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Some isomorphically polyhedral

arXiv:math/9304206v1 [math.FA] 1 Apr 1993Some isomorphically polyhedralOrlicz sequence spacesDenny H. LeungAbstractA Banach space is polyhedral if the unit ball of each of its finite dimensionalsubspaces is a polyhedron. It is known that a polyhedral Banach space has aseparable dual and is c0-saturated, i.e., each closed infinite dimensional subspacecontains an isomorph of c0.

In this paper, we show that the Orlicz sequence spacehM is isomorphic to a polyhedral Banach space if limt→0 M(Kt)/M(t) = ∞forsome K < ∞.We also construct an Orlicz sequence space hM which is c0-saturated, but which is not isomorphic to any polyhedral Banach space. Thisshows that being c0-saturated and having a separable dual are not sufficient fora Banach space to be isomorphic to a polyhedral Banach space.A Banach space is said to be polyhedral if the unit ball of each of its finite dimen-sional subspaces is a polyhedron.

It is isomorphically polyhedral if it is isomorphic to apolyhedral Banach space. Fundamental results concerning polyhedral Banach spaceswere obtained by Fonf [1, 2].Theorem 1 (Fonf) A separable isomorphically polyhedral Banach space is c0-saturatedand has a separable dual.Recall that a Banach space is c0-saturated if every closed infinite dimensional sub-space contains an isomorph of c0.

Fonf also proved a characterization of isomorphicallypolyhedral spaces in terms of certain norming subsets in the dual. In order to state therelevant results, we introduce some terminology due to Rosenthal [4, 5].

The (closed)unit ball of a Banach space E is denoted by UE.Definition Let E be a Banach space. (1) A subset W ⊆E′ is precisely norming (p.n.) if W ⊆UE′, and for all x ∈E, thereis a w ∈W such that ∥x∥= |w(x)|.

(2) A subset W ⊆E′ is isomorphically precisely norming (i.p.n.) if W is bounded and1991 Mathematics Subject Classification 46B03, 46B20, 46B45.1

(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.It is easy to see that W ⊆E′ is i.p.n. if and only if there is an equivalent norm||| · ||| on E so that W is p.n.

in (E, ||| · |||)′.Theorem 2 (Fonf) Let E be a separable Banach space. Then E is isomorphicallypolyhedral if and only if E′ contains a countable i.p.n.

subset.This paper is devoted mainly to the problem of identifying the isomorphically poly-hedral Orlicz sequence spaces. In §1, we prove a characterization theorem for isomor-phically polyhedral Banach spaces having a shrinking basis.This result is appliedin §2 to obtain examples of isomorphically polyhedral Orlicz spaces.

In §3, a non-isomorphically polyhedral, c0-saturated Orlicz sequence space is constructed.Sinceevery c0-saturated Orlicz sequence space has a separable dual, this shows that the con-verse of Theorem 1 fails, answering a question posed by Rosenthal [4].Standard Banach space terminology, as may be found in [3], is employed. If (en) isa basis of a Banach space E, and ||| · ||| is a norm on E equivalent to the given norm,we say that (en) is monotone with respect to ||| · ||| if |||Pkn=1 anen||| ≤|||Pk+1n=1 anen|||for every real sequence (an) and all k ∈NI .

Terms and notation regarding Orlicz spacesare discussed in §2.1A characterization theoremThis section is devoted to proving the following characterization theorem. Readersfamiliar with the proofs of Fonf’s Theorems will find the same ingredients used here.Theorem 3 Let (en) be a shrinking basis of a Banach space (E, ∥· ∥).

The followingare equivalent. (a) E is isomorphically polyhedral;(b) There exists an equivalent norm ||| · ||| on E such that (en) is a monotone basiswith respect to ||| · |||, and for all P anen ∈E, there exists m ∈NI such that|||∞Xn=1anen||| = |||mXn=1anen|||.2

Proof: Let (Pn) be the projections on E associated with the basis (en). The sequence(Pn) is uniformly bounded with respect to any equivalent norm on E.Also, (Pn)converges strongly to the identity operator on E, which we denote by 1.

