On Banach algebras, renewal measures and regenerative processes

On Banach algebras, renewal measures and regenerative

processes

Citation for published version (APA):

Frenk, J. B. G. (1985). On Banach algebras, renewal measures and regenerative processes. (Memorandum COSOR; Vol. 8513). Technische Hogeschool Eindhoven.

Document status and date: Published: 01/01/1985

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Memorandum COSOR 85-13

On Banach algebras, renewal measures and regenerative processes

by

J.B.G. Frenk

Eindhoven, The Netherlands

This paper contains the first chapter on Banach algebras

of a forthcoming monograph called "On Banach Algebras, Renewal

O. Introduction

This chapter is divided into three sections.

In section 1 an introduction to the theory of commutative Banach algebras will be given. The main result (Theorem 1.1.21) is the one-to-one correspondence between the set of all maximal ideals in a commutative Banach algebra V and the set of all homomorphism~ L: V + ~. All results in this section are known and can be found in the literature on this subject (cf. [NAI], [HIL], [RUD-1], (RUD-2], (RIC], [GEL]).

In section 2 attention will be paid to the Banach algebra V(~) of (weighted) complex valued summable sequences on the nonneg~tive integers and to some subalgebras of V(w). Most of the results in this section are known (cf. [GRU] , [ROG-2], [CHO]). However, the purpose of this section is to present simplified proofs and at the same time unify the proof techniques for the different poss"Z",;,le cases.

In section 3 the Banach algebra

seW)

of (weighted) complex measures, concentrated on the positive halfline, and some subalgebras of

SeW)

are considered. Among the most important and new results is a characterization of the space of homomorphisms on S(~) (Theorem 1.3.2) and the implication of this result (Theorem 1.3.4). These results fill an important gap in the reasoning of Rogozin (cf. (ROG-3], [ROG-5]) who derived similar results for a special case. The rest of the proofs and results are straightforward generalizations of the ideas used in section 2.

Finally, we like to mention that we will use the theorems in sections 2 and 3 to derive asymptotic results for the renewal sequence (Chapter 2), respectively the renewal measure (Chapter 3) in case the expectation of the waiting time distribution is finite.

- 2

-1.

Genera

I

properties

DEFINITION 1.1.1. (cf. [DUN]). A nonempty set V ~s called a comple~ algebra if V is a ring as well as a vectorspace over the complex field ~ with the property that for every x,y

E V

and a

E

C

the equality a(xy) = (ax)y

=

x(ay) holds.

I f in addition V possesses a complete norm 11·11 satisfying IIxyil ;;i lIx11l1yll for

all ~,y E V we call V a Banach algebra.

REMARK 1.1.2.

1. Since V is a vectorspace there exists a unique element

e

(the zero element) such that

e

+ x

= x for all x

E

V.

Also Ox = (l-1)x "" x - x

=

e

for all x E V, where 0 denotes the zero element of

C.

2. The multiplicative norminequality implies that the mapping (x,y) + xy is jointly continuous.

3. The existence of an element e, which satisfies xe

= ex

= x for all x E V

and lie I!

=

1, is assumed. This element is called the unit of V. (These conditions are not re6'"''i!rictive (cf. (HIL]).)

4. The opsration of multiplication in V is assumed to be commutative, i.e. xy

= yx for all x,y

E V.

DEFINITION 1.1.3. An element x E V is called invertible if there exists an element y E V satisfying yx

= e. (This element

y is unique and is denoted

by x -1.)

LEMMA 1.1.4. Suppose V is a commutative Banach algebra with unit.

1. For every x E V with IIxll < 1 the element e - x is invertibZe and

(lO (e-x)-l

=

I

n=O where xO := e. n x

2. For every· h E V and every invertib Ze x E V with II h"1I ;;; (lix -1/1) -1 the element x + h is invertible and it satisfies

PROOF. Since !Ix II < 1 and Ifxnll ~ IIxU n for every n E IN we obtain that sk :=

~=O

xn is a Cauchy sequence. Hence

lim sk

=

I

k-xo n=O

n x

k+l '

belongs to V. But (e-x)sk

=

e - x and so by the continuity of the multi-plication and the uniqueness of the inverse element we have

()O (e-x)-l

=

I

n=O

n x

This proves the first part of the lemma.

-1 A proof of the second part J.S given as follows. Since x + h = x(e + x h) and by assumption IIx -1 hll < 1 we obtain applying the first part that x + h

J.s invertible. Also

=

a:> )' (-L}"?(x-1 _{h)n Ullx}-l ll n;;;2 (lIx-111)3(lIhll)2 ~ 1 - !Ix -1 hI!

REMARK 1.1.5. By Lemma 1.1.4 the set G of all invertible elements is open

-1

and the mapping x + x (defined on G) is continuous.

DEFINITION 1.1.6. Suppose V is a commutative Banach algebra with unit. For x

E

V, denote by crV(x) the set of all A

E

C

such that x - Ae is not

invertible in V. This set crV(x) is called the spectrum of x and the comple-ments of crV(x) in

C

(=: pv(x» the resoZvent of x.

LEMMA 1.1.7. Let $ be a bounded Zinear functionaZ on a Banach aZgebra V

with unit. Take x E V and define f(A) := $«x-Ae)-l) for aZZ A E PV(x). Then f is anaZytic on the resoZvent of x and lim f(A)

=

O.

IAI-xo

PROOF. First note that the resolvent is not empty since e - A-1 x is invertible for [A[ > IIxll and hence x -Ae E G. This observation implies {A E

C:

IAI > IIxll} ~ PV(x). Taking A E PVex) and applying (1) with (A-~)e replacing h and x - Ae replacing x yields

-1 -1 -2

lI(x-~e) - (x-Ae) + (A-j.l)(x-Ae) II

4

-~or IA-~1 sufficiently small. Hence

, - 1 -1

lim n(x-~e) - (x-Ae)

~ - A (2)

By the definition of f and the continuity and linearity of ~ we obtain from (2) that

lim f(~) - f(A)

=

~ - A

-2

.p( (x-Ae) ) .

So f is analytic on the resolvent of x. Also

-1 - 1 - 1

Af(A)

=

.p(A(x-Ae) ) = .p«A x-e) )

and this implies by the continuity of the mapping x

~

x-I that lim Af(A)

=

.p(e). Hence

IAI-lim HA) ::; O.

IAI-THEOREM 1.1.8. For every x

€

V the set 0V(x) is nonempty and compaot.

IJ

PROOF. Suppose O'V(x) 'Is an empty set. Then it follows from Lemma 1.1.7

that for every .p € V* the function f: PVCx) ~

C

defined by f(A) := <P«x-Ae)-l) is entire (cf. [TIT], [KOD]).

Since lim fCA)

= 0, Liouville's theorem implies that <p{(x-Ae)-l) ::; 0 for

IAI-all A €

C

and <P € V*. Thus, by the Hahn-Banach theorem, -1

(x-Ae) =

e

for all A € C ,

an obvious contradiction, and so 0V(x) is not empty.

Since 0v(x) is bounded and the complement of the set G of invertible

ele-ments is closed, the compactness of 0V(x) follows easily. c

Before proving a key theorem ~n the general theory of Banach algebras we need the following definition.

DEFINITION 1.1.9. Let V be a Banach algebra and L a complex-valued func-tional (not identically zero) on V.

1. L is called an (algebra) homomorphism if (1) L is a linear functional, i.e.

(ii) L is a multiplicative functional, i.e.

L(xy)

=

L(x)L(y) for all x,y E V •

2. L is called an isometric isomorphism if ( i) L is an (algebra) homomorphism. (ii) L is onto, i.e. L(V)

= (::.

