Three Non-Linear Problems on Normed Spaces A

THREE NON-LINEAR PROBLEMS ON NORMED SPACES

A dissertation submitted to Kent State University in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy

Francisco J. Garc´ıa

February, 2007 Dissertation written by

Francisco J. Garc´ıa

B.S., University of C´adiz, Spain, 2000

M.S., University of C´adiz, Spain, 2004

Ph.D., University of C´adiz, Spain, 2005

M.A., Kent State University, 2006

Ph.D., Kent State University, 2007

Approved by

Richard M. Aron, Chair, Doctoral Dissertation Committee

Andrew Tonge, Members, Doctoral Dissertation Committee

Per H. Enﬂo,

Johnnie W. Baker, (Outside Person)

Paul S. Wang, (Graduate Representative)

Accepted by

Andrew Tonge, Chair, Department of Mathematical Sciences

Jerry Feezel, Dean, College of Arts and Sciences

ii TABLE OF CONTENTS

ACKNOWLEDGEMENTS ...... v

INTRODUCTION ...... 1

1 THE LINEABILITY PROBLEM FOR FUNCTIONALS ...... 7

1.1 Preliminaries ...... 7

1.2 Lineability of NA (X)...... 16

1.3 Lineability of X∗ \ NA (X)...... 21

1.4 Density of X∗ \ NA (X)...... 24

2 THE MINIMUM-NORM PROBLEM FOR TRANSLATIONS ..... 31

2.1 Preliminaries ...... 31

2.2 Minimum-norm elements and norm-attaining functionals ...... 42

2.3 Non-complete normed spaces having only norm-attaining functionals . . . . 45

2.4 Partial solutions ...... 48

3 THE BANACH-MAZUR CONJECTURE FOR ROTATIONS ...... 51

3.1 Preliminaries ...... 51

iii 3.2 Geometrical conditions ...... 58

3.3 Topological conditions ...... 60

3.4 Intermediate solutions ...... 63

BIBLIOGRAPHY ...... 68

iv ACKNOWLEDGEMENTS

There are several people to whom I should be giving thanks for my stay at Kent State

University: Richard Aron, Joe Diestel, Andrew Tonge, Artem Zvavitch, Per Enﬂo, and

Juan Seoane. I would like to use this opportunity to thank all of them very much: Thanks to Richard for accepting me as his student and for all the invitations to his house; thanks to

Joe and Artem for the recommendation letters that they wrote for me; thanks to Andrew for all the forms that he signed for me; thanks to Per for his excellent mathematical support; and, ﬁnally, thanks very much Juan for making my stay at Kent as good as possible.

Also, I want to give thanks to the Department of Mathematical Sciences at Kent State

University for the privilege of having been supported by the Graduate Program. Specially,

I wish to thank Virginia Wright (Secretary of the Math Department,) Misty Tackett (Sec- retary of the Math Graduate Students,) and Michelle Cordier (Secretary Assistant,) for attending me so well every time I needed their help.

Other people I would like to thank are my friends in the Math Department like Mienie,

Ramiro, Tom, Hongcheng, Jeﬀ, Terry, Brian, Antonia, Daniele, and Alejandro. Thanks to all of you guys, overall for the rides from and to Cleveland Hopkins International Airport.

Finally I mention all my friends and my family from my town. Brothers and sisters, thanks for always being there for me. Please, never change!

Francisco J. Garc´ıa

Febraury 2007

v GARC´IA, FRANCISCO J., Ph.D., February, 2007 PURE MATHEMATICS

THREE NON-LINEAR PROBLEMS ON NORMED SPACES (71 pp.)

Director of Dissertation: Richard M. Aron

In this dissertation, we will study the following three non-linear problems:

1. The lineability problem for functionals.

2. The minimum-norm problem for translations.

3. The Banach-Mazur conjecture for rotations.

As far as we know, all of them are currently open, and we believe that any approach to their solutions will constitute a work of great interest to the mathematical community. In this dissertation, we obtain progresses that lead to partial solutions of these problems. INTRODUCTION

The ﬁrst chapter of this dissertation is on the structure of the set of functionals on a Ba- nach space that attain their norm. The reason for this interest comes from an open problem concerning the lineability of the set NA (X) of norm-attaining functionals on a Banach space

X. Speciﬁcally, it is unknown if NA (X) always contains an inﬁnite dimensional, or even a 2-dimensional subspace. The paper [9] is an excellent reference about this problem. As we will see, from this problem arises another one corresponding to the complementary set,

X∗ \ NA (X) , of non-norm-attaining functionals on X. In concrete terms, for non-reﬂexive spaces X, is X∗ \ NA (X) always lineable, or dense? Our best results from this chapter are the following:

1. Every Banach space admitting an inﬁnite dimensional separable quotient can be equiv-

alently renormed to make the set of norm-attaining functionals lineable.

2. There exists a non-reﬂexive dual Banach space that cannot be equivalently dually

renormed to make the set of non-norm-attaining functionals even 2-lineable.

3. Every Banach space can be equivalently renormed to make the set of non-norm-

attaining functionals non-dense.

All the results presented in this chapter, unless explicitly stated, can be found in [2].

The second chapter consists of the study of norm-attaining functionals on non-complete spaces. In 1964, James proved that a Banach space is reﬂexive if and only if every functional is norm-attaining (see [28].) Besides, James showed in 1971 that the completeness hypothesis cannot be skipped, since he found a non-complete normed space on which every

1 2

functional is norm-attaining (see [29].) Afterwards, Blatter proved in 1976 that a necessary and sufficient condition for a normed space to be reflexive is that every closed convex set has a minimum-norm element (see [17].) This was the first time when minimum-norm elements appeared in the literature. However, they appeared again in 2005, when Aizpuru and the author generalized Blatter’s result by proving that a normed space is reflexive if and only if every bounded closed convex set with non-empty interior has a minimum-norm element

(see [4].) Here, the bounded closed convex sets with non-empty interior became important, and with this we have the following conjecture: A necessary and suﬃcient condition for a normed space to have only norm-attaining functionals is that every bounded closed convex set with non-empty interior can be translated to have a non-zero minimum-norm element.

Another interesting problem arising from this is to characterize the reﬂexive spaces that contain a proper dense subspace on which every functional is norm-attaining. Our best results from this chapter are the following:

1. A necessary and suﬃcient condition for a normed space to have only norm-attaining

functionals is that every closed convex subset with non-empty interior and non-empty

boundary can be translated to have a non-zero minimum-norm element.

2. If a non-complete norm space is so that every bounded closed convex subset of it with

non-empty interior can be translated to have a non-zero minimum-norm element, then

its completion cannot be rotund.

3. Every inﬁnite dimensional reﬂexive Banach space can be equivalently renormed to be

non-rotund and to not possess dense proper subspaces on which every functional is

norm-attaining.

All the results presented in this chapter, unless explicitly stated, can be found in [24]. 3

The third and last chapter is about a famous old problem. It has been always well known that every Hilbert space is transitive. On the other hand, it seems likely that

Banach already knew some examples of transitive Banach spaces that were not Hilbert.

Apparently, Mazur, who was working with separable spaces at that time, asked Banach for the existence of transitive and separable Banach spaces diﬀerent from `2. Banach came up

with no answer and this is how the Banach-Mazur conjecture was born: every transitive

and separable Banach space is Hilbert. It is believed that Mazur conjectured a positive

answer. An excellent reference is [12], and another good one is [18]. Basically, there are two

geometrical properties clearly diﬀerentiated: Rotundity and smoothness. Therefore, it is

natural to wonder whether a transitive and separable Banach space is rotund and smooth.

In 1932 (see [33]) Mazur proved that in every separable Banach space the set of smooth

points of the unit ball is a Gδ dense subset of the unit sphere. As a consequence, every

transitive and separable Banach space is smooth. Now, the question remains of whether a

transitive and separable Banach space is rotund. Our best results from this chapter are the

following:

1. If the unit ball of a smooth and separable Banach space is free of rotund points, then

the set of non-norm-attaining functionals on it contains a Gδ dense subset.

2. If a transitive and separable Banach space is so that the set of non-norm-attaining

functionals on it is not dense, then the space is rotund, the set of rotund points of the

unit ball of its dual is dense in the unit sphere of its dual, and the set of norm-attaining

functionals on it is open.

3. If the unit ball of a transitive and separable Banach space either has normal structure

or is dentable, then the space is rotund.

All the results presented in this chapter, unless explicitly stated, can be found in [5]. 4

As the reader may suppose, in much of this dissertation we will make use of various classical notions, such as smoothness, rotundity, etc., from the geometry of Banach spaces.

Now, we will brieﬂy describe the notation we have followed through the whole dissertation.

Assume that X denotes a topological vector space. Then:

1. If X is metrizable then BX (x, δ), UX (x, δ), and SX (x, δ) will denote the closed ball

of center x and radius δ, the open ball of center x and radius δ, and the sphere of

center x and radius δ, respectively.

2. If X is normable then BX , UX , and SX will denote the closed unit ball, the open unit

ball, and the unit sphere, respectively.

3. X∗ will denote the topological dual of X.

4. If M is any subset of X, then:

(a) span (M) will denote the vector subspace generated by M and span (M) will

denote the closed vector subspace generated by M.

(b) co (M) will denote the convex hull of M and co (M) will denote the closed convex

hull of M.

and the closure of M.

To ﬁnish the introduction, we will review some basic concepts from the geometry of

Banach spaces. Again, assume that X denotes a topological vector space and consider a

convex subset H of X. Then:

1. The slice of H determined by f ∈ X∗ \{0} and δ > 0 is deﬁned by slc (H, f, δ) =

{h ∈ H : f (h) ≥ sup (Ref (H)) − δ} . 5

2. A subset C of H is said to be a face of H if it is convex and veriﬁes that, for every

x, y ∈ H and every α ∈ (0, 1) with αx + (1 − α) y ∈ C, then x, y ∈ C.

3. A subset C of H is said to be an exposed face of H if there exists f ∈ X∗ such that

C = {x ∈ H : f (x) = sup (Ref (H))} .

4. A subset C of H is said to be a strongly exposed face of H if there exists f ∈ X∗ such

that C = {x ∈ H : f (x) = sup (Ref (H))} and for every open set U of H containing

C there is δ > 0 such that slc (H, f, δ) ⊆ U.

5. C is said to be a maximal face of H if it is a non-trivial face maximal among the

non-trivial faces of H.

6. A subset C of H is said to be a smooth face of H if it is a non-trivial face and has

non-empty interior relative to the boundary of H.

7. A point c ∈ H is said to be an extreme point of H if {c} is a face of H. We will let

ext (H) denote the set of extreme points of H.

8. A point c ∈ H is said to be an exposed point of H if {c} is an exposed face of H. We

will let exp (H) denote the set of exposed points of H.

9. A point c ∈ H is said to be a strongly exposed point of H if {c} is a strongly exposed

face of H. We will let exps (H) denote the set of strongly exposed points of H.

10. A point c ∈ H is said to be a rotund point of H if {c} is a maximal face of H. We

will let rot (H) denote the set of rotund points of H.

Now, assume that X is a normed space. Then:

1. A point x ∈ SX is said to be a smooth point of BX if every sequence (fn) ⊂ SX∗ n∈N 6

∗ such that (fn (x)) converges to 1 is ω -convergent. The set of the smooth points n∈N

of BX is denoted by smo (BX ).

2. A point x ∈ SX is said to be a strongly smooth point of BX if every sequence (fn) ⊂ n∈N

SX∗ such that (fn (x)) converges to 1 is convergent. The set of the strongly smooth n∈N

points of BX is denoted by smos (BX ).

Finally, if X is a smooth normed space, then the dual map of X is deﬁned as the function

∗ ∗ JX : X −→ X such that, for every x ∈ X, JX (x) is the unique element in X verifying

2 that kJX k = kxk and JX (x)(x) = kxk .

More background information and notation can be obtained in [19], [20], and [34]. We

also refer the reader to [3], [10], [11], and [26] for a wider perspective of these concepts.

Francisco J. Garc´ıa

February 2007 CHAPTER 1

THE LINEABILITY PROBLEM FOR FUNCTIONALS

1.1 Preliminaries

We begin by recalling the following relatively new concepts related to the “size” of subsets of Banach spaces.

Deﬁnition 1.1.1 (Gurariy, 1991) A subset M of a Banach space is said to be

1. n-lineable if M ∪ {0} contains an n-dimensional vector subspace;

2. lineable if M ∪ {0} contains an inﬁnite dimensional vector subspace;

3. dense-lineable if M ∪ {0} contains an inﬁnite dimensional dense vector subspace;

4. spaceable if M ∪ {0} contains an inﬁnite dimensional closed vector subspace.

In order to have a better perspective of these new concepts, we refer the reader to the papers in [7], [8], and [25], where it is proved that several pathological properties occur more often than one might expect in the sense described in the deﬁnitions above.

The main questions that we want to attack in this chapter are the following:

Problem 1.1.2 (Aron/Gurariy, 2004) Is the set of norm-attaining functionals on an inﬁnite dimensional Banach space always lineable?

7 8

The very ﬁrst results relative to this question and trying to answer it are presented now.

All of them appeared in [9].

Remark 1.1.3 (Bandyopadhyay/Godefroy, 2005) Let X be a Banach space. Then,

X ⊆ NA (X∗) and hence NA (X∗) is spaceable if X is inﬁnite dimensional. Observe that X

is reﬂexive if and only if X = NA (X∗).

Theorem 1.1.4 (Bandyopadhyay/Godefroy, 2005) Let X be a Banach space such that

∗ BX∗ is ω -sequentially compact. The following conditions are equivalent:

1. There exists an equivalent norm on X that makes NA (X) spaceable.

2. There exists an inﬁnite dimensional quotient of X which is isomorphic to a dual space.

Theorem 1.1.5 (Bandyopadhyay/Godefroy, 2005) Let X be an Asplund Banach space with the Dunford-Pettis property. Then, the closed vector subspaces of NA (X) are ﬁnite dimensional. In particular, X cannot be equivalently renormed to make NA (X) spaceable.

