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TOTAL SUBSPACES WITH LONG CHAINS OF

arXiv:math/9303206v1 [math.FA] 29 Mar 1993TOTAL SUBSPACES WITH LONG CHAINS OFNOWHERE NORMING WEAK∗SEQUENTIAL CLOSURESM.I.OstrovskiiAbstract. If a separable Banach space X is such that for some nonquasireflexiveBanach space Y there exists a surjective strictly singular operator T : X →Y then forevery countable ordinal α the dual of X contains a subspace whose weak∗sequentialclosures of orders less than α are not norming over any infinite-dimensional subspaceof X and whose weak∗sequential closure of order α + 1 coincides with X∗Let X be a Banach space, X∗be its dual space.

The closed unit ball and theunit sphere of X are denoted by B(X) and S(X) respectively. The term “operator”means a bounded linear operator.Let us recall some definitions.

A subspace M of X∗is said to be total if for every0 ̸= x ∈X there is an f ∈M such that f(x) ̸= 0. A subspace M of X∗is said tobe norming over a subspace L of X if for some c > 0 we have(∀x ∈L)( supf∈S(M)|f(x)| ≥c||x||).A subspace M of X∗is said to be norming if it is norming over X.

If M is notnorming over any infinite dimensional subspace of X then we shall say that M isnowhere norming.The set of all limits of weak∗convergent sequences in a subset M of X∗is calledthe weak∗sequential closure of M and is denoted by M(1). If M is a subspace thenM(1) is also a subspace.

This subspace need not be closed and all the more need notbe weak∗closed [M]. In this connection S.Banach introduced [B, p. 208, 213] theweak∗sequential closures (S.Banach used the term ”d´eriv´e faible”) of other orders,including transfinite ones.

For ordinal α the weak∗sequential closure of order α ofa subset M of X∗is the set M(α) = ∪β<α(M(β))(1).It should be noted that for separable X the notion of the weak∗sequential closureof order α coincides with the notion of the derived set of order α considered in [A],[M1], [M2].For the chain of the weak∗sequential closures we haveM(1) ⊂M(2) ⊂. .

. ⊂M(α) ⊂M(α+1) ⊂.

. .

.If we have M(α) = M(α+1) then all subsequent closures coincide with M(α). Theleast ordinal α for which M(α) = M(α+1) is called the order of M.1991 Mathematics Subject Classification.

Primary 46B20.Key words and phrases. Banach space, Total subspace, Weak∗sequential closure, Nowherenorming subspace.Tt bAMS T X

2M.I.OSTROVSKIIThe present paper deals with one of the aspects of the following general problem:how far from the norming subspaces can the total subspaces be and what is thestructure of Banach spaces whose duals contain total “very” nonnorming subspaces?There are many works devoted to this problem (see [B, p. 208–215], papers [A],[DJ], [M1], [M2], [O1], [O2] and papers cited therein). We recall only the resultswhich motivate us to carry out the present research.1.

There is a total subspace M of l∞= (l1)∗such that M(n) is nowhere normingfor all n ∈N [A].2. Let X be a nonquasireflexive separable Banach space.

Then, for every count-able ordinal α, there is a total subspace of order α + 1 in X∗[O1]. (V.B.Moscatelli[M1], [M2] obtained this result in the case when α is not greater than the firstinfinite ordinal.

Explicit construction of [M2] is useful in further investigation ofsuch subspaces.)3. Let X be a separable Banach space.

Its dual contains a total nowhere normingsubspace if and only if for some nonquasireflexive Banach space Y there exists asurjective strictly singular operator T : X →Y [O2].The main result of the present paper is the following.THEOREM. Let X be a separable Banach space such that for some nonquasire-flexive Banach space Y there exists a surjective strictly singular operator T : X →Y .

Then for every countable ordinal α there exists a subspace M of X∗such thatM(α+1) = X∗and for every ordinal β < α the subspace M(β) ⊂X∗is nowherenorming.Let us introduce some notation. For a subset A of a Banach space X, lin(A)and cl(A) are, respectively, the linear span of A and the closure of A in the strongtopology.

