Lecture Notes in Mathematics Editors: J.-M. Morel, Cachan F. Takens, Groningen B. Teissier, Paris
1757
3 Berlin Heidelberg New York Barcelona Hong Kong London Milan Paris Singapore Tokyo
Robert R. Phelps
Lectures on Choquet's Theorem Second Edition
123
Author Robert R. Phelps Department of Mathematics Box 354350 University of Washington Seattle WA 98195, USA E-mail:
[email protected] Cataloging-in-Publication Data applied for
The first edition was published by Van Nostrand, Princeton, N.J. in 1966 Mathematics Subject Classification (2000): 46XX ISSN 0075-8434 ISBN 3-540-41834-2 Springer-Verlag Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer-Verlag. Violations are liable for prosecution under the German Copyright Law. Springer-Verlag Berlin Heidelberg New York a member of BertelsmannSpringer Science+Business Media GmbH http://www.springer.de © Springer-Verlag Berlin Heidelberg 2001 Printed in Germany Typesetting: Camera-ready TEX output by the authors SPIN: 10759944 41/3142-543210 - Printed on acid-free paper
Preface
First
to
Edition
notes expanded version of mimeographed for a seminar 1963, at the prepared originally during Spring Quarter, of Washington. with to be read by anyone They are designed University of theorem the the Krein-Milman and Riesz a knowledge representation and measure theorem analysis theory implicit (along with the functional of these theorems). The only major theorem in an understanding which is is in of the measures" Section used without one on "disintegration proof
These
notes
are
revised
a
and
15.
inor helped, directly of these He has especially in the preparation benefited notes. directly, from the Walker-Ames of Washington lectures in the at the University G. of Professor from the and the at summer same Choquet, 1964, by stay P. A. Meyer. He has received institution during 1963 by Professor helpful comments from many of his colleagues, Professors N. as well as from who used the earlier in a seminar Rothman and A. Peressini, version at of Illinois. the University Professor he wishes J. Feldto thank Finally, of the unpublished the inclusion in Section material man for permitting and ergodic 12 on invariant measures. A note to the reader: of the theory are Although the applications the for needed interspersed they are never subsequent notes, throughout material. Thus, Sections 2, 5, 7, 9 or 12, for instance, may be put aside for later without reading (To omit them encountering any difficulties. off from its many and interesting however, would cut the subject entirely,
author
The
with
connections
indebted
is
other
parts
to
of
many
people
who
mathematics.)
R. R. P.
Seattle, March,
Washington 1965
Preface to Second Edition
delightful Belgian canal trip during a break from a Mons University conference in the summer of 1997, Ward Henson suggested that I make available a LaTeX version of this monograph, which was originally published by Van Nostrand in 1966 and has long been Ms. Mary Sheetz in the University of Washington out of print. Mathematics Department office expertly and quickly carried out the difficult job of turning the original text into a LaTeX file, providing the foundation for this somewhat revised and expanded version. I am delighted that it is being published by Springer-Verlag. On
a
Since 1966 there has been
a
great deal of research related
to
Choquet's theorem, and there was considerable temptation to init, easily doubling the size of the original volume. I decided against doing so for two reasons. First, there exist readable treatments of most of this newer material. Second, the feedback I clude much of
have received
the years has indicated that the small size of the first edition made it an easily accessible introduction to the subject, suitable for first
closely
one-term seminar
original text, but
merely suggestions and
have received
some newer
some
other
in the
results which more
summarized in the final section. It also
number of
in this
(of the type which generated it
This edition does include
related to the
terial is a
a
place).
over
recent
are ma-
incorporates
corrections to the first edition which I
the years. I thank all those who have helped me especially Robert Burckel and Christian Skau (who
over
regard, have surely forgotten the letters they sent me in the 70's) as well as my colleague Isaac Namioka. Of course, I'm the one responsible for any new errors. I am grateful to Elaine Phelps, who tolerated my preoccupation with this task (during both editions); her support made the work easier.
R. R. P.
Seattle, Washington December, 2000
Contents
Section
Page Preface
.
I
Introduction.
2
Application
gral
.
.
.
.
.
.
.
.
.
.
.
.
.
The Krein-Milman
theorem
representation of
the
4 5
6 7 8 9
10 11 12
13
.
.
.
.
.
.
Krein-Milman
as
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
to
.
.
.
.
.
theorems 15
Orderings
16
Additional
.
.
of
Index
.
.
.
.
.
.
.
.
.
.
.
.
.
.
dilations
and
Topics
References Index
.
.
.
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.
.
symbols .
.
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.
of .
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measures .
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9
com.
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13
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17
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25
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27
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35
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39
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47
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51
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65
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73
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79
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88
93
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1
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.
v
.
.
.
.
inte-
an
theorem
.
.
monotonic functions pletely theorem: The metrizable case Choquet's The Choquet-Bishop-de Leeuw existence theorem and Haydon's to Rainwater's theorems Applications A new setting: The Choquet boundary of the Choquet Applications boundary to resolvents The Choquet boundary for uniform algebras The Choquet boundary and approximation theory of representing measures Uniqueness of the resultant Properties map and ergodic to invariant measures Application A method for extending the representation theorems: Caps A different method for extending the representation .
14
.
.
theorem
.
3
.
.
.
.
.
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.
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.
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.
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.
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.
.
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.
101
.
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.
.
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.
.
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.
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.
115
.
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.
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.
.
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122,
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.
123
Introduction.
1
The Krein-Milman
representation
simplest
The
on
If space
X1i x
a
compact
E, and if
of
bination
x
is
of
finite-
a
gral
of X, then x is of X. Thus, there
points
i.e.,
-y is the Borel measure which equals of X which contains y, and equals 0 otherwise. let
Ej,
p
--
p(X)
and on
a
finite
exist
E,
what
we
EpjEj; regular 1. Furthermore, for any have f (x) (Epif (xi) =) then
p is
=
-
when
we mean
say that
we
Borel
a
p
I
on
=
fx
I such that
f dp.
"point
represents
mass" subset
functional
by 0, f
assertion
is
on
linear This
"inte-
x as an
any Borel
measure
com-
points
extreme
Abbreviating
continuous
vector
convex
and positive numbers p,.... with Ek/t, ) Xk i Pk 1 Wenow reformulate this representation of Epixi. For of X let Ey be the representation." any point y
at y,
integral
dimensional
element
an
extreme
subset
convex
...
=
an
7).
page
X is
as
example of a theorem of the type with which we will is the following classical result of Minkowski (see the
be concerned exercise
theorem
theorem
last
X,
.5,,, >
p
x.
Suppose that X is a nonempty compact subset of a lomeasure on X. cally space E, and that p is a probability (That is, M is a nonnegative regular Borel measure on X, with I-t(X) 1.) A point x in E is said to be represented by p if f (x) fX f dp for every continuous linear functional f on E. (We will sometimes write IL(f) for fX f d[t, when no confusion can result.) (Other ter"x is the "x is the resultant minology: barycenter of p, of p. ") DEFINITION. convex
=
=
"
The restriction existence
this
Later, Borel
of
sufficiently
guarantees we
that
will
measures
that want
suffice
E be
locally
to
consider
for
R.R. Phelps: LNM 1757, pp. 1 - 8, 2001 © Springer-Verlag Berlin Heidelberg 2001
in one
measures
the present.
is
convex
many functionals there is at most
simply
E* to
point on
to
insure
separate
represented other a-rings,
the
points; by p. but
the
Lectures
2
that
Note
for
each
fact
convex
X of
subset
finite
a
"supported"
If /-t compact Hau8dorff that /-t is supported
DEFINITION.
is
regular
nonnegative
a
X and S is
space
dimensional
by S if [t(X
\ S)
measure
Borel
Borel
a
Theorem
represented by Ex; the the above example by
out
X may be represented by a probability the extreme points of X. by
in
x
compact
a
trivially brought
X is
in
x
(and important)
interesting is that
point
any
Choquet's
on
space,
which
measure
of X,
subset
is
on
the
we
say
0.
=
problems which concern us: If X is a compact convex sub-set convex of a locally space E, and x i's does there exist element measure a an probability of X, /-t on X which is supported by the extreme points of X and which represents is it unique? x? If /-t exists, Choquet [17] has shown that, under the first that X be metrizable, the additional question hypothesis We may
has
affirmative
an
[9]
de Leeuw
In
of
then
additional the
above
Y be
Let
of all
Leeuw a
real-valued
I
JIL11.
=
E
space
convex
theorem
asserts
probability C(Y). By
C(Y)*
that
to
[t
a
on
than
is affirmative
integral seems
in
place
worthwhile the-
representation
language which we is quite natintegrals theorems of Choquet
the of
use
(in
theorem.
[28, (evaluation
points
consider vanish
/-t on
subset
topology)
of the and the
locally Riesz
a unique corresponds L(f) fy f dM for each f in Y is homeomorphic (via 442],
X there
Y such that
---
convex
weak*
its
theorem
of X which
measures
and
space,
each L in
embedding y of X, so we may
subsets
Riesz
into
the
compact
the natural Borel
an
It
(the
theorem)
Then X is
well-known
Bishop
C(Y) the Banach space and functions on Y (supremum norm), L on C(Y) such that functionals linear
=
measure a
of
artificial.
bit
Hausdorff
compact continuous
=
a
second
X).
exactly how the the Krein-Milman generalize
of all
set
first
introduction
instances
continuous
X the
on
general question
more
the
to the
of X.
property
allow
to
answer
make clear
also
Bishop-de
we
theorems
introduced; will
if
was
in these
It
geometrical
hypotheses example, the
two well-known
affirmative
an
answer
Krein-Milman
ural.
L(1)
the
and the
orem
and
while
certain
combination
convex
a
the
have shown that
to translate
have
on a
measures,
(without
formulate
answer,
depends
question Borel
now
p.
=
at
as
a
the
y))
with
the set of extreme
probability open set
measure
X
\ Y,
on
the
and hence
Section
3
Introduction
1.
supported by the extreme points of X. One need only recall functionals linear that E*, the space of weak* continuous on C(Y)*, L of form the functionals of all those -* consists L(f) (f precisely theorem of the in C(Y)) in order to see that this is a representation type we are considering. in the above paragraph There are two points which, it should be situation. of the general First, emphasized, are not characteristic of X formed a compact (hence a Borel) the extreme points subset; was unique. second, the representation (We will return to these It is clear that any probability measure /-I on points a little later.) functional linear Y defines on -4 a C(Y) which is in (by f fX f dl-t) next X. This fact is true under fairly as the general circumstances, linear from recall that a function shows. First, result one 0 space to another is affine A)O(y) provided O(Ax + (1 Ao(x) + (1 A)y) is
/-t
=
-
for
Suppose that
PROPOSITION 1.1 space
If
/-t
is
a
x
in
X
of [i)
is
Hf
=
of
compact subset
a
hull
PROOF. We want to show that
point
Y is
locally
a
of compact. E, and that the closed convex there exists then a on measure point unique Y, probability which is represented by /-t, and the function M --+ (resultant into X. weak* continuous an affine map from C(Y)*
convex
a
-
A.
any x, y and any real
fy
f (x)
such that
x :
M(f ) 1; n f Hf : f E
f (y)
to show that
fy
=
=
the compact
f dl-t
these E*
I
for
n X is
Y is
set
convex
f in E*. hyperplanes,
and
Since
nonempty.
X contains
f, let
For each
each
closed
are
X
we
want
X is compact, n
it
suffices
T:
then It
rates
this
by
-+
fi,
.
.
.
,
fn
E*,nHf,
in
Rn
by
Ty
=
(h (Y) f2 (Y))
g
Since
and
to show that
=
.
=
means -
Eaifi,
that
(a,p) then
Y C X and
nX
fn (0);
i
TX is compact and so that continuous, where E TX, (M (fl), p (f2),. p p, on Rn which strictly 0 TX there exists a linear functional functional the and a by TX; representing (a,, a2 p T is linear
suffices
If p
E
set
end, define
To this
is nonempty.
any finite
for
to show that
>
the
I-t(Y)
supf(a,Ty) last =
:
assertion
1, this
is
y E
Xj-
becomes
impossible,
If
we
fygdtL
sepa-
-
-
,
g in
> sup
and the first
(fn)).
/,t
.
,
define
convex.
an), E*
g(X).
part
of
Lectures
4
proof
the
measures sure
compact,
convergent then
the
f (xe)
show that
to
(f )
p (f
--+
of
points
X,
hypothesis
) y
for
f
each
X be compact
the closed
obtained
instance, by taking
simple,
but
space
show that
to
every
E*;
in
the
since
latter
x.
may be avoided
of
hull
convex
if E is
for
compact;
f (x)
=
=
that
E in which
spaces
suffices
it
x
subnet
I-t,3
=
The
--+
x,,
Since
resultants.
But if xp -+ y, say, to x. xp of x, converges subnet weak* to p, and hence corresponding M0 converges
separates
complete,
or
compact
a
if E is the
Banach space in its
a
those
in
always
is
set
locally topology
convex
weak
[28,
434].
p.
A a
Y converges weak* in C(X)* and denote their x respective x,,
on
Theorem
the net /_t, of probability to the probability mea-
that
and let
p,
X is
Suppose, next,
complete.
is
Choquet's
on
compact
set
useful, characterization in terms of can be given
of the closed
of
barycen-
and their
measures
hull
convex
ters.
PROPOSITION 1.2 space
convex
Y
if
and
represents
E.
if
only
there
exists
Y is
E is
in
a
probability f (x)
=
M(f
)
it follows and convex, in the there exists a net
X, x. Equivalently, (Ai' > 0, EAj'
closed
the
probability
EM'E' Z
Xi
.
/-t
measure
locally
a
hull
convex
there =
1, xi'
Y,
a
x
convex
in
on
sup
that
Y which
f (Y) is
hull
:!
on
of
X
Y which
represents
y,
directed
By
form
y,
set)
convex
converges =
which
each y,, by the probability the set of all probability the Riesz theorem,
with a weak*-compact Y may be identified and hence there exists a subnet pp Of Aa
then
Since f (X). if x Conversely,
of Y which
some
x,
sup
X.
in
of the
points
exist in
:!
We may represent
x.
of
compact subset
a
in
measure
closed
in
to
that
x.
PROOF. If /-t is a for each f in E*, is
Suppose A point x
converging
to
converges /_t,
measures
(in
is
E '-,M'x '
measure
subset
X
of the
on
C(Y)*, weak*
In particular, measure p on Y. topology of C(Y)*) to a probability to Y) in C(Y), each f in E* is (when restricted so lim f (y,3) lim f f d[tp f f dp. Since y, converges to x, so does the subnet fy f d1L for each f in E*, which completes the y,6, and hence f (x) proof.
