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A Banach space is c0-saturated if all of its closed infinite dimensional sub-

arXiv:math/9306210v1 [math.FA] 21 Jun 1993Some stability propertiesof c0-saturated spacesDenny H. LeungAbstractA Banach space is c0-saturated if all of its closed infinite dimensional sub-spaces contain an isomorph of c0. In this article, we study the stability of thisproperty under the formation of direct sums and tensor products.

Some of theresults are: (1) a slightly more general version of the fact that c0-sums of c0-saturated spaces are c0-saturated; (2) C(K, E) is c0-saturated if both C(K) andE are; (3) the tensor product JH ˜⊗ǫJH is c0-saturated, where JH is the JamesHagler space.Let E be a Banach space. Following Rosenthal [10], we say that a Banach space F isE-saturated if every infinite dimensional closed subspace of F contains an isomorphiccopy of E. In this article, we will be concerned with the stability properties of c0-saturated spaces under the formation of direct sums and tensor products.

In §1, weprove a result which implies that c0-sums of c0-saturated spaces are c0-saturated. In§2, it is shown that the tensor product E ˜⊗ǫF is c0-saturated if E is isomorphicallypolyhedral (see §2 for the definition) and F is c0-saturated.

As a corollary, we obtainthat C(K, E) is c0-saturated if and only if both C(K) and E are. Finally, in §3, weshow that JH ˜⊗ǫJH is c0-saturated, where JH denotes the James Hagler space [4].Standard Banach space terminology, as may be found in [7], is employed.The(closed) unit ball of a Banach space E is denoted by UE.

The space c00 consists ofall finitely non-zero real sequences. If (xn) and (yn) are sequences residing in possiblydifferent Banach spaces, we say that (xn) dominates (yn) if there is a constant K < ∞such that ∥P anyn∥≤K∥P anxn∥for all (an) ∈c00.

Two sequences which dominateeach other are said to be equivalent.A sequence (xn) in a Banach space is semi-normalized if 0 < inf ∥xn∥≤sup ∥xn∥< ∞. If A is an arbitrary set, |A| denotes thecardinality of A.

For an infinite set A, P∞(A) is the set of all infinite subsets of A.1991 Mathematics Subject Classification 46B20, 46B28.1

1Direct sums of c0-saturated spacesIn [10], it is stated without proof that c0-sums of c0-saturated spaces are c0-saturated.In this section, we prove a result which includes this as a special case. Let (En) bea sequence of Banach spaces, and let F be a Banach space with a basis (en).

TheF-sum of the spaces En is the Banach space (⊕En)F of all sequences (x(n)) such thatx(n) ∈En for all n, andP ∥x(n)∥en converges in F, endowed with the norm∥(x(n))∥= ∥X∥x(n)∥en∥.For convenience, we will say that a Banach space is p-saturated if it is ℓp-saturated(1 ≤p < ∞) or c0-saturated (p = ∞).Lemma 1 Let E, F be p-saturated Banach spaces for some 1 ≤p ≤∞, then E ⊕Fis p-saturated.Proof: It suffices to show that every normalized basic sequence in E ⊕F has a blockbasis equivalent to the ℓp-basis (c0-basis if p = ∞). Let (xn ⊕yn) be a normalizedbasic sequence in E ⊕F.If p = 1, and (xn ⊕yn) has a ℓ1-subsequence, then weare done.

Otherwise, by Rosenthal’s Theorem [8], we may assume that (xn ⊕yn) isweakly Cauchy. If p ̸= 1, using again Rosenthal’s Theorem, we may assume that both(xn) and (yn) are weakly Cauchy.

In both cases, by replacing the sequence (xn ⊕yn)with (x2n−1 −x2n ⊕y2n−1 −y2n) if necessary, we may even assume that both (xn) and(yn) are weakly null. If ∥yn∥→0, then a subsequence of (xn ⊕yn) is equivalent to asubsequence of (xn ⊕0).

But then the latter is a basic sequence in E, and hence has ablock basis equivalent to the ℓp-basis. Therefore, (xn ⊕yn) has a ℓp-block basis as well.A similar argument holds if ∥xn∥→0.

Otherwise, we may take both (xn) and (yn) tobe semi-normalized weakly null sequences. By using a subsequence, it may be assumedthat both are basic sequences.

Then (xn) has a ℓp-block basis (uk) = (Pnk+1j=nk+1 ajxj).Let (vk) = (Pnk+1j=nk+1 ajyj) be the corresponding block basis of (yn). If ∥vk∥→0, thenwe apply the same argument as above.

If ∥vk∥→∞, then uk∥vk∥⊕vk∥vk∥!is a semi-normalized block basis of (xn ⊕yn). Since ∥uk∥/∥vk∥→0, we may applythe argument above yet again to conclude the proof.

Finally, then, we may assumethat (vk) is semi-normalized.Then (vk) has a ℓp-block basis (tk). Let (sk) be the2

corresponding block basis of (uk) formed by using the same coefficients. Arguing asbefore, we may assume that (sk) is semi-normalized.

But since (uk) is a ℓp-sequence,so is (sk). Therefore, (sk ⊕tk) is a ℓp-block basis of (uk ⊕vk), and hence of (xn ⊕yn).Theorem 2 Let (En) be a sequence of p-saturated Banach spaces, and let F be a p-saturated Banach space with a basis.

Then E = (⊕En)F is p-saturated.Proof: For each x ∈E, write x = (x(n)), where x(n) ∈En for all n. Let (xk) bea normalized basic sequence in E. For each m ∈NI , let Pm be the projection on Edefined by Pmx = y, where y(n) = x(n) if n ≤m, and y(n) = 0 otherwise. If forsome subsequence (zj) of (xk), and some m ∈NI , (Pmzj)j dominates (zj), then (zj) isequivalent to (Pmzj)j.

But then the latter is a basic sequence in E1 ⊕. .

. ⊕Em, whichis p-saturated by Lemma 1 and induction.

Hence (zj), and thus (xk), has a ℓp-blockbasis, and we are done. Otherwise, for all m ∈NI , and every subsequence (zj) of (xk),inf{∥XjajPmzj∥: (aj) ∈c00, ∥Xajzj∥= 1} = 0.

(1)Let m0 = 0, and y1 = x1. Choose m1 such that ∥(1 −Pm1)y1∥≤1.

