The Mazur Intersection Property In Banach Spaces And Related Topics
Pradipta Bandyopadhyaya
Thesis submitted to the Indian Statistical Institute in partial fulfilment of the requirements for the award of the degree of Doctor of Philosophy CALCUTTA 1991 Acknowledgements
It appears to be customary to begin by thanking one’s supervisor. But I find it superfluous, particularly since Professor A. K. Roy means much more to me than merely my supervisor. After all, one normally doesn’t thank his parents for holding his hands when he is taking his first faltering steps, for teaching him to walk. Though I consider this entire thesis as a kind of joint venture, Prof. Roy has allowed me to pass much of it as my own. For the records, only the materials of Chapter 3 have so far appeared as a joint paper, which too he has permitted me to include in this thesis without any questions asked. I joined this Institute as a student more than a decade ago. The list of teachers who have taught me, nurtured me, encouraged me, made me what I am today, is naturally bound to read like the whole faculty list. To each of them I owe a lot and it is impossible for me to choose only few of them for special mention. The same is also true for the numerous friends and colleagues in the hostel, the division or outside. Each of them contributed in making this place my “home away from home”, gave me much-needed moral support. I thank them all. In the initial days of my research, I had good fortune of many fruitful discus- sions with Prof. D. van Dulst of University of Amsterdam and Dr. T.S.S.R.K. Rao of ISI, Bangalore, who were visiting our Institute at that time. My sincere thanks to both of them. Last year, I went to New Delhi to participate in a National Seminar and there I met Dr. D. P. Sinha of Delhi University, who soon became a personal friend. It is from one of his ideas that the materials of Section 5.3 of this thesis came up. It’s a pleasure to thank him. Finally, the manuscript of this thesis was typeset using LATEX. Prof. A. R. Rao introduced me to LATEX, and Mr. Joydeep Bhanja helped me tailor it according to my needs. Also, Prof. A. K. Adhikari of the CSU has kindly allowed me to use the Laser printer there. I am in their debt.
February, 1991 Pradipta Bandyopadhyaya
ii Contents
1 Introduction 1
2 The MIP for a Family of Closed Bounded Convex Sets 11 2.1. The Set-up and Main Result ...... 11 2.2. The MIP with Respect to a Norming Subspace F ...... 18 2.3. A Digression ...... 23
3 Bochner Lp Spaces and the `p Sums of MIP Spaces 26 3.1. The `p Sums ...... 26 3.2. Bochner Lp Spaces ...... 28
4 Miscellaneous Results 43 4.1. The Subspace Question ...... 43 4.2. The MIP and Farthest Points ...... 45 4.3. The MIP in Projective Tensor Product Spaces ...... 48
5 Exposed Points of Continuity and Strongly Exposed Points 62 5.1. The Counterexample ...... 62 5.2. A Characterisation Theorem ...... 64 5.3. A Characterisation of Banach Spaces Containing `1 ...... 67
Reference 69
iii Chapter 1 Introduction
In the first part of this chapter we explain in general terms the main theme of this thesis and provide a chapterwise summary of its principal results. The second part recapitulates some of the known notions and results used in the subsequent chapters. The numbers given in parentheses correspond to those in the list of references on page 69. S. Mazur [40] was the first to consider the following smoothness property in normed linear spaces, called the Mazur Intersection Property (MIP), or, more briefly, the Property (I) :
Every closed bounded convex set is the intersection of closed balls containing it.
He showed that any reflexive Banach space with a Fr´echet-differentiable norm has this property. Later, R. R. Phelps [42] provided a more geometric insight into this prop- erty by showing that (a) A normed linear space X has the MIP if the w*-strongly exposed points of the unit ball B(X∗) of the dual X∗ are norm dense in the unit sphere S(X∗). (b) If a normed linear space X has the MIP, every support mapping on X maps norm dense subsets of S(X) to norm dense subsets of S(X∗). (c) A finite dimensional normed linear space X has the MIP if and only if the extreme points of B(X∗) are norm dense in S(X∗). He also asked whether the sufficient condition (a) is also necessary. To date, this remains an open question.
1 Nearly two decades later, Phelps’ characterisation (c) was extended by J. R. Giles, D. A. Gregory and B. Sims [21] to general normed linear spaces, developing an idea due to F. Sullivan [51], and they proved, inter alia,
Theorem 1.1 For a normed linear space X, the following are equivalent : (a) The w*-denting points of B(X∗) are norm dense in S(X∗). (b) X has the MIP. (c) Every support mapping on X maps norm dense subsets of S(X) to norm dense subsets of S(X∗).
They also showed that in dual Banach spaces, the MIP implies reflex- ivity and considered the weaker property that every weak* compact convex set in a dual space is the intersection of balls (w*-MIP). Investigating the necessity of Phelps’ condition (a), they showed that it is indeed necessary if, in addition, the dual X∗ has the w*-MIP, or, X is an Asplund space. They now asked whether the MIP necessarily implies Asplund. To date, this also remains open. Notice that if X is separable and has the MIP, Phelps’ condition (b) (or, Theorem 1.1(c) above) implies that it has a separable dual and hence is Asplund. So one asks, is the MIP hereditary, i.e., inherited by subspaces ? The answer, unfortunately, is no. In Chapter 4, we give an example and discuss the subspace question in more detail. However, since the Asplund Property is invariant under equivalent renorming, a more pertinent question is whether the existence of an equiv- alent norm with the MIP is hereditary. Some discussions on MIP-related renorming questions may be found in [9], [47], [54] and [57]. However, in this work, we do not discuss any renorming problem but concentrate instead on some of the isometric questions that arise. Recently, there have appeared several papers dealing with similar inter- section properties for compact convex sets [54, 47] (called the Property CI), weakly compact convex sets [57] and compact convex sets with finite affine dimension [49]. In Chapter 2, we give a unified treatment of the intersection properties for these diverse classes of sets by considering the MIP for the members of
2 a general family — subject to some mild restrictions — of closed bounded convex sets in a Banach space and recapture all the known results as special cases. We also introduce a new condition of separation of convex sets which is a variant of the following :
Disjoint bounded convex sets are contained in disjoint balls and this apparently stronger condition turns out to be equivalent to the intersection property in all known cases. This strengthens the results of Zi- zler [56]. We should point out that our proofs in this chapter are usually modifications, refinements and adaptations to our very general set-up of ar- guments for particular cases to be found in [21], [47] and [54]. This chapter, which is essentially contained in [3], also provides much of the background for what follows in the later chapters. Whitfield and Zizler [55] have also defined the Uniform Mazur Intersection Property, a property somewhat stronger than the MIP. But in this thesis, we only briefly touch upon their work. In Chapter 3, we discuss the question of lifting the MIP and the CI from a Banach space X to its associated Bochner Lp space and their stability under `p sums, 1 < p < ∞. In particular, we prove that the `p sum of a family of Banach spaces has the MIP (or, the CI) if and only if each coordinate space has it; that the Bochner Lp space for the Lebesgue measure on [0, 1] always has the CI, while the MIP in X is equivalent to a weaker intersection property in the Bochner Lp space which turns out to be equivalent to the MIP if and only if X is Asplund. Most of these results have already appeared in print in [4]. In Chapter 4, we present a collection of partial results relating to various aspects of the MIP that raise more question than they answer. Apart from the subspace question mentioned above, in this chapter we discuss the rela- tion between the MIP and a farthest point phenomenon that was observed by K. S. Lau [35], but seems to have passed largely unnoticed since then. Lau had shown that in a reflexive space the MIP is equivalent to the following :
Every closed bounded convex set is the closed convex hull of its farthest points.
3 We extend this result to characterise the w*-MIP in w*-Asplund dual spaces using a result of Deville and Zizler [10]. As far as we know, besides the work of Tsarkov [53] — who has characterised the MIP in finite-dimensional spaces in terms of convexity of bounded Chebyshev sets — this is the only attempt at an intrinsic characterisation of the MIP. We also point out some new direction of investigation that is indicated by our result. In this chapter, we also discuss the MIP in projective tensor product spaces. Ruess and Stegall [46] have shown that the injective tensor product of two Banach spaces of dimension ≥ 2 never has the MIP. And Sersouri [48] has shown that in fact there is a two-dimensional compact convex set in
X ⊗ε Y that is not intersection of balls. The situation appears to be much more difficult for projective tensor product spaces. Here we show that the projective tensor product of two Banach spaces X and Y never has the MIP if X and Y are Hilbert spaces or are two-dimensional `p spaces for a large range of values of p. For this purpose, we characterise the extreme p q contractions from `2 to `2 and obtain their closure. The technique used is similar to [34]. In the process, we reestablish relevant special cases of the results obtained in [25, 26, 28]. Chapter 5 is somewhat independent of the rest. Here we discuss the following phenomenon : For a closed bounded convex set K in a Banach space X and a point xo ∈ K, the following implications are easy to establish :
xo strongly exposed =⇒ xo denting point ⇓ ⇓
xo exposed and PC =⇒ xo extreme and PC where by PC we mean a point of continuity (defined more precisely later).
A recent result (see Theorem 1.5 below) is that xo is denting if and only if xo is extreme and PC, i.e., the implication down the right hand side above is reversible. This naturally leads to the conjecture that the implication down the left hand side above, too, can be reversed, i.e., if xo is exposed and PC
(or, equivalently, xo is an exposed denting point) then xo is strongly exposed. We show that the conjecture is false by constructing a counterexample in the Banach space `1 and provide a characterisation of strongly exposed points
4 among points of continuity of a closed bounded convex set. As a corollary, we deduce that the conjecture is true for the points of weakly compact sets. We also show that the counterexample, in some sense, is actually typical of `1, i.e., we characterise Banach spaces containing `1 in terms of the validity of the above conjecture. We also briefly touch upon the necessity of Phelps’ condition (a) in the light of our characterisation of (w*-) strongly exposed points. This chapter is a revised version of [2].
Notations, Conventions and General Preliminaries
General reference to this work are the monographs [7], [13] and [20]. We work only with real Banach spaces. Unless otherwise mentioned, by a subspace we always mean a norm closed linear subspace. The closed unit ball and the unit sphere of a Banach space X will be denoted by B(X) and
S(X) respectively. For z ∈ X and r > 0, we denote by Br[z] (resp. Br(z)) the closed (resp. open) ball of radius r and centre z. For x ∈ S(X), D(x) = {f ∈ S(X∗): f(x) = 1}. The set valued map D is called the duality map and any selection of D is called a support mapping. For f ∈ S(X∗), we similarly define the inverse duality map as D−1(f) = {x ∈ S(X): f(x) = 1}. For a bounded set K ⊆ X, denote [K] = ∩{B : B closed ball containing K} and following Franchetti [19], call a bounded set admissible if K = [K]. In this terminology, a Banach space X has the MIP (resp. the CI) if every norm closed bounded (resp. norm compact) convex set is admissible. For K ⊆ X, f ∈ X∗ and α > 0, the set S(K, f, α) = {x ∈ K : f(x) > sup f(K) − α} is called the open slice of K determined by f and α. For A ⊆ X, denote by co(A) (resp. aco(A)) the convex (resp. absolutely convex) hull of A and by int(A), the norm interior of A. For A ⊆ X, f ∈ X∗, o ∗ ∗ kfkA = sup{|f(x)| : x ∈ A}, A = {f ∈ X : kfkA ≤ 1} and for B ⊆ X ,
A-dia(B) = sup{kb1 − b2kA : b1, b2 ∈ B}. For A1,A2 ⊆ X, dist(A1,A2) = σ inf{kx1 − x2k : xi ∈ Ai, i = 1, 2}. For A ⊆ X, A denotes the closure of
5 A for the topology σ. Whenever the topology is not specified, we mean the norm topology. We identify an element x ∈ X with its canonical imagex ˆ in X∗∗. Let Kc be the image of a closed bounded convex K ⊆ X under this canonical embedding of X in X∗∗. We denote by Kf the w*-closure of Kc in X∗∗. Kf is of course a w*-compact convex set.
For a measure space (Ω, Σ, µ) and A ∈ Σ, we denote by χA, the indicator function of A, and by µ|A the restriction of the measure µ to the σ-field of measurable subsets of A. Let X be a real Banach space, let F be a total subspace of X∗ (i.e., F separates points of X). Let σ denote the σ(X,F ) topology on X. For a closed bounded convex set K ⊆ X, recall that xo ∈ K is called σ (a) a σ-denting point of K if, for each ε > 0, xo 6∈ co (K \ Bε(xo)). (b) a σ-point of continuity (σ-PC) for K if the identity map, id :(K, σ) →
(K, norm) is continuous at xo.
(c) a very strong extreme point of K if, for every sequence {xn} of K-valued Bochner integrable functions on [0, 1] with respect to the Lebesgue measure, the condition
Z 1 Z 1 lim k xn(t)dt − xok = 0 implies lim kxn(t) − xokdt = 0. n→∞ o n→∞ o
(d) an σ-exposed point of K if there exists f ∈ F such that f exposes xo,
i.e., f(x) < f(xo) for all x ∈ K \{xo}. (e) a σ-strongly exposed point of K if there exists f ∈ F such that f
exposes xo and for any sequence {xn} ⊆ K, lim f(xn) = f(xo) implies n that lim kxn − xok = 0. n In the sequel, whenever σ is the weak topology, i.e., F = X∗, we omit the prefix σ while referring to the above concepts. The following lemma is very well-known and can be found in [8, Propo- sition 25.13] or more recently in [45, Lemma 1.3] :
Lemma 1.2 Let E be a locally convex space, K be a compact convex set in E and x ∈ K. The following are equivalent :
6 (a) x is an extreme point of K. (b) The family of open slices containing x forms a local base for the topology of E at x (relative to K).
In Chapter 5, we note an analogue of this lemma for exposed points. The following characterisation of a σ-denting point is an immediate con- sequence of the Hahn-Banach Theorem :
Lemma 1.3 Let X be a Banach space, F be a total subspace of X∗, K be a closed bounded convex set in X and x ∈ K. The following are equivalent : (a) x is a σ-denting point of K. (b) The family of σ-open slices containing x forms a local base for the norm topology at x (relative to K).
The following lemma is also well-known :
Lemma 1.4 Let K be a closed bounded convex set of a Banach space X. Then x∗∗ ∈ Kf is a w*-PC if and only if x∗∗ =x ˆ for some x ∈ K and x is a PC.
The following characterisation of a denting point of a closed bounded convex set in a Banach space is a consequence of the two lemmas above and is essentially contained in [38] and [39] :
Theorem 1.5 Let x be an element in a closed bounded convex set K of a Banach space. Then the following are equivalent : (a) x is a denting point of K.
(b)x ˆ is a w*-denting point of Kf. (c) x is a very strong extreme point of K.
(d) For any probability space (Ω, Σ, µ) and any net {xα} of K-valued Bochner integrable functions on Ω, the condition Z Z lim k xα(ω)dµ − xk = 0 implies lim kxα(ω) − xkdµ = 0. α Ω α Ω
(e) x is an extreme point of K which is also a PC.
7 Let X be a real Banach space, let F be a norming subspace of X∗ (i.e., kxˆkB(F ) = kxk, for all x ∈ X). Then B(X) is σ(X,F )-closed and B(F ) is w*-dense in B(X∗). We will need the following generalisation of Lemma 1.4 and Theorem 1.5 (a) ⇐⇒ (e) in the sequel.
Lemma 1.6 Let X be a Banach space, F be a norming subspace of X∗. ∗ ∗ ∗ ∗ (a) A point xo ∈ B(X ) is a w*-PC of B(X ) if and only if xo is a w*-PC of B(F ).
∗ ∗ ∗ (b) A point xo ∈ B(X ) is a w*-denting point of B(X ) if and only if ∗ xo is an extreme point and a w*-PC of B(F ).
We will also use the following well-known results :
Lemma 1.7 [42, Lemma 3.1] Let X be a normed linear space and ε > 0. If f, g ∈ S(X∗) are such that x ∈ B(X) ∩ f −1(0) implies |g(x)| ≤ ε/2, then either kf − gk ≤ ε or kf + gk ≤ ε.
Lemma 1.8 [20, p 205] Let K be a closed bounded convex subset of a Banach ∗ space X. Given f ∈ X \{0} and xo contained in a slice S of K determined by f, there exists ε > 0 such that whenever kf − gk < ε, there is a slice S0 0 of K determined by g such that xo ∈ S ⊆ S.
For a Banach space X, a function F : X −→ IR is said to be (a) Gateaux differentiable at a point x ∈ X if there exists an f ∈ X∗ such that
F (x + λy) − F (x)
lim − f(y) = 0 for all y ∈ B(X) and λ→0+ λ
(b) Fr´echetdifferentiable at x if the convergence is uniform over y ∈ B(X). In particular, if F is the norm, we have the following duality result :
Theorem 1.9 [20, Theorem 3.5.4] In a Banach space X (a) the norm is Gateaux differentiable (or smooth) at a point x ∈ S(X) if and only if D(x) is single-valued and in that case, x w*-exposes D(x) in B(X∗).
