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EXTRACTA

MATHEMATICAE

Volumen 33, Número 2, 2018

instituto de investigación de matemáticas de la

universidad de extremadura

EXTRACTA MATHEMATICAE

Vol. 34, Num. 2 (2019), 135 – 200

doi:10.17398/2605-5686.34.2.135

Available online October 3, 2019

The ξ,ζ-Dunford Pettis property

R.M. Causey

Department of Mathematics, Miami University, Oxford, OH 45056, USA

causeyrm@miamioh.edu

Received May 30, 2019 Presented by Jesús M.F. Castillo
Accepted July 31, 2019 and Manuel González

Abstract: Using the hierarchy of weakly null sequences introduced in [2], we introduce two new
families of operator classes. The first family simultaneously generalizes the completely continuous

operators and the weak Banach-Saks operators. The second family generalizes the class DP. We

study the distinctness of these classes, and prove that each class is an operator ideal. We also
investigate the properties possessed by each class, such as injectivity, surjectivity, and identification

of the dual class. We produce a number of examples, including the higher ordinal Schreier and
Baernstein spaces. We prove ordinal analogues of several known results for Banach spaces with

the Dunford-Pettis, hereditary Dunford-Pettis property, and hereditary by quotients Dunford-Pettis

property. For example, we prove that for any 0 ≤ ξ,ζ < ω1, a Banach space X has the hereditary
ωξ,ωζ-Dunford Pettis property if and only if every seminormalized, weakly null sequence either has

a subsequence which is an `ω
ξ

1 -spreading model or a c
ωζ

0 -spreading model.

Key words: Completely continuous operators, Schur property, Dunford Pettis property, operator
ideals, ordinal ranks.

AMS Subject Class. (2010): Primary: 46B03, 47L20; Secondary: 46B28.

1. Introduction

In [14], Dunford and Pettis showed that any weakly compact operator de-
fined on an L1(µ) space must be completely continuous (sometimes also called
a Dunford-Pettis operator). In [17], Grothendieck showed that C(K) spaces
enjoy the same property. That is, any weakly compact operator defined on a
C(K) domain is also completely continuous. Now, we say a Banach space X
has the Dunford-Pettis property provided that for any Banach space Y and
any weakly compact operator A : X → Y , A is completely continuous. A
standard characterization of this property is as follows: X has the Dunford-
Pettis Property if for any weakly null sequences (xn)

∞
n=1 ⊂ X, (x

∗
n)
∞
n=1 ⊂ X

∗,
limn x

∗
n(xn) = 0. Generalizing this, one can study the class of operators

A : X → Y such that for any weakly null sequences (xn)∞n=1 ⊂ X and
(y∗n)

∞
n=1 ⊂ Y

∗, limn y
∗
n(Axn) = 0.

By the well-known Mazur lemma, if X is a Banach space and (xn)
∞
n=1 is

ISSN: 0213-8743 (print), 2605-5686 (online)

https://doi.org/10.17398/2605-5686.34.2.135
mailto:causeyrm@miamioh.edu
https://www.eweb.unex.es/eweb/extracta/
https://creativecommons.org/licenses/by-nc/3.0/


136 r.m. causey

a weakly null sequence in X, then (xn)
∞
n=1 admits a norm null convex block

sequence. Of course, the simplest form of convex block sequences would be
one in which all coefficients are equal to 1, in which case the convex block
sequence of (xn)

∞
n=1 is actually a subsequence. The next simplest form of a

convex block sequence is a sequence of Cesaro means. A property of signifi-
cant interest is whether the sequence (xn)

∞
n=1 has a subsequence (or whether

every subsequence of (xn)
∞
n=1 has a further subsequence) whose Cesaro means

converge to zero in norm. A weakly null sequence (xn)
∞
n=1 having the prop-

erty that for every ε > 0, there exists k = k(ε) ∈ N such that for any
x∗ ∈ BX∗, |{n ∈ N : |x∗(xn)| ≥ ε} ≤ k is called uniformly weakly null.
A weakly null sequence has the property that each of its subsequences has
a further subsequence whose Cesaro means converge to zero in norm if and
only if it has the property that each of its subsequences has a further sub-
sequence which is uniformly weakly null. Schreier [21] produced an example
of a weakly null sequence which has no uniformly weakly null subsequence.
Schreier’s example showed that the convex combinations required to witness
weak nullity in Mazur’s lemma cannot be assumed to be Cesaro means, and
must occasionally be more complex convex combinations. Providing a quan-
tification of the complexity of convex combinations required to witness weak
nullity in Mazur’s lemma, Argyros, Merkourakis, and Tsarpalilas [2] defined
the Banach-Saks index, which provides canonical coefficients which measure
the complexity a given weakly null sequence requires to obtain norm null con-
vex block sequences. As described above, norm null sequences are 0-weakly
null, uniformly weakly null sequences are 1-weakly null, and for every count-
able ordinal ξ there exists a weakly null sequence which is ξ-weakly null and
not ζ-weakly null for any ζ < ξ. By convention, we establish that a sequence is
said to be ω1-weakly null if it is weakly null. Consistent with this convention
is the fact that for any 0 ≤ ξ ≤ ζ ≤ ω1, every sequence which is ξ-weakly
null is ζ-weakly null. The ordinal quantification assigns to a given weakly null
sequence some measure of how complex the convex coefficients of a norm null
convex block sequence must be.

Our notation and terminology follows the standard reference of Pietsch
[20]. We denote classes of operators with fraktur letters, A,B,I, etc. We
recall that for a given operator ideal I, the associated space ideal is the class
of Banach spaces X such that IX ∈ I. Given an operator ideal A,B,I, . . . ,
the associated space ideal is denoted by the corresponding sans serif letter,
A,B, I, . . . . The notion of quantified weak nullity defined in the preceding
section yields a natural generalization of the class DP. Given an opera-



the ξ,ζ-dunford pettis property 137

tor A : X → Y , rather than asking that every weakly null sequence in
(xn)

∞
n=1 ⊂ X and any weakly null sequence (y

∗
n)
∞
n=1 ⊂ Y

∗, limn y
∗
n(Axn) = 0,

we may instead only require the weaker condition that every pair of sequences
(xn)

∞
n=1 ⊂ X, (y

∗
n)
∞
n=1 ⊂ Y

∗ which are “very” weakly null, limn y
∗
n(Axn) = 0.

Formally, for any 0 ≤ ξ,ζ ≤ ω1, we let Mξ,ζ denote the class of all opera-
tors A : X → Y such that for every ξ-weakly null (xn)∞n=1 ⊂ X and every
ζ-weakly null (y∗n)

∞
n=1 ⊂ Y

∗, limn y
∗
n(Axn) = 0. We let Mξ,ζ denote the class

of all Banach spaces X such that IX ∈ Mξ,ζ. Then DP = Mω1,ω1 and Mω1,ω1
is the class of all Banach spaces with the Dunford-Pettis property. Note that
every operator lies in Mξ,ζ when min{ξ,ζ} = 0, since 0-weakly null sequences
are norm null. Thus we are interested in studying the classes Mξ,η only for
0 < ξ,ζ. Furthermore, one may ask for a characterization, as one does with
the Dunford-Pettis property, of Banach spaces all of whose subspaces, or all
of whose quotients, enjoy a given property (in our case, membership in Mξ,ζ).
We note that the classes M1,ω1 were introduced and studied in [16], while
the classes Mω1,ξ, were introduced and studied in [1]. The study of classes of
operators with these weakened Dunford-Pettis conditions rather than spaces
with these conditions is new to this work. Along these lines, we have the
following results.

Theorem 1.1. For every 0 < ξ,ζ ≤ ω1, Mξ,ζ is a closed ideal which is
not injective, surjective, or symmetric. Moreover, the ideals (Mξ,ζ)0<ξ,ζ≤ω1
are distinct.

In addition to generalizations of the Dunford-Pettis property, one may use
the quantified weak nullity to generalize other classes of operators. Two classes
of interest are the classes V of completely continuous operators and wBS of
weak Banach-Saks operators. Also of interest are the associated space ideals
V of Schur spaces and wBS of weak Banach-Saks spaces. The concepts be-
hind these classes are that weakly null sequences are mapped by the operator
to sequences which are “very” weakly null (completely continuous operators
send weakly null sequences to 0-weakly null sequences, and weak Banach-Saks
operators send weakly null sequences to 1-weakly null sequences). In [12], the
notions of ξ-completely continuous operators, the class of which is denoted
by Vξ, and ξ-Schur Banach spaces were introduced. These notions are weak-
enings of the notions of completely continuous operators and Schur Banach
spaces, respectively. An operator is ξ-completely continuous if it sends ξ-
weakly null sequences to norm null (0-weakly null) sequences. Heuristically,
this is an operator which sends sequences which are “not too bad” to se-



138 r.m. causey

quences which are “good.” In [3], the notion of ξ-weak Banach-Saks was intro-
duced. An operator is ξ-weak Banach-Saks if it sends weakly null sequences to
ξ-weakly null sequences. Heuristically, this is an operator which sends any
weakly null sequence, regardless of how “bad” it is, to sequences which are
“not too bad.” Of course, there is a simultaneous generalization of both of
these notions. For 0 ≤ ζ < ξ ≤ ω1, we let Gξ,ζ denote the class of operators
which send ξ-weakly null sequences to ζ-weakly null sequences. Along these
lines, we prove the following.

Theorem 1.2. For every 0 ≤ ζ < ξ ≤ ω1, Gξ,ζ is a closed, injective ideal
which fails to be surjective or symmetric. These ideals are distinct.

We also recall the stratification (Wξ)0≤ξ≤ω1 of the weakly compact oper-
ators. Note that, by the Eberlein-Šmulian theorem, an operator A : X → Y
is weakly compact if and only if every sequence in ABX has a subsequence
which is weakly convergent. Equivalently, A : X → Y is weakly compact
if and only if for any (xn)

∞
n=1 ⊂ BX, there exist a subsequence (x

′
n)
∞
n=1 of

(xn)
∞
n=1 and y ∈ Y such that (Ax

′
n − y)∞n=1 is weakly null. The classes Wξ,

0 ≤ ξ ≤ ω1, are analogously defined using our quantified weak nullity: The
operator A : X → Y lies in Wξ if and only if for any (xn)∞n=1 ⊂ BX, there
exist a subsequence (x′n)

∞
n=1 of (xn)

∞
n=1 and y ∈ Y such that (Ax

′
n − y)∞n=1

is ξ-weakly null. The class Wξ appears in the literature under the names
ξ-weakly compact operators and ξ-Banach-Saks operators. The former name
is due to the fact that Wω1 is the class of weakly compact operators, while
the latter is due to the fact that W1 is the class of Banach-Saks operators. In
this work, we use the former terminology.

We recall the basic facts of these classes and basic facts about operator
classes, including the quotients A ◦ B−1 and B−1 ◦ A, in Section 3. We note
that W0 is the class of compact operators, also denoted by K. The class
of weakly compact operators is denoted by W and Wω1 , and W1 denotes
the class of Banach-Saks operators. It is a well-known identity regarding
completely continuous operators that V = K ◦ W−1. It is also standard that
DP = W−1◦V = W−1◦K◦W−1. Rewriting theses identities using the ordinal
notation for these classes gives

Vω1 = W0 ◦W
−1
ω1
,

Mω1,ω1 = W
−1
ω1
◦K◦W−1ω1 .

We generalize these identities in the following theorem.



the ξ,ζ-dunford pettis property 139

Theorem 1.3. For 0 ≤ ζ < ξ ≤ ω1,

Gξ,ζ = Wζ ◦W−1ξ ,

Gdualξ,ζ = (W
dual
ξ )

−1 ◦Wdualζ .

For 0 < ζ,ξ ≤ ω1,

Mξ,ζ = (W
dual
ζ )

−1 ◦Vξ = (Wdualζ )
−1 ◦K◦W−1ξ .

The appearance of Wdualξ , rather than simply Wξ as it appeared in the
identities preceding the theorem are due to the fact that

W0 = K = K
dual = Wdual0 and Wω1 = W = W

dual = Wdualω1 ,

while Wξ 6= Wdualξ for 0 < ξ < ω1. This duality is known to fail for all
0 < ξ < ω1. The failure for ξ = 1 is the classical fact that the Banach-
Saks property is not a self-dual property, while the 1 < ξ < ω1 cases are
generalizations of this.

We say Banach space X is hereditarily Mξ,ζ if for every every closed sub-
space Y of X, Y ∈ Mξ,ζ. We say X is hereditary by quotients Mξ,ζ if for every
closed subspace Y of X, X/Y ∈ Mξ,ζ. In Section 2, we define the relevant
notions regarding `

ξ
1 and c

ζ
0-spreading models. We also adopt the convention

that a sequence which is equivalent to the canonical c0 basis will be called
a cω10 -spreading model. We summarize our results regarding these hereditary
and spatial notions in the following theorem. We note that item (i) of the
following theorem generalizes a characterization of the hereditary Dunford-
Pettis property due to Elton, as well as a characterization of the hereditary
ζ-Dunford-Pettis property defined by Argyros and Gasparis.

Theorem 1.4. Fix 0 < ξ,ζ ≤ ω1.

(i) X is hereditarily Mξ,ζ if every ξ-weakly null sequence has a subsequence

which is a c
ζ
0-spreading model.

(ii) X is hereditary by quotients Mω1,ζ if and only if X
∗ is hereditarily Mζ,ω1 .

(iii) If ξ < ω1, then X is hereditarily Mγ,ζ for some ω
ξ < γ < ωξ+1 if and

only if X is hereditarily Mγ,ζ for every ω
ξ < γ < ωξ+1.

(iv) If ζ < ω1, then X is hereditarily Mξ,γ for some ω
ζ < γ < ωζ+1 if and

only if X is hereditarily Mξ,γ for every ω
ζ < γ < ωζ+1.



140 r.m. causey

We also study three space properties related to the ξ-weak Banach-Saks
property, modifying a method of Ostrovskii [19]. In [19], it was shown that the
weak Banach-Saks property is not a three-space property. Our final theorem
generalizes this. In our final theorem, wBSξ denotes the class of Banach spaces
X such that IX ∈ wBSξ.

Theorem 1.5. For 0 ≤ ζ,ξ < ω1, if X is a Banach space and Y is a closed
subspace such that Y ∈ wBSζ and X/Y ∈ wBSξ, then X ∈ wBSξ+ζ.

For every 0 ≤ ζ,ξ < ω1, there exists a Banach space X with a closed
subspace Y such that Y ∈ wBSζ, X/Y ∈ wBSξ, and for each γ < ξ + ζ, X
fails to lie in wBSγ.

2. Combinatorics

Regular families. Througout, we let 2N denote the power set of N.
We endow {0, 1}N with its product topology and endow 2N with the Cantor
topology, which is the topology making the identification 2N 3 F ↔ 1F ∈
{0, 1}N a homeomorphism. Given a subset M of N, we let [M] (resp. [M]<N)
denote set of infinite (resp. finite) subsets of M. For convenience, we often
write subsets of N as sequences, where a set E is identified with the (possibly
empty) sequence obtained by listing the members of E in strictly increasing
order. Henceforth, if we write (mi)

r
i=1 ∈ [N]

<N (resp. (mi)
∞
i=1 ∈ [N]), it will be

assumed that m1 < · · · < mr (resp. m1 < m2 < ... ). Given M = (mn)∞n=1 ∈
[N] and F ⊂ [N]<N, we define

F(M) =
{

(mn)n∈E : E ∈F
}
,

F(M−1) =
{
E : (mn)n∈E ∈F

}
.

Given (mi)
r
i=1, (ni)

r
i=1 ∈ [N]

<N, we say (ni)
r
i=1 is a spread of (mi)

r
i=1 if

mi ≤ ni for each 1 ≤ i ≤ r. We agree that ∅ is a spread of ∅. We write
E � F if either E = ∅ or E = (mi)ri=1 and F = (mi)

s
i=1 for some r ≤ s.

In this case, we say E is an initial segment of F. For E,F ⊂ N, we write
E < F to mean that either E = ∅, F = ∅, or max E < min F. Given
n ∈ N and E ⊂ N, we write n ≤ E (resp. n < E) to mean that n ≤ min E
(resp. n < min E).

We say G ⊂ [N]<N is

(i) compact if it is compact in the Cantor topology,

(ii) hereditary if E ⊂ F ∈G implies E ∈G,



the ξ,ζ-dunford pettis property 141

(iii) spreading if whenever E ∈G and F is a spread of E, F ∈G,
(iv) regular if it is compact, hereditary, and spreading.

Given a regular set G ⊂ [N]<N, we let MAX(G) denote the members of G
which are maximal in G with respect to inclusion. We note that, since G is
regular, MAX(G) coincides with the set of members of G which are maximal
in G with respect to the initial segment ordering, and also coincides with the
set of isolated points of G in the Cantor topology.

Let us also say that G is nice if

(i) G is regular,
(ii) (1) ∈G,
(iii) for any ∅ 6= E ∈G, either E ∈ MAX(G) or E ∪ (1 + max E) ∈G.

Let us briefly explain why these last two properties are desirable. We wish to
create norms on c00 of the form∥∥∥∥ ∞∑

n=1

anen

∥∥∥∥
F

= sup

{ ∑
n∈F
|an| : F ∈F

}
.

In order for this to be a norm and not just a seminorm, we require that
(1) ∈F. The last condition is because we wish to have the property that any
M ∈ [N] can be uniquely decomposed into sets F1 < F2 < ... , where each
Fn ∈ MAX(F). If F is compact and M ∈ [N], then there exists a largest
(with respect to inclusion) F which is an initial segment of M and which lies
in F, but this F need not be a maximal member of F. To see why, let

F =
{
E ⊂ N : |E| ≤ 2

}
\{(1, 2)}.

This is compact, spreading, and hereditary, but the largest initial segment
of the set M = (1, 3, 4, . . . ) which lies in F is (1), which is not a maximal
member of F.

If M ∈ [N] and if F is nice, then there exists a unique, finite, non-
empty initial segment of M which lies in MAX(F). We let MF denote this
initial segment. We now define recursively MF,1 = MF and MF,n+1 =

(
M \

∪ni=1MF,i
)
F. An alternate description of MF,1,MF,2, . . . is that the sequence

MF,1,MF,2, . . . is the unique partition of M into successive sets which are
maximal members of F.

If F is nice and M ∈ [N], then there exists a partition E1 < E2 < ... of
N such that MF,n = (mi)i∈En for all n ∈ N. We define M

−1
F,n = En.



142 r.m. causey

Given a topological space K and a subset L of K, L′ denotes the Can-
tor Bendixson derivative of L consists of those members of L which are not
relatively isolated in L. We define by transfinite induction the higher order
transfinite derivatives of L by

L0 = L, Lξ+1 = (Lξ)′,

and if ξ is a limit ordinal,

Lξ =
⋂
ζ<ξ

Lζ.

We recall that K is said to be scattered if there exists an ordinal ξ such
that Kξ = ∅. In this case, we define the Cantor Bendixson index of K
by CB(K) = min{ξ : Kξ = ∅}. We recall the standard fact that every
countable, compact, Hausdorff topological space is scattered with countable
Cantor-Bendixson index.

For each n ∈ N ∪{0}, we let An = {E ∈ [N]<N : |E| ≤ n}. It is clear
that An is regular. Also of importance are the Schreier families, (Sξ)ξ<ω1 . We
recall these families. We let

S0 = A1,

Sξ+1 = {∅}∪
{ n⋃
i=1

Ei : ∅ 6= Ei ∈Sξ, n ≤ E1,E1 < · · · < En
}
,

and if ξ < ω1 is a limit ordinal, there exists a sequence ξn ↑ ξ such that

Sξ =
{
E ∈ [N]<N : ∃n ≤ E ∈Sξn+1

}
,

and (ξn)
∞
n=1 has the property that for any n ∈ N, Sξn+1 ⊂Sξn+1 . The existence

of such families with the last indicated property is discussed, for example, in
[11]. With the fact that Sξn+1 ⊂Sξn+1 ⊂Sξn+1+1, and equivalent, useful way
of representing these sets is

Sξ = {∅}∪
{
E ∈ [N]<N : ∅ 6= E ∈Sξmin E+1

}
.

Sometimes for convenience, we simply represent

Sξ =
{
E ∈ [N]<N : ∃n ≤ E ∈Sζn

}
,

where ζn = ξn + 1. In each instance, we use the notation which is most
convenient.



the ξ,ζ-dunford pettis property 143

Given two non-empty regular families F,G, we let

F[G] = {∅}∪
{ n⋃
i=1

Ei : ∅ 6= Ei ∈G,E1 < · · · < En, (min Ei)ni=1 ∈F
}
.

We let F[G] = ∅ if either F = ∅ or G = ∅.
The following facts are collected in [11].

Proposition 2.1. (i) For any non-empty regular families F,G, F[G] is
regular. Furthermore, if CB(F) = β + 1 and CB(G) = α + 1, then
CB(F[G]) = αβ + 1.

(ii) For any n ∈ N, CB(An) = n + 1.

(iii) For any ξ < ω1, CB(Sξ) = ωξ + 1.

(iv) If F is regular and M ∈ [N], then F(M−1) is regular and CB(F) =
CB(F(M−1)).

(v) For regular families F,G, there exists M ∈ [N] such that F(M) ⊂ G if
and only if there exists M ∈ [N] such that F ⊂ G(M−1) if and only if
CB(F) ≤ CB(G).

(vi) For ξ ≤ ζ < ω1, there exists n ∈ N such that n ≤ E ∈Sξ implies E ∈Sζ.

(vii) For all 1 ≤ ξ < ω1, S1 ⊂Sξ.

Item (vi) is sometimes referred to as the almost monotone property.

Lemma 2.2. Fix a countable ordinal γ.

(i) For any L ∈ [N] and δ < ω1, there exists M ∈ [L] such that for all
(ni)

∞
i=1 ∈ [M], G ∈Sδ, and E1 < E2 < ... , ∅ 6= Ei ∈Sγ,⋃

i∈G
Eni ∈Sγ+δ.

(ii) For any L ∈ [N], there exists M ∈ [L] such that for all (ni)∞i=1 ∈ [M]
and any E ∈ Sγ+δ, there exist E1 < · · · < Ed, ∅ 6= Ei ∈ Sγ, such that
(nmin Ei)

d
i=1 ∈Sδ and E = ∪

d
i=1Ei.

Remark 2.3. Both parts of Lemma 2.2 are strengthenings of Propo-
sition 2.1.



