Geometric aspects of Banach spaces: essays in honour of Antonio Plans

Using

(4.5),

(4.8)

and

the

monotonicity of |3/3(e) we obtain by the assumptions of the theorem i/p

i/p Y

Cj £ * 65(c 2 (l/k)

) * c£|3(c 2 (l/k)

l/p -l/q

Y

)=

l/q

) ss cc3 ii .

(1/k) 1 / PP

Y Y

D e n o t e fi

For ii

= c (1/k)

then

< c < n we have

k+l

k

A

rw J

Because

So,

, k=l,2,...;

p =q Y

P

l i m —— = 1 ,

we h a v e ,

30(e,B(X))

p = 0 ( eY)

Y

and,

P

Y

y

f o r e-»o

and

by Maurey-Pisier' s

p = 0(e Y ) .

56(c,B(X))

theorem,

we conclude

that

I

p

is Y

finitely represented by blocks of X.

Remark.

Clearly

3°|3O(e) < 33(e;B(X)) ^ 56(e;B(X)) * 5°5°(c). Therefore, if there exists c £ 1 such that 5°6°(c) < c^0^0(c) then the conclusion of Theorem 2.2 holds also for ^°/3°(c).

Milman & Perelson: Geometric moduli

33

5. FINITE DIMENSIONAL MODULI ££(e,B ) AND 33(e;B ) AND THE LOCAL STRUCTURE OF A SPACE.

In this section, we consider the finite dimensional

moduli

££(e;B ) = £ £ 3 (e;B ) and <5<5(e;B ) and compare their 1behavior with the o o o oo o o o type and cotype of the space in the spirit of section 3.

Theorem 5.1.

Let B be an infinite dimensional Banach space of type

and cotype 2 ^ q < oo. Then

]

VO p

sup
np

[

and

= 0 >^ P

JB

££ ( inf Jq; lim

q

= sup {pi B is of type p>,

)

JB

> 0 V^ q

= inf {q| B is of cotype q>,

For the proof ot Theorem 5.1 we will need the following known fact (see [Ml], Lemma, 5.1, or [M2], section 5.8).

Lemma A. Let B be an infinite dimensional Banach space. There exists an integer function L

c

n(k;o,C)

>B and any xeS(B),

such

that, for

any

n-dimensional subspace

there exists a k-dimensional subspace L

n

c

> L

k

n

and for any et 0 ^ e ^ C,

sup |||x+ey|| |y € S C L j j - inf |||x+cy|

There exists, of course, a finite dimensional version of this lemma which obviously implies, by definition definitic of moduli 8 (c;x) and £ (e;x) (see section 1), the following proposition

Proposition 5.2.

Let B be an infinite dimensional Banach space, L an

N-dimensional subspace of B and L

an n-dimensional subspace of L. For n

C < oo , a given e > 0 and a positive integer m, there exists a positive integer

N

=

N(m,n,C,©),

such

that,

if

N

>

N ,

there

exists

a

34

Milman & Perelson: Geometric moduli

m-dimensional subspace L c L such that for every x € S(L ), y € S(L ) and m

e € [ 0 , C ] we

(5.1)

n

m

have,

( l - © ) ( l + 5 ( e ; x ) ) ^ llx+eyll * ( l + e ) ( l + / 3 ( e ; x ) ) .

Remark. Use t h e p r o p o s i t i o n f o r C = 1 + 2 / e . Then, f o r c > C, we have

) > f l + - L - j ( e - l ) = e + l > 1+5 ( e ; x ) . Then

|x+ey[

Therefore, (5.1) holds for every e > 0. We are now going to show a finite dimensional

version of

estimate (3.5).

Proposition 5.3.Let B be an infinite dimensional Banach space. For all positive integers m,k and any e > 0, there exists an integer N(m,k,e) such that in every H-dimensional subspace L c B for N ^ N(m,k,e) k

a set {x > c S(L) such that for a =1 and

kf (5.2)

(1-©) n

r 1+5

L

^

l!

Ln=l

kf * (1+0) n

j

=2

f I

1+3m

l

n=l

la I

I

' ji I

1

^ "Un=l

^

) a x

JJ

n=l

J-1

11 II

; Va.x II

n n"

L

1

n= l

^

L nn

lI f

-

IITJ"1 a X

L )

^

L nn

n n11

k

ii

; ) a x

I I P " 1 a x ||

m

j"2[

a.€ (R, 2 z j ^ k,

-1"1

ia I

we can select

.

n n

JJJJ

k

Moreover, we can replace {x } Proof.

For a given e > 0 let us select the numbers {e

k-l

1+e

I

in (5.2) by every block set {y }

i=l

> 0 } _ , with

k-l

£ n (l+e ) 11

c S(B) of

i

and

1-e

^

n (1~© )• " i=l

i

Let

L c 0

B be

an

N -dimensional k

Milman & Perelson: Geometric moduli

subspace with N

35

large enough (we may compute t h i s

k

N

function of N(m;N

as a

k-iterated

k

2

; 1+ - ;e ) where N(m,n,C,e) is defined in Proposition

5.2). Let x € S(L ). Using Proposition 5.2, we select a subspace L c L , dim L = N

, such that for every y € L

(5.3)

(l-e1)(l+6m(|y||;xi))^ fx^yl * (1+e^ ( 1 + M ||y||; x ^ ).

We continue by induction. Suppose that the elements {x > were L

already

c L

n-l

constructed

with

. As in the first

n-2

subspace L c L n

, dim L

n-l

the appropriate

step,

= N

n

for every y € L and x € S(span

n

of subspaces

) one can choose a

n-1

= N(m;N ; 1+- ;e ), such

k-n+1

k-n

e

n

n

that

{x > n ) we have

n

(5.4)

for x € S(L

and N

k-n

sequence

j 1

(l-en)(l+Sm(||y|;x))

s |x+y| * (1+oJ ( 1 + Mflyfl;x ) ) .

Repeating t h e use of ( 5 . 4 ) we o b t a i n , f o r a =1,

c-1

I

)

x aa. x J

Lj=1

J J

,,rk-l

"

k

I^I

E

ax

ax

r

II) •_ \

L Lj=l

'

^ II

ax ;

a

I

; k>—a1 x

\

^m ~k~T

^

||

iirk-l

I

C

+

II r

j=l

+ |,

II) _ k —1

(1+G

ax

J

jXjH

* ... ^

a y

E

-

j=l

J M

—

)

^ (l+e) n L

This proves t h e r i g h t left

; [ax

1+ P

j=2 I

m

l y3"1 a x I

L

Ln=l

n n"

hand s i d e of ( 5 . 2 ) .

L nn n=l

.

lI J J

In a s i m i l a r

way we p r o v e t h e

hand s i d e of ( 5 . 2 ) . Note

that

by c o n s t r u c t i o n

we may r e p l a c e

{x}

i n ( 5 . 2 ) by

36

Milman & Perelson: Geometric moduli 1*1

every

block

sequence

iy« =

) J=n

a x

>

, o f { x }

(since

the

sequence

of

£+1 n

subspaces {L.}

is a decreasing sequence and span {y«}^_ c span {x } s+ ).

We return to the proof of Theorem 5.1. Let T),K > 0 be given. There exists a positive integer m such that

(5.5) and \fi $ (7)) ~ BBit))\ 'mm 00

'

< K.

The functions (B (A;x) and 8 (A;x) are Lipshitz on [0,1] x S(B). Then, by [Ml] (Theorem 5.1), for any N, there exists an N-dimensional subspace L c B such that for every A€ [0,1] the oscillation of the two fuctions £ U;x) m

and

5 (A;x) on S(L ) is small (say, less J m o Proposition 5.3 with this subspace L , then

than

a

&given

e>0). Use

j-i (5.6)

1+5 h ; m L

V a x I n n J n=l

£ (l-e)(l+

L

sup 5 ( T ) ; X ) ) fc ( 1 - e ) ( 1 + 5 5 ( T J ) ) . m m x€S(L ) m 0

Suppose that for a given k we select € in such a way that

(l-e) k ^ 1/2. Using similar arguments to those in section 3 together with the left hand side of (5.2) and (5.6) we obtain

t

5.7)

k+1 rp i V n (1+6 5 (77)) < -=- 3T (1 + (3T T))p) .|J mm [l-Q pj p

for some integer function t such that t k

p

> oo (if k—^co). k

Milman & Perelson: Geometric moduli

37

Assume that K has been chosen in such a way that (1-K) k £ 1/2.

