BEN GURION UNIVERSITY OF THE NEGEV

THE FACULTY OF NATURAL SCIENCES

DEPARTMENT OF MATHEMATICS

Master of Science Thesis

BOUNDARIES AND EXTREME

POINTS OF THE DUAL BALL

OF A POLYHEDRAL SPACE

by

Roi Livni

Supervisor: Prof. Vladimir P.Fonf

Beer Sheva, 2008

To my parents, Rachel and Haim Livni

Abstract

A Banach space X is called polyhedral if the unit ball of every finite dimensional

subspace V ⊆ X is a polytope. The main question in our investigation is thefollowing

Question 0.1. Let X be an infinite dimensional separable polyhedral Banach

Space. Can X be renormed so that extBX∗ will be countable.

This question arose from the following theorem

Theorem 0.1 ([4]). Every Polyhedral Banach space has a countable boundary.

By the Krein-Milman’s theorem the set of extreme points of the unit ball of the

dual is always a boundary for X. In [9] an example of a polyhedral Banach space

X, such that the set extBX∗ is uncountable was given. The spaces constructed

in [9] had the the following property

Property 0.1. The set extBX∗ may be covered by a sequence of norm compact

sets

In [6] and [3] it was proved that if extBX∗ can be covered by a sequence of

norm compact sets, then X can be renormed in such a way that extB(X∗,‖|·‖|) is

countable. The consideration above suggests the following conjecture (which was

open for several years)

Conjecture 0.1. If X is a separable polyhedral Banach space, then the set

extBX∗ can be covered by ∪∞n=1Kn where each Kn is norm compact.We found a new class of separable Banach spaces without this property. In

particular extBX∗ cannot be covered by a sequence of balls with ri → 0, 0 <ri < 1. Any such space is a counter example for the conjecture above. Besides

we proved that there are Banach spaces X such that for every subspace M ⊆ Xwith codimM < ∞, extBM∗ is uncountable.

iii

Acknowledgements

I would like to thank my supervisor Professor Fonf who beyond being my su-

pervisor was also one of my great teachers during my undergraduate studies. In

his course ”Functional Analysis” I learned to love the subject of Banach spaces.

Professor Fonf is an inspiring teacher and continuously challenges the students

with stimulating questions. His enthusiasm for the subject was infectious and I

am certain that without these exciting classes, I wouldn’t have chosen to study

Banach spaces. As my supervisor, Professor Fonf was tolerant to my questions,

and his door was always open. He knew how to direct me and at the same time

gave me the opportunity to conduct an independent research on my own terms.

I would also like to thank the administrative and academic staff of the Mathe-

matics Department, especially Mrs. Rutie Peled, who were always helpful and

pleasant. The teachers in the department for their support and what knowledge

I have in mathematics I owe to them.

Last but not least, I thank my fellow students in the department who were a great

company in this voyage and with whom I have enjoyed many conversations, some

very fruitful, on various mathematical problems.

i

Contents

Abstract iii

Contents ii

1 Introduction 1

1.1 Polyhedral Banach Spaces and the Set of Extreme points of the

Dual Unit Ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Boundaries in Polyhedral Banach Spaces 4

2.1 Minimal Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2 A Polyhedral Banach space without (∗∗) Property . . . . . . . . . 5

3 A Banach space with hereditary property A 73.0.1 The idea of construction . . . . . . . . . . . . . . . . . . . . 7

3.0.2 Construction . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4 A Banach space with a massive extBX∗ 14

Bibliography 19

ii

Chapter 1

Introduction

1.1 Polyhedral Banach Spaces and the Set of Extreme

points of the Dual Unit Ball

A convex body C in a finite dimensional Banach space V is called a polytope if

it is the intersection of finitely many closed half planes (recall that a half plane

is a set of the form {v ∈ V : f(v) ≤ 1} where f is a linear functional over V ). Areal, finite dimensional Banach space V is called polyhedral if the unit ball BVis a polytope. If BV is a polytope, there is a finite set of functionals {f1, ..., fn}such that

BV = {v ∈ V : |fi(v)| ≤ 1 i = 1...n},and

‖v‖ = max{|f1(v)|, .., |fn(v)|}.From the separation theorem it follows that BV ∗ = conv{±fi}ni=1. Hence, theset of extreme points of the unit ball of the dual must be finite. The converse is

obviously true: If extBV ∗ is finite, then BV is the intersection of finitely many

half planes.

In an infinite dimensional Banach space, a convex body C that is the intersection

of finitely many half planes is never bounded. Hence this definition is not inter-

esting in case of infinite dimensional Banach spaces. In [10] V. Klee introduced

the following definition of a polyhedral Banach Space

Definition 1.1. A Banach space X is called Polyhedral if the unit ball of every

finite dimensional subspace of X is a polytope .

