ISRAEL JOURNAL OF MATHEMATICS, Vol. 37, Nos. 1-2, 1980

ON THE RELATION BETWEEN SEVERAL NOTIONS OF UNCONDITIONAL STRUCTURE

BY

W. B. JOHNSON,* J. LINDENSTRAUSSt* A N D G. S C H E C H T M A N *

ABSTRACT

The unconditional structure of a space constructed by Kalton and Peck is investigated. A m o n g other things it is proved that even though this space has an unconditional decomposit ion into subspaces of dimension two, it does not have G.L. I .u .s t .

§1. Introduction

In [4] Kalton and Peck found a general way to construct spaces which fail the

three space property, i.e. spaces Z which for some 0 < p < oo contain a subspace

X such that both X and Z / X are isomorphic to lp while Z itself is not

isomorphic to Ip. For 1 < p < oo the space Z can be chosen to be a Banach space.

We will be interested mainly in a space denoted in [4] by Z2.

It was shown in [4] that, although the space Z2 has an unconditional (even

symmetric) decomposition into two dimensional subspaces, Z2 does not have a

complemented subspace which has an unconditional basis. In Section 3 we show

that Z2 does not have G.L.I.u.st . Since, by Proposition 1, every space X which

has an unconditional decomposition into k-dimensional subspaces, for some

k < 0% has the G.L. property (i.e., every absolutely summing operator from X

factors through an L~ space), Z2 answers part of question 1 in [7].

In [4], Kalton and Peck proved that every infinite dimensional complemented

subspace of Z2 contains an isomorphic copy of Z2. In Section 4 we show that

their proof actually shows that such complemented subspaces contain com-

plemented isomorphic copies of Z2. Since Z2 does not have G.L.I .u .s t . , this

result yields that no complemented subspace of Z2 has G.L. i .u .s t .

The result of Section 4 suggests that Z2 is a new prime space. (Recall that a

* Supported in part by NSF Grants MCS 76-06565 and MCS 79-03042. ** Supported in part by NSF Grant MCS 78-02194. Received December 5, 1979

120

Vol. 37, 1 9 8 0 UNCONDITIONAL STRUCTURE 121

space X is prime if every infinite dimensional complemented subspace of X is

isomorphic to X. The only known prime spaces are lp, 1 =< p --- ~ and co.) We

believe this not to be the case; in fact, in Section 5 we give some partial results

that suggest that Z2 is not isomorphic to its hyperplanes. If our conjecture is true,

then a hyperplane Y in Z2 would be a space such that Y is isomorphic to

Y O Y O Y but not to Y ~ ) Y .

The space Z2 of Kalton and Peck is the space of all sequences {(a,, b,)}~=~ of

couples of real numbers such that

]]{(a., b.)}:=, ][ = ( ~ l b 2)1/2+ (~=1 (a , -b , log (I b, [( ,~=1 bZ") -1/2))2) 1/2

is finite. The expression above is not a norm but is equivalent to a norm; that is

one of the main points in [4].

Define

e. = {(6..,, 0)}7=,, f, = {(0, 8,.,))7=1, n = 1, 2 , . . . .

It is easily checked that the sequence

e,,fl, e2,h, "'"

forms a basis for Z2. It is obvious that Z2= E~=~)[e , , ] ' , ] and the sum is

unconditional (even symmetric); [e. ]~=~ and Z2/[e, ]~=1 are both isomorphic to 12,

but Z2 is not isomorphic to a Hilbert space since [f,]~=l is isomorphic to the

Orlicz space l~20ogx)2. So Z2 is another solution to the three space problem. We

are not going to use this property of Z2. Two properties we are going to use are

(i) Z2 is reflexive (this follows from a computation of the dual space, see [4]),

(ii) Z2 does not contain l"~'s uniformly (by [1] Z2 has type 2 - e and cotype

2+ e for every e >0) .

In fact, it is even true that Z2 is super-reflective.