Using (5.5) and (5.7) we have r >,

t

(1+5 5 (7))) oo

^

t /p

^

T~T: ^^

(1 + C3T T)) ) . pj p

^1-e

Taking the t -th root from both sides of the above inequality we obtain (for t=t ) k

1/t

l/p

(1 + (3TT>) P ) P

*

l/t

* j^l-e Ui- 3Tpj Let t—>oo,

(5.8)

(1 + (3T T))p).

then

5 5 (TJ) ^ (3T TJ 0 0

p

and 5 6 (3T ) p .

^ p

The p r o o f

p

of

the

formula

for

(3(3 ( c )

is

similar.

We

obtain

f 1 1q ( 5

'

9 )

3^

and then take the lim inf. •

As a corollary of Theorem 5.1

(or, to be more exact, from

(5.8) and (5.9)), we obtain Theorem 2.3 stated in section 2 in the same way as we proved Theorem 2.1 at the end of section 3.

38

Milman & Perelson: Geometric moduli REFERENCES.

[Cl]

Clarkson, J.A. (1962). Orthogonality in normed linear spaces. Archiv Math. 4:4, 297-318. Day, M. M. (1941). Reflexive Banach spaces not isomorphic to [D] uniformly convex spaces. Bull. Amer. Math. Soc. 47:4, 313-317. [K] Kadee, M.I. (1956). Unconditionally convergent series in a uniformly convex space. Uspekhi Mat. Nauk 11:5, 185-190 (Russian) [Li] Lindenstrauss J. (1963). On the modulus of smoothness and divergent series in Banach spaces. Michigan Math. J. 10:3, 241-252. [Ml] Milman, V. (1971). Geometric theory of Banach spaces II. Geometry of the unit sphere. Uspechi Mat. Nauk, 26 no. 6, 73-149; Russian Math. Surveys 26, 6, 80-159. [M2] Milman, V. D. (To appear),The heritage of P. Levy in geometric functional analysis. Asterisque. [M3] Milman, V. D. (1967). Infinite-dimensional geometry of the unit sphere of a Banach space. Soviet Math. Dokl. 8, 1440-1444 (Translated from Russian). [MaPi] Maurey, B. & Pisier, G. (1976). Series des variables aleatoires vectorielles independantes et priorietes geometriques des espaces de Banach. Studia Math. 58, 45-90. [Mil] Milman, D.P. (1938). Certain tests for the regularity of spaces of type B. Dokl. Akad. Nauk SSSR 20, 243-246 (Russian). [MSch] Milman, V.D. & Schechtman, G. (1986). Asymptotic theory of finite dimensional normed spaces. Springer-Verlag. Lecture Notes in Mathematics 1200, 156 pp. [MShl] Milman, V.D. & Sharir, M. (1979). A new proof of the Maurey-Pisier theorem. Israel J. of Math. 33, 1, 73-87. [MSh2] Milman, V. D. & Sharir, M. (1979). Shrinking minimal systems and complementation of n -spaces in reflexive Banach spaces. p

Proceeding of London Math. Soc., (3) 39, 1-29. [Sm] Shmul'yan, V. L. (1940). On the differentiability of the norm of a Banach space. Dokl. Acad. Nauk SSSR 27, 643-648.

HUBERT SPACE REVISITED.

J. R. Retherford Louisiana State University Baton Rouge, LA 70803 Dedicated to Professor Antonio Plans

1. INTRODUCTION, Since some of the work of A. Plans concerns operators on Hilbert

spaces

(e.g.

[PI

1-5], [PB])

it

seems

appropriate,

in

this

"Festschrift" on the occasion of the retirement of Professor Plans, to look back at a few (of many) operator characterizations of Hilbert space. We will show that all of these characterizations follow easily from the summability property of the eigenvalues of nuclear operators on isomorphs of Hilbert space. This property is discussed below.

2. DEFINITIONS. Recall that an operator T from a Banach space X to a Banach space Y, here after written T : X —> Y, is nuclear provided there are (fn ) * in X , (y ) in Y such that

Tx = ) f (x) y L n n n=l

for each x € X, and

I « f n i \WJ < -• This concept is due to Grothendieck [G3] and Ruston [Ru]. On Hilbert

spaces the nuclear operators

"trace-class" operators if , where T e if means

coincide

with the

Retherford: Hilbert space

40

n=l Here (A v TT ) denotes the eigenvalues of the unique positive square root * n of TT . Moreover, it is readily seen that

1

2

2

where !f denotes the Hilbert-Schmidt class. That is, each nuclear operator on a Hilbert space is the composition of two Hi lbert-Schmidt maps and, conversely, the composition of two such maps is nuclear. The

generalization

of

this

fact

to

Banach

spaces

(with

applications) is the subject of this paper. The Hilbert-Schmidt operators have been generalized to Banach spaces

in the following

way

[Gl], [Pi3],

[S]: An operator

T: X—>Y

is

absolutely two summing, written T € TT (X,Y), provided there is a constant K such that

for all finite sets x ,...,x 1

in X. n

Pietsch [Pi3] has shown that for Hilbert spaces H , H

so TT is a natural generalization of the Hilbert-Schdmidt class. The class TT , defined by replacing 2 by p, 1

p < +oo, above,

p

has seen numerous applications in Banach space theory. The

famous

Grothendieck-Pietsch

factorization

theorem

[Gl]

[Pi2,5] asserts that T€ TT (X,Y) if and only if T admits the factorization

41

Retherford: Hilbert space

where K is a compact Hausdorff space, ji a normalized Borel measure on K, and j the inclusion mapping. It is easily seen that j € II (C(K), L (JI)). We will also have occasion to use the p-nuclear operators, in what follows: T is p-nuclear, 1 ^ p < +00, written T € N (X,Y) if p

Tx = V f (x)y L

n

n

(f ) in X , (y ) in Y, x in X where

and

< +00

" «

In

particular

V |f(y ) |q

Sup

llflhl

L

< +co , - + - =1.

n

P

q

n=l

1-nuclear

operators

are

just

the

nuclear

operators.

3. EIGENVALUES OF NUCLEAR OPERATORS ON HILBERT SPACES. Since we will be concerned with eigenvalues, we assume all our spaces to be complex Banach spaces. Let H be a Hilbert space and let T € N(H) = N(H,H) have eigenvalues (A (T)) (ordered by decreasing modulus n

and counting multiplicities). The classical Weyl inequality [W] asserts that

so

the

eigenvalues

of

an

arbitrary

nuclear

operator

are

absolutely

convergent. Surprisingly, Hilbert space is the only Banach space enjoying this property:

42

Retherford: Hilbert space

Theorem [JKMR]. Let X be a Banach space. The following are equivalent: (a) X is isomorphic to a Hilbert space; (b) each nuclear operator on X has absolutely summing eigenvalues; and

(c) N(X) = lyiyx). Since

(c) is critical

to our work we discuss

it

in some

detail. Firstly, the notation T € IT -IT (X) means there is a Banach space Y and operators U € iyx,Y), V € Secondly,

if T e

Y,X) with T = VU. IT -IT (X)

then

T

has

absolutely

summing

eigenvalues. To see this, following Pietsch [Pi5] we say that A and B are related operators if

for Banach spaces X and Y. The importance of this notion is that

AB

and

BA

have, in

this case, the same eigenvalues. Thus if

T e TI *II (X) we have for some

Banach

II (X,Y),

space

Y,

T

=

BA

and

A

€

B

€

II (Y,X).

By

Grothendieck-Pietsch factorization we thus have the following diagram:

* X

Now

Uiai/32)(J2a2^i)€

= 5P (L (fi ))

and so this operator

has, by the Weyl inequality, absolutely summing eigenvalues. Clearly this operator is a relative of BA, i.e. T has absolutely summing eigenvalus. Our strategy is thus the following: to show that a Banach space X is isomorphic to a Hilbert space we need only show that for Te N(X), the eigenvalues of T, U

(T)), satisfy

Retherford: Hilbert space

43

00

I IVT)I

< +M

-

n=l

To accomplish this we show, under various hypotheses, that T (or T ) is in

II • II . 2

2

To do this

we need the "big picture" provided by a famous

result of Grothendieck.

4. GROTHENDIECK FACTORIZATION. As usual we denote by £ , 1 ^ p < +oo the Banach space of P 00

II All = V IA |p " "p [ L ' n1 J

sequences A=(A ) satisfying n

< +oo, and by I

the space

oo

of sequences A = (A ) with IIAII = sup |A I1 < +oo. M n

" "oo

Finally c

^ I n

n

denotes the closed subspace of I

0

those sequences A = ( A ) w i t h l i m A = 0 . n

n-X»

m

consisting of

oo

n

Let f®y, f€ X , y€ Y denote the rank one operator x

f(x)y.

