Atomic Boolean Subspace Lattices and Applications

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MEMOIRS -i-VA

of the

American Mathematical Society Number 445

Atomic Boolean Subspace Lattices and Applications to the Theory of Bases S. Argyros M. Lambrou W. E. Longstaff

May 1991 • Volume 91 • Number 445 (second of 4 numbers) • ISSN 0065-9266

American Mathematical Society Providence, Rhode Island Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

1980 Mathematics Subject Classification (1985 Revision). Primary 46B15, 47C05; Secondary 47D30, 06E15. Library of Congress Cataloging-in-Publication Data Argyros, S. (Spiros), 1950Atomic Boolean subspace lattices and applications to the theory of bases/S. Argyros, M. Lambrou, W. E. Longstaff. p. cm. - (Memoirs of the American Mathematical Society, ISSN 0065-9266; no. 445) "May 1991, volume 91, number 445 (second of 4 numbers)." Includes bibliographical references. ISBN 0-8218-2511-9 1. Banach spaces. 2. Summability theory. 3. Bases (linear topological spaces) I. Lambrou, M. (Michael), 1952- . II. Longstaff, W. E. (William Ellison), 1945- . III. American Mathematical Society. IV. Title. V. Series. QA3.A57 no. 445 [QA322.2] 510s-dc20 90-7545 [515'. 732] CIP Subscriptions and orders for publications of the American Mathematical Society should be addressed to American Mathematical Society, Box 1571, Annex Station, Providence, Rl 029011571. All orders must be accompanied by payment. Other correspondence should be addressed to Box 6248, Providence, Rl 02940-6248. SUBSCRIPTION INFORMATION. The 1991 subscription begins with Number 438 and consists of six mailings, each containing one or more numbers. Subscription prices for 1991 are $270 list, $216 institutional member. A late charge of 10% of the subscription price will be imposed on orders received from nonmembers after January 1 of the subscription year. Subscribers outside the United States and India must pay a postage surcharge of $25; subscribers in India must pay a postage surcharge of $43. Expedited delivery to destinations in North America $30; elsewhere $82. Each number may be ordered separately; please specify number when ordering an individual number. For prices and titles of recently released numbers, see the New Publications sections of the NOTICES of the American Mathematical Society. BACK NUMBER INFORMATION. For back issues see the AMS Catalogue of Publications. MEMOIRS of the American Mathematical Society (ISSN 0065-9266) is published bimonthly (each volume consisting usually of more than one number) by the American Mathematical Society at 201 Charles Street, Providence, Rhode Island 02904-2213. Second Class postage paid at Providence, Rhode Island 02940-6248. Postmaster: Send address changes to Memoirs of the American Mathematical Society, American Mathematical Society, Box 6248, Providence, Rl 02940-6248. COPYING AND REPRINTING. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy an article for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication (including abstracts) is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Manager of Editorial Services, American Mathematical Society, P.O. Box 6248, Providence, Rhode Island 02940-6248. The owner consents to copying beyond that permitted by Sections 107 or 108 of the U.S. Copyright Law, provided that a fee of $1.00 plus $.25 per page for each copy be paid directly to the Copyright Clearance Center, Inc., 27 Congress Street, Salem, Massachusetts 01970. When paying this fee please use the code 0065-9266/91 to refer to this publication. This consent does not extend to other kinds of copying, such as copying for general distribution, for advertising or promotional purposes, for creating new collective works, or for resale. Copyright © 1991, American Mathematical Society. All rights reserved. Printed in the United States of America. The paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability. @ 10987654321

9594939291


TABLE OF CONTENTS

1.

Introduction and preliminaries

1

2.

Quasi-direct sums

9

3.

Strong rank one density property

20

4.

Meshed products

30

5.

Strong M-bases

43

6.

Selecting and slicing

69

7.

The double commutant

76

References

89

Addendum

93


ABSTRACT

Those families of (closed) subspaces of a (real or complex) Banach space that arise as the set of atoms of an atomic Boolean algebra subspace lattice, abbreviated ABSL, are characterized.

This characterization

is

used to obtain new examples of ABSL's including some with one-dimensional atoms. M-bases.

ABSL's with

one-dimensional

atoms

arise

precisely

from

strong

The strong rank one density problem for ABSL's is discussed and

some affirmative results are presented.

Several new areas of investigation

in the theory of ABSL's are uncovered.

IV


1.

Introduction and preliminaries

Throughout the following 'Banach complex Banach space, 'subspace'

space'

will

mean

closed

will

'operator' will mean bounded linear operator. a Banach space, the set of operators on set of subspaces of

X

is denoted by

X

either

linear

real

subspace

Also, the letter

is denoted by

X

or and

denotes

JB(X)

and

the

denote

the

set

£(X) .

Z C £(X)

As usual, for a non-empty subset of operators on

X

mean

we let

Alg #

which leave every member of £

invariant, that is

Alg £ = { T e 5B(X) : T M c M , for every M E £} . Dually, for a non-empty subset invariant subspaces of

d C S(X)

we let

Lat d

if

Z c Lat Alg 1

£ = Lat Alg £ .

for any

iSc^(X).

This notion is due to Halmos [9] ,

the earliest results in the present context:

d).

