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
C O N T E N T S
1 i n t r o d u c t i o n 3
1.1 Assumptions and Notations 3 2 r i e s z s pa c e s 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 a b s t r a c t c o n e s 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 t h e h au s d o r f f d i s ta n c e 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 c o n v e x i t y 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 s 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 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 u n i f o r m c o n v e x i t y 43
b i b l i o g r a p h y 48 i n d e x 51
1
I N T R O D U C T I O N
An ordered vector space E is a vector space endowed with a partial order which is ‘compatible’ with the vector space operations (in some sense). If the order structure of E is a lattice, then E is called a Riesz space. This order structure 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 an important role. The most natural example is the ‘standard machinery’, a tool which is frequently used in measure and integration theory.
The aim of this thesis is to provide a construction of a normed Riesz space, given a cone. This requires a definition of a cone outside the context of ordered vector spaces and will be given by a modification of the Grothendieck group construction.
Another goal is to find an example of a Riesz space which does not have a weak order unit. There are several existence results of Riesz spaces that do not have such units, but there are not many explicit examples.
1.1 a s s u m p t i o n s a n d n o tat i o n s
Before reading any further, the reader should be familiar with basic measure theory and functional analysis. However, most of the material will be developed 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) 6r}, 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 the closure of A;
5. The dual of a vector space X will be denoted by X∗.
2
R I E S Z S PA C E S
This chapter will provide an elementary introduction to the theory of Riesz spaces. Later on, this theory will be used to describe certain vector spaces which consist of sets.
Note that this chapter is in no way a complete introduction in the field of positivity. This chapter only provides the bare necessities in order to describe the results of this thesis. The interested reader can find a detailed exposition on this subject in the literature, e.g. in [5] and [2].
2.1 d e f i n i t i o n s
2.1.1 Definition. A partially ordered set(X,4)is a lattice if each pair of elements x, 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 is denoted by
sup{x, y} =x∨y and inf{x, y} = x∧y.
Note that both the supremum and infimum are unique, provided that they exist.
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+z4y+z for all z∈X;
2. αx4αy 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 k·ksuch that
04x4y =⇒ kxk 6 kyk and k|x|k = kxk
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 space or an ordered vector space(E,4). Additionally, the standard ordering on R will be denoted by the symbol6.
2.1.2 Example. Let 16 p6∞.
1. Let n ∈ N and consider Rn under the usual vector operations. Let x, y∈Rn and write x= (x1, . . . , xn)and y= (y1, . . . , yn). Then Rnis a Riesz space under the partial ordering4defined by
x4y ⇐⇒ xk6yk for k∈ {1, . . . n}.
By a similar reasoning, the sequence spaces c0and`pare Riesz spaces.
2. Let X be a topological space, then C(X), the space of continuous functions on X, is a Riesz space under the (pointwise) ordering4given by
f 4g ⇐⇒ f(x) 6 g(x) for all x∈X.
Note that by slightly modifying the previous ordering, the Lebesgue spaces are also examples of Riesz spaces. Suppose (X,Σ, µ) is an arbitrary measure space, then Lp(X,Σ, µ)is a Riesz space under the ordering4defined as
f 4g ⇐⇒ f(x) 6g(x) for almost all x∈X. « The previous examples indicate that Riesz spaces appear in a natural fashion when studying function spaces. However, not every function space is a Riesz space:
2.1.3 Example. Consider E = C1[0, 1], the class of continuously differentiable functions on the interval [0, 1] ⊆ R. Observe that E is an ordered vector space 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 1 1
2
1
Note that both f∧g and f∨g are not differentiable at x= 12. Therefore, E is
not a Riesz space. «
2.2 p o s i t i v e c o n e s a n d r i e s z h o m o m o r p h i s m s 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 elements of 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 map
T 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, then T is a Riesz homomorphism (or lattice homomorphism) if T preserves the lattice operations, 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 and only 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 does T[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 A exists. Then
sup(λA) =λsup(A) for all λ∈R+ and
inf(λA) =λinf(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 the negative 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 fromProposition 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−. «
2.3 a r c h i m e d e a n r i e s z s pa c e s
2.3.1 Definition. Let E be a Riesz space and x∈E. If
n∈Ninf
1 n ·x =0
for any x∈E+, then E is called Archimedean. « There is another way to define the Archimedean property, one that is frequently 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{1nx : n∈N} =0 for all x in E+.