144 r.m. causey

Proof. For both (i) and (ii), we induct on δ.
(i) For δ = 0, we can simply take M = L. Now suppose that the result

holds for δ and L ∈ [N] is fixed. By the inductive hypothesis, there exists
M ∈ [L] such that for any (ni)∞i=1 ∈ [M], E1 < E2 < ... , ∅ 6= Ei ∈ Sγ,
and G ∈ Sδ, ∪i∈EEni ∈ Sγ+δ. Now fix (ni)

∞
i=1 ∈ [M], E1 < E2 < ... ,

∅ 6= Ei ∈Sγ, and ∅ 6= G ∈Sγ+1. Let k = min G and note that we may write
G = ∪di=1Gi for some G1 < · · · < Gd, ∅ 6= Gi ∈ Sδ, nd d ≤ k. By the choice
of M, for each 1 ≤ j ≤ d, Fj := ∪i∈GjEni ∈ Sγ+δ. Since F1 < · · · < Fd and
min F1 = min Ek ≥ k ≥ d,

⋃
i∈G

Eni =
d⋃
j=1

Fj ∈Sγ+δ+1.

Now suppose that δ < ω1 is a limit ordinal. Let (δn)
∞
n=1, (βn)

∞
n=1 be the

sequences such that

Sγ+δ = {∅}∪
{
E : ∅ 6= E ∈Sβmin E

}
,

Sδ = {∅}∪
{
E : ∅ 6= E ∈Sδn

}
.

Now let us choose natural numbers p1 < p2 < ... and q1 < q2 < ... such that

γ + δn < βpn

and if qn ≤ E ∈Sγ+δn, E ∈Sβpn . By the inductive hypothesis, we may fix

M0 := L ⊃ M1 ⊃ M2 ⊃ . . . ,

Mn ∈ [N], such that for each n ∈ N, each (ni)∞i=1 ∈ [Mn], each E1 < E2 < ...
with ∅ 6= Ei ∈ Sγ, and each G ∈ Sδn, ∪i∈GEni ∈ Sγ+δn. Since each Mn
may be taken to lie in any infinite subset of Mn−1, we may also assume that
min Mn ≥ max{pn,qn} for all n ∈ N. Now write Mn = (mni )

∞
i=1 and let

mn = m
n
n. Note that m1 < m2 < ... . Let M = (mi)

∞
i=1. Fix (ni)

∞
i=1 ∈ [M],

E1 < E2 < ... with ∅ 6= E ∈ Sγ, and ∅ 6= G ∈ Sδ. Let k = min G and note
that G ∈Sδk. Let

S =
(
mk1,m

k
2, . . . ,m

k
k−1,nk,nk+1,nk+2, . . .

)
∈ [Mk].

Write S = (si)
∞
i=1 and note that since si = ni for all i ≥ k, H := ∪i∈GEni =

∪i∈GEsi. Since G ∈Sδk and S ∈ [Mk], H ∈Sγ+δk. Note that

min H ≥ nk ≥ min Mk ≥ max{pk,qk}.



the ξ,ζ-dunford pettis property 145

Since qk ≤ H ∈Sγ+δk, H ∈Sβpk . Since pk ≤ H ∈Sβpk , H ∈Sγ+δ.
(ii) Note that if M = (mi)

∞
i=1 and N = (ni)

∞
i=1 ∈ [M], then for any

∅ 6= E ∈ [N]<N, (ni)i∈E is a spread of (mi)i∈E. Thus if we reach the conclusion
when (ni)

∞
i=1 = M, this implies the result for all (ni)

∞
i=1 ∈ [M].

For δ = 0, we may simply take M = L. Suppose the result holds for δ and
fix L ∈ [N]. Choose M = (mi)∞i=1 ∈ [L] such that for any E ∈Sγ+δ, there exist
F1 < · · · < Fd such that E = ∪di=1Fi, ∅ 6= Fi ∈ Sγ, and (mmin Fi)

d
i=1 ∈ Sδ.

Now fix E ∈ Sγ+δ+1 and let k = min E. Write E = ∪lj=1Ej, E1 < · · · < El,
∅ 6= Ei ∈Sγ, and l ≤ k. We may recursively select F1 < · · · < Fn, ∅ 6= Fi ∈
Sγ and 0 = d0 < · · · < dl = n such that for each 1 ≤ i ≤ l, Ei = ∪

di
j=di−1+1

Fj

and Hi := (mmin Fj )
di
j=di−1+1

∈ Sδ. Note that min H1 ≥ min F1 = min E =
k ≥ l. Therefore E = ∪li=1Ei = ∪

n
j=1Fj and

(mmin Fj )
n
j=1 =

l⋃
i=1

(mmin Fj )
di
j=di−1+1

=

l⋃
i=1

Hi ∈Sδ+1.

Last, let δ < ω1 be a limit ordinal. Let (δn)
∞
n=1, (βn)

∞
n=1 be the sequences

such that

Sγ+δ = {∅}∪{E : ∅ 6= E ∈Sβmin E},

Sδ = {∅}∪{E : ∅ 6= E ∈Sδmin E+1},

and recall that Sδn+1 ⊂ Sδn+1 for all n ∈ N. Choose natural numbers p1 <
p2 < ... , q1 < q2 < ... such that for all n ∈ N, βn ≤ γ +δpn and qn ≤ E ∈Sβn
implies E ∈Sγ+δpn . Recursively select

M0 = L ⊃ M1 ⊃ M2 ⊃ . . .

such that min Mn ≥ max{pn,qn} and, with Mn = (mni )
∞
i=1, if E ∈ Sγ+δpn ,

there exist F1 < · · · < Fd such that ∅ 6= Fi ∈ Sγ, E = ∪di=1Ei, and
(mnmin Ei)

d
i=1 ∈ Sδrn . Let mn = m

n
n. Now fix ∅ 6= E ∈ Sγ+δ and let

k = min E. If k = 1, then E = (1), and we may write E = E1, E1 = (1) ∈Sγ,
(mmin E1 ) ∈Sδ. Assume 1 < k. Then E ∈Sβk, and E ∩ [qk,∞) ∈Sγ+δpk . Let
us choose F1 < F2 < · · · < Fd, ∅ 6= Fi ∈Sγ such that E∩[qk,∞) = ∪di=1Fi and
J := (mkmin Fi)

d
i=1 ∈ Sδpk . Since min F1 ≥ k, pk ≤ mk ≤ J ∈ Sδpk ⊂ Sδpk +1,

J ∈ Sδ. Then since H := (mmin Fi)
d
i=1 is a spread of J, H ∈ Sδpk ∩ Sδ.

If E ∩ [qk,∞) = E, this is the desired conclusion. Otherwise enumerate
E ∩ (1,qk) = (b1, . . . ,bt) and let Gi = {bi} for each 1 ≤ i ≤ t. Note



146 r.m. causey

that G1 < · · · < Gt < F1 < · · · < Fd, E =
(
∪ti=1Gi

)
∪
(
∪di=1Fi

)
, and

∅ 6= Gi,Fi ∈ Sγ. Let G = (mmin Gi)
t
i=1 and note that mk ≤ G and

|G| ≤ qk ≤ mk, so G ∈ S1 ⊂ Sδpk . Since 2 ≤ G < H and G,H ∈ Sδpk ,
G∪H ∈Sδpk +1. Since pk ≤ mk ≤ G,

(mmin Gi)
t
i=1 ∪ (mmin Fi)

d
i=1 = G∪H ∈Sδ.

`
ξ
1 and c

ξ
0-spreading models. Given a regular family F, a Banach

space X, and a seminormalized sequence (xn)
∞
n=1 ⊂ X, we say (xn)

∞
n=1 is an

`F1 -spreading model provided that

0 < inf
{
‖x‖ : F ∈F,x ∈ aco(xn : n ∈ F)

}
.

Here,

aco(xn : n ∈ F) =
{ ∑
n∈F

anxn :
∑
n∈F
|an| = 1

}
.

We say that a sequence (xn)
∞
n=1 is a c

F
0 -spreading model provided that

0 < inf

{∥∥∥∥ ∑
n∈F

εnxn

∥∥∥∥ : F ∈F, maxn∈F |εn| = 1
}

≤ sup
{∥∥∥∥ ∑

n∈F
εnxn

∥∥∥∥ : F ∈F, maxn∈F |εn| = 1
}
< ∞.

If F = Sξ, we write `
ξ
1 or c

ξ
0-spreading model in place of `

Sξ
1 or c

Sξ
0 . Note that

a weakly null `01 or c
0
0-spreading model is simply a seminormalized, weakly

null sequence.

Note that for a regular family F, the spreading property of F yields that
for any k1 < k2 < ... ,

inf
{
‖x‖ : F ∈F, x ∈aco(xkn : n ∈ F)

}
≥ inf

{
‖x‖ : F ∈F, x ∈ aco(xn : n ∈ F)

}
,

so that any subsequence of an `F1 -spreading model is also an `
F
1 -spreading

model. Similarly, every subsequence of a cF0 -spreading model is also a c
F
0 -

spreading model.

We are now ready to define the notions of ξ-weak nullity.



the ξ,ζ-dunford pettis property 147

Definition 2.4. For ξ < ω1, we say a sequence (xn)
∞
n=1 is ξ-weakly null

if for any subsequence (yn)
∞
n=1 of (xn)

∞
n=1 and ε > 0, there exist F ∈ Sξ and

y ∈ co(yn : n ∈ F) such that ‖y‖ < ε. We say (xn)∞n=1 is ξ-weakly convergent
to x if (xn −x)∞n=1 is ξ-weakly null. We say (xn)

∞
n=1 is ξ-weakly convergent

if it is ξ-weakly convergent to some x.

We say a sequence is ω1-weakly null, ω1-weakly convergent to x, or
ω1-weakly convergent if it is weakly null, weakly convergent to x, or weakly
convergent, respectively.

Remark 2.5. Note that if (xn)
∞
n=1 is a ξ-weakly null sequence in the Ba-

nach space X, then there exist sets F1 < F2 < ... , Fn ∈ Sξ, and positive
scalars (ai)i∈∪∞n=1Fn such that for each n ∈ N,

∑
i∈Fn ai = 1, and such that

limn‖
∑

i∈Fn aixi‖ = 0. We will use this fact often. However, we will also
often need a technical fact which states that the coefficients (ai)i∈Fn can come
from the repeated averages hierarchy. We make this precise below.

Remark 2.6. It follows from Theorem C of [2] that a weakly null sequence
fails to be ξ-weakly null if and only if it has a subsequence which is an
`
ξ
1-spreading model. From this it follows that if (xn)

∞
n=1 is a weakly null

`
ξ
1-spreading model, it can have no ξ-weakly convergent subsequence. Indeed,

since ξ-weak convergence to x implies weak convergence to x, the only x to
which a subsequence of (xn)

∞
n=1 could be ξ-convergent is x = 0. But if (xn)

∞
n=1

is an `
ξ
1-spreading model, all of its subsequences are, and so no subsequence

can be ξ-convergent to zero by the first sentence of the remark.

Let P denote the set of all probability measures on N. We treat each
member P of P as a function from N into [0, 1], where P(n) = P({n}). We
let supp(P) = {n ∈ N : P(n) > 0}. Given a nice family P and a subset
P = {PM,n : M ∈ [N],n ∈ N} of P, we say (P,P) is a probability block
provided that

(i) for each M ∈ [N], supp(PM,1) = MP,1, and

(ii) for any M ∈ [N] and r ∈ N, if N = M \ ∪r−1i=1 supp(PM,i), then
PN,1 = PM,r.

Remark 2.7. It follows from the definition of probability block that for any
M ∈ [N], (MP,n)∞n=1 = (supp(PM,n))

∞
n=1 and for any s ∈ N and M,N ∈ N,

and r1 < · · · < rs such that ∪si=1 supp(PM,ri) is an initial segment of N, then
PN,i = PM,ri for all 1 ≤ i ≤ s. This was proved in [12].



148 r.m. causey

Suppose that Q is nice. Given L = (ln)∞n=1 ∈ [N], there exists a unique
sequence 0 = p0 < p1 < ... such that (li)

pn
i=pn−1+1

∈ MAX(Q) for all n ∈ N.
We then define L−1Q,n = N∩ (pn−1,pn].

Suppose we have probability blocks (P,P), (Q,Q). We define a collection
Q∗P such that (Q∗P,Q[P]) is a probability block. Fix M ∈ N and for each
n ∈ N, let ln = min supp(PM,n) and L = (ln)∞n=1. We then let

OM,n =
∑

i∈L−1Q,n

QL,n(li)PM,i

and Q∗P = {OM,n : M ∈ [N],n ∈ N}.
In [2], the repeated averages hierarchy was defined. This is a collection

Sξ, ξ < ω1, such that (Sξ,Sξ) is a probability block for every ξ < ω1. We
will denote the members of Sξ by S

ξ
M,n, M ∈ [N], n ∈ N.

For ξ < ω1, we say a probability block (P,P) is ξ-sufficient provided that
for any L ∈ [N], any ε > 0, and any regular family G with CB(G) ≤ ωξ, there
exists M ∈ [N] such that

sup
{
PN,1(E) : E ∈G,N ∈ [M]

}
< ε.

It was shown in [2] that (Sξ,Sξ) is ξ-sufficient.
The following facts were shown in [12]. Item (ii) was shown in [2] in the

particular case that (P,P) = (Sξ,Sξ).

Theorem 2.8. (i) For ξ,ζ < ω1, if (P,P) is ξ-sufficient and (Q,Q) is
ζ-sufficient, then (Q∗P,Q[P]) is (ξ + ζ)-sufficient.

(ii) If X is a Banach space, ξ < ω1, (P,P) is ξ-sufficient, and CB(P) =
ωξ + 1, then a weakly null sequence (xn)

∞
n=1 ⊂ X is ξ-weakly null if and

only if for any L ∈ [N] and ε > 0, there exists M ∈ [L] such that for all
N ∈ [M], ‖

∑∞
i=1 PN,1(i)xi‖ < ε.

Remark 2.9. Since for each ξ < ω1, at least one ξ-sufficient probability
block (P,P) with CB(P) = ωξ + 1 exists, item (ii) of the preceding theorem
yields that if X is a Banach space and (xn)

∞
n=1, (yn)

∞
n=1 are ξ-weakly null

in X, then (xn + yn)
∞
n=1 is also ξ-weakly null. This generalizes to sums of

any number of sequences. The importance of this fact, which we will use
often throughout, is that if for k = 1, . . . , l, if (xkn)

∞
n=1 ⊂ X is a ξ-weakly null

sequence, then for any ε > 0, there exist F ∈ Sξ and positive scalars (ai)i∈F



the ξ,ζ-dunford pettis property 149

such that
∑

i∈F ai = 1 and for each 1 ≤ k ≤ l,∥∥∥∥∑
i∈F

aix
k
i

∥∥∥∥ ≤ ε.
That is, there is one choice of F and (ai)i∈F such that the corresponding linear
combinations of the l different sequences are simultaneously small.

Note that the preceding implies that for two Banach spaces X,Y and ξ-
weakly null sequences (xn)

∞
n=1 ⊂ X, (yn)

∞
n=1 ⊂ Y , for any ε > 0, there exist

F ∈Sξ and positive scalars (ai)i∈F summing to 1 such that∥∥∥∥∑
i∈F

aixi

∥∥∥∥
X

,

∥∥∥∥∑
i∈F

aiyi

∥∥∥∥
Y

< ε.

This is because the sequences (xn, 0)
∞
n=1 ⊂ X⊕∞Y and (0,yn)

∞
n=1 ⊂ X⊕∞Y

are also ξ-weakly null, as is their sum in X ⊕∞ Y .

Remark 2.10. Let X be a Banach space and let (xn)
∞
n=1 be ξ-weakly null.

Let (P,P) be ξ-sufficient with CB(P) = ωξ + 1. Then by Theorem 2.8(ii), we
may recursively select M1 ⊃ M2 ⊃ . . . , Mn ∈ [N] such that for each n ∈ N,

sup

{∥∥∥∥ ∞∑
i=1

PN,1(i)xi
∥∥∥∥ : N ∈ [Mn]

}
< 1/n.

Now choose mn ∈ Mn with m1 < m2 < ... and let M = (mn)∞n=1. Then for
any N ∈ [M] and n ∈ N, if F1 < F2 < ... is a partition of N into consecutive,
maximal members of P and Nj = N \∪

j−1
i=1 Fi for each j ∈ N, Nn ∈ [Mn]. By

the permanence property mentioned in Remark 2.7,∥∥∥∥ ∞∑
i=1

PN,n(i)xi
∥∥∥∥ =

∥∥∥∥ ∞∑
i=1

PNn,1(i)xi
∥∥∥∥ < 1/n.

Before proceeding to the following, we recall that for M ∈ [N] and a regular
family F, we let M|F denote the maximal initial segment of M which lies in
F. If F is nice, then M|F lies in MAX(F).

Lemma 2.11. Let X be a Banach space, (xn)
∞
n=1 ⊂ X a seminormalized,

weakly null sequence, and F a nice family.



150 r.m. causey

(i) (xn)
∞
n=1 admits a subsequence which is a c

F
0 -spreading model if and only

if there exists L ∈ [N] such that

sup

{∥∥∥∥ ∑
n∈M|F

xn

∥∥∥∥ : M ∈ [L]
}
< ∞.

(ii) If (xn)
∞
n=1 admits no subsequence which is a c

F
0 -spreading model, then

there exists L ∈ [N] such that for any H1 < H2 < ... , Hn ∈ MAX(F) ∩
[L]<N, ‖

∑
i∈Hn xi‖ > n for each n ∈ N.

Proof. (i) Assume there exists L ∈ [N] such that

sup

{∥∥∥∥ ∑
n∈M|F

xn

∥∥∥∥ : M ∈ [L]
}

= C < ∞.

By passing to an infinite subset of L, we may assume (xn)n∈L is 2-basic. If
F ∈F∩ [L]<N, there exists an infinite subset M of L such that F is an initial
segment of M|F, from which it follows that∥∥∥∥ ∑

n∈F
xn

∥∥∥∥ ≤ 2
∥∥∥∥ ∑
n∈M|F

xn

∥∥∥∥ ≤ 2C.
Thus

sup

{∥∥∥∥ ∑
n∈F

xn

∥∥∥∥ : F ∈F ∩ [L]<N
}
≤ 2C.

Then if ∅ 6= F ∈F ∩ [L]<N, (an)n∈F ∈ [−1, 1]F ,

∑
n∈F

anxn ∈ co
( ∑
n∈G

xn : G ⊂ F
)
− co

( ∑
n∈G

xn : G ⊂ F
)
⊂ 4CBX.

Now for any ∅ 6= F ∈F ∩ [L]<N and for any scalars (an)n∈F with |an| ≤ 1,∥∥∥∥ ∑
n∈F

anxn

∥∥∥∥ ≤
∥∥∥∥ ∑
n∈F

Re (an)xn

∥∥∥∥ +
∥∥∥∥ ∑
n∈F

Im (an)xn

∥∥∥∥ ≤ 8C.
If L = (ln)

∞
n=1, this yields the appropriate upper estimates to deduce that

(xln)
∞
n=1 is a c

F
0 -spreading model. The lower estimates follow from the fact

that (xln)
∞
n=1 is seminormalized basic.



the ξ,ζ-dunford pettis property 151

For the converse, suppose that (xrn)
∞
n=1 is a c

F
0 -spreading model and let

c = sup

{∥∥∥∥ ∑
n∈F

xrn

∥∥∥∥ : F ∈F
}
< ∞.

Let us choose 1 = s1 < s2 < ... such that sn+1 > rsn for all n ∈ N. Let ln =
rsn, L = (ln)

∞
n=1, and S = (sn)

∞
n=1. Fix M ∈ [L] and note that M = (rtn)

∞
n=1

for some (tn)
∞
n=1 ∈ [S]. Let M|F = (rtn)

k
n=1 ∈ F and note that (tn)

k
n=2 ∈ F.

Indeed, if tn−1 = si and tn = sj, then i < j and then

tn = sj ≥ si+1 > rsi = rtn−1.

Thus E := (tn)
k
n=2 is a spread of (rtn)

k−1
n=1 ⊂ (rtn)

k
n=1 ∈ F, and E ∈ F.

Therefore, with b = supn‖xn‖,∥∥∥∥ ∑
n∈M|F

xn

∥∥∥∥ ≤
∥∥∥∥xrt1

∥∥∥∥ +
∥∥∥∥ k∑
n=2

xrtn

∥∥∥∥ =
∥∥∥∥xrt1

∥∥∥∥ +
∥∥∥∥ ∑
n∈E

xrtn

∥∥∥∥ ≤ b + c =: C .
Therefore we have shown that

sup

{∥∥∥∥ ∑
n∈M|F

xn

∥∥∥∥ : M ∈ [L]
}
≤ C.

(ii) For each n ∈ N, let

Vn =
{
M ∈ [N] :

∥∥∥∥ ∑
i∈M|F

xi

∥∥∥∥ ≤ n‖
}
.

It is evident that Vn is closed, and in fact M 7→ ‖
∑

i∈M|F xi‖ is locally
constant on [N]. By the Ramsey theorem, we may select M1 ⊃ M2 ⊃ . . .
such that for all n ∈ N, either [Mn] ⊂ Vn or Vn ∩ [Mn] = ∅. By (i) and
the hypothesis that (xn)

∞
n=1 admits no subsequence which is a c

F
0 -spreading

model, for each n ∈ N, Vn ∩ [Mn] = ∅. Now fix l1 < l2 < ... , ln ∈ Mn,
and let L = (ln)

∞
n=1. Fix ∅ 6= H1 < H2 < ... , Hn ∈ MAX(F) ∩ [L]

<N.
For each n ∈ N, let Nn = ∪∞i=nHi ∈ [Mn] and note that Hn = Nn|F. Since
Nn ∈ [Mn] ⊂ [N] \Vn, ∥∥∥∥ ∑

i∈Hn

xi

∥∥∥∥ =
∥∥∥∥ ∑
i∈Nn|F

xi

∥∥∥∥ > n.



152 r.m. causey

For ordinals ξ,ζ < ω1 and any M ∈ [N], there exists N ∈ [M] such that
Sξ[Sζ](N) ⊂ Sζ+ξ and Sζ+ξ(N) ⊂ Sξ[Sζ] ([18, Proposition 3.2]). From this
it follows that for a given sequence (xn)

∞
n=1 in a Banach space X, there exist

m1 < m2 < ... such that

0 < inf
{
‖x‖ : F ∈Sζ+ξ,x ∈ aco(xmn : n ∈ F)

}
if and only if there exist m1 < m2 < ... such that

0 < inf
{
‖x‖ : F ∈Sξ[Sζ],x ∈ aco(xmn : n ∈ F)

}
.

This fact will be used throughout to deduce that if (xn)
∞
n=1 is an `

ζ+ξ
1 -spreading

model (or has a subsequence which is an `
ζ+ξ
1 -spreading model), then there

exists a subsequence of (xn)
∞
n=1 which is an `

Sξ[Sζ]
1 -spreading model. Similarly,

if (xn)
∞
n=1 has a subsequence which is a c

ζ+ξ
0 -spreading model, then it has a

subsequence which is a c
Sξ[Sζ]
0 -spreading model.

Corollary 2.12. Fix α,β,γ < ω1. Let X,Y be Banach spaces,
A : X → Y an operator, and let (xn)∞n=1 be a seminormalized, weakly null
sequence in X.