1

CHAPTER 1. INTRODUCTION 2

In [10], V. Klee showed that c0 is polyhedral. This is the simplest example

of a polyhedral Banach space. Note that c0 is a ”minimal” polyhedral Banach

space, i.e. any infinite dimensional polyhedral Banach space contains c0 (see [4]).

One of our main concerns will be to evaluate the ”size” of the set extBX∗ in an

infinite dimensional separable polyhedral Banach space. If dimX = ∞, then bythe Krein-Milman Theorem, the set extBX∗ is not finite. Hence, the minimal

cardinality of extBX∗ is countable.

Definition 1.2. For a set A ⊂ BX∗ we denote by A′ the set

A′ = {f ∈ BX∗ : f ∈ w∗ − cl(A/{f})}.

In [9] the following theorem was proved

Theorem 1.1. For a separable Banach space X, the following statements are

equivalent

1. For some equivalent norm on X, the set extB′X∗ is contained in the interior

of BX∗. In particular X is isomorphically polyhedral.

2. For some equivalent norm on X, the set extBX∗ is countable.

Corollary 1.1 ([5]). If X∗ = `1, then X is isomorphically polyhedral.

The following question naturally arises. If X is a separable polyhedral Banach

space, is the set extBX∗ always countable? This was answered in the negative in

[9]. Namely, the space c0 can be renormed in such a way that the set of extreme

points of the dual unit ball will be uncountable.

1.2 Boundary

Definition 1.3. Let X be a Banach space . A subset B ⊆ SX∗ is a boundaryof X if for any x ∈ X there is f ∈ B with f(x) = ‖x‖.

By the Krein-Milman theorem, the set extBX∗ is an example of a boundary. If

X is infinite dimensional, then the minimal cardinality of a boundary is countable.

It was proved in [9] that this is the case of separable polyhedral spaces. So the

following question is natural

Question 1.1. Let X be a separable polyhedral space. Is it possible to introduce

an equivalent norm ‖| · ‖| on X, such that extB(X∗,‖|·‖|) is countable?

CHAPTER 1. INTRODUCTION 3

Definition 1.4. We say that a set A ⊆ X∗ is r-norming if

supf∈A

|f(x)| ≥ r‖x‖,

for every x ∈ X.

Definition 1.5. We say that a Banach space X has property (∗) if there existsa 1-norming set B ⊂ SX∗ such that for each f ∈ B′ and x ∈ BX we have|f(x)| < 1.

Note that a Banach space X has property (*) iff it is isomorphic to a polyhe-

dral Banach space ([12],[7]).

Definition 1.6. We shall say that a Banach space X has property (∗∗) if thereexists a 1-norming set B ⊆ SX∗ such that B′ ⊆ intBX∗.

If a set B is 1 − norming, then from the separation theorem it follows thatBX∗ = convB

w∗ and from Milman’s theorem it follows that extBX∗ ⊆ Bw∗

=

B ∪B′. Hence, property (∗∗) implies that extBX∗ ⊆ B. It follows that a Banachspace X has property (∗∗) iff it satisfies the statements of Theorem 1.1 . Wewill mainly be concerned with the connection between property (*) and property

(**) in separable Banach spaces. Obviously (**) implies (*). We investigate the

following conjecture:

Conjecture 1.1. A separable Banach space has property (∗) iff it is isomorphicto a Banach space with property (∗∗).

We have the following result, in evidence of conjecture 1.1

Theorem 1.2 ([7]). If X is a separable Banach space with countable boundary

B, then X can be renormed to have property (*)

If X has property (∗∗), then extBX∗ ⊆ B. Hence, our conjecture can bereformulated:

Conjecture 1.2. Every separable Banach space X with a countable boundary

can be renormed so that the set extBX∗ is countable.

Since the set extBX∗ is always a boundary, the converse of conjecture 1.2 is

true and conjecture 1.1 and conjecture 1.2 are equivalent.

Chapter 2

Boundaries in Polyhedral Banach

Spaces

2.1 Minimal Boundary

Definition 2.1. Let X be a Banach space. A functional f0 ∈ SX∗ is called aw∗-exposed point of BX∗ if there is an x0 in SX such that f0(x0) = 1 > f(x0)

for each f ∈ BX∗, f 6= f0. Moreover f0 ∈ SX∗ is called w∗-strongly exposed if‖ · ‖ − lim fn = f0 whenever {fn} ⊂ BX∗ and lim fn(x0) = 1.