Theorem 1.1.6 (Bandyopadhyay/Godefroy, 2005) Let X be a Banach space with the

Radon-Nikodym property or an almost locally uniformly rotund norm. Then, we have that span (NA (X)) = X∗. In particular, if X is not reﬂexive but has the Radon-Nikodym property then X cannot be equivalently renormed to make NA (X) be a vector space.

Now we will present several results on concrete spaces giving partially solutions to the previous question.

We begin with spaces of continuous functions, but ﬁrst we need the following two lemmas involving compact Hausdorﬀ spaces and subspaces of `1. 9

Lemma 1.1.7 Let K be an inﬁnite compact Hausdorﬀ topological space. Then, there exists a sequence (lj) ⊆ K such that li 6= lj if i 6= j, and verifying one of the following j∈N conditions:

1. The sequence (lj) converges to some point ∞ ∈ K. j∈N

2. The set {lj : j ∈ N} does not contain isolated points.

Proof Firstly, note that if K is scattered then we deduce (see [32]) that K is sequentially compact. Therefore, we can assume that K is not scattered. It is known (see [32]) that

K = P ∪ D where P is closed and perfect (the perfect kernel of K) and D is open and scattered (the scattered kernel of K.) (Note that this decomposition holds not only for compact Hausdorﬀ spaces, but for any topological space.) Now, there exists a sequence

(lj) ⊆ P such that li 6= lj for i 6= j and the set {lj : j ∈ } does not contain isolated j∈N N points.

Lemma 1.1.8 There exists an inﬁnite dimensional vector subspace M of `1 such that every

(αi) ∈ M \{0} veriﬁes one of the following conditions: ∈N

1. The set {j ∈ N : Re (αj) > 0} is non-empty and ﬁnite and the set {j ∈ N : Re (αj) < 0} is inﬁnite.

2. The set {j ∈ N : Re (αj) > 0} is inﬁnite and the set {j ∈ N : Re (αj) < 0} is non- empty and ﬁnite. 10

Proof Let us consider the following elements of `1:   y = 1, −1, 1 , 1 , 1 , 1 , 1 , 1 ,...  1 23 24 25 26 27 28   1 1 1 1  y2 = 0, 0, 1, −1, 5 , 6 , 7 , 8 ,...  3 3 3 3   1 1  y3 = 0, 0, 0, 0, 1, −1, 47 , 48 ,... .  .  .    y = 0,..., 0, 1, −1, 1 , 1 ,...  k (k+1)2k+1 (k+1)2k+2   .  .

We will take M := span {yk : k ∈ N}. Indeed, assume that 0 6= y ∈ M and take λ1, . . . , λk ∈

K not all zero so that y = λ1y1 + ··· + λkyk. Then, y has the form ! λ1 λ1 λ1 λ2 λk (1.1) λ1, −λ1, + λ2, − λ2,..., + + ··· + ,... . 23 24 22k+1 32k+1 (k + 1)2k+1

Therefore, from equation (1.1) we can see that M is the desired vector space.

In the above proof we note the following remark.

Remark 1.1.9 If y ∈ span {yk : k ∈ N}\ span {yk : k ∈ N}, then y has the form

λ λ λ λ λ λ λ , −λ , 1 + λ , 1 − λ , 1 + 2 + λ , 1 + 2 − λ ,... , 1 1 23 2 24 2 25 35 3 26 36 3

which shows that the closure of M does not serve our purposes.

Theorem 1.1.10 Let K be an inﬁnite compact Hausdorﬀ topological space. Then, both

NA (C (K)) and C (K)∗ \ NA (C (K)) are lineable. If, in addition, K possesses a non-trivial convergent sequence, then NA (C (K)) is spaceable.

Proof In the ﬁrst place, let us consider the vector space V of all regular Borel measures µ, with bounded variation on K, such that there exists a ﬁnite set L ⊂ K with |µ| (K \ L) = 0. 11

(Observe that V is no other than the vector space generated by {δt : t ∈ K}.) Also notice P that if µ ∈ V and kµk = 1, then l∈L |µ ({l})| = 1. Now, deﬁne the continuous function f : L −→ [−1, 1] as  |µ({l})|  if µ ({l}) 6= 0, f (l) = µ({l})  0 if µ ({l}) = 0, for all l ∈ L. Since K is normal and L is closed, we deduce by Urysohn’s Lemma that there

˜ ˜ exists a continuous extension f : K −→ [−1, 1] of f. To ﬁnish, we have that f = 1 and Z X X fdµ˜ = f (l) µ ({l}) = |µ ({l})| = 1. K l∈L l∈L

In the second place, by using Lemma 1.1.7, we will ﬁrstly assume the existence in K

of a non-trivial convergent sequence. So, let L = (lj) ⊆ K be a convergent sequence to j∈N S∞ some point ∞ ∈ K and verifying that li 6= lj if i 6= j. We can set N = k=1 Nk where

the Nk’s are pairwise disjoint containing inﬁnitely many even numbers and inﬁnitely many S∞ odd numbers. Thus, we can write L = k=1 Lk where l ∈ Lk if and only if l = lj for some

j ∈ Nk. Now choose, for each k ∈ N, any regular Borel measure µk with bounded variation j on K, such that |µk| (K \ Lk) = 0 and the sign of each µk ({lj}) is (−1) , for j ∈ Nk. Let

P∞ n 0 6= µ ∈ W := span {µk : k ∈ } and consider a sequence ( λ µk) converging to µ, N k=1 k n∈N n where (λ ) ∈ c00 for every n ∈ . Since k k∈N N ∞ ! X n λk µk ({l}) −→ µ ({l}) as n → ∞ k=1 for every l ∈ L, we deduce that |µ| (K \ L) = 0. Now, since µ 6= 0, there exists l0 ∈ L such

P∞ n that µ ({l0}) 6= 0. There also exists k0 ∈ N such that l0 ∈ Lk0 and so ( k=1 λk µk)({l0}) = λn µ ({l }) for every n ∈ , which means that k0 k0 0 N

n µ ({l0}) λk0 −→ λk0 := as n → ∞. µk0 ({l0})

Then, if l ∈ Lk0 we have that µ ({l}) = λk0 µk0 ({l}). Therefore, the signs of µ ({l}) are P alternating for l ∈ Lk0 . Finally, assume that kµk = 1. Then l∈L |µ ({l})| = 1. Now if 12

R P f ∈ C (K), kfk = 1, and K fdµ = 1, we have that l∈L f (l) µ ({l}) = 1, which means |µ({l})| that f (l) = µ({l}) for all l ∈ L such that µ ({l}) 6= 0. However, this is impossible because

(f (lj)) converges to f (∞). Therefore, each non-zero measure µ ∈ W is, in fact, in j∈N C (K)∗ \ NA (C (K)).

In the third and last place, we will assume the existence in K of a non-trivial sequence with no isolated points. So, let (lj) ⊆ K be a sequence free of isolated points and verifying j∈N ∗ that li 6= lj if i 6= j. Let us consider in C (K) the inﬁnite dimensional vector subspace nP∞ o W := αjδl :(αj) ∈ M , where M is the inﬁnite dimensional vector subspace j=1 j j∈N constructed in Lemma 1.1.8. Observe that W is linearly isometric to M. Let µ ∈ W with

P∞ kµk = 1, so that µ is of the form αjδl with (αj) ∈ M. Since kµk = 1, we have that j=1 j j∈N P∞ j=1 |αj| = 1. We can assume, without loss of generality, that A = {j ∈ N : Re (αj) < 0} is non-empty and ﬁnite and B = {j ∈ N : Re (αj) > 0} is inﬁnite. Let f ∈ C (K, R) with R P∞ P∞ kfk = K fdµ = 1. Then, j=1 αjf (lj) = 1 = j=1 Re (αj) f (lj), and from here we

deduce that, if j ∈ A then f (lj) < 0, and if j ∈ B then f (lj) > 0. However, by taking into

account that lk is a cluster point of {lj : j ∈ N} for every k ∈ A, we have that there must

be an inﬁnite amount of natural numbers k ∈ N with f (lk) < 0, which is a contradiction. Therefore, each non-zero measure µ ∈ W is, in fact, in C (K)∗ \ NA (C (K)).

In view of Theorem 1.1.5 we have the following remark.

Remark 1.1.11 Given any compact Hausdorﬀ topological space K we have the equivalence of the following assertions (see [21] and [27]):

1. The space C (K) has the hereditary Dunford-Pettis property.

2. The space C (K) is Asplund. 13

3. The space K is scattered.

Therefore, if K is scattered then for every inﬁnite dimensional closed subspace X of C (K) we have that NA (X) is not spaceable.

We now turn to spaces of integrable functions.

∗ ∗ Lemma 1.1.12 Let (Ω, Σ, µ) be a σ-ﬁnite measure space. Then, an element x ∈ L1 (µ) (=

L∞ (µ)) attains its norm if and only if there is a measurable set A with µ (A) > 0 satisfying that |x∗ (t)| = kx∗k for all t ∈ A.

Proof First, if some subset A of positive measure satisﬁes the above condition, then the function θ :Ω −→ K given by  |x∗(t)|  ∗ if t ∈ A, θ (t) = x (t)  0 otherwise,

is measurable and clearly satisﬁes that kθk∞ ≤ 1. Since µ is σ-ﬁnite, there is a measurable

χB subset B ⊂ A such that 0 < µ (B) < ∞. Thus, the element g := θ µ(B) is a function in the unit ball of L1 (µ) and satisﬁes the inequality

Z Z Z ∗ ∗ ∗ ∗ ∗ (1.2) kx k∞ = x gdµ ≤ |x g| dµ ≤ kx k∞ |g| dµ = kx k∞ . Ω Ω Ω

∗ Conversely, assume that x 6= 0 attains its norm at g ∈ SL1 (µ). Then we have that

∗ ∗ ∗ both x and g verify equation (1.2). Hence we obtain that |x (t) g (t)| = kx k∞ |g (t)| almost everywhere. If we set C = {t ∈ Ω: g (t) 6= 0}, then C is measurable and µ (C) > 0.

∗ ∗ Therefore, there is a subset Z ⊂ C such that µ (Z) = 0 and |x (t)| = kx k∞ for all t ∈ A := C \ Z. 14

Theorem 1.1.13 Assume that (Ω, Σ, µ) is a σ-ﬁnite measure space such that L1 (µ) is

∗ inﬁnite dimensional. Then, both NA (L1 (µ)) and L1 (µ) \ NA (L1 (µ)) are spaceable. If, in

∗ ∗ addition, (Ω, Σ, µ) is atomless, then L1 (µ) \ NA (L1 (µ)) is dense in L1 (µ) .

Proof Let us ﬁrst show the spaceability of NA (L1 (µ)). Since L1 (µ) is inﬁnite dimensional, there exists a countable family (An) of pairwise disjoint measurable sets such that, n∈N

for every n ∈ N, µ (An) > 0. Now, consider the inﬁnite dimensional closed subspace P∞ M := αnχA :(αn) ∈ c0 . Observe that M is a closed subspace of L∞ (µ) linearly n=1 n n∈N

isometric to c0. Now, if (αn) ∈ c0 then there exists n ∈ such that |αn| = (αn) = n∈N N n∈N P∞ P∞ k n=1 αnχAn k, and if t ∈ An then | n=1 αnχAn (t)| = |αn|.

∗ Let us second show the spaceability of L1 (µ) \ NA (L1 (µ)). Since L1 (µ) is inﬁnite

dimensional, there exists a countable family (An) of pairwise disjoint measurable sets n∈N S∞ k k such that, for each n ∈ , µ (An) > 0 and An = Bn, where Bn is a sequence of N k=1 k∈N pairwise disjoint measurable sets of positive measure. Next, consider the vector subspace of L∞ (µ) deﬁned as

( ∞ ∞ ! ) X X M := x (n) α (k) χ k : x ∈ ` , Bn ∞ n=1 k=1 where α := (α (k)) is a ﬁxed strictly increasing sequence of real positive numbers such k∈N ∗ P∞ P∞ that kαk = 1. If x = x (n) α (k) χ k is an element of M, then ∞ n=1 k=1 Bn

∗ ∗ kx k∞ = sup {kx χAn k∞ : n ∈ N}

n ∗ o = sup x χBk : n, k ∈ N n ∞

= sup {|x (n) α (k)| : n, k ∈ N}

= sup {|x (n)| : n ∈ N}

= kxk∞ . 15

∗ ∗ S For 0 6= x ∈ M and t ∈ Ω, we clearly have that x (t) = 0 for all t∈ / An. If t ∈ An n∈N k ∗ for some n ∈ N, then there is k ∈ N such that x ∈ Bn and so we have that |x (t)| =

∗ ∗ ∗ |x (n) α (k)| < kxk∞ = kx k∞. Therefore for every t ∈ Ω, we have that |x (t)| < kx k∞,

∗ and thus x cannot attain its norm on L1 (µ).

∗ ∗ We now turn to density of L1 (µ) \ NA (L1 (µ)). We take x ∈ L∞ (µ) and r > 0. We will show that the ball centered at x∗ of radius r contains a functional that does not attain its norm. We can clearly assume that x∗ attains its norm and x∗ 6= 0, and hence there is

∗ ∗ a measurable set A ⊂ Ω with µ (A) > 0 such that |x (t)| = kx k∞ for all t ∈ A. Since µ S is atomless, we can write A = An, where (An) is a sequence of pairwise disjoint n∈N n∈N measurable sets such that µ (An) > 0 for each n ∈ N. Next, we choose a strictly increasing convergent sequence (rn) of real numbers such that rn ≥ 1 for all n ∈ and whose limit n∈N N r ∗ l satisﬁes l ≤ 1 + ∗ . We will denote by y the element of L∞ (µ) given by kx k∞

∞ ∗ X ∗ ∗ y = rnχAn x + (1 − χA) x , n=1 where the above convergence is pointwise. Then

∗ ∗ ∗ ky k∞ = max {ky χAk∞ , k(1 − χA) y k∞}

∗ ∗ ≤ max {sup {rn : n ∈ N} kx k∞ , kx k∞}

∗ = l kx k∞ .

