Weak$^*$ dentability index of spaces $C([0,α])$

0.15 MB·English

by  Petr Hajek

WEAK∗ DENTABILITY INDEX OF SPACES C([0,α])
PETRHA´JEK,GILLESLANCIEN,ANDANTON´INPROCHA´ZKA
Abstract. Wecomputetheweak∗-dentabilityindexofthespacesC(K)where
K is a countable compact space. Namely Dz(C([0,ωωα]))= ω1+α+1, when-
ever 0 ≤ α < ω1. More generally, Dz(C(K)) = ω1+α+1 if K is a scattered
9 compactwhoseheightη(K)satisfiesωα<η(K)≤ωα+1withanαcountable.
0
0
2

n
1. Introduction
a
J
The Szlenk index has been introduced in [20] in order to show that there is

6
no universal space for the class of separable reflexive Banach spaces. The general

idea of assigning an isomorphically invariant ordinal index to a class of Banach
]
A spaces proved to be extremely fruitful in many situations. We refer to [16] for a
F survey with references. In the present note we will give an alternative geometrical
. description of the Szlenk index (equivalent to the original definition whenever X
h
is a separable Banach space not containing any isomorphic copy of ℓ [12]), which
t 1
a stressesits close relationto the weak∗-dentability index. The later index provedto
m
be very useful in renorming theory ([12], [13], [14]).
[ Letusproceedbygivingtheprecisedefinitions. ConsiderarealBanachspaceX

1 andK aweak∗-compactsubsetofX∗. Forε>0weletV bethesetofallrelatively
∗
v weak -open subsets V of K such that the norm diameter of V is less than ε and
1 s K = K \ {V : V ∈ V}. Then we define inductively sαK for any ordinal α by
ε ε
68 sεα+1K =sε(SsαεK)ands∗αεK = β<αsβεK ifαisalimitordinal. We denotebyBX∗
theclosedunitballofX . WethendefineSz(X,ε)tobetheleastordinalαsothat
0 T
1. sαεBX∗ =∅,ifsuchanordinalexists. OtherwisewewriteSz(X,ε)=∞.TheSzlenk
index of X is finally defined by Sz(X) = sup Sz(X,ε). Next, we introduce the
0 ε>0
∗ ∗ ∗
notion of weak -dentability index. Denote H(x,t) = {x ∈ K, x (x) > t}, where
9
0 x ∈ X and t ∈ R. Let K be again a weak∗-compact. We introduce a weak∗-slice
: of K to be any non empty set of the form H(x,t)∩K where x ∈ X and t ∈ R.
v
∗
i Then we denote by S the set of all weak -slices of K of norm diameter less than ε
X and d K =K\ {S :S ∈S}. From this derivation, we define inductively dαK for
ε ε
r any ordinal α by dα+1K = s (dαK) and dαK = sβK if α is a limit ordinal.
a S ε ε ε ε β<α ε
We then define Dz(X,ε) to be the least ordinal αTso that dαε∗BX∗ = ∅, if such an
ordinal exists. Otherwise we write Dz(X,ε) = ∞. The weak -dentability index is
defined by Dz(X)=sup Dz(X,ε).
ε>0
Letus now recallthatit followsfromthe classicaltheory ofAsplund spaces(see
for instance [10], [9], [6] and references therein) that for a Banach space X, each
of the following conditions: Dz(X)6= ∞ and Sz(X)6= ∞ is equivalent to X being
an Asplund space. In particular, if X is a separable Banach space, each of the
∗
conditions Dz(X)<ω and Sz(X)<ω is equivalent to the separability of X . In
1 1
Date:July2008.
2000 Mathematics Subject Classification. 46B20,46B03,46E15.
Keywords and phrases. Szlenkindex,dentabilityindex.
Supported by grants: Institutional Research Plan AV0Z10190503, A100190801, GA CˇR
201/07/0394.
1
WEAK∗ DENTABILITY INDEX OF SPACES C([0,α]) 2
other words, both of these indices measure “quantitatively” the “Asplundness” of
the space in question. Moreover,these indices are invariant under isomorphism.
Itisimmediatefromthe definition,thatDz(X)≥Sz(X)foreveryBanachspace
X. Relying on tools from descriptive set theory, Bossard (for the separable case,
see[4]and[5])andthe secondnamedauthor([14]),provednon-constructivelythat
there exists a universal function ψ : ω → ω , such that if X is an Asplund space
1 1
with Sz(X)<ω , then Dz(X)≤ψ(Sz(X)).
