AN ABSTRACT OF THE THESIS OF
WILLIAM EDWARD MARGOLIS for the
(Name)
in
MATHEMATICS
(Major)
presented on
DOCTOR OF PHILOSOPHY
(Degree)
9212t-
<-
6"//Q 26
(Date)
TITLE: TOPOLOGICAL VECTOR SPACES AND THEIR INVARIANT
MEASURES
Signature redacted for privacy.
Abstract approved:
T. R. Chow
First, topological vector spaces are examined from a partial
order structure derived from neighborhood bases of the origin.
This
structure is used to produce a minimal vector norm for every
Hausdorff locally convex space.
Then, topological vector spaces are examined to find translation invariant measures with respect to which functions in the topo-
logical dual are integrable. It is shown that every conical measure
has a unique translationally invariant
on a locally convex space
E
extension to all of a(E),
the Riesz space generated by the real
valued continuous affine functions on
E. Invariant measures are
constructed, characterized, and extended. An integral representation on certain complete weak spaces is found.
Topological Vector Spaces and Their
Invariant Measures
by
William Edward Margolis
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
June 1971
APPROVED:
Signature redacted for privacy.
Assistant Professor of Mathematics
in charge of major
Signature redacted for privacy.
Acting Chairman of Department of Mathematics
Signature redacted for privacy.
Dean of Graduate School
Date thesis is presented
Typed by Clover Redfern for
June 26, 1970
William Edward Margolis
ACKNOWLEDGMENT
I wish to thank Professor T.R. Chow for his encouragement, patience, and sound suggestions.
I also wish to acknowledge Clover Redfern for her
excellent typing, which gave undeserved clarity to many
sections of this work.
TABLE OF CONTENTS
Page
Chapter
I.
INTRODUCTION
1
TVS PRELIMINARIES
6
6
8
TVS
Posets
POSET DESCRIPTIONS OF TVS E
is a TVS
is a LCS
is a Separable Space
Bornology
10
10
19
22
25
VECTOR VALUED NORMS
28
INVARIANT MEASURES
33
33
41
57
65
Riesz Space Representation
Characterizations of Invariant Measures
0(E) and Other Extensions
Integral Representation on Weak Spaces
BIBLIOGRAPHY
70
APPENDIX
Notational Index
72
72
TOPOLOGICAL VECTOR SPACES AND THEIR
INVARIANT MEASURES
I.
INTRODUCTION
Ina series of three papers, 1917-1920, P. J. Daniell initiated
the investigation of integration from the point of view of positive linear functionals
on a vector lattice (Riesz space) satisfying
U
"(L) if f1 (p) > f2 (p)
lim U(fn) = 0"
> ...
and
fn(p) =
0
for all
p,
(Daniell, 1917, p. 280). He also required that the
functions in the Riesz space be bounded. In his second paper, Daniell
(1918) produced the first examples of integrals for functions defined
on an infinite dimensional space which do not reduce to an infinite
series or to an integral over a finite number of dimensions. Banach
(1937) generalized Daniell's first example by considering as the
domain of the functions the unit ball in a separable Hilbert space.
The more standard approach to integration is the Lebesgue
theory: a countably additive set function on a sigma ring or sigma
algebra is given, the integral is defined on linear combinations of
characteristic functions, the integral is then extended by taking
sup s
or
inf's
over a set of previously defined integrals. Bochner
(1939) produced a Riemann integration theory with respect to positive
finitely additive set functions on an algebra of sets by considering
limits of partitions.
2
Tolstov (1962) showed that the Daniell approach to integration
theory lead to Lebesgue integrals when the function space included all
bounded Borel functions. The necessity for
that a positive linear functional
U
(L)
holding in order
on the bounded continuous func-
tions be represented by Lebesgue integration with respect to a positive Baire measure had been shown 11 years earlier by Glicksberg
(1951).
In the same paper, Glicksberg proved that for a completely
regular space E,
every positive linear functional on the space of
bounded continuous functions is represented by a Baire measure if
and only if E allows no unbounded continuous functions. This is
obviously not the case when E is a Hausdorff topological vector
space. This clarified the result of Hewitt (1950), who studied the
representation for positive linear functionals
I
on the space of all
continuous functions on a completely regular space. He discovered
(Hewitt, 1950, p. 168) that
I
would then be represented by a Baire
measure, with respect to which every continuous function is bounded
except on a set of measure zero. In a later paper (Hewitt, 1952), he
returned to the more general setting of Daniell: the investigation of
positive linear functionals
I
on some Riesz space
L
of functions.
By using the bounded functions in the Riesz space he found a finitely
additive measure which represented
I
for some functions in
L.
But he did not succeed in finding reasonable necessary and sufficient
conditions in terms of
I
alone for the representation to hold for all
3
bounded functions in
L
(Hewitt, 1952, paragraph 3.8).
Glicksberg's 1951 paper and Hewitt's 1952 paper highlighted the
difficulties in investigating Riesz spaces of unbounded functions.
The investigation of the Riesz space generated by the dual space
of a locally convex Hausdorff topological vector space was initiated
by Choquet (1962), who introduced the ideas of conical measures and
affine measures. Choquet (1969) proved that conical measures on the
complete weak spaces
E(X), X denumerable, were localizable by
Radon measures. He discovered rotationally invariant conical and
affine measures on pre-Hilbert spaces.
Umemura (1965) emphasized the importance of investigating
measures--not necessarily sigma additive--on function spaces, both
for the theory of stochastic processes and for quantum mechanics.
He proved the existence of a continuum of quasi-invariant Borel
measures on infinite-dimensional vector spaces, and asked about the
existence of essentially different invariant measures.
This thesis is concerned with investigating this question using
translation groups, from the Daniell point of view, and seeing how
the properties of the resulting measures depend upon properties of
the topological vector space structure. We first consider some new
properties of a topological vector space, and see how the measures
are affected by this structure.
Chapter II recalls and assembles the notation and basic facts
4
needed concerning topological vector spaces and partially ordered
sets. We introduce the concept of co-initialness and observe a simple
property of it.
Chapter III relates characteristics of the topology of a vector
space to certain properties of related posets. There is introduced
the equivalence relation
.e.
and the partial ordering
.a.
on the
equivalence classes, and classifies spaces according to lc-type.
Chapter IV uses
.e.
to show that every Hausdorff locally con-
vex space is (vectorially) normizable.
Chapter V begins with a representation theorem for Riesz
spaces. Although the theorem must be known to at least a few people,
neither its statement nor its proof seem to have ever been published.
A similar statement could probably be made for the other theorems in
the first section. In particular, the proof of Theorem 3 is too short
to be unknown.
Theorem 4 arose out of an attempt to simplify some
of the arguments in Hewitt's 1950 paper. The method of proof reap-
pears in Proposition 4; it also appears in Choquet's 1967 paper and
throughout Chapter 8 of his 1969 book. In the second section, The-
orem 6 was stated but not proved by Choquet. Theorem 13 was
announced by Courrege (Bony and Courrege, 1964). The other results
are all original. We show that no non-zero affine measure has an
extension which is translation invariant for any translation group.
We show that every positive invariant measure on a complete weak
5
space
E(X),
X denumberable, is a non-localizable extension of a
localizable conical measure. We show that invariant measures are
not zero as conical measures without being identically zero. We
show that although invariant measures are plentiful, they correspond
to no even finitely additive set function on a subset of the Baire sets.
We show that ever conical measure has a unique translation invariant
extension, and produce two characterizations (Theorems 15 and 17)
of invariant measures. Finally, for weak spaces E(X), X denumer-
able, we produce an integral representation.
6
II.
TVS PRELIMINARIES
TVS
The present section recalls and assembles notation and basic
facts from the theory of topological vector spaces which are used in
this paper. The reader who is familiar with the basic facts of the
theory may skip this section and refer to the Notational Index if confronted with any unfamiliar or unusual terminology or notations
appearing in the remainder of the paper. For further elucidation of
the assembled notations, the reader is referred to Choquest (1969,
Chapters 5, 6), Horvath (1966), or Schaefer (1966). The section ends
with some terminology from lattice theory.
We shall deal exclusively with topological vector spaces over
Let us recall that a topological vector space (TVS) is
the reals R.
a set E equipped with a real vector space structure as well as a
topology such that the vector space operations, addition EX E~E
and scalar multiplication R X E
E,
are both continuous.
denotes the vector space of all linear maps from E into
algebraic dual).
E'
F
(the
R
denotes the vector space of all continuous lin-
ear maps from E into
If
E*
R
(the topological dual).
is a second TVS and if
f
E
F
is an algebraic iso-
morphism which is also a homeomorphism, we shall call f
a
7
toplinear isomorphism and say that
F
and
E
are toplinearly
isomorphic.
A base for the neighborhood system at the origin is called a
be subsets of E.
0-neighborhood base. Let A, B
absorbs
in
{x}
for
R,
IsI
> t.
A
for all x in E.
sI
such that
provided there exists a t > 0
B
is called absorbing provided
A
The subset
<1.
0-neighborhood absorbs
We say A
B C sA for
A absorbs
sA C A
is called balanced provided
of
A
If
A.
E
A
is bounded provided every
is absorbing, the gauge of A,
g(x) = inf {t > 0:x E tA}.
If
E
has a 0-neighborhood base of convex sets, it is called
locally convex. We say that E is an LCS provided it is a locally
convex hausdorff TVS.
Suppose A is balanced, convex and absorbing. Then
is a semi-norm. We let
EA
denote the vector space
with the topology determined by the semi-norm
E
gA
furnished
gA.
A TVS is called locally bounded provided it possesses a bounded
0-neighborhood.
Suppose
Then the
E
is a vector space and
o-(E, F)-topology
spect to the set of maps
An LCS E
identical with
for
E
F
is a subspace of E*.
is the initial topology with re-
F.
is called a weak space provided its topology is
cr(E, E').
Notice that in general,
o-(E,E')
is a
8
weaker topology. On any vector space
weak topology is
X,
let E(X) = RX
furnished with the product
is a (complete) weak space whose dual E(X)' =R(X).
E(X)
Actually, it is well known that
is a complete weak space if and
E
is toplinearly isomorphic to some
only if E
a complete weak space, let
map i:E
E(X)
in
y
X
it is clear that the finest
cr(E,E*).
For any set
topology.
E
and
E,
Given
be a Hamel basis for 2.
X
given by point evaluation:
in
E(X).
Then the
i(y)(x) = x(y)
for
x
is the required toplinear isomorphism.-1/
E
Posets
We shall also have need of some elementary notions concerning
partial orderings. Suppose
ordered set) a subset
provided for every
u, v
w < u and w < v.
that
u<s
0 = u.
S,
T
for all
If
then
S
S
s
in
S,
(S, <)
of
is said to be left filtering for
S
there exists a
T,
in
If
then
is a poset (i.e., a partially
has an element
S
S
in
w
u
T
such that
is said to have a zero element
has an element w such that s < w for all
is said to have a unit element
We define
such
s
1 = w.
u V v = sup(u, v) = the zero element of the poset
{w ES:u <w and v < w},
when it exists. We define
1/See for instance, Choquet, 1969, Volume II, p. 58-61.
in
9
u A v = inf(u, v) = the unit element of the poset {w
S:w < u and w < v},
when it exists. A semilattice is a poset in which either of the following two conditions hold:
Every two elements has an
inf.
