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
.
.
.
.
.
.
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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.
101
.
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.
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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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.
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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; if f
Choquet's
on
Theorem
CM, f =54 0, then f If (0+) E K, so it to prove the theorem for elements of K. Now, K is a subset of the space E of all real valued infinitely differentiable functions on (0, oo), and E is locally convex in the topology of uniform converE
suffices
(of
gence
(0, oo).
functions
and all
show that
Wewill
Krein-Milman
theorem
points
of K
are
define
e-"ox
to
K is compact
applicable
is
precisely be the
derivatives)
their
topology, Furthermore, -+ e-x, 0 :!
to K.
functions
the
function
compact subsets
on
in this
x
(0, oo).]
It
of
that
so
the
the extreme
:!
a
[We
oo.
follow
will
easily homeomorphic to [0, oo] and is therefore compact. By the Krein-Milman theorem, to each f in K there exists a Borel probmeasure m on ex K which ability represents f The measure m can that
ex
zero
on
K is
.
be carried
to
a measure
f (x)
(x
>
f
0)
/-t
[0, oo]
on
continuous
and the
functionals
evaluation
E; these facts are easily comThe uniqueness assertion representation. is obtained of the Stone-Weierstrass theorem by a simple application to the subalgebra of Q0, oo]) generated by the exponentials. The first step in our proof is to show that K is a compact subset of E. The topology on E is the same as that given by the countable of family pseudonorms -+
bined
to obtain
Pm,n
(rn,n
(f )
theorem, following we
[47].]
=
together
given seen
must show that
and for
f E KI following
for
it
is
finite
for
Let
each
are
bounded above
a
>
on
by 1,
so
!
n
!
(n)
:
the
O
[a, oo) by
m,O
<
k <
and every
nj closed
and
a
on
"distribution
To show that
spaces"
it is bounded
SuPfPm,n(f) : f E KI Supf If (n) (X) I : M-1 < m
>
1.
It
is
clear
is
finite,
X
Tn'
that
the
fact.
this
-n
by induction.
suppose
n,
0 and
1(-I)nf
0 and each
PROOF. Weproceed
m and
establish =
exercises
K is closed.
n
:!
x
[This may be proved by Ascoli's use of the diagonal procedure, or by
show that
each
Kn
for
above
to
lemma will
LEMMA2.1
that
<
metrizable,
in the
each
suffices
this
E is
of E is compact. with repeated
easily
is
on
(X) I: m-1
Thus,
the outline It
(k)
SUpf If
1,2,3,...).
=
bounded subset
in
are
the desired
the assertion
f
Kj, n (nonnegative) G
=
0, 1,
Then
2 .....
functions
in
Kn
2 (n+l)(n/2)
The functions
in
for
Kn.
is true
Ko
are
Since
bounded the
func-
Section
K,,,+, the point in
tions at
are a.
[a/2, a], we see (a/2) f (n+1) (C) (a/2)
suffices
it
f(n) (a/2). (a) (applied at a/2),
This
-n2 (n+l)(n/2)
<
(')
such
on
that
with
the
shows 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
<
c
bound
the
theorem
a/2
with
c
-
establish
to
value
mean
exists
there that (n)
f hypothesis =
induction
nonincreasing, By applying the
11
Functions
to Monotonic
Application
2.
(n+l)
(C)
(n+l)
(a),
follows.
result
from the above may be obtained proof different This is on (0, oo). convergence by using the topology of pointwise functionals are also locally and, of course, the evaluation convex, is function that a completely It is known [82, p. 151] continuous. a certain if and only if it satisfies monotonic sequence of "iterated it is easily since these are defined pointwise, difference" inequalities; and the Tychonov product CMis closed in this topology, seen that theorem then yields compactness of K.]
[A
compactness
Our next LEMMA2.2
form f (x)
-
step is
to
identify
x
>
0,
of
points
The extreme
e-'x,
the extreme
0 <
a
:! -
K
are
of K.
points those
f of
functions
the
oc).
Suppose that f E ex K and that x0 > 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
<
:!
=
strictly
convex x
to
is open,
[: ,
We say
1.
is said
Al
convex
a
A)y]
-
whenever
is strict
real-valued
-f
if
[convex]
defined
if
and 0 < A
each real
semicontinuous
function
valued
affine
defining
and the
convex
lower
A)h(y)
concave
real
a
h is
=h
if
y and
be upper while it is
semicontinuous.
by A the set of all continuous affine functions on X. Note subspace of the Banach space C(X) and that A contains the constant A contains all functions functions. of the Furthermore, form x -+ f (x) + r, where f Cz E*, r is real and x E X, so that A the points contains of X. to separate sufficiently many functions Denote A is
that
a
If f
DEFINITION.
f (x)
--
inf
jh(x)
The function
following
useful
:
is
a
bounded
h E A and h >
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)
<
at
of
existence
point
some
the separafunctional
xi,
linear
continuous
a
from K, i.e., there (xi, I(xi)) From the fact that sup L(K) L(xl, I(xi)). it follows that L(O, 1) > 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 <
<
=
-
Y
f <
so
yields
f
the desired
of
X.
Then there
and is
in
in
it
uniformly,
hn
h2n is strictly
that
we
g)
+
Interchanging
g
if
x
a
metrizable
E, and that p
f and
of
zA
I hn Jno :j
set
C(X) then
y,
nonconstant
and it is
g
the
con-
of
element
represents
x0
X.
C(X) (and
hence
jhn1
of functions
Let a
A)
in A
sphere of
E2-'h 2; this limit f convex function strictly =
n
hn(X) 0 hn(Y) on
[x, y]
an
is dense in the unit
of X.
points
compact
x0 is
X which
on
Since X is metrizable, can choose a sequence
separates
is
X is
points
extreme
hence is in
(Indeed,
function
-
-
measure
1, and the
=
particular,
C(X).
affine
the
[11]).
Thus,
I I hn I I
such that
A;
by
that
space
convex
probability
a
(Bonsall
separable.
exists
is
supported
PROOF. is
f
-
Suppose
locally
a
Y
+ g
result.
THEOREM (Choquet).
-subset
f
g and hence
-
vex
g)
-
for
segment
and therefore
f
some
[x,yl.
n,
It
so
the
follows
convex strictly the subspace A + Rf of C(X) generated on [x, y].) by A and f. Now, from property it follows that the functional (c) above, on C(X) and by p(g) p, defined g(xo) (g E C(X)) is subadditive convex
on
is
Let B denote
=
satisfies
by
h +
p(rg) rf
-+
rp(g) h(xo) + rj(x0) =
if
r
>
0.
(h
Define
a
A,
real).
in
r
linear
functional We will
on
B
show that
Section
Choquet's
3.
15
Case
The Metrizable
Theorem:
that p, i.e., by the functional : - (h + rf ) (xo) for each h in A, r in R. If r 2! 0, h(xo) + rl(xo) h + rf, then h + rf by (b) and (c), while if r < 0, then h + rf is h + rf > 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<
2
1(y)
+ 1 2
RZ) 5 AX)
Ix
is
DEFINITION.
A(f)
for each
If f
M and A
A,
in
X, then for f (x) Consequently,
prove
the
concave;
f'
To
say. -
is
see
Ex,,,
1.
-
prove a
that
net in X
that
f(x)
probability
will
write
-
if
is
x
an
function
follows
from
pa(f)
-
the
from
point
a
! r, suppose that > r
:
/-t
E.
6 >
By
the compact
of
p
:
-
x, with
Exj.
To
1.4.
Exj;
we
that
Indeed, each
0 and for
-
con-
X.
definition
semicontinuous. to
This
p(f)
Proposition
supf/,t(f)
easily
is upper
point
extreme
=
converging
on
supff f dM
=
follows
f'(x)
-
A.
I(x)
X,
in
x
let it
/-t
f (x)
:
of X
such that
measures -
continuous
a
assertion
It
such that
are
f (x)
assertion,
first
we
f x, I
that
=
The second
show that
is
each
set
PROOF.
/-ta
we
If f
PROPOSITION 3. 1 vex
=
-
to note (and will be useful later) that interesting with the set of extreme points coincides f (x) I actually is a consequence of the next proposition.
It
I
implies
fact
last
This
of
definition
the
from
follows
=
distinct 1
>
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<
-
and therefore
p vanishes the We complete
that
are
=
=
=
=
if h (E A and h !
hand,
On the other
m(l) m(f)
I
M(f)
Furthermore,
M(g) for g in C(X). Since m(g) measure. p(l), so /-t is a probability < 1, so tt(f) < /t(j). I(xo). Now, f
M on X such that
measure
f'
is
suppose
f '(x,,)
each
must
a
weak* -compactness,
: ! r, choose
there
16
Lectures
exist
probability
a
converges g (x,3)
-+
f(x); f (x, r)
it
h(x) get f
'
as
p(h)
;5 f =
a
>
g(xa)
in
!
A,
I-t(f).
x
in
It
lim
r -,-.
f
I is closed (and convex) (b) (above), we conclude r
1/-t,31 1-to(g)
subnet
-
if h is in =
and
-
argument
hand,
/-z
weak* to M. If g is in A, then we see that g (x), ex Thus, [t > r. follows that f'(x) Since
f '(x)
:
measure
is
that
f'(x)
which f/-t,f -4 1,z(g); since '< M(f ) ILp (f ) of
=
sernicontinuous, R; using the same f < f. On the other
that <
Theorem
upper
in E
X, and h ! f, then for follows
Choquet's
on
x
any [t
h(x),
-
sx,
and from
we
this
have we
Suppose that cally convex
[9],
the
least
at
p be
Borel
vanish
can
measures representing that only p vanish no extreme points, then
obtain
contain
(Recall
be obtained.
that
generated by leads easily to an equivalent ported by" remains formally a
Borel
theorem
it
sets
are
always If, property. subsets
Baire
of
theorem
the members of the
of
the definition measure
result
this
Furthermore,
in which
A
is not
representation
a
by"
of extreme
set
this
the
on
the same, but the
is
no
"suplonger
measure.
Leeuw).
THEOREM (Choquet-Bishop-de
pact
the
a
on
"supported might require that
p with
compact Gj sets.)
the
u-ring
the Baire
defined
of
from
demands
one
X which
disjoint
is
drop the require-
measures
definition
which
set
allow
There
definitions.
and de Leeuw have shown that
Bishop
to
change
in
We can
the definition
An alternative
lo-
Leeuw
Borel
a
present
our
(i.e.,
measure
we can
Borel
every
but
possible however,
or
measures.
on
points,
Borel
a
under
get around this.
ways to
a-ring),
different for
is
a
Bishop-de set. Thus, the supported by the extreme
form
p is
measure
of
subset
convex
by examples
As shown
E.
points probability meaningless
two
ment that
compact
of X need not
"the
of X"
points
nonmetrizable
a
space
extreme
statement
are
X is
theorem
Leeuw existence
Choquet-Bishop-de
The
4
subset
convex
Then there
exists
of a locally a probability
and which vanishes
of
the set
on
points
extreme
of this
The rest
every
of
section
convex
Suppose that space E, and that
measure
Baire
subset
/t
on
of
X which
X which is
X is x0 is
a
com-
in X.
represents
disjoint
x0
from
X. is
devoted
mainly
to
the
proof
of this
theorem. DEFINITION. exX.
The set
be denoted
by
The set
of A
all
of
points of X will be denoted by on X will affine [convex] functions
extreme
continuous
[C].
R.R. Phelps: LNM 1757, pp. 17 - 24, 2001 © Springer-Verlag Berlin Heidelberg 2001
18
Lectures
subspace C
The
C) is a MaX(fl it
lattice
A
92)
-
A, functions; norm topology
C
contains
measures
on
X,
write
A -
This
relation
/-t and /,t
C(X). A -
A
-
A and p
y
if A(f)
imply
measures,
then
worth =
noting
f
f (x)
inffh(x)
=
A has its
may be
hope that points.
maximal
Suppose
f
order
is dense in
it
the
that
nonnegative
Borel
measures
reflexive;
the fact
that
C
-f
and
resultant
then
u
f I
infftt(h)
-
on
in C.
f
both
same
(with
indeed,
6.,;
C,
in
are
X.)
in
A
so
that
linear
same
probability
It
also
is
f
if
that
C is dense in
-
(E
-C,
well then
measure,
that
v
Wbe
A is
Suppose we
to
since
will
to
ordering.
the
to the
maximal
are
be called
f I ! I-t(f).
h !
with
"maximal
a
The fact
respect
measure",
that
if
A >- /-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)
=
<
A, <
f(x,r)
A < inf
separation
it
g(x)
+
E
convex
functional
L(J2).
follows
If
x
in
L
we
f
that for
E X
nonempty,
convex,
linear
23
x
:
from the closed
origin
continuous
sup
=
Theorem
of the usual
version
the a
and J2
compact,
are
extended
R and A in R such that E
X,r
These
by the equation and that and continuous, completes the proof. f
E
Leeuw Existence
is
on
define affine
X, which
Applications
5
this
In
section
plications
theorem
if the
nontrivial
two
results
the first
Y is
(n
f,,
,
suppose that
one,
1, 2, 3,..
--
If.1
that
states
)
be
to
seen
special
a
THEOREM(Rainwater
that
suppose sequence
and lim
fxnl f (xn)
x,
[66]).
(n
x,,,
=
converges -
f (x) for
weakly
E be
Let
)
1, 2, 3....
weakly
to
x
if
each extreme
C(Y).
in
f f,, (y) to
A
only f (y) for
if and =
points of the unit ball U f (y), then this result theorem.
a
are
ap-
-4
following
of the
case
nice
are
compact Hausdorff
bounded and lim
each y in Y. If we recall that the extreme of C(Y)* are the functionals of the form f is
a
functions
are
converges
uniformly
is
-
theorems
which
theorem.
f f,, I
sequence
Haydon's
and
Cho quet-Bishop-deLeeuw
f
and that
classical
present
we
of the
To introduce space
Rainwater's
to
normed linear elements
Of
E.
only if 1xn1 point f of the unit and
and
space
Then the is
bounded ball
U
of
E*.
Q denote the natural isometry of E into E**. If fxnl weakly to x, then for each f in E*, the sequence of real
PROOF. Let converges numbers
(Qxn)(f)
theorem
shows that
prove the converse,
(Qxn) Y)
---
is
bounded
fQxnf,
and hence the
jxnj, fQxnl is
hence
suppose that
(Qx) (f )
-
f (x)
for
each
f
is
uniform bounded
boundedness in
norm.
bounded and that in
ex
U, and that
f (Xn) g is
To =
an
element of U. It suffices to show that (Qxn) (9) arbitrary (QX) (9) Now, in the weak* topology on E*, U is compact (and convex) so by the Bishop-de Leeuw theorem there exists a a-ring S of subsets of U U and S such that I-t(U \ ex E S) a probability measure y on (with 0 and such that L(g) ex U) f L dl-t for each weak* continuous affine function L on U. In particular, (Qxn) (9) f QXn d1t and to Qx (Qx)(g) dl-t. Furthermore, Qx on U a.e. f QXnj converges f theorem f Qxn dl-t /-t, so by the Lebesgue bounded convergence and the is proof complete. f Qx dy, -
--
=
=
=
R.R. Phelps: LNM 1757, pp. 25 - 26, 2001 © Springer-Verlag Berlin Heidelberg 2001
Lectures
26
application
Our second
deals
[41]).
