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Total subspaces in dual Banach spaces which are not norming

arXiv:math/9303204v1 [math.FA] 29 Mar 1993Total subspaces in dual Banach spaces which are not normingover any infinite dimensional subspacebyM.I.Ostrovskii1. Introduction.1Abstract.

The main result: the dual of separable Banach space X contains a totalsubspace which is not norming over any infinite dimensional subspace of X if and onlyif X has a nonquasireflexive quotient space with the strictly singular quotient mapping.Let X be a Banach space and X∗be its dual space.Let us recall some basic definitions.A subspace M of X∗is said to be total if for every 0 ̸= x ∈X there is an f ∈Msuch that f(x) ̸= 0.A subspace M of X∗is said to be norming over a subspace L ⊂X if for some c > 0we have(∀x ∈L)( supf∈S(M)|f(x)| ≥c||x||),where S(M) is the unit sphere of M. If L = X then M is called norming.The following natural questions arise:1) How far could total subspaces be from norming ones? (Of course, there are manydifferent concretizations of this question.

)2) What is the structure of Banach spaces, whose duals contain total “very” non-norming subspaces?3) What is the structure of total subspaces?These questions was studied by many authors: [Al], [B, p. 208–216], [BDH], [DJ],[DL], [D], [F], [G], [Ma], [Mc], [M1], [M2], [O1], [O2], [P], [PP], [S1], [S2]. The obtainedresults find applications in the theory of Frechet spaces [BDH], [DM], [MM1], [MM2],[M2]; in the theory of improperly posed problems [O3], [PP, pp.

185–196] and in thetheory of universal bases [Pl, p. 31].The present paper is devoted to the following natural class of subspaces which arefar from norming ones. A subspace M of X∗is said to be nowhere norming if it isnot norming over every infinite dimensional subspace of X.If X is such that X∗contains a total nowhere norming subspace then we shall write X ∈TNNS.

This classwas introduced by W.J.Davis and W.B.Johnson in [DJ], where the first example of atotal nowhere norming subspace was constructed. In the same paper it was noted thatJ.C.Daneman proved that every infinite dimensional subspace of l1 is norming oversome infinite dimensional subspace of c0.

In [O2] a class of the spaces with TNNSproperty was discovered. A.A.Albanese [Al] proved that the spaces of type C(K) arenot in TNNS.

The problem of description of Banach spaces with TNNS propertyarises in a natural way.Our main result (Theorem 2.1) states that for a separable Banach space X we haveX ∈TNNS if and only if for some nonquasireflexive Banach space Y there exists asurjective strictly singular operator T : X →Y.11991 Mathematics Subject Classification: Primary 46B201

Section 3 is devoted to the proof of the auxiliary Theorem 2.4. Using the samemethod we are able to prove the following result (Theorem 3.1):A Banach space M is isomorphic to a total nonnorming subspace of the dual ofsome Banach space if and only if M∗contains a closed norming subspace of infinitecodimension.Thus the class of total nonnorming subspaces coincide with the class of Banachspaces that gives a negative solution to the J.J.Schaffer’s problem [Sc, p. 358] (see [DJ,p.

366]).Section 4 is devoted to several remarks concerning general (not necessarily separable)spaces, in particular, we show that Banach spaces with the Pelczynski property are notTNNS.Section 5 is devoted to an example of a nonquasireflexive separable Banach spacewithout the Pelczynski property and without TNNS too.We hope that our notation is standard and self-explanatory. For a subset A of aBanach space X, lin A, A⊥and cl A are, respectively, the linear span of A, the set{x∗∈X∗: (∀x ∈A)(x∗(x) = 0)} and the closure of A in the strong topology.

For asubset A of a dual Banach space X∗, w∗−cl A and A⊤are, respectively, the closure ofA in the weak∗topology and the set {x ∈X : (∀x∗∈A)(x∗(x) = 0)}. For an operatorT : X →Y the notation T|Z denotes the restriction of T to the subspace Z of X.I wish to express gratitude to V.M.Kadets for his valuable advice.2.

