Download Operators between C(K)

A.E. Sloof
Operators between C(K)-spaces
Bachelor’s thesis, June 26, 2014
Thesis advisor: Dr. M.F.E. de Jeu
Contents
1 Introduction
2
2 Banach lattices and Banach algebras
3
3 Useful results
7
4 Preparatory lemmas
9
5 The characterization theorems
12
6 Lattice and algebra automorphisms
16
1
Introduction
In this thesis we characterize two types of operators between C(K)-spaces: lattice and algebra homomorphisms. This characterization is nice, because something that seems to be rather big and incomprehensible, suddenly becomes simple and comprehensible.
With C(K) is meant the space of all continuous functions on a compact Hausdorff space K. One can give these spaces a special structure, namely the structure of a lattice or an algebra. In this thesis we consider the morphisms between
C(K)-spaces that respect these structures.
C(K)-spaces are not simply lattices and algebras. Because of the completeness
with respect to the norm ||f || = sup{|f (x)| : x ∈ K}, these spaces are also
Banach lattices and Banach algebras. We treat this in the second section. We
also give some more examples of Banach lattices and Banach algebras.
To prove the characterization theorems, we need some useful results, like The
Riesz Representation Theorem. We discuss these results in Section 3.
Before we formulate and prove the characterization theorems in Section 5, in
Section 4 we give the proof of some lemmas that are essential in the proof.
Finally, we consider the lattice and algebra automorphisms. The structure of
the groups of such automorphisms can be described completely, and this is done
in the final section.
At the end of this introduction, I would like to thank my advisor M.F.E. de Jeu
for his kind help and the friendly contact during my project.
2
2
Banach lattices and Banach algebras
In this section we will define Banach lattices and Banach algebras and their
morphisms. We will also give some examples. To define Banach lattices, we
need some preparation. First recall that a relation is an order relation if it is
reflexive, antisymmetric and transitive.
Definition 2.1. An ordered vector space is a vector space E over R equipped
with an order relation ≥, such that
1. If e ≥ f , then e + g ≥ f + g for all g ∈ E.
2. If e ≥ f , then αe ≥ αf for all α ∈ R, with α ≥ 0.
Definition 2.2. A vector lattice is an ordered vector space E such that for every
two elements e, f ∈ E the supremum e ∨ f exists in E.
Two vectors e, f ∈ E have a supremum g in E if g ≥ e and g ≥ f , and whenever
h is an upper bound of {e, f }, then h ≥ g. If the supremum of e, f ∈ E exists,
the infimum e ∧ f also exists, indeed e ∧ f = −(−e ∨ −f ). Before we go further,
we still need a short definition, that corresponds to our intuition.
Definition 2.3. The absolute value of a vector e ∈ E is given by |e| := e ∨ −e.
Now we can define a lattice norm.
Definition 2.4. A norm || · || on a vector lattice E is called a lattice norm
whenever |e| ≤ |f | in E implies ||e|| ≤ ||f ||.
A vector lattice equipped with a lattice norm is called a normed vector lattice.
Finally we come to our definition of a Banach lattice.
Definition 2.5. A Banach lattice is a vector lattice E with a lattice norm such
that E is complete.
Let us consider some examples.
Example 2.6. Let X be a nonempty set. Then B(X), the set of all real-valued
bounded functions defined on X, is a vector lattice under the ordering f ≥ g
whenever f (x) ≥ g(x) for all x ∈ X. We have (f ∨ g)(x) = max{f (x), g(x)} for
all x ∈ X. Under the norm ||f ||∞ = sup{|f (x)| : x ∈ X}, B(X) is a Banach
lattice.
Example P
2.7. Let `1 denote the vector space of all real sequences x = (x1 , x2 , ...)
∞
such that i=1 |xi | < ∞. With the pointwise algebraic operations and coördinatewise ordering (i.e., x ≥ y if x1 ≥ y1 , x2 ≥ y2 , ... for each x, y ∈ `1 ) `1 is a
vector lattice.
We have x ∨ y = (max{x1 , y1 }, max{x2 , y2 }, ...). Under the norm
P∞
||x|| = i=1 |xi | the vector lattice `1 becomes a Banach lattice.
Example 2.8. Let X be a topological space.
- Denote by Cc (X) the vector space of all continuous functions on X that
have compact support. In other words, a function f : X → R which is
continuous, belongs to Cc (X) if the set {x ∈ X : f (x) 6= 0} has compact
closure.
3
- Let Cb (X) be the space of all continuous bounded functions on X. So,
Cb (X) contains all continuous functions f : X → R such that sup{|f (x)| :
x ∈ X} < ∞.
- C0 (X) is the set of all continuous functions f : X → R such that for all
> 0 the set {x ∈ X : |f (x)| ≥ } is compact.