(i) If (Axn)
∞
n=1 has a subsequence which is an `

α+β
1 -spreading model and

(xn)
∞
n=1 has no subsequence which is an `

α+γ
1 -spreading model, then

there exists a convex block sequence (zn)
∞
n=1 of (xn)

∞
n=1 which has no

subsequence which is an `
γ
1 -spreading model and such that (Azn)

∞
n=1 is

an `
β
1 -spreading model.

(ii) If (xn)
∞
n=1 has a subsequence which is a c

α+β
0 -spreading model but no

subsequence which is a c
α+γ
0 -spreading model, then there exists a block

sequence of (xn)
∞
n=1 which is a c

β
0 -spreading model and has no subse-

quence which is a c
γ
0 -spreading model. If 0 < β, the block sequence is

also weakly null.

Proof. (i) We first assume supn‖xn‖ = 1. By passing to a subsequence,
we may assume without loss of generality that

0 < ε = inf
{
‖Ax‖ : F ∈Sβ[Sα],x ∈ abs co(xn : n ∈ F)

}
.

Let P = Sγ[Sα], P = Sγ ∗ Sα = {PM,n : M ∈ [N],n ∈ N}. As mentioned in
Remark 2.10, we may also fix L ∈ [N] such that for all M ∈ [L] and n ∈ N,∥∥∥∥ ∞∑

i=1

PM,n(i)xi
∥∥∥∥ ≤ 1/n.



the ξ,ζ-dunford pettis property 153

Now fix F1 < F2 < ... , Fn ∈ MAX(Sα), L = ∪∞n=1Fn and let yn =∑∞
i=1 S

α
L,n(i)xi =

∑
i∈Fn S

α
L,n(i)xi. It follows from the second sentence of

the proof that

ε ≤ inf
{
‖Ay‖ : F ∈Sβ,y ∈ abs co(yn : n ∈ F)

}
.

That is, (Ayn)
∞
n=1 is an `

β
1 -spreading model. It remains to show that (yn)

∞
n=1

has no subsequence which is an `
γ
1 -spreading model. To that end, assume

R = (rn)
∞
n=1, δ > 0 are such that

δ ≤ inf
{∥∥∥∥ ∑

n∈F
anyrn

∥∥∥∥ : F ∈Sγ, ∑
n∈F
|an| = 1

}
.

Now let En = Frn, N = ∪∞n=1En, S = (sn)
∞
n=1 = (min En)

∞
n=1 and note that,

by the permanence property,

zn := yrn =
∞∑
i=1

SαN,n(i)xi

for all n ∈ N. Now fix 1 = q1 < q2 < ... such that qn+1 > sqn. Let
M = ∪∞n=1Eqn and note that there exist 0 = k0 < k1 < ... such that for all
n ∈ N,

PM,n =
kn∑

j=kn−1+1

SγT,n(sqj )S
α
M,j

and (sqj )
kn
j=kn−1+1

∈ Sγ, where T = (sqj )
∞
j=1. Moreover S

γ
T,n(sqkn−1+1 ) → 0

since 0 < γ. We now observe that since sqj < qj+1, Gn := (qj)
kn
j=kn−1+2

is a

spread of (qj)
kn−1
j=kn−1+1

, which is a subset of a member of Sγ. Therefore, for
any n ∈ N,

δ
(

1 −SγT,n(sqkn−1+1 )
)
≤
∥∥∥∥ kn∑
j=kn−1+2

SγT,n(sqj )zj
∥∥∥∥

≤
∥∥∥∥ kn∑
j=kn−1+1

SγT,n(sqj )zj
∥∥∥∥ + SγT,n(sqkn−1+1 )

≤
∥∥∥∥ ∞∑
i=1

PM,n(i)xi
∥∥∥∥ + SγT,n(sqkn−1+1 )

≤ 1/n + SγT,n(sqkn−1+1 ).



154 r.m. causey

Since limn S
γ
T,n(sqkn−1+1 ) = 0, these inequalities yield a contradiction for suf-

ficiently large n.
(ii) We may assume without loss of generality that

sup

{∥∥∥∥ ∑
n∈F

εnxn

∥∥∥∥ : F ∈Sβ[Sα], |εn| = 1
}

= C < ∞

and that (xn)
∞
n=1 is basic. By Lemma 2.11 applied with F = Sγ[Sα], there

exists L ∈ [N] such that for all H1 < H2 < ... , Hn ∈ MAX(Sγ[Sα]) ∩ [L]<N,
‖
∑

i∈Hn xi‖ > n. We claim that for any F1 < F2 < ... , Fn ∈ MAX(Sα) ∩
[L]<N, (

∑
i∈Fn xi)

∞
n=1 fails to have a subsequence which is a c

γ
0 -spreading

model. In order to prove this, it is sufficient to prove that (
∑

i∈Fn xi)
∞
n=1 is not

a c
γ
0 -spreading model. To see this, simply observe that if F1 < F2 < ... , Fn ∈

MAX(Sα) ∩ [L]<N and (
∑

i∈Frn
xi)
∞
n=1 is a c

γ
0 -spreading model, this contra-

dicts the previous sentence, since Fr1 < Fr2 < ... also lie in MAX(Sα)∩[L]<N.
Seeking a contradiction, suppose that

sup

{∥∥∥∥ ∑
n∈E

∑
i∈Fn

xi

∥∥∥∥ : E ∈Sγ
}

= D < ∞.

Now fix 1 = s1 < s2 < ... such that for all n ∈ N, sn+1 > min Fsn. Let
T = ∪∞n=1Fsn and let H1 < H2 < ... be such that Hn ∈ MAX(Sγ[Sα]) and
T = ∪∞n=1Hn. Note that ‖

∑
i∈Hn xi‖ > n for all n ∈ N. Note also that there

exist 0 = k0 < k1 < ... such that Hn = ∪knj=kn−1+1Fsj , and these numbers are
uniquely determined by the property that (min Fsj )

kn
j=kn−1+1

∈ MAX(Sγ). As
is now familiar, we note that for each n ∈ N, En := (sj)knkn−1+2 is a spread of
a subset (min Fsj )

kn−1
j=kn−1+1

, so that En ∈Sγ. We note that for each n ∈ N,

n <

∥∥∥∥ ∑
i∈Hn

xi

∥∥∥∥ ≤
∥∥∥∥ ∑
i∈Fkn−1+1

xi

∥∥∥∥ +
∥∥∥∥ kn∑
j=kn−1+2

∑
i∈Fsj

xi

∥∥∥∥
≤ C +

∥∥∥∥ ∑
j∈En

xi

∥∥∥∥ ≤ C + D.
This is a contradiction for sufficiently large n.

Schreier and Baernstein spaces. If F is a nice family, we let XF
denote the completion of c00 with respect to the norm

‖x‖F = sup
{
‖Ex‖`1 : E ∈F

}
.



the ξ,ζ-dunford pettis property 155

In the case that F = Sξ, we write ‖·‖ξ in place of ‖·‖Sξ and Xξ in place of XSξ.
The spaces Xξ are called Schreier spaces. Note that X0 = c0 isometrically.

Given 1 < p < ∞ and a nice family F, we let XF,p be the completion of
c00 with respect to the norm

‖x‖F,p = sup
{( ∞∑

i=1

‖Eix‖
p
`1

)1/p
: E1 < E2 < ... , Ei ∈F

}
.

For convenience, we let Xξ,p and ‖·‖ξ,p denote XSξ,p and ‖·‖Sξ,p, respectively.
The spaces Xξ,p are called Baernstein spaces. For convenience, we let Xξ,∞
denote Xξ.

Remark 2.13. The Schreier families Sξ, ξ < ω1, possess the almost mono-
tone property, which means that for any ζ < ξ < ω1, there exists m ∈ N such
that if m ≤ E ∈Sζ, then E ∈Sξ. From this it follows that the formal inclu-
sion I : Xξ → Xζ is bounded for any ζ ≤ ξ < ω1. In fact, there exists a tail
subspace [ei : i ≥ m] of Xξ such that the restriction of I : [ei : i ≥ m] → Xζ
is norm 1. We will use this fact throughout.

It is also obvious that the formal inclusion from Xξ,p to Xζ,p is bounded
for any ζ ≤ ξ < ω1, as is the inclusion from Xξ,p to Xξ,q whenever p < q ≤∞.
Combining these facts yields that the formal inclusion from Xξ,p to Xζ is
bounded whenever ζ ≤ ξ. Furthermore, the adjoints of all of these maps are
also bounded.

The following collects known facts about the Schreier and Baernstein
spaces. Throughout, we let ‖ · ‖ξ,p denote the norm of Xξ,p as well as its
first and second duals.

Theorem 2.14. Fix ξ < ω1 and 1 < p ≤∞.

(i) ‖
∑n

i=1 xi‖ξ,p = ‖
∑n

i=1 |xi|‖ξ,p for any disjointly supported x1, . . . ,xn
∈ Xξ,p.

(ii) The canonical basis of Xξ,p is shrinking.

(iii) The basis of Xξ,p is boundedly-complete (and Xξ,p is reflexive) if and
only if p < ∞.

(iv) If p < ∞ and 1/p + 1/q = 1,∥∥∥∥ n∑
i=1

xi

∥∥∥∥
ξ,p

≥
( n∑
i=1

‖xi‖
p
ξ,p

)1/p
and

∥∥∥∥ n∑
i=1

x∗i

∥∥∥∥
ξ,p

≤
( n∑
i=1

‖x∗i‖
q
ξ,p

)1/q
for any x1 < · · · < xn ∈ Xξ,p and x∗1 < · · · < x

∗
n, x

∗
i ∈ X

∗
ξ,p.



156 r.m. causey

(v) The canonical basis of Xξ,p is a weakly null `
ξ
1-spreading model, while

every normalized, weakly null sequence in Xξ,p is ξ + 1-weakly null.

(vi) The space Xξ is isomorphically embeddable into C(Sξ).

Remark 2.15. Throughout, if E ∈ [N]<N, we will use the notation x∗ @ E
to mean that ‖x∗‖c0 ≤ 1 and supp(x∗) = E. It is evident that for any regular
family F, ⋃

E∈F
{x∗ : x∗ @ E}⊂ BX∗F .

Moreover, a convexity argument yields that for any y∗ ∈ BX∗F with supp(y
∗) ⊂

F ∈ [N]<N,

y∗ ∈ co
( ⋃
F⊃E∈F

{x∗ : x∗ @ E}
)
.

Finally, we note that if there exist x∗1 < · · · < x
∗
d and for each 1 ≤ i ≤

d, there exist li ∈ N, Ei,j ⊂ supp(x∗i ), and x
∗
i,j, j = 1, . . . , li, such that

x∗i,j @ Ei,j ⊂ supp(x
∗
i ), x

∗
i ∈ co(x

∗
i,j : 1 ≤ j ≤ li), and for each (ji)

d
i=1 ∈∏d

i=1{1, . . . , li}, ∪
d
i=1Ei,ji ∈F, then∥∥∥∥ d∑

i=1

x∗i

∥∥∥∥
X∗F

≤ 1.

Moreover, if we replace x∗i with aix
∗
i , where a1, . . . ,ad are such that |ai| ≤ 1

for each 1 ≤ i ≤ d, the resulting functionals a1x∗1, . . . ,adx
∗
d also satisfy the

hypotheses, so ‖
∑d

i=1 aix
∗
i‖X∗F ≤ 1 for any (ai)

d
i=1 ∈ `

d
∞.

Let us see why ‖
∑d

i=1 x
∗
i‖X∗F ≤ 1. Write x

∗
i =

∑li
j=1 wi,jx

∗
i,j where wi,j ≥ 0

and
∑li

j=1 wi,j = 1. Let I =
∏d
i=1{1, . . . , li} and for each t = (ji)

d
i=1 ∈ I, let

wt =
∏d
i=1 wi,ji and x

∗
t =

∑d
i=1 x

∗
i,ji

. Then x∗ =
∑

t∈I wtx
∗
t , wt ≥ 0, and∑

t∈I wt = 1. Therefore it suffices to show that ‖x
∗
t‖X∗F ≤ 1 for each t ∈ I.

But x∗t @ ∪di=1Ei,ji ∈F, and ‖x
∗
t‖X∗F ≤ 1 follows.

Proposition 2.16. Fix 0 ≤ γ,δ < ω1, and 1 < p ≤∞.

(i) If (x∗n)
∞
n=1 ⊂ X

∗
γ is weakly null and satisfies lim inf

n
‖x∗n‖∗γ < C, then

there exists a subsequence (x∗ni)
∞
i=1 of (x

∗
n)
∞
n=1 such that for any G ∈Sδ,

‖
∑

i∈G x
∗
ni
‖∗γ+δ < C.



the ξ,ζ-dunford pettis property 157

(ii) Suppose (xn)
∞
n=1 ⊂ Xγ+δ,p is weakly null in Xγ+δ,p, and for every β <

γ, limn‖xn‖β = 0. Then every subsequence of (xn)∞n=1 has a further
subsequence which is dominated by a subsequence of the Xδ,p basis.

(iii) If (xn)
∞
n=1 ⊂ Xγ+δ,p is a weakly null sequence such that lim supn‖xn‖γ >

0, then (xn)
∞
n=1 has a subsequence which dominates the Xδ,p basis.

Proof. (i) By passing to a subsequence, we may assume that (x∗n)
∞
n=1 is a

block sequence and supn‖x∗n‖ < C1 < C. By scaling, we may assume C1 = 1.
For each n ∈ N, let Sn = supp(x∗n). For each n ∈ N, it follows from convexity
and compactness arguments that for each n ∈ N, there exist dn, (x∗n,i)

dn
i=1, and

(En,i)
dn
i=1 ⊂Sγ ∩ [Sn]

<N such that x∗n,i @ En,i, and x
∗
n ∈ co(x∗i,n : 1 ≤ i ≤ dn).

By Lemma 2.2, there exist n1 < n2 < ... such that for any G ∈ Sδ and
E1 < E2 < ... , Ei ∈ Sγ, ∪i∈GEni ∈ Sγ+δ. Now we conclude that for each
G ∈Sδ, ‖

∑
n∈G x

∗
n‖γ+δ ≤ C1 = 1 using the facts contained in Remark 2.15.

(ii) By perturbing and scaling, we may assume (xn)
∞
n=1 ⊂ BXξ,p is a block

sequence. If γ is a successor, let γn + 1 = γ for all n ∈ N if γ is a successor.
If γ is a limit ordinal, let (γn)

∞
n=1 be such that

Sγ = {E : ∃n ≤ E ∈Sγn}.

For each n ∈ N, let εn = 2−n−2. Let mn = max supp(xn). We may recursively
choose 1 = k1 < k2 < ... such that for any n < l,

‖xkl‖γkn < εn/mkn.

By relabeling, we may assume kn = n.

Now by Lemma 2.2, we may fix (ni)
∞
i=1 such that if E ∈Sγ+δ, there exist

E1 < · · · < Ed, ∅ 6= Ei ∈ Sγ such that (nmin E)di=1 ∈ Sδ and E = ∪
d
i=1Ei.

Now let ri = nmi. We first consider the p = ∞ case. We claim that (xi)
∞
i=1 is

dominated by (eri)
∞
i=1 ⊂ Xδ. Fix (ai)

∞
i=1 ∈ c00∩S`∞ and let x =

∑∞
i=1 aixi and

y =
∑∞

i=1 aieri. Fix E ∈ Sγ+δ and write E = ∪
d
i=1Ei, where E1 < · · · < Ed,

∅ 6= Ei ∈ Sγ, and (nmin Ei)
d
i=1 ∈ Sδ. If γ = 0, we can take each Ei to be

a singleton. By omitting any superfluous Ei and relabeling, we may assume
that for each 1 ≤ i ≤ d, there exists j such that Eixj 6= 0.

As the following estimates involve many definitions, we say a word before
proceeding. For each Ei, our choice of the sequence (xi)

∞
i=1 will yield that

‖Eixl‖`1 will be essentially negligible for all vectors except the first one whose
support Ei intersects. Moreover, of all of the sets Ei which intersect the
support of xl, since the sets are successive, at most one of the sets can intersect



158 r.m. causey

the support of a later vector, so we can control the number of negligible pieces.
For each 1 ≤ i ≤ d, let ji = min{l : Eixl 6= 0} and J = {ji : 1 ≤ i ≤ d}. For
each j ∈ J, let Sj = {i ≤ d : ji = j}. For each j ∈ J, let sj = max Sj and
let Tj = Sj \ {sj}. Note that for each i ∈ Sj, Eixl = 0 for all l < j by the
minimality of j = ji. Note also that for each i ∈ Tj, Eixl = 0 for all l > j,
since

max Ei < min Esj ≤ max supp(xj) < min supp(xl).

Furthermore, since Esjxj 6= 0, Esj ∈ Sγ with min Esj ≤ mj. If γ is a limit
ordinal, then Esj ∈Sγmj , which means that for any k > j,

‖Esjxk‖`1 ≤ εk/mj ≤ εk.

If γ is a successor, then γ = γmj +1 and min Esj ≤ mj yield that Esj = ∪
q
i=1Fi

for some F1 < · · · < Fq, q ≤ mj, and Fi ∈Sγmj . Then for k > j,

‖Esjxk‖`1 ≤
q∑
i=1

‖Fixk‖`1 ≤ mn‖xk‖γmj ≤ εk.

In the case γ = 0, each Ei is a singleton, so we have the trivial estimate that
for i ∈ Sj and l > j, Eixl = 0. Therefore in each of the γ = 0, γ a successor,
and γ a limit ordinal cases,

∑
i∈Sj

‖Eix‖`1 ≤ |aj|‖Exj‖`1 +
∞∑

k=j+1

‖Esjxk‖`1 ≤ |aj| +
∞∑

k=j+1

εk.

Summing over i yields that

‖Ex‖`1 ≤
∑
j∈J

∑
i∈Sj

‖Eix‖`1 ≤
∑
j∈J
|aj| +

∑
j∈J

∞∑
k=j+1

εk ≤
∑
j∈J
|aj| +

∞∑
j=m(E)

∞∑
k=j

εk,

where m(E) = min{j : Exj 6= 0}. Now for each j ∈ J, fix some ij ∈
{1, . . . ,d} such that j = jij . Then j 7→ ij is an injection of J into {1, . . . ,d},
and (mij )j∈J is a spread of a subset of (min Ei)

d
i=1. Therefore T(E) :=

(rij )j∈J = (nmij )j∈J is a spread of a subset of (nmin Ei)
d
i=1 ∈Sδ, so T(E) ∈Sδ.

Therefore

‖y‖δ ≥‖T(E)y‖`1 =
∑
j∈J
|aj|.



the ξ,ζ-dunford pettis property 159

Collecting these estimates and recalling our assumption that (ai)
∞
i=1 ∈ S`∞,

we deduce that

‖x‖γ+δ ≤
∑
j∈J
|ai| +

∞∑
j=m(E)

∞∑
k=j

εk ≤ 2‖y‖δ.

This completes the p = ∞ case.
Now assume 1 < p < ∞. Fix E1 < E2 < ... , Ei ∈ Sγ+δ. Let x =∑∞
i=1 aixi, y =

∑∞
i=1 aieri as in the previous paragraph. For each i ∈ N, let

Ji =
{
j ∈ N : (∀i 6= k ∈ N)(Ejxk = 0)

}
.

Let J = ∪∞i=1Ji and I = N \J. Let us rename the sets (Ei)i∈I as F1 < G1 <
F2 < G2 < ... (ignoring this step if I is empty and with the appropriate
notational change if I is finite and non-empty). By the properties of I, for
each i such that Fi (resp. Gi) is defined, there exist at least two distinct indices
j,k such that Fixj,Fixk 6= 0 (resp. at least two distinct indices j′,k′ such that
Gixj′,Gixk′ 6= 0). From this it follows that, with

Ui = {j : Fixj 6= 0} and Vi = {j : Gixj 6= 0},

the sets (Ui)i are successive, as are (Vi)i. In particular, Fixj = Gixj = 0
whenever j < i. Observe that(∑

i∈J
‖Eix‖

p
`1

)1/p
=

( ∞∑
j=1

|aj|p
∑
i∈Jj

‖Eixj‖
p
`1

)1/p
≤‖(aj)∞j=1‖`p ≤‖y‖δ,p.

Now, arguing as in the p = ∞ case, for each i such that Fi is defined, if
m(Fi) = min{j : Fixj 6= 0}, there exists a set T(Fi) ∈Sδ such that

‖Fix‖`1 ≤‖T(Fi)y‖`1 +
∞∑

l=m(Fi)

∞∑
k=l

εk.

Furthermore, T(Fi) ⊂ {nmj : j ∈ Ui}, from which it follows that the sets
T(Fi) are successive, since the sets Ui are. From this, the triangle inequality,
and the fact that m(Fi) ≥ i for each appropriate i, it follows that(∑

i

‖Fix‖
p
`1

)1/p
≤
( ∞∑
i=1

‖T(Fi)y‖
p
`1

)1/p
+
∞∑
i=1

∞∑
l=m(Fi)

∞∑
k=l

εk

≤ ‖y‖δ,p +
∞∑
i=1

∞∑
l=i

∞∑
k=l

εk = ‖y‖δ,p + 1 ≤ 2‖y‖δ,p.



160 r.m. causey

A similar argument yields that

(∑
i

‖Gix‖
p
`1

)1/p
≤ 2‖y‖δ,p.

Therefore ( ∞∑
j=1

‖Ejx‖
p
`1

)1/p
≤ 5‖y‖δ,p.

Since E1 < E2 < ... , Ei ∈Sγ+δ were arbitrary, ‖x‖γ+δ,p ≤ 5‖y‖δ,p.
(iii) By passing to a subsequence and perturbing, we may assume (xn)

∞
n=1

is a block sequence in Xγ+δ,p and infn‖xn‖γ = ε > 0. We may fix a block
sequence (x∗n) ∈ ε−1BX∗γ biorthogonal to (xn)

∞
n=1. By (i), after passing to a

subsequence and using properties of the Xγ+δ,p basis, assume that

sup

{∥∥∥∥ ∑
n∈G

εnx
∗
n

∥∥∥∥
γ+δ

: G ∈Sδ, |εn| = 1
}
≤ 1/ε.

If p = ∞, note that for any (ai)∞i=1 ∈ c00,∥∥∥∥ ∞∑
i=1

aiei

∥∥∥∥
γ

= sup

{ ∑
n∈G
|an| : G ∈Sδ

}

≤ sup
{

Re

( ∑
n∈G

εnx
∗
n

)( ∞∑
n=1

anxn

)
: G ∈Sγ, |εn| = 1

}

≤ ε−1
( ∞∑
n=1

anxn

)
.