The notations and definitions which are not explained here can be found in [5]

or [6]. We just recall the definition of G.L. l .u .s t . A Banach space X has G.L.

(Gordon and Lewis) l.u.st if for every finite dimensional subspace E of X, the

inclusion operator i : E ~ X factors through a finite dimensional space U with

an unconditional basis in a uniform manner, i.e., if there exists a K such that for

every finite dimensional subspace E of X there exist a finite dimensional space

U and operators B : E - - - * U , A : U ~ X such that A B = i and

tl A II" II B II" x(u) < K. (x(U) is the unconditionality constant of U.)

§2. The following is a simple generalization of the fact, observed by Gordon

122 w. 13. JOHNSON ET AL. Israel J. Math.

and Lewis [3], that spaces with an unconditional basis have the property that

every absolutely summing operator from them factors through 1~.

PROPOSITION 1. Let X be a Banach space with an unconditional decomposi- tion X = ~ 7 = ~ O E , where sup, d im(E, )<o~. Then every absolutely summing

operator from X factors through l~.

PROOF. Let T : X - ~ Y be an absolutely summing operator. For each n

choose a basis x T . . . , x ~ n for En so that the vectors TxT/]lTxT[t, i =

1 , 2 , . . . , dim TE., form an Auerbach system in TEn (cf. [5, p. 16]), and so that

TxT= 0 for i > d i m TE,. Let {e~'}~"_-,,~=. be the unit vector basis of l~ and define

A : I I ~ Y and B : X ~ I ~ by

Ae~= TxT/I[TxTll (0 if Tx 7-- 0),

Bx '~ = [I Tx '~[l e '~.

It is easily checked that A and B are bounded and that A • B = T. Indeed, to

check the boundedness of B let K be the unconditionality constant of the

decomposition X = 2 , O E,, let M be the absolutely summing norm of T and

let a ;' be scalars. Then

n ~ l i = 1 n n = l i = 1

<= K M sup dimE~ a ~x ~ . [] . n = l iff i l

This shows that Z : has the G.L. property. The next proposition is the first step

in the proof that Z2 does not have G.L. l .u .s t .

PROPOSITION 2. Let X be a Banach space with the following properties:

(1) X is complemented in its second dual,

(2) X does not contain l"~'s uniformly,

(3) X has an unconditional finite dimensional decomposition

X = ~, E]) E~. n = l

Then X has G.L. l .u . s t if and only if there exists a Banach space Y with an k

unconditional basis {Y~,,}~"--Ln=~ such that (i) X is a subspace of Y and E. C [y , , , ]~ .

o n t o k

(ii) There exists a projection P : Y --~ X with P([y,..],~=~) = E,.

VoL 37, 1 9 8 0 UNCONDITIONAL STRUCTURE 123

PROOF. The if part is trivial. Assume now that X has G . L . I . u . st. By [2], we may assume that X is complemented in a Banach lattice L which does not

contain 12's uniformly. In particular L can be considered as a space of functions

on [0, 1] with L~ C L C LI and L~ dense in L (see [6] theorem 1. b. 14). A simple

perturbation argument shows also that one may assume that each E., n =

1 ,2 , . - . consists of simple functions. Thus, for each n one can find disjoint

functions YL.,'" ", Yk..o in L such that [y,,.]~Z~ D E..

Recall that R a d X is the subspace of LdX ) spanned by {ro(')'x}7=LxEx where r.(t)= sign sin2~m are the Rademacher functions. It is easily checked

that if O is a projection from L onto X then 0 defined by

0(~ '~ r . ( t )y.) = 2 r . ( t )Oy, with y. E L

is a bounded projection from R a d L onto R a d X (apply [6] theorem 1.f.14). Now, span {r. ( . ) E . }7-~ is complemented in Rad X; there are two ways to see

this:

(i) use a diagonal argument as in prop. 1. c. 8 in [5],

or

(ii) use the Maurey-Khinchine inequality (for {r. (t)r~, (s)}7.= =, see [6] theorem

1. d. 6) to show that {r. (')E.~}7~,.=1 is an unconditional f .dd. for RadX.