Under the pointwise ordering and the supremum norm (as in Example 2.6)
Cc (X), Cb (X) and C0 (X) are normed vector lattices. Here we have again (f ∨
g)(x) = max{f (x), g(x)} for all x ∈ X. Moreover, Cb (X) and C0 (X) are Banach
lattices. If X is not compact, Cc (X) is not in general a Banach lattice, because
it need not be complete. If X is compact, we have Cc (X) = C0 (X) = Cb (X),
and these spaces are all Banach lattices.
Now we move on to Banach algebras. First, we need the definition of an algebra
over a field.
Definition 2.9. An algebra over a field F is a vector space A over F that has
a multiplication defined on it such that for all α ∈ F and a, b, c ∈ A
1. the associative property holds: a(bc) = (ab)c;
2. the distributive properties hold: a(b + c) = ab + ac and (a + b)c = ac + bc;
3. the following property holds: α(ab) = (αa)b = a(αb).
Definition 2.10. A Banach algebra is an algebra A over the field R or C with a
norm || · || such that A is complete and such that ||ab|| ≤ ||a||||b|| for all a, b ∈ A.
Some books also require that A contains a unit element, but we do not. We will
consider some examples with and without unit element.
Example 2.11. Let C(K) be the space of all complex- or real-valued continuous
functions on a nonempty compact space K, equipped with the supremum norm.
Define multiplication in the usual way: (f g)(x) = f (x)g(x) for all f, g ∈ C(K)
and x ∈ K. This makes C(K) into a Banach algebra, which is also commutive.
The constant function 1 is the unit element.
Example 2.12. Let X be a Banach space. Then B(X) is the algebra of all
bounded linear operators on X, whereby the multiplication is composition. This
is a Banach algebra, with respect to the operator norm ||A|| = sup{||Ax|| : x ∈
X, ||x|| < 1}. The identity operator is its unit element.
Example 2.13. B(X) as in Example 2.6 with the usual multiplication is a
Banach algebra. The constant function 1 is its unit element.
Example 2.14. The space `1 , defined in Example 2.7, equipped with the pointwise multiplication is a Banach algebra. `1 does not contain a unit element.
Example 2.15. Just as in Example 2.8 we consider Cc (X), Cb (X) and C0 (X).
Besides R, we can also take C as field. These spaces are algebras when the multiplication is defined pointwise. They are also commutative. Moreover Cb (X)
and C0 (X) are Banach algebras. If X is not compact, C0 (X) and Cc (X) need
not have a unit element. If X is compact, the three spaces are equal and Cc (X)
is also a Banach algebra. We are then in the case of Example 2.11.
4
For more examples of Banach algebras, see [4, Chapter VII.1]
Now we want to consider morphisms between these two types of spaces.
Definition 2.16. Let T : X → Y be a linear map between two vector lattices
X and Y . Then T is a lattice homomorphism if T (x ∨ y) = T x ∨ T y for all
x, y ∈ X.
It is easy to see that T is positive. Let x ∈ X with x ≥ 0. Then x ∨ 0 = x,
hence
T (x)
= T (x ∨ 0)
= T (x) ∨ T (0)
= T (x) ∨ 0
≥ 0.
Just like the definition of a lattice homomorphism the definition of an algebraic
homomorphism is intuitively clear:
Definition 2.17. Let T : X → Y be a linear map between two algebras X and
Y . Then T is a algebra homomorphism if T (xy) = T (x)T (y) for all x, y ∈ X.
We do not require that T (1X ) = 1Y , when X and Y contain unit elements
denoted by 1X and 1Y . Note that the kernel of T is an ideal.
We will give two examples.
Example 2.18. Let Ω be a compact Hausdorff space. The real space C(Ω) and
R are algebras and lattices. Let ω ∈ Ω be an arbitrary element. Define
→ R
δω : C(Ω)
7→ f (ω).
f
Clearly, δω is linear. Moreover, δω (f ∨ g) = (f ∨ g)(ω) = max{f (ω), g(ω)} =
δω (f ) ∨ δω (g) for all f, g ∈ C(Ω). Hence δω is a lattice homomorphism. Finally,
δω (f g) = (f g)(ω) = f (ω)g(ω) = δω (f )δω (g) for all f, g ∈ C(Ω), thus δω is also
an algebra homomorphism.
The above example plays an important role in the proof of our theorems.
Example 2.19. Let T be the map given by
→ C([0, 1])
7→
f˜ : x 7→ xf (x) .
T : C([0, 1])
f
Then T is a lattice homomorphism, since
T (f ∨ g)(x)
= x · (f ∨ g)(x)
= x · max{f (x), g(x)}
=
max{x · f (x), x · g(x)}
=
(T (f ) ∨ T (g))(x)
5
for all f, g ∈ C([0, 1]) and x ∈ [0, 1]. But if we take f = g the constant function
1, and x = 21 , we find:
T (f g)(x) = x(f g)(x) =
1
1
1 1
6= = · = xf (x) · xg(x) = (T (f )T (g))(x).