(iii) L l.S an isometry, Le. IL(x)1

=

II xii for all x E V.

THEOREM 1.1.10 (Gelfand-Mazur). Suppose V is a commutative Banach aZgebra with unit in which every nonzero element is invertible. Then V is iso-morphic to the complexfieZd ~ (notation: V ~ C).

PROOF. Fix x

F

e

and choose A E 0V(x). (This is possible by Theorem 1.1.8.) Then x - Ae is not invertible, and so, by assumption, x - Ae

=

e

or equiva-lently x

= Ae. This proves that 0V(x) consists of precisely the element A,

which we denote by A(X). Define also A(e) := O. It is now obvious that the mapping L: V. + ¢ defined by L(x)

= A(x) is an isometric isomorphism.

IJ

DEFINITION 1.1.11.

..".

1. A subset I of the commutative Banach algebra V l.S called an ideal if

(i) I is a linear subspace of V.

(li) xy E I _{for all x Eland y E}

v.

2. An ideal I l.S said to be proper if I :f

v.

3. An ideal I is said to be nrIximaZ i f it l.S a proper ideal and there exists no proper ideal I' such that I c I ' and I :f I' .

REMARK 1.1.12. It is easy to verify that e i I if I l.S a proper ideal.

In order to prove that every proper ideal is contained in a maximal ideal we need the next result.

LEMMA 1.1.13 (Zorn's lemma). If in a partially ordered nonempty set X every linearly ordered subset has an upperbound in X then X contains a maxinrIZ eZement.

6

-LEMMA 1. 1 • 14. Every proper idea l is con tained in a mxinv. Z idea Z.

PROOF. The set of all proper ideals which contain a given proper ideal is

nonempty and partially ordered by inclusion. Since every linearly ordered subset K has the upperbound U {I: I E K} and U {I: I E K} is an ideal with

e

rt

u

{I: I E K} (hence U {I: I E K} is a proper ideal) and U {I: I E K}

contains the given one we can apply Lemma 1.1.13. This yields the stated result.

The following theorem is very helpful in identifying whether an ele-ment is invertible or not.

THEOREM 1.1.15. An element x E V is invertible if and only if x does not belong to any maximl ideal.

[J

PROOF. Suppose x E V. is invertible and belongs to a maximal ideal

\i.

Then

-1

e :::: xx E 1M and so 1M :: V. This proves that an invertible element does not belong to a maximal ideal.

Conversely, suppose the element x E V is not invertible and consider the set Vx := {yx: y E V} :""Then Vx is an ideal (use the commutative property of the multiplication) and e

rt

Vx (use x is not invertible).

Hence Vx is a proper ideal with x:::: ex E Vx and by Lemma 1.1.14 we arrive at the desired result.

The Gelfand-Mazur th~orem states that a Banach algebra V can be iden-tified with

C

in case every nonzero element in V is invertible. Therefore we like to reduce every Banach algebra V into a Banach algebra V' with the above property. This reduction can be carried out as follows.

Suppose I 1S a proper ideal in the complex algebra V with unit e and

consider for fixed x E V the coset II(x) := x + I := {x + y: y E I}. Since I is a linear subspace of V we obtain for x

1-xZ

rt

I that IT(x1)

n

IT(xZ)

=

0

and for x₁-x₂ E I that H(x

1) ; II(xZ)' The set of all cosets is then denoted by

ViI

(V modulo I) and one defines in this set in a consistent way (since I

is an ideal) as follows the operations of multiplication, addition and scalar multiplication (ef. [RUD-l ], [RUD-Z]) :

IT (x+y) := IT(x) + II (y) x,y (; V

II (Ax) := AIT(X) x

E

V. A E (::

II (xy) := IT(x)IT(y) x,y E V •

This makes ViI a aomplex algebra with unit nee).

LEMMA 1.1.16. Let V be a aommutative aomplex algebrs with unit e and I a proper ideal. Then every nonzero element in ViI is invertibZe if and only if I is a rraximal ideal.

PROOF. Suppose I is a maximal ideal and consider the set ViI. In this set the unit is nee) and so we have to prove for every nonzero element TI(x) (in VII) the existence of an element

v

E V such that n(v)n(x)

=

TI(e).

Take now an arbitrary nonzero elem~nt TI(x) E ViI and define

I' := {vx + y: v E V, Y E I} ~ I •

Then If is an ideal. Also I' contains I properly since x

=

ex +6 E I' and x I/. I. This implies by the maximality of I that I'

=

V and so there exists some v E V such that vx + y

= e. Hence IT(v)TI(x)

=

TI(Yx)

=

TI(e) and this proves the first part of this lemma.

If I is not a maximal ideal there exists by Lemma 1.1.14 a maximal ideal 1M containing I properly ~.p.d so we can find some Vo E V such that Vo E 1M and Vo I/. I. Hence TI(v

O) is a nonzero element in VII. Note also, since 1M ~s a maximal ideal, that TI(v)TI(v

O)

= IT(vv

O) ~ TI(e) for every v E V. Thus we have found a nonzero element n(v

O) (in VII) which is not invertibl,e in VII. C

In the preceding notes we looked at the algebraic properties of ideals and quotient spaces. We now pay attention to the topological properties.

LEMMA 1.1.17. Every rraximal ideal is alosed.

PROOF. It is easily verified that the closure of an ideal is again an ideal. Since by Remark 1.1.5 the nonempty set G of invertible elements is open and by Theorem 1.1.15 every invertible element does not belong to any proper

ideal we obtain that the closure of a proper ideal is again a proper ideal. By this observation and the maximality of 1M we get that the closure of 1M

equals 1M and so 1M is closed.

Consider the quotient space VII with I a proper and closed ideal and V,a commutative Banach algebra. Define now:

IlTI(x) II := inf {lIx+yil: y E I} .

8

-It follows that this defines a norm (the so-called quotient norm) on V/I which satisfies the following property.

LEMMA 1.1.18. If V is a commutative Banach algebr>a and I is a proper and

closed ideal, then V/I with norm Ii .li

q is a commutative Banach algebra.

PROOF. Cf. [RUD-2].

Combining the algebraic and topological properties the next important theorems hold.

THEOREM 1.1.19. If 1M is a maximal ideal in a commutative Banach algebra V

with unit, then V/1

M is isometrically isomorphic to the complex field ~. Also there exists for every x E V a unique nwnber x(~) E (: sHch that

x

=

x(~)e + y , y E ~ • (4)

The correspondence x + x(~) has the following properties for every x,y E'V

and a

E (:

.dJ$> (x+y) (1 M)

=

x(~) + y(~) (5) (ClX) (1 M)

=

Clx(IM) (6) (xy) (1 M)

=

x(~)y(IM) (7) e(I M)

=

1 (8) x(I M) =O~x E 1M (9) x(I M) E aV(x) (10) !x(IM) I ~ "xII • ( 11)

PROOF. Since by Lemmas 1.1.17 and 1.1.18 V/I

M is a commutative Banach algebra and every nonzero element IT(x) E V/I

M is invertible by Lemma 1.1.16 we can apply the Gelfand-Mazur theorem (Theorem 1.1.10). Hence for every x E V there exists some complex number x(I

M) such that IT(x)

= x(IM)IT(e).

The complex number x(~) is unique since IT(e) # 1M (zero element in V/I_M). Obviously, by the above observation we can find for every x E V and

corre-sponding complex number x(I

M) some y E 1M such that

x = x ( 1

M)

e + y • ( 12)

Using this representation we easily obtain (5) up to (9) and so we only have to prove (10) and (11).