THEOREM(Haydon
weak* compact convex Then K is the separable. be
(and
points
M
PROOF. Let
f fi I the
be
a
itself
hence is
supf Ilf 11: f
-
subset
dense
norm
K of
with
intersection
E*
hull
of
suppose
that
c
its
extreme
separable). E
of
Kj,
ex
For
K.
closed
the
Banach spaces.
convex
closed
norm
norm
Theorem
Banach space and let K such that ex K is norm
real
a
of
subset
a
arbitrary
with
Let E be
Choquet's
on
ball
each
0 and let
>
Bi denote
let
i,
c/3
of radius
centered
Thus, each Bi is weak* compact and convex and U Bi :) f be a point of K and let /-z be a maximal probability resultant r (M) measure on K with f Since U Bi is a weak* F,1. Let be have a positive n set, we /-t(U Bj) integer such that, if D Uni= 1 Bi, then p(D) > 1 3M Then p can be decomposed as A/-tl + (1 A)A2, where A p(D) and Al, A2 are probability A measures on K defined by at
fi.
ex
K.
Let
=
.
=
'
-
-
.
=
=
-
AA1
(If f
A =
r
=
92 be
1, let
(A)
=
(/-tl)
Ar
IIf
=
-
+
Ar
AID an
(I
(pj) I I
(I
and
arbitrary -
=
-
A)P2
=
probability
A) r (A2)
(1
-
-
Since
A) I I r (A2)II
1-tl(K\D)measure
(A2)
r
:5
E K '6 *
3M
K.)
on
we
M
Then
have 'E
=
3
supported by Ui=j Bi7 the point the hull of Un I Bi, which is weak* compact. convex r(pi) lies in 1. Hence r(/-tl) Eni= 1 Aigi, where gi E Bi, Ai > 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 <
-'6 3
+
-'6 3
+
LE 3
A
6
new
setting:
The
Choquet boundary
the Riesz representation theorem was reformuIntroduction, of the theorem as a representation Choquet type. Although the conclusion of the Riesz theorem is quite sharp (for each element of the convex set X under consideration there exists a unique repthe restrict its measure resenting supported by exX), hypotheses class of compact convex sets. to a very special In what application follows will other describe related we a family of sets things) (among which appears to be only slightly larger than that involved in the Riesz theorem, but which actually "contains" all the sets which inin the that subset of a locally terest sense convex us, every compact convex homeomorphic to a member of the family. space is affinely Y will denote a compact Hausdorff Throughout this section, and will the denote complexC,(Y) space, space of all continuous valued functions on Y, with (We continue to desupremum norm. note the space of real-valued functions continuous In on Y by C(Y).) order to work with the complex Banach space C,(Y), we recall some basic facts about any complex Banach space E. As usual, the dual of the locally is E itself, weak* topology with convex space E* in its each x E E defining linear functional a weak* continuous f -+ f (x) In looking at convex sets in (E*, weak*), it as on E*. one considers In the
lated
a
f
real -+
vector
Re f
(x),
space whose dual x
Cz
consists
of all
functionals
of the form
E.
Suppose that M is a linear subspace (not closed) of C(Y) (or of QY)) and that 1 C- M. The K(M) of M is the set of all L in M* such that L(1) DEFINITION.
=
necessarily state
I
=
space
JIL11.
then K(M) is a nonempty topology, and the results from a locally convex space, compact convex Note that the Riesz theorem dealt preceding sections are applicable. with the set K(C(Y)). In order to make full use of these results, If
M* is taken
in
subset
weak*
its
of
R.R. Phelps: LNM 1757, pp. 27 - 34, 2001 © Springer-Verlag Berlin Heidelberg 2001
Lectures
28
it
is necessary
There
is
not
Bishop
later)
it
we
be of
will
regular sure
Borel
A
JJ/_tJJ
help
ILI
-
Lf
f (Y);
JL(f
!
discs)
! in
if
K(C,(Y)) the
remarks
)
=
that
(0y) (f ) weak*
f (y), topology
that
represented
K(M), then it may be (complex form of the) there
y is
for
exists
each
f
M,
then
f on
K(M).
way:
weak*
[t'
=
p
o
0'.
topology),
LEMMA6.1
and that
a measure
"carry"
we can
p to
Since
a
-
II
Lf
real
to
func-
from
on
element
an
/-t
measure
in
we
measures
Hahn-Banach
r
such
axis, the
case,
:! is
of all
0 a
is
theorem. Y such
on
the
previous
and hence is
one-to-one,
compact
M is the
M is
that
a
K(M) equals
)
=
/Y on K(M) conjugate space a' represents
subspace
of C(Y)
the weak*
K(M). p (f ) for f
closed
If
of
subset
L (f
a measure
Suppose that Then
IIf
Thus,
extended
Y such that
on
easily
I E M.
the bounded set
measure.
follows
it
D be
0, let
intersection
follows
It
is,
Y, let Oy be the element of K(M) defined by Note that 0 is continuous from Y into the
in M.
and A is
!
Then
G D.
complex
M.
in
probability in
a
(that
0
[A simple proof f
If
nonnegative the complex by probability
in
even
M separates
L E K (M)
Lf is the
mea-
variation
!
L
0.
> 0.
the
a
:
the
total
the
in
points of Y, then homeomorphism, embedding Y as If
then
contains r
-
is contained
may be
f dl-t
If
and radius
a
al, i.e., f (Y) (which
follows
[t is
=
JIL11,
=
and hence p pi L goes as follows:
of
theorem
DEFINITION.
I
=
=
JL(f)
=
L E
fy
represents
complex plane which
f (Y)
by
Riesz
L (f
that
0)
Y such that
on
L and the
about
hull
0.] It K(M)
Y; indeed, By
a)J
-
convex
Lf
tionals
N) L(1)
cases
Riesz
A3 and A4
[427
D has center
and since
have
If
f
in the
disc
closed
the
a
i(A3 JIL11.
-
simultaneously, therepresentation then there exist nonnegative
of the
C,(Y)*,
in
pi,
complex
and
form
one
L is
A2 +
assume
r
of
If
above assertion
any closed
so
recall
to
of /-t equals ! 0 whenever
of the
the real
measures -
a
Leeuw have obtained
consider
C,(Y)*:
for
orem
will
general case, subalgebra of and interesting useful but
when M is
case
K(M). (as QY),
of
points
extreme
the
in
Theorem
exK(M).
of
characterizations
of the
be said
to
the
for
Bishop-de
and
Since
deal
great
a
be shown
will
descriptions
have
to
Choquet's
on
in to
the
in
obvious
M*
(in
its
L.
(or of C,(Y)) convex
hull
of
Section
O(Y)
A New Setting:
6.
the
above
assertion
supf(Ref)(y) Re supf L(f) : L EE C(Y)*, such that
adding
Suppose
DEFINITION.
QY))
and that
Oy boundary for which
is
The
1
E
extreme
an
K(M)
there
f
exists
M
in
< : L E K(M)j YJ < supfReL(f) JIL11 11 IlRef 11. We may assume (by Re f ! 0; the first to f) that term then =
a
=
contradiction.
subspace of C(Y) (or of M. Let B(M) be the set of all y in Y for We call B(M) the Choquet point of K(M). that
M is
is
a
linear
this notion An element is apparent: introducing if of and L if point only Oy for K(M)
for
extreme
an
=
theorem
B(M). [The "if" of B(M); on the (Section 1) imply
following on Y, at
least
definition
then
M.
reason
y in
false,
is
y (=-
:
constant positive Re and we have f II II
a
becomes
some
29
-
PROOF. If
L in
Choquet Boundary
The
"intrinsic"
part
of this
other
hand,
that
subspaces
comes
from
the
and Milman's
Lemma 6.1
We have the exK(M) O(Y).] of B(M) in terms of measures C
characterization
for
assertion
Mwhich
of Y.
points
the
separate
Suppose that M is a subspace of C(Y) (or of QY)) which separates the points of Y and contains the constant Then y is in the Choquet boundary B(M) of M if and functions. measure on Y such that only if M EY is the only probability f (y) each M. in f fy f dl-t for PROPOSITION 6.2
-
-
Suppose that Y, f (y) f f dy
PROOF. on
0'
=
is defined
1.4
implies
we
have
p [t
K (M).
distinct
(the
on
y'
that
means
ex
y E
o
Borel
Eoy
-
for
E.
0-1
E.
1
/-tj 1
+
M,
f yj
21
< 1.
of) K(M),
subsets
0-1. 0-1,
o o
Since
Oy
and since
I Al + 2 A2
It
follows
.
in
0
pi
that
tijyj
and A2 <
/-t
M'
=
/-t
o
exK(M),
Proposition homeomorphism B(M), so that Oy is
a
in K (M),
0 Ey. are distinct, 1, i.e., g 0 Ey.)
M, although
Since
some measure
measure
and the above relation
G
and A2 on Y which represent P2 (f ) for each f in M. Let /-t
=
EY- M
f
suppose y EY* Conversely, Then there exist distinct functionals
(00 Y 2 Al (f ) 2 f (y) for each f I-t(f) =
each
for
and suppose that in M. Then the
-
measures =
B(M)
and hence
them,
such that
then 1/11 1/12; 2 2 (Indeed, suppose +
=
yj
=h
6
Y
and hence
Lectures
30
(The
above characterization
quet boundary for
nition
subspaces
version
cated
1
=
2
follows:
as
f (0)
0
x
x,
exK(M);
O(Y) B(M)
where
=
Y may be
=
in
can
be
1 2
M, then by choosing
seen
that
p
=
C(Y) B (M).
E
=
=
it
defi-
this
use
have
we
example
(1) 1. Then f (x) for each f (f ) at
of the Cho-
to
[0, 1] and let M If Y \ 111. [Clearly, B (M) 2
Y
Let
+ 12 f
and /-t in Mwhich "peak" If
one
wishes
points, a more compliThis proposition makes it evident
An
Y.
=
definition
the
Theorem
do not separate
[9, 30].) C(Y) or C,(Y),
of
B(M)
constructed
( ')
some
If
authors.
is necessary
equivalently, f
actually
is
Mwhich
when M is all
that
by
used
Choquet's
on
:
functions
ex.]
subspace of C(Y) or of Cc(Y) and suppose 1 E M. A subset B of Y is said to be a boundary exists in B such that for M if for each f in M there a point x If there is a smallest closed I I f I I (= supf I f (y) I : y E Yj) I f (x) I closed a boundary which is contained in every boundary for M (i.e., it is called the ilov closed boundary), boundary for M. (For some Suppose
DEFINITION.
that
M is
a
=
-
examples,
illuminating
8.)
the end of Section
see
Suppose that Mis a sub-space of C(Y) (or QY)) wZth I c M. If f E M, then there exists y in the Choquet boundary B (M) such that I f (y) I I I f 11, i. e., B (M) is a boundary for M. PROPOSITION 6.3
=
PROOF. Let
Lo be
(for instance, mum) and let Ko
evaluation
The set an
face) L,
Ko
extreme
-
at
of
K(M),
Oy
for
L, which, is necessarily
some
y in
where
point
L in
ILo(f)I
such that
K(M)
If I
attains
its
L(f)
such that
==
=
Ilf 11
maxi-
Lo(f).
compact, and convex, hence it has subset (or since Ko is itself an extremal
weak*
is nonempty,
point
some
be the set of all
K(M)
of
any element
B(M),
an
and
extreme
point
If (y)l
ILj(f)I
=
of =
K(M). IL0(f)I
Hence =
Ilf 11.
Suppose that Mis a subspace of C(Y) (or QY)) and separates the constant which contains of Y. points functions Then the closure of the Choquet boundary is the Alov boundary for
PROPOSITION 6.4
M. PROOF. It a
closed
follows
boundary
from for
Proposition M.
It
remains
6.3 that to
the closure
show that
if
of B is
B(M) -a
is
closed
Section
A New Setting.
6.
then
not;
pose
M, then-B(M) there exists,y
for
boundary
Choquet Boundary
The
(and hence clB(M) c B). Sup\ B and hence a neighborhood
C B
in B (M)
show that U of y with U c Y \ B. We will such that sup I f (Y \ U) I < sup I f (U) 1; this will a
for
boundary
(for exK(M) in O(Y). due
that
oy
1
the
case) to Choquet. that O(U) is a (relative)
Using the definition M, we can find fl,...
c-
Re L(fi)
:
obvious.
We can
<
Ej
n
o(Y)
find
certainly
imply
that
a
B is not
proof
of the
is
Oy is an element of of Oy neighborhood topology and the fact
weak*
f,,
c
M
in
weak*
of the ,
f
exists
Note that
real
and
njL
c
M,
there
The remainder
contradiction.
a,
31
M and
in
O(U)
[This
-
0 such
>
E
isn't
immediately
number of functions
finite
that
gj in M
such that
0(y) If
we
lRe L(gj)
njL:
E
replace
by fj
each gj
(y)
Re gj
-
-=
gj
-
<
EJ
Re gj
(y),
-I
n
o(Y)
n
O(U).
c
first
the
intersection
above is
njL:
lRe fj I
so we
I
Oy
otherwise, i
I
=
I
.
obtain
of
.
,
.
O(Y) \o(U) adding
f
function
a
C
J, large
UKj
sufficiently
a
we
Re L(fi)
:
union
is
again
have
positive
separation
the
(J) < sup(Ref)(Y\U) sup Re f
constant
to
f
Re
ej,
with.] Ej n K(M) a
of elements
apply
<
started >
combination
we can
we
IL
L(fj)
-Re
:
the extreme
in Msuch that C
IL
does not contain
a convex
Oy 0 J,
Since
n.
be
-
J of their
hull
but
Ki
sets
convex
K(M),
<
the number of functions
convex
would
Re L(fj)
:
doubled
The
n.
subset
convex
njL
=
the compact
I....
=
ej
simply
have
Consider i
<
7
compact
point Oy; Li of Ki, theorem
(0y) (f ). Ref(y).
to
Since
By
<
get the desired
we
result. We next
locally
show that
convex
space is of the
If
PROPOSITION 6.5 vex
space
I E
M,
every
X is
E, then there
such that
X is
PROOF. Let Mbe those r,
where
f
is in
E*,
r
a
exists
nonempty compact form
compact a
affinely
convex
suitable subset
of
Y and a
locally
separating subspace M of C(X), homeomorphic with K(M).
functions
in R.
K(M) (for
subset
convex
Define
in
C(X)
of the form
0 from X to K(M)
9(x) as
=
before;
of
a
M). con-
with
f (x)+ it is
Lectures
32
that as
L is in
above,
it
that
a measure
By Proposition L. Ox
M.
follows
Theorem
=
find
we can
each g in
Choquet's
To see that O(X) 0 is affine. K(M), suppose and Hahn-Banach Riesz theorems the K(M); by using
checked
easily
on
that
p
p has
1.1,
L(g)
X such that
on
a
unique
=
resultant
p(g)
for
X;
in
x
=
foregoing discussion that we can carry probof function into -the context lems concerning measures representing has This been aptly latter boundaries. and setting Choquet spaces One advantage of the Leeuw setup." referred to as "the Bishop-de with which examples may the relative is Leeuw ease Bishop-de setup be constructed. [Another advantage is that it lends itself to the dismotivacussion of function algebras, which was Bishop's [81 original Section tion for proving see case of the a special Choquet theorem; from
follows
It
the
8.] this
Weconclude
section
with
a
form of the
theo-
representation
Bishop and de Leeuw. In order to do this (and observe that for separating subspaces Mthere purposes) of A >- p for measures A and /-t on Y, namely, definition is a suitable If we are given a A - p to mean that A o 0' define >- 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
<
1/A. Indeed, 11f 11 R,\1
--
-
1/A. This
is
trivial
if
A
=
A'
36
Lectures
on
Choquet's
Theorem
The operators
Given A, let MA Rx have the same range. RA[C(X)] be the range of RA. For any A and A', if f E C(X), then Rx f Ry f (A' A) RA(RA, f ), so RA, f E MA. Thus, MA'C MA-
3.