Thus any non-zero T€ N(X) can be written

T =

V f ® y L

and f

* 0, y n

n

with n

Y

l l f II lly II <

L

" n"

* 0 f o r each n. L e t

A

n

where, a n

n o

f

clearly (||gn|),

= pa

n

n

1/2

n n

( | | z n | ) € CQ

La n=l

n n

" n"

and write

" n"

A n

= a(3 n n

and (0 ) € I . Then 0

o

® y

n

= llf II lly II

n

> 0, (a ) € c

+co

" n"

1

fn

1/2

f

n

TT^-TT ® a

yn

-n—n-

y

_

= £ g

n n

® z

n

and

n

This yields the Grothendieck factorization theorem for nuclear operators

44

Retherford: Hilbert space

[Gl]: if T€ N(X,Y), then T admits the following factorization:

where K

and K

are compact and 5 is a diagonal nuclear operator. Indeed 00

let O ), (g ), (z ) n

n

be as above and let K (x) = (g (x),K ((*•))= V ? z 1

n

n

2

n

L

n n

and 5((? )) = (|3 £ ). n

where 6

n n

Observe that 6 admits a further factorization

and 6 1

are the diagonal mappings defined by (|3

show from the definition that 6

2

^

€ TT (c ,1 ) (even N (c tt )) and 8 1

7r (£ ,1 ).

). It is easy to

n

2

2

0

2

2

0

2

Or, if one desires, a deep result of Grothendieck

€ 2

[Gl] also

oo* 2

covers the above situation. This diagram is now produced in it's entirety. Our strategy outlined above reduces to the intellectual (?) game -chasing the diagram! Grothendieck Factorization for

K ,K ,6,6 ,6

have the meanings as above.

T€ N(X)

45

Retherford: Hilbert space

Main o b s e r v a t i o n

(Related

operators): ^

(6 K ) ( K 5

1122

)

has

the

same

eigenvalues as T. Actually, starting anywhere in the

diagram and going all the

way around yields an operator with eigenvalues the same as T. For example, K K 5

is an operator on

c

with eigenvalues the same as T.

5. APPLICATIONS. In what

follows

we

will

operators as needed. We will write

introduce

X € H

classes

to mean

of

spaces

X is isomorphic to a

Hilbert space.

I. Theorem of Cohen-Kwapien [C], [Kl: An operator T e D (X,Y) provided T Theorem: Let

X

€ II (Y ,X ).

be a Banach space. The following are equivalent:

(i)

X € H;

(ii)

iyx.Y) c D2(X,Y) for every Banach space Y;

(iii) D (Y,X) c n (Y,X) for every Banach space Y; and (iv)

Same as (i) or (iii) with Y = I .

Proof: ( i M i i ) . If

X

is isomorphic to a Hilbert space

T € II (H,Y), by Grothendieck-Pietsch factorization

-» Y

C(K)

and

H

we have

and

46 and

Retherford: Hilbert space j a € ^ (H,H ) 2 0

so

a*j*

€ 5P ( H , H ) 20

and so

T % II ( Y * , H ) . 2 m #

(i)^(iii) is very similar: If Te D2(Y,H) then T € iyH,Y ) so by what was Just shown T**€ n (Y**,H) so T€ n (Y,H) since T** extends T with Y 2 2 ** canonically imbedded in (ii )=>( i)

and

Y

(iii )=>( i):

factorization.

If

(ii)

Let

T€

N(X)

and

consider

Grothendieck

holds

SK € TI(X,£) c D ( X , £ ) . Also S € 11 2 2 2 2 2 D (£ yl ) so T € D - D (X). Thus T*€ II -TI (X*) and thus has absolutely 221 22 2 2 # converging eigenvalues. But, the eigenvalues of T

are exactly those of T,

so X € H. If

(iii)

holds

KSe 2 2

DU.X) 2 2

c

II(£,X) 2 2

and

so

(K 5 )(5 K )€ll -II (X) and T has absolutely converging eigenvalues. (iv)=»(i) is clear since

all we used above was the space I .

An isometric

version of I is valid. Indeed isometric

are valid for most of

the

results

theorems we give. To state such results

requires conditions on the various norms of the ideals involved. Since we have not

explicitly given these norms we will not give the isometric

II. Theorem of Lindenstrauss-Pelczynski [LP]: results. One of the remarkable results of Grothendieck is the fact that £(L , H), the space of all bounded linear operators from an L (^)-space to a Hilbert space, coincides with

IT (L ,H). This result is proved using

what is now called the Grothendieck inequality [G2], [LP].

Theorem. If

X has an unconditional basis and £(X,Y) = II (X,Y) then X is

isomorphic to Proof:

< a

e.g.

n II

H

n

n

( x ) f ( x ) | = K < +00. Let n

i n Y,

n

'

||y || = a " n"

with n

X

and assume ||x|| = l,

( a )€ c , 0 < a n

sup

0

" " n

2 | g ( y ) | ^ 2.

ll ll

[MR] for such a c o n s t r u c t i o n ) . First,

N

is isomorphic to a Hilbert space.

n

let

n

T x =

Y

sup I | f II II llxll =11 f 11=1 and c h o o s e (y )

" n"

n+l

and

Let (x ,f ) be an unconditional basis for

||f || s M. Let a

I

V ^ f ( x ) y . Then Lt

i

i

i

(/3 ) € £ , n

2

|(J3 ) || "

n "2

= 1. For each N £ 1,

let

< 1, (see

Retherford: Hilbert space

47

n

||T x|| *

sup

V |f(x)||

M|x||(Z|e | 2 ) 1 / 2 ( s u p l | g ( y ) | 2 ) 1 / 2 s 2M|x| 1

so each T

HglUl

'

is continuous. By hypothesis there is a constant c, such that N

N

N

[ |T N (f i (x)x i )| = [ 1^: n=l

i=l

i=l N

* c sup

j

| f ( x ) f ( x ) | ^ c K | x ||.

II f 11=1

n=l

Since (/3 )

was arbitrarily chosen from

the unit

ball of I , (f (x))€l .

n

2

Knowing (f (x))e I for x € X, allows us without the sequence (/3 ). Since

n

doesn't matter) and ||y | = a

2

to repeat the construction above

(a )€ c

n

i

(being positive

and monotone

0

we now obtain (f (x))€ t . Since ||x | = 1 it

follows that X is isomorphic to t . Now let

T € N(Y)

and go to the diagrams. By hypothesis

(since X is isomorphic to I ) K € n {I ,Y) c n (£ ,Y). 2

2 1

1 1

Thus

so T has absolutely summing eigenvalues. A word of explanation is Grothendieck

in order. While our proof avoids the

inequality, the construction

theorem [Dl]. Moreover the

used requires the Dvoretzky

proof of the eigenvalue result we are using

requires the deep result of Lindenstrauss-Tzafriri

[LT] that a space is

isomorphic

each

to

a

Hilbert

space

if

and

subspaces is complemented. This result in So

we've

replaced

one

deep

only

if

of

its

closed

turn uses[D2]. result

with,

essentially,

equally deep results! Still, diagram chasing's fun. Let's continue.

two

48

Retherford: Hilbert space

III. Theorem of Morre11-Retherford [MR]: We operators

write

T € T (X,Y) if there

is a Hilbert

space

H and

A € £(X,H), B € £(H,Y) such that T = BA, i.e., T factors

through a Hilbert space. A Banach space

X

is in the class

2)

if there is a constant

00

such that for each

n

there is a subspace X

of X and an isomorphism

n

n : X n

n

>£

(complex n-space with the sup-norm) such that

KiiO Theorem: The following

are equivalent for a Banach space X :

(i)

X € H ;

(ii)

m , Y ) c T (X,Y) for any Y€ 2) ; 2

oo

(iii) m , c Q ) c r 2 (X,c Q ). Proof: (I )=>( ii ) =»( iii) is trivial (as is it all). we have by hypothesis that K (iii)=*(i). Diagraming again, we there is a Hilbert space

Then

H

€ T (X,C Q ) S O

with

AK SB € N(H) and as before X € H. This shows we may remove the "p" in the title of [MR].

IV. Theorem of Gordon-Lewis-Retherford [GLR]: Here C(X,Y) denotes the closure of the finite rank operators from X to Y.

Theorem: The following are equivalent: (i)

X€ H;

(ii)

C(X,Y) c r (X,Y) for all Y;

(iii) C(X,c ) c r (X,c ); 0 2 0 (iv)

C(Y,X) c T (Y,X)

(v)

C U ,X) c r (£ ,X).

for all Y; and

Retherford: Hilbert space

49

Proof. That ( i )=*( ii M iii ) and (i )=>( iv)=>(v) is trivial. i). Let T € N(X) and proceed to the diagram. By hypothesis Ke

T (X,c ) so there are operators

A, B

to and from a Hilbert space H,

with K = BA. Now 6 B € if (H,£ ) and so also to D (H, I ), i. e. T € D -D (X). 2 2 2 2 2 2 2 (iv)=>(i). Again let Te N(X). Now K ^ r il ,X) and thus A,B,H exist with K

= BA factoring through H. Hence

A<5 € D (£ ,H) and so to II (£ ,H).