By w∗−limm→∞x∗m we denote the weak∗limit of the sequence {x∗m}∞m=1in the dual Banach space (if this limit exists). For a subset A of X∗, A⊤is the set{x ∈X : (∀x∗∈A)(x∗(x) = 0)}.Our sources for Banach space theory are [B], [S].At first we shall prove the theorem in the case when α is a nonlimit ordinal.

Inorder to do this we need the following result.Lemma 1. Let Y be a separable nonquasireflexive Banach space.

Then for everycountable ordinal γ there exist a subspace N of Y ∗and a bounded sequence {hn}∞n=1in Y ∗∗such that:A. If a weak∗convergent sequence {x∗m}∞m=1 is contained in N(β) for some β < γand x∗= w∗−limm→∞x∗m thenhn(x∗) = limm→∞hn(x∗m).(1)B.

There exists a collection {x∗n,m}∞n,m=1 of vectors in N(γ) such that for everyk, n ∈N we havew∗−limm→∞x∗n,m = 0;(2)(∀m ∈N)(hk(x∗n,m) = δk,n). (3)At first we shall finish the proof of the theorem in the case of nonlimit α withthe help of Lemma 1

TOTAL SUBSPACES3We apply Lemma 1 to γ = α −1 (this ordinal is correctly defined since α isnonlimit).Let N, {hn} and {x∗n,m} be as in Lemma 1.Let {s∗k}∞k=1 be a nor-malized sequence in X∗such that lin({s∗k}) is a norming subspace of X∗.Letc1 = supn ||hn||.Let νn > 0 (n ∈N) be such that P∞n=1 νn ≤1/(2c1).Wemay assume without loss of generality that T is a quotient mapping. In this caseT ∗: Y ∗→X∗is an isometric embedding.

Therefore we may (and shall) identifyY ∗with T ∗(Y ∗) ⊂X∗.Let R : Y ∗→X∗be given by the equalityR(y∗) = y∗+∞Xn=1νnhn(y∗)s∗n. (4)It is clear that(∀y∗∈Y ∗)((1/2)||y∗|| ≤||Ry∗|| ≤(3/2)||y∗||).

(5)Let M = R(N). We prove that for every β ≤γ we haveM(β) = R(N(β)).

(6)We use the transfinite induction. For β = 0 we have (6) by definition.

Let ussuppose that (6) is true for some β < γ and prove that M(β+1) = R(N(β+1)). Letx∗∈M(β+1), i.e.

x∗= w∗−limm→∞x∗m for some sequence {x∗m}∞m=1 in M(β). Lety∗m ∈N(β) be such that x∗m = R(y∗m).

Denote supm ||x∗m|| by c2. By (5) we have||y∗m|| ≤2c2 for every m ∈N.

Therefore by separability of X we can select a weak∗convergent subsequence {y∗m(i)}∞i=1 of {y∗m}∞m=1. Let y∗= w∗−limi→∞y∗m(i).

It isclear that y∗∈N(β+1). Since β < γ then by the assertion A of Lemma 1 we havelimi→∞hn(y∗m(i)) = hn(y∗)for every n ∈N.

By the definition of R it follows thatlimi→∞∞Xn=1νnhn(y∗m(i))s∗n =∞Xn=1νnhn(y∗)s∗n,where the limit is taken in the strong topology. Therefore we have:w∗−limi→∞R(y∗m(i)) = R(y∗).Hence, x∗= R(y∗) and M(β+1) ⊂R(N(β+1)).

The inclusion R(N(β+1)) ⊂M(β+1)follows immediately from (1), (4) and (6).The case of a limit ordinal β ≤γ is more simple:M(β) = ∪τ<βM(τ) = ∪τ<βR(N(τ)) = R(∪τ<βN(τ)) = R(N(β)).Therefore formula (6) is proved. In particular, we have M(γ) = R(N(γ)).