It is clear that each boundary B contains the (possibly empty) set w∗expBX∗

of all w∗-exposed points. If B = w∗ − expBX∗ then B is a minimal boundary.We also introduce the following notation. For a functional f ∈ SX∗ put

Γf = {x ∈ X : f(x) = 1}, γf = Γf ∩ SX .

The following result was obtained in [4]. For an easier proof see [8].

Theorem 2.1. Let X be a polyhedral Banach space with density character w.

Then the set B = w∗ − strexpBX∗ is a (minimal) boundary for X and moreoverfor each f ∈ B, intΓf γf 6= ∅ and cardB = w.

Recall that by Theorem 1.2, a separable polyhedral Banach space X can be

renormed so that X has property (∗) (see [7]). For the sake of completeness wepresent the proof.

4

CHAPTER 2. BOUNDARIES IN POLYHEDRAL BANACH SPACES 5

Proof. Let B = {±hi}∞i=1 be a boundary of X and let ²i → 0 be a sequence ofpositive numbers tending to zero. Define

V ∗ = conv ± (1 + ²i)hiw∗.

V ∗ is a w∗-closed convex set in X∗ and BX∗ ⊆ V ∗. Hence, ‖|x‖| = supf∈V ∗ |f(x)|is an equivalent norm on X and extV ∗ ⊆ ±(1 + ²i)hiw

∗. Let e be an accumulation

point of {±(1+²i)hi}, then e is also an accumulation point of B. Hence, ‖e‖ = 1.If e attains its maximum on x ∈ V then x ∈ BX but then we can find some hjsuch that hj(x) = 1. Hence, ‖|x‖| ≥ (1 + ²j)|hj(x)| > 1

2.2 A Polyhedral Banach space without (∗∗) PropertyAs mentioned before, in [9] an example was given of a separable polyhedral Ba-

nach space with uncountably many extreme points in the dual unit ball. This

Banach space is actually isomorphic to c0. We denote this Banach space by X

and mention some of its properties

Proposition 2.1. The set extBX∗ can be covered by a countable union of norm

compact sets. In particular the set extBX∗ = B ∪ K where B is a countableboundary of X and K is a norm compact set.

Corollary 2.1. The set extBX∗ can be covered by the union of a countable set

of balls f + ²iBX∗ such that ²i → 0, 0 < ²i < 1.

Proposition 2.2. There is a subspace Y ⊆ X of finite codimension in X, suchthat extBY ∗ is countable.

Each of these properties may be used to prove that X is isomorphic to a

polyhedral Banach space with (∗∗) property.

Theorem 2.2. Assume that there exists a sequence of real numbers ²i → 00 < ²i < 1 satisfying extBX∗ ⊆ ∪∞i=1B(hi, ²i). Then X can be renormed so thatthe set extBX∗ is countable.

The proof is similar to the proof of theorem 1.2 given in [12] and [7].

Proof. Let {hi} be a countable boundary and ²i a sequence of real numbersconverging to zero, satisfying the requirements in the theorem. Define

V ∗ = conv(±{(1 + 2²i)hi}∞i=1)w∗

.

CHAPTER 2. BOUNDARIES IN POLYHEDRAL BANACH SPACES 6

It is easily seen that V ∗ defines a dual norm on X. Denote this norm by ‖| · ‖|.By Milman’s theorem ([1], p. 103), extV ∗ ⊆ ±(1 + 2²i){hi}∞i=1

w∗. Assume that

e is an extreme point of V ∗ and e /∈ {±(1 + 2²i)hi}∞i=1. It follows that there isa sequence (1 + 2²in)hin →w

∗e. Since ²i → 0 we have that e is in {hi}∞i=1

w∗.

Hence e is in the unit ball BX∗ , hence also an extreme point of the unit ball. By

the Hahn-Banach theorem we can find a function F such that ‖|F‖| = 1 = F (e).But there exists an i such that ‖e − hi‖ < ²i. Hence, F (hi) ≥ 1 − ²i andF ((1 + 2²i)hi) > 1.

Theorem 2.3 ([2], p.162 [6]). Assume that there is a sequence {Ki}∞i=1 ofnorm compact sets such that

extBX∗ ⊆∞∪i=1

Ki.