(b) If x, y∈E+and nx4y for all n∈N, then x=0. « The next proposition provides infinitely many examples of Archimedean Riesz 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 4y for any n ∈ N. This means n·x 4 y 4 y+ for all n ∈ N, which implies 04n·x+4y+ for any n∈N. Hence
nkx+k = kn·x+k 6 ky+k for all n∈N.
Since ky+kis a finite real number, it follows thatkx+k = 0. On the other hand, x+=0 by definition ofi the norm. Therefore x4x+ =0. This shows E is Archimedean.
The previous proposition implies that each Riesz space fromExample 2.1.2 is Archimedean.
At this point a remark is in order. The Archimedean property from Definition 2.3.1 differs from the Archimedean property in the set of real numbers, which states that for any x, y∈Rthere is a n∈Nsuch that
|y| 6n|x|.
Note that this property is implied by bothLemma 2.3.2andDefinition 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 of Definition 2.3.1.
Indeed, let 04 f ∈C(0, 1). Then
∀x ∈R: 1nf(x) ↓0 in R, therefore
1
nf ↓0 in C(0, 1).
This shows that C(0, 1) is Archimedean. To prove that it does not satisfy the Archimedean property of the set of real numbers, consider f , g∈C(0, 1) given by f(x) = 1x and g(x) =1. Then there is no n∈Nsuch that f 6ng. «
8
2.4 o r d e r u n i t s
2.4.1 Definition. Let E be a Riesz space and let x∈E. The principal band generated by x∈ E, denoted by Bx, is the collection given by
Bx=ny∈E : |y| =sup
n∈N
|y| ∧n|x|o.
The principal ideal generated by an element x∈E+, denoted by Ex, is the set
Ex= {y∈ E : there is a λ>0 such that|y| 4λx}.
If there is an element e+ ∈E such that Be =E, then e is called a weak order unit 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 weak order unit. The converse statement is not true.
A natural question to ask is whether a Riesz spaces has any order unit at all (strong or weak). There are very few explicit examples of Riesz spaces that do not have a weak order unit. One of the goals of this thesis is to add a new example to that list.
Let’s state some known results. The proof of the next lemma is somewhat involved 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,`pdo not have a strong order unit.
2. Let (X,Σ, µ)be a measure space, then Lp(X,Σ, µ)does not have a strong order 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 a
strong order unit in L∞(X,Σ, µ). «
Proof. Since any f ∈L∞is essentially bounded, it follows that|f| 6 kfk∞. Therefore,
−kfk∞·1X6 f 6 kfk∞·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 function
f ∈C(K)is a strong order unit. «
Proof. Let f ∈C[K]be strictly positive. Consider α=min
x∈K f(x).
Note that α is well-defined since continuous functions on compacts sets attain a minimum. Moreover, α>0 and f 6α·1K. This implies f is a strong order unit.
2.4.5 Theorem. Let (X,Σ, µ) be a measure space and p ∈ [1,∞). Suppose (fn) is a sequence of Lp-functions which converges in norm to a f ∈Lp. If fn6 f , then
f =sup
n∈N
fn.
«
Proof. Define Fn = f −fnand consider the sequence(Fn). Then, by using the assumptions, it follows that Fn(x) >0 for all x∈X and
n→lim∞ Z
Fnpdµ= lim
n→∞kFnkpp=0.
By applying Fatou’s lemma, 06
Z
lim inf
n→∞ Fnpdµ 6lim inf
n→∞
Z Fnpdµ
= lim
n→∞
Z Fnpdµ
=0.
This means lim inf Fnp=0 almost everywhere.
On the other hand, note that for any n∈Nand any x∈X 06 inf
n∈NFn(x)p6 lim
k→∞inf
n>kFk(x)p=lim inf
k→∞ Fk(x)p. Hence,
n∈Ninf Fn(x) =0 almost everywhere on X. Therefore,
f(x) =sup
n∈N
fn(x)
for almost all x∈X.