2
4
2 2
Hence T is not an algebra homomorphism.
6
3
Useful results
In this section we will give some results that are useful for the next section.
Before we come to the first theorem, Urysohn’s lemma, we need the definition
of a normal space.
Definition 3.1. A topological space X is a normal space if, given any disjoint
closed sets E and F , there are open neighbourhoods U of E and V of F that are
also disjoint.
Theorem 3.2 (Urysohn’s lemma). [2, Theorem 10.5] For a topological space
X the following statements are equivalent:
1. X is a normal space.
2. Every pair of disjoint closed sets can be separated by a continuous function.
In other words: for every pair A, B of disjoint and closed subsets of X
there exists a continuous function f : X → [0, 1] such that f (a) = 0 for
all a ∈ A and f (b) = 1 for all b ∈ B.
Since a compact Hausdorff space is normal [5, Example 3.5.11], Urysohn’s lemma
is very useful to us.
Another result we will use is The Riesz Representation Theorem. First some
definitions.
Definition 3.3. Let Ω be a set. A collection A of subsets of Ω is called a
σ-algebra (in Ω) if it has the following properties:
1. Ω ∈ A;
2. If U ∈ A then U c ∈ A;
S
3. If (Un )n∈N ∈ A then n∈N Un ∈ A.
For every collection E of subsets of Ω, there is a smallest σ-algebra σ(E) which
contains E. σ(E) is called the σ-algebra generated by E and E is called a
generator of σ(E).
Definition 3.4. Let A be a σ-algebra in a set Ω and µ : A → [0, +∞] a function.
µ is called a measure on A if µ(∅) = 0 and if for every sequence (Un )n∈N of
pairwise disjoint sets from A
!
∞
[
X
µ
Un =
µ(Un ).
n=1
n∈N
Definition 3.5. Let Ω be a topological space. Denote by O the collection of its
open subsets. The Borel σ-algebra B(Ω) in Ω is the σ-algebra generated by O,
in other words B(Ω) = σ(O).
From now on let X be a locally compact Hausdorff space. A space X is locally
compact if each element of X has a compact neighbourhood.
7
Definition 3.6. A measure µ on the σ-algebra B(X) is called:
1. a Borel measure on X if µ(K) < ∞ for every compact K ⊂ X;
2. inner regular if for every A ∈ B(X)
µ(A) = sup{µ(K) : Kcompact ⊂ A};
3. outer regular if for every A ∈ B(X)
µ(A) = inf{µ(U ) : A ⊂ U open};
4. regular if it is both inner regular and outer regular.
Now we have enough luggage for the theorem.
Theorem 3.7 (The Riesz Representation Theorem). [2, Theorem 38.3] For
every positive linear functional
R F on Cc (X), there exists a unique regular Borel
measure µ such that F (f ) = f dµ holds for every f ∈ Cc (X).
The next theorem is about the support of a measure.
Theorem 3.8. [2, Theorem 31.12] Let µ be a regular Borel measure on X. Then
there exists a unique closed subset E of X with the following two properties:
1. µ(E c ) = 0;
2. If V is an open set such that E ∩ V 6= ∅, then µ(E ∩ V ) > 0.
The set E is called the support of µ and is denoted by supp(µ).
8
4
Preparatory lemmas
From now on let Ω and Q be compact Hausdorff spaces. We can see C(Ω) and
C(Q) as Banach lattices, as in Example 2.8. They are also Banach algebras
according to Example 2.11. Before we will describe the algebra and latticehomomorphisms between C(K)-spaces in the characterization theorems, we need
a lemma which characterizes the lattice homomorphisms from a C(Ω)-space to
the real numbers.
Lemma 4.1. A non-zero linear functional φ : C(Ω) → R is a lattice homomorphism if and only if there exists some c > 0 and some ω0 ∈ Ω, such that
φ = cδω0 , with δω0 as in Example 2.18.
In this case c and ω0 are both uniquely determined.
Proof. Let φ = cδω0 for some c > 0 and ω0 ∈ Ω. Clearly, φ is linear. Let
f, g ∈ C(Ω). Because c > 0, we have
φ(f ∨ g)
=
c · δω0 (f ∨ g)
=
c · (f ∨ g)(ω0 )
=
c · max{f (ω0 ), g(ω0 )}
=
max{c · f (ω0 ), c · g(ω0 )}
=
φ(f ) ∨ φ(g).
Thus φ is a lattice homomorphism.
