- Email: [email protected]
Three sequential properties of dual Banach spaces in the weak* topology
Topology and its Applications 190 (2015) 93–98
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
Three sequential properties of dual Banach spaces in the weak* topology Anatolij Plichko Department of Mathematics, Cracow University of Technology, Cracow, Poland
a r t i c l e
i n f o
Article history: Received 24 October 2014 Received in revised form 30 April 2015 Accepted 30 April 2015 Available online 16 May 2015
a b s t r a c t We present an example of a Banach space X in whose dual the weak* and weak* sequential closures of every bounded convex set coincide but weak* and weak* sequential closures of bounded sets do not necessarily coincide. We also answer a question of Gulisashvili on scalarly measurable functions. © 2015 Published by Elsevier B.V.
MSC: 46B26 54D55 Keywords: Non-separable Banach space Weakly* angelic dual Totally scalarly measurable function Markushevich basis
1. Introduction Let A be a subset of a topological space T . The sequential closure of A is the collection of limits of all convergent sequences in A. The set A is sequentially closed if A coincides with its sequential closure. A topological space T is called a Fréchet–Urysohn space (or Fréchet space) if the closure of any subset of T coincides with its sequential closure. A topological space T is said to be sequential if any sequentially closed subset A ⊂ T is closed. Of course, every Fréchet–Urysohn space is sequential. The converse statement is false. Let X be a real Banach space, BX be its unit ball, SX be its unit sphere and X ∗ be its dual. We say that a functional f ∈ X ∗ attains its norm if there exists an element x0 ∈ SX such that f = f (x0 ). A subset F ⊂ X ∗ is total if for every x ∈ X, x = 0, there is f ∈ F with f (x) = 0. A Banach space X has weak* angelic dual [2] if BX ∗ is angelic space in the weak* topology, i.e. if weak* and weak* sequential closures of every bounded subset of X ∗ coincide. Obviously, it is the same that every bounded subset of X ∗ is E-mail address: [email protected]. http://dx.doi.org/10.1016/j.topol.2015.04.017 0166-8641/© 2015 Published by Elsevier B.V.
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a Fréchet–Urysohn space in weak* topology [10]. A Banach space X has property E (of Efremov) [10, p. 352] if weak* and weak* sequential closures of every bounded convex subset of X ∗ coincide. Of course, a Banach space with weak* angelic dual has property E. In [10, p. 352], the question about reverse implication was posed. We answer this question in the negative. The example, which we present, answers also a question of Gulisashvili; for details see below. We use the following notations: conv(A) (conv(A)) denotes the convex (resp. closed convex) span of a set A. Body denotes a subset of a Banach space having a norm inner point. 2. Weak* angelicity and property E Many results of this section are, in fact, well known. We present them for sake of completeness. ∞ Lemma 1. ([3, Lemma 3.1.1]) Let X be a Banach space and (fn )∞ n=1 ⊂ BX ∗ . If for every f in conv(fn )1 we have f ≥ β for some 0 < β < 1 and limn→∞ |fn (x)| ≤ β 2 for every x in BX , then there exists an element g in conv(fn )∞ 1 which does not attain its norm.
∞ Proof. Let λk > 0, k=1 λk = 1. By the original James lemma [8], there exist α ∈ [β, 1] and a sequence ∞ gn ∈ conv(fk )∞ k=n such that k=1 λk gk = α and n ∀n,
k=1
∞ λk gk ≤ α 1 − β λk . n+1
Evidently, limk→∞ |gk (x)| ≤ β 2 for every x in BX . Take any x in BX and choose n so that |gk (x)| < β 2 ≤ αβ for all k > n. Then ∞
λk gk (x) <
n
1
Therefore, the functional g =
λk gk (x) + αβ
1
∞
λk ≤ α 1 − β
n+1
∞ 1
∞ n+1
λk
+ αβ
∞
λk = α .
n+1
λk gk ∈ conv(fn )∞ n=1 does not attain its norm. 2
Proposition 2. ([3, Th. 3.2.1]) Suppose a norm closed and total subspace E ⊂ X ∗ consists of norm attaining functionals and is weakly* sequentially dense in E ∗∗ . Then E ∗ is isometric to X under the natural duality (X, X ∗ ). Proof. We identify X with its natural image in E ∗ . Suppose the proposition is false. Then BX BE ∗ , hence [1, I, §3, Lemma 2] the norm closure of BX in E ∗ does not cover BE ∗ . So, there are β ∈ (0, 1) and e∗∗ ∈ SE ∗∗ such that e∗∗ (x) < β 2 for every x ∈ BX
(1)
and there is e∗0 ∈ BE ∗ such that e∗∗ (e∗0 ) > β. By condition of proposition, there exists a sequence (en )∞ n=1 in E such that for every e∗ ∈ E ∗ en (e∗ ) → e∗∗ (e∗ ) as n → ∞ ,
(2)
hence one can suppose that for each n en (e∗0 ) > β.
