Lectures on Choquet's Theorem 3540418342

Citation preview

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 .

.

.

.

.

.

.

.

.

.

of .

.

.

.

.

.

.

.

.

.

.

.

.

.

measures .

.

.

.

.

.

.

.

9

com.

.

.

.

.

.

.

.

.

.

13

.

17

.

25

.

.

.

.

.

.

.

27

.

35

.

39

.

.

47

.

.

.

51

.

.

.

.

65

.

.

.

.

73

.

.

.

.

.

79

.

.

.

88

93

.

.

.

.

.

.

.

.

.

.

.

.

.

.

1

.

.

.

.

.

.

.

.

.

.

.

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

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

101

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

115

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

122,

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

123

Introduction.

1

The Krein-Milman

representation

simplest

The

example

of

is

the

following

page

7).

be concerned exercise

on

If

X is

E,

space

of

bination X1i

)

...

compact

a

if

and

is

x

a

classical

as

integral

an

of

finite-

a

of X, Thus,

X.

x

dimensional

is

there

p,....

i

finite

a

exist

convex

=

the

vector com-

points

extreme

Ek/t,1

Pk with

will

we

(see

of Minkowski

then

numbers

positive

of

which

with

type

result

subset

element

points

of the

theorem

convex

an

extreme

Xk and

theorem

theorem

that

I such

We now reformulate this representation of x as an "inteEpixi. For any point y of X let Ey be the "point mass" gral representation." at y, i.e., the Borel is which I Borel subset measure on any equals -y of X which contains and otherwise. 0 equals Abbreviating y, .5,,, by let Borel measure on X, p > 0, EpjEj; then p is a regular Ej, p and p(X) 1. Furthermore, for any continuous linear functional f on E, we have is f (x) (Epif (xi) =) fx f dp. This last assertion x

=

--

=

-

what

cally

convex

M

is

A point

for

said

IL(f)

for "x

fX is

f d[t, the

The restriction of

existence

Later, Borel

we

will

measures

be

to

a

Borel

measure

represented

that want

suffice

X, I-t(X) by p if f (x) fX f dp can

locally

is

consider

for

the

R.R. Phelps: LNM 1757, pp. 1 - 8, 2001 © Springer-Verlag Berlin Heidelberg 2001

=

simply

E* to

point on

other

will

sometimes

(Other

resultant

the

is

in one

(We result.)

E.

convex

measures

present.

(That 1.)

X.

on

with

on

functional f on no confusion "x barycenter of p,

many functionals there is at most to

measure

lo-

a

=

when

E be

of

subset

compact

probability

"

that

sufficiently

guarantees

p is

x.

nonempty

a

linear

continuous

minology:

this

and that

represents

p

X is

regular

E is

in

x

every

write

that

nonnegative

a

that

say

we

Suppose space E,

DEFINITION.

is,

when

we mean

to

separate

represented a-rings,

of insure

terp.

")

the

points; by p. but

the

Lectures

2

that

Note

for

each

"supported"

If Hau8dorff supported

DEFINITION.

compact that

is

/-t

fact

is

/-t

by

space,

which

measure

is

of X.

regular

S is

X and

space

dimensional

probability

a

nonnegative

a

finite

a

Theorem

represented by Ex; the the above example by

out

X of

subset

convex

X may be represented by the extreme points

in

x

compact

a

trivially brought

X is

in

x

(and important)

interesting is that

point

any

Choquet's

on

Borel

a

by S if [t(X \ S)

Borel

measure

of X,

subset

on

the

we

say

0.

=

which concern us: If X problems sub-set is a compact convex convex E, and x i's of a locally space 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 under is it unique? x? If /-t exists, Choquet [17] has shown that, the first X be metrizable, that the additional question hypothesis We may

has

[9]

de Leeuw

the

of

a

to

translate

will

It

in these

also

Bishop-de Y be

Let

of all

of all

set =

I

=

JIL11. E

space

convex

theorem

asserts

probability C(Y). By

measure a

X is

C(Y)* [t

on

embedding y of X, so we may

Borel

subsets

of X which

Bishop

is

affirmative

integral

place

in

worthwhile

seems

the-

representation

which we language is quite natintegrals of Choquet theorems

the of

use

Krein-Milman

compact

(in

theorem.

---

consider vanish

L(f) 442], at y))

that

[28, (evaluation /-t on

topology)

X there

L in

p.

as

a

the

of the

subset

convex

weak*

its

Y such

and

than

measures

space,

theorem

natural

points

a

each

to

well-known

second

C(Y) the Banach functions on Y (supremum norm), L on C(Y) such functionals linear

continuous

that

Riesz

how the

the

Hausdorff

Then

It

into

the

exactly

generalize

compact

=

(the

theorem)

real-valued

continuous

X the

L(1)

a

a

an

artificial.

bit

instances

make clear

Leeuw

of

introduction

theorems

Krein-Milman

the

first

the

X).

on

was

of X.

general question

more

the

to

to

answer

property

allow

we

hypotheses the example,

combination

introduced;

and

if

answer

two well-known

and

ural.

the

affirmative

an

geometrical

shown that then

above

convex

have

the

have

additional

In

while

certain

on a

measures,

(without

the

answer,

depends

question

orem

formulate

affirmative

an

Borel

now

and

the

space

and that

locally Riesz

a unique corresponds fy f dM for each f in Y is homeomorphic (via =

with

probability open

set

the

set

of extreme

measure

X\

on

the

Y, and hence

Section

3

Introduction

1.

supported by the extreme points of X. One need only recall functionals linear that on C(Y)*, E*, the space of weak* continuous L of form the functionals of all those -* consists L(f) (f precisely theorem of the this is a representation to see that in C(Y)) in order type we are considering. in the above paragraph There are two points which, it should be situation. of the general not characteristic are First, emphasized, X formed of the extreme a a compact (hence Borel) subset; points the representation was unique. second, (We will return to these that It is clear measure points a little /-I on later.) any probability which is in functional linear Y defines on -4 a C(Y) (by f fX f dl-t) next X. This fact is true under fairly as the general circumstances, linear from function recall that shows. result one a 0 First, space to another is affine A)O(y) provided Ao(x) + (1 O(Ax + (1 A)y) is

/-t

=

-

for

x, y and

any

Suppose

PROPOSITION 1.1 space

If

/-t

is

a

x

in

X is

PROOF. a

point

Hf to

=

Y is

of

subset

compact

a

hull

We want such

x

fy

that

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 weak* continuous an affine map from C(Y)* into X.

convex

of [i)

-

A.

any real

f (x)

that

show that

fy

=

M(f ) 1; n f Hf : f E

f (y)

:

show that

to

the

f dl-t

these

=

E*

I

compact 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

to

T:

then

T is linear

It suffices

0

If p

to

this

means

g

Since

-

Eaifi,

-+

Rn

and

by

Ty

=

a

linear

TX; representing that

(a,p) then

Y C X and

>

the

I-t(Y)

.

.

,

fn

E*,nHf,

in

n X

last =

p

=

:

assertion

1, this

(M (fl),

on

is

y E

Xj-

becomes

impossible,

p

(f2),.

a

If

=

(a,,

we

.

convex.

(fn)).

/,t

.

strictly

Rn which

by

functional

supf(a,Ty)

and

TX is compact

functional the

fn (0);

i

so that continuous, where E TX, p,

exists

.

(h (Y) f2 (Y))

show that

TX there p and

rates

by

E

fi,

set

define

end,

To this

is nonempty.

any finite

for

show that

a2

define

fygdtL and the

,

sepa-

-

> sup

first

-

,

g in

an), E*

g(X). part

of

Lectures

4

proof

the

measures sure

X is

let

compact,

f (xe)

=

(f )

I-t,3

--+

hypothesis

The

E in which

spaces

(f )

X,

y

net

in

C(X)*

to

the

=

their

to

x.

M0 converges for each

every

-+

xp

to

p, since

E*;

in

say,

y,

and

hence

the

latter

x.

X be closed

may be

compact

of

hull

convex

avoided

compact

a

if E is complete, or if E instance, obtained by taking a Banach space in its

space

if

weak*

f

Since

show that

to

But

for

compact;

probability meaprobability

resultants.

suffices

it

x

Theorem

of

/_t,

respective

converges

f (x)

=

that the

the

--+

x,,

subnet p

of

points

separates

is the

weak

those

in

always convex locally topology [28, set

is

434].

p.

A simple, a

xp

that

weak*

of x,

corresponding

the

next,

denote

x

show that

to

subnet

convergent then

Y converges and x,,

on

and

p,

Suppose,

complete.

is

Choquet's

on

compact

but set

can

useful, be

of the

characterization

given

of

in terms

closed

of

barycen-

and their

measures

hull

convex

ters.

PROPOSITION 1.2 space

convex

Y

if

and

represents

A

E.

if

only

there

that

exists

Y is

E is

in a

closed

the

probability

of

subset

compact

a

in

convex

measure

/-t

on

a

locally

hull

X

of

Y which

x.

PROOF. If /-t is a each f in E*,

for is

Suppose point x

and

closed

probability f (x)

measure

=

it

convex,

M(f

follows

)

:! that

Y which

on

sup x

f (Y) is

in

:! X.

represents

x,

f (X). Conversely,

Since

sup

if

then X x

is

to X, there exists a net in the convex hull of Y which converges form of the there exist points Equivalently, E '-,M'x ' y, y, 1, xi' in Y, a in some directed set) which converges (Ai' > 0, EAj' each y,, by the probability We may represent to x. measure /_t, the set of all probability the Riesz theorem, EM'E' measures on By Xi of C(Y)*, subset with a weak*-compact Y may be identified convex Of the weak* and hence there exists a subnet (in Aa converging pp of C(Y)*) to a probability measure topology p on Y. In particular, lim f (y,3) to Y) in C(Y), each f in E* is (when restricted so lim f f d[tp f f dp. Since y, converges to x, so does the subnet hence and f (x) fy f d1L for each f in E*, which completes the y,6, proof.

in

x.

=

=

Z

.

=

=

=

Section

The

proposition

above

Milman

5

Introduction

1.

makes

Recall

theorem.

the

it

reformulate

to

easy

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. is the convex point of a compact convex subset X of a locally space which X is on measure by the supported of a probability barycenter of these closure of the extreme points of X. To prove the equivalence Let Y in X. is and that holds the former x two assertions, suppose of X; then x is in the closed of the extreme be the closure points of hull of Y. By Proposition convex 1.2, then, x is the barycenter the obvious If extend Y. we measure a probability way) p (in M on the second result. the desired to X, we get Conversely, suppose

of

subset

the

a

and

valid

is

assertion

by Proposition

1.2,

x

that

x

is in the

Then

X.

in

is

closed

(defining hull

convex

closed

of

Y

Y,

as

hence

above) in the

points of X. theorem now using measures any representation than by their closure) supported by the extreme points of X (rather Klee In theorem. of the Krein-Milman is a sharpening fact, [50] closed

has

shown

that

convex

in

a

(which

sense

of

subset

of its

closure

extreme

that

clear

is

compact the

of the

hull

convex

It

an

he makes

dimensional

infinite

such

For

points.

extreme

precise) sets,

almost

Banach the

then,

every is

space

Krein-

than the "point representation gives mass" representation. The problem of finding measures by the extreme points supported of X arises mainly from the fact that the set of extreme points need in case X is is avoided This difficulty set [9, p. 327]. not be a Borel result. as shown by the following metrizable, Milman

no

PROPOSITION 1.3

topological PROOF.

vector

Suppose integer

If

X is

then

space,

that

a

the

more

information

compact convex subset of points of X form a Gj set.

metrizable, the

extreme

topology

of X is

given

by

the

a

d,

metric

n Fn z), 2-'(y fx and each Fn is closed, that checked It is easily X, d(y, z) ! n-'j. if it is in if and some not extreme that Fn. a point x of X is only is of the extreme the an F,. points complement Thus, have the trivial measure that Recall we always representing Ex it is of then If extreme X. is of not for a point an x x X, point

and

for

each

!

I let

=

:

x

=

+

y and

z

in

Lectures

6

easily

that

seen

there

points of representing

other

measures.