8 (b) the norm is Fr´echetdifferentiable at x ∈ S(X) if and only if D(x) is single-valued and x w*-strongly exposes D(x) in B(X∗). Recall that a Banach space X is Asplund if every continuous convex function F : X → IR is Fr´echet differentiable on a dense Gδ subset of X, and a dual space X∗ is w*-Asplund if every continuous w*-lower semicontinuous ∗ convex function F : X → IR is Fr´echet differentiable on a dense Gδ subset of X∗. The following characterisation theorem is a classic [7] : Theorem 1.10 For a Banach space X, (i) the following are equivalent : (a) X is Asplund. (b) Separable subspaces of X have separable dual. (c) X∗ has the Radon-Nikod´ymProperty (RNP). (d) X∗ has the RNP with respect to the Lebesgue measure on [0, 1]. (ii) the following are equivalent : (a) X∗ is w*-Asplund. (b) X has the RNP. If X is a Banach space and (Ω, Σ, µ) a measure space, let Lp(µ, X) de- note the Lebesgue-Bochner function space of p-integrable X-valued functions 1 1 defined on Ω, 1 ≤ p < ∞ (see [14]). Recall (from [13]) that if p + q = 1 (1 < p < ∞), the space Lq(µ, X∗) is isometrically isomorphic to a subspace of Lp(µ, X)∗ and that they coincide if and only if X∗ has the RNP with respect to µ. Also recall (from [14] and [31]) that Lp(µ, X)∗ is isometrically isomorphic to the following two spaces : ∗ ∗ (1) Vq(µ, X ) = the space of all vector measures F :Σ −→ X such that the q-variation of F is finite, i.e.,
( q )1/q X kF (E)k kF kq = sup q µ(E): π is a finite partition of Ω < ∞, E∈π µ(E) (2) Lq(µ, X∗; X) = the space of all w*-equivalence classes of X∗- valued w*-measurable functions h, such that the real-valued function kh(·)k ∈ Lq(µ, IR).
9 Only the representation given by (1) will be used by us. [13, Chapter VIII] contains all necessary information on tensor product spaces. We just recall here that the dual of the projective tensor product, ∗ X ⊗π Y , of two Banach spaces X and Y , is L(X,Y ), the space of continuous linear operators from X to Y ∗.
10 Chapter 2 The Mazur Intersection Property for a Family of Closed Bounded Convex Sets
2.1. The Set-up and Main Result
Let X be a real Banach space, let F be a norming subspace of X∗ and let
A = {K ⊆ X : K is admissible}
Then A is a family of norm bounded, σ(X,F )-closed convex sets that is closed under arbitrary intersection, translation and scalar multiplication. Naturally, A contains all singletons and all closed balls. In this chapter, we denote the σ(X,F ) topology on X simply by σ. Taking A as our model, we let C be a family of norm bounded, σ-closed convex sets with the following properties : (A1) C is closed under arbitrary intersection, translation and scalar mul- tiplication. σ (A2) C1,C2 ∈ C =⇒ aco (C1 ∪ C2) ∈ C. (A3) C ∈ C, C absolutely convex and f ∈ F =⇒ C ∩ f −1(0) ∈ C. Note that (A1) implies that C contains all singletons. Examples : (i) C = {all closed bounded convex sets in X}, F = X∗ (ii) X = Y ∗, C = {all w*-compact convex sets in X}, F = Yb (iii) C = {all compact convex sets in X}, F = any norming subspace (iv) C = {all compact convex sets in X with finite affine dimension}, F = any norming subspace
11 (v) C = {all weakly compact convex set in X}, F = any norming subspace. Let F = {Co : C ∈ C}. Then F is a local base for a locally convex Hausdorff vector topology τ on X∗, the topology of uniform convergence on elements of C. Clearly, τ is stronger than the w*-topology on X∗ and weaker than the norm topology. ∗ Definitions : (1) Denote by Eτ , the set of all extreme points of B(X ) which are also points of continuity of the identity map, id :(B(X∗),w*) −→ (B(X∗), τ). (2) For C ∈ C, C absolutely convex, ε > 0 and δ > 0, we say that a point x ∈ S(X) belongs to the set MC,ε,δ(X) if
kx + λyk + kx − λyk − 2 sup < ε yC,0<λ≤δ λ
For C = B(X), we write MC,ε,δ(X) simply as Mε,δ(X). Let MC,ε(X) =
∪δ>0MC,ε,δ(X) and for C = B(X) write MC,ε(X) simply as Mε(X), or even as Mε whenever there is no scope of confusion. Similarly, MC,ε(X) will often be abbreviated as MC,ε. Notice that ∩ε>0Mε(X) gives the set of points of S(X) at which the norm is Fr´echet differentiable. τ (3) Hτ = ∩{D(MC,ε(X)) : C ∈ C, C absolutely convex, C ⊆ B(X) and ε > 0}.
Lemma 2.1 For any absolutely convex C ∈ C, C ⊆ B(X), x ∈ S(X) and δ > 0, let
kx + λyk + kx − λyk − 2 d1(C, x, δ) = sup yC,0<λ≤δ λ ∗ d2(C, x, δ) = C-dia S(B(X ), x,ˆ δ)
d3(C, x, δ) = C-dia [∪{D(y): y ∈ S(X) ∩ Bδ(x)}]
Then for any α, δ > 0, we have, 2α (i) d (C, x, α) ≤ d (C, x, δ) + 2 1 δ (ii) d3(C, x, δ) ≤ d2(C, x, δ)
(iii) d1(C, x, δ) ≤ d3(C, x, 2δ)
12 Proof : (i) Fix α, δ > 0. Put d1 = d1(C, x, δ) and d2 = d2(C, x, α). ∗ For each n ≥ 1, we can choose fn, gn ∈ S(B(X ), x,ˆ α) such that 1 1 kfn − gnkC > d2 − n . Choose yn ∈ C such that (fn − gn)(yn) > d2 − n . Then kx + δy k + kx − δy k − 2 f (x + δy ) + g (x − δy ) − 2 1 2α n n ≥ n n n n ≥ d − − δ δ 2 n δ 2α Thus, d1 ≥ d2 − δ . (ii) immediately follows from the observation that ∪{D(y): y ∈ S(X) ∩ ∗ Bδ(x)} ⊆ S(B(X ), x,ˆ δ). (iii) Let λ ≤ δ. Observe that for any y ∈ C ⊆ B(X),
x ± λy x ± λy
− x ≤ − (x ± λy) + λ = |1 − kx ± λyk | + λ kx ± λyk kx ± λyk = | kxk − kx ± λyk | + λ ≤ 2λ x + λy x − λy Let f ∈ D( ) and g ∈ D( ). Then kx + λyk kx − λyk kx + λyk + kx − λyk − 2 f(x + λy) + g(x − λy) − 2 = λ λ ≤ (f − g)(y) ≤ kf − gkC
And we immediately have :
Lemma 2.2 For any absolutely convex C ∈ C, C ⊆ B(X), x ∈ S(X) and ε > 0, the following are equivalent :
(i) x ∈ MC,ε(X) (ii) There is a δ ∈ (0, 1) such that C-dia S(B(X∗), x,ˆ δ) < ε (iii) There is a δ ∈ (0, 1) such that C-dia [∪{D(y): y ∈ S(X) ∩
Bδ(x)}] < ε.
Corollary 2.2.1 For any absolutely convex C ∈ C, C ⊆ B(X) and ε > 0,
MC,ε(X) is an open subset of S(X).
Proof : Follows immediately from the Lemma above and Lemma 1.8. Remark : In the case of C = B(X), Lemma 2.1 is quantitatively more precise than Lemma 1 of [55]. Now we have our main result :
13 Theorem 2.3 If X, F and C are as above, consider the following state- ments : + τ (a) F ⊆ IR Eτ
+ τ (b) F ⊆ IR Hτ
(c) If C1,C2 ∈ C are such that there exists f ∈ F with sup f(C1) <
inf f(C2) then there exist disjoint closed balls B1,B2 such that Ci ⊆ Bi, i = 1, 2. (d) Every C ∈ C is admissible. (e) For every norm dense subset A of S(X) and every support mapping τ φ : S(X) −→ S(X∗), F ⊆ IR+φ(A) . Then we have (a) ⇒ (b) ⇒ (c) ⇒ (d) ⇒ (e).
(For the reverse implications, see corollaries and remarks at the end of this section.)
Proof : (a) ⇒ (b) Enough to prove Eτ ⊆ Hτ .
Let f ∈ Eτ . Let C ∈ C, C absolutely convex, C ⊆ B(X) and ε > 0. We τ want to prove f ∈ D(MC,ε) . Let K ∈ C and 0 < η < ε. We may assume σ K ⊆ B(X). Let Ko = aco (K ∪ C). Then Ko ⊆ B(X) as B(X) is σ-closed.
Note that Ko ∈ C by (A2). Since f is an extreme point of the w*-compact convex set B(X∗) and id : (B(X∗), w∗) → (B(X∗), τ) is continuous at f, it follows from Lemma 1.2 that w*-slices of B(X∗) containing f form a base for the relative τ-topology at f. Thus there exist x ∈ S(X) and 0 < δ < 1 such that f ∈ S = S(B(X∗), x,ˆ δ) and Ko-dia(S) < η.
Now, by Lemma 2.2, x ∈ MKo,η ⊆ MC,ε and for any fx ∈ D(x), fx ∈
D(MC,ε) and fx ∈ S, so, kf − fxkK ≤ kf − fxkKo < η.
(b) ⇒ (c) Let C1,C2 ∈ C and f ∈ S(F ) be such that sup f(C1) < 1 inf f(C2). Let z ∈ X be such that f(z) = 2 (sup f(C1) + inf f(C2)) and 1 put ε = 12 (inf f(C2) − sup f(C1)) > 0. Then, inf f(C2 − z) > 5ε and inf(−f)(C1 − z) > 5ε. We may assume without loss of generality that z = 0, σ Ci ⊆ B(X), i = 1, 2 and kfk = 1. Let K = aco (C1 ∪ C2), then K ∈ C, K is absolutely convex and K ⊆ B(X).
14 By (b), there is λ ≥ 0 and g ∈ Hτ such that kf − λgkK < ε. If λ = 0, we have kfkK < ε and hence, infC2 f < ε, a contradiction. Thus, λ > 0. τ Now, g ∈ Hτ ⊆ D(MK,ε/λ) . So, we can find x ∈ MK,ε/λ and h ∈ D(x) such that kg − hkK < ε/λ. By definition, there is a δ > 0 such that kx + αyk + kx − αyk − 2 ε sup < yK,0<α≤δ α λ
λ Choose an integer n > εδ . The proof will be complete once we show that B1 = B(n−1)ε/λ [−nεx/λ] and B2 = B(n−1)ε/λ [nεx/λ] work.
Clearly, B1 and B2 are disjoint. Suppose, if possible, y ∈ C2 and y 6∈ B2. λ Then y ∈ K. Take α = nε < δ and observe that kx + αyk + kx − αyk − 2 kx + αyk − kxk x 1 = + k − yk − α α α α (n − 1)ε nε ε ≥ h(y) + − = h(y) − λ λ λ 2ε 1 ≥ g(y) − = [λg(y) − 2ε] λ λ 1 1 2ε > [f(y) − 3ε] ≥ [5ε − 3ε] = λ λ λ
This contradicts the fact that x ∈ MK,ε/λ. The other inclusion follows sim- ilarly once we note that K, and hence MK,ε/λ, is symmetric and h ∈ D(x) implies (−h) ∈ D(−x). (c) ⇒ (d) Since singletons are in C and every C ∈ C is σ-closed, (d) follows from (c). (d) ⇒ (e). (We adapt Phelps’ [42] arguments) Let A be a norm dense subset of S(X) and φ be a support mapping. Let f ∈ S(F ), K ∈ C and 0 < ε < 1. We may assume K ⊆ B(X) and further that K is absolutely convex and kfkK > 1 − ε/2 (Let x ∈ B(X) such that f(x) > 1 − ε/2. Let σ L = aco [{x} ∪ K]. Then L ⊆ B(X), L ∈ C and k · kL ≥ k · kK ). Let u ∈ K be such that f(u) > 1 − ε/2. Put u0 = εu/4 and D = K ∩ f −1(0). Then D ∈ C [by (A3)] and u0 6∈ D. By (c), there exists z ∈ X and r > 0 such that 0 D ⊆ Br[z] and ku − zk > r. 0 1 0 Let µ = ku − zk − r > 0. Put w = r+µ (ru + µz). Then kw − zk = r. 1 0 µ Put x = r (w − z) ∈ S(X). Let C = co[{u } ∪ Br[z]]. Let 0 < δ < (r+µ) .
15 rµ rµ If p ∈ Brδ[w], then kp − wk < (r+µ) , so, p = w + r+µ · y for some y ∈ X, r 0 µ kyk < 1. Thus, p = r+µ u + r+µ (z + ry). Now, z + ry ∈ Br(z) and hence, p ∈ int(C). So, Brδ[w] ⊆ int(C).
Let y ∈ Bδ[x] ∩ A and g = φ(y). Put v = ry + z. Clearly, kv − wk ≤ rδ, hence v ∈ int(C) and g(v) = sup g(Br[z]). Now, v ∈ int(C) ⇒ there exists 0 0 0 t ∈ (0, 1) and v ∈ Br(z) such that v = tu + (1 − t)v . Thus, g(v) = 0 0 0 tg(u ) + (1 − t)g(v ) < tg(u ) + (1 − t)g(v). Also, 0 ∈ D ⊆ Br[z] =⇒ 0 1 1 0 ≤ g(v) < g(u ) = 4 εg(u) ≤ 4 ε · kgkK . So, 0 < kgkK ≤ kgk = 1. Put 1 λ = 1/kgkK . Then sup λg(D) ≤ sup λg(Br[z]) = λg(v) < 4 ε · kλgkK = ε/4. By symmetry of D, kλgkD ≤ ε/4. Now, by Lemma 1.7 applied to the linear space sp(K), spanned by K, equipped with µK , the gauge or Minkowski functional of K, we have
f ε f ε
+ λg ≤ or − λg ≤ kfkK K 2 kfkK K 2
But u ∈ K and ε < 1 implies f(u)/kfkK ≥ f(u) > 1 − ε/2 > ε/2 and f λg(u) > 0. Thus, k − λgkK ≤ ε/2. Then we have kfkK
ε f ε ε ε
kf − λgkK ≤ + − f = + (1 − kfkK ) ≤ + = ε 2 kfkK K 2 2 2
Corollary 2.3.1 If in the set-up of Theorem 2.3, we have that the set
A = {x ∈ S(X): D(x) ∩ Eτ 6= ∅} is norm dense in S(X), then all the statement in Theorem 2.3 are equivalent.
Proof : We simply note that in this case there is a support mapping that maps A into Eτ and hence, (e) ⇒ (a).
Corollary 2.3.2 [54, 47, 49] In the case of examples (iii), i.e., C = {all compact convex sets in X}, F = any norming subspace, and (iv), i.e., C = {all compact convex sets in X with finite affine dimension}, F = any norming subspace, all the statements in Theorem 2.3 are equivalent and (c) can be reformulated as (c0) Disjoint members of C can be separated by disjoint closed balls.
16 Proof : In example (iii), τ is the bw* topology (see [16] for more on bw* topology) and in example (iv), τ is the w*-topology on X∗ and in both cases F τ = X∗, so we may as well take F = X∗. Further, as the bw* topol- ogy agrees with the w*-topology on bounded sets, in both the cases, Eτ = {extreme points of B(X∗)}. Clearly, in both cases A as in Corollary 2.3.1 is S(X) and so, in Theorem 2.3 all the statements are equivalent. Since members of C in both cases are σ-compact, (c) ⇐⇒ (c0). Remarks : 1. In example (v), i.e., C = {all weakly compact convex set in X}, F = any norming subspace, τ is the Mackey topology on X∗ (see [8] for further information) and again, F τ = X∗. In this case, we do not know whether any of the implications in Theo- rem 2.3 can be reversed. However, we note that (a) ⇒ (d) in Theorem 2.3 gives a weaker sufficiency condition for MIP for weakly compact sets than the one used in [57]. And in this case, too, (c) and (c0) of Corollary 2.3.2 are equivalent. 2. It seems unlikely that, in general, the implications in Theorem 2.3 can be reversed. Nevertheless, it appears to be an interesting and difficult problem to find conditions on X, F , and C under which this can be done. In particular, when is the assumption of Corollary 2.3.1 satisfied ? In this context, the work of Namioka [41] on neighbourhoods of extreme points may conceivably be relevant. However, there is yet another situation when the statements can actually be shown to be equivalent. And particular cases of this yield the character- isations of MIP and w*-MIP, i.e., examples (i) and (ii). This we take up in the next section. 3. Note that the subspace F ⊆ X∗ was assumed to be norming in order to ensure that σ-closure of norm bounded sets remain norm bounded which is implicit in (A2). However, if (A2) is satisfied, as in examples (iii), (iv) and (v), for any total subspace F , our results easily carry through with only minor technical modifications in the proofs.
17 2.2. The MIP with Respect to a Norming Subspace F
Our standing assumption in this section is that F is a subspace of X∗ such that
the set TF = {x ∈ S(X): D(x) ∩ S(F ) 6= ∅} is a norm dense subset of S(X)
(we shall write simply T when there is no confusion). Then F is necessarily norming. However, one can give examples (see below) of norming subspaces where this property does not hold. Let C = {all norm bounded, σ-closed convex sets in X}. We say that X has F -MIP if every C ∈ C is admissible. Examples : (i) F = X∗, T = S(X) and we have the MIP. (ii) X = Y ∗, F = Yb , T = D(S(Y )), which is dense by Bishop-Phelps Theorem [5], and we have the w*-MIP. ∗ Now, since B(X) ∈ C, τ is the norm topology on X and Eτ = {w*- denting points of B(X∗)}. We need the following reformulation of Lemma 2.2 :
Lemma 2.4 For x ∈ S(X), F, T as above and ε > 0, the following are equivalent :
(i) x ∈ Mε (ii) x determines a slice of B(F ) of diameter less than ε (iii) There exists δ > 0 such that
dia[∪{D(y) ∩ S(F ): y ∈ T ∩ Bδ(x)}] < ε.