Now suppose that 1 < p < ∞. Fix (ai)∞i=1 ∈ c00 and let x =
∑∞

i=1 aiei.
Fix E1 < E2 < · · · < En, Ei ∈Sδ and a sequence (bi)ni=1 ∈ S`nq , such that

‖x‖γ,p =
( n∑
i=1

‖Eix‖
p
`1

)1/p
=

n∑
i=1

bi

( ∑
j∈Ei

|aj|
)
.

Let y∗i =
∑

j∈Ei εjx
∗
j , where εjaj = |aj|, and let y

∗ =
∑n

i=1 biy
∗
i . Since

‖y∗i‖γ+δ ≤ ε
−1 and

∑n
i=1 b

q
i = 1, ‖y

∗‖γ+δ,p ≤ ε−1. Indeed, by Hölder’s in-



the ξ,ζ-dunford pettis property 161

equality, for any x ∈ c00, if I1 < · · · < In are such that supp(y∗i ) ⊂ Ii, then

|y∗(x)| ≤
n∑
i=1

bi|y∗i (x)| ≤ ε
−1

n∑
i=1

bi‖Iixi‖γ

≤ ε−1
( n∑
i=1

b
q
i

)( n∑
i=1

‖Iix‖pγ
)1/p

≤ ε−1‖x‖γ,p.

Moreover,

ε−1
∥∥∥∥ ∞∑
i=1

aixi

∥∥∥∥
ξ,p

≥ y∗
( ∞∑
i=1

aixi

)
=

n∑
i=1

bi

( ∑
j∈Ei

|aj|
)

= ‖x‖γ,p.

Let us recall that for any ordinals γ,ξ with γ ≤ ξ, there exists a unique
ordinal δ such that γ + δ = ξ. We denote this ordinal δ by ξ −γ.

We also recall that any non-zero ordinal ξ admits a unique representation
(called the Cantor normal form) as

ξ = ωε1n1 + · · · + ωεknk,

where k,n1, . . . ,nk ∈ N and ε1 > · · · > εk. Using the Cantor normal form
ξ = ωε1n1 + · · · + ωεknk, we define the least non-trivial part λ(ξ) of ξ by
λ(ξ) = ωε1 . For completeness, we let λ(0) = 0. We also note that if ζ ≤ ξ,
λ(ζ) ≤ λ(ξ).

For 0 < ξ, let ωε1n1 + · · · + ωεknk be the Cantor normal form of ξ. By
writing ωεn = ωε + · · ·+ ωε, where the summand ωε appears n times, we may
also uniquely represent ξ as

ξ = ωδ1 + · · · + ωδl

where l ∈ N and δ1 ≥ ···≥ δl. In this case, δ1 = ε1.

Theorem 2.17. Fix ξ < ω1 and 1 < p ≤ ∞. Fix a weakly null sequence
(xn)

∞
n=1 ⊂ Xξ,p. Let Γ = {ζ ≤ ξ : lim supn‖xn‖ζ > 0}.

(i) If p = ∞, then Γ = ∅ if and only if (xn)∞n=1 is norm null.
(ii) If p < ∞ and Γ = ∅, then either (xn)∞n=1 is norm null or (xn)

∞
n=1 has a

subsequence equivalent to the canonical `p basis.

(iii) If Γ 6= ∅ and γ = min Γ, then (xn)∞n=1 admits a subsequence which is
equivalent to a subsequence of the Xξ−γ,p basis. In particular, (xn)

∞
n=1

is ξ −γ + 1 weakly null and not ξ −γ weakly null.



162 r.m. causey

(iv) If p = ∞, then every subsequence of (xn)∞n=1 has a further WUC subse-
quence if and only if Γ ⊂{ξ}.

(v) If 0 < ξ, a weakly null sequence (xn)
∞
n=1 is ξ-weakly null if and only if

for every β < λ(ξ), limn‖xn‖β = 0.

Proof. First note that by the almost monotone property of the Schreier
families, if ζ ∈ Γ, then [ζ,ξ] ⊂ Γ.

(i) It is evident that limn‖xn‖ξ = 0 if and only if ξ /∈ Γ.
(ii) If ξ /∈ Γ, then let γ = ξ and δ = 0. By Proposition 2.16(ii), any

subsequence of (xn)
∞
n=1 has a further subsequence which is dominated by a

subsequence of the Xδ,p = `p basis. Then since every seminormalized block
sequence in Xξ,p which dominates the `p basis, either limn‖xn‖ξ,p = 0, or
(xn)

∞
n=1 has a seminormalized subsequence which dominates the `p basis, and

this subsequence has a further subsequence equivalent to the `p basis.
(iii) Let δ = ξ − γ, so that γ + δ = ξ. Proposition 2.16(ii) yields that

every subsequence of (xn)
∞
n=1 has a further subsequence which is dominated

by a subsequence of the Xδ,p basis. Since no subsequence of the Xδ,p basis is
an `δ+11 -spreading model, this yields that (xn)

∞
n=1 is δ + 1-weakly null. Since

γ ∈ Γ, Proposition 2.16(iii) yields the existence of a subsequence (yn)∞n=1 of
(xn)

∞
n=1 which dominates the Xδ,p basis, so (xn)

∞
n=1 is not δ-weakly null. Now

note that limn‖yn‖β = 0 for all β < γ, so by Proposition 2.16(ii), there exists
a subsequence (zn)

∞
n=1 of (yn)

∞
n=1 which is dominated by some subsequence

(xni)
∞
i=1 of the canonical Xδ,p basis. This sequence (zn)

∞
n=1 also dominates

some subsequence (xmi)
∞
i=1 of the canonical Xδ,p basis (where mi has the

property that zi = ymi). Now let us choose 1 = k1 < k2 < ... such that
mki+1 > nki for all i ∈ N and let ui = zki. Then (ui)

∞
i=1 is dominated by

some subsequence (xri)
∞
i=1 of the Xδ,p basis and dominates some subsequence

(xsi)
∞
i=1 of the Xδ,p basis, where s1 ≤ r1 < s2 ≤ r2 < ... . This is seen

by taking si = mki and ri = nki. But it is observed in [10] that two such
subsequences of the Xδ,p basis must be 2-equivalent, so (ui)

∞
i=1 is equivalent

to (eri)
∞
i=1 (and to (esi)

∞
i=1).

(iv) If Γ ⊂{ξ}, then by Proposition 2.16(ii) applied with γ = ξ and δ = 0,
every subsequence of (xn)

∞
n=1 has a further subsequence which is dominated

by the Xδ = c0 basis. Conversely, if ξ > γ ∈ Γ, then with δ = ξ − γ > 0,
(xn)

∞
n=1 has a subsequence which is an `

δ
1-spreading model. No subsequence

of this sequence can be WUC.
(v) Note that both conditions are satisfied if (xn)

∞
n=1 is norm null, so

assume (xn)
∞
n=1 is not norm null. If Γ = ∅, then p < ∞, and every sub-

sequence of (xn)
∞
n=1 has a further subsequence which is equivalent to the `p



the ξ,ζ-dunford pettis property 163

basis, which means (xn)
∞
n=1 is 1-weakly null, and therefore ξ-weakly null. Thus

both conditions are satisfied in this case as well.
It remains to consider the case Γ 6= ∅. Let γ = min Γ. Let us write

ξ = ωε1 + · · · + ωεk,

where ε1 ≥ ···≥ εk. Note that λ(ξ) = ωε1 . First assume that limn‖xn‖β = 0
for all β < λ(ξ), which means γ ≥ λ(ξ). Then if γ +δ = ξ, δ ≤ ωε2 +· · ·+ωεk.
By (iii), (xn)

∞
n=1 is δ + 1-weakly null, and

δ + 1 ≤ ωε2 + · · · + ωεk + 1

≤ ωε2 + · · · + ωεk + ωε1 ≤ ωε1 + · · · + ωεk = ξ

yields that (xn)
∞
n=1 is ξ-weakly null. Conversely, assume there exists β < λ(ξ)

such that lim supn‖xn‖β > 0. Then γ < λ(ξ). If γ + δ = ξ, then δ = ξ. By
(iii), (xn)

∞
n=1 is not ξ-weakly null.

Corollary 2.18. For any 0 < ξ < ω1 and any seminormalized, weakly
null sequence (xn)

∞
n=1 in Xωξ, (xn)

∞
n=1 has a subsequence (yn)

∞
n=1 which is

either equivalent to the canonical c0 basis or to a subsequence of the Xωξ
basis.

Proof. By Theorem 2.17(iv), every subsequence of (xn)
∞
n=1 has a further

WUC (and therefore equivalent to the c0 basis) subsequence if and only if
limn‖xn‖β = 0 for every β < ξ = λ(ξ). If this condition fails, then there
exists a minimum γ < ωξ such that lim supn‖xn‖γ > 0. Then if γ + δ = ωξ,
δ = ωξ, and (xn)

∞
n=1 has a subsequence equivalent to a subsequence of the

Xωξ basis.

Corollary 2.19. Fix 0 < ξ < ω1, 1 < p ≤ ∞, and let (xn)∞n=1 ⊂ Xξ,p
be weakly null. Then (xn)

∞
n=1 is ξ-weakly null in Xξ,p if and only if for every

γ < λ(ξ), limn‖xn‖γ = 0 if and only if every subsequence of (xn)∞n=1 has a
further subsequence which is WUC in Xλ(ξ).

Proof. This follows from combining Theorem 2.17 (iv) – (v).

In the sequel, we will need the following standard duality argument. As it
involves some non-trivial computation, we isolate it.

Proposition 2.20. Suppose that F is a spreading set of finite subsets of
N, (xn)∞n=1 ⊂ X is weakly null, (x

∗
n)
∞
n=1 is weakly null, inf |x

∗
n(xn)| ≥ ε > 0.



164 r.m. causey

(i) If

sup

{∥∥∥∥ ∑
n∈F

anx
∗
n

∥∥∥∥ : F ∈F, |an| ≤ 1
}

= C < ∞,

then there exists a subsequence (xkn)
∞
n=1 of (xn)

∞
n=1 such that

inf

{∥∥∥∥ ∑
n∈F

bnxkn

∥∥∥∥ : F ∈F, ∑
n∈F
|bn| = 1

}
≥

ε

2C
.

(ii) If

sup

{∥∥∥∥ ∑
n∈F

anxn

∥∥∥∥ : F ∈F, |an| ≤ 1
}

= C < ∞,

then there exists a subsequence (x∗kn)
∞
n=1 of (x

∗
n)
∞
n=1 such that

inf

{∥∥∥∥ ∑
n∈F

bnx
∗
kn

∥∥∥∥ : F ∈F, ∑
n∈F
|bn| = 1

}
≥

ε

2C
.

Proof. (i) First note that the condition

sup

{∥∥∥∥ ∑
n∈F

anx
∗
n

∥∥∥∥ : F ∈F, |an| ≤ 1
}
≤ C

passing to subsequences, since F is spreading. Fix (εn)∞n=1(0,ε) such that∑∞
n=1

∑∞
m=n+1 εm < ε/4. We may recursively choose 1 = k1 < k2 < ...

such that for all n < m, |x∗n(xm)|, |x∗m(xn)| < εm. Then for any F ∈ F and
(bn)n∈F , fix (an)n∈F such that |an| = 1 for all n ∈ F and∑

n∈F
anbnx

∗
kn

(xkn) =
∑
n∈F
|anbnx∗kn(xkn)| ≥ ε

∑
n∈F
|bn|.

Since ‖
∑

n∈F anx
∗
kn
‖≤ C by the first sentence of the proof,

C

∥∥∥∥ ∑
n∈F

bnxkn

∥∥∥∥ ≥
∣∣∣∣
( ∑
n∈F

anx
∗
kn

)( ∑
n∈F

bnxkn

)∣∣∣∣
≥
∑
n∈F

anbnx
∗
kn

(xkn) −
∞∑
n=1

∞∑
m=n+1

|bn||x∗km(xkn)|

−
∞∑
n=1

∞∑
m=n+1

|bm||x∗kn(xkm)|

≥ ε
∑
n∈F
|bn|− 2 max

n∈F
|bn|

∞∑
n=1

∞∑
m=n+1

εm ≥
ε

2

∑
n∈F
|bn|.



the ξ,ζ-dunford pettis property 165

(ii) This follows from (i) by considering (xn)
∞
n=1 as a sequence in X

∗∗.

Lemma 2.21. Fix 0 < ξ < ω1 and 1 < p ≤∞.

(i) If (x∗∗n )
∞
n=1 ⊂ X

∗∗
ξ,p is ξ-weakly null, then limn‖x

∗∗
n ‖γ = 0 for every

γ < λ(ξ).

(ii) If (x∗∗n )
∞
n=1 ⊂ X

∗∗
ξ,p is ξ-weakly null and (x

∗
n)
∞
n=1 ⊂ X

∗
λ(ξ)

is weakly null,

then limn x
∗∗
n (x

∗
n) = 0.

Proof. (i) Suppose not. Then for some γ < λ(ξ) and ε > 0, we may pass
to a subsequence and assume infn‖x∗∗n ‖γ > ε. We may choose a sequence
(x∗n)

∞
n=1 ⊂ BX∗γ ∩ c00 such that infn |x

∗∗
n (xn)| > ε. Since limn x∗∗n (e∗i ) = 0 for

all i ∈ N, we may, by passing to a subsequence and replacing the functionals
x∗n by tail projections thereof, assume that (x

∗
n)
∞
n=1 is a block sequence in

BX∗γ ∩c00. Then by standard properties of ordinals, if δ is such that γ +δ = ξ,
δ = ξ. By Proposition 2.16(i), we may pass to a subsequence once more and

asssume (x∗n)
∞
n=1 is a c

ξ
0-spreading model in X

∗
ξ , and therefore weakly null in

X∗ξ . By passing to a subsequence one final time and appealing to Proposition

2.20, assume (x∗∗n )
∞
n=1 is an `

ξ
1-spreading model. Therefore (x

∗∗
n )
∞
n=1 is not

ξ-weakly null. This contradiction finishes (i).

(ii) Also by contradiction. Assume (x∗∗n )
∞
n=1 ⊂ X

∗∗
ξ,p is ξ-weakly null,

(x∗n)
∞
n=1 ⊂ X

∗
λ(ξ)

is weakly null, and infn |x∗∗n (x∗n)| > ε > 0. By perturbing, we
may assume (x∗n)

∞
n=1 is a block sequence and there exist I1 < I2 < ... such

that Inx
∗
n = x

∗
n for all n ∈ N. Let (γk)∞k=1 ⊂ [0,λ(ξ)) be a sequence (possibly

with repitition) such that [0,λ(ξ)) = {γk : k ∈ N}. By (i), limn‖x∗∗n ‖γk = 0
for all k ∈ N. By passing to a subsequence and relabeling, we may assume
that for each 1 ≤ k ≤ n, ‖x∗∗n ‖γk < 1/n. Let xn = Inx

∗∗
n ∈ Xξ and note that

for each γ < λ(ξ), limn‖xn‖γ = 0. Indeed, if γ = γk, then for all n ≥ k,

‖xn‖γ ≤‖x∗∗n ‖γk ≤ 1/n.

Since Inx
∗
n = x

∗
n, |x∗n(xn)| = |x∗∗n (x∗n)| > ε. But by Corollary 2.19, some

subsequence of (xn)
∞
n=1, which we may assume is the entire sequence after

relabeling, is WUC in Xλ(ξ). But now we reach a contradiction by combining
the facts that (xn)

∞
n=1 is WUC in Xλ(ξ), (x

∗
n)
∞
n=1 ⊂ X

∗
λ(ξ)

is weakly null, and

infn |x∗n(xn)| > 0.



166 r.m. causey

3. Ideals of interest

Basic definitions. We recall that Ban is the class of all Banach
spaces and L denotes the class of all operators between Banach spaces. For
each pair X,Y ∈ Ban, L(X,Y ) is the class of all operators from X into Y .
Given a subclass I of L, we let I(X,Y ) = I∩L(X,Y ).

We say that a class I of operators is an operator ideal (or just an ideal)
provided that

(i) for any W,X,Y,Z ∈ Ban, C ∈ L(W,X), B ∈ I(X,Y ), and A ∈ L(Y,Z),
ABC ∈ I(W,Z),

(ii) IK ∈ I,
(iii) for each X,Y ∈ Ban, I(X,Y ) is a vector subspace of L(X,Y ).

Given an operator ideal I, we define the

(i) closure I of I to be the class of operators such that for every X,Y ∈
Ban, I(X,Y ) = I(X,Y ),

(ii) injective hull Iinj of I to be the class of all operators A : X → Y such
that if there exists Z ∈ Ban and an isometric (equivalently, isomorphic)
embedding j : Y → Z such that jA ∈ I(X,Z),

(iii) surjective hull Isur of I to be the class of all operators A : X → Y
such that there exist W ∈ Ban and a quotient map (equivalently, a
surjection) q : W → X such that Aq ∈ I(W,Y ),

(iv) dual Idual to be the class of all operators A : X → Y such that A∗ ∈
I(Y ∗,X∗).

We also let {I denote the class of operators such that for each pair X,Y of
Banach spaces, {I(X,Y ) = L(X,Y ) \I(X,Y ).

Each of I, Iinj, Isur is also an ideal.
Given two ideals I,J, we let

(i) I ◦ J−1 denote the class of all operators A : X → Y such that for all
W ∈ Ban and R ∈ J(W,X), AR ∈ I(W,Y ),

(ii) I−1 ◦ J denote the class of all operators A : X → Y such that for all
Z ∈ Ban and all L ∈ I(Y,Z), LA ∈ J(X,Z).

We remark that for any three ideals I1,I2,J,

(I−11 ◦J) ◦I
−1
2 = I

−1
1 ◦ (J◦I

−1
2 ),

so that the symbol I−11 ◦J◦I
−1
2 is unambiguous.



the ξ,ζ-dunford pettis property 167

We say an operator ideal is

(i) closed if I = I,

(ii) injective if I = Iinj,

(iii) surjective if I = Isur,

(iv) symmetric if I = Idual.

With each ideal, we will associate the class of Banach spaces the identity
of which lies in the given ideal. Our ideals will be denoted by fraktur lettering
(A,B,I, . . . ) and the associated space ideal will be denoted by the same sans
serif letter (A,B, I, . . . ).

We next list some ideals of interest. We let K,W, and V denote the class of
compact, weakly compact, and completely continuous operators, respectively.

For the remaining paragraphs in this subsection, ξ will be a fixed ordinal
in [0,ω1]. We let Wξ denote the class of operators A : X → Y such that
any bounded sequence in X has a subsequence whose image under A is ξ-
weakly convergent in Y (let us recall that a sequence (yn)

∞
n=1 ⊂ Y is said to

be ξ-weakly convergent to y ∈ Y if (yn −y)∞n=1 is ξ-weakly null). Note that
W0 = K and Wω1 = W. We refer to Wξ as the class of ξ-weakly compact
operators. This class was introduced in this generality in [6].

We let wBSξ denote the class of operators A : X → Y such that for
any weakly null sequence (xn)

∞
n=1, (Axn)

∞
n=1 is ξ-weakly convergent to 0 in

Y . Note that wBS0 = V, wBSω1 = L, and wBS1 is the class of weak
Banach-Saks operators. For this reason, we refer to wBSξ as the class of
ξ-weak Banach-Saks operators. These classes were introduced in this gener-
ality in [4].

We let Vξ denote the class of operators A : X → Y such that for any
ξ-weakly null sequence (xn)

∞
n=1, (Axn)

∞
n=1 is norm nul. It is evident

that Vω1 = V and V0 = L. These classes were introduced in this gener-
ality in [12].

For 0 ≤ ζ ≤ ω1, we let Gξ,ζ denote the class of operators A : X → Y
such that whenever (xn)

∞
n=1 is ξ-weakly null, (Axn)

∞
n=1 is ζ-weakly null. We

isolate this class because it is a simultaneous generalization of the two previous
paragraphs. Indeed, Vξ = Gξ,0, while wBSξ = Gω1,ξ. It is evident that
Gξ,ζ = L whenever ξ ≤ ζ. These classes are newly introduced here.

For 0 ≤ ζ ≤ ω1, we let Mξ,ζ denote the class of all operators A : X → Y
such that for any ξ-weakly null (xn)

∞
n=1 ⊂ X and any ζ-weakly null (y

∗
n)
∞
n=1 ⊂

Y ∗, limn y
∗
n(Axn) = 0. The class Mω1,ω1 (sometimes denoted by DP) is a



168 r.m. causey

previously defined class of significant interest, most notably because the asso-
ciated space ideal Mω1,ω1 is the class of Banach spaces with the Dunford-Pettis
property. As a class of operators, Mξ,ζ has not previously been investigated,
but the space ideals M1,ω1 and Mω1,ξ have been investigated in [16] and [1],
respectively.

Remark 3.1. Let us recall that the image of a ξ-weakly null sequence under
a continuous, linear operator is also ξ-weakly null, for any 0 ≤ ξ ≤ ζ ≤ ω1, any
sequence which is ξ-weakly null is also ζ-weakly null, and the 0-weakly null
sequences are the norm null sequences. From this we deduce the following:

(i) Gξ,ζ = L for any ξ ≤ ζ ≤ ω1.
(ii) Mξ,ζ = L if min{ξ,ζ} = 0.
(iii) For ζ ≤ α ≤ ω1 and β ≤ ξ ≤ ω1, Gξ,ζ ⊂ Gβ,α.
(iv) If α ≤ ζ ≤ ω1 and β ≤ ξ ≤ ω1, then Mξ,ζ ⊂ Mβ,α.

We next record an easy consequence of Corollary 2.12.

Corollary 3.2. For any 0 ≤ ζ,ξ ≤ ω1,

Gξ,ζ ⊂
⋂
α<ω1

Gα+ξ,α+ζ.

Proof. Suppose X,Y are Banach spaces, A : X → Y is an operator,
α < ω1, and 0 ≤ ζ,ξ ≤ ω1 are such that A ∈ {Gα+ξ,α+ζ. Then there exists a
sequence (xn)

∞
n=1 ⊂ X which is α + ξ-weakly null and such that (Axn)

∞
n=1 is

not α +ζ-weakly null. Note that ζ < ω1, since otherwise α +ζ = α +ω1 = ω1,
and (Axn)

∞
n=1 would be a non-weakly null image of a weakly null sequence.

If ξ = ω1, we deduce that A ∈ {Gξ,ζ, since (xn)∞n=1 is a ξ-weakly null se-
quence the image of which under A is not α+ζ-weakly null, and therefore not
ζ-weakly null. If ξ < ω1, we use Corollary 2.12 to deduce the existence of some
convex blocking (zn)

∞
n=1 of (xn)

∞
n=1 which is ξ-weakly null and the image of

which under A is an `
ζ
1-spreading model. Thus A ∈ {Gξ,ζ. Therefore

{Gα+ξ,α+ζ ⊂ {Gξ,ζ.

Taking complements and noting that α < ω1 was arbitrary, we are done.