In order to prove (10) we note that the element x - x(~)e belongs to 1

M. Hence x - x(IM)e is not invertible by Theorem 1.1.15 or equivalently

x(I_M) € 0V(x). Also, Ix(I_M)I ~ sup {IAI:

A E

0V(x)} and since for all

A E

~ with IAI > "xII the element x - Ae is invertible, i.e. A E PV(x), we obtain

Ix(I_M) I ~ IIxli. 0

REMARK 1.1.20. In the sequel we will use homomorphism instead of (algebra) homomorphism.

THEOREM 1.1.21. Let V be a commutative Banach a~gebra with unit. Then for every homomorphism L: V ~ t the set L+(O) := {x

E

V: L(x)

=

O} is a maxima~ ideaZ.

Conversely, for every maximal ideal 1M there exists a unique homomorphism L +

such that L (0)

=

1 M•

PROOF. Since L is a h~~omorphism the set L+(O) is an ideal and its

codimen-+

sion equals one. This implies directly that L (0) is a maximal ideal. Conversely, suppose the set ~ is a maximal ideal in V. Then by (12) there exists for every x

E

V a unique complex number x(I

_M)

such that x

=

x(IM)e +y for some y

El

_M'

From (5) up to (8) it is easy to see that the mapping L: x + x(I

M) defines a homomorphism. Also by (9), L+(O)

=

1M and hence we have found a homomor-phism satisfying the given property.

On the other hand this homomorphism is unique since every x

E

V has by (12)

a unique representation and L(e)

=

1 for every homomorphism L on V.

REMARK 1.1.22. By Theorem 1.1.21 we obtain immediately that there exists a one-to-one mapping T of the set 6(V) of all homomorphisms L: V + C onto the set of all maximal ideals in V. (TakeT: L + L+(O).)

LEMMA 1.1".23. Suppose V is a commutative Banach algebra with unit. Then for every x

E V

- 10

-PROOF. Suppose x € V fixed and let L: V + ~ be a homomorphism. Then +

x - L(x)e belongs to the set L (0) and this set is a maximal ideal by Theo-rem 1.1.21. Hence x·- L(x)e is not invertible by TheoTheo-rem 1.1.15 and this means by definition L(x) € 0V(x). So we have proved

Take now an arbitrary

A E

0V(x). Then by definition x - Xe is not invertible and this implies by Theorem 1.1.15 that x - Ae belongs to some maximal ideal

\i.

By Theorem 1.1.21 we can find a unique homomorphism L: V + C such that +

L (0)

=

1

M, For -this homomorphism we get L(x)

= X and so

LEMMA 1.1.24. Suppose V is a commutative Banach algebra with unit and let

L: V +

t

be a homomorphism. Then the operator norm of L equals one.

PROOF. By Lemma 1.1.23 we have L(x) € 0v(x) for every x € V. Since x - Ae is invertible for all h'·E t with I A I > IIxll, i.e. A

f/.

0v(x) J we obtain

IL(x)

I

~ !lxll.

Also L(e)

=

1, lie II

=

1, and this implies together with the above inequality that the operator norm

L(x)

11111 := sup { Ilxll x € V, IIxll :F O}

[J

equals one. [J

In order to discuss the next theorems we need the following observa-tions. A Banach algebra V with unit contains every polynomial

and in general every function

A(x)

=

I

n=O n

a x _n

• •• + x € V, k € IN

where A: t + t, defined by A(z)

=

l~ anzn, is an entire function. (Use the multiplicative norm inequality in V, the absolute convergence of the series \~ a zn for every z €

t

and the completeness of V.)

These examples are special cases of Theorem 1.1.28 as can be seen from Theorem 1.1.30. However, before mentioning these theorems, we introduce some well-known definitions.

DEFINITION 1.1.25 (cf. [RUD-2).

1. A curve in

C

~s a continuous mapping y of a compact interval

[a,S] c

lR

(a

< a) into

C.

We call

[a,S]

the parameter intervaL of y and denote the

range of y by y*, i.e. y* := {yet): t

€ [a,S]}.

2. A path is a piecewise continuously differentiable curve ~n ~.

It is now possible to define the integral ofa function A: ~ + V over the path y* in case A is continuous on an open set containing y*.

DEFINITION 1.1.26. Let A: ~ + V be continuous on an open set containing the path y*. Then

J

A(A)d>.. := y*

a

J

A(y(t»y'(t)dt

where y is the piecewi~e continuously differentiable curve with parameter interval [a,a] belonging to y*.

REMARK 1.1.27.

1. This definition does not depend on the parametrization of y*.(Use the substitution theorem.) Since the integral

fS

A(y(t»y'(t)dt is defined

a

in a similar way as the classical Riemann integral, i.e. as the limit of a Cauchy sequence {c

k} of which every element ck has the form k

L

A(y(t »y'(t )(t l-t )

n=O n n n+ n

we obtain by the completeness of V that f A(A)dA belongs to V.

y*

THEOREM 1.1.28. 'Let x € V be fixed and suppose

1. A: D +

C

is anaLytic in an open set D containing the compact spectrum 0vex)

= {L(x): L

€ 6(V)}.

and where If

we

define 12 -(i

=

1, ••• , n) n

I

h i k=1

J

d>' >.-z y*

=

n

r*

=

u

y~. k=1 k

o

1 n

J

-1

A(x) := 2~i

I

A(>.) (Ae-x) dA k=l

y* k

if

Z € 0V(x)

if

Z

t!.

D

:=

2~i

J

A(>') (Ae-x)-l d>'

r*

( 13)

then

1. A(x) E V

2. L(A(x»

=

A(L(x))

fOr

evepd homomorphism L on V.

PROOF. Note that>. + (>.e-x)-l is a continuous function on the open set pv(x). Hence by Remark 1.1.27 and condition (1) the first result follows immediately.

Observe also by the definition of a contour integral and the continuity of every L € ~(V) (Lemma 1.1.24) that

L(A(x)

=

2~i

I

A(>')L«>.e-x)-l)dA

=

2~i

f

A(A) (>'-L(x)-1 dA •

r*

r*

Hence, since the Cauchy formula holds (cf. [RUD-2]), we get

L(A(x» = A(L(x) .

REMARK 1.1.29.

o

1. In order for the Cauchy formula to hold we assumed the existence of a c.ontour

r*

satisfying condition (2) of Theorem 1.1.28. However, this condition is not restrictive since the first condition in Theorem 1.1.28 implies the existence of such a contour r~. (A constructive proof of this

will always use in (13) (unless stated otherwise) a contour

r*

con-structed along the lines of the proof of Theorem 13.5 of [RUD-2].

2. By Lemma 1.1.23 and Theorem 1.1.28 it is easy to see that aVCA(x»

=

= A(av(x»

for every A satisfying the conditions of Theorem 1.1.28.

THEOREM 1.1.30. Let x E V be fixed and A _x the collection of all analytic funotions A on some open set D containing av(x). Then the mapping A ~ ~Cx) satisfies the following properties:

1. A(z) _ 1 ~

hex)

=

e 2. A(z) _ z ~

hex)

=

x

5. If the sequence {A } converges uniformly to A in every oompaot subset of

n

D then the sequence {A (x)} converges in norm to A(x). n

PROOF. We will only prove (1), (2) and (4), since (3) and (5) are obvious. ,~

For the proof of (1) and (2) we note that the functions A are entire and so

by Cauchy's formula we can take for the contour in (13) the positively oriented circle r* := {l E

C:

III

=

UxU + 1}. On this circle (le-x)-l has

• ~oo -n-1 n

the expanswn Ln=O A x (Lemma 1.1.4) and so substituting this in (13) and using Cauchy's formula yields the desired results (1) and (2).