∗ ∗ ∗ ∗ ∗ In fact, rn kx k∞ = krnχAn x k∞ ≤ ky k∞ for all n ∈ N, so l kx k∞ ≤ ky k∞, and we

∗ ∗ have that ky k = l kx k . Since the sequence (rn) is strictly increasing and since the ∞ n∈N essential supremum is not attained at any measurable set with positive measure, y∗ does 16

not attain its norm at L1 (µ). Also,

∞ X ky∗ − x∗k = (r − 1) χ x∗ ∞ n An n=1 ∞ ∗ ≤ sup {rn − 1 : n ∈ N} kx k∞

∗ ≤ kx k∞ (l − 1)

≤ r.

∗ ∗ ∗ As a consequence, we have checked that y ∈ x + rBL1(µ) and y does not attain its norm.

Remark 1.1.14 If the measure space (Ω, Σ, µ) is σ-ﬁnite and has an atom A of ﬁnite

∗ ∗ 1 ∗ measure, then for all x ∈ L∞ (µ) such that kx − χAk∞ < 2 , we have that x is norm- ∗ attaining. As a consequence, L1 (µ) \ NA (L1 (µ)) is not dense.

Note that these previous results motivate the following question.

Question 1.1.15 Is the set of non-norm-attaining functionals on a non-reﬂexive Banach space always lineable?

1.2 Lineability of NA (X)

We will begin by providing positive results, in other words, suﬃcient conditions for the set NA (X) to be lineable (sometimes under renormings.)

In the ﬁrst place, we have the following result.

Proposition 1.2.1 Let X be an inﬁnite dimensional Banach space. Then, for every n ∈ N, X can be equivalently renormed so that NA (X) is n-lineable. 17

Proof Observe that X is isomorphic to a Banach space of the form Y = H ⊕2 M where

∗ ∗ ∗ H a Hilbert space of dimension n and M is a Banach space. Next, Y = H ⊕2 M and

H∗ ⊆ NA (Y ).

Theorem 1.2.2 Let X be a separable Banach space with a Markushevich basis (en) . n∈N Then:

1. If (en) is monotone, then NA (X) is lineable. n∈N

2. If (en) is a monotone and shrinking Schauder basis, then NA (X) is dense-lineable. n∈N

Proof

1. We will show that every vector in

( n ) X ∗ M = αkek : αk ∈ K for 1 ≤ k ≤ n and n ∈ N k=1

is norm-attaining, where (e∗ ) is the sequence of biorthogonal functionals associated n n∈N

to (en) . Notice that, since the basis is monotone, n∈N n n ! X X α e∗ = α e∗ | , k k k k span{e1,...,en} k=1 k=1

and hence every element of M is norm-attaining.

2. Note that, if (en) is a monotone and shrinking Schauder basis, then the sequence n∈N of biorthogonal functionals (e∗ ) in X∗ is a Schauder basis for the dual X∗, and n n∈N hence M is dense in X∗. 18

Observe that the previous theorem says that any example of an inﬁnite dimensional separable Banach space such that the set of norm-attaining functionals on it is not lineable is an example of a separable space admitting no monotone Markushevich basis. On the other hand, following the line proposed by Theorem 1.2.2, we can try to see what happens for Banach spaces admitting an inﬁnite dimensional separable quotient.

Theorem 1.2.3 Let X be a Banach space. Assume that M is a closed subspace of X.

Then:

1. If M is 1-complemented and NA (M) is lineable (spaceable) then NA (X) is lineable

(spaceable.)

2. If M is proximinal and NA (X/M) is lineable (spaceable) then NA (X) is lineable

(spaceable.)

Proof

1. Let p : X −→ X be a linear projection of norm 1 so that p (X) = M. If Y ⊆

NA (M) is an inﬁnite dimensional vector space then Z := {f ◦ p : f ∈ Y } is an inﬁnite

dimensional vector space contained in NA (X). Indeed, since kpk = 1 and every f ∈ Y

is norm-attaining, we have that kfk = kf ◦ pk and f ◦ p is norm-attaining for every

f ∈ Y (at the same point at which f is.) Finally, if Y is closed then Z is closed too.

Indeed, if (fn ◦ p) is a sequence converging to g, then (fn) is Cauchy (because n∈N n∈N

kfnk = kfn ◦ pk,) therefore it converges to some f ∈ Y , and hence g = f ◦ p.

2. Let Y ⊆ NA (X/M) be an inﬁnite dimensional vector space. Then we have that Z :=

{f ◦ p : f ∈ Y } is an inﬁnite dimensional vector space contained in NA (X), where p

denotes the quotient map from X onto X/M. Indeed, since M is proximinal and every 19

f ∈ Y is norm-attaining, we have that kfk = kf ◦ pk and f ◦ p is norm-attaining for

every f ∈ Y . Finally, if Y is closed then Z is closed too. Indeed, if (fn ◦ p) is a n∈N

sequence converging to g, then (fn) is Cauchy (because kfnk = kfn ◦ pk,) therefore n∈N it converges to some f ∈ Y , and hence g = f ◦ p.

Corollary 1.2.4 Let X be a Banach space. If X admits an inﬁnite dimensional separable quotient, then X can be equivalently renormed so that NA (X) is lineable.

Proof According to [35], if X admits an inﬁnite dimensional separable quotient, then X admits an inﬁnite dimensional quotient X/M with a Schauder basis. Now, according to [22],

X/M can be endowed with an equivalent norm |·| so that X/M has a monotone Schauder

basis. Next, according to [9, Lemma 2.4], there exists an equivalent norm k|·|k on X which

coincides with the original norm on M, whose quotient norm on X/M is a positive multiple

of |·|, and which makes M proximinal. Finally, X with the norm k|·|k has a proximinal

subspace M so that NA (X/M) is lineable (by Theorem 1.2.2,) and so, by Theorem 1.2.3,

NA (X) is lineable.

Now, we will try to approach the converse of the previous corollary. To do this, the following will be helpful (see [9].)

Lemma 1.2.5 (Bandyopadhyay/Godefroy, 2005) Let X be a Banach space. Then:

1. If Y ⊆ NA (X) is a closed separable subspace not containing `1 then X has a quotient

isomorphic to Y ∗. 20

∗ 2. If BX∗ is sequentially ω -compact then every closed separable subspace Y ⊆ NA (X)

does not contain `1.

∗ Theorem 1.2.6 Let X be a Banach space such that BX∗ is sequentially ω -compact. If

NA (X) contains an inﬁnite dimensional closed subspace with an unconditional basic sequence then X admits an inﬁnite dimensional separable quotient.

Proof Let Y ⊆ NA (X) be a closed separable subspace with an unconditional basic sequence. Then, we have three possibilities for Y :

1. c0 ⊆ Y . In this case, by Lemma 1.2.5, we have that X has a quotient isomorphic to

`1.

2. `1 ⊆ Y . This is impossible if we take into account Lemma 1.2.5.

3. There exists an inﬁnite dimensional reﬂexive Banach space R contained in Y . Then,

take any closed separable subspace W of R. Then, W is reﬂexive and it does not

∗ contain `1. Again by Lemma 1.2.5 we have that X has a quotient isomorphic to W

which is separable.

At this moment, we will concentrate on providing negative results, in other words, on the non-lineability of NA (X). As we will see, the hypothesis of smoothness will be crucial for the development of this section, and everything here will be based upon the following fact.

Remark 1.2.7 Let X be a smooth Banach space. If x∗ 6= y∗ ∈ X∗ \{0} are so that

∗ ∗ kx∗k = ky∗k = x +y , then x∗ + y∗ cannot be norm-attaining. As a consequence, every 2 21

element of NA (X) ∩ SX∗ is an extreme point of BX∗ .

A direct consequence of this remark is the next suﬃcient condition.

Theorem 1.2.8 Let X be a smooth Banach space. If X∗ does not contain rotund subspaces then NA (X) is not even 2-lineable.

Remark 1.2.9 Observe that the Banach space c0 almost veriﬁes the conditions of the pre-

vious theorem, because according to the Mazur Theorem (see [33]) we have that smo (Bc0 )

is a Gδ dense subset of Sc0 . Besides, c0 is also closed to verify the thesis of Theorem 1.2.8

since rot (B`1 ) = ∅.

1.3 Lineability of X∗ \ NA (X)

In this section we will show that the answer to Question 1.1.15 is negative. Nevertheless,

we will begin by presenting some positive results, in other words, by showing conditions for

X∗ \ NA (X) to be spaceable.

∗ ∗ Remark 1.3.1 Let X be a smooth Banach space. If x ∈ NA (X) ∩ SX∗ then x is not only

∗ an extreme point of BX∗ but a (ω -strongly) exposed point.

A direct consequence of this remark is the next “hint.”

Proposition 1.3.2 Let X be a smooth Banach space. If Y is a vector subspace of X∗ so

∗ that Y ∩ exp (BX∗ ) = ∅, then Y ⊆ X \ NA (X) ∪ {0}.

Lemma 1.3.3 Let X be a Banach space with a Schauder basis (en) . Then, the set n∈N

X \ {±en : n ∈ N} is spaceable. 22

Proof It can be checked that

span {e1 + e2, e3 + e4, e5 + e6,... } ⊆ X \ {±en : n ∈ N} .

Lemma 1.3.4 Let K be an inﬁnite Hausdorﬀ compact topological space. Let X be a rotund Banach space. Then, C (K,X) \ ext BC(K,X) is spaceable.

Proof Let us ﬁx an arbitrary s ∈ K. Then, consider the continuous linear operator

δs : C (K,X) −→ X

f 7−→ f (s) .

Since δs is a non-zero operator and C (K,X) is inﬁnite dimensional, we deduce that ker (δs) is an inﬁnite dimensional closed subspace of C (K,X). Now, according to [15],

ext BC(K,X) = {f ∈ C (K,X): kf (t)k = 1 for all t ∈ K} ,

therefore ker (δs) ⊆ C (K,X) \ ext BC(K,X) .

Theorem 1.3.5 Let X be a smooth Banach space. Assume that either one of the following conditions hold:

∗ 1. `1 ⊆ X .

2. There exists an inﬁnite compact Hausdorﬀ topological space K and a rotund Banach

space Y such that C (K,Y ) ⊆ X∗.

Then, X∗ \ NA (X) is spaceable. 23

Proof

1. In the ﬁrst place, observe that, if (en) denotes the ` -basis, then ` \{±en : n ∈ }∩ n∈N 1 1 N

exp (BX∗ ) = ∅. The result follows now from Proposition 1.3.2 along Lemma 1.3.3.

2. In the second place, in accordance with Lemma 1.3.4, C (K,Y ) \ ext BC(K,Y ) is

spaceable. Finally, by Proposition 1.3.2 we have that X∗ \ NA (X) ∪ {0} is spaceable.

Now, it is time to take care of the negative result, which actually solves Question 1.1.15.

Theorem 1.3.6 Let X be a Banach space such that X is a maximal subspace of X∗∗.

Then, X∗∗ \ NA (X∗) is not even 2-lineable.

Proof If Y is a vector space contained in X∗∗ \NA (X∗)∪{0}, then X ∩Y = {0}, therefore

Y has dimension at most 1.

In virtue of [34, Theorem 4.5.7] and [34, Theorem 4.5.9] we have the following remark.

Remark 1.3.7 The James Space J veriﬁes that J is of codimension 1 in its bidual J ∗∗,

and is isometrically isomorphic to J ∗∗. In particular, by Theorem 1.3.6, J ∗∗∗ \ NA (J ∗∗) is

not 2-lineable, since J ∗ is of codimension 1 in J ∗∗∗, therefore J ∗ \ NA (J ) is not 2-lineable

either.

We want to ﬁnish this section by presenting a generalization of Theorem 1.3.6.

Theorem 1.3.8 Let X be a Banach space such that X is a maximal subspace of X∗∗. Then

X∗ cannot be equivalently dually renormed to make X∗∗ \ NA (X∗) be 2-lineable. 24

Proof If k·k is an equivalent dual norm on X∗ then there exists an equivalent norm |·| on

X so that |·|∗ = k·k. Now, (X, |·|) veriﬁes the conditions of Theorem 1.3.6.

1.4 Density of X∗ \ NA (X)

In this section we pretend to take advantage of some geometrical techniques to keep studing the set of non-norm-attaining functionals in a topological way. In particular, we will begin by ﬁnding some suﬃcient conditions for the set X∗ \ NA (X) to be dense.

Lemma 1.4.1 Let X be a Banach space. Then,

[ {C ⊆ SX : C is a face of BX and diam (C) > 0} ⊆ cl (SX \ ext (BX )) .

Proof Let C be a non-trivial face of BX and x ∈ C. Then there is y ∈ C \{x} and so the

segment between x and y lies in C ⊂ SX . To conclude the proof, it suﬃces to observe that for every t ∈ (0, 1), tx + (1 − t) y is not an extreme point of BX .

Theorem 1.4.2 Let X be a smooth Banach space. Then,

[ {C ⊆ SX∗ : C is a face of BX∗ and diam (C) > 0} ⊆ cl (SX∗ \ NA (X)) .

In particular, if cl (SX∗ \ NA (X)) contains all rotund points of BX∗ , then SX∗ \ NA (X) is dense in SX∗ .