Every two elements has a sup.
that
t
T
is a poset and
If
(S, <)
T
is co-initial in
S
is a subset of
provided
sES
S
is left filtering and
also left filtering. For let
s
T
implies there exists a
and
is co-initial in
t
and
ment w E T
u < t.
By the co-initialness of
such that w < u
S
then
be elements of T.
is left filtering there exists an element
u<s
then we say
S,
such that t < s.
If
S
T
u
of
T
S
is
T
Since
such that
there exists an ele-
and therefore such that w < s
and
w < t.
The next chapter relates characteristics of the topology of a
TVS to certain properties of related posets. For a presentation of
other uses of posets to the investigation of TVS, see Birkhoff (1966).
10
III. POSET DESCRIPTIONS OF TVS E
E is a TVS
In this section let E be a TVS, and let
be a fixed
B
B we introduce
0-neighborhood base consisting of balanced sets. In
the following relations: for
provided
.e. V
U
is absorbed by
provided
and
U
for
V;
in
V
U
let
B,
and
in
V
U .a. V
let
B,
U .a. V and V .a. U.
Propostion 1. The following properties of elements U, V
of
are equivalent.
U .a. V
U
is a subset of
tV
for some
t > 0.
Proof.
implies (b): by the definition of absorbing.
implies (a): because the elements of
C tV C tsV for
IsI
> 1.
Therefore,
B
U C XV
and
for all
Q. E. D.
>t.
If
are balanced;
U C sV
and
V C tW then U C stW
V .a. W implies that U .a. W.
established, it is clear that
.e.
so that
U .a. V
With the transitivity of .a.
is an equivalence relation.
11
Corollary. The following properties of elements
B
U, V
of
are equivalent.
U .e. V
U C sV C tV for some s, t >0.
Let
U
[U] ={v E B:V .e. U}.
an element of B,
U .e. U'
If
U' .a. U
V .e. V'
and
induces a partial order on BI.e.
by
.a.
[U]
and
Thus,
.
in
V
[U] .a. [V]
Proof. It is clear that
[U], [V]
provided
for some
U .a. V
.a.
U
in
is a left filtering partial order.
(B/.e., .a. ) is a partial order. Sup-
are elements of BI.e.
But then [W] .a. [U]
and
Proposition 3. If
B/.e.
Thus,
which will continue to be denoted
neighborhood base, there exists W in
then
then since
U .a. V,
[V].
Proposition 2. B/.e.
pose
and
it is the case that U' .a. V'.
V .a. V'
and
be the canonical surjection, so that for
B/.e.
:B
[
B
Then since
B
B
is a
such that W CU fTh V.
[W] .a. [V].
Q. E. D.
is closed under finite intersections,
is a semilattice under A .
Proof. Let
[U], [V]
[U] A [V] = [U n V].
For,
be elements of B/.e.
U rm V .a. U and
.
We show that
U cm V .a. V
so
12
that [U cm V] .a. [U]
[W] .a. [U]
and
that We sU
and
V] .a. [V].
[U
Now suppose
Then W C sU and W C tV
[W] .a. [V].
Thus W .a. U rm V
tV C max(x, t)U rm V.
so
Q. E. D.
[W] .a. [U rm V].
Recall that a subset
provided that for all
that
so
[U]
T
of the poset
in
B/.e.
is co-initial
(B/.e. , .a.)
there exists
in
[V]
T
such
[V] .a. [U].
Notice that
itself is co-initial, so that the set of all
B/.e.
co-initial subsets of B/.e.
is nonempty.
2/
lc(B) = minfcardinality T :T C B/.e. is
are well-ordered, so that
exist a co-initial subset
lc (B)
The cardinals
is well defined. Indeed there must
such that
T
Let
lc(B) = the cardinality of T.
Proposition 4. Co-initial sets are left filtering.
Proof. This follows from Proposition 2, and the remarks at
the end of the preceding section.
Theorem 1.
If
balanced sets for E,
Proof. Let
B'
Q. E. D.
is another zero-neighborhood base of
then
T C B/.e.
lc(B) = lc(B1).
be co-initial in B/.e.
.
lc is a typewritten version of the small script aleph as written
in Hebrew.
13
co-initial in
MC
[U] E B/.e.
.
V C U, V E B'.
in
Let
m([U])
of that chosen
137.e.
is a neighborhood base, we may choose a
B'
Since
is constructed as follows: Let
V.
Then M is co-initial in B'/.e.
Since
.
B
Since
.
B1/.e.
For let
Let M m(T).
.
M is surjective, the cardinality of
m: T
Thus, for each co-initial subset T of BI.e-
there exists a co-initial subset M of 137-e.
B1
lc(B1) < lc(B).
Thus,
lc
B
and
and thus by the Schr6der-
lc(B) = 1c(31).
Q. E. D.
is independent of the choice of neighborhood base,
and so we may set lc(E) = lc(B),
of
of as low cardinality-
Since the argument with respect to
is symmetric, also lc(B) < lc(131),
Bernstein theorem,
be an element of
[W]
But then m([U]) .a. [U] .a. [W].
M < cardinality T.
Thus,
[U]
is a neighborhood base, there exists a U in
U C W.
such that
the equivalence class in
Performing this choice for every
defines a map m:B/-e-
B/.e.
137.e.
[la
equal
and consider
as a function
lc
E.
Theorem 2.
The following are equivalent statements about
lc(E) = 1.
The partially ordered set B/.e.
E
is locally bounded.
has a
0
element.
E.
14
Proof. (i) implies (ii): Since
co-initial subset
{[U]}.
This means that
[U]
[U] .a. [V]
So
is a
(ii) implies (i): The
initial subset for B/.e.
We show that
V
V'
V' C V.
U .a. V' .a. V and thus
origin in E.
U' C U.
0
element of B/.e.
Then there exists an element
[U] .a. [V]
But
U
is a
[U']
be an element of B/.e.
so that
be a bounded neighborhood of the
Then there exists an element
We show that
.
Then
0
U'
of
element of
B
such that
B/.e.
Let
U' .a. U .a. V so that
[U'] .a. [V].
Q. E. D.
The following are equivalent statements about E.
Theorem 3.
lc(E) =
1
or
lc(E) =
E
has a countable zero-neighborhood base.
E
is semi-metrizable.
Proof.
.
is absorbed by V.
U
implies (ii): Let
[V]
be the
[U]
is a bounded neighborhood of the origin in E. Let
U
such that
B
.
lc(E) = lc(B) = 1.
so
be any neighborhood of the origin.
of
B/.e.
B/.e.
in
[V]
element furnishes a singleton co-
0
implies (iii): Let
for all
element for
0
has a
lc(E) = lc(B), 13/.e.
(ii) implies (iii) and (iii) implies (ii) are well known
theorems.
Suppose
E
is semi-metrizable. Then there is a semi-metric
15
d :E X E
whose open balls form a base for the topology of E.
R+
{{x E E :d(x, 0) < n} :n E N} is a countable zero-neighborhood
Then
base. Now suppose that E has a countable zero-neighborhood base.
We sketch one possible construction of a semi-metric: Given a
countable neighborhood base
form a sequence
B = {Bn:n E N},
{Vn:n E N} of balanced zero-neighborhoods inductively as follows.
Choose as
any balanced zero-neighborhood such that
V1
Having chosen
such that
Vn
+
Vn-1,
C
Vn
neighborhood base,
d:EXE'R+
C B.
let Vn be any balanced zero neighborhood
Then {Vn:n E N} is a zero
Vn-1
Bn.
Vn
+
Vn
C
Vn-1,
and Vn\ {o}.
Define
by
inf
d(x, y) =
n EH
The semi-metric
F(N) = {H C N:H finite}.
(ii) implies (i): Suppose
a co-initial subset of Bi.e.
V C n U.
Let
[U]
generates
,
[U
n]}.
has a countable zero-
E
neighborhood base. Then clearly
that
d
For details see Schaefer (1966, p. 28-29).
the topology of E.
T = {[U11, [U2],
V}
2-n : x-y E
{
H E F(N) n EH
where
V1
1 < lc(E) <
Suppose
T
of finite cardinality.
There is some element V of
But then
[V] .a. [U.]
for all
1,
i
B
,n.
i=1
[Ui] .a. [U]
be an element of Bi.e.
so that
is
[V] .a. [U].
.
Then for some
Thus,
{[v]}
i
is co-initial so
such
16
lc(E) = 1.
(i) => (ii): Suppose
Then there exists a balanced
lc(E) = 1.
bounded zero-neighborhood
C
in
E.
B' =
So
a countable zero-neighborhood base. For if
neighborhood in E,
t >0.
some
So
U
for
U
t < n.
Suppose
C C tU for
lc(E) =
B'/.e. , where B'
{[Un]:n E N} be a co-initial subset in
and let
is a balanced zero-
then by the boundedness of C,
n-1C C
is
EN}
is the collection of all balanced zero-neighborhoods of E.
Then
B"
{m-1Un:n EN, m E N} is a countable zero-neighborhood base
for
E.
exists
that
For let
V E B'
U
be a balanced zero-neighborhood. Then there
such that V C U.
[U n] .a. [V]
so
[U n]
r-
U.
Thus
neighborhood implies there exists V E B"
suppose
m-1Un, r-1Us E B".
U C tU for some
and thus,
.a. [U],
So for m > t, m-1U n
tE R
Also, there exists n such
U
a zero-
such that V C U.
Now
We need only show that there exists
such that
q-1Ut C m-1Un
r-1Us to conclude that
B" is a zero-neighborhood base. By Proposition 4, there exists
q-1Ut E B"
] .a.] and
such that [Ut
[U]
.a. [U].
Thus, Ut C MUn
t
s
n
for some M, S E N. So (mM)1
U
C m1
SUs
Ut
Un
[Ut
and
and
q
(rS)-1Ut C
-1
Ut
base
rUs.
Let
q = max(mM, rS).
Then
C m-1Un
r-1Us.
Thus if lc(E) = 1 or
B".
lc(E)
,
there exists a countable
Q. E. D.
17
In the proof of the preceding theorem, we have shown the following:
<,
1 < lc(E)
if
cardinals which
lc(E)
1c(E) = 1
or lc(E)
0
so
2, 3, 4, etc. Are there any other
never equals
that lc(E)
then
avoids?
The answer to this question is no,
as shown by the following.
Theorem 4.
Suppose
is a set of cardinality
X
Then
>
there exists an LCS E such that lc(E) = cardinality of X.
Proof. Because of Theorems 2 and 3, we need only assume that
cardinality (X) >
tko.
that is,
E = E(X),
Let
equipped
RX,
with the product topology. We have seen in the preceding chapter that
E
is locally convex Hausdorff, and that its topology corresponds to
Let
cr(E,EI).