In order to prove (4) we have to show that the product of Al(x) and

A

2(x) equals

By the construction of 1'* (see Remark 1.1.29) it 1.S easy to verify that av(x) c int(r*) and cl(int(r*» c D. Applying again Theorem 13.5 of [RUD-2] with K

=

cl(int{r*» and n

= D, we can construct a contour r* with

1

r

7

c D - cl ( in t

(r

*) ) and

f

A1(1)(le-x) -1 dl.

r*

1

Hence the product of A

Since

-4>

J J

r* r*

₁

- 14

-(Ae-x)-1 - (~e-x)-l

=

(Ae-x)-l(~e-x)-l«~e-x) - (Ae-x»

=

=

(~-A)(Ae~x)-l(~e-x)-l

this implies that the product A

l(X)A2(X) equals the sum of

and

r*

r*

1

The first term in this sum equals zero (use A2(~)/(~-A) is analytic on cl(int

r*)

for every A E

r7),

while the second by Cauchy's formula equals

2~i

J A2(~)A,(~)(~e-x)-1 d~

•

r*

Finally, we conclude this section with the following remark.

-1

REMARK 1.1.31. If 0 ~ qv(x) (~x exists) then by (1) and (4) of Theorem 1.1.30 we obtain x-I

= A(x) with A(z)

=

liz.

2. The Banach algebra of complex-valued sequences on the nonnegative integers

Let 'i' be the set of all functions 1jJ: IN -+ [0,(0) satisfying ljI(O)

=

1 and

o <

1jJ(n+m) ~ ljI(n)1jJ(m) for all m.n E IN. It is known (cf. [HIL]) that each such ljI has' the property

lim (ljI(n» lin

= inf (ljI(n» lin

n+a> n~l

with 0

~

inf (l/I(n»l/n n~l

< 00

For each ~ E ~ let V(~) be the set of all complex-valued sequences {x(n)}:=o for which

I

~(n) !x(n)! < 00 •

n=O

Define in V(~) as follows the operations of addition, scalar mUltiplication and multiplication: (x + y) (n) :== x(n) + yen) (ax)(n) := ax(n) n (x

*

y) (n) :

=

L

x(n-k)y(k) k=O

Then the following result holds.

for all n E IN

for all a E

C.

n E IN

for all n E IN •

THEOREM 1.2.1 (d. [RUD-l J) . The space v(ljJ) with norm 1I-IlIP is a commutative Banach aZgebra (with unit) for every ljJ E ~.

PROOF. The only prope'nies of a commutative Banach algebra, which need verification, are the completeness and the (multiplicative) norm ,inequality. We start with the completeness.

Since V(ljJ) can be identified with the space Ll

(~)

with

~

a positive measure on the nonnegative integers defined by

~({n})

== ljJ(n) and Ll

(~)

(cf. [RUD-2])

is a complete space, we obtain the completeness of V(~).

The multiplicative norm inequality follows immediately by the observation that

eo n

L

ljJ(n)! (x

*

y)(n)! ::;

_1:

_1:

~(n-k)~(k) !y(n-k)x(k)! ==

n=O n=O k=O

00

=

L

~(k)!x(k)!.

L

~(n-k)!y(n-k)!. c

k=O n=k

REMARK 1.2.2. The unit in V(~) is given by e = {e(n)}:=O with e(O) :;:; 1 and e(n)

= 0 for all n

~ 1. Also, since IP(O) == I, we obtain lIelll/i == 1.

16

-THEOREM 1.2.3 (cf. [RUD-1]). Let ~ € f and suppose

C : = {z' € (;: I z I ;;; inf (

~

( n» 1 / n } • n~l

1~en L: V(~) + ( is a homomorphism if and onZy if there exists some z

€

C such that L(x) = I:=o x(n)zn for all x € V(~).

PROOF. Obviously for al1 z € C and x € V(~) we obtain I:=o Ix(n) Izn ,lii "xll~

and so the mapping L: V(~) +

C

with L(x)

= I:=o x(n)zn is well defined.

This mapping is clearly a homomorphism for all z

€

C.

It remains to prove that every homomorphism has this form. This can be done as follows. Consider the real valued sequence Xo

=

(0,1,0, •.. ). By the definition of multiplication in V(~) the real valued sequence x~* for n ~ equals 1 in the Cl+1) th component and 0 in the other components. Also,

0* ~

Xo

:=

e. Hence every complex valued sequence x in V(~), where x

= {x(n)}n=O'

has the representation x

=

L:=O

x(n)x~* and this implies that

n

Hx -

I

X(k)xk

_O

*"

+

0

k=O',;if> ~

(n +~) •

Thus by Lemma 1.1.24 we obtain for every homomorphism L: V(~) + t that

We now have to prove that L(x

O) € C. Since the operator norm of L equals one (Lemma 1.1.24) we get for every n € :IN that

.This implies IL(x O)/

result.

;;; inf

(~(n»

l/n and thus we have proved the desired

n~l

CJ

TIIEOREM 1.2.4. If x € v(~) for some ~ E ~ and

L:=o

x(n)zn ~ 0 for all z € C then x is invertible in V(~).

PROOF. The result follows immediately by applying Theorems 1.1.21, 1.1.15

REMARK 1.2.5. It is easy to verify by the compactness of C and the con-tinuity of the functioni(z)

=

I:=o

x(n)zn that the condition

I:=o

x(n)zn ~ 0 for all z E C is equivalent with the condition

inf

{I

I

x(n)zn,: z E

c}

> 0 • n=O

We now introduce the following class of commutative Banach algebras (with unit), which are connected with V(~).

Let

S

denote the set of all positive functions ~: ~ + (O,~) for which

2* ( )

sup)J n < ~

n;;:O ].I(n)

This class contains the so-called subexponential and other related sequences, discussed in [EMB-l].

For all ~

E

~ and ~

E

S

we define

V(l/J,].I) := {x

E

V(l/J):

Ii

ex) := sup Ix(n)1 _~(n) <

~}

,

~ _n;;;;O

~

-0'

:= {x

E

V(ljJ,].I): lim Ix(n)1

_{= O}}

V (1ji,].I)

].I(n)

.

n~

Clearly, V(Iji,~) ~s a vectorspace and it is easy to verify that

with

< ~

is a norm on V(l/J,].I).

THEOREM 1.2.6. VCl/J,].I) with the above norm is a commutative Banach aZgebra (with unit) for every l/J

E

~

and every

~

E

Sand VOCl/J,].I) is a closed sub-aZgebra, aZso with unit.

PROOF. Fix l/J

E

~, ].I

E

S

and let x,y

E

V(l/J,].I) be arbitrary. Then for every

n E ~ we have n I (x

*

y) (n) I ~

L

k=O I x(n-k) I'y(k) I n ~

Ii (x)Ii

(y) ].I Jl

I

k=O

- 18

-Hence

MP

(x)i? (y)

II ~ for every n E ::N

and this implies

P

(xy) :;i

MP

(x)

P

(y).

)l II II

Using this inequality it is inunediately clear that IIxyll :;; IIXU", lIyll,l. •

_ lJJ,1l ~,)l ~,ll

The completeness can be proven as follows. Let V(ll) be the set of all complex-valued sequences x for which

IIxll II Ixen) I := sup ll(n) n~O < co • co

This set is isomorphic to the Banach space ~ of bounded sequences and so it is a complete space.

