AN INVESTIGATION OF UNIFORMLY CONVERGENT

0‘.

«n

”WM"

POWER SERIES ON THE CLOSED, UN” DISK";  ;

Thesis for the Degree of Ph. D, ' "
MICHIGAN STATE UNIVERSITY ' '
LOUIS THURMAN RICHARDS 7

1970

{1113124121} L.
Michigan State '
University

    
     

mm}?

This is to certify that the

thesis entitled

AN INVESTIGATION OF UNIFORMLY
CONVERGENT POWER SERIES ON THE
CLOSED UNIT DISK

presented hg

Louis T. Richards

has been accepted towards fulfillment
of the requirements for

P C .
h D degree in Mathematlcs

fl7§5fl

/ I Major pfigr

Date July 17, 1970

 

0-169

ABSTRACT

AN INVESTIGATION OF UNIFORMLY CONVERGENT
POWER SERIES ON THE CLOSED UNIT DISK

By

Louis Thurman Richards

Two facts are immediately known about a given power series

with radius of convergence R3]:

(l) the series converges absolutely for Ill 1, and

(2) the series converges uniformly on izi1p l,

Included in the class of all such power series are two subclasses:
(1') those power series which converge absolutely on Izl=l, and
(2’) those power series which converge uniformly on iziij.

The class of all power series obeying (I') has been
extensively investigated“ However, the class of all power series
obeying (2') has not been adequately investigated;

After showing that U, the space of all power series obeying
(2'), is a Banach algebra, this paper investigates some of the func-
tional analysis properties of the space“ The investigator was
also interested in finding classes of functions,¢, such that the
composition of any power series in U with i would again be in U(

The following are typical results from the study:

Theorem: If {ak} is a sequence such that.‘Z:akzk is in U,

and if A is a complex number such that if 0, i # ak k = 0,l,2, 0.,

Louis T. Richards

then {§E%XJ is a sequence whose terms are coefficients of an element
in U.

Theorem: If X is a sequence space which is a Banach algebra
under coordinate—wise multiplication, and has a Schauder basis,
then prOjections into the coordinates are the only non zero homo-

morphisms on X.

Theorem: If ¢(z) - a + (l-a-e)z , 0<a<l, O<B<l, then
l-Bz

 

for any f in U, foo is in U.

AN INVESTIGATION OF UNIFORMLY CONVERGENT
POWER SERIES ON THE CLOSED UNIT DISK

By

Louis Thurman Richards

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements

for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

I970

r" J. ”M If
_..- 3. Q Q [5

/. g9 “ ~7/

T0 JEAN

ii

ACKNOWLEDGMENTS

The investigator wishes to acknowledge the suggestions
and guidance which Dr. William Sledd rendered in the preparation
of this manuscript. The investigator is also very thankful to
Miss Linda Taylor who typed the manuscript and proved to be

very helpful in proofreading the final copy.

TABLE OF CONTENTS

INTRODUCTION ........................... 1
CHAPTER ONE: UNIFORMLY CONVERGENT POWER SERIES 0N|z|_<_i ..... 5
l. The Banach Space U ................... 5
2. Some Spaces Equivalent to U ............... ll
3. The Banach Algebra U . . . . . . . . . . . . . . . . . . l6
4. The Dual Space ..................... 29
CHAPTER TWO: CONTINUOUS LINEAR OPERATORS ON U .......... 37
1. Concepts From Summability ................ 37

2. Bounded Linear Operators From A Banach Space X
(With Schauder-Basis) To U ............... 38
3. Karamata Type Operators on U .............. 43
4. Bajahski Type Operators on U .............. 6O
OPEN QUESTIONS .......................... 78
BIBLIOGRAPHY ............. i .............. 79

iv

" INTRODUCTION"
If Zanzn is a given powersseriesmwith radius of convergence
greater than or equal to 1, then
(l) Zanzn converges absolutely for Iz I< l . and
(2) Zanz" converges uniformlyon I zI :_ o <. I

Now if [anzn has radius ofconvergence .R> 1. then

(T ') Z anz" converges absolutely onl zl - l, and. ' hence
I n
(2 ) Zanz. converges uniformly on Iz Igl.

However, a series need not have radius of.convergence greater

than l in order to satisfy.(l'). .TheseriesE—z—g- has radius
n

1 l/n
of convergence l [since lim (—12) = I] . and it Is absolutely
n

convergent on i z I = l.

The space of all power series obeying (l') can easily be
identified with [1 . In fact, letting f be a typical power series
which obeys (l') and denoting the norm of f by Hf l'l -X|anl .
the map {anIT f is an isometric isomorphism from [1 onto this

space. The space of all power series obeying (l') is generally given as

an example of a Banach algebra, and theorems are proved about it
in books dealing with functional analysis and Banach algebras.

The space of all power series obeying 12') is not so very
well known. This paper will deal with the space of all power series
obeying (2') as a Banach algebra, its dual space, its Gelfand trans-
form, and continuous linear operators mapping'theispace into itself.

Kahane and Katznelson [l4] proved that the space of functions
satisfying (2') is not an algebra under pointwise multiplication.
Although the investigator was well into this paper before seeing
their results, the notation for the space is nearly identical [they
denoted the space of series obeying (2') by UT, and it is denoted

in this paper by U], and the norm used is the same:

(3) Hill = sup sup 1:6ka I -
p izlsl =0

The only mention which the investigator has seen of the space of
series obeying (2') as a Banach space occurred in the above paper.

It will be shown in this paper that U is a semi-simple commuta-
tive Banach algebra under coordinate-wise multiplication, and, hence
(3) is essentially the only norm that can be used.

Another author who has written about series obeying (2')

is Alpar [3], and he has proved the following three theorems:

Theorem l: Given a fixed point a (O<I&I< 1), then one can
always find a function fi(z), which is holomorphic in I2I<l and
which has an absolutely convergent power series such that the

power series defined by - 2::
i6 :2) = 11(2) .. bk“)zk
.- a 4

 

3

is not absolutely convergent on izi = 1.

Theorem 2: Let f1(z) =Z-akzk be holomorphic in |z|<l,
Zlak|<co, a (O<|a|< l) a fixed point, and lzll = 1, I22] =l

two points related by 22 _ a

 

Then the power series obtained by the transformation

mar-42:31.2“) =k(§:b~1)

is uniformly convergent on the circumference IzI=l and
- k _ - k
-Zakz - f2(zz) -E bk(a)z

Theorem 3: There exist functions fl(z) holomorphic in Iin l.

 

whose power series development Zakzk converges uniformly but not

absolutely on IzI = l and which are changed by the transformation

- O.
f;(—:—.—3-.:—) into a function f2(z) whose power series {bk(a)zk
is not absolutely convergent on the circumference izi = l for no
value of a (O<Ia|<l).

There are both propositions and theorems in this paper.
Propositions will refer to those mathematical truths whose proofs
are fairly elementary. Theorems will refer to those truths which
demand some care in proving. In each chapter, lemmas, theorems,

propositions, and corollaries have been numbered consecutively without

regard to their special characters.

Chapter I deals principally with the functional analysis
consequences of U being a Banach algebra. In order to facilitate

the investigation, u will be identified with three spaces:
a 2 ‘ = int 0 THE . .
i4) U.I {f(t) Zane . Zane converges uniformly }

(5) U2 -{ x = (an? : KaneInt is afunction in U1}
(6) U3 = {5(tm) =I§ake1kt :2:akeIkt is in U1) .
Under appropriate norm, all of these spaces are isomorphic-iso-
metric to U.
Chapter 11 deals with continuous linear operators on U.
The major theorem of Chapter II states that Karamata type functions,

i.e., functions of the type
¢(Z) = O. + (1 -a-B.) Z ; (0< C1<]), (0 <5 (I),
l - B 2

 

Operate on U under composition of functions. This result is used
to show that certain BaJanski [4] type functions also operate on

U,

The following usages and n0tations fhave been employed
throughput the investigation:
a) infinite series and sequences whose indices begin with O

or an appropriate positive integer have been written without
the index
I'M

b).1$ will denote the space of functions obeying (l')

c) B[X,Y] will denote the space of all continuous linear
operators from the Banach space X to the Banach space Y

d) X' will denote the dual space of X

e) coordinate-wise multiplication will be denoted by " *".

CHAPTER ONE

UNIFORMLY CONVERGENT POWER SERIES ON Rléjl

l. The Banach Space U

In this chapter. some basic theory concerning the structure of
the class of uniformly convergent power series on |z| ; I will be de-
veloped. It will be shown that this class forms a Banach space under
appropriate definitions of addition, scalar multiplication, and norm.
Furthermore, it will be shown that the space forms a semi-simple
Banach algebra under coordinate-wise multiplication. The Gelfand
transform of the space will also be investigated, and additional
information about its structure will be obtained. The dual space
will also be briefly investigated.

The goal of this section is to establish that the class.

(1) U = the set of all uniformly convergent power series on| qé=l,
is a Banach space under very natural conditions. Since each j[:akzk
which belongs to U defines a unique function, f, which is holomorphic

k, the

in [2|<J and whose power series devel0pment is precisely 2E:akz
elements of U will be denoted by these f's.

On the set U, addition is defined by the rule: if f1, f2 are
elements of U, then f1 + f2 is the element of U defined by

(f] + f2)(2) = f](z) + f2(z) for all [4; l.

6
If ais any complex number, and er, then of is the function de-
fined by (af)(z) - a-f(z) for all |z|:;l. The set U clearly becomes
a linear space over the complex numbers under these definitions.

If er, f(z) -:akzk, define IIfll by

n
(2) Ilfll = sup sup lSn(z)I . where Sn(z) = 'EILakzk.
n |zl_<_l k=0
Lemma 1.1: The function,|l ll, defined by (2), is a norm

on U.
3399:: Let f,g cU, deC, [C denotes the field of complex numbers].
It must be shown that:
i) ||f|| < (x) .
ii) ||f|| = 0 iff f is the zero function.
iii) llafll = Idl Ilfll.
in II he |l< Ilfll + ugh.

Since ii), iii), and iv) involve only direct calculations, only

i) will be verified.

Verification (i): By definition. Ilfll = sup su Is (z)I
n lzigl n

 

n
where Sn(z) . g akzk. Also Sn converges uniformly to f on lzl_<_l.

We note that Sn is an entire function for n= O,l,2,.... Hence,

su IS (z)l - Sn(c) . for some t satisfying ltl a T. New for
IZI<T " -

each n, choose a point on Izl a l where Sn attains its supremum and
call it tn. Since f is continuous on lzI;_ I, there exists M such
that sup lf(z)| < M. Since Sn converges uniformly to f on Illéj,

z_<_l

there exists N such that n>N implies ISn(z) - f(z)|<l for all lzlgj.

7

Hence, n,N implies that |Sn(z)I<l + If(z)| for all IZIéj. Therefore,

S = S . T+M=M', ”’f n>N. Hence, 5 5 IS 2 <00.
Tgp=1 | n(z)| I n(tn)|< i ngfi IZUEJ ( )I

It follows easily now that IIfII = sup sup 1 IS (z)I<09 since there
Ill 3
exists M" such that sag |S (c )I<M".
. n< n
In addition to proving that II II is a norm on U, the verifi-
cation of i) has also shown
3 f = su su '3 z '.
() |||l npmpfllnfll

Theorem l.2: The linear space U is a Banach space underII II
as norm.

3199:; Since (U,II II ) is a normed linear space, it suffices
to show that every Cauchy sequence in (U, II II) converges. Let tip}

be a Cauchy sequence in U. Then, giveneo-O, there is an N such that

 

i" k
(4) Irfp-quI=sup sup I [Cup-{)2 I<e ifp,q>N.
n IzI=l k=0 k

n
Hence, for any n, IEME - aEII 3; Hip — quI<c if p,q>N.
keO

Given m, we have,

Iap - an -éflfiap - an< Itai - an<5 if p, q>N, and therefore.
m m TKO k k'= k=-0 k

(5) ME - ails. 2e if p.q.>N..

