C O N E S , P O S I T I V I T Y A N D O R D E R U N I T S

w.w. subramanian

Master’s Thesis

Defended on September 27th 2012

Supervised by dr. O.W. van Gaans

Mathematical Institute, Leiden University

C O N T E N T S

1 introduction 3

1.1 Assumptions and Notations 3

2 riesz spaces 5

2.1 Definitions 5

2.2 Positive cones and Riesz homomorphisms 6

2.3 Archimedean Riesz spaces 8

2.4 Order units 9

3 abstract cones 13

3.1 Definitions 13

3.2 Constructing a vector space from a cone 14

3.3 Constructing norms 15

3.4 Lattice structures and completeness 16

4 the hausdorff distance 19

4.1 Distance between points and sets 19

4.2 Distance between sets 20

4.3 Completeness and compactness 23

4.4 Cone and order structure on closed bounded sets 25

5 convexity 31

5.1 The Riesz space of convex sets 31

5.2 Support functions 32

5.3 An Riesz space with a strong order unit 34

6 spaces of convex and compact sets 37

6.1 The Hilbert Cube 37

6.2 Order units in c(`p) 39

6.3 A Riesz space with a weak order unit 40

6.4 A Riesz space without a weak order unit 42

a uniform convexity 43

bibliography 48

index 51

1

1I N T R O D U C T I O N

An ordered vector space E is a vector space endowed with a partial orderwhich is ‘compatible’ with the vector space operations (in some sense). Ifthe order structure of E is a lattice, then E is called a Riesz space. This orderstructure leads to a notion of a (positive) cone, which is the collection of all‘positive elements’ in E.

In many applications positivity of normed (function) spaces plays animportant role. The most natural example is the ‘standard machinery’, a toolwhich is frequently used in measure and integration theory.

The aim of this thesis is to provide a construction of a normed Rieszspace, given a cone. This requires a definition of a cone outside the context ofordered vector spaces and will be given by a modification of the Grothendieckgroup construction.

Another goal is to find an example of a Riesz space which does not have aweak order unit. There are several existence results of Riesz spaces that donot have such units, but there are not many explicit examples.

1.1 assumptions and notations

Before reading any further, the reader should be familiar with basic measuretheory and functional analysis. However, most of the material will bedeveloped along the way.

Throughout this thesis, the following assumptions, conventions and nota-tions will be used without further ado:

1. The Axiom of Choice (AC) is accepted;

2. A sequence {xn : n ∈ N} in a topological space X will be denoted as(xn);

3. Let S be a metric space, x ∈ S and r > 0. Now define

B(x; r) = {y ∈ S : d(x, y) < r},B(x; r) = {y ∈ S : d(x, y) 6 r},B(x; r) = {y ∈ S : d(x, y) = 1};

4. If A is a subset of a topological space X, then the set A denotes theclosure of A;

5. The dual of a vector space X will be denoted by X∗.

3

2R I E S Z S PA C E S

This chapter will provide an elementary introduction to the theory of Rieszspaces. Later on, this theory will be used to describe certain vector spaceswhich consist of sets.

Note that this chapter is in no way a complete introduction in the field ofpositivity. This chapter only provides the bare necessities in order to describethe results of this thesis. The interested reader can find a detailed expositionon this subject in the literature, e.g. in [5] and [2].

2.1 definitions

2.1.1 Definition. A partially ordered set (X,4) is a lattice if each pair of elementsx, y ∈ X has a least upper bound (a supremum) and a greatest lower bound(an infimum). The supremum and infimum of any two elements x, y ∈ X isdenoted by

sup{x, y} = x ∨ y and inf{x, y} = x ∧ y.

Note that both the supremum and infimum are unique, provided that theyexist.

The maps

∨ : X× X −−→ X(x, y) 7−−→ x ∨ y

and∧ : X× X −−→ X

(x, y) 7−−→ x ∧ y

are the lattice operations on X.If, in addition, X is a vector space over R, then X is called an ordered vector

space if for all x, y ∈ X the following holds:

1. x + z 4 y + z for all z ∈ X;

2. αx 4 αy for all α > 0.

A Riesz space (or vector lattice) is an ordered vector space that is simultane-ously a lattice. A normed Riesz space is a Riesz space endowed with a norm‖·‖ such that

0 4 x 4 y =⇒ ‖x‖ 6 ‖y‖ and ‖|x|‖ = ‖x‖

for any x, y ∈ E. A complete normed Riesz space is called a Banach lattice. «

From this point on, the letter E will be used to denote either a Riesz spaceor an ordered vector space (E,4). Additionally, the standard ordering on Rwill be denoted by the symbol 6.

2.1.2 Example. Let 1 6 p 6 ∞.

1. Let n ∈ N and consider Rn under the usual vector operations. Letx, y ∈ Rn and write x = (x1, . . . , xn) and y = (y1, . . . , yn). Then Rn is aRiesz space under the partial ordering 4 defined by

x 4 y ⇐⇒ xk 6 yk for k ∈ {1, . . . n}.

By a similar reasoning, the sequence spaces c0 and `p are Riesz spaces.

5

2. Let X be a topological space, then C(X), the space of continuousfunctions on X, is a Riesz space under the (pointwise) ordering 4 givenby

f 4 g ⇐⇒ f (x) 6 g(x) for all x ∈ X.

Note that by slightly modifying the previous ordering, the Lebesguespaces are also examples of Riesz spaces. Suppose (X, Σ, µ) is anarbitrary measure space, then Lp(X, Σ, µ) is a Riesz space under theordering 4 defined as

f 4 g ⇐⇒ f (x) 6 g(x) for almost all x ∈ X. «

The previous examples indicate that Riesz spaces appear in a naturalfashion when studying function spaces. However, not every function spaceis a Riesz space:

2.1.3 Example. Consider E = C1[0, 1], the class of continuously differentiablefunctions on the interval [0, 1] ⊆ R. Observe that E is an ordered vectorspace under pointwise ordering. However, E is not a Riesz space.

Indeed, consider the functions f , g ∈ E given by f (x) = x and g(x) = 1− x.The figure below depicts the graphs of both f and g.

y

x

g(x) f (x)

0 112

1

Note that both f ∧ g and f ∨ g are not differentiable at x = 12 . Therefore, E is

not a Riesz space. «

2.2 positive cones and riesz homomorphisms

2.2.1 Definition. Let (E,4) be an ordered vector space. The set

E+ = {x ∈ E : x < 0}

is the positive cone of E. The members of E+ are said to be the positive elementsof E. The strictly positive elements of E are all the non-zero members of E+. «

2.2.2 Definition. Let E and F be Riesz spaces and T a linear map E→ F. The mapT is called positive if T[E+] ⊆ F+. «

2.2.3 Definition. Let T : E→ F be a linear map between Riesz spaces E and F, thenT is a Riesz homomorphism (or lattice homomorphism) if T preserves the latticeoperations, that is,

T(x ∨ y) = Tx ∨ Ty and T(x ∧ y) = Tx ∧ Ty

6

for all x, y ∈ E.A Riesz isomorphism is a bijective Riesz homomorphism. «

Note that any Riesz homomorphism is a positive map, since x ∈ E+ if andonly if x = x ∨ 0.

The next proposition lists some important properties of the lattice opera-tions on a Riesz space E. The proof is straightforward and will be omitted.

2.2.4 Proposition. Let E be a Riesz space and A ⊆ E a non-empty subset.

1. Let T : E → E be a Riesz isomorphism. If A has a supremum, then so doesT[A]. Furthermore,

sup T[A] = T(sup A) and inf T[A] = T(inf A).

2. If x, y ∈ E, then−(x ∨ y) = (−x) ∧ (−y).

3. Define λA = {λa : a ∈ A} for λ ∈ R+ and suppose that sup A and inf Aexists. Then

sup(λA) = λ sup(A) for all λ ∈ R+

andinf(λA) = λ inf(A) for all λ ∈ R+.

4. For any x, y ∈ E and all λ ∈ R+,

λ(x ∨ y) = (λx) ∨ (λy).

5. Define for x0 ∈ E, the collection A + x0 = {a + x0 : a ∈ A}. Then

sup(A + x0) = (sup A) + x0 and inf(A + x0) = (inf A) + x0,

provided that both sup A and inf A exist.

6. For all x, y, z ∈ E the identity

(x + z) ∨ (y + z) = (x ∨ y) + z

holds.

7. The lattice operations are distributive, i.e. the operations ∧ and ∨ satisfy

(x ∧ y) ∨ z = (x ∨ z) ∧ (y ∨ z) and (x ∨ y) ∧ z = (x ∧ z) ∨ (y ∧ z)

for all x, y, z ∈ E. «

2.2.5 Definition. Let E be a Riesz space and x ∈ E. The positive part x+ and thenegative part x− of x are defined by

x+ = x ∨ 0 and x− = (−x) ∧ 0.

The absolute value |x| of an element x ∈ X is given by

|x| = x ∨ (−x). «

The next proposition follows directly from Proposition 2.2.4:

2.2.6 Proposition. Let E be a Riesz space and let x ∈ E. Then

x = x+ − x− and |x| = x+ + x−. «

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2.3 archimedean riesz spaces

2.3.1 Definition. Let E be a Riesz space and x ∈ E. If

infn∈N

1n · x = 0

for any x ∈ E+, then E is called Archimedean. «

There is another way to define the Archimedean property, one that isfrequently used in the literature. The proof can be found in [2, p .14].

2.3.2 Lemma. Let E be a Riesz space, then the following statements are equivalent:

(a) inf{ 1n x : n ∈ N} = 0 for all x in E+.

(b) If x, y ∈ E+ and nx 4 y for all n ∈ N, then x = 0. «

The next proposition provides infinitely many examples of ArchimedeanRiesz spaces:

2.3.3 Proposition. Any normed Riesz space is Archimedean. «

Proof. Let E be a normed Riesz space and let x, y ∈ E such that n · x 4 yfor any n ∈ N. This means n · x 4 y 4 y+ for all n ∈ N, which implies0 4 n · x+ 4 y+ for any n ∈ N. Hence

n‖x+‖ = ‖n · x+‖ 6 ‖y+‖ for all n ∈ N.

Since ‖y+‖ is a finite real number, it follows that ‖x+‖ = 0. On the otherhand, x+ = 0 by definition ofi the norm. Therefore x 4 x+ = 0. This showsE is Archimedean.

The previous proposition implies that each Riesz space from Example 2.1.2is Archimedean.

At this point a remark is in order. The Archimedean property fromDefinition 2.3.1 differs from the Archimedean property in the set of realnumbers, which states that for any x, y ∈ R there is a n ∈ N such that

|y| 6 n|x|.

Note that this property is implied by both Lemma 2.3.2 and Definition 2.3.1,but it need not be its equivalent:

2.3.4 Example. Consider the space C(0, 1) and endow this space with the point-wise ordering. Observe that this space is Archimedean in the sense ofDefinition 2.3.1.

Indeed, let 0 4 f ∈ C(0, 1). Then

∀x ∈ R : 1n f (x) ↓ 0 in R,

therefore1n f ↓ 0 in C(0, 1).

This shows that C(0, 1) is Archimedean. To prove that it does not satisfythe Archimedean property of the set of real numbers, consider f , g ∈ C(0, 1)given by f (x) = 1

x and g(x) = 1. Then there is no n ∈ N such that f 6 ng. «

8

2.4 order units

2.4.1 Definition. Let E be a Riesz space and let x ∈ E. The principal band generatedby x ∈ E, denoted by Bx, is the collection given by

Bx ={

y ∈ E : |y| = supn∈N|y| ∧ n|x|

}.