(BAUER [4])

PROPOSITION 1.4 vex

subset

an

extreme

of locally point of a

probability

measure

for

that

I-t(D)

>

that

there

0 for

=

0 for

neighborhood that

M would

X

each

for /-tl IL

(K)

+

rp,

-

M(K)

=

K.) Thus, r-lp(B n K) B in

set

we

(I

that

see

r)A2,

-

X.

define

Let

which

=

be

xi

E K

x,

Borel

A2(B)

and

and

implies

1, then

=

the

that

the

=A

x, x

=

the

follows

0 for

every

compact

and

resultant

x

M, and A2

of on

(X \ K))

n

of /-ti;

since

Furthermore,

x.

(I

+

rx,

by show

neighborhood

resultant

hence

that

Suppose

>

r)-'M(B

-

only

to

jxj.

measures

(1

the

of D it

M(U n X)

that

/_t)

of

compactness

(If p(K)

< 1.

we can

=

1,

the

y of D such

point

< r

Borel =

D;

from

is

is

x

supported

M is

D C X\

D with

con-

Then

of X and

regularity

the

to

set

ex

show that

to

(due

mass

=

be in

by pl(B)

point

X.

U of y. Choose U to be a closed convex The set K is K U n X C X \ f xj.

and 0

convex,

extreme

compact

a

E

x

point x.

compact

such

the

X is

that

represents

an

suffices

each

some

if

We want

it

some

is

of y such

this,

is

x x.

that

E and

only

and

X which

that

M represents

f xj; I-t(D)

set

if

X

Suppose

space

convex

on

Suppose

PROOF. measure

the

are

Theorem

Indeed, the they have no

measures. representing characterized by the fact that

other

exist

X

extreme

Choquet's

on

-

r)X2,

a

contradiction. It

the

interesting

is

to

Krein-Milman

Propositions smallest

and

1.2

closed

of

closed in

the

convex

closure

of

there /-t

=

Indeed, exists Ex.

It

[28,

p.

of X which

convex

of

Milman's

implies

it

(Milman)

locally hull

PROOF.

1.2, 1.4,

a

1.4;

subset

PROPOSITION 1.5

subset

that

note

theorem

440] that

the

space,

an

that

that Z C

extreme

"converse"

easy

Y

=

a measure

follows

that

of

closure

X is

a

X is

ex

X,

and

points

of

compact that

X

cl Z and suppose x E exX. x; /,t on Y which represents x

E Y.

of

consequence

the

are

convex

X is

the

contained

Z.

let

to

X.

generates

Suppose

Then the

Z.

classical

is

By Proposition by Proposition

Section

1.

7

Introduction

example of a dimensional X of a finite E, in order compact convex subset space of integral the question to illustrate uniqueness concerning repreis X that is sentations. or more a generally, plane triangle, Suppose Y of E, that is, X subset the convex hull of an affinely independent Y is is a simplex. no point provided independent affinely (A set y follows It then Y in Y is in the linear by variety generated \ Jyj.) that Y is the set of extreme from the affine points independence of X has element of X, and that a by unique representation every To conclude

a

10.10)

(Proposition of X has will

us

to

not

a

It

(among

is

difficult

not

then

Section

of the

10

to

some

will

we

show

element

give

an

which "simplex" Choquet's uniqueness

of

notion

things)

other

the

to

simplex,

In

representations.

prove

return

of Y.

generalization

dimensional allow

if X is

that

such

two

we

of elements

combination

convex

infinite

introduction,

this

that for a metrizable which states compact convex set X theorem, of has X each point in a locally a unique convex representing space, if X is a if X of and the extreme measure only points by supported simplex. of the Krein-Milman In the next section an application we give to make some general Before theorem. doing this, it is worthwhile theof the various remarks representation applications concerning of that the objects to recognize It is generally not difficult orems.

form

interest

a convex

X of

subset

some

linear

space

One is then

E.

for topology and at the same time yields E which makes X compact sufficiently the assertion functionals linear so that "A repremany continuous the extreme x" has some content. sents points of Second, identify has by the extreme points" "p is supported X, so that the assertion faced

a

with

useful

two

problems:

find

First,

a

locally

convex

interpretation. EXERCISE

Carath6odory's

Prove a

compact

in X is

(Hint:

Use induction

X, there and

the

exists

latter

of

combination

a convex

a

set

form

sharper subset

convex

on

the

supporting has

an

of Minkowski's

n-dimensional

of at

most

dimension.

hyperplane

dimension

at

most

theorem:

space

E,

+ I extreme x

a

H of X with n

-

1.

If X is

each

points of boundary point

n

If

is

then

If

x

x

is

in an

Hn interior

x

X.

of

X,

Lectures

8

X, choose an extreme point the segment [y, z] for some boundary point

of

on

Choquet's

y of X and

point

z

of

X.)

note

Theorem

that

x

is

in

2

of the

monotonic

functions

A real

valued

f

function

theorem

Krein-Milman

Application

(0, oo)

on

is

said

to

be

to

completely

completely

mono-

and if f, f (1), f (2).... and nonThus, f is nonnegative (_ 1) f (n) > 0 for n (n). each of the functions [Some examples: is as (-l)nf increasing, fundamental a Bernstein ! S. e-x represenx-a and proved 0).] (a for such functions. theorem tation (See [821 for several proofs and We will prove the theorem only for bounded much related material). if

tonic

f

has

0)

derivatives

n

=

f 0, 1, 2.....

of all

=

orders

functions to unbounded the extension repre(with infinite functions; from this by classical follows [821. We arguments measures) senting of [0, oo) by [0, oo]. the one-point denote compactification THEOREM(Bernstein). on on

(0, co), [0, oo]

then

such

If f

there that

exists

/_tQ0, oo])

-

f W

=

(Note

that

the

converse

is

unique

a

true,

monotonic completely Borel measure nonnegative and for each x > 0,

bounded

is

I

f (0+)

and

p

00

e-ax

since

dp(a).

if

a

function

f

on

(0, oo)

can

under the integral then differentiation sign as above, represented monotonic. is that Moreover, follows it and is possible, completely f to the theorem dominated the convergence Lebesgue by applying that we see 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 results setting. who proved this and related the of sketch We start proof. by giving a funcmonotonic Denote by CMthe convex cone of all completely funcmonotonic tions f such that f (0+) < oo. (Since a completely allimit at 0 always this right-hand exists, tion f is nonincreasing, in CM those of set be the K Let convex f though it may be infinite.)

be

=

R.R. Phelps: LNM 1757, pp. 9 - 12, 2001 © Springer-Verlag Berlin Heidelberg 2001

=

10

Lectures

such

that

suffices

to

of the

f (0+)


0. =

For

0.

>

-

+

so

(X)

_

(_ I)nf

(n)

(X

+

X0)]

that

we

oo)

=

must

(so

+ b

< I

or

have that

and

Furthermore,

(_ I)nf

+

that

continuous a

b) f (0+)

! - 1.

0,

=

is

f (xo)

b

0, let

>

shown

case -ax

x

(n)

(X

(n)

f

+

X0)

U) (n) (X) (n)

the

oc).

whenever

a.

to

f (0+)

n

=

is

some

and =

ex

:! -

"iterated

of

of K.

points

Suppose that this implies extreme,

f (x)f (xo) implies

remains

1), u) (0+)

(_ 1) (f

f

Since

E

a

completely

is

it is easily pointwise, Tychonov product

f (x)f (xo).

-

this

e-cx

=

a

-

-

f

that

xo)

+

E K.

u

on

(f

f (x

=




x

the

is

K.]

of

identify

This

functionals

sequence

defined

are

(0, oo).

on

function

a

certain

topology,

compactness is to

that a

these

in this

step

-

151]

p.

since

CMis closed

that

seen

convergence the evaluation

satisfies

if it

only inequalities;

monotonic

[82,

known

is

if and

may be obtained

above

the

from

proof different of pointwise topology and, of course, convex,

compactness

by using the also locally

the

with

follows.

result

desired

on

that

such

a

(a/2)

(a/2) (_1)n+lf and the

(')

that

shows

>


0) of K into

(r

points,

extreme

By

what

we

for

some

a

is

have

and hence

Since

(and

extreme

proof

the

We now finish It

[0, oo]

into

also

is

f

m on

is

the

K such L

=

M

f (x)

--

T).

fex f

=

measure

v([O, oc]) on

I

the

function

e-clx

under

T,

exponentials clearly extreme),

are

for

map T:

K

bounded

are so

func-

e-a0

-+

a

--

[0, oo]

f (0+).

[0, oc].

Since

as

linear

A separates it

that

>

from

points in

fT-1(ex for

K)

x

on

Lx dm for each

T >

d(m

a

a

-+

(x

>

0)

and

is contin-

e-clx

/-t and

of the v

Stone-Weierstrass -

second

QO, oo]) generated

C[O, oo], p

m(TB)

T)

o

exists

combinations

the

=

0.

C-axdv(a)

0

of

so

o

there

function

linear

[0, oo], QO, oo]),

Lx

Suppose

0 the

K

functional

have

each

=

fex

==

linear

continuous

evaluation

[0, oo] by p(B)

we

subalgebra

functionals

is dense

e-ax,

of finite

of

the

B of

unique. f (x)

x

A be the

A consists

and

that

p is

For each

Let

=

each

f (x)

that

so

dy(a)

such

0, then

=

Lx dm

e-ax

L dm for

subset

L,,(Ta)

that

prove

on

functions;

functions

implies

to v

E,

Borel

Since

K

>

x

on

each

M on

fex

=

Now, if

E.

is continuous

mo

remains

)

L (f

(0

A.

0,

form

the

theorem

show that

to

00

these

all

0 and

proof of Bernstein's

that

on

Define

0.

uous

of this

>

r

functions

difficult

not

f W

It

each

of its

is nonconstant.

is of the

point

-a,x

hull

convex

which

one

extreme

image

for

closed

K

ex

since continuous; [0, oo] is compact, its image ex K By the Krein-Milman to representation theorem, K there Borel probability a regular corresponds measure

in ex

Lx (f )

(i.e.,

least

carries

K is

functional

>

is

T, T, is

Since

compact.

each

x

transformation

complete.

is

tions.

the

constant

nonnegative. the

the

it at

this

holds

this

is

consider

has

proved,

just

0,

>

extreme.

the

and therefore

latter

=

onto,

itself.

the

Theorem

by (T, f ) (x) f (rx). convex it combinations,

defined

itself

and preserves Since K is compact,

one-to-one, onto

nonincreasing, reverse inclusion,

Choquet's

on

v.

are

by same

equal

on

theorem

Choquet's

3

The metrizable

theorem:

case.

for theorem Choquet's representation X. This is actually of the general metrizable case a special Choquetbut short its is it Leeuw and Bishop-de proof theorem, quite gives us to introduce an opportunity some of the machinery which is needed this

In

section

in the

I Ah(x) that

(I

+

h is

h is

that

h is

The function

C.

that

A is

the

constant x

contains

if for

a

a

--

f (x) + sufficiently

The function

following f

r,

If f

jh(x) useful

is concave,

:

-f

is upper

Banach

to

which

>

is called

to

is open,

functions

C(X)

space

r

bounded

f

=h

x

said

is

[: ,

=

We say if convex

1.

strictly

Al


0,

then

semicontinuous,

g and

thenrf

If

-

j =rf.

Ilf

-

g1j,

14

Lectures

proofs

The manner

from

of most

from

fact

the

sertion

that

(b)

in

may be

K

is closed

and

proved

second

of

set

I(xi)




0.

(h

Define

a

A,

real).

in

r

linear

functional We will

show

on

B

that

Section

Choquet's

3.

functional

this

rl(xo)

h(xo)

+

then

h +

rf

is

dominated

:-

(h

h +

=

concave,

and hence

theorem,

then,

m(g)

that

=

j(x0)

M(h)

=

p(j),

!

-

there

>

exists

m(g) p(l), so

I(xo).