2.4.6 Theorem. Let(X,Σ, µ)be a measure space, then any strictly positive Lpfunction
is a weak order unit. «
Proof. Let f , g ∈ Lp such that f is strictly positive and g > 0. Then, by Theorem 2.4.5, it suffices to show
g=sup
n∈N
g∧ (n f).
Recall that the collection of step functions is dense in Lp. So for any ε>0 there is a step function s ∈Lp such thatkg−skp< ε. Since s is bounded, there exists a M∈Nsuch that M f >s almost everywhere. Furthermore,
g∧ (n f) −s∧ (n f)
p p=
Z
g∧ (n f) −s∧ (n f)
pdµ
for all n ∈ N. Let x ∈ X and note that if g(x) 6 n f(x), which means s(x) >n f(x). So in this case, g(x) −n f(x) 6n f(x) −n f(x) =0. Therefore, n f(x) −g(x) 6s(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=
Z
g∧ (n f) −s∧ (n f)
pdµ
= Z
g−s)
pdµ
= kg−skpp
<εp.
10
Hence, for all n>N,
kg−g∧ (n f)k 6 kg−skp+ ks−s∧ (n f)kp+ ks∧ (n f) −g∧ (n f)kp 6ε+0+ε.
This shows that g∧ (n f) →g (w.r.tk·kp) as n→∞. In addition,(g∧n f) 6g for any n∈N. Therefore, by applyingTheorem 2.4.5, the result follows.
3
A B S T R A C T C O N E S
The Grothendieck group construction is a tool used in abstract algebra that constructs 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 of K-theory.
This chapter provides a definition of a positive cone outside the context of 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 the Grothendieck group.
3.1 d e f i n i t i o n s
3.1.1 Definition. A non-empty setC 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), thenCis a monoid. The operation+is called the addition onC or simply addition. The element e from (M2) is called the unit element for addition onC and will be denoted by the symbol 0 for convenience. «
Observe that the unit element of a monoid is unique. Indeed, suppose that there are two elements e∈ C and u∈ Csuch 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 commutative
monoid under the usual addition. «
3.1.3 Definition. A commutative monoidC 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 onC. If no confusion can arise, the symbol·will be left out. «
3.1.4 Definition. LetCbe a cone. If there exists a partial ordering4onCsuch that a4b =⇒ λa4λb for all λ∈R+ and all a, b∈ C;
a4b =⇒ a+c4b+a for all a, b, c∈ C,
thenC is called an ordered cone. «
Any coneC with the property
a+b=e =⇒ a=b=e for all a, b∈ C
can be ordered by defining a6b if and only if there is a c∈ Csuch that a+c=b.
A routine verification shows that6defines a partial order that meets all the requirements ofDefinition 3.1.4. From now on, this order6will be called the standard order on a coneC.
3.2 c o n s t r u c t i n g a v e c t o r s pa c e f r o m a c o n e LetC be a cone and define the relation∼onC × Cby
(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 the Cartesian productC × 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 coneCcan be mapped into EC under the quotient map
q :C −−→ EC
a 7−−→ [a, 0]
This map preserves the cone structure ofC, but it need not necessarily be an embedding. 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 cancellation law, q is indeed injective hence an embedding.
From this point on, elements of the form[a, 0]are called positive elements of EC and elements of the form[0, b]are said to be the negative elements of EC. This convention ensures that the members ofC correspond to the positive elements of EC.
Observe that the comments above agree withProposition 2.2.6. That is, if c∈EC, then there are elements a, b∈ C such that c=a−b.
14
3.3 c o n s t r u c t i n g n o r m s
Given a coneC in a metric space S, it is possible to construct a norm on EC
under 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. LetCbe a metric cone. Assume that the metric d is translation invariant and 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 normk·kon EC, which induces d. « Proof. Let a, b, c∈ Csuch 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
k[a, b]k =d(a, b).
The mapk·kis well-defined. Indeed, suppose(a, b) ∼ (x, y), then there is a c ∈ C such that a+y+c= x+b+c. Then a+y = x+b, by using the cancellation law. Since d is translation invariant, it follows that
d(a, b) =d(a+x+y, b+x+y)
=d(x, y). Therefore,k·kis well-defined.