=
=
=
Section
proposition
The above Milman
5
Introduction
1.
makes it
Recall
theorem.
the
easy
reformulate
to
If
statement:
X is
the
compact
a
Kreinconvex
hull then X is convex convex locally space, is the following: Our reformulation Every of its extreme points. convex point of a compact convex subset X of a locally space is the which X is on measure supported by the of a probability barycenter closure of the extreme points of X. To prove the equivalence of these two assertions, suppose the former holds and that x is in X. Let Y of X; then x is in the closed of the extreme points be the closure of hull of Y. By Proposition convex 1.2, then, x is the barycenter the obvious If extend Y. we measure M on a probability way) p (in result. to X, we get the desired Conversely, suppose the second
of
subset
the
a
and that
valid
is
assertion
by Proposition
1.2,
x
closed
is in the
x
Then
X.
in
is
convex
closed
(defining hull
of
Y
as
above)
hence in the
Y,
points of X. theorem using measures now any representation supported by the extreme points of X (rather than by their closure) Klee [50] In fact, theorem. of the Krein-Milman is a sharpening closed It
of the extreme
hull
convex
that
clear
is
has shown that
in
compact
subset
the
convex
of its
closure
a
representation mass" representation. The problem of finding of X arises mainly from be
a
metrizable,
Borel as
set
topological
[9,
shown
PROPOSITION 1.3 vector
of
an
he makes
points. gives no
If
p.
327].
more
space,
a
then
information
supported difficulty
This
by
almost
every
Banach space is then, the Kreinthe
"point
the extreme
points
than
the set of extreme
that
by the following X is
such sets,
For
measures
the fact
precise)
dimensional
infinite
extreme
Milman
not
(which
sense
is avoided
points
need
in
X is
case
result.
metrizable, the extreme
compact convex subset of a points of X form a Gj set.
Suppose that the topology of X is given by the metric d, 2-'(y + z), y and z in fx : x integer n ! I let Fn and checked that each Fn is closed, It is easily X, d(y, z) ! n-'j. if it is in if and some Fn. not extreme that a point x of X is only is of the extreme the an F,. points complement Thus, measure Recall that we always have the trivial representing Ex of X, then it is for a point x of X. If x is not an extreme point PROOF. and for
each
=
=
Lectures
6
easily
that
seen
there
points of representing
other
subset
an
extreme
(BAUER [4])
of locally point of measure
M
f xj; I-t(D)
the
for
that
I-t(D)
>
that
there
0 for
=
0 for
the
is
extreme
point
an
We want
(due
suffices
D;
such
D with
set
from
the
only
of X and that
the
is
by
supported
M is
/_t)
of
DCX
show
to
\ jxj.
Suppose follows
of D it
M(U n X)
that
is
x
the
compactness
y of D such
point
con-
Then
ex
regularity
the
to
X.
mass
show that
to
compact
a
E
x
point x.
each compact
some
X is
represents
x
it
some
is
only if
and
x.
this,
that
X which
that
represents
set
if
X
Suppose
space E and that
convex
on
Suppose
PROOF. measure
are
a
probability
measures. Indeed, the representing characterized by the fact that they have no
measures.
PROPOSITION 1.4 vex
Theorem
other
exist
X
extreme
Choquet's
on
0 for
>
every
neighborhood U of y. Choose U to be a closed convex neighborhood of y such that K The set K is compact and U n X C X \ f xj. =
and 0
convex,
< r
M would be in
X
by pl(B) each
for /-tl IL
(K)
+
rp,
-
=
1,
(If p(K)
< 1.
K.) Thus, r-lp(B n K)
we can
Borel
=
M(K)
=
B in
set
that
we see
(I
-
r)A2,
X.
define
Let
which
=
(1
r)-'M(B
-
implies
that
x
n
=A
=
since
Furthermore,
x.
(I
+
rx,
of on
(X \ K))
of /-ti;
resultant
and hence x,
x
M, and A2
measures
be the
xi
E K
x,
Borel
A2(B)
and
1, then the resultant
=
r)X2,
-
a
contradiction. It
the
interesting
is
to
Krein-Milman
Propositions smallest
and 1.4;
1.2
closed
closed in
the
of
a
convex
closure
of
PROOF.
Indeed,
convex
of
440] that
the
an
easy
that
X,
Z C
extreme
to
consequence
of
ex
X is the
X.
generates
X is
space,
"converse" of
closure
Suppose that
Then the
Z.
classical
is
points
compact
a
convex
and that
X is
of
contained
X
are
Z.
cl Z and suppose x E exX. measure /,t on Y which represents x;
let
1.2, there exists a 1.4, /-t Ex. It follows =
p.
implies
it
(Milman)
locally hull
[28,
Milman's
of X which
subset
PROPOSITION 1.5
subset
that
note
theorem
Y
=
that
x
E Y.
By Proposition by Proposition
the
Section
1.
7
Introduction
example of a dimensional compact convex subset X of a finite space E, in order the question to illustrate uniqueness of integral concerning repreis X that is sentations. or more a generally, plane triangle, Suppose the convex hull of an affinely independent subset Y of E, that is, X Y is is a simplex. independent provided no point y affinely (A set It then follows in Y is in the linear variety generated by Y \ Jyj.) from the affine independence that Y is the set of extreme points of X, and that every element of X has a unique representation by To conclude
introduction,
this
10.10)
(Proposition
if
that
X is not
of X has two such
representations.
infinite
generalization
will
dimensional allow
us
to
prove
(among
return
of Y.
of elements
combination
a convex
we
a
It
is
then
Section
10
to
some
will
we
show
element
give
an
which "simplex" Choquet's uniqueness
of the notion
things)
other
difficult
not
simplex,
In
the
to
of
compact convex set X theorem, which states that for a metrizable of has X each point in a locally a unique convex space, representing if X is a if X of and the extreme measure supported only points by simplex. of the Krein-Milman In the next section an application we give to make some general it is worthwhile Before doing this, theorem. theof the various remarks concerning representation applications of that the objects to recognize It is generally not difficult orems. form
interest
a convex
subset
X of
some
linear
One is then
space E.
topology for E which makes X compact and at the same time yields sufficiently the assertion functionals linear so that "A repremany continuous the extreme sents x" has some content. points of Second, identify has "p is supported by the extreme points" X, so that the assertion faced
a
with
useful
problems:
two
find
First,
a
locally
convex
interpretation. EXERCISE
Prove a
Carath6odory's
compact
in X is
(Hint: and the
exists
latter
of
combination
a convex
Use induction
X, there
sharper subset
convex
a
set
on
the
supporting
form of Minkowski's an
n-dimensional
of at most dimension. at
space
+ I extreme x
most
a
H of X with n
-
1.
If X is
each
points of boundary point
n
is
then
E,
If
hyperplane
has dimension
theorem:
If
x
x
is
in an
Hn
x
X.
of
X,
interior
Lectures
8
X, choose an extreme point y of the segment [y, z] for some boundary point point
of
on
Choquet's
X and note z
of
X.)
that
Theorem
x
is
in
2
of the
monotonic
functions
A real
if
f
f increasing,
(n)
tonic
(_ 1)
n
has derivatives >
0 for is
as
(a
and e-x
x-a tation
f
function
valued
theorem
much related
n
=
!
0).]
(0, oo)
on
f 0, 1, 2.....
said
to
S. Bernstein Wewill
proved
a
(See [821 prove
the extension functions; senting measures) follows from this denote the one-point compactification
on on
(0, co), [0, oo]
then
If f
there
such that
exists
is
/_tQ0, oo])
-
f W
=
(Note
that
the
converse
I
is true,
classical
by
of
mono-
represen-
proofs
several
for
only
(with
for
infinite
arguments
and
bounded repre-
[821.
We
[0, oo) by [0, oo].
completely monotonic Borel measure p nonnegative and for each x > 0,
bounded and
unique
a
fundamental
the theorem
to unbounded functions
THEOREM(Bernstein).
completely
be
=
such functions.
material).
is
completely
to
of all orders and if f, f (1), f (2).... is nonnegative and nonThus, f functions (-l)nf (n). [Some examples:
0)
each of the for
theorem
Krein-Milman
Application
f (0+) 00
e-ax
since
dp(a).
if
a
function
f
on
(0, oo)
can
under the integral then differentiation sign as above, represented monotonic. that f is completely Moreover, and it follows is possible, to the theorem dominated convergence by applying the Lebesgue we see that p([O, co]), so f functions a _+ e-a/n f (0+) fo' dl-t The idea of the proof is due to Choquet [16, Ch. VII], is bounded.) in a much more general setting. results who proved this and related We start by giving a sketch of the proof. monotonic funcDenote by CMthe convex cone of all completely tions f such that f (0+) < oo. (Since a completely monotonic funcallimit at 0 always exists, this right-hand tion f is nonincreasing, Let K be the convex set of those f in CM though it may be infinite.)
be
=
R.R. Phelps: LNM 1757, pp. 9 - 12, 2001 © Springer-Verlag Berlin Heidelberg 2001
=
10
Lectures
f (0+)
such that
(_1)nf
>
(_I)n+l
(n)
(a/2)
(a/2) f
(a/2) (_1)n+lf and the desired
a
together
fact,
f
to
0,
of
points
The extreme
e-'x,
the extreme
0
0. For x > 0, let Suppose that we have shown that f (x)f (xo). f (x + xo) u(x) that u this implies 0, so that Since f is extreme, u E K. f f (x)f (xo) whenever x, xo > 0. Since f is continuous f (x + xo) 0 (the case a this oo) or on (0, oc), implies that either f have must we ae 0 e-cx for some a. Since -f'(x) f (x) Let b u E K. f (xo) (so that 0. It remains to show that f a b :! 1), and note that (f + u) (0+) 0 b) f (0+) + b < I and (I f (0+)] < f (0+) ! - 1. Furthermore, b[l f (0+) u) (0+) (f PROOF. =
-
=
-
=
=
=
=
=
=
-
(_ 1) (f n
and
+
(- 1) (f n
=
-
[(_ 1)nf
-
-
-
-
U) (n) (X)
-ax
(I
-
b) (_ I)nf
(n)
(X)
+
(_ I)nf
(n)
(X
(n)
f
+
X0)
U) (n) (X) (n)
(X)
_
(_ I)nf
(n)
(X
+
X0)]
+
b(- 1)
(n)
(X).
0
12
Lectures
(')
(- 1)'f
Since
(r
itself.
points,
extreme
By
what
for
some a >
(and
extreme
and therefore
proof
the constant
Wenow finish It
[0, oo]
into
transformation
closed
for
each
functions
of this
-a,x
0, all
>
r
0 and I
K
ex
hull
of its
is nonconstant.
is of the form e-clx
point
extreme
T, is
carries
convex
which
one
T,
Since
function
the
exponentials
T, are
extreme),
clearly
are
under
so
complete.
is
tions.
the
is
has at least
holds
the =
it
Theorem
nonnegative.
by (T, f ) (x) f (rx). convex combinations, it
just proved, 0, and hence the image this
is
consider
this
Since
is extreme.
the
defined
itself
have
we
the latter
reverse
onto, and preserves Since K is compact,
one-to-one, onto
nonincreasing, inclusion,
is
To prove the > 0) of K into
Choquet's
on
is
the
proof of Bernstein's
difficult
not
theorem
show that
to
the
for
map T:
bounded func-
a
e-a0
-+
from
since [0, oo] is compact, its image ex K continuous; is also compact. By the Krein-Milman representation theorem, to each f in K there corresponds Borel probability a regular measure m on
ex
K is
Lx (f ) x
>
(i.e.,
L
f (x)
=
is continuous
Define
0. M
mo
--
Now,
E.
on
T).
fex f
=
(0
measure
v([O, oc]) uous
these
on
v
on
f (0+). [0, oc]. Let --
and
E,
that
then
0,
that
so
subset =
Lx dm
e-ax
L dm for
=
dy(a)
e-ax,
for
A consists as
x
A be the
linear
>
of finite
functionals
Since A separates points implies that it is dense in
x
on
o
T
>
d(m
there
exists
a -+
(x e-clx
-
second
a
>
0)
and
is contin-
QO, oo]) generated /-t and
of the v
the Stone-Weierstrass p
m(TB)
T)
o
combinations
C[O, oo],
so
=
0.
C-axdv(a)
of
linear
[0, oo], QO, oo]), of
Lx
0 the function
subalgebra
functional
have
Suppose 0
linear
Lx dm for each K
[0, oo] by p(B)
K)
=
fex
==
each
p is
For each
evaluation
f (x) we
fT-1(ex
each continuous
the
B of
unique. [0, oo] such that f (x)
prove
functions;
functions A.
to
K
K
>
x
on
L,,(Ta)
Since
00
remains
fex
=
if
M on each Borel
f W
It
)
L (f
K such that
functional
v.
are
by same
equal
on
theorem
Choquet's
3
The metrizable
theorem:
case.
theorem for Choquet's representation of the case a special actually general ChoquetBishop-de Leeuw theorem, but its proof is quite short and it gives us which is needed to introduce an opportunity some of the machinery this
In
section
metrizable
in the
X.
we
will
This
is
main result.
Suppose that
I Ah(x) that
(I
+
h is
h is
h is
The function
C.
set
prove
-
Recall
0 < A < 1.
semicontinuous
for -h
if
that
if for
h[Ax
+
each x, y in C is convex, and h is called
a
on
(1
inequality
f
function
A, Jx: f (x)
is upper
1,
which
is called
(c)
f f
i.
If
g
while
on
X and
x
X,
E
let
f 1. the
upper
envelope
of
f,
has the
properties:
(a) f is concave, bounded, measurable). (b)
function
f and if f
is
(hence
and upper semicontinuous
concave
bounded, then f f +g--f +gifgEA. are
R.R. Phelps: LNM 1757, pp. 13 - 16, 2001 © Springer-Verlag Berlin Heidelberg 2001
and upper
+ g :5
Ifr
f
+
then
semicontinuous,
g and
If
>0, thenrf
-
j =rf.
Borel
Ilf
-
g1j,
14
Lectures
The
proofs
manner
from
from
of most of the above facts
fact
the
definitions.
the
that
follow
The second
functions
constant
Choquet's
on
in
straight-forward
a
assertion
(a)
in
affine.
are
Theorem
follows
The second
as-
If f is concave and up(b) may be proved as follows: in then the convex semicontinuous, per locally space E x R the set K I (x, r) : f (x) ! rj (i.e., the set of points below the graph of f) sertion
in
=
is closed
and
theorem
tion
f (xi)
convex.
If
asserts
the
I(xi)
0, and hence L(xi, f (xi)) < L(xi, I(xi)), the function h defined on X by h(x) A exists r if L(x, and is r) in A. Furthermore, The f < h and h(xi) < f (xi), a contradiction. second assertion in (c) again uses the fact that functions constant Since f < I I f 11, we have are affine: Furthermore I I f 11. L
E
on
strictly
R which
x
separates
A such that
exists
A
h + rf and hence h + rf By the Hahn-Banach concave, such that m on C(X) theorem, then, there exists a linear functional if h cz A, h(xo)+rl(xo) m(g) ! - g(xo) for g in C(X), and m(h+rf) > > If g E C(X) and g :! - 0, then 0 m(g), i.e., m is g(xo) r E R. continuous. is hence and functions By on nonpositive nonpositive the Riesz representation theorem, there exists a nonnegative regular
functional
this
dominated
is
B
on
=
=
.
=
Borel
A,
1 c
that
we see
h(xo)
m(h)
=
j(x0)
that
M(h)
=
p(j),
!
2
I-t(f).
It
f and consequently
then h !
,
tz(f)
=
,
/_t(j).
on
of S
the
x
-
f (x)
:
+ .1
(Y)
2
1
W
f
j(x) I f complement proof by showing that .6 is contained in the set + 1z, where y and z ly points of X. Indeed, if x 2 2 that of f implies the strict then of convexity X, points
of extreme
f W
- /-t,
of X than does
if
a
nonnegative such that
/j,
A
the is
chain
in
measure
an
Z.