(3)
By the Uniform Boundedness Principle, (en )∞ n=1 is bounded. So, we can assume each en ∈ BE , multiplying ∗∗ (if needed) this sequence and the element e by an appropriate scalar (and decreasing β).
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The formula (2) means, in particular, that en (x) → e∗∗ (x), as n → ∞, for every x ∈ X, i.e. the sequence ∗ ∗∗ ∗∗ 2 (en )∞ n=1 ⊂ E ⊂ X weakly* converges to the restriction e |X ; and, by (1), e |X < β . By (3), we have ∞ e ≥ β for every e in conv(en )1 . Therefore, all conditions of Lemma 1 are satisfied. Hence, there is an element in E which does not attain its norm. Contradiction. 2 Under additional assumption that the subspace E is norming, Proposition 2 was proved also in [4, Th. IV.4]. Let V be a weakly* compact convex subset of a dual Banach space X ∗ . A set B ⊂ V is called a (James) boundary of V if for every x ∈ X there exists vx ∈ B such that x(vx ) = supv∈V x(v) (see e.g. [3,4]). As usual, we identify an element of a Banach space with its image in the second dual. We say that a boundary B ⊂ V is large if conv B = V , and that a Banach space X admits large boundaries only if every convex weakly* compact subset V ⊂ X ∗ is the norm closed convex span of each its boundary. Lemma 3. Suppose X ∗ contains a convex weakly* compact subset V which does not coincide with the norm closed convex span of any of its boundaries B. Then X ∗ contains a symmetric convex weakly* compact body U which does not coincide with the norm closed convex span of any of its boundaries. Proof. Of course, we may assume B to be convex and norm closed. Let v0 ∈ V \ B and x∗∗ ∈ X ∗∗ be such that x∗∗ (v0 ) > supv∈B x∗∗ (v0 ). Take f ∈ X ∗ such that supv∈V x∗∗ (v + f ) > 0, moreover, such that x∗∗ (v0 + f ) > supv∈B x∗ (v). Then the (symmetric convex weakly* compact) body U = conv(BX ∗ ∪ (V + f ) ∪ (−V − f )) does not coincide with the norm closed convex span of its boundary B = BX ∗ ∪ (B + f ) ∪ (−B − f ) because of v0 + f ∈ U but v0 + f ∈ / conv B .
2
Lemma 4. Let a dual unit ball BX ∗ have a non-large boundary B. Then BX ∗ has a non-large boundary B which is a convex norm closed symmetric body. Proof. Since B is non-large, there is a functional x∗∗ ∈ SX ∗∗ such that supv∈B x∗∗ (v) = a < 1. Then the set B = {v ∈ BE ∗ : |x∗∗ (v)| ≤ a} has desired properties. 2 Lemma 5. ([3, Th. 3.1.4]) Let a Banach space E have a non-large boundary B in its dual ball BE ∗ . Then there exists a Banach space Y such that E isometrically imbedded as a total subspace into Y ∗ , consists of norm attaining functionals, but E ∗ = Y isometrically in the duality (Y, Y ∗ ). Proof. By Lemma 4, one may suppose B to be a norm closed symmetric body. The gauge function ||| · ||| of B is a (non-dual) norm in E ∗ . Denote by Y the (E ∗ , ||| · |||). Each element e ∈ E can be considered as a linear continuous functional on Y . So, we obtain the linear isomorphism I : E → Y ∗ onto a total subspace of Y ∗ . Since B is a boundary of BE ∗ , for any e ∈ E Ie = supy∈B Ie(y) = supy∈B y(e) = e . Hence, I is an isometry of E into Y ∗ and any functional from E attains its norm on Y . However, Y is not isometric to E ∗ , because B = BE ∗ . 2
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Remark 6. Lemma 5 has also the following modification [3, Th. 1.0.2]: Let a Banach space E have a boundary B of BE ∗ , whose norm closed linear span does not coincide with E ∗ . Then there exists a Banach space Y such that E is isometrically imbedded as a total subspace into Y ∗ and consists of norm attaining functionals, but E ∗ is not norm dense in Y . Proposition 7. ([3, Th. 3.3.8]) If a Banach space X is weakly* sequentially dense in X ∗∗ then X has large boundaries only. Proof. Suppose X has a non-large boundary. By Lemma 3, X ∗ contains a symmetric convex weakly* compact body U which does not coincide with the norm closed convex span of any of its boundaries. Let ||| · ||| be the gauge function of the polar U ◦ ⊂ X. By Lemma 5, there exists a Banach space Y such that E = (X, ||| · |||) is isometrically imbedded into Y ∗ , E consists of norm attaining functionals with respect to Y , but Y is not canonically isometric to E ∗ . On the other hand, the space E is isomorphic to X, hence the weak* sequential closure of E in E ∗∗ coincides with E ∗∗ . By Proposition 2, Y must be canonically isometric to E ∗ . Contradiction. 