
G -
so
f c Eo
(14.0.4)
strong neighborhood of
2. THEOREM.
x
= cA:
( i )A
i s closed i n
I n=m+l 1 tn(Aql
=
If(x)-G(x)l
i.e.
c f
5 1
+ Eo. Since Eo is a basic
0 the result follows.
T h e s e a r e e q u i v a l e n t c o n d i t i o n s on a m a t r i x
is
(ivl G
X',
( i i )'p
p-continuous,
+
X6
= c~
x',
A,
+ x6
(iii)G
= c Q X'.
That (iv) implies (ii), and (i) implies (iii) are by 14.2.3. That (ii) implies (i) is by 5.0.1. Finally (iii) implies (iv) by Theorem 1.
A remark of very little significance is that c o n t i n u o u s iff
zk = P,(z) p
p = 0,
p(Pk)
=
=
so
0
z
p(f) = f(z) = 0.
for all
Thus if
A
is
f
then
p-unique,
cannot be weak* continuous. 3.
Let
LEMMA.
t i n u o u s on
x
X
b e an
FK
space
3
bv.
Then
X
i s con-
X'.
See 3.2.1 for in
for if
i s weak"
p
since
1
6
B
X.
Let
E
(10.2.7).
=
{1-1(")1. NOW
if
f c
This is a bounded set EO
(14.0.41,
IX(~)(
llim f(l-l(n))/ 5 1 so X is bounded on a strong neighborhood of 0, hence is continuous (14.0.3).
=
14.3
Every c o r e g u l a r m a t r i x
THEOREM.
4.
X
Let 9.6.1 and
=
227
cA.
is
A
u-continuous.
This satisfies the hypothesis of Lemma 3 by
X(f) = p(f)X(A)
(9.6.8).
The result is immediate from
Lemma 3.
invariant.
X',
x
The c o n d i t i o n " u - u n i q u e and p - c o n t i n u o u s "
THEOREM.
5.
= c
Indeed
has t h i s p r o p e r t y i f f
A
XB
i s not dense i n
-
A'
Sufficiency: Using Theorem 1 and 14.2.3, X' # X B
so
X'.
is a closed proper subspace of
pL
is
3
Necessity:
X8 + G = L
by Theorem
2. 6.
tinuous.
p-continuous
B , E = {x} "lx'")},
exists
f c XB
<
1 limAx-f(x) I
+
If(x-x("')(.
such that +
1 f(x)-f(x("))
1 akxk
continuous.
X.
+ + 0
Let
n
as
A x(n)l
so
An
AB
m a t r i x must be
by Theorem 6 , hence
2E
A(x) = 0.
A
is not
p-unique.
Thus
p-continuous.
(13.2.14) i.e.
A = 0 p
= 0 which is
This is a contradiction.
A
was given earlier (13.2.25). In contrast, an p-unique
5
and so
+ m
The proof of Theorem 6 shows also that
be
0. There n llimAx - 1 akXk k=1 >
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{<:t
6
TI.
by 10.2.6,
10.2.7 and the rest is trivial.
Possibly
P
is strictly larger than
B
and the perfect part.
2 36
15.2
See Examples 5 and 9 .
-
a l l matrices
Note a l s o t h a t t h e d e f i n i t i o n of
covers
P
n o t j u s t t h o s e w i t h c o n v e r g e n t columns a s f o r t h e
other subspaces.
Then
Let
THEOREM.
4.
nip:
P =
be a m a t r i x w i t h c o n v e r g e n t columns.
A
i s a t e s t function for
f
A}.
From 1 5 . 1 . 5 . 5.
Let
EXAMPLE.
satisfy
W
#
Then
B.
f e c t p a r t of
cA.
be a l m o s t c o r e g u l a r o r more g e n e r a l l y
A
-
P = B.
If
is c o r e g u l a r , P
A
i s the per-
I t f o l l o w s from 1 5 . 1 . 4 , Theorem 4 and t h e Hahn-
P c
Banach theorem ( 3 . 0 . 1 ) t h a t
B.
The rest i s by Theorem 3 and
13.1.5. 6.
THEOREM.
A coreguZar m a t r i x
A
i s p e r f e c t i f f P = cA.
From Example 5 . EXAMPLE.
7. A
be a m a t r i x w i t h c o n v e r g e n t columns.
A
i s n o t r e p l a c e a b l e or n o t
t h e Hahn-Banach
Q = W = F = B = P 8.
If
Let (tA)x = 0.
trA.
A'
w i t h columns i n t c T
Then
Taking
A
P =
i s not
T
by 1 5 . 1 . 3 and
p-unique t h e n
x
Suppose t h a t a m a t r i x P.
and s o f o r
Then
has a
A
A i s of T y p e M .
x c P
w e have
t o be any one of t h e columns of
t(Ax) = A'
yields
0.
A c r u d e form of t h i s r e s u l t was g i v e n i n 3 . 3 . 9 . conservative
is i n
P
If
by 1 3 . 2 . 2 5 .
THEOREM (Cathy Madden).
right inverse
=
p-unique t h e n
theorem ( 3 . 0 . 1 ) .
-
t
Let
+ + B 3 B (c)
(Theorem 3 ) .
If
A
is
( 1 0 . 2 . 9 ) = E m , so any bounded s e q u e n c e (The more g e n e r a l assumption
3 . 3 . 8 i s c o v e r e d by u s i n g
Li.)
A c 0
of
15.2
Thus a
237
P
P-perfect matrix (i.e.
=
c ) A
with a right inverse
must be of Type M. EXAMPLE.
9. part of
X
n
Let
A'
f = 0
B
on
but
Since the perfect part
X, where f(u) B
2
X
=
u
=
W = F = B
P
=
=
f = limA we
In}.
Thus
the perfect part i s
3
perfect
B
-
# X.
The
B.
X.
We have to prove that (tA)x = t(Ax). n Now using Abel's identity (1.2.9), 1 (tA)kxk = 1 (tk-tk+l)xk k= 1 n = 1 tk(xk-xk-l).- tn+lxn = Un - Vn, say. Since Un + t(Ax) we k=l know that V = lim Vn exists and we have to prove that it is 0 . Taking
t c T, x
P #
cA. With
=
1 where
=
result follows when it is proved that Let
with
A
be the matrix of 13.5.2 with
A
c perfect part of
Ern
have
c
A conull t r i a n g l e
x = {n}
X.
6
yields that
lim n tn+l
exists and since
t
Q
6
this implies that lim n tn+l Now let
ere
C
y = Ax
so
is the
(C,1) matrix.
is regular.
that
y
6
c.
= 0
(1)
Vn -
1 Yk = ntn+Fn(Y) tn+l k=l This tends to 0 by (1) since C Then
u - u n i q u e by Example 7 .
This matrix i s
The matrix in Example 9 is a special case (6.3.2).
n
(rn = n)
A
of
R. DeVos has calculated that the result of Example 9 holds
for each such
A.
10. THEOREM.
Let
A
be a m a t r i x w i t h c o n v e r g e n t c o l u m n s
such t h a t a l l equipotent matrices are invariant f o r
u-continuous.
Then
P
is
A.
This is immediate from Theorem 4 and 15.1.6. 11.
EXAMPLE.
c o l u m n s and
P
=
Let
cA.
A
Then
be a r e v e r s i b l e m a t r i x w i t h n u l l A
is
p-unique.
(The assumption of
15.2
238
If not
null columns cannot be omitted, 4.4.4.) hence on
P
Thus x $ P
(Example 7 ) .
if
limA
0 on
=
$,
Ax = 1.
A converse for Example 11 holds: 12.
EXAMPLE (R. DeVos).
-
n u l l columns and
is
$ 3
u-unique i f f
that
z $
P
t(Ax) - (tA)x.
0
P
# 0. Thus A
is not
i s o f Type M .
for some
t
(Theorem 4) so
2
is a non-zero multiple of
f
A
Half is Example 11.
t(Az) # (tA)z
Now
be a r e v e r s i b l e m a t r i x w i t h
A
cA, e q u i v a l e n t l y
P = cA.
i.e.
Let
limA
6
Then
A
Conversely suppose
T. Let
f = 0 on
f(x) = 0
cA, hence
(15.0.5); non-zero since
f(z)
u-unique.
13. This suggests an extremely tantalizing question: must a m u l t i p l i c a t i v e t r i a n g l e be 14.
u-unique?
X = cA = aB has
AK
multiplicative matrix
so A M
has
X
is a
BK
are invariant.
P
AB, hence
M
of 4.4.4. Here =
X.
An equipotent
Example 11 shows
is shown in 5.2.7.
paradoxically that no s u c h =
A
Consider the non-p-unique triangle
c a n be r e v e r s i b l e , even though
cM
space, since all the conditions (except reversible) (Theorem 10 and 14.2.5).
On the other hand, a
p - u n i q u e r e p l a c e a b l e t r i a n g l e h a s an
equipotent multiplicative triangle.
(15.5.4)
What has been proved so far in the direction of the invariance
of
P? Sufficient conditions are: cA is a BK
15.1.6, 14.2.4.
W # B
This almost names P); A
is sufficient and
(Example 7 ; P
P
is named); A
is named); A has
AB
space (Theorem 4,
is coregular (Example 5 ; is not replaceable
(Theorem 3; P = cA).
ever Example 7 does not yield invariance of
How-
P for non-p-unique
matrices since this condition has not yet been proved invariant (15.4.9).
15.3
239
FACTORIZATION
15.3.
If 0
5 D
D
=
A,M
A.
<
M.
If
are positive integers with
>
A, M
EA + D
=
Without the last restriction we could take A, M
are matrices with
M = EA + D
to write
M
For example if
D
where
cM
=I
cA
is "small" (or
In general it is not to be expected that
A
the first column of
D
=
cD
D
E = MA-',
= 0.
0, for example if
Some valuable ones are given in 131
(D has small norm, is coercive, etc.) and 1581, (D m, e, d
is "large").
0. Such results are called f a c t o r i z a -
is
tion or q u o t i e n t theorems.
Now if
E = 0,
it is often possible
are triangles we may take
A, M
with
represent the
M
respectively, the equation
nth
EA + D
=
row of
= 0).
M, E, D,
is the same as
m
=
eA + d
for each n. This will follow from mx = e(Ax) + dx by setting x = 6 k . Thus the problem is to represent the row function mx (11.1.1) which is defined on
could take
e = 0, d
=
m
cA, and this is easy.
In fact we
but to obtain the l'smallnessll of
require a little more subtlety.
D will
(The next two results may be
omitted. ) 1. hlx)
=
LEMMA.
mx.
Let
be a
A
Then f o r
hlx)
x c X,
This is immediate from because 2.
columns. L
h
u-unique m a t r i x , X
4.4.3
elAx)
=
+
= cA,
m c X B,
d x , e c L, d
and 4 . 4 . 4 .
That
l.~ =
6
0
B uA.
is
has the representation mx. Let
COROLLARY. Let
c M 2 cA.
and t h e rows o f
D
be a
A
Then in
u-unique m a t r i x u i t h convergent
M = EA
+
D
w i t h t h e rows o f
E
in
B
uA.
Apply Lemma 1 to each row of
M
and set
x = 6
k
.
This result is given for comparison with the next one which is tailor-made for its application to proving invariance theorems.
240
15.3
One can see that Corollary 2 is not strong enough by considering A to be a non-AB
matrix and taking
M
=
satisfy Corollary 2 but not Theorem 3. THEOREM.
3.
columns, 2 = c A .
Let Let
for
(tD)x = t(Dx)
cM
t
be a
A
6
Then
E
0, D
=
(Take Y = c
A
=
will
in 12.5.3).
p-unique m a t r i x w i t h convergent
M = EA
Then
2 2.
1, x 6
A.
w
where
D
f
IIEll
<
m,
A'
D is that every
Thus the smallness condition on
x c uA
has
the associativity property. hn(x) = (Mx),.
Let
Then
{hn}
i s
pointwise convergent on
Z
hence equicontinuous (7.0.2). f
=
hn, X
=
In the construction of 4.4.1 with r w A , Y = c , we have Ihn(x)l 5 K 1 pi(x) + L ] ~ x Iwith ~~ i=l
K, L , r
independent of
seminorms of +
gnoA
y
c
6
are all the
gn
6
c',
hn = Fn
/)Y(,RI
5
as in 4.4.1.
c
nth row of
E
( M x ) ~ and
A
= pn
(enk)
=
is
limAx + en(Ax) + dnx
m
p-unique it follows that
and set
is because u(x)
=
x = gk
L 2 IIgnll =
t(Dx)
where
for
shows that
1
x c uA
enkl.
pn
M = EA + D. That
D, let
Turning to
t c
1
this quantity exists since
z
D
are the
hn (x) = = 0 and so finally
Itn(Wnl = ItnFn(x)I 5 K 1 P i x).]ltll,, hence Banach-Steinhaus theorem (1.0.4). Since uA has k u(x) = 1 u ( 6 )xk = (tD)x. The condition on
en, dn
D, respectively. Since also
and
Mx = E(Ax) + Dx. Taking
c
1 pi(x);
uA
hn(x)
<
{pnl
Ipn,hn) in 4.3.13.) Hence
Fn c u;Z, IFnCx)l 5 K
(1.0.2) so
IIEll
x c Z. Here
k has AK (4.3.8), Fn(x) = Fn(& )xk = (Dx), k dnk = Fn(& ) . Also gn(y) = p n lim y + 1 enkYk for y c c
Since where
and
(listed as
uA
with
~llyll, for
n
u AK
6
u;Z
by the
(4.3.8),
is related to bounded sections by
12.5.3 with Y = k m ; it is easy to see that so that this makes sense.
D
has bounded columns
15.3-15.4
THEOREM.
4.
The m a t r i c e s i n T h e o r e m 3 h a v e t h e p r o p e r t i e s
for
(EAIx = E(Ax)
241
x
6
and
Z
t
for
(tEly = t(Ey)
a, y
6
c c.
The first condition holds because in the course of proving Theorem 3 the identities Mx = E(AX) + Dx proved.
The latter gives
Mx
=
(EA)x + Dx
aE
B [a"]
am
2
and so
B [a;]
Y
c, using 10.2.9. The rest is by 12.5.3 with
= Ilm2
were
and the result follows.
m
The second condition holds because
M = EA + D
and
2
=
R
m
.