Z
nonnegative v
=
=
element
>-)
ordering a
X, then there
on
exists
p >.- A.
0 and let
have found
v
extreme
verified
E Z and hence a
which
A,
h E
:
points measures and convex by considering in the plane, say. This fact is what leads us maximal measure will be supported by the extreme
measure
respect
=
measures
"closer"
support
If
that
>
reference
a
LEMMA4.1
let
theorem,
A and /-t represent the if particular, they are
(In
the
a measure
heuristically on a triangle
to
PROOF.
with
such
further
functions
so
the
-
regular
and
i.e.,
6,
-
Since
partially
each
then
A.
A, h
h E
:
ordering;
without
a
if p
be concerned
to this
p,
[t(f),
g in
of X and contains
in
and hence
Wewill
then
A,
subspace they have
f,
g,
-
C(X). [Note (91 + 92) -1
ordering A + 91)
from the fact
comes
the
f
form
1
nonnegative
transitive
/-z is in
=
that
of the
Theorem
way:
M(f) for
=
f
A(f)
on
f
A if
Wenow
are
!
clearly
is
functional
+ 92
following
If
implies
/-t
C(X).
of
Note that
partial
MaX(fl
::::::
X in the
DEFINITION.
functions
all
C separates the points by the Stone-Weierstrass
-
constant
the
(of
the usual
under
911
-
C
-
Choquet's
on
/-t.
We may
in
fp
:
p
y in
>
0 and p
Z which
is
Al.
-
maximal
Then /-t will be a maximal then v >- A, measure and v >- /,t, Z.
To find
regard
a
maximal
Was
a
net
element
(the
of
Z,
directed
4.
"index
set"
in
exists
[to
compact
with
po
have yo c-
a
/_t(i) f p, I of
> 0 and
subnet
is contained
which
A(1)1.
--
19
there
Thus,
Wwhich
to converges from follows
W, hence po >- [q. and eventually /_tj /-t,, since /-to >- A, we is an upper bound for W; furthermore, a maximal element. Z. By Zorn's lemma, then, Z contains is any element
If [t,
it
in
that
of subnet
the definition
Thus,
0 and
topology.
the weak*
po in
!
/-to
IM
set
Theorem
Leeuw Existence
of Wthemselves)
the elements
being
weak*
the
Choquet-Bishop-de
The
Section
at maximal the idea of looking Bishop and de Leeuw originated from they used an ordering which differs slightly measures, although the one used here. The notion is applied in a very simple way: If xo
X, choose a maximal measure /,t such that /-t >- Ex, As noted of the maximality xo; it remains to show that above, p represents no extreme sets which contain on Baire that p vanishes /-t implies The first step toward doing this is contained in the following points.
is in
result.
If
PROPOSITION 4.2
/_t(j)
for
p is
f
functional
sublinear
L(rf)
=
p(rf),
f
p,
on
if
while
r
on
measure on
X, then I_t(f)
-
X.
the
and define
functional
linear
Rf by L(rf) C(X) by p(g) < 0, then 0 rf
subspace
one-dimensional
the
C(X)
in
maximal
function
each continuous
PROOF. Choose
a
=
rp(j).
I_t(g).
=
=
rf
-
L
Define If
r
_f
:!
>
the then
0,
(-rf)
+
on
=
Thus, p(rf). M(rj) :! p(rf) L(rf) and therefore on (by the Hahn-Banach theorem), there exists L' of L to C(X) such that L' ; p. If g ;S 0, then g ; 0, an extension 0 and hence there ! p(g) so V(g) [L(g) :! - 0. It follows that L' that V(g) v on X such measure v(g) for exists a nonnegative each g in QX). If g is convex, then -g is concave and -g -g, so < is Since v. maximal, /-t fL(-g), i.e., /,t p(!_-_g) p(-g) v(-g) therefore and [t(j), have must L(f) v(f) we I_t(f) v, /-t
rf
rf, Rf,
and hence
-
L < p
=
=
=
=
=
=
-
completes
which As result
will
is true.
following:
the
fx
we
:
I(x)
=
proof.
the
see
later
More
=
=
=
(Proposition
importantly,
If /-t f (x) 1, for is
a
f
in
the
converse
to
the above
implies proposition is by supported p measure, 3. C. As shown by Proposition 1, note
maximal
each
10.3) that
the
then
20
Lectures
each of these
strictly exX
fx
=
Herv6
[42]
:
do in the
of
inter-section if
(as
and the
proof
fo(x)J,
=
of X.
points
would have
in
If C contained
a
Choquet's
theorem)
be
complete.
would
f (x)
nonmetrizable
case
form f f (x) for each f in C, and the
all =
-sets
of
the
x
+
f (z)
!
2f (x)
=
21(x)
>
1(y)
+
that
is prove
I(x)
:
if
x
then
f (y)
Theorem
shown, however, that the existence of a strictly convex function that X is metrizable. on X implies About the
we can
Indeed,
lo(x)
we
Choquet's
has
continuous
best
fo,
function
convex
the extreme
contains
sets
on
1(z)
1 2
>
(y
+
f (y)
z),
y,
+
the
X is
ex
f (x) 1, f
=
C.
in z
E
X,
f (z),
i.e., 2f (x) f (y) + f (z) for each f in C. It follows that the same holds for any f in -C, hence for each element of C equality C. Since the latter is in dense have must subspace x C(X), we y z, i.e., x is an extreme point of X. --
-
=
To show that
any maximal
measure
y vanishes
on
the Baire
--
sets
from ex X, it suffices to show that p (D) 0 if D is disjoint a compact Gj set which is disjoint from ex X. (This is a consequence of regularity: If B is a Baire set and /-t is a nonnegative regular Borel then p(B) measure, supf I_t(D) : D c B, D a compact G61.) It will be helpful later if we merely assume that D is a compact subset of a Gj set which is disjoint from exX. To show that I_t(D) 0, we first lemma to choose a nondecreasing use Urysohn's sequence f fn I of continuous functions 1 ! fn < 0, fn (D) on X with I and 0 if x G exX. We then show that if /-t is maximal, then limfn(x) it is immediate from this that p(D) 0. To obtain 0; lim[t(fn) this "limit" result two technical lemmas. The first slightly requires of these is quite since it reduces the desired result to interesting, theorem for metrizable Choquet's X, using an idea due to P. A. fact that for each x in X, we will use the Meyer. (More precisely, there exists which is Ex supported by ex X. Since it is not /-t true that A can be extended to an element of in generally f every this is formally than the stated of Choquet's version E*, stronger theorem. See Proposition 4.5.) which
are
=
=
=
=
-
-
=
=
=
-
Suppose that f fnJ is a bounded sequence of concave with lim inf fn (x) ! 0 for on X, per semicontinuous functions in exX. Then liminf x fn(x) ! 0 for each x in X. LEMMA4.3
up-
each
Section
The Cho q uet-Bishop-de'Leeuw
4.
PROOF. Assume first
probability esis, lim inf f,, Since each f,,
measure
y
!
a.e.
0
is
X, choose By hypoth-
is in
x
21
supported by by Fatou's lemma, lim inf assertion and upper semicontinuous,
Ex which
-
M,
concave
f,,
3 shows that
Section
If
X is metrizable.
that
Theorem
Existence
ex
X.
f,,(x) [t (fJ.
that
so
=
inffh(x)
(b)
h E
:
a
0.
so
1,,,
=
is
A,
in
h >
f,, (x) > Thus, f,, I ! 0. Turning to the general case, suppose x is in X, and lim inf I-t(f,,) for each n choose hn in A such that hn : fn and h,, (x) < fn (x) + n-'. Let RN be the countable product of lines with the product topolN The function 0 Jhn(Y)J. 0 : X -+ R by 0(y) ogy and define f,, I
f [t (h)
inf
-
h E
:
!
A, h !
lim inf
=
is
affine
and continuous,
metrizable
of the
of R
projection x'
is
X',
in
the
is
7rn
W)
in
ex
X',
each x'
for
!
in
set
(x') (Ox)
lim inf 7rn
0 < lim
inf 7r,,
LEMMA4.4
decreasing
If
we
so
> 0
for
from the first
(x')
W
Assuming
is
/-t
=
0
is in
x
(and
exX;
in
bounded
X.
It
in
the
Since
functions.
uous
a
for
PROOF. Consider
Taking x' lim inf fn (x),
proof 0 (x),
we
=
on
conclude
we
obtain
completes
which
maximal
QX) each
x
in
ex
sequence
-1
:! -
the addition, by zero),
from
-1
1,,
of
:!
sequence so
Lemma4.3
X, and if f fnJ :! f,, :! 0 (n
that
that
Q,
we
have liM
ffnJ is limfn(x) lim In (x)
also
Thus, the
the we
Baire
in (X)
0
-
nondecreasing for
exists =
1, 2,...)
semicontin-
upper
concave
a non-
0.
=
=
completes
is
=
X, then lim I-t(f,,)
J1nJ
fn :!
above
follows
on
measure
such that
X. From the Lebesgue bounded convergence theorem 4.2 we have IL(f_n) 0; from Proposition limp(ln)
on
that
and continuous
each x' in X. =
in
If hn (y). X; by the
Since y is in ex X. > lim inf f,, (y) > 0,
of this
part
(0y) y.
affine
are
7rn
subset
coordinate" =
convex
shows that
have lim inf 7rn
lim inf hn
--
7rn
point
extreme
convex
"n-th
usual
and
compact an
compact
X, then
in
argument
X',
sequence
f,, (x)
and lim
in
has
a
be the
-xn
y is
is
proof.
the
if
Let is
it
O(X)
The functions
X.
ex
the metrizable that
0-'(x)
simple f,, (y),
a
.
R; if
onto
set
hn (Y)
-
R
theorem
Krein-Milman
x'
N
space N
X'
so
=
0 for
=
each
follows
it
each
tt(fn),
x
x
in
that
which
proof. have shown that
subsets
of X
\
any maximal ex
X.
meabare
We have also
on
shown
X vanishes
something
22
Lectures
different: slightly X contained of
A maximal
shows, closed
set
particular,
of a
generalizes
which
THEOREM (Bishop-de
(Indeed, such a set.) maximal
This
measure
and hence the
Gj subset
I_t(D)
since
theorem. Leeuw theorem
be
for
Leeuw).
convenient
Suppose
it
by any Choquet-Bishop-de
Choquet-Bishop-de more
0 if
=
important, supported
is
is
Theorem
any
on
showed that
we
the Krein-Milman
the
perhaps
can
vanishes
/-t
measure
exX,
contains
formulate
We next
X.
that
which
Leeuw theorem
manner
ex
subset
D is any compact in
\
X
in
Choquet's
on
that
X is
in
a
applications. compact
a
convex
of locally of X which is generated point x0 in X there exists
by S the a-ring of subsets space, by exX and the Baire sets. Then for each
1 such that
x0 and
subset
Borel
of X \
/-t
Baire
(ex X) As
x -*
A (X)
/-t
measure
/-t(exX)
on
S with
M(X)
there
exists
1.
=
Leeuw theorem
a
f
E be the
Ex,,.
in
/-t(exX) [Bi n ex X] we let I_t(S)
earlier, E*, r in
Hilbert x
U =
f (x)
not
R.
space
fxnl
=
Then
2 such that
To do
1.
--
this,
observe
that
any set
[B2 n (X \ ex X) ], where B, A(Bj), then /-t is well defined
and and
1.
=
remarked + r,
-
If
sets.
of sequences
by f (x) y in
we
f (x) Let
set
=
nonnegative
A which represents on the Baire x0 and which vanishes X. Weneed only extend A to a nonnegative measure
S and show that
are
a
ex
S in S is of the form
B2
and denote
Choquet-Bishop-de
the
measure
subsets on
/-t represents
By
PROOF.
[t
convex
a
f =
the
in
A is of the
following
is in A and
for
all
Ix, I f (0) x
2-n
form
example:
f2 in its weak topology,
such that
(x, y)
function
every
Consider
let
X be the
and define
0, but there
is
f no
on
X
point
in X.
example shows that the subspace M E*lx + R of C(X) the two notions may be a proper subspace of A. Nevertheless, "/t and x" X and a measure /-t on EX" "A represents (for a probability as the following point x in X) coincide, proposition implies. This
=
,
PROPOSITION 4.5 tions
is
tinuous
uniformly functions
PROOF. It that
subspace M (defined above) of affine funcdense in the closed subspace A of all affine conThe
on
is evident
g E A and
.5
X.
that >
0,
the space A is uniformly and consider the following
closed. two
Suppose subsets
of
Section
E
x
The
4.
f(x,r) g(x) + 61.
J1
R:
and
r
and
disjoint.
=
theorem
difference E
x
Choquet-Bishop-de
-
By
a
(obtained set
J2
-
:
x
on
g(x)J
=
sets
slightly by separating
Ji)
there
exist
L(JI) L(x, f (x)) g(x) < f (x)
=
0 and Eni= I Ai hI Let h Eni= I Aifi. This is a point of co (ex K) and I I r (pi) c/3Since
p,
is
a
probability
n
measure
_
_
=
-
Consequently,
IIf
-
h 11 :5
11 f
-
Thus, co(ex K)
Ar
(pi) I I
is
norm
+
(1
-
A) I I r (pj) I I
dense in K.
+
I I r (pj)
-
hII
- p o 0'.
rem
for
which
we
p
measure
Y and
on
choose
we can
In
is due to
later
a
of the
view
we
maximal
remarks
in
want
a
maximal
measure
Section
A' 4
on
measure
K(M)
(prior
to
the
with
A with
A'
-
Bishop-de
A p
o
-
p,
0'.
Leeuw
A' is supported by the compact set O(Y), hence is of the A >- p. To form A o 0' for a (maximal) measure A on Y such that A vanishes on the Baire subsets of Y \ B(M), we need only see that
theorem),
show that
it
vanishes
on
any compact
Gj subset
D C Y
\ B(M).
same is exK(M), O(Y) true of A O(D) U [K(M) \ O(Y)]. It follows that the complement is an F, in K(M), so A is a Gj of A is an F, in O(Y) and therefore Lemma in K(M) which misses exK(M). By the remarks following 0. A'(O(D)) 4.4) A' vanishes on A D O(D), hence A(D)
Now, O(D)
is
a
Gj
hence the
and it misses
in
=
=
THEOREMSuppose
which
separates
points
that
M is
a
and contains
subspace the
=
of C(Y) (or of C,(Y)) If L functions.
constant
M*, then there exists a complex measure p on Y such that L(f) 0 for any Baire set S in Y fy f dl-t for each f in M and p(S) which is disjoint from the Choquet boundary for M. =
Section
By applying
PROOF.
obtain for
a measure
each
K(M)
A New Setting:
6.
f
with
which
are
define
A
properties.
=
M.
in
>-
tti
disjoint Y1
-
A
Choquet Boundary
The
the Hahn-Banach =
Al -A2+i(A3
For each i
Ai.