N co N co

Now {x }N=O ~s a Cauchy sequence in V(IJi.ll) if and only if {x }N=O is a Cauchy sequence ~n V(ll) and V(tjI). Since Y(~) and V(lJJ) with norms respectively 11·11

co co _ II

and 1I~1I1Ji are complete spaces, we can find elements xl E V(lJJ) and x

2 E V(l.!) such that

and

N co

I t is clear that x (n) -+ x. (n) (i

=

1,2) as N t co for every n E ::N and so

~

x~ = x;. This imp lies .<Yfr

and lim IIxN -

x~1I

= 0 •

N-+co tjI,1l

It is easy to verify that yO(tjI,ll) is a closed subspace of YCtjI,ll) and so we only have to verify that yO(tjI,)l) is closed under multiplication.

-0 Let x

1,x2 E V (tjI,ll). Then clearly

and x. ~,n -0 x _1,n

*

_{x2 E V (lJJ,ll) ,n} (i = 1,2) for every n E IN •

Also, lim x.

= x.

(i

=

1,2) in yOCtjI.ll) and so by the joint continuity of n-+co ~, n ~

*

and the closedness of VO(tjI.ll) we obtain

REMARK 1.2.7. The only def iciency in Theorem 1.2.6 is that lie 11,1. :f: 1, 't',ll

where e is the (discrete) Dirac measure at zero. However, one can renorm V(lji,l-I) .with an equivalent norm II.II~

'j',ll such that lIell* l/J,ll

=

1. This can be carried out as follows:

Give any element x

E

V(lji,ll) the norm of the bounded linear transformation T _x (y)

= x * y. Then

IlxII~ 't',J.1 { Ilx

*

Ylltp } := sup _,J.1 y:f:e lIylll/J,il

and by the definition of Jlxll* and the multiplicative norm inequality. one l/J,ll

obtains

IIxll* S Oxll • 1ji,J.1 l/J,J.1

Obviously, lie II

*

l/J,ll

=

1 and by the above inequality the different norms are equivalent. (In the

·used. )

sequel we assume that-the equivalent norm

II-II:'

't',ll

"'iii>

is

Like ~n the case of V(lji) we are interested in the representation of every homomorphism on V(l/J,ll). It turns out that solving this problem is easy

-0

on the Banach subalgebra V (l/J,J.I). However, before stating the result, we need the following observation.

GO

REMARK 1.2.8. For every positive sequence {J.I(n)}n=O it ~s obvious that n+m

J.I(m)ll(n) S

I

J.I(k)ll(n+m-k) k=O

for all m;n E IN

This implies, in case J.I E S, that

for all m,n E IN •

M

Putting fen) := In J.I(n) we see that fen) is a subadditive function. This proves the existence of lim fen) (cf. [HIL]). Moreover, since lim fen) < GO,

/ n~ n n~ n

we have lim {J.I (n» 1 n > O.

n~

~

THEOREM 1.2.9. Let l/J E ~ and II E S. Then the following conditions are

20

-1. lim

(~(n)~(n»

lin

~

(the limit exists by the preceding remark and is

n

-< 00).

2. The mapping L:

VO(~,~)

+

~

is a homomorphism if and only if there exists

00 n -0

some z

E C

such that L(x)

= In=o x(n)z

for all x

E V

(~;~).

PROOF. Suppose lim

(~(n)~(n»

lin

~

1. Note first that

VO{~,~)

is a commu-n_

tative Banach algebra with unit.

Since the proof of (2) is very similar to the proof of Theorem 1.2.3 we will only discuss the differences.

Observe that

pix

n _{I x(k)xO} k*)

₌

sup Ix(k) I +0

k=O k~n+1 ~(k)

-0

for all x

E V

(~,~). and so

n

k*

lim IIx -

L

_x(k)xO "ljI,ll

= o ,

n~ k=O or equivalently lim IIx -

o .

This implies L(x) =

L

x(k)(L(XO»k k=O -0

for every x

E V

(~ill) •

It rema~ns to prove that A := L(x

O) E C. Suppose

:= inf

(~(n»

lin. n~ 1

In that case, define

and p :

=

n

x

(n) :

=

----:;p~­ Anljl(n) (~ P <

IA

I)

' b ' I ,.,. V() d I' Ix(n) I

o v~ous y, x € ~ an ~m ( )

=

__ -0 n-+<>o ]l n O. (Use lim n..,.."

(~(n)]l(n»

l/n

~

p < IXI.) Hence x € V (~,]l).

On the other hand,

n

L(X)

=

lim

L

X(k)Ak

=

n lim

I

n-+<>o k=O n""'" k=O

and this tends to infinity s~nce

p > inf

(~(n»)

lin:: lim (W(n) l/n •

n-+<>o

Now we have obtained a contradiction since ILex) I :.i IIxll,l. < "".

'1',11

The converse can be proved as follows. Suppose

lim

(~(n)l1(n»

l/n

₌

a • O < a < l .

n""'"

Then there exists some nO E N such that

(~(nh(n»

l/n ;;; a + ₂ 1 _< 1 Hence

for all n ii;;

00

L

k=n

_o

< ""

-0

for some constant C, depending on x E V (~,11). This implies that

co

L(x) :=

L

x(k)(a:l +

~)k w~

k=O

1 and

is a well defined mapping on

VO(~,]l).

Clearly L is a homomorphism and we have obtained a contradiction since --1-1 +

~

> lor, equivalently,

1 a+

(a+1 + !)w_O > Woe 0

For the proof of the next result we need the following simple lemma.

LEMMA 1.2.10. Let 11 E

S

and ~ E ~. Put xn

=

l[O,n]x (n

=

1,2, ••• ) for

00

implies that lim sup n-+<:o pE:N 22

-= 0

for every

x

E

V(~,~)

PROOF. Let x E V(l/J,~) be given. Trivially, lim Ilx-xn",,,

=

0, so all that

n-+<:o 'I'

must be proved is that

- 2*

lim P «x-x) )

=

0 • n-+<:o j..I n

I f M '"

P

(x), then for all p,n E IN we have Ix-x I (p) ;;i M(J.!-J..1 ) (p) and

}l n n

therefore

2* 2* 2 2*

I (x-x)

I

(p) ~ (Ix-x

I)

(p);;; M (J.!-~) (p)

n n n

Dividing by J..I(p) and using the assumption yields

REMARK 1. 2 • 11.

1. Let

a(n) := sup

pEJN

c

Then ~

E

S

~s equivalent with a(O) < 00, Since {a(n)}:=O is a nonincreasing

sequence, lim a(n) exists (and is finite) in case J.!

E

S.

n-+<:o • 2. Observe that since sup pEJN = sup p;;:2n+2 2* (I-\-J.!n) (p) J.!(p)

lji E '¥ and ,.... Suppose aLso THEOREM 1.2.12. Let )J E S. 2* lim sup' (]l-lln) (p)

°

ll{P) ""

.

n_ pElN

Then the foLZowing conditions are equivaZent.

1. lim (lji(n)jJ(n» lin

~

1 • n

-2. The mapping L: V(lji,ll) ~ ~ is a homomorphism if and only if there exists some z

E C

such that L(x)

=

I:=o

x(n)zn for aZl x

E

V(w,jJ).

PROOF. Suppose lim ()J(n)lji(n» 1/n

~

1 and let L: V(lji,ll)

~ ~

be given. Put

,.... -0 n-;4> -0

L := L/v (lji,)J). Then

L

is a homomorphism on V (lji,).I) and so by Theorem 1.2.9 there exists some z

E C

such that L(x)

=

1;=0

x(p)zp for all x

E

VO($,jJ). This implies in particular L(x ) _{. n}

=

\00 _~p=

°

x (p)zp for every n _n E IN. (Remember: xn := l[O.n]x.)