T e

£

We say that

terminology 'reflexive algebra' is also attributed.

to

Halmos

reflexive

is whom

the

found

one

Every atomic Boolean

subspace lattice on complex Hilbert space is reflexive. given below.)

of

d , that is

Lat d - { M e £(X) : T M c M , for every Obviously

denote the set

dual of

algebra

(Definitions

are

Throughout, the abbreviation ABSL is used for atomic Boolean

algebra subspace lattice. Aims of the present work include an investigation of how ABSL's arise, a study of their geometric properties and the provision

of

new

examples.

Also, density properties of the finite rank subalgebra of Alg £ are discussed. In the case of one-dimensional atoms, the question of the density of the finite rank subalgebra in the strong operator topology is shown to be

Received by editor June 2,

198 9

1


ARGYROS, LAMBROU and LONGSTAFF

2

equivalent to a long standing open problem in the theory of bases. Section 2 includes a characterization (Theorem 2.4) of those families {L }_, of non-zero subspaces of 7 I

X

which arise as the set of atoms of some

ABSL on X . A necessary and sufficient condition is simply that quasi-direct

sum

of

{L }

in

G.E. Kisilevskii [3]. For any ABSL Alg £

the

sense

of

M.S.

X

be the

Brodskii

and

J£ the set of finite rank operators of

is shown to be pointwise dense in Alg L ?

L9

are (abstract) complete lattices, a mapping

is called a complete

homomorphism

L 9

be a complete

sublattice

L„ .

of

the atoms of

If

L- is an atomic

(p(L-) are those

space

Let

elements

X . There exists

of atoms if and only PROOF:

Suppose that

if

Then

1

Y J

implies that the subspaces r

L C L' . Thus for disioint subsets AJ 7 /i 1 V. L and V4 (L') are orthogonal. In A± 7 A 2 /i

L and V (L') 1 7 -1 2 2^ 1 7 ^

are orthogonal and so are "


V L l 17 7

J

and

ABSL's AND THE THEORY OF BASES

Let x

x be a vector belonging to the left hand side of (1). Then

= y-L + y 2 = zi + z 2 yie

19

\

\

•

?i -

z

y

with

2

e

\

x

^

• zie

\

L

7

a n d

z

2

e

V

J2(L;)X

Thus i

=

z

2 - y2

e

(V

AL7>

n

(v

r\A(L;)X) -

(0)

•

So x = y + z where y - yx - z t E ( v ^ ) n ( v ^ y - v ^ ^ L 7

and

z - y2 - z2 e ^ a / ) n (v^a;)1) - v^ n ^a;) 1 , using Theorems 2.4 and 2.8.

This completes the proof.

m


•

3.

Strong rank one density property £

A subspace lattice strong

rank one density

on a Banach space X

property

is

said

to have

if the algebra generated by the

the

operators

of Alg £ of rank one is dense in Alg £ in the strong operator topology. algebra generated by the rank one operators of Alg £ finite sums of the form

2 R.

where each R.

most one, belonging to Alg £. Alg £, £

is

the

set

is an operator, of

The

of all rank at

Since this algebra is a (two-sided) ideal of

has the strong rank one density

algebra contains the identity

operator

property

on

in

e > 0

closure, that is, if and only if for every x-.xx of vectors of X 1 2 m described above such that

X

if and its

only

strong

if this operator

and every finite set

there exists a finite sum £ R. j

of the form

llx. - (E R.) x.ll < c , for i = 1,2, ... ,m.

11

l

j

Which subspace lattices

i"

£

have the strong rank one density property?

The answer is not known. This question was first raised for Hilbert spaces in [20] where it was distributive.

shown

that

£

such

This result was extended

are

necessarily

to normed

finite-dimensions complete distributivity complex, separable Hilbert space totally

is

spaces

also

ordered

completely

in

[15].

In

sufficient

[20].

On

subspace

lattices

and,

more generally, completely distributive commutative subspace lattices have the strong rank one density property, by Theorem 2] and by [11, Corollary 6.1]

and

the

Erdos

density

[17, Theorem

(actually, Erdos' theorem holds in non-separable

spaces

theorem

3] respectively as well).

every ABSL on a Banach space is completely distributive since a Tarski

[29] says

that

an

(abstract) complete

completely distributive if and only if it have the strong rank one density property?

is

Boolean

atomic.

Corollary 2.6

20


[5,

Does

Now

result of

algebra

is

every ABSL

says

that they


21

all have a somewhat weaker density property, but the full answer

is still

unknown.

the case

In fact, a very special case of this question, namely,

where all the atoms are one-dimensional, was raised by Ruckle in 1974 [27] and is still open (actually, in [27], the question is raised in a different way, but we show that it is an equivalent way). Much of what follows is concerned with the strong question for ABSL's.

Immediately

affirmative for ABSL's with

below

two atoms.

we show

rank

that

For complex,

one density

the answer separable

spaces this was first proved by Harrison [10]. His previously proof, using the spectral theorem, which leads to an even

is

Hilbert

unpublished

stronger

result

is postponed to a later section (Theorem 4.5). THEOREM 3.1. the strong PROOF: such that

Every ABSL on a Banach space with precisely

rank one density Let L

and M

atoms

has

property. be two non-trivial subspaces of a Banach space

L n M = (0) and L V M = X . Let

{L,M} . Let e > 0

two

£

be the ABSL

with

X atoms

and vectors

x-,x n ,...,x of X be given. We are to 1 I m show that there exists a finite sum F of operators of Alg £ of rank at most one such that llx. - Fx.ll < c , for i = l,2,...,m . 11 l

Let

i"

N be the finite-dimensional subspace spanned by

Assume for the moment that {y-\ »Yo» • • • 'Yi^

De a

{x..,x9,...,x } .