[ 6 ] , D81, [Sl].
Sources:
INCLUSION AND INVARIANCE
15.4.
The factorization theorem just proved will allow u s to complete the discussion of
v., P and the test functions in a satisfactory
way. DEFINITION.
1. i s
pA(M)
Let
A,
M
be m a t r i c e s w i t h
c M 2 cA.
i s c a l l e d c o r e g u l a r (mod A ) i f
p(limMlcA); M
Then
# 0,
pA(M)
o t h e r w i s e conull (mod A ) .
2.
x c cA
=
EXAMPLE. c
Take
(1.3.8).
A
I.
=
Thus
p
Then
x + tx
limMx = X(M)lim
I(M) = X(M)
i s conserva-
M
whenever
for
t i v e and s o c o r e g u l a r (mod I) = c o r e g u l a r .
3.
EXAMPLE.
(9.6.8) and so 4.
(mod A ) . (mod A ) ,
5.
A
M
i s
If
Let
A
be coregular.
i s not
i n particular
REMARK.
X(M)
uA(M)X(A)
=
is coregular iff it is coregular (mod A).
p-unique i f f A
Then
pA(A) # 0
p - u n i q u e and A
i s conull
i.e.
cM
3
i f f
eA
A
then
i s coregular M
i s conull
(mod A ) .
In the remainder of this section A
i s a matrix
We shall eschew the usual reference t o this remark - caveat lector.
w i t h convergent columns.
242
15.4
6. THEOREM (W. Beekmann).
+
f(xl = p l i m x M
If A is
is not
p-unique.
+
t(Mx)
ax.
Let
Then
(4.4.3).
0.
Fix some particular value of
M
is
so t h a t
= p-pAIMJ.
Now assume that
A
x c cA we have
B limMx = pA(M)limAx + u(Ax) + vx, u c L , v c cA
f c
ch
6
(so that it is
p
u-unique.) For
f
"(flcA)
p-unique both sides are
irrelevant whether
and
c M 2 cA
(4.4.3) and so for
CA
f(x) = p.uA(M)limAx + pu(Ax) + pvx + t(Mx) + ax = p-uA(M)limAx + g(x)
+ h(x), where g(x)
p(h) = 0 since
pu c
L
=
h(x) = pu(Ax) + (pv+a)x.
t(Mx),
and
a
so that
c c & C ci
The result will follow when it is shown that
+ t(Dx) with
0, for
p =
(since IIEll <
A
+ (tD)x.
(tE)(Ax)
=
01
8 tD c cA
- by
B + a c cA'
pv
p(g) = 0 .
g(x) = t [(EA+D)x]
applying 15.3.3 and 15.3.4 we have
Now, t[(EA)x]
=
This is a representation of (since
(tD)x
exists) and
Now
g c ci
tE c
8
direct calculation or by 8.3.l(d) with
= E'1).
The essence of Theorem 6 may be summarized in the formula
+ ci, where G is given in 14.1.1.
G[M] c G[A]
COROLLARY.
7.
matrix
M
If
A
is
p-unique f p - c o n t i n u o u s ) , t h e n e v e r y
which i s coregularlmod A ) i s
p-unique I p - c o n t i n u o u s ) .
The theorem is true (but vacuous by Remark 4) if p-unique. where that
f = lim, in Theorem 6 gives
Taking
Thus
p = pM(M).
A
pA(M) =
is
p =
1 and
M
is
pA(M)
p-unique.
is not
= p.pA(M)
Now suppose
p-continuous and, with no loss of generality, that
1. By Theorem 6, pM(f) = pA(flcA); now if
strong topology on topology on
A
ci
ch; it follows that
(14.0.5) and so
pM(f)
-+
f IcA
+
0
f
+
0 in the
in the strong
0.
This generalizes the earlier result that a coregular matrix
15.4
must be
u-unique and
in the
u-continuous (14.4.2 which, however, holds
setting).
FK
243
Of course a
u-unique and continuous con-
servative matrix need not be coregular(mod I) (13.5.1, 13.5.6). From these results it follows that various properties behave
properly with respect to inclusion; in particular, are invariant: 8. THEOREM. L e t is N
If
N.
M
c N 3 cM
conulllmod A )
.
A
g
be
g
10.
so
cNJ then
and
to
f c ch
f
=
limN\cM in
For the second part by the Hahn-Banach
is continuous on its domain by 4.2.5). f , replacing M
by
N.
This yields 0.
u-uniqueness is invariant.
u-unique and
(mod A ) (Example 4), hence consequently
is closed in
and so neither number on the right can be
pA(N)
9. COROLLARY.
Let
is conulllmod A )
cw
pA(N) = uM(N).uA(M).
Now apply Theorem 6 to = p
and
M
In p a r t i c u l a r , c o r e g u l a r l m o d A )
limM and extend
theorem (noting that
pA(M)
If
F o r the first part take
Theorem 6 which gives =
A'
are invariant.
Compare 9.3.6.
g
c
is c o r e g u l a r l m o d A )
is c o r e g u l a r l m o d A ) .
let
2
M
cM
=
cA. Then
A
is coregular
is coregular(mod A ) (Theorem 8) and
p-unique. (Corollary 7).
COROLLARY.
p-continuity
is invariant.
Like Corollary 9 o r by Corollary 9 and 14.3.5. non-u-unique matrix is
Note that any
u-continuous.
In some earlier results the hypotheses were that every equipotent matrix is
u-continuous (15.1.6, 15.2.10).
By Corollary 10
it is sufficient to assume that the matrix in question is tinuous.
u-con-
However, the conclusions of these theorems will now be
obtained without even this assumption:
15.4
244
11.
M
c M x cA.
Let
THEOREM.
is a t e s t f u n c t i o n f o r
i.e.
A
f
Then each t e s t f u n c t i o n
for
is a t e s t f u n c t i o n for
flcA
A.
I n p a r t i c u l a r t h e s e t of t e s t f u n c t i o n s i s i n v a r i a n t . By Thoerem 6 , p ( f I c A ) = 0 . 12.
P
THEOREM.
P(MI
cM 3 c A J t h e n
I f
In particular,
€'(A).
2
i s invariant. By Theorem 11 and 1 5 . 2 . 4 .
Thus
P
joins the other distinguish-
e d s u b s p a c e s i n b e i n g monotone ( 1 0 . 2 . 9 ) .
13.
may o e e u r e v e n i f Z = X"
Let
A
if
a l g e b r a i c d u a l of
X'
may c o n s i d e r
XC Z
Also
Let
c Z.
p
Ih Thus
6
X
For
THEOREM.
P
= cAJ
=
or
is
A
be t h e set
P + c Z
Z
be t h e
t h e set of a l l l i n e a r f u n c t i o n a l s .
i.e.
H = 4 0
p,
($*nuA)'
i s a maximal s u b s p a c e o f
f = 0
4
and t h i s is
or
P+.
$1.
on
T @p
topology of
or is a l l of
P+
*
i.e.
and
*
Since
(15.0.1,
Z.
Hence
P = P'nX
( 1 5 . 2 . 4 ) a n d , on
X, t h e weak
weak t o p o l o g y o f
X, t h e f i r s t s t a t e m e n t o f t h e t h e o r e m f o l l o w s
topology of
Z
w i t h t h e c l o s u r e t a k e n i n t h e weak t o p o l o g y . since
$
(5.0.2).
1c F
coincides with the
But t h i s i s t h e same,
i s a subspace, as t h e c l o s u r e i n t h e
FK
topology
For t h e s e c o n d p a r t c o n s u l t 1 5 . 2 . 7 w i t h 5 . 2 . 5 ; a l s o ob-
serve that i f
A
by 1 0 . 2 . 7 )
i s regular, 1 f hence
0
P = T O 1.
W e
where h ( f ) = f ( x ) .
x -+ h
1 5 . 0 . 2 ) w i t h t h e c l o s u r e t a k e n i n t h e weak
-4
Both c a s e s
p-continuous, otherwise let
b y t h e embedding
where
T @u.
=
i s a coregular conservative triangle.
Z : h ( f ) = 0 whenever p ( f ) = 0
P+ = H"
P
(5.2.1) but
1 6 P
(indeed
15.5
245
15.5. REPLACEABILITY
In this section we complete the "function as matrix" theorem 5.1.1 and give necessary and sufficient conditions for replaceability
.
1.
REMARK.
columns, X 2.
=
In this section
is a matrix with convecgent
A
cA'
THEOREM.
Let
f
6
(Remark I).
X'
M
is
v(fI # 0.
such
exists i f f
M
Necessity is because
ZimM
=
u-un-ique,
f.
If
A
Sufficiency in the second part is
The first part is by 5.1.4. by 5.1.1.
i s not
A
such t h a t
t h e r e e x i s t s an e q u i p o t e n t m a t r i x u-unique,
If
~ ( f )= pA(M) # 0 by 15.4.8 and
15.4.4. THEOREM (W. Beekmann, S-Y.
3. i f
A
u-unique it i s replaceable.
i s not
i s replaceable i f f there e x i s t s
#
With
Kuan).
f
f
u-unique i t
i s
A
with
X'
6
I f
a s i n Remark I,
A
=
on
0
$,
p(fl
0.
The first part is by 13.2.18.
For the second part, if
an equipotent multiplicative matrix, f
=
limM
is
M
satisfies the
requirements as in 15.4.8. Conversely if such
exists, A
f
is
replaceable by 5.1.1. 4. If
A
i s a
p - u n i q u e r e p l a c e a b l e t r i a n g 2 e i t has an e q u i -
potent multipZicative triangle.
in Theorem 3.
Then
f(x)
= 1-1
(Compare 15.2.14).
limAx + t(Ax)
5. A T [L]
D
a, then
A
a
=
0
be as u # 0.
so that the
(1.7.15.)
J . Boos has pointed out that i f
contains
f
(4.4.3) with
The procedure of 5.1.1 yields a triangle since proof ends with the definition of
Let
A
i s rep2aceabZe.
i s c o n s e r v a t i v e and
Here
ak
=
lim ank;
15.5
246
note that
A
T
-
f ( x ) = limAx Theorem 3.
by 8.3.8.
c (L:L)
t(Ax)
defines
f
The c o n v e r s e h o Z d s i f
sat s f y i n g t h e c o n d i t i o n s of
is
A
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 one t o one and r o w - f i n t e s i n c e w i t h Theorem 3, f ( x ) = Then
0 = f(6
k
p
limAx + t(Ax)
uak +
) =
T (A t ) k .
'I'
a = A t,
The p r o o f i s t h a t w i t h
as i n
f
(5.4.3, 5.5.3) w i t h
p
#
0.
The c o n v e r s e f c l i l s f o r a non-p-
u n i q u e t r i a n g l e a s may b e e a s i l y c h e c k e d f o r t h e m a t r i x o f 4.4.4. I d o n o t know what h a p p e n s i n o t h e r cases.
6.
answer i s : *:
iff
A
Define a matrix
14.4.6.
is n o t D
limD = 0
Now
0
u-unique.
such t h a t
so
D
The
M?
By Theorem 2 w i t h
a:
f = 0.
0
cD = cM by t h e d e v i c e shown i n
is not
p-unique.
A
Hence
is n o t
Note t h a t t h i s g i v e s a n a l t e r n a t e p r o o f o f 14.4.4.
(15.4.9). 7.
cA - cM f o r some
Under what c i r c u m s t a n c e s i s
COROLLARY.
u
p-unique i f f
With
A
a s i n Remark 1 , A
d o e s n o t b e l o n g t o t h e weak
*
i s repZaceabZe and c l o s u r e of
$.
By Theorem 3 and 15.0.3 w i t h t h e Hahn-Banach t h e o r e m (3.0.1) applied to
X'
#
. Note a l s o t h a t by
C o r o l l a r y 7 is a n i n v a r i a n t c o n d i t i o n .
15.4.6, any t w o p o s s i b l e
u's
( f o r e q u i p o t e n t m a t r i c e s ) are
l i n e a r l y dependent. 8.
COROLLARY.
a b l e , p - u n i q u e and Necessity:
W i t h t h e n o t a t i o n of Remark I , A u-continuous i f f
XB
Theorem 3 s a y s t h a t
pL
t h e r e s u l t by 14.3.2.
Sufficiency:
t i o n s h o l d by 14.3.5.
Hence
pL
i s replace-
i s n o t dense i n $'
$'.
and t h i s i m p l i e s
t h e second and t h i r d condi-
-
= XB
(14.3.2) ;P $'
and
A
is
r e p l a c e a b l e by Theorem 3. 9.
C o r o l l a r y 8 names t h e c o n j u n c t i o n o f t h e t h r e e p r o p e r t i e s
15.5-15.6
and so we are led to study XB
i s n o t d e n s e i n .'0
FK
If
replaceable, u-unique and (which are automatically
247
spaces X
X
=
cA, X
with the property ( * )
has
A
iff
is
For coregular matrices
u-continuous. u-unique and
(*)
u-continuous by 9.6.10 and
14.3.4) the condition ( * ) is equivalent to replaceability which, 1 @
in turn is equivalent to the condition
7
Thus there are two generalizations to ability for coregular matrices.
(13.2.16).
FK
I do not know whether these coneli-
tions are dependent for general coregular FK 10.
If
A
regular matrix 5.2.1, 4.2.4.
spaces of replace-
spaces.
is coregular and not replaceable, no stronger coB
can be replaceable. This is immediate from
Coregular may not be omitted, for example take
B = 0. 11. angle
A
(1.7.5).
EXAMPLE.
Mercerian t r i a n g l e
A
such t h a t
i s n o t conservatiue.
ATA-'
=
1
(5.2.5).
using the column limits of Then
A
cTA = cA
Compare this with
replaceable, X(A)
A
is consistent with
limits and value of
X.
and a c o r e g u l a r t r i -
T
Thus
(1.7.5).
Let
tk
=
ak.
since they have the same column
This implies that
is not conservative
be non-
i.e.
AM-'
is regular
(1.7.10) and so by Remark 10 it is not stronger than AM-lA-'
A
cA
Let 'M be Mazur's matrix (1.8.12) as the sequence t
M
cAT
(1.7.5).