By (1), there existsk2 ≥2, and y2 ∈span{xk : 2 ≤k ≤k2}, ∥y2∥= 1, such that ∥Pm1y2∥≤1/4. Thenchoose m2 > m1 so that ∥(1 −Pm2)y2∥≤1/4.

Continuing inductively, we obtain anormalized block basis (yj) of (xk), and (mj)∞j=0 such that∥yj −(Pmj −Pmj−1)yj∥≤1/jfor all j ≥1 (P0 = 0). Let vj = (Pmj −Pmj−1)yj.

Then (yj) has a subsequence equiva-lent to a subsequence of (vj). But, writing the basis of F as (en), it is clear that (vj)is equivalent to the sequence (Pmjn=mj−1+1 ∥vj(n)∥en) in F. Since F is p-saturated, weconclude that any subsequence of (vj) has a ℓp-block basis.

Thus, the same can be saidof (xk), and the proof is complete.2Tensor products of c0-saturated spacesFor Banach spaces E and F, let Kw∗(E′, F) denote the space of all compact weak*-weakly continuous operators from E′ into F, endowed with the operator norm. Theǫ-tensor product E ˜⊗ǫF is the closure in Kw∗(E′, F) of the finite rank operators that3

belong to Kw∗(E′, F). These spaces are equal if either E or F has the approximationproperty [11].

In this section and the next, we investigate special cases of the following:Problem: Is Kw∗(E′, F) (or E ˜⊗ǫF) c0-saturated if both E and F are?A Banach space is polyhedral if the unit ball of every finite dimensional subspace isa polyhedron. It is isomorphically polyhedral if it is isomorphic to a polyhedral Banachspace.

Our interest in isomorphically polyhedral spaces arises from the following resultof Fonf [2].Theorem 3 (Fonf) An isomorphically polyhedral Banach space is c0-saturated.A subset W of the dual of a Banach space E is said to be isomorphically preciselynorming (i.p.n.) if W is bounded and(a) there exists K < ∞such that ∥x∥≤K supw∈W |w(x)| for all x ∈E,(b) the supremum supw∈W |w(x)| is attained at some w0 ∈W for all x ∈E.This terminology was introduced by Rosenthal [9, 10] to provide a succint formulationof the following result of Fonf [3].Theorem 4 (Fonf) A separable Banach space E is isomorphically polyhedral if andonly if E′ contains a countable i.p.n.

subset.In this section, we consider the space Kw∗(E′, F) when one of the spaces E or Fis isomorphically polyhedral, and the other is c0-saturated.Note the symmetry inthe situation as Kw∗(E′, F) is isometric to Kw∗(F ′, E) via the mapping T 7→T ′. ForLemma 5, note that if x′ ∈E′ and y′ ∈F ′, the pair (x′, y′) defines a functional onKw∗(E′, F) by T 7→⟨Tx′, y′⟩.Lemma 5 Let E, F be Banach spaces, and let W and V be i.p.n.

subsets of E′ and F ′respectively. Then W × V is an i.p.n.

subset of (Kw∗(E′, F))′.Proof: It is clear that if both W and V satisy (a) of the definition of an i.p.n. set withconstant K, then W × V also satisfies it with constant K2.

Now assume that both Wand V satisfy part (b) of the definition. It is easy to see that (x′, y′) 7→⟨Tx′, y′⟩is acontinuous function on Ww∗×Vw∗, where both Ww∗and Vw∗are given their respectiveweak* topologies.

Since Ww∗×Vw∗is compact, there exists (w0, v0) ∈Ww∗×Vw∗suchthatsup(x′,y′)∈W ×V|⟨Tx′, y′⟩| =sup(x′,y′)∈Ww∗×Vw∗|⟨Tx′, y′⟩| = |⟨Tw0, v0⟩|.4

Now there exists v ∈V such that|⟨Tw0, v⟩|=supy′∈V|⟨Two, y′⟩|=supy′∈Vw∗|⟨Two, y′⟩|≥|⟨Tw0, v0⟩|.Similarly, there exists w ∈W such that|⟨Tw, v⟩|=|⟨w, T ′v⟩|=supx′∈W|⟨x′, T ′v⟩|=supx′∈Ww∗|⟨x′, T ′v⟩|≥|⟨Tw0, v⟩|.Combining the above, we see that |⟨Tw, v⟩| ≥sup(x′,y′)∈W ×V |⟨Tx′, y′⟩|. Since the re-verse inequality is obvious, the proof is complete.Lemma 6 Let (xn) be a non-convergent sequence in a c0-saturated Banach space F.There exists a normalized block (uk) = (Pnkn=nk−1+1 bnxn) of (xn) which is equivalent tothe c0-basis.Proof: Going to a subsequence, we may assume that infm̸=n ∥xm −xn∥> 0.ByRosenthal’s Theorem [8], we may also assume that (xn) is weakly Cauchy.

Let yn =(x2n−1 −x2n)/∥x2n−1 −x2n∥for all n. Then (yn) is a weakly null normalized block of(xn). Without loss of generality, we may assume that (yn) is a basic sequence.

SinceF is c0-saturated, (yn) has a normalized block basis (uk) equivalent to the c0-basis. As(uk) is also a normalized block of (xn), the proof is complete.Lemma 7 Let E, F be Banach spaces so that F is c0-saturated, and let (Tn) be anormalized basic sequence in Kw∗(E′, F).

For every x′ ∈E′, there is a normalizedblock basis (Sn) of (Tn), and a constant C, such that∥XanSnx′∥≤C supn |an|for all (an) ∈c00.5

Proof: There is no loss of generality in assuming that ∥x′∥= 1.Case 1 (Tnx′) converges.We may assume without loss of generality that ∥(T2n−1 −T2n)x′∥≤2−n for all n. Letǫn = ∥T2n−1 −T2n∥. Since (Tn) is a normalized basic sequence, ǫ ≡infn ǫn > 0.

Nowlet Sn = ǫ−1n (T2n−1 −T2n) for all n. Then (Sn) is a normalized block basis of (Tn).Furthermore, for any (an) ∈c00,∥XanSnx′∥≤ǫ−1 supn |an|X∥(T2n−1 −T2n)x′∥≤ǫ−1 supn |an|.Case 2 (Tnx′) does not converge.By Lemma 6, (Tnx′) has a normalized block (uk) = (Pnkn=nk−1+1 bnTnx′) which is equiv-alent to the c0-basis. Let Rk = (Pnkn=nk−1+1 bnTn).