We now have Ia?“ is a Cauchy sequence of complex numbers for each k.
p=0
Hence, there is an ak such that

(5) lim ap = a (Uniformly in k).
p k k

Now define f by f(2) = IZ:akzk. We claim that 2:akzk converges

for all Izl < T. To see this, we note that if {ak} is bounded, say
k k

by M, then for any Izol<l,2 Iakzo|< MZIZOI < 00, hence.

2::akzk converges for any |zI<l. To show that {ak} is bounded,

observe that {ap}OD is bounded for any p. According to 5,
k=0

given 5 = 13 there exists N such that p'>N implies that

- P' _
lakl Iak + ak ak

< pl + - pl
=Jak I lak ak |

pl
<Iak I + 1.

Since sup IaE I is finite, sup Iakl is finite.
k k
Thus f is a candidate for the limit function of fp. To see
that fp converges to f, observe that by (4),
n q k
sup sup IE (aE - ak)z |< e if p,q>N.
n IzI=l k=0

Fix 2, n, and p>N and let q+a9. Then

n
p
IZZL(a - a )zkl < 8-
k=0 k k

Since this is true for any Iinj. n, and p,

(7) Ipr-fII = sup sup IEEf(ap-a )zkl < e if p>N.
n |z|=l k=0 k k

It must now be shown that er. But,

m k m p k m k
| E .,akz | =| 2::(ak-a )z + 25::apz I
=n k=n k k=n

3

k m k
; Izaak-app I+Iz."_...akz I .
k k=n

k=n

Let e>0 be given, then, by (7), there is an N such that p>N implies
that

 

r p k
sup I}E:(a - a )z I < 5-.
r,IzI=l k=0 k k 3
Choose p>N and fix it. Then
m p k ”'1 p
sup I§::L(ak-ak)z I:; sup IZZ: (ak-a )z I
Iz|=l k=n |z|=] k=0
(“"1 )
+ sup I2,” a - ak z I
Izl=I K=O k
< E_ E = 2a
=.3 + 3. 3

Since fp e U, there existsan M such that m, n>m implies that

Therefore, if m, n > M, then

m k
sup IEZ: a z I < 6.
IzIz] k=n k

Hence, ch. This finishes the proof of Theorem 1.2.

10

Corollary 1.3: Define the collection {pk} by p (f) = ak.

 

k

where ak is the kth_coordinate of f, then the set {p } is an

equicontinuous family of functionals on U. Hence, 1: particular,
for each k, pk is a continuous linear functional on U.
3599:: It is obvious that each pk is a linear functional. It
follows from (4) and (5) that given c>O, we can choose a = 3-50
that if IIf—gII<6 = %-, then ka(f) - pk(g)I = Iak-ka < e for
each k.
Proposition 1.4: Let ek, k = 0.1.2,... be the functions
(2) = zk. Then {ek} is a Schauder-basis for U.

 

defined by ek
Proof: Let er, f(2) = 2::akzk. Let e>O be given. Since

er, there is an N such that

n k
sup sup ll 2:: akz I < e if p>N.

n>p lz - k=n
Therefore,
IIf-giaell = sup sup IVLazkI<e
tea M an |z|=i i336 k
Hence,
f = 2::akek.

It is interesting to note that one merely observed the
behavior of f at z=1 in some of the most crucial steps in the
proof of Theorem 1.2, and that the norm used on U is analogous to
the norm used on the space of convergent series. Hence, a natural
question to ask is, "If Zak converge, does {Zakzk converge uni-
formly on Izl;l?" Although Zak converges implies that fiakzk
converges uniformly on IzL;p<l, it does not imply uniform convergence

on IzL;l as the following simple example shows:

11

(8) 2::3-11E

k + 1

k k
The series defined by (8) converges; however, the series ELL—‘1 2
k+l

does not converge at z = -l; a fortiori,it cannot be uniformly con-
vergent on Iz lg] .

Since, the elements of U converge uniformly on Ingl, one
may naturally ask, "Does er imply that the radius of convergence
of f is R>1?" The answer to this question is, "No, there are elements
in U whose radius of convergence is 1." One way to see this is to
observe that 22%; converges absolutely, and hence, uniformly

on Izléj, but the radius of convergence is 1.
2. Some Spaces Equivalent To U

If X is a Banach space, then we will say that X is equivalent
to U provided there exists a continous linear operator A such
that it maps U one-to-one and onto X, and such that IIA(f)I IX= IIfI I.
That is, x is equivalent to U iff there existsan isomorphism be-
tween X and U which is also an isometry.

The purpose of this section is to develop some theory about
U by looking at different ways in which it is possible to describe
the class of uniformly convergent power series on IzL;l as a Banach
space while maintaining the norm which was defined by (2). Of
course, if X is equivalent to U, then any information which is ob-
tained about X can be easily translated via.A to information con-
cerning U. Since this can obviously be accomplished with no effort,

there will not be a need to specifically translate anything.

12
Since by (3), if f belongs to U, then IIfII = sup sup ISn(z)I,
n I2 =
the norm will be restricted to the unit circle T.

The spaces which follow are all Banach spaces under the norm

defined by (3).

(9) U1 = the set of all f such that f(t) = Zakeikt, and iake‘kt

=0
converges uniformly to f.
(10) U2 = the set of all x = {ak} such that 2::ake1kt is a

function in U].

Let N denote the non-negative integers with the discrete

topology. Let N denote the one-point compactification of N.

(11) U3 = the set of all functions, S, on TxN of the form

n . .
S(t,n) = Z ake1kt , and Zakelkt
k=0

is a function in U].

k
Define r: U +U] by r(f) = 9, where f(2) = 2::akz and

9(t) = f(eit)3 TiU‘I")U2 by T(g) = x, where g(t) .-. Zakelkt
and X = {6k}; and WIUTI U3, by WIQ) = S. where g(t) = ZELakeikt

and n -

S(t,n) = ii: a e1kt.
k=0 k

It is a trivial matter to check that FsT, and v are isometric

isomorphisms. Hence, the following proposition is stated without

proof:

Proposition 1.5: U1, U2, and U3 are equivalent to U.

13

Proposition 1.6: U], U2.,and U3 have Schauder-bases.

 

3399:; The proof follows immediately from Propositions
1.4 and 1.5.
From the mappings defined prior to Proposition 1.5, the
following facts are obVious:
'1. A Schauder basis for U1 is given by the set of
functions defined by ek(t) = elkt'
:2; A Schauder basis for U2 is given by the set of sequences
of the form ek = (0,0,...,O,l,0,0,...) with 1 in the
kth coordinate .

3. A Schauder basis for U3 is given by the set of
functions defined by

( ) 0 if k3 n
e t,n = ,
k elk’c if kf__n

Proposition 1.7: If {akle U

 

2 , then { ak}e U2, where ak
denotes the complex conjugate of ak.

Proof: Since {akle U , given n> 0, there is an N such that

2

sup IIE:,akeiktI < n if p,q > N.
t k=p

Choose p,q >N, and fix them. Then

sup IEE‘a'keiktI sup If; akeTikt I
t =p t k=p

W

supI I; akelkt
t =p

 

 

I

14

Since IE-I= IzI, Proposition 1.7 now follows from the last
equality.

It is well known that the complex series Zak + ibk
converges ( converges absolutely) iff the two real series
2:ak and IIbk converge (converge absolutely). The following
corollary follows directly from Proposition 1.7 and the fact
that U is a vector space over C; hence, it is stated without

2
proof:

 

Corollary 1.8: {ak + ibk} a U2 iff {ak} and {bk} eU2 .

U3 is an interesting space and seems to be the "natural"
space which one should use in investigating uniformly convergent
power series. Note that TxN’is a compact Hausdorff space, and
that U is at least a subset of B(TxN), where B(TxN) is the class

3
of bounded functions on TxN:

Theorem 1.9: U3 is a closed subspace of C(TxN), where

C(TxN) is the Banach algebra of continuous functions on TxN

 

with Sup-norm.

Proof: Since U3 is a Banach space under sup-norm on TxN,
and, therefore, is a closed subspace of B(T§N), it suffices to
Show that u is a subset of C(TXN).

3
Let Se U3, and let a be any open set in C. It must be
shown that S'][n] is open in TxN, where S'I[n] =[ (t,n): S(t,n)en} .
Case 1. S'1[n] = a ; whichis open, and we are done.

Case 2. (t0,n0)e S'I[o] with "o #09. In this case,

15

n
S(t ,n ) = 2:3:a e1kt° . Since S(t,n ) is continuous in t,
o o k=0 k 0

there is a neighborhood V 0 about to such that S(t,no)en for

t
all t a Vto. Hence, Vto x no is an open set about (t0,no) which

is contained within S'][o].

Case 3. (to,CX» e S'][n]; S(to,czbe Q . Since Q is open,
there exists a 6>0 such that Va = it: IS(to,0©) - c|<6} is
a subset of 9. Since S(t,0©) is continuous in t, there is an
open neighborhood Vto about to such that lS(t,00) — S(to,00)l< $52-
for all ttho. Since S(t,n) converges uniformly to S(t,6o),
there is an N such that, for all n>N, sup|S(t,ct» - S(t,n)l< g-.
A fortiori, |S(t,oo) - S(t,n)l< %-for all teVt and n>N. Finally,

0

if t e V and n>N, we have
to

IS<to.oo) - S(t,n)!s. |S(to.oo) -- S(two)! + IS(t.0o) - S(t,n)!

6 6
5‘ 7+735.
Hence, (t,n) c Vt x {n}"3 implies that S(t,n)e V6 which is a
0 n=N+l 11)
subset of 9. Therefore, Vt x {n} “‘ is an open neighborhood
0 n=N+l
about (to, a) which is contained in S‘1[Q].

From the above three cases, it follows that if Q is any
Open set in C, and if S a U3, then 5-][9] is Open in Txfil Hence,

U is a subset of C(Txfi).

3

Corollary l.lO: The norm,|| ll , originally defined on U

as sup sup lSn(z)l can now be redefined as
n Izlgl n 'kt
(l2) Ufll - max Iii: ake1 I, where maximum is used
(n,t) k=0

l6
in the sense that the value n =<z> is admitted.

Proof: Since TxN is compact, and since continuous functions

on compact sets attain their supremum, we have for SeU , there is

3

a p01nt (to,no) such that sgpn)|8(t,n)| = |S(to,no)|.

3. The Banach Algebra U

Since C(Txfi) is a Banach algebra under pointwise mul-
tiplication of functions, one is tempted to conclude that U3
is also a Banach algebra under pointwise multiplication. However,
this is not the case. Kahane and Katznelson [l4] proved that
U1 is not an algebra under pointwise multiplication; hence,
U3,U2, and U are not algebras under this definition of multiplication.
In fact, this seems to be an unnatural way to define multiplication
on U .

2

In this section, it will be shown that U2 is a Banach

algebra under coordinate-wise multiplication. This multiplication

corresponds to convolutions on U].

