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ADDITIONAL PROPERTIES OF WEIGHTED SHIFTS
presented by
James Lowell Hartman
has been accepted towards fulfillment

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Ph . D . degree in Mathematics

 

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ADDITIONAL PROPERTIES OF WEIGHTED SHIFTS

BY

James Lowell Hartman

A DISSERTATION

Submitted to
Michigan State university
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1981

ABSTRACT

ADDITIONAL PROPERTIES OF WEIGHTED SHIFTS

BY

James Lowell Hartman

In this paper we examine some preperties of weighted
shifts which were previously unknown. Some of these
properties are stimulated by properties of the unweighted
bilateral shift. In particular, we define Toeplitz
and Hankel operators for weighted shifts. Some of the
properties that hold for Toeplitz and Hankel operators
for the unweighted shift carry over for weighted shifts in
general. However. there are some striking differences
which we point out with several examples. Some pr0perties
for unweighted shifts carry over for weighted shifts with
only minor modifications. This happens mainly when the
weighted shift under consideration has a periodic weight
sequence. Thus we also prove some properties for weighted
shifts with periodic weight sequences. The material above is

found in Chapters I, II, III, and IV.

In Chapters V and VI our work takes a slightly different
direction. In Chapter V, we concern ourselves with

answering Question 11 in Allen Shields' survey article

James Lowell Hartman

on weighted shifts [23]. In particular, we give a
general sufficient condition for the ”analytic” projection

on L”(B) to be bounded.

In Chapter VI, we concern ourselves with spectral
sets. We give a new proof of von Neumann's Theorem which
says the closed unit disc is a spectral set for all
contractions. We then investigate what happens when we
replace the disc with an annulus. In particular, we
answer the last half of Question 7 in Shields' article
with an example. Finally, we examine what happens
when we restrict ourselves to operators on two-dimensional

spaces.

[23] Shields, A., Weighted Shift Operators and Analytic
Function Theory, Amer. Math. Soc. Surveys
13 (1974). 49-128.

ACKNOWLEDGEMENTS

First and foremost, I would like to thank my
thesis advisor Dr. Sheldon Axler for his encouragement
and guidance during the course of this research. I
greatly appreciate the time and energy he gave to me
and his willingness to discuss questions I had. He
has given me a sense for what is important mathematically
by steering me in the right direction when I was tempted

to divert my attention to tangential interests.

I am also greatly indebted to my wife Peggy who
has supported and encouraged me constantly during our
time at Michigan State University. Her unfailing love
kept me going at times when I felt it was not worth it.
Words cannot express the importance that’her life is to

mine.’

I would also like to thank Cindy Lou Smith for
typing this manuscript. I have great admiration for
anyone who is able to read my writing and put up with my
careless errors in the original draft. I thank her again
for the excellent job she did in taking my sloppy draft

and putting it into good form. I also would like to

ii

thank other members of the faculty and staff in the
Mathematics Department of Michigan State University for

their support during my time as a graduate student there.

Finally, I would like to thank my parents Lowell and
Doris Hartman for all their love and support. Their help
and guidance in my life has been invaluable. I would
also like to thank my wife's parents Art and Lydia Kursch.
our friends Tom and Doris Schumacher. and Gene and Jewel
Ratzlaff. Last, but certainly not least. I would
like to thank Pam Gorkin with whom I spent many hours

discussing the intricacies of graduate school.

iii

TABLE OF CONTENTS

Chapter Page
I. WEIGHTED SHIFT OPERATORS . . . . . . . . . 1
II. SHIFTS WITH PERIODIC WEIGHT SEQUENCES . . 8
III. TOEPLITZ OPERATORS FOR WEIGHTED SHIFTS . . 21
IV. HANKEL OPERATORS FOR WEIGHTED SHIFTS . . . 40
V. ANALYTIC PROJECTIONS FOR WEIGHTED SHIFTS . 59
VI. SPECTRAL SETS . . . . . . . . . . . . . . 74
BIBLIOGRAPHY . . . . . . . . . . . . . . . . .

iv

CHAPTER I

WEIGHTED SHIFT OPERATORS

Let H be a complex separable Hilbert space
with orthonormal basis {en : n E Z} . We denote the
set of bounded linear operators on H by 6(H). A
bilateral weighted shift T is a bounded linear Operator
on H that maps the basis vector en into a scalar
multiple wh of en+l' i.e. Ten = when+l' Weighted
shifts have been used generously through the years to
provide examples and counterexamples to questions in
operator theory. However, the first specific study of
weighted shifts was done only recently by R.L. Kelley
in his unpublished dissertation at the university of
Michigan in 1966. Since that time many other properties
of weighted shifts have been identified and examined.
Most of what is known about weighted shifts is compiled
in the survey article by Allen Shields, "Weighted Shift
Operators and Analytic Function Theory," [23]. This
article contains all of the basic facts one needs when
working with weighted shifts. It also contains a list
of unsolved problems concerning weighted shifts. We
will rely heavily on the material found in Shields'
article and use the notation develoPed there. The
article contains facts about both bilateral and unilateral
weighted shifts. We will restrict our attention here to
bilateral weighted shifts which are injective and

l

also whose weight sequences {wn : n E Z] consist only
of positive terms. That it is sufficient to do this is
pointed out by the following pr0position. Before that
prOposition, we should note that when we say weighted

shift, we mean an injective, bilateral weighted shift.

Proposition 1.1: (Shields,[23]) If T is a
weighted shift with weight sequence {wn}, then 7T
is unitarily equivalent to the weighted shift with

weight sequence {\wnll.

The assumption that T is injective guarantees
that there are no weights which are zero. Noninjective
shifts may be studied by considering them as a direct
sum of weighted shifts in which some of the summands may
operate on finite dimensional spaces. Thus from here on.
if T is a weighted shift with weight sequence {wn}, it
will be understood that wn > O for all integers n.
Given such a weighted shift, the following definitions

are made.

Definition 1.1: Let T be a weighted shift with

weight sequence {wh}. Then

n-l
i) B(n) = n wk if n > O
k=O
ii) B(n) = 1 if n = O
_l _1
iii) mm) = 11 wk) if n<O

Definition 1.2: Let T and B(n) be as in

Definition 1.1. Then

m A
L2(B) = [f = '2: f<n>znz

n=-m

A “1A 22
f(n) E C for all n and Z, \f(n)\ B (n)<(m}

n=-m

on
A
We note that the sum 2) f(n)zn is not taken in a
n=-m

literal sense at this point. It is taken in the formal
sense that L2(B) is a set of sequences indexed by the
integers with the summand f(n)zn indicating that f(n)
is the nth term of the sequence f. For f,g E L2(B)

we define an inner product

 

so

A
(f.g) = Z [f(n)g(n)52(n)

n=-m

With this inner product. L2(S) is a Hilbert space with
addition and scalar multiplication of vectors being
componentwise. The set {zn : n E 2} can be thought of

as an orthogonal basis of L2(B).
For f,g E L2(B) we define a formal product h = fg

”An A °°/\/\ ,
by h = Z) h(n)z where h(n) = Z) f(k)g(n—k) If all

n=-m k=—m

of the latter sums converge. We note that this mimics the
multiplication of analytic functions whose Laurent series
would be given as f and g are given. Now let

L°°(B) = {(9 e 142(3)sz e Lzm) for all f e LZUBH.

Then for m e Lm(B) we can define the linear map
M% :I?(B) 4 L2(B) by be = ¢f. Under these definitions

we have the following theorem.

Theorem 1.1: (Shields,[23]) For T 6 LF(B),
M.cp is a bounded linear operator on L2(B) and M2
is unitarily equivalent to the weighted shift T e 8(H).

Furthermore. under this unitary equivalence. {Mb =¢ 6 LP(B)}

corresponds to the commutant {T}’ = {S e 6(HJ :ST TS}

of T.

The theorem above says that a weighted shift, which
weightedly shifts an orthonormal basis, can also be thought
of as an unweighted shift of a weighted basis. This
follows from the equalities below.

°°/\ °°/\

Mz(f) = zf = Z f(n—l)zn = '23 f(n)z

n= -oo n=-m

n+1

The second equality comes from the identity:
as

;f(n) = Z) g(k)f(n-k) = f(n-l) .
k=-m

At times it is convenient to think of T strictly as
a weighted shift on‘ H. At other times. though, it
is helpful to think of T as M2 on L2(B). During
the first four chapters. I will basically think of a
weighted shift as M2 on L2(B). However, in the

last two chapters, I will sometimes think of them as

‘weighted shifts on unweighted spaces. Some further

notations, definitions, and facts are as below:

A
{f 6 L2(B) :f(n) = O for all n < O]

111
N

11>

ll

B(B) = {m E Lm(B) :M; = MW for some W e Lm(B)}

o(T) = {A 6 ¢ :T-—AI is not invertible}

H
*3
ll

sup{\z\ :Z G 0(Tll

For e e Lm(B). we let Hen” = “Man. When L”<e)
is endowed with this norm, it is a commutative Banach

algebra.

Definition 1.3: For f 6 L2(B) we define

 

—~ 2
f e L (B) by

 

A A
f(n) = f(-n)B(-n)/B(n)

From the definitions above, the following facts

are easy to verify.
2 . 2
l. H (B) is a closed subspace of L (B). (The
coefficient maps Tn :I?(B) 4 C given by

A n 2 .
rn(f) = f(n) = (f,z )/B (n) are continuous.)
2. {xI:i.e;¢} c B(B)c:IFKB)
3. For f e L2(B), HfH2 = (f,f)l/2 = WEN;

4. If T is invertible then o(T) =

{z 6 ¢ :r(T-1) g_lz\ g r(T)}. If T is

not invertible then o(T) = {z 6 ¢ :\21 g_r(T)}.

5. For cpELw(5)o ”CPH2=HMCP(1)H2 S ”McpHHJ-HZ S “CPU“, °

Also, we will let p : L2(B) » H2(B) be the
orthogonal projection of L2(B) onto its closed sub—
space H2(fl). This projection is described by the

formula:

m A an A
P( 2: f(n)z“> = 2:f(n)zln for f e Lzm) .
n=O

n=-m

* 'k
If we let WO(T ) be the point spectrum of T
*
(i.e. the set of eigenvalues of T ), then we have the

following theorem.

Theorem 1.2: (Shields,[23]) Let T be a weighted

 

shift represented as M2 on L2(B). Then the following
properties hold:

1 -l

a) If T is invertible and r(T- ) < lw‘ < r(T),

m . m A
then )‘w: L (B) 4 (1; given by Awhp) = Z cp(n)wn = cp(w)

n=-m
is a multiplicative linear functional on Lm(B). Thus
1 -l
)

lwlwll S Hell“, when r(T" < \w\ < r(T).

b) If w 6 nO(T*), then there exists kw 6 L2(B)

* -———— no
such that Mcpkw = cp(w)kw for all m E L (B).

'k
c) For w E nO(T ) and kw as above, we have

(D

e A n 2
(f,kw) = Z, f(n)w = f(w) for all f E L (B).

=-co
We note that this theorem implicitly says that

w A
I n U I
the series Z} f(n)w converges when w is as given.
n=-m

Before we go on to Chapter II, we should mention
a very important weighted shift. It is called the unweighted
shift because all of its weights are l's. For the
unweighted shift, L2(B) is the space of measurable
functions on 513 = {z E ¢ :lzl = 1] whose absolute
values are square integrable with respect to arclength
measure. ‘We will denote this by L2(a])). Furthermore,
L"(B) is the set of essentially bounded measurable functions
on BI). We will denote this by L?(613). (See Douglas,
[6]). We will refer to the unweighted shift and its
properties quite frequently. In particular, we will
use it as a model for some of our definitions and lines

of thought.

CHAPTER II

SHIFTS WITH PERIODIC WEIGHT SEQUENCES

We recall that {AI =1 6 ¢} C B(B) C Lm(B). One
may ask whether equality holds at either end of this
chain of inequalities. To answer this question, we

present the following lemmas and theorem.

. A
Lemma 2.1: If p 6 B(B) and ¢(N) # 0, then
B(N+k) = B(N)B(k) for every integer k.

Proof: $(N) = (ezk,zN+k)/BZ(N+k)

= (Mcpzk.zN+k)/BZ(N+1<)

= (2k.M;zN+k)/BZ<N+k)

*
= (2k,M zN+k)/B2(N+k) where M = M¢

W W

 

= $(-N) 52(k)/B2(N+k) for each k.

 

A
Thus 52(N+k)/B2(k) = $(-N)/¢(N) for all k. Letting
_ K _ 2 2 2
k — O we get ¢(-N)/¢(N) - B (N). Hence B (N+k)/B (k) =
52(N). Hence 32(N+k)/Bz(k) = B2(N) for every integer
k. This is the desired result since B(i) > O for

every integer i. Q.E.D.

We also note at this point that B(N+k) = B(N)B(k)

for all k implies B(-N) = 1/B(N).

Lemma 2.2: Assume there exists an integer N
such that B(N+k) = B(N)B(k) for all k. Then

wN+k = wk for all k (i.e. the weight sequence for

the weighted shift is periodic).

Proof: w = B(N+K+l)/B(N+k) = B(N)B(k+l)/B(N)B(k) =

N+k
B(k+l)/B(k) = wk. The second equality above holds because

of the assumption on the sequence {B(n) : n e Z} . Q.E.D.

Theorem 2.1: Let T be a bilateral weighted shift

 

with periodic weight sequence of least period N. Then
B(B) = {T e L”(B) :$(n) = O for all n which are not

integer multiples of N}.