I-t(f).

f

It

is

/-t

,

=

continuous.

i.e.,

1,




0, then

that

m(l) m(f)

I

M(f)

other =

that

g :! -

C(X)

m on

m(h+rf)

and

M on X such

Furthermore,

h(xo)

functional

linear

functions

we see

On the

and

.

a

C(X),

C(X)

measure

A,

1 c

exists

on

Riesz

Borel

=

nonpositive theorem, representation

nonpositive the

rf ) (xo) for each h in A, r in R. If r 2! 0, rf, by (b) and (c), while if r < 0, then h + rf is h + rf > h + rf h + rf By the Hahn-Banach

for g in

If g E

E R.

r

that

i.e.,

p,

+

there

g(xo)

!-

functional

by the

B

on

15

Case

The Metrizable

Theorem:

j(x) I f complement in the set .6 is contained that the proof by showing We complete where if X. of + 1z, x ly of extreme y and z Indeed, points 2 2 that of f implies distinct convexity are points of X, then the strict 1 1 .1 1 < f W< 2 (Y) + 2 W 2 1(y) + 2 RZ) 5 AX) p vanishes

that

of S

the

on

x

-

f (x)

:

=

-

=

-

to note (and will be useful later) that interesting with the set of extreme coincides points f (x) I actually of the next proposition. is a consequence

Ix

such

that

It is

If f

DEFINITION.

A(f) for

each

M and

A,

in

PROOF.

f (x)

f'

/-ta

we

To

see

Ex,,,

=

f (x)

is

1.

-

that

prove a

net

that

such

It

if

is

is

x

f(x)

>

r

from

:

the

from

a

that E.

point 6 >

By

the

This

p(f)

compact

x,

of

p

:

/-t

-

Exj;

we

that

Indeed, each

0 and for

Exj.

To

1.4.

definition with

-

con-

X.

Proposition

semicontinuous. to

-

point

supf/,t(f)

easily upper

on

supff f dM

=

extreme

=

! r, suppose

pa(f)

I(x)

f'(x)

in X converging

that

function

an

-

A.

follows

follows it

/-t

X,

in

x

let

that

-

each

measures -

continuous

a

assertion,

show

say.

is

write

assertion

first

f x, I

will

second

the

that

we

The

prove

concave;

for

then

X, Consequently, set

vex

are

If f

PROPOSITION 3. 1

probability

A

f (x)

:

of X

weak* -compactness,

f'

is

suppose

f '(x,,)

each

must

a

: ! r, choose

there

16

Lectures

exist

probability

a

converges

(x,3) f(x); f (x, r)

g

-+

it

as

h(x)

>

in

if h is in

p(h)

f ;5 f '

=

a

!

r

I

is closed

(b) (above), A,

I-t(f).

x

in

It

g(xa)

conclude

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 for


p

y in

measure

To find

:

v

Was

a

be

>- /,t,

maximal net

is a

p

maximal

then

v

element

(the

Al.

-

maximal

>-

of

directed

A, Z,

4.

"index

set"

being

weak*

the

in

exists

[to

po in

the

po is

have

yo c- Z.

is

used

here.

X, choose

in

above,

they used

an

The notion

is

maximal

a

represents

p

to converges from it follows

p vanishes

implies that The first points. /-t

that

/-t

>-

we

element.

maximal

at

from

If xo

As noted

Ex,

of

maximality

the contain

extreme

no

in the

following

X, then

I_t(f)

is contained

this

A,

slightly simple way:

very

a

which

sets

doing

toward

step

in

>-

differs

show that

to

Baire

on

looking

which

such

/,t

measure

of

idea

ordering applied

remains

it

xo;

the

po >- [q.

maximal

a

there

Thus,

Wwhich

Z contains

originated

de Leeuw

although

one

then,

lemma,

By Zorn's

and

measures,

subnet

a

A(1)1.

--

in W, If [t, is any element that eventually - /_tj and hence of subnet /-t,, since for bound an upper W; furthermore, /-to

Thus,

the

0 and

/_t(i) f p, I of

> 0 and

19

is contained

which

topology.

weak*

Bishop

!

/-to

IM

set

Theorem

Leeuw Existence

of Wthemselves)

elements

the

compact

with

definition

the

Choquet-Bishop-de

The

Section

result.

If

PROPOSITION 4.2

/_t(j)

for

each

L(rf)

=

p(rf),

on

rf, Rf,

an

extension

so

each

g in

v(-g) we




convex

About

that

(y

z),

+

f (y)

the

the

X is

ex

f (x) 1, f

=

a

theorem)

be

prove

I(x)

:

then

f (y)

If C contained

metrizable.

is x

Theorem

would

existence

case

the

in

proof

form f f (x) for each f in C, and the

all

Choquet's

of X.

points

extreme

however, that the that on X implies

do in the

inter-section

Indeed,

shown,

we

fo(x)J,

=

function

continuous

best

lo(x)

:

has

fo,

function

convex

exX

the

contains

sets

on

y,

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

the

on

Baire

--

sets

from ex X, it suffices to show that 0 if D is disjoint p (D) a compact from ex X. (This is a consequence Gj set which is disjoint 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 f fn I sequence 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 0. To obtain 0; lim[t(fn) p(D) this "limit" result two technical lemmas. The first slightly requires of these is quite it reduces since the desired result to interesting, theorem for metrizable Choquet's X, using an idea due to P. A. the fact that for each x in X, we will use Meyer. (More precisely, there exists which is Ex supported /-t by ex X. Since it is not that true A in be extended can to an element generally of f every this is formally than the stated of Choquet's version E*, stronger theorem. See Proposition 4.5.) which

are

=

=

=

=

-

-

=

=

=

-

LEMMA4.3 per x

Suppose

semicontinuous

in exX.

that

f fnJ

functions

Then liminf

fn(x)

is

a

bounded

sequence

X, with lim inf fn (x) ! 0 for each x in X.

on

of

concave

!

0

for

up-

each

Section

The Cho q uet-Bishop-de'Leeuw

4.

first

Assume

PROOF.

that

If

metrizable.

X is

Theorem

Existence

X, choose By hypoth-

in

is

x

21

a

measure by ex X. Ex which is supported probability y 0. Fatou's ! inf lim 0 so a.e. lemma, lim inf by f,, esis, M, assertion and upper semicontinuous, Since each f,, is concave (b) in h that E : that 3 shows Section A, h > 1,,, so inffh(x) f,,(x) f,, inf f [t (h) : h E A, h ! f,, I ! [t (fJ. Thus, lim inf f,, (x) > f,, I ! 0. Turning to the general x is in X, and lim inf I-t(f,,) suppose case, for each n choose hn in A such that hn : fn and h,, (x) < fn (x) + n-'. with the product of lines Let RN be the countable topolproduct N function The X R -+ 0 Jhn(Y)J. by 0(y) 0 : ogy and define and continuous, is affine so X' O(X) is a compact convex subset -

=

=

-

=

=

metrizable

of the

x'

is

of R

X',

in

the

is

in

W)

7rn

X',

ex -

hn (Y) x' in

for

each

the

metrizable

set

(x') (Ox)

0 < lim inf 7r,,

If

LEMMA4.4

decreasing

f,, (x)

for

7rn

first

from the

=

ex

inf

that

X.

Since

f,, (y)

>

and continuous

0, on

proof we conclude 0 (x), we obtain Taking which completes lim inf fn (x), of this

part

hn W

in

> lim

x'

x' in X.

each

lim inf

--

x

is

/-t

=

0

is

(and

in

It

=

Thus,

we

Baire

x

such in

ex

fn :!

:! -

that

-1

X,

then

J1nJ

sequence

-1

from

0; from

completes

QX) each

the

Lebesgue

the

the

for

on

measure

in

follows

From the

limp(ln)

in

maximal

the addition, above by zero),

exX;

bounded

X.

a

Since

functions.

uous

on

> 0

Consider

PROOF.

X.

so

sequence

and lim

in

X',

Assuming

y is

affine

are

in

y.

(x')

If hn (y). X; by the

=

=

proof.

the

if

lim inf 7rn

have

we

that

shows

argument

(0y)

convex

point

extreme

an

7rn

and

compact

The functions

X.

ex

lim inf 7rn

that

simple ! f,, (y),

is has

it

X, then

in

coordinate"

"n-th

usual

be the

-xn

y is

0-'(x)

set

a

Let

.

R; if

onto

theorem

Krein-Milman

x'

R

space N

projection

N

of

1,,

sequence so

Lemma 4.3

that

that

:!

X, and if f fnJ :! f,, :! 0 (n lim

I-t(f,,)

concave

Q,

we

ffnJ is limfn(x) lim In (x)

theorem convergence have 4.2 we IL(f_n)

bounded

Proposition

liM

1, 2,...)

=

in (X) for

0 for

it follows =

each

each

tt(fn),

x

subsets

shown

that

of X \

any ex

X.

maximal

We have

meabare

also

on

shown

x

in

that

which

proof. have

0

-

nondecreasing

exists =

non-

semicontin-

upper

also

a

0.

=

have

is

X vanishes

something

22

Lectures

different: slightly of X contained D is

A maximal

shows,

in

particular,

closed

set

which

that

formulate

We next

which

of

of

x0 in

that

1 such

Borel

[t

of X\

ex

S in S is of the

B2 /-t

Baire

are

(ex X) As

x -*

=

f (x)

+ r,

f

of sequences

by f (x)

Ex,,.

-

2 such

y in

This may be EX" and

point

If

sets.

A (X)

=

x

x

proper

=

tions

is

tinuous

shows

"A represents X) coincide,

g

exists

the

on

X) ],

a

Baire

measure

that

any set

where

B, and

/-t is well

defined

and

A is

of the

form

the

space

f2 in its

weak

such

on

(x, y)

x"

for

all

Ix, I f (0) x

following topology, 2-n

example: let

and

X be the

f

define

0, but there

is

on

no

the

the

point

M E*lx + R of C(X) subspace the two notions Nevertheless, "/t X and a probability measure on a /-t proposition following implies.

subspace in

X

in X.

the

(for

as

=

,

M (defined closed

subspace

above) A

of affine funcof all affine con-

X.

that >

that

in

of A.

The

.5

there

observe ex

function

every

Consider

that

dense

is evident

E A and

M(X)

nonnegative

this,

n

R.

subspace

uniformly functions

S with

vanishes a

[B2 (X \ A(Bj), then

is in A and

=

in

PROOF. It that

f

f (x)

PROPOSITION 4.5

=

not

fxnl

Then

that

U

of subsets for each

1.

Hilbert

example a

/-t(exX) [Bi n ex X] we let I_t(S)

on

convex

Then

a-ring

sets.

theorem

A to

To do

1.

a

1.

=

only extend

compact

a

the

/-t

in

applications.

X is

Leeuw

--

earlier, E*, r in

in

E be the

Let set

form

remarked

we

that

x0 and which

represents

We need

X.

S and show that

on

for

/-t(exX)

any

theorem.

convenient

Suppose

it

by Choquet-Bishop-de

Leeuw theorem

x0 and

A which

important, supported

be

more

0 if

=

since

Choquet-Bishop-de

Choquet-Bishop-de

the

measure

subsets

the

Krein-Milman

Leeuw).

represents

/-t

By

PROOF.

hence

locally convex space, and denote by S is generated by exX and the Baire X there exists a nonnegative measure

a

X which

point

is

Gj subset

I_t(D)

that

is

measure

and

the

the

THEOREM(Bishop-de

subset

exX,

perhaps

can

maximal

a

contains

This

Theorem

any

on

showed

we

of such

generalizes

Leeuw theorem

(Indeed, a set.)

X.

ex

subset

compact

any

manner

X\

in

vanishes

/-t

measure

Choquet's

on

the

space

0, and consider

A is the

uniformly following

closed. two

Suppose subsets

of

Section

E

x

The

4.

J1

R:

and

r

and

disjoint.

=

theorem

x

f(x,r) g(x) + 61. -

By

a

(obtained

difference E

Choquet-Bishop-de

R and

set

J2

-

x

:

These

on

E

X,r

=

sets

g(x)J

slightly by separating

Ji)

there

A in R such

exist

L(JI) L(x, f (x)) g(x) < f (x)

that

sup

=




let

/-z be a maximal Since U Bi is

let

'

-

measures

(If

r

Let

1.

