SOME APPROXIMATION PROPERTIES OF BANACH SPACES AND BANACH LATTICES

T. FIGIEL, W. B. JOHNSON^{∗} AND A. PEŁCZYŃSKI

Abstract. The notion of the bounded approximation property = BAP (resp. the uniform approximation property = UAP) of a pair [Banach space, its subspace] is used to prove that if X is a L_{∞}−space, Y a subspace with the BAP (resp. UAP), then the quotient X/Y has the BAP (resp. UAP). If Q : X → Z is a surjection, X is a L_{1}−space and Z is a L_{p}−space (1 ≤ p ≤ ∞) then ker Q has the UAP. A complemented subspace of a weakly sequentially complete Banach lattice has the separable complementation property = SCP. A criterion for a space with GL-l.u.st. to have the SCP is given. Spaces which are quotients of weakly sequentially complete lattices and are uncomplemented in their second duals are studied. Examples are given of spaces with the SCP which have subspaces that fail the SCP. The results are applied to spaces of measures on a compact Abelian group orthogonal to a fixed Sidon set and to Sobolev spaces of functions of bounded variation on ℝ^{n}.

To the memory of Lior Tzafriri

Introduction and notation

Typical problems studied in this paper are the following: Suppose that Q : X → Z is a surjection acting between Banach spaces (resp. J : Y → X is an isomorphic embedding). Assume that X is a nice space, for instance a Banach lattice. Having some information on Z (resp. on Y) we would like to obtain some information on ker Q (resp. on the quotient X/Y).

In Section 1 we introduce the notion of the bounded approximation property (resp. the uniform approximation property) for pairs consisting of a Banach space and a fixed subspace. As an easy application we

Keywords: Bounded approximation property; uniform approximation property; separable complementation property; Banach lattices; GL-lu.s.t.; complementation of a space in its second dual.

2000 Mathematics Subject Classification primary: 46B28, 46B42, 46B10, 46B26; secondary: 43A46, 46E35.

∗ Supported in part by NSF DMS-0503688.

obtain the existence of some new phenomena concerning approximation properties of some subspaces of c_{0}, l_{1} and S_{1} (cf. Corollaries 1.13 and 1.14).

In Section 2, influenced by a construction of Lusky [Lu], we apply that notion to show that a quotient of a L_{∞}−space has the bounded approximation property = BAP (resp. the uniform approximation property = UAP) provided that the kernel of the quotient map has the same property. By duality we obtain a corresponding criterion for subspaces of L_{1}−spaces. As a corollary we get that if Q is a surjection from a L_{1}−space onto a L_{p}−space (1 ≤ p ≤ ∞) then ker Q has the UAP. On the other hand it follows from a result of [J2] that ker Q fails GL-l.u.st.

In Section 3 we discuss an invariant of non-separable Banach spaces - the separable complementation property = SCP. A Banach space X has the SCP provided that every separable subspace of X is contained in a separable subspace which is complemented in X. The property goes back to [AL], who proved that weakly compactly generated = WCG spaces have the SCP. It was observed in [GS] that a weakly sequentially complete Banach lattice has the SCP. We prove that a weakly sequentially complete complemented subspace of a Banach lattice has the SCP. As a corollary we get that a dual Banach space of finite cotype with GL-l.u.st. has the SCP.

In Section 4 we isolate an isomorphic invariant of Banach spaces called property (k). It appears implicitly in [J2] and it is a modification of an invariant invented by S. Kwapień (cf. [KP, pp.142-144]). There is some analogy between property (k) and Grothendieck’s characterization [Gr1] of the Dunford - Pettis property. Combining property (k) with the lattice variant of the factorization theorem for weakly compact operators we give a criterion for kernels of quotient maps of Banach lattices to be uncomplemented in their second duals. At the end of the section we give an example which shows that the SCP need not be preserved when passing to subspaces. This answers a question posed in [PlY]. More examples of that sort are presented in Section 5.

In Section 5 we examine two important examples of spaces failing the SCP: the space of all Borel measures on a compact Abelian group orthogonal to an infinite Sidon set, and the Sobolev space of functions of bounded variation on an open subset of ℝ^{n}.

In Section 6 we mention open problems that are suggested by this research.

Terminology and notation. We employ the standard terminology and notation used in Banach space theory [LT, JL]. B_{X} denotes the open unit ball and S_{X} the unit sphere of a Banach space X. By cl A we denote the norm closure of a subset A of a normed space X. A (short) exact sequence of Banach spaces is a diagram

0 → Y →^{J} X →^{Q} Z → 0,

where X, Y, Z are Banach spaces, J : Y → X is an isomorphic embedding, Q : X → Z is a linear surjection, and J(Y) = ker Q. By L_{p}[0, 1], 1 ≤ p ≤ ∞, we denote the usual L_{p}−space on the interval [0, 1] with respect to the normalized Lebesgue measure.

1. The bounded approximation property of pairs

Definition 1.1. Let X be a Banach space, let Y ⊆ X be a closed linear subspace, let λ ≥ 1. The pair (X, Y) is said to have the λ − BAP if for each λ′ > λ and each subspace F ⊆ X with dim F < ∞, there is a finite rank operator u : X → X such that ||u|| < λ′, u(x) = x for x ∈ F and u(Y) ⊆ Y. If moreover for each λ′ > λ there is a function n → ϕ_{λ′}(n) for n ∈ ℕ such that in addition u can be chosen so that dim u(X) < ϕ_{λ′}(dim F) then the pair (X, Y) is said to have the λ − UAP. If Y = X we say that X has the λ − BAP (resp. λ − UAP).

X has the BAP (resp. UAP) iff X has the λ−BAP (resp. λ−UAP) for some λ ≥ 1.

We begin with some simple consequences of Definition 1.1.

Corollary 1.2. If a pair (X,Y) has the λ−BAP (resp. λ−UAP) then X, Y and the quotient space X/Y have the λ−BAP (resp. λ−UAP).

Proof. Let q : X → X/Y be the quotient map. Pick a finitedimensional subspace F̃ ⊂ X/Y. Then there is a subspace F ⊂ X such that dim F = dim F̃ and q(F) = F̃. Since (X, Y) has the λ−BAP, for every λ′ > λ there is a finite rank operator u : X → X satisfying the requirement of Definition 1.1. In particular u(Y) ⊆ Y which implies the existence of a unique ũ : X/Y → X/Y such that qu = ũq. Clearly ũ has finite rank, ∥ũ∥ ≤ ∥u∥ < λ′ and ũ(x̃) = x̃ for x̃ ∈ F̃. The argument for the λ−UAP is similar. □

Corollary 1.3. Let X be a Banach space, q : X → Z a quotient map, and let dim ker q < ∞. Then the assumption that X has the λ−BAP (resp. λ−UAP) implies that Z has the same property.

Proof. If a subspace F ⊆ X is finite dimensional then the subspace E := ker q + F is also finite dimensional. Thus the assumption on X implies that for each λ′ > λ there is a finite rank operator u : X → X with ∥u∥ < λ′ which is the identity on E. In particular u(ker q) = ker q.

Thus the pair (X, ker q) has the λ−BAP. Now we apply Corollary 1.2. The argument for the λ−UAP is similar. □

Corollary 1.4. Let X be a Banach space, Y a subspace of finite codimension.

(i) If the dual space X^{∗} has the λ−BAP then Y has the λ−BAP.

(ii) If X has the λ−UAP then Y has the λ−UAP.

Proof. (i). Let Y^{⊥} = {x^{∗} ∈ X^{∗} : x^{∗}(y) = 0 for y ∈ Y }. Clearly, dim Y^{⊥} = dim X/Y < ∞. The dual space Y^{∗} is naturally isometric to X^{∗}/Y^{⊥}. Thus, by Corollary 1.3 the space Y^{∗} has the λ−BAP. This yields that Y has the λ−BAP (see, e.g., [C, Proposition 3.5 and Theorem 3.3]).

(ii). X has the λ−UAP if and only if X^{∗} has the λ−UAP [H, M]. Thus the proof of (ii) reduces to the previous case. □

Lemma 1.5. Let Y be a closed linear subspace of the Banach space X and let λ < ∞. Suppose that for every finite dimensional subspace E of X and every ε > 0 there is a finite rank operator T : X → X such that T (Y) ⊆ Y, ∥T ∥ ≤ λ and ∥T x − x∥ ≤ ε∥x∥ for x ∈ E. Then the pair (X, Y) has the λ−BAP.

Proof. Fix numbers ε > 0, λ′ > λ and a subspace E ⊂ X with dim E < ∞. Pick a bounded linear projection P : X → X such that P (X) = E and P (Y) ⊆ Y. Let α > 0 be a small number to be specified later. Choose a finite rank operator T : X → X such that T (Y) ⊆ Y, ∥T ∥ ≤ λ and ∥T x−x∥ ≤ α∥x∥ for x ∈ E. Put S = T +P −T P = T +(I−T)P. Clearly, S(Y) ⊆ Y, Sx = x for x ∈ E and rank S ≤ rank T + rank P < ∞. Furthermore, one has ∥S∥ ≤ ∥T ∥ + α∥P ∥. Thus we will get the estimate ∥S∥ < λ′ if α has been chosen so that α∥P ∥ < λ′ − λ. □

Proposition 1.6. Let X be a Banach space and λ < ∞. Then X^{∗} has the λ−BAP if and only if the pair (X, Y) has the λ−BAP for each finite codimensional subspace Y ⊆ X.

Proof. Our proof of the “only if” part uses Lemma 1.5 and the full force of Proposition 3.5 in [C], which yields that if X^{∗} has the λ−BAP then X has the so called λ−duality bounded approximation property. The latter property means that for each ε > 0 and each pair of finite dimensional subspaces E of X and F of X^{∗}, there is a finite rank operator T on X with ∥T ∥ ≤ λ, ∥T x − x∥ ≤ ε∥x∥, for all x ∈ E and ∥T^{∗}x^{∗} − x^{∗}∥ ≤ ε∥x^{∗}∥, for all x^{∗} ∈ F.

Let λ′ > λ, ε > 0 and α > 0 be fixed (the number α will be specified later on). We also fix a subspace Y ⊆ X of finite codimension and let F = Y^{⊥}. Since F is a weak^{∗} closed subspace of X, there is a weak^{∗} continuous projection Q = P^{∗} from X^{∗} onto F.