Then ∥Rk∥≥∥Rkx′∥= ∥uk∥= 1.Let Sk = Rk/∥Rk∥for all k. Then (Sk) is a normalized block basis of (Tn), and∥XakSkx′∥= ∥Xak∥Rk∥uk∥≤C supk|ak|for all (ak) ∈c00, since (uk) is equivalent to the c0-basis and ∥Rk∥≥1.Theorem 8 Let E, F be Banach spaces so that E is isomorphically polyhedral and Fis c0-saturated. Then Kw∗(E′, F) is c0-saturated.Proof: Let (Tn) be a normalized basic sequence in Kw∗(E′, F).

It is easily seen thatG = [∪T ′nF ′] is a separable subspace of E, and that the sequences (Tn) and (Tn|G) areequivalent. Thus we may assume that E is separable.

Then E′ contains a countablei.p.n. subset W. Write W = (wm).

By Lemma 7, (Tn)∞n=1 has a normalized block basis(T (1)n ) such that∥XanT (1)n w1∥≤C1 supn |an|for some constant C1 for all (an) ∈c00. Inductively, if (T (m)n) has been chosen, let(T (m+1)n) be a normalized block basis of (T (m)n)∞n=2 such that there is a constant Cm+1satisfying∥XanT (m+1)nwm+1∥≤Cm+1 supn |an|6

for all (an) ∈c00. Now let Sm = T (m)1for all m ∈NI .

Then (Sm) is a normalizedblock basis of (Tn). Also, for all k, (Sm)∞m=k is a block basis of (T (k)n ).

Fix k, andwrite Sm =Pjmn=jm−1+1 bnT (k)nfor all m ≥k. Since (Sm) is normalized, and (T (k)n ) isnormalized basic, (bi) is bounded.

Therefore, by the choice of (T (k)n ),∞Xm=kamSmwk=∞Xm=kamjmXn=jm−1+1bnT (k)n wk≤Ck supm≥ksupjm−1

But by Lemma 5,W ×UF ′ is an i.p.n. subset of Kw∗(E′, F).

Applying Elton’s extremal criterion ([1], seealso [10, Theorem 18]), we see that [Sm] contains a copy of c0.Recall that a subset A of topological space X is dense-in-itself if every point of Ais an accumulation point of A. A is scattered if it contains no non-empty dense-in-itselfsubset.Corollary 9 Let K be a compact Hausdorffspace, and let E be a Banach space.

ThenC(K, E) is c0-saturated if and only if both C(K) and E are c0-saturated.Proof: The “only if” part is clear, since both C(K) and E embed in C(K, E). Nowassume that both C(K) and E are c0-saturated.

To begin with, assume additionallythat C(K) is separable. Then K is metrizable [12, Proposition II.7.5].

If K is notscattered, by [13, Theorem 8.5.4], there is a continuous surjection φ of K onto [0, 1].Then f 7→f ◦φ is an isometric embedding of C[0, 1] into C(K). This contradicts thefact that C(K) is c0-saturated.

Thus K is scattered. By [13, Theorem 8.6.10], K ishomeomorphic to a countable compact ordinal.

In particular, K is countable. HenceC(K)′ contains a countable i.p.n.

subset, namely, {δk : k ∈K}, where δk denotes theDirac measure concentrated at k. Therefore, C(K) is isomorphically polyhedral, andC(K, E) = Kw∗(C(K)′, E) is c0-saturated by Theorem 8.If C(K) is non-separable, as in the proof of the theorem, it suffices to show thatKw∗(G′, E) contains a copy of c0 for an arbitrary separable closed subspace G of C(K).However, Kw∗(G′, E) is isometric to Kw∗(E′, G), which clearly embeds in Kw∗(E′, F)7

for any closed subspace F of C(K) containing G. Take F to be the closed sublatticegenerated by G and the constant 1 function. By Kakutani’s Representation Theorem[12, Theorem II.7.4], F is lattice isometric to some C(H).

Note that C(H) is separablesince F is. Therefore, Kw∗(E′, F), which is isometric to Kw∗(F ′, E) = C(H, E), is c0-saturated by the above.

Since Kw∗(G′, E) is isomorphic to a subspace of Kw∗(E′, F),it contains a copy of c0.3The space JH ˜⊗ǫJHIn view of Theorem 8 in §2, it is interesting to consider spaces Kw∗(E′, F) whereboth E and F are c0-saturated, but neither is isomorphically polyhedral.In thissection, we investigate one such case. Namely, when E = F = JH, the James Haglerspace.In [6], it was shown that JH ˜⊗ǫJH = Kw∗(JH′, JH) does not contain anisomorph of ℓ1.

We show here that, in fact, JH ˜⊗ǫJH is c0-saturated. The proof usesElton’s extremal criterion for weak unconditional convergence, and the “diagonalizationtechnique” employed by Hagler to show that every normalized weakly null sequence inJH has a c0-subsequence.Let us recall the definition of the space JH, as well as fix some terms and notation.Let T = ∪∞n=0{0, 1}n be the dyadic tree.

The elements of T are called nodes. If φ is anode of the form (ǫi)ni=1, we say that φ has length n and write |φ| = n. The length ofthe empty node is defined to be 0.

For φ, ψ ∈T with φ = (ǫi)ni=1 and ψ = (δi)mi=1, wesay that φ ≤ψ if n ≤m and ǫi = δi for 1 ≤i ≤n. The empty node is ≤φ for allφ ∈T.

We write φ < ψ if φ ≤ψ and φ ̸= ψ. Two nodes φ and ψ are incomparable ifneither φ ≤ψ nor ψ ≤φ hold.

If φ ≤ψ, we say that φ is an ancestor of ψ, while ψ isa descendant of φ. For any φ ∈T, let ∆φ be the set of all descendants of φ.

If φ ≤ψ,letS(φ, ψ) = {ξ : φ ≤ξ ≤ψ}.A set of the form S(φ, ψ) is called a segment, or more specifically, a m-n segmentprovided |φ| = m, and |ψ| = n.A branch is a maximal totally ordered subset ofT.The set of all branches is denoted by Γ.A branch γ (respectively, a segmentS) is said to pass through a node φ if φ ∈γ (respectively, φ ∈S). If x : T →RI is afinitely supported function and S is a segment, we define (with slight abuse of notation)Sx = Pφ∈S x(φ).