B'
=S
`BS, n
F(X) = {SC X :S is finite}
B
B'
S, n
= {f
E
E F(X), n
E
N} where
and
E: -n1
< f(s) < + n.-1 for all
s
E
S}.
is a zero-neighborhood base for the topology of E.
if and only if
S
D S.
[B]
= {BAS, n :BA
A .e. B S,n }
5, n
S, n
{BA
BS, n
.a. BA A
If we set
S, n
B
= [BS , n
then
S, n
so
=
= {B
5,3
B' /.e. = {Bs : S E F(X)}.
cardinality (B'/.e.) = cardinality F(X).
Now
j
E
1\1}.
Clearly
But as is well known,
18
00
cardinality (F(X)) = cardinality (X). For let us write
F(X) =
i Fn(X)
n= 1
This is a
Fn(X) = {A C X :A has exactly n elements}.
where
countable disjoint union of sets of cardinality the same as that of
For clearly cardinality (X) < cardinality (Fn(X)) < cardinality
cardinality (X).
So indeed
X.
(X11)
cardinality (F(X)) = cardinality (X).
The next construction shows that by removing sets from
F(X),
we
X remains
are not in danger of lowering cardinality so long as
covered.
Let
T
be a co-initial subset of B'/.e.
value. Then clearly
= cardinality (X).
only show that
of minimal lc-
cardinality (T) < cardinality (B'/.e.)
So to conclude the proof of the theorem we need
cardinality (X) < cardinality (T).
Because in that
cardinality (X) = cardinality (T) = lc(B') = lc(E).
case
Let
E = {S C X :Bs
E
T}.
cardinality (Z) = cardinality (T).
Clearly
Also
Because by the co-initialness of T,
B
E
XE
S.
T
such that
Bs .a. [B{x}, 11
For each S eZ
Enumerate
S
so that
for all x
X
there exists
which is the case if and only if
construct a map fs: N
S=
E
, sn}.
X as follows.
Then let
19
1<j<n
if
fS(i)
if n < j
sr
Clearly
fs(N) =
Now we can consider the map
p:
S
X defined by
)< N
p(S, n) = f (n).
By (*) and (**)
p
is surjective.
So
cardinality (Z)).
cardinality (X) < cardinality (Z X N) = max(
But
0
< cardinality (X),
so
cardinality (X) < cardinality (Z) = cardinality (T).
Q. E. D.
E is an LCS
We shall now suppose in this section that E
that
B
is an LCS, and
is a fixed 0-neighborhood base consisting of convex balanced
sets and closed under finite intersection.
Proposition 1'.
The following properties of sets
U, V E B
are equivalent:
U .a. V
U C XV
for some
the identity map E
in
X
id
EV
is continuous, where
E
20
is the seminormable locally convex space on the set E
U, gu.
generated by the gauge of
(a) implies (b): by Proposition 1.
Proof.
implies (c): Suppose (b) holds. Consider the map
E
id
E
V
Because
.
g(x) < X for all x
bounded on the unit ball of E
,
in
id(U),
id
is
is a bounded linear
so that id
mapping of normed spaces, and thus (c) follows.
implies (a): if id
inV so
id(U) .a. V
is continuous then id(U)
and thus
U .a. V.
Corollary to the Proposition.
U, V
in
is bounded
Q. E. D.
The following properties of sets
are equivalent:
B
U .e. V
U
C
E
V C X2I-1
EV
for some
X.1, X2
in
as topological vector spaces.
Theorem 2'. The following are equivalent statements about E.
lc(E) = 1.
Bi.e.
E
is a semilattice under A
with
0.
is normable.
Proof. (iii) implies (ii): Suppose
E
is normable. Then the
norm is the gauge of a bounded zero-neighborhood, say
C .a. V
for every zero-neighborhood
V.
There exists
Thus
C.
U
C C
21
such that
Then
U E B.
[Ul
is the
of
0
Bi.e.
.
So by Propo-
sition 3, (ii) holds.
(ii) implies (i): Follows from Theorem 2.
(i) implies (iii): By the statement preceding Theorem 1, we
may choose a co-initial set of minimal cardinality, say {[C']}.
Let
be any zero-neighborhood, and
U
V CJ U.
[C] .a. [V],
Then
C' a. U so that
C'
V
in
B
such that
so that C' .a. V .a. U,
and thus
is bounded. Choose a zero-neighborhood W
which is balanced convex and absorbing, such that W C C'.
identity map E
neighborhood.
EW
is continuous because W is a zero-
The inverse map Ew
is bounded. Therefore
E
is continuous since W
is toplinearly isomorphic to the normed
E
Ew.
space
Q. E. D.
Recall that there exists nonlocally convex spaces
lc(E) = 1
let E
The
E
for which
but condition (iii) of Theorem 2' does not hold. For instance,
LP[O,
0 <p < 1.
where
Then
1
f(x) Pdx < 1}
U = {f E E :
0
is a bounded zero-neighborhood in the topology generated by the
pseudo-norm
1
f(x) Pdx.
f
0
22
So
lc(E) = 1.
empty. So
But
E'
is not locally convex. Indeed,
E
is
is not normable. Thus, Theorems 2 and 2' are
E
essentially different.
E is a Separable Space
This section will be concerned with separable spaces.
Theorem 5.
Suppose
lc(E) < 2
vector space. Then
Proof. Suppose
(D) =
is a separable Hausdorff topological
0
.
is dense in E,
D
and cardinality
be an open zero-neighborhood base. For each A P(D)
let
E
F(A) = {G
each
Then
X.
E
E
A
Let F
P(D) = {A:A C D} = the power set of D.
Let
.
0
E
:G
D = A}.
choose a
For
Let A = {A E P(D):F(A)
Uk E F(X).
Finally, let B' = {Uk:
E
cardinality (131) < cardinality (A) < cardinality (P(D)) =
Moreover,
Bi
is a zero-neighborhood base. For suppose
open neighborhood of the origin. Then there exists a
G
A}.
2°.
V
is an
Er such
GC GC V because Hausdorff topological vector spaces are regular. For X = G
D
there exists a
U
E 13/.
But
UC'TCGCV.
The first inclusion holds because any open set
X
not containing a point of
X
cannot meet
Uk,
section would be an open set without a point from
indeed, a zero-neighborhood base. Therefore
since their interD.
So
13°
is,
1
23
lc(E) = lc(B4) < cardinality (BI) < 2
".
Q. E. D.
That the conclusion of the previous theorem cannot be strengthened may be seen from the following example of a separable Hausdorff
locally convex space E,
Let
E = E( [0,1]).
with
lc(E) = 2
By Theorem 4,
so that E
lc(E) = 2 Ko,
is certainly nonmetrizable. Because it is possible to pass a poly-
nomial through any finite set of points in the plane, it is clear that the
continuous functions
C([0,1]) are dense in E.
Now
C([0,1])
is separable with respect to the stronger uniform topology, so a
fortiori is separable with respect to the weaker relative topology.
For the possibility of set theory of the real line without the
continuum hypothesis, see Geidel (1964). The following proposition
may present a stronger result than the preceding paragraph.
Proposition 5. Let X C [0,1] be such that
cardinality (X) > ,;()
Then E(X)
Hausdorff locally convex space with
of
is a nonmetrizable separable
lc
value equal to the cardinality
X.
Proof. Theorems 3 and 4 establish all but the separability of
the space. Let
Dx
Now
{f
D
be a countable dense subset of E( [0,1]).
Rx :f = glX for some g E D}.
We claim that
B = {Bs, n(f) :f E E(X), n E N, S C X finite}
Let
Dx = E(X).
is a base for the
24
topology of E(X),
BS, n(f)
Then since
D
B.
-1
for all s E S).
such that
Choose any T E E( [0, 1] )
-1
for all
s S.
Therefore dIX E
DX
Q. E. D.
dIX EB5, n(f).
and
71 S = f I S.
there exists a dED such
is dense in Ed 0, 1]),
Id(s) - 7(s)I < n
that
<n
= fg E E(X) : g(s) - f(s)
B5
So let
where
Theorem 6. Suppose
is separable if and only if
is a complete weak space. Then
E
is toplinearly isomorphic to some
E
where cardinality of X < 214°.
E(X)
Proof.
By our remarks in the preceding chapter and the pre-
vious theorems, it is sufficient to show that E(X) is separable
whenever cardinality (X) < 2
o.
Suppose
Then there exists an injection i :X
able dense subset in E( [0, 1] ).
DX =
Clearly,
B
S, n
Choose
cardinality (Dx) < cardinality D =
where
S
D
0
a count-
Let
E E(X): there exists d E D such that d
(co) C E(X),
BS, n(cp)
[0, 1].
cardinality (X) < 2
is finite,
= {f E E(X) : I f(s) - cp(s)1 < n
-1
n E N,
i = f}.
Consider any
EE (X),
and
for all s ES).
To conclude the proof, we need only find an element of
DX
lurking
25
in
B
S, n
There exists a d ED such that
(cp).
for all
I c(i(s)) - yo(s)I < n-1
dE
Dx
EB
s E S.
for all
id(s) - go(s)1 < n
and
So let
i.
d=d
sES
Then
so that
Q. E. D.
S, n (c0)
Bornology
Let
E
be a TVS.
We shall now consider the
E;
that is, on all subsets of E.
E by
relation on the power set of
We shall denote the power set of
P(E).
then A .a. B provided
If A, B E P(E),
B.
.a.
A .e. B provided
A .a. B and
B .a. A.
A
is absorbed by
We use
[
]
again
for the canonical surjection
[]:P(E)~ P(E)/.e.
Let
of
E.
Let
C P(E) be the set of all balanced zero-neighborhoods
p c P(E)
be the set of all bounded subsets of E.
Proposition 6. A subset
[X]
is a lower bound for
[y]
X
of
E
in the poset
is bounded if and only if
(P(E)/.e.
Proof. X is bounded iff X .a. V for all
[X] .a. [V]
for all [V] E [y].
VE
,
.a. ).
iff
Q. E. D.
26
[Y]
[G] E [y],
[13]
[X]
E
[13],
as an "upper segment. " That is,
sits in P(E)/.e.
[H]
E
P(E)/.e.
E
P(E)/.e.
[H] E [y].
as a "lower segment. " That is,
sits in P(E)/.e.
[y]
implies that
[G] .a. [H]
and
implies that
[Y] .a. [X]
and
[Y]
E [13].
We may define other notions from the theory of topological vec-
tor spaces in terms of this partial order. The subset A of
absorbing if and only if
A
is an upper bound for {[x]: x E}.
[A]
E
is bornivorous if and only if
[A]
is an upper bound for [P]. An
LCS E is bornological if and only if every balanced convex set
(P):[G]
satisfies the following property,
is an upper bound for
is
E
G
if and only if [G]
E [y]
[13].
Let us consider spaces
which satisfy
E
(P)
but which are
not necessarily locally convex.
Call a subset
for some
X > 0.
X
of
Taking
E
a
X = 2,
C-set provided X + X
C
XX
we see that every convex set is a
C-set. Call a TVS a C-space provided it has a zero-neighborhood
base of balanced C-sets. By Theorem 2 of Chapter III, if lc(E) = I
then
is a C-space.