On the other hand, by lemma 1.2.10 we obtain (L(x) - L(x »2

=

L«x-x )2*)

~

0

n -0 n

(n ~ 00) and so

n E IN)

lim L(x )

= L(x). This implies (since x

E

V ($,jJ) for every

n n n_ L(x) "" lim L(x ) _{. n} n

-= lim L(x )

_n

=

n-· lim

I

n - p=O

p=o

I

The other result can be proved In a similar way as Theorem 1.2.9.

x(p)z~ •

We shall now derive a result similar to Theorems 1.2.9 and 1.2.12 for yet another subalgebra of V($) and for a special choice of $

E

'£I.

CJ

DEFINITION 1.2.13. For r ~

jJ: IN ~ (0,1] satisfying

let S(r) denote the set of probability measures

( i) (ii) (iii) co ~(r) :=

L

r n ]len) n=O /*(p) lim \J(p) 2U(r) p -l _{1m ll(n+1)} ' !J(n) _{== r •} n -< 00

24

-REMARK 1. 2 • 14.

1. The members of S(r) are called r-subexponentiaZ (cf.

[EMB-lJ).

In the

special case r =, 1 we call them s'UbexponentiaZ instead of 1-subexponential.

2. It is possible to show that the probability measure ~ with ~(n)

=

Crnh(n) and h satisfying (i) (ii) (iii) (iv) h positive, h E L1(0,en) , . hen) hm h(n+1) = 1 , n-+<><> lim sup n-+<><> a~t~l h( [nt]) h(n) belongs to S(r). < en for some 0 < a < 1 ,

In particular, the probability measure ~ with

/3>l,rElR

are r-subexponential sequences.

3. An equivalent set oimconditions characterizing S(r) ~s:

(ii' )

(iiit)

J.l2*(n)

lim (n) exists

n-+<><> ].I

1_{lorn J.I(n+l)} · ].I(n)

₌

_{r •} n-+<><>

(cf. [CHO]).

In the next lemma we discuss another set of conditions characterizing S(r) (cf.

[EMB-l]).

LEMMA 1.2.15. Let].l: IN -+ (0, 1] be a probability measure satiSfying

(1) ( ii) (iii) 00

per)

:=

L

n=O n ].I(n)r < en ll(n) liminf ll(n+l) ~ r • n+<lO

Then necessariZy lim

~(~~~)

=

r (or equivalently

~

E S(r»).

n-+a>

PROOF. We only have to verify that lim pen)

=

r for r

=

t, since the

n-joa) \J ( n + 1)

general case r > 1 can easily be reduced to the case r

=

1 by putting

The proof for r

=

1 can now be carried out as follows. 'Since by definition \J(n) > 0 for every n E ~, one easily obtains

for every m ~

[I]'

Heney by (iii) and (ii) it follows

limsup J.I(n-l) <

1 (1 _

m~

J.I(k»)

\J(n) -

ilTiT

k 1

n+OQ

Letting m t OQ finally yields

and so lim J.I(n-l)

=

].J(n) n-k» 1 • for every m E ~ • c

For simplicity we write ~ (n) := rn and we define for all ].J

E

S(r) and r a

E

C

{x t:: V-( ) l ' x(n)" a} := ~

W

,IJ: ~m ~

=

r n4a> II (n) -a U V (W ,Il) aEC"" r

LE~~ 1.2.16.

1(W

_r,ll) is a closed subalgebra of V(wr,IJ) for all r ~ 1 and ~ E S(r).

PROOF. We omit the proof since a proof for a similar Banach algebra in a

26

-LEMMA 1.2.17. If ~ € S(r) for some r ~ 1, then

lim sup'

n-+<>o p€:JN

=

0 •

PROOF. Let e > 0 be arbitrary. Since ~

€

S(r) there exists an nO

= nO(e)

€

:IN such that

OG

I

rn~(n) & E •

n=no

An elementary computation shows for all p ~ 2no + 2 that

This implies using (1) and Definition 1.2.13 that

for all p € ]N with p ';;'..:/)9PO

= p(n

O)'

. ( )2* ( )2* .

S1nce ~-~ & ~-~ whenever m ~ n 1t follows that for n ~

m n we have sup p€]N 2* (~-p.n) (p) ~(p) sup p~2n+2 2e • (1)

REMARK 1.2.18. We do not know wheth~r the conclusion of Lemma 1.2.17 holds for every ~ € S (see also Remark 1.2.11).

LEMMA 1.2.19. If r

~

1 and

~

€ S(r), then lim (p.(n» l/n

=

r-1• (Hence the

n-+<>o

concZusion of Theorem 1.2.9 hoZds with 1/J :: I/J and]l € S(r).J r

PROOF. Setting hen)

= -

In(~(n)) it follows from 11m . ~(n) ( 1 ) : r that

n-+<>o ~ n+

lim h(n+l) - hen) = In r •

n-+<x>

o

Hence, in particular, h is bounded on each finite interval, so Lemma 1.12 of

We are now prepared for the proof of the announced analogue of Theorems 1.2.9 and 1.2.12 for the subalgebra ~(~r'~) of V(~r) (~E S(r».

THEOREM 1.2.20. Let

u

E S(r) for some

r

~ 1. Then the mapping L: V(~ ,~) + ~

f'o,J r

is a homomorphism if and only if there exists some z

E C

such that L(x)

=

~:=O x(n)zn for aZZ x

E

i($r'~)'

PROOF. -0

V (I/Jr'u)

- .... -0

Let L: V(~ ,~) +

C be given and put L

:= L/V (~ .~).

,.... r r

c VO(I/J

.~)

c V(I/J

.~).)

By Lemma 1.2.19 we obtain

-,..., r - .... r

lim

(I/Jr(n)~(n»

lin

n-+<x>

(Clearly,

and so applying Theorem 1.2.9 yields the existence of some z

E

C such that L(x):; ~ x(n)z n

n=O

On the other hand, we know by for every x E V(~ .~).

-0 '" r

Since x E V _n ($ _r .~) fo):;,. every _.

L(x) = lim L(x ) n n-+<x> 00

=

_L

x(p)zp p=O

=

-0

for all x

E

V (I/Jr'~)

Lemmas 1.2.10 and 1.2.17 that lim L(x )

=

n-+<x> n

n E IN we finally obtain that

co lim L(x ) = lim

_L

x (p)zp

₌

n n n-+<x> n-+<x> p=O for every x E V(tJ; .~) r v r

Clearly a mapping of this form defines a homomorphism.

L(x)

[J

From the general theory of commutative Banach algebras (section 1) the results stated in Theorems 1.2.9. 1.2.12 and 1.2.20 have the following

corollaries. (We do not repeat the assumptions here.)

(A) x E V -0 (I/J,~) and x invertible in V(I/J) + x -1 E V -0 (ljJ.~) (Theorem 1.2.9).

(B) x E V(tJ;,~) - and x invertible in v(tJ;) + x -1 E -V(I/J,~) (Theorem 1.2.12).

• 28

-These corollaries are special cases of the following results, which will be stated without proof. (Proofs of similar results in a more general setting will be given in the next section.)

Before stating these results we like to remind the reader of the following well-known facts (see also Theorem 1.1.28). Suppose V is a com-mutative Banach algebra with unit e and A is an analytic function on an open set D containing the compact spectrum 0V(x). If

r

is a contour surrounding 0V(x) in D ([RUD-2]) then the formula

~

_{1\(x) := 2ITi} 1

J

_{AO.) (I.e-x)} -1 _dA

r

defines an element of V.

Moreover, for every homomorphism L: V ~ ~ we have

L(A(X» == J\(L(x» •

In particular

THEOREM 1.2.21. Suppose ~ E ~, x E V(~) and A is anaZytia on an open set D

.jIII>

containing crV(~)(x).

n~

2. If~ in addition, lim sup

=

0 then x E V(~,ll) impZies

A

(x) E V(~ .11) • n~ pEN

We now sharpen en generalize (C).