=

K is

ex

hull

each

ball

compact

of K and

resultant

have

For

closed

weak*

is

suppose

K.

ex

the

/-t(U Bj) Uni= 1 Bi, then p(D) > 1 A/-tl + (1 A)A2, where

we

D

Bi point

each

and let

space

that

convex

Kj,

E

of

K of

with

intersection

at

subset

E* such

spaces.

separable).

norm

supf Ilf 11: f

-

dense

norm

of

Theorem

Banach

Banach

real

a

closed

norm

itself

is

E be

subset

the

arbitrary

with

Let

convex

K is

M

Let

PROOF.

deals

Choquet's

on

if as

by

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 in of hull lies which is weak* compact. convex Un I Bi, r(pi) 1. Hence r(/-tl) Eni= 1 Aigi, where gi E Bi, Ai > 0 and Eni= I Ai This of h is Let h and a point co (ex K) Eni= I Aifi. I c/3I I r (pi) Since

p,

is

a

probability

n

measure

_

_

=

-

Consequently,

IIf

-

Thus,

h 11 :5

11 f

-

co(ex K)

Ar

(pi) I I

is

norm

+

(1

-

dense

A) I I r (pj) I I in K.

+

I I r (pj)

-

h

II


-

p

measure

A'

0'.

o

maximal

a

A' is supported by the compact 0' for a (maximal) measure A

set

that

it

Now, O(D)

is

on

vanishes a

Section

in

the on

Baire

4

Gj in O(Y) and it misses

we

on

exK(M),

-

A p

Bishop-de hence that

B(M),

Gj subset

Y, namely, are given

A'

O(Y),

Y such

Mthere

A with

with

the

of Y \

subsets

any compact

If

on

measure

K(M)

on

/-t

(and

do this

to

subspaces

separating

measures

theo-

representation

In order

to

remarks

A vanishes

that

a

(prior

of the

view

with

observe

mean

Y and

choose

we can

show

Bishop

we

p to

-

measure

see

to

definition

suitable

form

section

purposes)

later a

In

is due

this

we

o

a

p,

0'.

Leeuw is

of the

A >- p. To need only

D C Y

hence

-

the

\ B(M). same

is

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. hence A(D) on A D O(D), A'(O(D)) 4.4) A' vanishes true

of A

=

=

THEOREMSuppose

which

separates

points

that

M is

and

a

contains

subspace the

=

of C(Y) (or of C,(Y)) If L functions.

constant

measure a complex M*, then there exists p on Y such that set 0 for any Baire fy f dl-t for each f in M and p(S) M. the which is disjoint for boundary Choquet from =

L(f) S in

Y

Section

PROOF.

obtain for

A New

6.

By applying A

a measure

f

each

K(M)

are

define

A

properties.

M.

in

with

which

Setting:

>-

tti

disjoint =

Y1

-

the =

Choquet

The

Hahn-Banach

Al -A2+i(A3 each

Ai.

We know that

i

we can

B(M), i (P3 N)

from

and

-

,

and Riesz

-A4)

For

A2 +

Boundary

we

on

find

a

yi(f)

=

a

theorems,

Ai(f)

measure

we

L(f)

that

maximal

vanishes

/-ti

get

Y such

33

=

measure

on

the

for

f

with

may

A(f) yi

Baire in

the

M. If

on

sets we

required

7

Applications

Let

X be

A > 0 there

family

is

(i.e.,

R,X ! 0

> 0

R.Xf

Ry families a

T,Tt

=

potheses,

=fo

[68]

for

is due

We first

definition 1.

For

a

2.

f

For

(Tt

If

C(X)

is

itself

into

:

t

>

(i.e.,

conditions

suitable

(x

defines

a

[55]

and

facts

is

the

convergence

[55]

None of the

shown

which

a a

hysemigroup

certain

papers

exposition

the

originally

elementary

some

this

on

was

(See subject.)

result.

follow

to

Under

section

0)

>

way from

of this

of this

X, A

in

resolvent. in this

content

needed

follow

us

and facts

given below, by Choquet.] easily from the

resolvent.

each

A

>

0, R,\

C(X), (1/A)Ilf 11, so JJR,\11

if

the

identity

processes.

from

under

then

information

Lion

prove

of

and

proof

are

to

of Markov

is obtainable

the

detailed

paragraph

which

following

that

A) Ry Rx.

-

operators

0),

and the

to

(A

e-"'(Ttf)(x)dt

resolvent

operators,

more

this

=

if the

each

00

C(X)

in

related

theorem

for

C(X) such 1/A. We call

-+

R)J

and

resolvent

study

>

1, Tt

=

every

of Markov

in

f

all

=

of Markov

semigroup T,+t, TtI

for

Rx

-

the

in

arise

(R,\ f ) (x) exists

a

0)

that

suppose

C(X)

:

resolvents

to

> 0:

H

[Such 0) is

0)

>

Rx !

f

whenever

RA(A

of operators all A, A'

and

space,

transformation

linear

a

for

valid

Hausdorff

compact

a

boundary

Choquet

of the

E

each

and follows

A and

from

:! -

A',

(*)

:-

I/A. R,\RX,

11f 11

hence

1, But RAI ==

otherwise.

R.R. Phelps: LNM 1757, pp. 35 - 38, 2001 © Springer-Verlag Berlin Heidelberg 2001

and

continuous

is

f

then

-

RyRA.

=

11R.J RJ

=


I-E

(V)

The

set

U

If I

E

Y with =

Ilf 11

point

y

The

point

y is

in

Ilf 11 y

the

:

Suppose following

IfI

0

6 >

any

:! -F

in

y,

=A

If (x)I

=

that

A is

there

y,

Ilf III a

assertions

C

S.

uniform algebra equivalent:

are

x,

there

f

exists

E A such

that

A such

that

Y\U. there

exists

f

E

-

Choquet

R.R. Phelps: LNM 1757, pp. 39 - 46, 2001 © Springer-Verlag Berlin Heidelberg 2001

exists

L

in

Condition

satisfies

The

fx

and

containing


0 such

S,

then

see

that

A(Yk)

and

the

from then

not;

JJx

'2

fYkJ

yll

1

+ 91

in S and

=&)

I

for


a

fk (x,,)

and

a

above

Theorem

I < 2-k-16. Thus, jjYk+1jj Yk+1jj < (I 2-k-1)6


afk(Xn)llllYk

-

1

CeXnll-

have

OXn) 0) and he proved the following Bernstein

,

THEOREM(Korovkin[521).

Suppose itself

Then

To show

that

the

Bernstein

(x

+

a)n

(x)

I

=

(1)

operators show that

we must theorem, to Ik for k 0, 1, 2, where binomial expansion

Korovkin's

=

Yk=O

Setting

(1)

with

utilizing

is

a

sequence

with the property from C[O, 1] into three the to f for f (x) functions uniformly to f for f uniformly every f Tnf I converges

operators verges

f Tnj

that

a

=

I

respect the

-

x

to

previous

shows x

(n)

Bn1 multiplying

twice,

Bn.[2

satisfy

the

E

[0, 1]

x

=

each

I for

by

X2 n2)

yields

identities

=

T2

I + n

R.R. Phelps: LNM 1757, pp. 47 - 50, 2001 © Springer-Verlag Berlin Heidelberg 2001

>

0,

of positive f Tnf I con-

kI k C[O, 1]. x

=

0, 1,

2.

.

Consider

the

Xka n-k.

k

that

E

P.

of hypotheses f Bn Jkj converges uniformly for

x

=

that =

ob-

(I

_

J2)

n.

settinga

Differentiating =

I-x

and

48

Lectures

for

each

n,

We won't in a

a

prove

for to

theorem

result

to

Hausdorff

Ma Korovkin

due in

is

linear

span,

( akin[73]).

THEOREM

and that

space

Mis

only if

the

[To

see

a

this

fTnj

is

of positive f for each

uniformly converges that f 1, X, X21 is a Korovkin set Korovkin set if and only if the same is

a

may

Then

does

holds

uniformly

sequence

a

We

theorem

converges to

that

assume

Mis

Suppose that X is linear subspace of C(X)

X.

that

fTjJ

X is

C(X).

of

subset

a

that

asserts

of Choquet boundary

points

separates

we

so

Suppose

Korovkin's

that

true

interested

we are

f Tnf I

on

of its

Mis

Theorem

T2.

to

since

[73].

that

whenever

C(X) such that M. theorem f E (Korovin's in C[O, 1].) Note that Mis true

itself,

C(X) provided it

[0, 1]

on

akin

and

space set

M, that is, provided f for each f E C(X)

operators

uniformly

converges

Korovkin's

general

more

compact

call

f BjT21

so

Choquet's

on

M is

B(M) for yield

indeed

Mis

linear

subspace. metrizable

compact

a

which

Korovkin

a

a

of

all

Korovkin's

I and

contains

C(X) if

in

set

and

X.

theorem

need

we

I only observe that for any xo E [0, 1], the polynomial (X XO)2 of the peaks at x0, so the latter point is in the Choquet boundary of and x 1, X2.] span that X and that Suppose, first, B(M) fTnj is a sequence of such that on C(X) 0 for all g E M. positive operators IITng gII Given f E C(X), we must show that f 11 IITnf 0; equivalently, show that we must of f I I Tn f subsequence f I I I has itself a every which converges to 0. For simplicity of notation, subsequence assume that and choose, for each n, f III is the initial f IITnf subsequence a point xn (E X such that _

_

=

-

-

-

IITnf By taking some on

x

a

(E X.

further Define

-:::::

I(Tnf)(Xn)

subsequence

we

sequence

jLnI

a

-

can

f (xn)lassume

of positive

that

xn

linear

-4

for

x

functionals

C(X) by Ln h

Since for

f 11

-

I

each

Cz Mwe n.

Thus,

have

=

(Tn h) (Xn),

Ln1 can

-+

1,

so

be considered

h E C (X). we

may to be

assume a

that

probability

Ln1

>

measure

0

Section

X and the

on

y,,

which

JIT,,,g

probability by hypothesis,

weak*

g1l

-

0

-+

has from

so

f y", I

subsequence

a

Now if g E M, of g and continuity p.

measure

a

49

the

inequalities

the

I (T., g) (X-J

WI : I (T., g) (X-J

g

-

1IT-k

:-
0, UN. Moreover, each

y G X

we

p

for

x

if

If (X)

that

Ex,

with

the

set

and

with

0 < g,,

nUn

f xj

9n(X)

1,


> then P n * base that and x X, x imply (-P) f 01, so y y if and in the * P P x are Furthermore, subspace y. y generated by P, then there exists z in P such that z ! x and z ! y, i.e., x and a

map X E)

is

to X under

&lf (x).

-+

x

P which

-

-

-

y have

bound

upper

an

bound

for

denote

this

and

x

y if

least

translation

P

in


1.

f

and

j(x)j

>

andE

is arealnumberr

Measures

From the

will

we

measure,

j(x)

- p. then A(g) p(g), so g E -C,

follows

_< A(9k)

function

to the

<

maximal

pffl

>

fj

if

=

p(f)

measure

p(j).

If




fI

=

in

C;

since

C

-

C is dense

in

maximal.

fact important about the set of all maximal we present measures. however, First, lattices. lemma concerning vector an elementary Suppose that P, and P2 are cones in a vector E, with P, c P2. Denote the space We say that induced by :! j and f is directed f inffh : h E H1. Indeed, Lemma 10.2) of the in then we have if this be true, as proof (just if /t h E HI for any M; in particular, inf f p (h) Ex, then p (f ) the to It h inf Jh(x) c HI then, remains, f (x). prove M(f) H is directed H. To see that about assertion downward, suppose A such that in that h, > f and h2 > f (hi in A); we want h subsets h > f and h :! hl, h2. To this end, define 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 and affine is Since semicontinuous, E X, r x hi (x) 1. f upper that of hi implies while the continuity and convex, J is closed Ji is of hull from the J is disjoint convex J3 JUJ2, Furthermore, compact. 0 and to the and J3 is compact. theorem, By separation (applied PROOF. It suffices

to prove

that

the

=

-

=

=

--

=

=

=

Section

Uniqueness

10.

closed

the

functional

linear

(but

Finally,

if

Proof

Choose

1-4f)

that

functional

homogeneous,

m(f

that

C

subspace shows the to

defined

f

m(l)

-

denote

Proposition whenever

which

and

g) I

-

subspace

IIf

defines

(c) 11. Thus, C (X) and

represents

of

that

the

next

are

COROLLARY10.8 on

particular,

If

=

that

/-t

X.

fx

:

and

In

As the

g) (x)

+

=

consider

and

This

positive-

is

From this

it follows

functional

m on

at

1

most

the

:

p

unique

on

C(X).

the

Since which

measure,

m(f) exj, i.e.,

=

-

maximal

=

I(x),

px >- p measure

of

of representing uniqueness The following extreme points.

of X \

by

of

X\

ex

X,

from the

X, then, it is I(x)J, f inC(X). f(x) that 10.3 implies Proposition

ex

E X

x

envelopes (Section 3) continuous on uniformly hence has a unique extension

hence

ex

particular, =

example

of

us

is

to

prove

mea-

easy

Choquet's

orig-

X.

nonnegative

a

subset

supported

is

PROOF. It is immediate

F, subset

I(x).

supjp(f)

ax is

problem by the

p is

compact

every

if

what

m is

to

ishes

H1.

h E

:

apply

(f

get

E X

x

linear

norm

supported 10.3 will enable corollary Proposition inal uniqueness for metrizable theorem which

Ih

g E C and that we

A

x.

