The Szlenk index of $L_p(X)$ and $A_p$

0.22 MB·

by  Ryan M. Causey

THE SZLENK INDEX OF L (X) AND A
p p
RYAN M. CAUSEY
Abstract. Given a Banach space X, a w∗-compact subset of X∗, and 1 < p < ∞, we
7 provide an optimal relationship between the Szlenk index of K and the Szlenk index of
1 an associated subset of L (X)∗. As an application, given a Banach space X, we prove an
p
0
optimal estimate of the Szlenk index of L (X) in terms of the Szlenk index of X. This
2 p
extends a result of Ha´jek and Schlumprecht to uncountable ordinals. More generally, given
n
a an operator A : X → Y, we provide an estimate of the Szlenk index of the “pointwise A”
J
operator A :L (X)→L (Y) in terms of the Szlenk index of A.
p p p
2
2

]
A
1. Introduction
F
.
h Throughout this work, X will be a fixed Banach space and K ⊂ X∗ will be a w∗-compact,
t
a non-empty subset. For1 < p < ∞, welet K denote thew∗-closureinL (X)∗ ofall functions
m p p
of the form gh ∈ L (X∗) ⊂ L (X)∗, where g : [0,1] → K is simple and Lebesgue measurable,
[ q p
1
and h ∈ B . Recall that these functions act on L (X) by hgh,fi = hg(̟),f(̟)ih(̟)d̟
1 Lq p 0
6v for f ∈ Lp(X). Note that if R > 0 is such that K ⊂ RBX∗, Kp ⊂ RRBLp(X)∗, so that Kp is
2 also w∗-compact. If K = BX∗, Kp = BLp(X)∗ by the Hahn-Banach theorem. If A : X → Y
2 is an operator, then there exists a “pointwise A” operator A : L (X) → L (Y) given by
6 p p p
0 (Apf)(̟) = A(f(̟)) for all ̟ ∈ [0,1]. Then if K = A∗BY∗, Kp = (Ap)∗BLp(Y)∗, which
.
1 follows from the Hahn-Banach theorem. Thus it is natural to examine what relationship
0
7 exists between K and Kp. In particular, one may ask what relationship exists between the
1
Szlenk indices ofthese sets. To thatend, weobtaintheoptimal relationship. Inwhat follows,
:
v
ω denotes the first infinite ordinal.
i
X
r Theorem 1. Fix 1 < p < ∞. Suppose that ξ is an ordinal such that Sz(K) 6 ωξ. Then
a
Sz(K ) 6 ω1+ξ. If K is convex, Sz(K ) 6 ωSz(K). If K is convex and Sz(K) > ωω,
p p
Sz(K) = Sz(K ).
p
Using the facts stated in the introduction that Kp = (Ap)∗BLp(Y)∗ if K = A∗BY∗, we
immediately deduce the following from Theorem 1.
Corollary 2. Fix 1 < p < ∞. If A : X → Y is an operator and K = A∗BY∗, then
Sz(A ) 6 ωSz(A), and if Sz(A) > ωω, Sz(A ) = Sz(A). In particular, A is Asplund if
p p p
and only if A is.
Applying Corollary 2 to the identity of a Banach space, we extend the result of H´ajek and
Schlumprecht from [8] to uncountable ordinals.
1
2 RYANM. CAUSEY
We recall that K is said to be w∗-fragmentable if for any non-empty subset L of K and
any ε > 0, there exists a w∗-open subset U of X∗ such that L∩U 6= ∅ and diam(L∩U) < ε.
We recall that K is w∗-dentable if for any non-empty subset L of K and any ε > 0, there
exists a w∗-open slice S of X∗ such that L∩U 6= ∅ and diam(L∩S) < ε. We recall that
a w∗-open slice is a subset of X∗ of the form {x∗ ∈ X∗ : Re x∗(x) > a} for some x ∈ X
and a ∈ R. As mentioned in [5], a consequence of Corollary 2 is that if Sz(K) 6 ωξ,
then Sz(K) 6 Dz(K) 6 ω1+ξ, where Dz(K) denotes the w∗-dentability index of K. Thus
Corollary 2 implies that K is w∗-dentable if and only if it is w∗-fragmentable.
In addition to considering the Szlenk index of a set, one may consider the ξ-Szlenk power
type p (L) of the set L, which is important in ξ-asymptotically uniformly smooth renorm-
ξ
ings of Banach spaces and operators. The concept of a ξ-asymptotically uniformly smooth
operator was introduced in [6], and further sharp renorming results regarding the ξ-Szlenk
power type of an operator were established in [4]. To that end, we have the following.
Theorem 3. For any ordinal ξ and any 1 < p < ∞, if 1/p + 1/q = 1, p (K ) 6
1+ξ p
max{q,p (K)}.