Main result.Our sources for Banach space basic concepts and results are [DS, LT, W]. The unitball and the unit sphere of Banach space X are denoted by B(X) and S(X) respectively.The term “operator” means a bounded linear operator and the term “subspace” meansa closed linear subspace.Let us recall some definitions.A Banach space X is called quasireflexive if its canonical image has finite codimensionin X∗∗.

The number dim(X∗∗/X) is called the order of quasireflexivity of X and isdenoted by Ord X.An operator T : X →Y is called strictly singular if the restriction of T to anyinfinite dimensional subspace of X is not an isomorphism.The main result of this paper is the following.2.1.Theorem. Let X be a separable Banach space.

Then X ∈TNNS if and onlyif for some nonquasireflexive Banach space Y there exists a surjective strictly singularoperator T : X →Y.Proof. Let us suppose that such an operator T exists.

We may assume without lossof generality that T is a quotient map. Then T ∗is an isometric embedding of Y ∗intoX∗.

The subspace M1 := T ∗(Y ∗) is nowhere norming because T is strictly singular.This space is also not total. Our aim is to find an isomorphism Q : X∗→X∗whichis a small perturbation of the identity operator and is such that under its action M1becomes a total subspace but remains a nowhere norming one.The space Y is nonquasireflexive, hence, by [DJ, p. 360] there exist a weak∗nullbasic sequence {yn} ⊂Y ∗, a bounded sequence {gn} ⊂Y ∗∗and a partition {In}∞n=1 of2

the integers into pairwise disjoint infinite subsets such thatgk(yn) = 1,if n ∈Ik;0,if n ̸∈Ik.Let us denote the vector T ∗yn by u∗n. The operator T ∗is weak∗continuous andisometric hence {u∗n}∞n=1 is a weak∗null basic sequence in X∗and in X∗∗there exists abounded sequence {v∗∗n }∞n=1 such thatv∗∗k (u∗n) = 1,if n ∈Ik;0,if n ̸∈Ik.Let {s∗k}∞k=1 be a normalized sequence spanning a total subspace in X∗.

Let theoperator Q : X∗→X∗be given byQ(x∗) = x∗+∞Xk=14−kv∗∗k (x∗)s∗k/||v∗∗k ||.It is clear that Q is an isomorphism. Let M = Q(M1).

We shall show that Mis a total nowhere norming subspace. Let 0 ̸= x ∈X and let k ∈N be such thats∗k(x) ̸= 0.

Since {u∗n}∞n=1 is weak∗null then we can choose n ∈Ik such that |u∗n(x)| <4−ks∗k(x)/||v∗∗k ||. We have(Q(u∗n))(x) = u∗n(x) + 4−ks∗k(x)/||v∗∗k || ̸= 0.Hence M is total.Recall that if U, V are subspaces of a Banach space, then the numberδ(U, V ) = inf{||u −v|| : u ∈S(U), v ∈V }is called the inclination of U to V .We shall prove that M ⊂X∗is nowhere norming.

Let us suppose that it is notthe case and let an infinite dimensional subspace L ⊂X be such that M is normingover L. By strict singularity of T there is no any infinite dimensional subspace of Xwith nonzero inclination to ker T. Using standard reasoning with basic sequences (see[Gu]) we can find in L a normalized basic sequence {zi} such that for some sequence{ti} ⊂ker T we shall have ||zi −ti|| ≤2−i, and moreover we may require that(∀n ∈N)( limi→∞s∗n(ti) = 0). (1)Let c > 0 be such that(∀x ∈L)(∃f ∈S(M))(|f(x)| ≥c||x||).In particular(∀i ∈N)(∃fi ∈S(M))(|fi(zi)| ≥c).By the definition of M we can find such y∗i ∈Y ∗thatfi = T ∗y∗i +∞Xk=14−kv∗∗k (T ∗y∗i )s∗k/||v∗∗k ||.3

