How Many Operators Do There Exist on a Banach Space?

HOW MANY OPERATORS DO THERE EXIST ON A BANACH SPACE?

TH. SCHLUMPRECHT

Abstract. We present some results concerning the following question: Given an inﬁnite dimensional Banach space X. Are there two normalized basic sequences (xn) and (yn), with lim infn→∞ kxn − ynk > 0, so that the map xn 7→ yn extends to a linear bounded operator between the linear span (xn) and the linear span (yn)?

1. Introduction Over the last years the following question caught the interest of an increasing number of researchers in the area of infinite dimensional Banach space theory and became for many of them one of the central problems. (Q1) Assume X is an infinite dimensional Banach space. Does there exist a linear bounded operator T on X which is not of the form T = λId + C, where λ is a scalar, C is a compact operator on X, and Id denotes the identity on X? Clearly, if E ⊂ X is a finite dimensional subspace and S : E → X is any linear operator, then S can be extended to a linear operator T : X → X of finite rank. Thus, there must be “many” finite rank operators on X, and thus “many” elements in the closure of the finite rank operators, which is, assuming X has the approximation property, the ideal of compact operators. Of course, the identity must also be a linear bounded operator. Therefore question (Q1) asks wheather or not there are always some other linear bounded operators defined on a Banach space, beyond the class of operators which must exist by elementary reasons. Note that a counterexample to (Q1) would be the first example of a Banach space X for which the invariant subspace problem has a positive answer: Does every operator T on X have a non trivial subspace Y (Y =6 {0} and Y =6 X) so that T (Y ) ⊂ Y ? Indeed, due to a result of V. I. Lomonosov [Lo], any operator on a Banach space which commutes with a compact operator C =6 0 must have a none trivial invariant subspace. A counterexample to (Q1) might also be a prime candidate for a Banach space on which all Lipschitz function admit a point of differentiability [Li]. Recall ([GM1] and [GM2]) that an infinite dimensional Banach space X is called a hereditarily indecomposable space (HI) if every operator T : Y → X, with Y being a subspace of X is of the form In(Y,X) + S, where λ is a scalar, In(Y,X) denotes the inclusion map from Y into X and S is a strictly singular operator, i.e. an operator which on no infinite dimensional subspace is an isomorphism into X. The first known example of an HI space, we denote it by GM, was constructed by T. W. Gowers and B. Maurey in 1993 [GM1]. Since then HI spaces with additional properties were constructed (e.g. [AD], [F], and [OS]). Although a counterexample to (Q1) does not need to be an HI space, these spaces are nevertheless natural candidates for counterexamples. In [AS] it was shown that the space constructed in [GM1] admits an operator (defined on all of GM) which is strictly singular but The research was supported by NSF. 1 2 TH. SCHLUMPRECHT

not compact. Already earlier it was shown in [Go1] that such an operator can be constructed on a subspace of GM. Given the seemingly easy task to write down on a concrete space a nontrivial operator (i.e. not of the form λ + compact) the proof in [AS] is quite involved. It would be interesting, but probably technically even harder, to establish the existence of non trivial operators defined on other known HI spaces. Since it seems to be hard to answer (Q1) already for specific spaces, the tools to give an answer (at least a positive one) in the general case are probably not developed yet. It also seems that (Q1) is the type of question which will not be answered directly, but by solving first some other, more structure theoretical questions. Following more along the line of the concept of HI spaces we turn therefore to the following “easier question” (for a positive answer) and ask whether or not it is always possible to define a non trivial operator on a subspace. (Q2) Assume X is an infinite dimensional Banach space. Is there a closed subspace Y of X and an operator T : Y → X, so that T is not of the form T = λIn(Y,X) +C, where λ is a scalar and C : Y → X is a compact operator? It is easy to see, that (Q2) can be equivalently reformulated as follows (Q3) Assume that X is an infinite dimensional Banach space. Does there exist two normalized basic sequences (xn) and (yn) so that (xn) dominates (yn) and so that kxn − ynk > infn∈Æ 0? We want to go one step further and formulate a more structure theoretical approach to our problem. We call a Banach space X with a basis (ei) a space of Class 1 if:

(C1) Every block basis of (ei) has a subsequence which is equivalent to some subsequence of (ei).

Remark. Note that the spaces `p,1≤p<∞, and c0 are clearly (C1). Actually in `p, 1 ≤ p<∞, and c0 all blockbasis are equivalent to each other, a property which characterizes the unit bases of `p,1≤p<∞, and c0 [Z]. Moreover, in [CJT] it was shown that Tsirelson’s space T (as described in [FJ]) as well as its dual T∗ (the original Tsirelson space defined in [T]) are spaces of Class 1. Recently the result of [CJT] was generalized to all finitely mixed Tsirelson spaces in [LT]. The reader unfamiliar with the usual notations and concepts (like mixed Tsirelson spaces) is referred to the last paragraphs of this section, where all the notions of this paper are introduced. Also note that every space of Class 1 must contain an unconditional basic sequence, an observation which follows immediately from the result in [Go1] which says that every infinite dimensional Banach space must either contain an infinite dimensional subspace with unconditional basic sequence or it must contain an infinite dimensional subspace which is HI. Of course, a space of Class 1 cannot contain an infinite dimensional subspace which is HI. We do not know of any elementary proof, i.e. a proof which does not use the above cited result of [Go1], of the fact that a space of Class 1 must contain an unconditional basic sequence. It seems that until the early nineties all known Banach spaces had subspaces which were of Class 1. Then, in 1991, the author [Sch] of this work constructed a space, nowadays denoted by S, which fails (C1) in an extreme way: In all infinite dimensional subspaces, spanned by a block, one is not only able to find two normalized blocks (xn) and (yn) having no subsequences which are equivalent to each other. But one can even find in each subspace OPERATORS ON BANACH SPACES 3

two normalized blocks (xn) and (yn), for which the map xn 7→ yn extends to a bounded linear and strictly singular operator between the span of (xn) and the span of (yn). Let us therefore deﬁne the following second class of Banach spaces. Let us call a Banach space X with a basis (ei) a space of Class 2 if:

(C2) Each block basis (zn) has two further block bases (xn) and (yn) so that the map xn 7→ yn extends to a bounded linear and strictly singular operator between the span of (xn) and the span of (yn). The main purpose of this paper is to establish criteria for a Banach space to be of Class 2 and thereby address the following problem. (Q4) Does every infinite dimensional Banach space contain a Banach space which is either of Class 1 or of Class 2? Question (Q4) might seem at first sight somewhat daring, let us therefore motivate it. If one believes (Q3) to be true, one could argue that experience shows us that positive results in Banach space theory are often derived from dichotomy principles. Therefore (C2) could be a candidate for the second alternative in a dichotomy result in which (C1) is the first alternative. Secondly, if one believes Question (Q3) to have a counterexample, one might try to look first for a counterexample of (Q4). Question (Q4) could have, contrary to Question (Q3), a counterexample with an unconditional basis, which probably will be easier to define. Then, starting with a counterexample to (Q4), one might ask for a modification of it to obtain a counterexample to question (Q3). Similarly, the key argument toward defining the first known HI space, was the definition of a space of Class 2. Finally note that a negative answer of (Q3) is equivalent to the statement that the following class of Banach spaces is not empty. A Banach space X with a basis (ei) is said to be a space of Class 3 if:

x z e kxn − znk > (C3) For any two block bases ( n) and ( n)of( i), with infn∈Æ 0, neither the map xn 7→ zn nor the map zn 7→ xn extends to a linear bounded operator between the spans of (xn) and (zn). In this paper we are interested in formulating criteria which imply that a given Banach space X is of Class 2. Therefore we want to find sufficient conditions for X to contain two seminormalized basic sequences (xn) and (yn) for which the map xn 7→ yn extends to a bounded linear and strictly singular operator. In Section 2 we discuss the following problem: Given two normalized basic sequences (xn) and (yn), which conditions should we impose on the spreading models of (xn) and (yn), in order to insure the existence of subsequences (˜xn) and (˜yn)of(xn) and (yn) respectively, so thatx ñ 7→ yñ extends to a strictly singular and bounded linear operator? The properties of spreading models can be quite different form the properties of the underlying generating sequences. For example, the spreading model of the basis of Schreier’s space (using the no- S S <∞ E≤ E} tations of Definition 1.6 this is the space ( 1), with S1 = {E ∈ [Æ ] :# min )is isometrically equivalent to the `1-unit vector basis. On the other hand Schreier’s space is hereditarily c0. Therefore we expect to need to impose rather strong conditions on the spreading models of (xn) and (yn) in order to conclude that for some subsequences

m , n Æ x 7→ y ( k) ( k) and it follows that mk nk extends to bounded linear and strictly singular

x k ∈ Æ → y k ∈ Æ map [ mk : ] [ nk : ]. The main result of Section 2 is the following answer to our question. 4 TH. SCHLUMPRECHT

Theorem 1.1. Let(xn) and (yn) be two normalized weakly null sequences having spreading models (en) and (fn) respectively. Assume that (en) is not equivalent to the c0-unit vector basis and that the following condition holds.

(1) There is a sequence (δn) of positive numbers decreasing to zero so that ∞ n X X a f ≤ δ a e .

i i max n ij i n∈Æ i=1 i1

Then there are subsequences (˜xn) and (˜yn) of (xn) and (yn) respectively, so that x˜n 7→ y˜n extends to a strictly singular and bounded linear operator. In order to formulate our second result, we need to recall the Cantor Bendixson index of

<∞ Æ subsets of [Æ ] , the set of ﬁnite subsets of . A⊂ <∞ A ∈A B ⊂ A A set [Æ ] is called hereditary if for all and all it follows that B ∈A. We call A spreading if A = {a1,a2,...ak}∈A, with a1

<∞

A ⊂ Æ χ consider the topology of pointwise convergence on [Æ ] (identifying with A, the characteristic function of A). A⊂ <∞ For an hereditary and compact [Æ ] we define (2) A(1) = {A ∈A|Ais accumulation point of A} ,N ∀n ∈ NA∪{n}∈A}. = {A ∈ A|∃N ⊂ Æ infinite <∞ A ∈A A Since A is compact in [Æ ] each is subset of a maximal element of and therefore we conclude that (1) A, A6 ∅. (3) A ´ if = Secondly it follows that A(1) is also compact and hereditary. We can therefore define A(α) (0) (β) for each α<ω1 by transfinite induction. We let A = A and assuming we defined A for all β<αwe put \ (1) (4) A(α) = A(β) if α = β +1, and A(α) = A(β) if α = sup β. β<α β<α <∞ α<ω A(α) Since [Æ ] is countable it follows from (3) that there is an 1 for which is empty and we define the Cantor Bendixson index of A by (α) (5) CB(A) = min{α<ω1 :A =∅}.

We note that CB(A) is a successor ordinal. Indeed, assume that α<ω1 is a limit ordinal and that A(β) =6 ∅ for all β<α. Then it follows from the fact that A(β) is hereditary that ∅∈A(β) for all β<α, and, thus, by (4) that ∅∈A(α) which implies that CB(A) >α. The strong Cantor Bendixson index of A is deﬁned as follows (cf. [AMT]): (6) CBS(A) = sup inf CB A∩[M]<∞

M⊂N,#M=∞

N⊂Æ,#N=∞ Remark. A N

Note that CBS( ) could be a limit ordinal. Indeed, let ( Sn)n∈Æ be a sequence

Æ A A A

of inﬁnite pairwise disjoint subsets of Æ whose union is . Take = n, with := n∈Æ {A ⊂ Nn ﬁnite : #A = n}. Then it follows that CBS(A)=ω. OPERATORS ON BANACH SPACES 5

<∞

On the other hand it is clear that for a successor ordinal γ and an hereditary A⊂[Æ ] we have that , N ∞∀M ⊂ N, M ∞ A∩ M <∞ ≥γ. CBS(A) ≥ γ ⇐⇒ ∃ N ⊂ Æ # = # = CB( [ ] ) Using the Cantor Bendixson index we can characterize spaces which have subspaces of Class 1 and quantify the property of not having a subspace of Class 1. More generally, we will use the Cantor Bendixson index to measure “how far away” two basic sequences are to each other, up to passing to subsequences”.

Deﬁnition 1.2. Assume that x =(xn) and z =(yn) are two seminormalized basic sequences. For c ≥ 1 we deﬁne

C <∞ <∞ ∃B∈ Æ , B A, x ∼ z }, (7) (x, z, c)={A∈[Æ ] : [ ] # =# ( n)n∈A c ( n)n∈B

by “(xn)n∈A ∼c (zn)n∈B ” we mean that (xn)n∈A and (zn)n∈B are c-equivalent (see end of A B γ<ω ∅ ∞ section), where on and we consider the order given by Æ .For 1let (put inf = ) (8) c(x, z, γ) = inf c ≥ 1 : CBS C(x, z,c) ≥ γ +1 (with inf ∅ = ∞) ∞ ∞ c ≥ ∃N ∀M ⊂N C x, z, c ∩ M <∞ ≥ γ = inf 1: ⊂Æ CB ( ) [ ] +1

(9) γ0(x, z) = sup{γ<ω1 :c(x, z,γ) < ∞}.

In the case that z is the unit basis of `1 we deﬁne B(x, c)=C(x, z,c), b(x, γ)=c(x, z, γ), and β0(x)=γ0(x, z) and for a Banach space X and β<ω1 we put

(10) B(β)=B(β,X) = sup{b(z,β):z=(zn)⊂BX semi normalized, weakly null} ( ) ∃(zn)⊂BX semi normalized and weakly null = sup b>0: . CBS(B(z,b))}≥β+1

(11) β0(X) = sup{β<ω1 :B(β)>0}

= sup{β0(z):z=(zn)⊂BX semi normalized and weakly null}

It is clear that β0(X) must be a limit ordinal. Proposition 1.3. (see proof after Corollary 4.9 in Section 4). Assume X is a Banach space with a basis (xn).If(zn)is another basic sequence for which γ0((zn), (xn)) = ω1 then a subsequence of (zn) is ismorphically equivalent to a subsequence of (xn) or to the spreading model of a subsequence of (xn). Moreover, X is of Class 1 if and only if for all blockbases (zn) of (xn) it follows that γ0((zn), (xn)) = γ0((xn), (zn)) = ω1. In Section 6 we will prove the following result, implying a condition for a space to be in Class 2.