For the converse, assume that φ : C(Ω) → R is a lattice homomorphism. By
definition, φ is linear and positive. Hence by the Riesz Representation Theorem,
R
there exists a unique regular Borel measure µ on Ω such that φ(f ) = f dµ for
all f ∈ C(Ω).
Now we will prove that supp(µ) contains exactly one element. Assume that the
support of µ contains two distinct points, say s and t. Because Ω is Hausdorff,
we can find two open sets U and V , such that s ∈ U , t ∈ V and U ∩ V = ∅.
Let T = Ω\U and S = Ω\V . Then T and S are closed. The singletons {s}
and {t} are also closed, because Ω is Hausdorff. According to Urysohn’s lemma,
there exists a continuous function f : Ω → [0, 1] with the following properties:
f (s) = 1 and f (a) = 0 for all a ∈ T . In the same way we can make a continuous
function g : Ω → [0, 1] such g(t) = 1 and g(a) = 0 for all a ∈ S. So we have
two functions f, g ∈ C(Ω) satisfying f (s) =Rg(t) = 1 and f ∧ g = 0, indeed
min{f (x), g(x)} = 0 for all x ∈ Ω. Moreover, f dµ > 0. To see this, recall that
F := f −1 (( 21 , ∞)) is an open set, because of the continuity of f . The element s
belongs to F , so supp(µ) ∩ F 6= ∅ and therefore µ(supp(µ) ∩ F ) > 0, according
to Theorem 3.6. Hence we have
Z
Z
1
f dµ ≥
f dµ ≥ µ(supp(µ) ∩ F ) > 0
2
supp(µ)∩F
9
In the same way, we find that
0
R
gdµ > 0. This implies
=
φ(f ∧ g)
=
=
min{φ(f ), φ(g)}
Z
Z
min{ f dµ, gdµ}
>
0,
which is a contradiction. Thus supp(µ) contains at most one element. Since
supp(µ) can not be empty, because φ is not the zero function, supp(µ) contains
only one element, say ω0 . Thus,
Z
Z
φ(f ) = f dµ =
f dµ = f (ω0 )µ({ω0 }) = µ({ω0 }) · δω0 (f ).
{ω0 }
So we find c = µ({ω0 }). This implies c > 0, because µ is non-negative. Finally,
ω0 and c are unique, because µ is unique.
We also need the characterization of the algebra homomorphisms from a C(Ω)space over R or C to respectively the real or complex numbers.
Lemma 4.2. Let C(Ω) be the real or complex vectorspace of functions. A nonzero linear functional φ : C(Ω) → F (with F = R if C(Ω) is real and F = C if
C(Ω) is complex) is an algebra homomorphism if and only if there exists a point
ω0 ∈ Ω such that φ = δω0 .
In this case ω0 is uniquely determined.
Proof. Let φ = δω0 for an ω0 ∈ Ω. As in the previous lemma, φ is clearly linear.
Furthermore, φ(f g) = φ(f )φ(g) for all f, g ∈ C(Ω), as we saw in Example 2.18.
So φ is an algebra homomorphism.
For the converse, assume that φ : C(Ω) → F is a non-zero algebra homomorphism. Let 1 be the constant function one on Ω. Because of the multiplicativity
we have φ(1) = φ(12 ) = (φ(1))2 . So φ(1) = 0 or φ(1) = 1. Assume φ(1) = 0.
Then φ(f ) = φ(f · 1) = φ(f )φ(1) = 0 for all f ∈ C(Ω), which is a contradiction
with the fact that φ is non-zero. Hence φ(1) = 1.
Let M = {f ∈ C(Ω) : φ(f ) = 0} be the kernel of φ. Denote by Nω the set
{f ∈ C(Ω) : f (ω) = 0}. Our claim is now that there exists a point ω0 ∈ Ω such
that M ⊂ Nω0 .
Let us assume that there is no such point. Then for each ω ∈ Ω we can find
a function fω in M which is not in Nω , in other words fω (ω) 6= 0. Since fω is
continuous, the set Vω = fω−1 (F\{0}) is open. We have ω ∈ Vω and fω (t) 6= 0
for all t ∈ Vω . Then (Vω )ω∈Ω is an open cover of Ω. By the compactness of Ω,
there exists a finite open subcover (Vωi )i=1,...,n of Ω.
Let us consider the function
x=
n
X
fωi fωi ,
i=1
where f (t) = f (t) for all t ∈ Ω. Let t ∈ Ω. There exists an j such that t ∈ Vωj ,
so fωj (t) 6= 0. We have fωi (t)fωi (t) ≥ 0 for all i, hence x(t) > 0. Therefore, we
10
have y =
1
x
∈ C(Ω) and
φ(x)φ(y) = φ(xy) = φ(1) = 1.