The principal ideal generated by an element x ∈ E+, denoted by Ex, is theset

Ex = {y ∈ E : there is a λ > 0 such that |y| 4 λx}.

If there is an element e+ ∈ E such that Be = E, then e is called a weak orderunit and if Ee = E, then e is a strong order unit. «

It is clear from the definition that any strong order unit is also a weakorder unit. The converse statement is not true.

A natural question to ask is whether a Riesz spaces has any order unit atall (strong or weak). There are very few explicit examples of Riesz spacesthat do not have a weak order unit. One of the goals of this thesis is to add anew example to that list.

Let’s state some known results. The proof of the next lemma is somewhatinvolved and will therefore be omitted, but can be found in Meyer-Nieberg

(see [15, p. 20]).

2.4.2 Lemma. Let p ∈ [1, ∞).

1. The Riesz spaces c0, `p do not have a strong order unit.

2. Let (X, Σ, µ) be a measure space, then Lp(X, Σ, µ) does not have a strongorder unit.

3. The dual of `∞ does not have a weak order unit. «

2.4.3 Proposition. Let (X, Σ, µ) be a measure space. Then the constant function 1X is astrong order unit in L∞(X, Σ, µ). «

Proof. Since any f ∈ L∞ is essentially bounded, it follows that | f | 6 ‖ f ‖∞.Therefore,

−‖ f ‖∞ · 1X 6 f 6 ‖ f ‖∞ · 1X for any f ∈ L∞.

Hence 1X is a strong order unit in L∞.

2.4.4 Proposition. Let K be a compact Hausdorff space. Then any strictly positive functionf ∈ C(K) is a strong order unit. «

Proof. Let f ∈ C[K] be strictly positive. Consider

α = minx∈K

f (x).

Note that α is well-defined since continuous functions on compacts setsattain a minimum. Moreover, α > 0 and f 6 α · 1K. This implies f is a strongorder unit.

2.4.5 Theorem. Let (X, Σ, µ) be a measure space and p ∈ [1, ∞). Suppose ( fn) is asequence of Lp-functions which converges in norm to a f ∈ Lp. If fn 6 f , then

f = supn∈N

fn.«

9

Proof. Define Fn = f − fn and consider the sequence (Fn). Then, by usingthe assumptions, it follows that Fn(x) > 0 for all x ∈ X and

limn→∞

∫Fp

n dµ = limn→∞‖Fn‖p

p = 0.

By applying Fatou’s lemma,

0 6∫

lim infn→∞

Fpn dµ

6 lim infn→∞

∫Fp

n dµ

= limn→∞

∫Fp

n dµ

= 0.

This means lim inf Fpn = 0 almost everywhere.

On the other hand, note that for any n ∈ N and any x ∈ X

0 6 infn∈N

Fn(x)p 6 limk→∞

infn>k

Fk(x)p = lim infk→∞

Fk(x)p.

Hence,inf

n∈NFn(x) = 0

almost everywhere on X. Therefore,

f (x) = supn∈N

fn(x)

for almost all x ∈ X.

2.4.6 Theorem. Let (X, Σ, µ) be a measure space, then any strictly positive Lp functionis a weak order unit. «

Proof. Let f , g ∈ Lp such that f is strictly positive and g > 0. Then, byTheorem 2.4.5, it suffices to show

g = supn∈N

g ∧ (n f ).

Recall that the collection of step functions is dense in Lp. So for any ε > 0there is a step function s ∈ Lp such that ‖g− s‖p < ε. Since s is bounded,there exists a M ∈ N such that M f > s almost everywhere. Furthermore,∥∥g ∧ (n f )− s ∧ (n f )

∥∥pp =

∫ ∣∣g ∧ (n f )− s ∧ (n f )∣∣p dµ

for all n ∈ N. Let x ∈ X and note that if g(x) 6 n f (x), which meanss(x) > n f (x). So in this case, g(x)− n f (x) 6 n f (x)− n f (x) = 0. Therefore,n f (x)− g(x) 6 s(x)− g(x).

This yields∣∣g(x) ∧ (n f (x))− s(x) ∧ (n( f (x))∣∣ 6 ∣∣g(x)− s(x)

∣∣ for all x ∈ S.

Then ∥∥g ∧ (n f )− s ∧ (n f )∥∥p

p =∫ ∣∣g ∧ (n f )− s ∧ (n f )

∣∣p dµ

=∫ ∣∣g− s)

∣∣p dµ

= ‖g− s‖pp

< εp.

10

Hence, for all n > N,

‖g− g ∧ (n f )‖ 6 ‖g− s‖p + ‖s− s ∧ (n f )‖p + ‖s ∧ (n f )− g ∧ (n f )‖p

6 ε + 0 + ε.

This shows that g∧ (n f )→ g (w.r.t ‖·‖p) as n→ ∞. In addition, (g∧ n f ) 6 gfor any n ∈ N. Therefore, by applying Theorem 2.4.5, the result follows.

11

3A B S T R A C T C O N E S

The Grothendieck group construction is a tool used in abstract algebra thatconstructs an abelian group from a commutative monoid in a universal way.This construction was develloped in the 1950s by Alexander Grothendieck

(see [3] and [7]) and has played an important role in the devellopment ofK-theory.

This chapter provides a definition of a positive cone outside the contextof ordered vector space and a tool to construct a Riesz space from the cone.The construction is based on a technique similar to the construction of theGrothendieck group.

3.1 definitions

3.1.1 Definition. A non-empty set C with a binary operation

+ : C × C −−→ C(a, b) 7−−→ a + b

such that

(M1) (a + b) + c = a + (b + c) for all a, b, c ∈ C;

(M2) There is a e ∈ C such that e + a = a + e = a for all a ∈ C;

(M3) a + b = b + a for all a, b ∈ C;

is said to be a commutative monoid. If the operation + satisfies only (M1) and(M2), then C is a monoid. The operation + is called the addition on C or simplyaddition. The element e from (M2) is called the unit element for addition on Cand will be denoted by the symbol 0 for convenience. «

Observe that the unit element of a monoid is unique. Indeed, suppose thatthere are two elements e ∈ C and u ∈ C such that

a + e = a and a + u = a

for all a ∈ C. Then, for a = e, it follows that u = u + e = e. Hence e = u.

3.1.2 Example. The collection of natural numbers including 0 is a commutativemonoid under the usual addition. «

3.1.3 Definition. A commutative monoid C equipped with a map

· : R+ × C −−→ C(λ, a) 7−−→ λ · a

such that

(SM1) 1 · a = a and 0 · a = 0 for all a ∈ C;

(SM2) λ · a + µ · a = (λ+ µ) · a for all λ, µ ∈ R+ and all a ∈ C;

(SM3) λ · a + λ · b = λ · (a + b) for all λ ∈ R+ and all a, b ∈ C;

is said to be a (positive) (convex) cone . The map · is called the scalar multiplica-tion on C. If no confusion can arise, the symbol · will be left out. «

13

3.1.4 Definition. Let C be a cone. If there exists a partial ordering 4 on C such that

a 4 b =⇒ λa 4 λb for all λ ∈ R+ and all a, b ∈ C;

a 4 b =⇒ a + c 4 b + a for all a, b, c ∈ C,

then C is called an ordered cone. «

Any cone C with the property

a + b = e =⇒ a = b = e for all a, b ∈ C

can be ordered by defining a 6 b if and only if there is a c ∈ C such that

a + c = b.

A routine verification shows that 6 defines a partial order that meets all therequirements of Definition 3.1.4. From now on, this order 6 will be calledthe standard order on a cone C.

3.2 constructing a vector space from a cone

Let C be a cone and define the relation ∼ on C × C by

(a, b) ∼ (x, y) ⇐⇒ there is a c ∈ C such that a + y + c = x + b + c.

Then, by a straightforward verification, ∼ is an equivalence relation on theCartesian product C × C.

Consider the quotient space

EC = C × C/∼.

In order to avoid confusing notation, an element of EC will be denoted by[a, b]. The quotient space EC is a vector space by defining the following(vector) operations

λ[a, b] = [λa, λb] for all λ ∈ R+ and all a ∈ C;

[a, b] + [x, y] = [a + x, b + y] for all a, b, x, y ∈ C;

[a, b] = −[b, a] for all a, b ∈ C.

The cone C can be mapped into EC under the quotient map

q : C −−→ ECa 7−−→ [a, 0]

This map preserves the cone structure of C, but it need not necessarily be anembedding. The cone operation + must satisfy

a + c = b + c =⇒ a = b for all a, b, c ∈ C.

This is called the cancellation law of the cone operation +.Indeed, suppose that q(a) = q(b) for a, b ∈ C, then [a, 0] = [b, 0]. By the

definition of ∼, there is a c ∈ C such that a + c = b + c. By the cancellationlaw, q is indeed injective hence an embedding.

From this point on, elements of the form [a, 0] are called positive elements ofEC and elements of the form [0, b] are said to be the negative elements of EC .This convention ensures that the members of C correspond to the positiveelements of EC .

Observe that the comments above agree with Proposition 2.2.6. That is, ifc ∈ EC , then there are elements a, b ∈ C such that c = a− b.

14

3.3 constructing norms

Given a cone C in a metric space S, it is possible to construct a norm on ECunder certain extra assumptions on the metric.

3.3.1 Definition. A metric space C =(C, d

)which is simultaneously a cone is

called a metric cone. «

3.3.2 Theorem. Let C be a metric cone. Assume that the metric d is translation invariantand homogeneous in both arguments, that is,

d(a, b) = d(a + c, b + c) for all a, b, c ∈ C;

λd(a, b) = d(λa, λb) for all λ ∈ R+ and all a, b ∈ C.

Then there is a norm ‖·‖ on EC , which induces d. «

Proof. Let a, b, c ∈ C such that a + c = b + c. Then, by translation invariance,

0 = d(a + c, b + c) = d(a, b).

This proves the cancellation law.Next, let a, b ∈ C and put

‖[a, b]‖ = d(a, b).

The map ‖·‖ is well-defined. Indeed, suppose (a, b) ∼ (x, y), then there isa c ∈ C such that a + y + c = x + b + c. Then a + y = x + b, by using thecancellation law. Since d is translation invariant, it follows that

d(a, b) = d(a + x + y, b + x + y)

= d(x, y).

Therefore, ‖·‖ is well-defined.The mapping ‖·‖ is indeed a norm on EC . Suppose that λ ∈ R+ and

a, b ∈ C, then, by homogeneity,

λ‖[a, b]‖ = λd(a, b) = d(λa, λb) =∥∥[λa, λb]

∥∥ =∥∥λ[a, b]

∥∥.

If λ 6 0, then∥∥λ[a, b]∥∥ = ‖|λ|[b, a]‖ = |λ|‖[b, a]‖ = |λ|d(b, a) = |λ|d(a, b) = |λ|

∥∥[a, b]∥∥.

Therefore, the norm is homogeneous.Let a, b, x, y ∈ C, then by translation invariance,

‖[a, b] + [x, y]‖ = ‖[a + x, b + y]‖= d(a + x, b + y)

6 d(a + x, a + y) + d(a + y, b + y)

= d(x, y) + d(a, b).