Then, X can be renormed so that the unit ball of the dual has countably many

extreme points

Remark 2.2.1. Theorem 2.3 may be considered as a corollary of theorem 2.2

Theorem 2.4. Assume X has a finite codimension subspace Y , such that Y

has property (∗∗) then X can be renormed so that the unit ball of the dual hascountably many extreme points

Proof. Recall that Y is c0 saturated and that if c0 is contained in a separable

Banach space Y then by the Sobczyk Theorem, it is complemented ([11]). If

X = V ⊕ Y , dimV < ∞ and Y = c0 ⊕ Z then,

X ∼= V ⊕ Y ∼= V ⊕ co ⊕ Z ∼= c0 ⊕ Z ∼= Y.

Definition 2.2. Let X be a separable polyhedral Banach space. We say that X

has property A if the set extBX∗ is uncountable. We say that X has hereditaryproperty A, if every subspace V ⊆ X with codimV < ∞ has property A.

Definition 2.3. Let X be a separable polyhedral Banach space and let C ⊆ BX∗.If C cannot be covered by a countable union of norm compact sets, we say that

C is massive.

Chapter 3

A Banach space with hereditary

property A

The main result of this chapter is the following

Theorem 3.1. There exists a separable polyhedral Banach space X with heredi-

tary property A (see definition 2.2)

3.0.1 The idea of construction

In [9] it was proven that if we take a sequence {λi}∞i=1 such that∑∞

i=1 λi = 1,

% ∈ (0, 1), a = 1λ1 and an =a

∑ni=1 λi

1−% ∑∞i=n+1 λi . Next we define

Am =

am

(m∑

i=1

λi

)−1 m∑

i=1

²kλke∗i : ²k = ±1

.

Then

U∗ = conv({e∗i } ∪∞⋃

i=1

Am

w∗

),

will be the dual unit ball of a polyhedral space X with countable boundary yet

uncountable extU∗. If we take a finite codimensional subspace of X this might

not be the case, i.e. X has a finite co-dimension subspace V such that if BV ∗ is

the unit ball of its dual then extBV ∗ is countable.

Proof. choose k such that∑∞

i=k λi <13a <

13 and take

V = span{ei}∞i=k.

7

CHAPTER 3. A BANACH SPACE WITH HEREDITARY PROPERTY A 8

Then,

extBV ∗ ⊆ {e∗i |V } ∪∞⋃

m=1

Am|V

w∗

.

For each f ∈ Am for some m, consider f as linear functional over V . We havethat

‖f‖V = ‖am(

m∑

i=1

λi

)−1 m∑

i=2

²kλke∗i ‖V ≤

a∑m

i=k λi1− %∑∞m+1 λi

<13· 2 < 2

3.

Hence,{∪∞m=1Am|V

}w∗ ⊆ 23BV ∗ and extBV ∗ = {e∗i |V }∞i=k.

To construct a polyhedral space such that all its finite codimension subspaces

will have uncountably many extreme points in the dual balls, we enlarge the sets

Am by ”shifting”, i.e. instead of taking Am we shall take a countable set Ãm:

Ãm =

am

(m∑

i=1

λi

)−1 m∑

i=1

²kλke∗i+n : ²k = ±1, n ∈ N

.

We claim that if λi and % are chosen in appropriate way then the set

a∞∑

i=1

²kλke∗i+n,

will be the extreme points of the dual ball, and further, in each finite codimen-

sional subspace, uncountably many will remain extreme points.

3.0.2 Construction

Let {ei}∞i=1 be the standard basis of c0 and {e∗i } the standard basis of `1. Fix% ∈ (0, 12) and λi = 12i . Let,

a =1λ1

and

an =a

∑ni=1 λi

1− % ∑∞i=n+1 λi.

Next define:

Am =

am

(m∑

i=1

λi

)−1 m∑

i=1

²kλke∗i+n : ²k = ±1, n ∈ N

.

Each Am is countable. The set

U∗ = conv∞∪

m=1Am

w∗

,

CHAPTER 3. A BANACH SPACE WITH HEREDITARY PROPERTY A 9

is the dual unit ball of an equivalent norm ‖| · ‖| on c0 (Note that {a1e∗i } ⊆ A1.Hence, λ1B`1 ⊆ U∗). Put X = (c0, ‖|·‖|) and denote by {hi}∞i=1 the set ∪∞m=1Am.To prove that {hi} is a boundary, it is enough to prove that each vector in {hi}′with ‖|h‖| = 1 does not attain its maximum.

Claim 3.0.1. every f ∈ {hi}′ with ‖|f‖| = 1 does not attain its norm.