A
Proof: Suppose T 6 8(5) and ¢(n) # 0. Then
Lemma 2.1 implies B(n+k) = B(n)B(k) for all k. Lemma 2.2

then implies wh+k = wk for every k. This implies

that n = mN for some integer m since the least period
of the weight sequence is N. Thus B(B) c E =

A
{w E Lm(B) :¢(n) = O for n not an integer multiple

of N}.

*
Now let m Q B. We will show that M1 = M5. This
k

'k —— ...
will be true if and only if M 2 = wzk = M:(w) for all

W

k. The reason for writing this in such a strange way

is that it is unknown whether W e LQ(B) implies

'V E Lm(fi). Also, we should mention that an injective

bilateral shift with a periodic weight sequence
k

is invertible. Hence Mz E 6(L2(B))

10

for every integer k. Continuing with the proof now,

we have:

(M* zk.z“)/Bz(n) (2k;¢zn)/Bz(n)

W

$(k-n) 32(k)/Bz(n>

(zk EznVBZm)

(M: T:zn)/I32(n)

A
((n-k)

7r—
wk-n) B(k-n)/B(n-k) .

A
Now if w(k-n) # 0 then k-n = mN for some integer m.

Hence
B2(k)/Bz(n) = {32(n+mN)/Bz(n) = 52mm
= B(mN)/B(-mN) = B(k-n)/B(n-k) -
So we have (M; 2k,zn) = (M: Eflzn) for all integers n.
This shows that M; 2k = $.zk for every integer k. Q.E.D.

Corollary 2.1: under the involution w 4‘T. B(B)

* a:
is a commutative C -subalgebra of L (B).

Corollary 2.2: For shifts with periodic weight
sequences with least period N, M n is normal if and

2
only if n is an integer multiple of N.

*
Proof: For n = kN, 2n 6 B(B). Thus M = M which

n '—n
. . * * . a . 2 z .
implies M n M n = M n M n Since L (B) is a commutative
z z z z *
Banach algebra. For the converse, we assume M n M n =
z z

'k
M n M n' This implies that
z z

ll

(zn+k'zn+m) = (M 2k,M nzm)
Z Z
= (M*n M nzk,zm)
Z Z
*
= (M n M n2k, m)
2 Z
= (M* 2k,M*nzm)
Zn 2
= (2k n 2m" n)B2 (MB2 (m)/Bz(k-n)B2(m-n)

B4(m)/B2(m-n) if m = k
0 if m # k

B2(m+n) if m = k
= . Thus we must

0 if m # k

(zn+k zn+m)

I

Also,

have B(m+n)B(m-n) 32(m) for all m. Hence
B(m+n)/B(m) = B(m)/B(m-n) for all m, which implies
B(m+n+l)/B(m+l) = B(m+l)/B(m-n+l) by using m4-1 instead
of m. Now dividing the latter equality by the former

one gets

[B(m+n+l)/B(m+n)][8(m)/B(m+l)] = [8(mwl)/B(m)][B(m-n)/B(m-n+1)]

or equivalently

w - w2/w
m+n m m—n

Now we prove by induction that wkn+m = (wn+m/wm) Wm

for all k 2_O. The equality is certainly true for k = O

and k = 1. So assume it is true for k‘g L. Then

12

w(1,+l)n+m = w(2n+m)+n

2

= wzn+m/w(z-l)n+m

21 2 L- -l
[(WMm/wm) (Wm) ] / [(wn+m/wm) Wm]

3+1
(wn+m/wm) wm °

Thus the identity is established for k = 1+1. By
induction then, we have wkn+m = “n+m/wm) wm for all
k 2.0.

If we now let m be fixed and let k get large,
we see that wn+m = wm since a periodic weight sequence
is bounded above and bounded away from zero. This is
true for all m and hence {wm} is periodic with

period \n‘. This implies that n = kN for some integer

k. Q.E.D.

Corollaryiz. 3: For m 6 6(B) and f 6 L2(B)

we have ¢f= ¢f.

an

A A
Z} c (k)f(n-k)
k=—w

/\
Proof: ($f)(n)

on

$7....
k2: m(-k) f(n-k)B(-k)/B(k

 

° A A
[B(-n)/B(n)][ Z) o(-k)f(n-k)

fi(n)B(-k)/B(k)B(-n)]

 

[m-nvemm Z $<-k)’f(n-k)

k=—m

B(n-k)/B(k-n)]

13

since $(-k) #’0 implies B(-k)B(n) = B(n—k) and
B(k)B(-n) = B(k-n). Thus (Ean) = [B(-n)/B(n)]

 

” A .e 7sr———' ,
I :3 ¢(-k) f(k-n)] = [B(-n)/B(n)]¢f(-n). This last term
k=-w

4: éh- 4>
is just ¢f(n), however. Thus (¢f)(n) = ¢f(n) for all

n implying Ef = 3?. Q.E.D.

Corollary 2.4: The equality B(B) = Lm(B) holds

 

if and only if all the weights are equal.

Proof: If 3(5) = L”(e), then 2 6 3(3). This
implies the weight sequence is periodic with period one.
Thus wk+l = wk = wO for every integer k. The converse

is immediate from the characterization of B(B) given

in Theorem 2.1. Q.E.D.

The following corollary follows immediately from

Theorem 2.1.

Corollary 2.5: The equality B(B) = {11 :1 6 ¢} holds

if and only if the weight sequence is not periodic.

We now recall that a von Neumann algebra of operators
in 8(H) is a selfadjoint algebra of Operators in 8(H)
which is closed in the weak operator t0pology. The
algebra B(B) is an abelian von Neumann algebra.
However, the following pr0position says that it is

not maximal.

14

Proposition 2.1: If {l1,:l 6 ¢} c B(B) g Lm(B).
then {Mb :¢ € B(B)} is not a maximal abelian von Neumann

algebra.

.grggf: If B(B) = (11 3A 6 ¢), we let S be any
nontrivial, selfadjoint Operator and consider the
von Neumann algebra generated by B(B) and S. This
will be abelian and properly contain B(B). If
B(B) # {11 =1 6 ¢}, let N > 1 be the least period
of the weight sequence. Let A E B(L2(B)) be given by

N) = z(k+1)N

A(zk for any integer k and A(zn) = 0

otherwise. Then a direct computation shows that

N (k‘1)N and A*(zn) = O for n y'kN.

(k—1)N

A*(zk ) = 32(N)z

kN

Hence, AA*(zkn) A(BZ(N)z ) = BZ(N)zkN = A*(Az )

* n * n .
and AA (2 ) = 0 AA (2 ) for n #‘kN. Thus A is

normal. Also for T E B(B), AM¢(zk) = A(¢zk)

m A
= A( Z) o(zN)zzN+k)
L=-w
o if ksimN
= a A
Z chN)zLN+k+N if k=mN
L=-o

Now if we look at M¢A, we see that M.A(zk) = 0 if
k+N)

k # mN. If k = mN, then M¢A(zk) M¢(z

= w zk+N

A
Z} o(LN)z

=—m

E N+k+ N

15

Hence AM¢(zk) = M¢A(zk) for all k, implying

AMw = MbA for all m E B(B). By Fuglede's theorem

(Rudin,[l7],Theorem 12.16), A* also commutes with

everything in {M¢::¢ e B(B)}. Hence the von Neumann

algebra generated by {Mb :w e B(B)] and A properly

contains {Mb =¢ e B(B)] since {A g {ME :w E B(B)}. Q.E.D.
We now discuss some properties of weighted shifts

which have periodic weight sequences.

Proposition 2.2: Let T be an injective bilateral

shift having a periodic weight sequence with least period

N-l l/N
N. Then B(n) = rno(n) where r = ( n w ) and
k=0 k

o(n) is periodic.

Proof: Suppose n 2_l and n = tNa—s where 0 g.8 < N.

n-l N—l t
Then B(n) = n w = ( n w ) (w w ---vr ) .
k=0 k k=0 k 0 1 5-1
N—l (tN+s)/N N-l s/N
=(HW) (WW°°°w )/(1'IW)
k=0 k 0 1 5-1 k=0 k

rn[(wowlo o . ws-l)/(wo . . o o wN_l)S/N] .

Since 0 g_s < N, the right half Of the product above is
a bounded sequence o(n) which is periodic and has

o(kN) = l for all k“; 0.

NOw suppose n = tNi-s where t < O and
-l

"N < S S 0. Then B(n) = ( II wk)-‘t‘ (w_1----wS)-l
N-l tN+S
_ —N- ‘1 S/N

= rn(w-1. o o ows)-1/(woo o o .wN-l)S/N .

16

Again since —N <53 g_O, the right half Of the product
is a periodic sequence o(n) with d(-kN) = l for all

k 2_o.

Thus B(n) = rnd(n) where o(n) is bounded. TO
see that o(n) is periodic overall we note that for all

integers k:

a<N+k)/a(k> sownrk/B(k)rN+k

B(N)B(k)/B(k)rN = Bow/r“ = 1. Q.E.D.

Proposition 2.3: For injective bilateral shifts having
periodic weight sequences with least period N and

r = B(N)1/N, we have the following:

i) r(T- ) = r(T) = r

ii) “T“ = max{w0,°'°':WN_l} and HT—lu =

min[w0.°"'.WN_l)

... -1 -1 -1 -1
iii) If N # 1, then HT H < r(T ) = r(T) < HTH

iv) There exists a constant L > 0 such that

L sup{B(k-n)/B(k) zk,2 Ol'g.inf{B(k-n)/B(k) =k 2,0}

for all n

°°A
v) f 6 L2(B) if and only if 23 f(n)rnzn E L2(aIn

—m
Proof: Parts i) and ii) follow from the corollary

to Proposition 7 in Shields [23] and the equation

r(T) = lim HTnHl/n. Part iii) then follows immediately

nam
from parts i) and ii).

17

For part iv) we consider

sup{B(k-n)/B(k) :k 2_O} sup{rk-no(k-n)/rko(k) :k 2.0]

r-n sup{o(k-n)/d(k) k.2 O}

r-n max{d(k—n)/O(k) N‘Z k 2_O}.

Likewise
inf{B(k-n)/B(k) :k 2 o] = r‘“ min{o(k-n)/d(k) : N 2 k 2 0}.

Now let 1 = min{d(k-n)/O(k) :N 2_k 2 O}/max{o(k-n)/O(k):
N 2_k 2_O). Since o(n) is bounded away from zero, we

have L > O. This constant L satisfies the desired

property.
To prove v), note that f E L2(B) implies
Z) lf(n)1232(n) < w. so Z) \f(n)(2r2n02(n) < w,

n=-m =—w
Using the boundedness of {o(n) : n 6 2} again, we see that

an A a A
Z) \f(n){2r2n < a. This implies Z} f(n)rnzn € L2(aIU .

n=-Q 11:-”

The converse is established by tracing this argument
backwards Since the statements are equivalent at each

point. Q.E.D.

Corollary 2.6: Let R.:I?(B) 4 L2(a])) be given by

m
A
R(f) = Z) f(n)rnzn. Then R is a similarity between
n=-m

M2 on L2(B) and a scalar multiple of the unweighted

shift.

18

Proof: The fact that R is bounded and invertible
follows from the boundedness of o(n): or one may appeal

to the closed graph theorem. For f 6 L2(aID) we have
-1 a A -n n
RMZR f = RM2( 2 f(n)r z )

n=-m

R( Z f(n)r-ner'l)

n=-m

m

A
Z, f(n)r

n=-w

-n n+1 n+1
r z

r( Z) f(n)zn+l) . Q.E.D.

=—m

Proposition 2.4: If m E Lm(B) and f e L2(B),

then R(cpf) = R(cp)R(f).

/\ n
Proof: R(pf)(n) r (of)(n)

r“ 23 $(k)?(n-k)

23 $(k)rk f(n-k)rn-k

a”~‘\~
[R(cp)R(f)](n) -
This is true for all n. Hence R(wf) = R(¢)R(f). Q.E.D.

Corollary 2.7: The map R is a Banach algebra

isomorphism between Lm(B) and L¢(al)).

19

3:22;; If w 6 LF(B). then Tf E L2(B) for all
f E L2(B). Then by Proposition 2.4, R(¢)R(f) 6 L2(BID)
for all f 6 L2(B). This says R(m) 6 L°(aIH since
{R(f) :f e L2(B)] = L2(a]D). The converse again is
achieved by tracing the argument above backwards. Thus
:9 e L°°(B) iff R(cp) e L°°(an) . We also have that

HR(C(>)R(f)H 2 g HR<cpf>H 2
L (BIN L (BIN

s.uRnuef(2
S.HRHHmeHfH2

$.HRHHR-1HHwHwHRfH 2 ,
L 61))

Hence nR<e>n . S.HRHNR-1HHmHm- Thus
L (BIN

R.:LP(B) 4 Lf(a]3) is continuous. By the Open mapping
theorem, so is R-l. The only thing left to verify is
that R(ww) = R(¢)R(w) for ¢,¢ 6 L”(B). This follows

from PrOposition 2.4. Q.E.D.

Corollary 2.8: Suppose T is a weighted shift
with periodic weight sequence. If ¢ E L@(B) and

O #’f E H2(B). then ¢f = 0 implies w = O.

2322:: If ¢f = 0 then R(mf) = R(¢)R(f) = 0.
But 0 # f E H2(B) implies O # R(f) e H2(al)). However,
R(m) e LF(a]D) and R(¢)R(f) = 0 implies R(o) = 0 from
the F. and M. Riesz theorem (see Douglas,[6]). Finally,

R(m) = 0 if and only if T = O. Q.E.D.

20

From the work above several questions may have
arisen. First Of all, we know that m e B(B) implies
3'6 B(B), and hence 6.6 Lm(B). At this point we may
ask the question; does a e Lm(B) imply 5’6 L?(B)?
The answer to this question is unknown. However, in
special cases one can answer the question affirmatively.
For example, if the weighted shift is rationally strictly
cyclic (see Shields, [23], p.101), then the answer is
yes. An affirmative answer is also Obtained if B(k) = rk
for all k or if B(k) = B(-k) for all k. For bilateral

shifts with periodic weight sequence the answer is determined

by whether o(—n)/O(n) is a multiplier on L?(a]D).