Therefore‖[a, b] + [x, y]‖ 6 ‖[a, b]‖+ ‖[x, y]‖,

which proves the triangle inequality.Lastly, note that ‖[a, b]‖ = 0 if and only if d(a, b) = 0. Hence, a = b and

consequently, [a, b] = [0, 0].

3.3.3 Corollary. Let (C, d) be a metric cone such that d is homogeneous and translationinvariant. Then there is an isometric embedding of C into EC . «

15

Proof. By the proof of Theorem 3.3.2, C has the cancellation law, so themap ψ : C → EC where ψ(a) = [a, 0] is a well-defined embedding. By usingthe definition of the norm on EC (from the proof of Theorem 3.3.2), ψ is anisometry.

3.3.4 Corollary. Let X be a vector space over R and C ⊆ X a metric cone in X such thatthe metric on C is homogeneous and translation invariant. Then there is a linearembedding ψ : EC → X.

If in addition, X is a normed space and d is the metric on C induced by the normon X, then ψ is an isometric embedding. «

Proof. For the first part, put ψ([a, b]) = a− b. The definition of the normon EC ensures that EC ⊆ X is a subspace. Explicitly,

EC = {a− b : a, b ∈ C}.

For the second part, observe that

‖a− b‖ = d(a, b) = ‖[a, b]‖ for all a, b ∈ C.

Therefore ψ is isometric.

The previous theorem and its corollaries, lead to the next definition:

3.3.5 Definition. A metric cone (C, d), is said to be a normed (convex) cone wheneverthere is a linear isometric embedding into some normed space (X, ‖·‖). «

3.4 lattice structures and completeness

3.4.1 Definition. An ordered cone C is a lattice cone if both a ∨ b and a ∧ b exist forany a, b ∈ C. «

Given an ordered cone C, it is possible to put an ordering on the associatedvector space EC . This is done by defining

[a, b] 6 [x, y] ⇐⇒ a + y 6 x + b.

It turns out that EC inherits the lattice cone structure of C:

3.4.2 Proposition. If C is a lattice cone, then EC is a Riesz space. «

Proof. Let a, b ∈ C. It suffices to show that [a ∨ b, b] is the supremum of[a, b] and 0.

Because a 6 a ∨ b, it follows that [a, b] 6 [a ∨ b, b]. Due to b 6 a ∨ b, itfollows that [a ∨ b, b] is positive. So [a ∨ b, b is an upper bound of [a, b] and 0.

It remains to show that [a, b] ∨ 0 is indeed the supremum of [a, b] and 0.To this end, let x, y ∈ C such that [x, y] > 0. Then there is a c ∈ C such that

(x, y) = (c, 0).

Suppose [a, b] 6 [c, 0], then a + 0 6 b + c. Since b 6 b + c, it follows thata ∨ b 6 b + c. Therefore,

[c, 0] = [b + c, b] > [a ∨ b, b],

which proves that [a ∨ b, b] ∨ 0 is the supremum of [a, b] and 0.

There is an expression for the absolute value of elements in EC :

16

3.4.3 Proposition. Let C be a normed cone. Then∣∣[a, b]∣∣ = [|a− b|, 0

]for all [a, b] ∈ EC . «

Proof. Note that [a, b] 6 [|a − b|, 0], since a + 0 6 b + |a − b|. Similarly,[b, a] = −[a, b] 6 [|a− b|, 0], because b + 0 6 a + |a− b|.

Let [x, y] ∈ EC and suppose that [x, y] > [a, b] and [x, y] > −[a, b]. Then

a + x 6 y + b and b + y 6 x + a.

This yieldsa− b 6 x− y and b− a 6 x− y.

Completeness of C does not imply that the associated vector space EC iscomplete:

3.4.4 Remark. Let C be a complete normed cone, then EC need not be complete.

Proof. Consider a normed space E which is not complete. Define the trivialordering

x 6 y ⇐⇒ x = y.

Then E is an ordered vector space and the only positive element is 0, whichmeans E+ = {0}. This is clearly a complete normed space.

Recall the following result from Banach space theory (see [14, p. 20] for aproof):

3.4.5 Lemma. Let X be a normed space. Then X is a Banach space if and only if everyabsolutely convergent series converges in X. «

3.4.6 Theorem. Let C be a normed lattice cone such that its norm is a non-decreasingmap. Then the associated normed space (EC , |||·|||) where

|||x||| = ‖|x|‖

is a Banach lattice whenever C is a Banach space. «

Proof. First note that EC is a lattice under its natural ordering.Let (xn) be a |||·|||-absolutely convergent sequence in X.Note that 0 6 x+n 6 |xn| for all n ∈ N. Therefore, because the norm on C

is assumed to be non-decreasing,

‖x+n ‖ 6 ‖|xn|‖ = |||xn||| for all n ∈ N.

Consequently, the series over all positive parts is absolutely convergent. So,by completeness of C, the series ∑ x+n is convergent. By a similar reasoning,the sum over all negative parts is also convergent.

Then, for N ∈ N sufficiently large,∣∣∣∣∣∣∣∣∣ ∑n6N

x−n −(

∑n∈N

x+n − ∑n∈N

x−n)∣∣∣∣∣∣∣∣∣ 6 ∣∣∣∣∣∣∣∣∣ ∑

n6Nx+n − ∑

n∈Nx+n∣∣∣∣∣∣∣∣∣

+∣∣∣∣∣∣∣∣∣ ∑

n6Nx−n − ∑

n∈Nx−n∣∣∣∣∣∣∣∣∣

6∣∣∣∣∣∣∣∣∣ ∑

n6N+1x+n∣∣∣∣∣∣∣∣∣+ ∣∣∣∣∣∣∣∣∣ ∑

n6N+1x−n∣∣∣∣∣∣∣∣∣

6 ∑n6N+1

|||x+n |||+ ∑n6N+1

|||x−n |||.

By letting N → ∞, the series ∑n6N xn converges in EC with respect to |||·|||.Therefore, EC is complete with respect to |||·|||, which means that it is aBanach lattice by Lemma 3.4.5.

17

4T H E H A U S D O R F F D I S TA N C E

The main goal of this chapter is to define a distance between subsets in aBanach space X = (X, ‖·‖). Since the power set of the entire space X can bequite ‘large’, it is hard to define a notion of distance on the whole power set.It is therefore necessary to make a restriction to a suitable subcollection ofthe power set instead.

4.1 distance between points and sets

In a metric space S, a distance is assigned to every pair of elements x, y ∈ S.

4.1.1 Definition. Let S be a metric space, x ∈ S and A ⊆ S. The distance from x toA is then given by

d(x, A) = infa∈A

d(x, a). «

Loosely speaking, the notion of distance between a point x ∈ S and asubset A ⊆ S is the distance from x to the ‘nearest’ point a ∈ A. There mightbe some difficulties when A is empty. Note that the distance between a pointand a non-empty subset is a non-negative element in R, by definition. Toavoid complications with the empty set, define inf∅ = ∞ so that d(x,∅) = ∞for each x ∈ S.

To illustrate this definition, consider the following example:

4.1.2 Example.

1. Consider R equipped with the Euclidean metric. Take x ∈ R, thend(x, Q) = 0, because Q is dense in R.

2. Consider R2 with the Euclidean metric. Take the point x0 = (−1, 2)and the line A = {(x, y) : y = x} = {(x, x) : x ∈ R}. Then

d(x0, A) = infy∈A‖x0 − y‖ = inf

x∈R

√(−1− x)2 + (2− x)2.

Put f (x) = (−1− x)2 + (2− x)2, then, by using the derivative of f , itfollows that the point in A nearest to x0 is (1/2, 1/2) ∈ A. Therefore,

d(x0, A) = ‖(−1, 2)− ( 12 , 1

2 )‖ =32

√2. «

4.1.3 Proposition. Let S be a metric space and let A ⊆ S be a fixed non-empty subset.Then the function

fA : S −−→ R+

x 7−−→ d(x, A)

is Lipschitz continuous (as a function of x) with Lipschitz constant 1. «

Proof. Let x, y ∈ S, then

d(x, A) 6 d(x, a) 6 d(x, y) + d(y, x) for all a ∈ A.

This yieldsd(x, A)− d(x, y) 6 inf

a∈Ad(y, A) = fA(y),

which givesd(x, A)− d(y, A) 6 d(x, y).

19

The map fA has another important property:

4.1.4 Proposition. Let S be a metric space and A ⊆ S a non-empty subset. If fA is notidentically zero, then fA(x) = 0 if and only if x ∈ A. «

Proof. Let A ⊆ S. Suppose that d(x, A) = 0. Then x ∈ A or x /∈ A. Notethat if x /∈ A holds, then x must be in the closure of A.

Indeed, let ε > 0. The definition of d(x, A) implies that there is a y ∈ Asuch that d(x, y) < ε. That is, y ∈ B(x; ε). So every ball with centre x andradius ε contains a point of A. This implies B(x; ε) ∩ A 6= ∅, which meansthat x is a limit point of A. Therefore, x ∈ A.

For the converse, assume x ∈ A. Then d(x, A) = 0. Indeed, let ε > 0then B(x; ε) contains a point y ∈ A. So d(x, A) < ε. Since ε was arbitrary,d(x, A) = 0.

4.2 distance between sets

Let S be a metric space and A, B ⊆ S. In the previous section the concept of‘distance from point to subset’ was introduced and discussed. This conceptcan be extended in order to find a suitable definition of distances betweensubsets of S.

The Hausdorff semi-distance

Definition 4.1.1 can be modified to ‘measure’ the distance between subsets inthe following way:

4.2.1 Definition. Let S be a metric space and A, B ⊆ S non-empty subsets. Thedistance from A to B or Hausdorff semi-distance is defined as

δ(A, B) = supa∈A

d(a, B) = supa∈A

infb∈B

d(a, b).«

Note that δ(A, B) may be infinite for unbounded sets A, B ⊆ S according tothe conventions from Definition 4.1.1. Since this might lead to some technicaldifficulties later on, it is therefore convenient (and sometimes necessary) toconsider a ‘suitable’ collection of subsets in S.

Recall that a subset A of a metric space S is said to be bounded if and onlyif A has finite diameter, that is,

diam A = supx,y∈A

d(x, y) < ∞.

The supremum of the empty set is defined to be −∞.

4.2.2 Definition. Let S be a metric space, then define the collection

H(S) ={

A ⊆ X : A is bounded, closed and non-empty}

.

This set is known as the hyperspace of all non-empty bounded, closed subsets of S.The set of all non-empty and compact subsets in S will be denoted by K(S).«

There is a relation between K(S) and H(S). Clearly, K(S) ⊆ H(S) for anymetric space S. Recall that a metric space S is said to have the Heine-Borelproperty, if any closed and bounded set is compact. If this happens to be thecase, then K(S) = H(S).

It turns out that δ is a pseudometric on H(S):

20

4.2.3 Proposition. Let S be a metric space and A, B, C ∈ H(S), then

1. δ(A, B) = 0 if and only of A ⊆ B.

2. δ(A, B) 6 δ(A, C) + δ(C, B). «

Proof. Let A, B, C ∈ H(S).

1. Note that if A ⊆ B, then δ(A, B) = 0. Conversely, suppose thatδ(A, B) = 0, then d(a, B) = 0 for all a ∈ A. Then a ∈ B, by using theresult of Proposition 4.1.4

2. By using the ‘ordinary’ metric of the underlying space S, for all a ∈ A,b ∈ B and c ∈ C the distance satisfies

d(a, b) 6 d(x, c) + d(c, b) 6 d(x, c) + d(c, B).