Proof According to Remark 1.2.7 we have that SX∗ \ NA (X) ⊇ SX∗ \ ext (BX∗ ), and hence cl (SX∗ \ NA (X)) ⊇ cl (SX∗ \ ext (BX∗ )). In view of Lemma 1.4.1, we have that 25

cl (SX∗ \ ext (BX∗ )) contains all the elements in SX∗ which are not rotund points. So the stated result follows from the assumption.

∗ Corollary 1.4.3 Let X be a smooth Banach space. If rot (BX∗ ) = ∅ then X \ NA (X) is dense.

Next remark shows that the assumption of smoothness in the above result cannot be

dropped. In fact, there are spaces with a dense set of smooth points in the unit sphere and

satisfying the second assumption that we imposed, but not the conclusion.

Remark 1.4.4 The unit ball of `∞ has no rotund points and S`1 possesses a dense set of smooth points. However, as we already remarked, NA (`1) has non-empty interior and so

S`∞ \ NA (`1) is not dense in S`∞ .

Now, what we will do in the ﬁnal part of this section is give another proof of the fact that every real Banach space can be equivalently renormed so that the set of non-norm-attaining functionals is non-dense (see [1] and [30].) We want to indicate that we did not know of the existence of these preceding proofs when we obtained ours.

Theorem 1.4.5 Let X be a Banach space. Then,

[ ∗ ∗ cl (SX \ NA (X)) ⊆ bdSX∗ (C): C is a face of BX .

In particular, we have the following:

1. If BX∗ possesses a smooth face, then SX∗ \ NA (X) cannot be dense in SX∗ . 26

2. If the union of the smooth faces of BX∗ is dense in SX∗ , then SX∗ \NA (X) is nowhere

dense in SX∗ .

∗ Proof In the ﬁrst place, let x ∈ SX∗ \NA (X) and assume that there exists a smooth face C

∗ ∗ of BX such that x ∈ intSX∗ (C). By the Bishop-Phelps Theorem, there is a norm-attaining

∗ ∗ functional y ∈ intSX∗ (C). Let x ∈ SX so that y (x) = 1. Now, intSX∗ (C) is contained in

∗ ∗ the set of all smooth points of BX . Since intSX∗ (C) is a convex set contained in SX , we deduce that x (x∗) = 1; that is, x∗ (x) = 1 which is a contradiction. In the second place, we

∗ have that given any two distinct faces C and D of BX , intSX∗ (C)∩intSX∗ (D) = ∅, and on

∗ the other hand, given any face C of BX with intSX∗ (C) 6= ∅, cl intSX∗ (C) = C. There-

S ∗ S ∗ fore, the sets bdSX∗ (C): C is a face of BX and intSX∗ (C): C is a face of BX

∗ S ∗ are disjoint. Since their union is SX , we have that bdSX∗ (C): C is a face of BX is closed.

1. By hypothesis, [ ∗ ∗ bdSX∗ (C): C is a face of BX 6= SX .

2. By hypothesis,

[ ∗ ∗ intSX∗ (cl (SX \ NA (X))) ⊆ intSX∗ bdSX∗ (C): C is a face of BX

= ∅.

Theorem 1.4.5 leads us to the idea that, for non-density of the set X∗ \ NA (X), it is

suﬃcient to ﬁnd an equivalent norm on X so that BX∗ has a smooth face. For this, the following result, which explains the relation between the faces of B L in terms of the X ∞ Y faces of BX and BY , will be helpful. 27

Theorem 1.4.6 Let X and Y be Banach spaces. Then:

1. If C is a maximal face of BX⊕∞Y , then C is either of the form BX × D, where D

is a maximal face of BY , or of the form E × BY , where E is a maximal face of BX .

Conversely, for D and E maximal faces of BY and BX respectively, both BX × D and

E × BY are maximal faces of BX⊕∞Y .

2. If C is a smooth face of BX⊕∞Y , then C is either of the form BX × D, where D

is a smooth face of BY , or of the form E × BY , where E is a smooth face of BX .

Conversely, for D and E smooth faces of BY and BX respectively, both BX × D and

E × BY are smooth faces of BX⊕∞Y .

Proof

1. In the ﬁrst place, assume that C is a maximal face of BX⊕∞Y . The set πX (C)×πY (C)

is a face of BX⊕∞Y as well as πX (C) and πY (C) are faces of BX and BY , respectively.

Let us see that πX (C)×πY (C) 6= BX⊕∞Y . Otherwise, keeping in mind that BX⊕∞Y =

BX × BY , we have that 0 ∈ πX (C) therefore there exists y ∈ BY so that (0, y) ∈ C.

Now, y ∈ πY (C) so −y ∈ πY (C) therefore there exists x ∈ BX so that (x, −y) ∈

1 1 1 C. Since C is convex, we deduce that 2 x, 0 = 2 (x, −y) + 2 (0, y) ∈ C. Next, 1 1 2 x, 0 ∞ ≤ 2 but C ⊂ SX⊕∞Y . As a consequence, πX (C)×πY (C) 6= BX⊕∞Y . This

means that πX (C) × πY (C) is a proper face of BX⊕∞Y . Since C ⊆ πX (C) × πY (C)

and C is maximal, we deduce that C = πX (C) × πY (C). By using again that

πX (C) × πY (C) 6= BX⊕∞Y , we deduce that either πX (C) 6= BX or πY (C) 6= BY .

Without loss, we can assume that πX (C) 6= BX . Then, πX (C) is a proper face of BX ,

so there exists a maximal face E of BX containing πX (C). Now, E × BY is a proper

face of BX⊕∞Y containing C. By maximality, they have to be equal. Conversely, both 28

BX × D and E × BY are proper faces of BX⊕∞Y , and according to what is proved

right above they have to be maximal.

2. As above, let us start by assuming that C is a smooth face of BX⊕∞Y . Since C is

maximal, C is either of the form BX × D, where D is a maximal face of BY , or of

the form E × BY , where E is a maximal face of BX . Since the projections are open,

we conclude the result. Conversely, both BX × D and E × BY are maximal faces of

B . Moreover, int (B × D) = U × int (D) and int (E × B ) = X⊕∞Y SX⊕∞Y X X SY SX⊕∞Y Y

intSX (E) × UY , so they are smooth faces.

Theorem 1.4.7 Let X be a real Banach space. Then, X can be equivalently renormed so that its new unit ball possesses a smooth face.

∗ Proof Consider x ∈ SX∗ to be a norm-attaining functional and consider a point a ∈ SX

∗ so that x (a) = 1. Now, it suﬃces to consider the equivalent norm on X given by kxk∞ =

∗ ∗ ∗ max {|x (x)| , kx − x (x) ak} for every x ∈ X. Notice that (X, k·k∞) = ker (x ) ⊕∞ Ra. Finally, we apply Theorem 1.4.6.

Corollary 1.4.8 Let X be a real Banach space. Then X can be equivalently renormed so that the set of non-norm-attaining functionals is not dense.

∗ ∗ Proof Let a ∈ SX and a ∈ SX∗ with a (a) = 1 and consider the equivalent norm on X

∗ ∗ ∗ given by kxk1 = |a (x)|+kx − a (x) ak for every x ∈ X. Notice that (X, k·k1) = ker (a )⊕1

∗ ∗ ∗ ∗ ∗ ∗ Ra. The dual norm of k·k1 is k·k1 = k·k∞, where kx k∞ = max {|a (x )| , kx − a (x ) a k} 29

∗ ∗ ∗ ∗ for every x ∈ X . Again, (X , k·k∞) = ker (a) ⊕∞ Ra . Finally, by applying ﬁrst Theorem 1.4.7 and then Theorem 1.4.5, we deduce the result.

Note that the idea used in Theorem 1.4.7 does not work for the complex case, because the unit ball of C does not possess any smooth face. Actually, we have these better results.

Lemma 1.4.9 Let X be a real Banach space. Then, X can be equivalently renormed to have a convex set in SX with non-empty interior relative to SX .

Proof Let f ∈ SX∗ be a norm-attaining functional. Then, consider the set B = BX ∩

−1 −1 1 −1 −1 f 2 , 2 . Clearly, B deﬁnes an equivalent norm on X for which both BX ∩f 2

−1 1 and BX ∩ f 2 are convex sets with non-empty interior relative to the boundary of B.

Observe that Lemma 1.4.9 provides a faster proof of Theorem 1.4.7. However, we still need Theorem 1.4.7 to obtain Corollary 1.4.8 because we cannot assure that the new norm given in Lemma 1.4.9 is always a dual norm. In fact, we have the following remark related to this topic (see [6].)

Remark 1.4.10 (Aizpuru/Garc´ıa-Pacheco, 2006) Let X be a real Banach space. If

X is not reﬂexive then there exists an equivalent norm on X∗ that cannot come from an equivalent norm on X.

Lemma 1.4.11 Let X be a complex Banach space. If C ⊂ SX is a convex set, then

intSX (C) = ∅. 30

Proof Assume that there are c ∈ C and r > 0 such that BX (c, r) ∩ SX ⊆ C. Now, choose

λ ∈ C \{1} with |λ| = 1 and |λ − 1| < r. Obviously, λc ∈ BX (c, r) ∩ SX , therefore λc ∈ C.

λc+c λc+c Since C is convex, we have that 2 ∈ C and hence 2 = 1, but this means that λ = 1, which is a contradiction.

Theorem 1.4.12 Let X be a complex Banach space. Let XR denote the underlining real

space. Then, XR can be equivalently renormed so that XR cannot be complexiﬁed.

Proof According to Lemma 1.4.9, XR can be equivalently renormed to have a convex set in the unit sphere with non-empty interior relative to the unit sphere. Now, by applying

Lemma 1.4.11, we deduce that this new norm cannot come from a complex norm. CHAPTER 2

THE MINIMUM-NORM PROBLEM FOR TRANSLATIONS

2.1 Preliminaries

In this chapter everything comes up to the following result (see [28].)

Theorem 2.1.1 (James, 1964) Let X be a Banach space. Then, X is reﬂexive if and only if every functional on X is norm-attaining.

James was asked for the possibility of removing the completeness hypothesis, and like this he came up with the following counterexample (see [29].)

Remark 2.1.2 (James, 1971) There exists a non-complete real normed space on which every functional is norm-attaining.

Later, Blatter realized the following (see [17].)

Remark 2.1.3 (Blatter, 1976) Let X be a normed space. If f ∈ X∗ \{0} and α > 0 then dist 0, f −1 (α) = α/ kfk, thus f is norm-attaining if and only if f −1 (α) has a minimum- norm element.

And then he came up with the following nice result (see [17].)

31 32

Theorem 2.1.4 (Blatter, 1976) Let X be a normed space. Then, X is complete if every closed convex set has a minimum-norm element. In particular, X is reﬂexive if and only if every closed convex set has a minimum-norm element.

With this theorem the minimum-norm elements appeared in the literature. And the next theorem improved Blatter’s result (see [4].)

Theorem 2.1.5 (Aizpuru/Garc´ıa-Pacheco, 2005) Let X be a normed space. Then, X is complete if every bounded closed convex set with non-empty interior has a minimum-norm element. In particular, X is reﬂexive if and only if every bounded closed convex set with non-empty interior has a minimum-norm element.

So, the point is to characterize the non-complete normed spaces on which every functional is norm-attaining. As shown above (Theorem 2.1.4,) we cannot use the existence of minimum-norm elements, but maybe we can use the existence of translations to obtain them. Following this line, here we present the conjecture on which we have been working.

Conjecture 2.1.6 Given a normed space, then:

1. If every functional is norm-attaining, then every closed convex set with non-empty

boundary can be translated to have a non-zero minimum-norm element.

2. If every bounded closed convex set with non-empty interior can be translated to have

a non-zero minimum-norm element, then every functional on X is norm-attaining.

Later on we will see more precisely the motivation to state this conjecture, which can be found in Theorem 2.2.2. Another problem related to this comes from the following result of James (see [29].) 33

Theorem 2.1.7 (James, 1971) Let X a non-complete normed space on which every functional is norm-attaining. Then, the completion Y of X is reﬂexive but not rotund.

This result motivates the following question.

Question 2.1.8 Given any non-rotund reﬂexive space, does it contain a proper dense subspace on which every functional is norm-attaining?

Before starting with the main results, let us see some basic ones related to translations and minimum-norm elements.

Theorem 2.1.9 Let X be a normed space and consider M to be a closed convex subset of X such that M has a maximum-norm element m. Then, we can ﬁnd a ∈ X such that m + a 6= 0 and m + a is a minimum-norm element of M + a.

Proof We can start by assuming that m 6= 0. Let us take

m a = − − m. kmk

If x ∈ M then

1 + kmk kx + ak = x − m kmk 1 ≥ |kmk kxk − (1 + kmk) kmk| kmk = |kxk − (1 + kmk)|

= 1 + kmk − kxk

≥ 1

= km + ak . 34

That is, m + a is a minimum-norm element of M + a.

The following theorems show that the converse to Theorem 2.1.9 does not hold, that is, the existence of non-zero minimum-norm elements does not imply that we can ﬁnd translations which map these non-zero minimum-norm elements into maximum-norm elements. Here, we can see the relations between translations, minimum-norm elements, and norm-attaining functionals.

Lemma 2.1.10 Let X be an inﬁnite dimensional normed space on which every functional is norm-attaining. Then, X can be equivalently renormed to be locally uniformly rotund

(and hence to have the Radon-Riesz property.) As a consequence, X contains a ω-null sequence of elements of norm 1.

Proof In the first place, remember (see [34]) that a Banach space has the Radon-Riesz property if on its unit sphere every ω-convergent sequence is convergent. Now, the completion of X is a reflexive Banach space, and every reflexive Banach space can be equivalently renormed to be locally uniformly rotund (see [20].) Finally, every reflexive Banach space has a ω-null sequence of norm 1 (see [31] and [36],) therefore due to the density of X in its

completion we deduce the existence of a ω-null sequence in SX .