Since X has the λ−duality bounded approximation property, we can choose a finite rank operator T on X with ∥T ∥ ≤ λ, ∥T x − x∥ ≤ α∥x∥, for all x ∈ E and ∥T^{∗}x^{∗} − x^{∗}∥ ≤ α∥x^{∗}∥, for all x^{∗} ∈ F. Put U = T + P − P T = T + P (I − T). Observe that U : X → X, rank U ≤ rank T +rank P < ∞, and U^{∗} = T^{∗}+(I^{∗}−T^{∗})Q. Thus for every x^{∗} ∈ F we have Qx^{∗} = x^{∗} and hence U^{∗}x^{∗} = T^{∗}x^{∗} + (I^{∗} − T^{∗})Qx^{∗} = x^{∗}. It follows that for y ∈ Y we have Uy ∈ Y, because for every x^{∗} ∈ F one has x^{∗}(Uy) = (U^{∗}x^{∗})y = x^{∗}y = 0.

Clearly, for each x^{∗} ∈ X^{∗} one has ∥(I^{∗} − T^{∗})P^{∗}x^{∗}∥ ≤ α∥Q∥∥x^{∗}∥, which implies that ∥U − T ∥ = ∥P (I − T)∥ = ∥(I^{∗} − T^{∗})P^{∗}∥ ≤ α∥Q∥. This yields ∥U∥ ≤ λ′, if α has been chosen so that α∥Q∥ ≤ λ′ − λ.

Using the identity x − Ux = x − T x − P (x − T x) = (I − P)(x − T x) which holds true for each x ∈ X, we obtain ∥x−Ux∥ ≤ α(1+∥Q∥)∥x∥ for every x ∈ E. If α is chosen so that α(1 + ∥Q∥) < ε, then we obtain the estimate ∥x − Ux∥ ≤ ε∥x∥ for every x ∈ E.

Now, thanks to Lemma 1.5 we can conclude that the pair (X, Y) has the λ′−BAP for every λ′ > λ. By Definition 1.1 this yields that the pair (X, Y) has the λ−BAP.

To prove the “if” part, let F ⊆ X^{∗}, dim F = n < ∞. Choose a finite dimensional E ⊂ X such that F ∩ E^{⊥} = {0}. Fix ε > 0. Put Y = F_{⊥} := {y ∈ X : fy = 0 for f ∈ F }. Since (X, Y) has the λ−BAP, there is an operator u : X → X such that u(Y) ⊆ Y, ue = e for e ∈ E, rank u < ∞ and ∥u∥ < λ + ε.

We claim that the operator u^{∗} : X^{∗} → X^{∗} satisfies the condition u^{∗}f = f for f ∈ F. For, observe that if f ∈ F, then u^{∗}f ∈ Y^{⊥}, because for each y ∈ Y we have (u^{∗}f)y = f(uy) = 0. Since dim F < ∞, F is weak^{∗} closed in X^{∗} and hence Y^{⊥} = (F_{⊥})^{⊥} = F by the bipolar theorem. It follows that g := u^{∗}f − f ∈ F. Observe that g ∈ E^{⊥} because for e ∈ E we have ge = (u^{∗}f)e − fe = f(ue) − fe = 0. It follows that g ∈ F ∩ E^{⊥} = {0} and hence u^{∗}f − f = g = 0. Since ∥u^{∗}∥ = ∥u∥ < λ+ε, this completes the proof that X^{∗} has the λ−BAP.

□

Remark 1.7. If X is a Banach space with the λ−UAP and Y ⊆ X is a closed subspace of finite codimension, then the pair (X, Y) has the λ−UAP. Indeed, since X^{∗} has the λ−UAP [H, M], the proof can just follow the lines of that of Proposition 1.6 (with appropriate estimates of the ranks of finite rank approximations used in the process).

Proposition 1.8. Let X be a Banach space and let Y ⊆ X be a closed subspace such that dim X/Y = n < ∞ and Y has the λ−BAP (resp.

Y has the λ−UAP). Then the pair (X, Y) has the 3λ−BAP (resp. 3λ−UAP).

Proof. Let ε > 0. Choose a subspace E of X to (1 + ε)-norm Y^{⊥} so that dim E ≤ f(n, ε). Define Q : E ⊕_{1} Y → X by

Q(e, y) = e + y.

Claim: Q is a 3(1 + ε)-quotient map.

To prove the claim observe first that the quotient map R : X → X/Y has the property that R(B_{X}) ⊆ (1 + ε)R(B_{E}). For, if that were false then there would exist an x ∈ B_{X} such that Rx ∉ (1 + ε)R(B_{E}). Let η ∈ (X/Y)^{∗} = Y^{⊥} be a separating linear functional with η(Rx) = 1. Then we would obtain that 1 > (1 + ε) sup{|η(Re)| : e ∈ B_{E}} ≥ 1, because E (1 + ε)-norms Y^{⊥}. This contradiction proves the inclusion.

Now, given any x ∈ X, we can find an e ∈ E such that Re = Rx and ∥e∥ ≤ (1+ε)∥x∥. Observe that x−e ∈ Y, because R(x−e) = Rx−Re = 0. Letting y = x−e, we have y ∈ Y and ∥y∥ ≤ ∥x∥+∥e∥ ≤ (2+ε)∥x∥. Since Q(e, y) = e + (x − e) = x and ∥e∥ + ∥y∥ ≤ (3 + 2ε)∥x∥, we have proved the claim.

Now suppose G ⊆ X is finite dimensional and G ⊇ E, where E is from the claim. We may and do assume that G ∩ Y ≠ {0}. Take a finite rank T : Y → Y with ∥T ∥ ≤ λ + ε and T y = y for y ∈ G ∩ Y. Define S on E ⊕_{1} Y → E ⊕_{1} Y by S(e, y) = (e, T y). Then ∥S∥ = ∥T ∥, S is a projection if T is, and S(e, y) = (e, y) if e ∈ E and y ∈ G ∩ Y. Since G ⊇ E, S is the identity on the kernel of Q and hence S induces an operator S̃: X → X defined by S̃x = e + T y whenever Q(e, y) = x. Then ∥S̃∥ ≤ 3(1 + ε)(λ + ε) and S̃ is the identity on G. □

Remark 1.9. Clearly S̃ is a projection if T is and is weak^{∗} continuous if X is a dual space, Y is weak^{∗} closed, and T is weak^{∗} continuous. If T_{1} : Y → Y is a finite rank operator commuting with T then S̃_{1} commutes with S̃ where S̃_{1} : X → X is constructed for T_{1} in the same way as S̃ for T.

Remark 1.10. We do not know whether in the statement of Proposition 1.8 one can replace 3λ by a smaller quantity. However, the number 3 which appears in the claim made in the proof of that proposition cannot be replaced by a smaller one, even in the simple case where Y = c_{0} ⊂ c = X and E ⊂ X is any subspace of finite dimension. We omit the easy verification of the latter fact.

Corollary 1.11. If X is a Banach space and Y has the λ−BAP for every finite codimensional subspace Y ⊆ X, then X^{∗} has the 3λ−BAP.

Proof. This follows from Proposition 1.6, because every pair (X, Y) has the 3λ−BAP by Proposition 1.8. □

Remark 1.12. If X is a Banach space with the BAP such that X^{∗} fails the BAP, then by Corollary 1.11 for every λ < ∞ there is a finite codimensional subspace Y_{λ} ⊂ X which fails the λ−BAP. This observation shows that one cannot replace X^{∗} by X in the statement of Corollary 1.4 (i). Also it leads to a somewhat simpler proof of the main result of [FJ]. In fact, by applying Corollary 1.4 (i) to some specific Banach spaces X we construct here some new examples with rather surprising properties.

Corollary 1.13. Each of the spaces c_{0} and l_{1} has a subspace Y which has the approximation property but Y fails the BAP.

Proof. In the case of c_{0} we use an old result of G. Schechtman and the second author of the present paper (see [JO], Corollary JS), who proved that there is a subspace X of the space c_{0} which has the BAP while X^{∗} fails the BAP. By Corollary 1.11, for each positive integer n we choose a finite codimensional subspace of X which fails the n−BAP. Obviously, all the X_{n}’s have the approximation property. Hence if Y is the c_{0}−sum of the sequence (X_{n}) then Y is isometric to a subspace of c_{0} and has the approximation property but Y fails the BAP.

Similarly, in the case of l_{1} we start with a subspace X of l_{1} which fails the approximation property found in [Sz1]. As shown in [J1], from the existence of such a space it follows that if we let Z be the l_{1}−sum of a dense sequence (X_{n}) of finite dimensional subspaces of X, then Z^{∗} fails the BAP and yet Z has the approximation property. We conclude as in the previous paragraph (now Y will be the l_{1}−sum of a suitable sequence of finite codimensional subspaces of Z). □

By the result of A. Szankowski [Sz2] the space B(H) fails the BAP. Since B(H) is the dual space of the trace class S_{1}, using again Corollary 1.11 we obtain

Corollary 1.14. Every finite codimensional subspace of the trace class S_{1} contains for each λ < ∞ a finite codimensional subspace Y_{λ} such that Y_{λ} fails the λ−BAP.

Of course, S_{1} has the 1−BAP and it is the simplest non commutative L_{1}-space. On the other hand, Corollary 1.4 implies that every finite codimensional subspace of a commutative L_{1}-space has the 1−BAP.

Remark 1.15. One may ask whether the analogue of Proposition 1.8 for quotient spaces is true. Namely, let Y be a Banach space with the λ−BAP. Consider those Banach spaces X for which there is a quotient map from X onto Y whose kernel is finite dimensional. When is it true that there exists μ < ∞ such that X has the μ−BAP for all those X? The answer is that if Y^{∗} fails the BAP then no such μ exists, while if Y^{∗} has the μ−BAP then X has the 3μ−BAP for all those X. The 3μ−estimate follows from Proposition 1.8, while the nonexistence can be demonstrated using the construction applied in the proof of that proposition.

2. Bounded and uniform approximation properties of quotients of L_{∞}−spaces (resp. subspaces of L_{1}−spaces).

The main result of this section is

Theorem 2.1. Let

0 → Y →^{J} X →^{Q} Z → 0

be an exact sequence of Banach spaces. Then

(a) If X is a L_{∞}−space and Y has the BAP (resp. Y has the UAP) then Z has the BAP (resp. Z has the UAP).

(b) If X is an L_{1}−space and the dual space Z^{∗} has the BAP (resp. Z has the UAP) then Y^{∗} and Y have the BAP (resp. Y has the UAP).

