A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
M. S OLVEIG E SPELIE Extreme positive functionals
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 7, n
o3 (1980), p. 505-509
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M. SOLVEIG ESPELIE
1. - Introduction.
In 1948 Gelfand and Naimark
using
thetechniques
ofrepresentation theory
for C*algebras
characterized thepositive homomorphisms
of acommutative Banach
*-algebra
A withidentity
as the extremepoints
ofan
appropriate compact
convex subset of thetopological
dual A’[2,
ch.VIII].
Bucy
and Maltese[1]
in 1966 gave a niceproof
of this resultusing only
theinherent
properties
of thealgebra
A. Thatis, they
were able to avoid thetechniques
ofrepresentation theory
which tend to obscure thegeometric picture, especially
in the commutative situation.In this note we characterize the
multiplicative
functionals for an ap-propriate
class of commutative Banach*-algebras
and obtain the Gelfand- Naimark result as a consequence. Since our results are based ongeometric properties
of certaincompact
convex subsets of the cone ofpositive
func-tionals,
as in[1],
we avoid the use ofrepresentation theory.
2. - Notation.
Let A denote a
complex
Banachalgebra
with isometric involution.We call such an
algebra
a Banach*-algebra.
Let A’ denote thetopological
dual of A. Then
is the usual cone
of positive functionals
defined on A.Further,
we letand is
multiplicative) .
Pervenuto alla Redazione il 28
Luglio
1979 ed in forma definitiva il 10 Dicembre 1979.506
The
nonempty
convex set P is weak*compact,
hence extP,
the set of ex-treme
points
ofP,
isnonempty. Finally,
we let extr K denote the union of the extreme rays of K.In Theorem 3 we
give
a sufficient condition on .K to insure that .K havean extreme ray.
3. - The main result.
In the work below we
always
assume that thealgebra
A is a Banach*-algebra
with dimensiongreater
than one. We first recall that a cone isproper if and
only
if it contains no line(i.e.,
if:t f
E K thenf
=0.)
As thefollowing
lemmaindicates,
there is arelationship, perhaps
wellknown,
between
A 2 = {xy Ix, y c- A}
and the cone K.LEMMA 1. The set A2 is total
if
andonly if
K is proper.PROOF. K is not proper
« there
exists f #
0 suchthat + f
e li« there exists
f
# 0 such thatf(xx*)
= 0 for all x E A« there exists
f
# 0 such thatf (A2) _ {01
« A 2 is not total in A.
In the above
proof
we have usedonly
the Hahn-Banach Theorem and the known fact that every ..,product
.... can beexpressed
as a linear combina- tion of elements of the formThus,
A2 total isequivalent
to elements of the form xx*total,
and the twoideas are used
interchangeably.
To obtain the desired
representation
of the extremal rays the restriction that .K be proper is necessary. Forinstance,
consider C2(2-dimensional
Hilbert
space)
endowed with themultiplication (x, y)(a, b)
==(xa, 0)
andinvolution
(x, y)*
=(x, fj)
for elements(x, y), (a, b)
E C2. Then the line L ={(Oy 2i):
2real}
lies in K and is notgenerated by
ahomomorphism.
Conversely,
thehomomorphism h(a, b)
= a does not lie on an extreme ray of the cone K.Although,
ingeneral, positive
functionals need not bereal, i.e., f(x*)
===
( f x),
we recall thatpositive homomorphisms
are real. Note that in thepreceding example
no element of the line L is real.Further,
if thealgebra
Ahas an
identity,
it is well known thatpositive
functionals are real. Thefollowing
lemmaanalogizes
thisresult,
LEMMA 2.
If
A 2 is total then every elementof
K is real.PROOF. Let
f
E.g,
a E A and since A2 istotal,
letlp.1
be a sequence in A withlim
pn = ac where each pn is a linear combination of elements of the form xx*. Thenf(p£)
=f ( pn )
andby
thecontinuity
off
it follows thatf
is real.
THEOREM
3.Suppose
A2 is total.If f =,p4-- 0
andf
EM,
thenf
lies on anextreme ray
of
K.PROOF. If
f E .M’
andf # 0
assumeO c g c f
withg # o.
To showf E
extr .K it is sufficient tofind ø
> 0 so thatf3f
= g. Theinequality
for all x,
YEA
shows that iff (x)
=0,
theng(xy*)
= 0.Thus,
foreach y
the linear functional
cp(0153)
=g(xy*)
is amultiple of /
and we writeSince A 2 is total and
g =A
0 there exists some x e A withg(xx*)
> 0 andconsequently by (1), f (x)
-=F 0. Thus define a continuous linear functional of yby a(y)
=g(xy)(f (x,))-1,
which isindependent
of x from(2).
Iff(y)
= 0then from
(1 )
it follows thata(y)
= 0 so a=== f3f
for somecomplex
constant(3.