Proof : (i) ⇒ (ii) ⇒ (iii) follows as easy adjustments of Lemma 2.1.
(iii) ⇒ (i) Let δ > 0 be as in (iii). Let do = dia[∪{D(y) ∩ S(F ): y ∈ 2 2 T ∩ Bδ(x)}] < ε. Choose δo > 0 such that δo + 2δo < δ and δo + 2δo/ε <
(1 − do/ε). Let y ∈ S(X), 0 < λ < δo. Then as in the proof of Lemma 2.1,
x ± λy
− x ≤ 2λ. kx + λyk
18
x+λy 2 x−λy 2 Find x1, x2 ∈ T such that kx+λyk − x1 ≤ λ and kx−λyk − x2 ≤ λ . Let f1, f2 ∈ S(F ) such that fi ∈ D(xi), i = 1, 2. Observe that kxi − xk ≤ 2 2 λ + 2λ ≤ δo + 2δo < δ, i.e., x1, x2 ∈ T ∩ Bδ(x). Thus kf1 − f2k ≤ do. Now, ! ! x + λy x + λy x + λy 2 0 ≤ 1 − f1 = f1 x1 − ≤ − x1 ≤ λ . kx + λyk kx + λyk kx + λyk
2 2 So, f1(x + λy) ≥ (1 − λ )kx + λyk. Similarly, f2(x − λy) ≥ (1 − λ )kx − λyk. So, we have
kx + λyk + kx − λyk − 2 f (x + λy) + f (x − λy) − 2(1 − λ2) ≤ 1 2 λ λ(1 − λ2) (f + f )(x) − 2 + λ(f − f )(y) + 2λ2 = 1 2 1 2 λ(1 − λ2)
kf1 − f2k + 2λ do + 2λ do + 2δo ≤ 2 ≤ 2 ≤ 2 1 − λ 1 − λ 1 − δo d + 2λ (since o is increasing in λ). Thus, 1 − λ2
kx + λyk + kx − λyk − 2 do + 2δo sup ≤ 2 < ε yS(X),0<λ≤δo λ 1 − δo by the choice of δo.
Definition : Let DF : T → S(F ) be defined by DF (x) = D(x) ∩ S(F ).
We say that DF is quasi-continuous if for every f ∈ S(F ) and ε > 0, there exists x ∈ T and δ > 0 such that y ∈ Bδ(x) ∩ T implies DF (y) ⊆ Bε[f]. ∗ Notice that for X = Y and F = Yb , DF is the inverse duality map. Now, we are in a position to prove :
Theorem 2.5 If X, F and T are as above, the following are equivalent : (a) The w*-denting points of B(X∗) are norm dense in S(F ).
(b) For every ε > 0, D(Mε) ∩ S(F ) is norm dense in S(F ).
(c) If C1,C2 ∈ C are such that there exists f ∈ F with sup f(C1) <
inf f(C2) then there exist disjoint closed balls B1,B2 such that Ci ⊆ Bi, i = 1, 2. (d) X has the F-MIP.
19 (e) DF is quasi-continuous. (f) For every support mapping φ that maps T into S(F ) and for every norm dense subset A of T, φ(A) is norm dense in S(F ).
(Observe that since Mε is open and T is dense, D(Mε) ∩ S(F ) 6= ∅ when- ever Mε 6= ∅.) Proof : (a) ⇒ (b) Let f be a w*-denting point of B(X∗) and ε > 0. As noted above f ∈ S(F ). Proceeding as in Theorem 2.3 (we may take K = B(X) and so, Ko = B(X)), for any 0 < η < ε, there is x ∈ S(X) and α > 0 such that f ∈ S = S(B(X∗), x,ˆ α) and dia(S) < η. Since T is dense in S(X), by Lemma 1.8, there is y ∈ T and δ > 0 such that f ∈ S0 = S(B(X∗), y,ˆ δ) and 0 S ⊆ S. Again as in Theorem 2.3, y ∈ Mε ∩ T and for any fy ∈ D(y) ∩ S(F ), kfy − fk < η. (b) ⇒ (c) ⇒ (d) ⇒ (f) Just a simplified version of the implication (b) ⇒ (c) ⇒ (d) ⇒ (e) in Theorem 2.3 where we replace K by B(X). (e) ⇒ (f) Follows from definitions.
(f) ⇒ (e) Suppose DF is not quasi-continuous. Then there exist f ∈ S(F ) and ε > 0 such that for all x ∈ T and δ > 0, there exists y(x, δ) ∈
Bδ(x) ∩ T such that DF (y) 6⊆ Bε[f]. Let A = {y(x, δ): x ∈ T , δ > 0}. Then A is dense in T and we can get a support mapping φ that maps T into S(F ) and φ(A) ∩ Bε[f] = ∅. (e) ⇒ (b) Let f ∈ S(F ) and ε > 0. Let 0 < η < ε/2. By (e), there exists x ∈ T and δ > 0 such that y ∈ T ∩ Bδ(x) implies D(y) ∩ S(F ) ⊆ Bη[f].
But then dia[∪{D(y) ∩ S(F ): y ∈ T ∩ Bδ(x)}] ≤ 2η < ε. So by Lemma 2.4, x ∈ Mε ∩ T and fx ∈ D(x) ∩ S(F ) implies kfx − fk < η.
(b) ⇒ (a) For n ≥ 1, let Dn = {f ∈ S(F ): f is contained in a w*-open 1 slice of B(F ) of diameter < n }. By Lemma 2.4, D(M1/n) ∩ S(F ) ⊆ Dn. Thus, for all n ≥ 1, Dn is norm open dense subset of S(F ) and by Baire
Category Theorem, ∩Dn is dense in S(F ). But, it is easy to see that ∩Dn = {w*-denting points of B(X∗)}. For completeness, we record in full the following immediate consequences, most of which are well-known results in the MIP literature.
Theorem 2.6 [21, Theorem 2.1 and 3.1] For a Banach space X
20 (1) the following are equivalent : (a) The w*-denting points of B(X∗) are norm dense in S(X∗). ∗ (b) For every ε > 0, D(Mε(X)) is norm dense in S(X ).
(c) Whenever C1,C2 are closed bounded convex sets in X with
dist(C1,C2) > 0, there exist disjoint closed balls B1,B2 such that
Ci ⊆ Bi, i = 1, 2. (d) X has the MIP. (e) The duality map is quasi-continuous. (f) Every support mapping maps norm dense subsets of S(X) to norm dense subsets of S(X∗). (2) the following are equivalent : (a) The denting points of B(X) are norm dense in S(X). −1 ∗ (b) For every ε > 0, D (Mε(X )) is norm dense in S(X). (c) Disjoint w*-compact convex sets in X∗ can be separated by disjoint closed balls. (d) X∗ has the w*-MIP. (e) The inverse duality map is quasi-continuous. (f) Every support mapping that maps D(S(X)) into S(X) maps norm dense subsets of D(S(X)) to norm dense subsets of S(X).
Proof : (1) Let dist(C1,C2) = δ > 0. Let K2 = C2 + Bδ/2B(0). Then C1 and K2 are disjoint closed convex sets and K2 has non-empty interior. Now, ∗ f ∈ X that separates C1 and K2 strictly separates C1 and C2. Thus (1) follows from Theorem 2.5. (2) Observe that {xˆ : x is a denting point of B(X)} = {w*-denting points of B(X∗∗)}. Now (2) is immediate from Theorem 2.5. Remark : Since disjoint closed balls always have positive distance, this result cannot be strengthened. Note that this Theorem and Corllary 2.3.2 considerably strengthen Corollaries on p 341 and p 343 respectively of [56]. In the following proposition, we collect without proof some observations regarding the MIP. Recall (from [22]) that a norming subspace of X∗ is minimal if it is contained in all norming subspaces.
21 Proposition 2.7 For a Banach space X (a) If the norm is Fr´echetdifferentiable at all x ∈ S(X), then X has the MIP and is Asplund. If the norm is smooth at all x ∈ S(X), then X has the CI. (b) [42] If X is finite dimensional, then X has the MIP (equivalently, the CI) if and only if the extreme points of B(X∗) are dense in S(X∗). In particular, a 2-dimensional space has the MIP if and only if it is smooth. (c) [42] If the w*-strongly exposed points of B(X∗) are dense in S(X∗), X has the MIP. (d) [21] If X has the MIP and the norm is Fr´echetdifferentiable on a dense subset of S(X), then the w*-strongly exposed points of B(X∗) are dense in S(X∗). In particular, this happens when X has the MIP and either X∗ has the w*-MIP or X is Asplund. (e) [51] If X is separable and has the MIP, then X is Asplund. (f) If X has the F -MIP, then F is the minimal norming subspace of X∗. Hence, if X has the MIP, then X∗ contains no proper norming subspace. In particular, X∗ has the MIP implies X is reflexive. (g) X has the MIP implies X∗∗ has the w*-MIP.
Remarks : 1. Apropos (d) above, does X has the MIP imply X is Asplund ? In this context, Sullivan [51] has shown that if for some ε > 0,
Mε = S(X), then X is Asplund. 2. As a strengthening of (f) above, he has also proved that if X is the range of a norm one projection in a dual space and for some 0 < ε < 1, ∗ D(Mε) is dense in S(X ), then X is reflexive. 3. On the other hand, Godefroy and Kalton [22] have proved that if X is the range of a norm one projection in a dual space and every closed bounded convex set in X is the intersection of finite union of balls, then X is reflexive. 4. Does the minimal norming subspace of the dual, if it exists, always satisfy our standing assumption of this section ? Does there exist a subspace
22 F of X∗ satisfying our standing assumption that is minimal with respect to this property ? 5. Is the converse of (g) above true ?
2.3. A Digression
In [21], a version of Lemma 2.4 was proved for X = Y ∗, F = Yb using the Bollob´as’estimates for the Bishop-Phelps Theorem (see [5] and [6]). Specifically, the authors of [21] used the fact that in this case the following holds : For every x ∈ S(X) and every sequence {fn} ⊆ S(F ) such (∗) that fn(x) −→ 1, there exists a sequence {xn} ⊆ T and fxn ∈ D(xn) ∩ S(F ) such that kxn − xk → 0 and kfxn − fnk → 0.
In fact, one can show that in this situation, the following stronger property holds (see [20]) :
2 For every x ∈ S(X), f ∈ S(F ) and ε > 0 with f(x) > 1 − ε (∗∗) there exists y ∈ T and fy ∈ D(y) ∩ S(F ) such that kx − yk ≤ ε and kf − fyk ≤ ε.
Using the fact that (∗∗) holds for F = X∗, one can show that (∗∗) also holds if F is an L-summand in X∗, i.e., there is a projection P on X∗ with P (X∗) = F such that for any f ∈ X∗, kfk = kP fk + kf − P fk. Here is a quick proof. Let x ∈ S(X), f ∈ S(F ) and 0 < ε < 1 be given such that f(x) > 1 − ε2. Then there exists y ∈ S(X) and y∗ ∈ D(y) such that kx − yk ≤ ε and kf − y∗k ≤ ε. Note that kf − P y∗k ≤ kf − y∗k ≤ ε < 1, so that P y∗ 6= 0. If y∗ = P y∗, we are done. Otherwise observe that
1 = ky∗k = kP y∗k + ky∗ − P y∗k and 1 = y∗(y) = P y∗(y) + (y∗ − P y∗)(y) P y∗ ! y∗ − P y∗ ! = kP y∗k (y) + ky∗ − P y∗k (y) ≤ 1 kP y∗k ky∗ − P y∗k
23 Thus, P y∗/kP y∗k ∈ D(y) ∩ S(F ), whence y ∈ T . Further
P y∗ ∗ ∗ f − ≤ kf − P y k + (1 − kP y k) kP y∗k = kf − P y∗k + ky∗ − P y∗k = kf − y∗k ≤ ε
Clearly, if (∗∗) holds for the pair (X,F ), it also holds for the pair (F, Xc). In particular, (∗∗) holds for each of the following : (1) X = C[0, 1], F = {discrete measures on [0, 1]}, (2) X = C[0, 1], F = {absolutely continuous measures on [0, 1]}, (3) X = L1[0, 1], F = C[0, 1]. So, in these cases, (∗) also holds and the proof of [21] can be used to prove Lemma 2.4. Another example of a situation in which (∗∗) is satisfied is furnished by Proposition 3.4 in the next chapter. However, one can construct examples (see below) to show that (∗∗) does not, in general, follow from the density of T in S(X). It would be interesting to know whether (∗) does. Also it would be interesting to find general suffi- ciency conditions for (∗∗) to hold which would cover at least the case X = Y ∗ and F = Yb . In particular, is the following obviously necessary condition also sufficient for (∗∗) to hold :
T is dense in S(X) and D(T ) ∩ S(F ) is dense in S(F ) ?
But these may be difficult problems. Example : Let X be a non-reflexive Banach space. Let F ⊆ X∗ be a norming subspace which is an L-summand in X∗. Let P be the corresponding ∗ L-projection. Let fo ∈ (I −P )(X ) such that kfok = 1 and fo does not attain ∗ its norm on B(X). Let F1 = F ⊕1 IRfo. Then F1 is a norming subspace of X 1 and fo ∈ S(F1). Let 0 < ε < 2 . Suppose there exists x ∈ S(X), g ∈ S(F1) such that kfo − gk < ε and g(x) = 1.
Now, g = f + αfo for some f ∈ F , α ∈ IR. We have
1 = kgk = kfk + kαfok = kfk + |α|
24 If α = 0, g = f and we have
ε > kfo − gk ≥ kP fo − P gk = kgk = 1
So, α 6= 0. Also, f = 0 implies g = αfo and so fo(x) = ±1, a contradiction as fo does not attain its norm. Thus, f 6= 0. But then,
f αx! 1 = g(x) = f(x) + αf (x) = kfk · (x) + |α|f ≤ kfk + |α| = 1 o kfk o |α|
This implies fo(αx/|α|) = 1. Again, a contradiction.
As noted earlier, the pair (X,F ) satisfies (∗∗), so TF is dense in S(X) and D(TF ) ∩ S(F ) is dense in S(F ). Now, clearly TF1 ⊇ TF , but the above shows D(TF1 ) ∩ S(F1) is not dense in S(F1). Consequently, though TF1 is dense in S(X), (∗∗) is not satisfied.
Also, interchanging the roles of X and F1, the above shows that though Xc is a norming subspace of F ∗, T = D(T ) ∩ S(F ) is not dense, i.e., our 1 Xb F1 1 standing assumption is not satisfied. Finally, we note that X = C[0, 1], F = {discrete measures on [0, 1]} and fo = λ|[0,1/2] − λ|[1/2,1] satisfies the hypothesis of the above example, where λ denotes the Lebesgue measure.
25 Chapter 3 Bochner Lp Spaces and The `p Sums of Banach Spaces with the MIP
3.1. The `p Sums
That the existence of an equivalent renorming with the MIP or the CI is p stable under co or ` sums for 1 < p < ∞ was established by A. Sersouri [47] using a renorming result for the MIP by R. Deville [9] and for the CI by himself [47]. We can, however, prove the following direct result for the `p sums, 1 < p < ∞.
Theorem 3.1 Let {Xα : α ∈ Γ} be a family of Banach spaces. Then the p space X = (⊕α∈ΓXα)`p with the usual ` -norm (1 < p < ∞) has the MIP
(resp. the CI) if and only if for all α ∈ Γ, the space Xα has the MIP (resp. the CI).
∗ ∗ 1 1 q Recall that if X is as above, X = (⊕α∈ΓXα)` , where p + q = 1. The following lemma is well-known :
Lemma 3.2 [50, p 154] Let {Xα} and X be as above. A point x = (xα) ∈
S(X) is an extreme point of B(X) if and only if for all α ∈ Γ, either xα = 0 or xα/kxαk is an extreme point of B(Xα).
[50] also contains similar characterisations of various other types of ex- treme points of B(X). In the dual situation, we can prove
26 ∗ ∗ ∗ Lemma 3.3 Let {Xα} and X be as above. A point x = (xα) ∈ S(X ) is ∗ ∗ a w*-denting point of B(X ) if and only if for every α, either xα = 0 or ∗ ∗ ∗ xα/kxαk is a w*-denting point of B(Xα).
We postpone the proof of Lemma 3.3 till the next section where similar results will be proved in a more general setting.