The other question involves Corollary 2.8. Can one
say that this corollary holds for all bilateral weighted
shifts and not just for those with periodic weight sequences?
Here again, one can say something about particular weighted
shifts. For example, if nO(T*) contains an Open annulus,
then one can answer yes since in this case the functions
w and f have Laurent series which converge on the annulus.
Thus w and f are analytic on this annulus. However,
it is not true that w0(T*) contains an Open annulus

for all weighted shifts T.

CHAPTER III

TOEPLITZ OPERATORS FOR WEIGHTED SHIFTS

Toeplitz Operators Tcp : H2(aD) 4 H2(a]D) for
w E L?(al)) have been studied quite widely in recent
years. (Douglas, [6]; Halmos, [8], or Sarason, [21]).
Some interesting properties have emerged from these
studies along with generalizations to ¢n and other
H2 spaces. (Gohberg, [7]; Abrahamse, [1]; or Devinatz,
[5]). The idea used here is that M2 on L2(aIn is
a weighted shift with all weights equal to 1. Then
{ME :¢ 6 Lm(a]))] is just the commutant {Mz]’ of M2
on L2(a])). For weighted shifts whose weights are not
all 1, we can then generalize the idea of a Toeplitz
Operator Tw.:H2(B) 4 H2(B) for m e Lm(B). This
follows from thinking of a weighted shift as M2 on L2(B).
The commutant Of T is then identified with [ME :m 6 LF(B)].

Hence we have the following definitions.

Definition 3.1: For n e L”(e). let T¢ e B(H2(B))

be given by Tm(f) P(¢f). Here P is the orthogonal

projection of L2(B) onto H2(B) given in Chapter I.

on 03 A
Definition 3.2: w'(e) = {a e L (e) :¢(n) = o

for all n < 0].

We first note that flp(B) is a closed subalgebra
of Lm(B). It is closed since Th :IF(B) 4 ¢ given by

A
Fn(m) = ¢(n) is continuous for each n. It is an

21

22

/\
algebra since if ¢,w 6 NP(B), then (mw)(n) =

Z) $(k)$(n-k) = 0 if n < 0. Thus $1 6 HQ(B).
k=O

At this point, one may ask which prOperties of Toeplitz
Operators on H2(a])) carry over for Toeplitz Operators
on H2(B). Many of the same prOperties do hold,

some with minor modifications. For example, if

m 6 Lm(a]D) then M; = M5. This property does not
hold for all w e L”(e) in general. It will hold,
though, if e e B(B). Hence some of the prOperties for
Toeplitz Operators on H2(B) will not hold for all

e E LF(B), only for some subset of LF(B). Also, we
may be able to show some properties hold for certain
classes of weighted shifts (e.g. those with periodic
weight sequences). However, there are some striking
differences which will be pointed out. We will now

examine prOperties of Toeplitz Operators on L2(B).
*
Proposition 3.1: If m E B(B). then Tcp = T5.

Proof: Suppose w 6 3(8), then 6'6 Lm(B) and

M; = M—. It is well known that A 6 B(L2(B)) and

o
S = PA] 2 implies that S* = PA*[ 2 . Thus
H (B) H (B)
* 'k
T = PM \ 2 = PM-[ 2 = T-. Q.E.D.
”Hus) c"B(B) “P

23

The "converse" of Proposition 3.1 is not true,

1

then it is not necessarily true that W = $3 To see

*
however. That is, if TT = T for some W e L°(B),

this, consider the following example.

Example 3.1: Let the weight sequence for the
weighted shift be given by i) wn = 1 for n # -1 and
ii) w_l =-%. Then B(n) = l for all n 2_O and
B(n) = 2 for all n < O.

For n,k 2.0, (Tzzk,zn) = l for )(= n-l and

(Tzzk.z") = o for k g n-l. Thus T;(zn) = P(z“'1) =

T _1(zn). However, '5 = z-lB(l)/B(-l) =-% z.1 implying
z

._ m _.l
2 e L (B) and T; — 2 T

Proposition 3.2: Let a: L”(e) 4 B(H2(B)) be
defined by 9(w) = Tm. Then @ is linear and contractive.

Moreover, Q)B(B) is *-linear and contractive.

Proof: It is easy to see that e is linear.
Also, by PrOposition 3.1 we need only verify that e is
contractive. This is easy, however, since for f e H2(B)

and $ 6 LF(B) we have:
HTcprZ = HP(CPf)))2 S ”CPfHZ S H‘PHODHfHZ -

The last inequality holds, of course, by the definition

0f Hme- We thus have HTgH S.Ume- Q-E-D-

24

We now have the following proposition which was
proved by Brown and Halmos [9] in the unweighted case.
We note that in the statement Of the proposition, we
say 'T 6 H2(B). This just says that $(n) = O for
n > 0. We use this instead of T'e MQ(B) since it is
unknown whether w E L”(B) implies Tie Lm(B) as

mentioned before.

Proposition 3.3: Let ¢,¢ 6 L?(B). Then

TwTo = Tww if and only if m E NQ(B) or T'é H2(B).

Proof: Suppose ¢ 6 ¥m(B). Then for f e H2(B)
it is easy to see that mf E H2(B). Thus wa = mf.
Then T T f = T (of) = P((¢¢)(f)) = T

W w W W
E'E H2(B). For k 2,0 let h = TWT$(zk). Then for

¢(f). Now suppose

n 2.0

(TwT¢(zk).z“)

(T ( $(2—k)zz).zn)
W o

L:

32(n)h(n)

A
((( z:e(z-k)z‘).z“)
L=0

2 °°A A
B (n)( Z)e(L-k)w(n-L))
i=0

a A A
B2(n)( Z3o(L-k)w(n-z)) .
L=n

If f = T (zk), then
WT
A
52(n>f(n)
Letting z’ = L-tk
32(n)?(n)

25

for “.2 O

= (((e)(zk).z“)
2
= B (n)(wo)(n-k)
2 ”AA
= B (n) 23 o(z)¢(n-k-£)
we get:
”A A
= 32(n) ,2: e(z'-k)((n-:')

l

B2(n) 22$(z’-k)$(n-2’)

£’=n
= 52(n)h(n)
k k . . .
Thus Tme(z ) = T¢m(z ) for all k 2_O. This implies
T - T since {zk :k 2_O] forms an orthogonal

w T ' (w 2
basis for H (B).

T A
Zcp(-£)
i=1

(*)

Now for the converse, suppose T¢T¢ = Tww and
to E'Np(B). Then for k,n 2 O, we must have Z) $(L-k)$(n-£) =
L=-w
” A A .
Z}¢(L-k)w(n-L) from the calculations above. Thus
{:9
$(L-k)®(n-L) = O for k,n 2_O or equivalently
£=-w
” A A .
Zim(-L-k)w(n+z) = O for all k,n 2 0. Taking k = O
L=l
and n = m we get

A
¢(m+z) O for all m.2 O .

26

m A A
Also letting L’ = zi-k we get Z) m(-z’)¢(n-k+z’) = O.
L’=k+1

Then for n = ki-m we have

m A A ,
(**) Z) w(-£)w(m&z) = O for all m.2 O, k 2_O .
L=k+l

k A A
Putting (*) and (**) together we get Z)¢(-z)¢(m+z) = O
£=l

for all m‘z O, k.2 1. Since ¢ g’y”(e) there exists
A
N > 0 such that ¢(-N) # 0. Now by the first part of

this prOpOSition TmeT N-l = TWT N-l = T N—l' SO
A 2 :pz 1].!th
we may assume that ¢(-l) # O. Letting k = l in our
A
last sum above, we get ¢(m+l) = O for all m“; 0.

Thus '3 E H2(B) as desired. Q.E.D.

In the case of the unweighted shift, it has
been shown (Douglas, [6]) that if T E Lw(al)), then
0(MT) C 0(T¢)' This inclusion is used to show that
T = M
1|:le n“,

inequalities: HM¢H = r(Mb) g_r(T¢) g_HT¢H g ”MT”.

 

 

. The proof is given by the chain of

Thus Q :IF(a])) 4 B(H2(al))) is an isometric *—homo-
morphism between L?(a])) and a closed subspace Of
B(H2(a]))). However, in the case when not all of the
weights are 1, it is not necessarily true that

r(Ma) = ”Mb”. This will be seen in Chapter 6. One
may still ask, though, whether there is a spectral
inclusion theorem (o(M¢) c o(Tw)). This property

is examined in the next two results and the example

following them.

27

Proposition 3.4: Let T, a weighted shift with
periodic weight sequence, be represented as M2 on
L2(B). If T E Lm(B) and To is invertible then

M is invertible.
CD

Proof: Since T is invertible there is a
constant C1 > 0 such that “TfH2.2 Hvanzlz cleH2
for all f 6 H2(B). Also, there is a constant c2 > O

* 2
such that ”Tmfuz‘z CZHfH2 for all f 6 L (B). Now
let n > O and consider

Q

( z:(§(k)(252<k-n))1/2
=o

Hz-anZ

co

/\
= [k (f(k)l232(k)B2(k-n)/Bz(k)]l/2
=0

” A
g,(k2:(f(k)(252(k))l/Zsupte(k-n)/B(k) :k 2_o}
:0

By PrOposition 2.3 there exists 1 > 0 such that

SUP{B(k-n)/B(k) zk 2.0} g.inf(B(k-n)/B(k) :k 2.01/1. Thus
uz‘“fn2 g_HfH2 inf(B(k-n)/B(k) zk 2.0}/2
g_HT¢fH2 inf{B(k—n)/B(k) :k 2_O}/£cl
g_Hz'nT fHZ/zcl
m
g.H2'n%ng2/zcl
g HM¢(z-nf)H2/‘C1 since Mz_nMhp = M¢M2_n .

Now since {z-nf :f 6 H2(B),n > O] is dense in L2(B),

we have Mcp is bounded below on L2(B).

28

We will now show that M; is bounded below.
The two conditions that M¢ is bounded below and M;
is bounded below then imply M is invertible
(Douglas, [6], p.84). TO prove M; is bounded below
we attempt to imitate the proof that M¢ is bounded

below. A new difficulty is encountered here since M -n
2

does not necessarily commute with M;. However, we
note that it is sufficient to use n = kN where N is

the period of the weight sequence and k is a nonnegative
integer. It is sufficient since {szNfz f e H2(B), k‘z O]

is also dense in L2(B). Now we have

-kN . -kN * . . -kN * 1 * -kN (

((2 fnz g ((2 Tcpf512/c21. g ([2 Mcpf,)2/£CZS (\Mwm f).lz/zc2 .
. -kN . . .

Noting that z e B(B), the last inequality is a

result of the following equation.

M M* M*——M* M*M-*—— M*M
z—kN m z-kN ¢ T z-kN T z-kN
*
Thus both M.CP and Mfi are bounded below. Q.E.D.

Thus we do have a spectral inclusion theorem for
shifts with periodic weight sequence. We note that a
key part Of the proof involved the existence Of a
constant 2 > 0 such that L sup{B(k-n)/B(k) :k.2 0] g
inf{B(k-n)/B(k) :k.2 O]. The following theorem shows
that this is a sufficient condition for a Spectral

inclusion theorem.

29

Theorem 3.1: Let T be an invertible

 

weighted shift for which there exists a constant
1, > 0 such that zsup{B(k-n)/B(k) :k 2 0} g
inf{B(k-n)/B(k) :k 2 o}. If cp e L°°(B) and Tcp

is invertible, then MT is invertible.

.2E22£= We again prove that both M.cp and M;
are bounded below. The proof that M.cp is bounded
below is exactly the same as in Proposition 3.4. We
note, however,that if the weight sequence is not
periodic, then B(B) = {XI =1 6 ¢]. Thus we cannot
use the same idea as in the last part Of Proposition 3.4.
TO show that M; is bounded below we will first show

*
that U = {M nf :f E H2(B), n'2 O] is dense in L2(B).
2

If g = 23 3002“ 6 13(5). then (m*n)'1(g) e H
=-n 2

2(B)

as shown by the following computation:

* '1 k -l k
((M n) (g).2 ) = (g,M nz )
z z
= (g,M _nzk)
z
" (gozk-n)

3(k-n)B2(k-n)

=0 if k<o.

* * -
This implies g = M n((M n) 1(g)) e U and shows that U
z z

is dense in L2(B).

30

NOw

” A
llM*anZ = (k2 [f (k+n) \ 2:32 (k+n)B2 (mm/{32 (R) )1/2
2 =—n

g (R): \f(k+n){232(k+n))l/zsup{B(k+n)/B(k) :k2 -n]
=-n

g Il’fH2 sup{B(k)/B(k-n) :k 2 o}

g HfH2(inf{B(k-n)/B(k) : k 2 o})'1

g{1f{!2(SUp{B(k-n)/B(k) :k 2 o})'1i"l

3 1|sz inf{B(k)/B(k-n) :k 2 om

3 )[Tng2 inf{B(k+n)/B(k) :k 2 -n]/Lc2
g HM:n T;f((2/:e2

_<. HM:n eggs/e,

t IIM;<M:nf)n2/:c2

We note that the first equality comes from a direct

” A
computation of Mjnf = Z) f(k+n)(B2(k+n)/B2(k))zk.
z =-n

*
We have hence verified that MT is bounded below on

Lzm).

Thus we have both M¢ and M; bounded below on

L2(B) which again implies that M¢ is invertible. Q.E.D.