L n N ^ (0) and M n N ?* (0) and let

basis of L n N

and let {y, - ,y, 9 , . . . ,y«) be a basis

of M n N .


22

ARGYROS, LAMBROU AND LONGSTAFF

The set

N

= {y-,y9,...,y«}

is linearly independent since if

2 A.y. - 0 1 X L then

2 A.y. = x

i

l

2 A.y.

k+i

x

l

But the vector on the left hand side of the equality belongs to other belongs to

M . So each belongs to

L n M = (0)

L

and the

and so

t

k

S A.y. - - S A.y. = 0 . -i i i i -i i i 1 k+1 It now follows that Now extend moment that

N

A. - 0 , for to a basis

I < n .

i - 1,2,...,£ .

l

{v*,yrt,...,Y }

Since every

x.

of

hL We build up this

F

F

e

,

denote

it

for y.'s

the it

by

as described earlier with

~ FyJI < e' , for

i - 1,2,... ,n .

in stages.

First observe that, for every span of

and assume

is a linear combination of

is enough to show that, for a suitably adjusted e' ,there is an operator

N

{y. : l < j < n

, j * i)

1 < i < k , y. UM

.

y. G v{y. : l < j < n , j / i ) v M = Then, for suitable scalars

does not belong to

For, suppose that v{y.

1 < i < k

the and

: l < j < n , j ^ i } + M .

{/i. : 1 < j < n , j # i)

and vector

would have y. = .S./i.y. + z ,

y. - .S.u.y. e N


z e M

we

23


Thus

(y. : k+1 < j < 1}

z G M n N . Since the latter has basis (A. : k+1 < j < 1}

are scalars

, there

such that I

y. — .E.u.y. = ,k ?1! A.y. y i

3*1."3 3

+

JJ

and this contradicts the linear independence of that y.

{y. : 1 < j

0 , every finite set

every set

y 1> y 9> ---,y 2 S. J

with

£

X

*

rank •k

one

be the

R e Alg £ it is

7Gr . So, since every

(L') ' s , we have that

E S. , with J

belongs

to

k

S. J

R

k

an operator

S(X ) , it is enough the

weak

operator

is reflexive, it is enough to show that *

*

x-,x~,...,x

of vectors of S. J

strong

I

of rank at most one, is a convex subset of

to show that the identity operator on

of the form

the

property.

has rank at most one so does

Since the set of finite sums of the form •k

7 r

for every

is the closed linear span of

density

{L }„ and let

(see Theorem 2.8).

* 1 i R (L') C (L')

. Additionally, if

rank one

also has

{ (L') }

be the ABSL on

easily verified that element of

with

X

*

of vectors of

X , there exists a

as earlier described, such that

|(x. - Fx.) y.l < e , for i = 1,2,...,m .


finite

X

'k

sum

and F

26


Now, by hypothesis, there exists a finite sum with each

R. E

Alg £

G

of the form

2 R.

of rank at most one such that

|x.(y. - Gy.)| < e , for i = l,2,...,m . But, for each

1 < i < m , |x

and so we may take

F =

i(yi ~ G

Gy

i}|=

= £ R. .

|(x

i " G*Xi}

y

i'

H

We defer until Section 4 (Theorem 4.3) the proof of the following: {L }

is the set of atoms of an ABSL with

the

strong

property on a Hilbert space, then, for every subset atoms

{L }.U{(L') }_. -y A *y 1 \A

also

has

this

rank

A C T ,

density

one

If

density

the ABSL

with

property.

The

one-dimensional version of this result is discussed in Section 5. We have already remarked that a subspace lattice X

0

and every finite set

a finite sum

F

x-,x0,...,x of vectors of X there exists 1 z m of operators of rank at most one of Alg J£ such that ||x. - Fx. II < e , for i = l,2,...,m. 11

l

l "

We have also remarked that distributivity.

this

vectors

property

The following result says that, in

latter condition, an operator the

density

x 1 ,x

...,x

F

implies the

complete

presence

of

the

of the required form exists provided that

satisfy

a

certain

strong

form

of

linear

independence.

THEOREM 3.4. Banach

space

M. n M. = (0) l

j

Let X .

£ If

whenever

be a completely

distributive

subspace

lattice

x,,xn,...,x are non-zero vectors of X and 1 z m i ^ Jj , where M. = n{L E £ : x. e L} , then I


l

on

a

for


every

e > 0

one of

there

Alg £

exists

such

a finite

sum

F

27

of operators

of

rank

at

most

that llx. - Fx.ll < € , for i - l,2,...,m . 11 1 i"

PROOF:

M. E £

By completeness

and by complete distributivity we have

M. n (.V.M.) = .V.(M. n M.) = (0) , for every

i = l,2,...,m . (Since

the

are linearly independent.)

x.'s Let

e > 0

be given.

x.E M. , this shows

For each

i

satisfying

in

particular

1 < i < m

,

that

complete

distributivity implies (see [19]) that M. = V{L E 1 : M. & L_} L E £ ,L

where, for every

is given by

L_ = v{K E £ : L d K) . Thus, as x.E M. , t h e r e e x i s t non-zero subspaces {L.. r 1

such t h a t {f..