2 A result, close in spirit to Proposition 7, was proved by Godefroy [4, Th. I.2]. Proposition 8. Let a Banach space X have large boundaries only, and suppose moreover that the dual ball BX ∗ is weakly* sequentially compact. Then X has property E. Proof. Let C be a bounded convex subset of X ∗ . Obviously, the weak* sequential closure C(1) of C is convex and norm closed. So, by definition of the property E, it is sufficient to show that C(1) forms a boundary for the weak* closure of C. Take any element x ∈ X. If a = supf ∈C x(f ) then there is a sequence (fn )∞ 1 in C ∞ such that x(fn ) → a as n → ∞. By weak* sequential compactness, (fn )1 contains a subsequence (fnk )∞ k=1 which weakly* converges to some f ∈ X ∗ . Evidently, f ∈ C(1) and x(f ) = a. Hence, C(1) forms a boundary for the weak* closure of C, which proves the proposition. 2 Let (Nγ )γ∈Γ be a maximal family of almost disjoint subsets of N which are infinite and have infinite complements. Denote by χγ the characteristic function of Nγ , γ ∈ Γ. Let X0 be the linear span of c0 ∪ {χγ }γ∈Γ . The norm on X0 is defined by k k x + ai χγi = max x + ai χγi i=1 i=1
, ∞
k i=1
1/2 |ai |2
,
(4)
where · ∞ is the usual supremum norm in ∞ , x ∈ c0 and γi = γj if i = j. The Johnson–Lindenstrauss space JL2 is the completion of X0 in the norm (4) [9]. Theorem 9. The space JL2 has property E. Proof. Since JL∗2 is isometric to 1 ⊕ 2 (Γ) [9], the bidual JL∗∗ 2 is weakly* angelic, hence JL2 is weakly* sequentially dense in JL∗∗ . On the other hand, JL has an equivalent Fréchet differentiable norm [9], so the 2 2 ball of JL∗2 in this equivalent norm is weakly* sequentially compact [7]. So, by Propositions 7, 8, JL2 has property E. 2 Since JL∗2 is not weakly* angelic [2], Theorem 9 answers the mentioned question of [10]. Question 10. Let BX ∗ be weakly* sequential. Does X have weakly* angelic dual? Is BJL∗2 weakly* sequential?
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3. Properties D and D We say that a Banach space X has property D if X ∗ is weakly* sequential with respect to subspaces, i.e. if every weakly* sequentially closed subspace of X ∗ is weakly* closed. This means that the ω1 th transfinite weak* sequential closure F(ω1 ) of each linear subspace F ⊂ X ∗ (where ω1 is the first uncountable ordinal) is equal to its weak* closure. We introduce this notion motivated by the following definitions. Let (Ω, Σ) be a measurable space and X be a Banach space. A function ϕ : Ω → X is scalarly measurable if f ◦ ϕ is measurable for every f ∈ X ∗ . The ϕ is totally scalarly measurable if the set F = {f ∈ X ∗ : f ◦ ϕ is measurable}
(5)
is total on X. A Banach space X satisfies the property D provided that (Ω, Σ) is a measurable space and ϕ : Ω → X is totally scalarly measurable imply that ϕ is scalarly measurable. Property D has been introduced by Gulisashvili [5,6] in connection with the Pettis integral in interpolation spaces. Gulisashvili has proved that the weak* angelicity of X ∗ implies the property D of X and naturally this gives rise to the problem of the reverse implication. We show that the space JL2 answers this problem negatively. First we present a generalization (with the complete proof) of a Gulisashvili lemma [5] which allows to construct Banach spaces without the property D. Given a Banach space X, a system (xγ , fγ )γ∈Γ , xγ ∈ X, fγ ∈ X ∗ , where Γ is some set of indices, is called M -basic if fβ (xγ ) = δβγ (the Kronecker δ) and (fγ )γ∈Γ is total on the closed linear span [xγ ]γ∈Γ . An M -basic system is said to be an + 1 -system if xγ = 1 for each γ and there is a constant c > 0 such that for all finite collections (aγ ) of positive scalars aγ x γ ≥ c aγ . γ
γ
Lemma 11. Let Γ be an uncountable set. If a (nonseparable) Banach space X has an M -basic + 1 -system (xγ , fγ )γ∈Γ , then X fails the property D. Proof. Let (Ω, Σ) be a measurable space such that Σ contains all singleton subsets of Ω and there exists some Γ ⊂ Ω with Γ ∈ / Σ. Such (Ω, Σ) exists whenever Ω is uncountable. Put ϕ(ω) =
xω 0
for ω ∈ Γ, for ω ∈ Ω \ Γ.