Theorem 2.1.11 Let X be an inﬁnite dimensional normed space on which every functional is norm-attaining. Assume that X has the Radon-Riesz property. There exists a bounded closed convex subset M with a non-zero minimum-norm element x ∈ M such that no translation exists which maps x into a maximum-norm element. 35

Proof In accordance with Lemma 2.1.10, we can pick a sequence (yn) in SX which is n∈N ω-convergent to 0. By taking a subsequence if it is necessary, we can assume without loss of generality that there exists f ∈ SX∗ such that Ref (yn) ≥ 0 for every n ∈ N. Let x ∈ SX

with f (x) = 1. For every n ∈ , let us denote yn + x by xn. Then, (xn) is ω-convergent N n∈N to x but none of its subsequences converge to x. Let us see that x is a minimum-norm

element of

M = co ({xn : n ∈ N} ∪ {x}) .

Take an element λx + λ1xn1 + ··· + λkxnk ∈ co ({xn : n ∈ N} ∪ {x}). Then

kλx + λ1xn1 + ··· + λkxnk k ≥ Ref (λx + λ1xn1 + ··· + λkxnk )

= 1 + λ1Ref (y1) + ··· + λkRef (yk)

≥ 1

= kxk .

Next, suppose we could ﬁnd a ∈ X so that x + a is a maximum-norm element of M + a.

Then, (xn + a) ω-converges to x + a and there exists a subsequence of (kxn + ak) n∈N n∈N which converges to kx + ak. Therefore, since X has the Radon-Riesz property, there exists

a subsequence of (xn + a) converging to x+a, in other words, there exists a subsequence n∈N

of (xn) converging to x, which is a contradiction. n∈N

The following theorems are more precise than the previous one (which, by the way, requires that X be inﬁnite dimensional.) To show them, we require the following lemma.

Lemma 2.1.12 Let X be a real normed space. Assume that x ∈ X and f ∈ X∗ are so that

δ := f (x) > 0 and t := kxk > 0. Then: 36

−1 1. If there exists r > 0 such that BX (x, r) ∩ f (δ) ⊆ SX (0, t), then there exists 0 <

0 0 −1 r ≤ r so that BX (x, r ) ∩ SX (0, t) ⊆ f (δ).

−1 0 2. If there exists r > 0 such that BX (x, r)∩SX (0, t) ⊆ f (δ), then there exists 0 < r ≤

0 −1 0 −1 r so that BX (x, r ) ∩ f ([0, δ]) ⊆ BX (0, t). In this situation, BX (x, r ) ∩ f (δ) ⊆

0 SX (0, t). Furthermore, r can be chosen to diverge to ∞ if r diverges to ∞.

Proof

δ 1. In the ﬁrst place, let us see that kfk = t . Obviously,

x δ f = t t

δ δ therefore kfk ≥ t . If y ∈ BX and f (y) > t then we can take 0 < α < 1 small enough to assure that f (α (ty) + (1 − α) x) > 0 and

α (ty) + (1 − α) x δ ∈ B (x, r) ∩ f −1 (δ) , f (α (ty) + (1 − α) x) X

which means that

α (ty) + (1 − α) x t = δ f (α (ty) + (1 − α) x) δ = kα (ty) + (1 − α) xk f (α (ty) + (1 − α) x) δ ≤ t, f (α (ty) + (1 − α) x)

and δ < f (α (ty) + (1 − α) x) ≤ δ, which is impossible. In the second place, choose

r t 0 < r0 ≤ min , . 4 2

0 0 With this choice of r we have that, if y ∈ BX (x, r ) then

δ δ f (y) ≥ f (x) − |f (y) − f (x)| ≥ δ − r0 ≥ . t 2 37

0 Now, take any y ∈ BX (x, r ) ∩ SX (0, t). Then

δy 1 − x = kδy − f (y) xk f (y) f (y) 1 = kδy − δx + δx − f (y) xk f (y) δ 1 ≤ ky − xk + |f (x) − f (y)| t f (y) f (y) δ t δ ≤ ky − xk + ky − xk f (y) f (y) t δ = 2 ky − xk f (y) ≤ 4 ky − xk

≤ r.

Therefore, δy ∈ B (x, r) ∩ f −1 (δ) ⊆ S (0, t) , f (y) X X in other words,

δy δ t = = t f (y) f (y) and f (y) = δ.

δ 2. In the ﬁrst place, let us see that kfk = t . Obviously, x δ f = t t

δ δ therefore kfk ≥ t . If y ∈ BX and f (y) > t then we can take 0 < α < 1 small enough to assure that α (ty) + (1 − α) x t ∈ B (x, r) ∩ S (0, t) , kα (ty) + (1 − α) xk X X which means that

α (ty) + (1 − α) x δ = f t kα (ty) + (1 − α) xk t = f (α (ty) + (1 − α) x) kα (ty) + (1 − α) xk t > δ, kα (ty) + (1 − α) xk 38

and t < kα (ty) + (1 − α) xk ≤ t, which is impossible. In the second place, choose r 0 < r0 ≤ . 2

0 −1 Let y ∈ BX (x, r ) ∩ f ([0, δ]) with kyk > t. Then

ty 1 − x = kty − kyk xk kyk kyk 1 = kty − tx + tx − kyk xk kyk t 1 ≤ ky − xk + |kxk − kyk| t kyk kyk t ≤ 2 ky − xk kyk < 2 ky − xk

≤ r.

Therefore, ty ∈ B (x, r) ∩ S (0, t) ⊆ f −1 (δ) , kyk X X in other words, ty t δ = f ≤ δ < δ, kyk kyk δ which is a contradiction. In the third and last place, since kfk = t , we have that

0 −1 BX (x, r ) ∩ f (δ) ⊆ SX (0, t).

Theorem 2.1.13 Let X be a real normed space. The following conditions are equivalent:

∗ ∗ −1 1. There exists a norm-attaining x ∈ SX∗ such that (x ) (1) ∩ BX has empty interior

relative to SX .

2. There exists a bounded closed convex subset M with non-empty interior so that M has

a non-zero minimum-norm element x ∈ M and cannot be translated mapping x into

a maximum-norm element. 39

∗ ∗ Proof Assume that 1 holds. We can pick x ∈ SX∗ and x ∈ SX such that x (x) = 1 and

∗ −1 ∗ −1 (x ) (1) ∩ BX has empty interior relative to SX . Let M = BX (x, 1) ∩ (x ) ([1, ∞)).

∗ −1 Let us set C := BX (x, 1) ∩ (x ) (1). Suppose that there is a ∈ X such that x + a is a

maximum-norm element of M + a. Then, kx + ak= 6 0 and M + a ⊆ BX (0, kx + ak). Let

us show that C + a ⊂ SX (0, kx + ak). If c + a ∈ C + a with c ∈ C, then, by assuming

that c 6= x, we can ﬁnd d ∈ C such that x ∈ (c, d). Now, x + a ∈ (c + a, d + a) which

means that kc + ak = kx + ak = kd + ak, because x + a is a maximum-norm element of

∗ −1 ∗ M + a. On the other hand, C + a = BX (x + a, 1) ∩ (x ) (1 + x (a)). Next, let us see that

1 + x∗ (a) < 0. Otherwise, pick y ∈ M such that x∗ (y) > 1. Now, ky + ak ≥ x∗ (y + a) >

1 + x∗ (a) = kx + ak, which contradicts the fact that x + a is a maximum-norm element of

M + a. Finally, since

∗ −1 ∗ BX (x + a, 1) ∩ (−x ) (−1 − x (a)) ⊂ SX (0, kx + ak) ,

we deduce, according to paragraph 1 of Lemma 2.1.12, the existence of a positive number

0 < r0 ≤ 1 such that

0 ∗ −1 ∗ BX x + a, r ∩ SX (0, kx + ak) ⊂ (−x ) (−1 − x (a)) ,

∗ −1 which is impossible since (x ) (1) ∩ BX has empty interior relative to SX .

Assume that 2 holds and consider M to be a bounded closed convex subset with non-

empty interior so that M has a non-zero minimum-norm element x ∈ M and cannot be

translated mapping x into a maximum-norm element. On the contrary, let us suppose that

−1 f (1) ∩ BX has non-empty interior relative to SX for all norm-attaining f ∈ SX∗ . Let

∗ ∗ ∗ x ∈ SX∗ verify that x (u) < x (m) for all u ∈ UX (0, dist (0,M)) and all m ∈ M. Clearly,

x∗ (x) = dist (0,M) = kxk, that is, x∗ is norm-attaining. Since M is bounded we can

∗ −1 consider a number K > 0 such that M ⊆ BX (0,K). Now, by assumption (−x ) (K) ∩

BX (0,K) has non-empty interior relative to SX (0,K), so we can take an interior point 40

∗ −1 z ∈ BX (z, r) ∩ SX (0,K) ⊆ (−x ) (K). By paragraph 2 of Lemma 2.1.12 there exists

0 0 ∗ −1 0 < r ≤ r such that BX (z, r ) ∩ (−x ) ([0,K]) ⊆ BX (0,K). Observe that we may assume that r0 > diam (M) by taking K large enough. Finally, consider the translated set

M + (z − x). If m ∈ M then

0 m + (z − x) ∈ BX (z, diam (M)) ⊆ BX 0, r ,

and

(−x∗)(m + (z − x)) = −x∗ (m) − x∗ (z) + x∗ (x) ≤ −dist (0,M) + K + dist (0,M) = K,

and hence m + (z − x) ∈ BX (0,K). And kx + (z − x)k = kzk = K, which means that

x + (z − x) is a maximum-norm element of M + (z − x) reaching a contradiction.

Corollary 2.1.14 Let X be a complex normed space. Then, there exists a bounded closed convex subset M with non-empty interior so that M has a non-zero minimum-norm element

x ∈ M and cannot be translated mapping x into a maximum-norm element.

Proof Let us consider XR, that is, the underlining real space of X. By Lemma 1.4.11

the unit sphere of XR is free of convex sets with non-empty interior relative to it, therefore Theorem 2.1.13 allows us to conclude the result.

−1 Lemma 2.1.15 Let X be a real Banach space with dim (X) > 1. Assume that f (1)∩BX

has non-empty interior relative to SX for every f ∈ NA (X) ∩ SX∗ . Then:

1. X is smooth. 41

2. exp (BX ) = ∅.

3. char (X) = card (NA (X) ∩ SX∗ ).

4. X cannot be separable.

Proof

1. Let x ∈ SX . We will show that x is a smooth point of BX . Otherwise, let f 6= g ∈ SX∗

f+g such that f (x) = g (x) = 1. We have that 2 ∈ SX∗ and

f + g −1 (1) ∩ B = f −1 (1) ∩ B ∩ g−1 (1) ∩ B 2 X X X −1 −1 = bdSX f (1) ∩ BX ∩ bdSX g (1) ∩ BX .

−1 f+g Therefore, 2 (1) ∩ BX cannot have non-empty interior with respect to SX .

−1 2. If x ∈ exp (BX ) then there exists f ∈ SX∗ such that {x} = f (1) ∩ BX . This

−1 contradicts the fact that f (1) ∩ BX has non-empty interior with respect to SX .

3. On the one hand, char (X) = char (SX ). On the other hand,

[˙ −1 ∗ SX = cl intSX f (1) ∩ BX : f ∈ NA (X) ∩ SX .

Finally, the map

−1 ∗ f ∈ NA (X) ∩ SX 7−→ intSX f (1) ∩ BX

is a bijection.

4. Assume that X is separable. Then we have that NA (X) ∩ SX∗ is countable. If Y

is a 2-dimensional subspace of X then SY ∗ is uncountable. By the Hahn-Banach

Theorem, we can extend all the elements in SY ∗ to an uncountable set contained in

NA (X) ∩ SX∗ , which is impossible. 42

Corollary 2.1.16 Let X be a real normed space with dim (X) > 1. Then:

1. If X is separable then there exists a bounded closed convex subset M with non-empty

interior so that M has a non-zero minimum-norm element x ∈ M and cannot be

translated mapping x into a maximum-norm element.

2. If X is not separable then at least it can be equivalently renormed to possess a bounded

closed convex subset M with non-empty interior so that M has a non-zero minimum-

norm element x ∈ M and cannot be translated mapping x into a maximum-norm

element.

Proof The results follows from Lemma 2.1.15 and from the easy fact that every real normed space can be equivalently renormed so that its new unit ball has an exposed point.

2.2 Minimum-norm elements and norm-attaining functionals

In this section we will provide a positive approach to Conjecture 2.1.6. The ﬁrst theorem we present in this section, a primary version of which appears for the ﬁrst time in [4] as a “key lemma,” will clarify why we try to characterize normed spaces having only norm- attaining functionals in terms of translations to obtain minimum-norm elements.

Deﬁnition 2.2.1 Let X be a normed space. Let H be a convex subset of X. A subset C of H is said to be a norm-attaining exposed face of H if C is an exposed face of H and the functional that determines C is norm-attaining. 43

Theorem 2.2.2 Let X be a normed space. Consider a closed convex subset M of X and

a point x ∈ bd (M). The following are equivalent:

1. There exists a ∈ X such that x + a is a non-zero minimum-norm element of M + a.

2. There exists a norm-attaining exposed face C of M containing x.

3. There exists a closed ball B such that x ∈ M ∩ B and M ∩ int (B) = ∅.

Proof Let us see that 3 implies 2. The Hahn-Banach Theorem allows us to deduce the existence of an element f ∈ SX∗ such that Ref (u) > Ref (m) for every u ∈ int (B) and

every m ∈ M. Since the interior of B is dense in B, we have that f deﬁnes a non-trivial exposed face in M. Let us see that f is norm-attaining. Let b be the center of B. Since x ∈ bd (M), we have that the radius of B is kx − bk. We will show that

x − b Re (−f) = 1. kx − bk

For every z ∈ UX (0, kx − bk), we have that Re (−f)(z + b) < Re (−f)(x), so Re (−f)(z) <

Re (−f)(x − b). Therefore

kx − bk = sup Re (−f)(UX (0, kx − bk)) ≤ Re (−f)(x − b) ≤ kx − bk .