¿From this equality we obtain||fi|| ≥(2/3)||T ∗y∗i ||.Using this inequality we obtain for every positive integer i thatc ≤|fi(zi)| ≤|fi(zi −ti)| + |fi(ti)| ≤2−i + |∞Xk=14−kv∗∗k (T ∗y∗i )s∗k(ti)/||v∗∗k ||| ≤2−i + (3/2)∞Xk=14−k|s∗k(ti)|.Using (1) and the boundedness of sequences {s∗n} and {ti} we arrived at a contradiction.Hence M is nowhere norming.Now we begin to prove the converse statement. We need the following result, whicheasily follows from the arguments of [DJ, p. 358].2.2.Lemma.

Let X be a separable Banach space and let N be a subspace of X∗such that the strong closure of the canonical image of X in N∗is of infinite codimension.Then N contains a weak∗null basic sequence {u∗n}∞n=1 such that for some bounded se-quence {v∗∗k }∞k=1 in X∗∗and some partition {Ik}∞k=1 of the positive integers into pairwisedisjoint infinite subsets we havev∗∗k (u∗n) = 1,if n ∈Ik;0,if n ̸∈Ik.It turns out that a total nowhere norming subspace need not satisfy the conditionof Lemma 2.2.2.3.Proposition. There exists a total nowhere norming subspace L of (l1)∗suchthat the canonical image of l1 is dense in L∗.Proof.

Let a Banach space X be such that X∗is separable, contains closed normingsubspaces of infinite codimension and does not contain subspaces isomorphic to l1. Wemay take e. g. X = (P ⊕J)2, where J is James’ space [LT, p. 25].We need the following definition.

Let a ≥0, b ≥0. We shall say that subset A ⊂X∗is (a, b)-norming if the following conditions are satisfied:(∀x ∈X)(sup{|x∗(x)| : x∗∈A} ≥a||x||);sup{||x∗|| : x∗∈A} ≤b.Let K ⊂X∗be a closed norming subspace of infinite codimension.

Let α : l1 →Kbe some quotient mapping. Hence the set α(B(l1)) is (c, 1)-norming for some c > 0.

Leta sequence {zi}∞i=1 ⊂X∗be such that its image under the quotient map X∗→X∗/Kis minimal. Let us introduce the operator β : l1 →X∗in the following way:β({ai}∞i=1) = (c/2)∞Xi=1aizi/||zi|| + α({ai}∞i=1).4

It is clear that β is injective and that the set β(B(l1)) is (c/2, 1 + c/2)-norming. Let(finite or infinite) sequence {yi}ki=1 ⊂X∗be such that its image under the quotientmapping X∗→X∗/clβ(l1) is minimal and that X∗= cl(lin({yi}ki=1 ∪β(l1))).

Let usrepresent l1 as l1⊕lk1 (or l1⊕l1 if k equals infinity) and define the operator γ : l1⊕lk1 →X∗in the following way:γ({ai}∞i=1, {bi}ki=1) = β({ai}∞i=1) +kXi=1biyi/||yi||.It is clear that γ is injective, its image is dense in X∗and that γ(B(l1)) is (c/2, 1 +c/2)-norming. Besides this γ is a strictly singular operator since X∗does not containsubspaces isomorphic to l1.Let L = γ∗(X) ⊂(l1)∗.

This subspace is total since γ is injective. Since γ(B(l1))is (c/2, 1 + c/2)-norming then γ∗|X is an isomorphic embedding.

Therefore the strictsingularity of γ implies that L is nowhere norming. On the other hand it is easy tocheck that the canonical image of l1 in L∗may be identified with γ(l1) and therefore isdense.

The proposition is proved.In order to make Lemma 2.2 applicable for our purposes we need the following result.2.4.Theorem. Let X be a Banach space and M be a total nowhere norming sub-space of X∗.