1

lj

L . . CM. ^ ( L . . ) l

lj

: 1 < Ji < n . )

of

i

ij

ij -

X

, for

i = l,2,...,n., i

such t h a t

f..

ij

EL..

ij

: 1 < Ji

r

However i t is not true that some subsequence

{P

"k

oo

the sequence identity.

{P }.

oo

}-

the of

of natural projections converges strong operator to the

Indeed Menshov [22] (see also Bari [1, p.354]) has

continuous function which is not the uniform limit of the sequence of partial sums of its Fourier series.

constructed a

any subsequence of (That

is,

the A 's

in (C6) cannot be replaced by ones.) The authors

thank

Professor

S. Pichorides

for

pointing

out the

reference to Menshov's example. Below (see remarks following Theorem where no subsequence of

the sequence

5.8)

we give

of natural

another

projections

strong operator to the identity operator, even though there

example converges

is a net of

natural projections so converging. The following corollary extends Corollary 3.1 of Markus [21]. COROLLARY 5.3. Let vectors {f }

{f }

X

be a reflexive

is a strong H-basis of

is a strong K-basis of

Banach space. X , then

its

If

the

family

biorthogonal

X


of

family

49


PROOF:

By the preceding theorem, {f }

is the set of atoms of

an ABSL

on X . Since ( ' ) X = , for every Theorem 2.8 shows that

{}

7 G T ,

is the set of atoms of an ABSL

The desired result now follows from Theorem 5.1. Clearly the condition

of

reflexivity

on

X

a

cannot be

omitted

from the

statement of the above corollary. A trivial example is any (Schauder) basis of I

. The span of its biorthogonal

sequence

cannot be

dense

I

in

since the latter is not separable. There is a dual version of the above corollary, coming 'down' from the dual.

Here reflexivity, as expected, is not required.

If

{f }

is a strong

M-basis

COROLLARY 5.4. and there {f } PROOF:

is a strong Let

M-basis "k

{f }

is a family of

of

of vectors

which is

X

of a Banach space

biorthogonal

~k

biorthogonal to

i(

natural isometry.

it,

then

X .

be a strong M-basis of X

The family biorthogonal to

to

X

{f }

It is clear that

(f }

~k~k

is

{?r(f )} {f }

where

n : X -> X

is complete

is the

and minimal.

We

establish condition (C2) of Theorem 5.1. * Let I c T and let x G n T ker f . We show that x G v_. T f . Let I 7 r\I 7 f G X satisfy f (v_v f ) - {0} and let e > 0 be given. By (C6) •k •* applied to {f }„ , there is a finite sum S A (n(£ ) ® f ) such that 7 T 7 7 7 || f* - (E A7(*(f7) ® f*)) f* 1 < e . Thus I f*(x) - E A^ f*(f )f*(x) I < e


50


But the sum inside the modulus signs is zero since, for f*(x) = 0

and for

7 e T\I

we have

f*(f ) = 0 .

So

7 e I

we have

| f*(x) | < e || x ||

and we deduce that

f (x) = 0 . An application of the Hahn-Banach theorem

gives

as required.

x G V .

f

Thus n_ ker f* c v f I 7 T\I 7 and since the reverse inclusion is obvious the proof is complete.

B

Notice that by using Theorem 2.10 instead of Theorem 2.8 in the of Corollary 5.3 we can recapture Lemma 3.1 of [21] which {f }„ 7 T

states

is a strong M-basis of a Hilbert space, then so is °

for every subset

A C T .

proof

that

if

*

{f }A U 7 A

{f }_,. . 7 r\A

Several other corollaries to the results of

previous sections can be obtained by using the equivalence M-bases and ABSL's with one-dimensional atoms proved in particular, the one-dimensional analogues of

our

between

Theorem

results

the

strong

5.1.

In

concerning

the

transmission of the (metric) strong rank one density property provide other corollaries. As summable

already

mentioned

M-bases.

Strong

Ruckle M-bases

[27]

calls

whose

strong

M-bases

corresponding

one-dimensional atoms, see Theorem 5.1) has the

strong

rank

1-series

ABSL one

density

property are precisely Ruckle's finitely series summable M-bases. example if

{f }

is a finitely series summable

Banach space, then its biorthogonal family

{f }

M-basis is

a

of

Hilbert space

H, {f }A U {f }„, A 7 A 7 r\A

H , for every subset

A C T

Thus for reflexive

finitely

summable M-basis of the dual space (this follows from Theorem "k

a

(with

3.3);

series on

a

is a finitely series summable M-basis of

(by Theorem 4.3) .


51


It is an open question whether or not every 1-series summable is finitely series summable.

In

other

words,

whether or not every ABSL with one-dimensional

it atoms

is

an

has

open the

M-basis question

strong

rank

one density property. £

Given a subspace lattice

on a Banach space

X , the following

question is stronger than the question of strong rank one density: Does there exist a uniformly bounded net of finite rank operators, each a finite sum of operators of AlgJ£ of rank at most one, converging to the identity operator in the strong operator topology? affirmative, then

X

Clearly (see [18,p.37]) if the answer is

has the bounded approximation property.

However, even

for an ABSL on a separable Banach space with the bounded approximation property the answer can be negative as we now explain. If Banach space, the unit ball of topology.