Let usshow that this equality implies that M(γ) is nowhere norming.Suppose that it is not the case. Let an infinite dimensional subspace L of X besuch that Mis norming over L

4M.I.OSTROVSKIIRecall that if U and V are subspaces of a Banach space X then the numberδ(U, V ) = inf{||u −v|| : u ∈S(U), v ∈V }is called the inclination of U to V .Since T is a strictly singular quotient mapping then X does not contain aninfinite dimensional subspace with non-zero inclination to ker(T). Using the well-known arguments (see [AGO], [G] or [R]) we can find a normalized sequence {zi}∞i=1in L such that for some sequence {ti}∞i=1 in ker(T) we have ||zi −ti|| < 2−i and,furthermore,(∀n ∈N)( limi→∞s∗n(ti) = 0).

(7)Let c > 0 be such that(∀x ∈L)(∃f ∈S(M(γ)))(|f(x)| ≥c||x||).In particular,(∀i ∈N)(∃fi ∈S(M(γ)))(|fi(zi)| ≥c).Since M(γ) = R(N(γ)) then we can find y∗i ∈N(γ) such that fi = R(y∗i ), i. e.fi = y∗i + P∞n=1 νnhn(y∗i )s∗n. By (5) we have ||y∗i || ≤2.

Furthemore, we havec ≤|fi(zi)| ≤|fi(zi −ti)| + |fi(ti)| < 2−i + |∞Xn=1νnhn(y∗i )s∗n(ti)| ≤2−i + c1||y∗i ||∞Xn=1νn|s∗n(ti)|.Using (7) and boundedness of the sequences {s∗n} and {ti} we obtain a contra-diction. Hence, the subspace M(γ) is nowhere norming.

Thus we prove that Msatisfies the second assertion of the theorem.It remains to prove that M(γ+2) = X∗.Let {x∗n,m}∞n,m=1 be the collection whose existence is asserted in Lemma 1. By (6)we have R(x∗n,m) ∈M(γ) for every m, n ∈N.

By (3) we have R(x∗n,m) = x∗n,m+νns∗n.Since the sequence {x∗n,m}∞m=1 is weak∗null and νn ̸= 0 then we have s∗n ∈M(γ+1)for every n ∈N, therefore lin({s∗n}) ⊂M(γ+1). Since the subspace lin({s∗n}) isnorming then by [B, p. 213] we have M(γ+2) = X∗.

The theorem is proved in thecase when α is nonlimit.Proof of Lemma 1. By [DJ, Theorem 2] Y contains a bounded away from 0 basicsequence {zn}∞n=0 such that the setkXi=jzi(i+1)/2+j∞∞j=0,k=jis bounded.

We may assume without loss of generality that ||zi|| ≤1 for everyi ∈N. Let Z =cl(lin({zn}∞n=0)).

It is easy to see that the following claims are true.1. The space Z∗∗may be identified with the weak∗closure of Z in Y ∗∗.2.

Every weak∗null sequence in Z∗has a weak∗null sequence of extensions toY

TOTAL SUBSPACES53. If we denote the canonical embedding of Z into Y by ξ then for every ordinalα and every subspace N of Z∗we shall have(ξ∗)−1(N(α)) = ((ξ∗)−1N)(α).

(In this connection see Lemma 1 in [O1]. )4.If z∗∗∈Z∗∗and y∗∈Y ∗then the value z∗∗(y∗) depends only on therestriction of y∗to Z.These claims imply that it is sufficient to prove Lemma 1 with Y ∗and Y ∗∗replaced by Z∗and Z∗∗respectively.Let us introduce some notation.

We shall write zji forz(j+i−1)(j+i)/2+j (j = 0, 1, 2, . .

. , i ∈N),biorthogonal functionals to the system will be denoted by ˜zn(˜zji ).

By the citedabove result of [DJ] we havesupj,m||mXi=1zji || = c1 < ∞.Therefore for every j = 0, 1, 2, . .

. the sequence {Pmi=1 zji }∞m=1 has at least oneweak∗limit point in Z∗∗.

Let us choose one of these limit points and denote it byfj. It is clear that ||fj|| ≤c1.We need the following result from [O1].Lemma 2.