This estimate is valid for each b ∈ B, so

d(a, B) 6 d(a, c) + d(c, B) 6 d(a, c) + δ(C, B).

for all a ∈ A and c ∈ C. Note that this estimate is valid for any c ∈ C.Therefore,

d(a, B) 6 d(a, C) + δ(C, B)

for any a ∈ A. By taking the supremum over all a ∈ A, the resultfollows.

Note that δ need not be a symmetric map:

4.2.4 Example. Let S = C and consider the subsets A = {z ∈ C : |z| 6 2} andB = {z ∈ C : |z − 4| 6 1}. Both of these sets are bounded, closed andnon-empty.

−2 4

−2

2

0

δ(B, A)

δ(A, B)

AB

Re z

Im z

Then δ(A, B) = 5 and δ(B, A) = 3, which yields δ(A, B) 6= δ(B, A). «

The previous example show that (H(S), δ) is not a metric space. However,this problem can be fixed, by ‘symmetrizing’ δ. This yields a metric on H(S).

21

The Hausdorff metric

4.2.5 Definition. Let S be a metric space and A, B ∈ H(S). The mapping

dH : H(S)×H(S) −−→ H(S)(A, B) 7−−→ δ(A, B) ∨ δ(B, A)

is the Hausdorff metric or Hausdorff distance on H(S). «

The word ‘metric’ from the previous definition needs some verification:

4.2.6 Theorem. dH is a metric on H(S). «

Proof. Since both A and B are bounded and non-empty, both δ(A, B) andδ(B, A) are finite. The triangle inequality follows from Proposition 4.2.3.Note that dH is symmetric by definition. By symmetry and Proposition 4.2.3,it follows that

dH(A, B) = 0 ⇐⇒ A ⊆ B ⊆ A.

If both A and B are closed, then they are equal to their respective closures,which means A = B.

Now that H(S) =(H(S), dH

)is indeed a metric space, one could wonder

whether there is an embedding of S into H(S).

4.2.7 Theorem. Let S be a metric space, then S can be isometrically embedded into H(S)by the map

J : S −−→ H(S)x 7−−→ {x}

Furthermore, the set J[S] is closed in H(S). «

Proof. Note that for any x, y ∈ S the distance between x and y satisfies

d(x, y) = dH({x}, {y}

).

This shows that S can be isometrically identified with J[S] ⊆ H(S).Let A ∈ J[S], then the distance from A to J[S] is 0 by Proposition 4.1.4.

Therefore, for any ε > 0, there is an element x ∈ S such that dH(A, {x}) < ε.This yields d(x, y) < ε for any y ∈ A. In addition, diam A 6 ε, because A isnon-empty. This shows that A must be a set that contains only one element,which means A ∈ J[S]. Therefore, J[S] contains all of its limit points, whichimplies J[S] is closed.

The following result will be used freely without mention in this thesis:

4.2.8 Proposition. Let S be a metric space and K, L ∈ K(S). Then there is an elementa ∈ A and b ∈ B such that

d(a, b) = dH(A, B). «

Proof. Assume without loss of generality that dH(A, B) = δ(A, B). Sincethe map fB : A→ R+ : x 7→ d(x, B) is continuous and A is compact, there isan element a ∈ A such that d(a, B) = dH(A, B).

Similarly, by continuity of the map x 7→ d(a, x) and compactness of B, itfollows that there is a b ∈ B such that

d(a, b) = infx∈B

d(a, x) = d(a, B) = dH(A, B).

22

The Hausdorff distance between non-compact set might not be attained:

4.2.9 Example. Consider the metric space `2 and consider the collection of unitelements U = {un : n ∈ N} where

un(k) =

{1 if k = n;0 otherwise.

Next, define A = {x} ∪U where x is the sequence {−1/n : n ∈ N}. Notethat both A and U are closed, bounded and non-empty subsets of `2, but notcompact. Observe that dH(A, U) = d(x, U), because δ(U, A) = 0.

On the other hand, by using Euler’s identity,

d(x, un) =(

1 + π2

6 + 2n

)1/2for all n ∈ N.

Therefore,

dH(A, U) = infun∈U

d(x, un) =(

1 + π2

6

)1/2.

But then,

d(x, un) >(

1 + π2

6

)1/2for all n ∈ N. «

4.3 completeness and compactness

This section will investigate sufficient conditions to ensure that H(S) iscomplete and K(S) is compact, for a given a metric space S.

4.3.1 Theorem. S is a complete metric space if and only if (H(S), dH) is complete. «

Proof. [⇐=] Suppose H(S) is complete, then the collection {{x} : x ∈ S} isclosed in H(S) by Theorem 4.2.7 and complete by completeness of H(S). Byutilizing the isometric map from Theorem 4.2.7, it follows that S is complete.[=⇒] Assume S is complete. Let (An) be a Cauchy sequence in H(S) and

put

A =⋂

m∈N

⋃n>m

An.

Note that A is closed by definition. The goal is to show that A is the limit of(An) and that A ∈ H(S).

Claim 1. Let ε > 0, then by the Cauchy property, there is a N ∈ N such that

dH(An, Am) < ε whenever n, m > N.

Then d(a, AN) 6 ε for any a ∈ A and A is bounded.

Indeed, note that δ(An, AN) < ε for all n > N (by the Cauchy property).Therefore ⋃

n>NAn ⊆ {x ∈ X : d(x, AN) 6 ε},

so A ⊆ {x ∈ X : d(x, AN) 6 ε}. Thus d(x, AN) 6 ε for all x ∈ A. Next, inorder to prove that A is bounded, take x, y ∈ A. Then there are a, b ∈ ANsuch that

d(a, x) 6 2ε and d(b, y) 6 2ε.

23

Then, by the triangle inequality,

d(x, y) 6 d(x, a) + d(a, b) + d(b, y) 6 4ε + diam AN .

Therefore diam A 6 4ε + diam AN , which implies that A is bounded. HenceA ∈ H(S), which concludes this claim.

Claim 2. Let ε > 0, then d(y, A) 6 ε for any y ∈ AN .

In order to prove the claim, it suffices to show that given ε > 0, there is anelement a ∈ A such that that d(y, a) 6 ε. This will be done by building aCauchy sequence (an) in X for each y ∈ A. Then, by completeness of X,there is an element a ∈ X such that an → a.

To this end, take y ∈ AN and put N = N1. By repeatedly applying theCauchy property, for each k ∈ N there is a Nk ∈ N such that Nk < Nk+1 and

dH(An, Am) < ε · 21−k whenever n, m > Nk.

Set a1 = y, then a1 ∈ AN1 . Now recursively take ak ∈ ANk such that

d(ak+1, ak) < ε · 21−k.

Observe that each ak can be chosen in this way, due to the ‘clever’ choice ofthe indices Nk. In addition, by repeatedly applying the triangle inequality,

d(ak+n, sk) 6n−1

∑`=0

d(ak+`+1, ak+`)

<n−1

∑`=0

ε · 21−k−`

6 ε · 21−k (1)

for all k, n ∈ N where n > 0. This proves that (an) is Cauchy in X and bycompleteness, it converges to some a ∈ X. Let n→ ∞, then by the estimatein (1),

d(a, ak) < ε · 22−k

for all k ∈ N and consequently,

d(a, y) = d(a, a1) 6 ε.

In order to show a ∈ A, note that a is defined as the limit of the sequence(ak) and each term of this sequence is contained in the corresponding setANk . Therefore a ∈ {ak : k > m} for all m ∈ N. Since k < Nk for all k ∈ N, itfollows that

a ∈ {ak : k > m} ⊆⋃

k>m

ANk ⊆⋃

n>mAn

for all m ∈ N. So a ∈ A, which finishes the proof of the claim.By combining the results of the previous two claims, dH(A, AN) 6 2ε. So

by using the triangle inequality,

dH(A, An) 6 dH(A, AN) + dH(AN , An) 6 ε for all n > N.

4.3.2 Corollary. If S is a complete metric space, then K(S) is complete. «

24

Proof. Let (Kn) be a Cauchy sequence in K(S). Since K(S) ⊆ H(S), thelimit of (Kn) exists in H(X) by Theorem 4.3.1. Therefore, by completeness ofS, it remains to show that the limit

K =⋂

m∈N

⋃n>m

Kn

is totally bounded.To this end, let ε > 0. By the proof of Theorem 4.3.1, there exists a N ∈ N

such thatK ⊆ {x ∈ X : d(x, KN) 6 ε/3}.

Furthermore, KN is compact by definition. This means there is a n ∈ N andx1, . . . xn ∈ AN such that

KN ⊆n⋃

k=1

B(xk; ε/2).

Take x ∈ K, then d(x, KN) 6 ε/3. Therefore there is a point y ∈ KN withd(x, y) < ε/2 and there exist k such that y ∈ B(xk; ε/2). Thus

d(x, xk) 6 d(x, y) + d(y, xk) 6 ε.

Therefore

K ⊆n⋃

k=1

B(xk; ε),

which means K is totally bounded.

4.3.3 Corollary. If S is a compact metric space, then K(S) is compact. «

Proof. Recall that any compact metric space is complete. Therefore, K(S) iscomplete by Corollary 4.3.2. It remains to show that K(S) is totally bounded.

Observe that S is totally bounded. So, given ε > 0, there is a n ∈ N andthere exist x1, . . . xn ∈ S such that

inf16k6n

d(x, xk) = min16k6n

d(x, xk) < ε for all x ∈ X.

Let K ⊆ X be a compact non-empty subset and define

B = {xk : d(xk, K)} < ε.

Then dH(B, K) < ε. Therefore, any K ∈ K(S) is at Hausdorff distance lessthen ε of a subset of the (finite) collection {xk : 1 6 k 6 n}. Hence K(S) istotally bounded and complete, which implies that K(S) is compact.

4.4 cone and order structure on closed bounded sets

If S =(S, d

)is a metric space, then H(S) is a partially ordered space of sets

under the inclusion relation ‘⊆’.By definition, the inclusion is a partial order on H(S). The addition in

this vector space is given by the Minkowski sums. Explicitly, let A, B ∈ H(S)then define

A + B = {a + b : a ∈ A and b ∈ B}.

Define the scalar multiplication on H(S) by

λA = {λa : a ∈ A}

25

for any λ ∈ R and A ∈ H(S). By definition, both addition and scalarmultiplication are well-defined.

Under these assumptions, H(S) is indeed an ordered vector space andan ordered cone (according to Definition 3.1.3). Moreover, inclusion is thestandard order as defined in section 3.1. From now on, the cone H(S) shallbe investigated in more detail.

Note that all of the previously defined operations and observations allhold for K(S) as well.

There is way to incorporate the Minkowski sums into the definition ofHausdorff distance:

4.4.1 Lemma. Let A, B ∈ H(X), then

dH(A, B) = inf{ε > 0 : A ⊆ B + εB and B ⊆ A + εB}. «

Proof. Let ε > 0 be given arbitrarily. Suppose A ⊆ B + εB, then d(a, B) 6 εfor all a ∈ A. Therefore

δ(A, B) = supa∈A

infb∈B

d(a, b) 6 ε

and by an analogous reasoning, δ(B, A) < ε. Hence dH(A, B) < ε.To show the other inequality, suppose that A is not contained in B + εB.