With
T
= M-l
A
i.e. the example
is concluded.
Sources:
[5],
1431.
15.6. MISCELLANY AND QUESTIONS. Once again we deal only with matrices with convergent columns.
248
15.6
1.
If
A
u
is a triangle it is very easy to compute
in
limAx + t(Ax), (4.4.5), namely define -1 g c c' by g(y) = f(A y) then p = X(g) = g(1) - 1 g(6 k ) = f(u) k - 1 f(v k ) where u is the sequence of row-sums of A-' and v
the expression
is the
f(x)
= p
column of
kth
A-I.
This fact can be used to give easy
proof f o r t r i a n g l e s of such results as 15.4.8.
equipotent matrix i s ueakly
p-unique iff e v e r y
A matrix i s
2. THEOREM (Cathy Madden).
u-unique.
Necessity is by invariance (15.4.9). Conversely if p-unique there is an equipotent matrix
M
This
is not weakly
3.
c l o s u r e of
If range
A
is not
= 0
limM
A[X]
M;
u-unique the matrix
and
X(g)
Let
f(x) = limAx + t(Ax)
all
x
so If
A
is not
A
X = c
M
0 (5.1.1)
A = 'M'
given in Theorem 2 has
M = A
by invariance
=
For
1.
for
y c c , g(y)
x c X.
Then
f(x)
g
6
c'
with
lim y + ty.
=
g(Ax)
=
= 0
for
u-unique.
is a triangle, A[X]
=
c
which is dense in
(Theorem 3 ) so the condition of Theorem 3 for some sufficient.
=
X c u(c",c'I-
p-unique i f f
If the conclusion is false, there exists
on
4.
is
Conversely we may assume
{O}.
(15.4.9).
g
A
f o r every equipotent
M[X]
with
u-unique.
A matrix
THEOREM.
M
A is not
M
c"
is not
In all the cases of non-u-uniqueness we have seen, it
was revealed by an equipotent matrix closure does not contain
M
with
M[X] c co
whose
X.
As a corollary we can find yet another sufficient condition for
p-continuity.
The condition is satisfied in two special
cases (14.1.3) in which however the domains are
BK
spaces
u-continuity is assured because
(14.2.4).
15.6
5.
IX
rp x 8 c
COROLLARY.
G, then
A
is
p-continuous.
c 1. A
=
A
We may assume that a sequence
{xnl
X
in
By hypothesis we have lim y + ty
for
is
p-unique.
such that
f c X'
For each
(15.0.4).
p
249
f(x)
y c c.
By Theorem 3, there is
g(Axn)
X(g)
+.
we shall prove that limAx + t(Ax).
= p
f(x n )
Then
rest is as in the proof of 14.3.3 with
=
g(Ax n )
E
=
for all f(xn)
Let +
+
g(y)
X(g)
g
c'
6
p(f). =
= p.
The
Ixn 1 , a (weakly)
bounded set. 6.
Here are some questions and suggestions for further study:
Is P
invariant for a matrix whose columns are not conver-
gent? What is the
FK
answered by 14.3.5.
equivalent for
p-uniqueness?
This may be
Matrix-bound conditions were given in 14.2.3,
and Theorems 2, 3. Solve the naming problem for functions and Give an
p-continuity, the set of test
P. FK
equivalent f o r coregular(mod A) analogous to
4.6.2.
7. To prove p-continuous
M
M
cM
such that
p-continuity f o r such that is a
A
BK
A
it is sufficient to find a
is coregular(mod M)
(15.4.7).
space will do (14.2.4).
Any
Theorem 3.5.1
looks promising in this direction; however the conditions do not
imply that Ax
A 2
=
{xn/n 1
(Also
A
is
is coregular(mod M). then
cM
n
m
L
=
cA
n
p-unique - otherwise
F o r example if Lm
A
Mx
=
{xn/n},
but
A is conull(mod M).
= 0
would do as well.)
8. Cathy Madden has suggested a study of what she calls
R-invariance in which properties common to
A, M
are studied if
250
15.6
CA = CM,
of p a i r s
WA
--
X, Y
(The
WM.
of
FK
FK
spaces with
To g i v e o n e example: (Compare 5.4.12)
Sources:
1471,
program would d e a l w i t h t h e p r o p e r t i e s
[Ell.
weak
X C Y, Y
an
u-uniqueness is
AK
space.)
R-invariant.
251
CHAPTER 16 SEQUENTIAL COMPLETENESS AND SEPARABILITY
FUNCTIONAL ANALYSIS
16.0.
1. Let
X
be a Banach space.
Then
X'
topology is sequentially complete 1801, Examples 9-3-8, The sup ( 4 . 0 . 9 )
2.
(A pseudometric is the same as a metric
d(x,y) = 0 is allowed.)
r80A],
Y'
convex space with separable FK
*
weak
space.
Let
4. (15.0.2).
X
=
{x
6
borhoods of
for a topology on
4-1-9;
*
iff
a finite set in
f(x)
compact set in
topology on
X
0
be a
X which
Theorem 12-5-13.
for all
X#
f c El.
Z, is a local base of neigh-
called
The set of all
f
E
a(X,Z).
Z.
[SO],
In this Examples
XI
Eo, E
an absolutely convex
Z is a local base of neighborhoods for a
called
topology such that
T(X,Z). =
Z.
6. With the topology #9-5-107.
-f
X
X
Definition 4-1-10.
Continuing 4 :
5.
weak
0
+
to
X: (f(x)l 2 1 for all
Eo, E
topology, x
[SO],
Let
Z a total subspace of
be a vector space and
The set of all 0
Y
Then each linear map from
E c Z, Eo
For
Y be a locally
Let
sequentially complete.
has closed graph is weakly continuous.
8-5-4,
#6.7.114.
(N.J. Kalton's closed graph theorem.)
3.
9-3-2.
of two pseudometrizable separable vector
topologies is separable. except that
*
with its weak
[80], .r(Lm
It is the largest locally convex Theorem 9-2-3. L),
Lm
is an
AK
space.
1801,
252
16.0-16.1
7. Let
X
be a locally convex sequence space
topology larger than the relative topology of
x
with
$
Then a set in
a.
r301, p . 1010,
is compact iff it is sequentially compact.
Theorem 6,
3
It follows that two such topologies with the same
convergent sequences have the same compact sets. X , Z be as in 4 , and let a subspace
8. Let
the property that Then
.c(Z,X)IV
sets E
=
T(V,X).
X, and
are in
9.
a(X,V)
X, Y
Let
and
o(X,Z)
have the same compact sets.
Eo
in
Z.
or
V
be locally convex spaces and
The topology of an
FK
T: X
+
a
Y
T is continuous when
Then
X, Y
[80] Example 11-2-5.
T(X,X'), T(Y,Y') respectively. 10.
Z have
This is clear from 5 in which now the
weakly continuous linear map. have
of
V
space
X
is
T(X,X').
[So],
Theorem 10-1-9, Example 10-1-10.
16.1.
SEQUENTIAL COMPLETENESS
The role of sequential completeness was pointed out by G. Bennett and N.J. Kalton in [I21 in which they proved Theorem 10 of this section. analysis
-
Their tools were the deepest methods of functional
most of their results are not presented here.
proof of Theorem 10 given here is due to the author Snyder.
The
and A.K.
Motivation for the result is contained in Remark 14 and
the next section. 1.
REMARK.
satisfying
2. ed
co c
In t h i s s e c t i o n
v
i s a v e c t o r s p a c e of s e q u e n c e s
c em.
DEFINITION.
V-Cauchy i f
V
A sequence
lim f(an)
{an)
of members of
e x i s t s for e a c h
f
E
P. i s c a l l -
V , where
ffa)
16.1
253
1 fkak.
means
LEMMA.
3. where
A sequence
Ian)
in
is
R
V-Cauchy i f f
cA
2 V
A = (a:).
x
For
E
V, x
c
E
iff
lim
k
k
is just the
nth
A
lim x(a n )
2) iff
1 xkan
exists
i.e.
r o w of
A.
(Definition
exists. an
4. Note that
Since
such A has convergent columns and IIAll < m . In particular n exists for each k and is norm bounded in R , ak = lim ak
V
co,
2
Ian}a
E
R
(1.3.7).
EXAMPLE.
5.
A
=
I
(Lemma 3).
f(an) + f(a) all
V
Let Here
for all
f
c.
=
Then
a
=
E
V , indeed for
V-Cauchy since
is
{6n}
(Remark 4 ) and it is false that
0
f
=
1, f(an) = 1
for
n. For the next definition refer to 16.0.4: 6. DEFINITION.
called a
flun)
EXAMPLE.
c
b e a s i n Remark 1 .
V
u(a,Vl
space i f
C
V-Cauchy i m p l i e s 7.
Let
f(a)
-f
c o J Bm
directly.
are
f o r a12
i s not a
LOJ
i s
A
=
Ian]
(Remark 4 1
f E V.
s p a c e by Example 5 .
C
I, cA
=
c
=
This is V, W(A) = co.
These are fairly easy to check
C spaces.
In the case of
V
i s sequentiaZZy complete i . e .
also a special case of Theorem 8 with However
Then
one uses 1.3.7 o r Example 11 o r 13.
co it follows from 16.0.1 o r by the next result and the monotonicity theorem 10.2.9 since W(co) = co; see also Example For
11 or 13. 8.
THEOREM.
every matrix
A
With
V
such t h a t
a s i n Remark I, V cA
3
V
is a
has t h e p r o p e r t y
C
space i f f V C WlAl.
254
16.1
Recall the definition of {an] so
be
in 5.6.1.
n A = (a,).
V-Cauchy and
V c W(A).
W
By 13.2.14,
Then on
A = 0
Sufficiency: Let
cA i.e.
V
V
3
x
(Lemma 3) and E
V
implies
x(a). Necessity: Let limAx = ax. This is the same as x(an) an be the nth row of A. By Lemma 3 and the definition, -f
x(a")
x(a)
+.
for all
x
V
E
(10.4.7) C F+(A)
F+(co)
V.
on
A = 0
V c F(A);
is sufficient to show that =
i.e.
Ry 13.2.14 it
this follows since
(10.2.9) and
V c Lm
V C cA.
The next result is an improvement of 6.1.2. Let
9. THEOREM. m a t r i x such t h a t
A
Of course
D
Let
=
cA
A - X(A)I.
3
be conuZl and c o n s e r v a t i v e and
X
Then
Wb(Xl.
D
Suppose it is coregular.
is conull, hence so is
n
X
cD
Q ( r ) (6.4.3) which we denote by
(4.6.6) which thus includes some
Y.
a
i s conuZZ.
A
is conservative also. Then
A
In the following calculation we make several applications of
n
the monotonicity theorem 10.2.9: Y Wb(XncD) c cD Since Y
il
Wb(X) c cD r l cA
(13.4.15)c
Wb(Y)
as in the proof of 3.5.4.
c
is conull (6.3.3), this is impossible by 6.4.5 o r 6.1.2.
10. THEOREM. is a
=
Em =
Let
be an
X
FK
space
3
c
.
Then
Wb(Xl
space.
C
To apply Theorem 8 assume that Y = z-l-X.
By 6.5.1, 6.5.2, Y
cA
2
Wb(X).
Let
z
E
Wb(X),
is conull and conservative. Next
we show that CAnz Let For
zsy
Y E
E
g
wb(y)l
Wb(X)
9, A.z
X'
E
wb(y)
and define g(2.Y) = f(Y)
f
F
=
1
(1)
Y'
by f(y) = g(2.y) k k f(6 )Yk = 1 g(6 )ZkYk
with yields (1) by the definition of
is conull and so
z
E
Wb(A)
by 12.1.6
A.
(6.5.1).
so that
By Theorem
(Y = c).
16.1-16.2
11.
C
is a
space; X
12. = c
EXAMPLE.
EXAMPLE.
space.
C
Let
=
Taking c
X = w
in Theorem 10 shows that
shows that
co
There exists no
is a
FK
C
space
9.-
space.
X
such that
Wb(X)
by Theorem 10 and Example 7. 13.
a
EXAMPLE.
255
cA
3
I in
Every i d e a l
such t h a t
9.-
This shows yet once more that
I, z
E
I
and
M
Then
= A.z.
M
Y
=
c) and a fortiori
z E W.
I 2 are
9.
is
co
spaces.
C
is coercive, hence
strongly conull (5.2.12). This shows that with
co,
m
z E S(A)
(12.1.7
The result follows by Theorem
8. 14.
Our discussion deduced the sequential completeness
theorem 10 from results which are essentially equivalent to the The approach of n 2 ] is to
bounded consistency theorem 5.6.8.
prove Theorem 10 directly by functional analytic methods and to deduce the bounded consistency theorem from it via Theorem 8 and 5.6.9. 15.
ever
By Example 7 , o(ll,c) is (16.0.1). Now
a(c',c)
(14.1.2) so
is not sequentially complete; howL
has codimension 1 in
c'
has o n e - d i m e n s i o n a l s e q u e n t i a l compZetion i n t h i s
9.
top0 logy. 16.
A simple minded attempt to extend these results to
spaces would lead to the conjecture that complete if
V
=
Wb(cA).
is sequentially
o(bs,V)
This is false even for
sc
A
=
0. However
the result of Theorem 9 can be extended; this is done in 1631.
16.2. SEPARABILITY
In this section we show that the
FK
program can be extended
16.2
256
to situations in which it has failed, provided that separability assumptions are made.
The contribution from functional analysis
is N.J. Kalton's closed graph theorem, 16.0.3. shows that this is a valid part of the
A
Let
1. THEOREM.
FK
The first result
program. is a s e p a r a b Z e
X = eA
be a m a t r i x t h e n
space.
FK
T1 be the topology of
Let
restricted to
uA
X.
Since
is separable and metrizable (4.3.8) the same is true of
subspace of
c
A[X]
hence is a separable metric space with
II.II,.
(X,T2)
topology for
is
X
2. EXAMPLE. m
. 3.
space.