THEOREM 1.2.22. Let 11 E S(r) for some r ~ 1 and Zet x E _,.., V(~ _r ,11). If 1\ is analytic on an open set 0 containing 0V(~)(x) then A(x) E X(~r'll).

More preciseZy, if _. x E _,..,

Vb(~

_r .11) then A(x) E _,....

Vd(~

_r ,11) with d

=

bJ\t(Im_o _n- r ll(n». n (A' denotes the derivative of A.)

Finally, we mention the following class of Banach algebras. Let S be the set of probability measures ].l: :N ~ (0,1] satisfying

For all ~ E ~ and ~ E S we now define 0:>

I

Ix(k) I k=n } s up -~'7"'( ':"'"[ n-.-o:>~)') - < 0:> n~O and 0:>

I

Ix(k) I lim

_k:-(7"':[:-n-.o:>')~)-

= O} •

n-?<iO

Clearly, V(~,~) is a vector space.

It is also easy to verify that II xii ,I, := IIxll + MP (x) is a norm on 'r,j.! l/J j.!

Moreover, if the multiplication is given by the convolution operation

*,

V(~,j.!)

is a Banach algebra (with unit) and

VO(~,~)

is a closed subalgebra of V(I/J,ll).

Now we mention without proof the following results. (The proofs of

these results follow the same lines as previous proofs.)

THEOREM 1.2.23. Let l/J ~ and \.l E S. Then the following aonditions are equivaZent:

1. lim

(l/J(n)~([n.co»)

l/n

~

1. (Note that this limit always exists.)

n-+oo

2. The mapping L: VOCI/J,\.l) +

C

is a homomorphism if and only if there exists

some z E C suah that L(x)

=

I:=o

x(n)zn for all x E

VO(l/J,~).

THEOREM 1.2.24. Let l/J E ~ and j.l E S. Suppose also

lim sup n-+<x> pEJN

2*

(j.1-\.ln) ([p,co»

]J([p,""» = 0 •

Then the foUowing aonditions al'e equivalent:

[ l/n .

1. lim (l/J(n)ll( n,o:»» ~ 1.

n-+oo

2. The mapping L: V(I/J,\.l) +

t

is a homomorphism if and onZy if there exists

30

-At last we introduce the following set of probability measures.

DEFINITION 1.2.25. For r ~ 1 let S(r) denote the set of probability measures

\1: IN -+ (0, 1] satisfying co (i) ~(r) :=

~

r n il(n) < co n=O 2* lim lJ ([n,oo»

₌

2~(r) ll([n,co») p-l-OO (ii) lim il([n,oo» ::: _r il( [n+1 ,00»

.

n-l-OO (iii)

If we define for all lJ E S(r), r ~ 1 and a E

C,

I

x(k)

lim k=n

=

a}

il([n,oo» n-l-OO

and

then it is easy to prove that _,.... V(~ _r ,il) (remember ~ _r (n) ,: r , n n ~ 0) is a commutative Banach algebra.

Before mentioning the next result we like to make the following remark.

REMARK 1.2.26. The above definition and results are only interesting for r

= 1, since for r

> lone can easily prove that S(r) = S(r) and _,.., V(~ _r ,lJ) :::

THEOREM 1.2.27. Let il

E

S(r) for some r ~ 1. Then the mapping L: V(~ ,ll) +

C

r

is a homomorphism if and only if there exists some z E C suah that

L(x)

=

r:=o

x(n)zn for all x E !(~r,\1).

REMARK 1.2.28. Most of the results in this section already appeared in

(CHO] and [GRU]. However, the purpose of this section was to unify the proofs for the 0-0 and limit-results.

Also we have stated the results and given the proofs since it is relatively easy to

identify every homomorphism

in the Banach algebra of complex valued sequences (in contrast to the more general Banach algebra in the next sec-tion).

Finally we like to remark that the important results (A), (B), (e) (similar

o

results also hold for the Banach algebras V(~,~), V (~,~) and _{,..., r} V(~ ,~)!) can also be derived from the more general setting in the next section.

3. The Banach algebra of complex measures concentrated at [0,(0)

Let 'i' be the set of all (Borel) measurable functions ~: [0,00) -+ (0,<:0) satis-fying

~(O)

=

1 and 0 < ~(x+y) ~ ~(x)~(y) for all x,y ~ 0 • It is known (cf. [HIL]) that each such ~ has the following properties:

-1 -1

~ and ~ are bounded on (e,e ) for every 0 < e < 1

and

inf In p(x)

=

lim In p(x) x ~,"", x

x>O x~

For each ~

€

'i' let S(~) be the set of all complex measures v on [0,00) for which

IIvlI~:=

J

!p(x)jvl(dx) < 00 •

a

The term 'measure' is to be taken here in the Bourbaki sense. Alternatively, a measure v is a complex valued set function on the bounded Borel sets which is countably additive, and therefore of bounded variation, on BC[O,a])

(= the Borel sets in [O,a]) for every a ~ O. The variation Ivl can of course be extended to a countably additive (generally non-finite) set function on B([O,<:o». It is a well-known fact (cf. [HIL]) that with the usual addition and scalar multiplication and with convolution as the product operation,

S(lji), with norm U.II!p is a commutative Banach algebra.

Our main goal in this section is to derive Theorem 1.3.4, an essential tool in later developments. For this we must first prove an integral repre-sentation theorem (Theorem 1.3.2) for the homomorphisms of the algebra S(!p).

32

-Special cases of this are treated in [HIL].

[SRE]

presents a ge~eral repre-sentation, but only in the case of bounded measures without weight func-tions. Our approach is essentially Sreier's. We start with the following simple result. PROPOSITION 1.3.1 • . If

. {OO

An T:=

2-,

O n. n=

A a nonnegative measure on

[O,OO)}

and if, for every

l/J E If

and

II E T,

then

(i) (H)

S (l/J) := {v

€

S(l/J) : v « ll} ,

II

S (l/J) is

a commutative Banach aZgebra with:unit;

II

S(l/J)'" U S (l/J).

llET II

PROOF. Let l/J € If and v

€

S(l/J) be arbitra~y. Then obviously

II := GO n

l:

M...

€ T O n! n= and 'V « II •

Hence, v E S (l/J) and (ii) is proved.

1.1 ".I¥>

• QQ n

To prove (i), let I.l

=

Ln=O

A In! € T be fixed. It is easily checked that

for every bounded Borel set B

(1.1

*

jJ)(B) ... ~ ~ An+k(B) _{L. L. n!k!}

n=O

k=O

from which it follows that

jJ

*

\l «jJ and ll'« \l

*

II • Hence for every pair 'V l ' v

2 E S (l/J) we have v 1 :Ie v2 « jJ

*

jJ « ll, so

. II

v

1

*

v2 E S\l(l/J). Since evidently v1 + v2 and aV1 (a E

C)

also belong to

S (l/J), we have thus shown that S (l/J) is an algebra. Also S (l/J) contains the

II 0 II II

unit e =: A of S(l/J) , the Dirac measure at O.