Then there exists an isomorphic embedding E : M →X∗such that E(M)is a also nowhere norming subspace and the closure of E∗(X) in the strong topologyhas infinite codimension in M∗.We postpone the proof until section 3.Let X ∈TNNS and let M ⊂X∗be a total nowhere norming subspace.Ap-plying Theorem 2.4 we find an embedding E : M →X∗such that N = E(M) isa nowhere norming subspace satisfying the condition of Lemma 2.2. Let {u∗n}∞n=1 ⊂E(M), {v∗∗k }∞k=1 ⊂X∗∗and {Ik}∞k=1 be sequences obtained by application of Lemma 2.2to N = E(M).We need the following definition [JR].A sequence {x∗n}∞n=1 ⊂X∗is called weak∗basic provided that there is a sequence{xn}∞n=1 ⊂X so that {xn, x∗n} is biorthogonal and for each x∗∈w∗−cl(lin{x∗n}∞n=1),x∗= w∗−limn→∞nXi=1x∗(xi)x∗i .By [JR, p. 82] (see also [LT,p.

11]) every bounded away from 0 weak∗null sequence inthe dual of a separable Banach space has a weak∗basic subsequence. Therefore, wemay select a weak∗basic subsequence {u∗n(j)}∞j=1 ⊂{u∗n}∞n=1.

Moreover, by argumentsof [JR] we may suppose that the intersection Ik ∩{n(j)}∞j=1 is infinite for every k ∈N.Let Y = X/(({un(j)}∞j=1)⊤) and let T : X →Y be the quotient map. The dual Y ∗may be naturally identified with w∗−cl(lin{u∗n(j)}∞j=1).

The space Y is nonquasireflexivebecause intersections Ik ∩{n(j)}∞j=1(k ∈N) are infinite. By the well-known propertiesof weak∗basic sequences [LT, p. 11] it follows that for some λ < ∞we haveB(Y ∗) ⊂λw∗−cl(B(lin{u∗n(j)}∞j=1)) ⊂λw∗−cl(B(E(M))).5

Therefore Y ∗is nowhere norming subspace of X∗. Hence T is strictly singular.

Theproof of Theorem 2.1 is complete.2.5. Corollary.

If X is a separable Banach space which contains a complementedsubspace Y with Y ∈TNNS then X ∈TNNS.Proof. The composition of a strictly singular surjection T : Y →Z and any projec-tion P : X →Y is a required surjection.2.6.

Corollary. For every separable Banach space X we have X ⊕l1 ∈TNNS.To prove this we need only to recall that there is a surjective strictly singular operatorT : l1 →c0 [LT, p. 75, 108].2.7.Remark.

There exists a space X for which X ∈TNNS but X does notcontain any subspaces isomorphic to l1. In fact, by the James-Lindenstrauss theorem[LT, p. 26] there exists a separable space Z for which Z∗∗/Z is isomorphic to c0 and Z∗∗∗is isomorphic to Z∗⊕l1.

Let X = Z∗∗. Then there exists a quotient map T : X →c0.This map must be strictly singular because if not then X must contain a subspaceisomorphic to c0 [LT, p. 53] but this contradicts the fact that X is a separable dualspace [LT, p. 103].

At the same time X does not contain subspaces isomorphic to l1because X∗is separable.3. Total nonnorming subspaces in dual Banach spaces.In this section we shall prove Theorem 2.4 and the following characterization of totalnonnorming subspaces.3.1.Theorem.

A Banach space M is isomorphic to a total nonnorming subspace ofthe dual of some Banach space if and only if M∗contains a closed norming subspace ofinfinite codimension.We need the following lemmas.3.2.Lemma [B, p. 39]. If U and V are Banach spaces and P : U →V is an operatorwith nonclosed image then the closure of P(B(U)) in the strong topology does notcontain interior points.3.3.Lemma [LT, p. 79].

If P : U →V is an operator with nonclosed image andF : U →V is a finite rank operator then the operator (P + F) has nonclosed image.3.4.Lemma. Let P : U →V be an operator with nonclosed image and let ε > 0.Then there exist a functional f ∈V ∗and an operator P1 : U →V such that f doesnot vanish on im P, the image of (P −P1) is one-dimensional, ||P −P1|| ≤ǫ andim P1 ⊂ker f ∩cl(im P).Proof.