Similarly, if γ ∈Γ, we define γ(x) = Pφ∈γ x(φ). A set of segments{S1, .

. .

, Sr} is admissible if they are pairwise disjoint, and there are m, n ∈NI ∪{0}such that each Si is a m-n segment. The James Hagler space JH is defined as the8

completion of the set of all finitely supported functions x : T →RI under the norm:∥x∥= sup( rXi=1|Six| : S1, . .

. , Sr is an admissible set of segments).Clearly, all S and γ extend to norm 1 functionals on JH.

Finally, if x : T →RIisfinitely supported, and n ≥0, let Pnx : T →RI be defined by(Pnx)(φ) =(x(φ)if |φ| ≥n0otherwise.Obviously, Pn extends uniquely to a norm 1 projection on JH, which we denote againby Pn.We begin with some lemmas on “node management”.Let φ and ψ be nodes.Denote by A(φ, ψ) denote the unique node of maximal length such that A(φ, ψ) ≤φand A(φ, ψ) ≤ψ. A sequence of nodes (φn) is a strongly incomparable sequence if(a) φn and φm are incomparable if n ̸= m,(b) no family of admissible segments passes through more than two of the φn’s.The first lemma is due to Hagler [4, Lemma 2].Lemma 10 (Hagler) Let (φn) be a sequence of nodes with strictly increasing lengths.Then there exists N ∈P∞(NI ) such that either (φn)n∈N determines a unique branch ofT, or (φn)n∈N is a strongly incomparable sequence.Lemma 11 Let (φ(n)) be a strongly incomparable sequence of nodes such that |φ(n)| <|φ(n + 1)| for all n. Then for all m ≥n ≥1, φ(m) ∈∆A(φ(n),φ(n+1)).Proof: Otherwise, there exist m ≥n ≥1 such that φ(m) /∈∆A(φ(n),φ(n+1)).

In par-ticular, note that m ≥n + 2. Let φ1 = φ(n), and let φ2 and φ3 be the ancestorsof φ(n + 1) and φ(m) respectively of length |φ(n)|.

Then φ1 ̸= φ2 since φ(n) andφ(n + 1) are incomparable. Also φ1 ̸= φ3 and φ2 ̸= φ3 since φ1, φ2 ∈∆A(φ(n),φ(n+1)),while φ3 /∈∆A(φ(n),φ(n+1)).

Let ψ1, ψ2 be nodes of length |φ(m)| which are ≥φ(n) andφ(n+ 1) respectively, and let ψ3 = φ(m). Then {S(φi, ψi) : i = 1, 2, 3} is an admissibleset of segments.

However, φ(n) ∈S(φ1, ψ1), φ(n+1) ∈S(φ2, ψ2), and φ(m) ∈S(φ3, ψ3),violating the strong incomparability of (φ(k)).9

Lemma 12 Let m ∈NI , and let (φ(i, j))∞i=1 be a strongly incomparable sequence ofnodes for 1 ≤j ≤m.Assume that |φ(i, j)| = |φ(i, j′)| < |φ(i + 1, j)| wheneveri, j, j′ ∈NI , 1 ≤j, j′ ≤m, and that for each i, {φ(i, j) : 1 ≤j ≤m} are pairwisedistinct. Then there exists k0 such that {φ(i, j) : i ≥k0, 1 ≤j ≤m} are pairwiseincomparable.Proof: Induct on m.If m = 1, there is nothing to prove.Now assume that thestatement is true for m −1 (m ≥2).

Let {φ(i, j) : i ≥1, 1 ≤j ≤m} be as given.Without loss of generality, we may assume that {φ(i, j) : i ≥1, 1 ≤j ≤m −1} arepairwise incomparable. First observe that if φ(i1, j1) < φ(i2, j2) for some j1, j2 ≤m,then A(φ(i2, j2), φ(i2 + 1, j2)) ≥φ(i1, j1).

Indeed, since A(φ(i2, j2), φ(i2 + 1, j2)) andφ(i1, j1) share the same descendant φ(i2, j2), they are comparable. Hence if the claimfails, A(φ(i2, j2), φ(i2+1, j2)) < φ(i1, j1).

But then the ancestor of φ(i2+1, j2) of length|φ(i1, j1)|, φ(i1, j1), and φ(i1, j2), are distinct. From this it is easy to construct an admis-sible set of segments which pass through all three nodes {φ(i1, j2), φ(i2, j2), φ(i2+1, j2)},in violation of their strong incomparability.

This proves the claim. In particular, underthe circumstances, Lemma 11 implies φ(i, j2) ∈∆φ(i1,j1) for all i ≥i2.

The remainderof the proof is divided into two cases.Case 1There exist j1 < m and i1, i2 ∈NIsuch that φ(i1, j1) ≤φ(i2, m).Note that since {φ(i1, j) : j ≤m} are pairwise distinct, we must have φ(i1, j1) <φ(i2, m). By the observation above, we obtain that φ(i, m) ∈∆φ(i1,j1) for all i ≥i2.However, for any i′ ≥i2, j < m, φ(i′, j) is incomparable with φ(i1, j1) by induction.Hence φ(i′, j) /∈∆φ(i1,j1).

Thus φ(i, m) and φ(i′, j) are incomparable whenever i, i′ ≥i2and j < m. This is enough to show that {φ(i, j) : i ≥i2, 1 ≤j ≤m} are pairwiseincomparable.Case 2For all j1 < m and i1, i2 ∈NI , φ(i1, j1) ̸≤φ(i2, m).Let I = {i : there exist i′ ∈NI , j < m such that φ(i, m) < φ(i′, j)}. Let i1 and i2be distinct elements of I.

Choose i′1, i′2 ∈NI , j1, j2 < m such that φ(ik, m) < φ(i′k, jk),k = 1, 2. By the observation above, φ(i, jk) ∈∆φ(ik,m) for all i ≥i′k, k = 1, 2.

Nowφ(i1, m) and φ(i2, m) are incomparable by assumption. Therefore, ∆φ(i1,m) ∩∆φ(i2,m) =∅.Combined with the above, we see that j1 ̸= j2.It follows that |I| ≤m −1.Now choose k0 such that i < k0 for all i ∈I.