We consider sures

inf

we can

(3)

The

a.

=

is needed.

of upper

M,,(f)

follows

It

,

=

given by a probability f in C, we have px(f)

is

Since, implies that

f

-+

a

g

-

for

Ex.

f

it is additive.

that

functional

px.

-

in

C of

-

functional

3.1 p

C

do what

9W.

Suppose C by f

property :!

linear

1, this by

+

-

-

continuous

a

for

(4) implies I(x) g(x)

C,

-

Ex;

Ax) (5):

I m(f

that

dense

we

g)

=

that from

-

continuous

a

inf L (J3)


0 and i E N otherwise. we have Ejcpaj, (since 0 in for I a so some Finally, -EiENai. E*) Eaj f f (X) and

hence

.

that

E* such

.

.

.

.

n

=

.

,

=

,

n

7

-

-

-

,

=

=

=

=

let

x

=

EiEPa-'ajyj

=

EiEN(-a)_1aiYi-

-

Since

these

are

convex

Section

Uniqueness

10.

combinations, measures

from last

we on

Choquet's step

have

X which

Representing

represented have

uniqueness proved

may be

decomposition

of

an

lemma and the

a

that

63

x

in

X is not

elementary that the points

more

fact

element

contained

support theorem

in

Measures

by

exX. a

different

two

It

follows

simplex. (This by using the

way yj

are

extreme.)

Properties

11

As

was

bility

of the

Proposition

in

seen

P(X)

measures

weak*

surjective, if jective

and

set

from

only properties

selection

the

theorem

plex. following

(i) (ii)

r

the

compact

uniqueness a simplex. map,

for

the

Suppose

inverse

map

theorem

that

r-1:

we

In this

including

the

theorem,

its

simple

re-

is still

know that we

and

is

r

prove

bi-

some

potentially

but

case.

compact

X

affine

measures

section a

proba-

the

X is

set

convex

metrizable

-*

Q(X)

X is

set

convex

exists

and

a

sim-

has

the

properties: -'

affin

is

For

each

e.

f

E

C(X)

the

is

Borel

r-1

is

(i)

PROOF.

function

real-valued x -+

(iii)

map from

resultant

Choquet-Bishop-deLeeuw of maximal probability

of this

PROPOSITION 11. 1 Then

Q(X)

if X is

and

additional useful

the

to

r

the

the

map

the

1.1,

onto

By

continuous.

striction

resultant

r-,(X)(f)

measurable. continuous

Since

Q(X)

if is

and

only if

convex

and

exX is

r

is

closed.

affine, by part

its

inverse

is affine.

(ii) f Choquet(3) theorem for we have each x E Meyer uniqueness f(x), r-'(x)(f) 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 of C(X) spanned C, the dense subspace by the convex functions. If f Cz C(X) is arbitrary, it is the uniform limit of a sequence from C of a sequence limit of Borel C, so that (1) is the pointwise measurable hence is itself Borel measurable. functions, Assume first

that

is convex;

then

=

-

-

R.R. Phelps: LNM 1757, pp. 65 - 72, 2001 © Springer-Verlag Berlin Heidelberg 2001

of the

Lectures

66

(iii) of

that

Then

there

X.

ex

1.4,

lim E,,

Ex,

To

C X.

The

=

y,

x

evaluation

at

dominating

a

x,,,

=

e,,,.

Thus, ex X,

C

x0

p >-

X is

in

net

probability

Since

p.

that

see

(hence

x0

and that

is continuous

exists

r-'(x,)

by Proposition

z),

r-'

Suppose

ex

p

2

exists

that

in the

+

closure

-+

xo

e,)

is

measure

=

.1

=

2

(y

+

represents

maximal

a

and

x0,

limr-'(x,)

that

(sy

Theorem

x, =

suppose

and there

simplex,

a

xo is

X with

r-'(xo)

measure

e,o)

Choquet's

on

measure

r-'(x,),

so

we

have

Ex"

Thus,

M

if

that

ex

is weak*

and

E, X is

=

[5]

hence

closed,

compact

=

y

that

To prove P (ex X). It

r

the

x.

z

=

A >-

-

Q(X)

then

so

r-'(x,,)

is in fact

an

note

converse,

follows

Q(X)

that

homeomorphism.

affine

X which of those characterizations given several and ex X is closed" "X is a simplex (which is why simplices satisfy the Note that called Bauer often with this property are simplices). shows that any Bauer simplex X can be identiproposition foregoing on the measures compact Hausdorff fied with the set Of all probability

Bauer

space

ex

has

X.

If X is not

X,

some

results

a

maximal

conditions

11.2)

(Proposition

way

11.4),

respectively.

DEFINITION.

By

a

Q(X)

X into

it

measure

give

which

affine

from

simplex,

is

for each to choose, possible We present x. having resultant

still

p,,

which

under and

in

The first

of these

selection

for

such

r(px)

that

a

map =

x

can

measurable

is due

the

this

r

for

to

H.

x

two

in

an

(Theorem Fakhoury [35]. way

we mean

each

be done

E

x

a

map

x

-+

px

E X.

P, and P2 are cones in real vector and that that P, is lattice-ordered 0 is an order-preserving, spaces, and positive additive If there exhomogeneous map of P, onto P2. another ists P, such that 0 o 0 is the identity map 0 from P2 into In particular, ordered. an then P2 is latticeon P2, if there exists onto or resultant the X, for its for affine selection map from P(X)

PROPOSITION 11.2

restriction

r:

Q(X)

Suppose

-+

that

X, then X

is

a

simplex.

Section

of the

Properties

11.

resultant

67

map

if x, y E P2, then x V y that to verify straightforward is and this is all that exists in fact, and is given, by 0[0(x) V 0(y)], about result the assertion this to obtain needed. To apply r, say, to the cones extends it and its selection first one by homogeneity TC. (As in Section 10, we have assumed P, 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.) It

PROOF.

is

=

--

=

The property

cinctly More

(ii)

in part

precisely,

will

we

Proposition

of

by saying

described

r-1

that

make

be

measurable."

more

suc-

terminology.

following

of the

use

could Borel

11.1

"weak*

is

0 from a compact Hausdorff function space X into measurable Borel Y be is said to a compact provided Hausdorff space sub-set X whenever is Borel subset U is an a of Y. of 0`(U) open real-valued on Y, functions family of continuous If A is a separating the real-valued measurable Borel is will that we if A-weakly 0 say each f E A on X, for function f o 0 is Borel measurable A

DEFINITION.

The lemma below metric

the

Using separating

LEMMA11.3

A is

that the

a

compact

metric

topology Since

which

Y is

of basic

each

a

is set

above

form.

The Rao

space,

any

of the

form

real

interval.

[67]

open

But

following and

there

then

open

0`(U)

by hypothesis,

is,

selection

exists

Suppose a

Borel

is

definitions,

above

compact

Thus, Y, we

n(fi

o

Borel

X is

measurable

the a

0)-'(Ii)

union E

A and

0`(U)

that

is

U has

that

and each

set

a

the

in the

of X.

independently

proved proof

below

metrizable map

fi

each

assume

can

subset

was

[791;

show

to

countable

a

where

Y, the weak topolgy.

initial

the

of Y is

q7 _jfj-'(Ij),

=

space

with

subset

open

in

on

Borel

a

of the

a

suppose

real-valued

Y coincides

theorem

that

if Y is

that

coincide.

functions 0: X -4 Y is function A-weakly Borel measurable.

Then

a

G. Vincent-Smith

THEOREM11.4

Then

U is

whenever

intersection

on

the

continuous

points

defines

it

sets

an

Borel

it

metric

open

Ii

space

A separates

Since

PROOF.

of

of

Y.

only if)

if (and

measurable

family

measurability

of

notation

fact)

standard

useful

two kinds

the

then

space,

(the

shows

x

compact --+

px

by

M.

is Rao's.

from

convex

X into

set.

the

68

probability

measures

represents

x,

all

probability

measures

follows,

(rather

L

x)

than

Theorem

(Section

strictly

convex

a

dense

for

ff-ln=l

generality fnf A fl) f2, closed subspaces i

-

1

-,

-

A(X)

fn

that

1,

Ao

for

c

is n

A,

the

linear

...

c

the

An-,

c

a

C

...

of

existence

of

existence

have

c

Choquet's

of

a

without

An of A(X)

span

An

and write

assume

can

We thus

1) 2) 3.

=

c

in

not

A(X)

of

as

We

which

argument

proof implies the

well

as

C(X) \ A(X).

in

X, the measure point of the set of

x.

space

of X

Theorem

E

As in the

metrizability fo in C(X)

function

x

induction

state

of X.

element

the W

sequence

in the

X to be the an

3),

of

loss

of notation

consider

will

we

each

I-tx is an extreme exX which represent

on

simplicity

For

PROOF.

for

Choquet's

that

and such

p.,

on

that,

X such

ex

on

Lectures

sequence

U

of

C(X)

An-, and fn-1 and such that their the of C(X). Let Sn denote union A0,, UAn is a dense subspace in the weak* of An (hence state So X), always considered space Define -* On: Sn Sn+j for each n > 0 as follows: topology. such

An

that

is the

linear

of

span

=

=

On(L)(g+Afn)=L(g)+Aan(L),

9EAn,

LESn,

AER

where

an(L)

=

inf

f L(h)

+

11h,

-

fn1j:

A,J.

h E

fnjj is continuous Now, for each h E An the map L -+ L(h) + 11h which infimum over all such h is upper semicontinous, so the on Sn, function implies that for fixed g E An, A G R, the real-valued -

L is Borel over,

measurable.

if A

0

On(L) (g


k >

and in-

0,

then

Ok(L)(g) On (L)(g)

L(g)

On(L)(g),

=

An. Let S,,, denote the state space of to define A,,,; the coherence just shown makes it possible property If X and X L follows: E -+ S(,, a map 0: as UAn, then g E A,,,, On (L) (g). Every g E An for some n > I and we can let 0 (L) (g) dense subspace of S,,, continuous element is uniformly on the A(,, hence admits extension to C(X), so we can a unique S,', identify while

=

if g E

=

=

the

with

probability

of

set

measures

L E X and g E A0, each of clear from the definition

whenever is

also

Borel

(ii),

the

for

surable

Borel

for

measurable

11.1

density

each

A,,,,

0 (L). (as in the proof of Choquet's By definition,

place

To

AL(A)

=

Now, if

g E A(X),

see

=

that

each

theorem)

OO(L)(fo)

inffL(g) then

shows

+

so

is

=

JJg g'

-

-

It

0 (L) (g) of Proposition

is

is Borel

that

L

proof we by ex X, supported

of the

AL is

will

AL(A)

to show that

Oo(L)(fo) foll: g +

=

g E

IIg

-

+

JJg

-

fo 11

=

L(g')

! inf

f L(h):

mea-

O(L)

-+

is

AL in

write it

suffices

=

AL(10)-

ao(L)

A(X). fo 11;

g'

moreover,

Consequently

L(g)

L(g) L.

map L -+

the

and Lemma 11.3

remainder

For the

measurable.

of

that

=

resultant

g E

C(X)

g E

0

has

As in the proof A,,,. L -+ O(L)(g) that implies

each

of

O(L)

measure

O(L)(g)

Since

X.

on

h E

A(X),

h >

fol

>

fo.