Remark 3.3. We remark that adding α on the left in the previous corollary
is necessary. The analogous statement fails if we add α on the right. For
example, for any 0 < ξ < ω1 and ζ < ω

ξ, the formal identity I : Xωξ → Xζ
lies in Gωξ,0 ∩{Gωξ+1,ζ.



the ξ,ζ-dunford pettis property 169

Examples. In this subsection, we provide examples to show the richness
of the classes of interest, wBSξ, Gξ,ζ, and Mξ,ζ. We note that wBS0 = V,
Gξ,ζ = L whenever ξ ≤ ζ, and Mξ,ζ = L whenever min{ξ,ζ} = 0. We typically
omit reference to these trivial cases.

Proposition 3.4. Fix 0 < ξ < ω1. Then for any subset S of [0,ξ) with
sup S = ξ, (⊕ζ∈SXζ)`1(S) ∈ wBSξ ∩

⋃
ζ<ξ {wBSζ.

Proof. By Theorem 2.14(v), if ζ < ξ, Xζ ∈ wBSξ. We will prove in
Proposition 3.15 that the `1 direct sum of members of wBSξ also lies
in wBSξ.

Theorem 3.5. For 0 ≤ ζ < ξ < ω1, the formal inclusion I : Xξ → Xζ lies
in Gξ,ζ ∩{Gξ+1,ζ.

Proof. Fix (xn)
∞
n=1 ⊂ Xξ ξ-weakly null. Then by Theorem 2.17 (v),

limn‖xn‖β = 0 for every β < λ(ξ). If ζ = 0, then ζ < λ(ξ) and limn‖xn‖ζ = 0.
Therefore (Ixn)

∞
n=1 is ζ-weakly null. If ζ > 0, then since λ(ζ) ≤ λ(ξ),

limn‖Ixn‖β = 0 for every β < λ(ζ), and Theorem 2.17(v) yields that (Ixn)∞n=1
is ζ-weakly null in this case. In either case, (Ixn)

∞
n=1 is ζ-weakly null, and

I ∈ Gξ,ζ. However, the canonical basis is ξ + 1-weakly null in Xξ and not
ζ-weakly null in Xζ, so I ∈ {Gξ+1,ζ.

It is well-known and obvious that every Schur space and every space whose
dual is a Schur space has the Dunford-Pettis property. The generalization of
this fact to operators is V,Vdual ⊂ DP. The ordinal analogues are also
obvious: For any 0 < ξ ≤ ω1, Vξ ⊂ Mξ,ω1 and V

dual
ξ ⊂ Mω1,ξ. Thus it is of

interest to come up with examples of members of Mξ,ω1 , or more generally
Mξ,ζ, which do not come from Vξ or V

dual
ζ .

Theorem 3.6. For 0 < ξ < ω1 and 1 < p ≤ ∞, the formal inclusion
I : Xξ,p → Xλ(ξ) lies in Mξ,ω1 ∩ {Mξ+1,1 ∩ {Vξ and the formal inclusion
J : X∗

λ(ξ)
→ X∗ξ,p lies in Mω1,ξ ∩{M1,ξ+1 ∩{V

dual
ξ .

Proof. It follows from Lemma 2.21(ii) that I ∈ Mξ,ω1 and J ∈ Mω1,ξ. Since
the canonical basis of Xξ,p ⊂ X∗∗ξ,p is ξ + 1-weakly null and the canonical basis
of X∗

λ(ξ)
is a c10-spreading model, and therefore 1-weakly null, I ∈ {Mξ+1,1 and

J ∈ {M1,ξ+1. Now if (γk)∞k=1 ⊂ [0,λ(ξ)) is such that [0,λ(ξ)) = {γk : k ∈ N},
we may select F1 < F2 < ... , Fi ∈ Sλ(ξ), and positive scalars (ai)i∈∪∞n=1Fn



170 r.m. causey

such that for each 1 ≤ k ≤ n,
∑

i∈Fn ai = 1 and ‖
∑

i∈Fn aiei‖γk < 1/n. Then
with xn =

∑
i∈Fn aiei, Theorem 2.17(v) yields that (xn)

∞
n=1 is ξ-weakly null

in Xξ,p ⊂ X∗∗ξ,p. Evidently (xn)
∞
n=1 is normalized in Xλ(ξ), so that I ∈ {Vξ

and J ∈ {Vdualξ .

Corollary 3.7. For any 0 ≤ α,β,ζ,ξ ≤ ω1, Gβ,α = Gξ,ζ if and only if
one of the two exclusive conditions holds:

(i) ξ ≤ ζ and β ≤ α (in which case Gβ,α = L = Gξ,ζ).
(ii) α = ζ < ξ = β.

Proof. It is obvious that (i) and (ii) are exclusive and either implies equal-
ity. Now suppose that neither (i) nor (ii) holds. Suppose ξ ≤ ζ and β > α.
Then IXα ∈ L∩{Gβ,α = Gξ,ζ∩{Gβ,α, and Gξ,ζ 6= Gβ,α. Similarly, Gξ,ζ 6= Gβ,α
if β ≤ α and ζ < ξ.

For the remainder of the proof, suppose that α < β and ζ < ξ. Now
suppose α < ζ. Then

IXα ∈ wBSα+1 ∩{wBSα ⊂ Gξ,ζ ∩{Gβ,α.

Similarly, Gξ,ζ 6= Gβ,α if ζ < α. Next assume ζ = α < ξ < β. Then if
I : Xξ → Xζ is the formal inclusion, I ∈ Gξ,ζ ∩ {Gβ,α. If ζ = α < β < ξ, we
argue similarly with the inclusion I : Xβ → Xα. Since this is a complete list
of the possible ways for (i) and (ii) to simultaneously fail, we are done.

Corollary 3.8. For any 0 ≤ α,β,ζ,ξ ≤ ω1, Mβ,α ⊂ Mξ,ζ if and only if
one of the two exclusive conditions holds:

(i) 0 = min{ζ,ξ} (in which case Mβ,α = L = Mξ,ζ).
(ii) 0 < ζ ≤ α and 0 < ξ ≤ β.

In particular, Mβ,α = Mξ,ζ if and only if min{β,α} = 0 = min{ξ,ζ} or
0 < α = ζ and 0 < β = ξ.

Proof. It is obvious that (i) and (ii) are exclusive, and either implies that
Mβ,α ⊂ Mξ,ζ.

Now assume that min{ζ,ξ} > 0. If min{α,β} = 0, Mβ,α = L 6⊂ Mξ,ζ,
since I`2 ∈ {M1,1 ⊂ {Mξ,ζ. If 0 < α,β and β < ξ, then let I : Xβ → Xλ(β) be
the formal inclusion. Then

I ∈ Mβ,ω1 ∩{Mβ+1,1 ⊂ Mβ,α ∩{Mξ,ζ.



the ξ,ζ-dunford pettis property 171

Now if 0 < α,β and α < ζ, let J : X∗
λ(α)
→ X∗α be the formal inclusion. Then

J ∈ Mω1,α ∩{M1,α+1 ⊂ Mβ,α ∩{Mξ,ζ.

The last statement follows from the fact that if Mβ,α = Mξ,ζ, then
either both classes must equal L, which happens if and only if min{β,α}
= 0 = min{ξ,ζ}, or neither class is L, in which case min{β,α}, min{ξ,ζ}
> 0. In the latter case, using the previous paragraph and symmetry, α = ζ
and β = ξ.

General properties. We will need the following fact, shown in [12].

Proposition 3.9. If X is a Banach space and (xn)
∞
n=1 ⊂ X isξ-weakly

null, then there exists a subsequence (xni)
∞
i=1 of (xn)

∞
n=1 such that the operator

Φ : `1 → X given by Φ
∑∞

i=1 aiei =
∑∞

i=1 aixni lies in Wξ(`1,X).

Remark 3.10. It follows that if Y is a Banach space and (y∗n)
∞
n=1 ⊂ Y

∗

is ξ-weakly null, there exist a subsequence (y∗ni)
∞
i=1 of (y

∗
n)
∞
n=1 such that the

operator given by Ψ : Y → c0 given by Ψy = (y∗ni(y))
∞
i=1 lies in W

dual
ξ (Y,c0).

This follows immediately from Proposition 3.9, since Ψ∗ : `1 → Y ∗ is given by
Ψ∗
∑∞

i=1 aiei =
∑∞

i=1 aiy
∗
ni

.

Remark 3.11. In the following results, we will repeatedly use the previ-
ously stated fact that a weakly null `

ζ
1-spreading model can have no ζ-weakly

convergent subsequence.

Theorem 3.12. Fix 0 ≤ ζ < ξ ≤ ω1. Then

Gξ,ζ = Wζ ◦W−1ξ and G
dual
ξ,ζ = (W

dual
ξ )

−1 ◦Wdualζ .

Consequently, Gξ,ζ is a closed, two-sided ideal containing all compact opera-
tors. Moreover, Gξ,ζ is injective but not surjective. Finally,

Gdual dualξ,ζ ( Gξ,ζ,

while neither of Gξ,ζ, G
dual
ξ,ζ is contained in the other.

Proof. Fix X,Y ∈ Ban and A ∈ L(X,Y ). First suppose that A ∈
Gξ,ζ(X,Y ). Fix a Banach space W and R ∈ Wξ(W,X). Fix a bounded
sequence (wn)

∞
n=1. By passing to a subsequence, we may assume there exists



172 r.m. causey

x ∈ X such that (x − Rwn)∞n=1 is ξ-weakly null, from which it follows that
(Ax−ARwn)∞n=1 is ζ-weakly null. Since this holds for an arbitrary bounded
sequence in (wn)

∞
n=1, AR ∈ Wζ. Since W ∈ Ban and R ∈ Wξ(W,X) were

arbitrary, A ∈ Wζ ◦W−1ξ (X,Y ).

Now suppose that A ∈ {Gξ,ζ. Then there exists a ξ-weakly null sequence
(xn)

∞
n=1 in X such that (Axn)

∞
n=1 is an `

ζ
1-spreading model. By Proposition

3.9, after passing to a subsequence and relabeling, we may assume the operator
R : `1 → X given by R

∑∞
i=1 aiei =

∑∞
i=1 aixi lies in Wξ(`1,X). But since

(ARei)
∞
i=1 = (Axi)

∞
i=1 has no ζ-weakly convergent subsequence, A ∈ {Wζ ◦

W−1ξ (X,Y ).

Next, suppose that A ∈ Gdualξ,ζ (X,Y ). Fix Z ∈ Ban and an operator
L ∈ Wdualξ (Y,Z). Then A

∗ ∈ Gξ,ζ(Y ∗,X∗) = Wζ ◦ W−1ξ (Y
∗,X∗) and L∗ ∈

Wξ(Z
∗,Y ∗), and (LA)∗ = A∗L∗ ∈ Wζ(Z∗,X∗). Thus LA ∈ Wdualζ (X,Z).

Since this holds for any Z ∈ Ban and L ∈ Wdualξ (Y,Z), A ∈ (W
dual
ξ )

−1 ◦
Wdualζ (X,Y ).

Now if A ∈ {Gdualξ,ζ (X,Y ), there exists (y
∗
n)
∞
n=1 ⊂ Y

∗ which is ξ-weakly

null and (A∗y∗n)
∞
n=1 is an `

ζ
1-spreading model. By the remarks preceding the

theorem, by passing to a subsequence and relabeling, we may assume the
operator L : Y → c0 given by Ly = (y∗n(y))∞n=1 lies in W

dual
ξ (Y,c0). But

since (A∗L∗ei)
∞
i=1 = (A

∗y∗i )
∞
i=1 is a weakly null `

ζ
1-spreading model, (LA)

∗ =
A∗L∗ ∈ {Wζ(`1,X∗). Thus LA ∈ {((Wdualξ )

−1 ◦Wdualζ )(X,Y ).
This yields the first two equalities. It follows from the fact that Wζ,Wξ

are closed, two-sided ideals containing the compact operators that Gξ,ζ is also.

It is evident that Gξ,ζ is injective, since a given sequence is ζ-weakly null if
and only if its image under some (equivalently, every) isomorphic image of that
sequence is ζ-weakly null. The ideal Gξ,ζ is not surjective, since Xζ ∈ {Gξ,ζ,
while Xζ is a quotient of `1 ∈ V ⊂ Gξ,ζ.

It is also easy to see that if A∗∗ ∈ Gξ,ζ, then A ∈ Gξ,ζ, so Gdual dualξ,ζ ⊂ Gξ,ζ.
If ζ = 0, note that `1 ∈ V ⊂ Gξ,ζ, but `∗∗1 contains an isomorphic copy of
`2, so that `

∗∗
1 ∈ {Gξ,0. This yields that G

dual dual
ξ,0 6= Gξ,0. Now if ζ > 0,

c0 ∈ wBS1 ⊂ Gξ,ζ. But `∞ = c∗∗0 ∈ {Gξ,ζ. In order to see that `∞ ∈ {Gξ,ζ,
simply note that `∞ contains a sequence equivalent to the Xζ basis, which is
ξ-weakly null and not ζ-weakly null.

Finally, let us note that if ζ = 0, `1 ∈ V ⊂ Gξ,ζ, while c0,`∞ ∈ {Gξ,0. Thus
neither of Gξ,0,G

dual
ξ,0 is contained in the other. Now suppose that ζ > 0. Then



the ξ,ζ-dunford pettis property 173

since X∗ζ,2 ∈ wBS1 ⊂ Gξ,ζ,

Xζ,2 ∈ Gdualξ,ζ ∩{Gξ,ζ and X
∗
ζ,2 ∈ Gξ,ζ ∩{G

dual
ξ,ζ .

Here we recall that Xζ,2 is reflexive. This yields that if 0 < ζ < ξ ≤ ω1,
neither of Gξ,ζ,G

dual
ξ,ζ is contained in the other.

Theorem 3.13. Fix 0 < ζ,ξ ≤ ω1. Then

Mξ,ζ = (W
dual
ζ )

−1 ◦Vξ = (Wdualζ )
−1 ◦K◦W−1ξ .

Consequently, Mξ,ζ is a closed, two-sided ideal containing all compact opera-
tors. Moreover, Mξ,ζ is neither injective nor surjective. Finally,

Mdualξ,ζ ( Mζ,ξ and M
dual dual
ξ,ζ ( Mξ,ζ.

Proof. It follows from the fact that Vξ = K ◦ W−1ξ , which was shown
in [12], that (Wdualζ )

−1 ◦ Vξ = (Wdualζ )
−1 ◦ K ◦ W−1ξ . We will show that

Mξ,ζ = (W
dual
ζ )

−1 ◦ K ◦ W−1ξ . To that end, fix Banach spaces X,Y and
A ∈ L(X,Y ).

Suppose that A ∈ L(X,Y ). Fix Banach spaces W,Z and operators R ∈
Wξ(W,X) and L ∈ Wdualζ (Y,Z). We will show that LAR ∈ K(W,Z). Seeking
a contradiction, suppose LAR ∈ {K. Note that there exists a bounded se-
quence (wn)

∞
n=1 ⊂ W such that infm 6=n‖LARwm −LARwn‖≥ 4. By passing

to a subsequence, we may assume there exist x ∈ X such that (x−Rwn)∞n=1 is
ξ-weakly null. Since ‖LARwm −LARwn‖≥ 4 for all m 6= n, there is at most
one n ∈ N such that ‖LAx−LARwn‖ < 2. By passing to a subsequence, we
may assume ‖LAx−LARwn‖≥ 2 for all n ∈ N. For each n ∈ N, fix z∗n ∈ BZ∗
such that |z∗n(LAx − LARwn)| ≥ 2. By passing to a subsequence one final
time, we may assume there exists y∗ ∈ Y ∗ such that (y∗−L∗z∗n)∞n=1 is ζ-weakly
null and, since (Ax−ARwn)∞n=1 is weakly null, |y

∗(Ax−ARwn)| < 1 for all
n ∈ N. Then (y∗−L∗z∗n)∞n=1 ⊂ Y

∗ is ζ-weakly null, (x−Rwn)∞n=1 is ξ-weakly
null, and

inf
n
|(y∗ −L∗z∗n)(Ax−ARwn)| ≥ inf

n
|L∗z∗n(Ax−ARwn)|− 1

= inf
n
|z∗n(LAx−LARwn)|− 1 ≥ 1.

This contradiction yields that Mξ,ζ ⊂ (Wdualζ )
−1 ◦K◦W−1ξ .



174 r.m. causey

Now suppose that A ∈ {Mξ,ζ(X,Y ). Then there exist a ξ-weakly null
sequence (xn)

∞
n=1 ⊂ X and a ζ-weakly null sequence (y

∗
n)
∞
n=1 ⊂ Y

∗ such
that infn |y∗n(Axn)| = 1. Using Proposition 3.9 and the remark following
it, after passing to subsequences twice and relabling, we may assume the
operators R : `1 → X given by R

∑∞
i=1 aiei =

∑∞
i=1 aixi and L : Y → c0

given by Ly = (y∗n(y))
∞
n=1 lie in Wξ(`1,X) and W

dual
ζ (Y,c0), respectively. But

LAR : `1 → c0 is not compact, since

|e∗n(LARen)| = |y
∗
n(Axn)| ≥ 1

for all n ∈ N. This yields that Mξ,ζ = (Wdualζ )
−1 ◦K◦W−1ξ .

Since `2 ∈ {M1,1 ⊂ {Mξ,ζ is a subspace of `∞ ∈ Mω1,ω1 ⊂ Mξ,ζ and a
quotient of `1 ∈ Mω1,ω1 ⊂ Mξ,ζ, Mξ,ζ is neither injective nor surjective.

Now suppose A ∈ Mdualξ,ζ (X,Y ). Now if (xn)
∞
n=1 ⊂ X is ζ-weakly null,

(y∗n)
∞
n=1 is ξ-weakly null, and j : X → X

∗∗ is the canonical embedding, then
(jxn)

∞
n=1 ⊂ X

∗∗ is ζ-weakly null. Since A ∈ Mdualξ,ζ (X,Y ),

lim
n
y∗n(Axn) = lim

n
A∗y∗n(xn) = lim

n
jxn(A

∗y∗n) = 0.

Thus A ∈ Mζ,ξ(X,Y ). This yields that Mdualξ,ζ ⊂ Mζ,ξ. To see that M
dual
ξ,ζ 6=

Mζ,ξ, we cite Stegall’s example [22], X = `1(`
n
2 ). This space has the Schur

property, and therefore lies in Mω1,ω1 ⊂ Mζ,ξ, while X∗ contains a comple-
mented copy of `2. The fact that X

∗ contains a complemented copy of `2 is
stated explicitly in [8]. Thus X ∈ {Mdual1,1 ⊂ {M

dual
ξ,ζ .

Next, we note that

Mdual dualξ,ζ = (M
dual
ξ,ζ )

dual ⊂ Mdualζ,ξ ⊂ Mξ,ζ.

To see that Mdual dualξ,ζ 6= Mξ,ζ, we make yet another appeal to Stegall’s example
and let Y = c0(`

n
2 ). Then Y

∗ = X has the Schur property, and therefore
Y ∈ Mω1,ω1 ⊂ Mξ,ζ. But

Y ∗∗ = X∗ ∈ {M1,1 ⊂ {Mξ,ζ.

Therefore Y ∈ {Mdual dualξ,ζ .

Direct sums. For 1 ≤ p ≤∞ and classes I,J, we say J is closed under
I-`p sums provided that for any set I and any collection (Ai : Xi → Yi)i∈I ⊂ I
such that supi∈I ‖Ai‖ < ∞, the operator A : (⊕i∈IXi)`p(I) → (⊕i∈IYi)`p(I) lies
in J. The notion of an ideal being closed under I-c0 sums is defined similarly.

We will use the following well-known fact about weakly null sequences in
`1 sums of Banach spaces.



the ξ,ζ-dunford pettis property 175

Fact 3.14. Let I be a set, (Xi)i∈I a collection of Banach spaces, and
(xn)

∞
n=1 =

(
(xi,n)i∈I

)∞
n=1

a weakly null sequence in (⊕i∈IXi)`1(I). Then for
any ε > 0, there exists a subset J ⊂ I such that |I\J| < ∞ and for all n ∈ N,∑

i∈J ‖xi,n‖ < ε.

Proposition 3.15. Fix 0 ≤ ζ < ξ ≤ ω1.

(i) The class Gξ,ζ is closed under Gξ,ζ-`1 sums.

(ii) The class Gξ,ζ is closed under Gξ,ζ-`p sums for 1 < p < ∞ if and
only if ζ > 0.

(iii) The class Gξ,ζ+1 is closed under Gξ,ζ-c0 sums.

(iv) The class Gξ,ζ is not closed under Gξ,ζ-c0 sums.

(v) The class Gξ,ζ is not closed under V-`∞ sums.

Proof. Throughout, let I be a set, (Ai : Xi → Yi)i∈I a collection of oper-
ators such that supi∈I ‖Ai‖ = 1. Let Xp = (⊕i∈IXi)`p(I), Yp = (⊕i∈IYi)`p(I),
and Ap : Xp → Yp the operator such that Ap|Xi = Ai. As usual, p = 0 will
correspond to the c0 direct sum.

(i) Assume Ai ∈ Gξ,ζ for all i ∈ I. Fix (xn)∞n=1 ⊂ X1 ξ-weakly null.
Write xn = (xi,n)i∈I and note that for each i ∈ I, (xi,n)∞n=1 is ξ-weakly null,
so (Aixi,n)

∞
i=1 is ζ-weakly null. Fix ε > 0 and M ∈ [N]. Using Fact 3.14, there

exists a subset J of I such that |I\J| < ∞ and supn
∑

i∈J ‖xi,n‖ < ε/2. Since
(Aixi,n)

∞
n=1 is ζ-weakly null, then there exists F ∈ Sζ ∩ [M]

<N and positive
scalars (an)n∈F summing to 1 such that∥∥∥∥ ∑

n∈F
anxi,n

∥∥∥∥
Yi

<
ε/2

1 + |I \J|
.

for each i ∈ I \J. Then∥∥∥∥A1 ∑
n∈F

anxn

∥∥∥∥ ≤ ∑
i∈I\J

∥∥∥∥ ∑
n∈F

anAixi,n

∥∥∥∥
Yi

+
∑
n∈F

an
∑
i∈I\J

‖xi,n‖Xi

< ε/2 + ε/2 = ε.

Since ε > 0 and M ∈ [N] were arbitrary, (A1xn)∞n=1 is ζ-weakly null.
(ii) Fix 1 < p < ∞. Since `p ∈ {Gξ,0 and K ∈ Gξ,0, Gξ,0 is not closed

under `p sums. It follows by an inessential modification of work from [3] that
for 0 < ζ < ω1, Gξ,ζ is closed under Gξ,ζ-`p sums. More specifically, let



176 r.m. causey

(xn)
∞
n=1 ⊂ BXp be ξ-weakly null and let vn = (‖xi,n‖Xi)i∈I ∈ B`p(I). Assume

(Apxn)
∞
n=1 satisfies

0 < ε ≤ inf
{
‖Apx‖ : F ∈Sζ,x ∈ co(xn : n ∈ F)

}
.