By the case assumption, φ(i, m) isincomparable with φ(i′, j) whenever i ≥k0 and j < m. This is enough to show that{φ(i, j) : i ≥k0, 1 ≤j ≤m} are pairwise incomparable.10

Lemma 13 Let (xn) be a bounded weakly null sequence in JH so that there is a se-quence 0 = j1 < j2 < j3 < · · · with xn ∈(Pjn −Pjn+1)JH for all n. Assume thatsupξ∈Γ |⟨xn, ξ⟩| ≤ǫ for all n. Then there is a subsequence (xnk) such thatsupξ∈Γ∞Xk=1|⟨xnk, ξ⟩| ≤7ǫ.Proof: For m, n ∈NI , letF(n, m)={φ ∈{0, 1}jn : there exists at least one branch ξ through φ with|⟨xn, ξ⟩| > ǫ/2m, and for all branches ξ through φ, |⟨xn, ξ⟩| ≤ǫ/2m−1}.Since (xn) is bounded, supn |F(n, m)| < ∞for all m. Let N1 ∈P∞(NI ) be such that(|F(n, 1)|)n∈N1 is constant, say, b1. Write F(n, 1) = {φ(n, 1, i) : 1 ≤i ≤b1} for alln ∈N1.

Choose N′1 ∈P∞(N1) so that for 1 ≤i ≤b1, (φ(n, 1, i))n∈N′1 is either a stronglyincomparable sequence or determines a branch. LetI1 = {i ≤b1 : (φ(n, 1, i))n∈N′1 determines a branch},and let Γ1 be the set of branches determined by some (φ(n, 1, i))n∈N′1 for some i ∈I1.Let L1 = {1, .

. .

, b1}\I1. By Lemma 12, there exists N′′1 ∈P∞(N′1) such that{φ(n, 1, i) : n ∈N′′1 , i ∈L1} are pairwise incomparable.

Finally, since Γ1 is finite, thereexists n1 ∈N′′1 such that |⟨xn1, γ⟩| ≤ǫ/2 for all γ ∈Γ1. Continue inductively to obtainn1 < n2 < · · ·, numbers b1, b2, .

. ., and sets I1, I2, .

. ., L1, L2, .

. ., and Γ1, Γ2, .

. .

so that1. for all k ≥m, |F(nk, m)| = bm,2.

for all k ≥m, F(nk, m) = {φ(nk, m, i) : i ≤bm}, where (φ(nk, m, i))∞k=mdetermines a branch if i ∈Im, and is a strongly incomparable sequence ifi ∈Lm = {1, . .

. , bm}\Im,3.

for all m, Γm is the set of branches determined by (φ(nk, m, i))∞k=m for somei ∈Im,4. for all m, {φ(nk, m, i) : m ≤k, i ∈Lm} are pairwise incomparable,5.

for all k, |⟨xnk, γ⟩| ≤ǫ/2k for every γ ∈Γ1 ∪. .

. ∪Γk.11

For all k, let G(k) = {0, 1}jnk\ ∪km=1 F(nk, m). Then {F(nk, 1), .

. .

, F(nk, k), G(k)} isa partition of {0, 1}jnk for each k. Fix ξ ∈Γ. Say ξ = (φn)∞n=0, where |φn| = n for alln ≥0.

Now let J0 = {k : φnk ∈G(k)}, and let Jm = {k ≥m : φnk ∈F(nk, m)} for allm ≥1. Then (Jm)∞m=0 is a partition of NI .

For all k such that k ∈Jm for some m ≥1,choose ik such that φnk = φ(nk, m, ik). For all m ≥1, let Jm,1 = {k ∈Jm : ik ∈Im},and let Jm,2 = {k ∈Jm : ik ∈Lm}.

Fix m ≥1 such that Jm,1 ̸= ∅.Case 1Jm,1 is infinite.In this case, there exists γ ∈Γm such that γ = ξ. Therefore,Xk∈Jm,1|⟨xnk, ξ⟩|≤∞Xk=m|⟨xnk, ξ⟩| =∞Xk=m|⟨xnk, γ⟩|≤∞Xk=mǫ2k =ǫ2m−1.Case 2Jm,1 is finite.Let k0 = max Jm,1. There exists γ ∈Γm such that φnk0 ∈γ.

Then, since φnk0 ∈F(nk0, m),Xk∈Jm,1|⟨xnk, ξ⟩|≤k0Xk=m|⟨xnk, ξ⟩| =k0−1Xk=m|⟨xnk, γ⟩| + |⟨xnk0, ξ⟩|≤∞Xk=mǫ2k +ǫ2m−1 =ǫ2m−2.Therefore, in either case, we haveXk∈Jm,1|⟨xnk, ξ⟩| ≤ǫ2m−2. (2)Now suppose for some m ≥1, there are distinct k1, k2 ∈Jm,2.

Then k1, k2 ≥m, andφnkl = φ(nkl, m, ikl), for some ikl ∈Lm, l = 1, 2. But then by choice, φnk1 and φnk2must be incomparable.

This is a contradiction since they both belong to the branchξ. Thus, for all m ≥1, |Jm,2| ≤1.

Now k ∈Jm,2 implies φnk ∈F(nk, m), and hence|⟨xnk, ξ⟩| ≤ǫ/2m−1. Consequently, for all m ≥1Xk∈Jm,2|⟨xnk, ξ⟩| ≤ǫ2m−1.

(3)12

Finally,Xk∈J0|⟨xnk, ξ⟩| ≤∞Xk=1ǫ2k = ǫ. (4)Combining inequalities (2)–(4), we obtainXk|⟨xnk, ξ⟩|≤∞Xm=1Xk∈Jm|⟨xnk, ξ⟩| +Xk∈J0|⟨xnk, ξ⟩|≤∞Xm=13ǫ2m−1 + ǫ = 7ǫ.For n ≥0, call a subset D of {0, 1}n ×{0, 1}n diagonal if whenever (φ1, ψ1), (φ2, ψ2)are distinct elements of D, then φ1 ̸= φ2, and ψ1 ̸= ψ2.Lemma 14 Let n ≥0, ǫ > 0, and let T : JH′ →JH be normalized.