Thus T € iyiyx). In [GLR] we also have characterizations of Hilbert space in terms of operators on X . These results can be obtained

as above by

dualizing the diagram. V. Theorem of Holub [Hi: An operator T is quasi-p-nuclear, 1 ^ p < +00, T € QN (X,Y), if p

t h e r e i s a sequence (f ) in X , V IIf ||p < +00 such that n

L " n" n=l

00

|Tx| * [ [|f i( x)| p ) 1/P n=l

for each x € X. This is equivalent to saying that if i:Y—> I ( D is an isometric imbedding of Y into a suitable I -space (the index set T may be 00

uncountable) then

IT € N {X,l ( D ) . p

00

Theorem: The following are equivalent: (i)

X € H;

(ii)

T € N (X,£ ) => T*€ N (I ,X*); 2 2 2 2 # #

(iii) T € N {X,l ) =» T € QN [I ,X ); (iv) (v)

T € QN (X,l ) =» T € N (£ ,X ); T € QN (X,£ ) =» T*€ QN (£ ,X*). 22 22

Proof: C l e a r l y for

Hilbert

N c QN c II

regardless

of t h e s p a c e s

i n v o l v e d and s i n c e

s pK a c e s H , H , n ( H , H ) = y ( H , H ) , a g6 l a n c e 1

representation shows that

2

2

1

2

2

1

2

a t t h e Schmidt

50

Retherford: Hilbert space

Thus,

y (H ,H ) = N (H ,H ) = QN (H ,H ) = II (H ,H ) 2

for Hilbert

1

2

2

1

2

2

1

2

2

spaces. So we certainly have

1

2

(i )=»( ii )=»( iii)

and

(i)=>(iv)=»

(v)=»(iii). Thus we need only show (iii) =» (i). Let T € N(X). Rushing once again to the diagram we (iii)

K*6*

€

see that by construction, 8 € N (c ,1 ) so by

QN 2 U 2 ,X*)c

TI U ^ X * ) .

Also

S*K*€

iyx*,£ 2 ),

i.e.

T €ll -TI (X ) so T has absolutely converging eigenvalues.

VI. Theorem of Rosenberger [R]: We end these Hilbert space characterizations with a result which requires a generalization of the Hilbert-Schmidt

operators to a

class of operators different from the absolutely two summing operators. For

T € £(X,Y)

let

a (T) = inf {|T-A|: rank A * n-1}.

On

Hilbert space

a (T) = X (/TT n

) [Pi4].

n

Let, f o r 1 s p < +oo,

X,Y) = s (X,Y) = IT I T € m,Y)

| £ C£(TX+CO|

Then clearly for Hilbert spaces H , H

S (H ,H ) = y (H ,H ). 2 However, S (X,Y).

1 2

2

1 2

for arbitrary Banach spaces X,Y it is not true that II2(X,Y) = These

properties of

ideas a (T)

are

due

to

Pietsch

[Pi4].

Using

multiplicative

it is easy to show that

S • S c S. 2 2 1 Also

[JKMR]

if T

€

S (X) then

the

eigenvalues

of

T

are

absolutely

Retherford: Hilbert space

51

summable. Thus (same idea as the strategy) to show that X € H i t i s enough to show that

T € N(X) implies

(under various hypotheses)

that

T € S•

S2(X).

Theorem: The following are

equivalent:

(i)

X €H;

(ii)

f o r a l l p , 1 s p < co, N ( X , £ ) c S ( X , i ) a n d N ( X * , £ ) c S ( X * , £ ) ; p

(iii)

there

is

a

p,

2

^

2

p

<

2

2

+oo such

p

that

2

N (X,£)

2

c

2

S (X,£)

and

N (X*,£ )c S (X*,£ ).

p 2 2 2 Proof: i)=>(ii). It is easily seen that N (H ,H ) c II (H ,H ) for Hilbert p

spaces H ,H . A. Pelczynski

1

2

p

1 2

[P] has shown t h a t in t h i s case

n (H ,H ) = n (H ,H ) p

for a l l

1

2

2

1

2

p, l^p<+oo. Thus if p^2

^"rV Clearly

i f lsp<2,

( i i )=>( i i i ) i s ( i i i M i ) .

=

c N (H H ) c P r 2

W V

N (H , H ) c n ( H , H )

L e t T e N(X). Again from But

m

5 € N (I tt ) and so, as above, o

o

t h ediagram

Nc N f o r p > 2 so by ( i i i ) 2p

2

s o (i)=»(ii).

trivial.

* 2

V ^ ' V = y2(Hi'V-

2

m

shown that a (T) = a (T ) for compact T n n S • S2(X) and so has absolutely

1

€ 1

€ S (X ,1 ). 2

m

6 K

#

6 K

5 K€ N (X,£ ).

2

2

S ( X , £ ) . Now 2

2

But, Hutton

[Hu] has

2

and so K 8 € S (£ ,X). Thus T€ 22 2 2

converging

eigenvalues.

Observe

that

reflexivity, used in [R], is not needed. 6. REMARKS. This concludes the applications of the eigenvalue theorem. Of course there are many more results scattered through the literature which can

be proved

by this

strategy

of diagram

chasing.

This

does not

necessarily detract from these results nor the theorems discussed here. Most

of these

results

were

obtained

when

we had the "complemented"

subspace problem" not the "complemented" subspace theorem". And remember, the eigenvalue theorem has its proof hidden in the complemented subspace

52

Retherford: Hilbert space

theorem. Really, what we have shown is that, unknown to

the various

authors, at the time, the eigenvalue behavior of nuclear operators on the given space unifies essentially all of the operator characterizations of Hilbert space.

BIBLIOGRAPHY [C] [Dl]

[D2]

[GLR] [Gl] [G2] [G3]

[H]

Cohen, J. (1970). A characterization of inner-product spaces using absolutely 2-summing operators. Studia Math. 38, 271-276. Dvoretzky, A. (1961). Some results on convex bodies and Banach spaces. Proc. Int. Symposium on Linear Spaces, Jerusalem 123-160. Dvoretzky, A. (1966). A characterization of Banach spaces isomorphic to inner product spaces. Proc. Colloquium Convexity, Univ. of Copenhagen 61-66. Gordon, Y.; Lewis, D.R. and Retherford, J.R. (1973). Banach Ideals of operators with applications. J. Funct. Anal. 14, 85-129. Grothendieck, A.(1955). Produits tensoriels et espaces nucleaires Memoir of the Am. Math. Soc. 16. Grothendieck, A. (1956). Resume de la theorie metrique des produits tensoriels topologiques. Bol. Soc. Math. Sao Paulo 8, 1-79. Grothendieck, A. (1951). Sur le notion de produit tensoriel topologique d'espaces vectoriels topologique, et une class remarquable d'espaces vectoriels liees a cette notion. C.R. Acad. Sci. Pris 233 1556-1558. Holub, J.R. (1972). A characterization of subspaces of L (ji). p

Studia Math. 42, 265-270. [Hu] Hutton, C.V. (1974). On the approximation numbers of an operator and its adjoint. Math. Ann. 210, 277-280. [JKMR] Johnson, W. B.; Konig, H. ; Maurey,B. and Retherford, J.R. (1979). Eigenvalues of p-summing and I -type operators in Banach p

[K] [LP] [MR] [P] [Pil] [Pi2] [Pi3] [Pi4]

spaces. J. Funct. Anal. 32, 380-400. Kwapien, S. (1970). A linear topological characterization of inner product spaces. Studia Math. 38, 277-278. Lindenstrauss, J. and Tzafriri, L. (1971) On the complemented subspaces problem. Israel J. Math. 9 , 263-269. Morrell, J.S. and Retherford, J.R. (1972). p-trivial Banach spaces. Studia Math. 48, 1-25. Pelczynski, A. (1967). A characterization of Hilbert-Schmidt operators. Studia Math. 28 355-360. Pietsch, A. (1963). Absolut p-summierende Abbildungen in lokal konvexen Raumen. Math. Nachr. 27, 77-103. Pietsch, A. (1967). Absolut p-summierende Abbildungen in normierten Raumen. Studia Math. 28, 333-353. Pietsch, A. (1969). Hilbert-Schmidt Abbildungen in Banach Raumen. Math. Nachr., 37, 237-245. Pietsch, A. (1970). Ide'ale von S -Operatoren in Banachraumen. p

[Pi5] [PI]

Studia Math. 38, 59-69. Pietsch, A. (1978). Operator Ideals. VEB Deutscher Verlag des Wissenschaften, Berlin. Plans, A. (1959). Zerlegung von Folgen im Hilbertraum in Heterogonal Systeme. Archiv der Mathematik, T.X. , 304-306.