=
=
-
the
Let Mdenote
of this
the remainder
(Even
-
of the operators R. Throughout M separates assume that points of
common range
section,
we
this, it would still be possible to formulate theorem analogous to the one below, but the a suitable would be unnecessarily statement The next theorem complicated.) the same as one originally is essentially proved by Ray [68], using a X.
different
if
did
we
a
resolvent
If
on
that
X is
C(X),
compact Hausdorff
a
and that
the
M is
that
space,
common
RA of the
range
RA.
operators
1.
assume
method.
THEOREMSuppose is
not
is
x
Choquet boundary
the
in
of M, then for
B
all
f
in
C(X), A (R),
lim,\,,,, 2.
If x is in X, there exists that, for each f in C(X),
for 3.
If
x
measure
px is
any Baire
set
is
A
-+
hence It
Indeed, ARAg-g
we can
follows
=
AR,\Rlf
ftx
on
that
px
measure
=fX
by B,
to
if g is in
write
f (x). X such
f dl-tx.
in the
sense
(A)
g
-Rjf
(1) holds,
M,
-
g1l
<
11RA11 JjRjf
0
Rjf 1-R,\Rlf
IIAR),g
E B.
x
g1l -+ 0 as in C(X), and some f -Rj Rx(Rif -f).
then
for
then
-
=
that
JJARAg
=
B.
conclusion
show that
oo.
\
=
Borel
f ) (x)
supported A CX
in X and the
PROOF. We first
regular
A (R),
lim,\--+oo The
a
f ) (x)
-
f 11
<
(I/A)IIR,f
-
f 11
--
0.
Section
Applications
7.
Choquet Boundary
of the
37
defined
The functional
X and A > 0.
Suppose, now, that x E C(X) by f -+ A(R,\f)(x) and takes I into 1, hence
to Resolvents
on
functions nonnegative on measure I-tx,,\ a probability X such that A(R,\f)(x) f f d1zx,,\ for each f in C(X). For each Ao > 0, let A(AO, x) be the closure (in the weak* topology of C(X)*)
nonnegative
is
there
on
exists
=
of
fpx,,\
family
nested
A01.
A >
:
For fixed
of nonempty
A(x).
intersection
the
x,
compact
if p E
g(x) for each g in M. Indeed, given e y(g) such that I A (R,\g) (x) gI g (x) I :! - I I ARxg =
-
-
A(AO, x),
p E
for
I-tx,,\
/\
some
A >
some
>
neighborhood then, particular,
A0. In
A0. It follows
that
I 1-t (g)
-
A(x), >
form
a
<
Ex, i.e.7 0, there exists AO > 0 e/2 for A ! A0. Since
then
/-t
of p contains
weak*
every
0)
>
and hence have nonempty
sets
show that
We next
A(Ao,x)(Ao
sets
jp,,,,\(g)
g (x)
< E
all
measure
a
p(g) I
-
for
-
E
6/2
<
for
> 0.
for M. The Suppose previous remark shows that if p E A(x), then M Ex, and from the that if of the Choquet boundary we conclude uniqueness property Hence if x E B and U is any weak* then A(x) x E B, fexj. have A(AO, x) C U for some AO, so that neighborhood of Ex, we must E U if A ! A0. Thus, px,A converges to Ex as A -4 oo, i.e., I-tx,,\ and equals f (x) for each f in for each x in B, lim A (R,\ f (x) exists now
that
E
x
boundary
B, the Choquet
-
=
C(X),
which
C(X) px(g) fx =
for
exists
and A
in
B,
y in
Suppose,
that
next,
there
theorem
(1).
proves
Suppose,
g
dyx.
=
that
now,
tt.,,
that
limA(Rj)(x)
to
1.
All
--
larger
this
theorem
as
the
a-ring
gx
E
on
Ex
Leeuw
f
Given
X.
gx(x) 11f 11 and,
M so that
=
=
-
=
dominated
yx(f).
proof of the Bishop-de generated by B and the =
ltx
=
Baire
functions and
convergence X is not metrizable,
If
Leeuw
the
ltx(B)
measure
ARAf; then
-
11f 11 JJARJ 11 < AIIR,\Il 11g,\11 what was A (R,\ f ) (y) -j ust proved. f (y), by I so gx -4 f Then px (B) X is metrizable.
Lebesgue
and the
a.e.
g,\
=
Now
(y)
g,\
a
let
0,
>
By the Choquet-Bishop-de
X.
x
maximal
theorem)
subsets involved
we can
apply
of are
the
extend
X,
so
implies
theorem
(as
we can
px to
that
px
measurable
the -
with
dominated
in
a-ring Ex and
respect
convergence
before.
f (x) show that x E B whenever lim A (R,\ f ) (x) for all f in C(X). Ex (and hence y - Ex); we Suppose, then, that p must show that p I-Ix (f ) Ex. The above proof that lim A (R,\ f ) (x) for any maximal measure I-tx such that is valid Thus, Ex. ltx It
remains
to
=
-
=
=
-
Lectures
38
we can
Since Ex
=
/-t,, px
suppose
(f ) -
=
that
/-t,,
lim A (R),
p >- ex,
and
is
a
maximal
f ) (x) the proof
=
on
measure
f (x) for all is complete.
f
Choquet's
satisfying in
C(X),
Theorem
p.,
-
we
have
ti.
8
The
By
a
boundary
Choquet
uniform
)
Hausdorff
any
algebras
uniform
QY) (Y compact subalgebra of C, (Y) which
algebra)
(or function
algebra
we mean
for
uniformly
closed
in
For points of Y. metrizable Y, the Choquet boundary of a uniform algebra A has a particularly simple description (Bishop [8]): It consists of the peak f points for A, i.e., of those y in Y for which there exists a function the
contains
A with
in
special
a
the
of
case
Suppose
DEFINITION.
We say that
there
Y)
main theorem
A is
a
uniform
is
Bishop
due to
of this
algebra
result
This
y.
section.
QY)
in
and that
neighborhood IIfII :! - 1, If(y)I
such that
of
U
any open
y and any
and
> 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
<
characterization
a
[9],
and separates
If (x) I
that
property
and de Leeuw
y E Y.
functions
constant
If (y)I
in A such that
QY)
(i) (ii)
The
and that
point
y
=
Ilf 11
y E Y.
The
satisfies
Condition
For each open set U containing lif 11 and If I < Ilf 11 in If (y) I For
each
x
E
If WI < If (Y) I (iv)
(V)
Y with =
Ilf 11
point
y
The
point
y is
in
=A
x,
the
If (x)I
=
that
A is
there
y,
Ilf III a
assertions
C
uniform algebra equivalent:
are
L y,
there
f
exists
E A such
that
A such
that
Y\U. there
exists
f
E
11
Choquet boundary B(A)
R.R. Phelps: LNM 1757, pp. 39 - 46, 2001 © Springer-Verlag Berlin Heidelberg 2001
exists
S.
-
Condition
satisfies
The
y
:
Suppose following
=
(iii)
G6 containing
a
fx
and
Leeuw).
THEOREM (Bishop-de in
S is
whenever
of A.
40
Lectures
(i)
PROOF.
that
U is
I
g,,
open set
an
of functions
(a)
llg,,+,
(b)
IIg,,II
(d)
Suppose,
for
sequence
f g,, I
2-n-1
<
the
Condition I and Suppose that y satisfies We will construct a sequence containing y. with the following properties:
2-n)
-
9nI
-
Theorem
2-n-1
-
3(1
-
Choquet's
2-1+1
<
3(1
<
19n+1
in A
g,,Il
-
(c) MY)
(ii).
implies
on
Y\U.
in
that
moment, to
converges
3 ply that Ilf 11 f gn + Ekm=n (9k+1 _-
f (y).
=
9k)
-
have
we
Moreover, we
,
By (a),
done this.
f
function
a
if
x
(b) Y\U, then,
A;
E
E
the
(c)
and
im-
writing
have 00
If WI
<
119nIj
E lgk+l(X)
+
-
9k(X) I
<
k=n
00
3(1
<
-
2
-n-1
)
E2
+
-k-1
< 3.
k=n
Weobtain I
the
implies
IgI1 3(1
that
If I
and
-
91)92)
3
4
2
3
2-1),
=
so )
...
2
Let 2
-
satisfies
g,
f
exists
Y\U. < 3 (1
by induction:
-
19kI
functionf
3(1 <
2
-
gi So
all
the
3(1
2 -k)
inAsuchthat
=
-2),
-k),
-
A such
in
9k have been chosen
9k(y)
Since
such that
there in
4
f9nj
sequence
+2
IIfIj
(b).
Also,
conditions.
satisfy a neighborhood the
-k-2
:!
Since
2
satisfies
relevant is
y E
IIf II
that
Q)[f(y)]-1f.
g,
to
there
Since
in
1,
V.
U, Condition I f (y) I >
If(y)I IgI(y) I
> 3
-
1 4
1
4) -
2
Suppose that
above four
conditions.
V of y,
V C
We can choose
> land If(y)l Ifj 4 thenh(y) -3.2
U,
another <
1
4
k
in
and (3-2 -k-1)[f (Y-)]-lf; -k-1 2 x in Y\V, Let < lh(x)l 9k + h; 9k+1 To check (b), properties (a), (c) and (d) are immediate. suppose < x c V; then 19k+1(X)I < lgk(X)I+lh(x)l 3(1-2 -k) +2 -k-2 +2 -k+l 2 -k-2). On the other hand, if x (z- Y \ V, then 19k+1 (X) I < 3(1 2-k-1 2 -k-1) +2 -k-1 2 -k < 3(1 3 2 -k-2) This ; 3(1 119kII + and the proof that (i) implies completes the induction (ii).. Y
IIhII
\
V.
:!
Define
2 -k+l
.
h
Also,
-
for
-
-
.
=
-
-
=
-
-
.
Section
(ii)
That
Algebras
Choquet Boundary for Uniform
The
8.
(iii)
implies
is immediate.
To
(iii)
that
see
41
(iv),
implies
II, suppose that S is a G6 set a decreasing sequence of open sets with containing JU,,j y. I S f. (y) nU,. For each n we will find fn in A such that I I f. I I function and IfnI < I in Y \ U,,. Once this is done, the E2-Ifn f II. Suppose, then, that n > 1 of Condition the properties will satisfy there exists and that x E Y \ U,,. By (iii), fx in A such that fx(y) in and I I a neighborhood Vx of x. By compactness of I I fx I I I fx I < number Y U,, we can choose a finite fx, of such functions k-1 Efx, then for which VX,, Vx, cover Y \ U,,. The function f,, has the required properties. To prove that (iv) implies (v), suppose that y satisfies Condition measure on Y 11. We will show that a -y is the only probability such that [t(f) 6.2, we can f (y) for each f in A; from Proposition conclude that y c B(A). Indeed, suppose that p is such a measure; 1 for any suffices to show that it that to see y(S) 1, M(f yj) If S is such a set, choose f in A such Gj set which contains y. :! that y E Ix : If (x)l jy(f)j if (y)I Ilf III C S; then Ilf 11 fS If Idy + fy\s If Idl-t < Ilf III-t(S) + fy\s If Idp. If p(Y \ S) > 0, then that
Condition
y satisfies
that
is,
be
Let
=
-
=
--
=
,
=
.
.
.
,
=
--
=
=
=
-
fy\s
that
If Idl-t < Ilf III-t(Y\S), (iv) implies (v).
If
M is
then Re M (the
subspace of
space
C(Y),
which
contradiction
(i),
(v) implies
To show that LEMMA8.1
a
M)
a
completes
simple
=
proof
the
lemma.
subspace of QY), in M) of functions
separating of real parts
a
and B (Re
need
we
=
M, separating
with is
a
1 E
B (M).
6.2 and the fact that for a real measure p Proposition on Y, p (Re f ) f (y) (Re f ) (y) for every f in Mif and only if y (f for every f in M. Wereturn to the proof that (v) implies (i): Suppose y E B(A) of y, and that 0 < < 1. B (Re A), that U is an open neighborhood < 0 1 and g Choose a function 1, g(y) g in C(Y) such that 0 :! g
PROOF. Use
=
-
-
=
=
in
Y \ U.
by
X and
f Proposition
that
=
Denote use
g
o
the weak* compact
Tietze's
0-'
3.1
on
that
theorem
extension
0 (Y)
(-f)(0y)
convex
C X. =
Since
(-f)(0y)
to
Oy
set
K (Re
obtain E -
ex
X,
-g(y)
f
A) in
it
(Re A)* C(X) such c
follows =
-1.
from
Now
42
Lectures
the
of continuous
space
4.5) isomorphic to therefore (- f ) (0y) there
(where
1 /6).
--
=
Let
f,
function
Ifil
easilythat
-
:!
I
inf h
closure
h (y)
equivalence
ik)
<
(iv)
and
1);
and
then
in Re A, and
each
(v) yields
>
follows
It
log (6
1) / log
-
A is closed
:! ,-
point the
If, I in
eh, Y\U. =
J
in
QY),
it
follows
Gj, and corollary.
of Y is
a
following
(Bishop [8]).
and A is If Y is metrizable Choquet boundary for A coincides
algebra in QY), then the of peak points for A.
that
h E Re A and there
Since
Ifil
f 1.
ho (y)
then
Since
A.
in
1-6
>
g and
-
-
Theorem
(by Proposition
X is
of the functions
E A.
is
metrizable,
between
COROLLARY8.2
ho
h + ik +
on
h G Re A, h !
:
(log J) (ho
=
exp(h 1, Ifi(y)l
if Y is
Note that
the
the uniform
Re A such that
k in
exists
the
J
functions
in Re A such that
ho
exists
affine
Choquet's
on
a
function with
the
set
Choquet boundary can be a proper subset of the Silov boundLet Y be the unit circle as by the following example: ary, in the and let : complex plane, A, be those functions fz IzI 11 in C,(Y) which are restrictions of functions f which are analytic in I z I < 1, continuous in I z I :! 1 and which satisfy f (1). f (0) from the maximum modulus principle It follows for analytic funcof Y except 1 is a peak point that every point tions for & since Y is metrizable, this shows that B (A,) Y while the ilov 11, boundary for A, is Y. We give a related for the term example which is the motivation Let Y "boundary:" fz : IzI :! 11 and let A2 be the set of all in C,(Y) which are analytic functions in IzI < 1. Then the Choquet with the ilov boundary coincides boundary and these equal the boundary fz: IzI 11 of Y. Y I I and let A3 be the restrictions let to Y Finally, f z : IzI in A2; then the Choquet and ilov of the functions boundaries equal Y (so this can happen for proper subalgebras of QY).) of Bishop's For a description of the foregoing material application to an approximation in the complex plane, see problem for functions The
shown --
=
=
=
=
=
Section It
16. is
boundary of
=
QY)
not
generally
which
true
(in
coincide are
the
not
that
the
metrizable
algebras.
peak points
Choquet subspaces M
and the
linear case) does (One inclusion for
hold:
The
Section
proofs
"(iii)
of
(iv)" (above)
boundary.)
quet
didn't
Now consider
(v)"
implies
peak point example: following
Bishopalgebra:
the
A is
an
Cho-
is in the
every
the
43
in
that
the fact
use
subspace,
linear
separating
"(iv)
and
implies
deLeeuw Theorem For any
Algebras
Choquet Boundary for Uniform
The
8.