A2 + i
-A4)
we can
B(M), (P3 N)
and
-
,
we
on
find
We know that
from
and Riesz
/-ti
yi(f) get
33
theorems,
a
maximal
vanishes =
Ai(f)
a measure
=
measure
on
the
for
f
with
may
we
L(f)
Y such that
A(f) yi
Baire
in
the
M. If
on
sets we
required
7
Applications
Let
X be
A > 0 there
R,X !
family
is
(i.e.,
0
> 0
R.Xf
of operators all A, A'
RA(A
Ry
T,Tt
families
for
f
all
potheses,
[68] in
for
more
which
Wefirst
1.
detailed
prove
of
For
a
0),
under
then
2.
f
For
that the
identity
is
and defines
a
proof
of this
t
>
(i.e.,
this
needed to follow
[55]
some
and
was
the
originally facts
elementary
>
0) hysemigroup
Under certain way from
is
section
(See subject.)
result.
on
X, A
in
resolvent. in this
of this
information
are
:
conditions
suitable
(x
a
a
convergence
the papers [55] and None of the facts
given below, by Choquet.] easily from the
exposition shown
us
which follow
resolvent.
each A >
0, R,\
is
and
continuous
f C(X), 11f 11 1, (1/A)Ilf 11, so JJR,\11 :! - I/A. But RAI
if
following
If (Tt of Markov processes. from C(X) into itself operators
is obtainable
the
is due to Lion
definition
C(X) such 1/A. We call
A) Ry Rx.
-
and the content
to
paragraph
this
(A
e-"'(Ttf)(x)dt
resolvent
related
R)J
if the
resolvent
-+ =
each
00
C(X)
of Markov operators, theorem
0)
C(X)
:
and
for
study
>
1, Tt
=fo
in
every
=
of Markov =
(R,\ f ) (x) exists
Rx
-
the
in
arise
semigroup T,+t, TtI
a =
a
that
> 0:
H
[Such 0) is
0)
>
Rx !
f
whenever
resolvents
to
and suppose
space,
transformation
linear
a
for
valid
Hausdorff
compact
a
Choquet boundary
of the
E
each A and
and follows
:-
then
from
A',
(*)
R,\RX,
-
==
otherwise.
R.R. Phelps: LNM 1757, pp. 35 - 38, 2001 © Springer-Verlag Berlin Heidelberg 2001
hence =
RyRA.
11R.J
=
RJ
I-E
IfI
0
6 >
:! -F
in
\ U; II-if,
Condition
f
0
x
satisfies:
for
f inA
exists
if
(for arbitrary
is the
that
y
I-if
Condition Y
which
If (y) I
nDmn. I
=
\
easily
=
Yk
=
S,
then
x
see
that
S
sm
To
f(x)
=
-+
A (Yk)
Dmn for
\ Dmn is
closed
0 such that
fl,
h (Xn) '2
sequence
fYkJ
JJx
-
yll
-
M
afk(Xn)llllYk
1
+
CeXnll-
have
OXn)
n
n
(Bn f ) W
=
Fk=O
(n) k
the
most
at
we
consider
k(1
_
f
for
by setting,
f (k/n)x
operators
to
example approximation
in
Bn from C[O, 1] into
operator
n
be for-
one
of the Weierstrass
1, define
degree
this,
can
of linear
sequence
a
theory
theory
approximation
in
of convergence of To illustrate operator.
mulated
approximation
and
X)n-k,
G
E
x
C[O, 1],
[0, 1].
which obto f proved that f Bn f I converges uniformly P. theorem. proof of the Weierstrass viously gives a constructive Korovkin [52] observed that each Bn is a positive operator (if f > 0, result. remarkable then Bnf > 0) and he proved the following Bernstein
,
Suppose that f Tnj is a sequence of positive from C[O, 1] into itself with the property that f Tnf I conoperators k x kI to f for the three functions 0, 1, 2. f (x) verges uniformly to f for every f E C[O, 1]. Then f Tnf I converges uniformly
THEOREM (Korovkin[521).
=
To show that
the
Bernstein
theorem, we must to Ik for k 0, 1, 2, where binomial expansion
show that
Korovkin's
=
(x
(1)
+
a)n
=
operators
(x)
I
Yk=O
Setting
(1)
with
utilizing
a
=
I
respect the
-
x
to
(n)
Bn1 multiplying
x
previous
twice,
Bn.[2
=
I
=
T2
by I
+ n
R.R. Phelps: LNM 1757, pp. 47 - 50, 2001 © Springer-Verlag Berlin Heidelberg 2001
[0, 1]
each
for
X2 n2)
yields
identities
E
x
the
.
Consider
the
Xka n-k.
k
shows that
hypotheses of f Bn Jkj converges uniformly
satisfy
for
x
=
=
(I
_
J2)
n.
settinga
Differentiating =
I-x
and
48
Lectures
for
each n,
Wewon't in a
a
result
Hausdorff
M a Korovkin
for
M, that
akin
and that
space
in
set
M is
is
we are
Suppose
a
fTjJ
interested that
X is
C(X).
of
subset
Korovkin's
that
true
Theorem
T2.
to
since
[73].
C(X) provided it
[0, 1]
on
itself,
theorem
due to
provided
is,
uniformly
converges
Korovkin's
prove
general
more
compact
call
f Bj T21
so
Choquet's
on
We
theorem
holds
uniformly
converges
of positive a sequence f C(X) fTnj C(X) such that f Tnf I converges uniformly to f for each theorem asserts that f 1, X, X21 is a Korovkin f E M. (Korovin's set in C[O, 1].) Note that Mis a Korovkin set if and only if the same is to
f
for
each
operators
linear
span,
so we
( akin[73]).
THEOREM
space and that
the
[To
M is
see
may
assume
that
a
Then M is
X.
this
that
Mis
Suppose that X is linear subspace of C(X)
of Choquet boundary B(M) for
points
separates
only if
is
on
of its
true
whenever
E
does indeed
a
yield
linear
subspace. metrizable
compact
a
which
Korovkin M is
a
of
all
Korovkin's
I
and
C(X) if
and
contains in
set
X.
theorem
need
we
I only observe that for any xo E [0, 1], the polynomial (X XO)2 peaks at x0, so the latter point is in the Choquet boundary of the span of 1, x and X2.] that B(M) X and that fTnj is a sequence of Suppose, first, such that IITng on C(X) 0 for all g E M. positive operators gII Given f E C(X), we must show that IITnf f 11 0; equivalently, we must show that of f I I Tn f f I I I has itself a every subsequence of notation, subsequence which converges to 0. For simplicity assume that f IITnf f III is the initial subsequence and choose, for each n, a point xn (E X such that _
_
=
-
-
-
IITnf By taking some x on
a
(E X.
further
-:::::
I(Tnf)(Xn)
subsequence
Define
a
we can
jLnI
sequence
-
f (xn)lassume
of positive
that
xn
linear
-4
for
x
functionals
C(X) by Ln h
Since for
f 11
-
I
each
Cz M we n.
Thus,
have
=
(Tn h) (Xn),
Ln1 can
-+
1,
so
h E C(X). we
be considered
may to be
assume a
that
probability
Ln1
>
0
measure
Section
which
converges
JIT,,,g
then
-
g1l
0
-+
I (T., g) (X-J
1IT-k
Ank
(g)
p(g)
-
of
continuity
f
subsequence
a
that
y
of open
neighborhoods
For each
we
Ex,
then
define
Tnf Clearly
each Tn is
g c M and y C-
E > 0, UN. Moreover,
y GX
we
p
1 :5 gW
for
we
g E
have
also
that
M,
is
implies that p This, together with
Ex,
=
the
f (Xnk) 1
If (X)
f (Xnk)l
-
that
x
if
with
0 < g,,
0, there exists A such that A and measure Ex li(f) A(f) < c. To this end, X by a finite number of closed convex neighborhoods cover Ui such that If (y) U, n X f (z) I < E/2 whenever y, z C= Ui n X. Let V, and let Vi 1. for i Then the Vi U n u > X) \ (V1 Vi-1) (Ui Borel subsets of X and for those i such that are pairwise disjoint 0 obtain measures Ai on X, supported probability p(Vi) =A we can n by Vi, by defining y(Vi)-1p(B Vi) (B a Borel subset of Ai(B) X). Let xi be the resultant of Ai; since Vi is a subset of the compact Define A set Ui n X, the latter must contain convex xi. Ep(Vi)6x,. If h is a continuous affine function A (h) on X then Ep (Vi) Ai (h) E fv h dl-t h (x), A (f so A Ex. Furthermore, p (h) p (f E [fv f dp p (Vi) f (xi) 6, and the fv [f f (xi) ] dl-t < EEp (Vi) proof is complete. follows
from
Proposition
3.1
that
we
need
-
-
-
=
-
...
-
=
=
=
,
,
-
-
-
-
Proof
that
1,
01 + Ce2
=
f(z)
alf(xi)
=
-
=
-
(1) implies and
f +
E
(2):
C. Let
a2f(X2).
Suppose z
-
aix,
Since
f
that
+ a2X2;
is
X1 X2 E X , a,, we
concave,
a2 >
O
want to show that
it
suffices
to
show
Lectures
58
that
is also
it
By Lemma10.6, 6,1. Suppose that
convex.
and /-t numbers exist
is discrete there
-
Eoj
such that
0 and
and ft
I
=
!
j
E jEj,
=
By applying the Decomposition jyj of X, we can choose zj in
jyj
and
z'j1
=
and hence
to
see
for
i
xi
(Use
of X.
1, 2, the right
=
=
f (Yj) so
=
over
supremum
Proof 10.7
=
aixi
(7ij
convex
for
represents
some
equal
sum
a
+
+
all
p(l).
that
Suppose
function
on
;-
/-z
-
(3):
that
f
E, gives
If
affine if p
3 l'Yljf
-
is
X and that
p is
EiEAj (i
+
E*,
1, 2)
X), =A 0,
r
follows
It
/-zi
that -
exi,
hand,
other
3 l-y2jf (Z2j)5
+
a2l(X2).
maximal
E
of elements
L in
measure
the desired
aixi, =
0, zij
On the
(Zlj)
J)
G
E iYi-
=
elements
1.)
to
discrete
all(xi)
aW2(f)
discrete
implies
LEMMA10.7
3-'172j%)
Z
=
-yijzij combination
=
L-'(r)
the =
=
(2) implies Since I is
that
(below)
tinuous
a
to
1(xi) yi(f) Ejai 1-yijf (zij). (yj) and for each j,
ailil(f)
:!
is
have
side
(0 3' 17ljZlj
f
Each zj
X C
>
E jf
p(f)
p(f)
that
coefficients
the
and therefore
y(f)
fact
X such that
J).
E
1-yijzij
Eiai
=
the
that
(j
z'j2
+
aixi
Lemma 10.1
then
-
+ a2X2
/-t
:
and p 6,; X in (j yj
sequence
i.e.,
Theorem
supf/-t(f)
=
p is discrete
finite
a
1(z)
have
we
Choquet's
on
Taking
the
conclusion.
and
f
C,
E
then
Lemma
and upper semicontinuous, I(x). Ex, then p(l) --
an
I-L
-
affine Ex..
upper
Then
(or lower) f (x) M(f ) =
semicon-
family H of all h in A such that downward and that f h > f is directed inffh : h E H1. Indeed, Lemma 10.2) we have of the in then (just if this be true, as proof if /t h E HI for any M; in particular, inf f p (h) Ex, then p (f ) h c HI inf Jh(x) f (x). It remains, then, to prove the M(f) H is directed about H. To see that assertion downward, suppose that h, > f and h2 > f (hi in A); we want h in A such that h > f and h :! hl, h2. To this end, define subsets J, J1, and J2 of J E x R as follows: f (x, r) : I (x, r) : x E X, r :! f (x) 1, Ji E X, r x hi (x) 1. Since f is affine and upper semicontinuous, that Ji is of hi implies J is closed and convex, while the continuity from the convex hull J3 of JUJ2, J is disjoint compact. Furthermore, to 0 and the and J3 is compact. theorem, (applied By separation PROOF. It
suffices
to prove
that
the
=
-
=
=
--
=
=
=
Section
Uniqueness
10.
closed
the
(but
much
E x R such that
X
J)
-
there
sup L (J)
by L(x, h(x))
simpler)
is affine
Measures
=
a
59
exists
continuous
a
inf L (J3)
- p whenever p that ax is the unique maximal measure Ex. It follows
we
-
denote
=
=
=
-
-
which
represents
x.
Weconsider
problem of uniqueness of representing The following sures which are supported by the extreme points. enable will to 10.3 us to prove corollary Proposition Choquet's inal uniqueness X. theorem for metrizable COROLLARY10.8 ishes
on
every
particular,
if
the
next
If
p is
ex
X.
fx
:
and
In
/-t
f(x)
of X
is
ex
I(x)J,
Proposition As the
\
by
supported
particular, =
nonnegative
compact subset
PROOF. It is immediate
F, subset
a
10.3
example
of
X
ex
X,
from the
X, then,
hence is it
is
ex
X, then
on
easy
orig-
X which
p is
maximal.
van-
In
then p is maximal.
hypothesis that p vanishes on every supported by every Gj containing supported by every set of the form Thus, p(f) =p(l) forf in QX),
f inC(X). implies that of
\
measure
mea-
p is
Mokobodzki
maximal.
(below)
will
show,
we
cannot
Lectures
60
hypothesis
weaken the
such that PROOF. imal
then
F, set,
an
/-t
y such
Corollary
that
THEOREM (Choquet).
of
for
each
locally
a
It
PROOF. exists
a
exists
Baire
set,
or
measure
a
1;
p
Section
know that
follows
it
convex
if
simplex which
/-t
measure
of
we
max-
from
M.
closed a
unique
a
4
A(exX)
X is
a
shown in
was
there
Then X is
space.
exists
a
unique
a
and hence A
unique the extreme points by
supported
and is
and
E,,
-
maximal,
convex
exX is
From Section
Ex.
-
A
if
exists
theorem,
Suppose that
in X there
x
the compact
on
1.
--
p
that
A is
that
10.8
vanishes
and
in X there
x
p(exX)
Suppose
1.
=
subset
simplex
a
each
By the Choquet-Meyer
measure
/-t(exX)
X is
and
E.,
-
"/-t
to
Theorem
X."
ex
If for
COROLLARY10.9 is
\
of X
subsets
Baire
corollary
in this
Choquet's
on
metrizable and
only if
represents
x
X. metrizable
for
3 that
convex) continuous function (strictly : x j(x) f (x) 1. Suppose that f
f
on
X there
X such
that
simplex; previous remark shows that ex X is a Baire set and Corollary to each x in X 10.9 yields Conversely, uniqueness. suppose that 1. We there corresponds measure I-tx a unique Ex with I-tx (ex X) that X is a simplex if we show that for each x in X can conclude then there is a unique maximal measure A Ex. But if A is maximal, A is supported ex X, hence (by f (x) I hypothesis) by Ix : j(x) X
ex
=
X is
=
then
a
the
=
-
-
=
A
-
px*
EXAMPLE(Mokobodzki) There exists
(i) (ii) (iii)
X is ex
a
compact
set X with
the
following
properties
simplex.
a
X is
a
Borel
There
exists
and
such that
v
convex
vanishes
on
a
set,
point
but not x
I-z(X \
every
Baire
a
Baire
in
X with
ex
X)
subset
=
set.
representing
two
0 and
of
X
\
v(X \ ex
X.
ex
X)
measures
p
1, but
v
=
Section
Uniqueness
10.
PROOF. Let yo which
is
Y be not
a
Gj,
a
of
Representing
Measures
Hausdorff
compact
a point containing measure on probability uncountable product of
space
and p a nonatomic take Y to be an
example, in Y which does intervals, yo any point and the base, neighborhood corresponding p Y.