The mapping k·k is indeed a norm on EC. Suppose that λ ∈ R+ and a, b∈ C, then, by homogeneity,
λk[a, b]k =λd(a, b) =d(λa, λb) = [λa, λb] = λ[a, b] . If λ60, then
λ[a, b] = k|λ|[b, a]k = |λ|k[b, a]k = |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, k[a, b] + [x, y]k = k[a+x, b+y]k
=d(a+x, b+y)
6d(a+x, a+y) +d(a+y, b+y)
=d(x, y) +d(a, b). Therefore
k[a, b] + [x, y]k 6 k[a, b]k + k[x, y]k, which proves the triangle inequality.
Lastly, note thatk[a, b]k =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 translation invariant. Then there is an isometric embedding ofC into EC. «
Proof. By the proof of Theorem 3.3.2, C has the cancellation law, so the map ψ : C →EC where ψ(a) = [a, 0]is a well-defined embedding. By using the definition of the norm on EC (from the proof ofTheorem 3.3.2), ψ is an isometry.
3.3.4 Corollary. Let X be a vector space over R andC ⊆X a metric cone in X such that the metric onC is homogeneous and translation invariant. Then there is a linear embedding ψ : EC →X.
If in addition, X is a normed space and d is the metric onCinduced by the norm
on X, then ψ is an isometric embedding. «
Proof. For the first part, put ψ([a, b]) =a−b. The definition of the norm on EC ensures that EC ⊆X is a subspace. Explicitly,
EC = {a−b : a, b∈ C}. For the second part, observe that
ka−bk =d(a, b) = k[a, b]k 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 whenever there is a linear isometric embedding into some normed space(X,k·k). «
3.4 l at t i c e s t r u c t u r e s a n d c o m p l e t e n e s s
3.4.1 Definition. An ordered coneC is a lattice cone if both a∨b and a∧b exist for
any a, b∈ C. «
Given an ordered coneC, it is possible to put an ordering on the associated vector space EC. This is done by defining
[a, b] 6 [x, y] ⇐⇒ a+y6x+b.
It turns out that EC inherits the lattice cone structure ofC:
3.4.2 Proposition. IfC 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, it follows 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∈ Csuch that (x, y) = (c, 0).
Suppose[a, b] 6 [c, 0], then a+0 6 b+c. Since b 6 b+c, it follows that a∨b6b+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. LetC 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+06a+ |a−b|.
Let[x, y] ∈EC and suppose that[x, y] > [a, b]and[x, y] > −[a, b]. Then a+x6y+b and b+y6x+a.
This yields
a−b6x−y and b−a6x−y.
Completeness ofCdoes not imply that the associated vector space EC is complete:
3.4.4 Remark. LetCbe a complete normed cone, then EC need not be complete.
Proof. Consider a normed space E which is not complete. Define the trivial ordering
x6y ⇐⇒ x=y.
Then E is an ordered vector space and the only positive element is 0, which means E+= {0}. This is clearly a complete normed space.
Recall the following result from Banach space theory (see [14, p. 20] for a proof):
3.4.5 Lemma. Let X be a normed space. Then X is a Banach space if and only if every
absolutely convergent series converges in X. «
3.4.6 Theorem. LetC be a normed lattice cone such that its norm is a non-decreasing map. Then the associated normed space(EC,|||·|||)where
|||x||| = k|x|k
is a Banach lattice wheneverC 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 06x+n 6 |xn|for all n∈N. Therefore, because the norm onC is assumed to be non-decreasing,
kx+nk 6 k|xn|k = |||xn||| for all n∈N.
Consequently, the series over all positive parts is absolutely convergent. So, by completeness ofC, the series∑ x+n is convergent. By a similar reasoning, the sum over all negative parts is also convergent.
Then, for N∈Nsufficiently large,
∑
n6N
x−n −
∑
n∈N
x+n −
∑
n∈N
x−n
6
∑
n6N
x+n −
∑
n∈N
x+n +
∑
n6N
x−n −
∑
n∈N
x−n 6
∑
n6N+1
x+n
+
∑
n6N+1
xn−
6
∑
n6N+1
|||x+n||| +
∑
n6N+1
|||x−n|||.
By letting N →∞, the series ∑n6Nxnconverges in EC with respect to|||·|||. Therefore, EC is complete with respect to |||·|||, which means that it is a Banach lattice byLemma 3.4.5.