Proof of Theorem 3.1 : (The MIP) : Let for all α ∈ Γ,Xα have the MIP. ∗ ∗ ∗ ∗ Let x = (xα) ∈ S(X ). Fix ε > 0. Let Γo = {α ∈ Γ: xα 6= 0}. For α ∈ Γo, ∗ ∗ ∗ ∗ xα/kxαk ∈ S(Xα) and since Xα has the MIP, there exists yα, a w*-denting ∗ ∗ ∗ xα ∗ ∗ point of B(X ), such that y − ∗ < ε. Define z = (z ) by α α kxαk α
∗ ∗ ∗ kxαkyα if α ∈ Γo zα = 0 otherwise
∗ ∗ ∗ ∗ By Lemma 3.3, z is a w*-denting point of B(X ) and kx − z kq < ε. ∗ ∗ Conversely, if X has the MIP, let αo ∈ Γ and xαo ∈ S(Xαo ). Fix 0 < ε < 1. ∗ ∗ The point x = (xα) defined by
∗ ∗ xαo if α = αo xα = 0 otherwise is in S(X∗) and hence there exists y∗, a w*-denting point of B(X∗), such ∗ ∗ that kx − y kq < ε. ∗ ∗ ∗ Clearly, kxαo − yαo k < ε < 1 and so, yαo 6= 0. Also, by Lemma 3.3, ∗ ∗ ∗ yαo /kyαo k is a w*-denting point of B(Xαo ). Now,
y∗ x∗ − αo ≤ kx∗ − y∗ k + |1 − ky∗ k | < ε + | kx∗ k − ky∗ k | αo ∗ αo αo αo αo αo kyαo k ∗ ∗ < ε + kxαo − yαo k < 2ε
Hence, Xαo has the MIP.
(The CI ) : Let for all α ∈ Γ, Xα have the CI. ∗ ∗ ∗ Let x = (xα) ∈ S(X ). Let K be a compact set in X. We may assume K ⊆ B(X).
Let πα : X −→ Xα be the co-ordinate projection. Then Kα = πα(K) is a compact set in Xα, for all α ∈ Γ. Fix ε > 0. Choose Γo ⊆ Γ, Γo finite, such
27 ∗ ∗ ∗ that kx χΓ\Γo kq < ε and for all α ∈ Γo, xα 6= 0. For α ∈ Γo, there exists yα, ∗ ∗ ∗ ∗ q an extreme point of B(Xα) and aα ≥ 0 such that kxα − aαyαkKα < εkxαk , ∗ ∗ ∗ ∗ where kzαkKα = sup{|zα(kα)| : kα ∈ Kα}. Define u = (uα) by
∗ ∗ aαyα if α ∈ Γo uα = 0 otherwise
∗ ∗ ∗ ∗ Clearly, uα/kuαk = yα is an extreme point of B(Xα). So, by Lemma 3.2, ∗ ∗ ∗ u /ku kq is an extreme point of B(X ) and for any k = (kα) ∈ K, we have
∗ ∗ X ∗ ∗ |(x − u )(k)| = | (xα − uα)(kα)| α∈Γ X ∗ ∗ X ∗ ≤ |(xα − aαyα)(kα)| + |xα(kα)| α∈Γo α6∈Γo X ∗ ∗ X ∗ q 1/q X p 1/p ≤ kxα − aαyαkKα + ( kxαk ) ( kkαk ) α∈Γo α6∈Γo α6∈Γo X ∗ q X p 1/p ∗ q ≤ ε kxαk + ε( kkαk ) ≤ ε[kx kq + kkkp] ≤ 2ε α∈Γo α6∈Γo
∗ ∗ ∗ ∗ since x ∈ S(X ) and k ∈ K ⊆ B(X). Hence kx − u kK ≤ 2ε. ∗ ∗ Conversely, let X have the CI. Let αo ∈ Γ, xαo ∈ S(Xαo ), Kαo ⊆ Xαo ∗ ∗ compact. Define x ∈ S(X ) as in the case of the MIP. Define K = {(xα): xαo ∈ Kαo and xα = 0 for α 6= αo}. Clearly, K is compact and for any ∗ ∗ ∗ ∗ z ∈ X , kz kK = kzαo kKαo .
Now, the CI in Xαo clearly follows from that in X. Remark : It follows from Proposition 4.1 of the next chapter that if the ∞ co or the ` sum of a family of Banach spaces has the MIP (or the CI), then each of them has the MIP (the CI). The converse is not true in general as ∞ the finite-dimensional spaces `n , n ≥ 1, fail the CI. See also Proposition 4.2 below.
3.2. Bochner Lp Spaces
Let X be a Banach space, (Ω, Σ, µ) a measure space, 1 < r < ∞. Let A ⊆ S(X). Define
28 n r X Mr(A) = {f ∈ S(L (µ, X)) : f is of the form f = xiχEi with n ≥ 1 i=1 arbitrary, Ei ∈ Σ and xi/kxik ∈ A for all i = 1, 2, . . . , n} ∞ r X Pr(A) = {f ∈ S(L (µ, X)) : f is of the form f = xiχEi with Ei ∈ Σ i=1 and xi = 0 or xi/kxik ∈ A for all i ≥ 1} and, r Nr(A) = {f ∈ S(L (µ, X)) : for almost all t ∈ Ω, either f(t) = 0 or f(t)/kf(t)k ∈ A}
r Note that Mr(A) ⊆ Pr(A) ⊆ Nr(A) ⊆ S(L (µ, X)). The following results are well-known :
(1) If A = {extreme points of B(X)} then Nr(A) ⊆ {extreme points of B(Lr(µ, X))}. (See, e.g., [52]).
(2) If A = {strongly exposed points of B(X)} then Nr(A) ⊇ {strongly r exposed points of B(L (µ, X))} ⊇ Mr(A) ([32, 23, 24]).
(3) If A = {denting points of B(X)} then Nr(A) = {denting points of B(Lr(µ, X))} [37]. A survey of similar results may be found in [50]. In the dual situation, we note that : ∗ (1) If A = {extreme points of B(X )} then Nq(A) ⊆ {extreme points of B(Lq(µ, X∗))}. In the following we prove a stronger assertion.
∗ (2) Any w*-denting point or w*-strongly exposed point of B(Vq(µ, X )), being a w*-PC, necessarily belongs to Lq(µ, X∗)) (see Proposition 3.4 below). And it is implicit in [32] that if A = {w*-strongly exposed ∗ points of B(X )} then Mq(A) ⊆ {w*-strongly exposed points of B(Lq(µ, X∗))}. Below, we prove by different methods, a similar result ∗ for w*-denting points of B(Vq(µ, X )), the proof being more involved than that for w*-strongly exposed points. But, first we note the following :
p q ∗ 1 Proposition 3.4 Let Y = L (µ, X) and F = L (µ, X ), 1 < p < ∞, p + 1 q = 1. Then every simple function in S(Y ) is in T , where T = {x ∈ Y : D(x) ∩ S(F ) 6= ∅}. So, T is dense in S(Y ), F is norming and we are in the
29 set-up of section 2.2. Moreover, the pair (Y,F ) satisfies (∗∗) as defined in section 2.3.
Pn Proof : Let x = i=1 xiχEi be any simple function in S(Y ) and φ : S(X) −→ S(X∗) be any support mapping. Define
n ∗ ! X p−1 xi Φ(x) = kxik φ ∗ χEi i=1 kxi k
q ∗ ∗ Then, Φ(x) ∈ S(L (µ, X )) ⊆ S(Vq(µ, X )) and hx, Φ(x)i = 1. Thus, Φ(x) ∈ D(x) ∩ S(F ). This proves the first part of the Proposition. Now, let x ∈ S(Lp(X)), f ∈ S(Lq(X∗)) and ε > 0 be such that f(x) > 1− ε2. Choose 0 < η < ε such that 0 < η[2(ε+1)−η] < f(x)−(1−ε2). Let z and g be simple functions in S(Lp(X)) and S(Lq(X∗)) respectively such that kx− zkp < η and kf −gkq < η. Refining the partitions if necessary, we may assume Pn that there is a finite partition {E1,E2,...,En} of Ω such that z = i=1 ziχEi Pn ∗ ∗ ∗ Pn ∗ and g = i=1 zi χEi where zi ∈ X, zi ∈ X . Now, g(z) = i=1 zi (zi)µ(Ei) > f(x) − 2η > 1 − (ε − η)2, by the choice of η. Now, consider the discrete 0 measure space Ω = {1, 2, . . . , n} with measure P , where P (i) = µ(Ei). Then z and g can be isometrically identified with elements of S(Lp(P,X)) and S(Lq(P,X∗)) respectively. But as P is discrete, Lp(P,X)∗ = Lq(P,X∗) and ∗ ∗ ∗ so (∗∗) is satisfied, i.e., there exists vectors (y1, y2, . . . , yn) and (y1, y2, . . . , yn) p q ∗ Pn ∗ in S(L (P,X)) and S(L (P,X )) respectively such that i=1 yi (yi)P (i) = 1 Pn p 1/p Pn ∗ ∗ q 1/q and [ i=1 kzi − yik P (i)] ≤ (ε − η) and [ i=1 kzi − yi k P (i)] ≤ (ε − η). Pn Pn ∗ p Put y = i=1 yiχEi and fy = i=1 yi χEi . Then y ∈ S(L (µ, X)), fy ∈ q ∗ S(L (µ, X )) and fy(y) = 1. Further kx − ykp ≤ (ε − η) + η = ε and kf − fykq ≤ ε.
The CI
Lemma 3.5 Let X be a Banach space, (Ω, Σ, µ) a measure space and 1 < q < ∞. Let A = {extreme points of B(X∗)} ⊆ S(X∗). Then ∗ ∗ (a) Mq(A) = Mq(S(X ))∩ {extreme points of B(Vq(µ, X ))}.
∗ ∗ (b) Pq(A) = Pq(S(X ))∩ {extreme points of B(Vq(µ, X ))}.
30 Proof : We prove (a), the proof of (b) being an easy adjustment. Pn ∗ Let F = i=1 xi χAi ∈ Mq(A). Suppose that there exist G1,G2 ∈ ∗ 1 B(Vq(µ, X )) with F = 2 (G1 + G2). We will prove that F = G1 = G2, i.e., for any E ∈ Σ with µ(E) > 0, n F (E) = G1(E) = G2(E). Fix any such E ∈ Σ. Put Ao = Ω \ (∪i=1Ai).
Then π = {Ao,A1,...,An} is a partition of Ω. For i = 0, 1, . . . , n, define c ∗ Ei1 = E ∩ Ai and Ei2 = E ∩ Ai. Put xo = 0.
Now, for any i = 0, 1, . . . , n and j = 1, 2 with µ(Eij) 6= 0, we have ∗ 1 xi µ(Eij) = F (Eij) = 2 [G1(Eij) + G2(Eij)] or, " # ∗ 1 G1(Eij) + G2(Eij) xi = 2 µ(Eij) So, " # ∗ 1 kG1(Eij)k + kG2(Eij)k kxi k ≤ 2 µ(Eij) and hence " q q # ∗ q 1 kG1(Eij)k + kG2(Eij)k kxi k ≤ q 2 µ(Eij) by the convexity of the map t −→ tq (q > 1). But then
n q X ∗ q 1 = kF kq = kxi k µ(Ai) i=0 n 2 X X ∗ q X ∗ q = kxi k µ(Eij) = kxi k µ(Eij) i=0 j=1 {(i,j):µ(Eij )6=0} q q 1 X kG1(Eij)k X kG2(Eij)k ≤ q µ(Eij) + q µ(Eij) 2 µ(Eij) µ(Eij) µ(Eij )6=0 µ(Eij )6=0 1 ≤ [kG kq + kG kq] ≤ 1 2 1 q 2 q So, we must have
" q q # ∗ q 1 kG1(Eij)k + kG2(Eij)k kxi k = q for all i, j with µ(Eij) 6= 0 2 µ(Eij) Then the strict convexity of the map t −→ tq gives
∗ kG1(Eij)k kG2(Eij)k kxi k = = whenever µ(Eij) 6= 0 µ(Eij) µ(Eij)
31 Thus,
∗ " # xi 1 G1(Eij) G2(Eij) ∗ = + whenever µ(Eij) 6= 0 kxi k 2 kG1(Eij)k kG2(Eij)k
∗ ∗ Now, the extremality of xi /kxi k implies
∗ xi G1(Eij) G2(Eij) ∗ = = kxi k kG1(Eij)k kG2(Eij)k whence ∗ G1(Eij) G2(Eij) xi = = whenever µ(Eij) 6= 0 µ(Eij) µ(Eij) So, n n X ∗ X F (E) = xi µ(Ei1) = G1(Ei1) = G1(E) i=0 i=0
Similarly, F (E) = G2(E). The reverse inclusion being obvious, this completes the proof.
Theorem 3.6 For any Banach space X, any finite non-atomic measure space (Ω, Σ, µ) and 1 < p < ∞, the space Lp(µ, X) has the CI.
∗ Proof : Let τ denote the topology on Vq(µ, X ) of uniform convergence on compact subsets of Lp(µ, X). By Proposition 3.4, B(Lq(µ, X∗)) is w*-dense ∗ q ∗ ∗ in B(Vq(µ, X )), so L (µ, X ) is τ-dense in Vq(µ, X ), and simple functions are norm dense in Lq(µ, X∗). Thus it suffices, by Lemma 3.5(a), to show that given any simple function F ∈ S(Lq(µ, X∗)), any compact subset K p of L (µ, X) and any ε > 0, there exists a function F1 ∈ Mq(A) such that ∗ ∗ kF − F1kK < ε, where A = {extreme points of B(X )} ⊆ S(X ). Now, since K is compact and simple functions are dense in Lp(µ, X), m there exists simple functions {g1, g2, . . . , gm} such that K ⊆ ∪i=1Bε/4(gi).
Suppose we have found a function F1 ∈ Mq(A) such that |(F −F1)(gj)| < ε/2 for all j = 1, 2, . . . , m. Then for any g ∈ K, there exists gj such that kg − gjkp < ε/4. So, |(F − F1)(g)| ≤ |(F − F1)(gj)| + |(F − F1)(g − gj)| ≤ 1 1 1 2 ε + (kF kq + kF1kq) · kg − gjkp ≤ 2 ε + 2 · 4 ε = ε, since kF kq = kF1kq = 1. Therefore we may as well assume K is a finite set {g1, g2, . . . , gm}. Fur- ther, passing to finer partitions if necessary, we may assume there exists a
32 partition A1,A2,...,An of Ω such that µ(Ai) > 0 for all i = 1, 2, . . . , n and each of the functions g1, g2, . . . , gm and F take constant values on each Ai, Pn ∗ i.e., the functions g1, g2, . . . , gm and F have the form F = i=1 αixi χAi with ∗ ∗ Pn xi ∈ S(X ) for all i = 1, 2, . . . , n and gj = i=1 xijχAi for all j = 1, 2, . . . , m, ∗ where some of the αi’s and xij’s may be zero and the αixi ’s and xij’s need not all be distinct. Then, n X ∗ F (gj) = αixi (xij)µ(Ai) for all j = 1, 2, . . . , m i=1 Pn Now, µ is finite implies kF k1 = i=1 |αi|µ(Ai) < ∞. Fix 0 < η < ε/kF k1. Since B(X∗) is w*-compact and convex, by the Krein-Milman Theorem we have :
for each i = 1, 2, . . . , n, there exists λik ≥ 0, k = 1, 2,...,N with PN ∗ ∗ ∗ ∗ k=1 λik = 1 and xi1, xi2, . . . , xiN , extreme points of B(X ) such that N ∗ X ∗ |(xi − λikxik)(xrs)| < η for all r = 1, 2, . . . , n, s = 1, 2, . . . , m. k=1
Since µ is non-atomic, for each i = 1, 2, . . . , n, there is a partition {Ai1,
Ai2,...,AiN } of Ai such that µ(Aik) = λikµ(Ai) for all k = 1, 2,...,N. Define n N X X ∗ F1 = αixikχAik i=1 k=1 Then n N n q X q X X q q kF1kq = |αi| µ(Aik) = |αi| µ(Ai) = kF kq = 1 i=1 k=1 i=1 ∗ and since each xik ∈ A, F1 ∈ Mq(A). Further, for all j = 1, 2, . . . , m
n N n N X X ∗ X X ∗ F1(gj) = αixik(xij)µ(Aik) = αixik(xij)λikµ(Ai) i=1 k=1 i=1 k=1 n N X X ∗ = αiµ(Ai)( λikxik)(xij) i=1 k=1 and so,
n N X ∗ X ∗ |(F − F1)(gj)| = | αiµ(Ai)[xi − λikxik](xij)| i=1 k=1
33 n N X ∗ X ∗ ≤ |αi| · µ(Ai) · |[xi − λikxik](xij)| i=1 k=1 n X ≤ η |αi|µ(Ai) = ηkF k1 < ε i=1
Remarks : 1. This shows that the CI is indeed much weaker than smoothness, as Lp(µ, X) is smooth if and only if X is (see [36]). 2. If the existence of an equivalent CI renorming is hereditary, it would follow from Theorem 3.6 that every Banach space admits an equivalent CI renorming. This was also noted by Sersouri [47], although in a different context.
The MIP
Coming to the question of the MIP, we need the following well-known result :
Lemma 3.7 [50, p 158] Let (Ω, Σ, µ) be a measure space, X a Banach space p and 1 ≤ p < ∞. For {fn} and f in L (µ, X), if kfnkp −→ kfkp and fn −→ f a.e. [µ], then kfn − fkp −→ 0.
Lemma 3.8 Let X be a Banach space, (Ω, Σ, µ) a measure space and 1 1 ∗ ∗ 1 < p < ∞, p + q = 1. Let A = {w*-denting points of B(X )} ⊆ S(X ). ∗ ∗ Then Pq(A) = Pq(S(X ))∩ {w*-denting points of B(Vq(µ, X ))}.