Take f ∈ (⋃∞m=1 Am)′ such that ‖|f‖| = 1 and assume to the contrary, thatthere is x ∈ BX such that f(x) = 1. If f ∈ A′m for some fixed m then this istrivial since A′m = {0}. Assume f ∈ (

⋃∞m=1 Am)

′ and f /∈ A′m for some fixed m.It is easy to see that f is of the form

f = a∞∑

i=1

²iλie∗i+n,

for some set of {²i}, ²i = ±1 and m ∈ N . Choose some x ∈ c0 such thatf(x) = ‖|x‖| = 1 and choose s > m so large that a ·max{|xk|}∞k=s+1 < %2 . Thenthe definition of ‖| · ‖|| implies

1 = f(x) = as∑

k=m

²kλkxk+m + a∞∑

k=s+1

²kλkxk+m

≤ aas

as

(s∑

1

λi

)−1 s∑

k=1

λk|xk+m|

s∑

1

λi +(

a ·maxk>s

|xk+m|) ∞∑

k=s+1

λk

<a

as·

s∑

1

λi +%

2

∞∑

i=n+1

λi < 1.

The last inequality is a result of the following equality:

a

an

n∑

i=1

λi + %∞∑

i=n+1

λi = 1.

Claim 3.0.2. {hi}∞i=1 is a countable boundary for X and X is polyhedral.

This is immediate from claim 3.0.1, the fact that extBX∗ ⊆ hiw∗

and the fact

that extBX∗ is a boundary.

Claim 3.0.3. The set E = extU∗ is uncountable

We shall prove that every

f = a∞∑

i=1

²iλie∗i+n,

CHAPTER 3. A BANACH SPACE WITH HEREDITARY PROPERTY A 10

is an extreme point. For simplification we will prove that

f = a∞∑

i=1

λie∗i+n,

is an extreme point and the general case follows from symmetry. For future use

we will prove a stronger statement

Proposition 3.1. Let (α1, ..., αn−1) be a set of numbers such that |αi| < 12 . Then

F = (α1, ..., αn−1,32,12,12, ...) ∈ `∞.

exposes f

We shall prove the proposition in steps. Denote

δ = minp 0.

It is easy to verify the δ > 0.

Lemma 3.1.1. ‖F‖ = |F (f)| = a

Proof. Define

xk = (α1, α2, ..., αn−1,32,12, ...

12︸ ︷︷ ︸

k

, 0, 0, 0, 0, 0) ∈ c0,

and note that xk →w∗ F ∈ `∞. It is enough to prove the ‖|xk‖| ≤ a. Take h inthe boundary

h = am

(m∑

k=1

λi

)−1 m∑

i=1

²kλke∗k+j ,

and consider three cases.

Case1 j < n− 1

|xk(h)| ≤ am(

m∑

i=1

λi

)−1

n−1−j∑

i=1

λk|αi+j |+ λn−j 32 +12

m∑

i=n+1−jλk

≤ am(1−δ).

case2 j > n− 1

|xk(h)| ≤ am(

m∑

i=1

λi

)−1 m∑

i=1

λk12

<am2

.

CHAPTER 3. A BANACH SPACE WITH HEREDITARY PROPERTY A 11

case3 j = n− 1

|xk(h)| ≤ am(

m∑

i=1

λi

)−1 (32λ1 +

12

m∑

i=2

λk

)=

= am

(m∑

i=1

λi

)−1 (λ1 + λ2 +

m+1∑

i=3

λi

)= am

2m

2m − 12m+1 − 1

2m+1

am2

(2m+1 − 12m − 1

).

Recall

an =a

∑ni=1 λi

1− %∑∞i=n+1 λi=

2(1− 12n )1− % 12n

<2(1− 12n )1− 1

2n+1

= 42n − 1

2n+1 − 1 .

Hence,

am2

(2m+1 − 12m − 1

)< 2

2m − 12m+1 − 1

(2m+1 − 12m − 1

)= 2 = a.

Lemma 3.1.2. Assume g ∈ BX∗ and |F (g)| = a then

g ∈ convBn−1w∗,

where

Bj = {amm∑

i=1

²kλkek+j : ²k = ±1,m ≥ 1}.

Proof. Define

H = conv ∪n−2j=1 Bjw∗

N = conv ∪∞j=n Bjw∗

V = convBn−1w∗

.

By using the definition of U∗ = BX∗ it is easy to see that conv(H ∪N ∪V ) = U∗therefore each g ∈ U∗ may be represented in the form

g = λ1h + λ2n + λ3v,

where λi > 0, λ1 + λ2 + λ3 = 1, h ∈ H, n ∈ N and v ∈ V . By our priorcalculations we have the following estimates:

CHAPTER 3. A BANACH SPACE WITH HEREDITARY PROPERTY A 12

1. |F (H)| < a(1− δ)

2. |F (N)| < a2So

|F (λ1h + λ2n + λ3v)| ≤

λ1|F (h)|+ λ2|F (n)|+ λ3|F (v)| ≤ λ1(a− δ) + λ2 a2 + λ3|F (v)| = a.