LetA={(φ, ψ) ∈{0, 1}n × {0, 1}n : there exists γ, ξ ∈Γ,φ ∈γ, ψ ∈ξ, such that |⟨TP ′nγ, P ′nξ⟩| > ǫ}.Then |D| ≤1/ǫ for all diagonal subsets D of A.Proof: Let D = {(φi, ψi) : 1 ≤i ≤k} be a diagonal subset of A. For each i, chooseγi, ξi ∈Γ such that φi ∈γi, ψi ∈ξi, and |⟨TP ′nγi, P ′nξi⟩| > ǫ.

Using the diagonal-ity of D, we see that (P ′nγi)ki=1 and (P ′nξi)ki=1 are both isometrically equivalent to theℓ∞(n)-basis. For 1 ≤i, j ≤k, let aij = ⟨TP ′nγi, P ′nξj⟩.

Define S, R : ℓ∞(k) →ℓ1(k)by S(b1, . .

. , bk) = (Pkj=1 aijbj)ki=1 and R(b1, .

. .

, bk) = (aiibi)ki=1 respectively.Then∥S∥≤∥T∥= 1, and ∥R∥≥kǫ.However, by [7, Proposition 1.c.8], ∥R∥≤∥S∥.Therefore, |D| = k ≤1/ǫ.Lemma 15 Let (Tn) be a normalized weakly null sequence in JH ˜⊗ǫJH such that thereis a sequence 0 = j1 < j2 < · · · with (Pjn −Pjn+1)Tn(Pjn −Pjn+1)′ = Tn for all n. Thenthere is a subsequence (Tnk) such thatX|⟨Tnkγ, ξ⟩| < ∞for all γ, ξ ∈Γ.13

Proof: Note that ∥Tn∥= 1 implies |⟨Tnγ, ξ⟩| ≤1 for all γ, ξ ∈Γ. For all m, n ∈NI , letA(n, m)={(φ, ψ) ∈{0, 1}jn × {0, 1}jn : there exist branches γ and ξ throughφ and ψ respectively so that |⟨Tnγ, ξ⟩| > 1/2m, and for all branchesγ and ξ through φ and ψ respectively, |⟨Tnγ, ξ⟩| ≤1/2m−1}.Fix n. Let B(n, 1) be a maximal diagonal subset of A(n, 1).

Then letC(n, 1) =[(φ,ψ)∈B(n,1){(φ1, ψ1) : φ1 = φ or ψ1 = ψ}.Inductively, if B(n, k) and C(n, k) have been chosen for k < m, let B(n, m) be amaximal diagonal subset of A(n, m)\ ∪m−1k=1 C(n, k). Then letC(n, m) =[(φ,ψ)∈B(n,m){(φ1, ψ1) : φ1 = φ or ψ1 = ψ}.It is easily seen that(a) ∪∞m=1B(n, m) is a diagonal subset of {0, 1}jn × {0, 1}jn,(b) for all m, B(n, m) ⊆A(n, m) ∩C(n, m),(c) for all k, ∪km=1A(n, m) ⊆∪km=1C(n, m).In particular, by (b) and Lemma 14, |B(n, m)| ≤2m for all m, n. Also, if (φ, ψ) ∈{0, 1}jn × {0, 1}jn\ ∪km=1 C(n, m), then (c) implies (φ, ψ) /∈∪km=1A(n, m).

Hence ifγ, ξ ∈Γ pass through φ and ψ respectively, then|⟨Tnγ, ξ⟩| ≤1/2k. (5)Now choose N1 ∈P∞(NI ) such that |B(n, 1)| is a constant, say b1, for all n ∈N1.

WriteB(n, 1) = {(φ(n, 1, i), ψ(n, 1, i)) : 1 ≤i ≤b1}for all n ∈N1. There exists N′1 ∈P∞(N1) such that for each i ≤b1, (φ(n, 1, i))n∈N′1 aswell as (ψ(n, 1, i))n∈N′1 are either strongly incomparable or determine a branch.

LetI1(φ) = {1 ≤i ≤b1 : (φ(n, 1, i))n∈N′1 determines a branch. }Let the set of branches so determined be denoted by Γ1(φ).

Define I1(ψ) and Γ1(ψ)similarly with regard to the sequence (ψ(n, 1, i))n∈N′1. Since (Tn) is weakly null, so are14

(Tnγ) and (T ′nξ) for all γ, ξ ∈Γ. Thus, because both Γ1(φ) and Γ1(ψ) are finite sets,Lemma 13 yields a set N′′1 ∈P∞(N′1) such thatsupξ∈ΓXn∈N′′1|⟨Tnγ, ξ⟩| ≤7for allγ ∈Γ1(φ)andsupγ∈ΓXn∈N′′1|⟨γ, T ′nξ⟩| ≤7for allξ ∈Γ1(ψ).Inductively, if N′′m ∈P∞(N) has been chosen, let Nm+1 ∈P∞(N′′m) be such that|B(n, m + 1)| is a constant, say bm+1, for all n ∈Nm+1.

For n ∈Nm+1, listB(n, m + 1) = {(φ(n, m + 1, i), ψ(n, m + 1, i)) : 1 ≤i ≤bm+1},choose N′m+1 ∈P∞(Nm+1) such that for each i ≤bm+1, (φ(n, m + 1, i))n∈N′m+1 as wellas (ψ(n, m + 1, i))n∈N′m+1 are either strongly incomparable or determine a branch. LetIm+1(φ) = {1 ≤i ≤bm+1 : (φ(n, m + 1, i))n∈N′m+1 determines a branch.

}Let the set of branches so determined be denoted by Γm+1(φ). Define Im+1(ψ) andΓm+1(ψ) similarly with regard to the sequence (ψ(n, m + 1, i))n∈N′m+1.

If γ ∈Γm+1(φ),ξ ∈Γ, and n ∈N′m+1, let φ and ψ be the nodes of length jn in γ and ξ respectively.Then φ = φ(n, m + 1, i) for some i. But (φ(n, m + 1, i), ψ(n, m + 1, i)) ∈B(n, m + 1).Therefore, (φ, ψ) ∈C(n, m + 1).

In particular, (φ, ψ) /∈∪mk=1C(n, k). By equation (5),|⟨Tnγ, ξ⟩| ≤1/2m.