For every vector g0 ∈Z∗∗of the form afj + zrs(a > 0, r ̸= j), everycountable ordinal α and every infinite subset A ⊂N such that j, r ̸∈A there existsa countable subset Ω(g0, α, A) of Z∗∗such that1) For a subspace K(g0, α, A) of Z∗defined by K(g0, α, A) = (Ω(g0, α, A))⊤wehave (K(g0, α, A))(α) ⊂ker(g0).2) All vectors h ∈Ω(g0, α, A) are of the form h = a(h)fj(h)+zr(h)s(h) with j(h), r(h) ∈A∪{j, r}, a(h) > 0 and for every h ̸= g0 from Ω(g0, α, A) we have j(h) ̸= r, r(h) ̸=r.3) If we denote by Q(b, g0, α, A) the intersection of the setb˜zrs + u, where u ∈lin({˜ztk}∞k=1,t∈A∪{j})(8)with K(g0, α, A) then the set (Q(b, g0, α, A))(α) contains all vectors of the form (8)which are in ker(g0).Let us introduce the functionals hn ∈Z∗∗by the equalities hn = f2n−1 (n ∈N)and the vectors x∗n,m by the equalities x∗n,m = ˜z2n−1m. It is clear that the vectors hnand x∗n,m satisfy equalities (2) and (3).Let {An}∞n=0 be a partition of the set of even natural numbers into pairwisedisjoint infinite sets.

Let εn,k > 0 (n, k ∈N) be such that P∞n,k=1 εn,k < ∞. Definethe family {gn,k}∞n,k=1 in the following way:gn,k = z2n−1k+ εn,kfj(n,k),where the mapping j : N × N →N is such that j(n, k) ∈An and j(n, k) ̸= j(n, l)for k ̸l

6M.I.OSTROVSKIILet {Dn,k}∞n,k=1 be a partition of A0 into pairwise disjoint infinite sets.The cases of limit and nonlimit γ will be treated separately.Let γ be a nonlimit ordinal and let Ω(gn,k, γ−1, Dn,k) be the sets whose existenceis asserted in Lemma 2. Let us define N ⊂Z∗by N = (∪∞n,k=1Ω(gn,k, γ−1, Dn,k))⊤.Let us show that N satisfies the conditions of Lemma 1.Let {x∗m}∞m=1 be a weak∗convergent sequence in N(β)(β ≤γ −1) and let x∗=w∗−limm→∞x∗m.

By Lemma 2 we have(∀n, k ∈N)(x∗m ∈ker(gn,k))(9)Since the sequence {zn}∞n=0 is a basis of Z, then their biorthogonal sequence {˜zn}∞n=0is a w∗-Schauder basis of Z∗[S, p. 155]. (It means that every vector z∗∈Z∗canbe represented as z∗= w∗−limk→∞Pkn=0 an˜zn, where an = z∗(zn)).Using (9) we can estimate some of the coefficients of the weak∗decompositionsof the vectors x∗m(m ∈N).

Precisely, if we denote supm ||x∗m|| by c2, then we obtain|x∗m(z2n−1k)| ≤εn,kc1c2.Therefore x∗m can be represented as u∗m+v∗m, where u∗m = P∞n,k=1 x∗m(z2n−1k)˜z2n−1k(wenote that this series converges unconditionally, therefore we need not indicate theorder of summability), and(∀n, k ∈N)(v∗m(z2n−1k) = 0)(10)Since the weak∗convergence implies the coordinatewise convergence for w∗-Schauderbases, then we can represent x∗in an analogous way, x∗= u∗+ v∗.We haveu∗= w∗−limm→∞u∗m.By (10) and by the definition of hn it follows that(∀m, n ∈N)(hn(v∗m) = 0);(11)(∀n ∈N)(hn(v∗) = 0). (12)Since the vectors {u∗m}∞m=1 and u∗are contained in the strongly compact setC = {∞Xn,k=1an,k˜z2n−1k: |an,k| ≤εn,kc1c2},then the weak∗convergence of {u∗m} to u∗implies the weak convergence of {u∗m}to u∗.

Therefore for every n ∈N we have limm→∞hn(u∗m) = hn(u∗). From hereby (11) and (12) we obtain limm→∞hn(x∗m) = hn(x∗).

Thus the assertion A ofLemma 1 is proved.In order to prove the assertion B it is sufficient to check that ˜z2n−1k∈N(γ).Let ˜z(t) = ˜z2n−1k−˜zj(n,k)t/εn,k(t ∈N). It is clear that ˜z(t) ∈ker gn,k and that˜z2n−1k= w∗−limt→∞(˜z(t)).