Hence φ(x) 6= 0. On the other hand, as φ(fω ) = 0 for each ω ∈ Ω, we find
φ(x) = 0, because ker(φ) is an ideal. This is a contradiction. Thus, there exists
some ω0 ∈ Ω such that M ⊂ Nω0 .
Now we shall show that Nω0 ⊂ M . To do this, let f ∈ C(Ω) satisfy f (ω0 ) = 0.
Consider the function u = φ(f )1 − f . Then φ(u) = φ(f )φ(1) − φ(f ) = 0, hence
u ∈ M . So u is also in Nω0 , thus u(ω0 ) = 0. This together with u(ω0 ) =
φ(f ) − f (ω0 ) = φ(f ), gives us φ(f ) = 0. Hence f ∈ M , and thus M = {f ∈
C(Ω) : f (ω0 ) = 0}.
Finally, let f ∈ C(Ω). Then f − φ(f )1 ∈ M because
φ (f − φ(f )1) = φ(f ) − φ(f ) · 1 = 0.
So f − φ(f )1 ∈ Nω0 , hence f (ω0 ) − φ(f ) = (f − φ(f )1)(ω0 ) = 0. In other words
φ(f ) = f (ω0 ).
The uniqueness of ω0 is a special case of the uniqueness of c and ω0 in Lemma 4.1.
11
5
The characterization theorems
Now we are ready to characterize the lattice homomorphisms between real C(Ω)spaces.
Theorem 5.1. A positive operator T : C(Ω) → C(Q) is a lattice homomorphism if and only if there exist a non-negative function w ∈ C(Q) and a mapping
τ : Q → Ω which is continuous on the set {q ∈ Q : w(q) > 0}, such that for
each f ∈ C(Ω) and each q ∈ Q we have
T f (q) = w(q)f (τ (q)).
Moreover, in this case, w = T 1Ω and the mapping τ is uniquely determined on
the set {q ∈ Q : w(q) > 0}.
Proof. Let T (f ) = w · (f ◦ τ ). T is well-defined, because T (f ) is continuous.
Indeed, if f is the constant function zero, T (f ) is also the zero function. So
assume that f is not the zero function and let q ∈ Q with w(q) > 0. Then τ
is continuous in q, and we know that w and f are continuous functions, hence
T (f ) = w · (f ◦ τ ) is continuous in q. So let q ∈ Q with w(q) = 0 and let > 0.
Since w is continuous, the set Uδ = w−1 ((−δ, δ)) is an open neighbourhood of q
for all δ > 0, such that |w(u)| < δ for all u ∈ Uδ . Let δ = ||f ||∞ . Note that we
can divide by ||f ||∞ , since f is not the zero function. Let u ∈ Uδ . Then
|T (f )(q) − T (f )(u)| = |w(q)f (τ (q)) − w(u)f (τ (u))|
= |w(u)f (τ (u))|
≤
|w(u)| · ||f ||∞
<
δ · ||f ||∞
= ,
hence T (f ) is continuous in q.
Clearly, T is linear. Because w ≥ 0, we have T (f ∨ g)(q) = (T (f ) ∨ T (g))(q),
just as in Lemma 4.1. Hence T is a lattice homomorphism.
For the converse, assume that T : C(Ω) → C(Q) is a lattice homomorphism.
The following step is important. Let q ∈ Q. We compose T with δq , so we get
δq ◦ T : C(Ω) → R
with (δq ◦ T )(f ) = δq (T (f )) = T (f )(q). Because T and δq are lattice homomorphisms and the composition of two lattice homomorphisms is again a lattice
homomorphism, δq ◦ T is a lattice homomorphism. Thus, by Lemma 4.1, there
exist a unique constant c = w(q) ≥ 0 and an ω0 = τ (q) ∈ Ω (only unique when
c > 0) such that
T (f )(q) = (δq ◦ T )(f ) = c · δτ (q) (f ) = w(q)f (τ (q))
for all f ∈ C(Ω). So for all f ∈ C(Ω) and q ∈ Q we have T (f )(q) = w(q)f (τ (q)).
Let f = 1Ω . Then
T (f )(q) = w(q)f (τ (q)) = w(q) · 1 = w(q),
12
thus w = T (1Ω ) and w ∈ C(Q).
We have already said that τ is unique on the set {q ∈ Q : w(q) > 0}, according
to Lemma 4.1. What remains is to prove that τ is continuous on {q ∈ Q :
w(q) > 0}.