Then
V
Let
x
-+
in
0
i: V
-+
The
X
3
be a
V
1 xiyi
+
0
and
space
C
i m p l i e s that
V
iff
FK
a separable
X
V C W(X).
o(V,L)
for each
y
This means that
16.0.4. E
The inc usion map
L.
this topology is smaller than the given topologies on
V
FK
X , this is by the definition of x
16.0.3, i
in
FK
has closed graph since it is continuous (hence has closed
X
because
V = X
Take
graph) when each space is given the relative topology of
(For
is a
co.
=
be given the topology V
Now the natural
program f a i l s f o r 1 6 . 1 . 8 .
FK
Let
THEOREM.
X.
T1VT2 (4.3.13) and is separable (16.0.2)
V q W(X)
Then
is separable.
on
(X,T1).
Now
It follows that
= L
II.lIA
T2 be the topology induced by
Let
wA
-+
0
in
w
1 xiyi
iff
is weakly continuous.
-f
0
space; for
for each
Now let
x
E
y
E
V
X.
V
it is
0.)
Then
V.
By
x(~) + x
as is obvious from the criterion for convergence in
just given; hence
x(~)
-+
x
weakly in
X
so
x
E
and
w
V
W(X).
4. The proof of Theorem 3 does not use the full force of the assumption that
co c V c
km.
It is sufficient to replace
co
by
16.2
257
In the proof of necessity in 16.1.8 the former assumption
@.
played a role, now seen to be unnecessary. ciency in 16.1.8 uses
V
co
2
The proof of suffi-
to conclude that
a
Q ; it is
E
essential there. 5.
Let
COROLLARY.
separable
FK
V
be a
space c o n t a i n i n g 1 .
C
space which i n c l u d e s
By Theorem 3 , 1
E
W.
m u s t be conull.
V
(Note that
Then a
co
included in the coregular separable space
is a
C
space which is
c ; some hypothesis is
required. ) The
FK
program may be applied to 3.5.5 which states that a
coercive matrix must be conull: 6.
X
Let
COROLLARY.
X
be a separable
FK
space
3
em.
Then
i s conull.
By Corollary 5 and 16.1.7. The same is true for 6.1.2: 7.
Let
COROLLARY.
s e p a r a b l e and
X 2 Y
n
Y
b e a conuZl s p a c e (or just
Qm
X
3
3
WblYll
c0’
If
then
X
X
is
i s conull.
By Corollary 5 and 16.1.10. It was pointed out in the earlier versions that these results fail for general (non-separable) FK 8.
a
C
Let
COROLLARY.
space.
Then
V
V
spaces.
be a c o n s e r v a t i v e
space which i s
i s n o t separable.
For Corollary 5 would otherwise imply that which is impossible since C
FK
Vc
(4.6.3).
Ilm
V
is conull
(Note that
co
is a
space so some hypothesis is required.) 9.
The results of this section hold under weaker hypotheses.
In 1123, p. 820, separability of
X
is replaced by the assumption
16.2-16,3
258
that
X
h a s no ( c l o s e d ) s u b s p a c e i s o m o r p h i c w i t h
f o r example t h e f o l l o w i n g b i g n e s s t h e o r e m : X = Y
n
space
L X
m
has a s u b s p a c e i s o m o r p h i c w i t h
am.
This yields,
i f
Y
i s conuZ2 t h e n
Em
for i f not, the
FK
T h i s is
i s c o n u l l by t h e improvement of C o r o l l a r y 7 .
i m p o s s i b l e as i n C o r o l l a r y 8. 10.
Snyder c a l l s a s e t
A.K.
implies t h a t
is c o n u l l .
A
i s pseudo-conull
For example t h e c o r e g u l a r s p a c e
(Corollary 7; take
t h a t c e r t a i n spaces
Wb
are p s e u d o - c o n u l l .
pseudo-conull
-
valent phrasing is t h e question:
answer i s n o for
Y
conull,
n
Ilm
BK
spaces.
t h i s is unsolved.
An e q u i -
does t h e r e e x i s t a separable
space?
FK
Theorem 5 s u g g e s t s
s p a c e which i n c l u d e s a
FK
s p a c e must be c o n u l l
Snyder h a s proved t h a t t h e
He h a s a l s o proved t h a t i f
is pseudo-conull
Lm
Theorem 16.1.9 shows
Y = w).
t h e c o n j e c t u r e t h a t any s e p a r a b l e
c o r e g u Z a r pseudo-conuZS
cA =I V
pseudo-conuZZ i f
V
Y
is
i n t h e weak s e n s e t h a t e v e r y
s e p a r a b l e s p a c e which i n c l u d e s i t i s c o n u l l .
T h i s is an e x t e n s i o n
of C o r o l l a r y 7 t o sc s p a c e s .
DENSE SUBSPACES OF
16.3.
!Lm
W e show t h a t a d e n s e p r o p e r s u b s p a c e o f bounded c o n v e r g e n c e domain. used i n t h e
FK
(Theorem 5 . )
Lm
T h i s r e s u l t is t h e n
p r o g r a m , Theorem 8 .
The f i r s t t h r e e r e s u l t s would b e t r i v i a l i f
m
1.
REMARK.
2.
LEMMA.
a A 3 D.
Then
In t h i s section
Let
A
D
D
2
co.
is a d e n s e s u b s e t o f
be a m a t r i x w i t h rows i n
L
I +-.
Let
un =
1 ar(n),k
k
and
{r(n))
m
L!
.
such t h a t
1 1 ~ 1 1 < -.
I f t h i s is f a l s e t h e r e is a sequence k1 l a r ( n ) , k
cannot be a
such t h a t
bnk = a r ( n ) ,kl'n'
16.3
Then of
B
(D:co).
E
But a l s o
LEMMA
3.
Let
a
DY = D B
B
B
so
(Rm:c0).
E
is a continuous map
This contradicts
B1 = 1.
= 9..
n
Y
and define
D
E
-
IIB[/ <
into itself, hence
Lm
259
by
A
(Ax)n
1
=
akxk.
By Lemma 2
k=1 IlAIl
<
which implies that
m
LEMMA.
4.
a
L.
E
I n Lemma 2 t h e a s s u m p t i o n on t h e r o w s of
may
A
be o m i t t e d .
A
The rows of
belong to
5. THEOREM. a matrix.
X
I f
Let
Y then
D
13
= w,
X
D Y , hence are in
YA n
Y =
W,
the rows of
6.
COROLLARY.
includes
If 5 that
Let
cA
Since
LI L m .
V
Em
Y
=
Lm
co
is a
IIAll
<
which
Em
it follows from Theorem
C
space (16.1.7), V C
V
V
b e a d e n s e s u b s p a c e of
L a c
W(A)
(16.1.8).
(16.0.41
]Im
which includes
h a v e t h e same c o m p a c t s e t s .
It is sufficient, by 16.0.7, to show that they have the same convergent sequences. This implies that
cA
5. This implies that
weakly in 8.
L.
Suppose that 2
V
-
Finally if
13
Let
afE,Lml
c , then
cA
space
and
is
A
space.
C
C
ofE,Vi
or
(1.3.11).
is a
7. LEMMA.
where
A
b e a d e n s e s u b s p a c e of
V
is a
V
X = Y
by Lemma 3.
L
is a matrix such that
A
Then
are in
Then
co.
(16.1.8) so
$.
A
If
is closed in
Lm
Lm,
Lm.
2
Lemma 2 covers the last case. by Lemma 2, so
or
coJ c
by Lemma 3.
.t
an + 0 in the first topology.
(16.1.3) so
IIanl[l -c 0
A
is coercive by Theorem
(1.7.18 ii) and so
an
+
0
The opposite implication is trivial.
THEOREM (G. Bennett and N . J .
Kalton).
Let
V
be a d e n s e
16.3
260
s u b s p a c e of 3
Then
V.
Let
which i n c l u d e s
IIm
X
x E
3
and
co
X
a separable
space
FX
tm.
x(~) + x
IIm.Then
1 6 . 0 . 8 and Lemma 7 ,
{x(")}
i: V
By
T ( V , ~ )Cauchy sequence. Now
is a
consider the inclusion map
~ ( t " , t ) (16.0.6).
in
+
X.
This is continuous, hence
has closed graph, when each space has the relative topology of u(V,t)
thus it has closed graph when the larger topologies
FK
the
topology of
X are used.
with these topologies and so
{x'")}
vergent.
It can only converge to
tinuous.
Thus, finally EXAMPLE. X
(*):
such that
and each space
2
zB
co
bs
+
since coordinates are con-
X.
E
Consider this property of an
Every separable space has
X B B = X, for example,
]Im,
(*)
since
X
is separable by 4.3.7. However if
which is a dense proper subspace of
X does not have that
X, hence con-
FK
space
i s t h e i n t e r s e c t i o n of a l l t h e s e p a r a b Z e
which incZude i t .
X
x
is continuous
is Cauchy in x
co
(*)
and
T(X,X')
16.0.9 and Corollary 6, i
(16.0.10). By 1 6 . 0 . 3 ,
9.
The latter is
w;
by Theorem 8. In D 1 ] ,
X: spaces
FK
as well as every B: z
=
X Em
is an
XB 1 ,
6
FK
(10.0.3d),
p. 31, it is shown
is another example.
The space
X mentioned in Example 9 shows that separable may
not be omitted in Theorem 8.
10. Lemma 3 , 8 . 6 . 2 and 8 . 6 . 7 suggest the following problem. I s t h e r e an
FK
(2)
Note that (1) and
X B = II?
space
X
3
$
with
X
XB c a
#
co
and
Ill X
C co,
implies ( 2 ) by 7 . 2 . 2 ;
if
X has AD, (2) implies (1) by 8.6.8. An example not satisfying (1) is given by letting
and
X
= L 0
E
E
be a quasicomplement for
(4.5.1). A much easier example takes
co
E
(10.0.3d) to be
261
16.3
the closure i n
[14A],
m
.t
of t h e p e r i o d i c s e q u e n c e s , X = .? 0 E .
Corollary 2 .
Sources :
I1 1 1,
I1 2 1.
See
This Page Intentionally Left Blank
263
CHAPTER 17 MAPS OF BANACH SPACES
FUNCTIONAL ANALYSIS
17.0.
i n 14.0.5. iff of
X, Y
Let
1.
Then
b e Banach s p a c e s , T
is dense i n
T
11-1-8.
Y.
*
closure of
T = { x : Tx = 01.
of t h e r a n g e o f
y E Y [80],
The s p a c e s c
3.
T' NT.
If T.
and
i s a l w a y s weak Here
Let
such t h a t xn + 0
5.
X
n
Theorem 11-1-6, L
+
weakly.
for all
0
is
c o n t i n u o u s ; NT"
lies i n t h e closure
y
Lemma 11-1-7,
#11-1-107.
h a v e no r e f l e x i v e c l o s e d s u b s p a c e s , [80], ##14-1-103, {xnl
104.
a bounded s e q u e n c e
i n a (norm) d e n s e s u b s e t .
f
Then
[80] #9-3-104. X, Y , T
With
*
T" [X"], t h e n
be a norrned s p a c e and
f(xn)
Corollary
d e n o t e s t h e n u l l s p a c e of
NT
except f o r t h o s e of f i n i t e dimension. 4.
11-3-4,
s 7.0.5.
C o n t i n u i n g 1:
t h e weak
as
( 4 . 0 . 6 ) , TI
is one t o one i f f t h e r a n g e
i s o n t o ; TI
[80], Theorems 11-3-1,
Some of t h i s
2.
B(X,Y)
i s a l i n e a r homeomorphism ( i n t o )
IIT'II = IITII; T
i s , and i f f T
T"
E
closed, i.e.
T[X]
there exists
z
E
as i n 1, s u p p o s e t h a t
is n o t c l o s e d i n Z
such t h a t
IIz/I
Y.
is n o t r a n g e
T
Then, f o r e a c h
= 1, IITzll
<
E.
E
[80],
> 0,
Lemma
11-3-6. 6. S
Let
i s weak
*
X
b e a Banach s p a c e a n d
closed i n
X"
iff S
S
a subspace of
is r e f l e x i v e .
A s h o r t e r b u t more a d v a n c e d p r o o f i s g i v e n i n
1321,
[80],
X.
Then
#11-3-118.
Lemma 1.
17.1
264
17.1. MATRIX MAPS OF
c
We now return to the study of conservative matrices and relate summability properties of a matrix to its behavior as a map from to
c
c. NOTE:
C o r e g u l a r and c o n u l l m a t r i c e s a r e now assumed c o n s e r v a -
t i v e as in Chapters 1-8.
1. LEMMA. g(y1
iff
Arc]
Let
A E
ty
for
=
r
b e c o r e g u l a r , g E c'. with
y E c
tsA
Sufficiency: g(Ax) = t(Ax) = (tA)x
y
=
Ax, x
E
c, g(y) =
(1.3.8) = p lim x + Yx
whenever
g(y) = 0
g(y) = u lim x . =
t(Ax). 2.
g
EXAMPLE.
3.
dense i n
A[c]c
(3.3.11.
0
(1.4.4). Necessity:
x
E
=
limAx + t(Ax) =
p
so
[lim x+ax] + t(Ax)
as in 4.4.9. The hypotheses is that c ; taking
Taking
Taking
0 on
=
x
p
on
g = 0
X(A) = 1. Now (3.2.1) g(y) = u lim y + ty
We may assume that if
=
Then
= 6k
u
x = 1 yields yields
gk
Let
x
(Ax),
co, but
=
yields =
so
Yk = 0
0 and so
g(y) = ty
tlA.
x
- xn-l, g(y)
lim Y.
=
Then
does not have the form of Lemma 1.
g
THEOREM. A c o r e g u Z a r m a t r i x i s of t y p e
M
i f f
~[c] i s
c.
From Lemma 1 and the Hahn-Banach theorem (3.0.1). 4. Arc]
THEOREM, A T a u b e r i a n m a t r i x
i s a c l o s e d s u b s p a c e of
maps these sets onto
A[c]
must be range c l o s e d i . e .
c.
One might expect this since A
A
c
and
is closed in
cA
(6.1.1)and
c ; indeed the proof is very
close to those of 3.4.4 and 6.1.1 - the difference is that the
final calculation takes place in
c
rather than
cA.
Assume that the conclusion is false and, without loss of generality, that
IIAII
= 1.