Furthermore, vectors ˜z(t) are of the form (8) with b =1, r = 2n −1, s = k and j = j(n, k). Therefore ˜z(t) ∈(Q(1, gn,k, γ −1, Dn,k))(γ−1)and ˜z2n−1 ∈(Q(1 gγ1 D))

TOTAL SUBSPACES7It remains to show that(∀r, s ∈N)(Q(1, gn,k, γ −1, Dn,k) ⊂(Ω(gr,s, γ −1, Dr,s))⊤)(13)For r = n, s = k it follows immediately from the definition of Q. Let (r, s) ̸= (n, k).Recall that every element of Ω(gr,s, γ −1, Dr,s) is a weak∗limit point of linearcombinations of z2r−1s, zj(r,s)t(t ∈N) and zpq(p ∈Dr,s, q ∈N) and that Q(1, gn,k, γ −1, Dn,k) consists of linear combinations of ˜z2n−1k, ˜zj(n,k)t(t ∈N) and ˜zpq(p ∈Dn,k, q ∈N).Our construction is such that the sets {2n −1, j(n, k)} ∪Dn,k and {2r −1, j(r, s)} ∪Dr,s intersect if and only if 2n −1 = 2r −1.

Since in this case we haves ̸= k, then we obtain (13).Thus we have finished the proof of Lemma 1 in the case when γ is a nonlimitordinal. Let us pass to the case when γ is a limit ordinal.

Let {γn}∞n=1 be anincreasing sequence of ordinals, for which γ = limn→∞γn . Let us introduce thesubspace N ⊂Z∗by the equalityN = (∪∞n,k=1Ω(gn,k, γn+k, Dn,k))⊤.We shall show that N satisfies all the conditions of Lemma 1.

Let {x∗m}∞m=1 bea weak∗convergent sequence in N(β) with β < γ and let x∗= w∗−limm→∞x∗m.Let i ∈N be such that γi−1 < β ≤γi(we let γ0 = 0). The definition of the setsΩ(gn,k, γn+k, Dn,k) implies that for those pairs (n, k) for which n + k ≥i we havex∗m ∈ker(gn,k), and, consequently, we have|x∗m(z2n−1k)| ≤εn,kc1c2.In the same time since ||z2n−1k|| ≤1 then(∀n, k ∈N)(|x∗m(z2n−1k)| ≤c2).Therefore we may argue in the same way as in the first part of Lemma 1 if we definethe set C in the following way:C = {∞Xn,k=1an,k˜z2n−1k: |an,k| ≤εn,kc1c2 if n + k ≥i and|an,k| ≤c if n + k < i}.The proof of Lemma 1 is complete.Thus in the case when α is a nonlimit ordinal the proof of the theorem is finished.Let us describe the changes which should be made in the proof of the theoremin the case when α is a limit ordinal.Let {αn}∞n=1 be an increasing sequence of nonlimit ordinals such that α =limn→∞αn.

Instead of Lemma 1 we shall use the following result.Lemma 3. For every separable nonquasireflexive Banach space Y there exists asubspace N of Y ∗and a bounded sequence {hn}∞n=1 in Y ∗∗such that:Anew.If a weak∗convergent sequence {x∗m}∞m=1 is contained in N(β) for someβ < α and x∗= w∗−limm→∞x∗m then we havehn(x∗) = lim hn(x∗)

8M.I.OSTROVSKIIfor those n for which β < αn.Bnew. For every n ∈N there exists a sequence {x∗n,m}∞m=1 in N(αn) such that theconditions (2) and (3) are satisfied.At first we finish the proof of the theorem with the help of Lemma 3.

Let {s∗k}∞k=1and {νn}∞n=1 be the same as in the first part of the theorem. Let R : Y ∗→X∗be defined by equality (4).

For every ordinal β < α we denote by W(β) the set ofnatural numbers n for which αn < β. It is clear that W(β) is a finite set.Let us show that for every ordinal β < α we haveM(β) = lin(R(N(β)) ∪{s∗k}k∈W (β)).