1
Recently,Raja[17]hasobtainedaconcreteexampleofsuchaψ,byshowingthat
Dz(X) ≤ ωSz(X) for every Asplund space. This is a very satisfactory result, but
it is not optimal, as we know from [8] that the optimal value ψ(ω) = ω2. Further
progressinthisareadependsonthe exactknowledgeofindicesforconcretespaces.
The Szlenk index has been precisely calculated for several classes of spaces, most
notably for the class of C([0,α]), α countable (Samuel [19], see also [8]). We have
Sz(C([0,ωωα]))=ωα+1, so it follows from the Bessaga-Pe lczyn´ski([3]) Theorem 1
below,thatthevalueoftheSzlenkindexcharacterizestheisomorphismclass([10]).
Computations of the Szlenk index for other spaces may be found e.g. in [2], [1],
∗
[11]. On the other hand, the precise value of the weak -dentability index is known
onlyforsuperreflexiveBanachspaces,whereDz(X)=ω ([13],[10]), andforspaces
∗
with an equivalent UKK renorming ([8]). For a detailed background information
on the Szlenk anddentability indices we refer the readerto [10], [15], [16], [18] and
references therein.
∗
The main result of our note, Theorem 2, is a precise evaluation of the w -
dentability index for the class of C([0,α]), α countable. These spaces have been
classified isomorphically by C. Bessaga and A. Pe lczyn´ski [3] in the following way.
Theorem 1. (Bessaga-Pe lczyn´ski) Let ω ≤ α ≤ β < ω . Then C([0,α]) is iso-
1
morphic to C([0,β]) if and only if β <αω. Moreover, for every countable compact
space K there exists a unique α<ω such that C(K) is isomorphic to C([0,ωωα]).
1
It is also well-known and easy to show that for α ≥ ω, C([0,α]) is isomorphic
to C ([0,α]) where C ([0,α]) = {f ∈C([0,α]):f(α)=0}. The aim of this note
0 0
∗
is to prove the next theorem. Note, as a particular consequence, that the weak -
dentability index gives a complete isomorphic characterization of a C(K) space,
when K is a metrizable compact space (similarly to the case of the Szlenk index).
Theorem 2. Let 0≤α<ω . Then Dz(C([0,ωωα]))=ω1+α+1.
1
Proof. We start by proving the upper estimate
Dz(C([0,ωωα]))≤ω1+α+1, (1)
The method of the proof is similar to [8], where a short and direct computation
of the Szlenk index of the spaces C([0,α]) is presented. Next lemma is a variant
of Lemma 2.2. from [8]. We omit the proof which requires only minor notational
changes.
Lemma 3. Let X be a Banach space and α an ordinal. Assume that
∀ε>0 ∃δ(ε)>0 dαε(BX∗)⊂(1−δ(ε))BX∗.
Then
Dz(X)≤α·ω.
We shall also use the following Lemma that can be found in [15].
Lemma4. LetX beaBanach spaceandL (X)betheBochnerspace L ([0,1],X).
2 2
Then
Dz(X)≤Sz(L (X)).
2
WEAK∗ DENTABILITY INDEX OF SPACES C([0,α]) 3
Thus, in order to obtain the desired upper bound we only need to prove the
following.
Proposition 5. Let 0≤α<ω . Then Sz(L (C([0,ωωα])))≤ω1+α+1.
1 2
Proof. For a fixed α < ω and γ < ωωα, let us put Z = L (ℓ ([0,ωωα))), together
1 2 1
with the weak∗-topology induced by L (C ([0,ωωα])) and Z =L (ℓ ([0,γ])) with
2 0 γ 2 1
∗
the weak -topologyinduced byL (C([0,γ])). We recallthat for a Banachspace X
2
∗ ∗
with separable dual, L (X ) is canonically isometric to (L (X)) .
2 2
Let P be the canonical projection from ℓ ([0,ωωα)) onto ℓ ([0,γ]). Then, for
γ 1 1
f ∈ Z and t ∈ [0,1], we define (Π f)(t) = P (f(t)). Clearly, Π is a norm one
γ γ γ
projection from Z onto Z (viewed as a subspace of Z). We also have that for any
γ
f ∈Z, kΠ f −fk tends to 0 as γ tends to ωωα.
γ
Next is a variant of Lemma 3.3 in [8].
Lemma 6. Let α < ω , γ < ωωα, β < ω and ε > 0. If z ∈ sβ (B ) and
1 1 3ε Z
kΠ zk2 >1−ε2, then Π z ∈sβ(B ).