Lemma l.ll: Let f, g e U , and define f*g by

 

1

2n
(l3) f*g(t) = 5%. I; f(t-u)g(u)du.
0

Then the function ffg e U].
Proof: It is well known--and it is easily proved via Fubini's
theorem--that the Fourier coefficients of h(t) = f*g(t) are given by

h(n) =cn= f(n)g(n) = anbn. How an = O for n<0 implies that cn = 0

l7

forin<0. Hence, to show that h 5U], it suffices to show that

Zakbkeikt is uniformly Cauchy.

ll

(14) liakbkeiktl =5 Ii: bkeikt 1 f(u)e'“‘“du I
k=p k=p 2n

 

 

TT

5.. ‘ ' |f(u)l If; bke‘kit-ull du
2 =1)
0

.5. sup lf(U)l suplzq: bke1k(t'”)| -
u u k=p

Since sup UEZbkeik(t'u)l = suplEE: bkeiku I , the right hand
u k=p u =p

side can be made arbitrarily small by choosing p sufficiently

large. Hence, it is immediate that h 2U];

Corollary l.l2: LEtALbe a complex Borel measure on T. Define

 

the sequence {ck} by

ck = Se'm am), k=0,l,....
T

Then {akckleU2 for all {akla U2.

Proof: Sinceinis a complex Borel measure an T, the total

variation of u,lul(T) is finite. Repeating the argument used in

I :akckelkulglul (T) sapliakeikw'w
=p t k=p

(14) yields

Corollary l.l2 follows immediately' from the above inequality.

l8
From Corollary l.lZ, one has the fact that if f eU1 and
g eLI, then f*g 2U]. This follows from the fact that the measure,
u , defined by I

u(E) = “J. 9(t) dt

E

is a complex Borel measure and that du = 9 dt.

 

Corollary 1.13: If f,g Eu], then IIf*gII;JIfII IIgII.

Proof: In (l4),let p = 0. Then

Iiakklbe H; sup|f(u Isup 129.;y be‘kti

.; IIfII IIqII
.3. = ..
Hence sup I k__ akbke 91ktl= IIf*qII ‘IIfII IIQII~
(q,t) <=0

 

Since it is obvious that U2 is a commutative ring under
addition and coordinate-wise multiplication, we have proved:

Ineg;§m_l.l4: U2 is a commutative Banach algebra under
coordinate—Wise multiplication.

Covollary 1.15 If {akla ”2’ then {Iakizla ”2°

 

Proof: Th1s follows immediately from Proposition l.7 and
Theorem I l4 since aiak = Iak I2

The converse Ef Corollary 1 I5 is not true since the sequence

 

defined by ak = does not belong to U2 while {IakI2}e U2.

19

Qefinition 1.1: If R is a commutative ring without identity,

 

‘then an ideal, I, of R is called a regular ideal provided there
exists UeR such that ux - XeI for all x in R.[u is called an identity
modulo I].

Definition 1.2: If R is a commutative ring, then it is

 

semi-simple iff the intersection of all its maximal regular ideals
is zero.

A well known result is that every regular maximal ideal in
a commutative ring R is the kernel of some non-zero homomorphism

from R to the complex numbers.

Theorem 1.16: If X is a space of sequences which is a

 

commutative Banach algebra under coordinate-wise multiplication
and has the set {ek} as a Schauder basis, then projections into
the coordinates are the only homomorphisms, and, therefore, X
'is semi-simple.

3399:; Since the system {ek} is a Schauder basis for X,
X contains the set of all finite Sequences.

Let h be anyznon-zero homomorphism of X into C. Then h is
a continuous linear functiona1,and h(x*y) = h(X)h(y) for all
x,y aX. Since x ={ak} , y = {b } can be written as

k
x drakek and y ==2:_bkek , we obtain

(15) ZakbkMek) = Zakh(ek) 2: bkh(ek)

for all x,y ax.
Let h(ek) = Ak , k = 0,1,... . Given n, let
xn = yn = (l,1,l,...,1,0,0,0,...), where 0 is in the n+p th_

coordinate for p = l,2,3,... . Hence’by (15)

20

 

n ,4. 2;
A = Iv A X
E0 k ‘56 k =0 k
_n_ n
Therefore, for any n, ‘1 Ak =0 or Ak = 1. Now Z: Ak - 0
IRE-6 k=0 k=0

 

for all n would imply that h was the zero homomorphism. "Hence,

E A = l.

:0 k

min {nz : A =1} .
.0 k

0 for k <N. It will be shown by induction

there existsn such that

Let

2
ll

Then A = l, and Ak

that A 0 for k >N.

Let x = y be the sequence defined by
IT ifk=N

ak=<’l/21'f kc N+l

 

L0 otherwise

. _ . 2
For the sequence x - y above, (15) yields -3 AN+1 = (AN+1) .
Since AN+1 must be -1 or 0, the above equality yields that it
must equal zero. Assume that AN+q = O for q <p. Then

AN+D must be -1 or 0. Let x = y be defined by

1 if k = N

ak = 1/2 if k= N+p

0 otherwise

21
Using (15) again, we obtain 1 + *l'AN+p = (1 + i-AN+p)2. Hence,

AN+p = 0. Since 4
0 if k f N
h(ek) =
1 if k = N ,
h(x) = PN(x) = aN. Projections into the coordinates are clearly

homomorphisms. It has now been shown that these are the only ones.

Define M by
k Mk=IX€X: ak30}.

Obviously, Mk is the kernel of P . Hence, Mk is a regular

k
maximal ideal for k = D,l,2,... . The intersection of all
regular maximal ideals is clearly the zero sequence. Hence X
is semi-simple. This completes the proof of Theorem 1.16.
It should be noted that one need only to have shown that
the set of projections was a subset of the set of homomorphisms
on X in order to have the fact that X was semi-simple. Hence}
the interesting part of the theorem is the fact that a characteri-
zation of the homomorphisms on this class of Banach algebras

is obtained.

Corollary:1.17: U2 is semi-simple.

 

Proof: The proof follows immediately from Proposition 1.4

and Theorem 1.16°

 

Corollary 1.18: The set 1% is an ideal in U2, but it is

not contained in any regular ideal.

Proof; If x,yeII, then x - 3/64. and if xa4 and yaUz, then
x*y = {akbk}€/1' Hence [1 is an ideal. To show that (1 is not

22
contained in any regular ideal, merely note that H/kflel,
and that no term of the sequence is zero; it now follows that
"1 is not in any regular ideal since all such ideals are con-

tained in maximal regular ideals.

From the fact that U2 is semi-simple, one can obtain more
information about the norm which has been used. In a paper on
F-H spaces, Nilansky and Zeller [25] gave a very short proof
of the fact that a commutative semiwsimple Banach algebra,A, has
an essentially unique norm under which A is a Banach algebra.
Hence, the norm, II II, defined on U2 is essentially the only

one that could be used.

Definition 1.3: In a commutative ring. an element x is

 

said to have an adverse y iff x + y - xy = 0.

It is a well known result that an element x in a com-
mutative ring has an adverse in the ring iff it is not an identity

modulo any regular maximal ideals. Using this fact, one obtains:*

Proposition 1.19: If {a }eU , ak f 1, for k =0,l,2,...,

k 2
a
then -——-—-l5——-— 5 U2.
ak - 1

Proof: For x = {ak} to be an identity‘mOdulo Mn’

X*y - y must be in Mn for all yeUZ. But this means that anbn 'bn =0

for all yaU , and this is true iff an = 1. Since ak # l for

2
any k, x is not an identity modulo any regular maximal ideal.

Hence, there is an element ch7 such that

X + y ‘ X*y = (09090909090-0)

23

Hence, ak + b - akb = 0 for k = 0,1,2,.... Therefore, y

k k
is given by the sequence

Let XEUZ. The Gelfand transform of x, denoted by’?, is a
function defined on, a ={ Mk : k = O, 1, ...} , the collection of

maximal regular ideals of U2 by
(16) 9(Mk) = n where n is a co-set of Mk and mm.

It is well known that if M is a regular maximal ideal in a
commutative Banach algebra X, then the quotient algebra, X/M , is
isomorphic to the field of complex numbers,C. In the particular

case X = U2 and M = Mk’ an isomorphism can be exhibited explicitly:

 

Proposition 1.20: Let Mk be a regular maximal ideal of U2.
k _ k .
Let IIn - {X8U2. ak - n}. Then Uz/Mk ={ 11n .neC }, and the
mapping 11k is an isomorphism ofg onto C
—> n 0
. n I //‘Ik

Proof: Elements x,y clearly belong to the same co-set of

Mk iff ak = bk, and it is equally clear that for any complex

numberiIthere is an XeU2 such that ak = n . Hence, it follows
that {US : ncC} is precisely the collection of co-sets of Mk'

From the way addition and multiplication are defined on Uz/Mk.

and from the fact that the co-sets of Mk are disjoint, it follows
that the mapping n: + n is an isomorphism.

 

24

The Gelfand transform of an element XeU2 can now be described

more completely by using Proposition 1.20:

(17) x(Mk) = ak where a is the kth_coordinate Of x.

k

Definition 1.4: Let T1 and T2 be two topologies on a set

X. Then T.l is coarser than T2 if T1 is a subset of T2.

Let X be a commutative Banach algebra without identity,
and let 1 denote the collection of all maximal regular ideals Of
X. Let T be the coarsest topology on i such that all the Gelfand
transforms are continuous on i. A subbase for the topology is

given by the sets
(18) Wm] for all Open O in c, and for all x.

This topology makes 1 a locally compact Hausdorff space. The
functions 2 have the property that given e>0, there is a compact
subset K.Of i such that |x(M)I<e for M not in K. (such functions
are said to vanish at infinity). Let Cb(fl) denote the collection
of all continuous functions on i-—with the above topology-~which

vanish at infinity.

Applying the above facts to U2 yields:

 

Proposition 1.21: QDGI) is isomorphic to the space of null

sequences.

‘Proof: Since (%(A) consists of all continuous functions
which vanish at infinity, it suffices to show that the topology which
is defined on.n must be the discrete topology. If the topology on

a.is the discrete topology, then n.is homeomorphic to the natural

25

numbers,N, with the discrete topology, and the continuous functions
on N which vanish at infinity is precisely the collection Of null
sequences. To show that the topology on A is the discrete topology,
it suffices to show that each Mk is Open. From (18)

{Hm =1 Mk: Q(Mk) am
will be an Open set--since it is a member of the subbase——for each
x in U2 and open set 9 in C. Let Mn ea . Then for en eUZ, let
9 be an open set in C such that 1 belongs to Q and 0 does not.

belong to 9. Then,

@n'IIO]

I Mk: €n(Mk)€Q}

= { Mk: akefl , 9n = {am} } by(17)

{M } ,
n

since en = (D,0,0,...,0,l,0,0,...) with l in the n th_ coordinate)
Hence, the coarsest topology on A for which all the RS are continu-
ous is the discrete topology.

One of the cases which one looks for in the Gelfand repre-
sentation of a commutative Banach algebra X is the case When
the collection of transforms equals C0(i). Lettinqfi2 denote
the collection of all Gelfand transforms of members of U2, (17)
’u‘

and Proposition 1.21 yield the fact that 2 is a proper subset

of CO(A).

Definition 1.5:' In a commutative Banach algebra,A, without

 

identity, the spectrum of an element XeA, o(x), is defined by

(19) 0(x) ={AcC: A # D, and (l/A)x does not have an adverseIL/‘Uli .

26
It is well known that the range of‘? is either identical to
the spectrum of x, g(x), or it is 3(x) with the value zero removed,
For U2, this means that g(x) = {a }. This yields the following

k
proposition which is an improvement upon Proposition 1.19:

 

Proposition 1.22: If {aKFeU2, and AcC,A # 0, and A? ak
for k = 0,1,2,..., then

' a
§:-——£———£} eU2 .
ak' A

Proof: x = {ak} , and Af 0, A # ak implies that A does
not belong to 0(X)e Hence. by (19), this means that (i;x)x has

an adverse in U2. This adverse is obviously given by

'___ii_<____
ak — A

Definition 1.6: The multipliers on U], M(U]), is the set of

 

all g such that the pointwise product, f(t)g(t), is in U1 for all f
in U].