By passing to a subsequence, we may assume vn → v = (vi)i∈I ∈ B`p(I)
weakly, and that vn is a small perturbation of v + bn, where the sequence
(bn)

∞
n=1 consists of disjointly supported vectors in BXp. We may fix a subset

J of I such that |I\J| < ∞ and
(∑

i∈J v
p
i

)
1/p

< ε/3. For k ∈ N, we may first
choose M = (mi)

∞
i=1 ∈ [N] such that Sζ[Ak](M) ⊂Sζ and let

un =
1

k

(n+1)k∑
j=nk+1

xmj.

If k was chosen sufficiently large, then

sup
n

(∑
i∈J
‖ui,n‖

p
Xi

)1/p
< ε/2.

By our choice of M, (Apun)
∞
n=1 also satisfies

ε ≤ inf
{
‖Aun‖ : F ∈Sζ,x ∈ co(xn : n ∈ F)

}
.

Since (Aixi,n)
∞
n=1 is ζ-weakly null, there exist F ∈ Sζ and positive scalars

(an)n∈F summing to 1 such that∥∥∥∥ ∑
n∈F

anAixi,n

∥∥∥∥
Yi

<
ε/2

1 + |I \J|

for each i ∈ I \J. We reach a contradiction as in (i).
(iii) Fix (xn)

∞
n=1 = ((xi,n)i∈I)

∞
n=1 ⊂ BX0 ξ-weakly null. Fix (εn)

∞
n=1 such

that
∑∞

n=1 εn < 1. Since for each i ∈ I, (Aixi,n)
∞
n=1 is ζ-weakly null, we may

recursively select F1 < F2 < ... , Fn ∈ Sζ, positive scalars (aj)j∈∪∞n=1Fn, and
finite subsets ∅ = I0 ⊂ I2 ⊂ . . . of I such that for each n ∈ N,

∑
j∈Fn

aj = 1 , max
i∈In−1

∥∥∥∥Ai ∑
j∈Fn

ajxi,j

∥∥∥∥ < εn and max
i∈I\In

∥∥∥∥ ∑
j∈Fn

ajxi,j

∥∥∥∥ < εn.



the ξ,ζ-dunford pettis property 177

Then since for each n ∈ N, ∪2nm=n+1Fm ∈Sζ+1 for each n ∈ N, we deduce that

sup
i∈I

∥∥∥∥A0 1n
2n∑

m=n+1

∑
j∈Fm

ajxi,j

∥∥∥∥
≤ max

{
max
i∈I\I2n

2n∑
m=n+1

∥∥∥∥ ∑
j∈Fn

ajxi,j

∥∥∥∥,
max

n<m≤2n

{
max

i∈Im\Im−1

1

n

∥∥∥∥Ai ∑
j∈Fm

ajxi,j

∥∥∥∥ + 2n∑
m6=l=n+1

∥∥∥∥Ai ∑
j∈Fl

ajxi,j

∥∥∥∥
}}

≤
1

n
+

∞∑
m=n+1

εm −→
n→∞

0.

(iv) For the ζ = 0 case, c0 = c0(K) yields that Gξ,0 is not closed under
Gξ,0-c0 sums. If ζ = µ + 1, let Fn = An[Sµ] and note that XFn is isomorhpic
to Xµ. If ζ is a limit ordinal, let (ζn)

∞
n=1 be the sequence defining Sζ and

let Fn = Sζn+1. In either the successor or limit case, Sζ = {E : ∃n ≤ E ∈
Fn}. Also, in both cases, XFn ∈ wBSζ ⊂ Gξ,ζ for all n ∈ N. Let xn =
(en,en,en, . . . ,en, 0, 0, . . . ), where (ei)

∞
i=1 simultaneously denotes the basis of

each XFn and en appears n times. Now fix ∅ 6= G ∈Sζ, let m = min G, and
note that G ∈ Fm. Fix (an)n∈G and note that the mth term of the sequence∑

n∈G anxn is
∑

n∈G anen, which has norm
∑

n∈G |an| in XFm. Thus (xn)
∞
n=1

is a weakly null, isometric `
ζ
1-spreading model. By (iii), (xn)

∞
n=1 is ξ-weakly

null (more precisely, we are using the fact that wBSζ+1, and therefore wBSξ,
is closed under wBSζ-c0 sums).

(v) Let En = [ei : i ≤ n] ⊂ Xζ,2, which lies in V. But, analogously to Ste-
gall’s example, `∞(En) contains a complemented copy of Xζ,2. More precisely,
let Z denote the subspace of `∞(En) consisting of those z = (

∑n
i=1 ai,nei)

∞
n=1

such that for all m < n ∈ N and 1 ≤ i ≤ m, ai,m = ai,n (that is, the se-
quences (ai,n)

∞
i=1 are each initial segments of a single scalar sequence (ai)

∞
i=1).

For x =
∑∞

i=1 aiei ∈ Xξ,2, let j(x) = (
∑n

i=1 aiei)
∞
n=1, which is an isometric

embedding of Xξ,2 into Z. Moreover, j is onto. Indeed, since the basis of Xξ,2
is boundedly-complete and if z = (

∑n
i=1 aiei)

∞
n=1 ∈ Z, then

sup
n

∥∥∥∥ n∑
i=1

aiei

∥∥∥∥
ξ,2

= ‖z‖`∞(En) < ∞,

and x :=
∑∞

i=1 aiei ∈ Xξ,2 is such that j(x) = z. Thus Z is isometrically iso-
morphic to Xξ,2. Let U be a free ultrafilter on N and for z = (

∑n
i=1 ai,nei)

∞
n=1 ∈



178 r.m. causey

`∞(En), let

Pz = weak lim
n∈U

n∑
i=1

ai,nei ∈ Xξ,2.

This limit is well-defined, since (
∑n

i=1 ai,nei)
∞
n=1 is bounded in the reflexive

space Xξ,2. Then Z is an isometric copy of Xξ,2 which is 1-complemented in
`∞(En) via the map jP . Since Xζ ∈ {Gξ,ζ, `∞(En), while each En is finite
dimensional and therefore a Schur space, we reach the desired conclusion.

Proposition 3.16. Fix 0 < ζ,ξ ≤ ω1.

(i) The class Mξ,ζ is closed under c0 and `1 sums.

(ii) The class Mξ,ζ is not closed under `p sums for any 1 < p < ∞.
(iii) The class Mξ,ζ is not closed under `∞ sums.

Proof. Item (i) follows from inessential modifications of the fact that the
class of spaces with the Dunford-Pettis property are closed under c0 and `1
sums, using Fact 3.14.

Item (ii) follows from the fact that `p = `p(K), 1 < p < ∞, does not lie in
M1,1, while K ∈ V.

Item (iii) again follows from Stegall’s example, which is an `∞ sum of Schur
spaces which contains a complemented copy of `2, and therefore does not lie
in M1,1.

4. Space ideals

Hereditary properties. Let us say a Banach space X is hereditarily
Mξ,ζ provided that any subspace Y of X lies in Mξ,ζ. For convenience, let us
say a sequence (xn)

∞
n=1 in a Banach space is a c

ω1
0 -spreading model provided

that it is equivalent to the canonical c0 basis.

Proposition 4.1. For 0 < ξ,ζ ≤ ω1, X is hereditarily M ξ,ζ if and only if
every seminormalized, ξ-weakly null sequence in X has a subsequence which
is a c

ζ
0-spreading model.

Remark 4.2. Since for ξ < ω1, a seminormalized, weakly null sequence
is either ξ-weakly null or has a subsequence which is an `

ξ
1-spreading model,

Proposition 4.1 can be restated as follows: For 0 < ξ < ω1 and 0 < ζ ≤ ω1, X
is hereditarily Mξ,ζ if and only if every seminormalized, weakly null sequence



the ξ,ζ-dunford pettis property 179

in X has a subsequence which is either an `
ξ
1-spreading model or c

ζ
0-spreading

model.

Remark 4.3. The proof below requires a result due to Elton concerning
near unconditionality. To the best of our knowledge, this result is only known
to hold for real scalars. We include a proof of the requisite result in the com-
plex case, which is an easy modification of what are now standard arguments
regarding partial unconditionality. We relegate the details of the complex case
to the final section of this work.

Proof of Proposition 4.1. Suppose that every normalized, ξ-weakly null
sequence in X has a subsequence which is a c

ζ
0-spreading model. Let Y be any

subspace of X. Suppose that (yn)
∞
n=1 ⊂ Y is ξ-weakly null, (y

∗
n)
∞
n=1 ⊂ Y

∗ is
weakly null, and infn |y∗n(yn)| = ε > 0. By passing to a subsequence, we may
assume (yn)

∞
n=1 is a c

ζ
0-spreading model. By Proposition 2.20 applied with

F = Sζ if ζ < ω1 and F = [N]<N if ζ = ω1, we may pass to a subsequence,
relabel, and find some C < ∞ such that

inf

{∥∥∥∥ ∑
n∈F

any
∗
n

∥∥∥∥ : F ∈F, ∑
n∈F
|an| = 1

}
≥

ε

2C
.

This yields that (y∗n)
∞
n=1 is not ζ-weakly null, and Y ∈ Mξ,ζ.

For the converse in the ζ < ω1 case, suppose that (xn)
∞
n=1 is a semi-

normalized, ξ-weakly null sequence in X having no subsequence which is a
c
ζ
0-spreading model. Assume that (xn)

∞
n=1 is a basis for Y = [xn : n ∈ N] and

let (x∗n)
∞
n=1 ⊂ Y

∗ denote the coordinate functionals. For M = (mn)
∞
n=1 ∈ [N],

let YM = [xmn : n ∈ N]. By hypothesis, there does not exist L ∈ [N] such that
(xn)n∈L is a c

ζ
0-spreading model. By [1, Theorem 3.9], there exists M ∈ [N]

such that for each L ∈ [M], (x∗n|YM )n∈L is not an `
ζ
1-spreading model. Then

(x∗n|YM )n∈M is ζ-weakly null in Y
∗
M . Since (xn)n∈M is ξ-weakly null in YM and

x∗n(xn) = 1 for all n ∈ M, YM ∈ {Mξ,ζ.
For the ζ = ω1 case of the converse, this is an inessential modification

of Elton’s characterization of the hereditary Dunford-Pettis property, with a
slight comment in the complex case to be discussed in the final section. For
the sake of completeness, we record the argument as given in [13, Page 28].
Suppose that (xn)

∞
n=1 ⊂ X is ξ-weakly null having no subsequence equivalent

to the c0 basis. By passing to a subsequence, we may assume (xn)
∞
n=1 is

basic with coordinate functionals (x∗n)
∞
n=1 and for any subsequence (yn)

∞
n=1 of



180 r.m. causey

(xn)
∞
n=1 and (an)

∞
n=1 ∈ `∞ \ c0, limn‖

∑n
i=1 aiyi‖ = ∞. Now if

Pk : [xn : n ∈ N] → [xn : n ≤ k]

denotes the basis projections, for any x∗∗ ∈ X∗∗, then

sup
n

∥∥∥∥ n∑
i=1

x∗∗(x∗i )xi

∥∥∥∥ ≤‖x∗∗‖sup
n
‖Pn‖ < ∞.

Therefore (x∗∗(x∗n))
∞
n=1 ∈ c0, and (x

∗
n)
∞
n=1 is ω1-weakly null. Since x

∗
n(xn) = 1

for all n ∈ N, [xn : n ∈ N] ∈ {Mξ,ω1 .

Remark 4.4. For each 0 ≤ ξ < ω1, Xωξ is hereditarily Mωξ,ω1 , since every
seminormalized, weakly null sequence in Xωξ has either a subsequence which

is an `ω
ξ

1 -spreading model or a subsequence equivalent to the canonical c0
basis.

In [5], for each 0 ≤ ξ < ω1, a reflexive Banach space Xω
ξ

0,1 with 1-uncondi-
tional basis was defined such that every seminormalized, weakly null sequence
has a subsequence which is either an `ω

ξ

1 -spreading model or a c
1
0-spreading

model, and both alternatives occur in every infinite dimensional subspace.
Thus Xω

ξ

0,1 furnish reflexive examples of members of Mωξ,1.

For 0 ≤ ζ,ξ ≤ ω1. Let us say that X is hereditary by quotients Mξ,ζ if
every quotient of X is a member of Mξ,ζ.

Theorem 4.5. Fix 0 < γ ≤ ω1. For a Banach space X, the following are
equivalent.

(i) X∗ ∈ Vγ.
(ii) X∗ is hereditarily Mγ,ω1 .

(iii) X is hereditary by quotients Mω1,γ.

(iv) X ∈ Mω1,γ and `1 6↪→ X.

Proof. (i)⇒(ii) Assume (i) holds. If (x∗n)∞n=1 ⊂ X
∗ is γ-weakly null, it is

norm null. Thus for any subspace Y of X∗, any γ-weakly null (yn)
∞
n=1 ⊂ Y ,

and any bounded sequence (y∗n)
∞
n=1 ⊂ Y

∗, limn y
∗
n(yn) = 0.

(ii)⇒(iii) Assume (ii) holds. For any quotient X/N of X, (X/N)∗ =
N⊥ ≤ X∗, so X/N ∈ Mdualγ,ω1 ⊂ Mω1,γ.

(iii)⇒(iv) Assume (iii) holds. If `1 ↪→ X, then `2 is a quotient of X, which
is a contradiction. Thus `1 6↪→ X. Since X is a quotient of itself, X ∈ Mω1,γ.



the ξ,ζ-dunford pettis property 181

(iv)⇒(i) Assume (iv) and ¬(i). Since X∗ ∈ {Vγ, there exists a seminor-
malized, γ-weakly null sequence (x∗n)

∞
n=1 in X

∗. Fix 0 < ε < 1
2

infn‖x∗n‖. For
each n ∈ N, we may fix xn ∈ BX such that x∗n(xn) > 2ε. By passing to a
subsequence and relabeling, we may assume that for all m < n, |x∗n(xm)| < ε.
Since `1 6↪→ X, we may also assume that (xn)∞n=1 is weakly Cauchy. Then with
y∗n = x

∗
2n and yn = x2n − x2n−1, (y

∗
n)
∞
n=1 is γ-weakly null, (yn)

∞
n=1 is weakly

null, and infn |y∗n(yn)| ≥ ε.

Distinctness of space ideals. We recall that, given an operator
ideal I, the associated space ideal I consists of all Banach spaces X such
that IX ∈ I. We showed in Section 3 that for any 0 ≤ ζ < ξ ≤ ω1 and
0 ≤ α < β ≤ ω1, Gξ,ζ = Gβ,α if and only if ζ = α and ξ = ζ. Our next goal is
to show that this is not true for the space ideals, due to the idempotence of
identity operators. We recall the result from [12] that a Banach space X lies in
Vζ for some ω

ξ < ζ < ωξ+1 if and only if X lies in Vζ for every ω
ξ < ζ < ωξ+1,

which is a consequence of considering blocks of blocks. We prove analogous
results below. We need the following result for blocks of blocks.

Proposition 4.6. Let X,Y,Z be operators, α,β,ζ countable ordinals,
and assume B ∈ Gβ+ζ,ζ and A ∈ Gα+ζ,ζ. Then AB ∈ Gα+β+ζ,ζ.

Proof. By Corollary 3.1, B ∈ Gα+β+ζ,α+ζ. Thus if (xn)∞n=1 is α + β + ζ-
weakly null, it is sent by B to a sequence which is α + ζ-weakly null, which is
sent by A to a sequence which is ζ-weakly null.

Corollary 4.7. For a Banach space X and ζ < ω1, let g ζ(X) = ω1 if
X ∈ Gω1,ζ, and otherwise let gζ(X) be the minimum ordinal ξ < ω1 such that
X ∈ {G ξ+ζ,ζ (noting that such a ξ must exist). Then there exists γ ≤ ω1
such that gζ(X) = ω

γ.

Proof. Note that gζ(X) > 0. Fix α,β < gζ(X). Then IX ∈ Gβ+ζ,ζ and
IX ∈ Gα+ζ,ζ. By Proposition 4.6, IX ∈ Gα+β+ζ,ζ. Thus we have shown that
if α,β < gζ(X), α + β < gζ(X). Since 0 < gζ(X) ≤ ω1, standard facts about
ordinals yield that there exists γ ≤ ω1 such that gζ(X) = ωγ.

For the following theorem, note that if ωξ < λ(ζ), then ωξ + ζ = ζ, so
Gωξ+ζ,ζ = L. This is the reason for the omission of this trivial case.



182 r.m. causey

Theorem 4.8. Fix 0 ≤ ζ < ω1 and ξ < ω1 such that ωξ ≥ λ(ζ). Then

∅ 6= {Gωξ+ζ,ζ ∩
⋂
η<ωξ

Gη+ζ,ζ.

Proof. It was shown in [12] that for any Banach space Y with a normalized,
bimonotone basis and 0 < ξ < ω1, there exists a Banach space Z (there
denoted by Zξ(EY )) such that Z has a normalized, bimonotone basis, Y is
a quotient of Z, Z ∈ ∩η<ωξVη, and if (yn)∞n=1 is an ω

ξ-weakly null sequence

in Y , then there exists an ωξ-weakly null sequence (zn)
∞
n=1 in Z such that

qzn = yn for all n ∈ N.
If ζ = 0, we consider Z as above with Y = c0. This space lies in

{Vωξ ∩
⋂
η<ωξ

Vη = {Gωξ,0 ∩
⋂
η<ωξ

Gη,0.

This completes the ζ = 0 case. For the remainder of the proof, we consider
ζ > 0.

Suppose that ξ = 0. Then since 1 = ωξ ≥ λ(ζ) ≥ 1, ζ is finite. Futhermore,
η + ζ = ζ for any η < λ(ζ), since the only such η is 0. Then X = Xζ is easily
seen to satisfy the conclusions. For the remainder of the proof, we assume
0 < ξ < ω1.

If λ(ζ) = ωξ, then for every η < ωξ, η +ζ = ζ. In this case, membership in⋂
η<ωξ Gη+ζ,ζ = Ban is automatic. In this case, Xζ ∈ {Gωξ+ζ,ζ is the example

we seek.
We consider the remaining case, 0 < ζ,ξ and λ(ζ) < ωξ. Note that this

implies that ζ < ωξ. We use a technique of Ostrovskii from [19]. If λ(ζ) is
finite, then it is equal to 1. In this case, let Y = c0. If λ(ζ) is infinite, then
it is a limit ordinal. In this case, let (λn)

∞
n=1 be the sequence used to define

Sλ(ζ). Let Y be the completion of c00 with respect to the norm

‖x‖ = sup
n∈N

2−n‖x‖λn.

Note that the formal inclusions I1 : Xζ → Xλ(ζ), I2 : Xλ(ζ) → Y are bounded.
The first is bounded by the almost monotone property. For n ∈ N and E ∈
Sλn, F = E ∩ [n,∞) ∈Sλ(ζ). Therefore for x ∈ c00,

2−n‖Ex‖`1 ≤ 2
−n((n− 1)‖x‖c0 + ‖Fx‖`1) ≤ n2−n‖x‖λ(ζ) ≤ 2−1‖x‖λ(ζ).

Let us also note that a bounded block sequence (xn)
∞
n=1 in Xζ is ζ-weakly null

if and only if limn‖x‖β = 0 for every β < λ(ζ) if and only if (I2I1xn)∞n=1 is



the ξ,ζ-dunford pettis property 183

norm null in Y . We have already established the equivalence of the first two
properties. Let us explain the equivalence of the last two properties. First, if
(I2I1xn)

∞
n=1 is norm null in Y , then for any β < λ(ζ), we can fix k such that

β < λk and note that

lim
n
‖xn‖β ≤ c lim

n
‖xn‖λk ≤ c2

k lim
n
‖xn‖Y = 0.

Here, c is the norm of the formal inclusion of Xλk into Xβ. For the reverse
direction, suppose (xn)

∞
n=1 ⊂ BXζ and limn‖xn‖β = 0 for all β < λ(ζ). Then

lim sup
n
‖I2I1y‖

≤ inf

{
max

{
lim sup

n

k∑
m=1

‖xn‖λm, sup
n>k

2−n‖I2‖‖I1‖
}

: k ∈ N

}
= 0.

Let i = I2I1 and let Z be as described in the first paragraph with this choice
of Y . Let q : Z → Y be the quotient map the existence of which was indicated
above. Let W = Z ⊕1 Xζ and let T : W → Y be given by T(z,x) = ix− qz.
Let X = ker(T). Since we are in the case ζ < ωξ, standard properties of
ordinals yield that for η < ωξ, η + ζ < ωξ. Suppose that (zn,xn)

∞
n=1 ⊂ X

is η + ζ-weakly null. Then since Z ∈ Vη, ‖zn‖ → 0. From this it follows
that (ixn)

∞
n=1 = (qzn)

∞
n=1 is norm null. Therefore (ixn)

∞
n=1 is norm null, and

(xn)
∞
n=1 is ζ-weakly null in Xζ. Therefore (zn,xn)

∞
n=1 is ζ-weakly null in X.

We last show that X ∈ {Gωξ+ζ,ζ. To that end, let us first note that the basis
of Y is λ(ζ)-weakly null. This is obvious if λ(ζ) = 1 and Y = c0. For the
case in which λ(ζ) is infinite, the space Y is a mixed Schreier space as defined
in [12], where it was shown that the basis of Y is λ(ζ)-weakly null. By the
properties of Z and q, since λ(ζ) ≤ ωξ, there exists an ωξ-weakly null sequence
(zn)

∞
n=1 in Z such that qzn = en, where (en)

∞
n=1 simultaneously denotes the

bases of Y and Xζ. Also note that (en)
∞
n=1 is ζ + 1-weakly null in Xζ. Since

ωξ + ζ ≥ ζ + ωξ ≥ ζ + 1,

(en)
∞
n=1 is ω

ξ + ζ-weakly null in Xζ. Therefore (zn,en)
∞
n=1 is ω

ξ + ζ-weakly
null in X. However, since (en)

∞
n=1 is not ζ-weakly null in Xζ, (zn,en)

∞
n=1 is

not ζ-weakly null in X. Therefore X ∈ {Gωξ+ζ,ζ.

Corollary 4.9. The classes wBSζ,Gωγ+ζ,ζ, ζ,γ < ω1,ω
γ ≥ λ(ζ), are

distinct.



184 r.m. causey

Theorem 4.10. The classes Gζ+ωγ,ζ, 0 ≤ ζ < ω1, 0 ≤ γ ≤ ω1, are
distinct.

Proof. We first recall that if ζ < ω1 and γ ≤ γ1 ≤ ω1, Gζ+ωγ1,ζ ⊂ Gζ+ωγ,ζ.
Thus the statement that these two classes are distinct is equivalent to saying
that the former is a proper subset of the latter.