Fix an integer u
>
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 <
u
= 01.
-.*.
Then
u- 1 Arc], hence is not closed in 0
E
0
>
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, <
<
2,
E
n2
0. Choose
= =
1 -. 2
xt = z1 n if 2 + maxIn: x1 # 0 1 , n
co.
for at most one value of
k ; there are infinitely many
5 IIxk - z k[ I m + k # 0 , xi if xi
IIAll
all
k , xi
(using 0
1
and
and divergent.
Let
=
(3.0.5) and
AX)^.
y
17.2. MAPS OF
T
c =
if no such
1) 5
k
=
matrix
exists.
Then
Ixi
since
=
k Xi k# 1 1 {akiXi:
k
0) =
i
c
Fix
THEOREM. A
Finally
for infinitely many i , so x is bounded k y = 1 Ax . Then Y E c since 1 ~ ~ A x k< ~ ~ w
n
and consider
=
2.
such
0
(Tx)~ = vn lim x + (Ax), k k PnfT(6 11 = (T6 )n.
PnrTx]
n
xi = 2 xn # 0 ) . k xn # 0
be a continuous linear map (called n o p e r a t o r " ) from
to itself.
ank
=
Yn = 1 (AX k )n is not Tauberian.
Ax
A
Thus
Let
=
(1)
Let
Choose z2 to satisfy (1) with u = n2, E = 114. Let 2 1 z2 if lznl > q , otherwise x2 = 0. Choose n3 = 2 + maxh: n n Continuing in this way we get xk satisfying: for each n,
=
E
Given
PnoT E c' where
i.e.
vn
=
X(PnoT)
T, t h e r e e x i s t a s e q u e n c e
w . i t h c o n v e r g e n t c o l u m n s and
11~11 <
m
(PnoT)(x)
v
E
(3.2.1),
Lm
such t h a t
and a
266
17.2-17.3
Tx = v Z i m x
+
The m a t r i x
Ax.
i s g i v e n by a m a t r i x i f f
The
1
is T6k
E
lankl = IIPnoTll 5 IITII. An example is
in which 3.
equivalently i f
- 1 ak.
c)
E
THEOREM. If
iff
Let
v
E
A F
r
iff
1
cA
E
X = X ( T ) = X ( 2 h oT).
T in Theorem 2.
expression for
+
X(Al.
(3.2.1), and the
X
ak
=
DEFINITION. F o r any v e c t o r s p a c e
X,
u E X,
T = u 8 g
X,
tensor
(u
lim ank
means
g)
If
a matrix.
T
Thus
is an almost matrix, T
T
=
v
8
=
lim(TG k )n.
a linear
g
Tx = g ( x ) u .
This operator has one dimensional range (unless u 6.
c,
X = lim(v+Al/
As usual
f u n c t i o n a l on
E
(1.3.6),
Then
i s an a l m o s t m a t r i x , X ( T ) = Z i m v
T
v
c.
This is clear from the definition of
5.
(-l)n rlim x-x,]
=
i s c a l l e d an a l m o s t m a t r i x i f
T
The equivalence is because
4.
(Tx),
+
i s c o n s e r v a t i v e i n Theorem 1 .
A
hence (since A 1
c, also (1.0.2) l X n l
n if k.
an., = (-l)n, ank = 0 if DEFINITION.
T
v = 0.
column of A
kth
n e e d n o t b e c o n s e r v a t i v e and
A
lim + A
is the sum of two operators on
or
g = 0).
A
where
is
c , one being
a one ( o r zero) dimensional operator, the other a matrix.
7. EXAMPLE.
i.e. g given by
Let
g
E
c'
have the special form
is a row function (11.1.1). Then
u 8 g
g(x)
bx,
=
A
is the matrix
ank = unbk.
17.3. THE DUAL MAP
The use of the dual map (14.0.5) as in this chapter was suggested by J. P. Crawford and R. J. Whitley.
T must be a matrix and
T':
(1 +
.t
For
X
=
Y
=
c0 '
is the transposed matrix.
We
17.3
T' for X
shall determine the form of
venient form is as a map from c'
is equivalent to
(3.2.1), tn
=
f(6");
identified with
by
II
T'.
Since
Particular values of h(1im)
=
6O,
h
hIT'(1im)I
The most con-
(X,tl,t2,. . . ) where
=
S
hoT'oh-l: II
=
AK, S
has
and members
X = x(f) +
a
may be
is a matrix and we
x
of
c
as
y
(x1,x2,x3, . . . ) .
are
h(Pk)
The zero column
.k
c.
=
It will be convenient to write members
(yo,y1,y2,. . . )
as
11
h(f)
Y
=
e ; this is available since
to
II
see 1.0.2. Thus
proceed to compute it. of
267
(Sno)
of
by (1) and this is
From the definition of
h
k 1. 1
for
= gk
S
is
hIT'(h-'6O)}
S6O =
h(1im oT) this is
=
by definition of
TI.
(X,al,a2,. . . )
where these
S6k = h(PkOT)
by (1)
numbers are given in 17.2.4. The
kth
(Snk), k 2 1, is
column
and a calculation similar to the preceding one. column is
(vk,akl,ak2,...)
For example
1. Thus
(PkoT)(6")
where v , A n (T6 )k = akn.
=
Thus the
kth
are given in 17.2.1.
S, which we identify with
TI
is given by the
matrix
. . .
x
v1
v2
v3
"1
"11
"21
"31
"2
"12
"22
a32
*
'
*
'
*
'
. . . . . . . . . . . . 2.
In abbreviated form, T'
explained in 17.2.1-4.
If
T
=
x (a
v
A ~ ) where the symbols are
is a matrix, T = A
and
T'
= A'
x o T). Since 9. has AK, the dual of a self-map (which must a A m be a matrix) is the transposed matrix and so T": ern + L! is
= (
17.3
268
1).
X
given by
("
embedding of y
E
x
E
It is very important to observe that the natural c
into
lies in
,trn
lim y
iff
(lim x,x1,x2,.. . )
is given b y
Ern
=
yl, in which case
y
=
2
and
with
x = (Y2,Y3,Y4,-4* The fact that
3.
r
is multiplicative on
X
trivial from Remark 2 since
(6.1.8) is
(AB)' = B'A'.
i s a l w a y s a m a t r i x and i s c o n s e r v a t i v e i f f
4. T"
a l m o s t m a t r i x , in which case
X(T")
=
i s an
T
X(A) = X(T) - lim v
by
17.2.4. If
A
i s a conull m a t r i x , A
morphism of
c
(onto a closed subspace).
5.
A'
and so
0
is
A'
cannot map
II
particular no Z i n e a r homeomorphism of
c a n n o t be a l i n e a r homeoFor the first row of
onto c
(See 17.0.1).
II. onto
In
can b e g i v e n
co
by a m a t r i x .
6. EXAMPLE. 0'
If
Tx = 0
Tx
T =
=
1 8 lim
then
- A
0 =
- xn- 1 so x
lim x of
T
Let
c
An a l m o s t m a t r i x w h i c h maps
(TX)= ~ lim x; for
= 0
is included in
(17.2.5) where
and co
T
one t o one o n t o
c
(Ax)n n
>
1, 0
is one to one.
and is all of
co
(xo
xn-1
=
(Tx),
=
= 0.)
=
Also the range
since for
y
E
co,
y where xn - y1 - Yn+1* 7.
No m a t r i x c a n have t h e p r o p e r t y of Example 6 .
a matrix would be conull; indeed multiplicative
For such
0 , and Remark 5
applies . 8. EXAMPLE. maps
c
onto
The matrix
co, while if
conull matrix and maps
c
A
given by
AX)^
onto
=
c.
(Ax)n
x1 + xZn
=
x
~ - ~x~~ -
- x ~ ~ A+ is ~ , a
~
17.3-17.4
269
A neater proof of 17.1.1 is now available: the conditioll on g Xt
is that 0
=
A'g
0, akto + (tA)k = 0.
to lim y + ty
17.4.
0; setting
=
=
ty.
g
=
Thus
(t0,tl,t2,. . . ) to = 0
and
this says
Q
E
tla
so
g(x) =
The converse is similar.
W-MATRICES
The study of matrix algebras can be carried out in a very general setting which makes certain summability facts quite transparent. If
1.
T: X
-f
Y
is a map of Banach spaces and
dual as in the preceding section, n o t e t h a t X
with its natural embedding in
X".
T"IX
=
T'
is the
identifying
T
This embedding is used to
prove results in the general setting.
They will hold for
X
=
c
but one must be careful to observe that when specializing the actual calculations to but a subspace of 2.
c
c , the embedding of in
Qm
w
Banack s p a c e and
in
is not
Qm
c
(17.3.2).
In t h i s s e c t i o n
REMARK.
c
X
i s a fixed non-reflexive
a f i x e d member o f
r,
X"\X,
are as i n
p
Definition 3.
3. of
DEFINITION.
T").
I'
=
rfX,wl
= IT E
BfX): w
The corresponding eigenvalue is denoted by
The meaning of the definition is that each continuous linear map from the preceding section; T such that 4.
i s an e i g e n v e c t o r
T"(w)
=
EXAMPLE.
E
X
r
to itself; T"
=
T
E
(TI)'
p(T).
B(X)
is a
is as in
iff there exists a number
p(T)
p(T)w. Let
w(f) = x(f) (3.2.1). w = 6 0 since f E c'
X = c, w = X.
This means that for
f
E
c'
Thinking of c" as Q m , this means that corresponds to {X(f),f(6'),f(62),...) E Q
17.4-17.5
270
w
so t h a t
picks out t h e f i r s t coordinate.
is a m a t r i x .
r(c,so)
Thus
s i t u a t i o n s ; see 1 7 . 5 . 2 .
r, t h e s e t
=
i.e. i f f
v = 0
T
of c o n s e r v a t i v e m a t r i c e s .
Because o f t h e f o r e g o i n g r e s u l t , members
are c a l l e d
r(X,w)
#-matrices.
W i t h t h e n o t a t i o n of Remark 2 ,
THEOREM.
5.
iff
p ( T ) = x ( T ) , b u t t h i s i d e n t i t y w i l l f a i l i n more g e n e r a l
Note t h a t
of
w
which is a m u l t i p l e o f
{x,v1,v2,,..)
From 1 7 . 3 . 2 , T"w =
subalgebra w i t h i d e n t i t y o f
and
BlX)
p
r
i s a closed
i s a (continuous) scalar
homomorphism. Let
h : B(X)
be g i v e n by
5 IIT"ll .IIwII = IITll .IIwII
E
r
implies
t i n u i t y is by 4.0.12.
Then
(17.0.1). p
i s con-
h Thus
Finally
I" = I
p = 0 , is a n i d e a l i n
To
r.
c l a s s i c a l case i n 6 . 1 . 9 and 1 7 . 3 . 3 .
so
I
r
=
is m u l t i p l i c a -
(AB)"w = A"p(B)w = p(B)p(A)w.
I t is t r i v i a l t h a t t h e s e t
f o r which
h ( T ) = T"w.
i s c l o s e d ; i t i s a s u b a l g e b r a and A,B
tive since
X"
11 h(T)II
tinuous since h - l [span w]
+
Con-
r.
E
w-matrices,
of c o n u l l
those
T h i s was n o t e d i n t h e The n e x t r e s u l t i s somewhat
u n e x p e c t e d , even i n t h e c l a s s i c a l case o f Example 3 . THEOREM.
6.
For
r0
T E B(X), A
i s a left ideal i n E
ro,
But n o t a r i g h t i d e a l
B(XI.
(TA)"w = T"p(A)w = 0 .
-
a s i m p l e example i s g o t t e n by t a k i n g
X = c , T = 1 8 l i m , A = 1 8 P1
(see 17.2.5).
a l s o f o l l o w s from g e n e r a l c h a r a c t e r i z a t i o n s i n
17.5.
AT = T.
This
1151.
W-ALMOST MATRICES If
+
Then
T
i s an almost m a t r i x , 1 7 . 3 . 2 shows t h a t
(a,vl,v2, ...)
where
a
= l i m
vn '
Setting
p = X-L,
= (X-L)A0
t h i s gives
17.5
p6O
+
where
is the natural embedding of
In the next definition we identify ing in
When
X".
271
X
=
c, X"
x
E
v
into
c" =
Rm.
X with its natural embedd-
is the second dual of
c
and not
R"
t17.3.2).
1. DEFINITION. W i t h t h e n o t a t i o n of 1 7 . 4 . 2 , = {T E
by
B ( X ) : T"w
E
and
T"w = p ( T ) w t x
i s ( n o t ) zero.
The f u n c t i o n a l
w 0 XI. T
T a = ra(X,wl
ra
on
p
i s defined
i s c a l l e d conull IcoreguZar) i f
ra
The members of
are called
plTl
w-almost m a t r i c e s .
This extends the concept of coregular and conull to conservative maps which are not matrices (in the setting X = c, w = 6 0 ) . The development beginning in Chapter 9 extended it to matrix maps which are not conservative. 2.
As pointed out before Definition 1, p(T) = X(T)
in the classical case.
p(T) = X(T")
Also
of these identities assume that
r c ra
3.
and t h e f u n c t i o n a l
In Definition 1, x 4.
T
=
T
0 whenever
=
X(A)
-
lim vn
by 17.3.4.
All
is an almost matrix. h a s t h e same meaning on b o t h .
p
is a
w-matrix.
THEOREM. W i t h t h e n o t a t i o n o f D e f i n i t i o n I ,
cZosed s u b a l g e b r a , w i t h i d e n t i t y , of
B(X)
and
p
ra
is a
i s a (continuous)
s c a l a r homomorphism.
With
h
as in 17.4.5, F a
is a subalgebra and =
p(T)S"(w) 5.
+ Sx
=
p
=
h-'[mw]
is closed (15.0.1); it
is multiplicative since
p(T)p(S)w
+ [p(T)y+Sx].
Similar calculations show that
matrices, i s a l e f t i d e a l i n
B(XI
(ST)"w
= S"
[~(T)w+x]
Continuity is by 4.0.12.
r z , the conull w-almost
as in 17.4.6.