S(X)

X

is a separable

is metrizable in the strong operator

Using this and the principle of uniform boundedness the question

in the preceding paragraph, for ABSL's with one-dimensional atoms on a 00

separable Banach space becomes: If {f }

is a strong M-basis is the identity

operator a strong operator limit of a sequence of finite rank operators (F }

of the form N

=

F

n

s

n

/ \

(n) A:

l

;

*

(f. ® f.) i

.

l

In [27] strongly

strong M-bases for which the answer is affirmative are called series summable. In [4] Crone, Fleming and Jessup give an example

(Example

4.24) of a separable Banach space

there

is a series summable strong M-basis

series summable. property.

Since

E

E

00

with a basis {e }_ 00

(f }-

in which

which is not strongly

has a basis it has the bounded approximation

For the ABSL with atoms (}

CO

, the answer to the question in the

f see ADDENDUM


52


preceding paragraph is negative. This example of [4] also shows that an ABSL with one-dimensional atoms on a Banach space with the metric approximation property need not have the metric strong rank one density property.

For, there is an equivalent norm

||| • ||| on E

such that

(E, III • |||) has the-metric approximation property (indeed, if the sequence of natural projections associated with the basis | | | x | | | = sup ||P x|| defines such a norm [18,p.2]). (E, III • |||) {f K summable.

Thus on

{P }- is

{e K

, then

In the Banach space

is still a strong M-basis which is not strongly series (E, |||« |||) the ABSL with atoms

{}.

does not

have the metric strong rank one density property. The following improves the example of [4] somewhat. An M-basis {f K of a Banach space X

is series

summable if and

approximation property and every nuclear map f (Tf ) - 0 (n € Z+) n n

T

only on X

if

X

has

the

satisfying

has trace zero [27, Theorem 1.2 D] .

EXAMPLE 5.5. There exists a separable Banach space

X

with the

approximation

property

and with a series summable M-basis but which possesses no strongly

series

summable M-basis whatsoever. We use the powerful constructions of [7,14],

To be specific,

in

[14]

00

Johnson constructs a sequence

{X K

of

finite-dimensional Banach

spaces

such that the space Y- (X1@X2©

...)(l

* has a basis and the metric approximation property but whose dual space fails the approximation property.

We proceed to define


X

following

Y the

53


example of

[7, p.199].

isomorphic copy

Y

By Theorem 1 of [7], for each n G Z

, Y

failing the n-approximation property.

has

Define

an

X by

X- (Y1@Y2® ...)f2 . This

space has

the

approximation

approximation property. operators on X topology.

property

of

strong operator to the identity.

{F K n 1

with

contradiction.) Hence X We now show that X

X

has

operators

the on

bounded rank

the

II F — " n

X

converging

{K }- was such a

approximation K

approximation

00

This is so because if

|| -» 0 n "

property, we would

finite get

a

admits no strongly series summable M-basis. has a series summable M-basis. For each

be an isomorphism. F

Consider the family

since

compact

sequence then choosing, by virtue of

let T : Y -» Y n n

the

converging to the identity operator in the strong operator

(A little more can be said:

operators

fails

In particular there is no sequence of finite

property, there is no sequence

rank

but

00

Let

{e.K j 1

be

a basis

{f.. : i,j G Z } of vectors of X

n G

of

Z

Y

given by

f. . - (0,0, . . . ^ T T 1 e.,0, . . .) i

ij

where the non-zero entry is at the

i-th

verified that the family biorthogonal to

j'

coordinate. {f..}

is

It can

easily be

"k

-f-

{f.. : i,JG Z } where

f*. - (0,0,...,0,T*e*,0,...) ij

•fc

where

{e.K

00

is the family biorthogonal to

M-basis is obvious. To show that S

ij

CO

{f..}

is

{e.K . That series

summable, let

= (S..) G S(X) be a nuclear map satisfying

Then

{f..}

f.. (Sf..) = 0 , for every i, j G Z + . 0 - T. e. (S..T. e.) - e. (T. S..T. e.) i J ii i J J 1 ii i J


is an

54


for all

i

for every

and

i j

00

so, as

{e.h J 1

i . Hence also trace S.. — 0

2.trace(S..) = 0 l n

. Thus

— 1

is a basis of

{f..} lj

is

Y , trace(T.S. .T. ) = 0 l n l '

for every

series

i , and so trace S =

summable

and we are

finished. Next we discuss a pointwise boundedness condition.

In

the

theorem, we use the fact that for an ABSL £ , Algi£ is abelian if £

if every atom of

biorthogonal constant

of

|| F || < K(x) belongs of

such

the form

and

PROOF:

that

for

for

M-basis every

every

operator

,

vector

(a finite

then

the

of

the set

closure

there

x G X there

X

with

exists

a finite

sum)

satisfying

identity

operator

of finite

exists

rank

a rank

on

X

operators

S A (f ® f ) . 7 7 7

We claim that there exists a constant

that for every

e > 0

F = S A (f* ® f ) For

only

of a Banach space

e> 0

* E A (f ® f ) 7 7 7

|| x - Fx || < e

to the strong

the form

. If

{f }

K(x) > 0 F

be a strong

{f }

family

operator

and

is one-dimensional [16].