Then for every γ ∈ Γ the function fγ ◦ϕ is measurable. Moreover, for each f in the annihilator ([xγ ]γ∈Γ )⊥ the function f ◦ ϕ is measurable. Hence, ϕ is scalarly totally measurable. On the other hand, the set A := conv(xγ )γ∈Γ is convex, closed and, since (xγ )γ∈Γ is an + 1 -system, does not contain 0. By the Hahn– Banach theorem, there is a functional g ∈ X ∗ such that g(x) > c > 0 for each x ∈ A. Then g ◦ ϕ is non-measurable. 2 By Lemma 11, the space 1 (Γ), with uncountable Γ, fails the property D. Since the space ∞ contains a subspace, isometric to 1 (Γ) with an uncountable Γ, it fails the property D. Denote by C[0, η] the space of continuous functions on the segment of ordinals [0, η]. The space C[0, η], with uncountable η, fails the property D. Indeed let xγ , be the characteristic function of the segment [0, γ] and fγ (x) := x(γ +1) −x(γ), x ∈ C[0, η], γ ∈ [0, η). It is easy to check that (xγ , fγ )γ∈Γ is an + 1 -system. So, one can apply Lemma 11. Proposition 12. E ⇒ D ⇒ D.
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Proof. E ⇒ D . Let F be a weakly* sequentially closed subspace of X ∗ . Then BF is weakly* sequentially closed, and because X has property E, is weakly* closed. By the Banach–Diedonne theorem [1, Ch. IV, §2.5], F is weakly* closed. D ⇒ D. If ϕ is totally scalarly measurable, then the set F , defined by (5), is a weakly* sequentially closed total subspace. Since X has property D , F is weakly* closed. As F is total, F = X ∗ , therefore X has property D. 2 This proposition and Theorem 9 answer the aforementioned question of Gulisashvili. We do not know whether E = D or D = D. Let us present arguments, why the property D can be equivalent to D. If X fails the property D , then there exists a weakly* sequentially closed subspace F ⊂ X ∗ which is not weakly* closed. Let Ω = X and Σ be the σ-algebra generated by functionals f ∈ F . Then, is each g ∈ cl∗ F \F (when cl∗ denotes the weak* closure) non-measurable? References [1] N. Bourbaki, Espaces Vectoriels Topologiques, Hermann, Paris, 1964. [2] G.A. Edgar, Measurability in a Banach space, II, Indiana Univ. Math. J. 28 (1979) 559–579. [3] N.M. Efremov, Duality conditions of a Banach space in terms of norm attaining linear functionals, Ph.D., Kharkiv State University, 1985 (in Russian). [4] G. Godefroy, Boundaries of a convex set and interpolation sets, Math. Ann. 277 (1987) 173–184. [5] A. Gulisashvili, Estimates for the Pettis integral in interpolation spaces and inversion of the embedding theorems, Dokl. Akad. Nauk SSSR 263 (1982) 793–798 (in Russian). English transl.: Sov. Math. Dokl. 25 (1982) 428–432. [6] A. Gulisashvili, Estimates for the Pettis integral in interpolation spaces with some applications, in: Lect. Notes in Math., vol. 991, 1983, pp. 55–78. [7] J. Hagler, F. Sullivan, Smoothness and weak* sequential compactness, Proc. Am. Math. Soc. 78 (1980) 497–503. [8] R.C. James, Reflexivity and the sup of linear functionals, Isr. J. Math. 13 (1972) 289–300. [9] W.B. Johnson, J. Lindenstrauss, Some remarks on weakly compactly generated Banach spaces, Isr. J. Math. 17 (1974) 219–230. [10] A.M. Plichko, D. Yost, Complemented and uncomplemented subspaces of Banach spaces, Extr. Math. 15 (2000) 335–371.