Retherford: Hilbert space [P12] [P13] [P14] [P15] [PB] [R] [Ru]

[S] [W]

53

Plans, A. (1961). Propiedades angulares de la convergencia en el espacio de Hilbert. Rev. Mat. Hisp.-Amer., T. XXI , 100-109. Plans, A. (1959). Propiedades angulares de los sistemas heterogonales. Rev. Acad. Ciencias de Zaragoza T. XV, 2. Plans, A. (1961). Resultados acerca de una generalizaci6n de la semejanza en el espacio de Hilbert. Coll. Math. XIII, 241-258. Plans, A. (1963). Los operadores acotados en relacion con los sistemas asintdticamente ortogonales. Coll. Math. XV, 105-110. Plans, A. and Burillo, P. (1975) Operadores Nucleares y Sumabilidad en bases ortonormales. Facultad de Ciencias, Zaragoza. Rosenberger, B. (1975). Approximationszahlen, p-nucleare Operatoren und Hilbertraum-charakterisierungen. Math. Ann. 213, 211-221. Ruston, A.F. (1951). On the Fredholm theory of integral equations for operators belonging to the trace class of a general Banach space. Proc. London Math. Soc. (2), 53, 109-124. Saphar, P. (1970). Produits tensoriels d'espaces de Banach et classes d'applications lineaires. Studia Math. 38, 71-100. Weyl, H. (1949). Inequalities between the two kinds of eigenvalues of a linear transformation. Proc. Nat. Acad. Sci. USA 35, 408-411.

PARTICULAR

M-BASIC

SEQUENCES IN BANACH SPACES.

Paolo Terenzi Dipart. di Matematica del Pol. di Milano, Milano, Italy. Dedicated to Antonio Plans on his 65 t h birthday.

Abstract. We are concerned with the existence of M-bases with weaker properties than the strong M-basis, moreover, with the possibility of decomposing an M-basis in a finite number of subsequences with more regular properties.

INTRODUCTION. In what follows B is a Banach space; for the definitions of biorthogonal system, minimal, uniformly minimal, M-basis, norming M-basis, strong M-basis, basis with brackets and basis, see [1] or [2] and [3]. Let (x ) be a sequence with B = [x ] (=span (x )); we shall n

n

n

consider the representation of the elements of B by means of the elements of (x ). n

We suppose (x ) minimal, with (x ,f ) biorthogonal and llx 11 = 1 n

for

every

n.

By

n

[4] there

exists

an

n

" n"

increasing

sequence

(q )

of

m m=0

integers, with q =0, such that for every x of B there exists 0 numbers for which

(1)

x = lim p oo

[

q P

r

q m

i

, P+l

"\

I I *."»*• ' I v.

m=lL

n=q

+1

m-1 where (q ) does not depend on x. m

We s h a l l say t h a t x of B i s

J

n = q +1 p

J

(a ) of n

Terenzi: M-basic sequences

55 m

[

-,00

V f (X) X L,

n

n I J

m=l

n=l m

(ii) bounded on

(x ) if it is not

lim

n

X f (x)x n

Li n=l

m o1o1 mm

= +».

n »

A panorama of the representation of B by means of (x ) is as n

follows

(the definitions

of

"middle

M-basis"

and

"quasi

basis

with

brackets" are new): a) (x ) uniformly minimal <

> lim f (x)=0 for every x of B;

n

n oo

oo

b) ( x j M-basis <

> £ |fn(xH > °

n for

every

x * 0

of B;

n=l

c) (x ) norming M-basis <

\ for every x of

n

|/| s K sup {|f(x)|;f€ span (f ) ||f| = l>, where K(2:l) is fixed; d) (x ) middle M-basis <

> every bounded

(on (x )) element of B is

n

n

regular on (x ); n

e) (x ) strong M-basis t

i every element of B is regular on (x );

n

n

f) (x ) quasi-basis with brackets i

? for every x of B there exist a

n

permutation (n) of (n) and an increasing sequence (p ) oo

integers, with p = 0 , such that x = ) *0 L m=0

f

p

of

m m=0 m

) \ L^

f~ (x) x~ n n

^ ~_ m-1

rf V g) (x ) basis with brackets < >x = ) ) where (q ) is the sequence of (1);

1 f (x) for every x of B,

00

h) (x ) basis < n

> x = V f (x)x for every x of B. u

n

n

n=0

We know that every separable B has an uniformly minimal norming M-basis, moreover we know that a separable B in general does not

56

Terenzi: M-basic sequences

have a basis with brackets; the existence of e) or f) is an open question. In ,fl we

consider the existence of the middle basis and of

other intermediate M-bases; in §2 we consider the basic subsequences of an M-basis and the possible decomposition in a finite number of subsequences with more regular properties; in §3 there are two open problems.

1. MIDDLE M-BASES AND QUASI-STRONG M-BASES. Firstly we prove the existence of the middle M-basis in the reflexive spaces. Theorem 1. If the dual is separable B has a middle M-basis. Proof. Let (x ) be an M-basis of B. Let (g ) be dense in the unit sphere n n # of B . Let q be the first integer such that g (x ) * 0, we set i

i qt

g

y = x f o r l ^ n ^ qM n

n

consider g , if (x ) 2

e g

In n>q

q >q such that g (x 2

y

n =V = x

x=

x

In

n

i(Xn]

- —-,

g (X ) 1 q

r- x q 1

for n > q ; 1

we consider g , otherwise there exists 3

2j_

)*0, then we set lq

2

for

V 1 *n * % • 2

x

, l

-

In 7

„

^ x

for n > q .

So proceeding we get (y ) and a not decreasing sequence {I ) n

n

of integers, so that for every m (2)

span (y ) m

n n=l

= span ( x ) m

n n=l

, (y ) « c g . n n>£ m

m, -1-

Terenzi: M-basic sequences Since

(y )

is

n

57

obviously

minimal

it

follows

that

oo

M-basis of B, indeed, by (2) if y € n [y ] 1

n n—m

' m=l

(y ) n

is

we have g (y) = 0 for every n

n, hence y=0. If (y ) is not strong M-basic, by [3] (p. 243, prop. 8.11, n

3) we have that there exist x of B and (n) = (n )u(n'), k

k

(n )n(n' ) = 0, so k

k

that setting Y'=[y , ], n

k

x + y € n ty 11

m=l

+ Y>

> llx + Y> II = i-

i

k^m

n k

"

"

Then there exist F of (B/Y*)* and f of B* so that

(3)

F(x+Y') = |F| = 1 = ||f| = f(x), where f(x) = F(x+Y') Vx € B.

Let (g ) be a subsequence of (g ) such that i

n P

(4)

||g - f I < 1/2P for every p.

We shall prove that x is not bounded on (y ), which will prove the thesis. n

Indeed suppose x bounded, there exist two increasing sequences (r ) and m

(t ) of integers such that, for every m,

(5)

[ uY fk (x) y = 1 ) f (x)ny k I l l II ^ nu w k=l

k

k=l

h a < + ». k

For every integer p £ 1 there exists

m(p)

(6)

p

x = V f ( x )y , + V P L nk \ L k

k=i

K

a y , Pk k

k=s

+i

p-i

s

p-l

= n' t

m(p)

> I . ||x - x|| < — . i

p

"H p

"

_p 2

with

58 By ( 2 ) ,

Terenzi: M-basic sequences (6),

(3),

(4) and (5) i t follows

that

m(p) lim

8.i

p oo

(x

= lim

J P

\

m(p) = lim P -*»

*

k.l t

l i m ||g. p->oo

fn,(x)

f||-

lim - ^ - = 0;

y

k

p

k

on the other hand, by (3), (6) and (4),

lim poo

I

i

p

-x)) |

(x ) | = lim | f ( x ) + f ( x - x ) + (g p ' ' p i p oo p £ 1 -

lim |f p oo

(x - x ) + (g P

p oo

P p

^ 1 - lim [|xp-x| - lim

- f )(x + (x - x ) ) |

!

+

|g t - f|[||x|

+

||x

r I + I r iixii + i i i = iUP

?p

i "

ppJJ

•

therefore x cannot be bounded on (y ), which completes the proof of th. I. n

Let us consider

another

intermediate

sequence between the

M-basis and the strong M-basis. We know that (x ) is strong M- basic «=> (x +[x ] n

n k

)

n' i=1 k=l i

every (n) = (n )°°_ u (n')°°_

is M-basic for

, (n )n(n')=0.