ξ
In the case that ξ > ω and p (K) 6 p, p (K) = p (K ), in showing that Theorem 3 is
ξ ξ ξ p
sharp in some cases.
The author wishes to thank P.A.H. Brooker, P. H´ajek, N. Holt, and Th. Schlumprecht for
helpful remarks during the preparation of this work.
2. L (X), Trees, Szlenk index, games
p
2.1. Trees, Γ , P , and stablization results. Given a set Λ, we let Λ<N denote the
ξ,n ξ,n
finite, non-empty sequences in Λ. Given two members s,t of Λ<N, we let sat denote the
concatenation of s and t, |s| denotes the length of s, s (cid:22) t means s is an initial segment of t,
and s| denotes the initial segment of s having length i. Given t ∈ Λ<N, we let [(cid:22) t] = {s ∈
i
Λ<N : s (cid:22) t}.
Any subset T of Λ<N which contains all non-empty initial segments of its members will
be called a B-tree. We define by transfinite induction the derivedB trees of T. We let
MAX(T′) denote the (cid:22)-maximal members of T and T′ = T \ MAX(T). We then define
T0 = T, Tξ+1 = (Tξ)′, and if ξ is a limit ordinal, Tξ = ∩ Tγ. We let o(T) denote the
γ<ξ
smallest ordinal ξ such that Tξ = ∅, provided such an ordinal exists. If no such ordinal
exists, we write o(T) = ∞. We say T is well-founded if o(T) is an ordinal, and T is ill-
founded if o(T) = ∞. For convenience, we agree to the convention that if ξ is an ordinal
ξ < ∞, and that ω∞ = ∞.
Given a B-tree T and a Banach space Y, we let T.Y = {(ζ ,Z )k : (ζ )k ∈ T,Z ∈
i i i=1 i i=1 i
codim(Y)}, where codim(Y) denotes the closed subspaces of Y having finite codimension in
Y. We let C denote the norm compact subsets of B and
X
T.X.C = {(ζ ,Z ,C )k : (ζ )k ,Z ∈ codim(X),C ∈ C}.
i i i i=1 i i=1 i i
THE SZLENK INDEX OF L (X) AND A 3
p p
We note that T.Y and T.X.C are B-trees. Furthermore, for any ordinal γ, (T.Y)γ = Tγ.Y
and (T.X.C)γ = Tγ.X.C. In particular, T.Y and T.X.C have the same order as T.
Given a B-tree T, a Banach space Y, and a collection (x ) ⊂ Y, we say (x ) is
t t∈T.Y t t∈T.Y
normally weakly null provided that for any t = (ζ ,Z )k ∈ T.Y, x ∈ Z . Given another
i i i=1 t k
B-tree S and a function σ : S.Y → T.Y, we say σ is a pruning provided that for every
s,s ∈ S.Y with s ≺ s , σ(s) ≺ σ(s ), and if s = sa(ζ,Z) and σ(s ) = ta(µ,W) for some
1 1 1 1 1
t ∈ T.Y, W 6 Z. If σ : S.Y → T.Y is a pruning and τ : MAX(S.Y) → MAX(T.Y) is such
that for every s ∈ MAX(S.Y), σ(s) (cid:22) τ(s), we say the pair (σ,τ) is an extended pruning,
and denote this by (σ,τ) : S.Y → T.Y.
For every ξ ∈ N and n ∈ N, a B-tree Γ was defined in [3] so that o(Γ ) = ωξn.
ξ,n ξ,n
Furthermore, a function P : Γ → [0,1] was defined so that for every t ∈ MAX(Γ ),
ξ ξ ξ
P (s) = 1. Furthermore, Γ is the disjoint union of Γ , n ∈ N. For convenience,
s(cid:22)t ξ ξ+1 ξ,n
Pwe define P : Γ → [0,n] by P (s) = nP (s). It follows from the definitions that
ξ,n ξ,n ξ,n ξ+1
Γ = Γ and P = P . For every ξ and every n ∈ N, there exist disjoint subsets Λ , ...,
ξ,1 ξ ξ,1 ξ ξ,n,1
Λ of Γ such that Γ = ∪n Λ . It follows from the facts regarding P discussed
ξ,n,n ξ,n ξ,n i=1 ξ,n,i ξ+1
in [3] that, with these definitions, for every ordinal ξ, every n ∈ N, every 1 6 i 6 n, and
every t ∈ MAX(Γ ), P (s) = 1. For any Banach space Y, we may define P
ξ,n Λξ,n,i∋s(cid:22)t ξ,n ξ,n
on Γ .Y and Γ .X.CPby letting
ξ,n ξ,n
P ((ζ ,Z )k ) = P ((ζ )k )
ξ,n i i i=1 ξ,n i i=1
and
P ((ζ ,Z ,C )k ) = P ((ζ )k ).