31

However, a spectral inclusion theorem does not
hold for all invertible weighted shifts. The following

example illustrates this point.

Example 3.1: Let T be the weighted shift with

 

weight sequence given as below:

i) wn = 1 if n‘z -1

ii) w =% if n<-l

n
k k .
Then for k‘z 0, HM _1H = 2 . It is also not
2
difficult to verify that HM _1\ 2 H = 1. Hence,
2 H (B)
r(M _1) = 2 from the first equality and r(T _1) g
z 2
HM _1\ 2 H g_l. Thus it is not possible that
2 H (B)
o(M -1) c o(T _1).
z 2

We now examine other conditions on Tcp which
imply something about the invertibility of MT. The
first result has been proven for the unweighted shift.

Its proof can be found in Douglas, [6].

Proposition 3.5: If T is a weighted shift with
periodic weight sequence, then either Ker Tcp = {O} or

* co
Ker Tcp = {O} for all T e L (B), ¢ # 0.

Proof: Suppose both Ker Tcp # {O} and Ker T; # {0].
Then there exist nonzero elements f,g 6 H2(B) such

* . —— 2
that T = O = T f. Since T = O we have H

32

where 113(3) = {h e H2(B) :hm) = 0}. Also Tczf = o

. . * n * n n
implies (Tmf,z ) = (wa,z ) = (f,¢z ) = O for all
. . ”9A 2
n‘z O. This last equation says Z) (k)¢(k-n)B (k) = O
k=O
for all n‘z O.

A A
Now let h 6 32(3) be given by h(k) = f(k)B(-k)B(k).
(We note that since the shift is periodic B(k)B(-k) is

bounded.) Now

an A A
k?) H(-k)¢(n+k)

/\
(wfi)(n)

 

A
k2 fi<k)(s(k)/B(-k))e(n+k)

a:

7‘ A 2
k2 f(k)cp(n+k)B (k)
:0

= O for all n g_O from the last line
of the previous paragraph. Thus $3.6 H3(B).

So R(cph) = R(cp)R(h) e Hgmm) and R(cpg) =

 

R(m) R(g) E Hg(a]D). This says R(¢)R(H)R(g) e Hé(al))

 

and R(cp)R(H)R(g) 6 H5513) since R(f)- 6 H2(a:D) and
R(g) e H2(a])). Now by Douglas, ([6], Corollary 6.7)
we have R(¢)R(h)R(g) = O which implies R(eg)R(h) = 0.
By the F. and M. Riesz Theorem if R(F) 7! 0, then
R(mg) = 0. Corollary 2.6 of this paper then implies
that T = O. This is a contradiction, and so we are

done. Q.E.D.

33

The next two corollaries have also been proven in
the unweighted case. Their proofs are also found in

Douglas [6].

Corollary 3.1: If 0 #’¢ 6 B(B) and Tcp has

 

closed range, then M¢ is invertible.

Proof: We may assume without loss of generality
that the weighted shift has periodic weight sequence.
If not, B(B) = {x1 :x 6 ¢] in which case the result

is trivial.

Now if ¢ 6 B(B) then T; = T5. Also by
Proposition 3.5, we may assume Ker Tcp = {0]. Then
since Tcp has closed range and Ker Tcp = {O}, we have
To is bounded below on H2(B). One can then show,
as before, that M¢ is bounded below on L2(B). That
is, there exists a constant c > 0 such that
Htprz 2 chHz for all f e L2(B). Then for f s 13(5),

we have:
(as), = ((5)2 = new, 2 em, = one), .

The second equality holds by Corollary 2.3. The above
*
chain shows that Mhp = M5 is bounded below on L2(B).

As before, we conclude that Mb is invertible. Q.E.D.

We recall now that S E B(H) is said to be
Fredholm if the range of S is closed and if both the

kernel Of S and S* are finite dimensional. .If S

34

is Fredholm, we define the index i(S) Of S by

i(S) = dim(Ker S) —dim(Ker 5*).

Corollary 3.2: If T is a weighted shift with

 

periodic weight sequence and T E L”(B), then Tea is

invertible if and only if T6 is Fredholm and i(TT) = O.

ggoofz It is easy to verify that if T is
invertible then Tfi is Fredholm and i(TT) = 0. SO
if Tcp is Fredholm and i(TT) = 0, then Ker Tcp = {O}
*
and Ker To = {O} by PrOposition 3.5. This implies
both Tcp and T; are bounded below on H2(B) since

they both have closed range. Thus Tcp is invertible. Q.E.D.

We now present the last result concerning B(B).

Proposition 3.6: Q: B(B) 4 6(H2(B)) is *-linear

and isometric.

‘ggogf: By Proposition 3.2, we need only show that
i is isometric on B(B). The C*—algebra B(B) is
commutative and hence r(Mb) = HM¢H for all ¢ 6 B(B).
Thus HMCPH = r(M”) g r(qu) g HTwH 3 ”Map“ for cp e B(B).
The first inequality holds from the spectral inclusion
theorem. The result is achieved since equality must

hold throughout. Q.E.D.

35

We now discuss some miscellaneous problems and
results concerning multiplication Operators, Toeplitz

Operators, and, N¢(B). First Of all we note that for

1

m ' A ' ' .
w 6 L (a), (”52 .23)/Bz(j) = o(j-i) = (M521+1.23*1)/52(3+1).

SO suppose L E B(L2(B)) is such that (sz,zj)/B2(j) =

l'zj+1

(Lz1+ )/B2(j+l). Is it then true that there is

w E L°(B) such that L = M”? The answer to this question

is given below.

Proposition 3.6: Let L E B(L2(B)) satisfy the

n+1 m+1
,2

equation (LG,zm)/B2(m) = (Lz )/B2(m+1). Then

there exists m E Lm(B) such that L = MT.

Proof: Since {Mcp 2:9 6 L°°(B)] = [Mz}’, the commutant

of M2 on L2(B), we only need to show that LMz = MzL.
This will be true if and only if (LMzzl,zJ) = (Mszl,zj)

for all integers i and j.

i 1

Now (LMzz .23) = (Lz1+ ,zj) and

i j _ i * j
(Msz ,2 ) — (Lz ,MzZ )

(in.zj'1)az(j)/52(j-1)

= (Lzl+l,zj) by the property of L given.

Thus (MszJ,zj) = (LMzzl,zJ) and we are done. Q.E.D.

36

Now note that for Toeplitz Operators Tcp with
m e L"(B), we have the same type Of prOperty. The
only difference is that (Twzi,zj)/Bz(j) =
(Twzi+l,zj+l)/B2(j+l) holds only for i,j 2_O. The
question here is like the question above. Suppose
S 6 B(H2(B)) satisfies the property above; is it
true that there exists T E Lm(B) such that S = TT?

The answer to this question is no and the solution

follows. We begin with the following proposition.

Proposition 3.7: Let T be an invertible weighted

 

shift with B(n) = l for n‘z O and sup{B(n) :n,< O] = a.

O
A so
Then there exists f 6 C(51)) such that Z} f(n)zn£' L (B)
2n nz'”

A . _.
where f(n) = I f(ele)e ine gfi .
O n

Proof: Suppose f 6 C(31)) implies

a A m a:
Z: f(n)zn 6 L (B). Let I :C(aID) 4 L (B) be the map

n=-w

co
A
that sends f to Z) f(n)zn. Then I is continuous

n=-a>

as a result of the closed graph theorem. The graph
B = {(f,If) :f 6 C(a]))] is closed since the coefficient

A
functional Tn which sends f to f(n) its continuous

on both C(51)) and LQ(B). Now since I is continuous,
we have:
em = Hz“), :11th 2 (manual, 2 (In for an n.

37

But this contradicts the assumption that
sup{B(n) :n < O} = m. Thus there must exist f E C(aIH

a A m
such that Z) f(n)zn E L (B). Q.E.D.

Corollary 3.3: Let T be an invertible weighted
shift with sup{B(n) :n < O] = o and B(n) = l for
n‘z 0. Then there exists S E B(H2(B)) such that
i+1 zj+l

(Szi,zj)/B2(j) = (Sz )/B2(j+l) for all i,j 2'0,

but S 7(Tcp for any T 6 Lw(B).

Proof: We note that H2(B) = H2(aIn . Let S

o o A
on H2(B) be defined by (821,23) = f(j—i) where

on A on
f 6 C(51)) but 73 f(n)zn f L (B). Then s e 5012(3))

n=-w

by Halmos, [8], Problem 194. But if S = M.cp for some
2 A A . .
¢ 6 L (B). then ¢(k) = f(k) for all k by picking

i,j 2 0 such that j-i = k. This equality implies

T A
Z) f(n)zn e Lm(B), a contradiction. Q.E.D.

=—w

Corollary 3.4: If T is an invertible weighted

 

shift with sup{B(n) :n < O] = w and B(n) = l for
all n 2.0, then {Tb :¢ 6 L°(B)] is not closed in

6012(5)).

Proof: Let f 6 C(61)) be such that

co n

A Q A a:
23 f(n)z“ 1 L (8). Then on(f) = k2: (1-§$1L>f<k)zker. (a).
=-¢ =-n

38

Also, on(f) 6 C(61)) and Hon(f)-fH 4 O as

Ian
n 4 m (Katznelson, [13], Thm. 2.11). Let S be as

in Corollary 3.3. Then

HT -SH S-H0n(f)-fHaI) (Douglas, [6], Prop. 7.4).

°n(f)

Hence S g {Tcp ‘T 6 IF(B)] implies this set is not

closed in B(H2(B)). Q.E.D.

We note that these last two corollaries show a
remarkable difference from the unweighted case,
B(n) = l for all n. In this case it is true that
if S 6 B(H2(B)) satisfies (Szi,zj) = (Szi+l,zj+l)
then S = T for some T e L°(B). This also says

W
that {Tcp ‘T e L?(B)] is closed in B(H2(B)).

In the last part of this chapter, we consider

. w _ _ a A n 2
the following. Let H (e) — {(1, — Z ¢(n)z : (:f e H (5)
n=0

all f e H2(B)] where the multiplication Hf is
defined as before. Then it can be shown (Shields, [23],

Thm. 3) that {Tz]’ = {M : W E Hw(B)]. The problem

we want to consider here is: what is the relationship
between ”9(5) and Hw(B) when considered as sets

of sequences? It is easy to verify that flp(B) C H”(B)
since flp(B) C LF(B). For the three cases: B(n) = l

for all n; B(n) = rn for all n; and the case where

the shift is rationally strictly cyclic, we have

VQ(B) HQ(B). However, in general, it is not true that

ir°°(B)

Hm(B) as the following example indicates.

39

‘lkl

Example 3.2: Let B(k) : 2 for all k.

Then it can be shown that w 6 Hm(B) if and only if

” A
w(z) = Z)¢(n)zn is a bounded analytic function on
n=0
{2 E C :\z{ < 1/2]. (Shields, [23], Theorem 10’).
m m k
Thus ([1: 22k 6 HQ(B). Also, g= Z z—k-z-kEL2(B).
k=O k=1

9 /\
If it were true that W E y (B) then (Hg)(O) would

be defined (and finite). However

/\ a) . on k
((thO) = 2’)<k)G(-k> = 2: l-ZT e e.
k=O k=l

Thus w E Hm(B) but W E'fim(B).

CHAPTER IV

HANKEL OPERATORS FOR WEIGHTED SHIFTS

The study of Hankel Operators is a natural out-
growth Of the study of Toeplitz operators. Again,
as in the case of Toeplitz Operators, Hankel Operators
for the unweighted shift have been studied quite
thoroughly. One area Of study has been to determine
which Hankel Operators are compact. We will be
considering this question for weighted shifts. We
will also point out during the course of the investigation
differences between the unweighted case and other cases.

We start with the following definitions.

Definition 4.1: Let T be an invertible weighted
shift represented as Mz on L2(B). For T e L°(B)
we define the Hankel Operator Hcp : H2(B) 4 L2(B) € H2(B)

with symbol T by

He“) = (1-P)(qpf) for all f e H2(B)

We recall that P is the orthogonal projection
of L2(B) onto H2(B) given in Chapter I. Thus l-P
is the orthogonal projection of L2(B) onto the orthogonal
complement L2(B)EH2(B) of H2(B) in L2(B). Also,
we note there is no loss in assuming T is invertible.
If T is not invertible then LP(B) = fim(B) (Shields,

[23]. P.68). We will show later that H) = O for all

40

41

W 6 Np(B). For the rest of this paper, the weighted

shift T will be invertible unless otherwise stated.

Definition 4.2: C(B) is the closed linear span

 

in Lm(B) of the (Laurent) polynomials.

Equivalently, f E L°(B) is in C(B) if and only

N A
if for every 6 > 0 there exists q = Z) q(k)zk such
=—N

that Hf-—qu < c. This definition is motivated by the
unweighted case. In that case, C(B) = C(31)); and
C(aIn is the closed linear span of the (Laurent)

polynomials in Lm(al)).

We now begin our study with some easy results

concerning Hankel Operators.

Proposition 4.1: The map T 4 HE is a contractive

linear map from La(B) into B(H2(B), L2(B)(E>H2(B)).

Proof: It is clear that the map is linear. The
proof that it is contractive is just like the proof for

Toeplitz Operators.
Hag), = \Hl-PHcprlzg new, 2 HTHJsz for an t e :f(n).
Thus New 2 Melt. Q.E.D.

Proposition 4.2: If E N”(B). then H = O.
T w

42

Proof: For w E Nm(B) and f E H2(B) we have
shown wf e H2(B). Thus H¢(f) = (l-P)(mf) = O for

all f t H2(B). Q.E.D.

Definition 4.3: For m e L°(B) we define the

 

distance d(¢,yp(B)) between m and flm(B) by

men/”(BM = inflllcp-(IH, z) e f(w) .