P∞ ∗ Proof : Let g = i=1 xi χEi ∈ Pq(A). Then by Lemma 3.5(b), g is an ∗ extreme point of B(Vq(µ, X )). So, by Lemma 1.6, it suffices to show that g q ∗ q ∗ is a w*-PC of B(L (µ, X )), i.e., to prove that if {gα} ⊆ B(L (µ, X )) is a w∗ net such that gα −→ g then kgα − gkq −→ 0. ∗ ∗ ∞ w w Put Eo = Ω \ ∪i=1Ei. Then gαχEo −→ gχEo = 0 and gαχΩ\Eo −→ gχΩ\Eo = g. Now, by the w*-lower semicontinuity of the norm we have :
1 = kgkq ≤ lim inf kgαkq ≤ lim sup kgαkq ≤ 1 α α and hence
lim kgαkq = 1 = kgkq α
34 Similarly, we have
lim kgαχΩ\E kq = 1 α o Now, q q q kgαkq = kgαχEo kq + kgαχΩ\Eo kq so that q q q lim sup kgαχEo kq = lim(kgαkq − kgαχΩ\Eo kq) = 0 α α
Thus, it suffices to show that kgαχΩ\Eo − gkq = 0. So, we may assume without loss of generality that (Ω, Σ, µ) is σ-finite, and for almost all t, g(t) 6= 0. P∞ ∗ q Fix ε > 0, choose N ≥ 1 such that i=N+1 kxi k µ(Ei) < ε. Put F = N ∪i=1Ei. We have,
q q q kgχF kq ≤ lim inf kgαχF kq ≤ lim sup kgαχF kq α α and, q q q kgχΩ\F kq ≤ lim inf kgαχΩ\F kq ≤ lim sup kgαχΩ\F kq α α Suppose q q kgχF kq < lim sup kgαχF kq α
Then there exists a subnet {gβα } of {gα} such that
q q q lim kgβα χF kq = lim sup kgαχF kq and lim kgβα χΩ\F kq exists. α α α Then
q q q q q 1 = kgαχF k + kgαχΩ\F k < lim kgβ χF k + lim kgβ χΩ\F k = lim kgβ k ≤ 1 q q α α q α α q α α q a contradiction. So, we have,
q q kgχF k = lim kgαχF k q α q
Similarly, q q kgχΩ\F k = lim kgαχΩ\F k q α q
35 Now,
q q q kgα − gkq = k(gα − g)χF kq + k(gα − g)χΩ\F kq q q ≤ k(gα − g)χF kq + [kgαχΩ\F kq + kgχΩ\F kq] q q−1 q q ≤ k(gα − g)χF kq + 2 [kgαχΩ\F kq + kgχΩ\F kq]
(by convexity of the map t → tq).
So, if k(gα − g)χF kq −→ 0, and ε > 0 is given, we can choose αo such that for all α ≥ αo,
q q q k(gα − g)χF kq < ε and kgαχΩ\F kq < kgχΩ\F kq + ε
q q−1 i.e., we have for α ≥ αo, kgα − gkq ≤ ε + 2 · 3ε.
Thus, it suffices to prove k(gα − g)χF kq −→ 0. So, again, without loss Pn ∗ of generality, we may assume µ is finite and g is of the form i=1 xi χEi ∈ q ∗ S(L (µ, X )) with {E1,E2,...,En} a partition of Ω.
Suppose now, that kgα − gkq 6→ 0. Then there exists εo > 0 and a subnet
{gβα } with βα ≥ α for all α, such that kgβα −gkq ≥ εo for all α. For notational ∗ q ∗ w simplicity, put Gα = gβα . Then Gα ∈ B(L (µ, X )) for all α, Gα −→ g and kGα − gkq ≥ εo for all α. As noted earlier,
lim kGαkq = 1 = kgkq (1) α
Fix A ∈ Σ, µ(A) > 0. Put Ai1 = A ∩ Ei and Ai2 = Ei \ Ai1. Fix Aij p such that µ(Aij) > 0. For any x ∈ X, xχAij ∈ L (µ, X), so, hxχAij ,Gαi −→ hxχ , gi, i.e., hx, R G dµi −→ hx, x∗iµ(A ). This implies Aij Aij α i ij
Z 1 w∗ ∗ Gαdµ −→ xi µ(Aij) Aij whence Z Z ∗ 1 1 kx k ≤ lim inf k Gαdµk ≤ lim inf kGαkdµ i α α µ(Aij) Aij µ(Aij) Aij " Z #1/q 1 q ≤ lim inf kGαk dµ (2) α µ(Aij) Aij
36 Thus, we have Z ∗ q q kx k µ(Aij) ≤ lim inf kGαk dµ for all i, j i α Aij Adding, we get
n 2 n 2 Z q X X ∗ q X X q 1 = kgk = kx k µ(Aij) ≤ lim inf kGαk dµ q i α i=1 j=1 i=1 j=1 Aij n 2 Z X X q q ≤ lim inf kGαk dµ = lim inf kGαk ≤ 1 α α q i=1 j=1 Aij So, Z ∗ q q kx k µ(Aij) = lim inf kGαk dµ for all i, j i α Aij
Suppose, for some io, jo, we have Z ∗ q q kxio k µ(Aiojo ) < lim sup kGαk dµ α Aiojo Then, Z Z ∗ q q X q kxio k µ(Aiojo ) < lim sup kGαk dµ − kGαk dµ α Ω Aij (i,j)6=(io,jo) Z X q = 1 − lim inf kGαk dµ α Aij (i,j)6=(io,jo) Z X q ≤ 1 − lim inf kGαk dµ α Aij (i,j)6=(io,jo) X ∗ q = 1 − kxi k µ(Aij) (i,j)6=(io,jo) making n 2 q X X ∗ q 1 = kgkq = kxi k µ(Aij) < 1 i=1 j=1 So, we must have Z ∗ q q kx k µ(Aij) = lim kGαk dµ for all i, j i α Aij Now, by (2) above and by the lim sup-version of (2) we have that for all i, j, Z ∗ kx kµ(Aij) = lim kGαkdµ i α Aij
37 Thus, Z n n Z Z X ∗ X kg(t)kdµ = kx kµ(Ai1) = lim kGα(t)kdµ = lim kGα(t)kdµ i α α A i=1 i=1 Ai1 A Since A ∈ Σ was arbitrary with µ(A) > 0, we have that for all A ∈ Σ with µ(A) > 0, Z Z lim kGα(t)kdµ = kg(t)kdµ α A A ∗ q w q whence, in L (µ, IR), kGα(·)k −→ kg(·)k. Now, since the space L (µ, IR) is uniformly convex and kg(·)k ∈ S(Lq(µ, IR)), kg(·)k is a w*-denting point of B(Lq(µ, IR)) and hence a w*-PC. So, we have Z q | kGα(t) − kg(t)k | dµ −→ 0 (3) Ω
Again, fix Aij such that µ(Aij) > 0. Then Z 1 w∗ ∗ Gαdµ −→ xi µ(Aij) Aij and Z 1 ∗ kGαkdµ −→ kxi k µ(Aij) Aij so R ∗ Gαdµ ∗ x Aij −→w i R kG kdµ kx∗k Aij α i ∗ ∗ ∗ Since xi /kxi k is a w*-PC of B(X ), this means
∗ R G dµ ∗ R G dµ xi Aij α xi Aij α − −→ 0 or − −→ 0 (as α −→ ∞) kx∗k R kG kdµ kx∗k µ(A )kx∗k i Aij α i ij i
Choose a net {εα} such that εα > 0 for all α and εα −→ 0. We have
Z G (t) Z G (t) α α dµ − ∗ dµ Aij kGα(t)k + εα Aij kxi k " # Z kx∗k − kG (t)k − ε i α α = Gα(t) ∗ dµ Aij kxi k(kGα(t)k + εα) Z 1 ∗ ≤ ∗ | kxi k − kGα(t)k − εα|dµ kxi k Aij ( Z )1/q 1 1 q ≤ ∗ µ(Aij) | kg(t)k − kGα(t)k | dµ + µ(Aij)εα kxi k µ(Aij) Aij ( Z )1/q µ(Aij) 1 q ≤ ∗ | kg(t)k − kGα(t)k | dµ + εα kxi k µ(Aij) Aij
38 Now, by (3) and as εα −→ 0, we conclude that
Z G (t) Z G (t) α α dµ − ∗ dµ −→ 0 Aij kGα(t)k + εα Aij kxi k whence x∗ 1 Z G (t) i α ∗ − dµ −→ 0 kxi k µ(Aij) Aij kGα(t)k + εα ∗ ∗ ∗ Now, xi /kxi k is a w*-denting, and hence denting, point of B(X ), so by Theorem 1.5(d), we get
1 Z x∗ G (t) i α ∗ − dµ −→ 0 (4) µ(Aij) Aij kxi k kGα(t)k + εα Now, Z ∗ kxi − Gα(t)kdµ Aij
Z x∗ G (t) G (t) G (t) ∗ i α α α = kxi k ∗ − + − ∗ dµ Aij kxi k kGα(t)k + εα kGα(t)k + εα kxi k " Z x∗ G (t) ∗ i α ≤ kxi k ∗ − dµ Aij kxi k kGα(t)k + εα Z ∗ # kGα(t)k | kxi k − kGα(t)k | + εα + · ∗ dµ Aij kGα(t)k + εα kxi k The first term of the sum goes to zero by (4) and the second by (3). Hence we have Z Z ∗ kxi − Gα(t)kdµ −→ 0 or kg(t) − Gα(t)kdµ −→ 0 Aij Aij
Now, since Aij and A were arbitrary, we have actually proved that Z kg(t) − Gα(t)kdµ −→ 0 Ω
Using this fact and (1), we can choose {αn} such that both Gαn (t) → g(t) a.e. [µ] and kGαn kq → kgkq. Hence, we have, by Lemma 3.7, kGαn −gkq → 0.
But this contradicts the choice of {Gα}. The converse follows immediately from Lemma 1.6, Lemma 3.5 and the ∗ P∞ ∗ ∗ ∗ w ∗ ∗ observation that if g = i=1 xiχEi and {yα} ⊆ B(X ), yα −→ xk/kxkk for ∗ def P ∗ ∗ ∗ ∗ w some k then gα = i6=k xi χEi + kxkkyαχEk ∈ B(Vq(µ, X )) and gα −→ g.
39 Proof of Lemma 3.3 : In the set-up of Lemma 3.8, put X = (⊕α∈ΓXα)`p , Ω = Γ, Σ = Power set of Γ and µ = counting measure. Then Lp(µ, X) = `p(X) and Lp(µ, X)∗ = `q(X∗) = Lq(µ, X∗). Now,
p X X p ` (X) = {((xαβ)α∈Γ)β∈Γ : kxαβk < ∞} α∈Γ β∈Γ
p Identify X with the subspace {((xαδαβ)α∈Γ)β∈Γ} ⊆ ` (X), where 0 if α 6= β δαβ = 1 if α = β
∗ ∗ q ∗ Then X gets identified with the subspace {((xαδαβ)α∈Γ)β∈Γ} ⊆ ` (X ). Ob- ∗ ∗ serve that with this identification, a net {xρ} ⊆ B(X ) is w*-convergent to ∗ ∗ ∗ q ∗ a point xo ∈ S(X ) if and only if the net {xρ} ⊆ B(` (X )) is w*-convergent ∗ q ∗ to xo ∈ S(` (X )). Now, Lemma 3.3 follows immediately from Lemma 3.8.
Lemma 3.9 Let X be a Banach space, (Ω, Σ, µ) a probability space and 1 < r < ∞. Let A ⊆ S(X). The following are equivalent : (a) A is norm dense in S(X).
r (b) Mr(A) is norm dense in S(L (µ, X)).
r (c) Pr(A) is norm dense in S(L (µ, X)).
r (d) Nr(A) is norm dense in S(L (µ, X)).
Proof : Since simple functions in S(Lr(µ, X)) are norm dense (a) implies
(b), and since Mr(A) ⊆ Pr(A) ⊆ Nr(A), (b) =⇒ (c) =⇒ (d). r Now, if x ∈ S(X), xχΩ ∈ S(L (µ, X)). So, by (d), there exists a se- quence {fn} ⊆ Nr(A) such that kfn − xχΩkr −→ 0. But then some sub- sequence {fnk } converges to xχΩ a.e. [µ]. Now, if t ∈ Ω is such that kfnk (t) − xχΩ(t)k −→ 0, then kfnk (t) − xk −→ 0. Since kxk = 1, for all sufficiently large k, kfnk (t)k= 6 0 and
f (t) nk − x −→ 0 kfnk (t)k
As fnk ∈ Nr(A), all such fnk (t)/kfnk (t)k ∈ A. Hence (a) follows.
40 Theorem 3.10 For any Banach space X, finite measure space (Ω, Σ, µ) and 1 1 1 < p < ∞ and p + q = 1, the following are equivalent : (i) X has the MIP (ii) Lp(µ, X) has the Lq(µ, X∗)-MIP.
Proof : Define Y , F and T as in Proposition 3.4. Suppose Lp(µ, X) has the F -MIP. Let A be a norm dense subset of S(X) and φ : S(X) −→ ∗ S(X ) be a support mapping. By Lemma 3.9, Mp(A) is norm dense p Pn in S(L (µ, X)). Let x = i=1 xiχEi ∈ Mp(A), define Φ(x) as in the proof of Proposition 3.4. Then, Φ can be extended to a support mapping p ∗ Φ:e S(L (µ, X)) −→ S(Vq(µ, X )). Now, Φe maps T into S(F ), Mp(A) is norm dense in S(Lp(µ, X)) and Lp(µ, X) has the F -MIP, so by Theo- rem 2.5, Φ(e Mp(A)) = Φ(Mp(A)) is norm dense in S(F ). Observe that q ∗ Mq(φ(A)) ⊇ Φ(Mp(A)), so that Mq(φ(A)) is norm dense in S(L (µ, X )). Now, Lemma 3.9 implies that φ(A) is norm dense in S(X∗). Since A was an arbitrary dense subset of S(X) and φ an arbitrary support mapping on X, Theorem 2.6 implies X has the MIP. Conversely, let X have the MIP. Let A = {w*-denting points of B(X∗)}. ∗ By Theorem 2.6, A is norm dense in S(X ). By Lemma 3.9, Mq(A) is norm q ∗ dense in S(L (µ, X )). And by Lemma 3.8, Mq(A) ⊆ {w*-denting points of ∗ p Vq(µ, X )}. Again by Theorem 2.5, L (µ, X) has the F -MIP.
Theorem 3.11 For any Banach space X, any finite measure space (Ω, Σ, µ) and 1 < p < ∞, the space Lp(µ, X) has the MIP if and only if X has the MIP and X∗ has the RNP with respect to µ.
Theorem 3.12 For any Banach space X, any finite measure space (Ω, Σ, µ) and 1 < p < ∞, X∗ has the w*-MIP if and only if Lp(µ, X)∗ has the w*-MIP.
Proof : As observed earlier, if A = {denting points of B(X)} then Np(A) = {denting points of B(Lp(µ, X))}. Now, the result is immediate from Theo- rem 2.6 and Lemma 3.9.
Corollary 3.12.1 Let X be a Banach space, let λ denote the Lebesgue mea- 1 1 sure on [0, 1], let 1 < p < ∞ and p + q . Then
41 (a) Lp(λ, X) has the CI. (b) X has the MIP if and only if Lp(λ, X) has the Lq(µ, X∗)-MIP. (c) Lp(λ, X) has the MIP if and only if X has the MIP and X is As- plund. (d) X∗ has the w*-MIP if and only if Lp(λ, X)∗ has the w*-MIP.
Remark : If X has the MIP implies X is Asplund — which is a long- standing conjecture — then we must have Lp(λ, X) has the MIP. However, in the light of our results, this would be tantamount to showing that if Lq(λ, X∗) ∗ is the minimal norming subspace of Vq(λ, X ), then it must be the whole of the latter space. But that seems unlikely.
42 Chapter 4 Miscellaneous Results
4.1. The Subspace Question
Phelps [42] noted that a two-dimensional space has the MIP if and only if it is smooth. Now, if the MIP were hereditary, a space X having the MIP would then have all its two-dimensional subspaces smooth and hence X itself would be smooth. However, Deville [9] sketches an argument to show that it is possible to construct a non-smooth norm on IR3 with the MIP and Tsarkov [53] has produced one such norm. Below, we produce a non-smooth norm on IR3 with the MIP that is much simpler than the one given by Tsarkov. Example : In IR3, consider the following set
U = {(x, y, z): |x| ≤ 1, |z| ≤ 1, y2 ≤ (1 − x2)(1 − z2)}
U is a closed bounded symmetric convex set with non-empty interior. Hence the Minkowski functional of U defines a norm on IR3 equivalent to the Euclidean norm. The norm turns out to be 1 k(x, y, z)k = [{(|x| + |z|)2 + y2}1/2 + {(|x| − |z|)2 + y2}1/2] 2 The corresponding pre-dual norm on IR3 with the above as the dual norm is given by
2 |a1| + |a2| if a ≤ |a1a3| 2 " #1/2 k(a1, a2, a3)k = (a2 + a2)(a2 + a2) 1 2 2 3 otherwise 2 a2 Observe that the surface of U is given by all those points where one of the defining inequalities is an equality, and that except for points on the
43 plane y = 0 (where the restricted norm is the `∞ norm on IR2) all the points on the surface are exposed and hence extreme points of U. So, by Phelps’ characterisation, the pre-dual norm indeed has the MIP. Moreover, the part of the sphere lying on the plane y = 0 is a square and hence, the pre-dual norm is not smooth. 1 And if we consider the subspace a2 = 0, the inherited norm is the ` norm on IR2, which clearly lacks the MIP. Observe further that the usual projection onto this subspace is a norm- 1 projection. So, neither complemented subspaces nor even those comple- mented by norm-1 projections necessarily inherit the MIP. We, however, have the following :
Proposition 4.1 If X has the MIP (resp. the CI) and a subspace Y of X is the range of an M-projection P, i.e., kxk = max{kP xk, kx − P xk} for all x ∈ X, then Y has the MIP (resp. the CI).