Y be the
Let
cirof the convex hull of two disjoint plane consisting The Y. functions affine valued on complex cles, four tangent points to the circles are in the Choquet boundary, but ReM. fact most easily seen by considering are not peak points-a for M are dense in the Choquet the peak points In this example, boundary for M, a fact which is true in general and is a corollary Banach spaces. result classical By a to the following concerning of the unit sphere of a Banach space E, we mean a smooth point 1, for which there is a unique f in E* such that point x, JJxJJ I f W. 11f 11
of the
subset
M be the
and let
=
=
-
(S. MAZUR) Let E plex) Banachspace and let S Ix : JJxJJ
be
PROPOSITION 8.3
=
of
Then the smooth
E.
the
PROOF. In
real
(in
space
by
denote
of
case
S)
way);
unchanged.
is
of
complex
a
usual
the
sm
points
S
the
a
form space, set
real
denote
the unit
will
we
it
as
(which
points sm S
com-
sphere of S.
consider
of smooth show that
Wewill
G6 subset
dense
a
(or
separable
11
--
is
a
a
we
countable
of dense open subsets of S; since S is a complete metric intersection conclusion. theorem will yield the desired space, the Baire category
f Xn I
dense sequence in S. S such that be those x in Dmn
Let let
f,g
in
sm S
satisfy
x
E S
=
Yk E S \
that It
follows
that
S. J
>
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
<
6 and
JJxJJ fi(yi)
g, in E* such that 1 Wewill rn+ 91 (Xn) -
in S and
fk,
compactness
corresponding
=
1
easilyverified
is
and n,
S,
in
1,
n,
m-1 whenever
m
Tn-
hence
that
suppose norm
k
such
one
1, 2, 3,
=
ball
of the unit
-
-
-
-
of E*
each set Dmn is dense in w can
imply =
It
A of
-
weak*
<
some
Y- Choose functions and A (Xn) A (Xn)
the
from
-q(xn) 11g1l.
f (Xn) =&)
I
m and
integers
positive
y E S \ Dmn- It remains to show that Suppose not; then for some m and n
and choose and
JJfJJ
Dmn and
A (Yk)
For
a
E*
if
that
be
11fill
proceed by functionals
choose
x
y in
Dmn.
=
=
11gill
induction
A,
A of
S and
Let y, =
=
y
gl(yi)
to define
a
norm one
44
Lectures
k m-1 +
11Y1 -Ykll < (1-2 -k)j, A (Yk) < Since fk(x,,,) gi(x,). 1, this will
Suppose
we
such that
define
Yk+1
JJYI
and
follows
that
fk+1 (Yk+l)
that
0
Yk+1
there
to
a
properties.
We
chosen
-
and
;!
contradiction.
0 is
>
fk (x,,)
and
a
above
Theorem
I < 2-k-16. Thus, jjYk+1jj Yk+1jj < (I 2-k-1)6 <
-
exist
(Yk+l)
9k+1
::::::
lead
where
Yk+1
JjYk
+
D ..... .so I
-
I I Yk
2-k)j
-
-
A (Yk)
::--
has the
aXn1j)
+
that
(I
<
A which
aXn)lllYk
+
to insure
Yk+1jj
-
chosen
(Yk
:`
enough
small
I
-:::::
have
Choquet's
on
fk+,, 9k+1 of norm fk+1 (Xn) > 7n_1 +
one
9k+1
be
to
I
::::
J.
It
such
(Xn)
Now,
11Yk+111 gk+I(Yk+l)
Since
JjYk These facts
-I M
proof
Mazur's of
ries
differentiable
+
(Xn)
>
-
M
afk(Xn)llllYk
1
+
CeXnll-
have
OXn)
show that
to
+
we
gk+I(Yk
:--
+ 9k+1
+
1 +
:!
A (Xn)
agk+l(Xn)-
gk+,(xn), !
(k
that
so
1)
+ g,
and
continuing
+
complete.
is
theorem that
:- gk+I(Yk),
aXnjj
investigations
guarantee ences
+
combine
fk+I(Xn) and the
I
:--:
[I
fk(Yk+l)
,
was
the first
into
those
any continuous
dense
on a
Gj
in
an
extensive
properties real-valued
subset.
See,
se-
of Banach spaces E which function convex on E is for
[62]
instance,
and refer-
therein.
COROLLARY8.4
Suppose that Y is a compact metrizable space and that M is a uniformly closed separating subspace of C,(Y) (or of Then the peak points C(Y)) which contains the constant functions. for M are dense in the Choquet boundary for M. points y in Y such that f (y) 11f 11 for some f of the unit sphere of M. It is immediate that every for M, and P will a peak point be dense in B(M)
PROOF. Let P be those smooth
point if
O(P)
Milman's hull
point
of P is
of
is
weak*
dense
If
in
in Section
theorem
O(P).
=
it
were
This exK(M). if 1) K(M) is
not,
we
could
latter
will
choose
g in
(by
be true
the weak* closed
M, 11g1l
convex =
1,
Section
8.
The
Choquet Boundary for Uniform
Algebras
45
Since the smooth points Reg (K (M)). dense in the unit are sphere of M, there would exist a uniformly For such a function this same inequality. smooth point f satisfying f however, the left side is I I f I I and the right side is at most I I f 11, a such that
,
contradiction.
sup
Reg (P)
< sup
The
9
boundary
Choquet
theorems
Many of the classical in terms
identity detail, obtaining
the
the
proof
Bernstein's
For each
theorem.
polynomials
of
>
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,,
<
nUn
9n(X)
1,
f xj -
each
n
on
f Un I
sequence
and for
I and
and
set
measure
a
a -
Korovkin
a
probability decreasing
p is
Choose
Ex.
=
M is
choose
101.
gn(X \ Un)
Tn by
A(f)9n
::=
a
positive
we can
Tng
-
(I
+
9n)f,
-
linear
=
operator
E
[g(x)
-
g]g,,,
C(X)Tn1
such that
Ig(x)
N such that
choose g
f
for
so
g(y) I
-
n
>
<
1.
If
c
for
N and any
have
shows that
IITng
-
gII
-
---*
19(X)
the
more
about
Mis
Korovkin-type
-
9(Y)19.(Y)
< 6,
being true for every g E M, that IITnf set implies f 11 -4 0 (Tn f ) (x) - f (x), so that p (f )
This
0.
Korovkin
a hypothesis for every f E C(X); in particular, f (x) for each f E C(X), that is,
that
[26].
g(x)
e,,.
+
suppose
-
Much
-
this
to
f (X)I
MOM 9(Y) I which
(X.")
Since
g E M. =
KTnkf)(Xnk) -
of
such that
n
weak*
show that
must
X with
C(X)
=
converse,
G
gn c:
f1l
-
x
e-
we
+ Ig
9(X)1)
-
g(x) for have p(g)
I
result.
the
X;
19(Xnk)
+
-4
RTnkf)(Xnk)
the desired
To prove
g (X-J
-
inequality
and the
'5
yields
g,
(9)
converges
1 1 Tnkf
ner
has
B(M) by hypothesis,
E
x
911
-
P"k
all
nk /Lnklf
IL
that
for
Since
Ex.
-
that
conclude
we now
I (T., g) (X-J
I : gW
-
:-<
so
to
inequalities
the
p
49
f y", I if Now measure p. g E M, of g and from the continuity
a probability so hypothesis, by
weak*
theory
approximation
and
f /-t,, I necessarily
X and the sequence
on
y,,
Choquet boundary
The
9.
-
=
y
=
Ex, which
theorems
was
to be shown.
may be found
in Don-
10
Uniqueness
The
question
one,
both
applications the clearly
in
What
asserted. those
representing
of uniqueness
specify
must
of
compact
we
representing and in the theory of
most is
X with
itself.
As
which
characterizes
point there and is supported
to each
that
the property
natural
a
always, one uniqueness is being
theorem
a
is
measures
which
within
context
would like
convex
measures.
the
point by the extreme points of X. Choquet has proved such a theorem result in the general for metrizable X, but there is no satisfactory On the other hand, Choquet and Meyer have characterized case. there corresponds a that to each point those X with the property maximal Since the which maximal measure point. represents unique measures are "supported" by the extreme points, it would seem that is taken in but the fact that "supported" this answers the question, An example difference. sense makes a considerable an approximate of maximal representing by Mokobodzki will show that uniqueness which of measures measures does not imply representing uniqueness exists
unique
a
vanish
section,
X in
real
the
affine
[convex]
poses,
hyperplane this
we
topology); X.] The ness
misses
[There
main
is
reason
naturally
when there that
X such
y
ex
X.
no
y
=
origin-this lost generality the hyperplane
III
of X is
doing
this
cone
there
If
X.
For
making
E
if
x
affinely
is that
III
X is
R.R. Phelps: LNM 1757, pp. 51 - 64, 2001 © Springer-Verlag Berlin Heidelberg 2001
contained
pur-
closed
a
throughout assumption,
this in E
x
R
(product
homeomorphic with of uniquethe question base of
(with
exists
in
be assumed
will in
P
present
our
X is contained
when X is the
only
ax.
study of a compact convex before, we denote by A[C]
on
the
a convex
and
As
E.
functions
studied is
E P if
that
for
to the
to assume that
may embed E as the image X x
is most
P, i.e., such in
which
section.
since
continuous
be convenient
will
it
\
then, we return locally convex space
In this set
a
represents
of X
subsets
Baire
on
that
measure
vertex a
in
unique a
a convex
at a
the
!
hyperplane
cone
origin) 0 and
x
which
52
Lectures
the
misses
X
origin,
lax
=
hand,
other
if
0 would
not
X),
so
E
x
there
f(x)
is also
this
0,
E
x
X is
XI
unique
take
P
by
generated P, then 0
representation
the
in
of X and the
functional
f
ThesetX'=
E X.
P, we can closed hyperplane
H in E which
X, where
=
On the
(otherwise
X
ax,
E and
fx'
E P:
theorem 0 such
>
f(x')
-01 the is
set
convex
form origin.
the
the
misses
0
X under
to
0,
>
a
separation
on
is
cone
Theorem
X.
form
affinely homeomorphic whenever a compact Thus, always assume that it is of
P which
&lf (x).
-+
x
the case; cone
cone
a
linear
continuous
a
> forallx
some
the
is
base for
a
a
certainly
is
by the compactness
base for
a
map X E) base for a
for
then
have
exists
that
!
a
:
Choquet's
on
a
Hn P
invariant Now, recall that a cone P in E induces a translation E: and if if on P. x If P has x partial ordering only y y E > > x then P n (-P) * base X, f 01, so that x imply y and y if and in the subspace P * P generated x y. Furthermore, y are by P, then there exists z in P such that z ! x and z ! y, i.e., x and -
-
-
y have
bound for
x
this
denote
translation
P. We say that z is the least upper and y if z < w, whenever w ! x and w ! y, and we least upper bound by x V y. It is easy to check that the
V
+
pair
have
lattice
Y1
A y
Y2
are
Z
X
(XI
=
of
a
z
k
if
two
w
-
z
is easy to
-
and
inequalities =
call
a
let
.
which
that
x
[Indeed, (X I
+
IfX
Y2)
V
V y in
TC
-
+
X2)
> x,
w
> y,
X).
-
x, y has
by is
a
A y;
x
vector
a
each pair
-
andy
(X2
must show that
we
if
is,
X
=XI-X2
(Y1
we
space
(that
X
is denoted
k
P
-
if the
suppose that
xVyinX. z
by
bound
P
(not necessarily
simplex
induced
note
=
in
upper bound of x, y in X -.k. + Y2) we have z > x; similarly,
(XI w
X
X
z
only if each pair
and
Finally,
x, y,
set
upper
lattice.
k,
-
for
convex
obvious),
bound
+ Y2) V (Y1 + X2)
WI+X2+Y2 -> W2+X2+Yl It
lattice
is the least
WI-W2
-
The first
earlier,
least
upper
elements
show that
W
is
we
a
has
=
will
If
k
bound
we -
-
T(
X has aleast
c
-
x
that
ordering
the
in
(definition -(-x V -y). if (and only ip TC is a lower
greatest
x,y
X
-
X,
a cone
lattice X
y in
x,
-
-
base of
vector
Equivalently, we
(x
=
the
-
each
z
P
of P guarantees If a + z) V (y + z)
invariance
(x y) compact) is 'k X- is a have
bound in
upper
an
+
=
Y2);
Since > y.
z
w
>
z.
imply that W1 + X2 + Y2 ! W2+ X1 + Y2 and invariance mentioned Using the translation
(WI
+ X2 +
verify
that
Y2)
-
being
[(W2 a
+ X2 +
simplex
Y1) is
V
an
(W2 + X1 + Y2)] "intrinsic"
>
0-1
property
Section
Uniqueness
10.
of
Representing
53
Measures
which misses the in a hyperplane X, that is, if X is contained in El, and if there exists situated a origin in E, if X, is similarly extended be then this X affine of onto X1, one-to-one map may map in the obvious way to a one-to-one, additive, order-preserving map of
which
T(
carries
only
if the other
that
the
in
case
ki,
onto
is
a
so
above definition X is finite
that
of these
one
of
a
section
coincides
simplex
if and
is lattice
cones
At the end of this
lattice.
will
we
the
with
show
usual
one
dimensional. that
each
point of X is represented by a unique maximal measure if and only if This result, X is a simplex. together with a number of equivalent is due to G. Choquet and P. Meyer [22], and our proofs formulations, follow theirs. uniqueness theorem for metrizable Choquet's original X is an easy corollary. the uniqueness Let us formulate portion of the Riesz represenof theorem in terms tation Suppose that Y is a compact simplices. Hausdorff space and let X be the compact convex set of all probathe Riesz As we noted in the Introduction, on Y. measures bility To each point of X there exists theorem can be formulated as follows: is supported measure which a unique by exX O(Y). representing of this
The main result
section
the
is
theorem
=
The
uniqueness fact
the
that
X is
on
lattice
in
a
fact,
but
we
X
set
one
of two
nonnegative measures continuous absolutely p f and Nikodym derivatives are
is defined
v),
a.e.
We need
one
"Decomposition
that
i.e.,
the
of
cone
not
v
to
respectively.
as
lower hence
v,
h
Let
a
bound A A p have Radon-
min(f,
=
important Lemma,"
(i)
(ii)
(x
+
lfx>
z)
Suppose A
(y
+
0, y !O
z)
which
that =
g) (this
=
is
concerning
assertion
(iii)
in
the lattices, following
the
lemma. LEMMA10.1
base and
well-known
then both A and
A+/-t;
=
of
no'nnegative
all this
prove
the greatest
A and p. Let with respect g,
consequence
a
measures
will
AAp hv.] technical result
and let
be
to
of probability
ordering. [We of defining way
the usual
recall
can
simplex,
a
Y has the
measures
is
assertion
be considered
(x
V is A
y)
a
vector
+
z
for
lattice. each x, y,
andz> 0, then
(X+y)Az<
(xAz)+(yAz).
z
in
V.