Since (en) isshrinking, (P ′n) converges to 1′ strongly as well. (a) ⇒(b).

By renorming, and using Theorem 2, we may assume that E′ contains ap.n. sequence (wk).

Fix sequences (ǫk) and (δk) in (0, 1) which are both convergentto 0, and so that (1 + ǫk)(1 −2δk) > 1 for all k. For each k, choose nk such that∥(1 −Pn)′wk∥≤δk for all n ≥nk. Define a seminorm ||| · ||| on E by|||x||| = supk(1 + ǫk) max1≤n≤nk |⟨Pnx, wk⟩|.

(1)Since (wk) ⊆UE′, |||x||| ≤2∥x∥sup ∥Pn∥. On the other hand, if x ̸= 0, choose k suchthat ∥x∥= |wk(x)|.

Then∥x∥=|wk(x)|≤|⟨x, P ′nkwk⟩| + |⟨x, (1 −Pnk)′wk⟩|≤|⟨Pnkx, wk⟩| + δk∥x∥.Thus|||x||| ≥(1 + ǫk)(1 −δk)∥x∥> ∥x∥. (2)Hence ||| · ||| is an equivalent norm on E. It is clear that (en) is monotone with respectto ||| · |||.

We claim that this norm satisfies the remaining condition in (b). To thisend, we first show that the supremum in the definition (1) is attained.

This is trivialif x = 0. Fix 0 ̸= x ∈E.

Choose k1 ≤k2 ≤· · · and (ji) , 1 ≤ji ≤nki for all i, so that|||x||| = limi (1 + ǫki)|⟨Pjix, wki⟩|.We divide the proof into cases.Case 1limi ki = limi ji = ∞.In this case, Pjix →x in norm. Thereforelim supi|⟨Pjix, wki⟩| = lim supi|⟨x, wki⟩| ≤∥x∥.Also, ǫki →0 as i →∞.

Thus, |||x||| ≤∥x∥, contrary to (2).Case 2limi ki = ∞, limi ji ̸= ∞.By using a subsequence, we may assume that ji = j for all i. Then|||x||| = limi (1 + ǫki)|⟨Pjx, wki⟩| ≤∥Pjx∥.3

Now choose k such that ∥Pjx∥= |⟨Pjx, wk⟩|. If j ≤nk,|||x|||≥(1 + ǫk)|⟨Pjx, wk⟩|=(1 + ǫk)∥Pjx∥> ∥Pjx∥,a contradiction.

Now assume j > nk, then∥(Pj −Pnk)′wk∥≤∥(1 −Pj)′wk∥+ ∥(1 −Pnk)′wk∥≤2δk.Hence∥Pjx∥=|⟨Pjx, wk⟩|≤|⟨Pnkx, wk⟩| + 2δk∥x∥≤(1 + ǫk)−1|||x||| + 2δk|||x|||.Therefore,|||x||| ≤∥Pjx∥≤((1 + ǫk)−1 + 2δk)|||x||| < |||x|||,reaching yet another contradiction. Consequently, we must haveCase 3limi ki ̸= ∞.By using a subsequence, we may assume that the sequence (ki) is constant.

Then it isclear that the supremum in (1) is attained.Now for any x ∈E, choose k so that the supremum in (1) is attained at k. Thenit is clear that |||x||| = |||Pnkx|||. (b) ⇒(a).

Let (ηn) and (ǫn) be sequences convergent to 0, with 1 > ηn > ǫn > 0 forall n. For each n, there is a finite Wn ⊆U(E,|||·|||)′ such that(1 + ǫn)−1|||x||| ≤maxw∈Wn |w(x)| ≤|||x|||(3)for all x ∈span{e1, . .

. , en}.