Theorem 1.4. Let X be a Banach space with a basis (ei) not containing c0. Assume that there is an ordinal γ ∈ [0,β0(X)] so that:

a) There is a semi normalized block basis y =(yn)⊂BX with β0(y) ≤ γ. b) infβ<γ B(β,X) > 0.

Then there is a seminormalized block basis (xn) in X and a subsequence (˜yn) of (yn) so that

x → y n ∈ Æ the map n 7→ y˜n extends to a strictly singular operator T :[xn :n∈ Æ ] [˜n : ]. 6 TH. SCHLUMPRECHT

Corollary 1.5. Assume that Z is a reﬂexive Banach space with a basis and each block subspace of Z satisﬁes the condition of Theorem 1.4. Then Z is of Class 2.

Remark. Note that in the statement of Theorem 1.4 either γ = β0(X) in which case it is clear that (a) is satisfied for all block bases. Or γ<β0(X) in which case it is clear that (b) is satisfied. Also note that the assumption in Theorem 1.4 says the following: On the one hand there is a block basis y which is “far away from the `1- unit vector basis, in the sense that given any ε>0 there is an α<γso that b(y, α) <ε, i.e. on sets of strong Cantor Bendixson index α the best equivalence constant between the `1 basis and y is at most ε. On the other hand if we let c = infβ<γ B(β,X), we can choose for any α<γand any ε>0 a block sequence z for which b(z,α) >c−ε. In other words the assumption in Theorem 1.4 states that we require the existence of blockbases which have “different `1-behavior”. But if one wants to attack Problem (Q4) one could start by assuming that our given Banach space X has no infinite dimensional subspace of Class 1 and therefore by Proposition 1.3 every block basis (yn) must admit a further block basis (zn) so that γ0((yn), (zn)) <ω1 or γ0((zn), (yn)) <ω1. So, instead of going the quite “uneconomical route” (as in Theorem 1.4) and comparing two blockbases to the `1-unit vector basis one should compare them to each other directly. Here exactly lies one major hurdle which one has to overcome before reaching further results. The proof of Theorem 1.1 as well as the proof of Theorem 1.4 make use of certain “partial unconditionality results” namely a result by E. Odell [O2] (see Theorem 2.2) and a result by S. A. Argyros, S. Mercourakis and A. Tsarpalias [AMT] (see Lemma 6.1 in Section 6). Both results give partial answers to the question, under which conditions a family of finite rank projections on certain subsets of a basis are uniformly bounded. In order to make advances on (Q4) we would need to address the following question which asks for generalizations of the results in [O2] and [AMT].

(Q5) Assume that X is a Banach space with a basis (en), γ ∈ [0,ω1), and assume that for all normalized block bases (xn)of(en) it follows that c((xn), (en),γ) < ∞ c((en), (xn),γ)<∞. Secondly, consider for a blockbasis (xn) and an r>0 the set <∞ kP k≤r , U((xn),r)= A∈[Æ ] : [xn:n∈A] P → x n∈A where [xn:n∈A] :[xn :n∈ Æ ] [ n : ] is the deﬁned to be the usual projection onto [xn : n ∈ A]. Now, does it follow that any block basis (xn) has a subsequence (zn) for which CBS(U((zn),r)) ≥ γ for some r>0? The result cited from [O2] gives a positive answer if γ = ω and the result cited from [AMT] gives a positive answer if we replaced in the assumption c((en), (xn),γ)byb((xn),γ).

To be able to prove Theorem 1.4 we will introduce in Section 3 a family (Fα)α<ω1 of subsets < ∞ α<ω of [Æ ] , having the property that for each 1 the strong Cantor Bendixson index is α + 1 (see Corollary4.9). The family (Fα) is deﬁned in a similar fashion as the Schreier

families (Sα)α<ω1 [AA], with one crucial diﬀerence: one obtains Fα+1 from Fα by adding to OPERATORS ON BANACH SPACES 7

each element of A ∈Fα at most one new element, therefore the step from Fα to Fα+1 is much smaller than the step from Sα to Sα+1. Feeling that this family (Fα) could be an important tool to analyse the combinatorics of blockbases we extensively discuss the properties of this family in the Sections 3, 4 and 5, in A⊂ <∞ particular we prove a result which, roughly speaking, says that any hereditary set [Æ ] ∞ F restricted to some appropriate M ⊂ Æ is equal to “a version of α” (see Theorem 4.10). Based on this result we will be able to prove Theorem 1.4 in Section 6. We will need the following notations and conventions. X For simplicity all our Banach spaces are considered to be over the ﬁeld Ê .If is a Banach space BX and SX denotes the unit ball and the unit sphere respectively. e c00 denotes the vectorspace of sequences in Ê which eventually vanish and ( i) denotes the usual vector basis of c00 and when we consider a Banach space X with a normalized basis, we thinkP of X being the completion of some norm on c00 with (ei) being that basis. For

x x x 6 } x E ⊂ Æ = Piei ∈ c00 we let supp(x)={i∈Æ : i =0 be the support of and if we let E(x)= i∈Exiei. We say that a basic sequence (xn) dominates another basic sequence (yn) if the map xn 7→ yn extends to a linear bounded map between the span of (xn) and the span of (yn), Pn Pn i.e. if the there is a c>0 so that for all (ai) ∈ c00 k i=1 aiyik≤ck i=1 aixik. We say that (xn) and (ynP) are c-equivalent,P c ≥ 1 and weP write (xn) ∼c (yn) if for any a ∈ 1 k n a x k≤k n a y k≤ck n a x k ( i) c00 it follows that c i=1 i i i=1 i i i=1 i i . x xn n∈ Æ The closed linear span of a subsequence ( n)n∈Æ of a Banach space is denoted by [ : ]. Let E be a Banach space with a 1-spreading basis (ei), i.e. m m X X

, m ∈ m a e a e Æ a ⊂ Ê n

and let (xn) be a seminormalized basic sequence. We say that (ei) is the spreading model of

x k a ⊂ Ê ( n) if for any k ∈ Æ and any ( i)i=1 it follows that k k X X ... a x a e . lim lim lim i ni = i i n1→∞ n2→∞ nk→∞ i=1 i=1 Recall [BL] that any seminormalized basic sequence has a subsequence with spreading model and that, using our notations in Definition 1.2, a seminormalized sequence (xn) has a subsequence whose spreading model is isomorphic to (ei) if and only if c((xn), (ei),ω)<∞). For a set M the set of all finite subsets of M is denoted by [M]<∞ and the set of all infinite ∞ subsets is denoted by [M]∞. We write M ⊂ N if M is an infinite subset of N. Definition 1.6. (Mixed Schreier and Tsirelson spaces) F <∞ For an hereditary, spreading and compact subset of [Æ ] containing all singletons of <∞ F S F [Æ ] we define the -Schreier space ( ) to be the completion of c00 under the norm defined by X (S(F)) kxk = sup |xi|, whenever x =(xi)∈c00. E∈F i∈E F ∞ <∞ Let ( n)n=1 be an increasing sequence of hereditary, spreading and compact subsets of [Æ ] and let (Θn) ⊂ (0, 1] be a non increasing sequence. The mixed Schreier space S((Θn), (Fn)) 8 TH. SCHLUMPRECHT

is the completion of c00 under the norm deﬁned by X

(S((Θn), (Fn))) kxk = sup sup Θn |xi|, whenever x =(xi)∈c00.

n∈Æ E∈F n i∈E

∞ ` Æ

If in addition ( n)n∈Æ is a sequence in increasing to , we deﬁne the mixed Schreier space ` S n , Fn , `n with the additional admissibility condition given by ( n)n∈Æ to be the space ((Θ ) ( ) ( )) with the following norm X (S((Θn), (Fn), (`n))) kxk = sup sup Θn |xi|, whenever x =(xi)∈c00. n∈Æ ` ≤min E,E∈F n n i∈E <∞ E

n n∈Æ n∈Æ (E ) F -adm. i i=1 n i=1

A ﬁnitely mixed Tsirelson space is a space T ((Θn), (Fn)) where (Θn) ⊂ (0, 1] and (Fn) are ﬁnite sequences.

2. Conditions on the spreading models implying the existence of nontrivial operators

Deﬁnition 2.1. For two normalized basic sequences (xn) and (yn) and an ε>0 we deﬁne X X

∆((xn),(yn))(ε) = sup aiyi :(ai)∈c00, max |ai|≤ε, and aixi ≤ 1 . i∈Æ

We say that (xn) strongly dominates (yn)if X (12) lim inf k xik = ∞ and

n→∞

A⊂Æ,#A=n i∈A

(13) lim ∆((x ),(y ))(ε)=0. ε&0 n n

Remark. Assume (xn) and (yn) are two normalized basic sequences.

a) If ∆((xn),(yn))(ε) < ∞ for some ε>0 (and thus for all ε>0) then (xn) dominates (yn), i.e. the map xn 7→ yn extends to a bounded linear operator. b) Note that in the case that (xn) is subsymmetric condition (12) means that (xn)is not equivalent to the c0 unit vector basis. c) No normalized basic sequence strongly dominates the unit vector basis in `1. We will need the following result by E. Odell on Schreier unconditionality. Theorem 2.2. [O2] Let (xn) be a normalized weakly null sequence in a Banach space and η>0. Then (xn) contains a subsequence (˜xn) so that

∞ X X , F ≥ F α x ≤ η α x . ∀ α ∈ ∀F ⊂ Æ ( i) c00 min # : i ˜i (2 + ) i ˜i i∈F i=1

If (xn) is a bimonotone sequence above factor (2 + η) could be replaced by (1 + η). OPERATORS ON BANACH SPACES 9

Remark. Assume that (xn) and (yn) are normalized basic sequence, that (xn) strongly dominates (yn) and that either (xn) is weakly null or 1-subsymmetric. By passing to subsequences (of (xn) and (yn) simultaneously) we can assume that (xn) has a spreading model (ei) and that the conclusion of Theorem 2.2 holds, i.e. that X X X∞ (15) k αieik∼2 k αixik≤3k αixik i∈F i∈F i=1 F ≥ F for all (αi) ∈ c00 and ﬁnite F ⊂ Æ , with min # . Since (xn) satisﬁes (12) (en) cannot be equivalent to the c0-unit basis and we deduce therefore that Xn (16) cn = k eik%∞. i=1

We deﬁne now the following basic sequence (˜y) which dominates (yn)by X∞ X∞ Xn k α y k k α y k 1 k α e k α ∈ . (17) i ˜i = i i + max √ max ki i whenever ( i) c00 n c k1<...kn i=1 i=1 n i=1 Proposition 2.3. Using the assumptions and notations as in above Remark it follows that a) (xn) strongly dominates (˜yn). b) For η>0it follows that X ` |α |≥η} α ∈ , k α y k≤ <∞. η = #{i ∈ Æ : i :( i) c00 i ˜i 1 η> x y Proof. In order to show (a) let 0. Since ( n) strongly dominates ( n) we√ can choose an ε0 0 / c <η/ > 0 so that ∆((xn),(yn))(ε ) <η/2. Secondly we choose n0 ∈ Æ so that 1 n0 12 and 0 we let ε = min(ε ,η/2Pn0). α n ≤ n If ( i) ∈ c00 with k αixik≤1 and max |αi|≤εwe deduce for n ∈ Æ , with 0, that n 1 X √ max k αieik≤nε ≤ η/2. c n≤k1<...kn n i=1 n ≥ n If n ∈ Æ , with 0, we deduce from (15) that

n n ∞ X X X 1 α e ≤ 2 α x ≤ 6 α x ≤ η/ , √ max ki i √ max ki i √ i i 2 c n≤k1<...kn c n≤k1<...kn c n i=1 n i=1 n i=1 P 0 which together with the choice of ε implies that k αiy˜ik≤eta and ﬁnishes the proof of the claim in (a). P α To show (b), let ( i) ∈ c00 so that k αiy˜ik≤1 and let η>0. We put I = {i ∈ Æ : 0 0 kαik >η}and let I the set of the d#I/2e largest elements of I. Note that #I ≤ min I which implies by our deﬁnition of (˜yn) and the fact that (ei) suppression 1-unconditional (thus 2-unconditional) that

X 1 X 1 X √ 1 ≥ αiy˜i ≥ √ αiei ≥ √ ηei ≥ η c#I0/2, c#I0 c#I0 i∈I0 2 i∈I0 √ c ≥ /η} which implies that `η ≤ 2 max{n ∈ Æ : n 2 and ﬁnishes the claim (b).