The space span{r . ( . )E.}7: , is clearly isomorphic to X and is contained in

span{r.(-)y,,.}~ZL~=~= Y so the only thing we still have to prove is that

{r.(-)y~,.}~"-_, ~=j is an unconditional basic sequence. But, by the

Maurey-Khinchine inequality

( f {I ~ ~, a,,.r. (t)y,.,,J{2)1/2 II(~ n (~i a,.,,y,..)2)~1{

= It(~dn ~i ~2~ny i2B)l[2Jl" []

REMARKS. (1) From the proof one gets that Y also does not contain ITs

uniformly.

(2) If sup. dim E. < o% one gets that sup. k~ < o0

§3. Given an operator T on R2 we consider it also as an operator on each of

the two dimensional spaces [e,,f,] in a natural way (using e,, f , as a basis). We

now define formally an operator 2P on Z2 by

n = l n = l

124 w.B. JOHNSON ET AL. Israel J. Math.

PROPOSmON 3. "F is bounded if and only if there exist ct and [3 so that

T(aen + bf.) = (aa + b[3 )e. + ba[.,

i.e., if and only if the matrix of T is o[ the [orm

(,~ /3). 0 a

PROOF. The if part is again very

"/~ is bounded and that the matrix

If y # 0 then

simple. T o prove the only if part assume that

of T is of the form

('~ /3). 3' 8

7"(~ a . e . ) = a ~ a.e.+y~,a.]:, . t l = l n = l n = l

Let {a.}7_, be such that ET=~aZ, = 1; then [IET=~a,e. ll<oo. This implies that

11~7~, a°f~ II is bounded , so we get that

(~1(a~logla. I)2)l/2<=K(~la2)'/2

which of course is false. Thus y = 0 and

For {b.}7~, such that ET=, b2. = 1 we get

:) "3 181+(~__ (cta.+/3b,-Sb, loglb. I = < J l ? ' J l ( l + ( ~ ( a , - b . loglb , I): ) .

If { b . } ~ are also chosen to satisfy

)2) 1,2 (~, (b. loglbn I _->N

and we choose a~ = b~ iog lb~ J we get

181+1o,-81" N- I/3 I--- 117"11 which is a contradict ion for large N unless a = 8. [ ]

VOI. 37, 1 9 8 0 UNCONDITIONAL STRUCTURE 125

We are ready now to prove the main result.

THEOREM 4. 2 2 does not have G.L.t .u.st .

PROOF. Assume that if has. By Proposition 3 it is enough to produce an

operator T on Z2 such that

(i) T ( e , , , f . l ) C [ e . , f . ]

and

(ii) i n fd (T , , , , . t . , , s pan{ (~ ~ ) } , . ~ ) > 0 .

(Tpf,o.r.i is considered as a matrix and the norm in (ii) is the operator norm.) Indeed, if such an operator exists it is easy, using the fact that Y,~)/5. is a "symmetric" decomposition and a simple compactness argument, to construct a

diagonal operator which does not satisfy the conclusion of Proposition 3.

We may assume that we are in the situation of Proposition 2, with E, = [e., f,],

n = 1 , 2 , . . . . The operator "/" that we shall use is of the form ]P(x)=

P(E,,.~A * x y , , . ( )y , , . ) for some set A (where the y,~. denote the system bi-

orthogonal to y,,.). It is clearly enough to find, for each n, a set A. C {1,..-, k.}

such that TA. defined by

i E A n

has the property

d (TA., span {( 0 ~)},~,~) _--> /z

for some absolute constant /z > 0, and all n. Fix n. For each 1_-< i =< k, define

7', * = P(y , . , (x )y i , , ) , x E [e. , f .] .