Lectures

70

f AL (h)

inf

pL(jfo)

hence

fo imply fo

that

since

It

=

inequality

follows

that

selection

of first

Baire

metrizable

holding

A is the

a

map

measure

the last

special

Mx

(using

into

the

He then

Suppose subspace -+

yx

result cases.

analytic

can

be

that

much

IIg

g +

fn I I

-

fn.

>

induction

obvious

an

X is

by longer

proof) to

measurability

which

certain

non-

properties.

metric

compact

a

there

that

measures

result

this

[76]

Talagrand

M.

maximal

extended

X,

x

and

px(B(A))

substantially

if X is

continuous then

at

and

space

=

1

con-

Borel E

X,

(where

for A).

improved

instance,

of the in

a

extreme

reasonable

boundary

For

and

extended

evaluation

represents

Choquet

2 since

or

of (real or complex) C(X) which Then there exists a points. separates from X to P(X) such that for each x

and x

A consists are

been

X, retaining

closed

I

A2 (fn)

has

mapping class.

k

proof.

he showed

constants

measurable

=

(fn)

Al

=

the

COROLLARY11.5 tains

we see

for

theorem

spaces

that

=

that

-

foregoing First, ways.

Cn and

7

=

-

AL (fn)

a

which

that

=

The

This

represent

and

Of AL

completes

argument

tant

of

set

that

L.

an[0n-1(L)](fn) 0n['0n-1(L)](fn) On(L)(fn) + jjg fn1j: g E Anj inflOn-l(L)(g) infIAL(g) + 119 fn1j: g E Anj inffAk(g)+Ilg-fnll:gEAnl>/-tk(fn))

=

last

is

of the

element

extreme

an

=

=

the

(10).

is obvious.

AL is

which

definition

the

=

B (A)

inequality

measures

=

is

AL

=

AL(fn)

exists

!

=

they Recalling

in two

AL (h)

that

Now, h E A(X) and h Thus, AL (A) > AL (A);

Suppose, then, /-t2 /-tl suffices It show to + 2AL Al A2these functionals are equal By hypothesis, A2 on A,,,. Al show will that are on we equal Ao. Assuming A., they are i.e., that AL(fn) Al(fn) /-t2(fn)equal on A,,+,,

that

It

SO

prove

measures

AL

the

i

last

Theorem

fo 1,

h >-

expression.

this

reverse

to

such

A(X)

on

the

remains

two

that

A

h >

fo,


r p(B) r p(C) holds Thus, v(B) v(C) v(B), so equality It follows that 0 and p(C) 0. Thus, for any throughout. I-t(B) < r r, f X : f (X) :! r I and T-'f x : f (x) :! r I =- f y : f (Ty) I differ number

let

r,

A

=

f (x)

:




(taking r) we

h(x)J

identity and by interchanging this

Suppose,

zero.

all

now,

that

over

the

unions

Ufx: g(x)

=

U[fx:r:' :h(x)j\fx:r :g(x)J] g

f

h

are

real

dense

have

=

to

g and

countable

=

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

PROOF. p

(A)

Let

f

A, i.e.,

such

functions T E

that

then

for

only by

f

=

f

>

f

that

g

and
0}, g) (x) (f g)d(p + v); it fA(f E

:

-

analogous

shows

argument

p and

(f

A

(f

(M

measures




a

proof

of this

1, let /-t,, be the n points (n-', kn-'), X of and the point n

represent

we

Hausdorff

extreme

points

on

S do not

form

for

fact measure --

sequence

a

closed

subset

special which assigns

1AnT

,

n

-

O(x)

mass

1.

converges

=

(x,

y +

X of T-invariant

set

case

from

homeomorphism

a

of the

the

0, 1, 2,...

T is

space,

the

k

T(x, y)

the

as

function

=

compact

measures

which

nonconstant be any continuous I x J into itself S by

T from

R and define

of S onto

0

Let

I into

circle,

J be the

let

and

=

of X. We will =

n-1

to

/-t,,

is

Then in the

x.

For

each an

weak*

each

of the

extreme

topology

78

Lectures

Lebesgue

to

measure

the

extreme

our

p is

T-'A

each

T

ergodic

p(A) in Tj.

in this

certainly

defines

goes

as

follows:

or

p(A)

=

So

Sl,,

Since

is

C

relate

to

the

notion

An invariant I for

set.

any

its

to

probability A in So

each

fA:

=

measure

ergodic

clearly

coincide

The two notions

sense.

in this

in measure" "ergodic be to measures ergodic probability this, measures;

if

0

probability

every

extreme

of

work

=

Theorem

the

of invariant one

0,

=

not

definitions

simply

definition

0(0)

Since

J.

p is

so

set

further

ergodic

x

other

two

if

for is

sense

X,

of the

points requires

Another

origins. measure

for

least

101

on

One of these

literature.

of course,

IL

J is in

x

at

are

the

=

f 01

on

There

A

measure

Choquet's

on

in

if

So S,, /-t(AAB) if T consists of a single T (or equals function instance, the semigroup let B generated by T)-simply nn-- , Uktn T-kA. More general on T which the of hypotheses equivalence guarantee the two notions are given by Farrell [36, Cor. 1, Theorem 3] and Lemma The Varadarajan [78, 3.3]. following simple example, due to shows that Farrell, they are not always the same. A in

each

there

exists

B in

such

that

This

0.

=

for

occurs,

=

n=

EXAMPLE S

Let

=

[0, 1]

fT,,T21,

x

where

[0, 1], let TI(x,,X2)

S be the =

Baire

(xi,xi),

of S and

subsets

T2(x,,X2)

the diagonal TI, T2 are continuous maps of S onto and So consists of S and the empty set. For any subset

(AATi-'A)

in D is

in

in

So, our

D.) Thus,

but

the

sense.

semigroup

point It

it follows

empty; and

invariant

support I in

n D is

is

generated

S,,

-

S.

every

such

masses

on

that

(In fact, measure

D are the

to note interesting is 7by simply 7-

any

takes

only that itself.

invariant

only ones

the

S, S,

D of

A of

/-z with

measure

every

(X2 X2)-

=

Then

let

support has

measure

the

values

which

are

0 and

ergodic

(noncommutative)

extending

for

A method

13

representation

the

theorems:

Caps

for

were

any

set

natural a

lead

of

elements to

in

satisfactory approach, involves is outlined

it

Choquet,

general

notion

for

admits

possible

are

be

involves

It

is

for

theorems

completely

no

lines

two

of

One of these

of interest.

measure"),

("conical

so

for the

base.

however,

are,

measure

approach

The other

such to

10,

cone,

convex

compact

a

seems

There

sections

Section

in

theorems

obtain

to

there

which of

closed

a

representation

which

is

As noted

set.

base

way to cone

due to

[19].

in

a

but of cones, of this nature.

result both

natural

convex

class

a more

as

whether

wonder

general

more

compact

in earlier

with

dealt

we

convex

regarded a

closed

a

a

be

can

results

these

of

elements

such

which

theorems

representation

The

replacing

which

the

notion

the scope to extend possible "cap"; by to the will be devoted This section theorems. of the representation consider the section, we only proper latter Throughout approach. K n cones f 01. (-K) K, i.e., of course, theorem" we mean, In using the term "representation which of measures than the mere existence points; represent more be these measures in by the supported some we require that, sense, the of a convex In the case only possible extreme cone, points. of an notion the introduce must and the is we extreme origin, point that

of "base"

makes it

this

of

=

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

=

-

an z

the

E

K,

x

of

extreme

E K

=A

K is

a

Since

0.

R+x

p may be said

ray

0 < A
0. By choosing a caps,

=

of p, if necessary, multiple positive that y E ex C, y =A 0. Then 1 ! If /-t 1. 0, so p(y) [I p(y)]

that

E

Then

is the

which

cone

of C.

PROOF. Since

C,

convex

0.

11 such that x f 01 C Ci, and


E z Eznxn (z, x) f zn I co E K, x : x 11 is 0, such that B Ix : x E K and (z, x) =

=

=

=

=

But

compact. hence

Xn

struct

C

Ix

E

.

,

.

0, z,-, ', 07 0,, and

for

cap

p(x)

K:

weak*

-

-

define

is

)

-

p,

on

ball

unit

z,,,

all

n,

--+

0, this

be compact.

by p(x)

K

(since of fl).

compact

compact

Since

B.

E

> 0 for

zn

B cannot

hence

K,

11

:!

that

shows

property

unbounded

K of the

with

.

universal

a

=

(0, 0,

=

is

sequence

first

the

it

To

the

con-

Then

EXn.

=

is

and

intersection

Since

p is

positive0, so p is

homogeneous, p(x) rj Ix Since the unit ball additive. and it is clearly lower-semicontinuous, is normed linear of the dual of a separable always metrizable space C is metrizable. that in the weak* topology, we see Finally, suppose K were metrizable. Since f, is weak* sequentially complete and K in itself. K is of second that conclude is closed, we could category relative to K. But K UnC, and C is closed and has empty interior if x c C, then x + an E K \ C and is weak* convergent (For instance, 2 at n and equals 0 of f 1 which equals to x where an is the element :!

is

compact

example

of

a

E K:

for

all

r

>

=

,

elsewhere.) Later, universal to

easy

K

(other

but

cap,

by

extreme

than

The

give

will

which base

a

points.

following

result measures

with

B, where

B is

Then K has

a

no

does

bounded

closed rays,

is

a cone

set

convex

hence

a

It

caps.

Take

caps:

extreme

have

not

of its

union

nontrivial

no

a

gives on

some

no

caps

concerning

information

unique-

caps.

If the cone K Conversely, simplex.

PROPOSITION 13.3

C is

the

nevertheless

is

cones

which

cone

f 01).

of maximal

K, then

an

closed

construct

generated

without

ness

we

is

a

if

lattice

each

if C

and

point

of

K is

is

a

cap

contained

of

Lectures

84

in

of

cap

a

K which

is

PROOF. A cap C of

p(x)

and

Co

cone

ifxo

(0,0)

p(x)

Assume,

follows:

as

follows

now,

that

of course, that that x0 ! 0 if and

lattice;

a

(x, r)

--

and

yo

=

-

-

-

-

only

p(x q that to show that !

zo

p(y that

!

if

r

w) p(z




yo

completes

wo.

the

proof. The

ing:

Is

remaining there

a

important

reasonably

question

large

class

concerning of

cones

caps

which

is the are

follow-

unions

of

Section

Extending

13.

Representation

the

Caps

Theorems:

85

in conhas led Choquet question [19] to investigate, of the class depth weakly complete cones. detail, (R. and Becker has extensive treatment an recently thorough given [7] of weakly complete and conical measures cones along with numer-

caps?

their

This

siderable

and

applications.)

ous

which

exhibit

PROPOSITION 13.4

of

each

cones,

Suppose

which

Since

PROOF. K is

a

of its n

cap

For

caps.

11 2, 3,..

=

For

each

n

homogeneous Kn: Pn(xn) verified

the

of the

that

p, is

of

intersection

there

exists

11.

of

sequence

a

to

exists

with

cap

a

convex

is

true

fl En.

(In

same =

of R+ with

closed

Cn

if

that

show a

a

show that

only

need

is

a

p

on

of Kn,

cap

functional

an

union

fl Cn

C of K with

cap

of

subcone

K is the

C

C.

additive,, semicontinuous, positiveCn f Xn Pn on Kn such that

lower

a

Define also

is

caps.)

its

nonnegative


0

If p E K, then (since gn / f) we have IgI : ' : b(g)f. En lim 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) P is compact since C C P that 11f [-b(g), b(g)] : g E Q,(Y)j; with the weak coincides in the product topology on topology (which that It is immediate in P. C is closed to show that C) it suffices

such

that

=

=

=

element

any

functional

-+

Co.(Y),

and

C is

/-t(gn)

We caps,

pointwise

increasing

the /-t

on

that

show

the

in

and

now

but

closed

in

not

an

have

a

measure.

a

follows

linear nonnegative Thus, we need only

from

fact

the

functions

continuous

on

that

p is

by

K defined

semicontinuous.

example a

is

This

K.

is lower

hence

exhibit

does

hence

of the

limit

of C is

closure

of

a

universal

cone

which

is

the

union

of its

cap.