4
T 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 a Banach space X= (X,k·k). Since the power set of the entire space X can be quite ‘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 of the power set instead.
4.1 d i s ta n c e b e t w e e n p o i n t s a n d s e t s
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 to A is then given by
d(x, A) = inf
a∈Ad(x, a).
« Loosely speaking, the notion of distance between a point x ∈ S and a subset A⊆S is the distance from x to the ‘nearest’ point a∈ A. There might be some difficulties when A is empty. Note that the distance between a point and a non-empty subset is a non-negative element in R, by definition. To avoid 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, then d(x, Q) =0, because Q is dense in R.
2. Consider R2with 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) = inf
y∈Akx0−yk = inf
x∈R
q
(−1−x)2+ (2−x)2.
Put f(x) = (−1−x)2+ (2−x)2, then, by using the derivative of f , it follows that the point in A nearest to x0is(1/2, 1/2) ∈A. Therefore,
d(x0, A) = k(−1, 2) − (12,12)k = 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) 6d(x, a) 6d(x, y) +d(y, x) for all a∈ A.
This yields
d(x, A) −d(x, y) 6 inf
a∈Ad(y, A) = fA(y), which gives
d(x, A) −d(y, A) 6d(x, y).
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 not identically 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. Note that 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∈ A such that d(x, y) <ε. That is, y ∈B(x; ε). So every ball with centre x and radius ε contains a point of A. This implies B(x; ε) ∩A6= ∅, which means that x is a limit point of A. Therefore, x∈ A.
For the converse, assume x ∈ A. Then d(x, A) = 0. Indeed, let ε > 0 then B(x; ε)contains a point y ∈ A. So d(x, A) < ε. Since ε was arbitrary, d(x, A) =0.
4.2 d i s ta n c e b e t w e e n s e t s
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 concept can be extended in order to find a suitable definition of distances between subsets of S.
The Hausdorff semi-distance
Definition 4.1.1can be modified to ‘measure’ the distance between subsets in the following way:
4.2.1 Definition. Let S be a metric space and A, B⊆ S non-empty subsets. The distance from A to B or Hausdorff semi-distance is defined as
δ(A, B) =sup
a∈A
d(a, B) =sup
a∈A
inf
b∈B
d(a, b).
« Note that δ(A, B)may be infinite for unbounded sets A, B⊆S according to the conventions fromDefinition 4.1.1. Since this might lead to some technical difficulties later on, it is therefore convenient (and sometimes necessary) to consider a ‘suitable’ collection of subsets in S.
Recall that a subset A of a metric space S is said to be bounded if and only if A has finite diameter, that is,
diam A= sup
x,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 byK(S).«
There is a relation betweenK(S)and H(S). Clearly,K(S) ⊆H(S)for any metric space S. Recall that a metric space S is said to have the Heine-Borel property, if any closed and bounded set is compact. If this happens to be the case, thenK(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) 6δ(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 that δ(A, B) =0, then d(a, B) =0 for all a∈ A. Then a∈ B, by using the result ofProposition 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) 6d(x, c) +d(c, b) 6d(x, c) +d(c, B). This estimate is valid for each b∈B, so
d(a, B) 6d(a, c) +d(c, B) 6d(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) 6d(a, C) +δ(C, B)
for any a ∈ A. By taking the supremum over all a ∈ A, the result follows.
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} and B = {z ∈ C: |z−4| 6 1}. Both of these sets are bounded, closed and non-empty.
−2 4
−2 2
0
δ(B, A) δ(A, B)
A
B
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).
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. dHis a metric on H(S). «
Proof. Since both A and B are bounded and non-empty, both δ(A, B)and δ(B, A) are finite. The triangle inequality follows from Proposition 4.2.3.
Note that dHis symmetric by definition. By symmetry andProposition 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 byProposition 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) <εfor any y∈ A. In addition, diam A6ε, because A is non-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, which implies 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 element a∈ 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). Since the map fB: A→R+: x7→d(x, B)is continuous and A is compact, there is an element a∈ A such that d(a, B) =dH(A, B).
Similarly, by continuity of the map x7→d(a, x)and compactness of B, it follows that there is a b∈B such that
d(a, b) = inf
x∈Bd(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 unit elements 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}. Note that both A and U are closed, bounded and non-empty subsets of`2, but not compact. 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+π62 +n21/2 for all n∈N.