This one dimensional operator must have the form

( a~ b, ) or ( 0 0 ) . a~a~ a~b~ a, b~

In the second case define ai = 1 (this wi]] simplify the notation). Since ~:~=~ T~ = I

and the sum is unconditional we get that k n k

Z a,b, = 1, Z I b , I<=K i=1 i=1

where K is the unconditionality constant of {y,..} times IlP[I.

126 W.B. JOHNSON ET AL. Israel J. Math.

Define

then B = {i, l a, I > 1/2K}

1 Y~ a,b, > ~ .

Clearly, for each i E B at least one of the following four possibilities occurs (with

/z an absolute constant):

(i) or,a, > /~ [a, bl I, (ii) - a,a, > I~ }a~b, [, (iii) a, - a,b, > /~ la, b, I ( - a,b, > i~ I ct,b~] in the second case),

(iv) a,b, - a, > p. let,b, I (ct,b, > ~ l a,b, I in the second case).

Thus, there exists A C B such that Ei~Aa,b~ > 1 / 8 and one of the four pos-

sibilities holds for all i E A simultaneously. It is easily seen that

s.an{( o ilEA 0 tlo)

is bounded away f rom zero (by an absolute constant) . [ ]

04. He re we prove

THEOREM 5. Every infinite dimensional complemented subspace of Z2 con- tains a complemented subspace isomorphic to Z2. In particular, no infinite dimensional complemented subspace of Zz has G.L. l .u . s t .

The proof depends heavily on theorems 5.1 and 6.5 of [4].

Let {x.}~=l deno te the unit vector basis of 12. For u = Z7=1 a.x., v = Z:=l b.x. we define (u, v) to be the sequence of couples {(a., b.)}~=~. We also define

= .=12 - a . log l a . Ix. + log , a .) .~=1 a.x.

(0 log 0 = 0) so that

Let {ok = Ej~a, aixj}~=l be a block basis of {xj}7=l normal ized in 12, i.e., the sets 2 Ak are disjoint and Ej~a aj - 1, k = 1, 2, • • .. It is easy to show that the sequence

{(vk, O),(F(vk), V~)}~=, is a basic sequence in Z2 and its basis constant does not

depend on the part icular choice of {v~}~=~. In [4] (proof of t heo rem 6.5) it is also

proved that this sequence is equivalent to the natural basis of Zz, namely

Vol. 37, 1980 U N C O N D I T I O N A L S T R U C T U R E 127

{e~,fk}~=~. Furthermore the proof of theorem 6.5 in [4] shows that any infinite

dimensional complemented subspace of Z2 contains a sequence which is a

perturbation, as small as we wish, of a sequence of the kind {(v~, 0), (F(v~), vk )}~=~. Thus, it is enough to show that any such sequence spans a complemented subspace.

By theorem 5.1 of [4], Z* is isomorphic to Z2. More precisely, let Z2 be the

space of all sequences {(a,, b,)}7=~ such that

2" ~ I/2 1/2 II{(a.,b.,}:=,ll, ( ~ b . ) + ( ~ ( a . + b . logOb, l ( ~ b Q - ~ / 2 ) ) ~)

is finite (notice the slight change from the definition of the "norm" in Zz). Then 22, with a norm equivalent to II" II~, is the dual space to Zz where the duality is given by

<(u, v), (w, z)) = (u, z ) + (v, w).