These subalgebras have a significant property, expressed in Theorem 7, which will follow from a preliminary lemma:
17.5
272
LEMMA.
6.
Let
U
T
and
and s u p p o s e t h a t Then
S
a =
ST
E
ra
(Definition 1).
with either
U
+ Sx.
we have
S"w
r
and
7. THEOREM. inverse closed) i n
p(T)
p =
S" bw+x] = pS"w
p(ST)/p
and if
S
E
r.
U
=
(ST)"w
Also
E
B(X)
coregular.
ST
or
T
S
r, x
y = 0. We have
=
+ y.
p(ST)w
=
In the
Setting
aw + (y-Sx)/p.
=
ra
are l e f t inverse closed (hence
B(XI.
This means, for example, that if then
Let
T must be coregular by Remark 5 and 17.4.6.
following calculation =
or
U.
E
First
(ST)"w
r
=
S
B(X), T
E
r , ST
E
The result is immediate from Lemma 6 since
=
I
I is a
coregular matrix. 8. Theorem 7 should not be confused with the more difficult
result that 9.
T: X
+
Y
r
is inverse closed in
EXAMPLE.
For X, Y
(6.1.9).
#
Banach spaces, a continuous linear map TI' [Xl'] c Y.
is called w e a k l y c o m p a c t if
It is trivial
from the definition that any weakly compact map from a conull almost matrix ( f o r any choice of
w.)
X
to
X
is
It is quite easy to
see that a c o n s e r v a t i v e m a t r i x d e f i n e s a w e a k l y compact map of
c
iff i t i s c o e r c i v e (1.7.18). 10.
p
conull for
x are both defined on Fa. We reserved the name
and p =
0. This is done for several reasons.
nice e x t e n s i o n of T"w
=
x
17.3.5
to
(Definition 1) = Tu
closed (17.0.2). So
T"w
=
for some u
T"u, T"
E
X
X
i s not a
homomorphism on
has the left inverse B
given by
if
T
T
is conull,
is range
is not one to one and
not a homeomorphism (17.0.1).] But the map Also
[If
w-almost m a t r i c e s .
Look at this
ra (By),
of 17.3.6 has
T
is
X =
since the map of 17.3.6 =
y1
- Y,+~.
0.
17.5
11.
since
The second matrix of 17.3.8 has no right inverse in It has the right inverse in
p = 0.
x + (L,xl,L, x2,L,x3,. . . )
where
B(c)
L = lim x.
12.
EXAMPLE.
Let
= g(x)f(a)
so
V
=
a
V'(f)
ij, in particular
V J J = ;E 8
@
g
=
f(a)g
(17.2.5). Then
v
matrix and is weakly compact.
It is a
#-matrix
Every
T
E
ra
is the sum of a
zero) dimensional operator. X'
with
w(h)
=
1
Then by Example 12, S"w
=
pw
iff
+ a
S
=
-
w(h)a
# ( g ) = 0.
#-matrix and a one ( o r
This generalizes 17.2.6.
and let
Hence
is a conuZZ w-almost
so
13.
<x,V'f>
V' = g 8 &.
i.e.
= u(g)a
E
given by
This shows that
V"W
h
ra
allows n o non-zero scalar homomorphism.
B(c)
=
273
T - a8 h
where
Choose T"w = pw+a.
= pw.
14. Almost matrices were first introduced in p8] in the way j u s t described.
They later appeared independently as the solution
of a problem concerning commuting operators. T E B(c)
which commutes with
For example, every
(C,1) must be an almost matrix.
See [54] p. 213. In [ Z O ] and articles cited there, the problem is studied as to whether, for various choices of
w, the algebras i"(X,w) are
isomorphic, and the same question for X
=
c , no for 16.
X
ra.
The answer is yes for
= L.
In 1291 the two algebras are introduced without use of
dual maps and in the more general setting of locally convex spaces which are not weakly complete.
For example
means of a weakly Cauchy and divergent net {T:(T-pI)(w)
+
0
weakly).
I'
w
may be defined by
in
X , namely
274
17.6
17.6. TAUBERIAN MAPS
1. DEFINITION. L e t c o n t i n u o u s and l i n e a r ; T w a'
X, Y
b e Banach s p a c e s and
i s called Tauberian i f
T: X
T"w @ Y
+ Y
whenever
x. This is behavior exactly opposite to that of a weakly compact
map, 17.5.9. 2. With the notation of 17.5.1,T
w, T
is not conull. 3.
A conservative almost matrix i s Tauberian i n the
EXAMPLE.
s e n s e of D e f i n i t i o n 1 w i t h Tauberian.
Necessity:
... ) .
- (x,v1,v2,
(17.5.2).
X = Y = c
In particular
This means that
Now let
x E cA
(T"Y)~. Thus T"y
(Also u = lim x.
6
E
vnwl + (Ax)n
=
follows that and so so
w
E
limAx
6
x
i.e. p ( T ) # 0
+
(x-p)u + limAx = xu + ax E
8 hence x
c.
6
This was the motivation for our definition of
(T"W),+~
Ax
8 does not contain T"6°
and so by hypothesis y
Suppose T"w
(17.5.2).
is
A
a m , u = (limAx-ax)/p, y = (u,xl,x2,. . . ) .
n
u, using 1.3.8.) Sufficiency: A is
i f f i t s matrix part
lim v #
Then by 17.3.2 (T"Y),+~ = vnu + (Ax), =
is Tauberianiff for every
E
= p
-+
E
6.
Let
x
=
(w2,w3,w4,.. . ) ,
(T"W)~ = xwl + ax.
limA x = pwl -+ ax. lim x + ax (1.3.8). Hence c
and
T
is coregular, (3.5.2), hence
Since vn
Then it
+ X-p
By hypothesis, x lim w
=
lim x
=
E
c
w1
(17.3.2).
By Example 3, Tauberian maps generalize Tauberian matrices. Their study has led to some interesting problems in general Banach space theory.
We shall do only enough of the general theory to
cover the classical characterization, Theorem 10. 4.
THEOREM.
reflexive.
The n u l l s p a c e of a T a u b e r i a n map m u s t be
17.6
If NT"
T"F
NT.
=
that
(NT
NT
F
= 0,
is weak
*
F
X", the hypothesis implies that
E
Ix:Tx = 01
=
275
T.)
is the null space of
closed in
X"
X
E
i.e.
It follows
(17.0.2) and so it is reflexive
(17.0.6).
The converse is obviously false since there are non-Tauberian triangles such as
(C,l).
These are one to one maps!
There is a
converse result for range closed operators:
iff
5.
THEOREM.
NT
i s reflexive.
Let
T
Half is by Theorem 4 . 17.0.2, T"w = Tx
weak
*
for some
closure of
so w-x
E
NT
be range c l o s e d .
Then
Conversely, let
w
x
NT"
E
X
so
(17.0.2). But
T
X",
E
w-x
E
NT
is weak
is T a u b e r i a n
T"w
E
Y.
By
which is the
*
closed in
XI'
In particular w e X.
NT.
T h e c o m p o s i t i o n of t w o T a u b e r i a n maps is T a u b e r i a n .
6.
LEMMA.
If
(TS)"w
7.
COROLLARY.
E
Z then
T"(S"w)
F
Z so
S"w
E
Y
and
w
E
X.
The r e s t r i c t i o n of a T a u b e r i a n map t o a c l o s e d
s u b s p a c e is T a u b e r i a n .
Let
X c
Y
composition of
and
T: Y
+
Z Tauberian. Then
T with the inclusion map.
T ( X is the
The latter is Tauberian
by Theorem 5 and the result follows by Lemma 6. 8.
DEFINITION. A s e m i - F r e d h o l m map i s a r a n g e c l o s e d map
whose n u l l s p a c e i s f i n i t e d i m e n s i o n a l .
Thus a semi-Fredholm map must be Tauberian, (Theorem 5 ) .
The
converse is false; for every map of a reflexive space is Tauberian (and weakly compact ) .
276
17.6
9. EXAMPLE.
Let
T: Q
a n y Banach s p a c e .
Y, Y
+
Tauberian i f f i t i s semi-Fredholm.
which
T
T[X]
=
T[R].
T: X
+
Y
with
is finite dimensional on
is Tauberian (Corollary 7 ) and one to one and such that Denoting
TIX
weakly in
=
1, Txn
+
is one to one since
it is sufficient to show that
If not there is a sequence
{xnl
in
xn
0
(17.0.5). We shall prove that
0
X , hence in
T
by
is range closed.
IIxnll
T' [Y']
T
has a closed subspace X
Q
is
T
Sufficiency is obvious by
Theorem 5. Conversely, the null space of by Theorem 4 and 17.0.3. Thus
Then
Q , and this contradicts 1 . 7 . 2 0 .
NT" = NT
is norm dense in
X'
+
First
as in the proof of Theorem 4. (17.0.1). Also
cxn,f> = <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
= p + =
X.
m = m = <mh,w>
=
=
+ h(x).
p
In the classical case this says that a multiplicative-m
x
matrix has
=
m.
(p
EXAMPLE.
5.
Let
X
r).
on
X
=
c , h = lim.
=
T
Then
E
M
T
iff
is
multiplicative in the summability sense i.e. lim Tx = m lim x x
E
c.
T
E
M
This is by Lemma 2. We may also take iff
T'h
=
xh
i.e.
iff
a
h
= 6
0
for
so that
E
0, from 17.3.1.
=
6. This formulation explains an old paradox in summability theory.
The following list of facts is at first glance somewhat
puzzling: (a)
The set of conull matrices in a left ideal in
(a') The set of multiplicative-0 in (b) (b')
B(c)
operators is a right ideal
B(c)
The set of matrices is left inverse closed in
M
The set
is right inverse closed in
We shall prove
these
two
B(c)
B(c).
pairs of facts in the general
setting and it will be seen that the puzzling shift from left to right in each pair is simply explained by the fact that matrices are defined in terms of
T"
are defined in terms of
TI; this with the facts that
S'IT", (ST)'
=
T'S'
(517.4) while multiplicative operators (ST)"
=
explains the shift from one side to the other.
Fact (a) is 17.4.6;Fact (b) is 17.5.7. To prove (a)' let
A
E
M
with
m(A)
To prove (b)' let (AB)'h
=
=
0, B
A
E
E
B(X).
M, B
B'(A'h) = B'(mh)
=
E
Then
B(X),
mB'(h),
(AB)'h
AB = I. hence
=
Then
B'(A'h) h
=
B'(h) = (I)h m
=
I'h
B'O = 0. =
since
17.7
m
279
m(A) # 0 by (a)'.
=
Every
7.
A
E
r
is the sum of a member of
r.
( o r zero) dimensional member of
Mr
and a one
This is analogous to 17.5.13
and the method is an abstraction of the trick used in 3.5.1, of which the present result is a generalization. Let Choose
x
will do.)
Let
- A"w
w(g)x Xw
and
A'h
-
h(x) = 1.
with
B
=
since w(g)
= 1.
g =
A - x
xh.
@
=
Also
(In the classical case h
B(c),
and
that
B e M
since
B'h
- Xh.
= lim, x =
since B"w = A"w
-
- X;A"w
=
=
A'h
- h(x)g
ra;
X(B) = X(A).
ra
is a maximal proper subalgebra
X
=
c
in both cases.
that there is no non-zero scalar homomorphism on
Mr
B(X)
It follows if
X
= c
(this was already observed in 17.5.11); this is true also for =
c
0'
X@ w
EP (
and Hilbert space. See D9],
[36].
[80], Remark 3-2-7) it is clear that
is a scalar homomorphism.
1
=
is the only non-zero scalar homomorphism on
is the only one on
p
r
B E
=
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 <
E,
is closed under multiplication.
E.
This is
immediate from continuity of multiplication (1.0.6). 3.
Let
z
belong to a Banach algebra with identity.
u ( z ) , the spectrum of
z , = {t: z-tl
a set of scalars.
t
If
E
u(z),
then
is not invertible}. It1 5
IIzll.
p9],
Then This is
Theorem
14.2.3. 4.
A weakly compact (17.5.9) member of
[SO], ##11-4-108, 11-4-203.
B(c)
must be compact
The spectrum of a compact map of a
Banach space is countable p3) p. 281 Th. 5 . 5 G .
5. Let
f
be a non-zero scalar homomorphism on a Banach
algebra with identity. Lemma 14.1.5.
Then
f(z)
E
u(z)
for each
z
1791,
282
18.1
18.1. TOPOLOGICAL DIVISORS OF
0
r
In this chapter we study the Banach algebras tive matrices, and
of conserva-
of conservative triangular matrices (1.5.4,
A
An extension to sc matrices is counter-indicated by 9.5.5.
1.5.5).
We saw a relation between Mercerian (or Tauberian) and invertible in
r,
(1.7.14) (or in
A
1.8.4).
I.D. Berg suggested the approach of this section.
The original
motivation was the observation of 18.2.4 which suggested the result of 18.2.5. There is a conceptual difference between a matrix
1.
x
the operator
For x
Ax.
+
c, A
E
r c B(c)
they are the same.
AB = AoB, the first being matrix multiplica-
This is expressed by tion.
E
and
A
See 1.4.4, 1.4.6. DEFINITION. L e t
2.
i s c a l l e d an
Then
z
each
E >
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
<
E.
r.
Then
is a bijection where
S
T-':
S +
X.
Then for all
11 TVll
is a closed
S
V
E
B(X),
is bounded away
is not an etz. E
(By 17.0.6 if
Ta = 0.) Let
+
which implies that
Sufficiency: Let IITall
T E
i s n o t a l i n e a r homeomorphism ( i n t o ) .
X. Let
IIVll = IIUTVll 2
let
g E X',
(17.2.5). Then f o r
y
> 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;
<
n 1 (-1) =
.
This gives a summability method, but the practical procedure was the converse: to define some sort of matrix or integral transform and use it to define a function (given by a series) outside the place where its original series converges Fejer's theorem on
See [88] pp. 145, 188.
(C,1) summabili y of Fourier series
(19.2.4) gives the Weierstrass approximat on theorem by approximat-
ing each trigonometric function by a segment of its power series expansion.
It shows also the totality of these functions:
if
n
I
f(x)t(x)dx
= 0
continuous) is
f o r every
t(x)
=
cos nx or sin nx
0 since its Fourier series is
then
f
(if
0. See also [88],
p. 154.