For
we
could
unit
itself.
with
measure
f (yo) Suppose
Let
C(Y)
Mc
61
have
not
product
be the
Lebesgue
f
of all
set
countable
a
of
in
C(Y)
fy
f dp. We first show that M separates points of Y. that A, and A2 are probability measures on Y, with Al(f) A2(f) for all f in M. (We will soon take A, and A2 to be 0 whenever f E C(X) and (/-t point masses.) Thus, (A, A2) (f ) are necessarily 0, so that these functionals proportional, Eyo)(f) i.e., there exists a real number r such that such that
=
--
-
-
=
(2)
A,
Since
p has
distinct
point
if y if A
Ey,
y is
in
A,
take
f y I, getting B (M)
masses,
then
yo,
has two
atoms,
no
A, (f
f yo 1. 0 (B (M))
A2
y I)
follows
r(p
-
Eyo)-
(2)
from
A, and A2 cannot be i.e., M separates points of Y. Furthermore, the Choquet boundary B(M) of M. Indeed, A, A2 E. in (2) and apply both sides to I 0, so A Ey. On the other hand, yo it
=
that
=
measures
Y\
=
=
representing
=
-
We let
X
=
(y
and
K (M);
eyo),
so
we can
from Section
6
conclude we
that
know that
0 (Y) \ f 0 (yo) I and that every maximal measure supported by the compact set O(Y), hence can be identified with a measure on Y. (We will use the same symbol for a measure on X which is supported measure on by O(Y) and for the corresponding measures on X; then Y.) Let Q, be the set of all maximal probability if A (ex X) 1. Indeed, a probability measure A is in Q, if and only the latter that A is 10.8. implies maximal, by Corollary property Suppose, on the other hand, that A is maximal but that A (ex X) < 1; since A is supported by O(Y), we must have a A(fyoj) > 0. Let A, (A aEyo) + ap; since each term of this sum is nonnegative, > Since 0. A, A, >- A. p Eyo, we have p >- Eyo and therefore A, 0 A, so A is not maximal, a contradiction. Clearly, from Q, onto X; Now, we know that the resultant map r is affine X
ex
on
-
-
X is
=
=
-
-
,
if
we
show that
it
is one-to-one,
we can
conclude
that
But if r(Al) simplex. r(A2), then (by definition) in that so f M, equation (2) applies to these two each necessarily vanishes on f yo 1, we conclude that a
for
X
AI(f)
=
(like =
measures. r
=
Qj) is AW) Since
0 and hence
Lectures
62
Al x
then,
A2- It remains,
-:::::
0(yo),
-
EO(y,,) represent
v
to
the assertions
prove
and consider
=
Choquet's
on
p
I-t(X \
(ii)
and
on
X.
a measure
as
X)
Theorem
(iii).
p(X \ O(Y))
Let
is clear
It
0, and subset of X ex X, 1. If v were positive that v (X \ ex X) on a Baire then it would be positive on some compact G6 subset D of X ex X, p and
that
v
x, that
ex
=
=
-
0(yo)
and therefore then be a
Gj
a
set.
X
ex
G6 Thus,
were
a
Baire
this
finite hull
exactly
n
of
n
extreme
PROOF. Wecan
Suppose
that
X is the
convex
linearly points.
(f, (x), unit
,
.
.
.
i-th
basis
R
in
vectors
cone
TX
and
X itself
Rn.
in
induces
a
that
X is
simplex by the Minkowski
a
points, have at
least
points. points
extreme aj,
n
all
i
E N otherwise.
n
Suppose
=
I
let
=
EiEPa-'ajyj
some
the
f
in =
-
it
of
cone
the
X is
E is
X has
-,T(.
-
since
xn;
,
-
sets a
=
independent, fn for fl, defined by Tx .
.
so
,
.
xi
of the
n
elements
in
simplex;
it
a
suppose
of its
hull
=
onto
that
extreme
X generates E, it must it has exactly n
show that -
Yn+j are distinct there exist numbers -
,
-
the
Partition
0.
P and
N, where
Ejcpaj,
have
EiEN(-a)_1aiYi-
TX is
proof,
the
7
=
hull
convex
points Y1 Y2, n-dimensional,
E*) Eaj
the
X generates
nonnegative
Since
=
the
is
E
basis
convex
will
we
the
X
linearly a
in Rn,
Eajyj
Then if
f (X)
for
Since
such that into
the
theorem.
that
+ I
TX is the
that
points;
zero,
through
x
a
extreme
of X.
I
is
Thus,
ordering To finish simplex.
and note
extreme
not
is
X1 7 X2 7
be
Choose
lattice
follows
that
and since
must
E.
spaces.
The map T : E -+ Rn and onto, and carries one-to-one
=
vector"
for
of "sim-
and
Points
points
and if
zero.
only if Equivalently,
if
is not
k spanned by
-
generality
extreme
n
equal
Jf
space
of
loss
of
dimensional
a
extreme
a they fi (xj) 6ij. fn (x)) is linear,
n
the
E, these points
space
"unit
This
Rn.
exactly of its
E* such that the
finite
would
that
the definition
proof that for
one
without
assume
hull
subset would
simplex independent points.
form
and hence
a
Then X is
n.
X has
n-dimensional
with
X)
ex
Suppose that
dimension
convex
(X \
the usual
PROPOSITION 10.10 has
v
section
with
coincides
then
set,
10(yo)l f yof X \ ex X,
=
the fact
Baire
every
on
O(Y)
But D n
O(Y), contradicting
to
vanishes
v
Weconclude
plex"
would be in D.
relative
we
0
so
a
-
Since
E P
i a
from
integers >
if
!
0,
(since
Finally,
-EiENai. these
ai
0 and
are
convex
Section
10.
Uniqueness
combinations, measures
from
on
we
have
X which
Choquet's
of
Representing
represented have support
uniqueness
theorem
Measures
an
element
contained that
63
x
in
X is not
by a
It
follows
simplex. (This using the
step may be proved in a more elementary way by lemma and the fact that the points yj are decomposition
last
different
two
exX.
extreme.)
Properties
11
As
resultant
Proposition
in
was seen
bility
of the
P(X)
measures
the
1.1, the
onto
map
map from
resultant
compact
convex
set
the
proba-
X is affine
and
weak* continuous.
By the Choquet-Bishop-deLeeuw theorem, its reof r maximal probability is still measures Q(X) and from the uniqueness theorem we know that r is bisurjective, if and only if X is a simplex. In this section jective we prove some additional of this but a properties simple map, including potentially striction
the set
to
useful
selection
theorem
PROPOSITION 11. 1
for
Suppose
r
(ii)
-'
affin
is
For each
that
e.
f
E
C(X)
r-1
(iii)
PROOF.
(ii)
is
(i)
function
real-valued
the
r-,(X)(f)
x -+
is Borel
case.
the compact convex set X is a simX -* Q(X) exists and has the map r-1:
Then the inverse plex. following properties:
(i)
the metrizable
measurable. continuous
Since
Assume first
if
and
only if
Q(X)
is
convex
that
f
is convex;
exX is
closed.
affine, its inverse is affine. then by part (3) of the Choquet-
and
r
is
we have r-'(x)(f) Meyer uniqueness f(x), for each x E X. Since the right side is upper semicontinuous, it is Borel measurwhenever f is in able, and it follows that (1) is Borel measurable, C C, the dense subspace of C(X) spanned by the convex functions. If f Cz C(X) is arbitrary, it is the uniform limit of a sequence from C that the is so of limit of Borel a sequence C, pointwise (1) measurable functions, hence is itself Borel measurable.
theorem
-
-
R.R. Phelps: LNM 1757, pp. 65 - 72, 2001 © Springer-Verlag Berlin Heidelberg 2001
=
Lectures
66
(iii) of
ex
Then there
X.
1.4,
lim E,,
To
Ex, C X.
=
y,
x
evaluation
a
net
=
e,,,.
(hence
x,
Thus, r-'(xo) X, suppose that p
and there
simplex,
a
X with
ex
2
exists
that
(sy a
+
-+
xo
is
=
=
.1 2
(y
+
e,) represents
maximal
measure
and
x0,
limr-'(x,)
ex
measure
e,o)
p >-
X is
in
x,,,
Theorem
in the closure
xo is
=
C
x0
probability
Since
p.
that
see
The
at x0
dominating
exists
Choquet's
and that
is continuous
r-'(x,)
by Proposition
z),
r-'
Suppose that
on
measure
r-'(x,),
so
we
have
Ex"
Thus,
M
if
that
closed,
compact
[5]
r-'(x,,)
and hence y
E, X is
=
ex
is weak*
=
Q(X)
then that
so
To prove P (ex X). It
x.
z
=
r
A >-
-
is in fact
note
converse,
Q(X) homeomorphism.
affine
an
the
follows
that
of those X which given several characterizations "X is a simplex and ex X is closed" (which is why simplices satisfy Note that the called Bauer simplices). with this property are often shows that any Bauer simplex X can be identiforegoing proposition on the compact Hausdorff measures fied with the set Of all probability Bauer
space
If
X,
ex
has
X.
X is not
some
results
a
measure
conditions
11.4),
respectively.
DEFINITION.
By
a
Q(X)
X into
p,,
11.2)
(Proposition
way
possible to choose, for each We present x. having resultant
is still
it
give
which
affine
from
simplex,
maximal
which
under and in
The first
of these
selection
for
such that
r(px)
the
a
this
can
map x
r
for
an
way
we mean a
each
E
two
(Theorem Fakhoury [35].
measurable
is due to H.
=
be done in
x
x
map
x
-+
px
E X.
P, and P2 are cones in real vector and that 0 is an order-preserving, that P, is lattice-ordered spaces, and positive additive homogeneous map of P, onto P2. If there exists another P, such that 0 o 0 is the identity map 0 from P2 into In particular, ordered. then P2 is latticeon P2, if there exists an affine selection for the resultant map from P(X) onto X, or for its
PROPOSITION 11.2
restriction
r:
Q(X)
Suppose
-+
that
X, then X
is
a
simplex.
Section
of the resultant
Properties
11.
67
map
that if x, y E P2, then x V y to verify straightforward exists in fact, and is given, by 0[0(x) V 0(y)], and this is all that is about r, say, the assertion to obtain needed. To apply this result extends it and its selection one first by homogeneity to the cones TC. P, (As in Section 10, we have assumed R+Q(X) and P2 in a hyperplane which that X is contained without loss of generality k misses the origin, so that R+X.) PROOF. It
is
=
--
=
The property
cinctly More
(ii)
in part
precisely,
we
will
Proposition
of
by saying
described
r-1
that
make
use
could
be
Borel
measurable."
11.1
is
"weak*
of the
following
more suc-
terminology.
0 from a compact Hausdorff space X into function a compact Hausdorff space Y is said to be Borel measurable provided 0`(U) is a Borel subset of X whenever U is an open sub-set of Y. on Y, functions If A is a separating family of continuous real-valued the real-valued measurable Borel is will that we if A-weakly 0 say function f o 0 is Borel measurable on X, for each f E A A
DEFINITION.
(the
The lemma below shows metric
then
space,
A is
that the
a
of
Using the notation separating family of
compact
metric
PROOF. Since
Since
which
Y is
of basic
each Ii Borel
space,
open sets
of the
But then
0`(U)
following
selection
THEOREM11.4
Then there
exists
Suppose a
Borel
that
definitions, suppose on functions 0: X -4 Y is Borel
a
of the compact space Y, the weak with the initial topolgy.
Y coincides
of Y is
q7 _jfj-'(Ij),
=
a
n(fi
o
Borel
X is
the a
measurable
show that
to can
assume
0)-'(Ii)
that
fi
E
union
A and
0`(U)
is
in the
of X.
by proved independently is Rao's. below proof
metrizable map
x
a
U has the
and each set
subset
was
[791;
countable
a
where each
Thus, Y, we
theorem
and G. Vincent-Smith
a
real-valued
any open subset
form
is, by hypothesis,
intersection
[67]
is
an
set
above
if Y is
that
coincide.
function A-weakly Borel measurable.
Then
points on
the
continuous
interval. open real whenever U is open in
is
The
it
A separates it defines
metric
a
above form.
Rao
Y.
if (and only if)
measurable
topology
space
measurability
of
LEMMA11.3
fact)
standard
useful
the two kinds
--+
compact
convex
from
X into
px
M.
set.
the
Lectures
68
probability
represents
x,
all
probability
measures
follows, L
(rather
a
dense sequence loss of generality
f A fl) f2,
ff-ln=l
fnsubspaces
i
-
1
-,
-
fn
that
A(X)
1,
Ao
for
c
is n
A,
of X
c
c
A(X)
An-,
the
as
We can linear
which
and write
c
An
c
a
without
An of A(X)
have
sequence
a
C
...
of
existence
assume
span
Wethus
1) 2) 3.
...
argument
proof of Choquet's implies the existence of
well
as
the
in
not
=
x.
space of
C(X) \ A(X).
in
W
X, the measure point of the set of
As in the
metrizability fo in C(X)
the
function
convex
of X.
element
an
Theorem
E
in the induction
X to be the state
consider
x) for (Section 3),
strictly
closed
of notation
than
Theorem
x
I-tx is an extreme exX which represent
on
simplicity
will
we
each
and such that
p.,
PROOF. For
for
that,
X such
on ex
measures
Choquet's
on
U
of
C(X)
An-, and fn-1 and such that their union A0,, UAn is a dense subspace of C(X). Let Sn denote the state X), always considered in the weak* space of An (hence So Define On: Sn -* Sn+j for each n > 0 as follows: topology. An
such that
span of
is the linear
=
=
On(L)(g+Afn)=L(g)+Aan(L),
9EAn,
LESn,
AER
where
an(L)
=
inf
f L(h)
+
11h,
-
fn1j:
h E
A,J.
fnjj is continuous Now, for each h E An the map L -+ L(h) + 11h which all such h is upper semicontinous, over on Sn, so the infimum function implies that for fixed g E An, A G R, the real-valued -
L is Borel over,
measurable.
if A
0
On(L) (g
k >
and in-
0,
then
Ok(L)(g) On (L)(g)
L(g)
=
On(L)(g),
An. Let S,,, denote the state space of A,,,; the coherence property just shown makes it possible to define If L E X and g E A,,,, X -+ S(,, as follows: a map 0: UAn, then On (L) (g). Every g E An for some n > I and we can let 0 (L) (g) dense subspace A(,, continuous element of S,,, is uniformly on the hence admits a unique extension to C(X), so we can identify S,', while
=
if g E
=
=
the
with
set
probability
of
measures
L E X and g E A0, each of clear from the definition
whenever is
also
Borel 11.1
(ii),
surable
Borel
for
measurable the
O(L)
measure
0
Since
X.
that
A,,,,
O(L)(g)
=
has resultant
L(g) It
L.
0 (L) (g) of Proposition
map L -+
the
As in the proof A,,,. L -+ O(L)(g) that implies
each g E
of
density
on
is Borel
is
mea-
each g E C(X) and Lemma11.3 shows that L -+ O(L) is For the remainder of the proof we will write AL in measurable. for
place of 0 (L). To see that each AL is supported by ex X, (as in the proof of Choquet's theorem) to show that AL(A) By definition,
AL(A)
=
Now, if
g E A (X),
=
OO(L)(fo)
inffL(g) then
=
+
so
is
JJg g'
-
-
Oo(L)(fo) foll: g +
=
g E
IIg
-
it
suffices
=
AL(10)-
ao(L)
A(X). fo 11;
g'
moreover,
Consequently
L(g)
+
JJg
-
fo 11
=
L(g')
! inf
f L(h):
h E
A(X),
h >
fol
>
fo.