Let

THEOREM 5.6.

following

n G Z X

such that

x G X

|| F || < K

and

> 0

with the

property

there exists a finite sum || x - Fx || < e .

let = {x G X : inf || x — Fx || = 0 , the infimum being taken over all

By J hypothesis

and every vector

K

F

of

the form

U- X = X . Now each n=l n

2 A (f ® f ) with || F || < n } . 7 7 7 " " X is closed. For, let n

00

{x, h c X k

> 1

F = S A

with

x, -> x

such that (f ® f )

and let

e > 0

be arbitrary.

|| x - x, || < €/2(n+l) . For this such that

| F || < n

and

|x

There exists x,

there exists

- Fx, || < e/2 . Then



|| x - Fx || ^ || x - x k

showing that

x G X

|| + || x k - F x k I

+

55

1 F || || x k - x || < e ,

. By the Baire category theorem there exist x G X ,

such that || y — x || < 8 implies y G X^ . ? Let K = 2N + N and let 0 * x G X and e > 0 be arbitrary.

N e Z

and

Clearly

8 > 0

y = -n—n- + x

each of the form

satisfies

|| y — x

2 A (f ® f ) with

|| < 8

|| F || < N

so there exist and

|| G

F and G

|| < N

such

that

I " - F> y I < i ' m ' Since

I — F and (I-F)

I —G (I-G)

M

«* I » " v*o I < I ' fe • M •

commute we h a v e x

|| - M

|| (H)(I-G)(y-x)

^ H
0 be given.

Choose

Fn , 1

of

IIFJI " 1" < Ko and

|| (I-F1)x1 || < 6 / (1 + K Q ) m . If

F- ,F„,...,F

step

L

Z

n

(1 < n < m) have been chosen, chose

F - (of the form

(considering b the vector II

(I

S A (f ® f )) such

that

for

the

II F , -

inductive

|| < K and

(I - F )(I - F -)...(I - F-)x n ) n n-1 1 n+1

- F n + l ) ( I - F n> ( 1 " Fn-1> ' • " (I " Fl>Xn+l » < e^1+Ko^


'

56


Then, for

1 < i < m we have

I ^-V^-W-^-Vi 1 £

ll(I-V(I-Fm-l)--- 0

i

if

projection

{f }..

and only if

and every finite

a natural

°°

set P

oP

if and only if {a K £ V n1

The sequence

(1 < p < co)

(£ )

defined

. Since

n ker :f = {{£.}-£ U : for some scalar A , £. = Aa. 1

THEOREM 5.7.

^0

(defined

that

|| x. -

i G

. Even more

of vectors P

x.

case,

of

||
1

|ai

'W |P ... +

+

|a £+1 |P

+ . . .+

l a l l P + •••

+

l a £ + ll P

l a l l P + •••

+

l a k + ll P

|a

-» 1

"k+11 l | P + - - -+

as

|a

k|P

k -> oo

so the Cauchy criterion fails. Let

e > 0

is a natural projection of

m

x = {£.}-, e tr

be arbitrary and let

vectors

x-,x 9 ,

P

such that ,x m

will

. We show that there

|| x — P x || < e . follow

from

this

(The general case case.)

Routine

calculation shows that

X - P

P

n

X || =

"

M

Kn+1

1

•*

n+1

J

2 |£.| P < eP/2 , for every n+1 J we claim that there exists an N > M such that

Now there exists For this

f |a 1 | P +...+|a n |P

M G I

such that


(1)

n > M .


(

P P a |a.| + ... i T . . .+-t-|a„| a_7

1

N

,.N l , P

|a

+

N

'W 0

••'+ lanl

n > 2 ,

1 e - a — n n a1 so it follows that

l'

in Theorem 5.7

is

a

basis

if +

•'•

+

|a

n|P

1 . Let

> r > 1 . Then, by the

there is a positive integer

N

r satisfy

definition

such that for every

of

limit inferior,

n > N we have

, **+! . ^

I -z— I > r •



So for

63

n > N we have

|a.| F + ... + |aJ

F

|a.| F + ... + |aJ N'

Kn+1

' a N + l ' P + ••• + ' a n ' P

"Vi"

1

n+11

"I r P + r 2p + "•+ r (n-N)p J which is uniformly bounded (the geometric progression is bounded by —1

D

(rr — 1)

) . Hence

oo

{f K n i

is a basis.

Conversely, suppose that lim sup lim sup fixed

a

n+l a n

a

n+l a n

e> 0

la i n+;i_.1 i
1 by the ratio test. We show that we cannot have 1 . On the contrary if this were the case, then for any there would be a positive integer

^ 1+e 'lan 'I , for every J

1

such that

n > N . Then, for n > N we would have

| a i | P + ... + |an|P

Kn+1

N

|aN|P +

iW l

•£(*) )b-m

)j_N+1

n+l-N


]•


64

Since v

n+l-N

it would follow that, for all sufficiently large

n

l * , r + ••• + I*

" *k

"W

This contradicts the fact that

|a

sup n

l|P

+

••'

+

|a

n|P

The two implications of the preceding For example, if n+1

lim sup basis

a9

(by

oo

- =1

and

a9

theorem

are

the

first part of the theorem). lim inf

n+1

exists, then the theorem shows that limit is strictly greater than one. In the case of i1 - « is also injective).