Proof : Let K be a closed bounded (resp. compact) convex set in Y . Since
X has the MIP (resp. the CI), there exist {xi}i∈I ⊆ X and {ri}i∈I with ri > 0 for all i ∈ I, such that K = ∩i∈I Bri [xi], where Br[x] = {z ∈ X : kz − xk ≤ r}. Let x ∈ K ⊆ Y , then for all i ∈ I, kx − xik ≤ ri so that kxi − P xik = k(x − xi) − P (x − xi)k ≤ kx − xik ≤ ri.
Claim : K = ∩i∈I {y ∈ Y : ky − P xik ≤ ri}.
Firstly, since kP k = 1, we have K = P (K) ⊆ P [∩i∈I Bri [xi] ⊆ ∩i∈I {y ∈
Y : ky − P xik ≤ ri} = RHS. Conversely, if x ∈ RHS, for all i ∈ I, kx − xik = max{kx − P xik, kxi − P xik} ≤ ri, as kxi − P xik ≤ ri. Thus x ∈ ∩i∈I Bri [xi] = K. Remark : All the known examples where the heredity of the MIP fails are non-smooth spaces. So, it is pertinent to ask : Is the property of being a smooth space with the MIP hereditary ? We note that no counterexample is possible in finite dimensions as smooth- ness is hereditary and implies the MIP in finite dimensions. Recall that a property (say, P) is called three space property if for a Banach space X and a subspace Y of X, Y has P and the quotient space X/Y has P together implies X has P.
44 Proposition 4.2 For any two Banach spaces X1 and X2, the space X =
X1 ⊕`1 X2 fails to have the MIP. And hence, the MIP is not a three space property.
∗ ∗ ∗ Proof : The dual of X is given by X = X1 ⊕`∞ X2 , the extreme points of ∗ ∗ ∗ ∗ ∗ B(X ) are of the form (x1, x2), where xi is extreme in B(Xi ), which are ‘far enough’ from norm 1 elements of the form (x∗, 0).
Observe that in this case, for Y = X1, X/Y = X2 and the above shows that even if both X1 and X2 have the MIP, X does not.
4.2. The MIP and Farthest Points
For a closed and bounded set K in a Banach space X, define
(i) rK (x) = sup{kx−yk : y ∈ K}, x ∈ X. rK is called the farthest distance map.
(ii) QK (x) = {y ∈ K : kx − yk = rK (x)}, x ∈ X. QK is called the anti-metric projection,
(iii) D(K) = {x ∈ X : QK (x) 6= ∅}, and
(iv) b(K) = ∪{QK (x): x ∈ D(K)}, i.e., b(K) is the set of all farthest points of K. Call a closed and bounded set K densely remotal if D(K) is norm dense in X. We note the following :
Lemma 4.3 [10, 35] A closed and bounded set K is densely remotal with respect to any equivalent norm if and only if K is weakly compact.
Lemma 4.4 [10, Proposition 3] If a dual space X∗ is w*-Asplund, then every w*-compact set in X∗ is densely remotal with respect to any equivalent dual norm.
The proofs of the following Lemmas and Theorem are essentially already contained in [35] and [17]. For the sake of completeness, however, we include the proofs which have been recast using our terminology.
45 Lemma 4.5 Let X be a Banach space and F be a norming subspace of X∗. If there exists a σ(X,F )-closed, norm bounded convex set K ⊆ X that is not admissible, then there exists a σ(X,F )-closed, norm bounded convex set K1 ⊆
X with non-empty norm interior (i.e., int(K1) 6= ∅) that is not admissible.
Proof : Let K be as above. Let xo ∈ [K] \ K. Since K is σ(X,F )-closed, there exists f ∈ F such that sup f(K) < f(xo).
Let B be a closed ball such that sup f(B) < sup f(K). Let K1 = σ(X,F ) co (K ∪ B). Then, K1 is σ(X,F )-closed, norm bounded convex set with int(K1) 6= ∅ and K ⊆ K1. Now, xo ∈ [K] ⊆ [K1] and sup f(K1) ≤ sup f(K) < f(xo), i.e., xo ∈/ K1.
Theorem 4.6 Let X be a Banach space and F be a norming subspace of X∗. Consider the following statements : (a) Any σ(X,F )-closed, norm bounded convex set in X is the σ(X,F )- closed convex hull of its farthest points. (b) X has the F -MIP (c) Any σ(X,F )-closed, norm bounded, densely remotal convex set in X is the σ(X,F )-closed convex hull of its farthest points. Then (a) =⇒ (b) =⇒ (c).
(The statements can be shown to be equivalent in some special cases. See corollaries below.) Proof : (a) =⇒ (b) Suppose there exists a σ(X,F )-closed, norm bounded convex set K ⊆ X that is not admissible. By Lemma 4.5, we may assume int(K) 6= ∅.
Let xo ∈ [K] \ K and yo ∈ int(K) ⊆ int([K]). Since [K] \ K is norm open in [K], there exists 0 < λ < 1 such that zo = λxo + (1 − λ)yo ∈ [K] \ K. Note that zo ∈ int([K]), and hence, so is any point of the form
(∗) αzo + (1 − α)x, α ∈ (0, 1], x ∈ K.
Let K1 = co(K ∪ {zo}). Then K1 is σ(X,F )-closed, norm bounded and convex. The proof will be complete once we show that b(K1) ⊆ K.
Let x ∈ X. Then B = BrK (x)[x] is a closed ball containing K, and so, contains [K]. Since each point of the form (∗) is in int([K]), it is in
46 int(B), i.e., its distance from x is strictly less than rK (x). Note that rK (x) ≤ rK1 (x) ≤ r[K](x) = rK (x). Thus, QK1 (x) ⊆ K. Since x ∈ X was arbitrary, b(K1) ⊆ K. (b) =⇒ (c) Let K be a σ(X,F )-closed, norm bounded, densely remotal convex set in X. Clearly, coσ(X,F )(b(K)) ⊆ K. Suppose there exists x ∈ K \ coσ(X,F )(b(K)). Since X has the F -MIP, there exists y ∈ X and r > 0 σ(X,F ) such that co (b(K)) ⊆ Br[y] and kx − yk > r. As K is densely remotal, 1 there exists z ∈ D(K) such that ky − zk < 3 (kx − yk − r). Let x1 ∈ QK (z). Then x1 ∈ b(K) and hence, kx1 − yk ≤ r. Now, since x ∈ K, we have
kx − yk ≤ kx − zk + kz − yk ≤ rK (z) + kz − yk = kx1 − zk + kz − yk 2 2 1 ≤ kx − yk + 2kz − yk < r + (kx − yk − r) = kx − yk + r 1 3 3 3 But this implies kx − yk < r, a contradiction that completes the proof.
Corollary 4.6.1 If X∗ is a w*-Asplund dual space, then X∗ has the w*-MIP if and only if every w*-compact convex set in X∗ is the w*-closed convex hull of its farthest points.
Proof : Since X∗ is w*-Asplund, by Lemma 4.4, every w*-compact set in X∗ is densely remotal and the result follows from Theorem 4.6.
Corollary 4.6.2 [35] If X is a reflexive Banach space, then X has the MIP if and only if every closed bounded convex set in X is the closed convex hull of its farthest points.
Remarks : 1. In studying the farthest points, the emphasis so far seems to be on finding conditions on the sets to ensure existence of farthest points. In view of Theorem 4.6, a natural question is can one give conditions on the norm to ensure that a ‘reasonably large’ family of sets admit farthest points? As far as we know, this line of investigation has rarely (see e.g., [18]) been pursued in the past. In particular, does the F -MIP imply every σ(X,F )- closed, norm bounded, convex set in X is densely remotal? Is the condition X∗ is w*-Asplund necessary in Corollary 4.6.1? In this context, notice that if the norm is strictly convex (respectively, locally uniformly convex), any farthest point of a closed bounded convex
47 set is also an extreme (resp. denting) point. So, if the MIP implies every closed bounded convex set admits farthest points, then strictly (resp. locally uniformly) convex spaces with the MIP necessarily have the Krein-Milman Property (KMP) (resp. the RNP) (see [7] or [13] for details). However, the space co, which does not have the KMP, admits a strictly convex Fr´echet differentiable norm (see e.g., [11]). Hence the answer must generally be no. Now, is the answer yes, if in addition, the space has the RNP? 2. As noted before, X has the MIP implies X∗∗ has the w*-MIP. And if it also implies X is Asplund, then by Corollary 4.6.1, every w*-compact convex set in X∗∗ is the w*-closed convex hull of its farthest points. Can this be proved directly?
4.3. The MIP in Projective Tensor Product Spaces
∗ For two Banach spaces X and Y , the dual of X ⊗π Y is L(X,Y ). Now, the extremal structure of such operator spaces are known only in some very special cases. See [29] or [34] for a survey. Here we only recall that the extreme contractions from a Hilbert space to another has been characterised, by Kadison [33] in the complex case (see also [30]) and by Grzaslewicz [26] in the real case, as :
Theorem 4.7 For Hilbert spaces E and F , T ∈ L(E,F ) is an extreme contraction if and only if T or T ∗ is an isometry.
As a corollary we deduce :
Theorem 4.8 For real Hilbert spaces E and F of dimension ≥ 2, E ⊗π F never has the MIP.
Proof : We simply note that the set of extreme contractions is closed and is not the whole of S(L(E,F )). Coming to 2-dimensional `p-spaces, some notations and preliminaries p first. For 1 < p < ∞, x = (x1, x2) ∈ `2 with kxk = 1, define
48 p−1 p−1 p−1 o x = (sgn(x1)|x1| , sgn(x2)|x2| ) and x = (−x2, x1). Notice that, in general, xp−1 is the unique norming functional of x and {x, (xo)p−1} is p p−1 o a basis for `2, and if p = 2, x = x and {x, x } is orthonormal. Assume o p−1 x1 ≥ x2 ≥ 0. Now, for r ∈ IR, denote by fp(x, r) = x + r(x ) and p Fp(x, r) = kfp(x, r)k . p−1 p p−1 p p Then Fp(x, r) = |x1 − rx2 | + |x2 + rx1 | . Clearly, Fp(x, r) = 1 + |r| if p = 2 or x2 = 0. Otherwise,
−p p p −p p p Fp(x, r) = x2 · |rx2 − x1x2| + x1 · |rx1 + x1x2| −p p −p p = x2 G(rx2 − x1x2) + x1 G(rx1 + x1x2) where G(u) = |u|p. Thus
∂ F (x, r) = G0(rxp − x x ) + G0(rxp + x x ) (1) ∂r p 2 1 2 1 1 2 and G0(0) = 0, G0(u) = p · sgn(u) · |u|p−1. Clearly, (1) also holds if p = 2 or x2 = 0. Now, G0(u) is an odd function, positive and strictly increasing for u > 0. Since the two arguments of G0 in (1) add up to r, we have if r > 0 (resp. r < 0), the one larger in absolute value is positive (resp. negative), and so, ∂ ∂r Fp(x, r) is positive (resp. negative), i.e., Fp(x, r) is strictly increasing (resp. decreasing) in r > 0 (resp. r < 0).
Further, if p 6= 2 and x2 6= 0 F (x, 0) = 1 p ∂ Fp(x, 0) = 0 ∂r ∂2 F (x, 0) = p(p − 1)(x x )p−2 ∂r2 p 1 2 (2) 3 ∂ p p F (x, 0) = p(p − 1)(p − 2)(x x )p−3[x − x ] 3 p 1 2 1 2 ∂r 4 ∂ F (x, 0) = p(p − 1)(p − 2)(p − 3)(x x )p−4[1 − 3(x x )p] ∂r4 p 1 2 1 2
49 q/p and so, if 1 < q < ∞, for Hpq(x, r) = [Fp(x, r)] , we have, Hpq(x, 0) = 1 ∂ Hpq(x, 0) = 0 ∂r 2 ∂ H (x, 0) = q(p − 1)(x x )p−2 ∂r2 pq 1 2 3 (3) ∂ p−3 p p Hpq(x, 0) = q(p − 1)(p − 2)(x1x2) [x1 − x2] ∂r3 ∂4 p−4 Hpq(x, 0) = q(p − 1)(x1x2) [(p − 2)(p − 3) ∂r4 p − 3(x1x2) {(p − 2)(p − 3) + (p − q)(p − 1)}]
p q Now, let 1 < p, q < ∞. Let T : `2 −→ `2, kT k = 1. Then there exists p x = (x1, x2) ∈ `2 such that kxk = 1 = kT xk. Let T x = y = (y1, y2). p−1 Let Ixy = {T : kT k ≤ 1,T x = y}. Then for any T ∈ Ixy,(T − x ⊗ y) annihilates x and so is of rank ≤ 1, whence (T − xp−1 ⊗ y) = xo ⊗ u, for p ∗ q−1 p−1 ∗ p−1 some u ∈ `2. Further, T (y ) = x , that is (T − y ⊗ x ) annihilates q−1 ∗ p−1 o q−1 q y , whence (T − y ⊗ x ) = (y ) ⊗ v, for some v ∈ `2. Combining, T must be of the form
p−1 o o q−1 Ts = x ⊗ y + sx ⊗ (y ) , for some s ∈ IR
In other words, Ixy = {Ts : s ∈ IR, kTsk ≤ 1}.
As in [34], pre- or post-multiplying by diag(sgn(x1), sgn(x2)), diag(sgn(y1), sgn(y2)) and permutation matrices, if necessary — each of which is an isometry — we may assume x1 ≥ x2 ≥ 0, y1 ≥ y2 ≥ 0.
Now, Ts(fp(x, r)) = fq(y, rs), and it follows from the above discussion q that for any r 6= 0, kTs(fp(x, r))k = Fq(y, rs) is strictly increasing in s ≥ 0 and strictly decreasing in s ≤ 0, and Fq(y, rs) is unbounded in s. Now, if r 6= 0, q/p q/p Fq(y, 0) = 1 = [Fp(x, 0)] < [Fp(x, r)] so, there exists unique s+(x, y, r) > 0 and unique s−(x, y, r) < 0 such that
q/p Fq(y, rs±) = [Fp(x, r)] (4)
And the quantity on the LHS becomes smaller or larger than the one in the
RHS according as |s| gets smaller or larger. Evidently, such s± also exist for
50 o p−1 (x ) , which we denote by fp(x, ∞). In fact, in this case, |s±(x, y, ∞)| = o p−1 o q−1 k(x ) k/k(y ) k. Notice that for fixed x, y, s± is a continuous function of r 6= 0 and elementary examples show that limr→0 s±(x, y, r) may not even exist. Let
∗ ∗∗ s (x, y) = inf{s+(x, y, r): r 6= 0} s (x, y) = lim inf s+(x, y, r) + + r→0 ∗ ∗∗ s−(x, y) = sup{s−(x, y, r): r 6= 0} s− (x, y) = lim sup s−(x, y, r) r→0 where lim infr→0 s+(x, y, r) = supε>0 inf{s+(x, y, r): |r| < ε} and ∗∗ lim supr→0 s−(x, y, r) = infε>0 sup{s−(x, y, r): |r| < ε}. Clearly, s− ≤ ∗ ∗ ∗∗ ∗ ∗ s ≤ 0 ≤ s ≤ s and T ∈ I if and only if s ≤ s ≤ s , i.e., T ∗ are − + + s xy − + s± end points of Ixy and hence are extreme. Also note that if Jxy = {s : Ts is ∗∗ ∗∗ contractive in a neighbourhood of x}, then s− = inf Jxy and s+ = sup Jxy. ∗ Now, either s± equals s±(x, y, r) for some r 6= 0 (including r = ∞), in ∗ which case s 6= 0 and T ∗ attain their norm on two linearly independent ± s± ∗ ∗∗ ∗∗ ∗∗ vectors, or s = s , in which case |s | < ∞, s ∈ J , in fact, T ∗∗ ∈ I . ± ± ± ± xy s± xy Note that T attains its norm on two linearly independent vectors if and only if T ∗ attains its norm on two linearly independent vectors. Moreover, any such T is exposed, and hence strongly exposed. Fix x, y.