Define a seminorm ρ on E byρ(x) = supn (1 + ηn) max1≤j≤n maxw∈Wj |⟨Pjx, w⟩|. (4)We will show that ρ is an equivalent norm on E, and the setW = {(1 + ηn)P ′jw : n ∈NI , 1 ≤j ≤n, w ∈Wj}4

is a countable p.n. subset of (E, ρ)′.

Then E is isomorphically polyhedral by Fonf’sTheorem (Theorem 2). Now let x ∈E.

By (b), there exists m such that |||x||| =|||Pmx|||. Hence, by (3), and the fact that (en) is monotone with respect to ||| · |||,|||x|||=|||Pmx|||≤(1 + ǫm) maxw∈Wm |⟨Pmx, w⟩|≤(1 + ηm) maxw∈Wm |⟨Pmx, w⟩|(5)≤ρ(x)≤2|||x|||.Thus ρ is an equivalent norm on E.Next we show that the supremum in (4) isattained.

Fix x ∈E. Choose sequences n1 ≤n2 ≤· · ·, (jk), and (wk) such that1 ≤jk ≤nk, wk ∈Wnk for all k, and ρ(x) = limk(1 + ηnk)|⟨Pjkx, wk⟩|.

First assumethat limk nk = ∞. Then ηnk →0.

Since (en) is monotone with respect to ||| · |||, wehave ρ(x) ≤|||x|||. But there exists k such that |||x||| = |||Pkx|||, and there is a w ∈Wksuch that |||Pkx||| ≤(1 + ǫk)|w(Pkx)|.

Thusρ(x) ≥(1 + ηk)|w(Pkx)| ≥1 + ηk1 + ǫk|||x||| > |||x|||,a contradiction. Therefore, limk nk ̸= ∞.

By going to a subsequence, we may assumethat (nk) is bounded. Using a further subequence if necessary, we may even assume itis constant.

Thus the supremum in (4) is attained. From this it readily follows thatthe set W is a p.n.

subset of (E, ρ)′. The countability of W is evident.

✷Remark The assumption that the basis (en) is shrinking is used only in the proofof (a) ⇒(b). If (en) is assumed to be unconditional and (a) holds, then (en) mustbe shrinking.

For otherwise E contains a copy of ℓ1, which contradicts (a) by Fonf’sTheorem (Theorem 1). Thus the assumption of shrinking is not needed if (en) is un-conditional.2Orlicz sequence spacesIn this section, we apply Theorem 3 to identify a class of isomorphically polyhedralOrlicz sequence spaces.

Terms and notation about Orlicz sequence spaces follow thatof [3]. An Orlicz function M is a continuous non-decreasing convex function defined fort ≥0 such that M(0) = 0 and limt→∞M(t) = ∞.

If M(t) > 0 for all t > 0, then it is5

non-degenerate. Clearly a non-degenerate Orlicz function must be strictly increasing.The Orlicz sequence space ℓM associated with an Orlicz function M is the space of allsequences (an) such that P M(|an|/ρ) < ∞for some ρ > 0, equipped with the norm∥x∥= inf{ρ > 0 :XM(|an|/ρ) < ∞}.Let en denote the vector whose sole nonzero coordinate is a 1 at the n-th position.Then clearly (en) is a basic sequence in ℓM.

The closed linear span of {en} in ℓM isdenoted by hM. Alternatively, hM may be described as the set of all sequences (an)such thatP M(|an|/ρ) < ∞for every ρ > 0.

Additional results and references onOrlicz spaces may be found in [3]. For a real null sequence (an), let (a∗n) denote thedecreasing rearrangement of the sequence (|an|).Theorem 4 Let M be a non-degenerate Orlicz function such that there exists a finitenumber K satisfying limt→0 M(Kt)/M(t) = ∞.

Then hM is isomorphically polyhedral.Proof: For all k ∈NI , letbk = infM(Kt)M(t): 0 < t ≤M−1(1k).Then limk→∞bk = ∞. Thus there is a sequence (ηk) decreasing to 1 such that ηk >(1 −b−1k+1)−1 for all k. Define a seminorm on hM by|||(an)||| = supk ηk∥(a∗1, .