E
Theorem 7. A C-space
C-set
G,
E
satisfies
if and only if for every TVS
every linear map f: E
is continuous.
F
F
(P)
for every balanced
such that
lc(F) = 1,
taking bounded sets into bounded sets
27
Proof. (=>): Suppose
balanced C-sets. Let
F.
is a C-space satisfying (P) for
be a balanced C-set zero-neighborhood in
V
Let A be a bounded set in E.
A .a. f-1(V).
[f-1(V)]
So
Then
bounded implies
f(A)
is an upper bound for
[p].
Also,
is a balanced C-set, since
f-1(V)
f
-1
(V) + f
-1
(V)
f
-1
(V+V) C f
Therefore, by (P), f-1 (V)
(<=): Now let
lc(E) = 1.
(V)
(V).
E
E
such that [V]
Then E V is a TVS such that
f: E
Ev be the identity map. Then f(V) = V
so that by assumption f is continuous. There-
Let
is bounded in E
V=f
(X.V) =
V be a balanced C-set of
is an upper bound for [p].
fore
E
E
Thus,
(P)
is satisfied for
V.
Q.E.D.
28
IV. VECTOR VALUED NORMS
For any LCS, we characterize its topology by considering its
image in a complete weak space.
Let
be an LCS,
E
convex sets, a subset
a zero-neighborhood base of balanced,
B
of
T
such that
B
initial set of minimal cardinality in Bi.e.
n:E
Consider the map
{[U] : U
E
T}
is a co-
.
defined by
E(T)
n(x) = (gU(x))U ET
Letting
r
= if
:T
E(T)
Considering
x, y E E(T)
with the Riesz space structure of F(T,R),
E(T)
see that E(T)
then
is the cone of positive elements in E(T).
x < y if and only if y - x E(T)+
E
for all
U
Proposition 1.
x.
x(U) < y(U)
it is clear that n(E) C E(T)+.
R : f > 0),
we
If
if and only if
T.
E
0
in
E
if and only if n(x.)
0
in
E(T).
Proof. Suppose
x.
1
0.
Let
e
E
T.
We want to find
implies that n(x.)G < E. There exists a J such
X.) <E.
that i > J implies x. EEG and therefore n(x.)
1iG
such that i > j
1
1
So
n(x.)
1
0.
29
Suppose now that n(x.)
We want to find
E.
U
E
be such that
T
G is a zero-neighborhood in
and
0
implies
such that i > j
j
U .a. G,
so that
X.0 C G
x.
n(x.)=
g U (x.) -" 0.
U
implies
implies
>J
Corollary. x.
x
g
U
k >0.
Therefore there exists J
implies x.
(x.) <
Let
G.
for some
such thatXi
~n(x.) 0
K
if and only if n(xi-x)
U C G. Q. E. D.
0.
Proposition Z. n is continuous.
Proof. Suppose x.
G
E
T,
g (x.)
G
x
But
gG(x).
Ig
Will show that for all
E.
in
(x.)-g (x)1 < g (xi -x)
G
corollary.
0
by the
Q. E. D.
Proposition 3.
n(x)
if and only if x = 0,
0
n()x) =
X.1n(x)
for all
E
R.
n(x+y) < n(x) + n(y)
Proof. (a) follows from the definition. Alternatively, notice
that Propositions 1 and 2 imply (a).
(b) follows from the definition of n
in
and scalar multiplication
E(T):
n(x)
= (gG(kx))G ET = (1k1 gG(x))G ET
=
(gG(x))G ET
=
n(x).
30
(c) follows from the fact that
(x+y) < g (x) + g (y)
g
the order structure of E(T).
and
Q. E. D.
Proposition 4. n is uniformly continuous.
Proof. Let
U
be a zero-neighborhood in
find a zero-neighborhood
n(x) - n(y)
E
G
in
contains a set If E E(T): I f(G)I
C T.
1
x- ye G implies
By definition of the weak topology on E(T),
U.
{G. :1 < i <
such that
E
Want to
E(T).
Let
< E,
1 < i < n} where
G = n G..
1
E>0
U
and
Then EG is the required
neighborhood.
For,
x-
implies
y E EG
gGi(x-y) < gG(x-y) < E
i = 1,
,n,
which implies that
= 1,
,n,
which implies that n(x) - n(y) E U.
Thus,
Ig
Gi
(x)-g
Gi
(y)1
<6
for
for
Q. E. D.
enjoys all the properties associated with a continuous
n
norm except real-valuedness. The following two theorems highlight
the relationship between n and continuous seminorms on E.
Theorem 1. Let
is, suppose
f E E(T)'
f
and
be a positive linear form on E(T). That
f(E(T)+) C R+.
Then
fon
is a con-
tinuous seminorm on E.
Proof. Let
since
E(T)1 = R
f
(T)
,
be a positive linear form on E(T).
Then
31
i=1
a. >0,
where
. (G.)E(T)R evaluation at G., G.
and
E
T.
Then
f
n(x) =
/aigG.(x).
i=1
So
fon
on
E.
is clearly an element of the cone of continuous semi-norms
Q. E. D.
A partial converse is the following.
Theorem 2. For every continuous semi-norm m on E,
there exists a continuous semi-norm
(i) g = f
x
E
o
n
for some
f E E(T)',
on
g
E
such that
and (ii) m(x) < g(x)
for all
E.
Proof. If m is a continuous semi-norm on E,
m(x) <a
with
a>0
and, we may assume,
then
(x)
G.
E
T.
For instance, see Horvath (1966 p. 97).) Let
(This is well-known.
32
i= 1
where
1.(G.)
is as in Theorem 1. Then clearly
f o n(x) =
i=1
is a semi-norm which satisfies (i) and (ii).
Q. E. D.
Notice that n is a scalar valued norm if and only if lc(E) = 1
if and only if E
is normizable.
33
V. INVARIANT MEASURES
Rie sz Space Representation
We shall assume the basic properties of Riesz spaces, which
can be found, for instance, in Bourbaki (1965, Chapter II) or even
Daniell (1917, Chapter 1). If L is a Riesz space,
L
= (L
is the algebraic dual of
If
)
EL
=
:
where
f(L+) C R+1,
L.
is a topological space, we let
E
designate s
x v y = sup(x, y).
the cone of positive elements of L.
x Ay = inf(x, y).
L+
real valued functions on E.
F(E,R)
be the set of all
F(E,R) is a Riesz space which is also
an algebra. Let B(E) be the subalgebra of continuous bounded functions.
Let
space of
be a TVS.
E
F(E,R)
Then h(E)
designates the Riesz sub-
generated by the elements of E', the topological
dual of E.
Theorem 1 in the present section will show us that
h(E) = {V a.
-
1= 1
1
V b.1 : a.,1 b.1 E El.
The elements of h(E)*+
are
1=1
called conical measures (Choquet, 1962, 1969).
Let
E' + R be the set of all continuous affine functions;
be the Riesz subspace of F(E,R)
b(E)
generated by
E' + R con-
sisting of all bounded functions. We shall see by Theorem 1
n
n
b(E) = { V a. - V b. E B(E) : a., b. E E` + R}.
1
b(E)
*+
1= 1
1= 1
are called affine measures.
The elements of
that
34
We shall be dealing with lattices generated by certain vector
subspaces of F(E,R).
It would be convenient to have a general
representation theorem.
Theorem 1. (General Representation Theorem).
Let R be
a Riesz space and S C R a vector subspace. Then the vector
lattice generated by
is
S
L(S)
{ V a.1
=
V
b.1 : a.,1 b.1 E s}.
V
-
1=1
1= 1
Proof. First it is shown that L(S)
n
V b., V a.
-
i=l
m
m _
n
V a.
j=1
i=1
is a vector space. Let
V 1;. E L(S).
-
j=1
3
3
Then
.
V a.
V b.
-
+
.
j=1
1=1
1=1
( V a.)
+
( V a.)
.
j=1
1 =1
V a.
- V 1-3-.
3
j=1
m
-
( V b.)
( V b. )
-
j=1
1=1
3
m
V(a.+Va.)V(b.+Vb.)
j
j=1
i=1
j=1
1
i=1
(n, m)
(a-4-a.)
V
(i,j)=.(1,1) 1
=
is in
L(S)
Suppose
since
X. >O.
S
Then
(n, m)
V
(i,j)=(1,1)
(b.+17).)
1
3
is closed under addition. Let
X. ER.
35
- V b.)
V
X V a. - X V b.
=X(
1=1
1=1
1=1
1=1
kb. E
V Xa.LV
1=1
1=1
since
S
(S)
is closed under scalar multiplication. Suppose
X < 0.
Then
X ( V a- i=1
.
V b.) = (4)( V b.
.
i=1
1=1
V a.)
1=1
= V (-X)bi - V (-)ai E L(S).
1=1
1=1
So
L(S)
is a vector space.
1
1
because aES=>a= Va- V OEL(S).
S C L(S)
Clearly,
1=1
V -y),
xAy=
Since
sup's
closed under
inf's,
and
1=1
in order to show that
L(S)
it is sufficient to show that
x, y E L(S) => xVy E L(S).
m
Let
x
V a.
y = V a.
V b.,
-
i=1
1=1
1
j=1
-
V b.
j=1
j
m
( V a.
i=1
- V
1
11
=
n
=
( V a.1
1=1
b.) V ( V a. - V b.)
1
m
n
-
m
_1 m
..._
V b. + V b.) V ( V a.) - V b.
i=1
1
j=1
=1
j=1
3]
3
=
is
36
n
m
i=1
j=1
m _
n
m
n
=i( V ai + V b.) V ( V a. + V b.)1- V b, - V b.
j=1
3
3
(n, m)
V
(n, m)
1(
aid- 1;.) V(
V
(i, j)=(1, 1)
3
(n, m, 2)
V
ai, 3, k
(i, j, k)=-(1, 1, 1)
i=1
j=1
1
i=1
3
m
I
n
a.+ bil- V b. - V b.
(i, j)=(1,1)
I
j=1
belongs to
L(S)
3
i=1
3
1
-VT-Vb
.
3=1
j
.
1=1
where
a.
1,
Because every
V s. = V s. -V 0
1
i=1
S.
E
.
the fact that
S,
L(S)
is a vector space shows that (4'4(4) be-
longs to L(5).
Q. E. D.
Theorem 2. Let
from E
I
whenever
1=1
1=1
F(E, R)
be the real algebra of all functions
Let A C F(E, R) be a real subalgebra. Let
be a positive linear functional on A; that is, I E A+. Supinto
R.
pose there exists an
and
f
in A
I(f) = 0.
Then I = 0.
Proof.
Let
(p
such that
f(x) > 1
for all x E E,
E A.
Then since for all n E N, 0 < (q-n)2 =2 - 2ncp + n2
0 < cp
2
2ncp + n f
so
we have
37
2n(p <
2
+ n2 f
1
2
n
7 f
Therefore
I(q,) <
1
I(ç)2 +
n
1
I(f) =
Since the above inequality is true for all
1(0 <0 for all
II A+ = 0,
Therefore
go
n
+ 2(p + 1 <
(co+1)2
0<
I((p2) + 2I((p) + I(f)
= cp2
E
2
)
N
in A.
by the positivity of
0<
I(cp
(*)
On the other hand,
I.