Then there exists an element a ∈ A such that a /∈ B + εB. Therefore

d(a, b) > ε for all b ∈ B,

so that d(a, B) > ε. Then, by taking the supremum of all a ∈ A, it followsthat dH(A, B) > ε. So if A is not contained in B + εB, then dH(A, B) > ε.This concludes the proof.

The expression of the Hausdorff distance from Lemma 4.4.1 will be usedwithout mention from this point on. The ‘power’ of this expression willbecome apparent in the next chapter.

In order to obtain a Banach lattice from the cone H(S), the cancellation law(of the Minkowski addition) must be verified. It turns out that the cancellationlaw holds in a very general setting, namely the setting of topological vectorspaces.

Recall that a subset A of a topological vector space X over a field F isbounded if for every neighbourhood N of the zero vector there exists a scalarλ ∈ F so that A ⊆ λN.

4.4.2 Theorem. Let X be a topological vector space, then

A + B ⊆ C + B =⇒ A ⊆ C,

for any non-empty subsets A, B, C ⊂ X such that B is bounded and C closed andconvex. «

Proof. Let N0 be a base of neighbourhoods of the zero-element in X. LetU0 ∈ N0 be a given and define a sequence (Vn) in N0 such that V0 + V0 ⊆ U0and

Vn + Vn ⊆ Vn−1 for all n > 1.

Now assume that A + B ⊆ C + B, then

A + B ⊆ C + B + V for any V ∈ N0.

26

ThereforeA + B ⊆ C + B + Vn for all n > 1.

Next, let a ∈ A, b1 ∈ B and let n ∈ N be a large, fixed number. Then thereexists c1 ∈ C, b2 ∈ B and v1 ∈ V such that

a + b1 = c1 + b2 + v1.

By proceeding inductively, there are ck ∈ C, bk+1 ∈ B and vk ∈ V such that

a + bk = ck + bk+1 + vk for k ∈ {1, . . . , n}.

Then, by adding up all the equations, rearranging the terms and using thetelescope sum on all the bk’s, it follows that

a =1n

n

∑k=1

ck +1n(bn+1 − b1) +

1n

n

∑k=1

vk.

Therefore, since C is convex and B is bounded, a ∈ C + V0 + . . . + Vn. Then,for n large and U0 ∈ N0,

a ∈ C + V1 + . . . + Vn ⊆ C + U0.

Hence A ⊆ C + U0 for all U0 ∈ N0, which means A ⊆ C.

If X is a normed space and the sets A, B and C from the previous theoremare convex, bounded, closed and non-empty, an alternative geometric proofcan be given by using a Hahn-Banach argument.

Recall the following separation theorem (see [21, p. 140] for a proof):

4.4.3 Theorem (Hahn-Banach separation theorem). Let X be a normed linear space andlet A, B ⊆ X be convex and non-empty subsets. If A is compact and B is closed,then there exists a functional φ ∈ X∗ and s, t ∈ R such that

φ(a) < t < s < φ(b) for all a ∈ A and b ∈ B. «

4.4.4 Lemma. Let X be a normed space and A, B, C ⊆ X be non-empty subsets such thatA and B are bounded, closed and convex and C is both closed and convex. Then

A + C ⊆ B + C =⇒ A ⊆ B. «

Proof. Let a ∈ A \ B, then the singleton {a} is both compact and convex.Note that B is closed and convex. Therefore, by the Hahn-Banach separationtheorem, there is a continuous linear functional φ : X → R which strictlyseparates {a} and B. In other words, there exists a t ∈ R such that

φ(a) < t < φ(b) for all b ∈ B.

Additionally, since C is bounded and φ is continuous, the set φ[C] is bounded.Therefore

φ(a) +φ[C] 6⊆ φ[B] +φ[C],

so that a + C 6⊆ B + C. Therefore A + C 6⊆ B + C.

The cancellation law of the Minkowski sums is a direct consequence ofTheorem 4.4.2:

27

4.4.5 Corollary. Let X be a topological vector space, then

A + B = C + B =⇒ A = C,

for any non-empty subsets A, B, C ⊂ X such that B is bounded and A and C closedand convex. «

Proof. Suppose A + C = B + C. Then A + C ⊆ B + C and A + C ⊇ B + C.So by using Theorem 4.4.2, A ⊆ B and A ⊇ B. Therefore A = B, whichproves the cancellation law.

The remainder of this section is devoted to showing the link between theordering on H(S) and the Hausdorff metric and the possibility of construct-ing a Riesz space from the cone H(S) by utilizing the results established inchapter 3.

4.4.6 Lemma. Let S be a metric space. Let (An) be a nested sequence in H(S) whereAn ⊇ An+1 for all n ∈ N and assume there is a number N ∈ N such that AN iscompact. Then

A =⋂

n∈NAn ∈ K(S).

«

Proof. Note that arbitrary intersections of closed sets are closed and a closedsubset of a compact set is compact. It remains to show that A is non-empty.

Pick for each n ∈ N an element an ∈ An. This provides a sequence (an)with a convergent subsequence

(ank

)k and with limit a, since every term after

aN is contained in the compact set AN .For each ` > N,

(ank

)k>`

is contained in the compact set A`. So a ∈ A`

and consequently, a ∈ A. This means A is non-empty.

4.4.7 Theorem (Order continuity of the Hausdorff distance). Let S be a metric space and{An} be a sequence in K(S) such that An ⊇ An+1 for all n ∈ N. Then

A =⋂

n∈NAn ∈ K(S)

and dH(A, An) ↓ 0. «

Proof. By Lemma 4.4.6, A ∈ K(S). For the other part, note that the sequence{dH(A, An) : n ∈ N} is non-increasing since An contains all Am for m < n.From Proposition 4.2.8, it follows that for any n ∈ N there is an an ∈ ANsuch that

d(an, A) = dH(An, A).

There is a subsequence (ank )k of (an) which converges to a point a ∈ A. Then

dH(

Ank , A)= d

(ank , A

)6 d

(ank , a

).

Let k→ ∞, then d(ank , a

)↓ 0.

In order to turn H(S) into a normed space, the Hausdorff distance needsto be homogeneous and translation invariant, according to the constructionof Theorem 3.3.2.

In order to obtain homogeneity of the Hausdorff distance, it is convenientto work with a normed space instead of a metric space. The main reason forthis assumption is the lack of linear structure of metric spaces.

28

For the remainder of this thesis, X =(X, ‖·‖

)will denote a normed space

over the field of real numbers. In particular, it follows that

dH(A, B) = supa∈A

infb∈B‖a− b‖ for A, B ∈ H(X).

This extra assumption yields the homogeneity of the Hausdorff distanceon H(X):

4.4.8 Proposition. Let X be a normed space, then dH is a homogeneous map on H(X). «

Proof. Let A, B ∈ H(X) and λ ∈ R, then, by using the homogeneity of thenorm,

dH(λA, λA) = supa∈A

infb∈B‖λa− λb‖ = |λ| sup

a∈Ainfb∈B‖a− b‖ = λdH(A, B).

Unfortunately, the Hausdorff distance is not necessarily translation invarianton every subset of H(X) or K(X):

4.4.9 Example. Consider X = R and take A = C = [−1, 1] and B = {−1, 1}. ThenA + C = [−2, 2] and B + C = [−2, 2]. Note that δ(B, A) = 0, since B ⊆ Aand δ(A, B) = 1. Then

dH(A, B) = 1 and dH(A + C, B + C) = 0,

by definition of the Hausdorff distance. This shows that dH is not translationinvariant. «

This last example proves that neither H(X) nor K(X) can be turned into anormed Riesz space (by using the construction of chapter 3). It turns outthat both H(X) and K(X) are too ‘large’. The construction is possible onthe space of convex, closed and bounded subsets of a normed space X thatcontain 0. This collection will be studied in more detail in the next chapters.

29

5C O N V E X I T Y

This chapter will provide additional conditions to insure translation invari-ance of the Hausdorff distance. The key is to consider a suitable subspace ofthe hyperspace of closed bounded non-empty subsets of a normed space X.This will be the space of convex, bounded and non-empty subsets of X. Itturns out that the ensuing Riesz space has a strong order unit.

5.1 the riesz space of convex sets

5.1.1 Definition. Let X be a normed space and A ⊆ X. The set of all convexcombinations of points in A, denoted by co(A) is said to be the convex hullof A. The set co(A) is called the closed convex hull of A and is defined as theclosure of co(A). «

5.1.2 Definition. Let X be a Banach space, then Conv(X) denotes the collection ofall non-empty, bounded, closed and convex subsets of X. «

For the remainder of this chapter, X will denote a normed space (unlessstated otherwise).

Note that Conv(X) is a metric cone when endowed with the Hausdorffmetric and the Minkowski sums. From this point on, the metric spaceConv(X) will be partially ordered by inclusion. The lattice operations ∨ and∧ are given by

A ∨ B = co(A ∪ B) and A ∧ B = A ∩ B for A, B ∈ Conv(X).

Note that A ∧ B might not always exist, as A ∩ B may be empty. To avoidthis possible complication, consider Conv0(X) instead, where

Conv0(X) = {A ∈ Conv(X) : 0 ∈ A}.

The collection Conv(X) inherits some structure from the underlying spacenormed space X.

5.1.3 Lemma. Let X be a normed space and A, B ⊆ X bounded, convex and non-emptysubsets. Then

co(A ∪ B) =⋃

λ∈[0,1]

[λA + (1− λ)B].«

Proof. Let A, B ⊆ X be bounded and convex. Define

L = co(A ∪ B) and R =⋃

λ∈[0,1]

[λA + (1− λ)B].

Note that R = {λa + (1 − λ)b : λ ∈ [0, 1], a ∈ A and b ∈ B}. With thisterminology, it remains to show that L = R.[⊆] Let x ∈ L. Note that if x ∈ A ∪ B, then, by definition of the convex

hull, x ∈ R.Suppose x /∈ A ∪ B, then there are elements x1, . . . , xn ∈ A ∪ B and scalars

λ1, . . . , λn in the unit interval that sum up to 1 such that

x =n

∑k=1

λkxk.

31

Define index-sets Λ1 and Λ2 by Λ1 = {k : xk ∈ A} and Λ2 = {k : xk ∈ B}and define

λ = ∑k∈Λ1

λk.

Then1− λ = ∑

k∈Λ2

λk.

Note that both λ and 1− λ are both non-zero, because x /∈ A∪ B. In addition,λ 6 1, since all λk are positive for 1 6 k 6 n and sum up to 1. Therefore, thesum over all elements of Λ1 cannot exceed 1.

By using that both A and B are convex, there is an element a ∈ A and anelement b ∈ B such that x = λa + (1− λ)b where

a =1λ ∑

k∈Λ1

λkxk and b =1

1− λ ∑k∈Λ2

λkxk.

This shows x ∈ R.[⊇] Let x ∈ R, then there is an element a ∈ A and an element b ∈ B such

that x = λa + (1− λ)b for λ ∈ [0, 1]. Then, by definition of the convex hull,x ∈ L.

5.1.4 Theorem. If X is a Banach space, then Conv(X) is complete. «

Proof. Let (An) be a Cauchy sequence in Conv(X). Since H(X) is complete(by Theorem 4.3.1) and Conv(X) is contained in H(X), there is a A ∈ H(X)such that

AndH−→ A as n→ ∞.

Because A is closed, bounded and non-empty by definition, it remains toshow that A is convex.