Therefore,
dH(A, U) = inf
un∈Ud(x, un) =1+π621/2. But then,
d(x, un) >1+π621/2 for all n∈N. «
4.3 c o m p l e t e n e s s a n d c o m pa c t n e s s
This section will investigate sufficient conditions to ensure that H(S) is complete andK(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}is closed in H(S)byTheorem 4.2.7and complete by completeness of H(S). By utilizing the isometric map fromTheorem 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∈Nsuch that dH(An, Am) <ε whenever n, m>N.
Then d(a, AN) 6εfor any a∈A and A is bounded.
Indeed, note that δ(An, AN) < εfor all n > N (by the Cauchy property).
Therefore
[
n>N
An ⊆ {x∈X : d(x, AN) 6ε},
so A ⊆ {x ∈ X : d(x, AN) 6ε}. Thus d(x, AN) 6 εfor all x ∈ A. Next, in order to prove that A is bounded, take x, y ∈ A. Then there are a, b ∈ AN
such that
d(a, x) 62ε and d(b, y) 62ε.
Then, by the triangle inequality,
d(x, y) 6d(x, a) +d(a, b) +d(b, y) 64ε+diam AN.
Therefore diam A64ε+diam AN, which implies that A is bounded. Hence A∈H(S), which concludes this claim.
Claim 2. Let ε>0, then d(y, A) 6εfor any y∈ AN.
In order to prove the claim, it suffices to show that given ε>0, there is an element a ∈ A such that that d(y, a) 6 ε. This will be done by building a Cauchy 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 the Cauchy property, for each k∈Nthere is a Nk∈Nsuch 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 ∈ANksuch that
d(ak+1, ak) <ε·21−k.
Observe that each ak can be chosen in this way, due to the ‘clever’ choice of the indices Nk. In addition, by repeatedly applying the triangle inequality,
d(ak+n, sk) 6
n−1
∑
`=0
d(ak+`+1, ak+`)
<
n−1
∑
`=0
ε·21−k−`
6ε·21−k (1)
for all k, n ∈ Nwhere n >0. This proves that(an)is Cauchy in X and by completeness, it converges to some a∈X. Let n→∞, then by the estimate in (1),
d(a, ak) <ε·22−k for all k∈Nand consequently,
d(a, y) =d(a, a1) 6ε.
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 set ANk. Therefore a∈ {ak: k>m}for all m∈N. Since k<Nkfor all k∈N, it follows that
a∈ {ak: k>m} ⊆ [
k>m
ANk ⊆ [
n>m
An
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) 62ε. So by using the triangle inequality,
dH(A, An) 6dH(A, AN) +dH(AN, An) 6ε for all n>N.
4.3.2 Corollary. If S is a complete metric space, thenK(S)is complete. «
24
Proof. Let (Kn) be a Cauchy sequence inK(S). SinceK(S) ⊆ H(S), the limit of(Kn)exists in H(X)byTheorem 4.3.1. Therefore, by completeness of S, 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 ofTheorem 4.3.1, there exists a N∈N such that
K⊆ {x∈X : d(x, KN) 6ε/3}.
Furthermore, KNis compact by definition. This means there is a n∈Nand x1, . . . xn ∈ANsuch 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 with d(x, y) <ε/2 and there exist k such that y∈B(xk; ε/2). Thus
d(x, xk) 6d(x, y) +d(y, xk) 6ε.
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, thenK(S)is compact. « Proof. Recall that any compact metric space is complete. Therefore,K(S)is complete byCorollary 4.3.2. It remains to show thatK(S)is totally bounded.
Observe that S is totally bounded. So, given ε>0, there is a n∈ Nand there exist x1, . . . xn ∈S such that
16k6ninf d(x, xk) = min
16k6nd(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 less then ε of a subset of the (finite) collection{xk: 1 6k6n}. HenceK(S)is totally bounded and complete, which implies thatK(S)is compact.
4.4 c o n e a n d o r d e r s t r u c t u r e o n c l o s e d b o u n d e d s e t s 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}
for any λ ∈ R and A ∈ H(S). By definition, both addition and scalar multiplication are well-defined.