We will show that the classes are distinct. Fix 0 ≤ ζ,ζ1 < ω1 and 0 ≤
γ,γ1 ≤ ω1. If ζ < ζ1,

Xζ ∈ wBSζ1 ∩{Gζ+ωγ,ζ ⊂ Gζ1+ωγ1,ζ1 ∩{Gζ+ωγ,ζ.

By symmetry, if ζ1 < ζ, Gζ+ωγ,ζ 6= Gζ1+ωγ1,ζ1 . Thus if ζ 6= ζ1, Gζ+ωγ,ζ 6=
Gζ1+ωγ1,ζ1 .

In order to complete the proof that the classes are distinct, it suffices to
assume that γ1 < γ ≤ ω1 and exhibit some Banach space Z ∈ Gζ+ωγ1,ζ ∩
{Gζ+ωγ,ζ. We first claim that it is sufficient to prove the case γ < ω1. This is
because if we prove that Gζ+ωγ,ζ ( Gζ+ωγ1,ζ whenever 0 ≤ γ1 < γ < ω1, then
for any 0 ≤ γ1 < ω1,

Gζ+ωω1,ζ = Gω1,ζ ⊂ Gζ+ωγ1+1,ζ ( Gζ+ωγ1,ζ.

Fix 0 < γ < ω1 and let (γn)
∞
n=1 be the sequence defining Sωγ . Fix a

sequence (ϑn)
∞
n=1 such that ϑ :=

∑∞
n=1 ϑn < 1. Given a Banach space E

with normalized, 1-unconditional basis, we define norm on [·] on c00 by letting
| · |0 = ‖ ·‖E,

|x|k+1,n = sup
{
ϑn

d∑
i=1

|Eix|k : n ∈ N,E1 < · · · < Ed, (min Ei)di=1 ∈Sγn
}
,

|x|k+1 = max
{
|x|k,

( ∞∑
n=1

|x|2k+1,n
)1/2}

,

[x] = lim
k
|x|k and [x]n = lim

k
|x|k,n.

Let us denote the completion of c00 with respect to this norm by Zγ(E). The
norm [·] on Zγ(E) satisfies the following

[z] = max

{
‖z‖E,

( ∞∑
n=1

[z]2n

)1/2}
.



the ξ,ζ-dunford pettis property 185

This construction is a generalization of a Odell-Schlumprecht construction.
We will apply the construction with E = Xζ. It is a well known fact of such
constructions that, since the basis of Xζ is shrinking, so is the basis of Zγ(Xζ)
(see, for example, [12]). It was shown in [12] that if (zn)

∞
n=1 is any seminor-

malized block sequence in Zγ(Xζ), then

(a) (zn)
∞
n=1 is not β-weakly null for any β < ω

γ,

(b) (zn)
∞
n=1 is ω

γ-weakly null in Zγ(Xζ) if and only if it is ω
γ-weakly

null in Xζ.

We will show that Zγ(Xζ) ∈ ∩β<ωγGζ+β,ζ, and in particular Zγ(Xζ) ∈
Gζ+ωγ1,ζ, while Zγ(Xζ) ∈ {Gζ+ωγ,ζ. This will complete the proof of the dis-
tinctness of the classes.

We prove that Zγ(Xζ) ∈ {Gζ+ωγ,ζ. As remarked above, the basis is shrink-
ing and normalized, and so it is weakly null. If it were not ζ +ωγ-weakly null,
there would exist some (mn)

∞
n=1 ∈ [N] and ε > 0 such that

ε ≤ inf
{

[z] : F ∈Sωγ [Sζ],z ∈ co(emn : n ∈ F)
}
.

But by Theorem 2.17, we may choose F1 < F2 < ... , Fi ∈ Sζ, and positive
scalars (ai)i∈∪∞n=1Fn such that

∑
i∈Fn ai = 1 and the sequence (zn)

∞
n=1 defined

by zn =
∑

i∈Fn aiemi is equivalent to the c0 basis in Xζ. But since

ε ≤ inf
{

[z] : F ∈Sωγ [Sζ],z ∈ co(emn : n ∈ F)
}
,

(zn)
∞
n=1 is an `

ωγ
1 -spreading model in Zγ(Xζ), contradicting item (b) above.

Therefore the canonical Zγ(Xζ) basis is ζ +ω
γ-weakly null. But it is evidently

not ζ-weakly null, and Zγ(Xζ) ∈ {Gζ+ωγ,ζ.
Now let us show that Zγ(Xζ) ∈ ∩β<ωγGζ+β,ζ. First consider the case

λ(ζ) < ωγ, which is equivalent to ζ + β < ωγ for all β < ωγ. In this case,{
ζ + β : β < ωγ

}
= [0,ωγ).

It therefore follows from property (a) above that

Zγ(Xζ) ∈
⋂
β<ωγ

Gβ,0 =
⋂
β<ωγ

Gζ+β,0 ⊂
⋂
β<ωγ

Gζ+β,ζ.

Let us now treat the case λ(ζ) ≥ ωγ. Write

ζ = λ(ζ) + µ



186 r.m. causey

and note that

µ + ωγ ≤ µ + λ(ζ) ≤ λ(ζ) + µ = ζ.

We claim that if (zn)
∞
n=1 is a seminormalized block sequence in Zγ(Xζ) which

is not ζ-weakly null in Zγ(Xζ), then there exists β < λ(ζ) such that

lim sup
n
‖zn‖β > 0.

To see this, suppose that for every β < λ(ζ), limn‖zn‖β = 0, but (zn)∞n=1
is not ζ-weakly null in Zγ(Xζ). Then, by Proposition 2.16(ii), by passing
to a subsequence and relabeling, we may assume (zn)

∞
n=1, when treated as

a sequence in Xζ, is dominated by a subsequence (emi)
∞
i=1 of the Xµ basis,

and (zn)
∞
n=1, when treated as a sequence in Zγ(Xζ), is an `

ζ
1-spreading model.

Since ζ ≥ µ + ωγ, we may, after passing to a subsequence again, assume

0 < ε ≤ inf
{

[z] : F ∈Sωγ [Sµ],z ∈ co(zn : n ∈ F)
}
.

We may select F1 < F2 < ... , Fi ∈ Sµ, and positive scalars (ai)i∈∪∞n=1Fn
such that

∑
i∈Fn ai = 1 and (

∑
i∈Fn aiemi)

∞
n=1 ⊂ Xµ is equivalent to the

canonical c0 basis (again using Theorem 2.17 as in the previous case). Since
(zn)

∞
n=1 ⊂ Xζ is dominated by (emi)

∞
i=1 ⊂ Xµ, (

∑
i∈Fn aizi)

∞
n=1 is WUC in Xζ.

But since

0 < ε ≤ inf
{

[z] : F ∈Sωγ [Sµ],z ∈ co(zn : n ∈ F)
}
,

(
∑

i∈Fn aizi)
∞
n=1 must be an `

ωγ
1 -spreading model in Zγ(Xζ), contradicting (b)

above. This proves the claim from the beginning of the paragraph. Now
suppose that (zn)

∞
n=1 is a weakly null sequence in Zγ(Xζ) which is not ζ-

weakly null. Then by the claim combined with Corollary 2.19, (zn)
∞
n=1 is not

ζ-weakly null in Xζ. After passing to a subsequence, we may assume (zn)
∞
n=1

is an `
ζ
1-spreading model in Xζ. Assume that

0 < ε ≤ inf
{

[z] : F ∈Sζ,z ∈ co(zn : n ∈ F)
}
.

Now fix n ∈ N and F ∈Sγn[Sζ] and scalars (ai)i∈F . By definition of Sγn[Sζ],
there exist F1 < · · · < Fd such that F = ∪dj=1Fj, ∅ 6= Fj ∈ Sζ, and
(min Fj)

d
j=1 ∈Sγn. Let Ei = supp(zi) and let Ij = ∪i∈FjEi. Since

min Ii = min supp(zmin Fj ) ≥ min Fj,



the ξ,ζ-dunford pettis property 187

(min Ij)
d
j=1 is a spread of (min Fj)

d
j=1, so that (min Ij)

d
j=1 ∈Sγn. Therefore

[∑
i∈F

aizi

]
≥
( ∞∑
k=1

[∑
i∈F

aizi

]2
k

)1/2
≥
[∑
i∈F

aizi

]
n

≥ ϑn
d∑
j=1

[
Ij
∑
i∈F

aizi

]

= ϑn

d∑
j=1

[ ∑
i∈Fj

aizi

]
≥ εϑn

d∑
j=1

∑
i∈Fj

|ai| = εϑn
∑
i∈F
|ai| .

Thus

0 < inf
{

[z] : F ∈Sγn[Sζ],x ∈ co(zn : n ∈ F)
}
.

From this it follows that (zi)
∞
i=1 is not ζ + γn-weakly null. Since this holds for

any n ∈ N and supn γn = ωγ, (zi)∞i=1 is not ζ + β-weakly null for any β < ω
γ.

Thus by contraposition, for any β < ωγ, any ζ + β-weakly null sequence in
Zγ(Xζ) is ζ-weakly null, from which it follows that Zγ(Xζ) ∈ ∩β<ωγGζ+β,ζ.
This completes the proof of the distinctness of these classes.

Remark 4.11. For ξ,η < ω1 and δ,ζ ≤ ω1 with η 6= ζ, the classes Gωξ+ζ,ζ,
Gη+ωδ,η are not equal. Indeed, if η < ζ, Xη ∈ Gωξ+ζ,ζ ∩ {Gη+ωδ,η. This is
because every sequence in Xη is η + 1-weakly null, and therefore ζ-weakly
null. However, the basis of Xη is η + 1-weakly null, and therefore η + ω

δ-
weakly null, but not η-weakly null. Now if ζ < η, either ωξ + ζ > ζ or
ωξ + ζ = ζ. If ωξ + ζ > ζ, Xζ ∈ Gη+ωδ,η ∩ {Gωξ+ζ,ζ. If ωξ + ζ = ζ, then
Gωξ+ζ,ζ = Ban 6= Gη+ωδ,η.

We next wish to discuss how the classes Gωξ+ζ,ζ can be compared to the
classes Gζ+ωδ,ζ. In particular, we will show that they are equal if and only

if ωξ + ζ = ζ + ωδ. If ζ = 0, then Gωξ+ζ,ζ = Vωξ and Gζ+ωδ,ζ = Vωδ . Then
Vmax{ωξ,ωδ} ⊂ Vmin{ωξ,ωδ}, with proper containment if and only if ξ 6= δ.

Now for 0 < ζ < ω1, write ζ = ω
α1n1 + · · · + ωαlnl l,n1, . . . ,nl ∈ N,

α1 > · · · > αl. Let us consider several cases. For convenience, let α = α1 and
n = n1.

Case 1: ξ < α. Then ωξ + ζ = ζ and Gωξ+ζ,ζ = Ban 6= Gζ+ωδ,δ.
For the remaining cases, we will assume ξ ≥ α, which implies that ωξ + ζ

> ζ.

Case 2: ωξ +ζ < ζ+ωδ. Then there exists β < ωδ such that ωξ +ζ = ζ+β.
Then the space Zδ(Xζ) from Theorem 4.10 lies in

{Gζ+ωδ,ζ ∩Gζ+β,ζ = {Gζ+δ,ζ ∩Gωξ+ζ,ζ.



188 r.m. causey

Case 3: ωξ + ζ = ζ + ωδ. In this case, of course Gωξ+ζ,ζ = Gζ+ωδ,ζ. By

considering the Cantor normal forms of ωξ + ζ and ζ + ωδ, it follows that
equality can only hold in the case that ξ = δ = α and ζ = ωαn, in which case
ωξ + ζ = ωα(n + 1) = ζ + ωδ.

For the remaining cases, we will assume ωξ + ζ > ζ + ωδ. Note that this
implies δ ≤ ξ. Indeed, if δ > ξ, then since we are in the case ξ ≥ α, it follows
that ωδ > ωξ,ζ. By standard properties of ordinals, ωξ > ωξ + ζ. Therefore
for the remaining cases, ωξ + ζ > ζ + ωδ and α,δ ≤ ξ.

Case 4: δ = ξ > α. Then the space Xωδ lies in {Gωξ+ζ,ζ ∩Gζ+ωδ,ζ. To see
this, note that since δ > α, ζ+ωδ = ωδ. Moreover, we have already shown that
any ωδ-weakly null sequence in Xωδ has the property that every subsequence
has a further WUC subsequence. Thus any ωδ-weakly null sequence in Xωδ is
1-weakly null, and Xωδ ∈ Gζ+ωδ,ζ. But of course the basis of Xωδ shows that
it does not lie in Gωξ+ζ,ζ ⊂ Gωδ+1,ωδ .

Case 5: ξ = α > δ. The space Zξ(Xζ), as defined in Theorem 4.10, lies in
{Gωξ+ζ,ζ ∩Gζ+ωδ,ζ. To see this, let us note that

Zξ(Xζ) ∈ {Gζ+ωξ,ζ ∩
⋂
γ<ωξ

Gζ+γ,ζ.

Since ξ ≥ α, ωξ +ζ ≥ ζ +ωξ, and Gωξ+ζ,ζ ⊂ Gζ+ωξ,ζ and Zξ(Xζ) ∈ {Gζ+ωξ,ζ ⊂
{Gωξ+ζ,ζ.Since ω

δ < ωξ, Zξ(Xζ) ∈ Gζ+ωδ,ζ.
Case 6: ξ > α,δ. Then the space Zξ(c0), as shown in [12], lies in wBSωξ ∩⋂

γ<ωξ Vγ. Furthermore, the basis of the space is normalized, weakly null.

Therefore the basis is ωξ-weakly null but not γ-weakly null for any γ < ωξ.
Therefore Zξ(c0) ∈ {Gωξ,ζ ⊂ {Gωξ+ζ,ζ. However, since α,δ < ξ, ζ + ωδ < ξ,
and Zξ(c0) ∈ Vζ+ωδ ⊂ Gζ+ωδ,ζ. Therefore Zξ(c0) lies in {Gωξ+ζ,ζ ∩Gζ+ωδ,ζ.

Case 7: ξ = α = δ. In this case, we can write ζ = ωαn + µ, where
µ = ωα2n2 + · · · + ωαlnl. Note that in this case, µ > 0, since otherwise
we would be in the case ωξ + ζ = ζ + ωδ. Then the space Xωα(n+1) lies in
{Gωα+ζ,ζ ∩Gζ+ωα,ζ. To see this, note that the canonical basis of Xωα(n+1) is

ωα(n + 1) + 1 ≤ ωα(n + 1) + µ = ωα + ζ

weakly null, but it is not ωα(n+ 1) = ωαn+ωα-weakly null, and therefore not
ζ-weakly null. Thus Xωα(n+1) ∈ {Gωα+ζ,ζ. However, if (xn)∞n=1 is ω

α(n + 1)-
weakly null, then by Theorem 2.17, every subsequence of (xn)

∞
n=1 has a further

subsequence which is dominated by a subsequence of the Xωαn basis. This
means (xn)

∞
n=1 is ω

αn + 1-weakly null. Since ωαn + 1 ≤ ωαn + µ = ζ and
ζ + ωα = ωα(n + 1), Xωα(n+1) ∈ Gζ+ωα,ζ.



the ξ,ζ-dunford pettis property 189

Our next goal will be to prove a fact regarding the distinctness of the space
ideals Mξ,ζ analogous to those proved above for the classes Gξ,ζ.

Remark 4.12. If ξ,η are ordinals such that ωξ + 1 < η < ωξ+1, then there
exist ordinals α,γ < η such that γ > 1 and α + γ = η. This is obvious if
ξ = 0, since since η > 2 is finite and we may take η = 1 + (η− 1) in this case.
Assume 0 < ξ. Then there exist n ∈ N and δ < ωξ such that η = ωξn + δ. If
n > 1, we may take α = ωξ(n−1) and γ = ωξ. Now if n = 1, then δ > 1, and
we may take α = ωξ and γ = δ.

Theorem 4.13. Fix 0 ≤ ξ < ω1 and 0 < ν ≤ ω1. Let X be a Banach
space.

(i) X is hereditarily Mµ,ν for some ω
ξ < µ < ωξ+1 if and only if X is

hereditarily Mµ,ν for every ω
ξ < µ < ωξ+1.

(ii) X is hereditarily Mν,µ for some ω
ξ < µ < ωξ+1 if and only if X is

hereditarily Mν,µ for every ω
ξ < µ < ωξ+1.

Proof. (i) Seeking a contradiction, suppose that X is hereditarily Mµ,ν
for some but not all µ ∈ (ωξ,ωξ+1). Let η be the minimum ordinal µ such that
X is not hereditarily Mµ,ν. Note that, since the classes Mµ,ν are decreasing
with µ and X is hereditarily Mµ,ν for some ω

ξ < µ < ωξ+1, it follows that
ωξ + 1 < η. We can write η = α + γ for some α,γ < η with γ > 1. Since X is
not hereditarily Mη,ν, there exists a seminormalized, η-weakly null sequence
(xn)

∞
n=1 in X which has no subsequence which is a c

ν
0 -spreading model. Since

α + 1 < α + γ, the minimality of η implies that X is hereditarily Mα+1,ν,
which means (xn)

∞
n=1 has a subsequence which is an `

α+1
1 -spreading model.

By Corollary 2.12(i), there exists a convex block sequence (yn)
∞
n=1 of (xn)

∞
n=1

which is an `11-spreading model and which is γ-weakly null. But since (yn)
∞
n=1

is an `11-spreading model, it can have no subsequence which is a c
ν
0 -spreading

model. Since γ < η, (yn)
∞
n=1 witnesses that X is not hereditarily Mγ,ν, con-

tradicting the minimality of η.

(ii) Arguing as in (i), let us suppose we have ωξ + 1 < η < ωξ+1 such
that X is hereditarily Mν,µ for every µ < η but X is not hereditarily Mν,η.
Then there exists a ν-weakly null (xn)

∞
n=1 ⊂ X which has no subsequence

which is a c
η
0-spreading model. Write η = α + γ, α,γ < η, γ > 1. By passing

to a subsequence, we may assume (xn)
∞
n=1 is a c

α+1
0 -spreading model. By

Corollary 2.12(ii), there exists a blocking (yn)
∞
n=1 of (xn)

∞
n=1 which is a c

1
0-

spreading model and has no subsequence which is a c
γ
0 -spreading model. Since



190 r.m. causey

(yn)
∞
n=1 is a c

1
0-spreading model, it is 1-weakly null, and therefore ν-weakly

null. But (yn)
∞
n=1 has no subsequence which is a c

γ
0 -spreading model. Since

γ < η, this contradicts the minimality of η.

Remark 4.14. The previous theorem yields that for a fixed 0 < ζ ≤ ω1 and
0 ≤ ξ < ω1, a given Banach space X may lie in {Mωξ,ζ ∩

⋂
η<ωξ Mη,ζ. That is,

the first ordinal η for which X fails to lie in Mη,ζ is of the form ω
ξ, 0 ≤ ξ < ω1.

But it also allows for X to lie in Mωξ,ζ and fail to lie in Mωξ+1,ζ. Let us make
this precise: For 1 ≤ ζ ≤ ω1, let mζ(X) = ω1 if X ∈ Mω1,ζ and otherwise let
mζ(X) be the minimum η such that X ∈ {Mη,ζ. Let m∗ζ(X) = ω1 if X ∈ Mζ,ω1 ,
and otherwise let m∗ζ(X) be the minimum η such that X ∈ {Mζ,η. Then the
preceding theorem yields that for any 1 ≤ ζ ≤ ω1 and any Banach space X,
there exists 0 ≤ ξ ≤ ω1 such that either mζ(X) = ωξ or mζ(X) = ωξ + 1, and
a similar statement holds for m∗ζ.

Contrary to the Gξ,ζ case, both alternatives can occur for both mζ and m
∗
ζ.

For example, for 0 < ξ < ω1, our spaces Zξ(c0) lie in
⋂
η<ωξ Vη, and therefore

lie in ⋂
η<ωξ

Mη,ω1 ⊂
⋂
ζ≤ω1

⋂
η<ωξ

Mη,ζ.

However, the basis of this space is ωξ-weakly null, and the dual basis is 1-
weakly null, so

Zξ(c0) ∈ {Mωξ,1 ⊂
⋂

1≤ζ≤ω1

Mωξ,ζ.

Thus for every 1 ≤ ζ ≤ ω1, mζ(Zξ(c0)) = ωξ. Since these spaces have a
shrinking, asymptotic `1 basis, they are reflexive. From this it follows that for
all 1 ≤ ζ ≤ ω1, m∗ζ(Zξ(c0)

∗) = ωξ. For the ξ = 0 case, mζ(`2) = m
∗
ζ(`2) = 1 =

ω0 for every 1 ≤ ζ ≤ ω1.
However, as we have already seen, for any 0 ≤ ξ < ω1, mζ(Xωξ) =

m∗ζ(X
∗
ωξ

) = ωξ + 1. This completely elucidates the examples with ξ < ω1.

For the ξ = ω1 case, we note that mζ(X) = ω1 if and only if X ∈⋂
η<ω1

Mη,ζ = Mω1,ζ, and a similar statement holds for m
∗
ζ.

Three-space properties. In [19], a Banach space X with subspace Y
was exhibited such that Y,X/Y have the weak Banach-Saks property, while
X does not. In [7], it was shown that Y,X/Y have the hereditary Dunford-
Pettis property, while X does not. In [9], it was shown that any Banach space
is a complemented subspace of a twisted sum of two Banach spaces with the



the ξ,ζ-dunford pettis property 191

Dunford-Pettis property. Therefore there exists a Banach space X containing
a complemented copy of `2 and a subspace Y of X such that Y and X/Y both
lie in Mω1,ω1 . Since `2 ∈ {M1,1 and X contains a complemented copy of `2,
X ∈ {M1,1. Thus Y,X/Y ∈ Mω1,ω1 , while X ∈ {M1,1. This implies that for
any 1 ≤ ξ,ζ ≤ ω1, the property Z ∈ Mξ,ζ is not a three space property.

We modify Ostrovskii’s example to provide a sharp solution to the three
space properties of the classes wBSξ.

Theorem 4.15. For any 0 ≤ ζ,ξ < ω1, any Banach space X, and any
subspace Y such that Y ∈ wBSξ, and X/Y ∈ wBSζ, X ∈ wBSζ+ξ.

For any 0 ≤ ζ,ξ < ω1, there exist a Banach space X with a subspace Y
such that Y ∈ wBSξ, X/Y ∈ wBSζ, and X ∈∩γ<ζ+ξ{wBSγ.

Proof. Assume Y ∈ wBSξ and X/Y ∈ wBSζ. Fix a weakly null sequence
(xn)

∞
n=1 ⊂ X and, seeking a contradiction, assume

0 < ε = inf
{
‖x‖ : F ∈Sζ+ξ,x ∈ co(xn : n ∈ F)

}
.