+ 2cp + f
So
since
co
E
2I((p),
A.
Therefore,
0 < I(9).
From (*) and (4":4),
(**)
it is clear that I(p) = 0.
Notice that if there exists any
inf g(x) > 0
and
g
Q. E. D.
in A
such that
I(g) = 0,
X EE
then the hypothesis is satisfied with
f
1
inf g
g-
38
The proof of the theorem is easier, of course, if it is the case
that A
consist only of bounded functions, or when A =
A+ - A+
which occurs when A is a Riesz space.
Notice that all we have used in the proof of the theorem is the
closure of the vector space A under the taking of squares. But
this characterizes an algebra, since
fg
(f+g)2-f2-g2
2
Theorem 3. Suppose A is a Riesz subspace of
is a Daniell integral.-3/ Then if I(f) = 0
IEA
such that
for all x E E,
f(x) > 1
Proof. Let
co
E
A+.
Therefore, it is clear that
zero pointwise.
I( co
When
cp
then
F(E,R) and
for some
fEA
I = 0.
(p(x) < n, nf(x) A (p(x) = (p(x).
(nf A co)
converges monotonically to
So
(nf A (p)) = I(q) - I(nf A (p)\ 0
I(nf A co) I I((p)
But,
I(nf A go) < I(nf) = nI(f) = 0.
3/Recall, this means that I satisfies the property: (pn A+,
decreases to zero pointwise, implies that I(p) \ 0.
E
q'
39
Therefore I(co) = 0,
so
II A+ = 0.
this implies that
A = A+ - A+,
That the hypothesis that
Since in a Riesz space
Q. E. D.
I = 0.
be a Daniell integral cannot be
I
relaxed may be seen from the following example.
the Riesz space generated by the affine func-
Let A = a(R),
That is,
tions in R.
co E A
if an only if
is a continuous
(I)
"piecewise linear" function: there exist real numbers
such that
y9I (-00, x1], cp [x
x 2], .
I
.
are all "linear" (affine). Let I:A
E A and
I(p) = a.
ca
I
[x, +00) =
That is,
constants, and yet
I
I
[x n,
f(t) = at + b,
is well defined,
0,
since, e.g.,
<xn
+00)
let
is the slope of the last piece of
I(y9)
easily verified that
yo
<
be defined as follows. If
R
where
[x, + 00)
and
1, xn]
[x
,
.
x1
E
*+
A+,
I(f) =
1
(P.
It is
is zero on the
I
for f(t) = t.
the previous theorem, it is not necessary to verify that
I
By
is not
Daniell.
A condition which ensures that an algebra be a Riesz space is
the following.
Theorem 4. Suppose
A and
B
are real algebras of real-
valued functions such that the set of positive elements of
closed under taking square roots. Let
homomorphism. Then
4,(A)
homomorphism. In particular,
L.p :A
B
is a Riesz space and
A
A, A+,
be an algebra
4,
is a lattice
is also a Riesz space. (Take
is
40
= identity map.)
Proof. It is sufficient to show that for all
0 I fl )
f
in
A,
Of) I
because in that case
4)(fVg) =
=
(f+g-/- If-g1)) = -}(iP(f)+00+ liP(f-g)
1
)
(4)(f)+Lp(g)+1 Lp(f)-Lp(g)1 ) = Lp(f)vip(g)
and
1
L1J(fAg) = LP(
So let
f
4i
Ifl
A+
If I 2
= N/
f2
E
)
= Lii(f)Atp(g).
A by hypothesis.
if and only if there exists h
in which case
h2
g
I Lp(f)-Lp(g) 1
2(Lp(f)+Lp(g)-
is an algebra homomorphism, it is order preis in
serving. For
Since
1
be an element of A.
Because
such that
(f+g If-g 1 )) =
f2,
Lp(If 1
)2
in
A+
4i(g) = Lp(hh) = kii(h)4i(h) = q(h)2 > 0.
=
So
Lp(f)2.
by the order preserving property of
LP,
1 Lp( I f 1)1 = 1 Of) 1
LP( I fl )
But
is positive. There-
fore
=
Notice that the only properties of
proof were that
Lp
ILp(f)I.
Lp
Q. E. D.
that were used in the
should be linear and preserve squares. Thus:
41
Theorem 5. Suppose
and
A
are real algebras of real-
B
valued functions such that the set of positive elements of A
closed under the taking of square roots. Let tp:A
transformation such that
p(f2)
is a Riesz space and
Lii(A)
qi
=
Of)z
for all
be a linear
B
f
A.
in
is
Then
is a lattice homomorphism.
Characterizations of Invariant Measures
Let
be a locally convex topological vector space. Let
E
be a positive measure-4/ on E.
Suppose
I
is invariant under some
translation operators, and every functional in
respect to
case that
I
sets of E,
that
I
in integrable with
E'
We shall study properties of this
I.
I
If it were the
I.
give rise to a finitely additive measure on a ring of subit would be sufficient to require that
I E (L(Er))
;
i.e.
be a positive linear functional on the Riesz space generated
by the continuous linear functionals (see Hewitt, 1952).
p E E,
If
for
define
T : F(E, R)
F(E, R)
by
T f(x) = f(x+p)
x E E.
In the remainder of this chapter, we shall consider a fixed subgroup
G
Let
taining
GO.
of E,
a(E)
be the smallest Riesz subspace of F(E,R)
E' and closed under the operation of translation in
4/See page 45.
conG;
that
42
is, such that p E G, f E a(E) implies that T f E a(E).
space certainly exists.
T F C F,
space,
a(E) =
Such a
C FC F(E, R), F a Riesz
{F
V p E G}.
Proposition 1. a(E) contains the constants.
Proof.
w E E'
Choose a
such that
ker w. This is
G
certainly possible by the Hahn-Banach theorem. Then w(p)
some
for
Without loss of generality, we may assume that
p E G.
w (p ) = 1.
0
w E E' C a(E) => T w E a(E).
Therefore,
Tw-wEa(E).
But
Thus,
1
E a(E).
Q. E. D.
Proposition 2. a(E) = L(EI+R),
of
so that the set is independent
G.
Proof. By the previous proposition,
a(E)
for all x E E.
(T w-w)(x) = w(x+p) - w(x) = w(p) = 1
E' + R C a(E).
is a Riesz space, this implies that L(ELFR) C a(E).
n
other hand, if
On the
n
V a.
i=1
Since
1
-
.
V b. E L(El+R),
1=1
1
where
a., b. E E' + R,
1
1
then
T(
P i=1
a.1
V b.)
i=1
1
T ( V a.)
P i=1
T( V b.)
P i=1
1
= V (T pa.) - V (T b.)
E L(El+R).
p
i= 1
1=1
43
E' C L(EI+R) C F(E,R),
Thus,
T L(ELFR) C L(El+R) for all p E G.
Therefore a(E)
is a Riesz space, and
L(E1-1-R)
a(E) C L(ELI-R).
So
L(El+R).
Q. E. D.
So by the General Representation Theorem,
a(E) = { V a. - V b. : a., b. E Et + R}.
1=1
1=1
We also have the following representation theorem for
a(E),
which
is modeled after Choquet (1969, Theorem 30.2(ii)).
Theorem 6. Every f E a(E)
has an affine decomposition:
E
may be written as a finite union of finite intersections of affine half
spaces
1
such that the intersection of two distinct
is a subset of an affine hyperplane, and for each
E.
f.
1 < i < m,
Ei,
E E' + R
such that
there exists
i
f I E. = f. I E..
1
1
1
Proof. First, supposing affine representations exist for
we show it true for f - g.
f, g E a(E),
a
flE. = f.IE.
1
and
1
1
1
n
E = v F.,
j=1
3
gnd
IF. = g.! F.
J
3
E..
= E. rm F..
13
3
1
j
f.,
g.j E E' + R,
1
(m, n)
E.
where
(i,j)=(1,1) li
Then E., F.j being finite intersections of affine
1
half spaces imply the same about each E...
E.. r,
where
J
satisfy the hypothesis. Let E =
E.,
F.j
I
1
m
E =i=1
v E.,
Let
C E. r- E., F. -m F.
Since
two distinct
E..
intersect
44
in either two distinct E.1 or two distinct
so that their inter-
F.,
section is in an affine hyperplane by hypothesis. Moreover,
f - g E..
13
f. -g. 3I E..13
and
Therefore, our initial
g.3 E E' + R.
f.
1
1
assertion has been proven. So by the previous proposition, we need
only prove the theorem for
i=1
proof is by induction on
a. E E' + R.
V a. where
f
The
1
The assertion is trivial for n=1.
n.
k+1
Suppose it has been proven for
n=k
f = V a..
and
k
Let
i=1
111
g = V ai have the decomposition E =UE., gIE. = a.IE.,
j=1
i=1
a.EE' + R.
Clearly
E. = E. n
J
J
Let
f = g V ak+1.
J
3
3
E E :ak+1(x)-a.(x) > 0},
J
J
E. = E.
fx E E :a k+1 (x)-a.(x)
J
J
r
{EA:
J
Let
J
:
E+.
{E7
9/}
SO.
J
3
3
E=
Each element of
is a finite intersection of affine half spaces.
Any two distinct elements of
meet in an affine hyperplane since
E. n E. C {x E E
and if
j1
j2,
:ak+1 (x) - a..(x) = 0)
E.+ n E.+J2 E._ n E.+
,
31
Ji
cm E.
J2
,
E.- n E.-
J2
31
are all subsets of E.
,
and
32
31
+
fIE.3 = a k+1
1
E.+ n E.
_
,
31
+
3
J2
45
Q. E. D.
1E7.
33
flE7
J
Notice that this decomposition need not be necessarily unique,
since the representation of f = V a.
b. in terms of the affine
i -V
i
.
i=1
functions
a.
Definition 4.
If
p.
E
a(E)
a(E), then
Let
p.
p.
*+
1=1
is not necessarily unique.
b.,
and
E a(E)
*+
is called a positive measure.
(T f) = i(f) for all p E G
and
and
is called a positive G-invariant measure.
m(E) = a(E)*+.
Let mG(E)
fp.
E a(E)
*+
:p.(T f) = p.(f) for all p E G, for all f E a(E)}.
Remarks. Since h(E) = L(EI) C L(El+R) = a(E) and
b(E)
a(E)
every positive measure is both a coni-
B(E) C a(E),
cal measure and an affine measure.
An example of a positive measure is the
the proof of Theorem 3. Notice that
prove a representation theorem for all
Theorem 7. Suppose
I
E
I E mR(R).
I
mG(E).
E
then
tivity of
I,
Then
0 < I(f) < 0.
-M < f < +M for some M E R.
I(-M) < I(f) < I(M)
so
We shall soon
MR(R)
Proof. It is sufficient to show that I(1) = 0.
f E b(E)
presented after
I
b(E) = 0.
Because if
Then by the posi-
-MI(1) < I(f) < MI(1).
So
46
Let w E E' be as in Proposition 1. That is,
some
p E G.