Then we say that (x ) is numerably strong M-basic if there n

exists a numerable sequence of partitions of (n)

Terenzi: M-basic sequences N

such

59

,=

that (x n^

+ [ x , ]°° )°° n t j j . i k=l

is

M-basic for

i= l,2,

Next theorem shows the existence of this intermediate type of M-bases.

Theorem II. Let

(x )

be a sequence of

B

with

[x ]

n

dimension and let

of infinite

n

(N ) be a numerable sequence of partitions of

N = (n ) i

ik k=l

u (n' )

ik k=l

(n)

, (n ) n (n' )=0 , i = l,2,... ik

ik

Then there exists (y ) of [x ] such that n

n

(i) (y ) is norming M-basis of [x ]' with (y ) n

n

for every m, where

n n—m

c span(x )

n n—q

m

(q ) is a non decreasing sequence of integers with m

q

>+oo; m

(ii) (y ) is numerably strong M-basic along the sequence n

i

of partitions of (n). Proof. By th. I of [5] we can suppose (x ) norming M-basic. n

Consider i=l. We have (x ) = x x

1

u (x

)_

n kt Ik " '1

u (x

)

n kt n' Ik k^t1

Set

y = x 1

i

, x

in

= x

n

;k

for k £ t ' , ;k oo

"lk

~

1

"lk

m=t 1

suppose that

(x

+ X') In

(N )

1 n=l

i s an M-basis of X 1

, with (x + X \ F ) In

1

In

60

Terenzi: M-basic sequences biorthogonal, (f

)c B

so that f (x) = F F (x+X') for forevery x of B

In

In

In In

1

and for every n. Consider f . n There exists an increasing sequence such

f (x

that

11

X

i,i

)* 0

for every

k

(p(l,l,k)) of integers

(otherwise

it would

be

n lp(l,l,k)

cf

111

), then we set

x

= x ln

n

lk

ik

for t £ k s p(l,l,l); *

moreover, for every k and for p(l,l,k)+l ^ i ^ p(l,l,k+l),

fi (x ) 11

n li

= X

X In

n li

li

77 / f (X 11

Then f (x' 11

ln

X n lp(l,l,k)

n

) = 0 for k > p(l,l,l); that is

ik

n

f

+ X'1

|x'

m = p(l,l,l) + lL

Consider

> )

.

If

(]

ln

s i x + X' I

'L^

lk

x'

m£p(l,l,l) •-

+ "lk

L I"

X' -"k^m

cF .

1J^2

j. F

U±

there

exists

increasing sequence (p(l,2,k)) _ of integers such that

f (x* n

* 0 for every k, hence we set

x

= x' In

lk

an

-1-

for p(1,1,1)+l < k ^ p(l,2,l);

1n lk

moreover, for every k and for p(l,2,k)+l ^ i ^ p(l,2,k+l),

)

ipd,2,k)

Terenzi: M-basic sequences f „ x

(x* 12

>

=

In l i li

x In

12

li

n lp(l,2,k)

such that

= span (x )

In n>l

;

n n>l

c span (x )

I n n^m

l

In n>l

span (x ) (x )

\ )

l n

So proceeding we get y u (x ) 1

)

In

7Z / * f (x

In l i

61

for every m, where (q ) i s a non

n n^q lm

ltn

decreasing sequence of integers with q — > +oo; lm

(x In + [xIn' ]°° )°° j=t* k=t Ik

lj

1

is

M-basic.

1

Consider i=2. We have

(X

ln)n>2=

(X

2

1,2

; x

2n' 2k

)

Ik

= Set y =x

ln

(X

=x

In' 2k

k ^ s U( X l n ' 1

Ik

.n2k K±e2U

(X

for k £ s

W

1

ln'2k W 2

2

Suppose that 00

x' = [x

]

ln

(x 2n

(f

2k^2

+ X')°° 2 n=l

p

, x = n 2

m=

x

S2Lln2k

21

2J

J = S 2

Jk^

M-basis of X with (x +X',F )°° biorthogonal, 2

2n

and for every n.

We have that

ln

2n

2

2n n=l

)°° c B* so that f (x) = F (x + X' ) for every x of B

2n n=l

Consider f .

-i

i 00

+ [x

2n

2

62

Terenzi: M-basic sequences (n

= (n(1)) ^ (i) U (n(2)) > (2) with

)

2k k^s

2k

2

k^s

2k

2

k^s

2

Consider

two

increasing

sequences

2

1

of

integers

(p

(2,l,k))°°_

(i =l,2)

so

that

f

21

(x (i) In

(i) 2p (2,l,k)

) * 0 for every k and for i=l,2

(maybe that one of these two sequences of integers does not exist, for example for i=l, then we shall set x

(1) = x

2n

ln 2k

(1) for k^s

and we

2

2X

shall consider only i=2). Now we set, for i=l,2,

x

s (l) s k s p (i) (2,l,l);

(i) = x (i) for 2n

2k

moreover, for every k and for p U) (2,l,k) + 1 ^ j s p ( i ) (2, 1, k+1),

f (8)

Then

X* (i) = X (i) n 2j °2j

f

(x*

) =

0

for

21

(X (1)) In X

f

21

large

(X (i) ) In 2p(2,l,k)

k;

moreover

(i) 2p(2,l,k)

n

(x*

+

[x' , ]°°_ t) ^

still M-basic because by (7) and (8) we have that

[*' . ] " _ .

(x* 2n

)^

= Ix , 1". . ;

c span(x

k^m

)^ In

for every m, where (q(l,m))

k2:q(l,m)

is a non decreasing sequence of integers with q(l,m)—> +00.

In the same way we consider f

and so on.

is

Terenzi: M-basic sequences

63

So proceeding we get (y , y ) u (x &

&

1^2

span (x

)

)

= span (x

2n n>2

(x

)

such that

2n n>2

)

;

In n>2

c span (x

2n n^m

)

for every m > 2, where (q

In n^q

) is a

2m

non decreasing sequence of integers with q —> +« ; 2m

(x

2n

+ [x Ik

(x

2n' lj

+ [x

^3

)°°

i s M-basic;

j=s* k=s 1 1

, ]°° ,) i s 2n

Consider i=3. We set y = x

]°°

J=S

2J

M-basic.

2

; then we have 2,3

= (x2n )k^r U (x2n )k^r' Ik

= (X *• x «

~

Set x

3n' 3k

= x

2n* 3k

Ik

U (X

'. «^

t

1

)

U IX

J

for k ^ r . Suppose that K 3

X' = [X 3

]

2n' k^r' 3k 3

(x + X')°° 3m

(f

1

)

^

. X = 3

M-basis of X

3 n=l

)°°

f|

'' m=r

tx

c B*

so that f

Consider f . 31

We have that

"

in

2n' 3j

(x) = F

]

j=r'

+ X' , F

3n

3n

and for every n.

tx

3k

with (x

3

3n n = l

+

2n

3

] 3

k^m

) biorthogonal,

3n

(x+Xl) for every x of B 3n

3

64

Terenzi: M-basic sequences

~r3

3k

~P3

3k k ^

k2:r3

(3)

~ ri

Ik k^rj

_

~P2

2k

k ^

,

P

F

~ 3

P

~ 3

~F2

~ 1

We consider again four sequences of integers (p

(3,l,k))°°_

(1=1,2,3,4) so that

f

31

(x

2n

(i) ) * 0 for (i) 3p (3,1,k)

e v e r y k and f o r

1=1,...,k

.

If some of these sequences of integers do not exist, then we shall proceed as for (p(i)(2,l,k)). It is now sufficient to set , for i=l,...,4,

x

r(i)-s k ± p(i)(3.1,l) ;

(i) = x (i) for

3n 3k

2n 3k

3

moreover, for every k and for (p(l}(3, 1, k) + l ^ j * p (l) (3,1,k+1),

f (x (D) 31

2n X

2n (U (i, 3

3p

Now the procedure is clear, for (f ^

)

(

(3,l,k)

and for i>3.

3n n>l

So proceeding we get (y ) as in the thesis (indeed, since (x ) is norming and since (y )

n n^m

n

n

c span (x )

n n^q

with q —-> +oo, (y ) is norming too). This m

m

n

completes the proof of th. II. We shall say that an M-basic sequence

(x ) is quasi-strong n

M-basic if (x ) has the possibility to become strong M-basic by means of n

the procedure of Th. II; that is if, after removing the not regular points along a numerable sequence of partitions of

(n), the sequence becomes

Terenzi: M-basic sequences

65

strong M-basic. About

the stability we point

out

that

a middle

strong) (quasi-strong) (strong) M-basic sequence keeps its sufficiently

"near" sequences:

it is sufficient

(numerably

properties for

to prove this for the

strong M-bases.

Proposition.