(c) Let 1 ≤ p ≤ ∞. If X is a L_{∞}−space (resp. X is a L_{1}−space) and Y is a L_{p}−space (resp. Z is a L_{p}−space) then Z has the UAP (resp. Y has the UAP).

Proof. (a) By equipping X with an equivalent norm [KP, Lemma 1.1] we can assume that J is an isometric embedding and Q is a quotient map. Suppose Y has the λ−BAP (resp. λ−UAP). By Corollary 1.2, it suffices to establish

(⋆) (X, Y) has the μ−BAP (resp. μ−UAP);

( μ ≥ 1 depends only on λ and X).

To verify (⋆) fix a finite-dimensional subspace F ⊆ X. Put Y_{1} = Y + F. Clearly Y_{1} is a closed subspace of X and dim Y_{1}/Y ≤ dim F < ∞. By Proposition 1.8, for every λ′ > λ there exists a finite rank operator u : Y_{1} → Y_{1} such that

∥u∥ < 3λ′, u(x) = x for x ∈ F, u(Y) ⊆ Y.

Moreover, in the case of UAP, dim u(Y_{1}) < ϕ_{λ′}(dim F).

Since X is a L_{∞}−space, there are c > 1 and a function f : ℕ → ℕ depending only on X such that for each finite-dimensional subspace F_{1} of X there exists a subspace E ⊆ X such that

E ⊇ F_{1}, (2.1) dim E < f(dim F_{1}), (2.2) (2.3) the Banach-Mazur distance d(E, ℓ_{∞}^{dim E}) ≤ c.

Now specify F_{1} := u(Y_{1}). By (2.1), we can regard u as an operator with values in E. Hence, by (2.3), u admits an extension ũ : X → E with ∥ũ∥ ≤ c∥u∥ < 3cλ′ because ℓ_{∞}^{dim E} is a 1-injective space. Clearly ũ(Y) = u(Y) ⊆ Y because ũ extends u. Clearly ũ is of finite rank because ũ(X) ⊆ E. By (2.2) it follows that, in the case of the UAP, dim E is controlled by dim F because in that case dim E ≤ dim F + dim u(Y_{1}) < dim F + ϕ_{λ′}(dim F).

(b) If X is a L_{1}−space then X^{∗} is a L_{∞}−space [JL, p.59]. The assumption that Z^{∗} has BAP allows us to apply the already established case (a) to the dual diagram

0 → Z^{∗} →^{Q^{∗}} X^{∗} →^{J^{∗}} Y^{∗} → 0

to conclude that Y^{∗} has the BAP. Then by [C, Proposition 3.5] we infer that Y has the BAP.

In the case of the UAP it is enough to assume that Z has the UAP. We use the result of Heinrich [H, M] that a Banach space has the UAP if and only if its dual has the UAP.

(c) By [PR], every L_{p}(μ) space has the UAP. By [LR], if 1 < p < ∞ then every L_{p}−space is complemented in some L_{p}(μ) and if either p = 1 or p = ∞ then some dual of every L_{p}−space is complemented in some L_{p}(μ). By [H] if the dual of a space has the UAP then the space itself has the UAP. Thus every L_{p}−space has the UAP for 1 ≤ p ≤ ∞. The desired conclusion follows now from assertions (a) and (b). □

Remark 2.2. Recently Szankowski [Sz3] proved that there exists a subspace X of L_{∞}[0, 1] (resp. C[0, 1]) which fails the approximation property while L_{∞}[0, 1]/X has the 1−BAP (resp. C[0, 1]/X has a basis). The result extends to L_{p}[0, 1] for p > 2, while for 1 ≤ p < 2 the dual result holds

Lusky [Lu, Corollary 3 (b)] proved

(L) If 0 → X →^{J} Y →^{Q} Z → 0 is an exact sequence of Banach spaces such that Y is a separable L_{1} space and Z is not reflexive and has a basis then X has a basis.

Our next result is a consequence of (L). It gives additional information for non-separable spaces. For the terminology and properties of abstract L− and M−spaces we refer to [K1, K2] and [LT, Vol II].

It holds for abstract L− and M−spaces over both real and complex scalars.

Proposition 2.3. Let 0 → X →^{J} Y →^{Q} Z → 0 be an exact sequence of Banach spaces. Assume that Y is isomorphic to an abstract L−space and Z is isomorphic to an abstract M−space with unit. Then every separable subspace of X is contained in a separable subspace of X with a basis.

Proof. We use the following well known facts.

(j) Every separable subspace of an abstract L−space is contained in a separable subspace which is an abstract L−space. (jj) Every separable subspace of an abstract M−space with a unit is contained in a separable subspace which is an abstract M−space with the unit.

(jjj) There is an a < ∞ such that if F is a separable subspace of Z then there is a separable subspace E of Y such that cl aQ(B_{E}) ⊃ B_{F}.

((jjj) is a simple consequence of the proof of the Open Mapping Principle.)

Let E be an infinite-dimensional separable subspace of X. Choose y_{0} ∈ Y so that Qy_{0} corresponds via an isomorphism to the unit of an abstract M−space isomorphic to Z. Using (j) - (jjj) we define inductively increasing sequences (E_{n})_{n=1}^{∞} and (Y_{n})_{n=1}^{∞} of subspaces of Y, and (F_{n})_{n=1}^{∞} and (Z_{n})_{n=1}^{∞} of separable subspaces of Z such that

E_{1} ⊃ J(E) ∪ {y_{0}}; (2.4) (2.5) Y_{n} is isomorphic to a separable abstract L−space; (2.6) Z_{n} is isomorphic to a separable abstract M−space; E_{n} ⊂ Y_{n}; F_{n} ⊂ Z_{n}; cl Q(Y_{n}) = F_{n}; (2.7) cl aQ(B_{E_{n+1}}) ⊃ B_{Z_{n}}. (2.8)

The isomorphisms in (2.5) (resp. in (2.6)) are restrictions of the same isomorphism of Y onto an abstract L−space (resp. of Z onto an abstract M−space).

Put

Y_{∞} := cl ⋃_{n=1}^{∞} Y_{n}; Z_{∞} := cl ⋃_{n=1}^{∞} Z_{n}; X_{0} := J^{−1}(J(X) ∩ Y_{∞}).

By (2.5) and the comment after (2.4)-(2.8), Y_{∞} is isomorphic to an abstract L−space. By (2.6) and the comment, Z_{∞} is isomorphic to an abstract M−space which has the unit because, by (2.4), Q(y_{0}) ∈ Z_{∞}.

Since the Y_{n}’s and the Z_{n}’s are separable so are Y_{∞}, Z_{∞} and X_{0}. It follows from (2.7) and (2.8) that

aQB_{Y_{∞}} ⊃ ⋃_{n=1}^{∞} aQB_{Y_{n}} ⊃ ⋃_{n=1}^{∞} aQB_{E_{n}} ⊃ ⋃_{n=2}^{∞} B_{Z_{n−1}} = ⋃_{n=1}^{∞} B_{Z_{n}}.

Since the sequence (B_{Z_{n}})_{n=1}^{∞} is increasing, cl ⋃_{n=1}^{∞} B_{Z_{n}} = cl B_{Z_{∞}}. Hence aQB_{Y_{∞}} is norm dense in B_{Z_{∞}}. Thus the restriction Q|Y_{∞} maps Y_{∞} onto Z_{∞}. Therefore ker Q|Y_{∞} = ker Q ∩ Y_{∞} = J(X) ∩ Y_{∞} = J(X_{0}).

Now we consider the exact sequence

0 → X_{0} →^{J|X_{0}} Y_{∞} →^{Q|Y_{∞}} Z_{∞} → 0.

Y_{∞} is isomorphic to a separable abstract L−space and thus by [K1] is isomorphic to a separable L_{1}(μ). The space Z_{∞} is isomorphic to a C(K) space on a compact metric K because Z_{∞} is isomorphic to a separable abstract M−space with unit [K2]. Obviously Z_{∞}, being infinite-dimensional, is non-reflexive. By a result of Vakher [V, MP], for every compact metric space K the space C(K) has a basis. Thus, by [Lu], X_{0} has a basis. Obviously X_{0} contains E. □

3. Banach spaces with the separable complementation

property and complemented subspaces of Banach lattices.

Definition 3.1. (cf. [C]) A Banach space X has the SCP (= separable complementation property ) if every separable subspace Y of X is contained in a separable complemented subspace Z of X. If always Z can be chosen to be the range of a projection of norm at most λ, we say that X has the λ-SCP.

A routine argument shows that a space with the SCP must have the λ-SCP for some λ < ∞.

Recall that the following classes of Banach spaces have the SCP (a) Abstract L−spaces [K1].

(b) Preduals of von Neumann algebras (Haagerup [GGMS, Appendix]).

(c) Weakly compactly determined spaces, in particular weakly compactly generated spaces, in particular reflexive spaces (cf. [DGZ, Chapt.VI]).

(d) Weakly sequentially complete Banach lattices [GS].

(e) Banach spaces with the commuting bounded approximation property (Casazza, Kalton, Wojtaszczyk [C, Theorem 9.3]).

It is not known whether a complemented subspace of a space with the SCP must have the SCP. However, we have

Proposition 3.2. A weakly sequentially complete Banach space which is isomorphic to a complemented subspace of a Banach lattice has the SCP.

Proof. By [LT, Vol.II,1.c.6] it is enough to show that if X is a weakly sequentially complete Banach lattice and Y is the range of a projection P : X → X then Y has the SCP. We need the following observations.

(i) A separable subspace of a Banach lattice is contained in a separable sublattice;

(ii) - a precise form of (d). If a Banach lattice X is weakly sequentially complete then every separable subspace of X is contained in the range of a contractive projection of X whose range is a separable sublattice.

(iii) A weakly sequentially complete Banach lattice is complemented in its second dual [LT, Vol.II,1.c.4 ].

Let E be a fixed separable subspace of Y. Using (i) and (ii) we construct inductively increasing sequences of separable subspaces of X, (E_{n}), (S_{n}), (G_{n}) and a sequence p_{n} : X → X of contractive projections with separable ranges such that

E_{1} = E, E_{n} ⊆ Y for n = 1, 2….; (3.1) (3.2) S_{n} is a sublattice of X and E_{n} ⊆ S_{n} for n = 1, 2, …; S_{n} ⊆ G_{n} := p_{n}(X) for n = 1, 2, …; (3.3) E_{n+1} := cl(P (G_{n})). (3.4)

Put F := cl(⋃_{n=1}^{∞} E_{n}) and S := cl(⋃_{n=1}^{∞} S_{n}). Clearly, F and S are separable and E ⊆ F ⊆ Y. Observe that P (S) ⊆ F, because for each n one has P (S_{n}) ⊆ P (G_{n}) ⊆ E_{n+1} ⊆ F.