ξ,n i i i i=1 ξ,n i i=1
We say an extended pruning (σ,τ) : Γ .X → Γ .X is level preserving provided that for
ξ,n ξ,n
every 1 6 i 6 n, σ(Λ ) ⊂ Λ .
ξ,n,i ξ,n,i
ThefollowingtheoremcollectsresultsfromTheorem3.3, Propositions3.2,3.3, andLemma
3.4 of [4].
Theorem 4. Suppose ξ is an ordinal and n is a natural number.
(i) If f : Π(Γ .n.X) → R is bounded and λ ∈ R is such that
ξ
λ < inf P (s)f(s,t),
ξ,n
t∈MAX(Γξ.n.X)X
s(cid:22)t
then there exist a level preserving extended pruning (σ,τ) : Γ .X → Γ .X and real
ξ,n ξ,n
numbers b ,...,b such that λ < n b and for every 1 6 i 6 n and every Λ ∋
1 n i=1 i ξ,n,i
s (cid:22) t ∈ MAX(Γ .X), b 6 f(σ(sP),τ(t)).
ξ,n i
(ii) If (M,d) is a compact metric space and f : Π(Γ .X) → M is any function, then
ξ,n
for any δ > 0, there exist x ,...,x ∈ M and a level preserving extended pruning
1 n
(σ,τ) : Γ .X → Γ .X such that for every 1 6 i 6 n and every Λ ∋ s (cid:22) t ∈
ξ,n ξ,n ξ,n,i
MAX(Γ .X), d(x ,f(σ(s),τ(t))) < δ.
ξ,n i
4 RYANM. CAUSEY
(iii) If F is a finite set and f : MAX(Γ .X) → F is any function, there exists a level
ξ,n
preserving extended pruning (σ,τ) : Γ .X → Γ .X such that f ◦ τ| is
ξ,n ξ,n MAX(Γξ,n.X)
constant.
(iv) For any natural numbers k < ... < k 6 n, there exists an extended pruning (σ,τ) :
1 r
Γ .X → Γ .X such that for every 1 6 i 6 r, σ(Λ ) ⊂ Λ .
ξ,r ξ,n ξ,n,i ξ,n,ki
2.2. The Szlenk index, Szlenk power type. Given a w∗-compact subset L of X∗ and
ε > 0, we let s (K) denote the set consisting of those x∗ ∈ L such that for every w∗-
ε
neighborhood V of x∗, diam(L∩V) > ε. We define the transfinite derivations
s0(L) = L,
ε
sξ+1(L) = s (sξ(L)),
ε ε ε
and if ξ is a limit ordinal,
sξ(L) = sζ(L).
ε ε
\
ζ<ξ
If there exists an ordinal ξ such that sξ(L) = ∅, we let Sz(L,ε) be the minimum such
ε
ordinal. Otherwise we write Sz(L,ε) = ∞. Since sξ(L) is w∗-compact, we deduce that
ε
Sz(L,ε) cannot be a limit ordinal. We agree to the conventions that ω∞ = ∞ and ξ < ∞
for any ordinal ξ. We let Sz(L) = sup Sz(L,ε). If B : Z → W is an operator, we
ε>0
let Sz(B,ε) = Sz(B∗BW∗,ε), Sz(B) = Sz(B∗BW∗). If Z is a Banach space, Sz(Z,ε) =
Sz(I ,ε) and Sz(Z) = Sz(I ).
Z Z
We recall that a set L ⊂ X∗ is called w∗-fragmentable if for any ε > 0 and any w∗-
compact, non-empty subset M of L, s (M) ( M. This is equivalent to Sz(L) < ∞. We say
ε
an operator B : Z → W is Asplund if B∗BW∗ is w∗-fragmentable, which happens if and only
if Sz(B) < ∞. We say a Banach space Z is Asplund if I is Asplund. These are not the
Z
original definitions of Asplund spaces and operators, but they are equivalent to the original
definitions (see [1]).
If Sz(K) 6 ωξ+1, then for any ε > 0, Sz(K,ε) 6 ωξn for some n ∈ N. We let Sz (K,ε)
ξ
be the smallest n ∈ N such that Sz(K,ε) 6 ωξn. We define the ξ-Szlenk power type p (K)
ξ
of K by
logSz (K,ε)
p (K) = limsup ξ .
ξ |log(ε)|
ε→0+
This value need not be finite. By convention, we let p (K) = ∞ if Sz(K) > ωξ+1. We
ξ
let pξ(A) = pξ(A∗BY∗) and pξ(X) = pξ(BX∗). The quantities pξ(X), pξ(A) are important
for the renorming theorem of ξ-asymptotically uniformly smooth norms with power type
modulus.