Lectures
54
(iii)
If Jxj
yj
PROOF. The fact
yields
ant
that
so
<
finite
are
such
of
sequences
EjEJYJ, EjEJZij
then
=
that
Theorem
xi
(i
=
nonneg-
there
exist
1)
C-
and
-
partial ordering in V is translation proof of (i). To see (ii), let u (x
that
the
invari+
=
+ y and
x
JI
E
J,
x
J)
E
immediate
an
u
E I
0,
=
j
:
of V, and if Eic:jxi
(i, j) (j EjEjzjj
!
jyj
and
elements
ative
zij
If
i E
:
Choquet's
on
u
0 :!
Since
z.
x,
we
have
<
u
y)
A z,
and
+ z,
x
On the other hand, yAz (x+y)A(x+z) x+(yAz). 0, :! and therefore so u z + (y A z), u [x + (y A z)] A [z + (y A z)] To A A + (x z) (y z). (the Decomposition Lemma), it is not prove (iii) hence
u
:
>
<
--
difficult
to
induction
use
order
to
case,
suppose
Z11
X1 A Y1 i Z12
=
reduce
X1 + X2 =
and Z12 A Z21
ative,
(i)).
=
Furthermore,
it
remains
to
But
Z21
tive.
(Z21
A
Z12)
Z22
> 0
X2)
of the
part
If f
f
fg
=
=
inf
infjp(g)
E :
g
g E
:
PROOF. Since
A c
A, h f I < inf h f f : h gives the first
1.
>
a
there let a
=
(y,
-
Z11
-
These zij
(xi
=
A
yi)
Y1 + Y2
=
Z21
=
SO
Z21
Y2
nonnegZ21 + Y2;
=
nonnega-
(Z12
Z21 A
(by
0
=
is
Z12
Z21
and let
zi,
Z11
this
prove
are -
-
-
=
therefore
Z21 A X2;
in I and in J in
f 1, 21. To nonnegative)
=
zil.
zil)
-
Z12 + X21
=
=
yj
-
X2
=
J
elements =
X1 + X2
Z22
function exists
=
E
downward, decreasing,
C(X), E Gj, GI for -C,
in g in
+
X2)
and hence
X2,
set
g E
from
we can
g
Gj;
and hence
gj, we
that
converges
we
g.
and g >
measure
GI
g E
proof.
a
(downward)
:
theorem
by >, /_z
inff
:!
f and
X.
on
h
minimum of two functions
the
9
include -C
g cz
:
a
:
h E
hand, for any g E G we have Taking the infimum on the right
<
must
downward; 92.
the
/_t(j)
show that -+
0.
sequence
pointwise
to
=
Since
G
in
if 91) 92 E
i.e.,
the last
To prove
I-t(g,,)
G such that
assume
infjg
is directed <
Ig
=
have
=
again,
it
of
consequence
directed
other
=
a
nonnegative
any
Since
G with
G
G is
we
gI
G, this :
we use
let
On the
assertion.
inff/_t(g) sequence Ignj =
since
h !
A,
lemma is
next
theory;
LEMMA10.2
is
zil)
A
=
-
from integration
p(f)
(xi
-
Z21 + Y2
A
(all
and Z21
zil,
I
case
Y1 + Y2
-
-
Z12 + X2
(Z21
the
to
show that
+
The last
Then
x,
the number of elements
on
proof
the
G,
assertion,
0.
Choose
G is directed
j9nj
is
a
(Borel
monotonically measurable
Section
Uniqueness
10.
bounded)
and
theorem,
gence
fx
then
not,
function
:
p(f') f'(x)
of
Representing
f,
with
,
=
j(x)j
>
1.
f
and
Measures
we
From the monotone
will
positive
has
fx
j(x)
conver=
If
0.
and hence there
measure, :
1)
M(f'-
show that
andE > Osuch that
is arealnumberr
55
>
rJ
set K of This latter set contains a compact measure. positive function K there is in and for each a point x positive g in measure, functions G such that g(x) < r we can find E. By compactness,
has
-
911, 92, with
9M in G such that for each x in KI there is a function g Let fk in G be less than or equal to the minimum < r e. '
1
I
...
(x) Ofgk,gl,--g
fk
while
-
91
; then
9k
on
k, which leads
to
As sition
on
M
<
X a
application 4.2, obtaining
\
K.
K, wehave fk Thus, A(fk) lemma,
a
useful
we
f
tion
if
for
and
on
X.
(MOKOBODZKI) p(j) for only if p(f )
f
E
each
A
gk-E,
Ep(K)
the
converse
of maximal
positive
measure
each continuous
-
Equivalently,
prove
characterization
PROPOSITION 10.3
,maximal
_< A(9k)
fl-E
-
for
each
contradiction. of this
an
<
p is maximal
if
and
to
Propo-
measures.
p
on
convex
only if p(f)
=
X is
func-
p(j)
C(X).
4.2, we need only show that if p(f) Proposition p(j) for each f in C, then p is maximal. Choose a maximal measure If A with A >- p. Then for f in C, A(!) p(j). A(f) > pffl p(g), so (by Lemma 10.2) g E -C, then A(g) PROOF. In view of
=
=
A(f) :! It
follows
C(X),
A
that --
inffA(g)
=
inffp(g)
A (f
)
=
y
:
g E
(f
for
g E
-C,g
-C,g
each
f
>
fj
<
p(j).
>
f I
=
in
C;
since
C
-
C is dense in
p and hence p is maximal.
fact important we however, present First, lattices. lemma concerning vector an elementary Suppose that P, and P2 are cones in a vector P, c P2. Denote the space E, with We say that induced partial orderings by :! j and <2, respectively. subcone of P2 if y E P1, x E P2, and x <_2 Y imply P, is an hereditary x E P1. We will
about
the
apply
set
of all
this
result
maximal
shortly
measures.
to
obtain
an
Lectures
56
If P2 is a lattice (in subcone of P2, then P, is
LEMMA10.4
the
hereditary
a
Choquet's
on
ordering
P,
and
(in
lattice
Theorem
an
<j).
ordering
the
is
Suppose that x, y are in P1, and let z be their greatest lower bound in P2. Then z <2 X, so z E P1, and we will show that if w :! j x and w :f j y, then w <1 z, so that z is the greatest lower bound in P1. Since P, C P2, we see that W : 2 x and w -"--2 y, hence that z 0 <2 W ! 2 Z- It follows w E P2 and that z w <--2 z; by the w E P1, so that w :! j z z. hereditary property, PROOF.
-
-
-
Q of all nonnegative maximal measures on X is a subcone of the cone P Of all nonnegative measures on X. The convex set Q, I I is a base for Q, and Q, y : [i E Q, I-t(X) is a simplex. The set
PROPOSITION 10.5
=
Suppose that A and /-t are maximal measures; by Propo10.3, (A + I_t)(f) (A + p)(1) for each continuous convex function f on X, so A + /-t is maximal. Similarly, rp is maximal if /-t E Q and r ! 0. Since Q, is the intersection of Q with the it is base for To a see that probability clearly Q. Q, is measures, in its natural a simplex, we must show that Q is a lattice ordering. the lemma the fact that P is a lattice) we need previous By (and in P. Suppose, then, that 0 < A < p only show that Q is hereditary and that [t E Q. Let A, be a maximal measure with A, - A. Then since /-t is maximal, we have A, + (/-t A) - A + (/-t A) /,t; A, A, E Q. A, so that A /,t /-t PROOF.
sition
=
=
-
=
-
Q,
We now know that
it
checked
that
this
onto.)
It
follows
is
function maximal
(1)"
is
measure
of the
tions
convex are
that
/-t with
if
subset
equivalent
of
is the
locally
each
Suppose convex
of p in X. (It and Lemma4.1 asserts
resultant
affine,
X itself
uniqueness
Meyer). a
is
x
closed; see from Q, onto X
function
affine
This
/-t
Choquet-Meyer
to
(not necessarily
simplex
function
i.e.,
one-to-one,
THEOREM (Cho quet-
pact
a
=
easily
that
is
we have an 11); furthermore, where -x by [t r(/-t), r(jL)
Section defined is
--
-
-
will in
be
X there
is the
simplex
a
exists
assertion
"(5)
if a
this
unique implies
theorem. that
X is
space E.
a
The
non-empty
following
comasser-
Section
Uniqueness
10.
(1)
X is
(2)
For each
(3) If
a
p is
a
measure
function
in
convex
(5)
For
each
x
that
px
E X there
supf IJV)
By
PROOF. which
is
a
:
If f
discrete
a
57
I(x)
then
on
and
x,
X.
if f
is
p(f).
=
g in
C(X),
f
is
unique
maximal
a
affine
is
resultant
X with
on
I
function
+ g
=
I
+ measure
such
I-tx
Ex.
-
LEMMA10.6 is
Measures
the
C(X),
f and
To prove that (1) recall that a bit; =
C(X),
in
maximal
For each
Ax) ft
f
convex
(4)
3.1
Representing
simplex.
convex
a
of
/-t
it
C(X),
measure
a
convex
for
then
and y
"discrete
finite
sharpen
need to
we
that
asserts
for
X and
in
x
Proposition f in C(X),
6-1.
,
E
(2),
implies
each
x
in
X,
I(x)
=supJ[z(f)
sxJ.
-
measure"
of course, a measure of measures of the form Ey. It
we
combination
mean,
only show the following: Given f in C(X), x in X, y a discrete Ex, and c > 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)
<
The
a.
=
A
do what is needed.
will
f
argument shows that semicontinuous,
and lower
=
I
inf
h
H1.
h E
:
apply what
we can
have
we
Choose
maximal
a
+ g) W
1-4f)
=
Proof
m(f
C
subspace
I m(f
shows that
-
m(l)
defines
a
(c)
Thus,
IIf
-
g 11.
C of C(X)
-
functional
g E C and that
(3)
(f
get
we
that -+
E X
x
x
E X
g) (x)
+
=
I(x).
and consider
This
is additive.
it
g) I
linear
continuous
that
and property
:!
,
9W.
Suppose C by f
in
g(x)
-
subspace C
the dense a
I(x)
-
C,
-
for
f
that from
Ex;
+
(5):
f (4) implies
and
g)
-
p
-
Ax)
=
defined
homogeneous, that
[t(g)
(4) implies
that
Suppose
measure
+
functional
the
to
on
f just proved to -f. Proof that (3) implies (4):
Finally,
if
J3
set"
on
L
h defined
similar
Representing
"difference
convex
functional
linear
function
of
is
From this
linear
functional
positiveit follows m on
the
envelopes (Section 3) uniformly continuous on hence has a unique extension
of upper mis
and of
norm
at
most
1
on
C(X).
Since
is given by a probability 1, this functional measure, which for have in we C, by px. Since, f I(x), px(f) m(f) that M,,(f) 3.1 implies : p Proposition supjp(f) exj, i.e., px >- 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 <
It
-+
On (L) (g
is clear
that
+
Afn)
On (L) (1)
=
1 for
each L.
fn1j:
h E
More-
0, then +
L(g)
Afn) +
=
L(g)
A[L(-A-1g)
+ A infIL(h)
+
11(-A-1g
+
-
11h
fn1j]
-
:5
jjg
+
Anf
Afnill
<
Section
so
Properties
11.
I I On (L) I I
that
11.3 that
G
On is Borel measurable composition maps On: So
On L
On (L)
1 and hence
=
X
E
On(L)(g)
-
duction,
On 0 on-1
=
=
So and
0
0n[On_1(L)](g)
Sn+ 1.
if
o
g E
from Lemma
consequently Sn+1 defined by
00,
n
S,,+,,
c
It follows
and -+
for
Ak+l,
is
too,
so,
0, 1, 2,....
=
O(L)
then
On-I(L)(g),
=
that
conclude
we
A,
E
g
01
o
...
69
map
each
each of the
If
of the resultant
each
n
L E X and
E
S,,,+,
>
1.
By
n
>
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,
<
:
this
last
SO
AL (h)
dominate
that
since
the
remains
A
i
!
AL
that
prove
Theorem
fo 1,
h >-
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
<
B
on
=
so
T-'A
-
-
=
=
=
=
-
=
=
=
=
=
-
only by valued set
set
of rational
fx:
M,
numbers
g(x)
By applying a.e.
Suppose,
of p-measure zero. functions. Then (taking a
>
h(x)J
r)
we
g and h
the
countable
unions
over
Ufx: g(x)
=
U[fx:r:' :h(x)j\fx:r :g(x)J] g
f
are
real
dense
have
=
to identity and by interchanging
this
all
now, that
=
f
,
and
h
f
>
=
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 <
1-ti(B)
p(A) =
<
I-t(B
that
measures
I for n
p is
on
S.
a
member
Then p is
of the
an
set
extreme
Xof all point of
ergodic. p is some
A)lp(A)
an
A in and
probability
invariant
S..
measure
Define
A2(B)
=
p(B \ A)/[I
-
p(A)];
and
76
Lectures
then
M, p
pi
ft(A)yj
=
(1
+
/_t(A))A2,
-
each pi
and moreover, each yj is invariant. measure, that p is invariant and that AAT-'A has /-t
is
a
[This
C,
probability the
uses
facts
together
zero,
measure
(C2AC3)
n
(Cl
=
C2)A(Cl
n
C3).]
n
To prove the converse, 0 or jL(A) suppose p(A) A in S,,, and suppose 2M = Al + A2, where Al, A2 =
probability =
order
theorem, Wewill
(as
yj
=
/-t
on
each
invariant
S,,
:D
Sp+jj,
12.2
above
results
locally
a
that
1 for
=
are
to
obtain
a
representation
P P under topology measures is compact. probability the method described by Choquet in convex
on
-
X of invariant
set
convex
use
the
must define
we
which the
apply
to
easily
S, by Corollary
on
p
follows
It
measures.
hence yj In
Theorem
identity
the
with
Choquet's
on
does
Feldman)
[17]. S be
Let
of Baire compact Hausdorff space, S the a-algebra subsets of S and let 7- be any family of continuous maps T of S into itself. Each T is measurable; T-1 carries the collection indeed, since a
of compact
S is contained in the a-algebra Gg subsets of S into itself, A such that T-'A The space of all finite E S. signed Baire
of all
measures
on
S
be identified
can
We will
restrict
difficult
to show that
a
linear
continuous
the
compact
itself
will to
induced
be closed
point
by
X will
theorem
shows,
family
7-
operation
(hence
the set
the
semigroup
generated
[23]).
Once
know that
extreme
yield
the
C(S)*
of
family
is not
which into
measures
maps
are
Weneed additional
of induced More
maps)
result.
is
carries
itself.
The the set of K
X continuous, hypotheses
is commutative
fixedif the
under
the
be nonempty if generally, mean (Day by T admits a left-invariant X will
X is nonempty, then and the existence and uniqueness
points following
itself
C(S).
It
poT`
map p -+
into
of
is, of course, precisely of affine transformations
the induced
compact.