A useful possibility is that, even if
1 xn
is convergent,
some matrix transformation will converge more quickly so that computation of the sum becomes feasible.
(It is easy to write
series whose sum would take inadmissible amounts of computer time to calculate directly.) expansions this way.
It is also possible to obtain asymptotic
See 1251, top of p. 96, [88],
p. 146.
A
recent reference is Math. Reviews May 1983 #40003. An application of summability due to A . K. Snyder is to showing that if one (ideal) point is removed from the Stone-Cech compactification of the rationals the result is countably compact. See [80A], Table 7, p. 339.
B. E. Rhoades [51],
[52]
has applications to fixed point
theorems and Markov chains. Classical theorems on convergence and interchange of summation can be obtained: each
n.
let
Then f o r any
1 ankl m, 1 ank converge f o r m J n k=I k t E L! the s e r i e s f tnank converges
sup
19.4-19.5
299
1 1. The first part is simply 8.4.1B which says n k (11:~s). For the second part let f(t) be the first sum.
and i s e q u a l t o
that
A'I' E
Then
f
is continuous on
1.0.4 and, since
has
11
by the Banach-Steinhaus theorem
11
AK, f(t)
=
1 tnf(6").
Another result of
this type is 9.6.5. An interesting elementary application was giver. by M. Aissen n [l]. By the binomial theorem (l+l/n)" = 1 ank(l/k!) where a nk k=O is easily calculated and seen to be in ( c s : ~ ) by 8.4.5B. Moreover lim x A
=
1 xn
for
x
E
cs.
Hence
lim(l+l/n)"
=
1 l/k!.
Applications to the study of harmonic functions and are given in [67],
Hp spaces
[68].
Inclusion theorems can be used t o prove that certain spaces are barrelled.
Suppose an
FK
space
(which does not need local convexity)
8.2.5 X
13
11.
X 2
Then by 4.2.4 is bounded in
This is sufficient to conclude that
barrelled subspace of
11,
[80]
fill2
X.
By
is a
Example 15-2-5.
Some impossibility proofs can be proved, for example no iff sequence {an} of sequences can be found such that x E 11n 1 akxk converges for each n. This is just the statement that 1"
#
uA
and is true because the latter space has
AK
(4.3.8). In
[ 7 5 ] it is deduced from 4.6.6 and 6.4.3 that the definition of
Cauchy sequence cannot be given by a countable set of sentences of a certain type.
19.5.
QUESTIONS
A s with any subject d i g n u s s u o s a l e the number of open ques-
tions can be multiplied without limit.
Some of these seize the
imagination because they are easy to state (1,3) or are deceptively easy looking (4,5). Some are so general that they seem easy to
19.5-19.6
300
disprove (2).
Some seem h o p e l e s s
(9,lO)
We s h a l l l i s t a few p r o b l e m s of s p e c i a l i n t e r e s t ( t c t h e a u t h o r ) and r e f e r t o o t h e r s i n t h e t e x t .
The f i r s t o n e r e a l l y h a s n o t h i n g
t o do with summability.
1. (8.6.8,
Suppose t h a t a n
AD
space
X
has
X B = k.
X = c0 ?
Must
16.3.10).
2.
I c o n j e c t u r e t h a t a l l matrices are
3.
I c o n j e c t u r e t h a t a m u l t i p l i c a t i v e t r i a n g l e must b e
unique
p-continuous.
(14.3) p-
(15.2.13) Find a t y p e
4.
t r i a n g l e n o t i n t h e c l o s u r e of G ( A )
M
(18.2.8).
5.
Must a l l ( o n e t o o n e ) non-Tauberian
boundary of t h e maximal g r o u p of
r?
matrices l i e i n t h e
(18.2.10)
6.
Must a o n e t o o n e m a t r i x h a v e a g r o w t h s e q u e n c e ?
7.
Must a m a t r i x w i t h c l o s e d i n s e t b e r e p l a c e a b l e ?
8.
Does t h e r e e x i s t a s e p a r a b l e c o r e g u l a r p s e u d o c o n u l l s p a c e ?
(5.5.6) (12.2.21)
(16.2-10)
9. c.
that
10.
F i n d c o n d i t i o n s on a m a t r i x
is a
BK
A
equivalent to t h e property
space.
C h a r a c t e r i z e s p a c e s of t h e form
cA among
FK
spaces.
(19.1.4)
11.
Characterize
A
such t h a t
c
is c l o s e d i n
cA.
(3.4)
O t h e r q u e s t i o n s o c c u r i n 9 . 5 . 2 ( c o n v e r s e ? ) , 9 . 5 . 1 0 , 10.3
(M(X)),
19.6
14.5.2, 14.5.6, 15.5.5, 15.5.9, 15.6.6.
HISTORY We h a v e spoken of t h e s e e d i d e a of s u m m a b i l i t y i n §l.l:
to
19.6
"sum" divergent series.
301
Ways of doing this were called summability
methods; some are described in Chapter 2.
Their number is legion
and their invention has stretched over the past 150 years. index of methods on p. 307 of [88] contains 99 entries.
The
Some
motivations f o r particular methods are indicated in 919.4.
In
1911, the first step towards abstraction came with the discovery
of conditions for regularity of an unspecified matrix (1.3.6). Thought of as a mapping theorem, this led to the investigation of spaces
(X:Y) as in Chapter 8. The first applications of
functional analysis were by Mazur in 1927 and Banach in 1932; the first is described in 13.1 and 3.3.6; the second is the proof of 1.3.6.
The latter, in particular, attracted great attention by
the extreme brevity of its proof in comparison with the classical construction used by Toeplitz 21 years earlier.
This is by now
an old story; the most classical of real and complex variabilists use uniform boundedness, closed graph and Gelfand representation. The rest of the history is as described (incompletely) in this book.
Triangles in Chapter 3, FK
spaces in Chapter 4 allow-
ing matrices of any shape and finally theorems about the spaces themselves.
See 14.1.
FK
A thread running from triangles
into more less pure functional analysis is described in [82A]. This history referred to is that of applications of functional analysis.
As just mentioned it is far from complete
- moreover
we have omitted all of the large area of summability in which methods of classical analysis predominate. [88]
and the Mathematical Reviews.
The reader may consult
A word of warning:
articles
written on the subject matter of this book are usually listed under Functional Analysis.
This Page Intentionally Left Blank
3 03
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42.
K. Knopp, G.G. Lorentz: Beitrage zur absoluten Limitierung. Arch. Math. 2(1949) 10-16.
43.
S-Y. Kuan: Some invariant properties on summability domains PAMS 64( 1977) 248-250.
BIBLIOGRAPHY
306
44.
S-Y. Kuan: On the inset of a convergence domain. 71(1978) 241-242.
44A. K. Kuratowski: Topologie. 45.
PAMS
Haffner, New York 1952.
B. Kuttner, L.H. Lawrence: On Mercerian sets.
JLMS 16(1977)
96-98. 46.
M.S. MacPhail, A. Wilansky: Linear functionals and summability invariants. Canad. Math. Bull. 17(1974) 233-242.
47.
C. Madden: A representation of Thesis 1978.
47A. I.J. Maddox:
ci.
Lehigh University Ph.D.
Elements of Functional Analysis.
47B. I.J. Maddox! Infinite Matrices of Operators. Notes #786, 1980. 48.
Cambridge 1970. Springer Lecture
W. Miesner: The convergence fields of N6rlund means.
PLMS
15(1965) 495-507. 48A. M.R. Parameswaran: On the reciprocal of a Indian Math. SOC. 20(1956) 329-331. 49.
A . Pietsch:
k-matrix.
Nuclear Locally Convex Spaces.
J.
Springer-Verlag
1969. 50.
L . A . Raphael:
into itself.
Matrices mapping the space of analytic sequences PJM 27(1968) 123-126.
B.E. Rhoades : 51. Fixed point iterations using infinite matrices. I. TAMS 194(1974) 161-176; 11. Constructive and Computational Methods for Differential and Integral Equations. Springer-Verlag Lecture Notes 430(1974) 390,294; 111. Fixed Points, Algorithms and Applications (Ed: S. Karamardian), Academic Press 1977, pp. 337-347. 51.
The convergence of matrix transformations for certain Markov chains. Stochastic Processes and their Applications 9(1979) 85-93.
53.
Triangular summability methods and the boundary of the maximal group. MZ 105(1968) 284-290.
54. B.E. Rhoades, A. Wilansky:
Some cornmutants in
B(c).
PJM
49(1973) 211-217. 55.
W.H. Ruckle:
Sequence Spaces.
Pitman 1981.
56.
W.H. Ruckle: FK spaces in which the sequence of coordinate vectors is bounded. CJM 25(1973) 973-978.
57.
W.H. Ruckle: Representation and series summability of complete biorthogonal sequences. PJM 34(1970) 511-528.
58.
D.C. Russell: Inclusion theorems for section-bounded matrix transformations. MZ 113(1970) 255-265.
BIBLIOGRAPHY
59.
W.L.C. Sargent:
307
On sectionally bounded BK spaces. MZ 83(1964)
57-66. 60. 61.
J.J. Sember: A note on conull FK spaces and variation matrices MZ 108(1968) 1-6. J.J. Sember, M. Raphael: The unrestricted section properties CJM 31(1979) 331-336.
of sequences. 62.
N.K. Sharma:
Spectra of conservative matrices.
PAMS 35(1972)
515-518. 62A. N.K. Sharma:
matrices.
Isolated points of the spectra of conservative PAMS 51(1975) 74-78.
A.K. Snyder: 63. Consistency theory in semiconservative spaces.
SM 5 1982)
1-13. 64.
Cech compactification and regular matrix summability DMJ 36(1969) 245-252.
65.
Heavy points in countable spaces.
66.
Generating heavy points with positive matrices.
JDAM 20(1967) 271-279. PAMS
19(1968) 973-975. 67.
Harmonic functions and regular matrix summability.
IJM
13(1969) 406-413. 68.
Sequence spaces and interpolation problems for analytic functions. SM 39(1971) 137-153.
A.K. Snyder, A. Wilansky 69. Inclusion theorems and semiconservative Rocky Mtn. J. of Math. 2(1972) 595-603. 70.
FK
spaces.
The Mazur-Orlicz bounded consistency theorem.
PAMS
80(1980) 374-376. 71.
Non-replaceable matrices.
MZ 129(1972) 21-23.
72.
M. Stieglitz, H. Tietz: Matrixtransformationen von Folgenraumen. Eine Ergebnisubersicht. MZ 1 5 4 ( 1 9 7 7 ) , 1-16.
73.
A.E. Taylor:
Functional Analysis.
74.
A.E. Taylor:
Theory of Functions and Integration.
John Wiley 1958. Blaisdell
1965. 74A. E.C. Tichmarsh.
The Theory of Functions.
Oxford 1939.
A. Wilansky: 75.
The Cauchy criterion.
AMM 64(1957) 469-471.
75A. Distinguished subspaces and summability invariants JDAM 12(1964) 327-350.
308
BIBLIOGRAPHY
76. Topics in Functional Analysis. #45. 1967.
77. Toplogical divisors of 78. Subalgebras of 632.
B(X).
TAMS 113(1964) 240-251.
PAMS 29(1971) 355-360; 34(1972)
79. Functional Analysis. 80.
0.
Springer Lecture Notes
Blaisdell 1964.
Modern Methods in Topological Vector Spaces. Hill 1978.
R.E. Krieger, Melbourne,
80A. Topology for Analysis. Florida 1983. 81. The p property of 21(1978) 371-380.
FK
82. Semi-Fredholm maps of
McGraw
spaces.
FK
Commentationes Math
spaces.
MZ 144(1975) 9-12.
82A. From triangles to separated inductive limits. 469-479.
SM 31(1968)
A. Wilansky, K. Zeller: 83.
The inverse matrix in summability.
JLMS 32(1957) 397-408.
84. Abschnittebeschrankte Matrixtransformationen. MZ 64(1956) 258-269.
85. Bounded divergent sequences, topological methods. 78(1955) 501-509; 80(1955) 386. 85A. Banach algebra and summability.
TAMS
IJM 2(1958) 378-385;
3(1959) 468. 86. K. Zeller: 55-70.
Abschnittskonvergenz in FK-Raumen.
MZ 55(1951)
87. K. Zeller: Faktorfolgen bei Limitierungsverfahren. MZ 56(1952) 134-151. 88. K. Zeller, W. Beekman: Springer-Verlag 1970.
Theorie der Limitierungsverfahren.
89. M. Zippin: The separable extension problem. 26(1977) 372-387.
Israel J. Math.