γ γ ε Zγ
Proof. Wewillproceedbytransfiniteinductioninβ. Thecasesβ =0andβ alimit
ordinalareclear. Nextweassumethatβ =µ+1andthestatementhasbeenproved
forallordinalslessthanorequaltoµ. Considerf ∈B withkΠ fk2 >1−ε2 and
Z γ
Π f ∈/ sβ(B ). Assuming f ∈/ sµ (B ) ⊃ sβ (B ) finishes the proof, so we may
γ ε Zγ 3ε Z 3ε Z
suppose that f ∈ sµ (B ). By the inductive hypothesis, Π f ∈ sµ(B ). Thus
there exists a weak∗3-εneigZhborhood V of f such that the diamγeter oεf V Z∩γsµ(B )
ε Zγ
is less than ε. We may assume that V can be written V = k H(ϕ ,a ), where
i=1 i i
a ∈R and ϕ ∈L (C([0,γ])). We may also assume, using Hahn-Banach theorem,
i i 2
T
that V ∩(1−ε2)1/2B =∅.
Zγ
Define Φ ∈ L (C ([0,ωωα)) by Φ (t)(σ) = ϕ (t)(σ) if σ ≤ γ and Φ (t)(σ) = 0
i 2 0 i i i
otherwise. Then define W = k H(Φ ,a ). Note that for f in Z, f ∈ W if
i=1 i i
∗
and only if Π f ∈ V. In particular W is a weak -neighborhood of f. Consider
γ
now g,g′ ∈ W ∩sµ (B ). TheTn Π g and Π g′ belong to V and therefore they
3ε Z γ γ
have norms greater than (1−ε2)1/2. It follows from the induction hypothesis that
Π g,Π g′ ∈sµ(B ) thus kΠ g−Π g′k≤ε. Since kΠ gk2 >1−ε2 and kgk≤1,
γ γ ε Zγ γ γ ′ γ ′
we also have kg−Π gk < ε. The same is true for g and therefore kg−g k < 3ε.
γ
This finishes the proof of the Lemma. (cid:3)
We are now in position to proveProposition5. For that purpose it is enough to
show that for all α<ω :
1
∀γ <ωωα ∀ε>0 sω1+α(B )=∅. (2)
ε Zγ
We will prove this by transfinite induction on α<ω .
1
Forα=0,γ isfiniteandthespaceZ isisomorphictoL andthereforesω(B )
γ 2 ε Zγ
is empty. So (2) is true for α=0.
Assume that (2) holds for α < ω . Let Z = L (C ([0,ωωα])). It follows from
1 2 0
Lemma 6 and the fact that for all f ∈Z kΠ f −fk tends to 0 as γ tends to ωωα,
γ
that
∀ε>0 sω1+α(B )⊂(1−ε2)1/2B .
ε Z Z
From this and Lemma 3 it follows that
∀ε>0 sω1+α+1(B )=∅.
ε Z
ByTheorem1weknowthatthe spacesC([0,γ]),C([0,ωωα]),andalsoC ([0,ωωα])
0
are isomorphic, whenever ωωα ≤γ <ωωα+1. Thus sω1+α+1(B )=∅ for any ε>0
ε Zγ
and γ <ωωα+1, i.e. (2) holds for α+1.
Finally, the induction is clear for limit ordinals. (cid:3)
WEAK∗ DENTABILITY INDEX OF SPACES C([0,α]) 4
In the rest of the note, we will focus on proving the converse inequality. Note
that it suffices to deal with the spaces C([0,ωωα]) where α < ω. Indeed, in case
α≥ω, our inequality (1) implies that
Dz(C([0,ωωα]))=Sz(C([0,ωωα]))=ωα+1.
Proposition 7. Let X,Z be Banach spaces and let Y ⊂X∗ be a closed subspace.
∗ ∗
Let there be T ∈ B(X,Z) such that T is an isometric isomorphism from Z onto
Y. Let ε > 0, α be an ordinal such that BX∗ ∩Y ⊂ dαε(BX∗), and z ∈ Z∗. If
z ∈dβε(BZ∗), then T∗z ∈dεα+β(BX∗).