To this end, let a, b ∈ A, λ ∈ [0, 1] and put x = λa + (1− λ)b. Then, forany ε > 0 there is a N ∈ N such that for any n > N,

An ⊆ A + εB and A ⊆ An + εB.

Note that An + εB is convex for all n > N, since it is a sum of convex sets.Therefore,

x ∈ An + εB + 2εB,

which yields x ∈ A.

Observe that all results of Section 4.4 are valid for Conv(X), since Theo-rem 5.1.4 implies that it is a closed subspace of H(X).

Note that the Hausdorff distance is still a homogeneous map on the ele-ments of Conv(X). Moreover, the Hausdorff distance is translation invariant.The proof of this assertion requires some tools from the theory of convexanalysis.

5.2 support functions

5.2.1 Definition. Let X be a normed space and let A ⊆ X be a fixed non-emptysubset. The map

hA : X∗ −−→ [0, ∞]φ 7−−→ sup

a∈Aφ(a)

is called the support function of A. «

32

For any A ⊆ X, the support function preserves lattice structure. The proofof the next proposition is available in [1, p. 292].

5.2.2 Proposition. Let X be a normed space and A, B ∈ Conv(X) and let hA and hBdenote the support function of A and B respectively. Then

1. hA∨B = hA ∨ hB and hA∧B = hA ∧ hB;

2. hA+B = hA + hB;

3. hλA = λhA for all λ > 0;

4. If A ⊆ B, then hA 6 hB. «

5.2.3 Proposition. There is a 1-1 correspondence between support functions and theelements of Conv(X). «

Proof. Let A ∈ Conv(X) and let hA be the support function of A. It sufficesto show that

A = {x ∈ X : φ(x) 6 hA(φ) for all φ ∈ X∗}.

[⊆] If x ∈ A, then

φ(x) 6 supa∈A

φ(a) = hA(φ) for all φ ∈ X∗.

[⊇] Suppose there is an element

x ∈ {a ∈ X : φ(a) 6 hA(φ) for all φ ∈ X∗} \ A.

Note that {x} is compact and A is closed and convex. In addition, {x} andA are disjoint. Therefore, by the Hahn-Banach separation theorem, there is afunctional φ ∈ X∗ and s ∈ R such that

φ(x) > s and φ(a) 6 s for all a ∈ A.

Thensupa∈A

φ(a) 6 s,

which yields hA(φ) 6 s. Hence

φ(x) 6 hA(φ) 6 s,

which leads to a contradiction.

By using support functions (given a fixed non-empty closed, boundedand convex set), it is possible to obtain another expression for the Hausdorffdistance:

5.2.4 Theorem. Let X be normed and A, B ∈ Conv(X), then

dH(A, B) = sup‖φ‖=1

∣∣hA(φ)− hB(φ)∣∣.

«

33

Proof. The case A = B is clear.Suppose A 6= B and let ε > 0 such that A ⊆ B+B(0; ε) and B ⊆ A+B(0, ε).

Then, for all φ ∈ X∗ with ‖φ‖ = 1,

hA(φ) 6 hB(φ) + ε and hB(φ) 6 hA(φ) + ε.

So |hA(φ)− hB(φ)| < ε. This means

sup‖φ‖=1

|hA(φ)− hB(φ)| 6 dH(A, B).

To show the other inequality, note that if

ε = sup‖φ‖=1

|hA(φ)− hB(φ)| > 0,

thenφ(a) 6 sup

b∈Bφ(b) + ε for every φ ∈ X∗ with ‖φ‖ = 1.

So for all η > 0 there is a b ∈ B with φ(a− b) 6 ε + η. Therefore

‖a− b‖ 6 ε + η.

This yields A ⊆ B + B(0; ε). An analogous reasoning yields B ⊆ A + B(0; ε).So dH(A, B) 6 ε.

If the space X is uniformly convex (see appendix), then the proof of theprevious theorem can be simplified:

5.2.5 Lemma. Let X be a uniformly convex normed space and A, B ∈ Conv(X). Let hAand hB denote the support function of A and B respectively. Then

dH(A, B) = sup‖φ‖=1

|hA(φ)− hB(φ)|.«

Proof. Note that the inequality 6 has been established in Theorem 5.2.4. Itremains to prove the other inequality.

Let ε > 0. There exists an element a0 ∈ A be at distance dH(A, B)− ε fromB. By using the Corollary A.10, there is a b0 ∈ B with minimal distance to a0.Now consider

φ =a0 − b0

‖a0 − b0‖.

Then ‖φ‖ = 1 and by a similar reasoning as in the proof of Theorem 5.2.4, itfollows that

hA(φ)− hB(φ) > dH(A, B)− ε.

The translation invariance of the Hausdorff distance with respect to theMinkowski sums follows directly from Proposition 5.2.2 and Theorem 5.2.4.Therefore, it is possible to construct a Riesz space from the metric coneConv0(X), by using the construction of Chapter 3.

5.3 an riesz space with a strong order unit

Let C = C(X) denote the Riesz space obtained from Conv0(X) by the con-struction in Chapter 3. By Theorem 5.2.4, the Hausdorff distance is translationinvariant and the Hausdorff distance is homogeneous in both arguments by

34

definition. Therefore, by Theorem 3.3.2, C is a normed space such that itsnorm induces the Hausdorff distance.

The Riesz space C has a strong order unit. The idea behind the proofis that any non-empty bounded set A is contained in a (closed) ball. Soby ‘stretching out the unit ball far enough’, the set A will eventually becontained in this ‘stretched ball’. The convexity-stucture ensures that the‘stretching-process’ is well-defined.

5.3.1 Theorem. The Riesz space C is Archimedean and he closed unit ball B is a strongorder unit in C. «

Proof. Let [A, B] ∈ C. If there there is a n ∈ N such that

−n[B, 0] 6 [A, B] 6 n[B, 0],

then the proof is complete.Suppose [A, B], [C, D] ∈ C such that n · [A, B] 6 [C, D] for all n ∈ N. Then

[A, B] 6 [0, 0], since C is Archimedean. Indeed, for any a ∈ A and d ∈ Dthere is a bn ∈ B, a cn ∈ C and an xn ∈ B such that

na + d = nbn + cn + xn.

This show that a ∈ B, because B is closed and both C and D are bounded.Hence the implication

nA + D ⊆ nB + C for all n ∈ N =⇒ A ⊆ B

holds true.Now take N ∈ N such that both A and B are contained in NB. This is

possible since both A and B have a finite diameter. Then for n > N and thefact that both A and B contain 0,

A ⊆ B + nB and B ⊆ A + nB = A + (−n)B.

Therefore, B is indeed a strong order unit in C.

35

6S PA C E S O F C O N V E X A N D C O M PA C T S E T S

The goal of this chapter is to show that for a given normed space X, that theRiesz space generated by the collection of all compact and convex subsets ofX containing 0, need not have a weak order unit.

6.1 the hilbert cube

6.1.1 Definition. Let X be a normed space, and define the collection

K(X) = {A ⊆ X : A is convex, compact and 0 ∈ A} ⊆ Conv(X). «

Observe that K(X) is a cone under the Minkowski addition. If the under-lying normed space is complete, then so is K(X):

6.1.2 Lemma. If X is a Banach space, then K(X) is a complete metric space. «

Proof. By combining the statements of both Corollary 4.3.2 and Theo-rem 5.1.4, the result follows.

The previous lemma implies that K(X) is a closed subspace of Conv0(X).Let c = c(X) denote the Riesz space obtained from K(X) (by using theconstruction in chapter 3). The lattice operations are given by

A ∨ B = co(A ∪ B) and A ∧ B = A ∩ B.

Note that c is closed under these lattice operations.Before studying order units on c, a remark is order. The idea behind the

proof of Theorem 5.3.1 will not work in c, because the unit ball is not compactin a general normed space. The ‘next best thing’ is to consider a set whichkeeps getting ‘thinner’ in each coordinate. The natural candidate to describethis behaviour in `p is the Hilbert cube.

For the remainder of this chapter, p is assumed to be a constant in theinterval [1, ∞).

6.1.3 Definition. Let α = (αn) ∈ `p be a fixed sequence. The subset given by

Hα = {(xn) ∈ `p : |xn| 6 |αn| for all n ∈ N}

is the Hilbert cube in `p. «

The Hilbert cube Hα is by definition non-empty, since it contains the zero-sequence. In addition, H is both convex and compact. The latter assertiondoes not follow directly from the definition. Its proof requires some resultsfrom topology:

6.1.4 Theorem. Let {[an, bn] : n ∈ N} be a collection of non-empty compact intervals inR, then the (countable) Cartesian product

∏n∈N

[an, bn]

is compact in the product topology. «

37

Proof. This is a direct consequence of Tikhonov’s theorem, but it is possibleto prove this theorem without using the Axiom of Choice (see [9, p. 28]).

6.1.5 Lemma. Let X, Y be topological spaces such that X is compact. Then for anycontinuous map f : X → Y, the set f [X] is compact in Y. «

6.1.6 Lemma. The Hilbert cube is a compact and convex subset of `p. «

Proof. Consider the (bijective) map

f : [−1, 1]N −−→ `p(xn) 7−−→ (αn · xn)

Note that Hα is the image of [−1, 1]N under f . By Theorem 6.1.4, [−1, 1]N iscompact (in the product topology). Therefore, by Lemma 6.1.5, it suffices toshow that f is continuous. Recall that

d(x, y) = ∑n∈N

2−n|xn − yn|

is a metric for the topology on [−1, 1]N.Let ε > 0. Since α ∈ `p, there is a N ∈ N such that

∑n>N+1

|αn|p <ε

2p+1 .

Let x, y ∈ [−1, 1]N and take

0 < δ = ε · 2−N

2‖α‖pp

.

If x, y ∈ [−1, 1]N with d(x, y) < δ, then

‖x− y‖pp < δ and |xn − yn| <

ε

2‖α‖pp

for all n 6 N.

Observe that |xn − yn| 6 2 for n > N, which yields

‖ f (x)− f (y)‖pp = ∑

n∈N|αn|p|xn − yn|p

= ∑n6N|αn|p|xn − yn|p + ∑

n>N+1|αn|p|xn − yn|p

<ε

2‖α‖pp

∑n6N|αn|p + 2p ∑

n>N+1|αn|p

<ε

2‖α‖pp· ‖α‖p

p + 2p · ε

2p+1

< ε.

So f is indeed continuous and therefore Hα is compact.To show that Hα is convex, let (xn), (yn) ∈ Hα and λ ∈ [0, 1]. Then,

|λxn + (1− λ)yn| 6 λ|xn|+ (1− λ)|yn|6 λαn + (1− λ)αn

= αn

for all n ∈ N.

38

6.2 order units in c(`p)

In order to find order units in a Riesz space, it suffices to only consider thepositive cone, by Definition 2.4.1. Hence for Riesz spaces as constructed fromhyperspaces, it suffices to consider the hyperspaces. By different choices, itis possible to get all variations of space with or without order units.

The aim of this chapter is to prove that the Hilbert cube is not a weakorder unit in c(`p).

By Definition 2.4.1, the Hilbert cube Hα ∈ c(`p) is a weak order unitwhenever any element A ∈ c(`p) can be expressed as

A = co[ ⋃

n∈N(nHα ∩ A)

]. (∗)

Intuitively, it is not possible to express every A ∈ c(`p) in this manner. Theproblem is that given a fixed `p-sequence (αn), there is an `p-sequence (βn)which converges at a slower rate. In other words, for any (αn) ∈ `p thereexists a (βn) such that βn/αn → ∞ as n → ∞. This observation can beused to show that for a given `p-sequence (αn) and a given A ∈ c(`p), theintersection nHα ∩ A can be empty for all n ∈ N.