Then
Corollary 1. If
I(1) = I(T w-w)
I
E
MG(E)
w(p) =
I(T w) - I(w) = 0.
for
1
Q. E. D.
is a Daniell integral. then I = 0.
Proof. This clearly follows from the preceding theorem cornbined with Theorem 3.
Corollary 2. If
Q. E. D.
I
E MG(E),
I
then
0,
cannot be
I
localized.
Proof. Recall,
is said to be localized if there exists a
I
compact set K C E and apositive Radon measure
that I(f)
pointwise then
fn
M+(K) such
E
a(E)
fn K
and
fn
monotonically converges to zero
monotonically converges to zero pointwise
uniformly by Dini's Theorem. Therefore, since Radon
fn I K \
measures are continuous for the uniform topology,
Thus,
e
for all f E a(E).
1.1.(flK)
Suppose
i.
Vfn)
(f 1K)0.
Therefore,
I
I.J.(fniK) \ 0.
would be a Daniell
integral. This is not allowed, by the previous corollary.
Q. E. D.
Remarks. Theorem 7 is actually a nonlocalizability condition.
Suppose
I
E
mG(E),
some compact set K.
and
f1, f2
Then
E a(E),
= f2
f11
I(f1) = I(f2).
continuous, and non-zero only on the compact
Because
K.
for
EK
f1
f2
is
It is therefore
47
bounded.
So
Therefore
f1 - f2 E b(E).
I(f1-f2) =
0
implies
I(f1) = 1(f2).
The previous theorem and its corollaries show that I E mG(E)
behaves differently than affine measures. The theorem states that
invariant measures are zero as affine measures. Hence no nontrivial
affine measure has an extension as a G-invariant measure for any
subgroup
G
of
E,
G
0.
Let us now examine the situation when E
space, so that E = E(X)
for some set
X.
is a complete weak
These spaces were
considered in detail in Chapter III. On these spaces the behavior of
I
is essentially different from its restriction Ilh(E) as a conical
measure. For
E = E(X),
h(E)
is Daniell (Choquet, 1969, The-
orem 38.13). Corollary 1 states that a G.-invariant extension of this
conical measure,
when
I = 0.
Ilh(E),
to all of a(E)
When cardinality
remains Daniell only
X <0, Ilh(E(X)) is localizable
(Choquet, 1969, Theorem 38.8). So Corollary 2 implies the following
Proposition 3. Let E
E(X)
where cardinality of X <
Then every positive invariant measure on E is a nonlocalizable
extension of a localizable conical measure.
Let us now return to the situation where E
convex TVS.
is any locally
Then Corollary 1 immediately shows that a non-zero
invariant measure is not an integral with respect to a o--additive
.
48
measure on a o--algebra of Baire sets in E.
Actually, the theorem
implies that a non-zero invariant measure cannot have an integral in
any of the usual ways--Lebesgue Theory or Riemann Theorywith
respect to even a finitely additive measure on an algebra of subsets
For if
E.
of
E,
in
is generated by a set function
I
and A
then
E
on an algebra
0 < I(xA) < I(1) = 0,
is the characteristic function of
A
p.
Thus
A.
where
XA
for all
p.(A) = 0
(see Hewitt, 1952).
E
The following is a more constructive proof of Corollary 1. Let
p E G,
p
by71P)
(X
Let
0.
F = Rp = {X.p
E F',
X
Banach theorem to a map
Tr
b
a
1
[
b-a (TT-a) A 1] V O.
,
X
E
and consider
R}
Tr1
:F
R
so it may be extended by the HahnTr
E
E'.
Clearly
For a < b
a
E
in
R,
let
a(ir).
for y in the closed half space
E E : Tr(y) < a} = Tr-1(-00, a]
for y in the closed half space
Tra(Y) =
E :Tr(y) > b} =
{y
71(y)-a
b-a
for
,
yE
Tr
-1
[a, b]
We shall construct fn a(E)+ such that
I(fn) = 0 for all n N. Let fn = Tr1-n
ingsince It -n < Tr -n-1
E
E
(+)
[b,+)
fn
.
/ 1 pointwise but
Then
fn
are increas-
49
For if Tr(y) < -n-1
then
0 < Tr1-n(y)< Tr1-(n+1)(y)
-n
-(n+1)
by (+)
0
-n-1 < Tr(y) < -n then
If
0=
1-n
-n
(y)
<
Tr1-(n+1)(y)
-(n+1)
Tr(y) + n =
-n+1 < Tr(y)
1
fn
-=
=
Tr
Tr
then
1-n
(y) < Tr1-(n+1)(y) = 1.
-n
-(n+1)
is, indeed, increasing.
when n> 1-x;
that is, when
so that
Tpfn = fn+1
1
1-n
= 1.
Tr(y)+n1-(n+1)(y)
-n (y) < -(n+1)
1
If
Tr(y)+n+1
=
then
-n < Tr(y) < -n+1
If
Thus,
so
11-(y) < -n
T
f
mp 1
=f
Let x E E.
1-n < x,
m+1
.
so
Then
fn
1.
fn
(x)
1
Also,
Thus I(fm) = I(Tmpfl) = 1(f1)
1
for all m E N.
Now let
gn = Trnn+1 = T-2npfn.
g. are decreasing and gn\60.
I(gn) \ 0.
But each
If
We can similarly show that the
I
were Daniell then
I(gn) = I(T_2npfn) = I(f1).
I(gn)\ 0 => I(f1) = 0 ==> I(1) = 0.
So
We now see how an invariant meas-
ure depends upon its restriction as a conical measure.
50
Theorem 8. Suppose
Proof. Since
I
mE(E)
E
is a Riesz space,
a(E)
So it is sufficient to show that
then I = 0.
II h(E) = 0
and
a(E)
a(E)
- a(E)
in order to conclude that
II a(E)+ = 0
I = 0.
Let
f
E
f = V a. - V b. where a., b.
a(E)+,
1
i=1
1
1
i=1
E
1
E' + R.
Always in a Riesz space
n
0<
n
g V aivV -a. = V (ai V -ai) = V I a.!.
V ai
i=1
i=1
i=1
i=1
1=1
So
o
< f < If! < IV ail +
+ V Ibi
I V bi I = V I ai I
i= 1
i= 1
i= 1
i= 1
Therefore,
0 <1(f) <
+I(Ibil)
a.
i=1
If=we
show that each
a.,
1
p. E
1
h(E)
and
Ia.!
x.,
y. E E,
1
1
Tx. a,
Ibi
=T
1
p.
Y..1
1
where
1
then the theorem will have been
proven. Because
0 < I(f) <T a.)
+ I(T
x.
i=1
1
+ I(pi) = 0
=
Yi
.
1
i=1
the last equality holding by the hypothesis on
I.
Without loss of
51
generality we shall show this only for al, assuming that al is
not constant. (For I is zero on constants. ) Consider the affine
hyperplane
3 ince
{x E E :
E1
Txal E E'.
0,
(x)
Tx(a1)(0) 1'
Choose some
0}.
a1(x)
Tx(a1)1 = Tx(a1) V - (Tx(al) E h(E).
-x,
x.
with
!Txal 1,
Theorem 9
Let
SO
But
ITx(a1)1 =
Q. E. D.
1
be a non-zero subgroup of
G
i
f
<x
E a(R)
< xn < +00
0<i <n
and
f
x1) =
0
1.
- 00
f.
, x.
I
xi]
1+1
where
[xi, xi
+1]
for
and
(p, q) E R+ X R+.
q) :a(E)
We will show that
0<
R+.
u°).
Define
<v
X
0.
i
1
fl[
[xn +
p
and functions
such that
1
n-1
.
f
2
b.,
Let
-.(z)
and R+
if and only if there exist real numbers
.
pose
Then
Recall that
at
1
mG(R)
R.
p E G.
1,
=
So
Proof. There exists
x
1
Txiall.
Tx al
al
there is a one to one correspondence between
1.(t)
x E El; then
R
p.( p, q)
by
aop + anq.
1.1.(p,q)(f)
is a well-defined func tion. For sup
has a second affine decomposition:
<y
<
. . .
<m, where
< ym < +co,
functions
g.
where gi(t) = c.t+d.,
J
J
52
1 <j <m-1
gj [Yj' Yj+11 = f [Yj' Yj+11
go
y1
= f (-00, y
gm [Ym, +3°) = f [Ym, +3*
Then
Therefore
YlA xi] = f0 ("c'
YlA xi] - go I
fl
c0
Y1
Axl
= a0.
Also,
gm I [YmV xn, +00) = fn ( -00, ymV xn]
fl [ymV xn, +00)
I
Therefore dm = an.
So
a0P
ancl
Clearly
cOP
F.s.(p, q) E
ant. Also, the map
and thus
dmq'
mG(R)
i : R+
q)(f) =(p, q)(f) for all
X
is well defined.
since it is positive, linear and invari-
R+
fE
1.1.(p, q)
mG(R)
mG(R).
is injective. For suppose
Let
x- ,x+:R
t >0
x+(t)
=
0,
t <0
t>0
t <0
Then
q = p.(p, q)(x
and
)
=
q)(x) = q
R by
53
=(p,
q)(x-
q)(x) = p.
)=
Thus, we need only show that
IE
with
p = I(x)
Let
mG(R).
-00 < X
<X
<
1
Fl
q
1
,
f-F
as before. Let F : R
1
R by
Tmp(a0x- )1 (-00, mp]
+00) = T
sp
+
(a
,
)1 [SP' +)
mp <
such that
m, s E Z
E a(R),
Suppose
=0
mp,Xsp
F [x sp
I(x+).
< +00, a., b.
n
(-00, mp] =
F I [X
where
and
is surjective . Let
[.1
< xn < sp.
Clearly, F E a(R).
(-00, mp] = bo + aomp
f-
is piecewise linear.
sp]
f - F I [sp, +00) = bn + asp
f - F, continuous on the compact set [mp, sp] is bounded there, so
f
F E b(R).
Therefore
by Theorem 13.
I(f-F) = 0
So
I(f) =
But
I(F)
I(T
= I(T
=
nrip
(a0 x) + T sp ( a n x+)
mp(a0x)) + I(T sp (an x+))
I(Tni(a
P
°
x-)) +
= I(a0x ) + I(a
= a0I(x ) + a
a0P
)
anci°
n
I(Ts(a x+))
p
n
x+)
I(x+)
Q. E. D.
(F).
54
We see from the theorem, that the positive cone of R- invariant
measures on
(p,q)
R
has two generators. Each
p(p.(1,0)) 4- q(p.(0, )).
So the set
generates
{p.(1,0), p.(0.1)))
the cone and is obviously independent.
Also from the Theorem, we see that y.(x, -x)
generates a
cone of invariants in m(R) - m(R) which vanish on
E + R.
This cone of measures illustrates the fact that the kernal of an
invariant measure need not be a lattice.
Theorem 10.
space,
p 1 0,
mG(E)
Let
E
pE G C Rp C E.
if and only if E
2
be a locally convex topological vector
Then the (linear) dimension of
is toplinearly isomorphic to
In general, the linear dimension of
R;
mG(E) > twice the linear
dimension of
Proof. By the previous theorem, it is sufficient to prove the
last statement..