Since S a separable weakly sequentially complete sublattice, the canonical image of S in S^{∗∗} is the range of a contractive band projection p_{S} : S^{∗∗} → S (by (iii)). We shall define a projection π from X onto F as the composition π = P p_{S}p, where p is defined as follows.

Let LIM be a fixed Banach limit, i.e., a positive linear functional on l_{∞} such that LIM(a) = lim_{n} a_{n} for each convergent scalar sequence a = (a_{n}) (cf. [B, II.3.3], [DS, II.4.22]). We define a linear operator p : X → S^{∗∗} by the formula (px)(s^{∗}) = LIM_{n} s^{∗}(p_{n}(x)) for every x ∈ X and s^{∗} ∈ S^{∗}.

Note that p_{S}p is a projection from X onto S, because for every fixed n and z ∈ S_{n} one has z ∈ G_{m} for m ≥ n. Hence p_{m}(z) = z for m ≥ n.

Thus p_{S}p(z) = z. Since ⋃_{n=1}^{∞} S_{n} is dense in S and ||p|| = sup_{n} ||p_{n}|| = 1, we infer that p_{S}p(z) = z for z ∈ S.

Now, if f ∈ F, then π(f) = P (p_{S}p(f)) = P (f) = f, because f ∈ Y. Also, if x ∈ X, then π(x) = P (p_{S}p(x)) = P (s), where s = p_{S}p(x) ∈ S. Since P (S) ⊆ F, it follows that π projects X down onto F. Thus the restriction of π to Y is the desired projection in Y with separable range containing E. □

For the sake of completeness we include

Proof of (ii). Let E be a separable subspace of a weakly sequentially complete Banach lattice X. Hence X is order continuous. Then, by a result of Kakutani (cf. [LT, Vol.II,1.a.9 ]) there is a sublattice X_{0} of X with a weak unit such that X_{0} ⊃ E and there is a contractive projection, say P_{0}, from X onto X_{0}. Since X_{0} has a weak unit and is order continuous, a result of Nakano (cf.[LT, Vol.II,1.b.16 ]) and the remark after the proof of [LT, Vol.II,1.b.16]) gives that X_{0} is weakly compactly generated. Thus, by a result of Amir and Lindenstrauss [AL], (cf. [DGZ, Chapt. VI, Theorem 2.5]), there is a contractive projection P_{1} : X_{0} → X_{0} whose range is separable and contains E. The desired projection is the composition P_{1} ∘ P_{0}. □

Remark 3.3. A Banach space X has the SEP=separable extension property provided that for every separable subspace E ⊂ X there exists an operator u : X → X with separable range such that u(x) = x for x ∈ E. Clearly the SCP implies the SEP. A slight modification of the proof of Proposition 3.2 shows that for weakly sequentially complete Banach lattices the SEP implies the SCP.

Recall [JL, section 8] that a Banach space E has finite cotype, equivalently E does not contain ℓ_{∞}^{n} uniformly, provided that

sup_{n} inf_{F ⊂E, dim F =n} d(F, ℓ_{∞}^{n}) = ∞.

Here d(·, ·) denotes the Banach-Mazur multiplicative distance. Recall [JL, section 9] that a Banach space Y has GL-l.u.st. provided that Y^{∗∗} is isomorphic to a complemented subspace of a Banach lattice.

Corollary 3.4. If a Banach space Y of finite cotype has GL-l.u.st. then Y^{∗∗} has the SCP. Moreover if Y is isomorphic to a dual Banach space then Y has the SCP.

Proof. By the local reflexivity principle Y^{∗∗} has finite cotype. Thus Y^{∗∗} does not contain subspaces isomorphic to c_{0} and hence is weakly sequentially complete [LT, Vol.II,1.c.4]. Remembering that Y has GLl.u.st., we infer that the weakly sequentially complete space Y^{∗∗} is isomorphic to a complemented subspace of a Banach lattice. Hence, by Proposition 3.2, Y^{∗∗} has the SCP. Moreover if Y is a dual Banach space then Y is complemented in Y^{∗∗}; again we use Proposition 3.2. □

4. Property (k)

Recall that (y_{m}) is a CCC sequence of (x_{n}) of elements of a linear space provided that there are a sequence (c_{n}) of non-negative numbers and a strictly increasing sequence of positive integers (n_{m}) such that

(4.1)  y_{m} = ∑_{n=n_{m}}^{n_{m+1}−1} c_{n}x_{n}, ∑_{n=n_{m}}^{n_{m+1}−1} c_{n} = 1 (m = 1, 2, …).

CCC - stands for “consecutive convex combinations”.

Definition 4.1. A Banach space X has property (k) provided that for every sequence (x_{n}^{∗}) ⊂ X^{∗} which weak^{∗} converges to 0 there exists a CCC sequence (y_{m}^{∗}) of (x_{n}^{∗}) such that for each linear operator u : L_{1}[0, 1] → X,

lim_{m} y_{m}^{∗}(uf_{m}) = 0

for every sequence (f_{m}) in L_{1}[0, 1] which weakly converges to 0 such that sup_{m} ||f_{m}||_{∞} < +∞.

Remark 4.2. We thank Professor Eve Oja who has pointed out that if X is a Banach space with the Radon-Nikodým property then X has property (k). For, by a result of Lewis and Stegall (cf. [DU, Chapt. III, Theorem 8, p. 66]) every operator u : L_{1}[0, 1] → X factors through ℓ_{1}. Hence if (f_{n}) ⊂ L_{1}[0, 1] is a weakly null sequence then lim_{n} ||uf_{n}||_{X} = 0. Therefore lim_{n} x_{n}^{∗}(uf_{n}) = 0 for every bounded sequence (x_{n}^{∗}) ⊂ X^{∗}. □

Remark 4.3. It follows from Sobczyk’s theorem [LT, Vol.I,2.f.5] that every weak^{∗} convergent to 0 sequence in the dual of a subspace of a separable space extends to a weak^{∗} convergent to 0 sequence in the dual to the whole space. Thus a subspace of a separable space with property (k) again has property (k). Clearly if X has property (k) then every complemented subspace of X has property (k). Therefore if X has the SCP and property (k) then every separable subspace of X has property (k).

Remark 4.4. In Definition 4.1 the requirement sup_{m} ∥f_{m}∥_{∞} < ∞ is superfluous in view of FACT. If (f_{n}) is a weakly-null sequence in L_{1}[0, 1] then for every ε > 0 there exists a weakly-null sequence (f′_{n}) ⊂ L_{1}[0, 1] such that sup_{n} ∥f′_{n}∥_{∞} < ∞ and sup_{n} ∥f_{n} − f′_{n}∥_{1} < ε. Proof (S. Kwapień). For f ∈ L_{1}[0, 1] let S_{m}(f) denote the m-th partial sum of the expansion of f with respect to the Haar basis (m = 1, 2, …). Since S_{m} is a conditional expectation with respect to the sigma field generated by the first m functions of the Haar basis, ∥S_{m}(f)∥_{1} ≤ ∥f∥_{1} and ∥S_{m}(f)∥_{∞} ≤ ∥f∥_{∞}. Since (f_{n}) is weakly-null, lim_{n} ∥S_{m}(f_{n})∥_{1} = 0 for m = 1, 2. . . .. Thus there is a non-decreasing sequence of indices (m_{n}) with lim_{n} m_{n} = ∞ such that lim_{n} ∥S_{m_{n}}(f_{n})∥_{1} = 0. Now given ε > 0 the equi-integrability of the set of elements of the weakly-null sequence (f_{n}) yields the existence of M < ∞ and (f̃_{n}) ⊂ L_{1}[0, 1] such that sup_{n} ∥f̃_{n}∥_{∞} < M and sup_{n} ∥f_{n} − f̃_{n}∥_{1} < ε/2. Put

f′_{n} := f̃_{n} + S_{m_{n}}(f_{n} − f̃_{n}) = f̃_{n} − S_{m_{n}}(f̃_{n}) + S_{m_{n}}(f_{n}) (n = 1, 2, …).

Note that (f̃_{n} − S_{m_{n}}(f̃_{n})) is a weakly-null sequence in L_{1}[0, 1] because sup_{n} ∥f̃_{n} − S_{m_{n}}(f̃_{n})∥_{∞} < 2M and lim_{n} S_{m}(f̃_{n} − S_{m_{n}}(f̃_{n})) = 0 for m = 1, 2, . . .. Thus (f′_{n}) is weakly-null. Finally ∥f_{n} − f′_{n}∥_{1} ≤ ∥f_{n} − f̃_{n}∥_{1} + ∥S_{m_{n}}(f_{n} − f̃_{n})∥_{1} < ε for n = 1, 2, . . .. □

Proposition 4.5.

(a) A separable subspace of a weakly sequentially complete Banach lattice has property (k).

(b) A weakly sequentially complete Banach lattice with a weak unit has property (k).

We need the following variant of the factorization theorem for weakly compact operators [DFJP].

Lemma 4.6. Let L be a Köthe function space on a probability measure space (Ω, Σ, μ) (cf. [LT, Definition Vol. II,1.b.17]). Assume that L as a Banach space is weakly sequentially complete. Then there is a reflexive Köthe function space R on (Ω, Σ, μ) such that

L_{∞} := L_{∞}(Ω, Σ, μ) ↪^{I_{∞,R}} R ↪^{I_{R,L}} L,

i.e. the set theoretical inclusion I_{∞,L} admits the factorization through the set theoretical inclusions I_{∞,L} = I_{R,L} ∘ I_{∞,R}.

Proof. Let Y be a function space on (Ω, Σ, μ) which contains constant functions and is a module over the algebra L_{∞}. Call a norm p on Y monotone provided that

p(ϕf) ≤ p(f)||ϕ||_{∞} (f ∈ Y, ϕ ∈ L_{∞}).

Then Y under the norm p is a Köthe function space on (Ω, Σ, μ).