Now, let us see that 2 implies 1. Let f be a norm-attaining functional deﬁning the

−1 exposed face C that contains x. We can assume that kfk = 1. Take y ∈ f (1) ∩ BX and consider a = −y − x. Let us see that x + a is a minimum-norm element of M + a. Let 44

z ∈ M, then

kz + ak ≥ |f (z + a)|

= |f (z) − 1 − f (x)|

≥ 1 + Ref (x) − Ref (z)

≥ 1

= kx + ak .

Finally, to see that 1 implies 3 it is enough to take B = BX (−a, kx + ak).

Now, it is time to state and prove the following characterization, which is based upon

Theorem 2.2.2 and could be considered as a positive approach to Conjecture 2.1.6.

Theorem 2.2.3 Let X be a normed space. The following conditions are equivalent:

1. Every functional on X∗ is norm-attaining.

2. Every closed convex subset of X with non-empty interior and non-empty boundary

can be translated to have a non-zero minimum-norm element.

Proof Assume that every functional in X∗ is norm-attaining. If M is a closed convex subset of X with non-empty interior and x is any point in the boundary of M, then the

Hahn-Banach Theorem allows us to deduce the existence of a functional f ∈ SX∗ such that

Ref (u) < Ref (x) for every u ∈ int (M). Taking into consideration that the closure of the

interior of M is M, it is clear that f deﬁnes a non-trivial exposed face in M, which will be

a norm-attaining exposed face by hypothesis. Now, with Theorem 2.2.2 we have completed

the proof of one implication. 45

Conversely, assume that every closed convex subset of X with non-empty interior and non-empty boundary can be translated to have a non-zero minimum-norm element. Let

−1 us take any functional f ∈ SX∗ . Consider the closed convex set M = Ref ([1, ∞)). By hypothesis, there are m ∈ M and a ∈ X such that m+a 6= 0 and m+a is a minimum-norm element of M + a. We have that M + a = Ref −1 ([1 + Ref (a) , ∞)). Since 0 ∈/ M + a we have that 1 + Ref (a) > 0. Therefore,

dist (0,M + a) = 1 + Ref (a) and Ref (m + a) = 1 + Ref (a), which means that

m + a Ref = 1. km + ak

Corollary 2.2.4 Let X be a Banach space. The following conditions are equivalent:

1. X is reﬂexive.

2. Every closed convex subset of X with non-empty interior and non-empty boundary

can be translated to have a non-zero minimum-norm element.

2.3 Non-complete normed spaces having only norm-attaining functionals

Remember from the Preliminaries that, according to the James Theorem (see [28],) the completion of a non-complete normed space on which every functional is norm-attaining must be reﬂexive. Furthermore, the completion of such spaces cannot be rotund, so no rotund reﬂexive space can have a proper dense subspace on which every functional is norm- attaining. The next proposition, which does not require any proof, is a little more precise about this. 46

Proposition 2.3.1 Let X be a non-complete normed space. The following conditions are equivalent:

1. Every functional on X is norm-attaining.

2. The completion Y of X is reﬂexive and X ∩ C 6= ∅ for every exposed face C of BY .

In particular, if X has only norm-attaining functionals then span (exp (BY )) ⊆ X and Y

cannot be rotund.

Corollary 2.3.2 Let X be an inﬁnite dimensional reﬂexive Banach space. Then X can

be equivalently renormed to not have proper subspaces on which every functional is norm-

attaining.

Proof The results follows from Proposition 2.3.1 and from the fact that every reﬂexive

Banach space can be equivalently renormed to be rotund (see [20].)

On the other hand, Proposition 2.3.1, which is a slight generalization of Theorem 2.1.7, motivates the Question 2.1.8. The next theorem answers negatively this question.

Lemma 2.3.3 Let X and Y be Banach spaces. Then,

co (SX × SY ) = BX⊕∞Y .

Proof It is enough to show that

co (SX × SY ) ⊇ SX⊕∞Y = (SX × BY ) ∪ (BX × SY ) . 47

Let (x, y) ∈ SX × BY . There are y1, y2 ∈ SY and α ∈ [0, 1] such that y = αy1 + (1 − α) y2.

Therefore,

(x, y) = α (x, y1) + (1 − α)(x, y2) , and (x, y) ∈ co (SX × SY ). Likewise, it can be proved that if (x, y) ∈ BX × SY then

(x, y) ∈ co (SX × SY ).

Theorem 2.3.4 Let X be an inﬁnite dimensional reﬂexive Banach space. Then, X can be equivalently renormed to be non-rotund and to not have proper subspaces on which every functional is norm-attaining.

Proof In the ﬁrst place, every reﬂexive Banach space can be equivalently renormed to be rotund (see [20],) therefore we can suppose that X is already rotund. Let f ∈ SX∗ be a norm-attaining functional and consider x ∈ SX such that f (x) = 1. Consider the non-rotund Banach space Y = Kx ⊕∞ ker (f). By Lemma 2.3.3, we have that

co (exp (BY )) = co exp (BKx) × exp Bker(f) = co SKx × Sker(f) = BY .

Proposition 2.3.1 concludes the proof.

From this theorem we ﬁnally obtain that, in the class of non-complete normed spaces, the fact of having only norm-attaining functionals is not a topological property.

Corollary 2.3.5 Let X be a non-complete normed space having only norm-attaining functionals. Then X can be equivalently renormed to have some functionals which are not norm-attaining. This equivalent renorming can be chosen to be rotund or non-rotund. 48

2.4 Partial solutions

In this section, we keep trying to obtain positive approaches to Conjecture 2.1.6. In virtue of the Bishop-Phelps Support Point Theorem (see [34, Theorem 2.11.9]) we have the following lemma.

Lemma 2.4.1 Let X be a Banach space. Let M be a closed convex subset of X with non-empty boundary. Assume either one of the following conditions:

1. There are a ∈ X and δ > 0 so that sup (Ref (M + a)) ≥ δ for all f ∈ SX∗ .

2. There are a ∈ X and f ∈ SX∗ such that Ref (M + a) = {0}.

Then, every x ∈ bd (M) is contained in an exposed face.

Proof In the ﬁrst place, the fact for a point in the boundary of a closed convex set to

be contained in a exposed face is preserved by translations. Therefore we can assume that

a = 0.

1. In accordance with the Bishop-Phelps Support Point Theorem there exist a sequence

(xn) ⊂ bd (M) converging to x and a sequence (fn) ⊂ SX∗ so that Refn (xn) = n∈N n∈N ∗ ∗ sup (Refn (M)). By the ω -compactness of BX , there exists a subnet (fni )i∈I that

∗ ∗ is ω -convergent to some f ∈ BX . Next, (fni (xni ))i∈I converges to f (x) and thus

Ref (x) = sup (Ref (M)). Let us show ﬁnally that f 6= 0. By hypothesis Refni (xni ) ≥

δ for all n ∈ N. Therefore Ref (x) ≥ δ.

2. Obviously, M = {x ∈ M : Ref (x) = sup (Ref (M))}. 49

Remark 2.4.2 If M is a closed convex subset of a normed space X with non-empty boundary and non-empty interior then there are a ∈ X and δ > 0 so that sup (Ref (M + a)) ≥ δ

for all f ∈ SX∗ . Indeed, it suﬃces to take a to be the opposite of the center of a closed ball

contained in M and δ the radius of this ball.

Lemma 2.4.1 together with Theorem 2.2.2 aﬀord the following result.

Theorem 2.4.3 Let X be a normed space. Assume that every functional on X is norm- attaining. Let M be a closed convex subset with non-empty boundary that veriﬁes either one of the following conditions:

1. There are a ∈ X and δ > 0 so that sup (Ref (M + a)) ≥ δ for all f ∈ SX∗ .

2. There are a ∈ X and f ∈ SX∗ such that Ref (M + a) = {0}.

Then M can be translated to have a non-zero minimum-norm element.

Proof It is suﬃcient to take into account that Lemma 2.4.1 may be applied to the com-

pletion Y of X to deduce that every x ∈ bd (M) is contained in an exposed face. Theorem

2.2.2 allows us to assure that these exposed faces are indeed norm-attaining exposed faces.

In the previous section we have seen that the completion of a non-complete real normed space on which every functional is norm-attaining cannot be rotund. Following this line, we present the following theorem.

Theorem 2.4.4 Let X be a non-complete normed space. If every bounded closed convex subset of X with non-empty interior can be translated to have a non-zero minimum-norm element, then the completion Y of X cannot be rotund. 50

Proof Assume that Y is rotund. Let y ∈ SY \SX and consider M = BY (y, 1/2)∩X. There is a ∈ X such that M + a has a non-zero minimum-norm element m + a with m ∈ M. We have that M +a = BY (y + a, 1/2)∩X and, since 0 ∈/ M +a, we deduce that ky + ak > 1/2.

Next, M + a is dense in BY (y + a, 1/2). Thus

dist (0,M + a) = dist (0, BY (y + a, 1/2)) .

Now, the rotundity of Y allows us to deduce that BY (y + a, 1/2) has a unique element of minimum-norm. Therefore

ky + ak − 1/2 m + a = (y + a) , ky + ak which implies that y ∈ X, and this is a contradiction. CHAPTER 3

THE BANACH-MAZUR CONJECTURE FOR ROTATIONS

3.1 Preliminaries

Let us recall the deﬁnition of transitivity for Banach spaces.

Deﬁnition 3.1.1 (Banach, 1932) Let X be a Banach space. Then:

1. The space X is said to be transitive if for every x, y ∈ SX there exists a surjective

linear isometry T : X −→ X such that T (x) = y.

2. The space X is said to be almost transitive if for every ε > 0 and for every x, y ∈ SX

there exists a surjective linear isometry T : X −→ X such that kT (x) − yk ≤ ε.

∗ ∗ ∗ ∗ 3. The space X is said to be ω -transitive if for every x, y ∈ SX∗ there exists a ω -ω

continuous surjective linear isometry T : X∗ −→ X∗ such that T (x) = y.

4. The space X∗ is said to be almost ω∗-transitive if for every ε > 0 and for every

∗ ∗ ∗ ∗ x, y ∈ SX∗ there exists a ω -ω continuous surjective linear isometry T : X −→ X

such that kT (x) − yk ≤ ε.

In 1932 Mazur and Banach had a conversation in a bar, and Mazur asked Banach for

an example of a transitive and separable Banach space diﬀerent from `2. This was the

beginning of the famous Banach-Mazur conjecture for rotations.

Problem 3.1.2 (Banach/Mazur, 1932) Is every transitive and separable Banach space a Hilbert space?

51 52

The main results trying to answer this question are given now. The ﬁrst one is given by

Mazur and appears in [33].

Theorem 3.1.3 (Mazur, 1933) Let X be a Banach space. If X is separable then smo (BX )

is a Gδ dense subset of SX . As a consequence, if X is transitive and separable, then X is

smooth and X∗ is almost ω∗-transitive.

Fifty years later Finet came up with this important result (see [23]).

Theorem 3.1.4 (Finet, 1986) Let X be a Banach space. If X is almost transitive and super-reﬂexive, then both X and X∗ are uniformly convex.

This theorem motivated Cabello to relax its hypothesis to keep obtaining the same thesis

(see [18].)

Theorem 3.1.5 (Cabello, 1999) Let X be an almost transitive Banach space. If X is

Asplund or has the Radon-Nikodym property, then X is super-reﬂexive.

Finally, the following result improved the two previous ones (see [12].)

Theorem 3.1.6 (Becerra-Guerrero/Rodr´ıguez-Palacios, 1999) Let X be a Banach space. The following conditions are equivalent:

1. The space X is almost transitive and super-reﬂexive.

2. There exists x ∈ SX such that

co {T (x): T : X −→ X surjective linear isometry} = BX

and the norm of X is Frech´etdiﬀerentiable at x. 53

According to the Mazur Theorem (see [33]) in order to assure the existence of smooth points it suﬃces to consider separable Banach spaces. Unfortunately, this hypothesis is not enough to assure the existence of rotund points as remarked as follows.

2 Remark 3.1.7 The space `∞ is separable and its unit ball is free of rotund points.

Note that the Banach space in the previous remark is not smooth. As we will see in this section, smoothness is crucial to obtain rotund points. A primary example is given by the following theorem, which shows that the unit balls of 2-dimensional real smooth Banach spaces have uncountably many rotund points.

Theorem 3.1.8 Let X be a 2-dimensional real Banach space. If X is smooth, then rot (BX )

is uncountable.

Proof Let C denote the set of all exposed faces of BX . Note that, in this situation, the

rotund points are exactly the exposed faces with null diameter. Due to the smoothness

of X, the exposed faces are pairwise disjoint, and thus there is a bijection between C and

SX∗ . Now, SX is separable and every exposed face with diameter greater than zero has

non-empty interior with respect to SX . Therefore there must be countably many exposed

faces with diameter greater than zero. This leaves uncountably many rotund points.

Let us continue by reviewing some geometrical concepts related to smoothness, which will be fundamental to study some topological properties of the set of rotund points, and can be widely consulted in [19].

Remark 3.1.9 Let X stand for a Banach space. Then a point x of the unit sphere of X 54

is said to be

∗ 1. a Y -smooth point of BX (where Y is a closed subspace of X ) if every f, g ∈ SY ∗ such

that f (x) = g (x) = 1 verify f = g;

2. a smooth-A point of BX (where A is a subset of SX ) if every f, g ∈ SX∗ such that

f (x) = g (x) = 1 verify f|A = g|A.