Given a w∗-compact subset L of X∗ and ε > 0, we let HL denote the set of Cartesian
ε
products n C such that C ∈ C for each 1 6 i 6 n and such that there exist (x )n ∈
i=1 i i i i=1
n C anQd x∗ ∈ K such that for each 1 6 i 6 n, Re x∗(x ) > ε.
i=1 i i
Q
THE SZLENK INDEX OF L (X) AND A 5
p p
2.3. The Szlenk index of K . Recall that for 1 < p < ∞, L (X) denotes the space of
p p
equivalence classes of Bochner integrable functions f : [0,1] → X such that kfkp < ∞,
where [0,1] is endowed with its Lebesgue measure. Recall also that if 1 < q < ∞R , L (X∗) is
q
isometrically included in L (X)∗ by the action
p
f 7→ hg,fi,
Z
for g ∈ L (X∗). We also recall that if ̺ : X → R is any Lipschitz function, then for any
q
f ∈ L (X), ̺◦f ∈ L .
p p
We note that the Szlenk index and the ξ Szlenk power type of K are unchanged by scaling
K by a positive scalar or by replacing K with its balanced hull. Moreover, for a positive
scalar c, (cK) = cK , which has the same Szlenk index and ξ-Szlenk power type as K . If
p p p
TK is the balanced hull of K, K ⊂ (TK) and Sz(K) = Sz(TK) ([3, Lemma 2.2]) so that
p p
Theorem 1, Corollary 2, and Theorem 3 hold in general if they hold under the assumption
that K ⊂ BX∗ is balanced. Therefore we can and do assume throughout that K ⊂ BX∗ and
K is balanced.
Let ̺ : X → R be given by ̺(x) = maxx∗∈K Re x∗(x). Since we have assumed K is
balanced, ̺(x) = maxx∗∈K |x∗(x)|. It is easy to see that for any 1 < p < ∞ and any
f ∈ Lp(X), k̺(f)kLp = maxf∗∈Kp Re f∗(f). Combining this fact with [4, Corollary 2.4] and
the proof of that corollary, we obtain the following.
Theorem 5. Fix 1 < p,α < ∞.
(i) If for every B-tree T with o(T) = ω1+ξ and every normally weakly null (f ) ⊂
t t∈T.Lp(X)
B ,
Lp(X)
inf k̺(f)k : t ∈ T.L (X),f ∈ co(f : ∅ ≺ s (cid:22) t) = 0,
Lp p s
(cid:8) (cid:9)
then Sz(K ) 6 ω1+ξ.
p
(ii) If there exists a constant C such that for every n ∈ N, every B-tree T with o(T) =
ω1+ξn, and every normally weakly null collection (f ) ⊂ B ,
t t∈T.Lp(X) Lp(X)
inf k̺(f)k : t ∈ T.L (X),f ∈ co(f : ∅ ≺ s (cid:22) t) 6 Cn−1/α,
Lp p s
(cid:8) (cid:9)
then p (K ) 6 α.
1+ξ p
Proposition 6. Suppose T is a non-empty B-tree. Suppose also that (C ) ⊂ C is fixed
s s∈T.X
and for s = (ζ ,Z )k ∈ T.X, let λ(s) = Z ∩ C . Suppose that S is a non-empty, well-
i i i=1 k s
founded B-tree and θ : S.X → T.X is a pruning. For s ∈ S.X, let s(s) = |s| λ(θ(s| )).
i=1 i
If ε > 0 is such that for every t ∈ S.X, s(s) ∈ HK 6= ∅, then for any 0 Q< δ < ε, any
ε
0 6 γ < o(S), and any s ∈ Sγ.X, s(s) ∈ Hsγδ(K) 6= ∅. Moreover, for any 0 < δ < ε,
ε
Sz(K,δ) > o(S).
Proof. We induct on γ. The base case is the hypothesis. Assume γ+1 < o(S) and the result
holds for γ. Assume s ∈ Sγ+1.X, which means there exists ζ such that sa(ζ,Z) ∈ Sγ.X
for all Z ∈ codim(X). Then for every Z ∈ codim(X), there exists Z > W ∈ codim(X)
Z
6 RYANM. CAUSEY
such that s(sa(ζ,Z)) ⊂ s(s) × B . From this and the inductive hypothesis, for every
WZ
Z ∈ codim(X), we fix x ∈ B , (xZ)|s| ∈ s(s), and x∗ ∈ sγ(K) such that Re x∗(x ) > ε
Z WZ i i=1 Z δ Z Z
and Re x∗(xZ) > ε for each 1 6 i 6 |s|. By compactness of s(s) × K with the product
Z i
topology, where λ(θ(s| )) has its norm topology and K has its w∗-topology,
i
∅ 6= {(xY,...,xY ,x∗ ) : Z > Y ∈ codim(X)} ⊂ s(s)×K.