Hence, λ1 = λ2 = 0 and g ∈ V

Lemma 3.1.3. If g ∈ BX∗ and |F (g)| = a then g ∈ conv ∪∞m=s+1 (Am ∩Bn−1)w∗

for each s ∈ N

Proof. Define

Ps = conv ∪sm=1 (Am ∩Bn−1)w∗

Qs = conv ∪∞m=s+1 (Am ∩Bn−1)w∗

.

Choose ps ∈ Ps, qs ∈ Qs and µs ∈ [0, 1] such that g = µsps + (1− µs)qs. Then,

a = |F (g)| ≤ µs|F (ps)|+ (1− µs)|F (qs)| ≤ a

Now

|F (ps)| ≤ max{|F (h)| : h ∈ Am ∩Bn−1 m ≤ s}.

For each h ∈ Am ∩Bn−1

|F (h)| ≤ am(

m∑

i=1

λi

)−1 (32λ1 +

12

m∑

i=2

λk

)< a.

The last inequality was evaluated in proposition 3.0.2 case 3. It follows that

µs = 0.

Proof of proposition 3.1. Let g ∈ BX∗ be such that F (g) = a. From proposition3.0.2, it is immediate that gm = 0 for each m < n. We claim that gk+n ≤ aλk forall k ∈ N. If not, there exists k ∈ N such that

gk+n − aλk =: 2² > 0.

Since an → a there exists s ∈ N such that∣∣∣∣∣∣a− am

(m∑

i=1

λi

)−1∣∣∣∣∣∣< ² foreach m > s.

CHAPTER 3. A BANACH SPACE WITH HEREDITARY PROPERTY A 13

Then, for each m > s, we have

gk+n − am(

m∑

i=1

λi

)−1λk ≥ gk+n − aλk −

∣∣∣∣∣∣a− am

(m∑

i=1

λi

)−1∣∣∣∣∣∣λk

> 2²− ²λk > ².

In other words, am(∑m

1 λi)−1λk < gk+n − ². But this implies that the k + n-th

coordinate of each element of ∪∞m=s+1Am ∩Bn−1 is smaller then gk+n − ², whichis in contradiction with g ∈ Qs.Since 32gn +

∑∞n+1

12gk = F (g) =

32aλ1 +

∑∞2 a

12λk, it must be that gk+n = aλk

and g = f

Corollary 3.1. Every finite codimension subspace, M of X, the unit ball of its

dual has uncountably many extreme points

Proof. Take {v1, .., vn} ∈ B`1 linearly independent such that

M = ∩ni=1 ker vi.

Since {vi}ni=1 are linearly independent WLOG we assume that there is a set ofcoordinates j1, .., jn such that v

jki = δi,k. Choose m sufficiently large so that

32|vmi |+

∞∑

n=m+1

12|vni | <

12,

for each choice of i ≤ n. Define F = (αk) ∈ `∞ as follows: for each i ≤ n

αji = −(

32vmi +

∞∑

n=m+1

12vni

),

and αk = 0 for every k 6= ji for some i ≤ n. Then by proposition 3.1

F = (α1, ., αm−1,32,12,12, ...),

exposes f = a∑∞

i=1 λie∗i+m . Since F (vi) = 0, F ∈ M∗∗ and it exposes f|M in

M∗ . Similarly each vector f =∑∞

i=1 ²kλie∗i+m is exposed.

Chapter 4

A Banach space with a massive

extBX∗

In this chapter we prove our main result which is the following

Theorem 4.1. There exists a separable polyhedral Banach space X, such that

for each sequence ²i → 0, 0 < ²i < 1 and a sequence of balls B(xi, ²i). It holdsthat extBX∗ * ∪∞i=1B(xi, ²i)

Fix % ∈ (0, 12) and a series of positive numbers {λi} such that λi = 12i ,

a =1λ1

,

and

an =a

∑ni=1 λi

1− % ∑∞i=n+1 λi.

Denote by Gm the set of injective, non-decreasing mappings from {1, .., m} to Nand G∞ the set of injective, non-decreasing mappings from N to N.

Next define:

Am =

am

(m∑

i=1

λi

)−1 m∑

k=1

²kλke∗g(k) : ²k = ±1, g ∈ Gm

.