Proposition 4.3: If m 6 Lw(B), then

 

1le!) _<. d(e.y°’(a)).

Proof: Let m E LF(B) and w E Nm(B). Then

Hw_w = HE by Proposition 4.2. Now by Proposition 4.1,
HH¢H = HHw_¢H g Hw-me. Since this is true for all
W 6 Nm(B), we have the result. Q.E.D.

The first three propositions and their proofs
are identical with ones for the unweighted case.
However, in that case it can be shown that
HHcpH = dues/”(BM for all :9 e L°°(B) (Nehari, [15]).
We will see that this is not true for all weighted
shifts when we study compact Hankel Operators. We
will begin this study after one more definition and a

theorem following it.

Definition 4.4: y°°(s)+C(s) = m+n : a; e VHS)

and w E C(B)).

43

For the unweighted case V”(B)4-C(B) has been
shown to be a closed subalgebra of L°(B) (Sarason,
[22]). We prove the same result for an arbitrary weighted

shift by using a theorem of Rudin, [19].

Theorem.4.1: flm(B)4-C(B) is closed in Lm(B).

 

Proof: We use the same notation as in Rudin, [19].
n

m A
For m E L (B) let on(o) = Z) (1-%§%)¢(k)zk be the
n

n'th Cesaro mean of m. Let T = {on :n.€ IN], y = C(B),
and. Z = flm(B). The latter two sets are then closed
subspaces of the Banach space Lw(B). Now for o E T,

f e y, g e z, and h e L”(e) we have the following

results:
i) o(h) e y (i.e. o(L°°(B)) c can)
ii) 0(9) e z (i.e. o(w”(e)) czy”(B))
iii) HoH g.l (i.e. sup{HoH :o 6 Q} < m).

Parts i) and ii) are easy to verify using the
definition of on. For part iii) see Shields ([23],
p.89). Rudin's theorem says we must only verify one
more thing to conclude that. ”p(B)4-C(B) is closed.

We must show that for each f e y and e > 0 there
exists 0 6 9 such that Ho(f)-—me < e. To see this
let f e C(B) and e > 0 be given. Then there exists

N
A
a (Laurent) polynomial p(z) = Z) p(k)zk such that
k=-N

Hf--pHco < e/3. Therefore,

44

ll0n<f>-fH gllon (f- p>iifl+uo (p)-pHm +Hp -f‘.lm or
non“) -me g ZHf-pll; Henna) —pHm (by part 111))
g 26/3+ Hon(p) —pHm ,

N
_ e JLL"
Now for n > N, on(p) p EZN n+1 p( (k)zk .
Using the results of Shields ([23], Prop.29), it can

be shown that

Homo) -pHm _<. lelmqw (qn(‘v7)( i‘g—Vg—L) where

N k
qn(W) = Z) [klw /(n+l). Thus

21/2
New in}

H0 (p)-p‘.‘ 3 HP!) (
n '00 an [an

N 1 2
g lelgz 231:2) //(n+l)
k=O

So there exists NO > 0 such that
Hon(p) -pHm < e/3 if n > NO. Finally if n 2 no,

then H°n(f)"wa < e. Q.E.D.

Proposition 4.4: If m e NQ(B). then

zncp e y”(a)+c(e) for all n.

Proof: For n 2.0, zn¢ E ”p(B). Thus we need

only consider the case when n < O. In this case,
-1 -

zncp = Z (p(k-n)zk+ Z cp(k-n)zk . Since 9 = Z‘ ¢/;;(k---n)zk
k=n k=O k=n

45

is in C(B), we see that Z}:p(k-n)zk = zn¢-g is
k=O

in .Mw(B). Thus zn¢ e z”(e)+-c(B). Q.E.D.

Corollary 4.1: 31mm) +C(B) is a closed subalgebra

of LQ(B).

Proof: By Theorem 4.1, flm(B)4-C(B) is closed.
Thus we need only show that it is a subalgebra of L°(B).
Because it is closed, it is sufficient to show that

pp 6 Mm(B)4-C(B) for all w E N”(B) and (Laurent)

polynomials p = Z) p(k)z . However, PT = Z) p(k)(z T)
=-N k=-N

and by Proposition 4.4 we know that 2km E ”p(B)4-C(B)

for all k. Hence po 6 u”(B)-tc(B). Q.E.D.

We will now discuss the set {w E LQ(B) :HE is
compact]. In the unweighted case, it is known that the
set above is exactly NQ(B)4-C(B) (See P. Hartman, [12]).
For weighted shifts in general this is not true as we
will see later. However, we will show that H is compact
for all T in. flm(B)4-C(B). We begin with th: following

proposition.

Proposition 4.5: If n > o and w e y”(e), then

H -n is a finite rank Operator and hence compact.

z T

46

-l

m A
Proof: If q = Z) $(n+k)zk and h = Z)¢(n+k)zk,
k=-n k=O
then q 6 C(B). w 6 Np(B), and z-nm = q+-m. Thus by
PropOSition 4.2, Hz-nm = Hq'

23 f(k)3(m-k) = o
k=O

A
for m < -n since q(z) = O for L < -n. Thus

2 /\
Now let f E H (B): then (qf)(m)

°° /\ k . .
qf = Z) (qf)(n)z , implying H -n (f) = (l-P)(qf) =
k=-n z w
'1 /\ 'k -n -n+1 -1
Z} (qf)(k)z . Hence {2 ,z ,---,z ] spans the
=-n
range of H -n . Q.E.D.
2 co

Corollary 4.2: If new”(e)+C(e) then Hcp

is compact.

Proof: Let T = qi-w where w e fim(B) and
q 6 C(B). Let qk e C(B) be (Laurent) polynomials

such that qu-qu 4 O as k 4 a. Then

\chp-H = HHq-quH S. liq-qkllm -+ 0 as k -v e .

qku

By PrOposition 4.5, Hq is compact. Also, it
k
is known that the collection Of compact Operators is

closed in the norm tOpOlogy on B(H2(B),L2(B) SH2(B))

(See Rudin, [17], Theorem 4.18). Thus Hfi is compact

since HH -H H 4 O as k 4 m and H is compact
w qk q

k
for all k. Q.E.D.

47

Before we continue with our discussion of compact
Hankel Operators, we need to take a short digression.
There are some general prOperties of Operators on Hilbert
space which we will need in our discussion. The results
we will give in this digression have been known for
many years, but for the sake of completeness, we will
include their proofs. The following result is due to

I. Schur. Its proof is found in Hardy, [10].

Theorem 4.2: (Schur) Let H and K be separable

Hilbert spaces with orthonormal bases {ej :j = O,l,2,---]

and {ei: i = -l,-2,---] respectively. Let A E B(H,K)

have matrix [aij] where aij = (Aej,ei). If bij = (fj'gi)
for fj' 9i 6 H with sup{HfjH:j = O,l,°°'] = M < w and
sup{HgiH :i = -l,-2,---] = N < m, then the Operator D

with matrix [aij bij] is in B(H,K). Furthermore,

(DH 2 HAHMN-

on

A °°A
Proof: Let fj = Z)f.(z)eL and gi = Z)gu(z)e

a i=o3J _1 i=0 1 2'
If x = Z3l e t H and y = z: B.e. e K then
k=O k k .=_a j j
e -l
(Dx,y)=Z ZlB.a. b
k=O j=-w k 3 3k 3k
5.2.x? (E?()())
= . a. L g. z
k=O j=_w k 3 )k-e=0 k 3
no 09-],

2(22i’iumAH)
.g. z a.
£=O k=O j=-e k k 3 3 3k

48

. _ a A _ ‘1 A
Now if hz — #ngk fk(z)ek and mi - jELQBj gj(z)ej
then (Dx,y) = (Ah ,m ). So
2:0 1 z
\(Dx.y)l g £§()((Ahz.m£)(
$120))Ai)))h£))))m£))
m 1/2 m 1/2 _
z:

23 ( 2) [1k] zlfk (2)12
L: O k=O

m 2
However, ZIthH
£=O

°° A
1.230!” (2( 2: \fk(£)]2)
i=0

2: A IZMZ
g k=O) k
_<. :42“tz
2)1/2
Thus ( 23Hh£ H) g.MHxH and similarly ( Z)Hm£ H2) 1/2 1
L=O 1:0

Therefore [(Dx,y)] g_HAHMNHxHHyH implying
HE" S. HAHMN-

           

We note that we should be a little more careful
in switching the order of summation. This can be done
by letting x and y be only finite linear combinations
Of the basis vectors. We would then get the norm inequality
on a dense set of vectors and be able to extend by

continuity. Q.E.D.

49

We now have the following two corollaries
which we will need in discussing compact Hankel
Operators. The operator An(A) defined below is
like the (n-l)th Cesaro mean of a function. If we
let Bk(A) be the Operator whose matrix entries
are those Of A above the kth cross diagonal and

n
zeros on or below it, then A (A) = ZJBk(A)/n.
n k=l

Corollary 4.3: (O'Donovan, [16]) Let A E B(H,K)

where H and K are as in Theorem 4.2. Let
aij = (Aej,ei) and let
0 if [11+j2 n

(An(A)ej.ei) =
aij(n-[i\-j)/n if \i[+j < n

Then An(A) e B(H,K), HAn(A)H g_HAH, and An(A) 4 A in

the strong Operator topology (SOT) as n 4 o.

n-)
Proof: For j = O,l,2,---,n-l, let f. = ( Z3e )/UH
'-—__' J _ L
1-1
and for j.2 n let fj = 0. Also for i = -l,-2,°°°,-n+1,
n
let gi = ( Z) eL)/Vn and for i g_-n let gi = O.
z=\i\+1

Then HfjH g_l, HgiH g_l, and it is easy to verify that

0 if \i[+j 2.n
(fj'gi) = '
(n-[i]-j)/n if \i[+j < n

50

Hence by Theorem 4.2, An(A) 6 B(H,K) and
HAn(A)H g_HAH. We note that in fact An(A) is a finite

rank operator.

To show convergence in the SOT we start with a
basis vector ej for j‘z 0. If e > O is given

then there exists an integer k < 0 such that

k -1
Z) \(Aej,ez)\2 < e2/2 since HAejH2 = Z) [(Aej,ez)[2 .

Lz—m 2:-”
Thus if n > [k]+—j+—l we have k > j-n+-1 and

2
HAn(A)ej -AejH

j—n 2 '1 . 2 2
e z [(Ae.,e£)\ + 23 (((z(+J)/n) \(Aewefil
=—cn J i=j‘n+l J
k-l -1

g 23 ((Ae..e ”2+ '2: ((\z[+j)/n)2\(Ae..e )(2
3 ‘ £=k 3 ‘

=—m

: 62/“ <(k\+j>21kmAH/n2.

Thus there exists NO > 0 such that

(DcHtj)2\k\HAH/h2 < e2/2 if n > NO. Therefore if

n > NO, then HAn(A)ej-AejH < 5. Or in other words

HAn(A)ej-AejH 4 O as n 4 a,

To show convergence in the strong Operator
topology we must show HAn(A)x-AxH 4 O as n 4 m

for each x e H. We have shown this for x = ej;

Q
j = 0,1,2,--- . SO let x = Z)l.e. e H and let e > o.
j=0 J J

51

Then there exists M > 0 such that

H :3 AjejH < S/BHAH- NOW let Y = Z) Aje.. Then
j=M+1 j=M+l

M
2: (inlAn(A)ej-Aejn + HAn(A)yH + HAyl

HAn(A>x - Axll g
3:0

g(M+l)HxH max{HAn(A)ej-AejH : O 3 j g M] + 5/2

By the part directly above we can make
max{HAn(A)ej-AejH :O g.j g.M] "small" by taking n
large. Thus for each x E H and e > 0 there exists

N > 0 such that HAn(A)x-AxH < e if n > N. Q.E.D.

Corollary 4.4: If A is compact then An(A)

 

converges to A in the norm topology on B(H,K) as

n4oo.

Proof: We first assume A is Hilbert-Schmidt and

let 6 > 0. Then there exists N > 0 such that

m

Z) HAe H2 < 22/9. We also note that for j‘z O
n

n=N+l

HAn(A)ejH g HAejH since )(An(A)ej'ei)) g_[(Aej,ei)\ .

as N
Thus if x = :31 e 6 H, we let x = Z) A e and
m k=O}: k l k=O k k
x = Z) x e . Then
2 k=N+1 k k

(Anon: -th g (Ansel -Ax1H+llAn(A)x2H+!le2H

N
_<. kgonkl HAn(A)ek -Aekll + (Anmmzn + \(szn

52

But HAn(A)x2H g k=§+l\lk\HAn(A)ekH

( E (2)1/2( :‘5 H ) ((2)1/2
A A (A e ‘
S k=N+1\ k k=N+l n k'
a: 2)l/2

3 MW 2 lHAekH

g Hxlle/3

The above inequalities also show that HAXZH g HxHe/3.

Therefore
HAn(A)x -AxH g [(N+l)max{HAn(A)ek-AekH : nggN] +3351] HxH .

However, by Corollary 4.3 there exists NO > 0 such that
max{HAn(A)ek-AekH :O>g_k g_N] < s/3(N+l) for all n > NO'
Thus if n > NO' HAn(A)x-AxH < eHxH for all x e H.

We note that NO does not depend on the vector x, only
on the Operator A and the 5 given. Thus

HAn(A)-AH < e if n > NO' This says An(A) 4 A in

the norm tOpOlogy if A is Hilbert-Schmidt.

However, the set of Hilbert-Schmidt operators
is norm dense in the set Of compact Operators. Thus
if A is compact and e > O is given, there exists a
Hilbert-Schmidt Operator B such that HA-BH < 5/8.