N

Let./

1 denote those elements of U] which have absolutely
convergent series. Kahane and Katznelson [14] have proved that

A

.2] is a proper subset of M(U]). Nevertheless, the following theorem

a’\.

yields a more direct proof that [I is a subset of M(U]).

lheggem 12;: 21 is a subset of M(U]),

Proof; Let f(t) =Zake1kt be in 1’ and g(t) =£bke1kt be
in U]. To show that f(t)g(t) is in U], it suffices to show that
given €>O, there is an M such that p,q>M implies that

sup IS(t;p,q)I = sup IZE::eInt a.b _ I < 6.
t t nap z:;; k n k

27

. av 11 “ . ~.’,_ ,r
S(t;p.q) = 2;, X ake‘ktbn_ke“l IV“
n=p keU

m

g 1
_-ake]kt ii b e‘n " e1Kt b eint

(:ucutjr 359-}

KP‘I 11'

7.
0

=1; +22.

11— '- k , "
2:} 1‘2“” al<eikt . bne1nt + ,f::vbakeikt21:£' eint
k-O ‘ ”M -“

n=p-k =Nr1 n- p1

.1312

Without loss of generality, one can as5ume that neither
f nor 9 is zero. Then there exists M1 Such that N>MI implies that
(Y)

Z: IakI< 513119;,

k:N+I

There exists M2 such that m‘M2 implies that

<:';1|r;l_-'~‘ ‘ .‘ . “ . I . V ,7“... ‘
sup ,r_ ,b e.nL. a/J Ii1 (,5 9I is ,f norm;
1 1'"? n 'l ' ' I ‘1
, l

Laacse M 4 mas1H,,M2)- Choose NO: M, and fix it Let M = M + N

Vt.- q’" f “.3310 -1
12“,,2 SUP 12:.“ Dnelny‘ gwms‘ 1'de
pilzk;q n-U k—p+1
All
1 I fits—"31"“ "
:_ 1 5'11 1......» IaK;
kepil
2/3 ,

sinep+1 M 7N0“ M .

28

Since N >M],
o

-k _ .
12:121. sup I bnemt-12i1ak1

N0 +1<k ép n=p-k k=NO+1
00
Q1911: 1a,)
k=NO+1
< 5/3 .

Now, if K<NO, then -k> -N0. Hence, p—k>p-NO. But
p-NO > M3 > M2. Hence, k<:NO implies that p - k > M2. Therefore,

-k .
sup sup IZEZ: bne1ntl< c/3IIfII].
t n=p-k

Q_k:NO

Hence,

N
n ”—9-

I}: I; 5UP supI bemtl‘t IaI

‘1 Oékého t n=p- '1 1256 k

;: sup sup I}E::kbn e1ntII IIfII1

QékéNo t n: p-k

 

6/3

. ‘T‘
Since S(t;p,q) = [L11 +'2E;2 i'EZT 9

SUP |S(t:p.q)I < e if p > M.
t

This completes the proof of Theorem 1.23.

29
4. The Dual space

Since U is isomorphic-isometric to U],U2, and U the dual

3’
space of U, U', is isomorphic-isometric to the dual spaces of U1,

U2, and Uj—denoted respectively by UI, Ué, and U3. That is, if X

is U], U , or U3 and r is an isomorphic-isometric mapping of U onto

2
X, then the map, 1, defined by

(20) w(L) = T where T is defined by T(f) = L(P(f))
is an isomorphic-isometric mapping of XI onto U'.

Proposition 1.24: The members of Ué are uniquely determined by

 

sequences {ck} which have the propertylz:akck converges for all {ak}

in U2.

Proof; Let L be a continuous linear functional on U2. Let x

be any member of U2. Since U2 has a Schauder basis, x = {Zakek .
Since L is a continuous linear functional, L(x) = Z:akL(ek) . Hence,
the sequence defined by L(ek) = Ck uniquely determines L. Conversely,
let 1Ck1 be a sequence such that 5::akck converges for all {akl in
U2. Now Pk(x) : ak isa_continuous linear functional on U2 for each
k. Hence, for each n,

n n

311“) = :Pk(ckx) = X: akck

k=0 k=0
is a continuous linear functional on U2. The assumption that 2:akck
converges for all Iakl in U2 implies that Sn converges weakly to S,
where S(x) = Xakck . Hence, S is a continuous linear functional.

An immediate result from Prop051tion 1.24 is that if 20 is such

that IzOI ;,1, then 1251 defines a continuous linear functional on LI,

2

(A1

O

K! awe C'WtinUOUS

C)

In Particular, the sequences (1,1,1, ,1 and {{-3)

linear functionals on U2,
For each fixed t, O étzé233 define the class B V.(t) by
(21) Bsz(t) = I 1CKI:§;ICK - e‘llckfig <cZ>r.
For t:0, the above class is simply the sequences Of bounded variations-
Eigpgsicignmlggg: If {ck aB.V.(t), then 11m cke'1K1 exists.

Progi: ickl eB V (t) implies that EEIck - e Itck+li « (I? 0

But
§:': .1i . _ ~lik‘l)t~ .1..2f1-. - -lt
,e .Ktck e tk+11 ,,ck e cK+],
-' ' -" + 1
Hence, 2:619 1m - ckfle 1“ 1’1) converges. But
'n ikt ’ lit
S a 2:: c e ' - c , e'IIK1" )
n kv‘O K k'iI
: _ -1(n+i)t
cO cn+1e
MEHLG, 12m cne Int Exists.
Now 11 C , lim cne'Int, then EC‘ » lim Icnl Since
'. 1 I ' ’, T -1nt 1
3K“: " .C1 :; ‘(ne - CI'

Elgpgsiglgflwlraéi B V.(t) is a subset of U' for each t.

2
Proof: Let {Ck} cB,V (t) Then for any n and any {dk1_U2
iii: . e QLL’ -1kt _ , -it , -1nt— .
kLO dkck LETE SKItIG [Ck Ck,]e I i cue Sn(t)1
k

where Sk(t) = .%:% amelmt , But lim cne‘lnt3n(t) eXIsts, and
l,—

31
:Sk(t)e“1kt[ck - Ck+l e'11] is aboslutely convergent. Hence,

_ , , , . ,1
2:3ka converges for each {akleU2. By Proposition 1.24, ickmUZ

Corollary 1 27: Let L = {ckch.V.(t). Then

 

IILII sup IL(.><)I

; ZICk - ck+]e’1tI + 11m IckI
Proof; From the proof of Proposition 1.26, it follows that
IL(X)I ; sip ISK(t)I(ZIck - ck+]e'1tI + lim IckI)
:: IIXII(:ICk " Ck+]e-1tI+ 11m ICkI)
Hence, Corollary 1.27 follows.

EQBOS_1t'OQ_l_,_2_8_ There BX‘lSt {Ck-1 EUZ SLICI‘I that 1ij? does

not belong to B V (t) for any t.

Proof: Since {(-l)kl and (l,1,l,l,...) are in U2, the sequence
‘Ckideilned by ck = l + (-1)le 1n U2, However, for any I,
m

m ,,, _ ._
I-ml _ l ', ‘ . . . ‘k‘f1 - ’
5—9 1C1. - €1.19”: '=2..-1[1 +1-11kJ-11 + 1-11 is it:
kvO k'IO

r\)

is

t1 :: 21m +1111)

k-=O

From the fact that [a is a subset of U2. the following proposit-

ion is eaSily obtained, and, is therefore stated without proof:

32

Proposition 1 29: As sets, Ué 15 a subset of m, where m

 

is the collection of bounded sequences.

Corollary 1.30: There exists f in U such that the derivative

 

of f, f', does not belong to U.

Proof: If E:akzke U implied that.2fkakzk'1e U, then

X kal< converges for all {a } in U Hence, by Proposition 1.24,

k 2’
{k} would be in U2, but this contradicts Proposition 1.29.

Corollary 1.30 is a prefectly natural result since there are
power series in U whose radius of convergence is precisely 1, and,

hence, have a singularity on izl = 1.

Theorem 1 31: {c 12Ué iff {akckleU

. . I ‘ . 1"
k for a 1 {ak ,U2.

2

 

Proof: Assume thatiakck}

(Z:akckzk converges on 125;] for all {akl U

€U2 for a11{akl,U2. Then the series

2. In particular, it
converges at z = 1. Hence, by PropOSition 1.24 {Ck}€Ué.
Now assume that L ; {ckirU . Let Ta }.U , To show that
2 k 2

—. 3' ,, -v- , - . _ v-e. - Tkt : H fi«._ 2, .
{dkck’ U2, it Suiilces to snow that E; dKLke 15 on formly Cauchy.
Let e15 be a point on T and fix it. Then y 4{ake'Ks} is in U2,

and

sup 13:; akeikse‘kti= sup i ii; ake‘kt j.
:p t

t K49

Hence,

. D. * s n .- ,
I L akcke1k I =| L f: ake1k°egi
kip k'D

(1
;11L11 11g akek 11 .

33

By choosing p sufficiently large, the right hand side of the inequal-
ity can be made as small as desired. Therefore,Z:akckelkt 15 uni-

formly Cauchy.

 

Corollary 1 32: If L = {ck}, M = {ukieu', then

* - ’ I
L M - {CkukitU2.

Proof: Let x = {akleU2. Then y ={akck1eU2. But
M(y) =zakckuk. Hence,:ak(ckuk) converges for all {ak}tU2.
Therefore, {ckukleU2.

Theorem 1.33: Ué is a commutative Banach algebra with identity

 

under coordinate—wise multiplication.

3599:; Let L = {ck} , M = {ukle ué. Let x = {akleU2.

Then y ={akckle u2. For fixed 5, {akeikSieu2. Hence,

n . n .
i§::: akcke1ks I = [L (S:: ake1k5eki I
k=0 k=0 /

;11L|| Ilfakekn

k=0

= IlLll I1XII.

Since the right-hand side is independent of n and s,

 

11y! llLli IIXII.

Hence, lM*L(x)l = IM(y)|;.llMi| llyi

' .
\
”—

 

;J1Mll llLll linl.

Therefore, iiM*LlI;JIM|| liLll.

The ring structure of U2 is obvious, and (l,1,l,...) is
the identity. Hence Ué is a commutative Banach algebra with

identity.

34
In order to find the precise dual space of U2, we will

examine the dual space of U3. From the fact that U5 and Ué

are equivalent, we will be able to deduce the manner in which the

sequence space, U2 , is generated.

Definition 1.7: Let B be the sigma—algebra generated by

 

the open sets of a topological space X. Let EeB. Then a
collection of sets, {En}, in B is said to partition E iff
E is equal to the union of the En, and the collection is
pair-wise disjoint. A complex Borel measure, u, on B is a

complex—valued function on B such that for each EeB,

00
u(E) = §;:; u(En) for every partition {En} of E,
n=0

and the above series always converges absolutely.

Definition 1.8: If v is a complex Borel measure on B,

 

the total variation of u is the finite, positive Borel measure In]

defined on B by

lul(E) = sup {:lu(En)| :{Enlis a partition of E1.

Definition 1.9: If u is a complex Borel measure on B,

 

then it is regular iff for every EeB,
M (E)
1u|(E)

inf{lul(V): E is a subset of V, and V is open}, and

sup{ipi(K): K is a subset of E, and K is compact}.