It is enough to show that if f ∈ Y then |f| ∈ Y and p(|f|) = p(f). To prove this, define

ϕ(x) = \begin{cases}|f(x)|/f(x) & for f(x) ≠ 0, \\ 1 & otherwise. \end{cases}

Note that ||ϕ||_{∞} = ||ϕ^{−1}||_{∞}. Thus the identities |f| = fϕ and f = |f|ϕ^{−1} imply p(|f|) ≤ p(f) and p(f) ≤ p(|f|). Thus p(f) = p(|f|).

Next observe that since L is weakly sequentially complete and L_{∞} is isomorphic to a C(K)−space, I_{∞,L} is weakly compact [DS, Theorem VI,§7.6]. Put W = I_{∞,L}(B_{L_{∞}}). Then W is a convex weakly relatively compact subset of L (In fact W is weakly compact). For each n = 1, 2, … let p_{n} be the gauge functional of the set 2^{n}W + 2^{−n}B_{L}. Define, for x ∈ L, p(x) = (∑_{n=1}^{∞} p_{n}^{2}(x))^{1/2}, let R = {x ∈ L : p(x) < ∞}. Clearly R is a function space on (Ω, Σ, μ) which is a module over L_{∞}. It is shown in [DFJP] that R under the norm p is a reflexive Banach space and we have the factorization I_{∞,L} = I_{R,L} ·I_{∞,R}. It remains to establish that R is a Köthe space on (Ω, Σ, μ), which reduces to verifying that the norm p is monotone. For some n ∈ ℕ pick f ∈ 2^{n}W + 2^{−n}B_{L}. Then f = g + h with ||g||_{∞} ≤ 2^{n} and ||h||_{L} ≤ 2^{−n}. Hence for every ψ ∈ L_{∞} with ||ψ||_{∞} ≤ 1 we have fψ = gψ + hψ with ||gψ||_{∞} ≤ 2^{n} and ||hψ||_{L} ≤ 2^{−n}. Thus fψ ∈ 2^{n}W + 2^{−n}B_{L}. Therefore p_{n} is monotone. Now, in a similar way, we infer that p is monotone. □

Proof of Proposition 4.5. Recall that every separable subspace of a Banach lattice is contained in a separable sublattice. Consequently, a Banach lattice is weakly sequentially complete iff every separable sublattice of X is weakly sequentially complete. Every separable lattice has a weak unit. Hence case (a) reduces to (b). Therefore one can assume that the lattice in question, say L, has a weak unit. Thus, by [LT, Vol II,Theorem 1.b.14], there exists a probability measure space (Ω, Σ, μ) such that L can be represented as a Köthe space on (Ω, Σ, μ) and the inclusion I_{∞,L} has a dense range. By Lemma 4.6, there is a reflexive Köthe space R on (Ω, Σ, μ) such that I_{∞,L} = I_{R,L}∘I_{∞,R}. Now we repeat an argument of [J2]. Since R is reflexive, so is R^{∗}. Thus the adjoint (I_{R,L})^{∗} : L^{∗} → R^{∗} is weakly compact, and it is injective because I_{R,L} has a dense range. Thus if (x_{n}^{∗}) ⊂ L^{∗} weak^{∗} converges to 0 then (I_{R,L})^{∗}(x_{n}^{∗}) converges to 0 weakly in R^{∗}. Therefore, by S. Mazur’s theorem, there is a CCC sequence (y_{m}^{∗}) of (x_{n}^{∗}) such that lim_{m} ||(I_{R,L})^{∗}(y_{m}^{∗})||_{R^{∗}} = 0. Thus also the sequence (|(I_{R,L})^{∗}(y_{m}^{∗})|) converges to 0 in the norm ||·||_{R^{∗}}. But |(I_{R,L})^{∗}(y_{m}^{∗})| = (I_{R,L})^{∗}(|y_{m}^{∗}|) because (I_{R,L})^{∗} is a lattice homomorphism. But then the only possible weak^{∗} cluster point in L^{∗} of (|y_{m}^{∗}|) is 0, so (|y_{m}^{∗}|) must converge weak^{∗} in L^{∗} to 0.

Next note that for every weakly sequentially complete Banach lattice L every operator u : L_{1}[0, 1] → L takes the unit ball of L_{∞}[0, 1], regarded as a subset of L_{1}[0, 1], into a lattice bounded subset of L (cf. [AB, p.24]). Thus for every weakly convergent to 0 sequence (f_{m}) ⊂ L_{1}[0, 1] with sup_{m} ||f_{m}||_{∞} < ∞ there is a positive x ∈ L such that |uf_{m}| ≤ x for m = 1, 2, . . .. Therefore lim_{m} y_{m}^{∗}(uf_{m}) = 0 because

|y_{m}^{∗}(uf_{m})| ≤ |y_{m}^{∗}|(|uf_{m}|) ≤ |y_{m}^{∗}|(x) (m = 1, 2 …)

and (|y_{m}^{∗}|) weak^{∗} converges to 0. □

Proposition 4.5 implies in particular that separable subspaces of abstract L−spaces have property (k). The next proposition extends this result to some preduals of von Neumann algebras.

Proposition 4.7. A separable subspace of the predual of a von Neumann algebra has property (k).

Proof. A precise version of a result of Haagerup [GGMS, Appendix] states that every separable subspace of the predual M_{∗} of a von Neumann algebra M is contained in a separable complemented subspace of M_{∗} which is itself the predual of a von Neumann algebra. Thus without loss of generality we can assume in the sequel that M_{∗} is separable. Let M_{∗}^{+} denote the positive cone of M_{∗}. Since M_{∗} is separable there is in M_{∗} a sequence (ω_{r}) whose elements form a dense set in M_{∗}^{+}. Now let (a_{n}) ⊂ M be a weak^{∗}-null sequence (= a sequence convergent to 0 in the σ(M, M_{∗})−topology). Then for every ω ∈ M_{∗}^{+} the sequence (a_{n}) converges weakly to 0 in the Hilbert space (M, ω). The space (M, ω) is the completion of M in the norm ∥·∥_{ω} where ∥g∥_{ω} = ω(g^{∗}g +gg^{∗})^{1/2} for g ∈ M. Thus applying Mazur’s theorem on convex combinations we construct a CCC sequence (b_{m}) of the sequence (a_{n}) so that for m = 1, 2, …,

(4.2)  ω_{r}(b_{m}b_{m}^{∗} + b_{m}^{∗}b_{m})^{1/2} < 1/m for r = 1, 2, …, m.

Next let u : L_{1}[0, 1] → M_{∗} be a bounded linear operator and let (f_{m}) ⊂ L_{1}[0, 1] be a weakly convergent to 0 sequence such that sup_{m} ∥f_{m}∥_{∞} < ∞. Then the sequence (uf_{m}) converges to 0 weakly in M_{∗}. Therefore, by [T, Theorem III.5.4 (iv)], there exists ω ∈ M_{∗}^{+} with the property that for any ε > 0 there exists δ > 0 such that if ω(bb^{∗} + b^{∗}b)^{1/2} < δ then |b(uf_{m})|∥b∥ < ε for m = 1, 2, . . .. Since (ω_{r}) is norm dense in M_{∗}^{+}, it follows from (4.2) that for some k large enough ω_{k}(bb^{∗} + b^{∗}b)^{1/2} < δ implies |b(uf_{m})| < ε for m = 1, 2 . . .. Therefore lim_{m} b_{m}(uf_{m}) = 0. □

The next example shows that in Proposition 4.5 the assumption on the existence of a weak unit is essential even for abstract L−spaces.

Example 4.I. Let T be the set of all strictly increasing sequences (n_{m})_{m=1}^{∞} of integers with n_{1} = 1. Define

X := (∑_{t∈T} L_{1}(μ_{t}))_{1},

where for all t ∈ T the measure μ_{t} is the normalized Lebesgue measure on [0, 1]. Then

X^{∗} = (∑_{t∈T} L_{∞}(μ_{t}))_{∞}.

Let (r_{j}) denote the Rademacher functions. Define (g_{n}^{∗}) ⊂ X^{∗} by

g_{n}^{∗}(t) = r_{j(n,t)},
where j(n, t) = m for n_{m} ≤ n < n_{m+1} (n = 1, 2, … t ∈ T).

Then g_{n}^{∗} → 0 weak^{∗} because, for all t ∈ T, g_{n}(t) → 0 weak^{∗} in L_{∞}(μ_{t}) = (L_{1}(μ_{t}))^{∗} as n → ∞ and sup_{n} ∥g_{n}^{∗}∥_{X^{∗}} = 1.

Suppose now that (h_{m}^{∗}) is a CCC sequence of (g_{n}^{∗}), say

h_{m}^{∗} = ∑_{j=n_{m}^{o}}^{j=n_{m+1}^{o}−1} c_{j}g_{j}^{∗}, ∑_{j=n_{m}^{o}}^{j=n_{m+1}^{o}−1} c_{j} = 1, c_{j} ≥ 0 (m, j = 1, 2, …).

Let t_{o} = (n_{m}^{o}) ∈ T. Define u : L_{1}[0, 1] → X to be identity on the t_{o} coordinate and 0 otherwise. Clearly r_{j(m,t_{o})} → 0 weakly in L_{1}[0, 1] as m → ∞, while h_{m}^{∗}(ur_{j(m,t_{o})}) = 1 for m = 1, 2, . . .. Thus X fails property (k).

Recall the Lindenstrauss Lifting Principle [L], [KP] (LLP) Let Q : X → V be a surjection (X, V −Banach spaces). As-

sume that ker Q is complemented in (ker Q)^{∗∗}. Then for every L_{1}−space Z, every linear operator u : Z → V admits a lifting ũ : Z → X, i.e. Qũ = u.

Corollary 4.8. Let

(4.3)  0 → W →^{J} X →^{Q} V → 0

be an exact sequence of Banach spaces. If W is complemented in W^{∗∗} and X has property (k) then V has property (k).

Proof. Let a sequence (v_{n}^{∗}) ⊂ V^{∗} weak^{∗} converge to 0. Then (Q^{∗}v_{n}^{∗}) weak^{∗} converges to 0. Since X has property (k), there is a CCC sequence (y_{m}^{∗}) of (Q^{∗}v_{n}^{∗}) which satisfies the requirement of Definition 4.1. Let (w_{m}^{∗}) ⊂ V^{∗} be a unique CCC sequence of (v_{n}^{∗}) such that Q^{∗}w_{m}^{∗} = y_{m}^{∗} for m = 1, 2 . . .. Since ker Q = W is complemented in W^{∗∗}, (LLP) implies that every u : L_{1}[0, 1] → V admits a lifting ũ : X → V. Since X has property (k), for every weakly convergent to 0 sequence (f_{k}) ⊂ L_{1}[0, 1] with sup_{k} ∥f_{k}∥ < ∞ one has

lim_{m} w_{m}^{∗}(uf_{m}) = lim_{m} w_{m}^{∗}(Qũ(f_{m})) = lim_{m} y_{m}^{∗}(ũ(f_{m})) = 0.