Proof. Byinductionwithrespecttoβ. Thecaseswhenβ =0orβ isalimitordinal
are clear. Let β = µ+1 and suppose that T∗z ∈/ dεα+β(BX∗). If z ∈/ dµε(BZ∗),
then the proof is finished. So we proceed assuming that z ∈ dµε(BZ∗), which by
the inductive hypothesis implies that T∗z ∈ dεα+µ(BX∗). There exist x ∈ X,
t > 0, such that T∗z ∈ H(x,t)∩dεα+µ(BX∗) = S and diamS < ε. Consider the
slice S′ = H(Tx,t)∩ dµε(BZ∗). We have hTx,zi = hx,T∗zi, so z ∈ S′. Also,
diamS′ ≤diamS <ε as T∗ is an isometry. We conclude that z ∈/ dβε(BZ∗), which
finishes the argument. (cid:3)
Let us introduce a shift operator τ : ℓ ([0,ω]) → ℓ ([0,ω]), m ∈ N, by letting
m 1 1
τ h(n)=h(n−m) for n≥m, τ h(n)=0 for n<m and τ h(ω)=h(ω).
m m m
Corollary 8. Let h∈dα(B ). Then τ h∈dα(B ) for every m∈N.
ε ℓ1([0,ω]) m ε ℓ1([0,ω])
Proof. Indeed, consider the mapping T :C([0,ω])→C([0,ω]) defined as
∗
T((x(0),x(1),...,x(ω))) = (x(1),x(2),...,x(ω)). Clearly, T = τ and the as-
1
sertion for m = 1 follows by the previous proposition. For m > 1 one may use
induction. (cid:3)
Definition 9. Let α be an ordinal and ε>0. We will say that a subset M of X∗
is an ε-α-obstacle for f ∈BX∗ if
(i) dist(f,M)≥ε,
(ii)foreveryβ <αandeveryw∗-sliceS ofdβε(BX∗)withf ∈S wehaveS∩M 6=∅.
It follows by transfinite induction that if f has an ε-α-obstacle, then f ∈
dαε(BX∗).
An(n,ε)-tree inaBanachspaceX isafinitesequence(x )2n+1−1 ⊂X suchthat
i i=0
x +x
2i 2i+1
x = and kx −x k≥ε
i 2i 2i+1
2
for i=0,...,2n−1. The element x is calledthe root ofthe tree (x )2n+1−1. Note
0 i i=0
that if (hi)2i=n+01−1 ⊂BX∗ is an (n,ε)-tree in X∗, then h0 ∈dnε(BX∗).
Define f ∈ℓ ([0,α]), for α≥β, by f (ξ)=1 if ξ =β and f (ξ)=0 otherwise.
β 1 β β
Lemma 10.
f ∈dω (B )
ω 1/2 ℓ1([0,ω])
Proof. In[7,Exercise9.20]asequenceisconstructedof(n,1)-treesinB with
ℓ1([0,ω])
roots
1 1
r =( ,..., ,0,...)
n 2n 2n
2n−times
whoseelementsbelongtoP = h∈B :khk =1, h(n)≥0, h(ω)=0 . We
| ℓ1({[0z,ω]) } 1
have r ∈ d2n (B ), and dist(f ,P) = 2. Finally, for every h ∈ P, every
n 1/2 ℓ1([0,ω]) (cid:8) ω (cid:9)
x∈C([0,ω]) and every t∈R such that f ∈H(x,t), there exists m∈N such that
ω
τ h ∈ H(x,t). Therefore the set τ r :(m,n)∈N2 is an 1-ω-obstacle for f .
m m n 2 ω
Thus f ∈dω (B ). (cid:3)
ω 1/2 ℓ1([0,ω]) (cid:8) (cid:9)
WEAK∗ DENTABILITY INDEX OF SPACES C([0,α]) 5
Proposition 11. For every α<ω,
fωωα ∈dω1/12+α(Bℓ1([0,ωωα])) (3)
Proof. The case α = 0 is contained in Lemma 10. Let us suppose that we have
provedtheassertion(3)forallordinals(naturalnumbers,infact)lessthanorequal
to α. It is enough to show, for every n∈N, that
f(ωωα)n ∈dω1/12+αn(Bℓ1([0,(ωωα)n])). (4)
Indeed, (4) implies
f(ωωα)n ∈dω1/12+αn(Bℓ1([0,ωωα+1])).
w∗
Since f(ωωα)n −→ fωωα+1 and f(ωωα)n −fωωα+1 = 2, we see that {f(ωωα)n : n ∈
N} is an 1-ω1+α+1-obstacle for f . That implies (3) for α+1.