Under these assumptions, H(S) is indeed an ordered vector space and an ordered cone (according toDefinition 3.1.3). Moreover, inclusion is the standard order as defined in section3.1. From now on, the cone H(S)shall be investigated in more detail.
Note that all of the previously defined operations and observations all hold forK(S)as well.
There is way to incorporate the Minkowski sums into the definition of Hausdorff 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) 6ε for all a∈A. Therefore
δ(A, B) =sup
a∈A
inf
b∈B
d(a, b) 6ε
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 follows that 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 fromLemma 4.4.1will be used without mention from this point on. The ‘power’ of this expression will become 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 cancellation law holds in a very general setting, namely the setting of topological vector spaces.
Recall that a subset A of a topological vector space X over a field F is bounded if for every neighbourhood N of the zero vector there exists a scalar λ∈Fso 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 and
convex. «
Proof. Let N0be a base of neighbourhoods of the zero-element in X. Let U0∈N0be a given and define a sequence(Vn)in N0such that V0+V0⊆U0 and
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
Therefore
A+B⊆C+B+Vn for all n>1.
Next, let a∈A, b1∈B and let n∈Nbe a large, fixed number. Then there exists 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 the telescope sum on all the bk’s, it follows that
a= 1 n
∑
n k=1ck+1
n(bn+1−b1) + 1 n
∑
n k=1vk.
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+U0for all U0∈ N0, which means A⊆C.
If X is a normed space and the sets A, B and C from the previous theorem are convex, bounded, closed and non-empty, an alternative geometric proof can 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 and let 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 that A 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 separation theorem, there is a continuous linear functional φ : X → Rwhich strictly separates{a}and B. In other words, there exists a t∈Rsuch 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+C6⊆B+C. Therefore A+C6⊆B+C.
The cancellation law of the Minkowski sums is a direct consequence of Theorem 4.4.2:
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 closed
and 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, which proves the cancellation law.
The remainder of this section is devoted to showing the link between the ordering 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 in chapter3.
4.4.6 Lemma. Let S be a metric space. Let(An) be a nested sequence in H(S)where An ⊇ An+1for all n∈N and assume there is a number N∈N such that ANis compact. Then
A= \
n∈N
An∈ K(S).
«
Proof. Note that arbitrary intersections of closed sets are closed and a closed subset of a compact set is compact. It remains to show that A is non-empty.
Pick for each n∈ Nan element an ∈ An. This provides a sequence (an) with a convergent subsequence ank
kand 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 inK(S)such that An⊇ An+1for all n∈N. Then
A= \
n∈N
An ∈ K(S)
and dH(A, An) ↓0. «
Proof. ByLemma 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 ∈ AN
such that
d(an, A) =dH(An, A).
There is a subsequence(ank)kof(an)which converges to a point a∈ A. Then dH Ank, A
=d ank, A
6d ank, a.
Let k→∞, then d ank, a
↓0.
In order to turn H(S)into a normed space, the Hausdorff distance needs to be homogeneous and translation invariant, according to the construction ofTheorem 3.3.2.
In order to obtain homogeneity of the Hausdorff distance, it is convenient to work with a normed space instead of a metric space. The main reason for this assumption is the lack of linear structure of metric spaces.
28
For the remainder of this thesis, X= X,k·k will denote a normed space over the field of real numbers. In particular, it follows that
dH(A, B) =sup
a∈A
inf
b∈B
ka−bk for A, B∈H(X).
This extra assumption yields the homogeneity of the Hausdorff distance on 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 the norm,
dH(λA, λA) =sup
a∈A
inf
b∈B
kλa−λbk = |λ|sup
a∈A
inf
b∈B
ka−bk =λdH(A, B).
Unfortunately, the Hausdorff distance is not necessarily translation invariant on every subset of H(X)orK(X):
4.4.9 Example. Consider X=Rand take A=C= [−1, 1]and B= {−1, 1}. Then A+C = [−2, 2] and B+C = [−2, 2]. Note that δ(B, A) =0, since B ⊆ A and 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 translation
invariant. «
This last example proves that neither H(X)norK(X)can be turned into a normed Riesz space (by using the construction of chapter 3). It turns out that both H(X)andK(X)are too ‘large’. The construction is possible on the space of convex, closed and bounded subsets of a normed space X that contain 0. This collection will be studied in more detail in the next chapters.