Let q ∈ Q such that w(q) > 0 but assume that τ is not continuous in q. That
means there exists an open neighbourhood V of τ (q) such that for all open
neighbourhoods U of q we have τ [U ] 6⊂ V . By Urysohn’s lemma there is a
continuous f : Ω → R with f (τ (q)) = 1 and f (ω) = 0 for all ω ∈ Ω\V . The
function T (f ) = w · (f ◦ τ ) is an element of C(Q), hence continuous. Let us
define another function:
g : {q ∈ Q : w(q) > 0}
→ R
w(q) · f (τ (q))
= (f ◦ τ )(q).
q 7→
w(q)
Clearly, g is also continuous. So g −1 12 , 32 is open in {q ∈ Q : w(q) > 0},
hence in X because {q ∈ Q : w(q) > 0} is open in X. Because f (τ (q)) = 1 and
w(q) > 0, we have q ∈ g −1 ( 21 , 32 ) . Moreover, because f is zero on Ω\V , it
follows that f −1 21 , 32 ⊂ V , and thus
1 3
1 3
−1
−1
g
,
= (f ◦ τ )
,
∩ {q ∈ Q : w(q) > 0}
2 2
2 2
1 3
= τ −1 f −1
,
∩ {q ∈ Q : w(q) > 0}
2 2
⊂
τ −1 (V ).
But that means that the image of g −1 21 , 32 under τ is contained in V . So
we have found an open neighbourhood U of q such that τ [U ] ⊂ V . This is a
contradiction, hence τ is continuous in q.
In C(K)-spaces we have the identity (f ∨ g)(x) = max{f (x), g(x)}. Therefore,
f ∨ g and f ∧ g satisfy also the identities
f ∨g =
1
1
(f + g + |f − g|) and f ∧ g = (f + g − |f − g|),
2
2
which we will use in the proof of the next theorem.
Theorem 5.2. Every algebra homomorphism between real C(K)-spaces is a
lattice homomorphism. However, the converse is false.
Proof. Let T : C(Ω) → C(Q) be an algebra homomomorphism. To prove that
T is a lattice homomorphism, it is enough to show that |T (f )| = T (|f |) for all
f ∈ C(Ω). Because then we have
1
(f + g + |f − g|)
T (f ∨ g) = T
2
1
=
(T f + T g + T |f − g|)
2
1
=
(T f + T g + |T f − T g|)
2
= Tf ∨ Tg
13
for all f, g ∈ C(Ω).
We know that T is positive. Indeed, if g ≥ 0, then
√
√
T g = T (( g)2 ) = (T ( g))2 ≥ 0.
Now if f ∈ C(Ω), we have T (|f |) ≥ 0, because |f | ≥ 0. We also have
|T (f )|2 = (T (f ))2 = T (f 2 ) = T (|f |2 ) = (T (|f |))2 ,
whereby we used that v 2 = |v|2 , for v = T (f ) and v = f and that T is an
algebra homomorphism.
So |T (f )(x)|2 = (T (|f |)(x))2 for all x ∈ Q. And since |T (f )(x)| ≥ 0 and
T (|f |)(x) ≥ 0, we have |T (f )(x)| = T (|f |)(x) for all x ∈ Q. So we find
|T (f )| = T (|f |).
For an example of a lattice homomorphism which is not an algebra homomorphism, see Example 2.19.
In the next theorem we will characterize the algebra homomorpisms.
Theorem 5.3. An operator T : C(Ω) → C(Q), with C(Ω) and C(Q) real or
complex spaces, is an algebra homomorphism if and only if there exists a clopen
subset V of Q and a mapping τ : Q → Ω which is continuous on V such that
for each f ∈ C(Ω) and each q ∈ Q we have
T f (q) = χV (q)f (τ (q)).
In that case χV = T 1Ω and T is uniquely determined on V .
Proof. Assume that T f = χV · (f ◦ τ ). T is well-defined, because T (f ) is
continuous, since T is a special case of Theorem 5.1. T is also linear. Moreover,
T (f · g)(x)
=
χV (x)(f · g)(τ (x))
=
χV (x)f (τ (x))g(τ (x))
=
(χV (x))2 f (τ (x))g(τ (x))
=
T (f )(x) · T (g)(x),
for all f, g ∈ C(Ω) and x ∈ Q hence T is an algebra homomorphism.
For the converse, we distinguish two cases:
1. Assume that T : C(Ω) → C(Q) is an algebra homomorphism between real
C(K)-spaces.
According to Theorem 5.2, T is a lattice homomorphism. So, by Theorem
5.1, there is a map τ : Q → Ω and a w ∈ C(Q) such that for all f ∈ C(Ω)
and q ∈ Q we have
T f (q) = w(q)f (τ (q)).
And we also know that w = T (1Ω ). When we use that T is an algebra
homomorphism, we get
w2 = (T (1Ω ))2 = T (1Ω 2 ) = T (1Ω ) = w
This implies that for all q ∈ Q we have w(q) = 0 or w(q) = 1. So w = χV
for a unique clopen subset V ⊂ Q. From Theorem 5.1 it follows that τ is
unique and continuous on V .