Therefore
(Ann) -AH s (\An<A>-An<a>n + (Age-Bu + HB—AH
g HAn(A-B)H + t/3 + HAn(B) - BH

g 26/34-HAn(B)-BH by Corollary 4.3 .

53

Since B is Hilbert—Schmidt there exists NO > 0
such that HAn(B)-BH < e/3 if n > NO by the first part

Of this corollary. Hence HAn(A)-AH < e if n > NO. Q.E.D.

We now relate the material above to the consideration

of compact Hankel operators.

Proposition 4.6: If w 6 LP(B) then HO (T) =
n

An+1(H¢) where on(¢) and An+1

(Hfi) are defined as

before.

“ k A k
Proof: From before on(¢) - ELn(l-%:+)¢(k)z .

An orthonormal basis for H2(B) is {Zn/B(j) :j = O,l,2,"°]

and an orthonormal basis for L2(B) 9 H2(B) is

{zi/B(i) :i = —l,-2,"°]. Thus if \i-j] = [i\4-j < n+-l,
then
j i . . _ j i . .
(H0n(w)z .z )/B(J)B(l) — (On(o)z .z )/B(J)B(1)
= on(e)(i-j)e(i)/a(j)
A - o
= o(i-j)[(n+1-Hl-)\)/(n+1)]B(i)/B(j)
A
= cp(i-j) [(n+l- {il - j)/(n+1)]B(i)/B(j) .
Similarly

<An+l(Hm)zj.zl)/B(i)a(j)

= (Hmzjsi) [(n+1- m - j)/(n+1)]/B(i)B(j)

A
o(i-j)[(n+l-\i\-j)/(n+l)]B(i)/B(j)

54

O ($)zj.zi)/B(i)B(j)
n

If \i-j\ 2 n+1 then (HO ((0)2121) e o = (An+l(Hcp)zj,zi):
n

Hence we must have H0n($) = An+l(H¢)' Q.E.D.

Theorem 4.3: Suppose there exists a constant

 

c > 0 such that HHwH.2 cd(¢.flm(B)) for all T E Lm(B).

Then Hcp is compact if and only if w e ”p(B)4~C(B).

Proof: We have already shown that if
e E ym(B)4-C(B) then Hcp is compact. So now we assume
H6 is compact. Then by Corollary 4.4 and PrOposition 4.6

HH -H¢H 4 O as n 4 m. Now by our hypothesis

On(o)
HHOn(¢)"H¢H = ”Hon(¢)—¢H 2.Cd(0n(o)-¢.Vm(B)). Thus

there exists {wn :n.= 1.2.°"] czym(B) such that
Hon(m)-tHn-¢H 4 O as n 4 m. However,

On(cp)+t)n t v°°(B)+C(B) since On(cp) e C(B). Now
by Theorem 4.1 we conclude that T e Nm(B)4-C(B)

since this space is closed. Q.E.D.

We note that this also gives us a proof that H
is compact if and only if m 6 flp(B)-tC(B) in the
unweighted case. This follows from the equation:
HHwH = d(¢,M”(B)). We now consider the problem of
knowing whether such a constant c (in Theorem 4.3)
exists for every weighted shift. The answer is no.

We illustrate this by using the following example.

55

Example 4.1: Let T be the weighted shift with

 

weight sequence as below:

i) wn = 1 if n 2_O

ii) wn = 1 if -[(k+l)2+l] < n g -[k(k+l)+l]

for k = O,l,2,°'°

iii) w =l otherwise
n 2

Then for m > O, I claim Hz-mHm = d(z_m,3/°°(B)) = 2m.
To see this pick an integer k > m. Then for T E ym(B)

and n = -[(k+l)2+l],

H (z-m+ cp)ZnH§ = Hz-mtn

((3 + usznni
2 Hz’mnll 3

Thus Hz‘mwnm 2 llz““""l(2/:'(z"u2 = strum/B(n) =

n-l -1
( H w ) - 2m Since this is true for all
k=n-m k

cp E flaw), we have d(z-m,U°(B)) 2 2m. But also

Hz-mH0° g_Hz-1H$‘g_2m implying the equality desired.

Now however,

lle-mH g Ile-m|H2(B)H

2112'”), since Hz’mzkllzgllz’mllz for R20
g B(-m)

$2m/2 .

56

The last inequality holds by a simple calculation.
Thus ||H _mH/d(z-m.1’°°(B)) 3 (firm 4 O as m 4 as,

2
Therefore no such constant as in Theorem 4.3 can exist.

One may ask at this point what is the set
{w E LP(B)‘:H¢ is compact} for Example 4.1? The
answer is given below.

For the weighted shift given in Example 4.1,

-1 -l -1 -1
r(T) = ”TH = 1 and r(T ) = HT H = 1/2. Now
by the remarks following the proof of Theorem 10'

(Shields, [23]) and Theorem 1.2 of this paper, f 6 L°(3)

Q
A
if and only if f = Z) f(k)zk is a bounded analytic
k=-m

function on the annulus A = {z 6 ¢ :%-< \21 < 1].
Also, HfHA = sup{\f(z)\ :z 6 A} g_Hme for all

f e L°(B). It is also easy to show that if f e C(B)

then f is continuous on {2 E q: : ‘2‘ = %}U {26¢ : \z\ =1].
This is done using (Laurent) polynomials and the norm
Q
A
inequality above. Now let f(2) = Z)f(k)zk be a bounded
k=O

analytic function on I) = {z 6 ¢ :\z\ < 1}. For

Q

-A _

z E A, let g(z) = f(fg). Then g(z) = 232 kf(k)z k
k=O

is a bounded analytic function on A. We recall that
{Zn/B(n) 3 n = 001020"'} and [zm/B(m) 3 m = '10'20'.°}
are orthonormal bases for H2(B) and L2(B) 9 H2(B)

respectively. we then have the following:

57

(ngn.zm)/B(n)B(m) = g(m-n)B(m) since B(n) = l for n'z O .

m 2 - ° 11- A 2 2
Thus 2 “H (27501))”. - Z ( L4 \g(m-n)\ B (m))
n=0 9 n=0 m=-a:
on Q A
= Z ( E \g(-m-n)\252(-m))
n=O m=l
C O A - -
= Z ( 23 \f(n+m)\22 2“ Zmazmn
n=0 m=l
_<. Z < 23 \?(n+m>\24'n) since
n=O m=1

m-m) g 2""/2

a _ O A
g 2 4 n( E \f(n+m)|2)
n=0 m=l

a -n .2
$.5Eg4 HfHA
.<. 211in -

This condition says Hg is Hilbert-Schmidt and
hence compact. In fact, the proof shows that if g
is a bounded analytic function for \z‘ > 1/2, then

Hg is compact.

. A k a
We now note that if ¢ = Z) ¢(k)z E L (B). then

R A k .
$1 = Z) ¢(k)z and $2 = 23 m(k)z are bounded analytic
k=O -~

functions for ‘z\ < 1 and ‘2‘ > 1/? respectively;

and hence are in LP(B). Thus H is

=H =H
w W1+¢2 W2
compact (in fact Hilbert-Schmidt) by the remark above.

Thus Em is compact for all ¢ 6 L°(B).

58

However, for some weighted shifts LQ(B) = ”p(B)-+C(B).
This is not true here. If g E C(B), then 9 is
continuous for \z‘ = %u There are, however, bounded
analytic functions on A which are not continuous for
\z} = 3n Such a function would be in L°(B) but not

in y"(B)+C(B).

CHAPTER V

ANALYTIC PROJECTIONS FOR WEIGHTED SHIFTS

Let T be an invertible weighted shift represented

as M2 on L2(B). Then, as before, L°(B) =

a: A
{¢ = k;; cp(k)zk':cpf e L2(B) for all f e L2(B)}.

m
A
If we consider m = Z) cp(k)zk as a formal Laurent
k=-w

series, then we can define the "analytic" projection

e>=lF(s) 4 L2(B) as below.

a
/\ co
Definition 5.1: For a = z) ¢(k)zk e L (a) we
k=-m

‘ define the "analytic" projection 0':L"(B) 4 L2(B) by

 

9(a) = E) $(k)zk
k=O
It is easy to see that. 6>:IF(B) 4 L2(B) is a
bounded linear map since H9(¢)H2 g_H¢H2 g_H¢HQ. The
problem we want to consider concerns the range of 9.
For which weighted shifts is the range of 0 contained
in L”(B)? By appealing to the closed graph theorem and
the continuity of the coefficient functionals Fn on
Lm(B). one can ask equivalently: for which weighted
shifts is 9 a bounded linear map from the Banach space
L°(B) to L”(s)? So 0 e B(LS(B)) if and only if
9(¢) 5 L”(B) for all m E L"(B). In what follows,

we will give a general sufficient condition for 9

59

60

to be bounded. We will then prove several corollaries
of this theorem. Finally, we will provide several
examples illustrating various aspects of this problem.
Before we give the theorem, however, we need the

following definition.

Definition 5.2: Let K C ¢ be compact and let
f be a function analytic on an open set containing K.

We then define the norm of f on K by
HfHK = sup{\f(z)\ :z e K]

We are now able to state the theorem.

Theorem 5.1: Let T be an invertible weighted

- -1
shift with r(T 1) < r(T). If there exists a constant

c > 0 such that HpH0° S-CHPH0(T) for all polynomials

NAk. .
p(z) = #2; p(k)z in z or if HqHa S-CHqu(T) for

k . -l

A -
all polynomials g(z) = q(-k)z in z , then

M42

1

9 is a bounded linear map from LQ(B) into Lm(B).

Proof: We first assume that up“co g CHPHO(T)

N
for all polynomials in 2. Let f = Z)f(k)zk and
-N

consider ”o(fmco g CH9(f)HO(T). At this point, we

note that o(T) = {z 6 ¢ :r(T-l) g_\z\ g_r(T)}.

61

_ -1

Since r(T 1) < r(T), there exists a constant d > O
I g g .

such that H9(f)dO(T) g-dif”o(T) (See Shields, [23],

p.81). Thus

H6<f>Hm s cHflleom s cdHfHO(T, s cdlifnm

The last inequality follows from the spectral mapping

theorem. For f, as above, f(O(T)) = o(Mf). Thus

anm) = may 3 HMf: = M,-
Now let f = Z) f(k)zk be any element of Lm(B).
k=-w
n . A k
Then on(f) = Z}[(n+l-§ki)/(n+l)]f(k)z and
k=—n
Gn(9(f)) = 9(On(f)). Hence

HoanHHm = Eié<on<f>>lim g cdil0n<f>Hm _<. cdnfnm .

The first inequality comes from the first part of this
proof and the second follows from the inequality
Hon(f)”e g Hf“... for all f e Lam). Thus

{M0n(€f) :n.= l,2,---) is a norm bounded sequence in

B(L2(B)) and hence must have a convergent subsequence
in the weak operator topology (WOT). Assume without

loss of generality that M in the WOT as

0n(0f) 4 S
n 4 o. Then S = M for some W E L°(B) since

W
LF(B) is closed in the WOT. Thus

62

(w.z£)/BZ(L)

A
w(£)

= lim (on(ef),z‘)/ez<i)
n-vco

= lim (Gn(9f))(L)

1140:

o if I,<O

A
lim [(n+l-H\)/(n+l)]f(£) if :20

11-9:

0 if z<o

A
f(n) if 2‘2 0

Thus W = 9(f) E LQ(B) and H9(f)flw g CdHwa for all
f e L°°(B).
The proof in the case where Hqu S-CHqu(T) is

very similar. Here, one shows that 91 = 1-0 is a

bounded linear Operator on Lw(B), and hence so is

 

‘9 = 1-91. Q.E.D.
-1 -1
Corollary 5.1: If r(T ) < r(T) and r(T) = “TH
or r(T-1) = HT-lH, then 9 is a bounded linear operator
on L°(B).
l N A k
Proof: Assume r(T) = HTJ. Then for p(z) = Z)p(k)z
k=O
we have

MPH“, = umpu = upmzm g npuK where K = {z e ¢= m = new.

63

The inequality is a result of von Neumann's Theorem
(von Neumann,[26]) and the maximum modulus theorem.
Now, since r(T) = ”T“, we have K c o(T). Thus
HPHK S HPHMT) implying that “pk. s HPHMT) for
all polynomials p in z. The result then follows

from Theorem 5.1.

On the other hand, if r(T-l) = HT-lH and
N
Q(Z) = Z} </§(-k)znk then
k=l
N A _ '
“qu = HkZl qc—k)(le)kn
N A k -l
g.uk:£q<-k>z (K, where K' = {z e e =\z1 = HT H3.
- II I "l -l "l -1
Letting K = {z 6 ¢ :\2\ = “T U = r(T ) } and

replacing z with 2.1. we get Hqu S.HqHK” S-HqH0(T)

since K” c o(T). Again Theorem 5.1 applies and we are

 

done. Q.E.D.

Corollary 5.2: If r(T-‘1)-l < r(T) and M2 is
similar to an Operator S such that r(S) = ”S” or
r(S-l) = “8.1”, then 9 is bounded.

Proof: Let M2 = RSR-l where R E B(L2(B)).

N
A
Assume r(S) = ”SH and let p(z) = Z)p(k)zk. Then
k=O

upu, = np<MZ>u = uhp<s>h'1n g.uhnuh'1nnp<s>u .

64

But Hp(8)H S-UPH0(T) since o(T) = 0(8) and
{z 6 ¢ :\21 = r(S) = HSH} c o(T). Thus

”PH” S.HRHHR-1HHPHO(T) and by Theorem 5.1 the result

follows.
The proof in the case r(S-l) = “8-1” is similar to
the case above, just as in Corollary 5.1. Q.E.D.

Before we present the third corollary, we need

the following definition.