If

(x ) is strong M-basic in B,

there exists

(e ) of

n

n

positive numbers such that every (y ) of B with fly - x II < c n

n"

n

for every n

n is strong M-basic. Proof. If (x ,f ) is biorthogonal suppose that n

n

= l/(2 n + 1 ||f ||) for every n.

||y - x II < e 11

n"

n

" n"

n

Following the techniques of the proof of the Krein-Milman-Rutman theorem (see [2] p. 84-99) it immediately follows, for every (a ) m

, of numbers,

n n=l

H I Ia * h I Ta y I s I I Ta * 2 £

L< n=l

n

n

IIn=l L. n

n II

^

In=1 L n n

If (y ) is not strong M-basic there exist (n) = (n ) u (n*), (n ) n (n')= 0 n

°

k

k

k

k

and y of B so that

p

r

_p

(10)

II y + iy niJ III = i ; III y - fI L) ap, kyn:,k + Y } II < ?L<--, \k yvk-' k=s

for every p where (s )

0

Set for every p

p

r

P

=

+1

is an increasing sequence of integers with s =0.

p p=0

x

p-1

) a x L Pk nik

+ k *—k=s +1 p-1

n

k

66

Terenzi: M-basic sequences

by (9) and (10) there e x i s t s x of [x ] so that n limx

= x

hence

,

^ ^ | | x + [ x

K

p-X»

]

II s 2 ;

k

(x ) would not be strong M-basic, which completes the proof of

the proposition.

2. BASIC SUBSEQUENCES AND DECOMPOSITION IN A FINITE NUMBER OF MORE REGULAR SUBSEQUENCES. Firstly we consider the from a sequence.

"extraction" of basic

subsequences

It is well known that every M-basic sequence has an

infinite basic subsequence. The presence of these basic subsequences is very "visible" in the norming and uniformly minimal case, as follows:

Theorem III. If (x ) is norming and uniformly minimal there exists an n

increasing sequence

(q )

of

integers with q =0

m m=0

( xn ) m

m=l

with

qm-1 + l ^ nm ^ q

such

that every

0 m

for every m is basic.

Proof. Since (x ) is norming we can choose the sequence (q ) of (1) so n

m

that, for every m. 1

n n>q

" "

m+1

Consider (x

n m

) with q

m-l

+1 ^ n

Let x € [x

q

) * llxll (1 - - ) for every x € [x ] m

d i s t (x, [x ]

^ q

mm

_m 2

n n=l

for every m.

] with |x| = 1, by (1) and since (x ) is uniformly m

minimal there exists a subsequence p ^ p(i) and for every i,

(p(i)) of (i) such that, for every

Terenzi: M-basic sequences

67

p

I

1

x - f V f (x) x + V L

I

n m

m=l

a x 1 | < -; f (x) n m m l p+1 L . J I 2

n m

m=q +1 P

P

I

P

x - y f (x) x + x I > I x - y f (x) x I fl - - 1 for L

n m

n

H

L

n

everym x € II [x ]II

n I I

m

n n>q

m

II

o

ij

J_ J

V

p+1

It follows that

I

P

-

P

r

-

x - m= ) f1 L

n

II

( x ) x n

m

II -

P+*

r

-

< x - )m=f 1

m"

L

"

n

q p+2

p+1

( x ) x m

V" n

- m=q > m

/ L

a +1

^ p+1 tn=q +1

II m

x

1

m . " 1 -

ro i 1 2-

Vi

< f I x - Y f (x) x - V a "-^p,,*1

.-1

21"1

2

'

1 - Il 2

for every i, that is x = y f (x) x , which completes the proof. L

n

n

We pass to consider the possible decomposition of an M-basis in a finite number of more regular subsequences. It is known that, if (x ) is norming M-basic, the sequence n

(q ) of (1) can be selected so that oo

U m= l

q

(x ) 2 m

n n=q +1 2m-l

oo

and

U m= 0

q

(x )

2m+1

n n=q +1 2m

are basic with brackets.

In the example of [4] we proved that in general an M-basic

68

Terenzi: M-basic sequences

sequence

(x ) cannot

be decomposed

in a f i n i t e

number of

basic

with

n

brackets subsequences (also if (x ) is quasi-basic with brackets). n

In a recent seminar at Zaragoza the following questions were raised: Question 1. Is it possible to decompose every M-basic sequence in 2 (or in a finite number of ) strong M-basic subsequences? Question 2. Does there exist in every M-basic sequence a strong M-basic (or basic) subsequence which is "maximal" (that is, by adding infinite elements the sequence loses its properties)? Next three examples give negative answers to question 1. 00

In what

follows

(x ) = U (x

)

is algebraic

basis of a

mn n=l

n

linear space L; (y ) is M-basis of a Banach space Y, with Y n L = 0. n

EXAMPLE 1.

Suppose that

(y )

is not strong

M-basic

and set for every

n P

U

P

(a )

m=l

mk k=l

of numbers

p

(11)

p

p

m-1

I y f y a x 1 1 = I y I y (a x +a x )+a x I \\ L [

L

m=l

mk mkj ||

|| L \\ L

k=l

m=l

mk mk km km

mm mmfl

k=l

m-1

p

= Im=l 7(1 k=l J ( | a |+|a f c | ) + | a | )y I || L \ || L

' mk '

' km'

' mm'

m ||

This defines a norm on span (x ), n

since, beeing

(y ) minimal, there exists n

moreover

(x ) is minimal n

(6 ) of positive numbers such n

that, for every (a ) m

of numbers,

n n=l

|a | 5 ;

ay

' n'

n

n n

hence for every

U m=l

(a ) p _ of numbers by (11) it follows that

69

Terenzi: M-basic sequences

L^L'mk m=l

k=l

m=l

1

kj

k=l

Moreover (x ) is M-basis of X = [x ], since otherwise there n

n

would exist x of X, ||x| = 1, an increasing sequence (p(s)) of integers and a sequence (a ) of numbers so that, for every s p(s +

m-l

f V (a x +a x )+a x 1 I < -i— ,

x m=p(s)+l

[

l_^ k=l

mk mk

km km

mm mmj

||

~s +l

hence for every i

p(s+l)

m-l

I (£••-".

. +a x mk

m=p(s)+l

p(s+l+l)

* I"

)a

km km

k=l

.

mm i mmj

m-l

)

)

/ I m=p(s+i)+l

/ k=l

(a x +a x mk mk


)+a x

km km

mm mmj

2s

therefore, setting for p(s)+l ^ m ^ p(s+l) and for p(s+i)+l ^ m ^ p(s+i+l)

k=l

by (11) i t follows that

p(s+l)

Y L^

m=p(s)+l

bm my > 1 - J— , I V b y - V bb yy I < p s + 1 II L^ m m Z - m mll m=p(s)+l

,

m=p(s+i)+l

which is impossible since (y ) is M-basic. n

We claim that (x ) cannot be decomposed in two strong M-basic

70

Terenzi: M-basic sequences

subsequences. We point out by

(11), that for every m and n, (x

)

and

mn n=l

(x

)

are not strong M-basic. Suppose (x ) = (x

mn m=l

n

(x t) strong M-basic; since n'

n

) k

u (x

k=l

)

n' k=l k

(y ) is not strong M-basic, by

with

(11) there

n

e x i s t s an i n f i n i t e s u b s e q u e n c e ( m ( i ) ) of (m) s u c h t h a t

U (x

)" m(i)n

therefore (x

n' k

fi

(x , ) ;

n=l

n ) cannot be strong M-basic, since every (x &

)

m(i)n n=l

is not

strong M-basic, for every i. This completes the example.

EXAMPLE 2.

As in example 1, where now we can suppose (by example 1) that

it is not possible to decompose (y ) in two strong M-basic subsequences. n

Therefore now, for every m and n, it is not possible to decompose (x ) mn n=l

and (x

)

in two strong M-basic subsequences.

mn m=l

Then, proceeding as in example 1, if (x )= (x with (x

n

)

u (x )

n k=l k

n

n k=l k

) strong M-basic, it is not possible to decompose (x

n k

k

) in two

strong M-basic subsequences. That is, it is not possible to decompose (x ) n

in three strong M-basic subsequences. So proceeding for every m there exists an M-basic sequence which cannot be decomposed in m strong M-basic subsequences. EXAMPLE 3. Suppose that, for every m,

(y

)

is M-basis of a Banach

mn n=l

space Y , so that

it is not possible to decompose

(y

) in m strong

mn

m

M-basic subsequences, moreover suppose that Y +Y + ...+Y + ...n L = 0. We 1 2 P

set for every

U m=l

(a

)

of numbers

mn n=l

\l[l \. *J | = I a

•

mn

y mn

m

Terenzi: M-basic sequences

71

Then (x ) i s M-basic of X = [x ] = V [x n

n

]°°

L

; but i t i s not

mn n=l

m=l possible to decompose (x ) in a finite number of strong M-basic subsequences n

Finally the next example gives negative answer to question 2.