90. A. Zygmund: Two notes on the summability of infinite series. Colloq. Math l(1947-8) 225-229.
309
INDEX
a
202
a
66
a-dual 65,195 ak = kth column of A ak = lim a n nk AB Matrix 202 B dua1,G 1 9 8 , 2 2 2 , 2 3 0 Criteria 211,212,215-217,230 227
1-1
Mercerian 214 Not AB 212,217 Perfect 202 Square 217
T 234 AB Space 160 Aspects 1 6 2 , 1 6 8 , 1 7 0 Closed 1 6 9 Criteria 1 7 1 228
y
Maps 1 9 8 Norm 169 Abel's identity,inequality 4 Absolutely convergent 3 9 , 8 9 Abstraction 210
AD
59
AK 172 Dual 1 0 6 , 1 0 8 Not AK 7 8 , 8 0 Admissible norm 93,143 AK (See AD) 59 AB 170 Aspects 8 0 , 1 9 3 , 1 9 4
Dual 106,107 Main Theorem 2 1 1 Monotone norm 104 Algebra (See Banach) 281-290 Algebraic dual 233 Almost coregular 218-220,227,234 Almost matrix 266-268,271-273 Analytic continuation 297,298 Applications 213,297-299 Approximate eigenvalue 289 Associative matrix 2 1 1 Associativity (See La) 7-9,154, 191-194
Asymptotic series 298 Attachment 6 2 , 6 8 , 8 7 , 9 7 , 2 1 9 b (See Reversible) B,B+ (See AB) 162 Aspects 1 6 2 , 1 6 6 , 1 6 8 Criteria 188 Growth 165,214 Formula 166 Inclusion 1 7 1 Norm 168 Not closed 1 7 1 , 2 1 5 Perfect part 2 0 1 T, Test functions 234 vs other subspaces 1 7 4 , 1 9 7 , 203,208,209,235,237
B(c),B(X) 273,279,281,283 B(X,Y) 52 Baire class 297 @-dual (See AB) 62,105
310
AK,BK 62,65 Closed(not) 65,158,167, 172,230,231 Inclusion 105,159 Multiplier 65 Norm,topology 62,65,159, sc 142 Second 105,172,174 vs other duals 80,106,107, 158,167,230,231 Banach algebra 2,52,281 Banach space 1 Banach limit 296 Banach-Steinhaus 1 Barrelled 165,299 Basis (See Weak) 211,212,293 BH space 55 Bigness (See Conull) 12,70, 90,100,102,156,212,258 Biorthogonal 293 BK 55 Bosanquet’s criterion 215 Bounded domain 7,258 Bounded consistency 43,88 220,255 Bounded divergent 47,48,90 Bounded sequences 17,49, 296,297 Bounded set 115 bs,cs (See Table, p. 116) 59, 65,139,142,143 bv,bvo (See Table, p. 116) 109-112,119,139,143 bvA (See YA) 150,196 c,co (See Perfect part, Table, p. 116) 1 BK 55,60 cA 291 C space 253 Closed 45,46,71 Complement 281
INDEX
Determining set 112 Dual 1,40,107,109 Dual map 267 Inclusion 120,138,174 Largest AD 139 Left inverse 283 Maps 268,281,283 Not closed 45,70,90,100 C space 253,257,259 X ( A ) (See Coregular) 6,lO Conull 155 Multiplicative 92,168 vsc 153 X(f) 1,40 Conull 71,146 Continuous 226 Fundamental formula 4.1,67,155 p 248 X(T) 266,272 cA ,ci (See AB,AK, Inequivalence, Perfect, YA) 3 BK 55,82,85,300 Bounded 7 Closed 7,64 Dual 66,82,199,291 FK 64 p 228 Possibilities 17,291,292 Separable 256,291 Ccr 31,217 (C,l), Ceskro (See Mercer) 7,30,31 AB 214,215 Commuting 24,273 Comparison 12,32 Fourier 298 Growth 11,44,214 Hausdorff 24,29,33 Maps 137 Monotone norm 105 idrlund 33
INDEX
Perfect 43 Regular 7 Spectrum 288 Cauchy sequence 299 c-like 197-199 Closed 64,65,233 Closed graph 51,252 Coercive 14 Almost coregular 219 Characterization 15,17 Conull 17,49,257,297 Separable 259-260 Spectrum 286 Weakly compact 272 Commuting 13,273,274 (not) Compact 150,251,252 Complements 69,70 Composition 8,9,282 Conservative
FK
13,
23,35,43,81,84
254
10,155
c closed 45 Inclusion, Invariance
Continuity (See p ) Coordinates 55,105,178 Embedding 55,159 Function 52,56 Matrix 57,229 Quotient 222 Row function 1 8 0 T , Weak 52,233,252 Conull matrix (Conservative in Chs. 1-8,17-19; vsc in Chs. 9-16) 10,155,271 Bigness 41,48,156,210 Coercive 17 Ideal 9 3 , 2 7 1 Inclusion, Invariance 41, 268
71,146
X
Domain 149 Inclusion 71,257,258 Miscellaneous 80,297 Two norm 94 W 87,99 Convergence domain = c A Convex closed 75 Convolution 32 Coordinates 40,55 Coregular matrix (Conservative in Chs. 1-8,17-19; vsc in Chs. 9-16)
Matrix 3,6 Sequence 27,223 Consistency (See Bounded)
Map
Miscellaneous 297 (Mod A) 231,243 Named 70 Perfect 63 Product 93 Conull space (1 c in Chs. 1-8 71,144 17-19; vsc in Chs. 9-16) Bigness 98,99
70,141
48,91,
311
41,48
68,156
LI
Mercerian 4 1 (Mod A) 241,243 Named 70 Tauberian 48 Coregular space (3 c in Chs. 1-8, 17-19; vsc in Chs. 9-16) 71, 145
Bigness 100,102 Domain 149 X 71,146 Inclusion 7 1 p
68,156
cs (See bs) cso 132 D
22,258
INDEX
312
FAK 1 6 1 , 1 7 3 Description 1 7 5 , 1 7 6 Main theorem 2 1 1 Maximal 175 aG 284-287 PMI 205 Determining set 112 FH 5 5 , 5 6 Direct theorems 97 Distinguished subspaces (See B , Finite norm (See @ ) F , P ,S , W) 160-164,187 Fixed point 298 Domain (See cA,uA,YA) 3 FK 5 5 , 8 9 Dominant diagonal 1 9 FK program 5 4 , 5 7 , 5 8 , 7 1 , 7 2 , 1 5 6 , 228,231 Dual (See f3,Y,f,c,cA,YA Fourier 294,298 reversible, row-finite) 1,107 Function as matrix 75,245 Dual map 221,267 Functional (See dual) 1 Fundamental 39
D,D(X) A 10 2 5
e
111
299
Existence domain = w A Equicontinuous 103 Equipotent 12,238 Equivalent 1 3 Extension 39,179-185
G (group) 284,285 G (c dual) 1 9 3 , 2 4 2 AB 1 9 8 , 2 2 2 Banach space 223 8-dual 222,226 c 222
F,F+ (See FAK) 161,162 Characterization 1 8 9 Closed(not ) 208 Formula 173 Inset 203,205 vs other subspaces 1 7 4 ,
r,ra
192,193,203,206,208 209,235,295
f-dual 105 AB 170 BK 108 Characterization 1 3 9 Inclusion 106,138 Norm 1 5 9 Quotient 107 Second 1 0 6 , 1 0 8 , 1 7 0 , 1 7 1 v s other duals 80,106-108 158,167,230
Factorization 239-241
9,269-273,289,290
Y-dual 6 5 , 1 0 5 - 1 0 7 , 1 5 8 , 2 3 1 Gauge 157 Generalized limit 2 , 3 , 2 9 6 , 2 9 7 Gliding hump 46 Graph 5 1 Growth sequence 5 8 , 8 6 Growth theorems 1 1 , 4 4 , 1 6 5 , 2 1 3 , 2 1 4 H 54 HU 22 Ha
31
Hp spaces 299 Hahn-Banach 39 Harmonic function 299 Hausdorff 22-32,196 Hilbert matrix 1 3 6 , 2 8 9 History 2 , 5 3 , 5 4 , 3 0 0
INDEX
313
HGlder 22,31,32,217 Hglder's inequality 111
Ell2 119,297,299 Lp (See Table p. 116) 111,139 L a (See Table p. 116) 1 I = Identity matrix 4 AK 251 Ideal 93,255,270,271 BK 55 Impossibility 299 C space 253 Improvement of convergence 298 Pseudoconull 258 Improvement of mapping 123,124 Quotient 291 Subspaces 99-101 In = Inset Inclusion 12,40,41,56,119,243 E i (See YA) 292 Inequivalence 140,196,211,231, L,Le,La 191-193,197,211,234 291-293 A 87,206-209 Inset 203 Left inverse Closed( not ) 208,209 Matrix 284,292 Internal 205 Reversible 84 Invariant(not) 205,206 Several 9 Maximal(not) 203,206,209 Tauberian 18,92,283,284
Intersection
48>59,60972,142,
146,260 Invariance 40,41,48,53,190, 243 Inverse (See Left,Right) 9, 12-14,19,84 Inverse closed 92,272
J matrix
220
K = scalars, R or C I, Kothe-Toeplitz (See a , B , Y ) Kobota-Petersen 36,37 L (See Table p. 116)
1
AK 59 BK 55 cA 291,292 Completion 255 Determining set 112 Dual 109 Inclusion 119,138 Subspaces 263 Q A (See YA) 195,196
Left topological divisor of 0 283 Lim 107 LimA 3,6 Local convexity Ltz 282,283
282,
119,297
M 22 u (Hausdorff) 22 p 41,225 Continuous 225-227,243 Formula 248 Invariant 225,227 242 246 space 228-229 u-unique 67 Characterization 248 Coregular 68,156 Invariant 225,227,243 Replaceable(not) 79,208,210, 246 Reversible 82,83,237 @
INDEX
314
Weakly
84,248 P 236-238 pA 241,242 M(X),M(X,Y) 64,65,160,169
Mapping (See Table p. 116)
Stronger 247 Ngrlund 32-38,44,287,289 Norm 1,5,40 Not closed (See 6,c) 166,208 Null space 275
14,231
Markov chain 298 0 , o 11,13,21 Matrix (See Continuity) 8,9,282 0; 95 Maximal group 284,285 w (See Table p. 116) Maximal subspace 75,221 BK 58 Mazur's matrix 20,215 Continuity 51 Mazur-Orlicz 88,220,255 Dual 140 Mercerian matrix 14,19,30,36, FK 51,55,60
3
42,92,285
f-dual 140 Maps 75,85 Subspaces 58
Mercerian theorems (See AB) 30,41,214,216,276,285
Mercer set 31 Mercer's theorem 30,216 Metric 89 MI',Mra 277 Moment problem 27 Monotone 164,244 kmotone norm 104,168 Multiplication (See Attachment, Product ) XY X-Y
3 62
Multiplicative Function 92,168,270,271 Matrix 3,277,278 Operator 277-279 Multipliers (See M(X).) NT 275 (N,p) (See Ngrlund) Naming program 53,70,79 Non-replaceable Example 78 Conull 204 p-unique 208
P 236
uA
3,62,293
n(r),nb(r) 95 BK 96 Conull 98 Inclusion 98,99 Subspaces 198,217 0 matrix 220 Onto 17,268 Open map 103 Operator (See Table p. 116) 265, 282
Oscillation (See n , V )
P 235-238,244 Pn = nth coordinate $,$1 (See Perfect part) Q
3,51,79
5,8,101
Perfect 41,72,201 AB 202 Consistency 81 Conull 63 Type M 42,81,84,86,202 Perfect part 41,42,72,201 P-Perfect 237
INDEX
PMI 205,211 Polar 221,251 Polynomial matrix 32-38 Principal diagonal 19 Problems 299 Product (See Multiplication) Programs (See FK,Naming) Pseudoconull 258 Pseudometric 251
cA 17 Matrix 292 Onto 268 Reversible 83 R-invariance 249,250 Row finite 3,8,61,85-86,195 Row function 179-183 Row sums of inverse 12 S
Q
4,38 Growth 11,44 Hausdorff 24 Ngrlund 33 Sections 182 Subspaces 214 Toeplitz basis 294 Type M 43 vs(C,l) 12 Questions 299 Quotient 107,157,222 p 269-273 Range closed 75,263 Reflexive 89,101,263 Regular FK space 77,209 Matrix 4,6 Sequence 27 Replaceable (See non-) 76 By triangle 238,245 Characterization 207,245 F = B 204 FK 247 Inset 209 246 Representation (See Dual) 199 Reversible 82-84,219 Right inverse (See Type M)
315
161,163,170,211,213
SA (See YA)
o(X,Y)
251 u ( z ) = spectrum of z SAX 87,176,211 sc = semiconservative Scalars = R or C Scalar homomorphism 52,273,279, 281 Schauder basis 177 Schnee and Knopp 31 Section 59 Sectional boundedness = AB Semiconservative 141,147,151, 152,174 Semi-Fredholm 275 Seminorms 51 Separable 89,233,255-258 Sequentially complete 251,255 Sgn 4 Silverman-Toeplit z 7 Smallness 49 Smoothing 172 Spectrum 281,287-290 Stone-Cech 296,298 Strong topology 221 Stronger 12 Strongly conull 79,80 Submatrix 101 Sum 221,233 Summable 3
INDEX
316
aup 5 2 , 2 5 1 Superdiagonal 8 1 T
251,252
T 234 2bv,2-, 2p ( See Two-norm)
Uniform convergence
6,15,16,3.7,79
Uniqueness (See Invariance)
V
56
156,252
Variational 145 Variational semiconservative
145,
151,153
2N
297
Tauberian map 274-277 Talkrim matrix 1 8 c closed 4 6 , 9 1 Characterization 276 Coregular 48 G
285
Hausdorff 30 Left inverse 1 8 , 9 2 , 2 8 3 , 2 8 4 Rtz 283 Norlund 38 Range closed 264 Tauberian Theorems 1 1 , 1 3 , 2 2 Test functions 234,235,244 Toeplitz basis 294-296 Topological divisor of 0 282 Topological vector space 2 Totally decreasing 26 Transpose 123,124 , 1 2 5 , 2 3 1 Triangle,Triangular 7,55,245 Trigonometric functions 294,298 Two-norm 93,95,143,144 Type M (See Perfect) 42 Density 264 Not important 8 1
P
233-235
Right inverse 44,236
V-Cauchy 252,253 Very conull 218-220 vsc = Variational semiconservative W,Wb (See Two norm) AB 1 7 0 Aspects 163,198 C space 254 Characterization 190 Closed 208,210 Conul1,Coregular 99,210 Dense subset 258,259 Inclusion 260 Main Theorem 2 1 1 Pseudoconull 258 vs other subspaces 192,206, 235,255,295
w-almost matrix 270-273,277 Weak( ly) Absolutely summable 1 3 8 , 1 3 9 AK (See SAK) Basis 177,178 Bounded 115 Cauchy 1 6 1 Closure 75 Compact 2 7 2 , 2 8 1 Convergent 1 7 , 1 8 7 , 1 9 4 - 1 9 7 , 263
Unbounded sequences 4 5 , 4 7 , 7 0 , 72,214
Unconditional convergence 140 Unconditional sectional boundedness 1 2 0 Uniform boundedness 1
p-unique 8 4 , 2 5 0 Sequentially complete 197 Topology 2,185 Weierstrass approximation 298 w-matrix 2 7 0 , 2 7 2 , 2 7 3 , 2 7 7
INDEX
X-continuous
180
317
11
bv
u
107
I
42,75,221,233
log > 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