Definition 5.3: The numerical radius w(A) for

A E B(L2(B)) is given by

w(A) = sup{\(A£.f)( :f 6 13(5) and ”5H2 = l}

- -1
Corollary 5.3: If r(T 1)

1)

< r(T) and

 

l

r(T) = w(T) or r(T- = w(T- ), then 9 is a bounded

linear operator on L”(B).

A
Proof: Assume w(T) = r(T) and p(z) = p(k)zk.

W

k

Then p is a bounded analytic function on
{2 6 ¢ :\2\ g_w(T)}. Also, it is well known that
“A“ g 2w(A) for all A e B(L2(B)). If

A
9(2) = p(Z) -p(0). then

65

Hmm=\W+gWW.

A
3 H9”... + \P(0)\

2w .
_ is de
_ IO p(r(T)e )2wr(T) °

A
s. \\9(Mz)\‘.+HpHO(T) since p(O)

Thus Hpr g 2w<g<MzH+ upnmy

We now apply Theorem 4 of Berger and Stampfli, [3].

This is a mapping theorem for the numerical range and

says:

w(g(Mz)) g_HgHB where B

{2 E T :\z‘ g w(T)] .

Thus Hp”OD $.2H9HB4'HPH6(T)

since r(T) = w(T) and HgHB = HgHB, g-H9H0(T)' where

B’ = {z 6 ¢ :\2\ = w(T)} c o(T). Finally,

hmgzm-Qmmwpwmmmgswmw,

and Theorem 5.1 applies again.

Also, as before, the case where w(T-l) = r(T-1)
. . . . N A -k .
15 Similar uSing g(z) = Z) q(-k)z instead of
k=l
p(Z).

Q.E.D.

66

-l
l) < r(T) = t

Corollary 5.4: Let r = r(T-
and A(r,t) = [z 6 ¢ :r g_\z\ g_t]. If there exists
a constant c > 0 such that Hwa S-CHfHA(r,t) for
all f E L¢(B) then 9 is a bounded linear Operator

on L°(B).

Proof: The proof is obvious. If the norm

inequality holds for all f E L°(B). then it certainly

holds for all polynomials p is z° Q.E.D.

This ends the set of corollaries to Theorem 5.1.
We will finish this chapter with four examples related
to Theorem 5.1. The first example is used to answer
the following question. Do the hypotheses of Theorem 5.1
imply the hypotheses for Corollary 5.4? That is, does
Ilpuco g CHpHo(T) for all polynomials p in 2 imply
HfHa g CHf”A(r,t) for all f e L”(B)? The answer to

this question is no, as illustrated below.

Example 5.1: Let T be the weighted shift with

weight sequence as below:

. _ 1_ .
i) wn — 2 if n 2.0

ii) wh = 1/4 if n = -k(k+1)/2 k = 1,2,3,...

iii) wn = 1/3 if -(k+l)(k+2)/2 < n < -k(k+1)/2 k==1,2'...
Then it is easy to show that r(T) = “T” = 1/2 and
1/4 = HT-IH-l < r(T-1)— = 1/3. Since r(T) = “TH we have

“pH” g_HpHC(T) for all polynomials p in 2 by

von Neumann's theorem.

67

Now for k < 0, let 2(k) be the number of
times 1/4 appears in {w_l,w_2,---,wk]. Then

L(k) 4 w as k 4 -m. Also, let qk(z) = 2k for k < O.

-k . .
Then quHO(T) = 3 ' but ‘iqki‘m 2 “qu2 = B(k) and

-k-z(k>4i(k)

B(k)=3 . Thus

i(k) 4 m as k 4 -w

quHm/quHmT) 2 (4/3)

Hence there can be no constant c > 0 such that

Hen“, geueuom for an e e Lam.

In all of the previous situations we have taken
r(T-'1)_1 < r(T). We have used this condition so that
o(T) contains an Open annulus. We were then able to
use the boundedness of the "analytic" projection on
the space of bounded analytic functions on this annulus.
The condition r(T-l).l < r(T) is not necessary,

however, as the following example shows.

Example 5.2: Let T be the weighted shift
represented as M2 on L2(B) where B(n) = \n\+-l.

The weight sequence {wn} for T is then given by

wn = B(n+l)/B(n) = (|n+l\4-l)/(‘n‘+l)

-l
l) = 1.

For this weighted shift r(T) = r(T-
This weighted shift is known to be rationally strictly
cyclic (See Shields, [23], p.101). For rationally

strictly cyclic weighted shifts, L2(B) = LQ(B) as

68

formal Laurent series and the norms on these spaces
are equivalent. Hence, the projection O is bounded

since 0 : Law) 4 L2(B) is bounded.

Again, we note that we have usually taken

 

 

- -l
r(T 1) < r(T). This, of course implies that
_ -1
“T 1H < HT . One might ask whether the condition
HT-IH-l < “TH is sufficient for the boundedness of

9. The answer to this question is no and is illustrated

by the following example.

Example 5.3: Let T be the weighted shift

with weight sequence as below:

i) wn = 1 if n # -1
ii) wn = 1/2 if n = -l,
. . 1 ‘ -lI-1 o
For this shift HT” = l and ”T H = 1/2. This

shift, though, is similar to the unweighted shift. Thus
the "analytic" projection on L”(B) will be bounded if
and only if it is bounded for the unweighted shift.
However, it is known that the "analytic" projection

for the unweighted shift is unbounded (See Rudin, [18],
Prob. 9, Chapter 14). Thus the analytic projection on

Lm(B) is unbounded.

Our last example concerns the hypotheses of

- 1
Theorem 5.1 again. If r(T 1) < r(T), is the

condition that {pHco gthpHG(T) for all polynomials

69

p in z or iqflOD g_chHO(T) for all polynomials

q in z"1 necessary for 0 to be bounded on L”(

B)?
The answer here is again no. We use the following

example to show this.

Example 5.4: Let T be the weighted shift with

 

weight sequence given below. For k = 1,2,3,--- let

i) wn = 1 if n = (k(k+1)/2)-l
ii) wn = 1/2 if n > o and n ;! (k(k+1)/2) -1
iii) wn = 1/3 if n < O and n # -[(k(k+l)/2)-l
iv) wr1 = 1/4 if n = -[(k(k+l)/2) -1] .
- -1 .. -
Then HT 1“ = 1/4 < r(T 1) = 1/3 < % = r(T) < HT” = 1.

For k > 0, let 2(k) be the number of 1's appearing

in {w "wk-1}' For k < 0, let L(k) be the

0'...

number of 1/4's appearing in {w ,w_2,---,wk}; and

-1
let 1(0) = 0. Then

2£(k)-k

a) B(k) = if k 2.0

-k-z(k)4z(k)

b) B(k) 3 if k < O .

We note that 1(k) 4 m as either k 4 m or k 4 —m,

For k.2 o, szHm 2_szH2 = B(k) = and

nzkum, = 24‘. hence. uz“1\.,/uz“um,2 for

k > 0. Similarly for k < 0, “2k“... 2 szH2 = 3-£(k)-k4£(k)

-k L(k)

and szHO(T) = 3 . Hence szHm/HZRHO(T)‘2 (4/3)

for k < 0. Thus there can be no constant C > 0 such

70

that either ”pun g_CHpHO(T) for all polynomials p in
z or ”q“ g_CHqHO(T) for all polynomials q in 2-1.
G
For this weighted shift, however, the "analytic" projection
0 is bounded on Lm(B). To prove this, we will show
that T is a rationally strictly cyclic weighted shift.
We do this by verifying the condition in PrOposition 32
of Shields, [23]. (Note that this can be extended to

bilateral shifts according to the remarks after

Proposition 36 of Shields, [23].)
To simplify notation let B(n,k) = B(n)/B(k)B(n-k).

Q
Thus we want to show that sup{ Z) 132(n,k) : n E Z} is

=-m

finite. For n 2.0,

°° 2 ‘1 2 n 2 ° 2

2 B (n,k) = 2 B (n,k)+ E B (n.k)+ 2‘ B (n,k)

=-w =-m k=O k=n+l

’1 2 '1 -n Mn) k Mk) 'Mk) n-k - ( -k)
Z) B (n,k) = Z) 4 4 9 9 16 4 4 3 n

—l
= 23 (4/9)’k(9/16)“k)4““)”Lin-k)

-l

g_ Z} (ll/9)-k since for k < O,
k=-w

L(k)‘2 O and L(n-k) 2.1(n).

-1 so
So 2 32(n,k) g )3 (4/9)k = 4/5. Also,
k=-m k=1

71

23 B2(n.k)
k=n+l
= E 4-n 4£(n) 4k 4-uk) 9n—k 9!,(n-k) l6-L(n-k)
k=n+l

C3

k=n+l

g, :3 (4/9)k-n since for k 2_n, L(k)‘2 L(n)
k=n+1

and L(n-k)‘2 O .

So 2) 52(n,k) g_ 23(4/9)k 34/5° The other term for
k=n+l k=l
n > O is Z) 32(n,k) = Z) 4L(n)-L(k)-L(n-k)' which

n n
k=O k=O

we will deal with a little later. First we will see what

happens if n < O. For the case when n < O,

°° 2 “‘1 2 O 2 ° 2

2 B (n,k) = 2 B (n.k)+ E B (n,k)+ 23B (n,k).
k=—m =—cp k=n k=1

n-l 2

Z) B (n,k)

=—m

gig 9-n 9-£(n) 161(n) 9k 92(k) 16-‘(k) 4n-k 4-£(n-k)

k=-w
n'l -k+n £(k)-L(n) -L(n-k)
= 23 (4/9) (9/16) 4
k=-m
n-l —k+n
g_ (4/9) since for k < n, L(k) 2.2(n)

and £(n-k) 2.0

gal/5 .

72

Similarly Z 32(n,k) g 4/5, but again
k=1

0 o
23‘ B2(n,k) = 2(16/9)1(n)-£(k)-£(n-k).
k=n k=n

One can show that for k 2 O, Mk) = [ —-%-+./%+ 2k ]

where [ °] is the greatest integer function. Thus
fifi-lgz(k)g¢2¢k+l for k20. So

2% 4L(n)-;c(k)-£(l’1-k):43 §(4./'§)¢E—fi-jn-k

k=O k=O
Let O< r< l and o(n,k) =Jk+Jn—k—Jn for

ngg [% ]. Then by symmetry

gflk+JHZk-A/ES 2mg” r0(n,k)
k=O k=O

One can now show that o(n,k) g o(n+l,k) by a direct
computation. Also if n is even, then o(n,n/2) =

([21le which tends to an as n 4 on. If

S = 2 2:) rd . then SmlgSn+2ra(n+l'[(n+l)/2])

_ _ a, _
Thus S g S +2r(‘/2-']')“/n and Zr(“/2-l)~ffi< co.
n+1 n n=O

Hence sup{Sn : n = O,l,°°°} < co. This says

n
sup{ 2 41(n)-2(k)-L(n-k) : n = O,l,2,°-°] < ca. Similarly
k=O

0
one can show sup{kz} (16/9)L(n)—L(k)-L(n-k) : n = —1,—2'°°°]<co.
=n

73
Now putting all the parts together, we have

N
sup{ 2‘, B2(n,k) : n e Z} < co. Thus T is a rationally

strictly cyclic weighted shift. Hence Lm(B) = L2(B)

and the norms are equivalent. Thus 9 is bounded on

L (B).

We end this chapter with a conjecture. We have

- -1
seen that the condition r(T 1) < r(T) is not necessary

for the boundedness Of 9. However, we have used it

extensively in most of the results in this chapter.

1-

Is r(T- ) l < r(T) a sufficient condition for 0

to be bounded? Another condition, stronger than

_ l
r(T 1) < r(T), which might be sufficient is to

*
require that WO(T ) contain an Open annulus.

CHAPTER VI

SPECTRAL SETS

Let T be a bounded linear operator on a

N
A
Hilbert space H. If p(z) = Z) p(k)zk, then we can
k=O
N A k
define p(T) e B(H) by p(T) = Z)p(k)T where
k=O
T0 = 1, the identity Operator on H. Also, if

h(z) = (z--)‘)-1 and A g o(T) then we can define

MT) 6 B(H) by mm = (T-ifl.

Putting these two
things together, if q(z) = pl(z)/p2(z) is any rational
function with poles (the zeroes of p2(z)) off o(T),

then q(T) = pl(T)p2(T)-l

E B(H). One can investigate
certain relationships between the function q (which

is analytic on a neighborhood of o(T)) and the Operator
q(T) E B(H). In this chapter, we want to consider the
relationship between ”q(T)” and HqHK where K is

a compact set containing o(T). Thus we have the

following definitions.

Definition 6.1: Let K c ¢ be compact. Then
Rat(K) is the set of rational functions with poles

off K.

Definition 6.2: Let T e B(H). A compact set
K containing o(T) is said to be a spectral set
for T if and only if Hf(T)H g HfHK for all

f 6 Rat(K).

74

75

Definition 6.3: Let T E B(H) and let c > 0.
A compact set K containing o(T) is said to be a
c-spectral set for T if and only if Hf(T)}g_chHK

for all f 6 Rat(K).

We note at this point that it is not possible for

Hf”o(T) to be greater than “f(T)

 

‘. From the spectral
mapping theorem, we have HfHo(T) = r(f(T)) g_Hf(T)H.
Thus in most cases we will have c 2_l since we will
consider sets K not much "larger" than o(T). Two

questions are raised by Definition 6.2.

i) If T 6 B(H) what kind of sets K containing
the o(T) can be chosen so that K is a

spectral set for T?

ii) What conditions can be placed upon the Operator
T e B(H) so that o(T) is a spectral set

for T?