1 |s| Y
\
Z∈codim(X)
Fix (x ,...,x ,x∗) lying in this intersection. Obviously x∗ ∈ sγ(K). Moreover, for any
1 |s| δ
w∗-neighborhood V of x∗, there exists Z ∈ codim(X) such that ker(x∗) ⊂ Z and x∗ ∈ V,
Z
whence
diam(sγ(K)∩V) > kx∗ −x∗k > Re (x∗ −x∗)(x ) = Re x∗(x ) > ε > δ.
δ Z Z Z Z Z
This implies x∗ ∈ sγ+1(K). It is obvious that Re x∗(x ) > ε for all 1 6 i 6 |s|. This shows
δ i
that s(s) ∈ Hsγδ+1(K) and completes the successor case.
ε
Finally, assume γ < o(S) is a limit ordinal and the result holds for all ordinals less than
γ. Fix s ∈ Sγ.X and let s(s)×K be topologized as in the successor case. By the inductive
hypothesis, for all β < γ, there exists (xβ,...,xβ ,x∗) ∈ s(s)×K such that x∗ ∈ sβ(K) and
1 |s| β β ε
for all 1 6 i 6 |s|, Re x∗(xβ) > ε. By compactness of |s| λ(θ(s| )) ×K,
β i i=1 i
(cid:0)Q (cid:1)
{(xµ,...,xµ ,x∗) : µ > β} =6 ∅.
1 |s| µ
\
β<γ
Clearly any (x ,...,x ,x∗) lying in this intersection is such that x∗ ∈ sγ(K) and for any
1 |s| δ
1 6 i 6 |s|, Re x∗(x ) > ε. This shows that s(s) ∈ Hsγδ(K) and completes the induction.
i ε
We have shown that for any 0 < δ < ε, Sz(K,δ) > o(S). If o(S) is a limit ordinal, we
deduce that Sz(K,δ) > o(S) since Sz(K,δ) cannot be a limit ordinal. If o(S) is a successor,
say o(S) = ξ +1, then there exists a length 1 sequence (ζ) ∈ Sξ. For every Z ∈ codim(X),
s((ζ,Z)) = W ∩C for some W ⊂ Z. The first part of the proof yields that for each
Z θ((ζ,Z))
Z ∈ codim(X), there exists x ∈ W ∩C ⊂ W ∩B and some x∗ ∈ sξ(K) such that
Z Z θ((ζ,Z)) Z X Z ζ
Re x∗(x ) > ε. Arguing as in the successor case, we deduce that any w∗-limit of a subnet
Z Z
of (x∗) lies in sξ+1(K), whence Sz(K,δ) > ξ +1 = o(S).
Z Z∈codim(X) δ
(cid:3)
2.4. Games. SupposeT ⊂ Λ<N isawell-founded, non-empty B-treeandE ⊂ MAX(T.X.C)
is some subset. We define the game on T.X.C with target set E. Player I first chooses
(ζ ,Z ) ∈ Λ ×codim(X) such that (ζ) ∈ T and Player II then chooses C ∈ C. Assuming
1 1 1
(ζ ,Z )n ∈ T.X and C ,...,C ∈ C have been chosen, the game terminates if (ζ ,Z )n ∈
i i i=1 1 n i i i=1
MAX(T.X). Otherwise Player I chooses (ζ ,Z ) ∈ Λ×codim(X) such that (ζ )n+1 ∈ T
n+1 n+1 i i=1
and Player II chooses C ∈ C. Since T is well-founded, this game must terminate after
n+1
finitely many steps. Suppose that the resulting choices are (ζ ,Z )n and C ,...,C ∈ C.
i i i=1 1 n
We say that Player II wins if (ζ ,Z ,C )n ∈ E, and Player I wins otherwise.
i i i i=1
THE SZLENK INDEX OF L (X) AND A 7
p p
A strategy for Player I for the game on T.X.C with target set E is a function ψ : T′.X.C∪
{∅} → Λ×codim(X) such that if ψ((ζ ,Z ,C )n−1) = (ζ ,Z ), (ζ )n ∈ T. We say ψ is a
i i i i=1 n n i i=1
winning strategy for Player I provided that for any sequence (ζ ,Z ,C )n ∈ MAX(T.X.C)
i i i i=1
such that (ζ ,Z ) = ψ((ζ ,Z ,C )i−1) for every 1 6 i 6 n, (ζ ,Z ,C )n ∈/ E.