We shall now show that SjJ(l/J) is complete by proving that it is isometrically isomorphic to Ll(jJ). Indeed, let v E S (l/J) be given. Define

II

PV(B)

:=

Jl/J(X)V(dX) B

Since

IIvll1/l =

I

1/I(x)lvl(dx) < a:> ,

o

the measure p has finite variation Ip 1([0,00» = IIvll,l.* Also, since v « 1.1,

v V ~

we have p «~. Therefore the Radon-Nikodym derivative v

dp 1

h

:= ~

E L

(~)

v

dll

exists and Uvllljl

=

IIh} l ' The map <Pll 1

S (1jI) 3 v - h E L (\.I)

II v

·is therefore a linear isometry. Since for every h E L1(1l) the measure

v := (h/1jJh.l is in S (1/1) and h

=

h ,~ is also surjective. This proves (i). 0

\.I v ~

Our next result gives an integral representation for the homomorphisms of the algebra S(IjI) (cf. [SRE]) •

.

"""

THEOREM 1.3.2.

Let

T

be the set of measures defined in Proposition 1.3.1

and Let

1/1

E

~

be fixed. Suppose that

{g : II

E

t}

is a coLLection of Borel

II

measurable complex functions on

[O,a:»

satisfying the following properties,

for all

II E T:

(i) g (0)

=

1 and Ig I ~ ljJ II

a.e.

II II

(H) g (x+y)

=

g (x)g (y) II x _II

a.e.

~ II II

(iii) _\.1

1 « llZ ~ gill

=

gllz III

a.e.

Then the formulas (one for each

II E T)

a:>

<L,v>

=

I

gll(x)v(dx)

o

(1)

defines a homomorphism

L

on

S(1/I)

=

Il~T Sll (lji).

Conversely, for each

homo-morphism Lon

S(lji)

there exists a coLlection

{g : \.I E T}

of compZex Borel

II

measurabZe functions on

[0,00)

satisfying

(1), (ii)

and

(iii)

above such that

(1)

holds. Moreover, the collection

{g : II

E T}is unique in the sense that

II

each

g is

determined

II

a.e.

II

34

-PROOF. Let L be a homomorphism of S(~)· U S (~). For each ~ E Tt set

~ET ~ .

L := Lis ( ). In the proof of Proposition 1.3.1 we have defined a

surjec-Jl ~ lJ! 1 . . . *. 1 . ,

tive isometry ~ : S (~) + L (~). S1nce the adJo1nt t 1S a so surJect1ve

~ ~ ~

(cL [LUE]. [KAN)) and since

Ll(~)*

may be identified with

L"\~),

there exists an f E L""(Jl) such that L

=

t*f • Observe also that IIf II

=

1 since

~ ~ ~Jl ~go

ilL II ;: 1 and 41* is isometric. Thus, with h = 41 v and g :;: ~ • f t we have

Jl ~ v Jl Jl ~ 00 <L.,v> = «=f.,v>

=

<f.,".v> •

oJ

f.(x)hv(x).(dx)

=

go

=

I

f~(x)~(x)v(dx)

a

(v E S (~» • ~ (2)

We shall now prove the properties (i), (ii) and (iii) for the functions g

~

(:"~.f).

Jl (i) is immediate since IIf II Jl<» ;: 1 and 1 ;: <L ,e> ;: g ~ ~ (0). To

prove (ii), let v .E SJl(~) 'and Bl'B2 E B([O,CIO» be arbitrary. Then

since L is a homomorpn~sm. Applying (2) to both members of (3) yields

~

II

g\l(x+y)v(dx)v(dy);:

If

g\l(x)g~(y)v(dx)v(dy)

B1XB

2 B1XB2

and therefore, more generally,

II

gp(x+y)v(dx)v(dy) ;:

Jf

gJl(x)g~(y)V(dx)v(dy)

B B

for all

B E B([O,oo)

x

[0,(0».

We conclude from this that

v x V a.e.

Since the function

~

is bounded on (e,e-1) whenever

a

< e < 1 and since

~(O) ;: 1, the measures l{O}U(l/n,n)P belong to SJl(~) for every nEE.

Applying (4) to v ;: l{O}U(l/n,n)~ and letting n +~, yields (ii). For the proof of (iii), let 1l

1,1l2 E T with III « 112 be given. Then

(3)

(4)

""

. J

g~

(x)v(dx)

=

J

g (x)v(dx)

o

1 0 ~2

Taking v

=

l{O}U(l/n,n)·~l and arguing as in the proof of (ii), we get that g~l

=

g~2 ~1 a.e.

This completes the proof of the second part of the theorem. That, conversely, given a collection {g : ~ E T}, the formulas (1) define a homomorphism, is an easy matter which we leave to the reader. Finally, to see that g is ~.

~

a.e. de~ermined for each ~ € T, observe that this is so for f J since $* is

~ ~

injective.

As an illustration of the preceding theorem we mention the following well-known result (cf. [HIL]).

PROPOSITION 1.3.3.

Let

~ E ~

be fixed and let

L(~) := {v E S(~): v « i},

where

1 is

the Lebesgue measure. Then

(i)

L(~) is

a commutative Banach algebra (without unit);

(ii)

there

is

a

1-1

correspondence between the set of

homomo~hisms L

on

L(~)

and the halfpZane

{c E ¢::m.ec

~

inf 1n t/J(x)},

given by

. ..,.. x>O x "" <L,v>

=

J

exp(cx)v(dx)

o

PROOF. Set ~ := 00 tn

I -, .

n=O n.

Then )l ( T. Since 1 is translation invariant, we have 1 * 1 « 1, hence

"" ""n

\' - , « "" • l.. n. n=l (5) o

It easily follows from this that S (t/J)

=

[e] + L(t/!) (e is the Dirach measure

~

at 0 and [e] denotes the span of e) and that L(t/J) is a closed subalgebra of S (t/J). Thus (i) is a consequence of Proposition 1.3.1.

~ .

For the proof of (ii), note that the homomorphisms of L(~) are the restric-tions of those of S (t/J), except for the one with kernel L(~). From Theorem

~

1.3.2 we know that for every homomorphism L of S (~) there exists a function

gag with g(O)

=

1 and lJ.

19l

~ ljJ 1..1 a.e. such that CD 36

-and g(x+y)

= g(x)g(y)

lJ. x l..l a.e.

<L,v>

=

J

g(x)v(dx)

o

(v

E S

(ljJ» • 1..1

Since g(O)

=

ljJ(O)

=

1 and r:=l 1n/n! « 1, (6) is equivalent to

Igl :.i ljJ 2, a.e. and g(x+y)'" g(x)g(y) 1 x 1 a.e.

(6)

(7)

Now it is clear that the function g

=

1{0} corresponds to the trivial homo-morphism on L(ljJ). We may therefore assume that 1({g ~ O}) > O. We then have

co

<L,vO>

=

J

g(x)vO(dx)

~

0

o

for some Vo € L(ljJ) •

For every x E [O,eo) let us denote by e _x the Dirac measure at x. It is not _. difficult to show that ex * Vo E L(ljJ) for all x !;: 0 (lle

x

*

vO"ljJ < CD follows

from ljJ(x+y) :.i ljJ(x)ljJ(y) Yx,y) and that the function

~

is continuous (see [HIL]). It follows that

x + _<L,ex*vO>

=

J

g(y) (ex*v_O) (dy)

=

J

g(x+y)vO(dy)

o

0

is continuous. From the fact that g(x+y) ... g(x)g(y) 1 x 2, a.e., we infer

that <D

J

g(x+Y)VO(dy)

=

o

CD g(x)

J

g(y)VO(dy)

o

for 1 a.a.x. Since <L,v

O> f 0 it follows that g coincides 1 a.e. with a

con-tinuous function. Observe next that changing g on an l-null set does not effect (7). This is clear from the proof of Theorem 1.3.2 and can also be seen directly in this case, using t *1 « 1. We may therefore assume that g is continuous. But then g(x+y)

= g(x)g(y) holds everywhere and it follows

that g is of the form g(x)

= exp(cx) for some c E

C (this is well known and