Each Am is countable. The set

U∗ = conv∞∪

m=1Am

w∗

,

14

CHAPTER 4. A BANACH SPACE WITH A MASSIVE EXTBX∗ 15

is the dual unit ball of an equivalent norm ‖| · ‖| on c0 (Note that {a1e∗i } ⊆ A1hence a1B`1 ⊆ U∗). Put X = (c0, ‖| · ‖|) and denote by {hi}∞i=1 the set ∪∞m=1Am.To prove that {hi} is a boundary, it is enough to prove that each vector f in{hi}′ with ‖|f‖| = 1, doesn’t attain its supremum on U .

Claim 4.0.4. Every f ∈ {hi}′ with ‖|f‖| = 1 doesn’t attain its norm.

Take f ∈ (⋃∞m=1 Am)′ such that ‖|f‖| = 1. If f ∈ A′m for some fixed m then

f = am(m∑

i=1

λi)−1n∑

k=1

²kλke∗g(k),

for some n < m and g ∈ Gn.

‖|f‖| = ‖|am(m∑

i=1

λi)−1n∑

k=1

²kλke∗g(k)‖| =

‖|am(∑m

i=1 λi)−1

an(∑n

i=1 λi)−1an(

n∑

i=1

λi)−1n∑

k=1

²kλke∗g(k)‖| ≤

a1−% ∑∞i=m+1 λi

a1−% ∑∞i=n+1 λi

< 1.

Assume f ∈ (⋃∞m=1 Am)′ and f /∈ A′m for some fixed m. It is easy to see thatf is of the form

f = a∞∑

k=1

²kλke∗g(k),

for some set of {²k} ²k = ±1 and g ∈ G∞ or

f = an∑

k=1

²kλke∗g(k),

for some g ∈ Gn. If f is of the first form, the same argument as the one usedin claim 3.0.1 may be used to prove the f doesn’t attain its norm. If f is of the

later form, then ‖|f‖| < 1.

Claim 4.0.5. {hi}∞i=1 is a countable boundary for X and X is polyhedral.

This is immediate from claim 4.0.4, the fact that extBX∗ ⊆ hiw∗

and the fact

that extBX∗ is a boundary.

Claim 4.0.6. The set E = extU∗ is uncountable

We shall prove that

f = a∞∑

i=1

λke∗g(k),

for some g ∈ G∞ is an extreme point.

CHAPTER 4. A BANACH SPACE WITH A MASSIVE EXTBX∗ 16

Definition 4.1. Given g ∈ G∞ and f = a∑∞

i=1 λke∗g(k). We define

Uj =

am

(m∑

i=1

λi

)−1 m∑

k=1

²kλke∗g′(k) : ∀i ≤ jg′(i) = g(i)

.

Before proving the claim we prove a proposition.

Proposition 4.1. If f = 12f1 +12f2 and f1, f2 ∈ U∗ then

f2, f1 ∈ ∩∞j=1convUjw∗

Proof. Denote

δ = infm≥0

{2−

∑m+1i=1 λi∑mi=1 λi

}> 0.

The proof is by induction. Assume f2, f1 ∈ convUn−1w∗. Define

H = conv ∪g(n)−1j=1 Bjw∗

∩ convUn−1w∗,

N = conv ∪∞j=g(n)+1 Bjw∗ ∩ convUn−1w

∗

and

V = convBg(n)w∗ ∩ convUn−1w

∗= convUn

w∗,

where

Bj =

am

(m∑

i=1

λi

)−1 m∑

i=1

²kλke∗gx(k)

: ²k = ±1,m ∈ N gx ∈ Gm ∧ gx(n) = j .

Note that Un−1 ⊆ H ∪N ∪ V . Let {ci}∞i=1 be the sequence of natural numbers,not in the image of g, ordered by their natural order. Define F = (F )j ∈ `∞ asfollows

(F )j =

0 j = ci

0 j = g(k), k < n

32 j = g(n)

12 else

.

Take h ∈ Un−1,

h = am

(m∑

k=1

λi

)−1 m∑

i=1

²kλke∗gh(k)

,

and consider three cases

CHAPTER 4. A BANACH SPACE WITH A MASSIVE EXTBX∗ 17

Case1 h ∈ H

|F (h)| ≤ am(

m∑

i=1

λi

)−1 (32λn+1 +

12

m∑

i=n+2

λi

)≤ amλn

(m∑

i=1

λi

)−1 (32λ1 +

12

m∑

i=2

λi

)=

amλn

(m∑

i=1

λi

)−1 (λ1 + λ2 +

m+1∑

i=3

λi

)≤ amλn·(2−δ) ≤ amλn−1(2−δ) ≤ a2n−1−δ

a

2n.