. .

, a∗k, 0, . .

. )∥,(6)where ∥· ∥is the given norm on hM.

It is clear that ||| · ||| is an equivalent norm onhM, and that (en) is a monotone basis with respect to ||| · |||. It suffices to show that||| · ||| satisfy the remaining condition in part (b) of Theorem 3.

We first show that if(an) is a positive decreasing sequence in hM, then there is a k such that∥(an)∥≤ηk∥(a1, . .

. , ak, 0, .

. .

)∥. (7)Assume otherwise.

There is no loss of generality in assuming that ∥(an)∥= 1. ThenP M(an) = 1 andPkn=1 M(ηkan) ≤1 for all k. In particular, note that the secondcondition implies ak ≤M−1(1/k) for all k, since ηk ≥1 and (an) is decreasing.

Nowchoose m such that ∥(0, . .

. , 0, am, am+1, .

. .

)∥≤K−1. ThenP∞n=m M(Kan) ≤1.

Also6

M(Kan) ≥bmM(an) for all n ≥m. Therefore,1=XM(an)=m−1Xn=1M(an) +∞Xn=mM(an)≤η−1m−1m−1Xn=1M(ηm−1an) + b−1m∞Xn=mM(Kan)≤η−1m−1 + b−1m<1,a contradiction.

Hence (7) holds for some k. Now for a general element (an) ∈hM,choose m such that ∥(an)∥= ∥(a∗n)∥≤ηm∥(a∗1, . .

. , a∗m, 0, .

. .

)∥.Note that sincelimk ηk∥(a∗1, . .

. , a∗k, 0, .

. .

)∥= ∥(an)∥, the supremum in equation (6) is attained, say,at j. Then choose i large enough that a∗1, .

. .

, a∗j are found in {|a1|, . .

. , |ai|}.

With thischoice of i,|||(a1, . .

. , ai, 0, .

. .

)||| ≥ηj∥(a∗1, . .

. , a∗j, 0, .

. .

)∥= |||(an)|||by choice of j. Since the reverse inequality is obvious,|||(an)||| = |||(a1, .

. .

, ai, 0, . .

. )|||,as required.

✷3A counterexampleTheorem 5 Let M be a non-degenerate Orlicz function. Suppose there exists a se-quence (tn) decreasing to 0 such thatsupnM(Ktn)M(tn)< ∞for all K < ∞.

Then hM is not isomorphically polyhedral.Proof: Suppose that hM is isomorphically polyhedral . By Theorem 3 and the remarkfollowing it, one obtains a norm ||| · ||| on hM as prescribed by part (b) of the theorem.Fix α > 0 so that |||x||| ≤α ⇒∥x∥≤1.

Choose a sequence (ηk) strictly decreasing to7

1. Let n1 = min{n ∈NI : η1|||tne1||| ≤α}.

If n1 ≤n2 ≤· · · ≤nk are chosen so thatηk||| Pkj=1 tnjej||| ≤α, then ηk+1||| Pkj=1 tnjej||| < α. Hence{n ≥nk : ηk+1|||kXj=1tnjej + tnek+1||| ≤α} ̸= ∅.Now definenk+1 = min{n ≥nk : ηk+1|||kXj=1tnjej + tnek+1||| ≤α}.

(8)This inductively defines a (not necessarily strictly) increasing sequence (nk) satisfyingηk|||kXj=1tnjej||| ≤α(9)for all k and the minimality condition (8). In particular, ||| Pkj=1 tnjej||| ≤α for all k,so ∥Pkj=1 tnjej∥≤1 by the choice of α.

Therefore Pkj=1 M(tnj) ≤1 for all k. For allK < ∞and all k ∈NI ,kXj=1M(Ktnj) ≤supmM(Ktm)M(tm)kXj=1M(tnj) ≤supmM(Ktm)M(tm) .Consequently, P∞j=1 M(Ktnj) < ∞for all K < ∞. Hence x = P∞j=1 tnjej convergesin hM.