Similarly, |⟨γ, T ′nξ⟩| ≤1/2m for all γ ∈Γ and ξ ∈Γm+1(ψ). ByLemma 13, there is a set N′′m+1 ∈P∞(N′m+1) such thatsupξ∈ΓXn∈N′′m+1|⟨Tnγ, ξ⟩| ≤7/2mfor allγ ∈Γm+1(φ),(6)andsupγ∈ΓXn∈N′′m+1|⟨γ, T ′nξ⟩| ≤7/2mfor allξ ∈Γm+1(ψ).

(7)Pick n1 < n2 < · · · such that nm ∈N′′m for all m, and letD(m) = {0, 1}jnm × {0, 1}jnm\m[k=1C(nm, k).For all m, {C(nm, 1), . .

. , C(nm, m), D(m)} is a partition of {0, 1}jnm × {0, 1}jnm.

Fixγ, ξ ∈Γ. We proceed to estimatePm |⟨Tnmγ, ξ⟩|.

For all m ∈NI , let φm and ψm be15

the nodes of length jnm in γ and ξ respectively. Define J0 = {m : (φm, ψm) ∈D(m)},and Jk = {m ≥k : (φm, ψm) ∈C(nm, k)} for all k ≥1.

Note that {J0, J1, J2, . .

.} is apartition of NI .

If m ∈J0, then equation (5) yields |⟨Tnmγ, ξ⟩| ≤1/2m. Consequently,Xm∈J0|⟨Tnmγ, ξ⟩| ≤Xm∈J012m ≤1.

(8)Now fix k ≥1 and m ∈Jk. Then m ≥k and hence nm ∈Nk.

Thus B(nm, k) islisted as {(φ(nm, k, i), ψ(nm, k, i)) : 1 ≤i ≤bk}. But (φm, ψm) ∈C(nm, k).

Hencethere exists 1 ≤im ≤bk such that either φm = φ(nm, k, im) or ψm = ψ(nm, k, im). LetJk(φ) = {m ∈Jk : φm = φ(nm, k, im)}, and let Jk(ψ) = Jk\Jk(φ).

Since nm ∈N′k aswell, we may further subdivide these sets into:Jk,1(φ)={m ∈Jk(φ) : im ∈Ik(φ)},Jk,2(φ)=Jk(φ)\Jk,1(φ),Jk,1(ψ)={m ∈Jk(ψ) : im ∈Ik(ψ)},Jk,2(ψ)=Jk(ψ)\Jk,1(ψ).Now {φ(nm, k, im) : m ∈Jk,2(φ)} is a subset of the branch γ. But it is also containedin the union of the strongly incomparable sequences (φ(n, k, i))n∈N′k, i /∈Ik(φ).

Hence|Jk,2(φ)| ≤2bk. Recalling equation (5), we obtainXm∈Jk,2(φ)|⟨Tnmγ, ξ⟩| ≤Xm∈Jk,2(φ)12k−1 ≤4bk2k .

(9)Similarly,Xm∈Jk,2(ψ)|⟨Tnmγ, ξ⟩| ≤4bk2k . (10)For any m ∈Jk,1(φ), φm belongs to a branch in Γk(φ).

Let ˜γ be a branch in Γk(φ)such that M(˜γ) = {m ∈Jk,1(φ) : φm ∈˜γ} is non-empty. If M(˜γ) is finite, let m0 beits maximal element.

Then φm0 belongs to both branches ˜γ and γ. Therefore,Xm∈M(˜γ)|⟨Tnmγ, ξ⟩|=Xm∈M(˜γ)\{m0}|⟨Tnm˜γ, ξ⟩| + |⟨Tnm0γ, ξ⟩|≤supξ∈ΓXn∈N′′k|⟨Tn˜γ, ξ⟩| + |⟨Tnm0γ, ξ⟩|≤72k−1 +12k−1 = 162k16

by equations (6) and (5) respectively. On the other hand, if M(˜γ) is infinite, then ˜γand γ coincide.

Thus the term containing Tnm0 may simply be omitted from the aboveinequality. Now since |Γk(φ)| ≤bk, we obtainXm∈Jk,1(φ)|⟨Tnmγ, ξ⟩| ≤16bk2k .

(11)Similarly,Xm∈Jk,1(ψ)|⟨Tnmγ, ξ⟩| ≤16bk2k . (12)Combining equations (8)–(12), we see thatX|⟨Tnmγ, ξ⟩| ≤∞Xk=0Xm∈Jk|⟨Tnmγ, ξ⟩| ≤1 +∞Xk=140bk2k .To complete the proof, it remains to show that P bk/2k < ∞.

Fix m ∈NI . By property(a) of the sets B(n, k), ∪mk=1B(nm, k) is a diagonal subset of {0, 1}jnm ×{0, 1}jnm.

Hence{φ(nm, k, i) : i ≤bk, k ≤m} are all distinct, as are {ψ(nm, k, i) : i ≤bk, k ≤m}. Bythe definition of B(nm, k), one can choose, for any i ≤bk, k ≤m, branches γk,i andξk,i, passing through φ(nm, k, i) and ψ(nm, k, i) respectively, so that |⟨Tnmγk,i, ξk,i⟩| >1/2k.

Keeping in mind the assumption on (Tn), we see that this inequality remainsvalid if γk,i and ξk,i are replaced by δk,i = P ′jnmγk,i and ζk,i = P ′jnmξk,i respectively.Since (δk,i) and (ζk,i) are isometrically equivalent to the ℓ∞(Pmk=1 bk)-basis, the mapS : ℓ∞(Pmk=1 bk) →ℓ1(Pmk=1 bk)S(bk,i)bk mi=1k=1 = (Xk,i⟨Tnmδk,i, ζk′,i′⟩bk,i)bk′ mi′=1k′=1has norm ≤∥Tnm∥= 1. Then [7, Proposition 1.c.8] implies that the “diagonal” of Salso has norm ≤1.

But this meansmXk=1bk2k ≤mXk=1bkXi=1|⟨Tnmδk,i, ζk′,i′⟩| ≤1.Since m is arbitrary,P bk/2k converges, as required.In order to apply Elton’s extremal criterion, we need the following. For convenience,we call an element of JH′ of the form P ′mγ, where m ≥0, and γ ∈Γ, a m-∞segment.17

Lemma 16 Let W be the collection of all elements in JH′ of the form Pri=1 aiSi,where r ∈NI , max |ai| ≤1, and there exist m, n, with n possibly equal to ∞, so that{S1, . .

. , Sr} is a set of pairwise disjoint m-n segments.