Finally, (c) follows easily from (b). £ 10 TH. SCHLUMPRECHT

Lemma 2.4. Let (ei) be a suppression 1-unconditional, 1-subsymmetric and normalized basis. For a normalized, basic and weakly null sequence (yn) the following statements are equivalent. a) (yn) has a subsequence (zn) which is strongly dominated by (ei). b) (yn) has a subsequence (zn) having a spreading model (fi) which is strongly dominated by (ei). c) (yn) has a subsequence (zn) for which there is a sequence (δn) ⊂ (0, ∞), with δn & 0, if n %∞, so that

∞ n X X αizi ≤ max δn max αj ei , whenever (αj) ∈ c00. j

d) (yn) has a subsequence (zn) for which there is a sequence (εn) ⊂ (0, ∞),with εn & 0, n %∞ ` if , and a subsequence ( n) of Æ so that

∞ `n X X αizi ≤ max εn max αj ei , whenever (αj) ∈ c00. n≤j

Proof. (a)⇒(b) Assume (zn) is a subsequence of (yn) which is strongly dominated by (ei). z f By passing to a further subsequence we can assume that ( n) has a spreadingP model ( i), k k α k ≤ ε k k α e k≤ being subsymmetric. If ε>0 and (αi)i=1 ⊂ Ê , with ( i) ∞ , and i=1 i i 1it follows that k k X X α f ... α y ≤ ε , i i = lim lim i ni ∆((ei),(yi))( ) n1→∞ nk→∞ i=1 i=1 which implies (b). (b)⇒(a) Let (fi) be the spreading model of a subsequence (ui)of(yi). We first note that it is enough to show that for a fixed η>0 we need to find a subsequence (vi)of(ui) and ε ε η ε <η an = ( ) so that ∆((ei),(yi))( ) . Then a standard diagonal argument using a sequence (ηn) decreasing to 0 will show that a subsequence of (ui) is stronglyP dominated by (ei). For a fixed η>0 we first choose (εn) ⊂ (0, 1) so that εnn<η/2 and so that

P n∈Æ ε <ε/ e ` {` ∈ Æ ∆(ei),(fi)( n) 4. Since ( n) satisﬁes condition (12) it follows that n = min : Pn∈Æ k ` i=1 eik≥1/εn+1} is ﬁnite for every n ∈ Æ . By passing again to a subsequence of we can assume that `n `n X X α f ∼ α u , (18) i i 2 i ki i=1 i=1

n ≤ k `n

X X∞ X X∞ X

αiui ≤ αiui ≤ nεn + αiui

n=1 n=1

i∈Æ,ε <|α |≤ε n≤i,ε <|α |≤ε n+1 i n n+1 i n

X∞ X X∞ ≤ nε α f ≤ nε ε <η, n +2 i i n +2∆((ei)(fi))( n)

n=1 n≤i,εn+1<|αi|≤εn n=1 OPERATORS ON BANACH SPACES 11 P (for the third inequality note that since k αieik≤1 it follows that #{i : |αi|≥εn+1}≤`n) which ﬁnishes the proof of (b)⇒(a). (a)⇒(c) Assume that (yi) is strongly dominated by (ei). Since we could replace (yn)by (˜yn) as in the remark after Theorem 2.2 and apply Proposition 2.3 we can assume that (yn) satisﬁes the properties (a) and (b) of Proposition 2.3. After a renormingP we canP also assume that (ei) dominates (yi) with constant 1, i.e. that we have k αiyik≤k αieik ∞ for all (αi) ∈ c00. Choose a strictly to 0 decreasing sequence (εi)i=0 with ε0 = 1 and

i+1 −i−1 ε ≤ /i i ∈ Æ ∆((ei),(yi))( i2 ) 2 and deﬁne for 0 n X o K ` {i |α | >ε } α ∈ , k α y k≤ . i = εi+1 = max # : i i+1 :( i) c00 i i 1

δ i ∈{ , ,...K } δ /n i ∈{K i ∈ Æ Finally, choose for i =2∆((ei),(yi))(1) if 1 2 0 and i =1 if n−1+ ,K 1 n−1 +2,...,Kn} for some n ∈ Æ . P α ∞ Assume now that ( i) ∈ c00 with k i=1 αiyik = 1. Therefore there must be an n ∈ Æ 0 for which ∞ X X α y ≥ −n−1 −n−1 α y . i i 2 =2 i i ,ε <|α |≤ε i=1 i∈Æ n+1 i n We deduce that

∞ X X α y ≤ n+1 α y i i 2 i i

i=1

i∈Æ,ε <|α |≤ε n+1 i n

X X α n+1 i =2 αiei yi

i∈Æ,ε <|α |≤ε i∈Æ,ε <|α |≤ε  n+1 i n n+1 i n X  

with c = k αieik

i∈Æ,ε <|α |≤ε n+1 i n

X ≤ n+1 α e ε n+1 2 i i ∆((ei)(yi))( n2 )

i∈Æ,ε <|α |≤ε  n+1 i n  X X  −n−1

Note that c = k αieik≥k αiyik≥2

i∈Æ,ε <|α |≤ε i∈Æ,ε <|α |≤ε n+1 (i n n+1 i n XKn 2∆((e ),(y ))(1) if n =0 ≤ α e · i i max ji i j1

Kn k X X = δK max αjiei ≤ max δk max αjiei . n j

which concludes this part of the proof. ⇒ z `˜ {` δ > −k2 } (c) (d) Let ( n) and (δn) be given as in (c). For k ∈ Æ , we let k = max : ` 2 . 12 TH. SCHLUMPRECHT

We can assume that `˜k ≥ k, by passing to a slower decreasing sequence δ˜k, if necessary. We also can assume that (xi) is a monoton basis. α Then it follows from (c) for ( i) ∈ c00 that there is an ` ∈ Æ so that

∞ ` X X α z ≤ δ α e i i ` max ji i j1

Therefore the claim follows in the case that ` ≤ `˜1 if we put ε1 = δ1 and `1 = `˜1.If`>`˜1

we can ﬁnd a k ∈ Æ so that

`˜ ˜ X∞ Xk+1 X∞ X`k 2 2 α z ≤ −k α e α y > −(k−1) α e . i i 2 max ji i and i i 2 ji i j1

Thus we conclude   ˜ X∞ X`k+1 Xk−1 2 α z ≤ −k  α e α e  i i 2 max ji i + ji i j1

˜ X`k+1 X∞ 2 2 2 ≤ −k α e −k (k−1) α z . 2 max ji i +2 2 i i k≤j1

˜ X∞ X`k+1 2 2 2 − −k (k−1) α z ≤ −k α e . 1 2 2 i i 2 max ji i k≤j1

−k2 −k2 (k−1)2 and ﬁnishes the proof of our claimed implication if we choose εk =2 /[1 − 2 2 ] and `k = `˜k+1. ⇒ z (d) (a) Assume ( n), (εn) and (`n) are chosen as required in (d). Let η>0, choose n0 ∈ Æ ε ≤ η/ ε<η/ n so that n0 2 and choose 2 0.P If now (αi) ∈ c00, maxi |αi| <ε, and k αieik≤1 then it follows from condition (d) that

∞ `n `n X X X αizi ≤ max εn αj ei ≤ ηn0 + max εn αj ei <η, ,n≤j <...j i n≥n ,n≤j <...j i n∈Æ 1 0 1 ` i=1 `n i=1 n i=1 ε <η

which implies that ∆(ei),(zi)( ) and proves the claim and ﬁnishes the proof of the Lemma. £ OPERATORS ON BANACH SPACES 13

Proof of Theorem 1.1. Using Lemma 2.4 (c)⇒(a) for the sequences (fi) and (ei) and then Lemma 2.4 (b)⇒(a) for the sequences (yi) and (ei) we can assume, by passing to a subsequence of (yn) if necessary, that (ei) strongly dominates (yi) and by passing to the sequence (˜yn) as defined in the remark after Theorem 2.2 we can assume that (a) and (b) of Proposition 2.3 are satisfied. Using Lemma 2.4 we can again pass to a subsequence, still denoted by (yn) for which we ˜ δ & ` %∞ n %∞ find sequences (δn) ⊂ (0, ∞) and `n ⊂ Æ , with n 0 and n for , so that

k X X α y ≤ 1 δ˜ α e , α ∈ i i max k max ni i whenever ( i) c00. k k≤n1

By using Theorem 2.2 and passing to a subsequence of (xn), if necessary, we can assume that

k X X X α y ≤ δ α x ≤ α x (19) i i max k max ni ni 3 i i k k≤n1

whenever (αi) ∈ c00.Thus(xn) dominates (yn) and in order to show that the formal identity I : xn 7→ yn extends to a strictly singular operator, let (un) be a seminormalized block of x ( n), write un, for n ∈ Æ as

Xkn Xkn (n) (n) un = αi xi, and let vn = I(un) αi yi

i=kn−1+1 i=kn−1+1

Using Theorem 2.2 and the fact that (en) has to satisfy (12), we can assume that (n) limn→∞ maxkn−10 so that X X (n) (n) 0

(20) 1 = ai fi ≥ c ai ei ≥ c .

i∈Æ i∈Æ

Now we let (xn) be the basis for the Schreier space associated to (ei), i.e. the norm deﬁned by n X X a x k a e a ∈ .

(21) i i = max ij j whenever ( i) c00 n∈Æ n≤i1<...in j=1 14 TH. SCHLUMPRECHT

As in the original Schreier space (where (ei) is set to be the `1-basis) it is easy to see

e x x x n ∈ Æ that ( i) is a spreading model of (Pi). Let (ñ) be a subsequence of ( n). For define m a(n) u mn a(n)x ku k≥c0/c n = max supp( i ) and n = i=1 i ˜mn+i. Then n and, again, as in the original Schreier space, we can show that a subsequence of un is equivalent to the c0-unit basis. Since F does not contain a copy of c0 we deduce that the mapx ñ 7→ fn can not be extended to a linear bounded operator. In general the condition that a subsymmetric and normalized basis (ei) strongly dominates another basis (fi) is much stronger than the condition that (ei) dominates (fi) without (ei) being equivalent to (fi). But in the case of E = `1, we have the following. Proposition 2.5. Assume (yn) is a normalized basic weakly null sequence. (yn) has a subsequence which is strongly dominated by `1 if and only if (yn) has a subsequence whose spreading model is not equivalent to the unit vector basis of `1.

Proof. By Lemma 2.4 it follows immediately that (yn) is not strongly dominated by `1 if it has a subsequence whose spreading model is equivalent to the unit vector basis of `1. To show the converse we need to show by Lemma2.4 that if a 1-subsymmetric basis (fi) is not strongly dominated by `1 then (fi) is equivalent to the unit vector basis of `1. f ∗ F f∗ Let n, n ∈ Æ , be the coordinate functionals on . Then ( n) is a 1-subsymmetric basis f ∗ of the closed linear span of ( n). P

δ ∞ (n) > n ∈ Æ y a f ∈ f By assumption there is a 0 0 andP to for each aP sequence n = i=1 i i [ i :

i ∈ (n) ∞ (n) ∞ (n) ≤ a < /n i ∈ Æ a k a f k≥δ Æ ] with 0 i 1 , for P, i=1 i = 1, and i=1 i i 0.

n ∈ ∗ ∞ (n) ∗ ∗ (n) y β f ∈ f i ∈ Æ ≤ β i ∈ Æ PFor each Æ choose n = Pi=1 i i [ i : ] with 0 i , for , ∞ (n) ∗ ∗ ∞ (n) (n) k i=1 βi fi k = 1, and yn(yn)= i=1 ai βi ≥ δ0/2. (n) (n) Letting cn =#{i:βi ≥δ0/4},n=1,2... it follows from the conditions on (ai ) that X∞ δ / ≤ a(n)β(n) ≤ c 1 δ / , 0 2 i i n n + 0 4 i=1 Pk

c nδ0 ∗ δ0 ∗ ≥ f n k ∈ Æ k f k≤ thus n 4 . Since ( ) is 1-subsymmetric it follows for all that 4 i=1 i 1 ∗ and thus that fi is equivalent to unit basis of c0, from which we finally deduce that (fi)is ` equivalent to the 1-basis. £ The following proposition describes another situation in which Theorem 1.1 applies. Its proof can be compiled from the techniques in [AOST], Section 3. Nevertheless, the proof is still quite technical, and since Theorem 1.4 provides a generalization, we will not give a proof here. Before we can state the result we need the following Definition from [AOST]. Definition 2.6. Let (xi) be a 1-subsymmetric basic sequence. The Krivine set of (xi) is the p ≤ p ≤∞ ε> n ∈ set of ’s (1 ) with the following property: For all 0 and Æ there exists

m ∈ m n λ ⊂ Ê a ⊆ Ê Æ and ( k) , such that for all ( i) , k=1 1 n 1 X k a n k ≤ a y ≤ ε k a n k , ε ( i)i=1 p i i (1 + ) ( i)i=1 p where 1+ i=1 Xm yi = λkx(i−1)m+k for i =1,...,n k=1 and k·kp denotes the norm of the space `p. OPERATORS ON BANACH SPACES 15

The proof of Krivine’s theorem [K] as modified by Lemberg [Le] (see also [Gu], remark II.5.14), shows that for every 1-subsymmetric basic sequence (xi) the Krivine set of (xi)is non-empty. It is important to note that our definition of a Krivine p requires not merely n that `p be block finitely representable in [xi] but each `p unit vector basis is obtainable by means of an identically distributed block basis. Proposition 2.7. Assume X is a Banach space containing a normalized basic sequence (xn) which has a spreading model (ei) which is not equivalent to the `1-unit vector basis, but whose Krivine set contains the number 1. Then there is a normalized basic sequence (zn) in X which strongly dominates (xn).

3. The transfinite family (Fα) and some of its basic properties

In this section we discuss a well ordered family (Fα)α<ω of subsets of the finite subsets of S Æ . Its definition is similar to the definition of the Schreier family ( α)α<ω [AA]. The Schreier set of order α, Sα, corresponds to our set Fωα , in the sense that they have the same Cantor Bendixson index. α<ω A α

For every limit ordinal 1 we consider a sequence of sets ( n( ))n∈Æ so that for each n ∈ A α ,α Æ n( ) is a ﬁnite subset of [0 ) and A α α. (22) An(α) ⊂ An+1(α) for, n ∈ Æ , and lim max n( )= n→∞ A α α α<ω

We call ( n( ))n∈Æ the sequence approximating . If for every limit ordinal (we write

α∈ ω A α α A α

n n∈Æ,α∈Lim(ω ) Lim( 1)) ( n( ))n∈Æ is a sequence approximating , we call the family ( ( )) 1 an approximating family. A <∞ α<ω n α ,α∈Lim(ω ) Fα ⊂ Æ 1 Given an approximating family ( ( ))n∈Æ 1 the sets [ ] , , are deﬁned by transﬁnite recursion as follows

(23) F0 = {∅}. <∞ β<α Assuming for some 0 <α<ω1 the sets Fβ ⊂ [Æ ] , with , are already deﬁned we proceed as follows. F ,E ∈F ∪ {∅} α β (24) α = {n}∪E:n∈ Æ β if = + 1 and [ <∞ E ∈ F α ∈ ω (25) Fα = E ∈ [Æ ] : β if Lim( 1)

β∈Amin E (α)

We say that the transfinite family (Fα)α<ω is defined by the approximating family A α ,α∈Lim(ω ) ( n( ))n∈Æ 1 . We first state some elementary properties of our family (Fα).