By Lemma 3.2 the closed convex set cl(P(B(U))) does not have interior pointsin the subspace V0 = cl(imP) ⊂V . Therefore there exists a functional f0 ∈S(V ∗0 ) suchthat(∀v ∈P(B(U)))(|f0(v)| ≤ε/2).It is clear that f0 does not vanish on im P. Let v0 ∈V0 be such that f0(v0) = 1 and||v0|| ≤2.

Let us define an operator P1 : U →V by P1(u) = P(u) −f0(P(u))v0. Let fbe any continuous extension of f0 onto the whole V .

It can be directly verified that P1and f defined in such a way satisfy all the requirements of Lemma 3.4.Let X be a Banach space and let M be a subspace of its dual. Every element of Xmay be considered as a functional on M. So there is a natural map of X into M∗.

Weshall denote this map by H.6

3.5.Proposition.Let X be a Banach space and let M be a total nonnormingsubspace in X∗. Then there exists an isomorphic embedding E : M →X∗such thatthe closure of the E∗(X) in the strong topology is of infinite codimension in M∗andthe difference (E∗|X −H) is a nuclear operator.Proof.

In our case the map H is injective because M is total and is not an isomorphicembedding because M is nonnorming. By the open mapping theorem the image of His nonclosed.Let us apply Lemma 3.4 to P = H and ε = 1/4, and denote the obtained functionalby f1 and the obtained operator by H1.By Lemma 3.3 the operator H1 also hasnonclosed image.

Applying Lemma 3.4 to P = H1 and ε = 1/8 we find functional f2and operator H2. We continue in an obvious way.We have ||Hi−1 −Hi|| < 2−i−1.

Therefore the sequence {Hi}∞i=1 is uniformly conver-gent. Let us denote by R its limit.The operator (R −H) is nuclear and satisfies the inequality||R −H|| < 2−1(2)It is clear that the set H(B(X)) is (1,1)-norming.By this and by inequality (2)we obtain that the set R(B(X)) is (1/2,3/2)-norming.

Moreover we have cl(imR) ⊂T∞i=1 ker fi. The sequence {fi}∞i=1 is linearly independent because it is constructed insuch a way that fi+1 does not vanish onTik=1 ker fk.

Therefore cl(imR) is of infinite codi-mension in M∗. Let us introduce an operator E : M →X∗by (E(m))(x) = (R(x))(m).This operator is an isomorphic embedding because the set R(B(X)) is (1/2,3/2)-norming.It is easy to see that the restriction of E∗onto X coincides with R.Therefore(E∗|X −H) is nuclear and cl(E∗(X)) is of infinite codimension in M∗.

The proof iscomplete.Proof of Theorem 2.4. A nowhere norming subspace is of course nonnorming.

Sowe can apply Proposition 3.5. It should be noted that for nowhere norming M theoperator H is strictly singular.Let E : M →X∗be the operator constructed in Proposition 3.5.

We need only tocheck that E(M) is nowhere norming. But this follows immediately from the fact thatE∗|X = R = H +(R−H) is strictly singular as a sum of two strictly singular operators[LT, p. 76].Proof of Theorem 3.1.

The necessity follows immediately from Proposition 3.5 andthe fact that cl(E∗(X)) is norming subspace in M∗.Let us suppose that M is a Banach space for which there exists a closed normingsubspace V ⊂M∗of infinite codimension. Let {zi}∞i=1be a normalized basic sequencein M∗/V and let m∗i ∈M∗(i ∈N) be such that ||m∗i || ≤2 and Q(m∗i ) = zi, whereQ : M∗→M∗/V is a quotient map.Let X = V ⊕l1.

Let us define an operator H : X →M∗byH(v, {ai}∞i=1) = v +∞Xi=1(ai/i)m∗i .It is clear that this operator is injective but is not an isomorphic embedding.7

The restriction of H∗onto M is an isomorphic embedding because V is a normingsubspace. The subspace H∗(M) ⊂X∗is total because H is injective and is not normingbecause H is not an isomorphic embedding.