□

Proposition 4.9. If a Banach space V contains a complemented subspace isomorphic to c_{0}, then V fails property (k).

Proof. Let w : c_{0} → V be an isomorphic embedding onto a complemented subspace of V. Then there is a operator p : V → c_{0} such that pw : c_{0} → c_{0} is the identity. Let (e_{n}) (resp. (e_{n}^{∗})) denote the unit-vector basis of c_{0} (resp. of ℓ_{1}). Then (e_{n}^{∗}) weak^{∗} converges to 0 in ℓ_{1} regarded as the dual space of c_{0}. Hence (p^{∗}(e_{n})) weak^{∗} converges to 0 in V^{∗}. Let (y_{m}^{∗}) be an arbitrary CCC sequence of (p^{∗}(e_{n}^{∗})). Define Δ_{m} ⊂ c_{0} for m = 1, 2, … by Δ_{m} = ∑_{n=n_{m}}^{n_{m+1}−1} e_{n} where (n_{m}) is the sequence of positive integers satisfying (4.1). Define u : L_{1}[0, 1] → V by

uf = w( ∑_{m}(∫_{0}^{1} ff̄_{m}) Δ_{m}) for f ∈ L[0, 1]

where (f_{m}) is any orthonormal sequence in L_{2}[0, 1] with ∥f_{m}∥_{∞} = 1 for m = 1, 2, …. Clearly (f_{m}) converges weakly to 0 in L_{1}[0, 1] and it is easy to see that uf_{m} = wΔ_{m} for m = 1, 2, … and ∥u∥ = ∥w∥. We have for m = 1, 2, …,

y_{m}^{∗}(uf_{m}) = ( ∑_{n=n_{m}}^{n_{m+1}−1} c_{n}p^{∗}e_{n}^{∗})(wΔ_{m})

= ( ∑_{n=n_{m}}^{n_{m+1}−1} c_{n}e_{n}^{∗})(pwΔ_{m})

= ( ∑_{n=n_{m}}^{n_{m+1}−1} c_{n}e_{n}^{∗})(Δ_{m}) = ∑_{n=n_{m}}^{n_{m+1}−1} c_{n} = 1.

Thus V fails property (k). □

Note that ℓ_{∞} contains c_{0} but has property (k) because in ℓ_{∞}^{∗} every weak^{∗} convergent sequence converges weakly (cf. [Gr1]).

Corollary 4.10. Suppose that we are given an exact sequence (4.3). Assume that X has property (k) and V contains a complemented subspace isomorphic to c_{0}. Then W is not complemented in W^{∗∗}, in particular W is not isomorphic to a dual Banach space. □

Applying results on property (k) and some results of previous sections to Banach lattices we get

Corollary 4.11. Suppose that we are given an exact sequence (4.3). Assume that V contains a subspace isomorphic to c_{0} and either

(i) X is a separable subspace of a weakly sequentially complete

Banach lattice, or

(ii) X is a complemented subspace of a weakly sequentially com-

plete Banach lattice with a weak unit, or

(iii) X is a separable subspace of a Banach space of finite cotype

with GL-l.u.st. Then W is not complemented in W^{∗∗}, in particular W is not isomorphic to a dual Banach space. □

For (iii) note that a Banach space of finite cotype with GL-l.u.st. is isomorphic to a subspace of a weakly sequentially complete lattice.

If in (4.3) X is a weakly sequentially complete Banach lattice without a weak unit and V contains an uncomplemented subspace isomorphic to c_{0} then sometimes W is not complemented in W^{∗∗} (see Example 4.II), and sometimes W is a dual Banach space (see Example 4.III). Example 4.II. If Γ is a set of cardinality continuum then there is a surjection Q : ℓ_{1}(Γ) → ℓ_{∞}. Then W = ker Q is not complemented in W^{∗∗}.

Proof. Define u : L_{1}[0, 1] → c_{0} ⊂ ℓ_{∞} by uf = (∫_{0}^{1} ff̄_{n}dx) for f ∈ L_{1}[0, 1] where (f_{n}) is any orthonormal uniformly bounded sequence. Then u does not lift to ℓ_{1}(Γ) because (f_{n}) weakly converges to 0, while ∥uf_{n}∥_{∞} = 1 for n = 1, 2, … and in ℓ_{1}(Γ) every weakly convergent sequence converges in norm. Thus in view of (LLP), W is not complemented in W^{∗∗}. □

In the last example of this section the kernel of the surjection is a dual Banach space. The example shows that the SCP may fail when passing to subspaces. This answers in the negative a question posed in [PlY]. Example 4.III. Let Q : L → c_{0}(Γ) be a surjection where L is a separable lattice of finite cotype and Γ is an infinite set. Then Q^{∗∗} : L^{∗∗} → ℓ_{∞}(Γ) = c_{0}(Γ)^{∗∗} is also a surjection. By Proposition 3.2, L^{∗∗} has the SCP because L^{∗∗} has the same cotype as L; hence it is a weakly sequentially complete lattice. Clearly ker Q^{∗∗} is a weak^{∗}-closed subspace of L^{∗∗}, hence it is a dual Banach space, in particular ker Q^{∗∗} is complemented in its second dual.

Proposition 4.12. ker Q^{∗∗} fails the SCP; precisely if Γ_{0} ⊂ Γ is a countable infinite set, L_{0} is a separable subspace of L such that Q(L_{0}) = c_{0}(Γ_{0}) and E is a separable subspace of ker Q^{∗∗} with ker Q ∩ L_{0} ⊂ E, then there is no projection from ker Q^{∗∗} onto E.

Proof. Let X = cl(L_{0} + E). Then X is a separable subspace of the weakly sequentially complete Banach lattice L^{∗∗}. Therefore, by Proposition 4.5(a), X has property (k). We show that the quotient space X/E contains c_{0}. Let (δ_{j}) be the unit-vector basis of c_{0} = c_{0}(Γ_{0}). Clearly c_{0}(Γ_{0}) can be regarded as a subspace of c_{0}(Γ). Choose ξ_{j} ∈ L_{0} so that Q(ξ_{j}) = δ_{j} for j = 1, 2, . . .. Identifying c_{0}(Γ) with its canonical image in ℓ_{∞}(Γ) and L with its canonical image in L^{∗∗} we infer that Q^{∗∗}(ξ_{j}) = δ_{j} for j = 1, 2, . . .. Since Q(L_{0}) = c_{0}(Γ_{0}) and Q^{∗∗}(L^{∗∗}) = ℓ_{∞}(Γ), there are positive constants a and b such that for arbitrary n = 1, 2, … and scalars t_{1}, t_{2}, …, t_{n} one has

a∥ ∑_{j=1}^{n} t_{j}δ_{j}∥_{c_{0}(Γ_{0})} ≥ inf_{ξ∈ker Q∩L_{0}} ∥ ∑_{j=1}^{n} t_{j}ξ_{j} + ξ∥ ≥ inf_{ξ^{∗∗}∈E} ∥ ∑_{j=1}^{n} t_{i}ξ_{j} + ξ^{∗∗}∥

= ∥ ∑_{j=1}^{n} t_{j}ξ_{j}∥_{X/E} ≥ inf_{ξ^{∗∗}∈ker Q^{∗∗}} ∥ ∑_{j=1}^{n} t_{j}ξ_{j} + ξ^{∗∗}∥ ≥ b∥ ∑_{j=1}^{n} t_{j}δ_{j}∥_{ℓ_{∞}}.

Thus (qξ_{j}) is equivalent to the unit-vector basis of c_{0}, where q : X → X/E denotes the quotient map. Therefore, by Corollary 4.11(i), E is not complemented in E^{∗∗} and hence is not isomorphic to a complemented subspace of any dual space, which implies that there is no projection from ker Q^{∗∗} onto E (cf. [D, L]). □

Remark 4.13. Analyzing the proof of Proposition 4.12 we obtain If L̃ is a subspace of ker Q^{∗∗} such that ker Q ⊂ L̃ and L̃ is complemented in (L̃)^{∗∗} then L̃ fails the SCP.

5. Application to Sidon sets and Sobolev spaces of functions of bounded variation

5.1. Spaces of measures on a compact Abelian group orthogonal to a fixed Sidon set. Let G be a compact Abelian group, Γ its dual. Let L_{1}(G) denote the space of all complex-valued functions on G absolutely integrable with respect to the Haar measure of G. Let M(G) denote the space of all complex-valued regular Borel measures μ on G with bounded variation var(μ) with the norm ∥μ∥ = var(μ(G)).

For S ⊂ Γ put S^{⊥} = Γ \ S. Put

L_{1}^{S^{⊥}}(G) = {f ∈ L_{1}(G) : ∫_{G} f(g)γ^{−1}(g)dg = 0 for γ ∈ S};
M^{S^{⊥}}(G) = {μ ∈ M(G) : ∫_{G} γ^{−1}(g)dμ = 0 for γ ∈ S}.

Clearly L_{1}^{S^{⊥}}(G) can be regarded as subspace of M^{S^{⊥}}(G). Less obvious is the following fact.

Proposition 5.1. There is an isometrically isomorphic embedding Υ : M^{S^{⊥}}(G) → (L_{1}^{S^{⊥}}(G))^{∗∗} and a contraction P : (L_{1}^{S^{⊥}}(G))^{∗∗} → M^{S^{⊥}}(G) such that P ∘ Υ = id_{M^{S^{⊥}}(G)}; in other words, M^{S^{⊥}}(G) is isometrically isomorphic to a complemented subspace (via a contractive projection) of the second dual of L_{1}^{S^{⊥}}(G).

Outline of the proof. Let (Φ_{U})_{U∈O} be an approximate identity of the convolution algebra L_{1}(G), where O is the set of neighborhoods of the unit of G; 0 ≤ Φ_{U} ∈ L_{1}(G); ∫_{G} Φ_{U}dg = 1; Φ(g) = 0 for g ∈ G \ U.