C(S)* C(S)*.
be nonempty. The Markov-Kakutani for instance, that X will be nonempty
of composition. we
the
on
T, the induced
measures
T Since
and therefore
that
insure
for
points
space
topology
probability
K of
set
the dual
the weak*
transformation
probability
common fixed
into
to
for each T in
convex
X of invariant
set
of
ourselves
with
we
know that
theorems
it
apply
has to
Section
Application
12.
THEOREMIf continuous
S is
from probability
probability
Hausdorff S, then
space
S into
functions
of 7--invariant
set X
compact
a
Ergodic
and
to Invariant
m(f
)
fX f
-
measures
-subsets
the Baire
m on
measure
Baire
each
dm for
and 7-
f
a
each element
to
77
Measures
on
family of p of the
S there
of
X such that
in
C(S)
exists
a
for each Baire subset B of X which contains no ergodic measures. If the ergodic measures form a Baire subset or G6 then the measure m is unique. subset of X (e.g., if S is metrizable),
m(B)
and
=
0
set
X of 77-invariant
T.
Downarowicz
measures
[27]
is
a
proved
has
metrizable,
when S is
that
theorem
from this
is immediate
It
the
metrizable).
simplex (necessarily fact the interesting
that
for
any
space S and the set of T-invariant
metric
a compact 8implex homeomorphism T of S onto itself for which on S is homeomorphic to K. measures affinely probability If, in the existence theorem above, the extreme points of X were theoclosed in X, then it could be proved using the Krein-Milman To see that Leeuw theorem. of the Choquet-Bishop-de rem in place due to Choquet an example we reproduce exX need not be closed, is implied by Downarowicz's [171. (The existence of such an example than that is far more complicated but the proof of the latter result,
K there
metrizable
of the
exists
a
example below.)
simple
EXAMPLE
[0, 1] R(mod 1). -1
Let
line
I into
O(x)).
Then S is
probability >
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 <
of
1),
K
of
set
to
the =
generate
if whenever
then
y,
z
form
R+y if x
E p.
by exr K the union of the extreme rays of K; this whose proofs are straightforuseful has the following descriptions, Wedenote
R.R. Phelps: LNM 1757, pp. 79 - 87, 2001 © Springer-Verlag Berlin Heidelberg 2001
p. E p
set
80
Lectures
on
Choquet's
Theorem
ward:
Suppose x E K; then x implies that y, z E R+x. :! denotes * Suppose that *
linear y
(for
Ax
-
A !
base B
a
0) (so
B in
of K if
and
p intersects
exB
Bn
=
called
If
K is
of
cap
a
x
<
y
point),
B in
an
exr
z
by
induced K if
(y,
E
z
K
on
K) the
only
and
if
x.
with
set
a convex
one
y +
=
extreme
B and each
0
then p is
extreme
an
ray
Thus,
of B.
point
K.
exr
DEFINITION. K is
exactly
if
ordering 0 :!
B is
only
of K is in
x
whenever that
ray of K intersects
only if
partial
the
An element
K.
-
some
If K has
*
K
space
if and
E exr K
closed
a
K
convex
provided
cone,
a
C
nonempty subset
C is compact,
of
and K \ C is
convex,
convex.
If K has
a
[O,r]B=fAx
set
K such that
(Thus, cap of
is
of
then for any r ! 0 the compact base B, for instance, :0:-< A
if K has
a
K.)
Un=1 nC, 00
K
compact
Note that
B,
then
C
cap of K.
universal
a
[0, I]B
=
is
a
universal
any cap of K
contains necessarily 0, and if K, subcone of K and C is a cap of K, then C n K, is a cap a closed of this notion comes from the following K1. The usefulness fact.
PROPOSITION 13.1 x
C is called
then
base
is
If C is point of C,
extreme
an
Suppose
PROOF.
a
ly 2
of
cap
then
x
the
lies
closed
on
an
convex
extreme
cone
ray
of
K and K.
1z, 2
where y, z E K \ R+x. Since at least a convex, one of these points, say y, is in C. Since C is compact and y : 0, 1 < A0 maxjA : Ay E C1 < oo. If A > A0, then Ay C and [letting we have z,\ A(2A I)-'z] that
x
=
K \ C is
+
=
--
x
=
A'(Ay)
+
convex
and
follows
that
z.\0
point
extreme
(I
Ay
-
A')zx, \ C,
C- K
where 0 we
<
must
E C and hence
x
A'
-
-
(2A)-1
< 1.
have zA Cz C for =
AO(Aoy)
+
(I
-
Since
K
each A >
AO)z),,,
\
C is
Ao.
is not
It an
of C.
proof shows that if y, z E K and 21 y + 21 z E C, then y, z E R+C. The following useful fact is an immediate consequence of this remark: If C is a cap of the cone K, and if y, z E K and y+z E R+C, This
then
y,
z
E
R+C.
Section
Extending
13.
The above
rem
point
we
given an extremely useful
K is
K is the
its
could
for the elements
have
extreme
f
--
x
also
state
of such cones,
(ii)
on
and
K with
lower
p, is
additive
p(x)
integral
an
but
we
will
that
:!
K is
in
cone
a
Then K is
caps.
theo-
representation this
postpone
until
we
if
11,
where p is
and
which is
closed
a
K
following
the
if
only
an
convex
C is
A
cone.
and
compact
extended
valued
real
properties: and 0 :5
semicontinuous
p:5
oo;
positive-homogeneou8.
and
if and only if
The cap C is universal
C is
Suppose that
PROOF.
of
convex
its
of caps-a
Suppose a cap of
K is
p is
closed
a
union
description description which will follow. the examples constructing
in
E K
x
:
function
(i)
of
C
subset
of the
extension
an
rays.
alternative
PROPOSITION 13.2
C
of
hull
convex
At this
Suppose that and that
space
convex
closed
to
81
theorem.
THEOREM (Choquet).
the
Caps
Theorems:
immediately
leads
proposition
Krein-Milman
locally
Representation
the
of
cap
a
finite
p, is
K;
valued. then
0
C and the
E
infjA > (defined by p(x) (or gauge) lower semicontinuous, 0 : x E ACJ) is nonnegative, positive homoge:5 11. It is easily verified that since Ix : p(x) neous, convex, and C K \ C is convex, p, is additive on R+C (and p +oo on K \ R+C). p, of C
functional
Minkowski
=
=
=
after
The remark
p(y
and on
in
+
z)
(i)
and
On the
(ii),
and if
it
In
is
order
theorem
union a
base.
Since
for
of its
C
verified
easily is a cap. Finally, homogeneity of then
the
=
fx
:
x
assertion
p and the definition to see how to formulate a
cone
K
(in
let
us
first
caps,
a
locally
see
follows
that
if p is a functional :5 E K and p(x)
C and K
that
last
< oo;
hand,
other
it
\
C
are
described
as
11
is
convex,
is
immediate
from
of
a
universal
cap.
the
if y, z E K p is additive
shows that
proposition
p(y), p(z)
then
< oo,
of K.
all
preceding
the
compact,
the
positive
Choquet-Bishop-de
convex
space
how it is formulated
E) for
Suppose, then, that K has a base B and that x of a base is a base, we can multiple any positive
Leeuw
which
is
a cone
E
C
that
so
K,
x
assume
the
with
0
0.
that
Lectures
82
x
follows
It
c B.
that
of
resultant
is the
x
Choquet's
on
maximal
a
Theorem
p
measure
on
B, and this measure is unique if B is a simplex. Now, as noted at of Section the beginning 10, we can assume that B f y : y C- K and f (y) where linear functional is continuous It a on E. f 11, follows and positive that f is continuous, additive, homogeneous on < and C is E : a cap containing x. K, f (y) K, fy y [0, 1]B 11 If B is metrizable, then p is supported by the extreme points of B; it is supported otherwise, by exB in the sense defined in Section 4. An analogous result is valid if K is the union of its caps. =
-
=
=
THEOREM Suppose that
K is
of
E
its
fy
and that
caps, y E
:
K,p(y)
Furthermore, which
<
ex
C\
represents
x
supported
then [i
is
points
of C.
x
a
K,
11
some
the
contain
R+x,
p. so
of its
p(x)
If
we can
-
C,
apply
we can
C,
fy
=
exists y E
:
cap C
a
K,p(y)
=
11.
=
Lemma 10.7
A
=
:
y
E
C,p(y)
any
caps,
to
C
G
x
assume
p(x)
>
we can assume
p(y)
>
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
<
=
=
=
=
=
=
measures
is
We now
that
the set
immediate.
give an example which will show, among other C, above need not be compact (and hence is not
things, a
base).
EXAMPLE K be the
Let space
f,
of
convex
absolutely
cone
of all
summable real
nonnegative
sequences
sequences.
Topologize
in K
the
by
Section
the weak*
real
Extending
13.
topology
sequences
have
a
induced
which
converge
0).
83
of the space co (of all closed, does not
the dual
as
to
and is not
compact base,
universal
j
on
Caps
Theorems:
Representation
the
Then K is
metrizable,
but
has
it
metrizable
a
cap.
Indeed, it is clear that K is closed since it is the polar set of the K set of nonnegative f ynj : Eynxn ! 0 ly sequences in co: ! 01. If K had a compact base B, then whenever xn -+ 0, x, (as remarked at the beginning of Section 10) there would exist a B H n K. H such that weak* closed hyperplane Thus, there would exist z Eznxn > 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
<
s
-
r
w) p(z
-
and hence In order
:!
r
-
to prove
the
in
p(y)
C0,
in
by
zo
13.1, are
and
-
-
!
x0
p(x
=
-
t; similarly,
from
!
-p(x
p(x z) + p(z p(z w)
z);
-
w) w),
<
-
to
s
zo
that
suppose
r
t and
-
conclude
we
p(y
>
then
wo,
<
-
of
remains
!
yo
-
-
It
zo.
wo and
that
converse,
Since
K.
definition
the
yo
t, which is equivalent
partial
lattice.
a
if q is the minimum of q) E C0, and we need
(z,
=
is
p(z)
so
that
mean
w)
the
-
z)
-
t,
wo.
each
of
point
It suffices to show that cap which is a simplex. A y E K. Choose a cap C of K which is a simplex
a
the
finite.
element
implies
this
Let xo
by hypothesis
(z, r);
=
C0,
E
-
-
:! - q
-
if x, y E K, then x and which contains and
p(x p(x z)
Since
r,
similarly,
zo;
inequalities
two
Co
A y in < r
x
=
is immediate
!
xo
so
(w, t)
=
w)
K is contained
Proposition
q,
-
wo
t.
It
x0 A yo.
=
The first
wo. -
z)
-
to
zo
-
z
<
-
-
-
E K
=
=
let
=
have
-
Co
cone
fx: x p), then
C
functional
must show that
(y, s) are in C0, x z E K, we p(x) p(x z) +p(z) similarly, p(z) < s p(y z). It follows r p(x z) and s p(y z), then zo If
if the
(x, r) E Co if and only This E Cxf1j. (x1r,1) K and c x G C, i.e., x1r < r. only if x E K and p(x)
xo =
of course, that that x0 ! 0 if and
now, that
only
if and
If wewrite
follows:
as
Theorem
lattice.
a
additive
means,
follows
It
r.
simplex
a
appropriate
>
orr
assertion <
in ExRis
the
may be described
=
latter
if
:!
then K is
K is
a cone
by Cxf1j
generated
simplex,
a
Choquet's
on
then
z
E
x
+ y.
that
and y
x
(x, p(x)),
=
yo
x0 A yo exists
K, and
we
As noted
will
=
the
R+C,
(y, p(y));
Co.
in
after
in
are
then
Denote
show that
z
x
of
p(x)
x0, yo
this
=
proof
hence
are
element
A y.
Since
< It remains to z E K. zo :! x0 and zo yo, we have x z, y showthatifw E Kandx-w, E K. Since y-w (EK, thenz-w x (x w) + w E R+C, we have x w and w in R+C; similarly, -
=
y
It
-
w
-
-
E
follows
-
R+ C. If that
zo
we
!
let
wo
wo and
=
(w,
hence
p (w)), z
-
then w
E
xo
!
> wo and yo wo.
K, which completes
the
proof. The
ing:
Is
question concerning caps remaining important a reasonably large class of cones which
there
is the followare
unions
of
Section
Extending
13.
Representation
the
Caps
Theorems:
85
in concaps? This question has led Choquet [19] to investigate, siderable the class of weakly complete cones. depth and detail, (R. and Becker [7] has recently thorough treatment given an extensive of weakly complete cones and conical measures along with numerhere to proving ourselves We restrict two results ous applications.) of "well-capped" which exhibit two major classes cones.
their
Suppose that Kn C En is a sequence of convex Then the same is true is the union of its caps. of cones, of any closed subcone of the product K 11 & C E fl En. (In product of R+ with particular, any closed subcone of the countable the is union its itself of caps.) PROPOSITION 13.4
which
each
=
=
PROOF. Since
the
of
intersection
a
cap with
a
closed
of
subcone
subcone, we need only show that K is the union if Cn is a cap of Kn, of its caps. For this, it suffices to show that then there exists a cap C of K with n 11 2, 3,.. fl Cn C C. For each n there exists a lower additive,, semicontinuous, positivethat functional such on Kn Cn homogeneous nonnegative f Xn E Pn E2-n Pn(xn)It is easily Kn: Pn(xn) < 11. Define p on K by p(x) verified that p, is also an extended real-valued function nonnegative If of the is which is additive and positive On projection homogeneous. of the lower semicontinuous limit E onto En, then p is the increasing EN functions lower hence semicontinuous. is 0 Thus, if n=12 Pn On, if C x E K: p(x) 11, then C is closed in E. Furthermore, 2n C then for each which 2n is C fl E C, x Cn, n, so Pn (xn) < I for each if then hence E x compact. Finally, 11 Cn, Pn (xn) n, p(x) < I so x E C and the proof is complete. K is
a
cap of the
=
.,
=
=
-n
If
Y is
a
locally
space
(no topology)
which
have compact
sures ous
on
way)
induced
Y which with
by C
...
are
C,,,,(Y); (Y).
compact of
all
support. finite
on
we
The
Hausdorff
will
M(Y)
The space
compact
sets
consider
following
Q,,(Y)
space,
is in
M(Y)
result
was
denotes
functions
real-valued
continuous
signed
of all
(in
duality in the
Y
mea-
the obvi-
weak
shown to
the on
us
topology by P. A.
Meyer. PROPOSITION 13.5
Hausdorff
Suppose
space and that
K is
that a
Y is
weakly
a
locally
closed
compact, subcone
of
o-compact the
cone
of
Lectures
86
all
nonnegative
PROOF. Since
Y is
M(Y).
in
measures
countable
a
Choquet's
on
of
Then K is the union of compact
union
sets,
Theorem
its
caps.
we can
write
UY, where Yn is compact and Yn c int Yn+j for each n. Choose I on Yn, fn 0 on C,, (Y) such that 0 :! fn ! - 1, fn fn Y \ int Yn+j. a 0; we will construct Suppose now, that /-to E K, /-to Y
==
in
=
cap C which
Mo. Choose
contains
f anI
sequence sequence of a
such that
Since the 1nonnegative Eantto(fn) the positive), f po (fn) I is nondecreasing (and eventually As For /-t in K, let p(p) is convergent. Eantt(fn).
and
=
numbers
K,
C
f
The function
linign.