EXAMPLE Let and

let

considered

s

be the

E

=

to

of all

space

s* be the

real

sequences

of

space

s.

in the

As is well

product known,

topology E

can

be

with the nonzero finitely sequences, E. E E a x Eanxn, Topologize by (a, x) s, topology defined by s and let K be the closed convex

be the

correspondence E by the weak*

dual

space

defined

of all

=

Section

Extending

13.

nonnegative

of all

cone

containing

C

and

suppose

that

A

verify is

ysuch

thaty=Oonl. that

To

Ix

K, p(x) by 6n the

Denote it

is

K does

:! -

C

x

:

that

Since

1.

p(Jn) (a, Xn)

=

n,

We conclude

for

criterion

I

weak*

have

to

p(Jn)




x

1, 2,3,...

convex

-54

of J

are

that

0

which

By

that

K is

locally

U of 0 such

=

in

[0, 1]J.

(-K)

F;

B, then

I [0, n]B

for

I

exists

we

can

some

f

=

Kn

locally

0 from is

a

base

a

Then K has

a

cone

that

x : f (x) (relative

0 c

linear

continuous

J, i.e.,

2b

for

K and

=

the

inf is

ex

exists

the

we

>

compact,

0.

a

F be

closed same

is

points

extreme

conclude

f

functional

f (J)

to

J be the

K,

nJ, K),




:

compact. cap,

object,

an

Suppose

space

Ik

=

is

hence

PROPOSITION 13.6 convex

J

a

E C.

universal

a

with

section

It

n.

and

which

Jnp(Jn)J

=

cone

a

01;

-=

define

we can

C be those

=

such

were

this

K,

E

Let

< YkX-1 k

have

not

C is not

so

Xk

(and closed)

universal,

a

:

Ik

=

EkEJ

sequence

C is Let

> 0.

I

x

K \ C is

C is bounded

see

If

87

to straightforward C x convex Furthermore, of all of E consisting dimensional subspace E J; it Ify E C,thenO A 0, nonnegative there exists on X, measures a sequence supf Anj of nonnegative Borel subsets EA" S,, of X, such that p disjoint ported by pairwise hull of and Of (Kn) < E for each n, where Kn is the closed convex Sn

If

LEMMA14.1

p is

a

=

-

PROOF.

X, v 7 0, Of (K)

on

that

for the

A in

If

to

SA

the

closed

Z is

A

Since

an

measure,

the

sets

the

Lebesgue

EA,,

converges

If A

=A

A

-

Sx, Sy

then

In the

Borel

a

Let

out.

X with

on

Sx of

subset

Z

partially necessarily

by

there

SA (A Mo

set

Cz

MO)

is

countable;

dominated to

the

EAn, then

are

exists

restriction we

can

a

say

A

apply

of p to

then

set,

it so

maximal

disjoint

pairwise

Mo

=

E.

simply

step,

restriction

ordered


- A

metrizable

compact A and A

if

and

regular

are

only

convex

if

there

a

general

probability

Borel exist

of

subset

a

dilation

T

TA.

of this

depends

theorem

on

result

of Cartier

with a classical use result together metrizability), of measures. disintegration With X as above, the space F we consider C(X)* x C(X)*, the of the weak* with itself. product using topology Thus, F is a linear functional L on F locally convex space, and every continuous on

does

not

the

=

is of the

form

L(a, 0) for will

some

pair

be interested

=

of functions in two

a(f)

-

0(g),

(a, 0)

E F

this f g in C(X). Throughout subsets J and K of particular F, ,

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 v >- E, implies

that

It is

(since and

the

closed

of the

(a, ) imply (I)

of all

compact

1,

and is itself

=

if L !

K, i.e., 0

on

(1)

a

we see

that

P,

set

0

on

a

is

certainly

J,

there

assume

so

J

(1)

H 1

=

of the

B C Kn H

for

base

L !

L E F* and

a

subset

Thus a

Its

Indeed, hyperplane

convex

be true

X,

K.

the

compact.

compact

convex.

E K and

K n H is

that

will

K whenever on

(a, ) closed

Ex

-

P, into

not

for

v

combination

convex

J is

base

compact

hence

This

L > 0

B , then

a

set

Since

1.

=

is clear

It

K n H.

the

K n H is

P1,

x

Since

a

from

K n H of K with

which

compact. B

is

J is

continuous

is

mass

intersection

convex

show that L !

for =

K; furthermore,

v) graph).

a

Since

in F.

cone

convex

C

of

its

to

not

hull

convex

subset

a

J

point B, however,

is

masses

closed

a

that

(resultant

-+

v

K is see

homeomorphic

J is

of point is

map

we

K;

we

will

if B generates 0 on B. Now, if

f

exist

,

C (X)

g in

will f (x) L(E,, v) v(g) 6,,; show that L (a, ) a (f ) (g) ! 0 whenever (a, 0) E K. Recall for that each x in X, 7(x) ej. (Proposition 3.1) supf v(g) : v It follows that V(x) :! f (x), so that g < y :! f Thus, 0 (g) < 0 ( ) :! affl; and a(y) from Lemma 10.2 we know that 0(y) < a(g) and hence L(a, 0) > 0.

such

that

!

-

0 whenever

v

we

-

-

=

-

.

The

following

Proposition

proposition

now

(CARTIER)

PROPOSITION 15.1

only if

and

is

an

immediate

of

consequence

1.2

there

exists

a

to

the

(A,p)

An element

nonnegative

measure

of F

J which

on

is

in

K

if

represents

(A, p). We now return that exists

for

X is metrizable a

each

and that

nonnegative L in F*.

A(f)

proof

p >- A.

m'

measure

This

-

means

p(g)

=

of the

on

that

fi [f (x)

Assume, By the above proposition, theorem

J such for

-

itself.

that

fj

each

(f, g)

v(g)]

dm'(Ex,

in

L dm'

C(X)

v).

=

x

then, there

L(A, M) C(X),

Lectures

96

(x, v) : x (x, v) from

S

Let

(Ex, 1/) m' to the

measure

a

above

(a)

Equation carried We

now

a

p.

58].

[13,

measures

Suppose

that

measure

on

function probability

0.

(i)

Y.

measures

For

each

h in

on

X, denote

the

the

C(Y),

the

X,

support

function

carry

can

0, f

--

0 in

S which

on

P,

onto

spaces, m

Ax(h)

x ---

is

X.

is

a

of

0 nonnegative that

image of m under X into -+ Ax from properties:

x function following

a

g

=

disintegration

metrizable

0'

Y, with

function

we

x

on

and that

exists

the

measure

of X

theorem

compact

Theorem

v), v).

dm(x,

Y onto

mo

-

there

Then

of the

are

from A

Let

dm(x,

probability projection

a

case

Y and X

function

continuous

a

is

natural

special

state

Since

-

a

v

=

m

the

I

Ex

-

fS f (x) fS (g)

=

(g)

that

shows A under

onto

S is

)

A (f p

v

homeomorphism, By alternately choosing that for all f, g in C(X),

S.

we see

(a) (b)

P1,

C-

v

J onto

m on

equation,

X,

E

Choquet's

on

Borel

is

is

the the

measur-

able.

(ii)

For

each

x

(iii)

For

each

h in

We apply

natural introduced A

measures

on

resultant

the

of S onto and is

0',

the

(2)

P1. which

resultant

m(h)

as

exists

P, of the

It

remains

image to

P,

=

S C Xx and

m

three

that

image

IL

Ax

-

=

fS

v

means

(f ) dA,, (y, v),

measures

that implies the probability We let T', be

natural

projection properties (1) TA. The fact that Tx

Tx satisfies

that

of

the

the

(a)

properties.

of Ax under

and

0-'(x).

P1, let 0 be

A be the

noted,

have

C(X), TX(f)

in

A., from X into

---

prove

of the

Y

let

we

above

dilations,

define in

x

the

in

Let

and

as

contained

is

fX Ax(h) dA(x).

=

X,

Then,

S, satisfying

of Ax

follows:

of S onto

there

so

the

C(Y),

result

this

projection previously.

mo

=

in

the

that

for

each

f

in

Section

Orderings

15.

(y, v)

Since

implies

S

in

f,

functions

this

of Measures

and Dilations

v

sy,

-

we

see

that

97

for

continuous

affine

becomes

T-W

fS f (y)

-

dA , (y, v).

supported by 0-'(x) f (x, v) : v ExI, and hence Tx (f ) x. f (x), i.e., Tx represents Property (2) of dilations follows from (**) and property to show that TA, (i). Finally, I.L in for that we must verify C(X), g Ax

We know that

is

=

-

=

=

p(g) By (* *), Tx (g) function

fS (g)

v

(iii)

dm (y,

v

(b),

we

v)

that

see

fX

dAx (y, v). implies that =

=

From

=

fS (g)

=

C(S),

h in

(TA)(g)

=

the

Tx (g) dA (x). h (y,

Since

v)

fX JS (g) dAx (y, v)) fX Tx (g) dA (x). p(g),

equals

side

(g)

defines

a

dA (x)

v

left

v

=

and

proof

the

is

complete. We next

define

considers

Loomis

ordering orderings;

the

several

[56].

p > A of Loomis

the

present

one

is

(Actually, "strong"

his

ordering.) If

on X, a subdivision measure nonnegative that measures on X such a finite of f pil of nonnegative each subdivision We say that Epi. f Ail of A y ft > A if for there exists a subdivision Ai for each i. f pil of p such that pi of this and relation its other to group descriptions ordering (For

DEFINITION. IL is

p is

a

set

=

-

see representations, [57] and [56].) In the following X and theorem,

15.1

of Cartier.

Note

THEOREM(Cartier-Fellmeasures

on

(a)

p >- A.

(b)

There

(A, p).

X, then

exists

a

that

X is

not

J

are

the

assumed

same

to

as

in

Proposition

be metrizable.

If A and /-t are nonnegative Meyer [15]). the following assertions are equivalent:

nonnegative

measure

m on

J which

represents

Lectures

98

(c)

Choquet's

on

Theorem

M > A.

Proposition

PROOF.

(b) holds,

and

f Ail

let

Radon-Nikodym

shows

15.1

(a) implies

that

be any

(b).

Suppose

By

of A.

subdivision

choose

that

of the

means

Borel

measurnonnegative 1. Define able functions fiA and Efi Ifil on X such that Ai for each J Borel functions fi (x) by gi (E:,:, v) (Ex, v) in J, Jgi I on each measure 15.1 again, and let mi By Proposition mi has gim.

theorem

we can

=

=

=

=

(vi, /zi) in the cone K. If (and if we carry the measure Proposition 15.1) we see that a

resultant

we

assertion

Similarly,

=fS

(f )

'Ji

since

(A, A),

represents

m

AM

=

dm(x,

f (x) fi (x)

fs f (x)

P),

we

dm(x, P),

to

the

for

f

f

of this

S defined

set

after

in C (X).

deduce

for

definition

the

use

m

that

C(X).

in

that A where 7r is the this m 0 7_1 means 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

we

noted

=

,

-

=

=

,

fs (f fi so

vi(f)

that

(Ai, pi)

E K

implies

in the

V1, V2,.

6 >

proof ..

0,

pi

f

is

-'_

A(f).

a

we can

Vn such

(c)

that

carry

that

out

the

m

the

for

a

o 7r

-') (x),

vi

/-t

=

But

ElLi,

so

that then, Suppose, to want on X; we

disjoint

union to

Vi is

as

was

of Borel nonzero

measure probability x in Vi. Thus, A

each

Ai.