Proposition 3.1. Assume that (Fα)α<ω1 is the transﬁnite family associated to an approxi- An α ,α∈Lim(ω ) mating family ( ( ))n∈Æ 1 . α<ω <∞ a) For 1,Fα is hereditary, spreading and compact in [Æ ] . b) For α<ω1 it follows that F ,n < E, E ∈F ∪ {∅} α+1 = {n}∪E:n∈ Æ min and α E ∈ <∞ E 6 ∅,E\{ E}∈F ∪ {∅}. = [Æ ] : = min α α ≤ β<ω c) For 1 there is an m ∈ Æ so that <∞ <∞ Fα ∩ {m, m +1,...} ⊂Fβ ∩ {m, m +1,...} . 16 TH. SCHLUMPRECHT

Proof. We can prove (a) by transfinite induction for all α<ω1 while (b) follows from the fact that Fα is hereditary and spreading. To show (c) we fix α and prove the claim by transfinite induction for all β>α. Assuming the claim is true for all γ<β.Ifβ=γ+ 1 the claim follows since Fγ ⊂Fγ+1.Ifβis a limit ordinal and if (An(β)) is its approximating sequence we proceed as follows. n ∈ β(n) A β >α First we choose Æ so that = max n( )) , and, using the induction hypothesis ` ∈ we can find an Æ so that <∞ <∞ Fα ∩ [{`, ` +1,...}] ⊂Fβ(n) ∩[{`, ` +1...}] .

Secondly we observe from the deﬁnition of Fβ it follows that <∞ <∞ Fβ(n) ∩ [{n, n +1...}] ⊂Fβ ∩[{n, n +1...}] . m {`, n} Therefore the claim follows by choosing = max . £

In the definition of approximating families we allow the sets An(α) to have more than one element, contrary to the definition of the Schreier families (see [AA]), because we want to ensure that the transfinite families are directed. Proposition 3.2. k ∈ A(k) α Æ n ,α∈Lim(ω ) Assume for every ( ( ))n∈Æ 1 is an approximating family F (k) defining the transfinite family ( α )α<ω1 . S

k ∈ (k) k (i) (k) n ∈ Æ α ∈ ω B α A α G For each Æ , and Lim( 1) define n ( )= i=1 n ( ) and let α , α<ω B (k) α 1 be defined using the approximating family ( n ( )).S n ∈ α ∈ ω B α n A(i) α G Further more define for Æ and Lim( 1) n( )= i=1 n ( ) and let ( α)α<ω1 be the transfinite family defined by (Bn(α)). Then it follows for all α<ω1 Sk (i) (k)

a) i=1 Fα ⊂Gα for k ∈ Æ . k ∈ F (k) ∩ {k, k ,...} <∞ ⊂G b) For all Æ it follows that α [ +1 ] α.

Proof. By transﬁnite induction. £ ∞

<∞

N {n i ∈ Æ } ⊂ Æ n %∞ i%∞ Deﬁnition 3.3. Let A⊂[Æ ] and = i : , i ,if . <∞ a) We call the set A∩[N] the restriction of A onto N. N b) We call the family A = {ni : i ∈ E} : E ∈A the spreading of A onto N. N Using the deﬁnition of (Fα), we obtain the following recursive description of Fα . ∞ Proposition 3.4. N N n n %∞ α<ω Assume ⊂ Æ , =( i), i and 1. Then a) If α = β +1 FN = {n}∪F :n∈N and F ∈FN ∪ {∅} α β <∞ = F ∈ [N] \ {∅} : F \{min F }∈Fβ ∪ {∅}.

b) If α ∈ Lim(ω1) and (An(α)) is the sequence approximating α, then [ N <∞ N Fα = F ∈ [N] : F ∈ Fβ . β∈A (α) min{i:ni∈F } ∞

Proposition 3.5. M,N m ∈ Æ Assume ⊂ Æ and 0 so that a) #(M ∩ [1,m0]) ≤ #(N ∩ [1,m0]) b) M ∩ [m0, ∞) ⊂ N ∩ [m0, ∞) OPERATORS ON BANACH SPACES 17

M <∞ N <∞ Then it follows for α<ω1 that Fα ∩ [{m0,m0 +1,...}] ⊂Fα ∩[{m0,m0 +1,...}] .

Proof. We prove the claim by transﬁnite induction on α<ω1, using at each inductionstep

Proposition 3.4 (a) or (b). £

∞ N n n %∞ β<α<ω ` ∈ Æ Proposition 3.6. If N ⊂ Æ , =( i), i , and 1 then there is an so that {ni:i≥`} N Fβ ⊂Fα . F ∩ {`, ` ,...} <∞ ⊂ Proof. Using Proposition 3.1 (c) we can choose ` ∈ Æ so that β [ +1 ] <∞ Fα ∩[{`, ` +1,...}] . Thus it follows (for the ﬁrst ”⊂” recall that Fβ is spreading) {ni:i≥`} F = {ni+`−1 : i ∈ E} : E ∈Fβ ⊂ {nj :j∈E}:E∈Fβ, and E ≥ ` β N ⊂ {nj : j ∈ E} : E ∈Fα, and E ≥ ` ⊂Fα ,

which ﬁnishes the proof. £

Proposition 3.7. An α Bn α

,α∈Lim(ω ) n∈Æ,α∈Lim(ω ) Assume that ( ( ))n∈Æ 1 and ( ( )) 1 are two approximating families deﬁning the transﬁnite families (Fα)α<ω1 and (Gα)α<ω1 respectively. ∞ ∞ α<ω M ⊂ N GM ⊂F For 1 and N ⊂ Æ there is an so that α α.

Proof. We proof the claim by transﬁnite induction on α<ω1. Assume that the claim is true for all β<α.Ifαis a successor it follows immediately that the claim is true for α.

Assume α = sup γ = sup max Bn(α) = sup max An(α).

γ<α n∈Æ n∈Æ

m i ∈ Æ γ ∈ B α Using Proposition 3.1(c) we can choose an i ∈ Æ for each , so that for all i( ) <∞ it follows that Fγ ∩ [{mi,mi +1,...}] ⊂Fα. ∞ ∞ Secondly, using the induction hypothesis, we can ﬁnd N ⊃ M1 ⊃ M2 ... so that for k ∈ γ ∈ B α GMk ⊂F all Æ and all k( ) it follows that γ γ. Since we can make sure that

G ⊂F k ∈ Æ γ ∈ B α min Mk ≥ mk, for k ∈ Æ , it follows that γ α for all and k( ). If we ﬁnally M M let be a diagonal sequence of the i’s we deduce the claim from Proposition 3.5. £ ∞

From the property that Fα is spreading it is easy to see that for any L ⊂ Æ it follows that L <∞ Fα ⊂Fα∩[L] . If one is willing to change the approximating family the converse becomes true.

Proposition 3.8. Let (Fα)α<ω1 be a transfinite family which is defined by an approximating ∞ A α ,α∈Lim(ω ) L ⊂ Æ family ( n( ))n∈Æ 1 , and let . B α ,α∈Lim(ω ) Then there is an approximating family ( n( ))n∈Æ 1 defining a transfinite family (Gα) for which it follows that for any α<ω1 <∞ L (26) Fα ∩ [L] ⊂Gα.

Proof. We write L as L = {`1,`2,...... } with `1 <`2,<.... For a limit ordinal α<ω1 and i ∈ B α A α F ∩ L <∞ ⊂GL an Æ we put i( )= ì( ) and show by transfinite induction that α [ ] α, G B α ,α∈Lim(ω ) where the family ( α) is defined based on the approximating family ( i( ))n∈Æ 1 . Assuming the claim to be true for all β<αthe claim follows immediately from Proposition <∞ 3.4 (a) for α if Sα is a successor. If α is a limit ordinal we observe that for an F ∈Fα∩[L]

<∞

F ∈ F ∩ L n F i ∈ Æ ` n we have β∈An(α) β [ ] with = min . Choosing 0 so that i = we 18 TH. SCHLUMPRECHT

deduce from the induction hypothesis that [ [ [ <∞ L L F ∈ Fβ ∩ [L] ⊂ Gβ = Gβ

β∈An(α) β∈An(α) β∈Bi(α)

which implies by Proposition 3.4 (b) the claim. £

4. The transfinite family (Fα) is universal

The main goal in this section is to prove that the family (Fα) which was introduced in section 3 is universal (see Theorem 4.10 for the precise statement). In the following Deﬁntion we are using the transﬁnite family (Fα) to measure the com- A⊂ <∞ plexity of hereditary sets [Æ ] . As we will see later this measure is equivalent to the Cantor Bendixson index introduced in Section 1.

Definition 4.1. A α ,α∈Lim(ω ) Consider an approximating family ( n( ))n∈Æ 1 defining a transfinite

family (Aα)α<ω1 . ∞

P <∞ α<ω A⊂ Æ A α P Let ⊂ Æ , 1, and [ ] be hereditary. We say that is -large on ,iffor ∞ ∞ N all M ⊂ P there is an N ⊂ M so that Fα ⊂A. A,P {α A α P } A P Moreover, we call Á( ) = sup : is -large on the complexity of on . Remark. Since the notion α-large depends on (Fα) which depends on the choice of the approximating family, we should have rather used the notion Fα-large instead of α-large. But in Corollary 4.4 we will show that the property of being α-large is independent to the underlying approximating family. For the results up to Corollary 4.4 we consider the approximating family and its transﬁnite family to be ﬁxed. Proposition 4.2. A,P P (Stabilization of Á( )with respect to ). ∞ ∞

A⊂ <∞ P ⊂ Æ α Á A,P Q ⊂ P Assume [Æ ] is hereditary, and let 0 = ( ). Then there is a so ∞ L A,L α that for all ⊂ Q it follows that Á( )= 0. ∞ ∞ A,P <α Q ⊂ P N ⊂ Q Proof. Since Á( ) 0+ 1 we deduce that there is a so that for all it ∞

F N 6⊂ A N ⊂ Q Á A,N <α follows that α0+1 . This implies that for all we have ( ) 0+ 1, and thus

∞

A,N ≤ α Á A,N ≥ I A,P N ⊂ P Á( ) 0. On the other hand it is clear that ( ) ( ) for all , which

implies the claim. £ ∞

<∞ α<ω A⊂ Æ Proposition 4.3. Let P ⊂ Æ , 1, and [ ] hereditary. a) If β<αand if A is α-large on P , then A is β-large on P . b) If α = β +1, then ∞ ∞ A is α-large on P ⇐⇒ ∀ Q ⊂ P ∃ L ⊂ Q so that  ∞ ∞ ∀` ∈ L ∀M ⊂ L ∃N ⊂ M N (∗) F ⊂A|` := {E : {`}∪E∈A,`

c) If α ∈ Lim(ω1) and (An(α)) is the approximating sequence for α it follows that A α P ⇐⇒ ∀ n ∈ A A α P is -large on Æ is max n( )-large on [ ⇐⇒ ( by part (a) )∀β<α Ais β-large on P ] OPERATORS ON BANACH SPACES 19

A,P <ω α<ω Together with (a), (c) implies that if Á( ) 1 it follows that the set of all 1 for A α P , A,P which is -large on is the closed interval [0 Á( )].

Proof. (a) follows immediately from Proposition 3.6. ∞ ∞ L For (b) ”⇒” let A be α-large and Q ⊂ P . Then there is an L ⊂ Q with Fα ⊂A. Since α = β + 1 it follows from Proposition 3.4 that L L Fα = {`}∪E:E∈Fβ,`∈L and `

∞ L Therefore it follows for all ` ∈ L that Fβ ⊂A|` and therefore it follows for any M ⊂ L that M Fβ ⊂A`. ∞ ∞ In order to prove “⇐” of (b) assume that M ⊂ P . We need to find N ⊂ M so that N Fα ⊂A. By assumption we find an L =(ì)⊂M satisfying (∗). n By induction we can choose 1

(27) ni+1 ∈ Ni for i =1,...k−1 (28) # Ni ∩ [0,ni+1] ≥ i + 1 for i =1,...k−1 F Ni ⊂A| i ,...k (29) β ni, for =1 .