Now, by
our choice of x,so
# > 0. Thus,
Finally,
whenf (x)
= 0 from(I)
we conclude that(3)
holds and theproof
is
complete.
COROLLARY 4.
If 11 f 11 = I
andf
E .M- thenf
E ext P.An
application
of theCauchy-Schwarz inequality
forpositive
functionalsyields
thefollowing
lemma which will be used in theproof
of Theorem 6.LEMMA 5.
Suppose
A2 is total andI
E K.Then, if f(xyx*)
= 0for
all x,y E.A. it
follows
thatf
= 0.508
PROOF. From the
hypotheses
onf
it followsthat If(abed) 12 : f(aa*) . .f(b(cd)(cd)*b*)=O
for alla, b, c,
d E Aand,
if A4 =(abcd :
a,b,
c, d EAl
then
f (A4)
= 0. Since everyproduct
xy E A2 is the limit of a sequence of elements from the span ofA4,
thecontinuity of f,
thetotality
of A 2 and the aboveinequality imply
thatf = 0.
The
following "square
root lemma " for Banach*-algebras
will be usedin Theorem 6. Let z E A with
Ilzll 1,
then inA1,
thealgebra
withidentity adjoined,
the element(e
-zz*)i
= a exists and a = a*. Since A is a maximal ideal inA1,
forany x E A
it follows that(xa) (xa)* = xaa* x* = x (e - zz*) x*
=xx* -
(xz)(xz)*
E A.For the first time in our work we now
require
that thealgebra
A becommutative. The
necessity
of thishypothesis
in Theorem 5 is indicatedby
thefollowing example.
Let S? denote the
algebra
of 2 X 2 matrices(with complex entries)
withthe usual matrix
multiplication.
The norm of a matrix Z in Q isgiven by
the sum of the absolute values of its entries and
Z*
is the usualconjugate transpose.
Thetopological
dual 0’ isagain
the set of 2 X 2matrices,
thenorm of an element of Q’ is the maximum of the absolute values of its en-
tries. The
positive
cone K c .Q’ is the setThe
following elements,
none of which aremultiples
ofmultiplicative
func-tionals, generate
extreme rays of K:THEOREM 6. Let A be a commutative Banach
*-algebra,
with A 2total.If f
lies on an extreme ray
of
K then there exists a constant3 >
0 suchthat (3f
ismultiplicative.
PROOF. If
f
= 0 the resultfollows,
hence we assumef
# 0 andf
E extr K.Choose z c- A with
11 z 1
and z # 0 and defineg E A’ by g(x)
=f (xzz* )
for x E A. Then
g E K
and(f - g) c- K
since frompreceding remarks,
foreach x E A there
exists y
e A such that xx* - xx*zz* =yy*.
Sincef
c extr Kand K is proper, there exists a constant
a(z) _>- 0
so thata(z)f
= g.Therefore,
from a routine
computation,
it follows that for everyZ E A, z 71-- 0,
thereexists
oe(z) > 0
such thatConsequently
fory, z E A (z # 0)
it follows thatFrom Lemma 5 and the fact that
f
0 there exist x, z E A withf (xzz*)0
and
hence,
from(4)
we havea(z)
> 0 andf (x)
=A 0.Further,
sincewe conclude that
f (zz* )
> 0.Finally,
y from(5)
and the fact thatf (zz* )
> 0 and«(z)
> 0 it follows thatf ( yy*)
= 0 if andonly
if«(y)
= 0 forall y
e A.Let a =
(a(z))-lf(zz*) which,
from(5)
isindependent
of z for«(z)
> 0.Then
With a fixed as
above,
it follows that(7)
is also valid whenf(yy*)
= 0 sincethen, (6) implies
thatf (xyy*)
= 0 for every x E ft. Thus( 7 )
holds for all x,YEA.
With elements of the form
yy*
total in A we conclude thatf (x) f (y)
==- a f (xy ) for x, YEA
andconsequently,
thata-If
ismultiplicative.
For a commutative
algebra,
we note that whenever A contains an ap-proximate identity
boundedby
one, the extremepoints
of P areexactly
the
homomorphisms. Further,
when A has anidentity
of norm one weobtain the well known result that the extreme
points
ofP, - {f > 0: f (e)
=1-1
are
exactly
themultiplicative
elements of the set[1].
In fact extP1
=-
(extr K)
r1Pi
sincePi
is the intersection of K and thehyperplane
c- A: f (e) = 11 (see, [3,
p.337].) >
REFERENCES
[1] R. S. BUCY - G. MALTESE, A
representation
theoremfor positive functionals
oninvolution
algebras,
Math. Ann., 162 (1966), pp. 364-367.[2] I. GELFAND - D. RAIKOV - G. SHILOV, Commutative Normed
Rings,
Chelsea,New York, 1964.
[3] G. KÖTHE,
Topological
VectorSpaces -
I,Springer-Verlag,
New York, 1969.Howard