Clearly |||x||| = limk ||| Pkj=1 tnjej||| ≤α. We claim that in fact |||x||| = α.Otherwise, suppose |||x||| = β < α.

Since (en) is monotone with respect to ||| · |||,||| Pkj=1 tnjej||| ≤β < α for all k. By the convergence of x, limj tnj = 0. So one canfind i such that |||tniej||| ≤α −β for all j. Then|||iXj=1tnjej + tniei+1||| ≤|||iXj=1tnjej||| + |||tniei+1||| ≤β + α −β = α.By the minimality condition (8), ni+1 = ni.

Similarly, we see that nj = ni for allj ≥i. This contradicts the convergence of x and proves the claim.

But now, by (9),|||Pkj=1 tnjej||| < α = |||x||| for all k, contradicting the choice of the norm ||| · |||. ✷We now construct an Orlicz function M satisfying Theorem 5 while hM is c0-saturated.

We begin with some simple results which help to identify the c0-saturatedOrlicz sequence spaces.8

Proposition 6 Let M be a non-degenerate Orlicz function. Then the following areequivalent.

(a) hM is c0-saturated;(b) hM does not contain an isomorph of ℓp for any 1 ≤p < ∞;(c) for all q < ∞,sup0<λ,t≤1M(λt)M(λ)tq < ∞.Proof: Clearly (a) implies (b). If (a) fails, let Y be an infinite dimensional closedsubspace of hM which contains no isomorph of c0.

By [3, Proposition 4.a.7], Y has asubspace Z isomorphic to some Orlicz sequence space hN. Then hN contains no iso-morph of c0.

By [3, Theorem 4.a.9], hN contains an isomorph of some ℓp, 1 ≤p < ∞.Hence Y contains a copy of ℓp, and (b) fails. The equivalence of (b) and (c) also followsfrom [3, Theorem 4.a.9].

✷Proposition 7 Let (bn)∞n=0 be a decreasing sequence of strictly positive numbers suchthatsupm,nbm+nbnKm < ∞for allK < ∞.Define M to be the continuous, piecewise linear function such that M(0) = 0,M′(t) =(bnif2−n−1 < t < 2−n, n > 0b0if2−1 < tThen M is a non-degenerate Orlicz function so that hM is c0-saturated.Proof: It is clear that M is a non-degenerate Orlicz function. For all n ≥0, 2−n−1bn ≤M(2−n) ≤2−nbn.

HenceCq ≡supm,nM(2−m−n)M(2−n) 2mq ≤2 supm,nbm+nbn(2q−1)m < ∞for any q < ∞. Now if λ, t ∈(0, 1], choose m, n ≥1 such that t ∈(2−m, 2−m+1],λ ∈(2−n, 2−n+1].

Then λt ∈(2−m−n, 2−m−n+2]. If m ≥2, thenM(λt)M(λ)tq ≤22q M(2−(m−2)−n)M(2−n)2(m−2)q ≤4qCq.9

If m = 1, then t > 2−1. ThereforeM(λt)M(λ)tq ≤t−q ≤2q.Thussup0<λ,t≤1M(λt)M(λ)tq < ∞,and hM is c0-saturated by the previous proposition.

✷Theorem 8 There exists an Orlicz function M such that hM is c0-saturated but notisomorphically polyhedral. In particular, a c0-saturated space with a separable dual isnot necessarily isomorphically polyhedral.Proof: It is well known that every c0-saturated space hM has a separable dual.

Thusthe second statement follows from the first. Let α0 = α1 = α2 = 1, and let αj = (e/j)jfor j ≥3.

Then (αj) is a decreasing sequence. Choose a decreasing sequence (cj)∞j=0of strictly positive numbers such that cj+1 ≤αjα2j2cj for all j ≥0.