Additionally, if A ⊆ X is a convex set which contains 0, then

A ⊆ λA for all λ > 1.

Indeed, let a ∈ A, then write

1λ · (λ · a) + (1− 1

λ ) · 0.

This shows that a ∈ λA.Combining all these observations yields sufficient material to disprove (∗).

Disproving (∗)

6.2.1 Theorem. There exists an element A ∈ c(`p) such that (∗) does not hold. «

Proof. Take a sequence (βn) in `p such that βn/αn → ∞ as n → ∞ andconsider

A = {λβ : λ ∈ [0, 1]}.

Then A ∈ K(`p) and

nHα ∩ A = {0} for all n ∈ N.

This shows that (∗) does not hold.

The previous theorem shows that the Hilbert cube is not a weak order unitin c(`p).

There is another cone in which the Hilbert cube is a weak order unit. Theproof requires the Baire Category theorem. There are several versions of thistheorem in circulation. The statement of the version that will be used in thistheses is presented below. The proof can be found in [17, p. 296].

The Baire Category theorem

6.2.2 Definition. A nowhere dense set in a topological space X is a set whose closurehas empty interior. «

39

In the setting of metric spaces there is another characterisation of nowheredense sets:

6.2.3 Proposition. Let S be a metric space and A ⊆ S a closed subset. Then A is nowheredense if and only if there is no open ball which is contained in A. «

Proof. Assume there is no open ball contained in A. This assumption is norestriction, for if there is an open ball contained in A, then A is not nowheredense.

Let U ⊆ S be a non-empty open subset. Then A is not contained in U,since U contains an open ball and A does not. Let x ∈ U such that x /∈ A.Because A is closed, there exists r > 0 such that B(x; r) does not intersect A,that is,

B(x; r) ∩ A = ∅.

Since U is open, there exists s > 0 such that B(x; s) ⊆ U.Next, take R = min(r, s). Then B(x; R) is contained in U and B(x; R) does

not intersect A. Therefore, A is nowhere dense.

The next proposition presents a common example of a nowhere dense setin a normed space of infinite dimension.

6.2.4 Proposition. Let X be an infinite dimensional normed space. Then any non-emptycompact subset of X is nowhere dense. «

Proof. Suppose K is not nowhere dense. Then there is an element x ∈ Kand a r > 0 such that B(x; r) ⊆ K. Define s = r · (eπ − π)/42, then s < r.

If X is infinite dimensional, then the closed ball centered at x with radiusr is not compact. For if it were, the closed ball B(x, s) would be compact aswell since it is a closed subspace of K. This contradicts the fact that closedballs in infinite dimensional normed spaces are not compact.

6.2.5 Theorem (Baire Category theorem). Any non-empty complete metric space is notequal to a countable union of nowhere-dense closed sets. «

6.3 a riesz space with a weak order unit

It turns out that the Hilbert cube is a weak order in a certain subspace ofK(`p).

6.3.1 Definition. Let K ∈ K(`p), x ∈ `p and y ∈ K. If |xn| 6 |yn| for all n ∈ Nimplies x ∈ K, then K is a set of type (A). The collection of all K ∈ K(`p) oftype (A) will be abbreviated by A(`p). That is,

A(`p) = {K ∈ K(`p) : K is of type (A)}. «

6.3.2 Example. For a fixed `p-sequence α, the Hilbert cube Hα is a set of type (A),according to the results of the previous section. «

6.3.3 Theorem. Let α ∈ `p be a fixed sequence. The Hilbert cube Hα is a weak order unitin A(`p), that is,

K = co⋃

n∈N[(nHα) ∩ K] for all K ∈ A(`p).

«

40

Proof. [⊆] Let K ∈ A(`p) and a ∈ K. Define for all N ∈ N the sequence

aN = (a1, . . . , aN , 0, . . .).

This sequence is contained in (nHα) ∩ K for n sufficiently large. Therefore,

aN ∈ co⋃

n∈N[(nHα) ∩ K] for all N ∈ N.

And since aN → a with respect to the p-norm, it follows that

a ∈ co⋃

n∈N[(nHα) ∩ K].

[⊇] Let n ∈ N, then (nHα) ∩ K ⊆ K. This implies⋃n∈N

[(nHα) ∩ K] ⊆ K.

Since K is convex and closed, it follows that

co⋃

n∈N[(nHα) ∩ K] ⊆ K.

Therefore, Hα is a weak order unit in A(`p).

Note that the Hilbert cube need not be a strong order unit in A(`p). Thisfollows from the fact that for any fixed `p-sequence α, there is a sequencewhich converges at a faster rate.

6.3.4 Proposition. For p = 2 and α = {1/n : n ∈ N}, the Hilbert cube Hα is not astrong order unit in A(`p). «

Proof. Consider the set

K = {x ∈ `2 : |xn| 6 log(n)/n}.

Note that K contains the zero-sequence. Because the sequence (βn) whereβn = log(n)/n for all n ∈ N is square summable, it follows that K is compactand convex. Therefore K ∈ A(`2).

Now note that the sequence (βn) is contained in K, but not contained inkHα for all k ∈ N. Therefore, Hα is not a strong order unit.

6.3.5 Proposition. A(`p) does not have a strong order unit. «

Proof. Suppose K ∈ A(`p) is a strong order unit. If z ∈ `p, then

{x ∈ `p : |xn| 6 |zn| for all n} ∈ A(`p).

Therefore, there exists a λ ∈ R+ such that

{x ∈ `p : |xn| 6 |zn| for all n} ⊆ λK.

This yields z ∈ λK and therefore 1λz ∈ K. This shows that for any z ∈ `p

exists a number n ∈ N such that z ∈ nK. Hence⋃n∈N

nK = `p.

Observe that each nK is compact and therefore closed and `p is a completemetric space. By applying the Baire category theorem, not every nK isnowhere dense. This means there is a natural number m such that mK is notnowhere dense. In other words, mK does not have nonempty interior. Thiscontradicts Proposition 6.2.4.

This concludes the proof.

41

6.4 a riesz space without a weak order unit

Consider the collection

A(`p) = {K ∈ K(`p) : there is an A ∈ A(`p) such that K ⊆ A}.

This collection is a cone and the Riesz space generated by it does not have aweak order unit.

6.4.1 Theorem. A(`p) does not have a weak order unit. «

Proof. Let H ∈ A(`p). Then there is an A ∈ A(`p) such that H ⊆ A. Thenthere is a sequence z ∈ `p such that

A = {x ∈ `p : |x| 6 |z|}.

Therefore |x| 6 |z| for all x ∈ H.Now take y ∈ `p such that yn/xn → ∞ as n→ ∞ and define

K = {λy : λ ∈ [0, 1]}.

Then K ⊆ {x : |x| 6 |y|}, so K ∈ A(`p). Then

nH ∪ K = {0} for each n ∈ N.

Thereforeco

⋃n∈N

nH ∩ K = {0} 6= K,

which shows that H is not a weak order unit.

42

AU N I F O R M C O N V E X I T Y

This is a self-contained chapter which explains the structure on uniformlyconvex normed spaces. It can be used to simplify certain proofs presented inthe previous chapters.

Let X be a normed space over a field F. The closed unit ball in X, definedas

B(X, ‖·‖) = {x ∈ X : ‖x‖ 6 1},

will be denoted by B if no confusion can arise.

A.1 Definition. Let X be a normed space over F. Then X is called uniformly convexif for all ε > 0 there is a δ > 0 such that∥∥∥ x + y

2

∥∥∥ 6 1− δ

for all x, y ∈ B such that ‖x− y‖ > ε. «

Loosely speaking, the center of a line segment inside the unit ball must lie‘deep inside’ the unit ball unless the segment is ‘short’.

Note that uniform convexity is a property of the norm, not a property ofthe underlying vector space. It can happen that a vector space is uniformlyconvex when endowed with some norm ‖·‖, but not uniformly convex whenendowed with another norm, even when the two norms are equivalent!

A.2 Example. Let n > 1 then (Rn, ‖·‖1) is not uniformly convex, but (Rn, ‖·‖2)is.

PutB1 = B(Rn, ‖·‖1) and B2 = B(Rn, ‖·‖2).

For the first part, consider x = (1, 0, . . . , 0) and y = (0, 1, 0, . . . , 0) and takeε = 2. Then x, y ∈ B1 and ‖x− y‖1 = 2. On the other hand,∥∥∥ x + y

2

∥∥∥1= 1

2 · 2 = 1 > 1− δ

for any δ > 0. This shows that (Rn, ‖·‖1) is not uniformly convex.To prove that (Rn, ‖·‖2) is uniformly convex, let x, y ∈ Rn and write

x = (x1, . . . , xn) and y = (y1, . . . , yn). Then

∥∥∥ x + y2

∥∥∥2

2−∥∥∥ x− y

2

∥∥∥2

2=

14

[ n

∑k=1

(xk + yk)2 +

n

∑k=1

(xk − yk)2]

=14

n

∑k=1

(2x2k + 2y2

k)

=12

(‖x‖2

2 + ‖y‖22

). (∗)

Let ε > 0 and take δ > 0 such that

(1− δ)2 = 1− ε2

4.

43

Let x, y ∈ B2 such that ‖x− y‖ > ε, then, by (∗),∥∥∥ x + y2

∥∥∥2

2=‖x‖2 + ‖y‖2

2−∥∥∥ x− y

2

∥∥∥2

2

6 1−∥∥∥ x− y

2

∥∥∥2

2

6 1− ε2

4= (1− δ)2.

This shows that (Rn, ‖·‖2) is indeed uniformly convex. «

The previous example can be generalized:

A.3 Theorem. Let (X, Σ, µ) be a positive measure space, then Lp(X, Σ, µ) is uniformlyconvex for 1 < p < ∞. «

The proof of this theorem is quite involved and will not be presented inthis thesis. The interested reader can find a proof in the literature (see [4] or[8]).

Recall the following result from linear algebra:

A.4 Lemma (Parallelogram law). Let X be an inner product space over F. Then, forx, y ∈ X, the norm on X satisfies

‖x + y‖2 + ‖x− y2‖ = 2 · ‖x‖2 + 2 · ‖y‖2. «

A.5 Theorem. Let X be an inner product space over F. Then (X, ‖·‖) is a uniformlyconvex normed space. «

Proof. Let ε > 0 and take δ > 0 such that

1− δ =√

1− ε2

4.

Let x, y ∈ B such that ‖x− y‖ > ε, then∥∥∥ x + y2

∥∥∥2= 2 ·

∥∥∥ x2

∥∥∥2+ 2 ·

∥∥∥y2

∥∥∥2−∥∥∥ x− y

2

∥∥∥2

6 1− ε2

4= (1− δ)2.

Therefore, X is uniformly convex.

It turns out that any uniformly convex Banach space is reflexive. The proofof this statement requires a result from the theory on the weak topology. Theproof is omitted, but can be found in e.g. Dunford and Schwartz (see [6,p. 424]).