By passing to a subsequence, we may assume

ε ≤ inf
{
‖x‖ : F ∈Sξ[Sζ],x ∈ co(xn : n ∈ F)

}
.

Since (xn + Y )
∞
n=1 is weakly null in X/Y , it is ζ-weakly null. Thus there

exist F1 < F2 < ... , Fi ∈ Sζ, and positive scalars (ai)i∈∪∞n=1Fn such that∑
i∈Fn ai = 1 and ‖

∑
i∈Fn aixi + Y‖ < min{ε/2, 1/n}. For each n ∈ N, we fix

yn ∈ Y such that ‖yn−
∑

i∈Fn aixi‖ < min{ε/2, 1/n}. Since (xn)
∞
n=1 is weakly

null, so are (
∑

i∈Fn aixi)
∞
n=1 and (yn)

∞
n=1. Since Y ∈ wBSξ, there exist G ∈Sξ

and positive scalars (bn)n∈G such that
∑

n∈G bn = 1 and ‖
∑

n∈G bnyn‖ < ε/2.
Since ∪n∈GFn ∈Sξ[Sζ],

ε ≤
∥∥∥∥ ∑
n∈G

∑
i∈Fn

bnaixi

∥∥∥∥
≤
∥∥∥∥ ∑
n∈G

bnyn

∥∥∥∥ + ∑
n∈G

bn

∥∥∥∥yn − ∑
i∈Fn

aixi

∥∥∥∥ < ε/2 + ε/2 = ε,
and this contradiction finishes the first statement.

Now if ζ = 0 = ξ, let X be any finite dimensional space and let Y = X.
If ζ = 0 and ξ > 0, let (ξn)

∞
n=1 be any sequence such that supn ξn + 1 = ξ.

Let X = (⊕∞n=1Xξn)`1 and let Y = X. If ξ = 0 and ζ > 0, let (ζn)
∞
n=1 be any



192 r.m. causey

sequence such that supn ζn + 1 = ζ. Let X = (⊕∞n=1Xζn)`1 and let Y = {0}.
Each of these choices is easily seen to be the example we seek in these trivial
cases.

We now turn to the non-trivial case, ξ,ζ > 0. Fix (ξn)
∞
n=1 such that if ξ is

a successor, ξn + 1 = ξ for all n ∈ N. Otherwise let (ξn)∞n=1 be the sequence
such that

Sξ =
{
E ∈ [N]<N : ∃n ≤ E ∈Sξn

}
.

Let (ζn)
∞
n=1 be chosen similarly. Let Im,nXζ+ξm → Xζn be the canonical

inclusion, which is bounded, since ζ + ξm ≥ ζ > ζn. Let am,n = ‖Im,n‖−1.
For each m ∈ N, let Zm = (⊕∞n=1Xζn)`1 and let Z = (⊕

∞
m=1Zm)`1 . Define

Jm : Xζ+ξm → Zm by Jm(w) = (2
−nam,nIm,nw)

∞
n=1. Note that ‖Jm‖ ≤ 1.

Now let W = (⊕∞m=1Xζ+ξm)`1 and define S : W → Z by letting S|Xζ+ξm = Jm.
Note that ‖S‖≤ 1. Let q : `1 → Z be a quotient map. Let X = `1 ⊕1 W and
define T : X → Z by T(x,w) = qx + Sw. Then T is also a quotient map, and,
with Y = ker(T), X/Y = Z. Since ζn < ζ, Xζn ∈ wBSζ. Since wBSζ is closed
under `1 sums, Zm and Z lie in wBSζ. Fix γ < ζ +ξ and note that there exists
m ∈ N such that γ ≤ ζ + ξm. Since X contains an isomorph of Xζ+ξm, the
basis of which is not ζ + ξm-weakly null, X in{wBSγ. It remains to show that
Y ∈ wBSξ. To that end, fix a weakly null sequence ((xn,wn))∞n=1 ⊂ Bker(T).
Then xn → 0, and Txn → 0. From this it follows that Swn → 0. Seeking a
contradiction, assume that

0 < ε = inf
{
‖z‖ : F ∈Sξ,z ∈ co((xn,wn) : n ∈ F)

}
.

By passing to a subsequence, we may assume ‖xn‖ < ε/2 for all n, so that

ε/2 ≤ inf
{
‖w‖ : F ∈Sξ,w ∈ co(wn : n ∈ F)

}
.

Since (wn)
∞
n=1 ⊂ W is weakly null, there exists k ∈ N such that for all n ∈ N,

∞∑
m=k+1

‖wn,m‖Xζ+ξm < ε/4,

where wn = (wn,m)
∞
m=1. Since Swn → 0, it follows that for all m ∈ N,

Jmwn,m →
n

0. In particular, for every β < ζ and m ∈ N, limn‖wn,m‖β = 0.
By passing to a subsequence k times, once for each 1 ≤ m ≤ k, we may
assume (wn,m)

∞
n=1 is dominated by a subsequence of the Xξm basis. For this

we are using Proposition 2.16(ii). Since ξm < ξ, (wn,m)
∞
n=1 is ξ-weakly null

for each 1 ≤ m ≤ k. From this it follows that there exist F ∈ Sξ and



the ξ,ζ-dunford pettis property 193

positive scalars (an)n∈F such that
∑

n∈F an = 1 and for each 1 ≤ m ≤ k,
‖
∑

n∈F anwm,n‖ζ+ξm < ε/4k. Then

ε/2 ≤
∥∥∥∥ ∑
n∈F

anwn

∥∥∥∥ ≤ k∑
m=1

∥∥∥∥ ∑
n∈F

anwm,n

∥∥∥∥
ζ+ξm

+
∑
n∈F

an

∞∑
m=k+1

‖wm,n‖ζ+ξm

< ε/4 + ε/4 = ε/2 ,

a contradiction.

5. Partial unconditionality

In this section, we give the promised modification in the complex case of
the cited result of Elton required for our proof of Proposition 4.1.

Lemma 5.1. Fix k ∈ N and suppose we have vectors (u1, . . . ,uk−1,v1,
v2, . . . ) ⊂ SX forming a normalized, weakly null, monotone basic sequence.
For any C,ε > 0, there exists a subsequence (wj)

∞
j=1 of (vj)

∞
j=1 such that for

any T ⊂ {1, . . . ,k − 1}, any n ∈ N, and m0 < · · · < mn, any functional
x∗ ∈ BX∗ such that ∣∣∣∣x∗

(∑
j∈T

uj

)
+ x∗

( n∑
j=1

wmj

)∣∣∣∣ ≥ C,
there exists y∗ ∈ BX∗ such that∣∣∣∣x∗

(∑
j∈T

uj

)
+ x∗

( n∑
j=1

wmj

)∣∣∣∣ ≥ C,
|y∗(uj) −x∗(uj)| ≤ ε for all j ≤ k, and |y∗(wm0 )| ≤ ε.

Proof. We prove only the k > 1 case, with the k = 1 case following by
omitting superfluous parts of the k > 1 case.

For L ∈ [N], U ⊂ B
`k−1∞

, T ⊂ {1, . . . ,k − 1}, and n ∈ N, let A(T,U,n,L)
(resp. B(T,U,n,L)) denote the set of x∗ ∈ BX∗ such that, with L = (l0, l1,
l2, . . . ), ∣∣∣∣x∗

(∑
j∈T

uj

)
+ x∗

( n∑
j=1

vlj

)∣∣∣∣ ≥ C



194 r.m. causey

and (x∗(uj))
k−1
j=1 ∈ U (resp.∣∣∣∣x∗

(∑
j∈T

uj

)
+ x∗

( n∑
j=1

vlj

)∣∣∣∣ ≥ C,
(x∗(uj))

k−1
j=1 ∈ U, and |x

∗(vl0 )| ≤ ε). Now for a fixed T ⊂ {1, . . . ,k − 1} and
U ⊂ B

`k−1∞
, let An denote the set of those L ∈ [N] such that if A(T,U,n,L) 6=

∅, then B(T,U,n,L) 6= ∅. Let An = ∩n∈NAn. We claim that for any N ∈ [N],
there exists L ∈ [N] such that [L] ⊂A. We prove this by contradiction. Note
that since membership in An is determined by properties of the n + 1-element
subsets of a given set, An is closed. Since A is an intersection of closed sets,
it is also closed, and therefore Ramsey. Therefore if the claim were to fail,
there would exist some L ∈ [N] such that [L] ∩A = ∅. Write L = (l1, l2, . . . ).
For 1 ≤ q ≤ p, let Lp,q = (lq, lp+1, lp+2, . . . ) and note that, since Lp,q ∈
[L] ⊂ [N] \ A, there exists np,q ∈ N such that A(T,U,np,q,Lp,q) 6= ∅ but
B(T,U,np,q,Lp,q) = ∅. For each such p,q, fix x∗p,q ∈ A(T,U,np,q,Lp,q). Fix
np = min{np,q : q ≤ p} and qp ≤ p such that np,qp = np. By monotonicity
of the basis, there exists x∗p ∈ BX∗ such that x∗p(uj) = x∗p,qp(uj) for all j < k,
x∗p(vj) = x

∗
p,qp

(vj) for all j ≤ lnp, and x∗p(vj) = 0 for all j > lnp. Note that

∣∣∣∣x∗p
(∑
j∈T

uj

)
+ x∗p

( np∑
j=1

vlp+j

)∣∣∣∣ =
∣∣∣∣x∗p,qp

(∑
j∈T

uj

)
+ x∗p,qp

(np,qp∑
j=1

vlp+j

)∣∣∣∣ ≥ C
and (

x∗p(uj)
)k−1
j=1

=
(
x∗p,qp(uj)

)k−1
j=1
∈ U.

Now note that since np,q ≥ np = np,qp for all 1 ≤ q ≤ p and x∗p(vlj ) = 0 for
any j > np, for each 1 ≤ q ≤ p,∣∣∣∣x∗p

(∑
j∈T

uj

)
+ x∗p

(np,q∑
j=1

vlj

)∣∣∣∣ =
∣∣∣∣x∗p
(∑
j∈T

uj

)
+ x∗p

( np∑
j=1

vlp+j

)∣∣∣∣ ≥ C,
and (x∗(uj))

p−1
j=1 ∈ U. Since B(T,U,np,q,Lp,q) = ∅, it must be the case that

|x∗p(vlq )| ≥ ε. Now if x
∗ is any weak∗-cluster point of (x∗p)

∞
p=1, |x

∗(vlq )| ≥ ε for
all q ∈ N, contradicting the weak nullity of (vj)∞j=1. This gives the claim.

Now let T1, . . . ,Tr be an enumeration of the subsets of {1, . . . ,k− 1} and
let U1, . . . ,Us be a partition of B`k−1∞

into sets of diameter not more than
ε. By repeated applications of the claim from the preceding paragraph, we



the ξ,ζ-dunford pettis property 195

may choose N = L0 ⊃ ··· ⊃ Lrs = L such that if j = (k − 1)r + (i − 1)
with 1 ≤ k ≤ s and 1 ≤ i ≤ r, then for any M ∈ [Lj], if for some n ∈ N,
A(Tk,Ui,n,M) 6= ∅, then B(Tk,Ui,n,M) 6= ∅. Then L has the property
that for any M ∈ [L], if A(Tk,Ui,n,M) 6= ∅, then B(Tk,Ui,n,M). Write
L = (lj)

∞
j=1 and let wj = vlj . Now suppose T ⊂{1, . . . ,k−1}, x

∗ ∈ BX∗, and
m0 < m1 < · · · < mn are such that∣∣∣∣x∗

(∑
j∈T

uj

)
+ x∗

( n∑
j=1

vlmj

)∣∣∣∣ =
∣∣∣∣x∗
(∑
j∈T

uj

)
+ x∗

( n∑
j=1

wmj

)∣∣∣∣ ≥ C.
Pick k such that T = Tk and i such that (x

∗(uj))
k−1
j=1 ∈ Ui. Fix any mn+1 <

mn+2 < ... such that mn+1 > mn and let M = (lmj )
∞
j=0 ∈ [L]. Then x

∗ ∈
A(Tk,Ui,n,M), so that B(Tk,Ui,n,M) 6= ∅. Now fix y∗ ∈ B(Tk,Ui,n,M).
By definition of B(Tk,Ui,n,M),∣∣∣∣y∗

(∑
j∈T

uj

)
+ y∗

( n∑
j=1

wmj

)∣∣∣∣ =
∣∣∣∣y∗
(∑
j∈T

uj

)
+ y∗

( n∑
j=1

vlmj

)∣∣∣∣ ≥ C
and |y∗(wm0 )| = |y∗(vlm0 )| ≥ ε. Since (y

∗(uj))
k−1
j=1, (x

∗(uj))
k−1
j=1 ∈ Ui, it follows

that

max
1≤j<k

|y∗(uj) −x∗(uj)| =
∥∥∥(y∗(uj))k−1j=1 −(x∗(uj))k−1j=1∥∥∥`k−1∞

≤ diam(Ui) ≤ ε.

Since this holds for any n ∈ N and m0 < · · · < mn were arbitrary, we
are done.

Corollary 5.2. Let (xj)
∞
j=1 be a normalized, weakly null, monotone

basic sequence.

(i) For any C,ε > 0, there exists a subsequence (yj)
∞
j=1 of (xj)

∞
j=1 such that

for any pairwise disjoint, finite subsets G,H of N and scalars (aj)j∈H
such that ‖

∑
j∈G yj‖≥ C,∥∥∥∥∑

j∈G
yj +

∑
j∈H

ajyj

∥∥∥∥ ≥ C −ε maxj∈H |aj|.
(ii) For any sequences (Cn)

∞
n=1, (εn)

∞
n=1 of positive numbers, there exists a

subsequence (yj)
∞
j=1 of (xj)

∞
j=1 such that for any n ∈ N, any pairwise



196 r.m. causey

disjoint subsets G,H of N such that ‖
∑

j∈G yn‖ ≥ Cn + 2n, and any
scalars (aj)j∈H,∥∥∥∥∑

j∈G
yj +

∑
j∈H

aj

∥∥∥∥ ≥ Cn − (n + εn) maxj∈H |aj|.
Proof. (i) Fix positive numbers (εj)

∞
j=1 such that

∑∞
j=1

∑∞
k=j εk < ε.

Let L0 = N and apply the k = 1 case of Lemma 5.1 with (vj)∞j=1 = (xj)
∞
j=1,

C = C, and ε = ε1 to find M1 ∈ [N] satisfying the conclusions of Lemma 5.1.
Let r1 = min M1 and L1 = Mn \ {r1}. Now suppose that for some k > 1,
r1 < · · · < rk−1 and L0 ⊃ ··· ⊃ Lk−1 ∈ [N] with min Lk−1 > rk−1 have been
chosen. Apply the k case of Lemma 5.1 with uj = xrj , (vj)

∞
j=1 = (xj)j∈Lk−1 ,

C = C, and ε = εk to find Mk ∈ [Lk−1] satisfying the conclusions of Lemma
5.1. Let rk = min Mk and Lk = Mk \ {rk}. This completes the recursive
construction of r1 < r2 < ... .

Let yj = xrj . Now fix a finite subset G of N such that ‖
∑

j∈G yj‖ ≥ C.
Fix x∗0 ∈ BX∗ such that ∣∣∣∣x∗

(∑
j∈G

yj

)∣∣∣∣ ≥ C.
We may use the conclusions of Lemma 5.1 to find x∗1,x

∗
2, . . . such that for each

k ∈ N and for each j < k, |x∗k(yj)−x
∗
k−1(yj)| ≤ εk,

∣∣∣x∗k(∑j∈G yj)∣∣∣ ≥ C, and if
k ∈ N\G, |x∗k(yk)| ≤ εk. We explain how to choose x

∗
k assuming x

∗
k−1 is chosen.

If k ∈ G, we simply let x∗k = x
∗
k−1. If k = 1 + max G, we use monotonicity to

deduce the existence of x∗1+max G such that x
∗
1+max G(yj) = x

∗
max G(yj) for all

j ≤ max G and x∗1+max G(yj) = 0 for all j > max G. We then let x
∗
k = x

∗
1+max G

for all k > 1 + max G. Now suppose that k /∈ G and k < max G. Fix n ∈ N
and some m1 < · · · < mn such that G ∩ (k,∞) = {m1, . . . ,mn}. Fix any
mn < mn+1 < ... . Now note that, since (rk,rm1,rm2, . . . ) ∈ [Mk] and∣∣∣∣x∗k−1

(∑
j∈G

yj

)∣∣∣∣ =
∣∣∣∣x∗k−1

( ∑
j∈G∩[1,k]

yj

)
+ x∗k−1

( n∑
j=1

xrmn

)∣∣∣∣ ≥ C,
the properties of Mk yield the existence of some x

∗
k ∈ BX∗ such that∣∣∣∣x∗k

(∑
j∈G

yj

)∣∣∣∣ =
∣∣∣∣x∗k
( ∑
j∈G∩[1,k]

yj

)
+ x∗k

( n∑
j=1

xrmn

)∣∣∣∣ ≥ C,
|x∗k(yj) −x

∗
k(yj)| ≤ εk for all j < k, and |x

∗
k(yk)| ≤ εk.



the ξ,ζ-dunford pettis property 197

Now note that the previous recursion yields x∗ = x∗1+max G ∈ BX∗ such
that ∣∣∣∣x∗

(∑
j∈G

yj

)∣∣∣∣ ≥ C.
Furthermore, for any j < max G such that j /∈ G,

|x∗(yj)| ≤ |x∗j (yj)| +
1+max G∑
k=j+1

|x∗k(yj) −x
∗
k−1(yj)| ≤

∞∑
k=j

εk.

For j > max G, x∗(yj) = 0. Now fix any set disjoint from H and any scalars
(aj)j∈H. Then∥∥∥∥∑

j∈G
yj +

∑
j∈H

ajyj

∥∥∥∥ ≥
∣∣∣∣x∗
(∑
j∈G

yj

)∣∣∣∣− maxj∈H |aj|
∞∑
j=1

∞∑
k=j

εk ≥ C −ε max
j∈H
|aj|.

(ii) Recursively select L1 ⊃ L2 ⊃ . . . such that (xj)j∈Ln is the sequence
obtained by applying (i) with C = Cn + n and ε = εn. Fix l1 < l2 < ... ,
ln ∈ Ln, and L = (ln)∞n=1. Let yj = xlj . Suppose that n ∈ N, G ⊂ N are such
that G is finite and ‖

∑
j∈G yj‖≥ Cn + 2n. Fix H ⊂ N \G finite and scalars

(aj)j∈H. Note that∥∥∥∥ ∑
j∈G∩(n,∞)

yj

∥∥∥∥ ≥ Cn + 2n− n∑
j=1

‖yj‖≥ Cn + n.

By the properties of (yn+j)
∞
j=1 obtained from the conclusions of (i),∥∥∥∥ ∑

j∈G∩(n,∞)

yj +
∑

j∈H∩(n,∞)

ajyj

∥∥∥∥ ≥ Cn + n−εn maxj∈H |aj|.
Now∥∥∥∥∑

j∈G
yj +

∑
j∈H

ajyj

∥∥∥∥ ≥ Cn + n−εn maxj∈H |aj|−
n∑
j=1

‖yj‖− max
j∈H
|aj|

n∑
j=1

‖yj‖

≥ Cn − (n + εn) max
j∈H
|aj|.



198 r.m. causey

Proposition 5.3. (Johnson) If (xn)
∞
n=1 is a normalized, weakly null se-

quence having no subsequence equivalent to the canonical c0 basis, then there
exists a subsequence (yn)

∞
n=1 of (xn)

∞
n=1 such that for any r1 < r2 < ... ,

sup
t

∥∥∥∥ t∑
j=1

yrj

∥∥∥∥ = ∞.
Since the complex version of the preceding result can be easily obtained

from the real part by splitting coefficients into real and imaginary parts, we
omit the proof.

Corollary 5.4. Let (xj)
∞
j=1 be a normalized, weakly null sequence with

no subsequence equivalent to the canonical c0 basis. Then there exists a
subsequence (yj)

∞
j=1 of (xj)

∞
j=1 such that for any (bj)

∞
j=1 ∈ `∞ \ c0,

sup
t

∥∥∥∥ n∑
j=1

bjyj

∥∥∥∥ = ∞.
Proof. By passing to a subsequence and passing to an equivalent norm, we

may assume that (xj)
∞
j=1 is monotone basic. We may pass to subsequences

twice and assume that for any r1 < r2 < ... ,

sup
t

∥∥∥∥ t∑
j=1

yrj

∥∥∥∥ = ∞,
a property which is retained by all subsequences. We may also let Cn = n

2

and εn = 1 and assume that for any n ∈ N and pairwise disjoint, finite subsets
G,H of N such that ‖

∑
j∈G yj‖≥ Cn + 2n and scalars (aj)j∈H,∥∥∥∥∑

j∈G
yj +

∑
j∈H

ajyj

∥∥∥∥ ≥ Cn − (n + εn) maxj∈H |aj|.
We prove that this sequence (yj)

∞
j=1 has the desired property.

Fix (aj)
∞
j=1 ∈ B`∞\c0. We may select r1 < r2 < ... and a non-zero number

a with |a| ≤ 1 such that
∑∞

j=1 |a − arj| < 1. By multiplying the sequence
(aj)

∞
j=1 by a unimodular scalar, we may assume a is a positive real number.

By monotonicity, supt‖
∑t

j=1 ajyj‖ = limt‖
∑t

j=1 ajyj‖ = limt‖
∑rt

j=1 ajyj‖.



the ξ,ζ-dunford pettis property 199

In order to reach the conclusion, it is sufficient to define Gt = {r1, . . . ,rt} and
Ht = {1, . . . ,rt}\Gt and show that

∞ = lim
t

∥∥∥∥ ∑
j∈Gt

yj +
∑
j∈Ht

aj
a
yj

∥∥∥∥.
Indeed, from this it follows that

∞ = −1 + lim
t

∥∥∥∥ ∑
j∈Gt

ayj +
∑
j∈Ht

ajyj

∥∥∥∥
≤−1 + lim

t

∥∥∥∥ rt∑
j=1

ajyj

∥∥∥∥ + t∑
j=1

|arj −a| ≤ lim
t

∥∥∥∥ rt∑
j=1

ajyj

∥∥∥∥.
Note that for each t, maxj∈Ht |

aj
a
| ≤ 1/a. For each n ∈ N, by the properties

of (yj)
∞
j=1, there exists t0 so large that for all t ≥ t0,∥∥∥∥ ∑

j∈Gt

yj

∥∥∥∥ > Cn + 2n,
so that ∥∥∥∥ ∑

j∈Gt

yj +
∑
j∈Ht

aj
a
yj

∥∥∥∥ ≥ Cn − (n + εn)/a = n2 − n + 1a .
Since this holds for any n ∈ N and limn n2 − n+1a = ∞, we are done.

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	Introduction
	Combinatorics
	Ideals of interest
	Space ideals
	Partial unconditionality