Let U be an ultrafilter on O such that {V ∈ O : V ⊆ U} ∈ U, for U ∈ O. For μ ∈ M^{S^{⊥}}(G) define Υμ ∈ (L_{1}^{S^{⊥}}(G))^{∗∗} by

Υμ(x^{∗}) = LIM_{U∈U} ∫_{G}(Φ_{U} ⋆ μ)ϕ_{x^{∗}}(g)dg for x^{∗} ∈ (L_{1}^{S⊥}(G))^{∗},

where LIM_{U∈U} denotes the limit with respect to the ultrafilter U, ϕ_{x^{∗}} ∈ L_{∞}(G) = (L_{1}(G))^{∗} is a norm preserving Hahn-Banach extension of the functional x^{∗} and ”⋆” stands for convolution.

To define P first observe that M^{S^{⊥}}(G) is naturally isometrically isomorphic with the dual space of the quotient C(G)/(M^{S^{⊥}}(G))_{⊥} where

(M^{S^{⊥}}(G))_{⊥} = {f ∈ C(G) : ∫_{G} f(g)dμ = 0 for μ ∈ M^{S^{⊥}}(G)}.

Define the isometrically isomorphic embedding

J : C(G)/(M^{S^{⊥}}(G))_{⊥} → (L_{1}^{S^{⊥}}(G))^{∗}

as follows: if [ψ] is a coset of a ψ ∈ C(G) then J([ψ]) ∈ L_{1}^{S^{⊥}}(G)^{∗} is defined by J([ψ])(f) = ∫_{G} ψ(g)f(g)dg for f ∈ L_{1}^{S^{⊥}}(G). Remembering that (C(G)/(M^{S^{⊥}}(G))_{⊥})^{∗} = M^{S^{⊥}}(G) we put

P := J^{∗} : (L_{1}^{S⊥}(G))^{∗∗} → M^{S^{⊥}}(G),

in other words P x^{∗∗} is the restriction of the functional x^{∗∗} acting on L_{1}^{S^{⊥}}(G)^{∗} to the subspace J(C(G)/(M^{S^{⊥}}(G))_{⊥}). We omit a routine verification that Υ and J are isometrically isomorphic embeddings and the proof of the identity P Υ(μ) = μ for μ ∈ M^{S^{⊥}}(G). □ Next we list several properties of the spaces M^{S^{⊥}}(G) and L_{1}^{S^{⊥}}(G) for S being Sidon sets. The properties (vi) and (vii) below were established with different proofs in [J2, Theorem 2.1 and Corollary 2.9].

Theorem 5.2. Let S ⊂ Γ be an infinite Sidon set in the dual Γ of a compact Abelian group G. Then

(i) The spaces L_{1}^{S^{⊥}}(G) and M^{S^{⊥}}(G) have the UAP.

(ii) If Γ is countable then L_{1}^{S^{⊥}}(G) has a basis.

(iii) Every separable subspace of L_{1}^{S^{⊥}}(G) (resp. M^{S^{⊥}}(G)) is contained in a separable subspace of L_{1}^{S^{⊥}}(G) (resp. M^{S^{⊥}}(G)) with a basis.

(iv) M^{S^{⊥}}(G) fails the SCP.

(v) M^{S^{⊥}}(G) fails the commuting bounded approximation property. (vi) M^{S^{⊥}}(G) fails GL-l.u.st. (vii) L_{1}^{S^{⊥}}(G) fails GL-l.u.st.

Proof. Recall that S is a Sidon set iff the map Q : f → (∫_{G} f(g)γ^{−1}(g)dg)_{γ∈S} is a surjection from L_{1}(G) onto c_{0}(S) (resp. the map μ → (∫_{G} γ^{−1}(g)dμ)_{γ∈S} is a surjection from M(G) onto ℓ_{∞}(S) (cf. [HR, Theorem 37.4]). Of course, L_{1}(G) and M(G) are L_{1}−spaces. Thus applying Theorem 2.1 (c) we get (i) because c_{0}(S) and ℓ_{∞}(S) are L_{∞}−spaces.

(ii) is due to Lusky and it follows directly from (L) (see Section 2). (iii) follows from Proposition 2.3 because for S infinite c_{0}(S) is iso-

morphic to the space of all continuous functions on the one-point compactification of the discrete set S, while ℓ_{∞}(S) is isometrically isomorphic to the space of all continuous functions on the Stone-Čech compactification of S.

(iv) L_{1}(G) is a Banach lattice of cotype 2. By Proposition 5.1, M^{S^{⊥}}(G) ⊂ ker Q^{∗∗} = (L_{1}^{S^{⊥}}(G))^{∗∗}. Moreover M^{S^{⊥}}(G) is complemented in its second dual because it is the dual of C(G)/(M^{S^{⊥}}(G))_{⊥}. The desired conclusion follows from Proposition 4.12 and Remark 4.13. To apply Remark 4.13 we put L̃ = Υ(M^{S^{⊥}}(G)), where Υ is defined in the proof of Proposition 5.1.

(v) Combine (iv) with the result of Casazza, Kalton and Wojtaszczyk mentioned in Section 3.

(vi) Combine (iv) with Corollary 3.4. Note that M^{S^{⊥}}(G) ⊂ M(G) has cotype 2.

(vii) If a Banach space X has GL-l.u.st. then all complemented subspaces of the even duals of X have GL-l.u.st. Thus (vii) follows from (vi) and Proposition 5.1. □

5.2. Sobolev spaces of functions of bounded variation. Let Ω ⊂ ℝ^{n} be an open non-empty set. For a measurable complex-valued f on Ω put

||f||_{BV} = ∫_{Ω} |f|dλ_{n} + |Df|(Ω),

where |Df| is a positive regular measure of bounded variation defined for open Ω′ ⊂ Ω by

|Df|(Ω′) = sup | ∫_{Ω′} f div ϕdλ_{n}| = sup | ∫_{Ω′}(f ∑_{j} \frac{∂ϕ_{j}}{∂x_{j}})dλ_{n}| ,

λ_{n} denotes the n−dimensional Lebesgue measure and the supremum extends over all infinitely many times differentiable complex valued functions ϕ = (ϕ_{j})_{j=1}^{n} with compact support from ℝ^{n} into the unit ball of the n−dimensional complex Hilbert space.

Put

BV (Ω) = {f : Ω → ℂ - measurable : ||f||_{BV} < ∞}.

BV (Ω) is a nonseparable Banach space. It consists of absolutely integrable functions whose distributional partial derivatives exist and are measures with total bounded variation. The Sobolev space W_{1}^{1}(Ω) is isometrically isomorphic with the subspace of BV (Ω) consisting of those functions whose distributional gradient is a ℂ^{n}-valued measure absolutely continuous with respect to λ_{n}.

Theorem 5.3. If Ω ⊂ ℝ^{n} is an open non-empty set, then

(a) BV (Ω) isometrically embeds into an abstract L_{1}−space - the Cartesian power (M(Ω^{n}))^{n+1}, where M(Ω) is the space of all complex-valued regular Borel measures on Ω with bounded variation.

(b) BV (Ω) is a dual Banach space.

(c) If n ≥ 2 then BV (Ω) fails the SCP.

Proof. (a) is obvious.

For (b) observe that M(Ω)^{n+1} is the dual space of the space C_{0}(Ω)^{n+1} where C_{0}(Ω) is the space generated in the uniform norm by all continuous complex-valued functions on Ω with compact support. The space BV (Ω) is isometrically isomorphic to a weak^{∗}−closed subspace of M(Ω)^{n+1} (cf.[PW]).

The proof of (c) requires several steps. First, it routinely reduces to the case Ω = ℝ^{n}. Indeed, an open non-empty set Ω contains a closed cube. If I^{n} denotes the interior of the cube then there is a linear extension operator from BV (I^{n}) into BV (Ω). Therefore BV (Ω) contains a complemented subspace isomorphic to BV (I^{n}); the operator of restriction of functions on Ω to I^{n} composed with the linear extension is the desired projection. Finally, by a decomposition technique (cf.

[PW, Proof of Theorem 7.1]) one shows that BV (I^{n}) is isomorphic to BV (ℝ^{n}). Thus BV (Ω) contains a complemented subspace isomorphic to BV (ℝ^{n}). To complete the proof of (c) it suffices to establish

Proposition 5.4. If n ≥ 2 then BV (ℝ^{n}) fails the SCP. More precisely, if E is a separable subspace of BV (ℝ^{n}) which contains W_{1}^{1}(ℝ^{n}) then E is not complemented in BV (ℝ^{n}).

We need several facts about traces.

Let b ∈ ℝ; x = (x(j))_{j=1}^{n} ∈ ℝ^{n}. Put

H_{b}^{−} = {x ∈ ℝ^{n} : x(n) < b}; H_{b}^{+} = {x ∈ ℝ^{n} : x(n) > b};
H_{b} = {x ∈ ℝ^{n} : x(n) = b}.

In the sequel we identify L_{1}(H_{b}, λ_{n−1}) with L_{1}(ℝ^{n−1}).

Write f ∈ CV^{b}(H_{b}^{−}) provided that f ∈ BV (H_{b}^{−}) and there is δ = δ(f) > 0 such that the restriction f|{x : −δ + b < x(n) < b} is λ_{n} a.e. equal to a C^{∞}−function uniformly continuous with all its partial derivatives. Note that if f ∈ CV^{b}(H_{b}^{−}) then f has a unique extension - denoted also by f - to a C^{∞}−function on H_{b}^{−} ∪ H_{b}. The definition of CV^{b}(H_{b}^{+}) is similar.

General Gagliardo-Nirenberg Theorem [PW], [AFP]. There is a unique contractive linear operator Tr_{b}^{σ} : BV (H_{b}^{σ}) → L_{1}(ℝ^{n−1}), where σ = + or σ = −, such that if f ∈ CV^{b}(H_{b}^{σ}) then Tr_{b}^{σ}(f) = f|H_{b}. Moreover:

(i) if f ∈ W_{1}^{1}(ℝ^{n}) then Tr_{b}^{−}(f|H_{b}^{−}) = Tr_{b}^{+}(f|H_{b}^{+}) =: Tr_{b}(f);

(ii) Tr_{b}(W_{1}^{1}(ℝ^{n})) = L_{1}(ℝ^{n−1}).