Lectures
70
inf
pL(jfo)
hence
fo imply fo It
h >
fo,
-
Now, h E A (X) and h Thus, AL (A) > AL (A);
expression.
inequality
reverse
to
h E A (X),
f AL (h)
Choquet's
on
(10).
is obvious.
AL is
of the
element
extreme
an
of
set
Suppose, then, /-t2 /-tl suffices It show to + 2AL Al A2these functionals that AL are equal By hypothesis, A2 on A,,,. Al show will that are on we on A(X) equal Ao. Assuming A., they that they are equal on A,,+,, i.e., that AL(fn) Al(fn) /-t2(fn)that Of definition the Recalling AL we see probability
measures
such
two
are
which
=
represent
and that
measures
that
L.
7
=
=
=
=
AL(fn)
an[0n-1(L)](fn) 0n['0n-1(L)](fn) On(L)(fn) + jjg fn1j: g E Anj inflOn-l(L)(g) + 119 fn1j: g E Anj infIAL(g) inffAk(g)+Ilg-fnll:gEAnl>/-tk(fn)) -
=
-
=
It
last
inequality
follows
that
holding
AL (fn)
argument completes
foregoing ways. First,
The in two
a
selection
of first
Baire
exists is
metrizable
A is
tains
the
mapping
the
measure
B (A)
is
the
This
last
tant
special
Mx
x
are
2 since
or
A2 (fn)
and
fn I I
-
fn.
>
induction
obvious
an
by much longer
has been extended
(using
Suppose subspace
a
extended
that
X is
of (real or complex) Then points.
from
X to
P(X)
evaluation
represents
at
which
certain
non-
to
C(X) there
for
such that
x
and
properties.
metric
compact
a
there
that
measures
measurability
and separates yx
proof)
result
this
reasonable
[76]
Talagrand
M.
maximal
the extreme
into
He then
-+
IIg
g +
px(B(A))
and
space
which
exists
a
each =
1
x
con-
Borel E
X,
(where
Choquet boundary for A). result cases.
Cn and A consists which
I
(fn)
X, retaining
closed
map
k
proof.
the
class.
constants
measurable
Al
=
he showed
spaces
a
for
theorem
COROLLARY11.5 that
=
=
=
=
the
=
analytic
can
For
be
instance,
of the in
improved
X,
continuous then
substantially
if X is
there
a
functions exists
on
a
impor-
in certain
connected
open subset of of X the closure
selection
x
--+
p.,
as
Section
11.
Properties
of the resultant
function complex-valued that, for each continuous the ilov boundary of A, the map X E) x -4 px(f) is analytic. for instance, [80] and references therein. above such
71
map
f
on
See,
Application
12
S be
Let
S
set,
a
functions
surable
S
T:
A
finite
nonnegative
T-invariant)
are
this
that
in the
S
general elementary
mea-
(or
be invariant
as
below).
probability obtained generalization
the
p is
the
/-t-invariant It is easily p
ergodic
if
definitions
other
by Proposition
of
12.4
=
a =
or
p(A)
"ergodic"
below.
A of S is said to be
[By
AAB
family of (A \ B) (B \ A).] or more simply by Sl,. by S,,(T), of S.
sub-o-ring 0
first.
aspect
each T in T.
0 for
extreme
illuminates
Choquet-Bishop-
The
U
be denoted
S,, is M(A)
that
seen
of the which
theoretic
S. An element
on
difference
will
sets
be used to prove a J. Feldman [37]
via the
measure
/_t(AAT-'A)
if
symmetric
all
the
a measure
(mod p),
invariant
In
measures
We treat
un-
probability average" "integral 1956, Choquet [17]
Subsequently, description
type. theoretic
that,
state
invariant an
could
theorem
of this
measure
(and theorem).
result
which
every
of the set of invariant
measure
of
have
A E S. to
literature
(definition
representation
his
Suppose that
the
on
an
we mean
we
each T in 7- and A in S.
for
many theorems
theorem
de Leeuw
are
p(A)
=
measures
observed
points
S is said
S has
on
ergodic
gave
family
a
each T in 7-
S whenever
on
p
and 7-
S,
for
hypotheses on S, S and T, a unique representation
der suitable
fairly
E
measures
if
There
of
T-1A
measure
lt(T-'A)
measure
S and
-+
S, i.e.,
S into
from
ergodic
of
of subsets
o-ring
a
and
invariant
to
in
Wewill
I for
--
the
We call
each A in
literature; this
discuss
an
ours
again
at
invariant
[There
S,,. is
motived end of
the
section.] Now, the
forms
a
set
convex
of all cone
invariant
P, which
R.R. Phelps: LNM 1757, pp. 73 - 78, 2001 © Springer-Verlag Berlin Heidelberg 2001
nonnegative generates
the
finite linear
measures
space
P
on -
S P.
74
Lectures
Furthermore, a
the
base for
we
that we
do this
will
prove
and that
dvldp ft
f
for
proof
all
say).
,
p and
absolutely
is
v
by
Then
is
v
with
on
P
P; and
To this
end
although
the
S, that
respect
if and only if f
invariant
on
Sion).
measures
continuous
is
-
lattice
a
Feldman,
him to M.
are
v
P is
-
Theorem
measures
measures.
to
is attributed
that
P
ergodic
the
are
(due originally
lemma
Suppose
LEMMA12.1
invariant,
of X
points
basic
elementary
present
probability no topology
of course, defined we show that First,
later.
the extreme a
X of invariant
set
convex
We have,
P.
Choquet's
on
p is
to A
=
f
o
(with
T
a. e.
T in T
f
PROOF. If
each such T
v
f
=
we
o
T
all
T in
T, and if A
E
S,
then
for
T in
T.
have
(T-'A)
fT-IA f dp fT-IA f oTdp fA f d(y T-') fA f dl-t
=
=
o
=
To prove a real
p for
a.e.
the
=
suppose
converse,
v
fx
T-'
o
v
=
=
v
for
(A). some
rf, let B \ A v(B) -rp(B) fB (f 0. Moreover, r)dp > 0 and we have equality if and only if p(B) v(C) Now, v(B) v(T-'A) fc f dy :! rp(C). v(T-'A n A) v(A) v(T-'A n A) v(C), and similarly, p(B) p(C). > r r p(C) Thus, v(B) p(B) v(C) v(B), so equality holds It follows that I-t(B) 0 and p(C) 0. Thus, for any throughout. < r :! :! and r I r I =- f y : f (Ty) r, f X : f (X) T-'f x : f (x) I differ Given
C
and let
=
number r, let A Then f -r A\T-'A.
=
>
f (x)
:
0
=
f
o
T
r
o
!
T,
we
h(x)l
we see
conclude
that the
f < f proof.
o
T
Section
Application
12.
If
COROLLARY12.2
Sl,+,,
then p
v
invariant
are
Measures
measures
75
and p
on
v
=
S.
on
v
=
p and
Ergodic
and
to Invariant
dvld(p + v). We will have for all A in S if fA f d (p + v) v (A) fA g d (p + v) for all p (A) if f such A, i.e., (p + v). Now f and g are S-measurable g a.e. in fact, Indeed, if functions on S, and, they are S,+, measurable. f
Let
PROOF.
dpld(p
=
+
v),
g
=
=
=
-
T E
that
T, f
and I
then T
o
v
If
=
=
(-oo,
=
p,
implies
Lemma 12.1
and p + v are invariant, (p + v). f and g o T g a.e. and then (f o T)-'(1) f -'(I) r),
since
is
r
number
(f '(1))
differ
a
T-1
=
real
is (p + v) measure zero (their symmetric difference E S,,+,. Thus, f of f x : f (x) :A f (Tx) 1) and hence f '(I) a subset If A g) (x) > 0}, f x : (f g) is S,,+,-measurable. (and similarly vA and hence 0 then A E Sl,+, g)d(p + v); it pA fA(f that f < g a.e. follows (p + v) and an analogous argument shows f > g a.e. (p + v).
only by
of
set
a
-
-
=
The
PROPOSITION 12.3
suffices
to
produce
to
negative the proof
a
order
PROOF. In it
S is
on
measures
invariant
(in
Of
P
cone
lattice
its
P is
greatest
lower
measures
finite
all
p and v. we have
nonnegative
invariant
its
ordering,
ordering).
own
show that a
-
-
a
lattice
in
own
bound in P for
f f f
Let
f
any two
non-
and g be defined as in o T and g = g o T a.e.
Corollary A g a.e. T in T, hence (f A g) o T (M + v). Since measures lower bound p A v for two nonnegative the usual greatest is defined (f A g) (M + v), Lemma12.1 implies that p A v by p A v It follows is invariant. easily that p A v is the greatest lower bound induced by P, so P is a lattice. of p and v in the ordering
(p
+
v)
12.2;
of
for
all
=
Suppose
PROPOSITION 12.4
and
probability only if p
is
PROOF.
Suppose
that
invariant X
that
if
0
Let
a
itself,
a
n
we
represent =
=
and the on
of this
extreme
from
for
measure --
the
which
0, 1, 2,...
sequence
1AnT
a
(x,
y +
homeomorphism
of the set X of T-invariant
points
S do not form
fact
k
space,
T is
the
as
function nonconstant be any continuous I x J into itself T from S by T(x, y)
compact Hausdorff
1, let /-t,, be the points (n-', kn-'), point of X and the
n
which
0
measures
proof
circle,
J be the
and let
R and define
of S onto sketch
=
a
closed
special assigns
case
subset
O(x)
mass
n-1
of X.
Wewill
x.
For each
=
to
each of the
Then /-t,, is an extreme in the weak* topology converges ,
n
-
1.
78
Lectures
Lebesgue
to
measure
measure
f 01
on
There
x
least
at
are
IL
of
of course,
further
Another
origins. A
T-'A
=
our
for
sense
definition
ergodic
p is
measure
for
if
each T
ergodic
is
S,,
each A in
X,
x
certainly
p is
so
0(0)
Since
J.
0,
=
not
definitions
Theorem
probability
every
in this
extreme
set.
"ergodic measure" in these simply defines the ergodic measures to be the set of invariant probability measures; this,
the extreme
points requires
101
two other
One of
the literature.
on
J is in
Choquet's
on
p(A) in Tj.
in this
there
of
work if
one
goes
as
follows:
or
p(A)
=
So
Sl,,
=
0
Since
C
relate
the
any
its
to
probability
each A in
So
f
=
A
measure
ergodic
in
clearly
coincide
if
/-t(AAB)
So such that
B in
notion
An invariant I for
The two notions
sense.
exists
is to
This
0.
=
for
if T consists of a single T (or equals function instance, the semigroup generated let B by T)-simply nn-- , Uktn T-kA. More general of hypotheses on T which guarantee the equivalence the two notions Farrell are given by [36, Cor. 1, Theorem 3] and Varadarajan [78, Lemma3.3]. The following simple example, due to shows that they are not always the same. Farrell, occurs,
=
n=
EXAMPLE Let
S
=
[0, 1]
fT,,T21,
[0, 1], let TI(x,,X2)
x
S be the
where
=
Baire
of S and let
subsets
(xi,xi),
T2(x,,X2)
=
Then
TI, T2 are continuous maps of S onto the diagonal and So consists of S and the empty set. For any subset
(AATi-'A)
n D is
in D is invariant
support I in
in
So, our
in
and
D.) Thus,
but the sense.
semigroup
empty;
point It
is
generated
S,,
it -
every
follows S.
(In fact,
such
masses on
that
any
D are the
interesting by 7- is simply to
note
invariant
every
measure
takes
only that
7- itself.
/-z with
measure
only ones
the
(X2 X2)A of
support
measure
the values
which
are
S, S,
D of
has
0 and
ergodic
(noncommutative)
:
A method
13
extending
for
representation
the
theorems:
Caps
were
for
these
of
elements
any such set
results of
elements
can
be
lead in
compact
regarded
we
a
base for
way to cone
it
As noted closed
a
sections
Section
in
admits
compact base.
a
so
for the
representation
which
10,
cone,
convex
theorems
possible
is
in earlier
with
dealt set.
convex
as
natural
convex
whether
to wonder
natural
a
a
closed
a
which
theorems
representation
The
It
such theorems
to obtain
is
for
be no completely general of two lines result are, however, satisfactory these of One interest. of which are approach, both due to Choquet, notion of measure ("conical involves measure"), which a more general The other approach involves in [19]. is outlined replacing the notion to extend the scope this makes it possible of "base" by that of "cap"; will be devoted to the This section theorems. of the representation we consider only proper latter Throughout the section, approach. K n (-K) cones K, i.e., f 01. theorem" we mean, of course, In using the term "representation which of measures the mere existence points; represent more than be these measures in by the supported some we require that, sense, the In the case of a convex cone, only possible extreme points. the notion of an and we must introduce is the origin, extreme point
of cones, but there There of this nature.
class
a more
seems
to
=
extreme
ray.
DEFINITION.
R+x
=
jAx
:
A ray p of a A : 01, where
Ax, A > 0, any nonzero A ray p of K is said to be and x Ay + (1 A)z, (y,
y
=
-
x
E
K,
element
=
an z
cone
convex
x
of
extreme
E K
=A
K is 0.
a
Since
R+x
p may be said
ray
0 < A
0, and
is
s,,
-
ly
=
:
y E
K, p(y)
0, then the compact set C would
conclude
a
By choosing a 1. Suppose p(x) 0.
=
y
=
probability that
p(p)
p(y) [y/p(y)]
+
measure =
p(x)
on
1.
=
11; since p, is lower semicontinuous, I this is a Borel set, and if p(A) > 0, then p(p) f p dlt contradiction < a + + 1, f 4 p dp fc, p, dl-t p (A) p (Ci) p (C) The assertion maximal which shows that 1. concerning p(Ci) Let
fy
E
=
multiple of p, if necessary, positive that y E ex C, y =A 0. Then 1 ! 0, so p(y) = 1. If /-t [I p(y)] -
which is the union
cone
Then there
measure /-t on C probability is supported by C1. If /-t is a maximal measure, (in an appropriate sense) by the nonzero extreme
appropriate ray
:
convex
0.
11 such that x f 01 C Ci, and
PROOF. Since K is the union
for
closed
x
0 for every f zn I E co with (z, x) E K, x : x 11 is 0, such that B Ix : x E K and (z, x) =
=
=
=
=
But
compact. hence
Xn
sequence struct
C
Ix
.
.
,
.
0, z,-, ', 07 0,,
-
-
)
-
E
K, define
11
:!
is
p,
on
Since
B.
ball
all
by p(x)
K
it
the
is
Since
0, this To
con-
Then
EXn.
=
and
n,
--+
z,,,
be compact.