The form of the Riesz Representation theorem which we need
states:
If X is a compact Hausdorff space.'Unyito each continuous

linear functional L on C(X) there corresponds a unique complex

35

Borel measure, u, such that

(21) L(f) = J f dp , fOY‘ all feC(X),
X

and llLll = |u|(X)~

By the Hahn-Banach Theorem, if LeU3, then there exists a
continuous linear functional H on C(TxN) such that L(S) = H(S)
for all S in U . Since TXN'is a compact Hausdorff space, there

3
exists a unique complex regular Borel measure, u , such that

H(S) = j S dp
TxN'

Hence, for each L in U', L can be represented by a complex

regular Borel measure on the Borel sets in TxN: Moreover,

L can be represented by one whose total variation on TxN is equal
to the norm of L. However, it is not true that L can be represented
by a unique Borel measure since the extension of L to H is not

unique.
Now the system {ek} defined by
0 if n<k

ek(t,n) =
eikt if n» k

is a Schauder basis for U3. If L c u; , and se U3, then there is

a complex regular Borel measure, u , on TxN such that

36

ek(t,m) du(t,m) .

7‘.
II
0

Txfi'

Finally, if L = {ck} belong to Ué, then it is immediately seen
that there is a complex regular Borel measure on TXN which

generates the c by the rule

k

(22) C ek(t,m)du(t, m) and

Txfi'
conversely, any complex regular Borel measure will generate

a sequence--when defined by (22)--which belongs to U2

CHAPTER TWO

CONTINUOUS LINEAR OPERATORS ON U

1. Concepts From Summability

Let A=(ank) be an infinite matrix. A sequence x = {xk} is

said to be A - limitable provided the sequence y = {yn} defined

by yn =k:ankxk converges. If every convergent sequence is

A - limitable, then the matrix A is said to be conservative. A
conservative matrix A = (ank) is clearly a linear operator on the
Banach space of convergent sequences. A conservative matrix is

called a regular matrix provided A - limit x = lim xn for every
”+00

convergent sequence x = {xn}. Toeplitz' Theorem gives the following
necessary and sufficient conditions for a matrix to be conservative:

6x3
1') sup Elam] < (D

Tl

ii) 1im ank = ak exists for each k,
n+oo

C!)
iii) 1im E 0 ank = a exists.
n =

If ak = O for each k and a=1, then the above conditions become
necessary and sufficient for A to be regular. Toeplitz' Theorem also

shows that a conservative matrix is a bounded operator, i.e.,

'27

38

“PIE 6 kal:= suplx I supfila k]
nk=o" k k nlk=0n

Let a be a holomorphic function in |z]<R, R>l. Then by

taking powers of a, [<i>(z)]n =f: ankzk, a matrix A= (ank) is
k=0
obtained where a00= 1, and 30k = 0 for k= 1, 2, 3, .... If

¢(z) = “ +(1'a'8)z , then the resulting matrix is called a Karamata

l-Bz

 

matrix. ‘
Bajanski [4] has proved that if
i) ¢> is holomorphic in |zI<R, R>l
ii) |o(z)|<l for Izlgl. 2f 1
iii) ¢(1) s l, and
iv) Re Af 0, where
q>(z) - za= iP A(z--1)p +ol(1)(z-l)p as z-rl, Afo, and a=¢>’(1),
then the matrix defined by [<I>(z)]n =§ank 2k is regular. In this
paper, he also has shown that necessary and sufficient conditions
for a Karamata matrix to be regular (for real a,B) are o<l, e< 1,
and a+B>0, or a=8=0.
Notation 1: If X and Y are Banach spaces, let B[X,Y] denote the
Banach space of all continuouslinear operators from X to Y.
2. Bounded Linear Operators From A Banach Space
X (With Schauder-Basis) To U.

Theorem 2.1 In order for an operator A to belong to B[X,U]

 

(where X is any Banach—space with a Schauder-basis (e0,ej ,...)) it

is necessary and sufficient that

39
i) A is uniquely determined by a matrix (ank) satisfying
ii) fk(z) = SEE ankzn belongs to U for each k, and
n=0

iii) Ln = {ank}cr) belongs to X’ for each n and the continuous
n=0

linear functionals defined by Fp t: Z eint Ln satisfy
’ n=0

I] I ((29.

sup lle,t X

DJ
Proof: Assume that A belongs to B[X,LU. Let x belong to X.

Then x can be written as x =5::; xk tk. Since A is continuous and
k=

linear, we have A(x) = SEE Xk A(ek). But .A(ek) belongs to U for
k=0

each k. Hence,

. _ _ n .
(l) A(ek) - if? anken where en(z) - z , and we obtain

n=0

(2) A(x) =§ x fa e . Since A(x) belongs to
k=0 “ n=0 "k n

U, we have
629

(3) A(x) =§bnen

We will show that b =§§E a xk. Without loss of generality, we
m k=0 mk

can assume that l||A1|| = supll ||A(x)||U f 0. Given e>0, there
x =1
is an N such that r2>N implies that llfifi xkekllx <
k=r

nuul'

40

Hence r>N implies that

11E x... A(ékiuu =11 M X, ékmu
k=r k=r

s111A111 Hf X-kék11X< .
k=r

- do
Hence, : xk 2 anan converges in U-norm to
= n=0

Pm<A<x>> = Pug xi we.)
k=0 n=0
" E
= Pm(11m 2:- X‘k ‘ anken)
woo k=0 k=O
to
= 1im : XkPm (E: anken)
r+m k=0 “:0
623
“‘Z: x'k amk
k=0
Hence, b =§ a X , and
m k=0 mk k

' CD
(4) ‘ A(X) =6 en;bankxk

41
We now have that A is completely determined by the matrix (ank) and

fromll), we have 1’ = M? ) = E a e belongs to U for each k.
k k “=0 nk n

Moreover, fromI4L we have that ankxk must converge for each n
k=0
and arbitrary x in X.

Hence, lfl= { amg‘29 is a continuous linear functional on X for each
k=0

n. Now let p and t be given. Let x be an arbitrary member of X.

 

Then
62> <27
I|A(x)IIU =II2: an 2: ankkuI
”=0 k=0 U
“' .. if
n
= e o a x
In; k=0 "k kI
(by Corollary 1.10 where p'may beoo).
Hence,
° t ' t
IIA("mu ' t em 0 Honk-IE: em LnMI 3 I F11.t(")I°
n=0 n=0
Hence, IIIAIII = SUP“. IIA(X)II ;. 'SUP IFp’t(X)I = IIFp’tIIo

IIXII-l 1|X|1=l

Since |IIA|I|<GD, and p and t are arbitrary, we have

511 F 1’. A <m,
a?" pJ11, =111111

Hence, conditionsi), ii) and iii) are necessary.
Assume that (ank) is a matrix which satisfies ii) and iii).

(X?
To show that EEC en 25: ankxk belongs to U for each x in X, we
n=0 k=0

42

need to show that.£ ann (x) converges uniformly on Ingl.
n=0

But this is equivalent to showing that 2::Le‘”th (x) converges
n=0

uniformly in t for each x. Hence, we need to show that the continuous
linear functionals {Fp,t} converges weakly and uniformly in t.
As a consequence of the Uniform Boundness Principle {Fp t} will

converge weakly and uniformly in t, provided

(ek) a :0 emt Ln (ek) =2)“: ankemt converges
n: n=0

uniformly in t for each k, and

(5) PM

(5) sup ||F ll .< 00
M M X

ao
But conditionslSland(6)are satisfied by (ank). Hence, 2::enLn(x) is

in U for each x in X. Letting A denote the operator defined by (ank),
we clearly have A(x1 + x2) = A(x1) + A(xz) and A(zx) a aA(x).
To show that A belongs to B[X,U], it suffices to show that A is

bounded.
But IIA(x)IIU=II2:)erm)IL(xI
= sup | 2% ean (x)|
Mi n= 0
= sup IF (X)|
M M
2.. sup lle’tllx. IIXLIIX.

p.t

43
Hence, A is bounded, and IIIAIII = Sup IIFp,tIIX"
Therefore, conditions ii and iii are sufficient.
For X = i], conditions ii and iii become fk = E anken 1s

n=0
in U for each k and sup IIfkllU <00. This follows from the fact
k

that C30
_ int
Fp’t(x) 'iG Z ank xk
n=0 k=0
an int
= Z xk anke
k=0 n=0
_ int
and lleJl I}. - :up I: anke | . Hence,
1 n=0

sap lle’tlIII = sip llfkllU .

Hence, if an f in U is chosen such that IIf||,;l and if we define

k

ank=an k=1, 2, 3, ... where f=2:anen, ano = 0 for all n,

then the matrix (ank) is a continuous linear operator from ,é:

into U. Herelfk(z) = EEi afizn, and since U is a Banach algebra,
n=0

we have app IIfk II = sup Ilfk ,;1. Where fk denotes f*f*f*...*f.

 

 

3. Karamata Type Operators on U.

If a is holomorphic in IzlsR, R>l, and if |¢(2)1; 1 for
IzL; 1, then for f belonging to U, we can consider the composition

of f with o, f(a). f(¢)will certainly be holomorphic in 121 < l.

44

and will be continuous on |zI= l. The power series coefficients

for fo¢ will be given by

 

 

CWO
= 1 2: .. 1 .
2—51' aka (2) zn+1 z
k=0
Ill-=1
c2:
- _;L__ k
_ 22::ak 2 H1 a (z) zn+1 dz
k=O
|Z|=1

. _ _l_ k l
Letting ank - 2111 fo (z) ;071— dz, we have that
121 =1

205‘

k=0

b" = ankak
Hence, if we define F by r(f) = foo, we can ask if1~ belongs to
B[lJ,U].

The following lemmas whose proofs are well known will be used
extensively:

Lemma 2.2: If 0;}<l, then for any pip

; (k) =—--—-k: .
(1-t)P+1 = P p p:(k-p):

Lemma 2.3: If m and n are nonnegative integers with mgp,

45

m m m+1
+ = ,
then (n) (n_]) ( n 1
Lemma 2.4: Abel summation:
m m-l
Z akbk = Ambm - Apbp+1+ : Ak(bk-bk+])
k=p+l k=p+l
k
where Ak = ar .
r=0

Notation 2: [x] will denote the greatest integer which is less

than or equal to x.

Theorem 2.5: If 4(2) =cx+(1-a)2, 0<o<l, then the operator

 

r, defined by r(f) = fo¢, belongs to B[U,U].

Proof; For each k, we have r(ek) =(o +(1'a)el)k = ak(z)

k k kt k'n n n

which certainly belongs to U. Now ch (2) = ED (”)0‘ (l-a) z ,
n:

(k)Il-'—9>nok if n3

n 01
Hence, ank =
0 if n>k
(“0 aka k k , *n
Now a a = 2 a Z—EE'
k=0 ”k k k=n (n) O. X ('1 j

00
_ ‘n k
= (1—9) 2 (:) a ak is absolutely convergent

a k=n
for each f = 2E:Iakek in U. Hence, the linear functional

00
L = {ank} belongs to U. We must now show that the continuous
k=0

46

P
linear functionals F - eIntL satisf s F .
M £6 n y pvgil ID.tl|<ao

Let f belong to u. f - Zakek.

Then Fp t(f) - if:_e1nt(1'a)nf k-n (: )aka

n-O
Since the expression on the right is absolutely convergent, we can

change the order of summation to obtain:

FP( "f )éoaka o”(2:6 ( W461?“ k-p+1 kmib ( :)[;ae ”In
.£1 + £2
12211 -1§O\Ikfiilg%lflkak 1

_. a1_aitk
125.311,,“(1611

Now,

allfil

Let m 2 [%:TJ . Then we write
2k§§1akm n-o £(n)I1—-Gelt]"+ 1;: aka krill (ME—“*3“

I 221+ :22 ., Ifm-p, then 221.0.

If map+1. then IZZI l a lap.” ap+l % (p'H) [1%1371
n- h

47

Hence, if m=p+l,

+l p+1 +1 ~ _ n
I221! ;|lfll up 2:00,, ) [J—l = um.
n=0 “

If m>p+l, then we write 5:2] as

M
.571

_:_2llf||.