=

construction

Ai of A

restriction

< e

(a). function

same

X as

in X of the

I

d (m

i.e., Emi implies

convex

to write

(x)

Ai(f),

implies

continuous

letting xi be the resultant have If (x) we will f (xi) -

A(ffi)

Ai, and

of Lemma 10.6 ,

=fX (f fi)

v)

=

-

show

to

/-t > A and that show that I_t(f)

Given

Jsffidm

=

p > A. It remains

dm(x,

7r) (x, v)

o

used sets

and,

Ai/Ai(X), =

E Ai,

and

Section

Orderings

15.

therefore

choose

we can

implies

The latter

of

resultant

consequently f :! f (xi)

A(f) Since

this

proof

is

that

I_tj1ttj(X). tt(f)

+

=

EAj(f)

is

true

this

a

A.

nonnegative

maximal,

almost

(xi)

E

0,

>

E/uj

=

Aj(Vj) and pj(f)1tzj(X)

eA(X)

+

:!

conclude

we

with

only

I-t(f)

if

dense

an

and /_tj that xi is

1.t(f)

and hence

EA(X).

+

that

and

hand,

other

6],

the

(xi),

! f

On the +

Aj.

-

Ai and the

-

tt

measure

T be

Let

=

/,t(y) of

with

conclude

every

f

C(X);

then

for

and hence

>

-

(Section that

Tx,

for

3)

almost

is maximal

that

for

-

the

such

respect

10.3

Tx (yn fn) dA (x). Now, Yn f,, A, for each n. It follows that Since

proposition

Suppose that and that on X, dilation

a

Proposition

subset

interesting

concern-

measures.

everywhere

from

Recall

countable

a.e.

convex,

/,t

EAj(Vj)f(xj). Ai(f) :! A(Vi) [f (xi)

(MEYER [57])

p >- A.

and

that =

!

section

with

PROOF.

fx

any

and maximal

A is

measure,

if

f

is

EAi(Vi)f

:!

PROPOSITION 15.2

is

Aj(X)

=

99

complete.

ing dilations

TX

1-ii(X)

that

so

for

We conclude

that

pi

EtLj(f)

=

such

measures

Since

Vi,

6 on

of Measures

and Dilations

for

map

f

all

T. (f

a.e.

x,

A.

-+ .,

Y

C(X).

each so

all is

TA

n,

n,

Tx(fn) for

Then

is maximal

/-t

=

f fnj

/-L(yn

Tx (yn

have

uniformly

/-t.

=

Let 0

we

TX (y)

maximal

a

A.

a measure

in

0,

is

/-t

that to

metrizable,

X is

a

fn) fn) T (fn) a. -

.,

continuous, each

be

-

f

in

=

0

=

e.

we

C (X),

Topics

Additional

16

Much of the

material

these

in

(other

notes

than

the

applications)

presented by Choquet [19] at the 1962 International of Mathematicians, and the paper Congress [22] 'by and very concise of treatment Choquet and Meyer gives an elegant the main parts of the theory. Bauer's lecture decontain notes a [6] tailed which starts from the very beginning, development using (as do Choquet and Meyer) his "potential theoretic" to the approach of extreme existence functions on a compoints via semi-continuous book [57] covers XI of Meyer's deal a great pact space [3]. Chapter of ground. He shows, other that the entire of things, subject among maximal measures of an abstract as a special case may be viewed of "theory balayage." A number of books and monographs have apon this subject peared since the 1966 first edition of these notes (which appeared in in 1968 [63]). Russian translation Among these have been Gustave Erik M. Alfsen Choquet [20] (1969), [1] (1971), Yu. A. agkin [73] contained

is

(1973), (1980) (without

the

in

outline

[53] (1975), In [21], [61] (1980).

S. S. Kutateladze and

Phelps

proofs)

of related

results

L. Asimow and

Choquet through

has

obtained

A. J.

given

Ellis a

[2]

survey

1982.

book by R. Becker a superb 245-page [7] (1999) contains which in many respects where these starts up-to-date exposition leave off. His emphasis notes on convex cones (rather than compact and conical to potential convex measures permits applications sets) statistical decision theory, theory and other topics where capacities,

The

the

cone

of interest

The

ough;

474-page in particular,

elements

(sets

[83],

the

not

admit

a

RNP;

non-compact see

G. Winkler

base.

compact

monograph by Bourgin his Chapter 6 covers

of certain

with In

does

convex

[14]

extraordinarily representations integral

thor-

of Banach

spaces

is

subsets

for

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

applications

view towards

a

(not necessarily

sets

convex

in Banach

probability,

in

Choquet's

on

Theorem

spaces,

[14])

in

as

statistical

mechanics

dimensional

convexity

and statistics.

A survey, in

appears

The

proofs, Fonf-Phelps-Lindenstrauss with

of this

rest

related

been

have

infinite

[38].

will

section

which

topics

of

some

be devoted

brief

to

body of these

from the

omitted

of

descriptions notes.

POTENTIAL THEORY

representation Integral and the tential theory,

play

theorems

Choquet potential

(or axiomatic)

theorem

role

important

an

in

of considerable

is

po-

in

use

use Unfortunately, that it would resubject deeply regard quire far more time and space then we are willing to spend in order to self-contained. What which is even moderately given an exposition harmonic functions and do is sketch some facts we can concerning theorems classical of the show how one representation integral may A much more theorem. of the Choquet be viewed as an instance book [7]. complete treatment may be found in Becker's subset of Euclidean nLet Q be a bounded, connected, open

abstract

this

is

(n

space

2)

!

harmonic

are

and Q.

in

Then

induces

a

Let

E

set

H

=

ordering

and on

E.

the

linear

convergence

H is

a

closed

convex

x0 be any metrizable compact

Let

Jh

=

:

h E

==

In a

of the

view cone,

0 :!

u

we

: :-

h,

see u

of this

property,

usually

referred

=:f

(X)

h(x)d1_t(h) (Section

characterization h lies

that

harmonic, the to

as

(X

on

implies

13)

extreme

an u

=

extreme

nonnegative harmonic

which

Q;

in

base

convex

theorems, measure

then

for

then,

M on

the

Q).

of extreme ray

Ah for

minimal

E

subsets

cone

point

H.

U

generated

space

compact

on

H, h(xo) 11 is a and uniqueness existence the cone By Choquet's exists a unique to each u in H there nonnegative extreme points h of X such that

X

h > 0 which

functions

of all

H be the

-

of uniform

metrizable

E is

lattice

in

H be the

let

by H, with the topology of Q.

theory.

imbedded

so

in

its

some

harmonic functions.

elements

of H if A

> 0.

of if

only

and

Because

functions

are

Section

Additional

16.

Topics

103

theorem to have any sigrepresentation of course, concrete one must nificance, description give a reasonably if Q is the open of the minimal functions. harmonic For instance, and if x0 ball of radius at the origin, r > 0 and center 0, then the h in X extreme i.e., a function points come from the Poisson kernel; with for where if and only if h is extreme some y IIy II r, P.

order

In

for

the

above

=

=

Py(x)

easily seen that the the boundary of the sphere It

is

(and

hence

we

could

IIX112 ylln

2

r'-2r

=

=

jjX

Py

map y -

exX,

onto

used

have

the

above

The

can

to

an

proof exposition Notes

fact

of this that

tained

in

sphere

induces

extreme, =

A

and

the

closure

related

they

a

of

by

Milman's

Y, and Y affine Since

are.

given

is

for

the

theorem). on

The

boundary a unique

the exists

r).

theorem

be ob-

can

provided one Py. An elementary

theorem, by

Holland[43].

F.

may be found

results

then

one-to-one

all

been

is compact

in

Since

itself, we

con-

of the

rotation

map of X onto

exX is nonempty,

Lecture

the

exX is

theorem,

closed.

An

if

one

conclude

Py

that

Y.

POSITIVE be

n

Herglotz

from

theorem


0

THEOREM

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, < of real and If (t) I f (0) for all t in G. If a function f is a character of G into the group of all complex numbers G (i.e., a homomorphism definite. of modulus Suppose that G is locally 1), then f is positive of and all P let be the continuous definite cone compact positive

whenever

tj,

numbers.

It

functions

on

f

of those

t,,

is

G.

in L'

ff In the

f

cap

that

in

the

are

every

Then

(G)

g(s

P

+

K with

(t)

< 1.

positive =

of

eralization

and

line

finite

nonnegative

a

Since

the

each

a

The

classical

points

subset

(g

E

L'(G)).

set

K

X of

f

dlt (X)

X(t)

p

a

on

the

on

-4

e'xt

set,

it

can

it follows

for

form

G has the

This

(where

t

form

G, and

characters.

of Bochner closed

f

consisting points of this

cap,

extreme

nonzero

function

of the

form

universal

a

definite

theorem is

a

characters

measure

character

extreme

and has

continuous

f W for

ds dt > 0

K is closed

I If 11

of the

as

satisfy

t)g(s)f

(essentially) continuous

be considered

can

which

topology,

weak*

of those

of G and

elements

are

is

G is some

actually

be

a

gen-

the

real

real

x).

proved

to use the Stonepossible theorem determined Weierstrass to show that by f. [t is uniquely has a close with group representations, This result connection in a definite function each continuous since 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 representation essentially integral representaof tion of G (and hence every representation G) is a "direct integral" further of irreducible For see details, representations. [33] and [58]. result which can be used to It is worthwhile to sketch a simple the The K of show that the extreme the characters. set are points in what follows which are left unproved facts 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

adjoining

identity

the

(which doesn't Rudin[70]. Suppose and

e

functionals in A. If f

all

x

for

all

Any

(e

+

Eix),

Define

the

-+

x

identity convex of all and f (x*x) > 0 for f (x)f (y) f (xy)

1

=

in

with

K be the

Let

f (e) satisfy point of K,

A which extreme

an

of A is

element

Ei's

then

set

-

fourth

form.

functional

linear

x

roots

A

on

4

i=1

Ic-,i

(e+Eix)*

hence

we

JJx*xJJ

that

f (x*xy).

f [(Xy)*(xy)]

-

4

assume

by g(y)

of the

of elements 1

F, unity);

-

of

We may also

g

g(y*y)

combination

identity

the

are

is of that

x

linear

a

polarization

the the

where that

assume

is

(consider

form x*x

on

may be found

*-algebra

Banach x*.

proof

A related

A.

x, y in

PROOF.

f

of the

commutative

a

involution

L'(G). involution)

algebra

group

continuity

A is

that

105

the

to

assume

continuous

linear

Topics

Additional

16.

may < 1.

For any y,

0

and

(f JJx*xJJ

since

g)(y*y)

-

< 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 we have Af for some A ! 0. From by K. Since f is extreme, g and that I conclude A we f (e) g (y) g (e) g (e) f (y) for all y, the proof. i.e., f (x*xy) f (x*x)f (y) for all y which completes =

-

-

=

=

=

=

=

It

is

even

g

(x)

and

=

0

lg(X)12

-




0 and

need

compact

image

the

X is

represen-

set

in

any

set

it will

set,

images

homothetic

generally

little

bear

plane,

triangle images will "look" homothetic another will be that is, it image of X. This theorem of the following version is the two-dimensional the avoids which D. G. Kendall [49] has given a proof assumption.

blence

to

X.

intersection

space

E is

of

any

or

another

two

a

in the

However, if X is any of two of its homothetic

THEOREM16.3

A compact

simplex homothetic homothetic

if

and

convex

only if

of

concept

convex

of

sim-

a

a

topo-

of

the

E E.

x

of two different convex

integral

or

We first

A homothetic

intersection

the

be

E.

where

aX + x,

orderings

to

geometric.

Suppose

16.2

DEFINITION

While

no

reference

dimensional

of infinite

characterization

is

homothetic

logical form

elegant

an

the

subset

X

intersection

images of X is either image of X.

of

a

a

of X resem-

nontrivial like

just

X;

observation of

Choquet.

compactness

topological

vector

(ax+x)npx+y) empty,

a

single

point

Lectures

110

characterization

This

leads

of a theorem generalization for a decreasing sequence

simplices

is

of finite

family

a

x, y G X and

for

(x

aX)

+

1,

n

directed

(y

X)

+

=

of

following

of the

proved simplices.

who

in 1952

it

family

directed

is

and let

that

(x

+ aX) n

a

0 < -yX

zX and

a

7xX.

+

zx

simplices

assume

X E X there

every

Theorem

of

com-

simplex.

a

PROOF. Let X be Then

[12],

downward

a

Choquet's

proof

dimensional

of

Let 0 < a,