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2. Assume that T : C(Ω) → C(Q) is an algebra homomorphism between
complex C(K)-spaces.
Let q ∈ Q. Just as in the proof of Theorem 5.1, we compose T with δq , so
we get
δq ◦ T : C(Ω) → C
with (δq ◦ T )(f ) = δq (T (f )) = T (f )(q). Because T and δq are algebra
homomorphisms and the composition of two algebra homomorphisms is
again an algebra homomorphism, δq ◦T is a algebra homomorphism. Thus,
by Lemma 4.2, if δq ◦ T 6= 0, there exists a unique ω0 = τ (q) ∈ Ω such
that
T (f )(q) = (δq ◦ T )(f ) = δτ (q) (f ) = f (τ (q)),
for all f ∈ C(Ω). So for all f ∈ C(Ω) and all q ∈ Q: if δq ◦ T 6= 0, then
T (f )(q) = f (τ (q)).
From (T (1Ω ))2 = T (1Ω 2 ) = T (1Ω ), it follows that T (1Ω ) = χV for a
unique clopen subset V ⊂ Q.
Now let q ∈ V . Then (δq ◦ T )(1Ω ) = T (1Ω )(q) = χV (q) = 1, so δq ◦ T 6= 0.
Therefore, we have
T (f )(q) = f (τ (q)) = χV (q)f (τ (q))
for all f ∈ C(Ω).
If q 6∈ V , then χV (q) = 0, so T (f )(q) = T (1Ω · f )(q) = (T (1Ω )T (f ))(q) =
χV (q)T (f )(q) = 0. Choose a τ (q) ∈ Ω arbitrary, then certainly T (f )(q) =
χV (q)f (τ (q)). So we have
T (f )(q) = χV (q)(f (τ (q))
for all q ∈ Q and all f ∈ C(Ω).
Finally, the continuity of τ on V is a special case of the continuity of the
map τ on the set {q ∈ Q : w(q) > 0} in Theorem 5.1.
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6
Lattice and algebra automorphisms
In this section we will consider the automorphisms of C(K)-spaces. Therefore
we introduce some notation. Denote by LatAut(C(Ω)) the group of lattice automorphisms T : C(Ω) → C(Ω). Further, let AlgAut(C(Ω)) be the group of
algebra automorphisms of C(Ω). Finally, Homeom(Ω) is the group of homeomorphisms τ : Ω → Ω. Recall that a continuous map is a homeomorphism if it
is bijective and its inverse is also continuous. In these groups the operation is
composition.
The following theorem will characterize the lattice automorphisms. For this we
need the group G = {f ∈ C(Ω) : f > 0}. This is a group under multiplication.
First we will consider the map
α : Homeom(Ω) →
τ
7→
Aut(G)
ατ : g 7→ g ◦ τ −1 .
α is well-defined, since ατ is a surjective and injective homomorphism. Moreover, α is a homomorphism, because
α(τ ◦ σ)(g) = g ◦ (τ ◦ σ)−1 = g ◦ σ −1 ◦ τ −1 = α(τ )(g ◦ σ −1 ) = (α(τ ) ◦ α(σ))(g),
for all τ, σ ∈ Homeom(Ω) and g ∈ G. This gives us the semi-direct product
G oα Homeom(Ω) with the following multiplication:
(g, τ ) ∗ (h, σ) = (g · ατ (h), τ ◦ σ) = (g · (h ◦ τ −1 ), τ ◦ σ).
Theorem 6.1. The map
φ : G oα Homeom(Ω)
→ LatAut(C(Ω))
(g, τ ) 7→
T : f 7→ g · (f ◦ τ −1 )
is a group isomorphism.
Proof. We will show that φ is a homomorphism. First, we have:
φ((g, τ ) ∗ (h, σ))(f )
= φ((g · (h ◦ τ −1 ), τ ◦ σ))(f )
=
(g · (h ◦ τ −1 )) · (f ◦ (τ ◦ σ)−1 )
=
(g · (h ◦ τ −1 )) · (f ◦ σ −1 ◦ τ −1 ).
And second, we have:
φ((g, τ )) ◦ φ((h, σ))(f )
=
φ((g, τ ))(h · (f ◦ σ −1 ))
=
g · ((h · (f ◦ σ −1 )) ◦ τ −1 )
=
(g · (h ◦ τ −1 )) · (f ◦ σ −1 ◦ τ −1 ),
for all (g, τ ), (h, σ) ∈ Goα Homeom(Ω) and f ∈ C(Ω). So φ is a homomorphism.