Case2 h ∈ N

|F (h)| ≤ am(

m∑

i=1

λi

)−1 m∑

k=n

λk12

< amλn−1

(m∑

i=1

λi

)−1 m∑

k=1

λk12

<12

a

2n−1.

Case3 h ∈ V

|F (h)| ≤ am(

m∑

k=1

λk

)−1 (32λn +

12

m∑

k=1

λn+k

)=

= amλn−1

(m∑

i=1

λi

)−1 (λ1 + λ2 +

m+1∑

i=3

λi

)= amλn−1

2m

2m − 12m+1 − 1

2m+1

=am2n

(2m+1 − 12m − 1

).

Recall

an =a

∑ni=1 λi

1− %∑∞i=n+1 λi=

2(1− 12n )1− % 12n

<2(1− 12n )1− 1

2n+1

= 42n − 1

2n+1 − 1 .

Hence,

am2

(2m+1 − 12m − 1

)< 2

2m − 12m+1 − 1

(2m+1 − 12m − 1

)= 2 = a.

Note that F (f) = a2n−1 . Since Un−1 = conv(H ∪N ∪ V ), then by our estimates

supg∈convUn−1w

∗F (g) ≤ F (f). Hence, f1(F ) = f2(F ) = a2n−1 . Next choose three

positive numbers α1 + α2 + α3 = 1, h ∈ H, n ∈ N and v ∈ V such that

f1 = α1h + α2n + α3v.

Again by our prior estimates

|F (H)| < a2n−1

− δ a2n

and

|F (N)| < 12

a

2n−1.

Hence, α1 = α2 = 0 and f1 ∈ convV w∗.

CHAPTER 4. A BANACH SPACE WITH A MASSIVE EXTBX∗ 18

Proof of claim. Now assume f = 12f1+12f2. By our prior result, f1 ∈ ∩∞i=1convUi

w∗ .

It is easy to see that eg(n)(convUm

w∗)≤ am (

∑mi=1 λi)

−1 λg(n). Hence f1(eg(n) ≤aλg(n) and the same holds for f2. Since f(eg(n)) = aλg(n) It holds that f1(eg(n)) =

f2(eg(n)) = f(eg(n)). It is also easy to see that for every ci f1(eci) = f2(eci) = 0

and f1 = f2 = f .

Claim 4.0.7. The set extU∗ cannot be covered by a countable union of balls

BX∗(xi, ²i)∞i=1 with 0 < ²i < 1 tending to zero.

Assume otherwise. Denote E ={

a∑∞

i=1 λie∗g(i) : g ∈ G∞

}and let

E ⊆ ∪BX∗(xi, ²i) ²i → 0.

Since BX∗ ⊆ 2B`1 it holds that

E ⊆ ∪B`1(xi, 2²i).

For each ball i choose a representative yi ∈ B`1(xi, 2²i) . Next note that forany two vectors y1, y2 ∈ E, if gy1(n) 6= gy2(n) then ‖y1 − y2‖ > 12n . Choose m0sufficiently large so that for m0 < m it holds that2²m < 14 . Choose n0 sufficiently

large so that if gx(1) > n0 then

max{2²1, .., 2²m0} < ‖x− yj‖,

for each j ≤ m0 (Note that this is possible since 2²i < 2 and E ⊆ 2S`1). Denote byG0 the set {1, 2, ...n0}. Choose m1 > m0 sufficiently large such that if m1 < mthen 2²m < 18 . Denote by G1 the set {gym0+1(1), ..., gym1 (1)}. By our remarkabove, if x ∈ E and gx(1) /∈ G1 then ‖x − yj‖ > 12 for m0 < j ≤ m1. Hence,x /∈ ∪m1i=m0+1B`1(xi, 2²i). Continue to choose inductively mn and Gn such that

1. for every mn < m, ²m < 12n+1 .

2. Gn is finite.

3. if gx(n) /∈ Gn then x /∈ ∪mni=mn−1+1B`1(xi, 2²i).

Choose mn+1 so that for mn+1 < m it holds that 2²m < 12n+2 . Denote by Gn+1the set {gymn+1(n + 1), . . . , gymn+1 (n + 1)}. For every x ∈ E and mn < j ≤ mn+1if gx(n+1) /∈ Gn+1 then ‖x−yj‖ > 12n+1 > 4²j and x /∈ ∪

mn+1mn+1

B`1(xi, 2²i). Define

a1 = maxG0 ∪G1 + 1 and Define an to be max{∪ni=0Gn, a1, . . . , an−1}+ 1. Nextdefine g ∈ G∞ to be g(n) = an, then x =

∑∞i=1 λie

∗g(i) /∈ ∪∞i=1B`1(xi, 2²i).

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