Case(I) : (i) p = 2 and either q = 2 or y2 = 0; (ii) q = 2 and either p = 2 or x2 = 0; (iii) p 6= 2 6= q and x2 = 0 = y2.
q/p Ts is a contraction ⇐⇒ Fq(y, rs) ≤ [Fp(x, r)] for all r ⇐⇒ 1 + |rs|q ≤ [1 + |r|p]q/p for all r [1 + |r|p]q/p − 1 ⇐⇒ |s|q ≤ for all r 6= 0 |r|q
Note that the RHS ≡ 1 if p = q and is strictly decreasing (resp. increasing) in |r| for q > p (resp. q < p). ∗ ∗∗ So, if p = q, s ≡ ±1, and hence, s = s = ±1 and T ∗ are isometries. ± ± ± s± And, if p 6= q, the infimum of the RHS over r 6= 0 yields ∗ q 1 if q > p |s±| = 0 if q < p
51 ∗ So, if q < p, s± = 0, hence Ixy = {T0}, and x ⊗ e if p = 2 > q 1 ∗ T0 = e1 ⊗ y if p > q = 2 ∗ e1 ⊗ e1 if p > q, p 6= 2 6= q is an extreme contraction which attains its norm at precisely one point. ∗ And if q > p, s± = s±(∞) = ±1 with T±1 attaining its norm at both x o p−1 and (x ) . It is interesting to note that if p 6= 2 6= q, T1 in this case is the identity operator. In each of the following cases our basic aim is to find extreme contractions that do not attain their norm on two linearly independent vectors. Then nec- ∗ ∗∗ ∗∗ essarily s± = s± . So, we first assume |s± | < ∞. If we reach a contradiction, ∗∗ ∗ ∗∗ we conclude that s = ±∞, whence s 6= s and T ∗ is not of the desired ± ± ± s± ∗∗ type. Otherwise, we check whether s± ∈ Jxy. If it does not, we are again done. And if it does, we further check whether T ∗∗ gives a contraction. If s± it does not, we are done, and if it does, we get an extreme contraction. We then check whether it attains its norm in any direction other than that of x and only if it does not, we get an extreme contraction of the desired type. This is exemplified in the analysis below. ∗∗ To calculate s± , let {rn} be a sequence of real numbers such that rn → 0 ∗∗ ∗∗ and s±(rn) → s± . Since |s± | < ∞, {s±(rn)} is a bounded sequence. Now, by (4),
q/p Fq(y, rns±(rn)) = [Fp(x, rn)] (5)
Case (II) : q 6= 2, y2 > 0 and either p = 2 or x2 = 0. 2 In this case, subtracting 1 from both side of (5), dividing by rn and taking limit as n −→ ∞, we get by L’Hospital’s rule and (2) that LHS −→ 1 2 q−2 ∗∗ 2 q(q − 1)s (y1y2) , where s = s± and 0 if p > 2 RHS −→ ∞ if p < 2 q if p = 2 2 ∗∗ So, if p < 2, we have a contradiction, whence s = ±∞, so that T ∗ is not ± s± ∗ ∗∗ ∗ of the desired type, and if p > 2, s± = s± = 0, Ixy = {T0} and T0 = e1 ⊗ y is extreme, and clearly of the desired type.
52 If p = 2, we have
2 q−2 s (q − 1)(y1y2) = 1 (6)
Now, if the s given by (6) belongs to Jxy, we must have
q/p Fq(y, rs) ≤ [Fp(x, r)] for all small r 6= 0 (7)
By (2), the Taylor expansion of the LHS above around r = 0 is given by 1 1 1 + q(q − 1)r2s2(y y )q−2 + q(q − 1)(q − 2)r3s3(y y )q−3[yq − yq] 2! 1 2 3! 1 2 1 2 1 + q(q − 1)(q − 2)(q − 3)r4s4(y y )q−4[1 − 3(y y )q] + ··· 4! 1 2 1 2 while that of the RHS is given by 1 1 1 + qr2 + q(q − 2)r4 + ··· 2 8 Comparing the two, we see that the coefficients of 1, r and r2 on both sides are equal, whence the inequality (7) for small r implies the corresponding inequality for the coefficient of r3 on both sides, which, for r > 0 and r < 0, leads to the following equality : 1 s3q(q − 1)(q − 2)(y y )q−3(yq − yq) = 0 (8) 6 1 2 1 2
q q 2 Combining equations (6) and (8), we have y1 = y2 = 1/2, s = 1 (q−2)/q (q−1) 4 . But again the equality in (8) forces the inequality of the coef- ficient of r4, i.e., 1 1 q(q − 2) ≥ s4q(q − 1)(q − 2)(q − 3)(y y )q−4[1 − 3(y y )q] 8 24 1 2 1 2 (q − 2)(q − 3) or 3(q − 2) ≥ (q − 1)
∗∗ Now for q < 2, this leads to a contradiction, so that s± 6∈ Jxy. And for q > 2, by [28, Proposition 1], Ts with these parameters is a contraction that attains its norm only in the direction of x. So for q > 2, Ts with these ∗ ∗∗ parameters is of the desired type and for q < 2, s 6= s , so that T ∗ is not ± ± s± of the desired type.
53 Case (III) : p 6= 2, x2 > 0 and either q = 2 or y2 = 0. ∗ ∗∗ This is the situation dual to case (II) and hence, if q < 2, s± = s± = 0, p−1 Ixy = {T0} and T0 = x ⊗ e1 is extreme and of the desired type; if q > 2, ∗ ∗∗ s 6= s , so that T ∗ is not of the desired type, and if q = 2, then for p < 2, ± ± s± ∗ ∗∗ q (2−p)/p 1/p s = s = ± (p − 1)2 , x = x = (1/2) , T ∗ is of the desired type, ± ± 1 2 s± and for p > 2, T ∗ is not of the desired type. s± Case(IV) : p 6= 2 6= q and x2 > 0, y2 > 0. 2 In this case too, subtracting 1 from both side of (5), dividing by rn and taking limit as n → ∞, we get by (2) and (3)
q−2 2 p−2 (q − 1)(y1y2) s = (p − 1)(x1x2) (9)
∗∗ where s = s± . Now,
" p−2 #1/2 " p #1/2 (p − 1)(x1x2) (p − 1)(x1x2) y1y2 s = ± q−2 = ± q · (q − 1)(y1y2) (q − 1)(y1y2) x1x2
" p #1/2 (p − 1)(x1x2) y1y2 Put α = q , then s = ±α , whence (q − 1)(y1y2) x1x2
p−1 o o q−1 Ts = x ⊗ y + sx ⊗ (y ) p−1 p−1 q−1 q−1 x1 y1 x2 y1 y1y2 y2 x2 −y2 x1 = p−1 p−1 ± α q−1 q−1 x1 y2 x2 y2 x1x2 −y1 x2 y1 x1 xpy xpy y yq y yq 1 1 2 1 1 2 − 1 2 x x x x = p1 p2 ± α 1 q 2q x y x y y y y y 1 2 2 2 − 2 1 2 1 x1 x2 x1 x2 y y 1 (xp ± αyq) 1 (xp ∓ αyq) x 1 2 x 2 2 = 1 2 (10) y2 p q y2 p q (x1 ∓ αy1) (x2 ± αy1) x1 x2 To continue, in this case, by (2), the Taylor expansion of the LHS of (7) around r = 0 is given by 1 1 1 + q(q − 1)r2s2(y y )q−2 + q(q − 1)(q − 2)r3s3(y y )q−3[yq − yq] 2! 1 2 3! 1 2 1 2 1 + q(q − 1)(q − 2)(q − 3)r4s4(y y )q−4[1 − 3(y y )q] + ··· 4! 1 2 1 2
54 while by (3), that of the RHS is given by 1 1 1 + q(p − 1)r2(x x )p−2 + q(p − 1)(p − 2)r3(x x )p−3[xp − xp] 2! 1 2 3! 1 2 1 2 1 + q(p − 1)r4(x x )p−4[(p − 2)(p − 3) − 3(x x )p · 4! 1 2 1 2 {(p − 2)(p − 3) + (p − q)(p − 1)}] + ···
So, if s ∈ Jxy, by similar arguments, we must have
3 q−3 q q p−3 p p s (q − 1)(q − 2)(y1y2) (y1 − y2) = (p − 1)(p − 2)(x1x2) (x1 − x2) (11) and
4 q−4 q p−4 s (q − 1)(q − 2)(q − 3)(y1y2) [1 − 3(y1y2) ] ≤ (p − 1)(x1x2) · p p [(p − 2)(p − 3){1 − 3(x1x2) } − 3(p − q)(p − 1)(x1x2) ] (12)
Eliminating s from (9) and (11) and using the fact that x, y are unit vectors, we get
(q − 2)2 " 1 # (p − 2)2 " 1 # q − 4 = p − 4 (13) (q − 1) (y1y2) (p − 1) (x1x2)
Also, dividing (12) by the square of (9), we get
(q − 2)(q − 3) " 1 # (p − 2)(p − 3) " 1 # q − 3 ≤ p − 3 − 3(p − q) (14) (q − 1) (y1y2) (p − 1) (x1x2)
Notice that if p = q, we get from (13) that xi = yi, i = 1, 2, whence from
(9), s = ±1, and from (11), y1 = y2 = x1 = x2 for s = −1. Also, (14) is consistent with (13) and equality holds. Now in (10), α = 1, and so,
1 0 0 1 T1 = and T−1 = 0 1 1 0 and hence, clearly, are isometries. 1 Now, let p 6= q. From (13) and (14), writing p = A, we get (x1x2)
(p − 2)2(q − 3) (q − 2)(q − 3) (p − 2)(p − 3) (A − 4) + ≤ (A − 3) − 3(p − q) (q − 2)(p − 1) (q − 1) (p − 1)
55 or "(p − 2)2(q − 3) (p − 2)(p − 3)# − · A (q − 2)(p − 1) (p − 1) 4(p − 2)2(q − 3) 3(p − 2)(p − 3) (q − 2)(q − 3) ≤ − − − 3(p − q) (q − 2)(p − 1) (p − 1) (q − 1) "3(p − 2)2(q − 3) 3(p − 2)(p − 3)# "(p − 2)2(q − 3) = − + (q − 2)(p − 1) (p − 1) (q − 2)(p − 1) (q − 2)(q − 3)# − − 3(p − q) (q − 1)
Simplifying we get (q − p)(p − 2) 2q(q − p)(pq − p − q) · A ≤ (p − 1)(q − 2) (p − 1)(q − 1)(q − 2)
So, if (a) 1 < q < 2 < p < ∞, or, (b) 1 < p < q < 2, or, (c) 2 < p < q < ∞, we have 2q(pq − p − q) 2(q − 2)(pq − p − q + 2) A ≤ i.e., A − 4 ≤ (15) (p − 2)(q − 1) (p − 2)(q − 1)
And if (d) 1 < p < 2 < q < ∞, or, (e) 1 < q < p < 2, or, (f) 2 < q < p < ∞, we have 2q(pq − p − q) 2(q − 2)(pq − p − q + 2) A ≥ i.e., A − 4 ≥ (16) (p − 2)(q − 1) (p − 2)(q − 1)
p p p Now, (x1x2) = 1/A and x1 + x2 = 1, so 0 < 1/A ≤ 1/4, i.e., A ≥ 4 or A − 4 ≥ 0. But since (pq − p − q + 2) = (p − 1)(q − 1) + 1 is always positive for 1 < p, q < ∞, if 1 < q < 2 < p < ∞, i.e., in case (a) above, we reach a contradiction at this point, whence T ∗ is not of the desired type. s± Summarising the results so far, we have the following
p q Theorem 4.9 For 1 < p, q < ∞, an operator T : `2 −→ `2 with kT k = 1 is an extreme contraction (i) [26] for p = q = 2, if and only if T is an isometry. (ii) [28] for p = 2 6= q, if and only if T satisfies one of the following (a) T attains its norm on two linearly independent vectors.
56 (b) T is of the form x ⊗ e if q < 2 T = i x ⊗ y + sxo ⊗ (yo)q−1 if q > 2
q 1 where x is any unit vector and, in the second case, |yi| = 2 and s = ±√ 1 2(q−2)/q. (q−1) (iii) [28] for p 6= 2 = q, if and only if T satisfies one of the following (a) T attains its norm on two linearly independent vectors. (b) T is of the form e ⊗ y if p > 2 T = i xp−1 ⊗ y + sxo ⊗ yo if p < 2
p 1 where y is any unit vector and, in the second case, |xi| = 2 and q s = ± (p − 1)2(2−p)/p. (iv) [25] for p = q, if and only if T satisfies one of the following (a) T attains its norm on two linearly independent vectors. (b) T is of the form e∗ ⊗ y if p > 2, y y 6= 0 T = i 1 2 p−1 x ⊗ ej if p < 2, x1x2 6= 0
(v) for 1 < q < 2 < p < ∞, if and only if T satisfies one of the following (a) T attains its norm on two linearly independent vectors. p−1 (b) T = x ⊗ y with x, y unit vectors and x1x2y1y2 = 0.
Remark : In the discussion preceding Theorem 4.9, the conditions (b) and (e) are dual to (c) and (f) respectively. And in the cases (b) and (c), the inequality (15) implies v u 1 p 1 u(q − 2)(pq − p − q + 2) ≤ x ≤ 1 + t (17) 2 1 2 q(pq − p − q) while in cases (e) and (f), the inequality (16) implies v u 1 u(q − 2)(pq − p − q + 2) p 1 + t ≤ x < 1 (18) 2 q(pq − p − q) 1
57 Now, in cases (b), (c), (e) and (f), we have from (11) that for s < 0, both p p q q sides of (11) must be 0, i.e., we must have x1 = x2 = 1/2 = y1 = y2. But ∗ then in cases (e) and (f), we have a contradiction. So, in these two cases, s− gives extreme contractions not of the desired type. Also in case (d), (16) is always satisfied and (11) implies that for s > 0, p p q q x1 = x2 = 1/2 = y1 = y2. Now combining Propositions 1 and 2 of [28], we p p q q see easily that for x1 = x2 = 1/2 = y1 = y2 for both s > 0 and s < 0, we have a contraction that attains its norm only in the direction of x, i.e., we get extreme contractions of the desired type. Thus, we are left with the following cases unsolved : p (1) Case (b) with x1 satisfying (17) for s > 0 and x1 = 1/2 for s < 0 with
y1 given by (13).
(2) Case (e) with x1 satisfying (18) for s > 0 with y1 given by (13). p (3) Case (d) with x1 > 1/2 and s < 0 with y1 given by (13). 1 1 p q Notice that for p + q = 1, (p < 2), (13) gives xi = yi , i = 1, 2, whence in this case, Ts — as given by (10) — has a comparatively simpler form. We proceed now to find the closure of the extreme contractions in the cases described in Theorem 4.9. In case (i), the set of extreme contractions is clearly closed. Also, in the cases (ii), (iii) and (v), the set of operators of the form (b) is clearly closed. And in case (iv), the closure of the set of operators of the form (b) contains only the operators ei ⊗ ej, i, j = 1, 2 in addition. Let us consider the set of operators of the type (a) in cases (ii), (iv) and
(v). Let {Tn} be a sequence of operators of the type (a). Let Tn −→ T in operator norm. Let xn = (xn1, xn2) be such that kxnk = 1 = kTnxnk. Let
Tnxn = yn = (yn1, yn2). Then Tn is of the form
p−1 ∗ o o q−1 Tn = xn ⊗ yn + s±(xn, yn)xn ⊗ (yn)
∗ where s±(xn, yn) is as in our earlier discussion. For notational simplicity, ∗ write s±(xn, yn) = ±sn. Passing to a subsequence, if necessary, assume xn → x = (x1, x2), yn → y = (y1, y2) (by compactness of the unit balls of p q `2 and `2), and all the sn’s have the same sign, without loss of generality, positive.
58 Clearly, kT k = 1 and T x = y, whence T is of the form
T = xp−1 ⊗ y + sxo ⊗ (yo)q−1
p−1 o o q−1 Also, as Tn −→ T , sn = kTn − xn ⊗ ynk/kxnk · k(yn) k −→ kT − p−1 o o q−1 x ⊗ yk/kx k · k(y ) k, i.e., {sn} is convergent. Clearly, sn → s. o p−1 Now, since Tn is of the type (a), there exists zn = xn + rn(xn) with rn 6= 0 ∈ IR such that kTnznk = kznk. Again we may assume all rn’s are of the same sign, in particular positive and rn → r, where 0 ≤ r ≤ ∞. If o p−1 0 < r < ∞, zn → z = x+r(x ) and kT zk = kzk, i.e., T is also of the type o p−1 o p−1 (a). Also, if rn → ∞, let un = zn/kznk. Then un → u = (x ) /k(x ) k and kT uk = 1, so that T again is of the type (a).
Now, suppose rn → 0. Then from kTnznk = kznk we have
q/p Fq(yn, rnsn) − [Fp(xn, rn)] = 0 (19)
For (ii), if p = 2 and q < 2, since Tn is of type (a), we have yn1yn2 6= 0 for all n. And if q > 2, we have two possibilities; either there is a subsequence for which yn1yn2 = 0, or, eventually yn1yn2 6= 0. In the first case, we restrict ourselves only to that subsequence, and we have, by case(I), sn = 1 for all n, o whence s = 1. Also, y1y2 = 0. So, T = x ⊗ ei + x ⊗ ej (i 6= j), and it is clear that T is of type (a) (see case (I)). And in the second case, we assume 2 yn1yn2 6= 0 for all n. Then dividing (19) by rn and taking limit as n → ∞, we get by L’Hospital’s rule
2 q−2 (q − 1)s (y1y2) − 1 = 0 (20)
If q > 2, for y1y2 = 0, this leads to a contradiction, whence either rn 6→ 0
(in which case we get an operator of type(a)), or y1y2 6= 0. In the latter ∗∗ case, (6) and (20) coincides, i.e., we have s = s+ (x, y). Now, our analysis in q q ∗∗ case (II) shows that only for y1 = y2 = 1/2, s+ gives a contraction (which is an extreme contraction of type (b)). And in every other case, we run into a contradiction, i.e., we must have rn 6→ 0.
And if q < 2, for y1y2 = 0, (20) makes sense only if s = 0. In that case,
T = x ⊗ ei, which, by case (I), is an extreme contraction of the type (b).