We will use Abel summation on Z212. Letting

k
-k kl-aitn
bkw E] (n)E—e} ,

we obtain ZAZIZ ‘ Smbm ‘ Spbp+l + ‘L‘ZlZl

where

-- m-1 k i k: _a-Z;k+1 k+1 kn ~. n7
Z2121 7': Skfi (:1) a afln‘ n )L; j“
k= +l n=p+l Wp+
111:1 k k i-a it n k k+l {‘11 it k+l
=2...st {IE—e (n)-a(n)- 1...sk(<-a)e>
k=p l n=p+l k=p+l

2121] ' '2l212 '

48

1-0L t
lsmbml (nj‘a ‘]n I

II

M

Q

3
3
ll

Fla

"A

lsmla'" g: (:)[1§;9—]n
|Sm| since :4: ( m)(:l—9"—f= (1 #14111")

;llfl|-

Similarly,

Ispbpfll; llfll.

m' k+l
‘szm léllfllVg] (1‘0‘) ,

We observe that (E) ; Mk:

 

) ‘iff k+l ;

l-a ‘

Now in 2212“ , m-l _>__ k. Hence, _>____ m_>__ k+l. BUt l" 2212]]:

we have mp. Hence, 11-1?) k+l. Therefore,

;]I(¥eit)n((:)-a(k;1))l hp” #‘FF‘WH k” (:9)

k
V"- lZPLn k .. -a k)
£51“ ) (MM) (1 )(n)

k 1
:(Mn (")-fl1nl'1?(:)

n=p+l an-1 "-1

-a pH a
glp) (k) _(1- 2 .

a p a

Thus,

I“. |
#21211

EZ:2l2ll

H“

49

I ]_ p+l m-l k m-l
élIfHL—Sl :: aim-5::
a k=p+l p k=p+l
; “fl—21211|+ lZ:21212'
_ +1 m'] k
sufHfl—alp 2:: at H
aP k=p+l P

l CK) k

'_ +

sufuflpflp Ema"
a k=p p

= llfll by Lemma 1.

lsmbml + lspbp+l| +l::2]2]l

3||f||~

é l7:211' + I2:212]

5llfll.

we observe that Abel summation, with

ak ZR; (:)(l;3elt)n , gives us
n=0

0.

Viki. k
Srbr'Sb++/ .50,
m m I k=m+l k

n=0 0

(l-a)

p , . ~ -
:(kee‘t)”(<k) - OUT)

n

We will show that the last sum on the right converges absolutely

and since the sum on the left converges, we must have lim Srbr

existing.

Hence,

2::22 = lim Srbr ' Smbm+1

r¥w

+ :22] 7

50

 

 

5‘3 P ~ .n
“' T" k l-o 11; k k+l
where =4___ SO‘Z:(—e)[()'°‘( ”-
221 k=m+l k n=0 °‘ " n
In E , k;m+l. Hence, k+l ;m+2 >—P—+i > n.
221 l-a l-OL
Hence,
p _ ' _n + pfi _ n
:: Mia—"12”) ||(k) - a<kn‘)l = 2(153) [(k - a “bl
n=0 n n=0 n n
P gi-e)"+‘ k l-a ” k
E n (l'ififlmfl
n=0 a a
.. ‘l-Glp+] (k)
up D
. +1
HEnce,|E—221l;llfll 9:31” Z (kMk
up k=m+1 p
- n+1 k
ennui—31 2:( )ak
a k=p
= IIfII.
iiirgsrbrn; :93 |er Ibrl
. t Y‘ -a n
; IIfII sup e'£:( )(10, )
rtm n=0 n

ellfli.

In like manner, ismbm+1 I; IIfII. Hence, IZZZI <=-3|lf|l.

Hence. l22|=|2:21| ӣ2221

; 8||fll«

Finally,

.

lF (flL; IZZZil +|3322l 9ilfll-

(1

Pet

Hence, lle tllé: 9, Since p, and t are arbitrary, we have

SUP HF ll: 9-
p,t p’t '—

Hence, r belongs to Bl:U,U].
In connection with the proof of Theorem 2,5, we observe

that our linear functionals, L , are also given by

 

n
(7) Ln(fl = %fii 3”f(o(zl) zn+l dz.
l¢(2)l=l

6x3
This follows from the fact L (f) = 2:: a ka = b where b is
n k=O n k n n

the n-th coefficient in the power series expansion of foe. In

 

' r " ‘ - _w-O. =9!!—
fact, letting w-¢(zl, we obtain 2 — H(w) — TZT’ dz 1da ,
and integrating around lw|= l, we have

= l i gl-aln
Ln(f) 25?. f(w) (w-a)“+] dw.
|wj=l
By the calculus of residues, we have
. - n
Ln(f) = £1112. lin f(n)(w)
n, w + a
673 CT)
1'3 n k a Since (n) J = k! a k-n
(T) EHW < f new 12

which is precisely the value that we should obtain.

Theorem 2,6: If i(z) = $%E%%5-, 0<Bcl, then the operator A,

defined by A(f) = fol, belongs to B[u,u].

 

52

k
Proof: We have A(ek) = 1%91) belongs to U for each k»
' l

= ‘l. )‘k(z) ll d2 t
Zn+1

 

 

1-
= A .. = W =

Let w (2). Then z-r(w) l-B+Bw , dz (l-B+Bw dw ,
Hence, for n:;l

l
k n-l
= l- w(L3$w)
ank yrs]. ‘3 WM] dw

le=l

 

a00=l an0=0 n=l,2,....

Using the calculus of residues, we obtain

_ _k
(E-l)8"(133) n;kn#0k#0
a = 0 n < k
"k l n=0,k=0
0 n:; l, k = 0

"V
—J

for n

Let q and t be given, and let f be an element of U.

Changing

c......./

L2

Let m

is lost.

53

. n-l
73mm)" 2: in" )(Lixki‘a
_: k=0 k B

n-l

a0 + k+l

 

-‘| C]
l-8 k+l n l
' a + ( ) "”‘ ( )(eelt)
k=0 8 k+l fiétli
= a + "1(1é5>'<+‘a 2:09 ("We 1t)"
O k=0 k+1 n=k+l k
'1 l- k+l €§§;L -l n
' _EEL ak+l‘4;””‘ (n )(Be1t)
k=0 n=q+l k
/ /
- ao iii} ' Z:é
q-i - it
= a + (l B)e k+l
0 =0 l-Beit k+l
q 1-3 elt
- :t‘ ). k

l:

O

+l /
[$jB‘J. If m = q+l, thenZ:2 remains as it is and nothing

For' m > q+l, we write Zig' as

54

/ ’22}.
5: Lieeit)"k f:( n-l magma

2 n=q+ k+1
00
k+la
+ Emit)" in" ‘)< ‘ )-—§— 1
n= m
- 2.2 +23 '
We write 2:: as
// m
n=q+l k= 0
m—l n- -l +
- ‘2‘:(ee‘t)n z:( )( ——->"‘ak+1
n=q+l k=q k 8
/II
= :2 - :2
Hence,
- — W ”-
q,t(f)‘>-i ‘2—2 +22 ‘2—3
But,
__A0
F f + = F f . Th refore,
q,t( ) )‘2 m-l,t( ) e
Fm-l,t (f) = 21+: 23

IZ1|=|23.‘a[U'B)e:t lkl ; urn.

k=0 l-Be1

 

55

We use Abel summation on the inner sum of 2'52. Letting

g n ‘| B k+l
bk ( wig—i .
we obtain
n- -l n- -2
z:a‘kflbk ‘ sn bn -l ' quq + 2:: Sk+l(b k ' bk+l)
ksq k=q
= 1-8 n- -l +l
sn(—_ 8 -qs (q >(1-'— 88W

n-2
l-B k” n-l
% Sk+l(_B—) I: ( k )'J(k+1l]
Hence, we can write ”:2 as

2:2 = :21 ' 2:22 +57:23'

IEle‘l‘Egéeitw s<15~8>nleufllzner

.=lllq+

m-l .
| Zzz| |$q(-]-g—B—)Q+] 2: (n-1)(Be1t)n l

n=q+] q

llfll( 'lgB)q+l£(nq)lB n
n= q+l

IIA

We observe that

 

(nil); 1&8 (:3) iff (k+l); (l-B)n.

In :23, n_<___m-l 1%” l =¥—:—,

Hence, q+e _>_= (l-3)n. But in 2:23, k;q. Hence, (k+l) ;q+8 ; (l-B)n.

56

Therefore,

m-l n
I‘ZZZ3I; llfllz: e"21[(‘—%E)k+1(";1)-(—§—1Blk+2(fl;l)]

n=q+l k=q
m—l m-l
=- Ilf|l[( (lg—W”: (""i e" - 5: 0—8)").
n=q+l q n=q+l

Finally, we obtain

l1::2lé=|2221| +l2322| +IZ:23l

; 2llf|| <l_- BW12: Wis"1
n= -q+l q

; 2|lf|l

+1 m-2
since (“B‘WH :(qn1=Mq Z(n)sn < 1.

n= -q+l Bq n=q q —
We use Abel summation on the inner sum of 2:5. Letting

b “’1) (1%)“ .

k=(k
we obtain
-l

_ (l- k+l n— l Bk+2 n l
Eakflbk ' quq-l ‘ a(M‘s—B) +:::Sk”[ (lg-B-l (k )J'e'“ B) (k+l)]‘

.0

Therefore, we can write :3 as

23 =Z3i 5232 +233

. l;£_q it n n-l
I23“ lsq( B) g(fle ) (Ci-l),
JV" “5'” in)“ 8 (l3:
[£3zl = [30(1-7— 88):?(Beitwl
n=m
1|!le (lr-éiz:6"
n=m
InZ33, we have (l-3)n > k+l. Hence,
C7<> q-2
—— k+2 -l l- k+l -l
'533l1llfllfi8néoHLg-l (2+1) ' (7'5) (nk ))

- "2:; -1 n i- '
' IIfII ((lflqr%1(3_1)e -(—155)ge")

Finally, we have

l23l11231|+|23zl+ |>333|
e2ilfu< if:

:lefll
Hence, [Fm_1.t(f)|f_|21l +If?" +l23l
5.5llfll.
We now have that for any positive integer q,
+l
lFm-1,t(f)|:~. 5||f|l. where m=[%;é~].

58
We now show that there exists an integer h such that if q is any

integer, then there are at most h integers between
= 9:21 = ELTE .
m1 [ l-B] and m2 [ l-BJ

Let h = [l—l—J + 2. Then
l-B
- .1- 9+1.

;[___]+2+9__-

1-3

= __.. + + qjl.

[l-B] l 1‘8
,__.1__. +2.41
=l-B l-3
= 32

l-B
z.[*9:gJ = m2. Hence,
—' l-B

hi; m2 - m1 and h is independent of q.

Now for any q and associated m, we have

lFm,t( )‘l [Fm-1t(f)+WUmFE-f-O-(IHIJHI—é—Byflakfll
; lFm_]’t(f)l + 8m|lf|l .m:] (In?) (155m
;5l|f|| + llfll (l-B)
= IIfII (5 + (i-eii.

'Fm+1.t(f)' ‘ lFm,t(f) + (8 ”WW1 :O(E)(l§fi>k+1 k+ll

Therefore,l

+l - .Jn k+l
IFm,1{f)l_ Fm t(iiI + em ||f||>_.. i2” 0 meg)

_.||f|l( (l- 8 )) + llfl|( l- -B)
llfll (5 + 2(1-8)) .

A

Continuing in this manner, we obtain
IFm,h_gfiI :. Hill is + [Tl—8.] Ii-eii .