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
Three sequential properties of dual Banach spaces in the weak* topology Anatolij Plichko Department of Mathematics, Cracow University of Technology, Cracow, Poland
a r t i c l e
i n f o
Article history: Received 24 October 2014 Received in revised form 30 April 2015 Accepted 30 April 2015 Available online 16 May 2015
a b s t r a c t We present an example of a Banach space X in whose dual the weak* and weak* sequential closures of every bounded convex set coincide but weak* and weak* sequential closures of bounded sets do not necessarily coincide. We also answer a question of Gulisashvili on scalarly measurable functions. © 2015 Published by Elsevier B.V.
MSC: 46B26 54D55 Keywords: Non-separable Banach space Weakly* angelic dual Totally scalarly measurable function Markushevich basis
1. Introduction Let A be a subset of a topological space T . The sequential closure of A is the collection of limits of all convergent sequences in A. The set A is sequentially closed if A coincides with its sequential closure. A topological space T is called a Fréchet–Urysohn space (or Fréchet space) if the closure of any subset of T coincides with its sequential closure. A topological space T is said to be sequential if any sequentially closed subset A ⊂ T is closed. Of course, every Fréchet–Urysohn space is sequential. The converse statement is false. Let X be a real Banach space, BX be its unit ball, SX be its unit sphere and X ∗ be its dual. We say that a functional f ∈ X ∗ attains its norm if there exists an element x0 ∈ SX such that f = f (x0 ). A subset F ⊂ X ∗ is total if for every x ∈ X, x = 0, there is f ∈ F with f (x) = 0. A Banach space X has weak* angelic dual [2] if BX ∗ is angelic space in the weak* topology, i.e. if weak* and weak* sequential closures of every bounded subset of X ∗ coincide. Obviously, it is the same that every bounded subset of X ∗ is E-mail address: [email protected]. http://dx.doi.org/10.1016/j.topol.2015.04.017 0166-8641/© 2015 Published by Elsevier B.V.
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a Fréchet–Urysohn space in weak* topology [10]. A Banach space X has property E (of Efremov) [10, p. 352] if weak* and weak* sequential closures of every bounded convex subset of X ∗ coincide. Of course, a Banach space with weak* angelic dual has property E. In [10, p. 352], the question about reverse implication was posed. We answer this question in the negative. The example, which we present, answers also a question of Gulisashvili; for details see below. We use the following notations: conv(A) (conv(A)) denotes the convex (resp. closed convex) span of a set A. Body denotes a subset of a Banach space having a norm inner point. 2. Weak* angelicity and property E Many results of this section are, in fact, well known. We present them for sake of completeness. ∞ Lemma 1. ([3, Lemma 3.1.1]) Let X be a Banach space and (fn )∞ n=1 ⊂ BX ∗ . If for every f in conv(fn )1 we have f ≥ β for some 0 < β < 1 and limn→∞ |fn (x)| ≤ β 2 for every x in BX , then there exists an element g in conv(fn )∞ 1 which does not attain its norm.
∞ Proof. Let λk > 0, k=1 λk = 1. By the original James lemma [8], there exist α ∈ [β, 1] and a sequence ∞ gn ∈ conv(fk )∞ k=n such that k=1 λk gk = α and n ∀n,
k=1
∞ λk gk ≤ α 1 − β λk . n+1
Evidently, limk→∞ |gk (x)| ≤ β 2 for every x in BX . Take any x in BX and choose n so that |gk (x)| < β 2 ≤ αβ for all k > n. Then ∞
λk gk (x) <
n
1
Therefore, the functional g =
λk gk (x) + αβ
1
∞
λk ≤ α 1 − β
n+1
∞ 1
∞ n+1
λk
+ αβ
∞
λk = α .
n+1
λk gk ∈ conv(fn )∞ n=1 does not attain its norm. 2
Proposition 2. ([3, Th. 3.2.1]) Suppose a norm closed and total subspace E ⊂ X ∗ consists of norm attaining functionals and is weakly* sequentially dense in E ∗∗ . Then E ∗ is isometric to X under the natural duality (X, X ∗ ). Proof. We identify X with its natural image in E ∗ . Suppose the proposition is false. Then BX BE ∗ , hence [1, I, §3, Lemma 2] the norm closure of BX in E ∗ does not cover BE ∗ . So, there are β ∈ (0, 1) and e∗∗ ∈ SE ∗∗ such that e∗∗ (x) < β 2 for every x ∈ BX
(1)
and there is e∗0 ∈ BE ∗ such that e∗∗ (e∗0 ) > β. By condition of proposition, there exists a sequence (en )∞ n=1 in E such that for every e∗ ∈ E ∗ en (e∗ ) → e∗∗ (e∗ ) as n → ∞ ,
(2)
hence one can suppose that for each n en (e∗0 ) > β.