(p, GI)
X
E
R+.
Let
I
be a Hamel basis for
Define
p.(x;p,q)(f)
where
l(p, q)
p.(x; p, q) E m(E)
p.(p, q)(fl Rx)
E.
Let
x E I,
by
for all
f
E a(E),
is the invariant measure of the last theorem. It is
clear that
{p(x; 0, 1) : x E
{p.(x; 1,0) : x E
55
q)E m Rx (E). We shall
see by Theorem 15 in the next section that this is enough to insure
are linearly independent.
that
Also,
y.(x; p,
Modulo this result, the theorem is proved.
1.1.(x;p, q) E mG(E).
Q. E. D.
We shall now examine the existence of F-invariant measures
where
m
E
is any finite dimensional subspace of E.
F
is n-dimensional, we may consider
F
and
(Rn)
If
p.
as
Rn
mF(F)
in
and define
J.
E m (E) by
(fl F).
11(f)
Therefore, it is sufficient to consider m
(Rn).
Rn
We sketch the construction of OS; p) c m n(Rn) where
(e
S
is an ordered basis for
, e)
n
,
1
Let
f E a(E).
f.
1
1
Let
1
1
t1
E
R
and
p E (R+)1.1.
f has an affine decomposition E
E.,
E E' + R.
R+
be defined by
n
ti
'
i=1
where
t1 =
inf {t >0:E. n
E. (m
t1
(te1 +Re2
+ ... +Ren)
(se1+Re2+ ... +Ren)
is well defined, since each E.
ci
projects onto
implies
for all s > t}.
Rel
as an
56
interval. Having defined
t.
1 < j < n,
for some
n
.
v tj
t+1 =
define
1=1
,
+j 1
where
+. .+Re)
(tlel+ ...+t.e.+te.
3+1
n
33
tij+1 = inf{t > 0 :E.1
implies that E.
(ti el+ . . . +t .e .+ se.3+1 +. . . +Ren)
33
for all s > t}
Thus we may define inductively
Considering
t1, t2,
fi t1e1+
en_
, tn.
+
Ren
as an element of
it is clear from the construction of tn that f is affine on
n-ia(R),
That is
[tn, +00).
fit e +... +tn-1 en-1
= fj itlel+
for some affine
affine on
R11
+tn- len- 1
+00)en
f.
continuous
implies that
a, b E Rn
Then define
+ [t,
n
in the decomposition of f.
f.
f.(x) = (x, a) + b
where
+ [t,n +00)en
and
(
for x
in
R
is e.g., the usual inner product.
14S; p)(f) = (p, a).
We claim that
p. = 1.1.(S;p) E m
(R11).
Rn
Let
57
x=
+
xlel +
xnen R11.
It is easy to verify that
E
For here the value of
p.(T
f)
x.e.
1 1
for
p(f)
,n.
i = 1,
depends only upon the behavior of
p.(f)
and our construction reduces to that shown in Theorem 9.
fl Re.,
So
p.(Tx
1
+x e f) = p.(Tx
e 1+
nn
Fi(T
1
e (Tx
1
2e2+
+x e f))
nn
f)
xe+
22 ...+xen
n
p.(Tx e
f) = p(f).
nn
The other properties that
p.
must have, its well-definedness,
linearity, may all be verified as in Theorem 9.
0(E) and Other Extensions
In the remainder of this chapter we shall show that every invariant measure has a unique invariant extension to a subspace of
F(E, R)
Let
which contains
0(E)
a(E)
properly.
be the vector space
a(E) + B(E) where
the subalgebra of all continuous bounded functions in
Theorem 11.
Proof. Since
B(E)
is
F(E,R).
0(E) is a Riesz space.
0,(E)
is a vector space, it is sufficient to show
58
that
f1
in
implies that
f+ = f V 0 E 01(E).
f- = (-f) V 0 E q(E)
Then
f = h + b,
0,(E)
so that
Ifl E 01(E).
Letting
h E a(E), b E B(E),
f VU
(h+b) V 0 = (h V -b) + b.
So it is sufficient to show that h V -b E 0,(E)
Since
m < -b < M.
there exists
-b E B(E),
m, M E R
such that
Then hVm < hV (-b) < V M Now
{-M-m,
h(x) <m <M
M-h,
m <_ h(x) <M
0,
m < M < h(x)
[hV M - h V m](x) =
But when m < h(x) < M,
..._
0 < M - h(x) < M-m.
So that always
0 <hVM- hVm <M-m.
h V (-b) = h V m + [h V (-b)-hv m],
with hVm E a(E)
and
0 < h V (-b) - hV m < M-m so that
(h V -b) - (h V m) E b(E).
Thus
Theorem 12. Every
G-invariant functional
I
E
MG(E)
I E 0,(E)
Proof. Let f E o,(E).
h v (-b) E 0,(E).
Q. E. D.
has a unique extension to a
*+
So
.
f = h + b where
h E a(E)
and
b E B(E).
Define
T(f) = I(h).
We shall show that T is well defined.
59
Suppose
is also equal to h' + b' where h' E a(E), b' E B(E).
f
Then
h-
=b-
E B(E).
So m < h - h' < M for some constants m, M.
0 = I(m) < I(h) - I(h') < I(M) = 0,
so
But
Thus,
I(h) = I(h') = T(f).
I
is well defined.
It is clearly linear.
Suppose f=h+bE
b' E B(E)
We shall find h'
01(E)+.
Then, since
f = h' + b'.
such that
a(E)+,
I E a(E)
);(4.
,
= I(h') >0.
Since
for some M E R.
h + b > 0, h > -b > -M
Then clearly,
b' = b - M.
So let h' = h + M,
h' E a(E)+, b' E B(E)
Thus, we have shown that 7 E
f = h' + b'.
and
01(E)*+.
To conclude the proof, we need only show the uniqueness. Suppose
12
are G-invariant functionals in
Illa(E) = 121 a(E).
01(E)
We shall show that1 = I2
*+
and
We can use the
proof of Theorem 13 to conclude that
I1
Let f=h+bE 01(E),
1B(E) =
I2
1 B(E) = 0
h E a(E), b E B(E).
11(f) = I1(h+b) = 11(h) + 11(b)
11(h)
I(h) =
(h)=
I2
Then
I2(h) + I2(b)= I2
(h+b)= I2(f).
Q. E. D.
60
We may employ the argument used in the proof of Theorem 11
to show that
h(E)
for
h(E) + b(E) is a Riesz space. For if we substitute
a(E)
and
mutatis mutandis.
for
b(E)
the proof goes through,
B(E),
The sum of two Riesz spaces is not, in general,
a Riesz space; however, it is so when one of the Riesz spaces consists of bounded functions.
Theorem 13. a(E) = h(E)
Proof. E' C h(E),
b(E).
R C b(E)
so
E' + R C h(E) + b(E) C a(E)
L(El+R) C L(h(E)+b(E)) C L(a(E))
But since
h(E) + b(E)
and
are Riesz spaces,
a(E)
L(E'+R) C h(E) + b(E) C a(E) = L(El+R)
So
a(E)
h(E) + b(E).
Let
X E h(E)
Then x
b(E)
agrees with some elements of E'
through the origin of E.
But
x
on polyhedral cones
bounded implies x
agrees with
the zero functional. Thus
h(E) rm b(E)
0.
Q. E. D.
61
d E E,
Theorem 14. If f E a(E) and
then
Tdf - f E b(E).
n
n
Proof. It is sufficient to show
f E a(E)
n
Tdf - f = (Td (\/v
i=1
f
n
implies
= V h.,
= i=1
V h.i
f
h. E
i=1
i
(E)
Because if that is the case, then
n
k''
-V
i=1
with
1
n
h.)i - \ V/ h.) 1
i=1
i
i=1
1<
_i _< n.
h.I E E' + R,
for
h.) - V
thatbTd(V
(Td
h., k.
1
n
n
i=1
i=1
E E' +R and then
1
\/v k. i- \ /vk.)i E b(E).
So let
h. E Et + R.
i=1
x E E.
Let
Suppose
for all
ha.(x) >h.(x)
Td hP
(x) >
Td
h.(x)
i
for all
(A)
a.
i
p
(B).
ha(x) - TdhPoo = ha(x) - h (x) - h (d) + h(0)
This expression is greater than or equal to
by (A),
ha(x) - hp(x) > 0.
On the other hand, the expression is less
than or equal to ha(0) - ha(d)
Tdhp(x) - Tdh a(x) > 0
-h (d) + h(0) because,
because, by (B),
so that
h(x)
+ h(d)
- h (0)
< h (x) + h (d) - h (0).
a
a
a
Thus
h(0) - h (d) < h (x) P
-a
Td hp
(x) < h (0) - ha(d).
Therefore,
A h.(0) - h.(d)
<
1
i=1
1
V
h. -T,u i=1
V h.
i=1
1
<V
h.(0) - hi(d).
i=i
62
So
T f - fE a(E)
b(E).
B(E)
Theorem 15. Let
p.
Q. E . D .
E m(E).
Then the following statements
are equivalent:
p.
P.
for some subgroup
E mG(E )
G
of E,
0.
G
e mE (E
p.lb(E) = 0.
p.(1) = 0.
Proof. (b) =-> (a) and (c) => (d) are trivial.
(c) and
(d)
(a) => (c) have been demonstrated in Theorem 7.
But (c) => (b) follows from Theorem 14.
Q. E. D.
Thus any G-invariant measure is automatically E-invariant,
and so we may refer to such measures simply as 'invariant measures.'
Theorem 16. Every conical measure has a unique extension to
an invariant measure.
Proof.
I(h4-b)
Oh)
Let
p.
E h(E)
.
Define
I: a(E)
for all h E h(E), b E b(E).
well defined on
Clearly
a(E).
that by Theorem 15,
I
I I h(E)
R
by
By Theorem 13.
p.
and
is an invariant measure.
II b(E) = 0
is
I
so
0. E. D.
Notice that this yields an alternative proof to Theorem 8. Combining the results of Theorem 12 with Theorem 15 we have
63
Theorem 17. Let
p.
E 0,(E)
*+
Then the following statements
.
are equivalent:
p.
J.
is G-invariant on
01(E).
E mE(E).
p.lb(E) = 0.
PdB(E) =
p.(1) = 0.
Theorem 16 shows that all conical measures can become trans-
lation invariant at very little extra cost. That extra cost is spelled
out in the remarks following Theorem 7.
We close with some observations on the extension problem. We
have extended invariant measures to positive linear functionals on
01(E).
Can we extend them much further? Theorem 2 immediately
implies that no nontrivial invariant measure may be extended to a
positive linear functional on all of C(E).
(This, by the way, shows
us that the results of Choquet (1967) and Schwarz (1964) are not
applicable to the study of invariant measures. )
Let
0(E) = {f E C(E)+:f < g for some g E cp,(E)} where
C(E) C F(E,R) is the Riesz space of all continuous functions. Let
0(E) = 0(E) - 0(E).
For
E = R,
continuous functions which are
x.