5
C 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 of the 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. It turns out that the ensuing Riesz space has a strong order unit.
5.1 t h e r i e s z s pa c e o f c o n v e x s e t s
5.1.1 Definition. Let X be a normed space and A ⊆ X. The set of all convex combinations of points in A, denoted by co(A)is said to be the convex hull of A. The set co(A)is called the closed convex hull of A and is defined as the
closure of co(A). «
5.1.2 Definition. Let X be a Banach space, then Conv(X)denotes the collection of all non-empty, bounded, closed and convex subsets of X. « For the remainder of this chapter, X will denote a normed space (unless stated otherwise).
Note that Conv(X)is a metric cone when endowed with the Hausdorff metric and the Minkowski sums. From this point on, the metric space Conv(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 avoid this possible complication, consider Conv0(X)instead, where
Conv0(X) = {A∈Conv(X): 0∈ A}.
The collection Conv(X)inherits some structure from the underlying space normed space X.
5.1.3 Lemma. Let X be a normed space and A, B⊆X bounded, convex and non-empty subsets. 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 this terminology, 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=1λkxk.
Define index-setsΛ1and Λ2 byΛ1 = {k : xk ∈ A}andΛ2 = {k : xk ∈ B} and define
λ=
∑
k∈Λ1
λk. Then
1−λ=
∑
k∈Λ2
λk.
Note that both λ and 1−λare both non-zero, because x /∈ A∪B. In addition, λ61, since all λkare positive for 16k6n and sum up to 1. Therefore, the sum over all elements ofΛ1cannot exceed 1.
By using that both A and B are convex, there is an element a∈ A and an element b∈B such that x=λa+ (1−λ)b where
a= 1
λ
∑
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 (byTheorem 4.3.1) and Conv(X)is contained in H(X), there is a A∈H(X) such that
An dH
−→A as n→∞.
Because A is closed, bounded and non-empty by definition, it remains to show that A is convex.
To this end, let a, b ∈ A, λ ∈ [0, 1]and put x =λa+ (1−λ)b. Then, for any ε>0 there is a N∈Nsuch that for any n>N,
An ⊆A+εB and A⊆ An+εB.
Note that An+εBis 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), sinceTheo- rem 5.1.4implies 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 convex analysis.
5.2 s u p p o r t f u n c t i o n s
5.2.1 Definition. Let X be a normed space and let A⊆ X be a fixed non-empty subset. The map
hA: X∗ −−→ [0,∞]
φ 7−−→ sup
a∈A
φ(a)
is called the support function of A. «
32
For any A⊆X, the support function preserves lattice structure. The proof of 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 hB
denote the support function of A and B respectively. Then 1. hA∨B=hA∨hBand hA∧B=hA∧hB;
2. hA+B =hA+hB; 3. hλA =λhAfor all λ>0;
4. If A⊆B, then hA6hB. «
5.2.3 Proposition. There is a 1-1 correspondence between support functions and the
elements of Conv(X). «
Proof. Let A∈Conv(X)and let hA be the support function of A. It suffices to show that
A= {x∈X : φ(x) 6hA(φ)for all φ∈X∗}. [⊆]If x∈A, then
φ(x) 6sup
a∈A
φ(a) =hA(φ) for all φ∈X∗.
[⊇]Suppose there is an element
x∈ {a∈X : φ(a) 6hA(φ)for all φ∈X∗} \A.
Note that{x}is compact and A is closed and convex. In addition,{x}and A are disjoint. Therefore, by the Hahn-Banach separation theorem, there is a functional φ∈X∗and s∈Rsuch that
φ(x) >s and φ(a) 6s for all a∈ A.
Then
sup
a∈A
φ(a) 6s,
which yields hA(φ) 6s. Hence
φ(x) 6hA(φ) 6s, which leads to a contradiction.
By using support functions (given a fixed non-empty closed, bounded and convex set), it is possible to obtain another expression for the Hausdorff distance:
5.2.4 Theorem. Let X be normed and A, B∈Conv(X), then dH(A, B) = sup
kφk=1
hA(φ) −hB(φ) .
«