The sets of Y -smooth points of BX and smooth-A points of BX will be denoted, respec-

Y ∗ tively, by smo (BX ) and smoA (BX ). Furthermore, if A is a ω -dense subset of SX∗ ,

X then smoA (BX∗ ) ⊆ smo (BX∗ ); and if A is a ω-dense subset of SX , then smoA (BX ) =

smo (BX ). Finally, in 1966 (see [38]) Phelps obtained (by using the Baire Category Theo-

rem) the following results for a Banach space X:

1. For every a ∈ SX and every m ∈ N, the set PX (a, m) of all x ∈ SX such that

|f (a) − g (a)| < 1/m for every f, g ∈ SX∗ with f (x) = g (x) = 1, is always an open

dense subset of SX .

T 2. If A ⊆ SX , then smoA (BX ) = {PX (a, m): a ∈ A, m ∈ N}. In particular, if A is

countable then smoA (BX ) is a Gδ dense subset of SX .

As a consequence, Phelps deduced the Mazur Theorem: Every separable Banach space has a Gδ dense subset of smooth points in its unit sphere.

Now, we are ready to state the following lemma and theorem, upon which we will base several of our main results in the next section.

Lemma 3.1.10 Let X be a Banach space. Then: 55

1. If X is smooth, then a necessary and suﬃcient condition for a point x ∈ SX to be a

rotund point of BX is that JX (x) is an X-smooth point of BX∗ .

X 2. If X is separable, then smo (BX∗ ) is a residual subset of SX∗ .

Proof

1. Assume that JX (x) is an X-smooth point of BX∗ . Let y ∈ SX such that k(x + y) /2k =

1. Then, JX (x) = JX (y), thus x (JX (x)) = y (JX (x)) = 1. Since JX (x) is an X-

smooth point of BX∗ , x = y. Similarly, if x is a rotund poind of BX then JX (x) is an

X-smooth point of BX∗ .

∗ 2. Let (an) be a ω -dense sequence in SX∗ and consider A = {an : n ∈ }. Since A n∈N N

is countable, according to Phelps’ results, smoA (BX∗ ) is a Gδ dense subset of SX∗ .

∗ X Finally, by taking into account that A is ω -dense in SX∗ , smoA (BX∗ ) ⊆ smo (BX∗ ).

A direct consequence of Lemma 3.1.10 is the following theorem.

Theorem 3.1.11 Let X be a separable and smooth Banach space. If BX is free of rotund

points, then SX∗ \NA (X) is residual in SX∗ , and hence NA (X)∩SX∗ is of the ﬁrst category in SX∗ .

Note that Theorem 3.1.11 looks like the following one (see [16, Theorem 3.5.5 and

Problem 3.5.6].)

Theorem 3.1.12 (Bourgain/Stegall, 1983) Let X be a Banach space. If X is separable and BX is not dentable then NA (X) is of the ﬁrst category. 56

Finally, the next theorem provides a version of the Mazur Theorem for the set of rotund points.

Theorem 3.1.13 Let X be a Banach space. Then:

1. If X is reﬂexive, separable, and smooth, then the closure of rot (BX ) contains the set

exps (BX ) of all strongly exposed points of BX , and therefore, co (rot (BX )) = BX .

2. If X is reﬂexive, separable, and strongly smooth, then rot (BX ) is a Gδ subset of SX .

∗ 3. If X is reﬂexive, separable, strongly smooth, and JX |SX : SX −→ SX is open, then

rot (BX ) is a Gδ dense subset of SX .

Proof

1. Let x be a strongly exposed point of BX . By taking into account that X is reﬂexive

and separable, we deduce that X∗ is also separable. Therefore we can ﬁnd a se-

quence (JX (xn)) of smooth points of BX∗ converging to JX (x). For every n ∈ , n∈N N

|1 − JX (x)(xn)| = |JX (xn)(xn) − JX (x)(xn)| ≤ kJX (xn) − JX (x)k and therefore

(JX (x)(xn)) converges to 1. Since x is a strongly exposed point of BX ,(xn) n∈N n∈N converges to x. Now, let us observe that, according to paragraph 1 of lemma 3.1.10,

xn ∈ rot (BX ) for every n ∈ N. Finally, the fact of co (rot (BX )) = BX follows from the fact that every reﬂexive Banach space has the Radon-Nikodym property, that is,

co (exps (BX )) = BX (see [17].)

2. According to paragraph 1 of Lemma 3.1.10 and keeping in mind that X is reﬂexive,

−1 ∗ rot (BX ) = JX (smo (BX∗ )). Now, again X is separable, therefore smo (BX∗ ) is a Gδ

∗ dense subset of SX . Finally, since JX |SX is continuous, we deduce that rot (BX ) is a

Gδ subset of SX . 57

3. We have already proved that rot (BX ) is a Gδ subset of SX . Thus, it just remains

to show that it is dense. Let U be an open subset of SX . Then, JX (U) is open in

SX∗ , therefore there exists x ∈ U such that JX (x) ∈ JX (U) ∩ smo (BX∗ ). Finally, by

applying again paragraph 1 of lemma 3.1.10, x ∈ rot (BX ).

The next remark shows that reﬂexivity, separability, and strong smoothness are not suﬃcient conditions to assure that the set of rotund points of the unit ball is a dense subset of the unit sphere.

Remark 3.1.14 Suppose that X is a 2-dimensional real Banach space which is smooth but not rotund. Then, rot (BX ) cannot be dense in SX because rot (BX ) ⊆ ext (BX ) and ext (BX ) is closed.

Now, the next proposition will show an equivalent condition to the fact that the restric- tion to the unit sphere of the dual map is open.

Proposition 3.1.15 Let X be a smooth Banach space. Let us consider on SX the equivalence relation deﬁned by R = {(x, y) ∈ SX × SX : JX (x) = JX (y)}. Let us deﬁne on SX /R the metric d ([x]R , [y]R) = kJX (x) − JX (y)k. Then:

1. JX |SX is continuous if and only if d-topology is contained in the quotient topology τR.

2. JX |SX is open if and only if the d-topology contains the quotient topology τR.

Proof First, notice that the quotient set SX /R is exactly the set of all exposed faces of

−1 BX , that is, for every x ∈ SX ,[x]R = JX (x) (1) ∩ BX . Therefore, if X is rotund then the 58

quotient map p : SX −→ (SX /R, τR) is an homeomorphism, and if X is reﬂexive then the map q :(SX /R, d) −→ SX∗ deﬁned as q ([x]R) = JX (x), is a surjective isometry.

1. The strongly smoothness of X is an equivalent condition to the fact that the quotient

map SX −→ (SX /R, d) is continuous, therefore, it is equivalent to the other fact that

the quotient topology τR contains the d-topology.

2. Let us assume that the d-topology contains the quotient topology τR and let A be

an open set in SX . Now, p (A) is open in (SX /R, d), so JX (A) = q (p (A)) is an

∗ open set in SX . Conversely, suppose that JX |SX is open and let A be an open set in

−1 −1 (SX /R, τR). Now, p (A) is open in SX , so JX p (A) is open in SX∗ , and ﬁnally,

−1 −1 A = q JX p (A) is open in (SX /R, d).

The previous theorem allows us to state the following characterization of strongly rotund and strongly smooth Banach spaces.

Corollary 3.1.16 Let X be a Banach space. The following conditions are equivalent:

1. X is strongly rotund and strongly smooth.

∗ 2. X is smooth and JX |SX : SX −→ SX is an homeomorphism.

3. X is reﬂexive, rotund, and the norm topology coincides with the topology on SX given

by the metric d (x, y) = kJX (x) − JX (y)k.

3.2 Geometrical conditions

Here we will use geometrical properties such as normal structure and dentability, that

can be widely consulted in [14] and [16], respectively. 59

Theorem 3.2.1 Let X be a transitive and smooth Banach space. Assume that BX has

normal structure. Then, X is rotund.

Proof Suppose that X is not rotund. Since X is transitive and smooth, the exposed faces

of BX are pairwise disjoint are isometric to each other, and hence they all have the same

diameter strictly greater than 0. Therefore, if C is any exposed face of BX , then for every

x, y ∈ C we have that sup {kx − ck : c ∈ C} = sup {ky − ck : c ∈ C}. As a consequence, diam (C) = sup {kx − ck : c ∈ C} for every x ∈ C, which contradicts that BX has normal structure.

Corollary 3.2.2 Let X be a transitive and separable Banach space. If BX has normal structure, then X is rotund.

Theorem 3.2.3 Let X be a transitive and smooth Banach space. Assume that BX is dentable. Then, X is rotund.

Proof Assume that X is not rotund. Again, the exposed faces of BX are pairwise disjoint, isometric to each other, and have the same diameter ε > 0. By hypothesis BX possesses a slice of diameter strictly less than ε, but this slice must contain an exposed face, and this is contradiction.

Corollary 3.2.4 Let X be a transitive and separable Banach space. If BX is dentable, then

X is rotund. 60

Recently, we have known of the following much stronger result, apparently not published yet (see [40].)

Theorem 3.2.5 (Talponen, 2006) Let X be a Banach space. Assume that BX is dentable and the set of all x ∈ SX such that

co {T (x): T : X −→ X surjective linear isometry} = BX

has non-empty interior relative to SX . Then X is almost transitive and uniformly rotund.

3.3 Topological conditions

In this section we will establish a topological condition on the set of norm-attaining

functionals of norm 1 in order to obtain the desired property of rotundity.

Remark 3.3.1 Let S be a metric space. Then:

1. The Hausdorﬀ distance (see [13]) between two bounded closed subsets C and D of S

is deﬁned as

dH (C,D) = sup {dist (c, D) , dist (C, d): c ∈ C, d ∈ D} .

2. This metric is complete if and only if S is complete.

3. If S is a normed space, then the subset composed by all bounded closed convex subsets

of S is dH-closed.

4. The Sierpinski Lemma (see [39]) says that a subset A of S is a Gδ subset of S if A is

topologically complete (that is, A can be endowed with a complete metric topologically

equivalent to the original.) 61

Remark 3.3.2 Given a separable and smooth Banach space X, we will take S to be SX∗

∗ with the metric given by the ω -topology and A to be {JX (x): x ∈ SX } (= NA (X) ∩ SX∗ .)

Now, by the smoothness, there exists a bijection between the norm-attaining functionals of

norm 1 and the exposed faces of BX . Next, all the exposed faces are bounded closed convex subsets therefore we can think of the Hausdorﬀ distance, that is, we will endow A with the

following metric: If x, y ∈ SX then

−1 −1 d (JX (x) , JX (y)) = dH JX (x) (1) ∩ BX , JX (y) (1) ∩ BX .

According to the Sierpinski Lemma, if d is complete and the d-topology coincides with the

∗ ∗ ω -topology, then A is a ω -Gδ subset of S, therefore it is a Gδ dense subset of SX∗ .

In the ﬁrst place, observe that next proposition assures the existence of many smooth

Banach spaces for which the metric d is not complete.

Proposition 3.3.3 Let X be a separable and smooth Banach space. Then:

1. If X is rotund then d is complete.

2. If the set of rotund points of BX is not closed in SX , then d is not complete.

Proof

1. It suﬃces to note that d (JX (x) , JX (y)) = kx − yk for every x, y ∈ SX .

2. Let (xn) be a sequence of rotund points of BX convergent to an element y ∈ SX n∈N

which is not a rotund point. For every n, m ∈ N, d (JX (xn) , JX (xm)) = kxn − xmk,

therefore the sequence (JX (xn)) is Cauchy. Finally, the sequence ({xn}) is a n∈N n∈N sequence of exposed faces converging, in the Hausdorﬀ metric, to the closed bounded 62

convex set {y}, which is not an exposed face. This shows that (JX (xn)) is not n∈N convergent.

Theorem 3.3.4 Let X be a smooth and separable Banach space. Then, the ω∗-topology is

contained in the d-topology. Furthermore, if both topologies coincide then d is a complete

metric.

Proof Let (JX (xn)) be a sequence in A that is d-convergent to some element JX (x) in n∈N

A. There exists a sequence (yn) in SX which is convergent to x and such that JX (yn) = n∈N

JX (xn) for all n ∈ . Now, (JX (yn)(x)) converges to 1, therefore by the smoothness N n∈N ∗ of X we deduce that (JX (xn)) is ω -convergent to JX (x). Next, assume that both n∈N

topologies coincide. If (JX (xn)) is a d-Cauchy sequence in A then there exists a bounded n∈N −1 closed convex subset C in SX which is the dH-limit of the sequence JX (xn) (1) ∩ BX . n∈N −1 Now there is x ∈ C such that C ⊆ JX (x) (1) ∩ BX . Following a similar reasoning as right

∗ ∗ above, we deduce that (JX (xn)) is ω -convergent to JX (x). Since the ω -topology and n∈N

the d-topology are the same, we have that (JX (xn)) is d-convergent to JX (x). Finally, n∈N ∗ by applying the Sierpinski Lemma, we deduce that A is a ω -Gδ subset of SX∗ , therefore it

is a Gδ dense subset of SX∗ .

Finally, we present the partial result, which is a direct consequence of Theorem 3.3.4 and Theorem 3.1.11.

Corollary 3.3.5 Let X be a transitive and separable Banach space. If the ω∗-topology coincides with the d-topology then X is rotund and NA (X) ∩ SX∗ is a Gδ dense subset of 63

SX∗ .

3.4 Intermediate solutions

Here we want to present suﬃcient conditions for a transitive and separable Banach space to be close to Hilbert spaces. Overall, we will take into account Theorem 3.1.11. The

ﬁrst two results deal with the possibility for a transitive and separable Banach space to be reﬂexive.