S 0 -T I

Hence

A

r

h+2

«2'

2 ?2

3 ? 3''"'

;

is injective with dense range. Define

subspaces L. (i = 1,2,3) by L

= {(x,0,0) : x e £2 © £2 }

L 2 = {(x,Ax,0) : x G l2 © l2 ) L 3 - {(x,Ax,Ax) : x G I2 © I2}

and

By the preceding lemma, (i) holds. Define linear manifolds by K± - {(x,0,0)

:x G i

© (0)

K 2 - {(x,Ax,0) : x G G(T) } , and

K 3 - {(x,Ax,Ax) : x G G(S+T)]

where for any operator

B , G(B) denotes the graph of B .


K. (i = 1,2,3)

As

G(T)

and

ABSL's

AND THE THEORY OF BASES

G(S+T)

are closed, each

K.

75

is a subspace.

Clearly

(ii) holds. We show that (iii) holds by showing that (K

V K ) n (K

V K ) * K

show that the vector e

and appealing to Theorem 2.4.

f = ((0,e ) , (0,0) , (0,0)) of H , where

- (1,0,0, ...) G I1 , belongs to n G 2

+

In fact we

(K V K ) n (K V K ) but not to ¥L . n

2

x G I by x ~ .£- e. where {e.K is the J n n j-1 j j 1 2 usual orthonormal basis of I . Clearly lim Tx - e. and lim Sx = 0 . J n n 1 n n Now, as A(x ,Tx ) = (Sx ,0) , we have n n n ((-x 0) , (0,0) , (0,0)) + ((x .Tx) , A(x Tx ) , (0,0)) n n n n n For

-

((0,Txn) , (Sxn,0) , (0,0))

->

((0,ei) , (0,0) , (0,0)) - f .

But, for every f

define

n , the left hand side is a vector in K- + K9

so

G Knv K0 . Similarly, as A(x ,(S+T)x ) = (Sx ,Sx ) we have 1 z n n n n ((-x ,0),(0,0), (0,0)) + ((x ,(S+T)x ), A(x ,(S+T)x ), A(x ,(S+T)x )) n n n n n n n = ((0,(S+T)x^), (Sx ,Sx ), (Sx^,Sx^)) n n n n n

so, as before, f e ^ clearly

v K3 .

Hence

f € (^

f € K- . This completes the proof.

We remark that, with

S and

T

V K2> n (^

V K 3 ) , but

B

as above,

M - {(0) , G(0) , G(S) , G(S+T) , I2 © I2 } is a medial subspace lattice on I M V N = I

2

2 © £

for every pair

both different from

2

© I

M, N

2

; that is, M n N = (0) and

of distinct non-zero elements of M

2 2 £ © £ . This may provide some insight into why the

above example 'works'


7.

The double commutant

Von Neumann's celebrated double commutant theorem states that operator closed

a weak

*-subalgebra of the set of operators on a complex Hilbert

space, containing the identity operator, is equal to its double commutant. For example, if £

is an ABSL with mutually orthogonal atoms, it

to see that Alg £

is self-adjoint. As it is also weak

the past

few years

operator

closed,

(Alg £)" = Alg £ .

and contains the identity operator, we have In

is easy

there has been

increasing

interest

in

non-self-adjoint algebras of operators. The ABSL's we study are defined on Banach spaces so self-adjointness (of their Alg's) is ruled out. But even if we restrict ourselves

spaces, Alg £

to Hilbert

self-adjoint (it is so if and only if equivalently, for each M G £ happen, for an ABSL

£

M

we also have

with Alg £

- Alg £ . For example the ABSL

between

L

and M

- £(X)

(see [15]).

for

K~ G

£ = {(0),L,M,X}

£)

with

M

still

a closed

(Alg £)" = Alg £ (see [15]).

On the

is a non-closed sum (equivalently, if the angle

is zero) we have the strict inclusion Alg £ c (Alg £)"

£

we

have

L'

are

non-zero angle for every

L G £ , L ^ (0), X

by Corollary 7.2 below).

Perhaps surprisingly we show (Theorem

LG#,L?*(0)

;

(Alg £)"

L + M

double commutant property, (Alg £)" = Alg £ , only if L and

£

£

E

It may

The above remarks seem to suggest that for an ABSL

there is an ABSL

in general

every

non-self-adjoint, that

(rather than just dense) sum satisfies other hand, if L + M

= M'

is not

L

and

at a

(the converse of this is true

on a separable Hilbert space

, H , the subspaces

the

L'

H

7.5)

that

such that, for every

are at a

76


zero

angle and

77


(Alg £)"

yet

= Alg J£ .

orthogonality of

L

In other words we can

and

go to

the

other

L' , so as far as possible from

extreme

self-adjointness

of Alg £ , and still retain the double commutant property.

In

passing

we

mention that the construction below gives us a new way of obtaining ABSL's. The following is a special case of Theorem 5.4 of [15]. We needing parts of the proof as well as the notation introduced

shall

be

therein,

so

we give a summary of its proof, only elaborating on the parts we shall use.

Let

THEOREM 7.1. unique sublattice PROOF:

1

ABSL M on of

he an ABSL on a Banach space X

such that

X . There exists

(Alg £)" = Alg M.

This

M

a

is a

£ .

As in the proof of Theorem 5.1 of [15], if

restriction of

T

to each atom of £

equivalently, each atom of

iS

T

every operator

and let

A

multiple

are elements of

closed linear spans of the atoms of T

a

of

£

they

£

(and

contain).

Let

be the corresponding eigenvalue.