(3)
By the Uniform Boundedness Principle, (en )∞ n=1 is bounded. So, we can assume each en ∈ BE , multiplying ∗∗ (if needed) this sequence and the element e by an appropriate scalar (and decreasing β).
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The formula (2) means, in particular, that en (x) → e∗∗ (x), as n → ∞, for every x ∈ X, i.e. the sequence ∗ ∗∗ ∗∗ 2 (en )∞ n=1 ⊂ E ⊂ X weakly* converges to the restriction e |X ; and, by (1), e |X < β . By (3), we have ∞ e ≥ β for every e in conv(en )1 . Therefore, all conditions of Lemma 1 are satisfied. Hence, there is an element in E which does not attain its norm. Contradiction. 2 Under additional assumption that the subspace E is norming, Proposition 2 was proved also in [4, Th. IV.4]. Let V be a weakly* compact convex subset of a dual Banach space X ∗ . A set B ⊂ V is called a (James) boundary of V if for every x ∈ X there exists vx ∈ B such that x(vx ) = supv∈V x(v) (see e.g. [3,4]). As usual, we identify an element of a Banach space with its image in the second dual. We say that a boundary B ⊂ V is large if conv B = V , and that a Banach space X admits large boundaries only if every convex weakly* compact subset V ⊂ X ∗ is the norm closed convex span of each its boundary. Lemma 3. Suppose X ∗ contains a convex weakly* compact subset V which does not coincide with the norm closed convex span of any of its boundaries B. Then X ∗ contains a symmetric convex weakly* compact body U which does not coincide with the norm closed convex span of any of its boundaries. Proof. Of course, we may assume B to be convex and norm closed. Let v0 ∈ V \ B and x∗∗ ∈ X ∗∗ be such that x∗∗ (v0 ) > supv∈B x∗∗ (v0 ). Take f ∈ X ∗ such that supv∈V x∗∗ (v + f ) > 0, moreover, such that x∗∗ (v0 + f ) > supv∈B x∗ (v). Then the (symmetric convex weakly* compact) body U = conv(BX ∗ ∪ (V + f ) ∪ (−V − f )) does not coincide with the norm closed convex span of its boundary B = BX ∗ ∪ (B + f ) ∪ (−B − f ) because of v0 + f ∈ U but v0 + f ∈ / conv B .
2
Lemma 4. Let a dual unit ball BX ∗ have a non-large boundary B. Then BX ∗ has a non-large boundary B which is a convex norm closed symmetric body. Proof. Since B is non-large, there is a functional x∗∗ ∈ SX ∗∗ such that supv∈B x∗∗ (v) = a < 1. Then the set B = {v ∈ BE ∗ : |x∗∗ (v)| ≤ a} has desired properties. 2 Lemma 5. ([3, Th. 3.1.4]) Let a Banach space E have a non-large boundary B in its dual ball BE ∗ . Then there exists a Banach space Y such that E isometrically imbedded as a total subspace into Y ∗ , consists of norm attaining functionals, but E ∗ = Y isometrically in the duality (Y, Y ∗ ). Proof. By Lemma 4, one may suppose B to be a norm closed symmetric body. The gauge function ||| · ||| of B is a (non-dual) norm in E ∗ . Denote by Y the (E ∗ , ||| · |||). Each element e ∈ E can be considered as a linear continuous functional on Y . So, we obtain the linear isomorphism I : E → Y ∗ onto a total subspace of Y ∗ . Since B is a boundary of BE ∗ , for any e ∈ E Ie = supy∈B Ie(y) = supy∈B y(e) = e . Hence, I is an isometry of E into Y ∗ and any functional from E attains its norm on Y . However, Y is not isometric to E ∗ , because B = BE ∗ . 2
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Remark 6. Lemma 5 has also the following modification [3, Th. 1.0.2]: Let a Banach space E have a boundary B of BE ∗ , whose norm closed linear span does not coincide with E ∗ . Then there exists a Banach space Y such that E is isometrically imbedded as a total subspace into Y ∗ and consists of norm attaining functionals, but E ∗ is not norm dense in Y . Proposition 7. ([3, Th. 3.3.8]) If a Banach space X is weakly* sequentially dense in X ∗∗ then X has large boundaries only. Proof. Suppose X has a non-large boundary. By Lemma 3, X ∗ contains a symmetric convex weakly* compact body U which does not coincide with the norm closed convex span of any of its boundaries. Let ||| · ||| be the gauge function of the polar U ◦ ⊂ X. By Lemma 5, there exists a Banach space Y such that E = (X, ||| · |||) is isometrically imbedded into Y ∗ , E consists of norm attaining functionals with respect to Y , but Y is not canonically isometric to E ∗ . On the other hand, the space E is isomorphic to X, hence the weak* sequential closure of E in E ∗∗ coincides with E ∗∗ . By Proposition 2, Y must be canonically isometric to E ∗ . Contradiction. 