Theorem 3.4.1 Let X be a transitive and separable Banach space. Assume that the set

X∗ \ NA (X) is not dense. Then, X is rotund, NA (X) is open, and the set of rotund points of BX∗ is dense in SX∗ . In particular, if NA (X) is a vector space then X is reﬂexive.

Proof By Theorem 3.1.11 we have that X has to be rotund. Now, we will show that

JX (x) is a rotund point of BX∗ as well as a smooth point for every x ∈ SX . If no x ∈ SX is such that JX (x) is a rotund point of BX∗ then, according to Corollary 1.4.2, we deduce that SX∗ \NA (X) is dense in SX∗ . Finally, since NA (X) has non-empty interior, we deduce by the transitivity of X that NA (X) is open. Let us suppose now that NA (X) is a vector space. If X is not reﬂexive then we can take f ∈ X∗ \ NA (X). Now, f − NA (X) is an open dense set, therefore f − NA (X) ∩ NA (X) 6= ∅, that is, there are g, h ∈ NA (X) such that f − h = g and hence f = g + h and NA (X) cannot be a vector space.

Theorem 3.4.2 Let X be a Banach space. Then, X is transitive and reﬂexive if and only if X∗ is ω∗-transitive. 64

Proof Assume that X is transitive and reﬂexive. Let C and D be exposed faces of BX∗ .

Our objective is to show the existence of a ω∗-ω∗ continuous surjective linear isometry from

∗ X onto itself mapping C onto D. Since X is reﬂexive, there exist x, y ∈ SX verifying

−1 −1 C = x (1) ∩ BX∗ and D = y (1) ∩ BX∗ . Now, the transitivity of X allows us to deduce

the existence of a surjective linear isometry T : X −→ X such that T (y) = x. Let us

see that T ∗ (C) = D. For every f ∈ C, T ∗ (f)(y) = (f ◦ T )(y) = f (T (y)) = f (x) = 1,

therefore T ∗ (f) ∈ D and T ∗ (C) ⊆ D. Let us see indeed that T ∗ (C) = D. Since X is

reﬂexive we can ﬁnd extreme points in BX , therefore the transitivity of X implies that X

is rotund. Now, by using again the reﬂexivity of X, we have that X∗∗ is rotund, so X∗

∗ is smooth. Then, C is a maximal face of BX∗ , so is T (C). This fact directly implies

the equality T ∗ (C) = D. Finally, to see that X∗ is ω∗-transitive, notice that, because of

∗ ∗ the reﬂexivity of X, BX∗ has exposed points. Conversely, assume that X is ω -transitive.

∗ ∗ In ﬁrst place, X is rotund because BX∗ has extreme points and X is transitive. As a

∗ ∗ consequence, X is smooth. Next, let f ∈ SX∗ and consider a point x ∈ SX and a ω -ω

∗ ∗ continuous surjective linear isometry T : X −→ X such that T (f) = JX (x). Now, there

exists a continuous linear operator S : X −→ X such that S∗ = T . Then, f (S (x)) =

∗ ∗ (f ◦ S)(x) = S (f)(x) = T (f)(x) = JX (x)(x) = 1. Since kSk = kS k = kT k = 1,

S (x) ∈ BX and f is norm-attaining. As a consequence, X is reﬂexive. Finally, to see that

X is transitive, it suﬃces to realize that X∗ is transitive and reﬂexive, therefore X∗∗ is

transitive, and so is X.

Observe that in [18, Corollary 2.4] it is already proved that the dual of every transitive and reﬂexive Banach space is transitive, and therefore ω∗-transitive. However, in Theorem

3.4.2 we present a new proof. 65

The following results deal with the possibility for a transitive separable reﬂexive Banach space to be a Hilbert space. Particularly, the next theorem is already known and can be obtained more generally from combining some results that appeared in [23] and [18]. Our intention is to provide a more direct proof by taking advantage of the following theorem

(see [3].)

Theorem 3.4.3 (Aizpuru/Garc´ıa-Pacheco, 2005) Let X be a Banach space. If X is strongly exposed and strongly smooth then it is locally uniformly rotund.

Theorem 3.4.4 Let X be a transitive and separable Banach space. If X is reﬂexive then

both X and X∗ are locally uniformly rotund and strongly smooth.

Proof In the ﬁrst place, by Theorem 3.4.2 we have that X∗ is transitive, separable, and

reﬂexive. Now, every reﬂexive Banach space has thee Radon-Nykodim property (see [16])

therefore exps (BX ) , exps (BX∗ ) 6= ∅. By transitive, exps (BX ) = SX and exps (BX∗ ) =

∗ SX∗ , in other words, both X and X are strongly exposed. Next, by duality between

strongly exposed points and strongly smooth points, we have that smos (BX ) = SX and

∗ smos (BX∗ ) = SX∗ , in other words, both X and X are strongly smooth. Finally, by

applying Theorem 3.4.3 we have that both X and X∗ are locally uniformly rotund.

Remark 3.4.5 Let X be a smooth Banach space. For every x ∈ SX , we will let γX (x)

denote the set of all y ∈ SX such that JX (x)(y) = JX (y)(x). Also, we will set γX = T {γX (x): x ∈ SX }. We have that X is a Hilbert space if and only if γX = SX . We will 66

also consider the continuous odd function

φX (x): SX −→ K

y 7−→ φX (x)(y) = JX (x)(y) − JX (y)(x).

−1 Clearly, γX (x) = φX (x) (0).

Lemma 3.4.6 Let X be a smooth Banach space. Let n ∈ N be an arbitrary ﬁxed natural

number. Assume that for any diﬀerent points x1, . . . , xn ∈ SX , the set γX (x1)∩· · ·∩γX (xn)

is non-empty and connected. Then, for any diﬀerent points y1, . . . , yn+1 ∈ SX , the set

γX (y1) ∩ · · · ∩ γX (yn+1) is non-empty.

Proof Let y1, . . . , yn+1 ∈ SX be diﬀerent points. We have that φX (yn+1) |γX (y1)∩···∩γX (yn) is a continuous odd function deﬁned on a symmetric non-empty connected set. Therefore, by

the Intermediate Value Theorem there is z ∈ γX (y1)∩· · ·∩γX (yn) such that φX (yn+1)(z) =

0, and hence z ∈ γX (y1) ∩ · · · ∩ γX (yn+1).

Theorem 3.4.7 Let X be a transitive and separable Banach space. Assume that:

1. The space X is reﬂexive, the dual mapping of X is ω-ω continuous, and the family

{γX (x): x ∈ SX } veriﬁes the ﬁnite intersection property.

2. There exists a dense sequence (en) in SX such that for every n ∈ we can ﬁnd n∈N N

zn ∈ γX (e1) ∩ · · · ∩ γX (en) such that (zn) does not ω-converge to 0. n∈N

Then X is a Hilbert space.

Proof Observe that, in virtue of the transitivity of X, in order to show that X is a

Hilbert space it will be enough to prove that γX 6= ∅. On the other hand, since JX is ω-ω 67

continuous, each γX (x) is ω-closed with respect to SX . Now, by the reﬂexivity of X, we can assume without any loss that (zn) is ω-convergent to some z ∈ BX \{0}. By the n∈N T ω-ω continuity of JX , we deduce that z/ kzk ∈ {γX (en): n ∈ N} = γX . BIBLIOGRAPHY

[1] Acosta, M. D.; Galan,´ M. R. New characterizations of the reﬂexivity in terms of the set of norm-attaining functionals. Canad. Math. Bull. 41 (1998), 279-289.

[2] Acosta, M. D.; Aizpuru, A.; Aron, R.; Garc´ıa-Pacheco, F. J. Functionals that do not attain their norm. Bull. Belg. Math. Soc. Simon Stevin (to appear)

[3] Aizpuru, A.; Garc´ıa-Pacheco, F. J. Some questions about rotundity and renormings in Banach spaces. J. Austral. Math. Soc. 79 (2005), 131-140.

[4] Aizpuru, A.; Garc´ıa-Pacheco, F. J. Reﬂexivity, contraction functions and minimum-norm elements. Studia Sci. Math. Hungar. 42 (2005), no. 4, 431-443.

[5] Aizpuru, A.; Garc´ıa-Pacheco, F. J. Rotundity in transitive and separable Banach spaces. Quaest. Math. 30 (2007), 1-12.

[6] Aizpuru, A.; Garc´ıa-Pacheco, F. J. A note on L2-summand vectors in dual spaces. Preprint.

[7] Aron, R.; Gurariy, V.; Seoane-Sepulveda,´ J. B. Lineability and spaceability of sets of functions on R. Proc. Amer. Math. Soc. 133 (2005), 795-803.

[8] Aron, R. M.; Perez-Garc´ ´ıa, D.; Seoane-Sepulveda,´ J. B. Algebrability of the set of non-convergent Fourier series. Studia Math. 175 (2006), no. 1, 83-90.

[9] Bandyopadhyay, P.; Godefroy, G. Linear structure in the set of norm-attaining functionals on a Banach space. Preprint.

[10] Bandyopadhyay, P.; Huang, D.; Lin, B.-L.; Troyanski, S. L. Some Generalizations of Locally Uniform Rotundity. J. Math. Anal. Appl. 252 (2000), 906-916.

[11] Bandyopadhyay, P.; Lin, B.-L.; Some properties related to nested sequence of balls in Banach spaces. Taiwanese J. Math. 5 (2001), 19-34.

68 69

[12] Becerra-Guerrero, J.; Rodr´ıguez-Palacios, A. Transitivity of the norm on Banach spaces. Extracta Math. 17 (2002), no. 1, 1-58. [13] Beer, G. Topologies on closed and closed and convex sets. Mathematics and its Applications 268, Kluver Academic Publishers, Dordrecht, 1993. [14] Belluce, L. P.; Kirk, W. A.; Steiner, E. F. Normal structure in Banach spaces. Pacific J. Math. 26 (1968), 433-440. [15] Bogachev, V. I.; Mena-Jurado, J. F.; Navarro-Pascual, J. C. Extreme points in spaces of continuous functions. Proc. Amer. Math. Soc. 123 (1995), no. 4, 1061-1067. [16] Bourgin, R. D. Geometric aspects of convex sets with the Radon-Nikodym property. Lecture Notes in Mathematics 993, Springer-Verlag, Berlin, 1983. [17] Blatter, J. Reflexivity and the existence of best approximations. Approximation Theory, Proc. Internat. Sympos., Univ. Texas, Austin, Tex., Aca- demic Press, New York, (1976), 299-301. [18] Cabello-Sanchez,´ F. Maximal symmetric norms on Banach spaces. Math. Proc. R. Ir. Acad. 98A (1998), no. 2, 121-130. [19] Deville, R.; Godefroy, G.; Zizler, V. Smoothness and Renormings in Banach Spaces. Pitman Monographs and Surveys in Pure and Applied Mathematics 64, Long- man, Brunt Mill, 1993. [20] Diestel, J. Geometry of Banach spaces-selected topics. Lecture Notes in Mathematics 485, Springer-Verlag, Berlin-New York, 1975. [21] Diestel, J. A survey of results related to the Dunford-Pettis property. Proceedings of the Conference on Integration, Topology, and Geometry in Linear Spaces (Univ. North Carolina, Chapel Hill, N.C., 1979), 15-60, Contemp. Math., 2, Amer. Math. Soc., Providence, R.I., 1980. [22] Diestel, J. Sequences and series in Banach spaces. Graduate Texts in Mathematics 92, Springer-Verlag, New York, 1984. [23] Finet, C. Uniform convexity properties of norms on a super-reflexive Banach space. Israel J. Math. 53 (1986), no. 1, 81-92. 70

[24] Garc´ıa-Pacheco, F. J. Minimum-norm elements and norm-attaining functionals. Preprint. [25] Garc´ıa-Pacheco, F. J.; Palmberg, N.; Seoane-Sepulveda,´ J. B. Lineability and algebrability of pathological phenomena in Analysis. J. Math. Anal. Appl. 326 (2007), no. 2., 929-939. [26] Giles, J. R. Strong diﬀerentiability of the norm and rotundity of the dual. J. Austral. Math. Soc. 26 (1978), 302-308. [27] Habala, P.; Hajek,´ P.; Zizler, V. Introduction to Banach spaces. Matfyzpress, 1996. [28] James, R. C. Characterizations of reﬂexivity. Studia Math. 23 (1964), 205-216. [29] James, R. C. A counterexample for a sup theorem in normed spaces. Israel J. Math. 9 (1971), 511-512. [30] Jimenez-Sevilla,´ M.; Moreno, J. P. A note on norm-attaining functionals. Proc. Amer. Math. Soc. 126 (1998), 1989-1997. [31] Josefson, B. Weak sequential convergence in the dual of a Banach space does not imply norm convergence. Ark. Mat. 13 (1975), 79-89. [32] Kuratowski, K. Topology Vol. I. Academic Press, New York-London, 1966. [33] Mazur, S. Uber¨ konvexe mengen in linearem normieten Raumen¨ . Studia Math. 4 (1933), 70-84. [34] Megginson, R. E. An Introduction to Banach Space Theory. Graduate Texts in Mathematics 183, Springer-Verlag, New York, 1998. [35] Mujica, J. Separable quotients of Banach spaces. Rev. Mat. Univ. Complut. Madrid 10 (1997), no. 2, 299-330. [36] Nissenzweig, A. W ∗ sequential convergence. Israel J. Math. 22 (1975), no. 3-4, 266-272. 71

[37] Phelps, R. R. Extreme points in functions algebras. Duke Math. J. 32 (1965), 267-277.

[38] Rolewicz, S. Metric linear spaces. Mathematics and its Applications 20, Reidel, Dordrecht, 1985.

[39] Sierpinski, W. General Topology. Dover Publications, Inc, Mineola, New York, 2000.

[40] Talponen, J. On assymptotic transitivity in Banach spaces. Preprint.