Since we have at most h points between m1-l and mZ-l, m1+h-l:m2—l.
1
Hence, we have IFm1+k,t(f)l :_|]f[] (5 + k(l-3)) where 1EF§IT=§J .

I _ .
Hence) k f-T:§' and k(l e) :_l. Therefore, for any integer p and
given t, IF (f)! g_6llfll. Finally, we have sup ||F ||< 6

pat p,t pgt '—
and Theorem 2.6 is proved.

 

Corollary 2.7: If i(z) = a +(1-G'B)Z , 0*d<l, 0<B<l, then

 

l-Bz
the Operator'T defined by T(f) = fay b910ngs to BEU,U].
BEQQEE Let ¢(z) = a+(]-a)z W2) (l-B)z
-82

Then T and A are members of B[U,U]. Since B[U,U] is a Banach algebra
under compositions, we have hot belongs to B[U,U], but

Ir (iii A not i

H

(f0 ¢ )oA

f0 (m 0A )

But ¢ (x (2)) e 0+ (l-d) gl-Blz

l-Bz

= a+ (l-a-Blg =‘¥(Z)

l-Bz

Hence 1‘: B[U,U}

6O
4. Bajanski Type Operators 0n U

The goal of this section is to show that certain Bajanski
type functions operate on U under composition of functions. The
method used consists of comparing a given Bajanski type operator
with a Karamata type operator.

Theorem 2.8: If

 

) ¢ is holomorphic in Izl < R, R>l

) l¢(z)l <1 for lzlf__l zafl

iii) ¢(l) = l
)

iv Re Aio, where
¢(z) - zY = -A(z-l)2 + 0(1) (2.1)2 as 2+ 1,
Y = ¢'(l)
v) «v'm > o. ¢"m > o. and (©‘(l))2 < mu + My ,

then the operator A defined by A(f) = f0¢ belongs to B[U,U].

One should note that the first four hypotheses serve to
identify the Bajanski type functions,and that the Karamata type
functions with 0<a< l, O<8<l, are included in the Bajanski type
functions. The fifth hypothesis is added for comparison purposes
with Karamata—type functions.

It will facilitate the proof of Theorem 2.8 to first prove
a few lemmas.

Lemma 2.9: If ¢ obey the hypotheses of Theorem 2.8, then
¢ has a local inverse W in a neighborhood of z=l, such that for
lel sufficiently small,

We”) - elel2 ; k(64 + (t-T)2)

where T= arg(©él@) and k is a constant independent of 9 .

6l

Proof: Since a'(l)>0, a local inverse to a, w, exists in
a neighborhood of z=l, and v'(l)>0. One can clearly choose a

neighborhood N about z-l such that

Re w'(z)> 2:51).) 0 if ZEN.

Let g be a number such that °(C) is in N. Integrating along

the line segment Joining °(C) and z in N, one obtains

Z

(2) - CI: 8;: '(E)d€
@(

Let 5= <15(C) + A(24(4)). then d5 = (z-¢(C))dA and

l

 

 

lw(z) - CI = '2 - ¢(€)l 5*"(¢(C) + A(2 - ¢(¢))d*
0
l
.1. '2 - 4>(C)l ISRe 9"Xm
O
; i'é‘) Iz - «>(c)l.
Hence,
(8) |W(Z) - Clz ;.Clz - ¢(C)l2 where C = Eiill- is

2
independent of C.

Let g(t) = leit - ¢(€)l2, and let T denote the point where g

attains a minimum.

62

Since

leit - ¢(c)l.; l-|®(c)| .
it is clear that 1= arg ¢(C). Let C = eie. Then,

l - l¢(ei9)|= 1-|<I>(e"ell2
l + |¢(ele)|

From hypothesis iv) of Theorem 2.8 ,

¢(Z) = iY - A(z-l)2 + o(l)(z-l)2 as z+l

‘6'
t3
II

__I

- A [l + (2-1)]‘Y(z-1)2 + o(l)(z-l)2 as 2 +1

= l - A(z-l)2 + o(l)(z-l)2 as 2+1 .

 

Hence,
|<I>(el9)l2 =I1- A(ele-i)2 + ...|2 as 6+0
=|1+-m2-+..l2 ase+0
= l+2 Re A92 + larger powers of e .
Hence,
l - [g(ele)| = l ' l¢(el9)|2
l + |o(ele)|

= -622 Re A + large powers of e .
l + |o(e‘9)|

 

Therefore ,

- ie
1 - leée )l s -Re A as e» 0 (Re A < 0)-
6

 

'6
Hence, 1 ' |¢(91 )l is bounded in a neighborhood

e2

 

63

of 9 =0. Hence, there exists C1 > 0 such that

1 - |¢(ele)L; c162.

Hence,

(9) g(T) = (l - |<I>(ele)|)2 ;.cfe4.
Expanding g in a neighborhood of t=T, one obtains
(10) g(t) = g(.) + 91§11-<t-r)2 ,
where T is between t and T, and

g”(T) = 2Re®(el9) cos T + 2 Im ¢(ele) sin T.
For ele in N, there exists n> 0 such that

(ll) g”(T);p > 0 independent of 9.

From 8, 9, l0, and ll, one obtains the existences of a k>0

such that
2> k(e4 + (t-t)2.

_—
—.

[g(elt) _ ele

 

Before the next Lemma is stated, observe that for

h(z) = 0 +](1'§;§)Z , the power series development is given in

a neighborhood of z=l by

 

6x3
h(z) = 1 + <A+B> 2i; Bk"<z-i)k .
k:

where A = l-a-B , B = ‘TQE
l-B —

u _ = 1-0 h”(l) - = Bl'a)
Hence, h (l) — A+B 'TTE , ‘_TZ_— B(A+B) l-B 2 .

64

 

Lemma 2.l0: If a, b>0 with _E§.< l, then there exist real
a+
numbers a and 8 such that
i) 0<a<l, 0<B<l and
ii) 15%”, and £9.19). =b.
(1-6)2
2 b
o = - a = .__..__
Proof. Let a l W, B a+b .

If e is a function which obey the hypotheses of Theorem 2.8,

then from Lemma 2.l0, there exists a, B, such that

- 0L + (l'G'BLZ
h(z) - l - Bz

satisfies h(l) = ¢(l) = l; h'(l) = ¢'(l), and h"(l) = ¢"(l).

Also, the local inverse to o,v, in a neighborhood of z=l satisfies

v(l) = l, v'(l) = 6.}T7., and v"(l) = - Talllll3 .

If H is the local inverse to h in a neighborhood about z=l,

H(l) = H’(l), H'(l) = Ll/‘(l), and H"(l) = Kl"'(l).

 

 

Therefore,
(12) I H(eit) - v(elt)l = 0(t3) as t+0
(13) I H'(eit) - v'(eit)| = 0(t2) as t»0
Lemma 2.ll: If I is a function of e such that %’= 0(1)
as a +0, then
b
2 d
4 t t 2 = 0(l) as a +0.
8 + (Z-T)

65

Proof: Let u = t-r, then

 

 

 

 

b b-t b;t b s ’
4 t2 dt (N 4. 21' Jill—__— + (12-94) (Ill 4
9 + (t'T)2 u2 + 94 u2 + 6
6-1 a-t a-t
b-T
du = (b-a) . 0(l) as a +0.
-T
b-T
udu (b-T)2 + ,4
2T = Tlog = 0(l) as 6 +0.
“2 + e4 (a-r)2 + e4
a-r
b-T
2 _ 94 dU 4[ tan “1(b-T _ tan-l a'T J.
a-r
Since tan'lx is bounded as x+so, and since %_.= 0(l) as 6+0,

the last integrand is 0(l) as 6 +0.

Corollary_2.l2: If r is a function of e such that
= 0(l) as 6 +0, then

 

b
ltln dt . = 0(1) as 6 +0, for n>l.
[e4 + (t-rrfz =

a

66

 

 

Proof:
b b
n - t dt
4|ti dt 3: é sup Itlnl 4| l ,2}: ,
[6 +(t-r)21 a;t.<.b [6 + (t-T)]
a a

sup Itln'1 . 0(1). and the Cauchy-Schwarz inequality

 

 

6365b

yields

6 2

2
Itldt < dt 4t dt 2 0(1)
[64 + (t-rizfé e + (t-r)

a a a

as a + 0

Lemma 2.13: If T is a function of 9 such that %-= 0(l)

 

as 9+0, and c>0 a constant independent of a, then

3 - ct2
I(p,6) = LfiL~Pe p dt = 0(l) as e-+0, p-+CK).

[64 + (t-srr
a

 

Proof: If 0 is between a and b,

0 2 . » 2
_ t3pe'Cpt dt + t3pe’CPt dt

[64+ (t-r >212 0[e4 + (t... >211:

 

 

I(p.e)~

a

I](p,£» + 12(p,e».

67

In I](P,8)s let

t2
”= 4 21/
[6 +(t-T)]i

 

dv = -pte'Cpt2dt.

 

 

Then

4 2

2t[e +(t~r)J-t2(t-T) l -c t2
du 2 dt and v = -—e p

[84 + (t-T) ]3/2 2:

Hence, 2
2 -cpt 0
l 2c[e4 + (t-T) ]2

1 e-pctz 2t[64 + (t'T)2] - t2(t-T)
2C [64 + (t-t)2]3/2
a

2
aZe-cpa

.- 2ch64 + (a-1 )leE

 

I]](p96).

The first expression on the right is clearly bounded as 8 +0,

p-sx; The integral on the right can be expressed as

0

 

 

 

2 dt
_ 1 - at? t2(t+r) ]__ -pct 2t
” 7‘5 [e4 + (a. 1213/2 2C [e4+<t« 1212
a a
=LI (o )- LI ( 1
2C 111':63 2c 112 W9
0
. I’ddt , _
NOW [1112(p,6)] _<_ 2gb); + (tr-“0231,; = 0(l) 515 9+0. p ‘* Cc»
a

by Corollary 2.l2.

68

tzlt-T' dt

 

“111(1),” Li

a

Assume that a<T<0.
Then

0
t2 lt-Tldt

Ld+<enfiwz
a

[64 + (t- T)2]3/2

If this does not prevail then nothing is lost.

T

 

 

o
- _:?lt_lidt + tzit-Tldt
[$+(eo%”2 .[d+<vnfi”2
a
t2(t-T) dt t2(t-T)dt

 

 

td+<vofiwz+ 94+(en%”2

T

1

"Il11(e) T 11)1(8)‘

 

 

 

It suffices to examine 1111(6).
Let u =t2 dv = t'T dt , , Then, du = 2tdt
[94 + (t'flylfl?
v - '1 H
— 4 2 L ence,
[u + (t- t) l 2
0
11 2 t d1;
1 (£9 = 'r + 2 °
“1 6—2 ['94 + (ta-t)2]1/2
T
T2
Now, -?-= 0(1) as 3+0, and
e
0

 

O
t m < 5
ei+<vohi
T

by Corollary 2.12.

_rLtLAdt

4 1/2 as 9+0
6 +usr1

T

69

Hence, |I111(p,e)| = 0(1) as 6 +0, p+ CK).
Hence, II]](p,e)| = 0(1) as a +0, p + CZ)-
Therefore, |I](p,e)| = 0(1) as a +0, p+av.

In like manner,
|12(p,e)| = 0(1) as a +0, p+aD.

This proves Lemma 2.13.

Lemma 2.14: If e obey the hypotheses of Theorem 2.8, then

 

r = arg ¢(ele) obeys §-= 0(l) as 6 +0.
Proof:
¢(rele) = u(r,e) + iv(r,e), and u and v are

continuously differentiable. Moreover, v(l,0) = 0.

arg (¢(eie)) = tan'1 XXlJEQ-.

u(l,e)
By L'Hopitals rule,
lim %-tan-] :(l’: = 3!-(1,0).
9+0 1
Hence, -% = 0(1) as 6+ 0.