∞ Indeed, choose n1 = `1 and, using the property (∗) (with ` = n1) we find N1 ⊂ L so that F N1 ⊂A| n ∈ N ∩{n ,n ,...} β n1. Then we choose an 2 1 1 +1 1 +2 , large enough in order to ∞ ∗ ` n N ⊂ N F N2 ⊂A| satisfy (28), and apply ( ) again (with = 2) to find an 2 1 with β n1.We continue in that way. } {ni:i∈Æ Now we claim that Fα ⊂A, which would finish this part of the proof. Indeed, if } E {n ,n ,...n }∈F{ni:i∈Æ n

}∩ ,n {n ,n ...n }∈F i2−1 i {n i∈ Æ 2 =# i : [0 i2]. Therefore, by Proposition 3.5 i2 i3 ik β , and N N N F i2−1 ⊂F i1 {n ,n ...n }∈F i1 thus, since β β , it follows that i2 i3 ik β , which implies by (29) that {ni ,ni ...ni }∈A|n {ni ,ni ...ni }∈A 2 3 k i1 and thus 1 2 k . ∞ The claim (c)“⇒” follows from (a). In order to show (c)“⇐” let L ⊂ P . Using the ∞ ∞ ∞ L Ni assumption and part (a) we ﬁnd ⊃ N1 ⊃ N2 ⊃ ... so that Fγ ⊂Afor all i ∈ Æ and ∞ γ ∈ Ai(α). Then choose N =(ni) to be a diagonal sequence of (Ni)i=1 in such a way that i ∈ n N i N for Æ i is in i and is at least as big as the -th element of i. F N ⊂A E ∈FN E ∈FN γ ∈ A α It follows that α . Indeed, let α and, thus, γ0 for some 0 i0 ( )

where ni0 = min E. Note that by the choice of N we have that i0 =#(N∩[1,ni0]) ≤

#(Ni0 ∩ [1,ni0]) and N ∩ [ni0 , ∞) ⊂ Ni0 ∩ [ni0 , ∞), and we deduce from Proposition 3.5 that N E ∈FN ∩{n ,n ,...} <∞ ⊂F i0 ∩{n ,n ,...} <∞ ⊂A γ0 [ i0 i0 +1 ] γ0 [ i0 i0 +1 ] , which ﬁnishes the

proof. £

Now we can conclude that the property of being α-large for a set A does not depend on the choice of the approximating sequences. ∞

Corollary 4.4. A⊂ <∞ P ⊂ Æ α<ω α For an [Æ ] , and 1, the property of being -large does not depend on the choice of approximating family one has chosen for deﬁning the sets F β. 20 TH. SCHLUMPRECHT

Proof. Assume that (Gα) is a transfinite family defined by another approximating family. ∞ FN We will show by transfinite induction that if α ⊂Afor some N ⊂ Æ then there is an ∞ M M ⊂ N so that Gα ⊂A. Assume that our claim is true for all β<αfor some α<ω1 and ∞ N FN ⊂A let ⊂ Æ be such that α . If α = β + 1 the claim follows from the induction hypothesis and Proposition 4.3 part (b). α If = supβ<α β the claim follows from applying Proposition 4.3 part (a) and (c). £ ∞

<∞ P ⊂ Æ

Corollary 4.5. For A⊂[Æ ] and it follows that

A,P ≤ Á A| ,L . (30) Á( ) sup sup ( p )+1 p∈P ∞ L⊂P A,P

Moreover if Á( )is a limit ordinal then it even follows that

A,P ≤ Á A| ,L (31) Á( ) sup sup ( p ) p∈P ∞ L⊂P A,P α Proof. Put α0 = Á( ). If 0 is a successor our claim follows directly from Proposition 4.3 part (b). If α0 is a limit ordinal and β<α0 arbitrary (and thus β +1<α0) we conclude ∞ p ∈ P L A| ,L ≥β from Proposition 4.3 part (b) that there is a and an ⊂ P so that Á( p ) .

β<α ∞ α ≤ Á A| ,L £ Since 0 was arbitrary it follows that 0 supp∈P supL⊂P ( p ). ∞

<∞ P ⊂ Æ Corollary 4.6. Assume that for A⊂[Æ ] and we have that ∞ Nα (32) ∀α<ω1∃Nα ⊂P Fα ⊂A. ∞ Then there is an L ⊂ P so that [L]<∞ ⊂A. ∞ A⊂ <∞ L ⊂ P Therefore, if an hereditary set [Æ ] has the property, that for no it follows <∞ that [L] ⊂A, the complexity of A must be some countable ordinal α0. A P ω A,P ω We will say that the complexity of on is 1 and write Á( )= 1 if (32) is satisﬁed.

Proof. We first show that there is an n ∈ P , so that (32) holds for A|n (instead of A). Indeed, ∞ n ∈ P α F N 6⊂ A| N ⊂ P otherwise we could find for each an n so that αn n for all . Letting ∞ N α = sup an we deduce that Fα 6⊂ A|n for any n ∈ P and any N ⊂ P . By Proposition 4.3 (b) this would contradict (32) for α + 1. Note that n could have been chosen out of any given cofinite subset of P . We can iterate this argument and produce a strictly increasing sequence (ni) ⊂ P so that

k ∈ A| {A ⊂ Æ {n ,...n }∪A∈A} for every Æ (32) holds for n1,n2,,...nk = : 1 k holds. This

{n <∞ } ⊂A £ implies that that [ i : i ∈ Æ ] .

∞

Á F ,L α Corollary 4.7. For α<ω1 and L ⊂ Æ it follows that ( α )= . L <∞ F ,L ≥ α β>α Proof. Since Fα ⊂Fα∩[L] it is clear that Á( α ) . Assume that for some ∞ N {n(1) (1) F N ⊂F and some = 1 ,n2 ...}⊂ Æ it follows that β α. By Propostion 4.3 (a) we can assume that β = α + 1 and we claim that it would follow N \ N γ<β that there is a family (Nβ)α<β<ω1 of inﬁnite subsets of Æ , with γ β being ﬁnite, if , so that

Nβ (33) Fβ ⊂Fα. OPERATORS ON BANACH SPACES 21

∞ L <ω ⊂F Using Corollary 4.6 this would imply that for some L ⊂ Æ so that [ ] α, contradicting the compactness of Fα. We will show the existence of Nβ by transﬁnite induction of β>α.Forβ=α+1Nβ (γ) (γ) ∞ exists by assumption. If β = γ + 1 and if Nγ = {n1 ,n2 ,...} ⊂ Æ is as in (33) we choose N (α+1) } FNγ ⊂F F Nγ ⊂F γ+1 = {n (γ) : i ∈ Æ and observe that (note that α+1 since γ α) n γ+1 i n o n o Nγ+1 (α+1) (α+1) Nγ F = {n (γ) : i ∈ E} : E ∈Fγ+1 = {n : m ∈ F } : F ∈F γ+1 n m γ+1 n i o (α+1) Nα+1 ⊂ {nm : m ∈ F } : F ∈Fα+1 = Fα+1 ⊂Fα.

If β ∈ Lim(ω1) and if (Ai(β)) is the sequence approximating β we ﬁrst note that we can choose ∞ ∞ S

⊃ M ⊃ M ... i ∈ Æ F ⊂F inﬁnite subsets Æ 1 2 so that for all it follows that γ∈Ai(β) γ α. (β) } n(β) i M i , ... Then we choose Nβ = {ni : i ∈ Æ with i being the -th element of i, for =1 2 .

We deduce then the claim in this case from Proposition 3.5. £

Using our results on the family (Fα)α<ω we can now show the relation ship between Á and the Cantor Bendixson index. Lemma 4.8. A⊂ <∞ Assume that [Æ ] is hereditary and compact. a) Deﬁne the expansion of A by Ep(A)= {n}∪A:A∈A ∪ {∅}. (Note that Fα+1 = Ep(Fα) for α<ω1 and that F ∈ Ep(A) if and only if F = ∅ or F \{min F }∈A.) (α) For α ≤ CB(A) it follows that Ep A(α) = Ep(A) .

` ` ∞ n ∈ Æ b) Assume that there is a sequence ( n) ⊂ Æ with limn→∞ n = and for <∞ hereditary and compact sets An ⊂ [{`n,`n +1,`n +2,...}S] so that αn = CB(An) strictly increases to some α<ω1, and assume that A = An. Then it follows that A α CB( )= +1. n ∈ α<ω A| (α) A(α) | c) For any Æ and any 1 it follows that n = n α A ≥ A(1) Proof. (a) First assume that = 1 and, thus, that CB( ) 1. Clearly, Ep as well as (1) <∞ F 6 ∅ Ep(A) contain ∅ as an element. For F ∈ [Æ ] , = , we observe that ∞ F ∈ A (1) , N> F ∀n ∈ NF∪{n}∈ A Ep( ) ⇐⇒ ∃ N ⊂ Æ min max Ep( ) ∞ ⇐⇒ ∃ N , N> F ∀n ∈ N F ∪{n} \{ F ∪{n}} ∈ A ⊂ Æ min max min ∞ ⇐⇒ ∃ N , N> F ∀n ∈ N F \{ F } ∪{n}∈A ⊂ Æ min max min ⇐⇒ F \{min F }∈A(1) ⇐⇒ F ∈ Ep A(1) . For general α ≤ CB(A) the claim now follows easily by transﬁnite induction. α >β ∅∈A(β) ⊂A(β) (b) Note that for β<αwe can choose m ∈ Æ so that m and thus m . Since α is a limit ordinal we deduce that ∅∈A(α) and, thus, that CB(A) ≥ α +1. m ∈ On the other hand note that for any Æ it follows that m ∞ [ (αm) [ (αm) (αm) <∞ A = Ai ∪ Ai ⊂ [{`m,`m +1,...}] i=1 i=m+1 T (α) <∞

and, thus, that A ⊂ [{`m,`m+1,...}] ={∅}, which implies that CB(A) ≤ α +1. m∈Æ 22 TH. SCHLUMPRECHT

<∞

F∈ Æ \ {∅} F>n To prove (c) let ﬁrst α = 1 and n ∈ Æ .For [ ] , min , it follows that ∞ F ∈ A| (1) , M>n∀m∈M {n}∪F ∪{m}∈A n ⇐⇒ ∃ M ⊂ Æ min (1) (1) ⇐⇒ { n }∪F ∈A ⇐⇒ F ∈ A |n

For general α we conclude the claim by transﬁnite induction. £ ∞

<∞ P ⊂ Æ Corollary 4.9. For α<ω1, a hereditary and compact A⊂[Æ ] , and an it follows that ∞ (34) A is α-large on P ⇐⇒ ∀ Q ⊂ P CB(A∩[Q]<∞)≥α+1. ∞ N Proof. In order to show “⇒” it is enough to observe that CB(Fα )=α+ 1 for any N ⊂ which follows by transfinite induction on α using (a) (in the successor case) and (b) (in the case of limit ordinals) of Lemma 4.8. We also show “⇐” by transfinite induction on α<ω1 and assume that the implication is α<α true for all ˜ . ∞ A⊂ <∞ Q ⊂ P Assume that [Æ ] is compact and hereditary so that for all it follows that CB(A∩[Q]<∞)≥α+1. If α = β + 1, it is by the induction hypothesis and by Proposition 4.3 enough to show: ∞ ∞ ∞ Claim. ∀Q ⊂ P ∃L ⊂ Q ∀` ∈ L ∀K ⊂ L, ` < min K CB(A∩[K]<∞)≥β+1. ∞ ∞ Assume the claim is not true and choose a Q ⊂ P so that for all L ⊂ Q there is an ` ∈ L ∞ and a K ⊂ L so that CB(A∩[K]<∞)≤β. ∞ We first put L1 = Q and then choose `1 ∈ L1 and K1 ⊂ L1, with min K1 >`1 so that ∞ <∞ CB(A|`1 ∩ [K1] ) ≤ β. Then we let L2 = K1 and choose an `2 ∈ L2 and a K1 ⊂ L2 with <∞ min K2 >`2 so that CB(A|`2 ∩ [K2] ) ≤ β. We can continue in this way and eventually ∞ ∞ get a strictly increasing sequence L =(ì) and (Li) with Q = L1 ⊃ L2 ⊃ ... so that

A| <∞ ∩ L ≤ β i ∈ Æ CB( ì [ i+1] ) . Thus it follows for each that A| ∩ L <∞ (β) A| ∩ {` ,` ,...} <∞ (β) ⊂ A| ∩ L <∞ (β) ∅, ì [ ] = ì [ i+1 i+1 +1 ] ì [ i+1] = which implies by Lemma 4.8 (c) that A(β) ∩ [L]<∞ must be finite and, thus, that CB(A∩[L]<∞) ≤ β+ 1, contradicting the assumption. This proves the claim and the induction step in the case that α is a successor. ∞ If α is a limit ordinal and CB(A∩[Q]<∞)≥α+ 1 for all Q ⊂ P it follows that for all β<αwe have CB(A∩[Q]<∞) ≥ β, and, thus, by our induction hypothesis that A is β-large P β<α A α P on , for all , which implies, by Proposition 4.3 (c), that is -large on . £ Using the equivalence of the strong Cantor Bendixson index and the concept of α-largeness and using Corollary 4.6 we can prove Proposition 1.3 of Section 1.

Proof of Proposition 1.3. Assume that X is a Banach space with a seminormalized basis (xi)

and let (zn) another seminormalized basic sequence with γ0((zn), (xkn )) = ω1. By Corollary 4.9 we deduce for 1 ≤ c<∞and γ<ω1 that n o ∞ c z C z , x ,c γ (( n), (xn),γ) = inf c ≥ 1:∃N⊂Æ (( n) ( n) )is -large on N < ∞.

Since c((zn), (xn),γ) is non decreasing in γ we deduce from the uncountability of [0,ω1) that there is a c0 > 0 so that c((zn), (xn),γ) ≥ c0 for all γ<ω1. But this means that for any OPERATORS ON BANACH SPACES 23

∞

γ<ω1 the set C((zn), (xn),c0/2) is γ large on some set Nγ ⊂ Æ . From Corollary 4.6 we ∞ L <∞ ⊂C z , x ,c / deduce therefore that there is an L =(`i)⊂Æ so that [ ] (( n) ( n) 0 2). By z passing to a subsequence of ( n) we might simply assume that L = Æ . n ∈ m(n)

We are now in the position to state and prove Theorem 4.10 concerning the universality of the transfinite families. Theorem 4.10. A⊂ <∞ α Let [Æ ] be not empty and hereditary, and assume that 0 = A,P <ω B α Á 1 n ,α∈Lim(ω ) ( ) . Then there is an approximating family ( ( ))n∈Æ 1 defining the trans- ∞ finite family (Gα)α<ω1 , and there is an L ⊂ P so that GL ⊂A∩ L<∞ ⊂G ∩ L<∞. (35) α0 [ ] α0 [ ]

Proof. Let (Fα) be a transfinite family being chosen a priori. We will prove the claim by transfinite induction for all α0 <ω1. If α0 = 0, we deduce that L = {` ∈ P : {`}6∈A}is infinite and, thus, since A is hereditary and not empty it follows that <∞ A∩[L] ={∅} = F0.

A⊂˜ <∞ ˜ Á A,P <α α ≥ Assume the claim to be true for all hereditary [Æ ] with ( ) 0, where 0 1.