2 (cid:13) ωωα+1 (cid:13)
In order to prove (4) we will(cid:13)proceed by induc(cid:13)tion. The casen=1 follows from
the inductive hypothesis as indicated above, so let us suppose that n=m+1 and
(4) holds for m.
Define the mapping T :C([0,(ωωα)n])→C([0,ωωα]) by
(Tx)(γ)=x((ωωα)m(1+γ)), γ ≤ωωα
∗
A simple computation shows that the dual map T is given by
g(ξ), if γ =(ωωα)m(1+ξ), ξ ≤ωωα
∗
(T g)(γ)=
(0 otherwise
Clearly, T∗ is an isometric isomorphism of ℓ ([0,ωωα]) onto rngT∗. We claim that
1
Bℓ1([0,(ωωα)n])∩rngT∗ ⊂dω1/12+αm(Bℓ1([0,(ωωα)n])). (5)
∗
Note that the set of extremal points of Bℓ1([0,(ωωα)n])∩rngT satisfies
ext(Bℓ1([0,(ωωα)n])∩rngT∗)⊂{fγ,−fγ :γ =(ωωα)m(1+ξ), ξ ≤ωωα}
By the inductive assumption and by symmetry, f(ωωα)m and −f(ωωα)m belong to
dω1/12+αm(Bℓ1([0,(ωωα)n])). It is easy to see that more generally, fγ and −fγ belong
to dω1/12+αm(Bℓ1([0,(ωωα)n])), whenever γ = (ωωα)m(1+ξ), ξ ≤ ωωα. Thus we have
verified that
ext(Bℓ1([0,(ωωα)n])∩rngT∗)⊂dω1/12+αm(Bℓ1([0,(ωωα)n])),
and the claim (5) follows using the Krein-Milman theorem.
Thistogetherwiththeinductiveassumption(3)allowsustoapplyProposition7
(with ℓ ([0,(ωωα)n]) as X∗, C([0,ωωα]) as Z, and rngT∗ as Y) to get
1
f(ωωα)n =T∗fωωα ∈dω1/12+αn(Bℓ1([0,(ωωα)n])).
(cid:3)
To finish the proof of Theorem 2, we use that for every Asplund space X,
Dz(X)=ωξ for some ordinal ξ (see [15, Proposition3.3], [10]). Combining Propo-
sition 11 with (1) we obtain
Dz(C([0,ωωα]))=ω1+α+1
for α < ω. For ω ≤ α < ω , we use that ω1+α+1 = ωα+1 = Sz(C([0,ωωα])) =
1
Dz(C([0,ωωα])), which finishes the proof. (cid:3)
OurnextpropositionisadirectconsequenceofTheorem2,Lemma4andPropo-
sition 5.
WEAK∗ DENTABILITY INDEX OF SPACES C([0,α]) 6
Proposition 12. Let 0≤α<ω . Then Sz(L (C([0,ωωα])))=ω1+α+1.
1 2
Our main result can be extended to the non separable case as follows.
Theorem 13. Let 0 ≤ α < ω . Let K be a compact space whose Cantor derived
1
sets satisfy Kωα 6=∅ and Kωα+1 =∅. Then Dz(C(K))=ω1+α+1.
∗
Proof. The upper estimate follows from the separable determination of the weak -
dentability index when it is countable and from Theorem 2 (the argument is iden-
tical to the one given for the computation of Sz(C(K)) in [14]).
On the other hand, since Kωα 6=∅, we have that Sz(C(K)) ≥ωα+1 (see [14] or
Proposition7in[15]). ThereforethereisaseparablesubspaceX ofC(K)suchthat
Sz(X)≥ωα+1. By considering the closed subalgebra of C(K) generated by X, we
may as well assume that X is isometric to C(L), where L is a compact metrizable
space. SinceSz(C(L))≥ωα+1,itfollowsfromTheorem2thatDz(C(L))≥ω1+α+1
and finally that Dz(C(K))≥ω1+α+1. (cid:3)
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WEAK∗ DENTABILITY INDEX OF SPACES C([0,α]) 7
MathematicalInstitute,CzechAcademyofScience,Zˇitna´ 25,11567Praha1,Czech
Republic
E-mail address: [email protected]
Universit´edeFrancheComt´e,Besanc¸on,16,RoutedeGray,25030Besanc¸onCedex,
France
E-mail address: [email protected]
Charles University, Sokolovska´ 83, 186 75 Praha 8, Czech Republic and Universit´e
Bordeaux 1,351cours dela liberation,33405,Talence,France.
E-mail address: [email protected]