2 A result, close in spirit to Proposition 7, was proved by Godefroy [4, Th. I.2]. Proposition 8. Let a Banach space X have large boundaries only, and suppose moreover that the dual ball BX ∗ is weakly* sequentially compact. Then X has property E. Proof. Let C be a bounded convex subset of X ∗ . Obviously, the weak* sequential closure C(1) of C is convex and norm closed. So, by definition of the property E, it is sufficient to show that C(1) forms a boundary for the weak* closure of C. Take any element x ∈ X. If a = supf ∈C x(f ) then there is a sequence (fn )∞ 1 in C ∞ such that x(fn ) → a as n → ∞. By weak* sequential compactness, (fn )1 contains a subsequence (fnk )∞ k=1 which weakly* converges to some f ∈ X ∗ . Evidently, f ∈ C(1) and x(f ) = a. Hence, C(1) forms a boundary for the weak* closure of C, which proves the proposition. 2 Let (Nγ )γ∈Γ be a maximal family of almost disjoint subsets of N which are infinite and have infinite complements. Denote by χγ the characteristic function of Nγ , γ ∈ Γ. Let X0 be the linear span of c0 ∪ {χγ }γ∈Γ . The norm on X0 is defined by k k x + ai χγi = max x + ai χγi i=1 i=1
, ∞
k i=1
1/2 |ai |2
,
(4)
where · ∞ is the usual supremum norm in ∞ , x ∈ c0 and γi = γj if i = j. The Johnson–Lindenstrauss space JL2 is the completion of X0 in the norm (4) [9]. Theorem 9. The space JL2 has property E. Proof. Since JL∗2 is isometric to 1 ⊕ 2 (Γ) [9], the bidual JL∗∗ 2 is weakly* angelic, hence JL2 is weakly* sequentially dense in JL∗∗ . On the other hand, JL has an equivalent Fréchet differentiable norm [9], so the 2 2 ball of JL∗2 in this equivalent norm is weakly* sequentially compact [7]. So, by Propositions 7, 8, JL2 has property E. 2 Since JL∗2 is not weakly* angelic [2], Theorem 9 answers the mentioned question of [10]. Question 10. Let BX ∗ be weakly* sequential. Does X have weakly* angelic dual? Is BJL∗2 weakly* sequential?
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3. Properties D and D We say that a Banach space X has property D if X ∗ is weakly* sequential with respect to subspaces, i.e. if every weakly* sequentially closed subspace of X ∗ is weakly* closed. This means that the ω1 th transfinite weak* sequential closure F(ω1 ) of each linear subspace F ⊂ X ∗ (where ω1 is the first uncountable ordinal) is equal to its weak* closure. We introduce this notion motivated by the following definitions. Let (Ω, Σ) be a measurable space and X be a Banach space. A function ϕ : Ω → X is scalarly measurable if f ◦ ϕ is measurable for every f ∈ X ∗ . The ϕ is totally scalarly measurable if the set F = {f ∈ X ∗ : f ◦ ϕ is measurable}
(5)
is total on X. A Banach space X satisfies the property D provided that (Ω, Σ) is a measurable space and ϕ : Ω → X is totally scalarly measurable imply that ϕ is scalarly measurable. Property D has been introduced by Gulisashvili [5,6] in connection with the Pettis integral in interpolation spaces. Gulisashvili has proved that the weak* angelicity of X ∗ implies the property D of X and naturally this gives rise to the problem of the reverse implication. We show that the space JL2 answers this problem negatively. First we present a generalization (with the complete proof) of a Gulisashvili lemma [5] which allows to construct Banach spaces without the property D. Given a Banach space X, a system (xγ , fγ )γ∈Γ , xγ ∈ X, fγ ∈ X ∗ , where Γ is some set of indices, is called M -basic if fβ (xγ ) = δβγ (the Kronecker δ) and (fγ )γ∈Γ is total on the closed linear span [xγ ]γ∈Γ . An M -basic system is said to be an + 1 -system if xγ = 1 for each γ and there is a constant c > 0 such that for all finite collections (aγ ) of positive scalars aγ x γ ≥ c aγ . γ
γ
Lemma 11. Let Γ be an uncountable set. If a (nonseparable) Banach space X has an M -basic + 1 -system (xγ , fγ )γ∈Γ , then X fails the property D. Proof. Let (Ω, Σ) be a measurable space such that Σ contains all singleton subsets of Ω and there exists some Γ ⊂ Ω with Γ ∈ / Σ. Such (Ω, Σ) exists whenever Ω is uncountable. Put ϕ(ω) =
xω 0
for ω ∈ Γ, for ω ∈ Ω \ Γ.