Then W is an i.p.n. subset ofJH′.

Consequently, W × W is an i.p.n. subset of JH ˜⊗ǫJH.Proof: The second assertion follows from the first by Lemma 5.

If Pri=1 aiSi ∈W, andx ∈JH, then|rXi=1aiSix| ≤rXi=1|Six| ≤∥x∥.To complete the proof, it suffices to show that for all x ∈JH, there is a collection{S1, . .

. , Sr} of disjoint m-n segments (n possibly = ∞) such that ∥x∥=Pri=1 |Six|.Let x ∈JH be fixed.

For each j, choose an admissible collection of mj-nj segmentsAj such that∥x∥= limjXS∈Aj|Sx|. (13)If (mj) is unbounded,PS∈Aj |Sx| ≤∥Pmjx∥→0 as j →∞.

Hence x = 0, and theresult is obvious. If (nj) is bounded, then so is (mj).

Without loss of generality, wemay assume that both (mj) and (nj) are constant sequences with finite values, say,m and n. But as there are only finitely many sets of admissible m-n segments, thelimiting value in (13) is attained, and the claim holds. Finally, we consider the casewhen (mj) is bounded and (nj) is unbounded.

Going to a subsequence, we may assumethat (mj) has a constant value, say m, and nj →∞. Then, for each j, |Aj| ≤2m.Using a subsequence again, we may assume that |Aj| = r for some fixed r for all j. Foreach j, write Aj = {S1(j), .

. .

, Sr(j)}. Choose a subsequence (jk) such that (Si(jk))kconverges weak* to some Si for every 1 ≤i ≤r.

It is easy to see that {S1, . .

. , Sr} isa collection of pairwise disjoint m-∞segments.

From equation (13), we deduce that∥x∥=Pri=1 |Six|, as desired.Lemma 17 Let (Tn) be as in Lemma 15, then [Tn] contains a copy of c0.Proof: Let W be as in Lemma 16. Choose a subsequence (Tnk) as given by Lemma 15.Then we have P |⟨Tnkw, v⟩| < ∞for all (w, v) ∈W × W. But by Lemma 16, W × Wis an i.p.n.

subset of JH ˜⊗ǫJH. Hence Elton’s extremal criterion [1] assures us that[Tnk] contains a copy of c0.18

Theorem 18 The space JH ˜⊗ǫJH is c0-saturated.Proof: For all m ∈NI , letEm = Kw∗(JH′, (1 −Pm)JH)andFm = Kw∗((1 −Pm)′JH′, JH).Both Em and Fm are isomorphic to a direct sum of a finite number of copies of JH,hence they are c0-saturated by Lemma 1. Let (Sn) be a normalized basic sequencein JH ˜⊗ǫJH.

If there exist a subsequence (Rn) of (Sn) and m ∈NIsuch that (Rn)is dominated by ((1 −Pm)Rn ⊕Rn(1 −Pm)′) ∈Em ⊕Fm, then (Rn) is equivalentto a sequence in Em ⊕Fm, which is a c0-saturated space by Lemma 1. Thus [Rn],and consequently [Sn], contains a copy of c0.

Otherwise, for all m ∈NIand everysubsequence (Rn) of Sn,inf(∥Xnan(1 −Pm)Rn∥+ ∥XnanRn(1 −Pm)′∥: (an) ∈c00, ∥XanRn∥= 1)= 0.Also, it is clear that for any T ∈JH ˜⊗ǫJH, limn(1 −Pn)T(1 −Pn)′ = T in norm. Us-ing these observations and a standard perturbation argument, we obtain a normalizedblock basis of (Sn) which is equivalent to some sequence (Tn) satisfying the hypothesesof Lemma 15.

By Lemma 17, [Tn] contains a copy of c0. Thus, so does [Sn].Hagler [4] proved that, in fact, JH has property (S): every normalized weakly nullsequence has a c0-subsequence.

Coupled with the absence of ℓ1, property (S) impliesc0-saturation. In [5], it was also shown that property (S) implies property (u).

Thus,we may askQuestion: Does JH ˜⊗ǫJH has property (S) or property (u)?References[1] J. Elton, Extremely weakly unconditionally convergent series, Israel J. Math.40(1981), 255-258. [2] V. F. Fonf, On a property of Lindenstrauss-Phelps spaces, Funct.

Anal. Appl.13(1979), 79-80 (translated from Russian).19

[3] V. F. Fonf, Polyhedral Banach spaces, Matematicheskie Zametki 30(1981), 627-634 (translated from Russian). [4] James Hagler, A counterexample to several questions about Banach spaces , StudiaMath.

60(1977), 289-308. [5] H. Knaust and E. Odell, On c0-sequences in Banach spaces, Israel J. Math.67(1989), 153-169.

[6] Denny H. Leung, Embedding ℓ1 into tensor products of Banach spaces, FunctionalAnalysis (eds. E. Odell and H. Rosenthal), Lecture Notes in Math., vol.

1470,Springer-Verlag, Berlin, 1991, 171-176. [7] Joram Lindenstrauss and Lior Tzafriri, “Classical Banach Spaces I, SequenceSpaces”, Springer-Verlag, Berlin, 1977.

[8] H. Rosenthal, A characterization of Banach spaces containing ℓ1, Proc. Nat.

Acad.Sci. (U.S.A.), 71(1974), 2411-2413.

[9] H. Rosenthal, Class Notes, Topics course in analysis, University of Texas at Austin. [10] H. Rosenthal, Some aspects of the subspace structure of infinite dimensional Ba-nach spaces, Approximation Theory and Functional Analysis (ed.

C. Chuy), Aca-demic Press, 1990. [11] W. Ruess, Duality and geometry of spaces of compact operators, Functional Anal-ysis: Surveys and Recent Results III, (eds.

K.-D. Bierstedt and B. Fuchssteiner)Math. Studies, no.

90, North-Holland, Amsterdam, 1984, 59-78. [12] H. H. Schaefer, “Banach Lattices and Positive Operators”, Springer-Verlag, Berlin,1974.

[13] Z. Semadeni, “Banach Spaces of Continuous Functions I”, Monografie Matematy-czne, PWN-Polish Scientific Publishers 55, Warszawa, 1971.Department of MathematicsNational University of SingaporeSingapore 0511E-mail(bitnet) : matlhh@nusvm20

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