A.6 Theorem. Let X be a normed space, then B is σ(X∗∗, X∗) dense in the closed unitof ball X∗∗. «

A.7 Theorem (Milman-Pettis). Any uniformly convex Banach space X over F is reflex-ive. «

44

Proof. Suppose X is a uniformly convex Banach space which is not reflexive.Let B∗∗ denote the closed unit ball in X∗∗ and consider the isometric dualitymaps

J : X → X∗∗ and J∗: X∗ → X∗∗∗.

By Theorem A.7,

B∗∗ = J[B]wk∗

,

where J[B]wk∗

denotes the closure of J[B] in the σ(X∗∗, J∗[X∗]) topology.Since X is not reflexive, J[X] 6= X∗∗ and so there must be a ψ ∈ X∗∗ that isnot in J[B] with ‖ψ‖ = 1.

Since J is an isometry, J[B] is (norm) closed in X∗∗. So

d(ψ, J[B]) = 2ε

for some given ε > 0. On the other hand, ψ ∈ J[B]wk∗

. So

ψ ∈ U ∩ J[B]wk∗

for any σ(X∗, J∗[X∗]) neighbourhood U of ψ.Next, let φ ∈ X∗ such that ‖φ‖ = 1 and |ψ(φ)− 1| < δ. Note that this

follows from the fact that ‖ψ‖ = 1. Put

U = {ψ ∈ X∗∗ : |ψ(φ)− 1| < δ},

then|ψ1(φ)−ψ2(φ)| < 2δ

for all ψ1,ψ2 ∈ U ∩ J[B]. Additionally,

‖ψ1 +ψ2‖ > |ψ1(φ) +ψ2(φ)|= |2 +ψ1(φ)− 1 +ψ2(φ)− 1|> 2− 2δ,

for all ψ1,ψ2 ∈ U ∩ J[B]. Therefore, ‖ψ1−ψ2‖ < ε) for all ψ1,ψ2 ∈ U ∩ J[B]since X is uniformly convex. Now fix ψ1, then

U ∩ J[B] ⊆ ψ1 + ε(U ∩ J[B]

wk∗).

Note that U ∩ J[B]wk∗

is closed so ψ ∈ U ∩ J[B]wk∗

. But this contradicts theassumption d(ψ, J[B]) = 2ε. Hence, X is reflexive.

The Milman-Pettis theorem was proven independently by Milman [16]and Pettis [18]. Shortly after publishing these papers, Kakutani came upwith a simplified proof (see [12]). About twenty years later, Ringrose [20]published a shorter proof.

A.8 Lemma. Let X be a uniformly convex normed space over F and let (xn) be a sequencein X such that

(1) ‖xn‖ → 1 as n→ ∞.

(2) For ε > 0 there is a N ∈ N such that∣∣‖xn + xm‖ − 2∣∣ 6 ε

for all n, m > N.

45

Then (xn) is a Cauchy sequence.In particular, if X is Banach space, then (xn) is convergent. «

Proof. Assume that xn ∈ B for all n. Let ε > 0, then there is a N ∈ N suchthat ‖xn − xm‖ < ε for all n, m > N. This is possible by using (2). So if thesequence (xn) is contained in B, then the sequence is Cauchy.

If not all of (xn) are in the unit ball, assume (without loss of generality and(1)) that ‖xn‖ 6= 0 for all n ∈ N. Next, define the sequence (yn) by setting

yn =xn

‖xn‖for all n ∈ N.

Then (yn) satisfies (1) and since the terms of (yn) are in B for all n, thereasoning from the first part yields that (yn) is a Cauchy sequence. And then(xn) is Cauchy due to (1).

A.9 Theorem. Let X be a uniformly convex Banach space and C ⊆ X a non-emptyclosed and convex set. Then there is a unique x ∈ C such that

‖x‖ = infz∈C‖z‖. «

Proof. Note that if 0 ∈ C, take x = 0 and the result follows. So assume that0 /∈ C.

Put d = inf{‖z‖ : z ∈ C}, then d > 0 by definition of the norm and thefact that 0 /∈ C. Let (xn) be a sequence in C such that ‖xn‖ → d as n → ∞.Define a sequence (yn) by putting

yn =xn

dfor all n ∈ N.

Then, by the triangle inequality,

‖yn + ym‖ =∥∥∥ xn + xm

d

∥∥∥ 6 ‖xn‖+ ‖xm‖d

for all n, m ∈ N. Next, since C is convex, d 6 12‖xn + xm‖ for all n, m ∈ N.

Therefore,

‖yn + ym‖ =∥∥∥ xn + xm

d

∥∥∥=

2d

∥∥∥ xn + xm

2

∥∥∥> 2 · 2

‖xn + xm‖· ‖xn + xm‖

2

= 2

for all n, m ∈ N. These estimates prove that the sequence (yn) satisfy theconditions of Lemma A.8, which means (yn) is a Cauchy sequence. Notethat is implies that (xn) is also a Cauchy sequence. Since X is Banach and Cis closed, there is a x ∈ C such that ‖xn − x‖ → 0 as n→ ∞. Since ‖xn‖ → d,it follows that ‖x‖ = d. This proves existence.

To prove uniqueness, let ε > 0 and assume there are x, y ∈ C such that‖x‖ = ‖y‖ = d and ‖x − y‖ = ε. Since X is uniformly convex, there is aδ > 0 with ∥∥∥ x + y

2

∥∥∥ 6 (1− δ) · d < d.

On the other hand, the fact that C is convex yields

x + y2

> infz∈C‖z‖ = d.

This is clearly a contradiction. Hence x = y, which establishes uniqueness.

46

A.10 Corollary (Nearest point projection). Let X be a uniformly convex Banach spaceover F and C ⊆ X be non-empty closed and convex. Let y ∈ X \ C, then there is aunique x ∈ C such that

‖y− x‖ = infz∈C‖y− z‖. «

Proof. Set D = C− y = {z− y : z ∈ C}. This set is non-empty, closed andconvex. By applying Theorem A.9, there is a unique x ∈ D such that

‖x‖ = infz∈D‖z‖

= infv∈C‖v− y‖.

Take u = x + y, then u ∈ C and

‖u− y‖ = ‖x‖ = infv∈C‖v− y‖.

This implies‖y− u‖ = inf

v∈C‖y− v‖.

A.11 Corollary. Let X be a uniformly convex normed space over F. If φ ∈ X∗ is anon-zero linear functional, then there is a unique x ∈ X with ‖x‖ = 1 such thatφ(x) = ‖x‖. «

The proof of this corollary requires some results from functional analysis.

A.12 Definition. Let X be a normed linear space over R and H ⊆ X. Then H is ahyperplane if and only if there is a non-zero φ ∈ X∗ and a λ ∈ R such that

H = {x ∈ X : φ(x) = λ}. «

Recall the following consequence of the Hahn-Banach theorem (see [23,p. 107-108]):

A.13 Theorem. Let X be a reflexive Banach space over F and let φ ∈ X∗. Then there isan element x ∈ X with ‖x‖ = 1 such that

φ(x) = ‖φ‖. «

At this point there is enough material available to prove Corollary A.11.

Proof (Corollary A.11). Consider for a non-zero linear functional φ ∈ Xthe collection H = {x ∈ X : φ(x) = ‖φ‖}. Then 0 /∈ H and according toDefinition A.12, H is a hyperplane in X. In addition, H is convex (by linearityof φ) and closed since φ is continuous. Therefore, by Theorem A.9, there is aunique x ∈ H such that

‖x‖ = infz∈H‖z‖ > 0.

Since φ(x) = ‖φ‖, it follows that ‖x‖ > 1.Next, since X is uniformly convex, it is reflexive due to Theorem A.7. By

applying Theorem A.13, there is a y ∈ X with ‖y‖ = 1 such that φ(y) = ‖φ‖.This means y ∈ H and ‖y‖ 6 ‖x‖. Additionally x = y, since x was theunique element in H with minimal norm. Hence ‖x‖ = 1.

47

B I B L I O G R A P H Y

[1] C. D. Aliprantis and K. C. Border, Infinite dimensional analysis,Springer, Berlin, third ed., 2006.

[2] C. D. Aliprantis and O. Burkinshaw, Locally solid Riesz spaces withapplications to economics, vol. 105 of Mathematical Surveys and Mono-graphs, American Mathematical Society, Providence, RI, second ed.,2003.

[3] A. Borel and J.-P. Serre, Le théorème de Riemann-Roch, Bulletin de laSociété Mathématique de France, 86 (1958), pp. 97–136.

[4] J. A. Clarkson, Uniformly convex spaces, Transactions of the AmericanMathematical Society, 40 (1936), pp. 396–414.

[5] E. de Jonge and A. van Rooij, Introduction to Riesz spaces, vol. 78 ofMathematical Centre Tracts, Amsterdam, 1977.

[6] N. Dunford and J. T. Schwartz, Linear operators Part I, John Wiley &Sons Inc., New York, 1988.

[7] A. Grothendieck, Classes de faisceaux et théorème de Riemann-Roch,vol. 255 of Lecture notes in Mathematics, Springer-Verlag, 1971.Mimeographed notes, Princeton 1957, Théorie des Intersections etThéorème de Riemann-Roch (SGA6).

[8] H. Hanche-Olsen, On the uniform convexity of Lp, Proceedings of theAmerican Mathematical Society, 134 (2006), pp. 2359–2362.

[9] H. Herrlich, Axiom of choice, vol. 1876 of Lecture Notes in Mathematics,Springer-Verlag, Berlin, 2006.

[10] S. C. Hille, Fundamentals of Nonlinear Analysis, Lecture Notes, 2009.

[11] V. Istratescu, Fixed point theory, Mathematics and its applications, D.Reidel Publishing Company, Dordrecht, Holland, 2002.

[12] S. Kakutani, Weak topologies and regularity of Banach spaces, Proceedingsof the Imperial Acadamy Tokyo, 15 (1939), pp. 196–173.

[13] R. Larsen, Functional analysis: an introduction, Marcel Dekker Inc., NewYork, 1973. Pure and Applied Mathematics, No. 15.

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[19] H. Rådström, An embedding theorem for spaces of convex sets, Proc. Amer.Math. Soc., 3 (1952), pp. 165–169.

[20] J. R. Ringrose, A note on uniform convex spaces, Journal of the LondonMathematical Society, 34 (1959), p. 92.

[21] B. P. Rynne and M. A. Youngson, Linear functional analysis, Springer Un-dergraduate Mathematics Series, Springer-Verlag London Ltd., London,second ed., 2008.

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50

I N D E X

absolute value, 7

Banach lattice, 5

boundedset, 20

topological vector space, 26

cancellation law, 14

cone, 13

convex, 13

lattice, 16

normed, 16

ordered, 14

positive, 13

standard order, 14

convex hull, 31

closed, 31

distancebetween sets, 20

Hausdorff, 20, 22

to a subset, 19

Hausdorff metric, 22

Heine-Borel property, 20

Hilbert cube, 37

hyperplane, 47

hyperspace, 20

lattice, 5

homomorphism, 6

operations, 5

metric cone, 15

monoid, 13

commutative, 13

operation, 13

unit, 13

negative part, 7

nowhere dense set, 39

order unitstrong, 9

weak, 9

ordered vector space, 5

positivecone, 6

elements, 6

map, 6

part, 7

strict, 6

principalband, 9

ideal, 9

Rieszhomomorphism, 6

isomorphism, 7

space, 5

Riesz spaceArchimedean , 8

normed, 5

scalar multiplication, 13

set of type A, 40

support function, 32

uniformly convex space, 43

vector lattice, 5

51