P

EXAMPLE 4. If (x ) and (y ) are as above, set now for every n

n

U

(a

)

mk k=l

m= l

of

numbers

m=l

k=l

Set X = [x ], it is easy to see that (x ) is an M-basis of X. n

Let (x

n

) be a strong M-basis (basic) subsequence of (x ), with

n

n

(xn ) = U (xp(m)q(k) )°° ; k=l k

m=l

it cannot be (p(m))=(m) otherwise by (12) (x

) would not be strong

n

k

M-basic, since (y ) is not strong M-basic. n

Therefore if m € (p(m)) we have that (x

n

) u (x-, ) k

mk

k=i

is

still strong M-basic (basic). Hence (x ) does not have a "maximal" n

strong M-basic (basic) subsequence.

3. OPEN PROBLEMS. About the middle M-bases: Problem 1_. Does there exist in every separable Banach space a middle M-basis?

This

is a weaker

problem

than

the

existence

of

the

strong

M-basis. About the quasi-strong M-basis: we don't know an example of an M-basis which is not quasi-strong M-basis, hence.

72

Terenzi: M-basic sequences

Problem 2. Is every M-basis a quasi-strong M-basis ? By th. II a positive answer to problem 2 implies positive answer to existence of the strong M-basis.

REFERENCES. [1] [2] [3] [4] [5]

Lindenstrauss, J. &Tzafriri, L. (1977). Classical Banach spaces I. Sequence spaces. Springer-Verlag 1977. Singer, I. (1970). Bases in Banach spaces I. Springer-Verlag. Singer, I. (1981). Bases in Banach spaces II. Springer-Verlag. Terenzi, P. (1984). Representation of the space spanned by a sequence in a Banach space. Archiv. der Mathematik 43, pp. 448-459. Terenzi, P. (1987). On the theory of fundamental norming bounded biorthogonal systems in Banach spaces. Transactions of the A.M.S. (299) 2, p. 497-509.

BEHAVIOUR OF SEMhFREDHOLM OPERATORS ON A HUBERT CUBE Alvaro A. Rodes Usan. Departamento de Matematicas. Universidad de Zaragoza.

A Antonio Plans, mi maestro y amigo.

0. INTRODUCTION be

Let H be a real separable Hilbert space. Let B, £j, o

the

set of bounded linear, compact and nuclear operators on H respectively. There

are

several

characterizations

through

absolute

summability of distinct types of operators of B. For instance

A € 3

if and only if 3(e ) orthonormal basis (o.n.b. ) of H,

1

n

£ ||AeJ< co [Ho] Y |Ae ||< oo for every o.n.b. of H if and only if A=0 [Bu]

Ae 3if inf j £ |AeJ; (ej (o.n.b.) of H j = £ |AeJ< « where (e ) denotes the orthonormal basis of singular vectors of A (i.e n

*

eigenvectors of (A A)

1/2

).

[Go-Ma]

If we change the role of the unit sphere (support set of all the orthonormal basis) by that of the Hilbert cube, we obtain the Macaev's ideal £

defined by

A€ 3; £ 1/n jAeJ < oo | [Go-K] Since A is linear, Y l / n I"I Aen"|| =LV "IA 1/n e IL so the Macaev's L n" ideal

is

made

up

of

all

the

compact

operators

which

give

absolute

74

Rod€s: Semi-Fredholm operators

summability over the vectors (e /n) which are precisely the central points n

of the faces of the Hilbert cube associated to the o.n.b. of singular vectors

Q(e ) = { x€ H; x = V x e , |x I * 1/n \ n

(^

L

n

n

' n'

J

In this paper we have started to study the set of operators of B which give absolute summability on vector sequences of a Hilbert cube.

The

set

of

o.n. bases

can be

considered

as

the set of

intersections of the unit sphere of H with distinct complete

orthogonal

systems of rays (r ) of H, i.e.

(e ) = S n (r ) n

n

(we take only one of the two intersection vectors) However there are as many Hilbert cubes as ordered o.n. basis. For a fixed Hilbert cube Q, the relative position of a complete orthogonal system of rays (r ) depends obviously on the set of rays; there are even n

complete orthogonal systems of rays which intersect Q only in the null vector,

and

in this case every operator

of B gives clearly

absolute

summability on the intersection vectors of (r )n Q. The problem will be non trivial only when (r ) n Q * 0, n

for ininitely many values of n.

The simplest case (we shall concentrate our attention on it) is when the rays (r ) intersect Q in the central points of its faces, that n

is for the orthogonal system r =[e ], ne IN. (We denote with [ ] the closed n

n

linear span). We only consider the Hilbert cubes whose faces central points are intersections with these rays. Such cubes can be denoted by

Rod6s: Semi-Fredholm operators

75

i
being


| x Iss 1 / n } . ' n1

Our goal is thus studying operators A of B with the condition

If

A

verifies

(*)

we shall

say that

A

gives

absolute

summability over r n
n

^ (Q,A) = inf jY ||A 1/n e , J f
^ (Q,A) = sup |^ ||A 1/n e / J,
1. THE BEHAVIOUR OF OPERATORS WITH RESPECT TO Q

In [Ro.l] the compact operators are characterized by

A€ 3 if and only if VQ, 3c, £ (Q.a.A) < »

This condition is equivalent to

A€ ,3 if and only if VQ, £ (Q,A)=O,

which can be proved easily by convenient choices of permutations o\

In [Ro.2] we have given the following characterization of semi-Fredholm operators

76

Rodes: Semi-Fredholm operators A* 4>+ if and only if 3 Q, £ (Q, 1,A)< oo ($ +

stands for the set of closed range,and finite

dimensional kernel operators).

As in the former case it is easily proved

A* $ + if and only if 3 Q, £ (Q,A)= 0

From now on we shall pay attention to the supremum over the permutations of a given o.n.b.

sup « V IA 1/n e . A,

instead of the infimum as we did before.

We give a new characterization of operators which are not semi-Fredholm, by the existence of o.n.b. which make the former supremum arbitrarily small, that is

inf J sup iy IA 1/n e , J , or permutation of IN V; (e )o.n.b. of H I = O ,

A * $ + if and only if inf j £ (Q.A); Q Hilbert cube 1 = 0 To prove this assertion we have to show first that for a fixed operator A, with non closed range or else with infinite dimensional kernel, and a given e>0, there exists a o.n.b. of H, (e ) verifying n

sup < V ||A 1/n e , J|,
77

Rod6s: Semi-Fredholm operators

On the other hand if A€ $ , it must be proved that for every o.n.b. (e ) +

n

and for every
I» A 1 / n e
in certain finite dimensional

subspaces of H. Such changes are

expresed in the following form

2P)

M :(e ; n=l P

where M

(u ; n=l,

2P)

n

is a 2px 2 P matrix obtained by induction

p

V2

|l

-l|

V2 N

11

I" I p p N -N

1

p

P

N

N 1 p

p+1

*

N

P

_

±1

~N

Numerical lemma: a) lim 2~ p/2 V 1/n = 0 p

n=i

b) Let (a ;n€ IN) and (b ; n€ (N) be n

two non increasing

n

sequences of positive real numbers. Then

sup -I V a b . .; c permutation of IN> = V a b .

Proposition 1.(Operators with non closed range). Let A be a bounded linear operator with non closed range. Then

78

Rod6s: Semi-Fredholm operators

inf | V (Q,A); Q Hilbert cube 1 = 0

Proof. - By [PI] there exists an o.n.b. of H, (e ) such that n

|A e j

> 0.

Let e>0. Take a sequence of positive real numbers (c ; n€ IN) n

s.t. V c < e. We construct a new o.n.b. of H (v ;n€ (N) performing changes U

n

n

in blocks. Block 1 (B. 1). For a given c >0, 3p € IN s.t.

C

-p /2

2

' I

1/n

< 21*1

which is always possible by the numerical lemma a) For p € IN, 3 a € R, a >0

s.t.

a < 1

Take now 2 a-l elements of the basis (e ), le1, e1, . . . , e1 n \ 1 2 Pa J 2

1A ej| < a , (1 = 1

2 ^l) > 0)

Perform

in

[e,e,e,...,e 1 1 2

-1

(this is always possible since ||A e |

the

], given by p

I such that

change

of

o.n.b.

the

subspace

Rod6s: Semi-Fredholm operators M :(e . e 1 ^ 1 , . . . ^ 1 1 1 2* p v

1

79

)

> (u , 1'

P

1 2-1

u

) P

2

1

-