AG Alg n, + 1 and such a projection

on

{f

1

> m,

{f

+ yfe) II > k . We : n > m, +

be such that y,

1}

may

suppose

(by perturbing

belongs to the linear

: ni,+l < n < IL - } . This completes the inductive step.

Now define oo

M - .v. (< f .,..., f >) i=l n.+l' m. l

and let

oo

{g }-, be

a

renaming

(preserving

l

order) of


the

f 's

not

80


appearing in M , that is, of f ,f -,..., f ,f -,..., f ,... . nm-+l n? nu+1 n~ 00

Since the subspaces 5.1) on X , so do M' =

CO

Vg

{) 1

form the set of atoms of an ABSL (Theorem

{M} U { : n G Z } . In this new ABSL we have n

, moreover

M

and M'

are at zero angle.

k G Z + , x k G M , y k G M' , I x k + y k || < 1 unboundedness of the projection onto M

and

Indeed, for every

|| x k || > k , showing the

along M' .

To summarize, we have constructed an ABSL

on

X

with atoms

M

and

the one-dimensional subspaces M == (n G Z ) , such that & r n n M + M' ^ X , yet for every other atom M we have M + M' = X . J

J

n

n

n

For a specific instance of the above situation one may take for n 2 example fn — 1=1 .En e. in £ (see Theorems 5.7,i 5.8)/ , and put r l » r CD 00 M = .V-f0. , and M - (n G Z ) . Here M' - .V-f0. and one can 1=1 2i-l n 2n 1=1 2i

show that the angle between M

and M' is zero; for example, it can easily

be shown that f

lim n

2n

«f 2n «

f

2n+l

+

=1 1

n" (2n (v^ST ~ V2^)

iif 2 n + 1 n

2ST J = °

In the proof of the next theorem we shall be needing the following observation.

LEMMA 7.4.

If

I U J = Q n [ 0 , l ] x

n

G l , yJ

n

I

and

, then

G J ( n G Z )

J there such

are

non-empty

disjoint

exist

sequences

that

l i m x - l i m yy n n n n

sets oo

{x }-

and


with oo

{y }-

wi th

81


PROOF:

The hypothesis

connectedness of

[0,1] gives oo

find sequences

I u J = 0 .

2 This is a quadratic form in a and b with the coefficient of a positive (recall that 0 < r,s < 1) so it is sufficient to show that its discriminant

A

is non-positive. Here

1 2 2 2 2 T- A — (4—rs) cos (r-s) — (4—r ) (4—s ) 2 2 2 = 4 (r—s) — (4-rs) sin (r-s) 2 2 2 < 4 (r—s) — 3 sin (r-s) . But (with the obvious meaning when 9 = 0) sin 9/9

is decreasing on [0,1] , 2

so

sin 9/9 > sin 1 > 2/3

0 < 9 < 1 . Hence

for

2

4(9-9

9

sin

< 0

(9 G [0,1]) and taking

9 = | r - s | we get j A < 0 , as required.

authors thank A. Lewis

for

ideas

leading

to

(The

this

simple

proof of

on its domain enables us to

extend

it

inequality (*).) The continuity of whole of H

without

extension also by

T

spoiling

continuity . If we

T , we show that

r G Q n [0,1] , f G L

T G (Alg £)'

and A G Alg £

ATf . As the latter is true for every V{L

: r G Q n [0,1]} - H , we have

we have f G L

denote

to the

this unique

. Indeed, if Af G L

, so

and for every

TAf

-

rAf -

r, and as

AT - TA .

We are now in a position to show that 7.1 there is a unique ABSL M on H with

(Alg £)" = Alg &

.

By Theorem

(Alg £)" = Alg M , so we

only

have to show that M = ft . The proof of Theorem 7.1 shows that every eigenspace of operator

T, constructed above, belongs to £

span of the atoms it contains.

the

specific

and so is the closed

It follows that, for every


linear

r G Q n [0,1] ,

85


L

—

(L ) „

for

r G Q n [0,1] (with notation as in the proof of Theorem

7.1).

It

is an eigenspace of T . Hence, since

every

now follows that M = £

T G (Alg £)' , L

and the proof is complete. a

A pertinent problem is to find necessary and sufficient conditions for an ABSL £

to satisfy (Alg £)" = Alg £ . The above theorem and what

shows that this is quite a difficult problem.

Notice

that

Corollary 7.2

gives a sufficient condition which is not necessary, by Theorem necessary condition (Theorem 7.6(2)) is given below but this not sufficient.

THEOREM 7.6.

(1)

£

X

with

be an ABSL on a Banach space (Alg£)"

For each finite

set

L- + Ln + ... + L L Z n (2)

PROOF:

If

condition is

(Alg 0 , applying

T

(nGZ+)

.

there exists

CO

00

{x }- C M , {y ) 1 C M '

ynii

n

n

A^x — z -» 0 . 2 n n

'n

llx — y II -> 0 . For each n G 2

ii

T(y © 0) = z © 0 and J

©0-y

^ = fJl I

There

T(0 © y) - 0 © z . Since M

are at a zero angle there exist sequences

let

z G X n

T(0 © Jy ) = 0 © z n

gives

A-x — z -» 0 . A similar argument using

n

. Since

A-x © 0 - z 0©x

©0->0,so

— 0 © y ->0 gives t h a t

Hence

|A X -A 2 | = |