59 ∗∗ And for y1y2 6= 0, we again have s = s+ (x, y) and our analysis in case (II) shows that this case always leads to a contradiction, whence rn 6→ 0. So, in both the cases, the closure of the set of operators of the type (a) contains at most operators of type (b), and therefore, the set of extreme contractions is closed. Since case (iii) is just the dual of case (ii), the set of extreme contractions, in this case too, is closed.
In (iv), i.e., if p = q, by duality, it suffices to consider p > 2. Since Tn is of type (a), we have three possibilities; either there is a subsequence for which both xn1xn2 = 0 and yn1yn2 = 0, or, there is a subsequence for which xn1xn2 6= 0 and yn1yn2 = 0, or, eventually both xn1xn2 6= 0 and yn1yn2 6= 0. In the first case, we again restrict ourselves only to that subsequence, and we have, by case(I), sn = 1 for all n, whence s = 1. Also, x1x2 = 0 and y1y2 = 0. Now again by case (I), T is of type (a). 2 In the second and the third case, dividing (19) by rn and taking limit — through a subsequence if necessary — as n → ∞, we get in the second case
p−2 (x1x2) = 0 and in the third case p−2 2 p−2 (y1y2) s = (x1x2)
So, in the second case, y1y2 = 0, and we have a contradiction unless x1x2 = 0. And in that case, T is of the form diag(1, s) upto isometric factors of signum or permutation matrices. Now, T is a contraction for −1 ≤ s ≤ 1 and is extreme (in fact, an isometry) only for s = ±1.
In the third case, if x1x2 = 0, we get a contradiction unless y1y2 = 0 or ∗ s = 0. If y1y2 6= 0, s = 0 = s±(x, y), whence T is extreme. And if y1y2 = 0, we get the conclusions as in the second case. If x1x2 6= 0, y1y2 = 0 leads to a ∗∗ contradiction (and therefore, rn 6→ 0), and if y1y2 6= 0, s = s+ (x, y), so that T is an isometry and hence is of type (a). Thus, in case (iv), the closure of the operators of type (a), and so the closure of extreme contractions, can have at most operators of the form diag(1, s), |s| < 1, upto isometric factors of signum or permutation matrices
60 as additional elements. However, we do not have precise description of the closure.
In case (v), i.e., if 1 < q < 2 < p < ∞, since Tn is of type (a), we must have xn1xn2yn1yn2 6= 0 for all n. And a similar argument leads to
2 q−2 p−2 s (q − 1)(y1y2) = (p − 1)(x1x2)
If x1x2 6= 0 the only situation that does not lead to any contradiction — either immediate or to the fact that T is a contraction — is both y1y2 = 0 ∗ and s = 0. And in that case, s = 0 = s±(x, y), so that T is extreme.
And if x1x2 = 0, we must have s = 0, in which case, by cases (I) and (II), ∗ s = 0 = s±(x, y) and T is extreme. Thus in this case too, the set of extreme contractions is closed.
p q Theorem 4.10 In each of the following cases of 1 < p, q < ∞, `2 ⊗π `2 lacks the MIP : 1 1 (i) p and q are conjugate exponents, i.e., p + q = 1. (ii) Either p or q is equal to 2. (iii) 2 < p, q < ∞.
p q p q0 1 1 Proof : The dual of `2 ⊗π `2 is L(`2, `2 ), where q + q0 = 1 and the closure of extreme contractions in each of the above cases does not contain norm 1 operators of the form xp−1 ⊗ y, where x and y are unit vectors with x1x2y1y2 6= 0. Remark : The fact that operators of the above form do not belong to the closure of extreme contractions in any of these cases seems to suggest that this is a general phenomenon. It is possible that this is happens in higher dimensions as well. Can one give a proof of this without precisely characterising the extreme contractions ? What seems to be required is a more tractable necessary condition for extremality, or, for belonging to the closure of extreme contractions.
61 Chapter 5 Exposed Points of Continuity and Strongly Exposed Points
In this chapter, we make the following change in our notations. For a Banach space X and A ⊆ X, by Ao, the polar of A, we mean Ao = {f ∈ X∗ : f(x) ≤ 1 for all x ∈ A} (i.e., we do not take the absolute value of f(x)). If K is a closed bounded convex set in X and f ∈ X∗, we will say that ∗ f ∈ X supports K at xo ∈ K if f(xo) = M(K, f) and we let S(K, xo) = ∗ {f ∈ X : f(xo) = M(K, f)}. Note that S(K, xo) is a w*-closed convex cone with vertex 0. We may sometimes abbreviate S(K, xo) by S(xo) when there is no scope of confusion.
Call xo ∈ K a relatively exposed point of K if for each x ∈ K \{xo} there exists fx ∈ S(K, xo) such that fx(x) < fx(xo). Call xo a vertex of K if for each x ∈ X \{xo} there exists fx ∈ S(K, xo) such that fx(x) 6= fx(xo), or, in other words, S(K, xo) separates points of X. We can also define w*-relatively exposed points or w*-vertices similarly. Obviously, a (w*-) vertex is (w*-) relatively exposed.
5.1. The Counterexample
1 1 Let {δn : n ≥ 1} denote the canonical basis of ` . Recall that ` as a dual space has a w*-topology induced on it. Let
∗ 1 1 1 K = cow [{ δ + δ : n ≥ 2} ∪ { δ − δ : n ≥ 2}] n 1 n n2 1 n n From Milman’s Theorem (see [43, p 9]) and the metrizable version of Cho- quet’s Theorem ([1] or [43]), it follows that x ∈ K if and only if x is of the
62 form X 1 X 1 1 x = αn( δ1 + δn) + βn( 2 δ1 − δn) n≥2 n n≥2 n n P with αn, βn ≥ 0 and n≥2(αn +βn) ≤ 1 (see [7, Example 3.2.5] for the details ∞ 1 ∗ in a similar situation). Observe that (−1, 0, 0,...) ∈ co ⊆ ` = (` ) exposes ∞ 1 0 = (0, 0,...) ∈ K. If f = (an) ∈ ` supports K at 0, then n a1 ± an ≤ 0 for 1 n ≥ 2, from which it follows that a1 ≤ 0 and |an| ≤ n |a1|,(n ≥ 2) and we 1 1 see therefore that (an) ∈ co. Since f( m δ1 + δm) = m a1 + am → 0 as m → ∞, 1 S(K, f, α) = {y ∈ K : f(y) > −α} contains m δ1 + δm for all sufficiently 1 large m, whence diam[S(K, f, α)] ≥ k m δ1 + δmk > 1, showing that 0 is not strongly exposed. 1 1 1 1 ∞ Next, for m ≥ 2, let fm = (−1, − m2 ,..., − m2 , − m , − m ,...) ∈ ` where 1 the mth coordinate onwards of fm has the value − m and consider the slice 1 Sm = {y ∈ K : fm(y) > − m3 } determined by fm. Clearly, 0 ∈ Sm and if
X 1 X 1 1 x = αn( δ1 + δn) + βn( 2 δ1 − δn) ∈ Sm n≥2 n n≥2 n n we must have
m−1 2 m−1 2 1 X m + n X m + n X m − n X n − m − 3 < − αn · 2 − αn · − βn · 2 2 + βn · 2 m n=2 m n n≥m mn n=2 m n n≥m n m whence it follows that
m−1 2 1 X n + 1 m + n X n + 1 m + n 2 > αn · · + αn · · m n=2 n m(n + 1) n≥m n n + 1 m−1 2 X n + 1 m − n X n − m + βn · 2 · − βn · 2 n=2 n m(n + 1) n≥m n m−1 m−1 X n + 1 X n + 1 m − 1 X n + 1 1 ≥ αn · + αn · + βn · 2 − n=2 n n≥m n m + 1 n=2 n 4m on using successively the following easily verified inequalities : m2 + n (i) for all m ≥ 1 and n ≤ m, ≥ 1. m(n + 1) m + n (ii) for all m ≥ 1, ≥ 1. n + 1 m2 − n m − 1 m2 − n (iii) for all n ≤ m, ≥ , since is decreasing in n. m(n + 1) m + 1 m(n + 1)
63 n − m 1 (iv) for all m, n ≥ 1, ≤ . n2 4m Thus,
m−1 m + 1 m + 4 m + 1 X n + 1 X n + 1 · 2 ≥ αn · + βn · 2 m − 1 4m m − 1 n≥2 n n=2 n m−1 X n + 1 X n + 1 ≥ αn · + βn · 2 n≥2 n n=2 n
Now, ! α β β n + 1 n + 1 X n n X n X X kxk = + 2 + αn − ≤ αn · + βn · 2 n≥2 n n n≥2 n n≥2 n n≥2 n (m + 1)(m + 4) X n + 1 (m + 1)(m + 4) m + 1 ≤ 2 + βn · 2 ≤ 2 + 2 4m (m − 1) n≥m n 4m (m − 1) m 5(m + 1) 15 = ≤ for all m ≥ 2. 4m(m − 1) 4m
Hence, diam(Sm) ≤ 15/2m → 0 as m → ∞ and we conclude that 0 is a denting point of K. 1 Remarks : 1. Let Fo be a w*-cluster point of { n δ1 + δn : n ≥ 2} in ∞ Kf. Then Fo 6= 0 since Fo(fo) = 1 where fo = (1, 1,..., 1,...) ∈ ` . As ˆ S(0) ⊆ co, we must have Fo ∈ ∩{ker(f): f ∈ S(0)} and hence 0 is not w*- relatively exposed in Kf. This observation proved useful in the formulation (d) of Theorem 5.3 in the next section.
2. Since S(0) ⊆ co, one in fact has S(0) = {(an) ∈ co : a1 ≤ 0 and 1 1 |an| ≤ n |a1|, n ≥ 2}. It is now easy to see that S(0) separates points of ` , i.e., 0 is a vertex of K.
5.2. A Characterisation Theorem
We need the following lemma :
Lemma 5.1 Let K be a compact convex set in a locally convex space E. Then ∗ (a) xo ∈ K is exposed by f ∈ E if and only if {S(K, f, α): α > 0}
forms a local base for the relative topology on K at xo.
64 (b) A relatively exposed point xo ∈ K is exposed if and only if xo is a
Gδ point.
Proof : (a) The sufficiency is obvious and the necessity follows from the compactness of K. ∗ (b) [We follow [1, p 119]] If xo is exposed by f ∈ E , then {xo} = 1 ∩n≥1S(K, f, n ) is clearly Gδ. Conversely, let xo be relatively exposed and {xo} = ∩n≥1Wn, where Wn is open in K. Now, an easy compactness argument shows that for any compact ∗ subset F of K with xo 6∈ F , there exists f ∈ E such that f ∈ S(xo) and f(y) < f(xo) for all y ∈ F .
Since for all n ≥ 1, Wn is open in K, by the above argument, there exists fn ∈ S(xo) such that fn(y) < fn(xo) for all y ∈ K \ Wn. It is now esay to P 1 show that f = n≥1 2n fn exposes xo.
Theorem 5.2 Let K be a w*-compact convex set in a dual Banach space X∗ ∗ and xo ∈ K be a w*-PC. The following are equivalent : ∗ (a) xo ∈ K is w*-strongly exposed. ∗ (b) xo ∈ K is w*-exposed. ∗ (c) xo ∈ K is w*-relatively exposed.
Proof : Follows from Lemma 5.1 and the fact that a w*-PC is a w*-Gδ point.
Theorem 5.3 Let K be a closed bounded convex set in a Banach space X and xo ∈ K be a PC. The following are equivalent :
(a) xo ∈ K is strongly exposed.
(b)x ˆo ∈ Kf is w*-strongly exposed.
(c)x ˆo ∈ Kf is w*-exposed.
(d)x ˆo ∈ Kf is w*-relatively exposed.
∗ o ∗ ∗ (e) X /sp(S(xo)) = π[(K − xo) ] where π : X −→ X /sp(S(xo)) is the quotient map and the closure is with respect to the norm topology.
65 Proof : (a) ⇐⇒ (b) is immediate, while (b) ⇐⇒ (c) ⇐⇒ (d) follows from Theorem 5.2 and Lemma 1.4. (d) ⇐⇒ (e) : Using definitions, simple separation arguments and polar calculations `ala the Bipolar Theorem, each of the statement below is easily seen to be equivalent to the next :
(1)x ˆo is w*-relatively exposed in Kf.
(2) 0 is w*-relatively exposed in Kf1, where K1 = K − xo. ˆ (3) Kf1 ∩ [∩{ker(f): f ∈ S(xo)}] = {0}.
oo o (4) K1 ∩ [sp(S(xo))] = {0}, ··· (∗) ˆ ⊥ o (since ∩{ker(f): f ∈ S(xo)} = [sp(S(xo))] = [sp(S(xo))] and Kf1 = oo K1 ) o o (5) [K1 ∪ sp(S(xo))] = {0}. o ∗ (6) co[K1 ∪ sp(S(xo))] = X . o ∗ This last condition implies that [sp(S(xo)) + K1 ] = X which in turn means that ∗ o X /sp(S(xo)) = π(K1 ). ··· (∗∗) oo o Conversely, suppose that (∗∗) holds and that φ ∈ K1 ∩ [sp(S(xo))] . We get immediately o ∗ ∗ (a) φ ∈ [sp(S(xo))] = [X /sp(S(xo))] , and
o (b) M(K1 , φ) ≤ 1. o But (a) and (b) together give M(π(K1 ), φ) ≤ 1 and using (∗∗), we see that φ = 0, thus (∗) holds.
Corollary 5.3.1 If K is weakly compact then xo ∈ K is strongly exposed if and only if xo is exposed and PC.
Proof : Follows immediately from Theorem 5.2 or from Theorem 5.3, once we observe that if K is weakly compact, Kf = Kc and thatx ˆo ∈ Kf is w*- exposed if xo ∈ K is exposed.
Corollary 5.3.2 (a) Let K be a closed bounded convex set in a Banach space
X. xo ∈ K is strongly exposed if and only if xo is contained in slices of K determined by functionals from S(K, xo) of arbitrarily small diameter.
66 ∗ ∗ (b) Let K be a w*-compact convex set in a dual Banach space X . xo ∈ ∗ K is w*-strongly exposed if and only if xo is contained in w*-slices of K ∗ determined by functionals from S(K, xo) ∩ X of arbitrarily small diameter.
Remarks : 1. Note that the proof of (d) ⇐⇒ (e) in Theorem 5.3 does not use the fact that xo is a PC.
2. If xo is a PC andx ˆo ∈ Kf is a w*-vertex, or equivalently, xo is a PC and ∗ sp(S(xo)) is dense in X , it follows from Theorem 5.3, that xo is strongly exposed. But, if xo ∈ K is a strongly exposed point with a unique (upto scalar multiples) exposing functional — this happens for instance for any 2 boundary point of the unit ball of the Euclidean space IR — then sp(S(xo)) ∗ is 1-dimensional and thus X 6= sp(S(xo)), i.e., this condition is, in general, not necessary. And since 0 in our example is a vertex (see Remark 2), the weaker condition that xo is a PC and xo ∈ K is a vertex is no longer sufficient for xo to be strongly exposed.
3. As pointed out earlier, ∩ε>0Mε = {x ∈ S(X) : the norm is Fr´echet differentiable at x} and by Theorem 1.9, D(∩ε>0Mε) = {w*-strongly exposed points of B(X∗)}. We note the following consequence of Corollary 5.3.2(b) :
∗ {w*-strongly exposed points of B(X )} = ∩ε>0D(Mε).
∗ Now, if X has the MIP, for each ε > 0, D(Mε) is dense in S(X ). And the necessity of Phelps’ condition (a) reduces to the question whether the intersection of these dense sets is also dense. Naturally, if each of these sets was a Gδ, the answer would be yes. So one asks :
Does the quasi-continuity of the duality map force the image of
Mε, an open set, to be Gδ ?
5.3. A Geometric Characterisation of Banach Spaces Containing `1
Let us say that a closed bounded convex set K in a Banach space X has Property (P) if every exposed PC in K is strongly exposed; and that a
67 Banach space X has Property (P) if every closed bounded convex set K in X has Property (P). It follows from the counterexample and Corollary 5.3.1 above that (i) If X has the Property (P) then X does not contain a copy of `1. (ii) If X is reflexive, then X has the Property (P). The monograph [15] is an excellent introduction to the theory of Banach spaces not containing `1. Here we show that a weaker version of the Property (P) is equivalent to X not containing `1, while among a certain class of Banach spaces, Property (P) implies reflexivity. Specifically we prove :
Theorem 5.4 Let X be a Banach space. (a) X does not contain a copy of `1 if and only if every norm bounded, weakly sequentially complete convex set in X has Property (P). (b) If X is weakly sequentially complete, then X is reflexive if and only if X has Property (P).
Proof : (a) Since `1 is weakly sequentially complete (see [16]), so is any closed convex subset of it, in particular, so is the set K in our example of Section 5.1. Note that if X contains `1, the set K above can be identified as a norm bounded, weakly sequentially complete convex subset of X which, by Section 5.1, lacks Property (P). Conversely, if X does not contain a copy of `1 and K ⊆ X is weakly se- quentially complete then K is weakly compact. Indeed, if {xn} is a sequence in K, then by Rosenthal’s `1 Theorem (see [15] or [12, Chapter XI]), it has a weak Cauchy subsequence which, by weak sequential completeness, is weakly convergent. And hence, by Eberlein-Smulian Theorem, K is weakly compact and, by Corollary 5.3.1, K has the Property (P). The proof of (b) is similar. Remark : Can one relax the assumption of weak sequential completeness in any of the two statements above ?
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