( ( - , . ) on the right denotes the inner product in 12.) Now, the sequence {( - F(v~ ), v~ ), (vk, 0)}~=, in Z2 is biorthogonal to {(v~, 0), (F(v~-), v,, )}~=~ in Z2. Put

P(u, v) = ~ [((u, v), ( - F(v~ ), vk )) (VE, O) + ((U, V), (V~, 0)) (F(vk), Vk )]. k=l

Then, in order to complete the proof, it is enough to notice that the sequence {( - F(vk ), v~ ), (vk, 0)}~=1 is equivalent to the sequence {[k, ek }~=1 in 22 which is the biorthogonal system to {ek, fk}~=, in Z2. This, again, follows from the proof of theorem 6.5 in [4]. []

05. R e m a r k s and a conjecture

We first remark that the index 2 (in Z2) does not have any special role; all the results here carry over to Zp for any 1 < p < oo. The changes in the proofs are formal except in the proof of Theorem 5; there the projection should be changed to

P(u, v) = ~ [((u, v), ( - F ( (p - 1)- 'v *), v ~))(v~, O) + ((u, v), (vk, O))(F(v~ ), Vk )l k= l

where

v ~ = ~ / aj I~-'sign ajxj. i~Ak

128 w.B. JOHNSON ET AL. Israel J. Math.

The rest of this section is devoted to a discussion regarding the following

conjecture.

CONJECTURE. The space Z2 is not isomorphic to any of its hyperplanes.

It is proved in [4] that the quotient mapping Q from Z2 onto Z2/[e,]7=, is

strictly singular and hence Q T is compact on [e,]7=~ for every bounded opera tor

T on Z2. Thus every bounded opera tor on Z2 "a lmos t" maps [e,]7=t into itself.

This suggests the following problem.

PROBLEM. Does every bounded operator on Z2 have a perturbation, by a

strictly singular operator , which carries [e,]7=1 into itself?

An affirmative answer to this problem would imply that Z2 is not isomorphic

to its hyperplanes. To see this, suppose that T is a bounded opera tor on Z2

which carries [e,]7=1 into itself. T then induces a bounded operator , 7 ~, on

Zz/[e,]~=~, which, like [e,]7=~, is a Hilbert space. If the Hilbert space [e.]7=~ and

Zz/[e,]7=~ are identified in the natural way, one checks that the results of [4]

yield that Tit,.l~= ~- 2P is strictly singular. On the other hand, if S is a bounded

opera tor on a Banach space Z, S X C_ X for some subspace X of Z, and ,,~ is the

induced opera tor on Z / X . Then it is routine to prove that S has finite Fredholm

index if and only if both Six and S do, in which case i ( S ) = i ( S i x ) + i ( S ) .

Therefore the opera tor T on Z2 cannot have odd Fredholm index (and hence

cannot carry Z2 isomorphically onto a hyperplane).

We conclude with a discussion of the well known decomposition problem for

Banach spaces: if X ~ Y O U and Y ~ X @ V, must X ~ Y? The answer is

unknown even when X - ~ X O X . A negative answer to the decomposit ion

problem would be obtained if Zz is not isomorphic to its hyperplane Y. Indeed,

Z2 is isomorphic to its square, because Zz is a symmetric sum of a two

dimensional spaces. Thus Y @ Y ~ Z 2 - ~ Z 2 ~ ) Z z and Y ~ Y @ Y ~ ) Y.

REFERENCES

1. P. Enflo, J. Lindenstrauss and G. Pisier, On the "three space problem", Math. Scand. 36 (1975), 199-210.

2. T. Figiel, W. B. Johnson and L. Tzafriri, On Banach lattices and spaces haoing local unconditional structure with applications to Lorentz [unction spaces, J. Approximation Theory 13 (1975), 395-412.

3. Y. Gordon and D. R. Lewis, Absolutely summing operators and local unconditional structures, Acta Math. 133 (1974), 27-48.

4. N. J. Kalton and N. T. Peck, Twisted sums of sequence spaces and the three problem problem, to appear in Trans. Amer. Math. Soc.

Vol. 37, 1980 UNCONDITIONAL STRUCTURE 129

5. J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces L Sequence Spaces, Springer Verlag, I977.

6. J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces IL Function Spaces, Springer Verlag, 1979.

7. G. Pisier, Some results on Banach spaces without local unconditional structure, Compositio Math. 37 (1978), 3-19.

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