Yn+,,
int
on
To this
cap.
a
f
p(tt)
Eok'=n,
gn +
=
gn
this
I:nk= jakfk
will
and let
positive Furthermore,
ak.
Let
C is
show that
and it
strictly
is
We will
semicontinuous; -
function
positive-homogeneous.
and
11.
:!
p is lower end, define
and that
compact is
Cz K and
tt
:
additive
nonnegative,
p is
1/-t
=
Ean
series a
=
on
an > 0
prove
f
-
weakly
Eanfn
is continuous,
if
-
since
C,,,, (Y),
g E
C
that
then
a some n, so b(g) > 0 \ int Yn+l If p E K, then (since such that f) we have IgI : ' : b(g)f. gn / lim En Thus, if p E C, then p(l_t). liM/-t(gn A(f) k= jakA(A) < : b(g). It follows for any g in C,,,,(Y) we have p(g) b(g)lt(f) since P is compact that C C P 11f [-b(g), b(g)] : g E Q,(Y)j; with the weak topology coincides on in the product topology (which that P. It immediate is in closed that is C suffices show it to C)
g
there
for
0 in Y
=
number
exists
=
=
=
element
any
functional
on
increasing
the -+
/-t(gn)
limit
in
does not
have
a
linear nonnegative need we only Thus,
from
follows
functions
the on
fact
that
p is
K defined
by
semicontinuous.
example of
an
a
a measure.
continuous
and hence is lower
but
This
K.
of the
of C is
closure
and hence is
closed
We now exhibit caps,
pointwise
the
Co.(Y),
C is
show that
/-t
in
universal
a cone
which
is the
union
of its
cap.
EXAMPLE Let
s
be the
space
of all
real
sequences
in the
product
topology
space of s. As is well known, E can be with the considered to be the space of all finitely nonzero sequences, defined by (a, x) Eanxn, a E s, x E E. Topologize correspondence
and let
E
=
s* be the
dual
=
E
by the weak* topology
defined
by
s
and let
K be the closed
convex
Section
containing
of E.
elements
follows:
as
x
Representation
the
nonnegative
of all
cone
C
Extending
13.
Let
I
J has n elements. and suppose that that A 0 for k in I and EkEJ is
Ix
K, p(x) by 6n the
Denote it
is
p(Jn) > (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
ysuch thaty=Oonl. follows
:
K \ C is
C is convex, of the finite
that subset
a
x
< YkX-1 k
=
verify
If
Ik
=
Caps
Theorems:
for
I
have nonempty
we can some =
f
Kn
interior
(as
assume
in
in E* such that
x : f (x) (relative
<
to
nJ, K),
On the
K; locally compact. hand, suppose that K is locally compact. Then there exists a convex neighborhood U of 0 such that U n K is compact. Let F be of K with the boundary of U and let J be the closed the intersection and their
union
is
that
follows
it
K is
other
convex
true
of J
of F.
hull
By
of J. are
Since
Milman's
contained
in
0 J. Thus, there which strictly separates B that f -'(b) n K that
0
=
contained
in
[0, 1]J.
the U n K is compact (and convex), the theorem (see Section extreme 1),
F;
J C K and 0 c
since exists
a
0 from is
a
linear
continuous
J, i.e.,
base for
2b
=
inf
K and is
ex
K,
we
>
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
<
disjoint.
are
induction
where A is the
MO)
Cz
dominated
If
to
Sx, Sy
then
partially necessarily
SA (A set
Lebesgue
the
measures
SA, then Of (K),)
of
"inductive"
and there
applicable
=A A',
order
=
an
of /-t
hull
convex
JAI,
be p and let M B. If we partially SA Z is
has been carried
restriction
Z is nonempty:
to
that
Assume,
of B.
hull
convex
measure.
E M and
The collection
closed
properties:
M is
p
A, A'
If
set
is any nonnegative measure B of positive v measure such
step" Mof nonnegative
sets
If KA is the closed
(iii)
Borel
v
"induction
of all
Z be the collection
following
a
where K is the
< e,
for the moment, that the
is
if
that
series =
(above)
S),,,,. to
Lectures
90
v
p
=
A, obtaining
-
positive
/,t
measure)
convex
hull
of B.
Since
we are
led to
US,
from It
of all
f
=
Y is
x
g)
\
measure
v
the
The set
the fact
and from
g has at least If S C nonempty.
g is
of
point
Y, then
we
disjoint
B is
of Mo.
f and
v,
to
J be the J
by
affine
it
closed
any
follows
continuity
the
and let
g,
(for
closed
of
closed
complement of
f
Y is
that
one
(hence the
Given
and let of
restriction
el.
>
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
<
V
write
we can
-
by regularity
measure.
and hence
Vk for which Og(Vk) Ix : x E V, (n
sets
convex
convex,
some
-
g is bounded
function
and
of
Y, and V of J \ Y of positive a compact convex subset V for V. in is since the x Oxg < Now, f bounded,
clearly,
measure;
locally point
and E is
a
convex
Jo; clearly complete the proof hull
of
K C
V,
we need Oxg < E for x in K. In fact, to : show that Let the be hull & convex J, only Og(K) [= Of (K)]. of JO; then J, C Vk, so Og(Jl) If x, y are in K (i.e., in the < .6. closure of JI), then there exist neighborhoods of (in J) U, Uy x, y, for which Og(Ux) < E, Og(Uy) < E. It follows from respectively, the triangle that inequality g(y) I < 3E, and the proof of the lg(x) lemma is complete.
so
-
We now finish
that
f
measures
E) A
=
Ali
E/-tk+A
set
Kk for
the
resultant
the
(1),
satisfies A2 i
...
proof
and that i
Of
6 >
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
<
convex
Xk be -
E
Section
f (-Ek)
Ak (f for
Different
14.
each k.
IJAII(A/IIAII) IIAIIy, so 11-t(f)
+
for
-F
=
f (X)
-
=
f (x) I=
Representation
Theorems
91
I I Ak I I f (Xk) I I Ak (f 611 Pk I I since A/IIAII; 1 11AJAk + p 14MIJ + IIAII, we have X EIIAkIIXk + 5-7111411f (Xk) + IIAIIf (y). Thus, Thus,
each k.
y be the and I IIAII
Let
that
of the
Extension
resultant
of
=
=
=
11 [Ak(f)
-
(Xk)]
IIAkIIf
+
/N(f)
-
II/NjIf
(Y)l
-EIIAkII+IA(f)-IIAIIf(Y)I <
Since
this
holds
Choquet [18] theorem
pointwise
fails
for
limit
an
of
+
each
has
for
E
a
211AII
E >
sup
0, p(f)
=
<
e(l
+ 2 sup
f (x).
an
function
of second
of functions
Baire
of first
example, but omit the proof A proof may be found in [1, second Baire class.
will
describe
If 1).
example which shows that
given affine
sequence
If I
the
class
Baire
that p.
the
(i.e., class).
the function
above
the we
is of
20]
EXAMPLE Borel measures probability p on [0, 1]. Each measure p in X admits the unique decomposition into its purely atomic and atom-free parts, and we let f (p) be the norm of the atomic f (p) part of p. (Equivalently, E ,c :[O,lj p(jxj).) If p It is easily function checked that f is a bounded affine on X. of X, then p is a point is an extreme point mass and f (p) 1; I on the image in X of [0, 1]. Let A be Lebesgue f consequently, then f (A) 0. On the other hand, A can be measure on [0, 1]; carried to a measure v on X; then v is supported by the extreme in X is A. But v(f) points of X and its resultant 10 f (A). Let X be the compact
convex
set of all
=
=
=
=
has Using Choquet's theorem, G. Mokobodzki [69, Appendice] of the that fact an function of elementary proof given every affine first Baire class on a compact convex set X is, in fact, the pointwise limit on X. of a sequence of continuous affine functions
and dilations
Orderings
15
If X is /_t, A
are
a
compact
1,t(f)
that
A(f)
!
subset
convex
nonnegative
measures
for
of
of
on
X,
measures
locally
a
A to
-
/-t
f
function
convex
E, and if
space
have defined
we
each continuous
convex
on
mean
For
X.
ordering has long been of interest A in statistics; it is used to defined "comparison of experiments." in terms of dilations characterization (defined below) has been given and and Polya for one dimensional spaces, by Hardy, Littlewood, dimensional finite for by Blackwell [101, C. Stein, and S. Sherman The general case has been proved by P. Cartier [15], based in spaces. part on the work of Fell and Meyer; this is the proof we present below. different An entirely approach has been carried out by Strassen [75]. dimensional
finite
There
spaces
ordering,
another
is
E, this
denoted
by p proof
by Loomis [56] in the course of his uniqueness theorem, and which is of interest The second theory of group representations. tion is the proof that /-t > A if and only if p duced
measures
P, denote
let
We will on
A
X.
the
mapping
of all
set x
-->
> A which
of the
was
Choquet-Meyer with
in connection
of this
main result >-
the sec-
A.
regular
T__ from
intro-
probability
Borel
X into
P, is called
a
dilation
(1)
The
measure
(2)
For
each
the
the
resultant
defined
of Section
language
There as
C(X),
in
the
x,
for
each
x
function
valued
real
in X.
x
-+
Tx(f)
is
measurable.
Borel
(In
f
T,, represents
11, T is
a
measurable
Borel
selection
for
P, into
P1,
map.) is
a
natural
follows:
If
extension
A E
P1, let
R.R. Phelps: LNM 1757, pp. 93 - 100, 2001 © Springer-Verlag Berlin Heidelberg 2001
of T to
a
TA be the
map from measure
obtained
(via
94
Lectures
theorem)
the Riesz
on
from the bounded linear
(TA) (f )
=
fX
Choquet's
functional
Tx (f ) dA (x),
f
E
Theorem
defined
by
C(X).
taking A Ex in (*) shows that T(Ex) T, so that (modulo the homeomorphism x -+ 6,,) this is a genuine We can picture the measure extension; it, too, is called a dilation. out" or "dilating" the unit mass at x. Condition Tx as "spreading (2) says that this should be done in a reasonable way as we change from one point and (*) says that TA is the measure to another, obtained dilations. It is by taking the A-average of these individual not surprising, that TA should have its support "closer" to the then, of X than does A: extreme points Since
Ty
Ey for
,
all
y,
=
If p
Indeed, Tx -
suppose e.,
immediate
from
of this
result
f
that
have T. ,,
we
(*)
is
>-
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).
<
easily seen that the map y - Py is a homeomorphism from is compact the boundary of the sphere onto exX, so that the latter It
is
(and
hence
could
we
have used the
theorem
Krein-Milman
for
the
theorem). portion integral representation obtained by carrying result, p to a measure on the boundary If y E X, there exists of the sphere, is Herglotz's theorem: a unique measure p on f y : I I y I I r I such that probability of the above
existence
The
final
=
U(X) The above
tained can
as an
fact
in
sphere
induces
is extreme, =
n
=
the
closure
they
a
I
of
Y, and Y
one-to-one
all
is
An
by
Py. Holland[43].
An
F.
in
theorem,
closed.
be ob-
Since
the
Lecture
exX is
rotation
con-
of the
if one Py map of X onto itself, that exX is nonempty, we conclude
affine
Since
are.
Milman's
can
provided one elementary
theorem,
may be found
results
by
given
r).
theorem
Y of functions
has been
then
<
Herglotz
the
set
and related
Schober[74]. if (*) holds,
tained
exX
for
of this
Note that
shows that
application the ex X equals
this
Notes of G.
(11XII
Py(x)dp(y)
of the Krein-Milman
show that
proof of exposition
sketch
f
=
Y.
POSITIVE DEFINITE FUNCTIONS AND BOCHNER's THEOREM
A
be
complex valued if positive- definite
function
f
on
an
Abelian
n
E AiTjf i,j=l
(ti
-
tj)
> 0
group
G is said
to
Lectures
104
Choquet's
on
Theorem
A,, are complex A,,... then f (0) is easily seen that if f is positive definite, < real and If (t) I f (0) for all t in G. If a function f is a character of the group of all complex numbers G (i.e., a homomorphism of G into definite. of modulus 1), then f is positive Suppose that G is locally of and all P let be the continuous definite cone compact positive
whenever
tj,
numbers.
It
functions of those
t,,
is
on
f
G. Then P
cap
that
f the
every
g(s
+
in
K with
(t)
< 1.
positive =
a
eralization
line
finite
nonnegative of
a
The
is
points
subset
(g
E
L'(G)).
nonzero
function
f
dlt (X)
X (t)
p
of the
form
universal
a
characters
theorem
and each character
a
definite
measure
classical
Since the extreme
and has
continuous
f W for
ds dt > 0
K is closed
I If 11
as
of the set K
satisfy
t)g(s)f
(essentially) continuous
be considered
can
which
topology,
In the weak*
are
(G)
in L'
ff of those
of G and
elements
are
a
on
f
G, and
on
the characters.
of Bochner form
closed
extreme
X of
t
set,
e'xt
it
can
for
follows
it
G has the form
This
(where
-4
consisting points of this
cap,
is
a
gen-
G is the
real
real
x).
some
actually
be
proved
possible to use the Stonetheorem to show that [t is uniquely determined Weierstrass by f. has a close connection with group representations, This result in a definite function since each continuous on G defines, positive of G, and the canonical unitary representation way, a continuous irreducible the The above characters to representations. correspond shows that every cyclic representaessentially integral representation of G) is a "direct tion of G (and hence every representation integral" further of irreducible and For see details, representations. [58]. [33] result which can be used to It is worthwhile to sketch a simple The of the set K are the characters. show that the extreme points facts which are left unproved in what follows may be found in [58, called The proof is a "symmetric 10, 301 (where a "*-algebra" ring"). of this result due to J. L. Kelley and R. L. Vaught[48]. is essentially of course, obtained It is applied, to the commutative by *-algebra
by
the
Krein-Milman
theorem.
It
is also
Section
16.
adjoining
identity
the
(which doesn't Rudin[70]. Suppose e
A is
that
all
x
for
all
f
PROOF.
Any
(consider
(e
where the
assume
Define
that
identity of all > 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)
-
<
easier
to
prove
every
=
-
=
=
=
=
=
=
=
g(X*X),
so
g(x)
ANOTHERAPPLICATION TION
that
=
=
and hence
lg(X)12
element of K multiplicative to prove that Indeed, if 2f g + h, g, h in K, it suffices h (x) whenever f (x) h (e) 0 (since 1 f (e) g (e) 0, then 0 f 2f (x*)f (x) h). But if f (x) g 0 so + Furthermore, g(x*x) g(x*x) h(x*x), h(x*x).
even
=
0 and
OF
(similarly)
h(x)
=
0.
CHOQUETBOUNDARIESAND FUNC-
ALGEBRASTo APPROXIMATIONTHEORY
Lectures
106
[8] original
Choquet's
on
Theorem
special case of the Choof function algebras (and his "peak quet theorem in the context of the Choquet boundary in this case) was appoint" description of continuous complex plied to a theorem concerning approximation of the rational certain subset valued functions on a compact plane by the of this theorem, and direct functions. Wewill give the statement reader to Wermer's monograph [81] for a survey of this and related Bishop's
concerning
result
a
results. Let Y be
a
compact subset of the complex plane,
with
empty inte-
subalgebra C,(Y) consisting complex valued functions on Y which can be uniformly approximated which have poles in the complement of Y. functions on Y by rational It is easily verified that A is uniformly closed, contains the constants, and separates points of Y. Let B C Y be the Choquet boundary for Let A be the
rior.
of those continuous
of
A. THEOREM (Bishop).
(i)
Y
\
(iii)
B
=
following
assertions
Lebesgue
B has two-dimensional
measure
A
MAXIMAL MEASURE
if /-t is a maximal probability whenever exX C B C X and B is
We know that =
This
set.
theorem
1,
result
proof)
or
the Baire
be extended
can
Choquet
of
if B is
true
K-Suslin
a
if B is
and the K-Suslin
sets,
need not
K-Suslin
the
meaning of "I_t(B)
sets
as
any maximal that I_t(K)
follows:
1.]