Lemma 2.15: If e obey the hypothese$ of Theorem 2.8, and

 

if it| is sufficiently small, there exists a constant c>0 such
that the inverse function to ¢,11, obeys,
- 2
|H(e‘t)| ; eCto

Proof: Let y = ¢'(l). Then

z‘”- (1 + ('2-111‘”

II
__J
+
l—l

A
N.

I

._a
V

+

l-A
A

70

Observe that from hypothesis v) that ®'(1)>0, ¢"(1)>0, and
(annzs «(1) £3211 < ¢'(1)+<I>"(1).

Hence, 2
1.8T. C a ¢"(]l+¢"(1)_'(¢i(ll) > 0.

1 2(¢'(l))3
H(z) - z”Y . 1 + H'(1)(z-1) + flléll (2-1)2 + ....

- [1 + H'(l)(z-l) + %.H'(1)(H'(1) - l)(z-l)2+...]

.. H“(i) - (H'llllz + H'(l) (2-1)2 + .
2

H"(l)-(H'(l))Z+H'(l)=- Jill—L 1 + 1
(«mm (M1112 rm

 

.. -<1>"(1) - Nu} (<1>'(1))2
wufi

‘ZC-l o

H(eit) - eit/Y = - c1(elt-l)2 + 0(1 ) (elt-l)2 as t+ 0 .

 

it . .
fl$§-l - 1 - C](e1t-l)2 + o(l)(e1t-l)2 as t+ 0.
elt/Y
it 2 . (1' )2 (it)3 2
Since (e -l) = [1t+ 2 + 3! + ....] ,

|H(elt)l 3,1 + (111:2 + ...

C] 2
2_.t

2W 4
Now, £Clt)4=l+§lt2+(%—L)2%+... o

;,1 + for It] sufficiently small.

7l

For kl sufficiently small,

c 4
l + E%- 2 > 1 + Zl-tz + (5%)2 £2'+ , since
C] C] 2 t2 C] 3 t4

since 2(3-1)2 53- + 2(——) ——- + ... can
4 2 4 31
be made arbitrarily small by choosing t sufficiently small.
c 2
Hence, letting c = 51., one obtains |H(elt)] ; eCt for
ltl sufficiently small.

Proof of Theorem 2.8: Since |o(z)|<l for |2L;l, zfl,

 

a curve r’ can be chosen such that r’ surrounds the unit circle
and touches it only at z=l, and r’ is inside the set of points
where |¢(z) |= 1 (2 f D. and f¥{2:|¢(z)|=ll for 2 close to 1.

Since A(ek) = ek is in U for each k, it suffices to show

 

 

that
p . '
k6 1 f(<1>(2))
suplG(f,p,e)| = sup 2 e1 ——-.— d
p,e p,0 k=0 2T1 zk+1

is finite for each er.

It will facilitate the following discussion to assume that
we have chosen a neighborhood, N, about z=l so small that all the
assertions which follow hold.

Denote the part of the curve, F’, inside 11 by -y, and the

part outside by r. In N, y = {z:|¢(z)l=l}.

7" '6 '1 .
G(f p-l, e) = —l-TSM%:(9—1—)kdz + J— ———f(¢(z)) E)Z_‘Z(9]-e-)kd2.
’ 2nl Z k=0 2 2W1 Z k=0 2
F Y

72
Hence , G(f,p-l,e) = Ir + ly.

Now, there exists 5 >0 such that |z|>1 + don r. Hence,

Let w=e(z) for 2 on y, so that z=v(w) [assuming that N is small
enough so that a has v as a local inverse in N], and y is mapped

onto an arc, c, of the unit circle. Hence,

pe)dw.

 

1Y 2 %‘j‘ ...JLQLL. (1-
" c

v (w)- We :p(w)
Clearly if |61>n >0, then

(15) sup |Iy|<w.
Ps|9|>n

Hence, only 9'5 sufficiently close to zero need be considered.

01+ (Fol-B);
l-BZ

satisfies h(l) = ¢(l), h'(l) = ¢'(l) and h"(l) = ¢"(l).

By Lemma 2.l0, there exist 6,8 such that h(z) =

By Corollary 2.7,

 

<

sup
1329

1 el n6
21—1“ ME dz

2 =0 zn
|h(z)}- =1

 

 

Let y' denote the part of the curve |h(z)|=l inside of N, and let

T denote the remainder. Then one easily obtains

SUp <

The

())
5%l-Uyfl: (z 22::9 ----d2

Y}

 

 

73

Without loss of generality, it can be assumed that the mapping

w=h(z) maps y' onto the arc of the unit circle c. One then

 

 

 

 

obtains
' P0
(l6) sup i,— W1“ (W ”Vi-N )dw < co.
m 2“ va-e‘e M)
c
Now write IY as
IY = I] + I2 , where
I we I
1=1 f(w)[ Hm Jl-e )- ”W l- ———)]dw

 

 

W(W) ' ei8\ wp(w) H(w)-e1.a Hp(W )

w )H'w ___Bi_)dw .
Hp(w )

From( 6) Csup|12I<C19
0.9
[Notation: For the remainder of the proof, it will be convient to

denote all constants which are independent of e by 0(l)]

From (12) and (l0), we have
We”) - H(e‘tN; o<1>1t3a
w (eit) - waging; 0(l)t2.

From Lemma 2.9,

 

 

1 §= 0(1) « where
|H(e1t) - elel [64 + (m1 >721"é
1] = arg h(elS); and
l 2; 0(1)

 

 

74

where T2 = arg ¢(eie).
[In order to have adequate space, in the remainder of the proof,
functions in the integrands will be written without the variable in

those cases where possible. For example we will write v(e1t) as v.]

 

 

 

 

Let w = eit; c being given by §i§;b° Write I] as
Il ‘ In ”12
where
b
_ l eltf[v'-H'l em8
111' Er? ~e "7 )dt
H - e1 H
a
and
b 1 _ el#3 1 _ eipa
p ——v—
112 = 1 e‘tfiy‘ "—L§_ - H '9 dt
2“ v - e1 H - e1
a
Now
b
' ' ”’9 tZdt
IIlll L sup (f(elt) l - e it ) 4 2 ,fi
I We ) [e +(t-w1>1
a
b
2
= 0(1) t dt 1 .
[e4 + mpg]1
a
By Corollary 2.l2
(17) sup |111| < CZ) .

Welsh

75

 

 

 

 

 

Write I12 as
112 = I121 + I122 Where
1__ it . H ' V
112] - ZTT e fl] (w-eie (H-eie) dt , and
a
b
ipe ' I l .____)‘l
I .-.. ———e 6"th "—'=— _ ‘ dt
122 2 1T (vp [W-em] HpLH'ew]
a
Now write I122 as
= + here
I122 I1221 I1222 w
b
in, it . H - W
11221 = e2 n §L193__ i . dt , and
\y (W -e 87(H‘e1e)
a
b
le eitf ’ 1 1'-
e W -
_ d .
I1222 2n H91“? Hp) t
a
Now b
ltltzdt

 

< o 1 f it ’ it ’ 1
|__._ ( ) S‘lépl (e ) W (e )l [8 4+(t—11)zj/2[84+.(t12)2]/2
a
b

ltl tzdt 1
[e4 + (t- wzfi [.4 + (”2W

0

 

= 0(1)

' a

76

 

 

 

 

But
b b b
JtJ t2 dt < t2 dt :4 d:
[e4 + (t-rpfit [e4 + (t-12)2];‘ " e4 + (is-1,72 e +(t-TZ)
a a
From the proof of Lemma 2.ll,
(l8) sup |I |< ct).
polelén 121
In like manner, we have
(19) su II km.
p.letsn 1221
By Lemma 2.15, there exists c>0 such that
2
|H(e1t)|2- eCt
and
2
|v(e1t)L; eCt for each point in N.
Then
b .2. 1
.t .t |t3|E2f IHIKIvIPH'k
.11222|:. 0(l) sup |f(e1 )l H“(e1 )l -*h "fiv— 2 1 2 dt .
t [6 +(t-r >1/
a 1
But
b b

2
|t3l ' p 1 |t3lPe-C(p+l)t

EEZ _ dt;= dt
[94+ (191921"2 k=l |H|k|v|p+1 k a [e4 + (t-,1)231/2

 

 

 

a

b 2
< lt3lpe'Cpt dt.
3

 

= 0(1) as p+oo ,9 +0)

77

by Lemma 2.13. Hence.

(20) su I I |<a3 .
p.|6l’:.n 1222

From (20), (l9), (l8), (l7), and (l6), we obtain

50 II |<OO.
Pafe :0 Y

Finally, from (15) and (14), we obtain

sup . lG(f°.p-l.e)| <60.
p9 6

OPEN QUESTIONS

The aim of this section is to present a series of questions
concerning U, its dual space and operators on U:
(1) As sets, “I is a proper subset of U2, and U2 is a proper
subset of ‘2' Does there exist p such that l<p<2 and I; is a

subset of U2 or U a subset of (L?

2

2
k
that there exists n>0 such that Ickl> n for k = 0,l,2,... Sufficient

(2) If {ck} e Uz', then a necessary condition for {%—J a U is

conditions for {%_J 5 U2' when {ck} a U2' are (l) there exists
k

n>0 such that Ickl> n for k = 0,1,... and (2)[cJ is of bounded

variation. Establish necessary and sufficient conditions for a

sequence to have an inverse in U2

(3) A question related to (2) is "What are the homomorphisms on UZ'?”

(4) If f belong to U, with partial sums Sn(z) obeying,

inf (sn(z)|:§>o. will %-belong to U?-

n,(z(;}
(5) A more difficult question is ”If f belong to U, If|>0 on

lzlgj, will %-belong to U?"

(6) Given o(0<|o|<l), will fo¢ be in U for all f in U, where

 

78

BIBLIOGRAPHY

BIBLIOGRAPHY

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2. L. Alpar, "Remarque sur la sommabilite des series de Taylor
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the Mathematical Institute gj_the Hungarian Academy of Sciénces
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3. L. Alpar, "Sur certaines transformees des series de ouissance
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4. B. Bajanski, "Sur une classe generals de procedes de sommation
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5. S. Banach, Theorie Des Operations Lineaires Monografje
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6. J. Clunie and P. Vermes, "Regular Sonnenschein Type Summability
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7. J. Clunie, "On Equivalent Power Series", Acta Mathematica,
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8. V. Cowling, "Summability and Analytic Continuation”, Proc.
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9. P. Dienes, The Taylor Series, Oxford at the Clarendon Press
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l0. G. Goffman and G. Pedrick, First Course in Functional
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ll. G. Hardy, ”A Theorem Concerning Taylor's Series"
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l2. G. Hardy, Divergent Series, Oxford University Press (l949).

 

13. J. D. Hill , "Summability of Sequences of 0's and l's",
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80

14. J. Kahane and Y. Katznelson, "Sur Les series de Fourier
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15. L. Loomis, Ag_1ntroduction to Abstract Harmonic Analysis,
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17.2. Nehari; Conformal Mappings, McGraw- Hill Book Company,
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18. A. Peyerimhoff, Lecture Notes on Summability, Winter and
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19. G. Piranian, A. Shields, and J. Wells, "BOunded Analytic
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20. C. Rickart; General Theory of Banach Algebras, D. Van
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21. W. Rudin, Real and Complex Analysis, McGraw- Hill Book
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23. W. Sledd, "Regularity Conditions for Karamata Matrices",
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24. P. Turan, "A Remark Concerning the Behavior of a Power
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