(since of fl).
compact
0 for
>
zn
and hence B cannot
cap for
p(x)
E K:
shows that
property
unbounded
universal
a
=
(0, 0,
=
is
first
the
intersection
positivehomogeneous, Ix E K: p(x) :! rj is compact for all r > 0, so p is Since the unit ball additive. and it is clearly lower-semicontinuous, normed linear of the dual of a separable space is always metrizable metrizable. C is in the weak* topology, we see that Finally, suppose K were metrizable. Since f, is weak* sequentially complete and K in itself. conclude that K is of second category is closed, we could relative interior has to K. closed and is C But K and empty UnC, (For instance, if x c C, then x + an E K \ C and is weak* convergent to x where an is the element of f 1 which equals 2 at n and equals 0 K of the
with
weak*
compact
unit
p is
=
,
elsewhere.) Later,
we
cap,
easy to
construct
K
generated
without
(other
but
by
than
base
a
points.
following
result
B,
with
the
where B is
a
no
does not
of its
union
nontrivial
no
Then K has
C is
a
gives
measures
on
some
bounded closed rays,
is
a cone
set
convex
hence
a
It
caps.
Take
caps:
extreme
have
no
caps
concerning
information
unique-
caps.
If the cone K Conversely, simplex.
PROPOSITION 13.3
K,
cones
which
a cone
is nevertheless
f 01).
of maximal
then
example of
an
which
closed
extreme
The ness
give
will
universal
is
a
if
lattice
each
and
point
of
if C
is
a
cap
K is contained
of
Lectures
84
in
of
cap
a
K which
is
PROOF. A cap C of
p(x)
and
Co
cone
ifxo
(for
(0,0)
p(x)
Assume,
alattice.
Oandr-'xo
K is
lattice;
a
we
xo
(x, r)
--
and yo
=
-
-
-
only
show that
p(x
q that
show that
zo
!
p(y that
!
if
w) p(z
(a, Xn)
Let
0.
=
n,
a
for
I
this
10)
f (x)
0 for
>
is
function
suitable
a
C
that
suppose
cap,
for
and Xn
s
with
section
have
to
a cone
K has
Section
=
E
p.
Suppose that
x
1, 2,3,...
a
B
that
-54
=
fx
K.
which
K is Kn
gives
:
f (x)
closed
a
(-K)
locally
a
topological
convex
101.
=
in
cone
a
Then K has
a
compact.
B, then
I [0, n]B
=
The sets
compact,
C; then
c
contradiction.
a
result
base
compact Kn
0 in
are
a
p(jn)-ljn
=
compact base.
a
space E such that base if and only if K is
PROOF. If
01i
K such
y in
hence compact.
object,
an
convex
compact
n
Jnp(Jn)J
=
PROPOSITION 13.6
locally
cap
Xk >
:
E C.
universal
a
such
were
It
n.
=
C is not weak* compact,
so
We conclude criterion
I
Ik
=
a
sequence which is 0 except in the n-th place, where C is universal, p(Jn) < oo, and since C is compact,
Since
1.
:! -
J
C be those
and
have
K does not
C
x
:
that
see
define
we can
01;
-=
Let
(and closed)
C is bounded
that
To
Xk
K,
E
87
to straightforward C x convex Furthermore, of all dimensional subspace of E consisting E J; it Ify E C,thenO A
compact,
0.
points
conclude
f
functional
f (J)
is
same
It
on
E
follows
since
it
is
A different
14
method
extending
for
representation
the
theorems
probability point x of X,
When we say that X "represents" a
a
for each continuous for
a
larger it
uous)
functions.
affine
affine
the
which
holds
functions the
are
for
locally resultant class
x,
first
proof which follows hypotheses, namely,
The
(1) of
f
The function
f
to any
f
If
tainly
has
inal
If
limit
of
we
p is
i.e.,
it
affine
holds
for
functions
(but
of continuous
a
compact
convex
not
of
subset
a
a
category
theorem
dense set
of
to
on
M, it
only
Borel
measurable,
it
is bounded.
that
f
is
not
Let
bounded.
assume
asserts
one
restriction
that
to
a
X with Baire
(1)
restriction
of
continuity.
point
it
compact
of first
follows
f
stated
and the
consequence
function so
that
following:
the
any
A classical a
than
functions,
of continuous
sequence
points of continuity f. In order to
suffices
then
weaker property
a
need
and its measurable, of first Baire class. again
hypoth ses
respect
show that those
of X has at least
Borel
a
X is
uses
affine,
is
compact subset
is the
of X is Baire
will
semicontin-
X.
on
in the
f (x)
X.
on
if
we
lower
measure on probability f (x) for each affine function f of first
E and =
(or
sequence
a
=
extending the show that this latter equality For instance, 10.7 Proposition
class,
of
limit
[18]).
p(f)
then
section Baire
I-t(f)
that
set
convex
One way of
X.
semicontinuous
In this
functions
space
convex
f
on
be to
upper
of
THEOREM(Choquet
mean, of course,
of functions.
pointwise
affine)
necessarily
would
class
showed that
we
function
theorems
representation holds
affine
compact
on a
IL
measure
Baire
from
the
is
cer-
subset
of the class
orig-
with integrable if a function to prove that f satisfies (1), of of be and a f, point continuity suppose y S nce X is compact, we can find a net x,,
R.R. Phelps: LNM 1757, pp. 88 - 92, 2001 © Springer-Verlag Berlin Heidelberg 2001
know that
is
Section
and
Different
14.
point
a
X such that
in
x
of the
Extension
-+
x,,
Representation
f f (x,) I
and
x
U of y such open neighborhood U and choose 0 < t < 1 such that ty + (I
Choose
u,
=
leads
ty
+
to
a
(I
t)x,
t)x
-
tf (y)
--
is
unbounded.
is
bounded
(I
t)f (x,,),
-
on
Eventually,
E U.
+
89
this
contradiction.
We next
introduce
Of (A)
X, let
f (u,,)
Since
E U.
f
that
an
-
Theorems
inff Of (U)
f (A)
sup
=
U open,
:
f (A),
inf
-
the
and for
If A C
Oxf
X, let
in
x
f:
of
oscillation
U1.
E
x
for
notation
some
then measure on X and E > 0, nonnegative there exists a sequence f Anj of nonnegative measures on X, supBorel subsets S,, of X, such that p EA" disjoint ported by pairwise and Of (Kn) < E for each n, where Kn is the closed convex hull of Sn
If
LEMMA14.1
p is
a
=
-
show below
PROOF. We will
X, v 7 0, that Of (K)
then
on
there
(i)
three
Each A in
positive
(ii)
this
the
A
-
Since
the
sets
the
measure,
EA,,
converges
=A
A
A
-
In the
Borel
a
Let
out.
X with
on
Sx of
subset
by
Z
to
Mo is countable;
EAn,
the
then
say
convergence
restriction we can
a
pairwise
are
then
A
apply
it
is
v
of p to B and easily verified lemma is
maximal
element
Mo in Z.
disjoint
and of
set,
Mo
=
so
Onf-
It
positive
induction
step
p
from
follows
theorem, say, that the of p to US, where S,, the
take
Zorn's
ordered exists
E.
simply
step,
restriction
inclusion
that
step.
(i.e., zero)
K is
maximality
the
v
Theorem
measure
assume
induction
Since
Y is
where
of the
of
support
Oxg
p
< s,
Of (K) certainly
the
prove
of
positive
contradiction
Denote
J,
c-
x
:
convex.
J
set
open sets
function
valued
to
of S.
hull
convex
a
B of
set
we can
S be the closed
let
union
Y
then,
remains,
0,
e >
Borel
a
such that
Choquet's
on
real that
J, the would have J c Y; in
From the definition S \ Y is nonempty. of S it follows consequently, that any neighborhood of any point of S \ Y has positive measure.
Since
closed
Y is
convex
-neighborhood
hence there
exists
sets
of the
will
cover
form
V.)
At least
Let
of S
we can
\
choose
Y which
< e.
1)6
it
of these
one
must
K be the
a
closed
misses
as a
finite
union
of
(For instance, finitQly many < n an nel, 2g(x) integer,
v measure, Vk has positive of set v J0 compact positive
sets
contain
closed
An and A with
and the support
which
of the
(0f)(Kk) Ak- It follows
theorem.
By disjoint
0.
Suppose that M 6, choose we can lemma, -
the
supports
such that
in a compact Of Yk is contained Let Ak < E. 14111MIJ and let =
that
Xk E
Kk and hence f (Xk)
I I AI I
-
that
=
TA, then
a
ft
(f )
=
that
Cartier
function
convex
Tx, (f )
(TA) (f )
following
Polya-Blackwell-Stein-Sherman-
A.
>-
/j.
continuous so
E,
is the
section
=
!
>
f f
theorem ,
f (x) dA of
under
A (f
=
Since
X.
for
all
).
It
x.
is
The main
Hardy-Littlewoodthat hypothesis
the
X be metrizable. THEOREM Suppose that a
locally
measures
on
such that
p
a
compact metrizable
of
subset
convex
Borel probability A and A are regular T if and only if there exist a dilation
TA.
=
proof of
The
(which
X is
space and that X. Then p >- A
convex
does not
this
depends on a general together with metrizability),
theorem
use
result a
of Cartier
classical
result
of measures. disintegration X as above, we consider the space F C(X)* x C(X)*, using the product of the weak* topology with itself. Thus, F is a and linear continuous functional L on F locally convex space, every on
the
With
is of the
=
form
L(a, 0) for will
some
pair
=
of functions
be interested
in two
a(f)
-
0(g),
(a, 0)
E F
f g in C(X). Throughout this subsets J and K of F, particular ,
section
defined
we as
Section
Orderings
15.
of Measures
and Dilations
95
follows: K
=
f(A,y):A !0,p !0and/-o-Aj
J
=
f(E,,,V):xEX,V,6xj-
easily verified implies v >- E, It
that
is
(since
the
map
K is
we see
and J is
homeomorphic
of point
masses
closed is
of the
subset
a
(a, ) imply (I)
of all
compact
for =
1,
and is itself
is
we see
that
P,
set
from
Since the
J is
Since
1.
=
K n H is
a
(a, ) closed
compact
X, Its
convex.
Indeed, hyperplane
J
K.
the
E K and
a
(1)
H 1
=
of the
subset
convex
P1, hence is compact. is clear that K n H is
Ex
-
P, into
not
compact base for
v
combination
a convex
set
K n H of K with
(1)
a
a
J is
is continuous
mass
intersection
which
convex
v) graph).
a
Since
in F.
K; furthermore,
of
its
to
not
cone
convex
J C
point B, however,
is
hull
convex
closed
(resultant
-+
v
a
that
Thus B C K n H
x
base for
K; we will K n H. This will show that B be true if B generates certainly ! if K L F* L and whenever L ! 0 on B. Now, if 0 on E K, i.e., > L ! 0 on B then L 0 on J, so assume there exist f g in C(X) ! such that whenever 0 v we will f (x) L(E,, v) v(g) 6,,; It
compact.
a
=
,
,
-
-
show that
L (a,
)
a
(f )
(g)
-
!
(a, 0) E supf v(g)
0 whenever
K.
Recall
: v ej. 3.1) that for each x in X, 7(x) < < :! :! that so 0( ) f (x), f Thus, 0 (g) y V(x) g and a(y) :! affl; from Lemma 10.2 we know that 0(y) < a(g) and hence L(a, 0) > 0.
(Proposition
It
follows
=
that
.
following 1.2 Proposition The
proposition
only if
there
is
now an
(CARTIER)
PROPOSITION 15.1 and
-
exists
a
immediate
(A,p)
An element
nonnegative
measure
on
consequence
of F
is
in
of
K
if
J which represents
(A, p). to the
We now return that
X is metrizable
exists
for
a
and that
nonnegative
each L in F*.
A(f)
proof of p >-
A.
m'
on
measure
This
-
means
p(g)
=
fi
Assume, then, there By the above proposition,
the theorem
that
[f (x)
itself.
J such that
for
-
fj
L dm'
C(X)
each
(f, g)
v(g)]
dm'(Ex, v).
in
=
x
L(A, M) C(X),
Lectures
96
(x, v) : x (x, v) from
S
Let
(Ex, 1/) m' to the
a measure
above
we see
(a)
Equation carried
shows that
We now state measures
a
[13,
m is
function probability
(i)
dm(x,
the
function carry
we can
0, f
=
0 in
--
v), v).
dm(x,
probability projection
Theorem
measure
of X
of the theorem
P,
x
S which
on
disintegration
on
is
X.
onto
of
58].
p.
Y.
A
Let
measures
For each h in
on
compact metrizable
are
Y onto
X,
0'
denote
mo
-
Then there
0.
a
case
Suppose that Y and X function from on
Since
-
a
v
the natural
continuous
measure
I
Ex
-
fS f (x) fS (g)
=
special
a
S is
=
p (g)
A under
onto
)
A (f
v
homeomorphism, By alternately choosing g that for all f, g in C(X),
S.
(a) (b)
P1,
C-
v
J onto
m on
equation,
X,
E
Choquet's
on
exists
a
and that
x function following
C(Y),
the
X,
support
function
is
a
image of m under the the -+ Ax from X into properties:
the
the
Y, with
0 is nonnegative that
spaces, m
Ax(h)
x ---
is
Borel
measur-
able.
(ii)
For each
(iii)
in
x
C(Y),
For each h in Weapply
the
result
this
as
projection previously.
mo
=
measures
on
resultant
the
of S onto and is
0',
(2)
the
P1. which
resultant
there
so
exists
S, satisfying in
It
the
and let
as x
contained
is
we
=
S C Xx
P1, let 0 be
m and A be the
have
above three
the
that
probability T',
We let
properties.
the
measures
(a) implies
noted,
A., from X into
---
0-'(x).
in
fX Ax(h) dA(x). Let Y
X,
Then,
introduced
=
follows:
of S onto
natural A
m(h)
of Ax
be
P, of the image of Ax under the natural projection to prove that Tx satisfies the properties (1)
remains
define in
P,
dilations, of the
and that
image
of
Ax
IL
-
means
C(X), TX(f)
=
fS
v
TA.
(f ) dA,, (y, v),
The fact that
for
that each
Tx
f
in
Section
Orderings
15.
(y, v)
Since
f,
functions
implies
S
in
this
of Measures
and Dilations
v
sy,
-
that
we see
97
for
continuous
affine
becomes
T-W
fS f (y)
-
dA , (y, v).
supported by 0-'(x) f (x, v) : v ExI, and Tx (f ) f (x), i.e., Tx represents x. Property (2) of dilations follows from (**) and property to show that TA, (i). Finally, I.L that for g in C(X), we must verify Ax
We know that hence
is
=
-
=
=
p(g) By (* *), function
Tx (g)
h in
fS (g)
v
(iii)
dm(y,
v
(b),
we see
fX
dAx (y, v). implies that
v)
=
=
From
=
fS (g)
=
C(S),
(TA)(g)
=
that
the
left
Tx (g) dA (x). Since
h (y,
v)
fX JS (g) dAx (y, v)) fX Tx (g) dA (x). p(g),
equals
(g)
defines
a
dA (x)
v
side
v
=
and the
proof
is
complete. We next Loomis
define
considers
ordering p several orderings; the
[56].
> A of Loomis
the
present
one
is
(Actually, "strong"
his
ordering.) If
on X, a subdivision measure nonnegative that set f pil measures on X such of IL is a finite of nonnegative We say that ft > A if for each subdivision Epi. f Ail of A y there exists a subdivision Ai for each i. f pil of p such that pi of this ordering and its relation to group (For other descriptions
DEFINITION.
p is
a
=
-
and [56].) see [57] representations, In the following theorem, X and 15.1
of Cartier.