=
proof
of either
and B is
an
If B is
measure
F,
The
[i there
a
is
the
containing
sets,
exists
X.
for
F,-set
that
I_t(B)
ex
it a
a
is still
complete
[Since
make clear
result X C
K, with 1 if
can
be
then for
B,
K C
measures =
of
means
family.
have to
set and
X, then or an F,
of sets contains
largest
we
of maximal
fact ex
an
set
by instance,
Choquet-Meyer
K-Suslin
only property
theorem set
The
V'
on
sets
set) (see [57]
be Borel
=
Baire
of these classes form the
sets
a
For
capacities.
set
measure
general
more
(= K-analytic [6]; each
set
K-Borel
a
to
abstract
on
the
formulated
zero.
Y.
THE SUPPORTOF
p(B)
equivalent:
are
(Y).
A= C,
(ii)
The
B,
such
used in the
p is
maximal
The K-Borel
sets
closed
family
countable
under
necessarily a a-ring,
not
closed
be
not
simplest
the
are
members of the smallest is
107
Topics
16. Additional
Section
differences). the
between
lies
set,
Corollary form: If X is following then for each x in X there
p
Ex and
is
immediate
to
the
-
and countable
unions
10.9
that
M(exX)
(in
I
=
OTHEREXTENSIONS OF
a
Borel
exX is
a
need It
sets.
K-Suslin such that
it
measure
and
sharpened
be
can
unique
a
which
and the
uniqueness simplex and if
above).
described
sense
THE
sets
on
exists
the
family,
the
(but
intersections
The
Baire
are
the compact sets
which contains
under
They
describe:
to
KREIN-MILMAN
AND
MINKOWSKI
THEOREMS convex 13.6, if a cone K in a locally Proposition then K no line, is locally compact, and contains space is closed, it admits a compact base. Thus, local compactness makes possible we theorem to (proper) to extend the Krein-Milman cones, provided by "extreme rays" in the statement of the replace "extreme points" this is true even under the weaker hypothesis theorem. (Of course, that What if we drop the hypothesis that K is the union of its caps.) this in two results K be a cone? Klee [51] has obtained direction,
As shown in
one
which
and
one
a
which
theorem
on
dimensional
We first
theorem.
the Krein-Milman
extends
finite
sets
require
definition. An extreme
property
the
]x, y[C if its
ray of
that
X and
]x,
If space,
and
set X is
an
p C X with
open half-line
in p whenever the open segment JXI Y[ is contained closed be said to is A set intersects linearly y[ p.
of the extreme X is
a convex
each line
with
intersection
union
ex
Minkowski's
extends
of X.
rays
a
locally
if
X contains
compact no
is closed. Klee's
closed
line,
Welet
convex
exr
X denote
the
following: locally convex closed convex hull of
results
are
of
subset
then X is the
the
a
X U exr X.
If tains
X is no
a
line,
linearly
closed
then X is the
finite
dimensional
convex
hull
SOME TOPOLOGICAL PROPERTIES
of
convex
set
exX U exr X
which
con-
-
OF THE
SET
the set
X of extreme
OF
EXTREME
POINTS As
of
a
was
shown in the
compact
convex
Introduction,
subset
X of
a
locally
ex
convex
space will
points form
a
Lectures
108
Gj
on
Choquet's
Theorem
general case, however, ex X need not set. even be a Borel Choquet has proved the following Nevertheless, results: convex If X is a compact convex subset of a locally space, then ex X is a Baire space in the induced topology. that a (Recall the intersection of any topological space T is a Baire space provided set if X is metrizable.
In the
of T is dense in
of dense open subsets
sequence an
interesting
in
[25,
application
to
C*-algebras,
and its
This
proof
result
has
may be found
355].
p.
R. Haydon [41] has shown that if T is metrizable is a separable space, complete with
T.)
topology),
its
then
there
exists
a
a
in
Polish some
compact
(that
is, T compatible simplex X
space
metric convex
homeomorphic to T. M. Talagrand Finally, [77] has proved a general result which imof that if the set extreme points of a compact convex set is plies K-analytic (that is, it is the continuous image of a Kj subset of a compact Hausdorff space), then it is a Borel set of a special form.
such that
ex
X is
THE POULSENSIMPLEX.
of extreme
points of a simplex need not is implied be closed (as, for instance, by Haydon's theorem quoted the is intuition to gained from looking at the finiteabove) contrary The fact
dimensional
the
that
Even
case.
example constructed X such
that
ex.
set
X is
by
counter-intuitive
more
E. T.
dense
X.
homeomorphism, this simplex It has other interesting is unique.
-
strauss,
Olsen and Sternfeld
THEOREM 16.1
phism, Poulsen
(ii)
a
(i)
[65]
Poulsen
in
It
now
turns
called
properties,
is
of
the
simplex
metrizable
a
that,
out
Poulsen
the
remarkable
as
shown
up to
affine
simplex by Linden-
-
[54]:
Uniqueness:
unique compact metrizable
There
is, simplex
up to
X with
affine
homeomor-
cl exX
=
X
(the
simplex).
homeEvery compact metrizable simplex is affinely Universality: the Poulsen simplex. omorphic to a face of For extreme two points s, and S2 of the (iii) Homogeneity: any Poulsen simplex there is an affine automorphism of the simplex which More carries to if F, and F2 are two closed proper generally, si S2faces of the Poulsen simplex and if p is an affine homeomorphism from F, onto F2 then W can be extended to an affine automorphism
of
simplex.
the POU18en
simplex
The Poulsen
in
appears
surprisingly
a
for
mechanics,
In statistical
situations.
109
Topics
16. Additional
Section
wide
instance,
it
of
variety
arises
as
the set
under certain measures) which are invariant the extreme are ergodic states. points actions; (dense) Details [46] and D. [45], Israel-Phelps may be found in R. Israel in Olsen's books [71, 72] as well as Ruelle's [59]. A proof is survey result: Let Z be the -set of all sketched in [38, Sec. 3] of the following sequences of integers and Q f 0, 1 JZ be the set of all doubly-infinite Q is its a and compact Hausdorff zeros product topology) ones, 80 (in natural the T is of Q onto iself, shift map (continuous) If space.
(probability
of states
its
natural
--
on Q forms measures probability of all T-invariant that is, the ergodic measures are dense in X. the Poulsen simplex,
then
the
X
set
A GEOMETRICALCHARACTERIZATIONOF SIMPLICES There
is
plices which but tations,
makes
purely image.
vector
will
space
again be
blence
a
We first
a >
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
<
set
x
(_3 , ).
point
OX)
C zx +
Similarly,
aXo
+
for
-yxX,
it
for
every
follows
that
X E X
every
Then
[0, I]Xo.
-
Since
that
so
Choose any X0 E X.
cluster
,k.
C E+
that
7xX.
the
to
simplices
assume
zx +
n
of
we
i- + =
i +
simplex whose extreme points form a closed Bauer simplex. D. A. Edwards [32] has shown that of Bauer simplices. "is" intersection an simplex.
([32])
If
X is
decreasing sequence X, simplices such that X is affinely a
a
D
metrizable
simplex,
X2
D
D
...
homeomorphic
Xn to
then D
...
there
of
have
X
a
THEOREM16.4 ists
proved simplices.
who
aX) n (y + OX), which implies that (y + OX) and therefore (x + aX) n (y + OX)
that
metrizable
belongs
7
C
n
Recall
X)
zX
=
have
we
ag)
+
+
f ZX 'YX IXEx (x + aX)
Thus the set X E X
directed
X E X there
every n
+
1,
following
of the
downward directed
a
Theorem
simplex.
a
PROOF. Let X be <
Choquet's
c
X. set
any
ex-
Bauer
nXn.
FACES AND EDWARDS' SEPARATION THEOREM
of
The notion or
pealing
is
called
Ay
+
(I
For
-
an
a convex
easily
appears
(also
set
visualized
called and
an
"extreme"
geometrically
ap-
in many contexts.
16.5
example,
entire
extreme
is
A nonempty convex subset F of a convex set X a face of X provided y, z E F whenever x E F and x A)z for some 0 < A < 1.
DEFINITION
the
which
concept
of
"face"
a
subset")
"extremal
set.
point.
the
It
faces
is
a
triangle always,
that, equally clear
is clear It
of
that
and sides, vertices face of X is an a one-point the set of points where an are
its
16. Additional
Section
affine
real-valued
Topics
ill
function
continuous
X attains
on
its
maximum is
face.
a
role in Alfsen's exposition [1] (although important Asimow of and Ellis that in and [2]; explicitly they are defined) face. Faces in particular, they make use of the notion of a "split" theorem, see, for instance appear in the proof of the Krein-Milman [70, p. 74]. Families of face-like sets appear in the proof of Bauer's maximum principle [20, Vol. II, p. 102] where the latter is shown to Here is the statement: theorem. the Krein-Milman imply Faces
play
an
not
(BAUER)
PROPOSITION 16.6 tion
a
on
compact
uniquely)
necessarily This
convex
result
is
at
an
A
convex
X attains
set
of
(not
X.
D. A.
step in motivating
one
maximum value
its
point
extreme
func-
upper semicontinuous
Edwards'
separation
[31]:
theorem
16.7 THEOREM
(EDWARDS) If f
functions continuous affine
convex
This
can
on
simplex
the
function
h
be considered
f :5
g,
f
<
of the
extension
an
upper semicontinuous
are
X such that
on
as
and -g
X with
then
there
exists
a
h < g.
following
classical
analysis. (See, [34, p. 88].) on functions Suppose that f and -g are upper semicontinuous < Then there exists that Y such a the compact Hausdorff f g. space < < h such that Y h continuous on f function g. Edwards' theorem for Bauer simplices, To see that this implies and f and -g are upper semisuppose that X is a Bauer simplex Restrict functions continuous convex on X. f and g to the compact classical result the to obtain Hausdorff a ex X, apply space Y identification with the X of the Y h and function use on separating funch as an affine continuous measures on Y to realize probability shows that f :5 h < g on all tion on X. Bauer's maximum principle for
from real
result
instance
,
=
of X.
UNIQUE REPRESENTATIONSIN As noted from
the
K(M),
in Section
study
of
6, there
arbitrary
where the latter
THE COMPLEXCASE.
is
no
compact is the state
loss
in
convex
space of
generality sets a
to
sets
subspace
in
switching
of the form Mof
C(Y)
112
Lectures
(or C,(Y)),
for
that
assume
Y.
In the
complex
6
produced,
for
ti
measure
Hausdorff
compact
some
the
M contains case,
the main
M*, Y, but nothing
on
[44]
Hustad
showed
senting
measure
that
fact,
before, points
the
we
of
in Section
"boundary"
complex
about
said
was
the
theorem
representation representing
a
Theorem
As
Y.
space
and separates
constants
each L E
Choquet's
on
norm
0.
of /-t.
boundary reprel ttll JIL11. (An exposition of Hustad's result in [64].) is contained This leads. naturally of to the question of such To formulate uniqueness norm-preserving representations. such a result definition. requires the following DEFINITION
if
every
or
the
fl,
of
Let
THEOREM16.9
C,(Y)
of complex
Assume that
Y.
a
is
M be
a
to be
a
simplexoid
simplex. as is simplexoid, an octahedron,
a
of
version
weak*
in its
set K is said
convex
K is
compact simplex.
a
choose
=
face of
proper
infinite-dimensional
ball
can,
one
which
A compact
16.8
closed
Obviously, unit
p for
in
octahedron
an
namely,
the
topology. closed
subspace
functions
on
a
continuous M contains
1 and
of
the
separates
complex
space
compact metric the points of Y.
space
the
Then
represented by a complex Borel measure A on Choquet boundary B(M) of M satisfying JJ/_tJJ JIL11, and each L E M* has a unique such representing and measure only if the if unit ball Bm. of M* is a simplexoid. each L c M*
be
can
the
=
This
mulated the
(due
result
to
R. Fuhr
and the
proved for nonmetrizable
and
exposition
[39])
author
can
compact spaces Y;
be reforsee
[39]
for
which
or
[64].
in
SETS. CHOQUETTHEOREMSFOR NONCOMPACT
There there
is
exists
a
of
class
subsets
convex
version
of
of Banach spaces Choquet's existence and
strong uniqueness theorem, namely, the class of all bounded closed convex sets having the Radon-Nikod#m property (RNP). As the name suggests, the RNP is defined in terms of Radon-Nikod m derivatives of certain vectorvalued measures; see Bourgin A [14] or Diestel-Uhl [24] for details. a
characterization
definition
notes)
can
of sets
since
it
be
given
is
with
the
RNP
closely
related
to
in terms
of the
(which
the
following
we
subject notion.
will
take
matter
as
the
of these
16. Additional
Section
113
Topics
A point x in a closed convex subset X of the Banach space E is said to be a strongly exposed point if there exists such that x1l -4 0 f (x) supf f (y): y E Xj and llx,, f E E* 16.10
DEFINITION
=
of
-
closed
convex
every
of
hull
X
a
exposed points.
strongly
its
of
Banach space E bounded closed convex subset
subset
convex
RNPprovided
to have the
X is the
f W
-+
A closed
16.11
DEFINITION is said
f (Xn)
Xn E X and
whenever
-
E, this yields of a Banach space with the RNP. Examples of spaces the definition dual Banach space and any separable with the RNP: Any reflexive Note that
(e.g.,
space
for
since
f
1).
we
The space L,
comprehensive
a
The set
ex
X is
ex
Borel
X of extreme
lies
a
the
in
measurable
universally is, for with completion that
and
convex
the
measure
f
p
(G.
separable
RNP. For on
does not
of this
points
of
(even [14],
for
any
X
=
RNP. See
have the
[24]
property. a
bounded closed
separable X in the separable
a
but
with
Borel
regular
respect
case
at
case,
respect
the
p
RNP)
the
to
measure
p of the
to
noncompact
with
on
least,
regular X, ex X
a-algebra
of Borel
of X.
subsets
THEOREM16.12
the
measurable
always
measures,
[0, 1]
treatment
Banach space need not be Borel measurable X of
subset
excluded
haven't
each
A.
EDGAR) Suppose
a
bounded closed
the Banach space E and that X has Borel probability exists a regular E X there
subset x
X is
X such that
of
p(ex X)
=
I
and
f (x)
=
fX
f dl-t for
each
E E*.
A uniqueness and
a
nonseparable
theorem version
(due
to
of this
R. D.
Bourgin
theorem
and G. A.
may be found
in
Edgar) [14].
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Lectures
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Index
of
on
symbols
Page A
13,18
B(M)
29
C
17
i
13
K(M)
27
0
28
X
VY
52
rl_l
15
>-
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,