Then for every γ ∈ Γ the function fγ ◦ϕ is measurable. Moreover, for each f in the annihilator ([xγ ]γ∈Γ )⊥ the function f ◦ ϕ is measurable. Hence, ϕ is scalarly totally measurable. On the other hand, the set A := conv(xγ )γ∈Γ is convex, closed and, since (xγ )γ∈Γ is an + 1 -system, does not contain 0. By the Hahn– Banach theorem, there is a functional g ∈ X ∗ such that g(x) > c > 0 for each x ∈ A. Then g ◦ ϕ is non-measurable. 2 By Lemma 11, the space 1 (Γ), with uncountable Γ, fails the property D. Since the space ∞ contains a subspace, isometric to 1 (Γ) with an uncountable Γ, it fails the property D. Denote by C[0, η] the space of continuous functions on the segment of ordinals [0, η]. The space C[0, η], with uncountable η, fails the property D. Indeed let xγ , be the characteristic function of the segment [0, γ] and fγ (x) := x(γ +1) −x(γ), x ∈ C[0, η], γ ∈ [0, η). It is easy to check that (xγ , fγ )γ∈Γ is an + 1 -system. So, one can apply Lemma 11. Proposition 12. E ⇒ D ⇒ D.
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Proof. E ⇒ D . Let F be a weakly* sequentially closed subspace of X ∗ . Then BF is weakly* sequentially closed, and because X has property E, is weakly* closed. By the Banach–Diedonne theorem [1, Ch. IV, §2.5], F is weakly* closed. D ⇒ D. If ϕ is totally scalarly measurable, then the set F , defined by (5), is a weakly* sequentially closed total subspace. Since X has property D , F is weakly* closed. As F is total, F = X ∗ , therefore X has property D. 2 This proposition and Theorem 9 answer the aforementioned question of Gulisashvili. We do not know whether E = D or D = D. Let us present arguments, why the property D can be equivalent to D. If X fails the property D , then there exists a weakly* sequentially closed subspace F ⊂ X ∗ which is not weakly* closed. Let Ω = X and Σ be the σ-algebra generated by functionals f ∈ F . Then, is each g ∈ cl∗ F \F (when cl∗ denotes the weak* closure) non-measurable? References [1] N. Bourbaki, Espaces Vectoriels Topologiques, Hermann, Paris, 1964. [2] G.A. Edgar, Measurability in a Banach space, II, Indiana Univ. Math. J. 28 (1979) 559–579. [3] N.M. Efremov, Duality conditions of a Banach space in terms of norm attaining linear functionals, Ph.D., Kharkiv State University, 1985 (in Russian). [4] G. Godefroy, Boundaries of a convex set and interpolation sets, Math. Ann. 277 (1987) 173–184. [5] A. Gulisashvili, Estimates for the Pettis integral in interpolation spaces and inversion of the embedding theorems, Dokl. Akad. Nauk SSSR 263 (1982) 793–798 (in Russian). English transl.: Sov. Math. Dokl. 25 (1982) 428–432. [6] A. Gulisashvili, Estimates for the Pettis integral in interpolation spaces with some applications, in: Lect. Notes in Math., vol. 991, 1983, pp. 55–78. [7] J. Hagler, F. Sullivan, Smoothness and weak* sequential compactness, Proc. Am. Math. Soc. 78 (1980) 497–503. [8] R.C. James, Reflexivity and the sup of linear functionals, Isr. J. Math. 13 (1972) 289–300. [9] W.B. Johnson, J. Lindenstrauss, Some remarks on weakly compactly generated Banach spaces, Isr. J. Math. 17 (1974) 219–230. [10] A.M. Plichko, D. Yost, Complemented and uncomplemented subspaces of Banach spaces, Extr. Math. 15 (2000) 335–371.