0(E) corresponds to the set of
0(x),
that is, which are 'big oh of
The extension theorem of Choquet (1969, p. 289) shows that each
invariant measure has an extension to a positive linear functional on
64
0(E). However such an extension need not be either unique or invariant, contrasting the situation in Theorem 12. We do have the following
Proposition 4. Let
tive linear functional
h > f2.
Then
h
mEE.
Suppose
I
extends to a posi-
on a vector space M where
I
a(E) C MC F(E,R).
IE
Then if f E M, T(f) >0
and
h E F(E, R),
M.
Proof. Assume that
h E M.
0 < f2 - 2nf + n2
SO
2nf<f2 +n2 <h+n2
2f < n -1 h + n
SO
0 < 2I(f) < n-11(h) + I(n)
n1
I(h)
for all n E N.
This cannot happen.
Let
V
Q. E. D.
{M:M is a subspace of F(E,R), a(E) C M,
and
every I E mEE has a positive linear extension
onto
V
is clearly a non-empty poset under set inclusion. It is easy to
65
verify that
is inductively ordered. Therefore, by Zorn's
V
lemma we can find a maximal element
M of
The following
V.
two facts may be verified.
f E M and
If
0 < g < f imply g E M.
F(E,R)+,
gE
0< fn
< g,
f
n
then
g
M if and only if there exists
+00
E M such that I(f)7
n
for some
I E mE(E).
Integral Representation on Weak Spaces
In this section, we consider operations on invariant measures,
other than convex combinations.
Let
E
linear map. Let
p,
E mE(E).
It is clear that
q(p.)(f) = 1.1.(f q).
q(p.)(1)
be LCS's, and let q:E
F
and
p.(loq) = p.(1) = 0.
ment of mF(F).
Define
qE
F be a continuous
R by
q(p.):a(F)
a(F)+
.
However,
Thus, by Theorem 15,
is an ele-
q(p.)
We see that the projection of an invariant measure
is an invariant measure.
Let
define
A be a convex cone in an LCS E.
p.A:a(E)
R
f E a(E)+.
For
p.
E mE(E)
we
by
p.A(f) = inflp.(co):
for
For
f E a(E)
E a(E), (p > f on A)
let p.A(f) = p.A (f+) - p.A(f
).
That
66
follows from the Riesz decomposition
is an element of a(E)
as well as from Proposition 11.2.1 of
property applied to a(E)
See Proposition 30.8 of Choquet (1965) in
Bourbaki (1965, p. 26).
which he applies the Riesz decomposition property to h(E)
in order
to prove a very similar proposition.
We say that
f
on X
0
then
X,
Notice that if
p.(f) = 0.
f> 0
then if f E a(E),
on
is carried by X provided if f E a(E)
p.
on X then
.i
p.(f) >0.
and
is carried by
X,
Because if f> 0
then
0=
1-1-(f+-f)
= 11(f)- 1-1.(f),
So that
0 < p.(f+)
Clearly,
p.
carried by A.
mE(E).
A
< p..
=
Also,
.
pA
< p.
We can now show that
if and only if
p.A
p.
is
is an element of
For
0 < p. (1) < p.(1) = 0.
A
Thus, the restriction of an invariant measure to a convex cone re-.
mains invariant.
Let
P
P(E) = {p.(x; 1,0) : x E E},
from the proof of Theorem 10 that
in
a(E).
where it will be recalled
p.(x; 1,0)(f) = p.(1,0)(f Rx)
We shall always assume that P
for
carries the initial
67
topology defined by the set of maps
f(i)= On- Notice that
by
f :P
f E a(E)} where
{f
p.(x; 1,0)
F1(x;1.0)Rx
since
R
p.(x; 1,0)
is carried by Rx.
Proposition 5. Let E be an LCS. Then for every compact
K C E,
p:K
positive Radon measure
P
v
on
K,
and continuous map
the positive linear functional
p(x)dv(x)
=
xEK
defined by
p(x)(f)dv(x)
p.(f) =
is an invariant measure on
Proof
x
p(x)(f)
E a(E)
fl
4.*
Clearly
E.
f op is continuous on
is within the domain of v.
p(x)(f)dv(x) =
p(x) E
so that
Also, it is clear that
by the positive linearity of each
is also clear since each
K,
p(x).
The invariance of
mE(E):
p(x)(T f)dv(x).
Q. E. D.
The converse of this proposition holds for certain weak spaces.
68
Theorem 18. Let E be a complete weak space with
lc(E) <
p.
Then there exists a compact
E mE(E).
a positive Radon measure
K C E,
p:K
and let
0'
on
v
and a continuous map
K,
such that
P(E)
p(x)dv(x).
p. =
x EK
Proof.
b E b(E).
and
h(E)
Since
Radon measure
h(E) by H(h+b) = h where h E h(E)
H:a(E)
Let
By Theorem 13, this projection is well-defined.
is localizable, there exists a compact K CE and a
v
on K such that
h(x)dv(x)
p.(h) =
for all
h
For each x
in h(E).
measure evaluation-at-x,
p(x)
ex0H
for
x
For let f = h + b.
let ex be the conical
for all h in h(E). We
K
ex(h) = h(x)
claim that ex0H = p.(x; 1,0)
by
E
in
and that the map p:k
K
Then
is eventually constant along
equals the slope of
h
defined
is continuous.
p,(1,0)(h Rx)
p.(x; 1,0)(h+b) = p.(1,0)(h+blRx)
since b
P(E)
R+x.
on the ray R+x,
But
p.(1,0)(hlRx)
which equals
h(x),
by
69
identification of Rx with R
in the construction of p.(p, q).
Therefore ex° H = 1.1.(x; 1, 0). To show that p is continuous, we
need only show that for all f in a(E) the map f p:K R is
continuous, where f p(x) = h(x)
since all of
h(E)
for
f = h + b.
But this is clear
are continuous functions.
But then
H(f)(x)dv(x) = p.(H(f)) = p.(f)
P(x)(f)dv(x) =
Lcx
x EK
so that
p(x)dv(x)
Q. E. D.
This theorem clarifies Proposition 3 and its preceding remarks.
70
BIBLIOGRAPHY
The Lebesgue integral in abstract spaces.
In: Theory of the integral, by S. Saks. 2d rev. ed. New
York, Dover, 1964. p. 320-330.
Banach, Stefan.
1937.
Birkhoff, Garrett. 1966. What can lattices do for you? In: Trends
in lattice theory, ed. by J. C. Abbott. New York, Van Nostrand
Reinhold, 1970. p. 1-40.
Bochner, S. 1939. Additive set functions on groups. Annals of
Mathematics 40:769-799.
Bony, Jean-Michel and Philippe Courrege. 1964. Mesures sur les
espaces vectoriels topologiques faibles, mesures de Radon et
compactifies affines. Comptes Rendus des Seances de
l'Academie des Sciences 259:3158-3161.
Bourbaki, N. 1965. Integration. 2d rev. ed. Paris, Hermann,
282 p. (Elements de mathematique, bk. 6, Actualites
Scientifique et Industrielles no. 1175)
Choquet, Gustave. 1962. ktude des mesures coniques. Canes con-
vexes saillants faiblement complets sans an6ratrices
extremales. Comptes Rendus des Seances de l'Academie des
Sciences 255:445-447.
1967. Cardinaux 2-mesurables et canes faiblement
complets. Annales de l'Institut Fourier 17:383-393.
1969.
Lectures on analysis. New York, Benjamin,
3 vols.
Daniell, P. J. 1917-8. A general form of integral. Annals of
Mathematics, ser. 2, 19:279-294.
1918.
Integrals in an infinite number of dimensions,
ibid., 20:281-288.
1919-1920. Further properties of the general
integral, ibid., 21:203-220.
71
Glicksberg, Irving. 1952. The representation of functionals by
integrals. Duke Mathematical Journal 19:253-261.
Godel, Kurt. 1964. What is Cantor's continuum problem? In:
Philosophy of mathematics, ed. by Paul Benacerraf and Hilary
Putnam, New Jersey, Prentice-Hall, p. 258-273.
Hewitt, Edwin. 1950. Linear functionals on spaces of continuous
functions. Fund amenta Mathematicae 37:161-189.
Integral representation of certain linear
functionals. Arkiv For Matematik 2:269-282.
1952.
Horvath, John. 1966. Topological vector spaces and distributions.
Reading, Addison-Wesley, 449 p.
Schaefer, Helmut. 1966. Topological vector spaces. New York,
Macmillan, 294 p.
Schwarz, Lanrent. 1964-5. Les mesures de Radon dans les espaces
topologiques arbitraires. Cours de Monsieur le Prof.
Schwarz. Paris, 19 numb. leaves.
Tolstov, G.P. 1962. An abstract integral investigated by S. Banach.
Matematieheskii Sbornik, new ser. , 57:319-322. (Translated
in American Mathematical Society Translations, 2d ser. , vol.
80, 1965. p. 151-155)
1966. Differentiation and integration in abstract
spaces. Ibid. , 71:420-422. (Translated in American Mathematical Society Translations, 2d ser. , vol. 80, 1969. p. 175179)
Umemura, Yasuo. 1965. Measures on infinite dimensional spaces,
Publications of the Research Institute for Mathematical
Sciences of Kyoto University, ser. A, vol. 1, no. 1, p. 1-47.
APPENDIX
72
NOTATIONAL INDEX
Terms or symbols
Hints or definition
Page of
initial
appearance
the field of real numbers
6
TVS
topological vector space
6
E'
topological dual
6
algebraic dual
6
*
toplinear isomorphism
7
absorbs
7
balanced
7
bounded
7
gA
LCS
EA
gauge of A
7
locally convex Hausdorff TVS
7
(E, gA)
7
locally bounded
7
o-(E, F)
7
weak space
cr(E, E')
7
E(X)
(RX, cr(RX, R(X)))
8
po set
partially ordered set
8
uVv
sup (u, v)
8
uAv
inf (u, v)
9
semilattice
9
73
Terms or symbols
Hints or definition
Page of
initial
appearance
co-initial
9
left filtering
9
is absorbed by
.a.
.e.
10
10
in Bi.e.
11
lc (B)
12
lc(E)
13
B
S, n
24
(f)
open sets
25
bounded sets
25
26
(P)
C-set
x+xC
26
C- space
26
n(x)
28
33
33
F(E, R)
33
B(E)
33
h(E)
33
conical measures
b(E)
h(E)
33
33
"74
Terms or symbols
affine measures
Hints or definition
b(E)
*+
Page of
initial
appearance
33
E' + R
33
L(S)
34
Daniell integral
38
translation by p
41
a(E)
41
affine decomposition
43
positive measure
a(E)
*+
45
45
m(E)
45
mG(E)
localization
46
Op, q)
51
p.(x; p, q)
54
55
(S; p)
01(E)
a(E) + B(E)
57
C(E)
continuous functions
65
0(E)
big oh of E
65
q(i)
projection of p.
65
restriction to A
65
A
carried
66
75
Terms or symbols
P(E)
f
Hints or definition
{p.(x; 1,0) : x
E
E}
Page of
initial
appearance
66
67