i i j j j j=1 i i i i=1
A strategy for Player II for the game on T.X.C with target set E is a function ψ defined
on the set
{((ζ ,Z ,C )n−1,(ζ ,Z )) :(ζ ,Z ,C )n−1 ∈ {∅}∪T.X.C,(ζ ,Z ) ∈ Λ×codim(X),
i i i i=1 n n i i i i=1 n n
(ζ )n ∈ T}
i i=1
and taking values in C. We say ψ is a winning strategy for Player II provided that for
any sequence (ζ ,Z ,C )n ∈ MAX(T.X.C) such that C = ψ((ζ ,Z ,C )i−1,(ζ ,Z )) for all
i i i i=1 i j j j j=1 i i
1 6 i 6 n, (ζ ,Z ,C )n ∈ E.
i i i i=1
Proposition 7. [5, Proposition 3.1] For any non-empty, well-founded B-tree T and any
E ⊂ T.X.C, either Player I or Player II has a winning strategy for the game on T.X.C with
target set E.
Proposition 8. Suppose that Player II has a winning strategy for a game on T.X.C with tar-
get set E. Then there exists (C ) ⊂ C such that for every t = (ζ ,Z )k ∈ MAX(T.X),
s s∈T.X i i i=1
(ζ ,Z ,C )k ∈ E.
i i t|i i=1
Proof. Fix a winning strategy ψ for Player II in the game. We define C by induction on |s|.
s
We let C = ψ(∅,(ζ,Z)). If |s| = k +1, C has been defined for every 1 6 i 6 k, and
(ζ,Z) s|i
a
s = s| (ζ,Z), we let C = ψ(s| ,(ζ,Z)).
k s k
(cid:3)
For the next proposition, if h ∈ L (X) is a simple function, we let h be the function in
p
L (X) such that h(̟) = 0 if h(̟) = 0 and h(̟) = h(̟)/kh(̟)k otherwise.
p
Proposition 9. Let ξ be an ordinal, n a natural number, and let T be a B-tree with o(T) >
ω1+ξn. If ψ is a strategy for Player I for some game on Γ .X.C, then for any 1 < p < ∞,
ξ,n
any δ > 0, and any normally weakly null (f ) ⊂ B , there exist s = (ζ ,Z )k ∈
t t∈T.Lp(X) Lp(X) i i i=1
MAX(Γ .X), ∅ = t ≺ t ≺ ... ≺ t ∈ T.L (X), g ∈ co(f : t ≺ u (cid:22) t ), h ∈ B ,
ξ,n 0 1 k p i u i−1 i i Lp(X)
and C ∈ C such that for every 1 6 i 6 k,
i
(i) h is simple,
i
(ii) range(h ) = C ⊂ B ,
i i Zi
(iii) kg −h k < δ,
i i Lp(X)
(iv) (ζ ,Z ) = ψ((ζ ,Z ,C )i−1).
i i j j j j=1
Remark 10. For a B-tree S on Λ and s ∈ S, we let S(s) denote those non-empty sequences
u ∈ Λ<N such that sau ∈ S. An easy induction argument yields that for any ordinals ξ,ζ,
Sξ(s) = (S(s))ξ for any ordinal ξ. From this it follows that s ∈ Sξ if and only if o(S(s)) > ξ.
8 RYANM. CAUSEY
Furthermore, another easy induction yields that if (Sξ)ζ = Sξ+ζ, from which it follows that
if o(S) > ξ +ζ, o(Sξ) > ζ. Therefore if s ∈ Sξ+ω, o(Sξ(s)) > ω.
Proof of Proposition 9. We first note that if Z ∈ codim(X), L (X)/L (Z) is either the zero
p p
vector space or isomorphic to L , and therefore has Szlenk index not exceeding ω. As
p
explained in [5], this means that for any B-tree T with o(T) > ω, any δ > 0, and any
normally weakly null (f ) ⊂ B , there exist t ∈ T.L (X), g ∈ co(f : ∅ ≺ s (cid:22) t),
t t∈T.Lp(X) Lp(X) p s
and h ∈ B such that kg −hk < δ. Moreover, by the density of simple functions,
Lp(Z) Lp(X)
we may assume this h is simple.