Now we will prove that φ is surjective. Let S ∈ LatAut(C(Ω)). According to
Theorem 5.1 for all f ∈ C(Ω) we have S(f ) = w · (f ◦ σ) for w ∈ C(Ω) a nonnegative function and σ : Ω → Ω a continuous map. This map S has an inverse,
16
say T with T (f ) = v · (f ◦ τ ), such that T ◦ S = idC(Ω) = S ◦ T . This means
g
=
(S ◦ T )(g)
= S(v · (g ◦ τ ))
= w · ((v · (g ◦ τ )) ◦ σ)
= w · ((v ◦ σ) · (g ◦ τ ◦ σ))
= w · (v ◦ σ) · (g ◦ τ ◦ σ)
for all g ∈ C(Ω). If we take g = 1, we find
1 = g(x) = w(x) · v(σ(x)) · 1 = w(x) · v(σ(x)),
1
for all x ∈ Ω, thus w(x) = v(σ(x))
, and because w is non-negative, we have
w(x) > 0 for all x ∈ Ω.
Now we get (S ◦ T )(g) = g ◦ τ ◦ σ for all g ∈ C(Ω).
We know from Theorem 5.1 that idC(Ω) is of the form idC(Ω) (f ) = v · (f ◦ τ ) for
v = idC(Ω) (1) = 1 and a map τ which is unique on the set {ω ∈ Ω : v(ω) > 0} =
Ω. So τ need to be idΩ . But S ◦ T = idC(Ω) , and S ◦ T corresponds with the
continuous map τ ◦ σ, so τ ◦ σ = idΩ . Likewise, σ ◦ τ = idΩ . Hence τ and σ are
bijections and have continuous inverses. So these maps are homeomorphisms.
Finally, we find φ((w, τ )) = φ((w, σ −1 )) = S. Hence φ is surjective.
For the injectivity we have to show that ker φ = {0}, since φ is a homomorphism.
Assume that φ((w, τ )) = idC(Ω) . So idC(Ω) (f ) = w · (f ◦ τ −1 ). We just saw that
w = idC(Ω) (1) = 1 and τ −1 need to be idΩ , so τ = idΩ . Hence ker φ = {0}.
In the next theorem we will see that looking at the algebra automorphisms on
C(Ω) is actually the same as looking at the homeomorphisms on Ω.
Theorem 6.2. The map
φ : Homeom(Ω) → AlgAut(C(Ω))
τ 7→ T : f 7→ f ◦ τ −1
is a group isomorphism.
Proof. First, φ is a group homomorphism, because φ has the same structure as
α in the proof of Theorem 6.1.
Now we will prove that φ is surjective. Let S ∈ AlgAut(C(Ω)). According to
Theorem 5.3 for all f ∈ C(Ω) we have S(f ) = χW · (f ◦ σ) for W a clopen subset
of Q and σ : Ω → Ω a continuous map. This map S has an inverse, say T with
T (f ) = χV · (f ◦ τ ), such that T ◦ S = idC(Ω) = S ◦ T . This means
g
=
(S ◦ T )(g)
=
S(χV · (g ◦ τ ))
=
χW · ((χV · (g ◦ τ )) ◦ σ)
=
χW · ((χV ◦ σ) · (g ◦ τ ◦ σ))
=
χW · (χV ◦ σ) · (g ◦ τ ◦ σ)
17
for all g ∈ C(Ω). If we take g = 1, we find
1 = g(x) = χW (x) · (χV (σ(x)) · 1) = χW (x) · χV (σ(x)),
for all x ∈ Ω, thus χW = 1. So S(f ) = f ◦ σ for all f ∈ C(Ω).
Hence for each algebra automomorphism M there is a continuous map µ such
that M (f ) = f ◦ µ. Moreover, this map is unique, by Theorem 5.3.
For idC(Ω) this continuous map has to be idΩ . But S ◦ T = idC(Ω) , and S ◦ T
corresponds with τ ◦ σ, so τ ◦ σ = idΩ . Likewise, σ ◦ τ = idΩ . Hence τ and σ are
bijections and have continuous inverses. So these maps are homeomorphisms.
Finally, we find φ(τ ) = φ(σ −1 ) = S, thus φ is surjective.
By the injectivity of φ is meant that for each algebra automorphism M there
is a unique continuous map µ such that M (f ) = f ◦ µ. This uniqueness is a
special case of the uniqueness of τ and V in Theorem 5.3.
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References
[1] Y.A. Abramovich, C.D. Aliprantis, An Invitation to Operator Theory. American Mathematical Society, Providence, 2002.
[2] C.D. Aliprantis, O. Burkinshaw, Principles of Real Analysis. Third Edition,
Academic Press, San Diego, 1998.
[3] H. Bauer, Measure and Integration Theory. Walter de Gruyter, Berlin, 2001.
[4] J.B. Conway, A Course in Functional Analysis. Second Edition, SpringerVerlag, New York, 1990.
[5] V. Runde, A Taste of Topology. Springer Science+Business Media, New
York, 2008.
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