This completes the proof of Theorem 3.1.3.6. Corollary.

If M is a total subspace of X∗and M is quasireflexive then X isquasireflexive and Ord(X) = Ord(M).Proof. It is known [CY] that Ord(X∗) = Ord(X) and that the order of quasire-flexivity of a subspace is not greater than the order of quasireflexivity of the wholespace.It is well-known and is easy to see that the duals of quasireflexive spaces do notcontain norming subspaces of infinite codimension.Therefore by Theorem 3.1 thesubspace M ⊂X∗is norming.

Hence X is isomorphic to a subspace of M∗. Using theabovementioned result we obtain Ord(X) ≤Ord(M∗) = Ord(M).Using the abovementioned result once more we obtain Ord(M) ≤Ord(X∗) =Ord(X).

The proof is complete.3.7.Remark. By [DJ, p. 355] nonquasireflexivity of X does not yield the existencein X∗of an infinite codimensional norming subspace.

Therefore there exist nonquasire-flexive spaces which are not isomorphic to total nonnorming subspaces.4. Remarks on the nonseparable case and on spaces withthe Pelczynski property.Theorem 2.1 and Corollaries 2.5 and 2.6 are not valid in nonseparable case.

In orderto prove this let us show that the space X = l1 ⊕l2(Γ) does not have TNNS propertyif card(Γ) > 2c.Let M be a total subspace in X∗. Then 2card(M) ≥card(X) ≥card(Γ).

Conse-quently card(M) > c. Therefore M contains a set of functionals of cardinality greaterthan c, whose restrictions to l1 coincide. Therefore the intersection of M with the sub-space of X∗which vanishes on l1 is an infinite dimensional subspace in {0} ⊕l2(Γ).

Ifwe “transfer” this subspace into X then we shall obtain a subspace over which M isnorming.Problem. Characterize TNNS in the nonseparable setting.At the moment it is known [Al] that C(K) ̸∈TNNS for every compact K.4.1.Proposition.

Let Banach space X be such that every strictly singular operatorT : X →Y is weakly compact. Then X ̸∈TNNS.Proof.

Evidently it is sufficient to consider the case when X is nonreflexive. Let Mbe a total nowhere norming subspace in X∗.

Let XM be the completion of X under thenorm||x||M = sup{|f(x)| : f ∈S(M)}.Let T : X →XM be the natural embedding. The operator T is strictly singular becauseM is nowhere norming.

Hence T is weakly compact.The subspace M ⊂X∗may be considered also as a subspace of (XM)∗. Moreoverthe restriction of T ∗: (XM)∗→X∗onto M is an isometry.The operator T ∗is weakly compact by V.Gantmacher’s theorem [DS, VI.4.8].

There-fore M is reflexive, hence the subspace M is weak∗closed in X∗by the M. Krein–V. Smu-lian theorem [DS, V.5.7].

Since M is a total subspace of X∗we obtain M = X∗. Thiscontradiction completes the proof.8

4.2.Remark.For separable spaces this proposition follows immediately fromTheorem 2.1.The conditions of Proposition 4.1 are satisfied by spaces with the Pelczynski property.Let us recall the definition.A Banach space X has the Pelczynski property if for every subset K ⊂X∗that isnot relatively weakly compact there exists a weakly unconditionally convergent seriesP∞n=1 xn in X such thatinfn supx∗∈K x∗(xn) > 0.This property was introduced by A.Pelczynski in [Pe] (under the name “property(V)”). In the same paper it was proved that for any compact Hausdorffspace S thespace C(S) has property (V ).

For other spaces with the Pelczynski property see [W,pp. 166-172].The fact that spaces with the Pelczynski property satisfy the conditions of Proposi-tion 4.1 follows from the next proposition.4.3.Proposition [W,p.

172]. Suppose X has the Pelczynski property.