(14)We shall prove this with the aid of transfinite induction. For β = 0 (14) followsimmediately from the definition.

Let us suppose that (14) is valid for some β < αand prove the analogous equality for β + 1. Let x∗∈M(β+1), i.e.x∗= w∗−limm→∞x∗m, where x∗m ∈M(β)=lin(R(N(β)) ∪{s∗k}k∈W (β)).

It is clear that the lastspace is a subspace of cl(R(N(β)))⊕F, where F is some subspace of lin({s∗k}k∈W (β)).Therefore, we may write x∗m = u∗m + v∗m, where u∗m ∈cl(R(N(β))) and v∗m ∈F.It is clear that the sequences {u∗m}∞m=1 and {v∗m}∞m=1 are bounded and that wemay suppose without loss of generality that u∗m ∈R(N(β)). Let u∗m = Ry∗m.

By(5) the sequence {y∗m} is also bounded.So we can find a sequence {m(i)}∞i=1of natural numbers such that the sequences {v∗m(i)}∞i=1 and {y∗m(i)}∞i=1 are weak∗convergent. Let v∗and y∗be corresponding weak∗limits.

By Lemma 3 we havehn(y∗) = limi→∞hn(y∗m(i)) for all n for which αn > β. Therefore,limi→∞Xn̸∈W (β+1)νnhn(y∗m(i))s∗n =Xn̸∈W (β+1)νnhn(y∗)s∗n,where the limit is taken in the strong topology.Since the set W(β + 1) is finite then we may without loss of generality as-sume that the sequence {Pn∈W (β+1) νnhn(y∗m(i))s∗n}∞i=1 is strongly convergent. LetPn∈W (β+1) ans∗n be its limit.

It follows thatx∗= R(y∗) −Xn∈W (β+1)νnhn(y∗)s∗n +Xn∈W (β+1)ans∗n + v∗∈lin(R(N(β+1)) ∪{s∗n}n∈W (β+1)).Therefore M(β+1) ⊂lin(R(N(β+1)) ∪{s∗n}n∈W (β+1)). The converse inclusion followsimmediately from Bnew and the induction hypothesis.If β is a limit ordinal and (14) is proved for all ordinals less than β then (14)follows immediately from the following fact: W(β) = W(τ) for some τ < β.

So(14) is proved for all ordinals which are less than α.It is not hard to check that the linear span of the union of a finite-dimensionaland a nowhere norming subspaces is nowhere norming. Therefore by (14) and thearguments of the first part of the theorem M(β) is a nowhere norming subspace forevery β < α.The proof of the equality M(α+1) = X∗is the same as in the first part of thetheorem.Proof of Lemma 3.

We repeat the arguments of the proof of Lemma 1 up to thepassage where Awas presented in the form A = ∪∞DNow we continue

TOTAL SUBSPACES9in the following way. Let {αn}∞n=1 be the increasing sequence of nonlimit ordinalsintroduced above.

Let Ω(gn,k, αn−1, Dn,k) be the sets whose existence is guaranteedby Lemma 2. Let us introduce the subspace N ⊂Z∗by the equalityN = (∪∞n,k=1Ω(gn,k, αn −1, Dn,k))⊤.The assertion Anew of Lemma 3 is proved in the same way as the assertion A ofLemma 1.

The only distinction is that we have relation (9) not for all natural nbut only for n ∈N\W(β + 1). This do not prevent to finish the proof since theassertion Anew concerns only those hn for which n ∈N\W(β + 1).The proof of the assertion of Bnew is the same as the proof of the assertion B ofLemma 1.References[A]A.A.Albanese, On total subspaces in duals of spaces of type C(K) or L1, preprint (1991).

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[O1]M.I.Ostrovskii, w∗-derived sets of transfinite order of subspaces of dual Banach spaces,Dokl. Akad.

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I, Springer-Verlag, Berlin and New York, 1970.Mathematical Division, Institute for Low Temperature Physics and Engineering,47 Lenin avenue, 310164 Kharkov, UKRAINEE-mail address: ostrovskii%ilt.kharkov.ua@relay.ussr.eu.net

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