A⊂ <∞ Á A,P α P Let [Æ ] be hereditary with ( )= 0. By passing to a subsequence of ,if ∞ A,L α L ⊂ P necessary, we can assume that Á( )= 0 for all (we are using Proposition 4.2). Since A is not α0 + 1-large on P we deduce from Proposition 4.3 part (b) that there is a ∞ Q ⊂ P so that

∞ (36) ∀M ⊂ Q ∃m ∈ M A|m is not α0-large on M. 24 TH. SCHLUMPRECHT

A| ,M <α We start by applying (36) to M1 = Q and ﬁnd an m1 ∈ M1 for which βm1 = Á( m1 1) 0

(recall that by Proposition 4.3 the set of ordinals α for which A|m1 is α-large is a closed inter- (1) Bn γ ,γ∈Lim(ω ) val). By the induction hypothesis we can find an approximating family ( ( ))n∈Æ 1 (1) <∞ which defines a transfinite family (Gγ )γ<ω1 and an M2 ⊂ M1 so that A|m1 ∩ [M2] ⊂ (1) <∞ G ∩ [M2] . βm1 A| β M Since m1 is m1 -large on 1 (which does not depend on the choice of the approximating (1) M2 family) we also can require that G ⊂A|m1. βm1 ∞

m i ∈ Æ Mi ⊂ Æ By repeating this argument we ﬁnd an increasing sequence ( i)i∈Æ , sets , for , (i) β ⊂ ,α0 Bn γ m i∈Æ ,γ∈Lim(ω ) a sequence of ordinals ( i ) [0 ), and approximating families ( ( ))n∈Æ 1

(i) i ∈ Æ deﬁning transﬁnite families (G )γ<ω1, for i ∈ Æ , so that for all

∞ (37) mi ∈ Mi,Mi+1 ⊂ Mi, and mi < min Mi+1, G(i) Mi+1 ⊂A| ∩ M <∞ ⊂G(i) ∩ M <∞, (38) mi [ i+1] [ i+1] βmi βmi

Putting M = {m1,m2,...} we deduce from (37) and (38) that for all i ∈ Æ G(i) {mi+1,mi+2,...} ⊂A| ∩ M<∞ ⊂G(i) ∩ M<∞. (39) mi [ ] [ ] βmi βmi

(for the second “⊂” recall that we deﬁned A|m in such a way that A|m ⊂ [{m +1,m+ ∞ ,...} <∞ ˜ A| , M˜ β 2 ] ) which implies that for any m ∈ M and any M ⊂ M we have Á( m )= m and, thus, by Corollary 4.5, it follows that . (40) sup βm +1=α0, for all k ∈ Æ m∈M,m≥k

To finish the proof we distinguish between the case that α0 is a successor and the case that α0 is a limit ordinal. If α0 = γ + 1 we deduce from (40) that the set L˜ = {m ∈ M : βm = γ} is infinite. Using B α ,α<ω Proposition 3.2 (b) we can find an approximating family ( n( ))n∈Æ 1 so that for any α<ω G(i) ∩ {i, i ...} <∞ ⊂G G 1 and any i ∈ Æ it follows that α [ +1 ] α, where ( α) is defined by B α ,α<ω ( n( ))n∈Æ 1 . We therefore deduce that <∞ <∞ A∩[L˜] = {`}∪E :`∈L,˜ ` < E and E ∈A|` ∩[L˜] ∪ {∅} ⊂ {`}∪E ∃i∈ ` m ∈L,˜ ` < E, E ∈G(i) ∩L˜<∞∪{∅} : Æ = i [ ] (by (39)) γ ˜ ˜ <∞ ˜ <∞ ⊂ {`}∪E :`∈L, ` < E and E ∈Gγ ∩[L] ∪ {∅}=Gα0 ∩[L] (since i≤mi).

α ∈ ω B α ,α<ω If 0 Lim( 1) we also deﬁne the approximating family ( n( ))n∈Æ 1 as in Proposition

3.2 (b), but add in the case of α = α0 the ordinals βm1 +1,βm2+1,...βmn+1 to the set Bn(α0) B α B α ,α<ω (still denoting it n( 0)). The transfinite family defined by ( n( ))n∈Æ 1 is denoted by (Gα). We put L˜ = M = {m1,m2,...}. <∞ Now if E ∈A∩[L˜] ,E=6 ∅, we write E = {m}∪F with m = mn = min E and (n) F ∈A|m ⊂G Gα α<ω n βmn (by (39)). From the definition of the family ( ) 1 we conclude that F ∈G E ∈G n ≤ m ≤ E β ∈B α βmn and thus βmn+1. Since n and since mn +1 n( 0) we deduce ˜<∞ that E ∈Gα0 ∩[L] . OPERATORS ON BANACH SPACES 25

Therefore we derive in both cases (α0 being a successor and α0 being a limitordinal) that ˜ <∞ ˜<∞ (41) A∩[L] ⊂Gα0 ∩[L] .

On the other hand it follows from the deﬁnition of α0 and from Proposition 4.3 that A is α0-large (which by Corollary 4.4 does not depend on the transﬁnite family). We can ∞ L ⊂ L˜ GL ⊂A therefore chose an so that α0 which implies that GL ⊂A∩ L<∞ ⊂G ∩ L<∞ α0 [ ] α0 [ ]

and ﬁnishes the proof of Theorem 4.10. £ <∞ A We will have to apply Theorem 4.10 not only for one A⊂[Æ ] but for a sequence ( n) simultaneously. Therefore we need the following reformulation. ∞ Corollary 4.11. P ⊂ N A

Assume we are given a , a sequence ( `)`∈Æ of nonempty and

<∞ ` ⊂ Æ hereditary subsets of [Æ ] , an increasing sequence ( k) , and a sequence of ordinals (α`) so that

∞

A ,Q , ` ∈ Æ Q ⊂ P. (42) α` ≥ Á( ` ) whenever and ∞ Then there is a transﬁnite family (Gα)α<ω1 and K ⊂ P , K = {k1,k2,...}, (ki) strictly n ∈ ` ≤ ` increasing, so that for all Æ and n A ∩ {k ,k ,...} <∞ ⊂GK. (43) ` [ n n+1 ] α` Proof. We ﬁrst use Proposition 4.2 and an easy diagonalization argument to assume that

∞

A ,Q Á A,P Q ⊂ P Á( ` )= ( ` ) for all (note that (42) stays valid if we pass to subsequences). i ∈ A ` ≤ ` For Æ we then apply Theorem 4.10 to each of `, i, in order to get an approximat- (i) (i) ∞ Bn γ ,γ∈Lim(ω ) Gα α<ω Li ⊂ P ing family ( ( ))n∈Æ 1 with associated transfinite families ( ) 1 and an ∞ ` ≤ ` L ⊂ L L P A ∩ L <∞ ⊂G so that for any i it follows that i i−1 (with 0 = ) and ` [ i] S α`. n ∈ γ ∈ ω B γ n B(j) γ Now define for Æ and Lim( 1) as in Proposition 3.2, i.e. n( )= j=1 n ( ), let (Hα) be the transfinite family associated to Bn(γ), and Let K = {k1,k2 ...}be a diagonal L ` ≤ ` sequence of the i’s. Then we deduce that for any m ∈ Æ and any m it follows that <∞ <∞ <∞ A` ∩ [{km,km+1 ...}] =A` ∩[Lm] ∩[{km,km+1,...}] ⊂G` ∩ {k ,k ,...} <∞ ⊂H ∩ K∞. α` [ m m+1 ] α` [ ] G <∞ K By Proposition 3.8 we then choose ( α)α<ω1 so that Hα ∩ [K] Gα , for α<ω1. £

5. Some Consequences of Theorem 4.10 Using Theorem 4.10 we deduce the following generalization of Ramsey’s theorem for ﬁnite sets. ∞

<∞ P ⊂ Æ Corollary 5.1. Let F⊂[Æ ] be hereditary and let . ∞

If F = A∪B with A and B also being hereditary then there is a Q ⊂ P so that

A,Q ,Á B,Q Á F,Q (44) max(Á( ) ( )) = ( ) 26 TH. SCHLUMPRECHT

Proof. By passing to an inﬁnite subsequence of P and using Proposition 4.2, if necessary, we ∞ can assume that there are ordinals α0,β0 and γ0 so that for all P˜ ⊂ P

˜ ˜ ˜

A, P ,β Á B,P, γ Á F, P . α0 = Á( ) 0=( )and 0 = ( )

We need to show that max(α0,β0)=γ0.

Assume that this is not true and, thus, assume that max(α0,β0)<γ0.

A,P Á B,P By Theorem 4.10 and the fact that Á( ) and ( ) are stabilized in the sense of ∞ <∞ Proposition 4.2 we ﬁnd a transﬁnite family (Gα)α<ω1 and an L ⊂ P so that A∩[L] ⊂Gα0 ˜<∞ A,L ≤

and B∩[L] ⊂Gβ0 and therefore, it would follow from Corollary 4.7 that Á( ) G ,L α ,β <γ £ Á( max(α0,β0) ) = max ( 0 0) 0 which is a contradiction. <∞

We introduce the following ”addition” of subsets of [Æ ] < AtB {A∪B A∈A,B ∈B} Definition 5.2. For A, B⊂[Æ ] we define = : F tF ,N α β Remark. At first sight one might believe that Á( α β )= + . But this cannot be true since on one hand addition on the ordinal numbers is not commutative, on the other AtB BtA A, B⊂ <∞ hand it is clear that = , for any [Æ ] . For example it is easy to see that Fω+1 tFω =2ω+1=26 ω=ω+1+ω. One could define the following ”commutative addition of ordinal numbers”:

α t β F tF ,Æ . = Á( α β ) It might be interesting to determine the properties of this binary operation and compare it with the addition of ordinal numbers. Nevertheless it is easy to prove by transﬁnite induction on β that for all α, β < ω1 it follows that

(45) Fα+β ⊂FαtFβ.

Proposition 5.3. Assume that α<β<ω1 and that γ<ω1 F ∩ {m, m ,...} <∞tF ⊂F ∩{m, m ,...} <∞tF a) There is an m ∈ Æ so that α [ +1 ] γ β [ +1 ] γ

∞

Á F tF ,N < Á F tF ,N b) For any N ⊂ Æ it follows that ( α γ ) ( β γ ) Proof. (a) follows immediately from Proposition 3.1(c). To prove (b) deﬁne δ = ∞ ∞ F tF ,N M ⊂ N L ⊂ M Á( α γ ). Let . By Proposition 4.3 (last part) there is an so that L Fδ ⊂FαtFγ. Then note that L L Fδ+1 = {{`}∪D:D∈Fδ and ` ∈ L} <∞ <∞ ⊂{{`}∪A∪G:A∈Fα∩[L] ,G∈Fγ ∩[L] `∈L} <∞

⊂{A˜∪G:A˜∈Fα+1 ∩ [L] ,G∈Fγ}⊂Fα+1 tFγ.

F tF ,N <δ ≤ Á F tF ,N ≤ Á F tF ,N Thus, by (a), δ = Á( α γ ) +1 ( α+1 γ ) ( β γ ) which ﬁnishes

the proof. £

Proposition 5.4. Assume that (Fα)α<ω1 and (Gα)α<ω1 are two transﬁnite families and let ∞

N α<β Á F tF ,N Á G tG ,N ⊂ Æ and be such that ( α β ) and ( α β ) are stabilized in the sense

˜ ˜

F tF ,N Á F tF ,N Á G tG ,N Á G tG ,N of Proposition 4.2, i.e. Á( α β )= ( α β ) and ( α β )= ( α β )

for all N˜ ⊂ N.

F tF ,N Á G tG ,N Then it follows that Á( α β )= ( α β ). OPERATORS ON BANACH SPACES 27

Proof. Write N as N = {n1,n2,...}, with ni %∞. By Proposition 3.7 we can ﬁnd an M =

{m M M ,m ,...}⊂N m %∞ G ⊂F G ⊂F L {m i ∈ Æ } 1 2 , i , so that α α and β β. Deﬁne = ni : .

M M

G tG ,N Á G tG ,L ≤ Á F tF ,L Á F tF ,N Then it follows that Á( α β )= ( α β ) ( α β )= ( α β ), which

ﬁnishes the proof by symmetry. £

Proposition 5.5. (Cancellation Lemma) α, β < ω <∞ Let 1 and consider a map Ψ:FαtFβ →[Æ ] with the following property: ∞

<∞

N ⊂ Æ F tF ⊂B Á B,N ≤β There exists a hereditary B⊂[Æ ] and an so that Ψ( α β) and ( ) . C⊂ <∞ {A \ A A∈F tF } If [Æ ] is hereditary and contains the set Ψ( ): α β then there is ∞ C,M ≥α an M ⊂ N so that Á( ) .

∞ M C,M <α Proof. Assume that our claim is not true and that for all ⊂ N it follows that Á( ) . First, by applying Proposition 4.2 and passing to a subsequence of N, we can assume that ∞

˜ ˜

B, N Á B,N ≤ β N ⊂ N Á( )= ( ) for all . By applying Proposition 4.2 a second time we can ∞

α ˜ ˜ C, N Á C,N <α N ⊂ N β Á B,N ≤β also assume that 0 = Á( )= ( ) , for all , and define 0 = ( ) . Using Theorem 4.10 (since we are in the stabilized situation we can apply it simultaneously ∞ to C and B as in the proof of Corollary 4.11) we obtain a transfinite family (Gγ) and a M ⊂ N <∞ <∞ <∞ so that B∩[M] ⊂Gβ0 and C∩[M] ⊂Gα0 and thus (BtC)∩[M] ⊂Gβ0 tGα0. On the other hand it is clear that Fα tFβ ⊂CtB. Now, using Proposition 4.2 we first ∞

˜ ˜

G tG ,L I G tG ,L Á G tG ,M L ⊂ L ﬁnd a subset L ⊂ M so that Á( β0 α0 )= ( β0 α0 )= ( β0 α0 ) for all . ∞

F tF ,K Á F tF ,K Á F tF ,L Then we pass to a subset K ⊂ L so that and Á( α β )= ( α β )= ( α β ), ∞ for all K˜ ⊂ K.