Let ψ be a strategy for Player I for a game on Γ .X.C. Let T be a B-tree with o(T) =
ξ,n
ω1+ξn and define γ : Γ .X∪{∅} → [0,ωξn] by letting γ(t) = max{µ 6 ωξn : t ∈ (Γ .X)µ}
ξ,n ξ,n
for t ∈ Γ .X and γ(∅) = ωξn. Let s = t = ∅. Now assume that for some k ∈ N and
ξ,n 0 0
all 1 6 i < k, s ∈ Γ .X, ζ ∈ [0,ωξn], Z ∈ codim(X), t ∈ T.L (X), g ,h ∈ B , and
i ξ,n i i i p i i Lp(X)
C ∈ C have been chosen such that for all 1 6 i < k,
i
(i) h is simple,
i
(ii) s = (ζ ,Z )i ,
i j j j=1
(iii) t ≺ t ≺ ... ≺ t ,
0 1 k−1
(iv) t ∈ (T.L (X))ωγ(si),
i p
(v) (ζ ,Z ) = ψ((ζ ,Z ,C )i−1),
i i j j j j=1
(vi) g ∈ co(f : t ≺ u (cid:22) t ),
i u i−1 i
(vii) kg −h k < δ,
i i Lp(X)
(viii) range(h ) = C ⊂ B .
i i Zi
If s is maximal in Γ .X, we let s = s , and one easily checks that the conclusions
k−1 ξ,n k−1
are satisfied. Otherwise let (ζ ,Z ) = ψ((ζ ,Z ,C )k−1) and s = sa (ζ ,Z ). Let u be
k k j j j j=1 k k−1 k k k−1
the sequence of first members of the pairs of t and let U denote the proper extensions of
k−1
uk−1 in Tωγ(sk). Then (ftak−1u)u∈U.Lp(X) ⊂ BLp(X) is normally weakly null and o(U) > ω by
the remark preceding the proof, so that the previous paragraph yields the existence of some
u′ ∈ U.L (X), g ∈ co(f : t ≺ u (cid:22) ta u′), and some simple function h ∈ L (Z ) such
p k u k−1 k−1 k p k
that kg −h k < δ. Let t = ta u′. In order to apply the remark before the proof, we
k k Lp(X) k k−1
note that since s ≺ s , γ(s ) > γ(s )+1. Since
k−1 k k−1 k
ωγ(s ) > ω(γ(s )+1) = ωγ(s )+ω,
k−1 k k
the remark preceding the proof applies. Note that C := range(h ) ⊂ B . This completes
k k Zk
the recursive construction. Since Γ .X is well-founded, eventually this process terminates.
ξ,n
The resulting s = (ζ ,Z )k ∈ MAX(Γ .X) clearly satisfies the conclusions.
i i i=1 ξ,n
(cid:3)
3. Definition of an associated space and two games
3.1. The associated space and its properties. If E is a vector space with seminorm
k·k, we say a sequence (e )n in E is 1-unconditional provided that for any scalars (a )n
i i=1 i i=1
THE SZLENK INDEX OF L (X) AND A 9
p p
and any (ε )n ∈ {±1}n, k n ε a e k = k n a e k. Recall that for 1 < p < ∞, a vector
i i=1 i=1 i i i i=1 i i
space E with seminorm k·Pk which is spannPed by the 1-unconditional basis (e )n is called
i i=1
p-concave provided there exists a constant C such that for any (f )n ⊂ L ,
i i=1 p
n n
k f e k 6 Ck kf k e k .
i i Lp(E) i Lp i E
X X
i=1 i=1
The smallest such constant C is denoted by M (E).
(p)
Given x ∈ span(e : 1 6 i 6 n), where (e )n is a Hamel basis for the seminormed
i i i=1
space E, we write x = n a e and supp(x) = {i 6 n : a 6= 0}. We say the vectors
i=1 i i i
x ,...,x ∈ span(e : 1 6Pi 6 n) are disjointly supported if the sets supp(x ), ..., supp(x )
1 n i 1 n
are pairwise disjoint.
For 1 < β < ∞, we say that an unconditional Hamel basis (e )n for a seminormed space
i i=1
E satisfies an 1-lower ℓ estimate provided that for any m ∈ N and any disjointly supported
β
elements (x )m ⊂ E,
i i=1
m m
kx kβ 1/β 6 k x k.
i i
(cid:0)Xi=1 (cid:1) Xi=1
Theorem 11. [7, Theorem 1.f.7] Fix 1 < β < p < ∞. There exists a constant C′ = C′(β,p)
such that if (e )n is a 1-unconditional basis for the seminormed space E which satisfies a
i i=1
1-lower ℓ estimate, then E is p-concave and M (E) 6 C′.
β (p)
For the remainder of this section, T is a fixed, non-empty B-tree.
Fora non-empty set J, we let c (J)bethespan ofthecanonical Hamel basis (e ) inthe
00 j j∈J
space of scalar-valued functions on J, where e is the indicator of the singleton {j}. We let e∗
j j
denote the coordinate functional to e . Given x ∈ c (J), we may write x = a e . Then
j 00 j∈J j j
we define |x| to be |a |e . A suppression projection is an operator P frPom span(e∗ : j ∈
j∈J j j j
J) into itself such tPhat there exists a subset F of J such that P a e∗ = a e∗.
j∈J j j j∈F j j
For 0 < φ < θ < 1, let P P
k
N = {0}∪ θ e∗ : t = (ζ ,Z ,C )|t| ∈ T.X.C,1 6 j < ... < j 6 |t|,
θ,φ,T n X t|ji i i i i=1 1 k
i=1
k
Z ∩C ∈ HK ⊂ span(e∗ : t ∈ T.X.C).