Note that
THEOREM(Cartier-Fellmeasures
on
(a)
p >- A.
(b)
There
(A, p).
X,
exists
then
a
the
X is
not
J
are
the
same as
in
Proposition
assumed to be metrizable.
If A and /-t are nonnegative Meyer [15]). assertions are equivalent: following
nonnegative
measure
m on
J which
represents
Lectures
98
(c)
Choquet's
Theorem
M > A.
PROOF.
Proposition
(b) holds,
and let
Radon-Nikodym Borel
functions
and let
mi
f Ail theorem
(a) implies
shows that
15.1
By Proposition
gim.
=
Similarly,
since
=fS f (x) fi (x) m
=
dm(x,
(A, A),
represents
AM
Suppose
that
fs f (x)
Efi
and
Define
1.
=
fi (x) for each (Ex, v) in J, again, each measure mi has
=
15.1
(vi, /zi) in the cone K. If assertion (and if we carry the measure Proposition 15.1) we see that (f )
fiA
=
resultant
'Ji
(b).
of A. By means of the be any subdivision choose we can nonnegative Borel measur-
Ifil on X such that Ai Jgi I on J by gi (E:,:, v)
functions
able
a
on
m to the
P),
of this
S defined
set
after
in C(X).
f
deduce that
we
dm(x, P),
for
definition
the
we use
f
for
C(X).
in
where 7r is the this means that A m 0 7_1 earlier, of S C X x P, onto X. Since the fi are bounded natural projection that A (f fi) Borel functions, it follows (m o 7r 1) (f fi) for each f in n. Now, for each f in C(X) and each i, we have 1, 2,... C(X), i
As
noted
we
=
,
-
=
=
,
fs (f fi so
o
vi(f)
that
(Ai, pi)
Jsffidm
=
implies
E K
dm(x,
7r) (x, v)
pi
A(ffi)
=
Ai,
-
=fX (f fi)
v)
(x)
d (m
o 7r
Ai(f),
and
i.e., Emi implies
m
-') (x),
vi
Ai.
=
/-t
=
But
ElLi,
so
p > A.
It
remains
/-t > A and that
show that
I_t(f)
Given
6 >
in the
V1, V2,.
letting we
will
proof ..
xi
,
0,
(c) implies
show that
to
f
is
-'_
A(f).
a
continuous
we can
carry
convex
out
of Lemma10.6 to write
Vn such that
If (x)
-
the X
f (xi) I
in X of the < e
for
Suppose, then, that function on X; we want to
same as a
construction
disjoint
Ai of A
the restriction
be the resultant
have
(a).
probability x in Vi.
each
union to
Vi is
as was
of Borel nonzero
measure
Thus,
A
used sets
and,
Ai/Ai(X),
=
EAi,
and
Section
Orderings
15.
therefore
choose
we can
implies
The latter
that
I_tj1ttj(X). consequently tt(f) f :! f (xi) + 6 on Vi, of
resultant
A(f) Since
this
proof
is
=
EAj(f)
is
true
dilations
is
a
only
for
a.e.
A.
=
is convex,
(xi) 0,
+
=
-
Aj(Vj) pj(f)1tzj(X)
eA(X)
:!
conclude
we
E/uj
and /_tj Aj. and that xi is the
/,t
EAj(Vj)f(xj). :! A (Vi) [f (xi)
Ai(f)
E >
any
p >-
if
with
section
f (xi),
!
On the
other
6],
+
1.t(f)
hand,
and hence
EA(X).
+
that
and
Ai and the
-
tt
an
A.
from
I-t(f)
dilation
a
everywhere
with
Proposition =
/,t(y) of
C(X);
then
conclude and hence
-
that
for
3)
the
almost
Tx, is maximal
that
every
-
(Section
concern-
map
f
all
T. (f
a.e.
x,
A.
-+ .,
f
to
TA
C(X).
Y
is
n,
for
/-t.
Then
is maximal
f fnj
/-L(yn Tx (yn =
Tx(fn)
uniformly TX (y)
/-t
Let
each n, 0 > 0, so we have all
maximal
a =
A.
for
for
is
/-t
a measure
in
metrizable,
X is
such that
respect
10.3
for
Tx (yn fn) dA (x). Now, Yn f,, A, for each n. It follows that Since
proposition
Suppose that and that on X,
measure
Let T be
almost
interesting
measures.
(MEYER [57])
dense subset
countable
fx
f !
that
nonnegative
with
maximal,
and
=
EAi(Vi)f
:!
this
PROOF. Recall
if
Aj(X)
and maximal
A is
measure,
TX
1-ii(X) EtLj(f)
so
PROPOSITION 15.2 that
pi
Since =
such that
measures
99
complete.
Weconclude
ing
of Measures
and Dilations
fn) fn) T (fn) a. -
-
.,
continuous, each
be
f
in
a =
0
=
e.
we
C(X),
Topics
Additional
16
Much of the
material
these
in
notes
(other
than
the
applications)
presented by Choquet [19] at the 1962 International and the paper [22] 'by Congress of Mathematicians, Choquet and Meyer gives an elegant and very concise treatment of the main parts of the theory. Bauer's lecture notes [6] contain a detailed development which starts from the very beginning, using (as do Choquet and Meyer) his "potential theoretic" approach to the of extreme points via semi-continuous existence functions on a compact space [3]. Chapter XI of Meyer's book [57] covers a great deal of ground. He shows, among other things, that the entire subject of maximal measures may be viewed as a special case of an abstract of "theory balayage." A number of books and monographs on this subject have appeared since the 1966 first edition of these notes (which appeared in in 1968 [63]). Russian translation Among these have been Gustave Choquet [20] (1969), Erik M. Alfsen [1] (1971), Yu. A. agkin [73] contained
is
(1973), (1980) (without
the
in
outline
[53] (1975), L. Asimow and A. J. Ellis [2] In [21], Phelps [61] (1980). Choquet has given a survey proofs) of related results obtained through 1982. 245-page book by R. Becker [7] (1999) contains a superb
S. S. Kutateladze and
The
up-to-date
which exposition leave off. His emphasis
notes convex
sets)
theory,
capacities,
the
cone
and conical
does not
starts
many respects
on convex
measures
statistical
of interest
in
cones
(rather
permits applications theory and other
decision admit
a
where
than
compact
potential topics where to
compact base.
474-page monograph by Bourgin [14] is extraordinarily his Chapter 6 covers integral ough; in particular, representations The
elements
(sets
with In
[83],
of certain the
RNP;
non-compact see
G. Winkler
these
convex
subsets
thor-
for
of Banach spaces
below). has focussed
R.R. Phelps: LNM 1757, pp. 101 - 114, 2001 © Springer-Verlag Berlin Heidelberg 2001
on
the
Choquet ordering
and
Lectures
102
noncompact with
a
Theorem
in Banach spaces, as in [14]) statistical mechanics in probability,
applications
view towards
Choquet's
(not necessarily
sets
convex
on
and statistics. of infinite proofs, Fonf-Phelps-Lindenstrauss [38].
A survey, in
appears
with
of this
The rest
related
will
section
be devoted
which have been omitted
topics
dimensional
some
to
from the
brief
convexity of
descriptions
body
of these
notes.
POTENTIAL THEORY role in potheorems play an important representation Integral of considerable is theorem the and tential use in Choquet theory, its use in abstract Unfortunately, potential theory. (or axiomatic) that it would rethis regard is so deeply imbedded in the subject then to far and time we are more spend in order to willing quire space self-contained. What which is even moderately given an exposition harmonic functions and facts do sketch is some we can concerning theorems may show how one of the classical representation integral A much more of the Choquet theorem. be viewed as an instance complete treatment may be found in Becker's book [7]. of Euclidean nLet Q be a bounded, connected, open subset
(n
space are
2)
!
harmonic
Q.
in
with
of Q.
Then E is a
the
Let
E
H be the
-
ordering
on
E.
Let
linear
generated
space
compact subsets
on
convex
be any
x0
h > 0 which
functions
of all
convergence and H is a closed
metrizable
lattice
H
=
set
of uniform
topology
by H, induces
H be the
and let
point
which
cone
in
Q;
then
compact H, h(xo) 11 By Choquet's existence and uniqueness theorems, then, a unique to each u in H there exists nonnegative measure M on the h of X such that extreme points
X
Jh
=
the
:
cone
h E
a
of the
view cone,
0 :!
u
we see
: :-
a
metrizable
convex
base for
H.
U
In
is
==
h,
u
of this
property,
usually
referred
(X)
=:f
h(x)d1_t(h) (Section
characterization h lies
that
harmonic, the to
as
(X
on
implies
13)
extreme
an u
=
Ah for
extreme
nonnegative
minimal
harmonic
E
Q).
of extreme ray
elements
of H if
some
harmonic functions.
and
A > 0.
only
of if
Because
functions
are
Section
16. Additional
order
In
nificance, of the
for
above
103
points
extreme
come
if and
is extreme
Py(x)
for
some
jjX
IIyII
y with
IIX112 ylln
2
r'-2r
=
P.
=
sigdescription
concrete
a
the
=
from the
only if h
have any
to
reasonably if Q is For instance, if and at the origin, 0, x0 Poisson kernel; i.e., a function
> 0 and center
r
theorem
representation
of course, one must give minimal harmonic functions.
of radius
ball
the
Topics
(11x1l
=
r,
open
then
the
h in X
where
r).
0 for
with
convex
set
f (x*x) f (x)f (y) f (xy)
1 and
=
in
then
-
A.
x, y in
Eix),
extreme
an
form x*x +
satisfy f (e) point of K,
A which
may be found
Let K be the
x*.
-+
x
proof
A related
*-algebra
Banach
commutative
on
is
of the
continuity
a
L'(G). involution)
algebra
group
involution
functionals in A. If f
105
the
to
assume
and continuous
linear
Topics
Additional
of A is
element
Ei's
form.
functional
the linear
combination
identity
the fourth
are
is of that
x
linear
a
polarization
the
roots
Wemay also
g
g(y*y)
x
A
on
by g(y)
Ic-,i
i=1
(e+Eix)*
hence that
assume
f (x*xy).
f [(Xy)*(xy)]
-
4
4 F, unity);
of
of the
of elements 1
-
we
JJx*xJJ
may < 1.
For any y,
0
and
(f since
g)(y*y)
-
JJx*xJJ
< I
f [y*y(e
=
implies
e
-
x*x
(
00
z*
z
-
=
E n=O
x*x)]
-
=
12 n
! 0,
where
z*z,
=
f (y*y z*z)
) ( -X*X)n
E A.
Thus, f g), where g and f g are in the cone generated g + (f Af for some A ! 0. From by K. Since f is extreme, we have g and that I conclude A we f (e) g (y) g (e) g (e) f (y) for all y, i.e., f (x*xy) f (x*x)f (y) for all y which completes the proof. =
-
-
=
=
=
=
=
It
is
is extreme. g (x)
=
0
2f (x*x)
-
compact
X is
a
0 and
x
need the
compact
represen-
set in any set
homothetic
a
a
topo-
of
the
set,
bear little
generally
it will
images of X
in the plane, if X is any triangle homothetic of two of its images will "look"
However,
to X.
of
concept
convex
is
sim-
E E.
of two different convex
integral
or
image of X
A homothetic
E.
intersection
the
orderings
to
geometric.
where
aX + x,
reference
Suppose that
16.2
DEFINITION
While
no
dimensional
of infinite
characterization
is
homothetic
logical form
elegant
an
a
resem-
nontrivial
X; observation This X. of image is, theorem of Choquet. of the following version is the two-dimensional D. G. Kendall [49] has given a proof which avoids the compactness assumption. intersection
that
it
will
be another
THEOREM16.3 space E is
a
if
of
any two
homothetic
or
another
homothetic
convex
subset
only if the images of X image of X.
and
like
homothetic
A compact
simplex
just
X
intersection is
either
of
a
topological
vector
(ax+x)npx+y) empty,
a
single
point
Lectures
110
characterization
This
leads
to
a
on
simple
proof
of a theorem of Borovikov generalization [12], for a decreasing of finite dimensional sequence
of
THEOREMThe intersection
pact simplices
is
family
a
Let 0 < a,
x, y G X and
Then for
(x for
aX)
X C
X0,
(y
(x zx
(x
aX)
+
n
7xX
+
is called
(y
+
09)
(x
+
a
in 1952
it
family
of
com-
is
a
has
a
(y
zX and
compact
+
and let
nXEXX.
(x
+ aX) n
a
0 < -yX
(y
X) : 0.
+
1
-
18
>
98
Choquet's
Theorem
123
Index
K-Borel
ilov
Completely
89
set,
boundary,
Barycenter,
(of
Bauer
Bauer's
73
-
Dilation,
41
Discrete
54
simplex,
maximum principle,
theorem, 5 Bernstein's theorem, 8 approximation Bishop's
93
Bauer's
Bishop-de
orem,
theo-
Cartier-
Fell-
peak point
the-
32
theorem, 7 Meyer theorem, 81
Choquet boundary, 25 Choquet's theorem (metrizable
case), Choquet's
13
uniqueness
47
measure,
Edwards'
of measures,
80
theorem,
separation
theorem,
50 Leeuw Choquet-Bishop-de 15 theorem, Choquet-Meyer uniqueness theorem, 46
Ergodic
measure,
Extreme
Bishop-de Leeuw representation theorem, 20 Bochner's theorem, 86 Boundary, 25 ilov boundary, 26 Choquet boundary, 25 Cap, 65 Carath6odory's
77
93
89
Leeuw
43
Disintegration
54
(for
lemma
lattices),
cone),
rem,
12
Decomposition
I a
12
Convex function,
90
space,
Bauer,
Concave function,
15
set,
Base
8
26
function,
Baire
function,
monotonic
59
65, 90
ray,
Function 12 affine, completely
monotonic,
8
12
concave,
12
convex,
class, 73 oscillation of, 74 86 definite, positive 76 second Baire class, 12 upper envelope of,
first
Baire
upper
Function
semicontinuous, algebra, 28, 32
Hardy-Littlewood, rem,
al
et
78
Haydon's theorem, subcone, Hereditary theorem, Herglotz's Homothetic image,
22
Invariant
59
measure,
45 85 92
12
tbeo-
124
Lectures
Klee's
theorems, 90 Korovkin's theorem, 38 Krein-Milman theorem,
Poulsen
42
Least
upper
tice),
lat-
29
measure,
theorem,
Mazur's
a
16
Measure 47 discrete, 59 ergodic, 59 invariant, maximal, 16 I representing, Milman's theorem, 6 Minkowski's theorem, 1, 7 Mokobodski's example, 50 Mokobodzki's theorem, 45
Oscillation
of
a
function,
74
Peak
points, 32 theory, Poulsen simplex, Potential
84 91
95 Radon-Nikody'm property, 21 theorem, Ray (of a cone), 65 Representing measure, I 29 Resolvent, 1 Resultant, Riesz representation theorem,
Rainwater's
2,
RNP,
24
95
Simplex, Bauer
6
simplex,
54
(of
Support
of
Uniform
algebra,
Universal
cap,
12
91
36
a
a
measure),
measure,
81
I
32 66
Upper envelope, 12 Upper semicontinuous
36
Theorem
23
space,
Subdivision
42
Markov operator, Maximal
(in
bound
simplex,
94 Simplexoid, Smooth point,
2
State
Lattice,
Choquet's
on
function,