For convenience,set sn = Pnj=1 j for all n ≥1. Now define b0 = c0, b1 = c1, and bsn+k = cn+1/αn+1−kwhenever n ≥1 and 1 ≤k ≤n + 1.

We first show that the sequence (bj) satisfies theconditions in Proposition 7.Claim 1(bj) is a decreasing sequence.One verifies directly that b0 ≥b1 ≥b2. If n ≥1 and 1 ≤k ≤j ≤n + 1,bsn+k =cn+1αn+1−k≥cn+1αn+1−j= bsn+jsince (αm) is decreasing.

Finally,bsn+1+1 = cn+2αn+1≤α2(n+1)2cn+1 ≤cn+1 = bsn+n+1for all n ≥1. This proves Claim 1.Claim 2bm+n ≤αmbn for all m ≥0, n ≥2.Express n = si + k, m + n = sj + l, where 1 ≤i ≤j, 1 ≤k ≤i + 1, and 1 ≤l ≤j + 1.If i = j, then l −k = m. Moreover, i + 1 −k ≥max{l −k, i + 1 −l}, from which itfollows that αi+1−k ≤αl−kαi+1−l.

Therefore,bm+n =ci+1αi+1−l≤αmci+1αi+1−k= αmbn.10

Now consider the possibility that j > i. Note first thatm = (m + n) −n ≤sj + j + 1 −(si + 1) ≤sj + j ≤2j2.Hence αm ≥α2j2.

Using Claim 1 and the properties of the sequence (cj), we obtainbm+n≤bsj+1=cj+1αj≤α2j2cj≤αmci+1=αmbsi+i+1≤αmbn.Claim 3supm,nbm+nbnKm < ∞for allK < ∞.First observe that for i ≥1, 1 ≤k ≤i + 1, and K < ∞,bsi+kKsi+k=ci+1αi+1−kKsi+k≤α2i2ciKsi+i+1≤c0α2i2Ksi+i+1→0as i →∞. Hence (bmKm)m is bounded.

Therefore supn=1,2 supm bm+nKm/bn < ∞.On the other hand, using Claim 2,supn≥2 supmbm+nbnKm ≤supm αmKm < ∞by direct verification.Define the function M using the sequence (bj) as in Proposition 7. Using Claims1 and 3, and the proposition, we see that hM is c0-saturated.

To complete the proof,it suffices to find a sequence (tn) as in Theorem 5. We claim that the sequence (tn) =(2−sn) will do.

Clearly (tn) decrease to 0. Fix m ∈NI .

For all n > m,bsn−m = bsn−1+(n−m) = cnαm.HenceM(2mtn)=M(2−sn+m)≤bsn−m2sn−m=cnαm2sn−m=2m+1αmcn2sn+1=2m+1αmbsn2sn+1≤2m+1αmM(tn)11

whenever n > m. Therefore,supnM(2mtn)M(tn)< ∞for all m ∈NI . ✷The obvious question to be raised is how to characterize isomorphically polyhedralhM in terms of the Orlicz function M. We suspect that the condition given in Theorem4 is the correct one.

It can be shown that if lim inft→0 M(Kt)/M(t) < ∞for all K < ∞,then for any sequence (ηk) decreasing to 1, the norm given by equation (6) does notsatisfy part (b) of Theorem 3.References[1] V. P. Fonf, On a property of Lindenstrauss-Phelps spaces, Funct. Anal.

Appl.13(1979), 79-80 (translated from Russian). [2] V. P. Fonf, Polyhedral Banach spaces, Matematicheskie Zametki 30(1981), 627-634(translated from Russian).

[3] Joram Lindenstrauss and Lior Tzafriri, “Classical Banach Spaces I, SequenceSpaces”, Springer-Verlag, Berlin, 1977. [4] H. Rosenthal, Class notes, Topics course in analysis, University of Texas at Austin.

[5] 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.Department of MathematicsNational University of SingaporeSingapore 0511e-mail(bitnet) : matlhh@nusvm12

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