Y ji ji φ o t
i=1
For 0 < φ < θ < 1 and 1 < α < ∞, let
k k
M = a g :g ∈ ∪∞ N ,a > 0, aα 6 1,supp(g ) are pairwise disjoint .
θ,φ,α,T i i i n=1 θn,φn,T i i i
nX X o
i=1 i=1
Note that the set M is closed under suppression projections.
θ,φ,α,T
We define the seminorm k·k on c (T.X.C) by
θ,φ,α,T 00
kxk = sup{f(|x|) : f ∈ M }.
θ,φ,α,T θ,φ,α,T
10 RYANM. CAUSEY
Claim 12. Fix 1 < α < ∞ and 0 < φ < θ < 1. For any t ∈ T.X.C, (e )|t| is 1-
t|i i=1
unconditional and satisfies a 1-lower ℓ estimate in its span, where 1/α+1/β = 1.
β
Proof. Note that 1-unconditionality is obvious. Fix x ,...,x ∈ span(e : 1 6 i 6 |t|)
1 n t|i
with disjoint supports. That is, there exist pairwise disjoint subsets S ,...,S of {1,...,|t|}
1 n
such that x ∈ span(e : j ∈ S ). Then there exist g ,...,g ∈ M such that for each
i t|j i 1 n θ,φ,α,T
1 6 i 6 n, g (|x |) = kx k . Since M is closed under suppression projections, we
i i i θ,φ,α,T θ,φ,α,T
mayassume thatsupp(g ) ⊂ S foreach1 6 i 6 n. Thenif(a )n aresuch that n aα = 1,
i i i i=1 i=1 i
a > 0, and n a kx k = ( n kx kβ )1/β, g := n a g ∈ M Pand
i i=1 i i θ,φ,α,T i=1 i θ,φ,α,T i=1 i i θ,φ,α,T
P P P
n n n n
k x k > g x = a g (|x |) = kx kβ )1/β.
i θ,φ,α,T i i i i i θ,φ,α,T
Xi=1 (cid:0)(cid:12)Xi=1 (cid:12)(cid:1) Xi=1 (cid:0)Xi=1
(cid:12) (cid:12)
(cid:3)
Claim 13. Fix 1 < α < ∞ and 0 < φ < θ < 1. For any t = (ζ ,Z ,C )k ∈ T.X.C, any
i i i i=1
sequence (x )k ∈ k Z ∩C , and any sequence (a )k of non-negative scalars,
i i=1 i=1 i i i i=1
Q
k k
1
̺( a x ) 6 k a e k .
i i θ −φ i t|i θ,φ,α,T
X X
i=1 i=1
Proof. We recall that if C ∈ C, C ⊂ B by the definition of C. With t, (x )k ∈ k Z ∩C ,
X i i=1 i=1 i i
and (a )k as in the statement, fix x∗ ∈ K such that Re x∗( k a x ) = ̺Q( k a x ).
i i=1 i=1 i i i=1 i i
For all j ∈ N, let B = {i 6 k : Re x∗(x ) ∈ (φj,φj−1]}. NPote that for evePry j ∈ N,
j i
θj e∗ ∈ N , so
i∈Bj t|i θj,φj,T
P
k k
φj−1 a = φ−1(φ/θ)j(θj e∗ )( a e ) 6 φ−1(φ/θ)jk a e k .
i t|i i t|i i t|i θ,φ,α,T
X X X X
i∈Bj i∈Bj i=1 i=1
Then
k ∞ ∞ ∞ k
̺( a x ) 6 a Re x∗(x ) 6 φj−1 a 6 φ−1(φ/θ)jk a e k
i i i i i i t|i θ,φ,α,T
X XX X X X X
i=1 j=1 i∈Bj j=1 i∈Bj j=1 i=1
k
1
= k a e k .
θ−φ i t|i θ,φ,α,T
X
i=1
(cid:3)
Corollary 14. Fix 1 < p,α,β < ∞ with 1/α+1/β = 1 and β < p. Let C′ = C′(β,p) be the
constant from Theorem 11. Suppose that ξ is an ordinal, n is a natural number, ε > 0, and
0 < φ < θ < 1 are such that Player I has a winning strategy in the game with target set
t ∈ MAX(Γ .X.C) : k P (s)e k > ε .
n ξ,n X ξ,n s θ,φ,α,Γξ,n o
s(cid:22)t