Finally we deduce the following chain of inequalities

F tF ,K ≤ Á CtB,K ≤ Á G tG ,K Á F tF ,K Á( α β ) ( ) ( α0 β0 )= ( α0 β0 )

(the last inequality follows from Proposition 5.4). But on the other it follows from Proposition

F tF ,K > Á F tF ,K ≥ Á F tF ,K £ 5.3 that Á( α β ) ( α0 β ) ( α0 β0 ) which is a contradiction.

6. Proof of Theorem 1.4 We now turn to the proof of Theorem 1.4. We will first restate it in an equivalent form using the equivalence of the Cantor Bendixson index and the concept of α-largeness. Recall the definitions of B(z,b), B(X, β), β0(z) and β0(X) in Definition 1.2 for X being a Banach space, z =(zn)⊂Xseminormalized, b>0 and β<ω1. Using Corollary 4.9 we get n o ∞ Bz,b β N , (46) b(z,β) = inf b ≥ 1:∃N⊂Æ ( )is -large on

b ≥ B <∞ z ∼ `A− } where for 1 and (z,b)={A∈[Æ ] :( n)n∈A b 1 unit vector basis . We secondly want to replace in the deﬁnition of b(z,β) the set B(z,b) by a somewhat more convenient set. We will need the following special case of a result from [AMT] (see also [AG] Lemma 3.2).

Lemma 6.1. Assume (xn) is a weakly null and semi-normalized sequence in a Banach ∞ ∞

X δ> N ε > n ∈ Æ M ⊂ Æ space .Let 0, ⊂Æ and n 0, for . Then there exists an , 28 TH. SCHLUMPRECHT

M = {m1,m2,...} so that for all ﬁnite F ⊂ M the following implication is true: ∗ ∗ ∗ (47) If there is an x ∈ BX∗ , with x (xm) ≥ δ for all m ∈ F, then there is a y ∈ BX∗ ,

∗ ∗ , m 6∈ F. y x ≥ δ m ∈ F |y x | <ε i ∈ Æ with ( m) for all and ( mi ) i for all with i For the sake of being self contained and reader-friendly we present a proof of this special case of the above cited result of [AG].

∞

m ∈ Æ L ⊂ N m

Clearly the claim of the Lemma follows if we can accomplish such a choice of mk’s and Lk’s. Assume for some k ≥ 1 we have chosen Lk−1 (recall: L0 = N) and m1 <...mk−1. We define mk = min Lk−1 and L, F, n } . L = {L ⊂ Lk−1 : ∀F ⊂{1,...k}∀n∈Æ ( ) satisfies (48) It is easy to see that L is closed in the pointwise topology and we can apply Ramsey’s theorem. In the case that there is an L ∈Lso that [L]∞ ⊂Lwe are done. We have to show that the alternative in Ramsey’s theorem leads to a contradiction. ∞ ∞ ∞ Assume that there is an L˜ ⊂ Lk−1 so that [L˜] ∩L=∅. Thus for any L = {`1,`2...}⊂L˜ ∗ ∗ F F x x ∈ B ∗ there is an = L ⊂{1,2...k} and an n = nL ∈ Æ so that there exists an = L X with x∗ x >δ i ∈ F, x∗ x >δ i , ,...n (49) ( mi ) for all and ( ì ) for all =2 3 ∗ but for any y ∈ BX∗ satisfying (49) there must be either an i ∈{1,...k}⊂Fwith |y∗ x | >ε |y∗ x |≥ε ( mi ) i or ( `1 ) k+1. We first use again Ramsey’s theorem to assume without loss of generality that the sets FL ∞ do not depend on L.ThusFL=Ffor all L ⊂ L˜. ∞ Fixing for a moment such an L = {`1,`2 ...} ⊂ L˜ we let j0 = max{j ∈ [0,k]:j6∈ F }. ∗ ∗ If j0 = 0 (meaning F = {1, 2 ...k}) we put zL = xL and observe that we must have ∗ |zL(x`1 )|≥εk+1.Ifj0≥1 we apply the fact that our induction hypothesis is true for j0 − 1 ∗ and are able to find a zL ∈ BX∗ satisfying (49) and secondly |z∗ x | <ε i ∈{ , ,...k}\F (50) L( mi ) i if 1 2 |z∗ x | >ε and, thus, we also must have L( `1 ) k+1 (apply the induction hypothesis to the set

L = {mj0 ,...mk,`2,`3,...}⊂Lj0−1,F∩[1,j0 −1] and n + k − j0 + 1). OPERATORS ON BANACH SPACES 29

˜ ˜ ˜ x∗ |x∗ x |≥ L {` , ` ,...} m ∈ Æ ˜ We write = 1 2 and claim that for any there is a m so that m( ì ) εk+1, for i =1,2,...m. This would be a contradiction to the assumption that (xi) is weakly null. For each j =1,2,...m we find an nj so that the triple (L˜j,F,nj) does not satisfy (48) ˜ ˜ ˜ ˜ m with Lj = {`j, `m+1, `m+2 ...}. Choose j0 so that nj0 is the maximum of (nj)j=1. Then ∗ ∗ choose xm = zL (where zL is defined as above). j0 ∗ We observe that xm satisfies (49) with respect to all of the Lj’s. Secondly it satisfies (50) ∗ |x x˜ | >ε j , ,...m and thus it must follow that m( `j ) k+1 for all = 2 . This finishes the proof of

the claim and, thus the proof of the Lemma. £ We will slightly reformulate Lemma 6.1.

Corollary 6.2. Assume (xn) is a weakly null and semi-normalized sequence in a Banach ∞ X δ> N ε> space .Let 0, ⊂Æ and 0. ∞ M M {m ,m ,...} F ⊂ M Then there exists an ⊂ Æ , = 1 2 so that for all ﬁnite the following implication is true: ∗ ∗ ∗ (51) If there is an x ∈BX∗ , with x (xm)≥δ for all m∈F, then there is also a z ∈BX∗ ,

∗ ∗ , m 6∈ F. z x ≥ δ − ε m ∈ F z x , i ∈ Æ with ( m) for all and ( mi )=0 for all with i

Proof. After passing to a subsequence of xn we can assume that there is for each n ∈ Æ ∗ ∗

∗ n ∈ Æ an xn ∈ (2/kxnk)BX , with xn(xm)=δ(n,m), whenever n, m ∈ Æ . Choose for ∞ −n

εn =2 ε/(2 + sup kxnk) and apply Lemma 6.1 in order to obtain M =(mi)⊂ Æ . n∈Æ ∗ ∗ ∗ If F ⊂ M is ﬁnite and x ∈ BX∗ so that x (xi) ≥ δ, wheneverPi ∈ F , then we let y ∈ BX∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

be as prescribed in (47) and let z =˜z/kz˜kwherez ˜ = y − m∈M\F y (xm)xm. £

We introduce notations similar to B(z,b), b(z,β), B(β,X), β0(z) and β0(X).

Deﬁnition 6.3. Let x =(xn) be a seminormalized sequence in a Banach space X.Fora>0 we put <∞ ∗ ∗ A ∃x ∈B ∗∀i∈Axx≥a. (52) (x, a)= A∈[Æ ] : X (i)

For α<ω1 we let ∞ a Ax, a α N (53) (x, α) = sup a ≥ 0:∃N⊂ Æ ( )is -large on ∞ a ≥ ∃M FM ⊂Ax, a = sup 0: ⊂ Æ α ( ) ,

(54) α0(x) = sup{α<ω1 :a(x, α) > 0},

(55) A(α, X) = sup{a(z, α):z⊂BX seminormalized and weakly null}, and

(56) α0(X) = sup{α<ω1 : A(α, X) > 0} = sup{α0(z):z⊂BX seminorm., weakly null}.

Lemma 6.4. Let x be a seminormalized sequence in a Banach space X with α0(x) <ω1. Then there is subsequence y =(yn)of x with the following properties. α<ω 0 0 a) For all 1 and all a

Moreover if β0 <α0(x)(and, thus a(x, β0) > 0) and if 0 <ηa(x, β0) − η. 30 TH. SCHLUMPRECHT

Proof. We ﬁrst note that if (u)=(un) is almost a subsequence of v =(vn) (i.e. for some

n ∈ Æ u v a u, α ≤ a v, α

0 it follows that ( n0+i)i∈Æ is a subsequence of ) then ( ) ( ). In particular this means that α0(u) ≤ α0(v) and therefore it is enough to ﬁnd a subsequence of x which satisﬁes (a), (b) and (c) for all α<α0(x). Claim. Let y =(yn) be a subsequence of x and let α<α0(x). Then there is a subsequence 0 0

z of y so that for all a a(y, α)−η (this proves the part of our claim starting with “moreover” if we let α = β0). Writing now the interval [0,α0(x)) as a sequence (αn) and applying successively the above claim to each αn, we obtain by diagonalization a subsequence y of x so that (a) and (b) of our statement are satisﬁed. It is also clear that a(y, α) is decreasing in 0 α. Let α be a limit ordinal and a

α 0 a

Proposition 6.5. Let x =(xn)be a weakly null, and normalized sequence in a Banach space X and c>η>0. Then there is a subsequence y of x so that

A y, c η ⊂B y, 1 ⊂A y, c − η (57) 2 + c

and, thus it follows that

1 1 1 1 (58) a(x, α) ≤ ≤ a(x, α) and A(X, α) ≤ ≤ A(X, α), 2 b(x, α) 2 B(X, α) (59) α0(x)=β0(x), and α0(X)=β0(X)

Proposition 6.5 will follow from Corollary 6.2 and the following simple observation. OPERATORS ON BANACH SPACES 31

,k·k n∈ Æ n Lemma 6.6. Assume E =(Ê ), is an -dimensional normed space for which e n n the unit vector basis ( i) of Ê is a normalized basis. Define: ( i=1 ) ∗ ∗ x (ei)≥c if i ∈ A c1 = max c ≥ 0:∀A⊂{1,2...n}:∃x ∈BE∗ ∗ x (ei)=0if i 6∈ A n Xn Xn o c c ≥ cB n ⊂ B ∗ c k a e k |a | 2 = max 0: `∞ E and 3 = min i i : i =1 i=1 i=1 c ≥ c c ≥ 1 c Then it follows that 1 2 = 3 2 1. Proof. It is clear that c1 ≥ c2. To show that c2P≥ c3 we first observe that by the maximality ∗ ∗ n ∗ ∗ ∗ of c2 we can find an x ∈ SX∗ of the form x = i=1 xi ei ∈ SE∗ (ei being the i-th coordinate |x∗| c i ,...n x x ∈S x∗ x functional)P so that i P= 2 for =1P . Then choose =( i) E so that ( )= ∗ ∗ xixi =1.Thus1= xixi ≤c2 |xi|≤c2/c3, which implies theP claimed inequality.

c n n n ≥ c c ≥ c / a ∈ Ê |a | In order to show P3 2 and 3 1 2 let ( i)i=1 with i=1 i = 1. First we ∗ n ∗ ∗ ∗ ∗ x x e ∈ B ∗ |x | c x a ≤ i ≤ n can choose an =P i=1 i i EPwith i = 2 and sign( i ) = sign( i), for 1 . k n a e k≥x∗ n a e ≥ c c ≥ c This proves thatP i=1 i i ( i=1 i i) 2, which implies 3 P 2. Secondly, we can n + n assume that i=1 ai ≥ 1/2 and a simliar argument implies that k i=1 aieik≥c1/2, and c thus 3 ≥ c1/2. £ Proof of Proposition 6.5. Assume c>η>0 to be given. Using Corollary 6.2 we can ﬁnd a subsequence y =(yn)ofxso that A(y, 2c + η) ⊂ A˜(y, 2c), and (trivially) A˜(y, c) ⊂A(y, c), where we put for r>0 <∞ ∗ ∗ ∗ A˜ ∀B ⊂ A∃x ∈B ∗ x y ≥r i∈B x y i6∈B} (y, r)={A∈[Æ ] : X ( i) if and ( i)=0 if . Secondly we deduce from Lemma 6.6 that X X ˜ ∞ k ayk≥c |a|} ⊂ A˜ y, c , A(y, 2c) ⊂{A∈[N] :∀(ai)i∈A ⊂Ê i i i ( ) i∈A i∈A

proving the claim (note that 1/c3, where c3 as deﬁned in Lemma 6.6 is the smallest c so that E c `n is -equivalent to 1 ). £ We now can restate Theorem 1.4 as follows. Theorem 6.7. Let X be a Banach space with a basis not containing c0. Assume that there is an ordinal β0 ∈ [0,α0(X)] so that the following two conditions hold. a) There is a seminormalized weakly null sequence y ⊂ BX with a(y, β0)=0.

b) infγ<β0 B(X, γ) > 0. Then there is a seminormalized block basis (xn) in X and a subsequence (˜yn) of (yn) so that

the map xn 7→ y˜n extends to a strictly singular and linear bounded operator

→ y n ∈ Æ . T :[xn :n∈ Æ ] [˜n : ] In order to prove Theorem 6.7 we will need several Lemmas. Lemma 6.8. Let (Fα) be a transﬁnite family and x =(xn)be a weakly null and seminormalized sequence in a Banach space X satisfying the conclusions of Lemma 6.4. Let 1 ≤ α<ω1 and assume that a(x, α) > 0. For any η>0there is a subsequence z of x so that: ∗ ∗ A ∈F ∗ \ A For α,orA∈FαtFα, there is a zA ∈ BX so that zA(xm)=0if m ∈ Æ and ∗ ∗ zA(xm) ≥ a(x, α) − η,orzA(xm)≥(a(x, α) − η)/2,ifm∈A, respectively. 32 TH. SCHLUMPRECHT

Proof. Let 1 ≤ α<ω1, with a(x, α) > 0, and η>0. Put a = a(x, α). From the definition a ·, · M F M1 ⊂Ax, a − η/ of ( ) it follows that there is an 1 ⊂ Æ so that α ( 4) and applying ∞ Corollary 6.2 we find an M2 ⊂ M1 so that (51) holds for δ = a − η/4 and ε = η/4. In M2 particular this means that our claim holds for all A ∈Fα . Secondly we deduce that M2 M2 Fα tFα ⊂A(x, (a − η/2)/2). Since each A ∈FαtFα is the disjoint union of two M2 M2 ∗ elements of Fα, it also follows that for any A ∈Fα tFα we find a zA ∈ BX∗ so that ∗ ∗ zA(xm)=0ifm∈M2\Aand zA(xm) ≥ (a(x, α) − η)/2, if m ∈ A.