Peetre’s Theorem [Pee], [PW]. The operator Tr_{b}^{−} : W_{1}^{1}(H_{b}^{−}) → L_{1}(ℝ^{n−1}) has no bounded lifting, i.e. there is no linear operator V : L_{1}(ℝ^{n−1}) → W_{1}^{1}(H_{b}^{−}) such that Tr_{b}^{−} ∘V = Id_{L_{1}(ℝ^{n−1})}.

Theorem on extending traces [PW]. There is a bounded linear operator

Λ_{b}^{−} : BV (H_{b}^{−}) → W_{1}^{1}(H_{b}^{+})

such that

Tr_{b}^{−}(f) = Tr_{b}^{+}(Λ_{b}^{−}f) ∀f ∈ BV (H_{b}^{−}),

and there is also Λ_{b}^{+} : BV (H_{b}^{+}) → W_{1}^{1}(H_{b}^{−}) with similar properties.

Proof of Proposition 5.4. Let E ⊃ W_{1}^{1}(ℝ^{n}) be a separable subspace of BV (ℝ^{n}). It suffices to construct a separable subspace F ⊂ BV (ℝ^{n}) such that (j) E ⊂ F; (jj) E is uncomplemented in F.

To this end observe that there exists b ∈ ℝ such that

(5.1)  |Df|(H_{b}) = 0 for all f ∈ E.

Indeed if b_{1} ≠ b_{2} then H_{b_{1}} ∩ H_{b_{2}} = ∅. Thus for every f ∈ E the set A_{f} = {b ∈ ℝ : |Df|(H_{b}) ≠ 0} is countable. Hence if S is a countable dense set in the separable space E then the set ⋂_{f∈S}(ℝ\A_{f}) is uncountable. It is easy to see that every b in the intersection satisfies (5.1).

Next fix b ∈ ℝ satisfying (5.1) and put

{ \atop W_{1}^{1}(H_{b}^{−})} := \{g ∈ BV (ℝ^{n}) : g|H_{b}^{−} ∈ W_{1}^{1}(H_{b}^{−}) and g|(ℝ^{n} \ H_{b}^{−}) ≡ 0\};

F := the closure in BV (ℝ^{n}) of E + { \atop W_{1}^{1}(H_{b}^{−})}.

Define the operator Q : F → L_{1}(ℝ^{n−1}) by Qf = Tr_{b}^{−}(f|H_{b}^{−}) −

Tr_{b}^{+}(f|H_{b}^{+}). Then Q(F) = L_{1}(ℝ^{n−1}), because Q(F) ⊇ Q({ \atop W_{1}^{1}(H_{b}^{−})}) ⊇

L_{1}(ℝ^{n−1}), by the General Gagliardo-Nirenberg Theorem.

Lemma 5.5. ker Q = E.

Proof. If e ∈ E then |De|(H_{b}) = 0. Thus by [PW, Lemma 3.1.1] (special case k = 1, Ξ_{a} = ℝ^{n}), given ε > 0 there is an f ∈ BV (ℝ^{n}) such that ∥f − e∥_{BV (ℝ^{n})} < ε and for some δ > 0 the restriction f|{x : |b−x(n)| < δ} is equal to a C^{∞}−function λ_{n} a.e. Clearly Tr_{b}^{−}(f|H_{b}^{−})− Tr_{b}^{+}(f|H_{b}^{+}) = 0. Hence | Tr_{b}^{−}(e|H_{b}^{−}) − Tr_{b}^{+}(e|H_{b}^{+})| < 2ε. Letting ε ↓ 0 we get Tr_{b}^{−}(e|H_{b}^{−}) − Tr_{b}^{+}(e|H_{b}^{+}) = 0. Therefore e ∈ ker Q.

To prove that ker Q ⊆ E, fix f ∈ ker Q and let ε > 0. Pick g ∈

{ \atop W_{1}^{1}(H_{b}^{−})} and e ∈ E with ||f − g − e||_{BV (ℝ^{n})} < ε. Since ||Q|| ≤ 2,

2ε > ||Q(f − g − e)||_{L_{1}(ℝ^{n−1})} = ||Qg||_{L_{1}(ℝ^{n−1})} = || Tr_{b}^{−}(g)||_{L_{1}(ℝ^{n−1})}.

Note that g|H_{b}^{−} ∈ W_{1}^{1}(H_{b}^{−}) because g ∈ { \atop W_{1}^{1}(H_{b}^{−})}. Thus using the fact that CV^{b}(H_{b}^{−}) is dense in W_{1}^{1}(H_{b}^{−}) (cf. [PW, Lemma 3.1.1]) we infer that there exists a < b with 2ε > ||g|{a < x(n) < b}||_{W_{1}^{1}({a<x(n)<b})}. Hence, by the extension formula for W_{1}^{1} functions (cf. [A]), there exists e^{∗} ∈ W_{1}^{1}(ℝ^{n}) which extends g|{a < x < b} with ||e^{∗}||_{W_{1}^{1}(ℝ^{n})} ≤ Cε, where the constant C is independent of ε. Define g^{∗} ∈ W_{1}^{1}(ℝ^{n}) ⊂ E by

g^{∗}(x) = \begin{cases}g(x) & for x ≤ a, \\ e^{∗}(x) & for x > a. \end{cases}

Then ∥g −g^{∗}∥_{BV (ℝ^{n})} ≤ Cε. Thus ∥f −g^{∗} −e∥_{BV (ℝ^{n})} < (2+C)ε. Hence, letting ε ↓ 0, we infer that f ∈ E. □

We now complete the proof of Proposition 5.4. Assume that (jj) is false, i.e., there exists a bounded projection P : F →^{onto} E. Define S : L_{1}(ℝ^{n−1}) → F by Sg = f − P f whenever Qf = g. Then S is well defined, because ker Q = E by Lemma 5.5 and S is bounded, by the Open Mappping Theorem. Clearly, QSg = Q(f − P f) = Qf = g for g ∈ L_{1}(ℝ^{n−1}).

Let Λ_{b}^{σ} : BV (H_{b}^{σ}) → W_{1}^{1}(H_{b}^{−σ}) be trace preserving operators and let R_{σ} denote the operator of restriction of functions to H_{b}^{σ}, where σ = ∓. Define U : F → W_{1}^{1}(H_{b}^{−}) by

Uf = Λ_{b}^{+}Λ_{b}^{−}R_{−}f − Λ_{b}^{+}R_{+}f (f ∈ F).

We shall show that V = US lifts Tr_{b}^{−} (contrary to Peetre’s Theorem). First note that given f ∈ F the properties of Λ_{b}^{σ} imply

Tr_{b}^{−}(Λ_{b}^{+}R_{+}f) = Tr_{b}^{+}(R_{+}f),
Tr_{b}^{−}(Λ_{b}^{+}Λ_{b}^{−}R_{−}f) = Tr_{b}^{+}(Λ_{b}^{−}R_{−}f) = Tr_{b}^{−}(R_{−}f),

and hence Tr_{b}^{−}(Uf) = Qf.

Now recall that if g ∈ L_{1}(ℝ^{n−1}) then Sg ∈ F and QSg = g. Hence

Tr_{b}^{−}(USg) = QSg = g.

This contradiction completes the proof of Proposition 5.4. □

Remark 5.6. BV (ℝ^{n}) has the BAP and every separable subspace of BV (ℝ^{n}) is contained in a separable subspace with a basis [ACPP]. We do not know whether BV (Ω) has the same property for arbitrary open Ω ⊂ ℝ^{n}.

6. Open problems

Problem 6.1. If Y is a subspace of X such that X, Y and X/Y all have the BAP, must the pair (X, Y) have the BAP?

We do not know the answer to Problem 6.1 even when X is reflexive.

Problem 6.2. If X is a L_{p} space, 2 < p < ∞, and Y is a subspace of X with the BAP (resp. UAP), must X/Y have the BAP (resp. UAP)? Equivalently, if X is a L_{q} space, 1 < q < 2, and Y is a subspace of X such that X/Y has the BAP (resp. UAP), must Y have the BAP (resp. UAP)?

Problem 6.3. Does every weakly sequentially complete Banach space with GL-l.u.st. have the SCP?

Problem 6.4. Does every complemented subspace of a space with the SCP have the SCP? Does X have the SCP if X^{∗∗} does?

Proposition 3.2 and Corollary 3.4 give positive answers to special cases of Problem 6.3 and Problem 6.4.

Remark 6.5. The projective tensor product of spaces with property (k) need not have (k). We are indebted to Professor Eve Oja for the following two examples.

The first one is the space Z = L_{1}([0, 1], ℓ_{∞}) of Bochner integrable ℓ_{∞}-valued functions. Since Z contains a complemented subspace isomorphic to c_{0} (cf. [E]), it fails (k), by Proposition 4.9. Note that L_{1}([0, 1]) ˆ⊗ℓ_{∞} = Z (cf. [Gr2, Chap. I §2.2, p. 59]), and both L_{1}([0, 1]) and ℓ_{∞} have (k).

The second example is provided by the space X = L_{λ}[ℓ_{2}] constructed in [BP], where λ > 1. Namely, X has (k) by Remark 4.2, while X ˆ⊗X fails (k) by Proposition 4.9 and Sobczyk’s theorem. Indeed, it was proved in [BP] that X is a separable L_{∞,λ}−space with the following properties: (i) X has the RNP, (ii) X contains an isometric copy E of l_{2}, (iii) X ˆ⊗X contains an isomorphic copy Y of c_{0}. □

Problem 6.6. Does a predual of a σ−finite von Neumann algebra have property (k)? (A von Neumann algebra is said to be σ−finite if it admits at most countably many mutually orthogonal projections.)

Problem 6.7. Does BV (ℝ^{n}) have the UAP?

A well known and important problem is whether X^{∗} has the metric approximation property whenever X^{∗} has the BAP; see [C, Theorem 3.7]. We do not know the answer to the following two special cases of this problem.

Problem 6.8. Does BV (ℝ^{n}) have the metric approximation property?

Problem 6.9. Does M^{S^{⊥}}(G) have the metric approximation property for every Sidon set S ⊂ Γ?

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Tadeusz Figiel, Institute of Mathematics, Polish Academy of Sciences, ul. Abrahama 18, 81825 Sopot, Poland

E-mail:  t.figiel@impan.gda.pl

William B. Johnson Department of Mathematics, Texas A&M University, College Station, TX 77843, USA.

E-mail:  johnson@math.tamu.edu

Aleksander Pełczyński, Institute of Mathematics, Polish Academy of Sciences, Śniadeckich 8 Ip; 00956 Warszawa, Poland

E-mail:  olek@impan.gov.pl