Then forevery operator T : X →Y that is not weakly compact there exists a subspace X1 ⊂Xsuch that X1 is isomorphic to c0 and the restriction of T onto X1 is an isomorphicembedding.The spaces which have no TNNS property need not have the Pelczynski propertyand need not satisfy the conditions of Proposition 4.1.A corresponding example is given by James‘ space J. Let us recall its definition [LT,p.

25]. The space J consists of all sequences of scalars x = (a1, a2, .

. .

, an, . .

.) for which||x|| = sup 2−1/2((ap1 −ap2)2 + (ap2 −ap3)2 + .

. .

+ (apm−1 −apm)2)1/2 < ∞where the supremum is taken over all choices of m and p1 < p2 < . .

. < pm; andlimn→∞an = 0.It is easy to see that the operator T : J →c0 which maps every sequence from Jonto the same sequence in c0 is a non-weakly compact strictly singular operator.

On theother hand any total subspace in J∗is norming over J. This follows from well-knownproperties of quasireflexive spaces.In fact there are nonquasireflexive spaces of this type as is shown in the next section.5.

A nonquasireflexive separable Banach space without thePelczynski property whose dual does not contain nowherenorming subspaces.5.1.Theorem. There exists a nonquasireflexive Banach space X ̸∈TNNS and suchthat there exists a strictly singular non-weakly compact operator T : X →c0.Proof.

Let X = (P∞n=1 ⊕J)2. The unit vectors in J we shall denote by {ei}∞i=1.

It isknown [LT, p. 25] that {ei} is shrinking basis of J, therefore its biorthogonal functionals{e∗i }∞i=1 form a basis of J∗.It is clear that vectorsen,j = (0, . .

. , 0, ej, 0, .

. .

).9

(where ej is on the n-th place) after any numeration preserving order in sequences{en,j}∞j=1 form a basis of X. We need the following two lemmas about X and its dual.5.2.Lemma.

Every weakly null sequence {xm}∞m=1 in X for which inf ||xm|| > 0contains a subsequence equivalent to the unit vector basis of l2.5.3.Lemma. Every infinite dimensional subspace of X∗contains a subspace iso-morphic to l2.This lemmas easily follows by well-known arguments (see [An] and [HW]).Let us consider an operator T : (P ⊕J)2 →c0 = (P ⊕c0)0 defined byT(x1, .

. .

, xn, . .

.) = (x′1, .

. .

, x′n, . .

. ),where (xi) is a sequence of elements of J and (x′i) is a sequence of elements of c0 withthe same coordinates.

It is clear that T is a continuous operator. It is not weaklycompact because for any n ∈N the sequence (T(Pkj=1 en,j))∞k=1 does not have limitpoints in weak topology.

At the same time the operator T is strictly singular becauseby Lemma 5.2 the space X does not contain subspaces isomorphic to c0.Let us suppose that X ∈TNNS. Then by Theorem 2.1 there exists a surjectivestrictly singular operator T : X →Z where Z is a certain Banach space.

ConsequentlyZ∗is isomorphic to a subspace of X∗. By Lemma 5.3 the space Z∗contains a subspaceisomorphic to l2.

Let us denote this subspace by U. Let R be the quotient map R :Z →Z/U⊤.

The space (Z/U⊤)∗may be in natural way identified with w∗−clU. SinceU is reflexive then by M.Krein-V.Smulian theorem [DS, V.5.7] we have w∗−cl(U) = U.Therefore Z/U⊤is isomorphic to l2.

Let {ui}∞i=1 be sequence in Z/U⊤equivalent to theunit vector basis of l2. By Lemma 2 of [GR] we can find in X a weakly null sequence{xi}∞i=1 for which {RTxi}∞i=1 is a subsequence of {un}.

By Lemma 5.2 the sequence {xi}contains a subsequence {xni}∞i=1 which is equivalent to the unit vector basis of l2. Therestriction of RT onto the closed linear span of {xni}∞i=1 is an isomorphism.

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