Choosing finally z to be the subsequence defined by M3, will finish the proof. £ Using Lemma 6.8 successively for different α’s and the appropriate choices of η we conclude from a simple diagonalization argument the following Corollary. Corollary 6.9. Let x =(xn)be a weakly null and semi-normalized sequence in a Banach X ` space satisfying the conclusions of Lemma 6.4, let ( k) ⊂ Æ be strictly increasing, and let α ` ≤ ` ( k) ⊂ [0,α0(x)). Then there is a subsequence z so that for any k ∈ Æ and any k it follows that: A ∈F ∩{k, k ,...} <∞ A∈ F tF ∩ {k, k ,...} <∞ For α` [ +1 ] ,or [ α` α`] [ +1 ] , there is a ∗ ∗ ∗ ∗ z ∗ \A z x ≥ a z, α / z x ≥az,α / A ∈ BX so that zA(xm)=0if m ∈ Æ and A( m) ( `) 2,or A( m) ( `) 4, if m ∈ A, respectively. Lemma 6.10. Let x =(xn)be a weakly null and semi-normalized sequence in a Banach space X satisfying the conclusions of Lemma 6.4. Let (δk) be decreasing sequence in (0, 1), δ with k ≤ 1/(k +1), for any k ∈ Æ . Then there is a subsequence z of x and a sequence of ordinals αk which increases to α0(x) k ∈ so that for each Æ 1 δ ≤ a z,α ≤ δ a) 2 k ( k) k b) αk is of the form αk =˜αk+k(here we identify positive integers with finite ordinals of the same cardinality). k ∈ β {β a x, β ≤ δ } δ ≤ / k Proof. We define for Æ k = min : ( ) k . Since k 1 ( + 1) it is easy to see that βk ≥ (k + 1). If βk, is a successor, say βk = γk + 1, then a(x, γk) ≥ δk If βk is a limit ordinal we deduce from the continuity that a(x, βk)=δk, and we let γk = βk. a x, γ ≥ 1 a x, γ ≥ 1 δ γ ≥ k It follows from (45) that ( 2 k) 2 ( k) 2 k. Since k we will be able to find α an k between βk and 2γk which is of the form as required in (b). £

Lemma 6.11. Assume that x =(xn)is a weakly null and seminormalized sequence in a Banach space X satisfying the conclusion of Lemma 6.4. Let αk be ordinals increasing to

α α α α k α <ω k∈Æ 0(x) so that for each k ∈ Æ k can be written as k =˜k+ for some ˜k 1.For let εk = a(z, αk). a) There exists a transﬁnite family (Gα) and a subsequence z =(zn)of x so that for all (ai) ∈ c00 it follows that ∞ X X X aizi ≤ 8 εk−1 max |ai|. A∈Gα k=1 k i∈A

b) Let (Fα) be a transﬁnite family. Then there is a subsequence z =(zn)of x so that for all (ai) ∈ c00 it follows that

X 1 X aizi ≥ max εk |ai|. ,A∈F k∈Æ α 8 k i∈A OPERATORS ON BANACH SPACES 33

∞ k ∈ Æ Proof. From our assumption on x it follows that for all Q ⊂ Æ and all it follows that A x, ε ,Q ≤ α Á( ( 2 k) ) k. Therefore we can apply Corollary 4.11 to obtain a transﬁnite family ∞ G k %∞ i%∞ ` ( α)α<ω1 and a K = {k1,k2,...}⊂ Æ , i ,if , so that for all A x, ε ∩ {k ,k ,...} <∞ ⊂GK. (60) ( 2 `) [ ` `+1 ] α` z x ` Putting =( ki) this implies that for all <∞ (61) A(z, 2ε`) ∩ [{`, ` +1,...}] ⊂Gα . P ` ∗ Let (ai)∈c00. We ﬁnd a z ∈BX∗ which norms aizi and (ε0 =sup kzik) and deduce that X∞ X X X ∗ ∗ ∗ (62) z aizi = z (zi)ai + z (zi)ai ∗ ∗ i=1 k=1 2εk

` ` ⊂ Æ

In order to prove part (b) we ﬁrst choose 0 = 0 and ( k)k∈Æ fast enough increasing so that for all (ai) ∈ c00 it follows that X ε X `k

(64) aizi ≥ sup |ai|.

A⊂Æ,#A=k 4 i∈A After passing to a subsequence we can assume that the conclusion of Corollary 6.9 is satisﬁed.

a <∞ + ∈ k ∈ Æ ` ≤ ` A ∈F ∩ {k, k ,...} A {i ∈ A a ≥ Let ( i) c00, , k and α` [ +1 ] . Letting = : i − + ∗ ∗ 0} and A = A \ A we can choose zA+ and zA− as prescribed in Corollary 6.9 and deduce that X∞ X∞ X∞ X 1 ∗ 1 ∗ 1 (65) aizi ≥ zA+ aizi − zA− aizi ≥ ε` |ai|. i=k 2 i=k 2 i=k 4 i∈A Therefore we obtain ∞ h ∞ k−1 i X 1 X X

aizi ≥ max aizi + aizi k∈Æ i=1 2 i=k i=1 h X Xk−1 i ≥ 1 ε |a | ε |a | max max ` i + `k−1 i

8 k∈Æ `∈(`k−1,`k], A∈F ,A≥k i∈A i=1 α` 1 X 1 X

≥ max max ε` |ai| = max ε` |ai|, proving part (b).

k∈Æ `∈Æ,A∈F 8 `∈(`k−1,`k] 8 α` A∈F i∈A i∈A α`

34 TH. SCHLUMPRECHT £

Lemma 6.12. Assume that u =(un)and v =(vn)are two weakly null and seminormalized sequences in a Banach space X satisfying the conclusions of Lemma 6.4 such that 1 (66) a(u, α) < a(v, α). 5

Let x = u + v =(un+vn). Then there is a subsequence z of x so that (67) Fα ⊂A z,a(v, α)/4 , in particular a(y, α) ≥ a(v, α)/4 for all subsequences y of z.

Proof. By passing to a subsequence we can assume that the conclusions of Lemma 6.8 are 1 ∗ v α η a v, α A ∈F tF v ∈B ∗ satisfied for , , and = 10 ( ). In particular we find for each α α,a A X v∗ x ≥ a v, α − η / 9a v, α m ∈ A which has the property that A( m) ( ( ) ) 2= 20 ( ), whenever , and ∗ vA(xm) = 0, if m 6∈ A. A ∈F tF ψ A {i∈A v∗ u <−1av, α } For α α we define ( )= : A( i) 5 ( ) , and note that

1 1

A u, a v, α , Æ ≤ α F tF ⊂Au, a v, α Á Ψ( α α) ( 5 ( )) and ( ( 5 ( )) ) (using (66)). Secondly we note {A \ A A∈F tF } A v u, 1 a v, α that Ψ( ): α α contains ( + 4 ( )). This means that we can apply ∞

Á A v u, a v, α ,N ≥ α N ⊂ Æ Proposition 5.5 and derive that for some we have ( ( + 4 ( )) ) and ∞ M ⊂ N F M ⊂Av u, 1 a v, α thus that there is an so that α ( + 4 ( )) which implies the claim for

the subsequence of v + u deﬁned by M. £

Lemma 6.13. For α<ω1,(αn)⊂[0,ω1) strictly increasing, (Θn) ⊂ [0, 1] being non ` ` ∞ S F decreasing, and ( n) ⊂ Æ , with limn→∞ n = it follows that the Schreier spaces ( α),

S((Θn), (Fαn )) and S((Θn), (Fn), (`n)) (recall Deﬁnition 1.6) are hereditarily c0, assuming

that in the case of the space S((Θn), (Fαn )) the sequence (Θn) decreases to 0.

Proof. We will proof by transﬁnite induction on α<ω1that the spaces S(Fα), S((Θn), (Fαn )) and S((Θn), (Fn), (`n)), where αn % α, is hereditarily c0 (of course the claim for the second and third space is vacuous if α is not a limit ordinal). Assume the claim is true for all β<α.Ifαis a successor, say α = β +1 it is clear that the norm on S(Fβ) and S(Fβ+1) are equivalent, and, thus, the claim follows from the induction hypothesis. α S F , F , ` If is a limit ordinal then ( α) is a special case of the space ((Θn) ( αn ) ( nS)). Indeed, A α A α we let Θn = 1, for n ∈ Æ , and if n( ) is the approximating sequence we write n( )as {α } ` {` α ∈ A α } a strictly increasing sequence n : n ∈ Æ and put n = min : n `( ) . Therefore we only need to consider the norms ( P sup sup Θn |xi| or kxk = n∈N F ∈Fαn ,`n≤FP i∈F . n |xi| supn∈N supF ∈Fαn Θ i∈F

Let X be the completion of c00 under k·k, and let, without of generality, Y be the closed subspace spanned by a normalized block (yn). Either there is a further block (zn) and an n k·k z 0 ∈ Æ so that the norm is on the closed subspace spanned by ( n) equivalent to the k·k S F norm αn0 on ( αn0 ). Then our claim follows from the induction hypothesis. Or we can choose a normalized block (zn)inY and increasing sequences (kn), (Pmn)in Æ , with k z a e ∈ n

k ∈ Æ z| a e [`k,∞) = i≥k i i) kz k < −n, k ≤ k (68) n αk 2 whenever n

(69) max suppzn ≥ `n (if we assume admissibility) and 1 = kznk =Θm kznkα  n mn kz k Θk n αk  −n (70) respectively < 2 , whenever k ≥ kn+1 kz k  Θk n [k,∞) αk

(for (70) we are using either the (`n)-admissibility condition or the fact that Θn & 0, if n %∞).

Finally it is easy to see that (68), (70)and (69) imply that (zmn ) is equivalent to the c0-unit

basis. £ We are now ready for the proof of Theorem 6.7.

Proof of Theorem 6.7. We choose a seminormalized and weakly null sequence y ⊂ BX with the property that a(y, β0) = 0 and (by passing to a subsequence) assume that y satisﬁes the conclusion of Lemma 6.4. We let n o 1 1 (71) c = min , inf B(X, γ) 5 2 γ<β0

Using first Lemma 6.10 we find an increasing sequence of ordinals (αk) ⊂ [0,α0(y)) so that, 1 c/ 3k ≤ a y, α ≤ c/ 3k after passing to a subsequence, we can assume that 2 ( 2) ( k) ( 2) and that α k is the k-th successor of another ordinal, for all k ∈ Æ . Using secondly Lemma 6.11(a) we can assume, after passing again to a subsequence, that there is a transfinite family (Gα)α<ω1 so that for any (ai) ∈ c00 it follows that X X∞ X X 3k−3 16 3k (72) aiyi ≤ 8 (c/2) max |ai|≤ max c |ai| A∈G 3 ,A∈G α c k∈Æ α k=1 k i∈A k i∈A (where the second inequality is an easy application of the inequality of Holder). We now choose (infinitely or finitely many) sequences x(1), x(2) in X as follows. (1) (1) We first choose x ⊂ BX weakly null, seminormalized, so that a(x ,α1) >c.By Lemma 6.4 (using also the part which starts with “moreover” for α1) we can assume that the conclusions of Lemma 6.4 are satisfied. Now we consider the following two cases: k ∈ a x(1),α ≥ c3k Case 1: For all Æ it follows that ( k) . In that case we are done for the moment and do not continue to choose other sequences in X. (1) 3k1 Case 2: There is a smallest k1 (necessarily bigger than 1) for which a(x ,αk1)

(1) 5 3k1 (1) 3k1 (73) y ⊂ c BX with a(y ,αk )≥5c c 1 and deﬁne (x(2)(n)) = x(2) = x(1) + y(1). Using now Lemma 6.12 we ﬁnd a subsequence

x ,n(2) x(2) x ,n n∈Æ ( (2 i ))i∈Æ of =( (2 )) so that

(2) 3k1 a x ,n ,αk ≥c . (74) (( (2 i ))i∈Æ 1) Once again we can consider two cases, namely k ∈ k ≥ k a x ,n(2) ,α ≥c3k Case 1: For all Æ , 1, it follows that (( (2 i )) k) . 36 TH. SCHLUMPRECHT

(2) k k a x ,n ,α |

N (1) (2) n ⊃ n ... ≤ j<` j ∈ Æ subsequences of = Æ =( i ) ( i ) , for 1 + 1 so that for any , 1 ≤ j<`+ 1, it follows that 5 y(j) ⊂ c3kj B (75) ( ) c X satisfying the conclusions of Lemma 6.4 and (j) 3k (76) a((x(j, ni )),αk)≥c , whenever kj−1 ≤ k

(where k0 = 1 and k` = ∞ if `<∞). Moreover (76) holds for all subsequences of x(j),n(j)

( i ))i∈Æ . x x j, n(j) ` ∞ x xi, n(i)

Now we choose to be the diagonal sequence of ( i ), if = (i.e. i =( ( i ))i∈Æ ) or we choose x = x(`, n(`)i )if`<∞.

k ∈ j ∈ Æ ≤j<` k ≤ k

Using Lemma 6.11 again we can assume (by passing to subsequence) that for any (ai) ∈ c00 X X 1 3k (78) aixi ≥ max c |ai|. ,A∈G k∈Æ α 16 k i∈A

Combining (72) and (78) we showed that for any (ai) ∈ c00 it follows that 3 X X X c 3k aiyi ≤ max c |ai|≤16 aixi . ,A∈G k∈Æ α 16 k i∈A

Thus the mapping xi 7→ yi extends to a linear bounded operator T which factors through a space which is hereditarily c0 (by Lemma 6.13). On the other hand we assumed that X c does not contain 0, therefore T must be strictly singular. £ References

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