In this chapter we will concentrate upon the
first question. However, in some cases we may limit
our study to weighted shifts. In order to provide
some type of answer to the second question, we note
that if T E B(H) is normal then o(T) is a spectral
set for T (Rudin, [17], p.309). In particular,

0(T) is a spectral set for T if T is hermitian.

76

Returning to the first question, we note that if
T e B(H) then C(T) c [z 6 ¢ : \2( 3 HT“). This
containment may not be strict as there are some
Operators with o(T) = {z 6 ¢ :\2\ g_HTH]. The
question of whether [2 6 ¢ :\2\ g_HTH] is a spectral
set for T was answered in 1951 by von Neumann [26].

It is sufficient, by scaling, to consider only operators

T with ”TH = 1. In this case, we let

16 =:[z 6 ¢ :\21 g_l], 61) = [z 6 ¢ :\2\ = l] and

I) = 33\BID. we will give a new proof of von Neumann's
theorem using some of his ideas. In place of one of

his arguments, we will use the solution Of the Caratheodory-
Schur problem suggested in the section on Hankel operators
in Sarason's VPI notes [21]. The result needed is

stated and proved below.

N A
Lemma 6.1: Let p(z) = Z) p(k)zk be a polynomial.
k=O

Then for each n > 0, there exists a set

[d1,°-- )] c I), a constant A 6 ¢ and a function

“Mn

h analytic on a neighborhood of ‘13 such that:
i) [M g MPH-5

ii) Hth s ZHPH-fi
i(n) _ _1
iii) p(z) = x k2l (z-ai)(1-diz) 4-2 h(z),

Proof: Let T, the unweighted shift (wn = l

for all n), be represented as M2 on L2(aIH . Let

f(2) = z-(n+l)p(z) and consider the Hankel Operator Hf.

77

Since f 6 C(31)), Hf is compact. Thus if B is the
closed unit ball of H2(BID), then Hf(B) is compact.
Therefore there is gl 6 B such that HHf ng = HHfH = 1-

Also there is a function h e Hm(aIH such that

anhuco = “Hf” = A- Now Hf(gl) = Hf(g) where
n+1 A k m
g = Z) gl(k)z since for m < 0 (H (g ),z ) =
k=O f 1
m G A A n+1 A A m
(fgl.z ) = Z gl(k)f(m-k) = Z 91(k)f(m-k) = (Hfg.z ).
k=O k=O
Thus h = HHfH = HHngz = li(1-P>l(f-h>91‘)2 _<. H<f-h)9‘.)2 g
Hf—hHmHgHZ g_x. Hence equality must hold throughout.
-1
This implies that Hfg = (f-—h)g = Z) bkzk.
=-(n+l)

From the string of inequalities above, we must have

\f-—h‘ = 1 almost everywhere on 31). Also,
zn+leg = zn+l(f-h)g is a polynomial and hence in
m n+1 1 2
H (51)). We note that z (f-h)g = [ H (z-ai)][ H (Z-A.)]
i=1 i=1 1
where [oi :i = l,°°',Nl] c I) and {xi :1 = 1,"',N2] c ¢\I).
It is known that zn+1(f—h)g has a unique inner-outer

factorization. Since zn+1(f—h)g has only a finite
number of zeroes in I), the right hand factor is an
outer function, and the left hand factor is bounded away
from zero on all) , the inner factor of zn+l(f-h)g

must be a finite Blaschke product (see Rudin, [18],

Th. 17.17). However, zn+1(f-h)/X is an inner function
and hence must be a finite Blaschke product b.

n+1

Thus p(z) = 2 f = xb(z)+-zn+1h(z). Also,

78

1 = “Hf“ g_HfHal) = HPHBID' Consequently,
\h<z>\ = um) ->.z"“"“h<z>\ .<. Hemen + m _<. 2“P“5]D’

Thus parts i), ii), and iii) are satisfied. Q.E.D.

We now state and prove von Neumann's theorem.
We use a complex function theory argument. A purely
Operator theoretic proof can be given using unitary

power dilations (see Halmos, [8]).

Theorem 6.1: If T E B(H) and “TH = 1, then

I) is a spectral set for T.

Proof: First let a E I) and let 0 < r < l.
1

We now want to consider the operator A = (rT-o)(l-4arT)-
If f e H and g= (l-Er'rflf, then
:2
“Af“ = (Af,Af)
= ((rT-G)g.(rT-a)g)
2 2 2 :2
= r “Tg“ -2Re (ag,rTg)+ ‘0‘ “g“
Also
Hfflz = ((l-ErT)g.(1-ErT)g)

“g“2 — 2 Re (ag,rTg) + ‘o‘2r2“Tg“2

Hence HfH2-HAfH2

ngzu- M2) —r2nTgu2<1- m2)

<1- \e‘2>wg!12-rznhg)21

2_0 since ‘a‘ < l and rHTgH g_HgH-

79

Thus “Af“ g “f“ for all f E H implying “A“ g l.

N A k
Now let p(z) = Z p(k)z and let
k=O

[ol,°--,a“n)], 1 and h be as in Lemma 6.1. Then for
O < r < l
L(n) — -1 n+1 n+1
p(rT) = X 1'1 (rT-ak)(l-o.krT) +r T h(rT) .

k=l
Thus “p(rT)“g ‘1‘+ rn+l“h(rT)“ by the triangle inequality
and the first part of this proof.

Now since h 6 HwUD) , h(rz) = Z, h(n)rnzn.
h=0

w A n n a A n
Hence h(rT) = Z h(n)r T . Thus “h(rT)“ g 2:, ‘h(n)‘r g
h=0 n=0

“1'1“” 230 rn g “h“m/(l-r). We now have
n:

upon); _<. w + )h‘am rn+l/<1-r> g. HpH-fi + zlapu—firm/u-r)

by Lemma 6.1 and the maximum modulus theorem. This

is true, however, for all n > 0. Hence
“P(I'T)“ _<_ “p“ffi for all r between 0 and 1

Letting r increase to l and using the continuity
Q:f the “p(rT)“ as a function Of r, we get
H p(T)“ g “p“fi for all polynomials p. The polynomials

are dense in Rat(f) , though. Thus we are done. Q.E.D.

80

Now let T E B(H) be invertible. Then we know
that o(T) is contained in the annulus
[z 6 ¢ :“Tml“-1 g_‘z‘ g “T“]. It will be sufficient
(as in the case of the disc) to concern ourselves
only with operators T for which “T“ = 1 and

o < “T-l“ 1 < 1. (We note that if “T“ = l = HT’IH-l.
then T is unitary and hence “f(T)“ S-HfHoI) for
all f E Rat(aIU (Rudin, [l7],Th. 12:13). In order
to simplify our writing, we introduce the following

notation: A(r) = [z 6 ¢ :r g ‘z‘ g_l]. We now state

a result which appears in Shields ([23], Prop.23).

Theorem 6.2: Let T E B(H) be such that “T“ = 1

 

and “T-l“ = r where 0 < r < 1. Then there exists
a constant cr such that “f(T)“ g_cr“f“A(r) for all

f 6 Rat(A(r)).

This theorem says the annulus A(r) is a
cr-spectral set for all T e B(H) satisfying the norm
requirements above. However, this theorem (as stated)
is unsatisfying in the sense that it does not tell us
anything about Cr‘ Can we take cr = 1? If not,
how large must cr be? In 1967, J. Williams, [28],
showed that we cannot take cr = l for all T E B(H)

satisfying the norm requirements. In the result by

81

Shields, cited above, it is shown that the constant

 

cr = 24-J(l+r)/(l-r) will suffice. However, we will

show that a slightly smaller constant works. The
question of determining the best constant for cr

is still Open.

PrOposition 6.1: Let T E B(H) be such that

HT“ = 1. \(T‘ln'l = r. Then (emu s hrnfnm) for

.. -1
all r e Rat(A(r)) where kr = min[3+47r lln(l6(l—r2)) ,

 

2+./(1+r)/(1-r)] .

Proof: Let f e Rat(A(r)) be uniquely decomposed

as follows:
f(z) = ¢l(z)-tw2(z); $1 is analytic on IL
$2 is analytic on [2 e‘¢ :‘z‘ > r]

and ¢2(z) 40 as ‘2‘ 4 n.

Then f(T) = ¢1(T)+¢2(T). Also “cp1(T)“ g “cplnam
by von Neumann's theorem and “¢2(T)“ g_“¢2“ral) where
r51) = [z 6 ¢ :‘z‘ = r]. The second part follows by

. -l .
uSing von Neumann's theorem for T and the maXimum

modulus theorem. Now “szuralD g “eraD + “cpl“raD g

”sum, + “‘Pluron -

Using the Cauchy Integral Formula, we get

lirr
r4]
1m

be‘

im

Si

he

82

‘cp1(rw)‘ = —1—“ f(z)(z-rw)_ldz‘ for all w e an).

2F 81)

g Menage!) \z-rwrl 1%.?

S.“f“A(r)‘aI)‘l-rz‘-l l%%i since ‘2‘ = 1

for z 6 31)

2V . -1
. 19 QE
3 MN ‘l-re ‘
A(r); 2W
2W -l/2
g “f“A(r)“' (l-2r cos 6+r2) %%
0
27’ —l 2 ..
Now ‘ (l-2r cos e+r2) / 53-9- = 21r 1k(r) where
0 w
w/2 _
'k(r)= [ (l-rzsin29) l/Zde from Ryshik, [20]. However,

0

1im (k(r)4~% ln 16(l-r2)) = 0 by Byrd [4]. Thus
r41-

k(r) g ln(16(l-r2))'l if 1 > r > rO for some rO

between 0 and 1. Thus
_ —1
“wl“ra]D g “f“A(r)(2n 11n(l6(l-r2)) if r0 < r <‘l ,
implying
- -1
MW)” g “f“A(r)(l+27T lln(16(l-r2)) if r0<r<l .
Similarly

”CleaD S HfHaJD + “42““) S. “f“A(r) + “COZHI'BJD by the

maximum modulus theorem. Putting both of these together we get:

works

int

11!?
r»;

res
is
Th

5!

In

83
Hf<T>H g Helmn + “co2(T)“

g “cpl“ap + Meaghan
S. “fHA(r) + 2HCPZ'i r51)

_ -l
g “f“A(r)(3+41T lln(16(l-r2)) ) if r0 < r < l

 

If 0 < r g.ro then the constant 2-tj(l+r)/(l-r)

works. Q.E.D.

 

The estimate above is better than 2-tJ(1+r)/(l-r)
-1
in the sense that 1im [(l-r)l/zln(l6(l—r2)) ] = 0.
r41'

This estimate is still "bad", however, in the sense that

lim (3-1-47r-11n(16(l-r2))ml

r41"

) = a. We now give three

results for Operators T E B(H). The first result

is obtained by placing restrictions on the Operators.
The other two results are obtained by considering
specific types of functions and investigating the norm

inequality for this type of function.

Proposition 6.2: Let [T :r0 < r < l] c B(H)

 

r

-l
'1‘

be a net of operators such that “Tr“ = l and “Tr = r.

Suppose there is a constant M > 0 such that “Tr-n“ g_M

for all n‘z 0 and for all r between and 1.

r
0
Then for r0 < r < l, “f(Tr)“g_M“f“A(r) for all

f e Rat(A(r) ).

Risk

is dens

let p

Hence

that
is :
difi
oi

the

be

84

Proof: The proof of this proposition is easy.

N
It is known that {q 6 Rat(A(r)) :q(z) = Z] g(k)zk]

=-N
. . N A )r
is dense in Rat(A(r)). For q(z) = Z) q(k)z we
=-N
let p(z) = znq(z). By von Neumann's theorem
“p(Tr)" _<. Hp”am = ‘lql‘ap s “q!|A(r)
Hence
-N . -N‘ u -
“q(Tr)“ = HT} P(Tr)“ $.HTr ““PCTr)H $.MHqHA(r)

Q.E.D.

One could also prove this theorem by noting

that if Tr and. T1-1 are power bounded, then Tr

is similar to a unitary operator Ur’ The only
difficulty would then come from computing the norms
Of the Operators providing the similarity. We now have

the two results concerning types of functions.

A
Proposition 6.3: Let q(z) = Z) q(k)zk E Rat(A(r))
k=-co

be such that g(k) 2_0 for all R. Then “q(T)“ $.2“q“A(r)

. ‘ -1 -1
for all T E B(H) With “T“ = l and “T “ = r.

G
A
Proof: Since q(1)2 Z q(k) and
-1 k=O

A k
q(r) 2 Z) q(k)r we have

85

“q(T)“ _<. If: (Gmmwk)

-1

° A
_<. ‘23 $(k)rk+ 2 q(k)
k=-a k=O
g_q(r)4-q(1)
g_2“q“A(r) . Q E D

Proposition 6.4: Let q(z) = z-Ni-on where

M,N 2 o and o 6 ¢. If T e B(H) with “T“ = l and

HT'lu'l = r then (emu g 22!q2\A(r,.

M+N

Proof: We see that q(z) = z-N(l4-dz ).

By the maximum modulus theorem, we have

(gum, = maxuxquam .ueurw)
= max[l+ (o‘,r’N(i+ ‘o‘rM+N)}
2%((1+‘O‘)+(r_N-l- ‘a‘rMH
Now
“q(T)“ g (T‘Nu+ mm”) s r‘N+ m s when) - 9-“-

In the next part of this chapter we restrict our
attention to invertible bilateral weighted shifts. The
question Of whether we can take Cr = l is question 7
in Shields, [23]. He actually has two parts in his

question. The first part asks whether sup{cr : 0< r < l] < an

86

when we are only considering weighted shifts. We

answer the second part of question 7 using the following

example. We state beforehand that the answer is no:

one cannot take cr = l for weighted shifts.

Example 6.1: Let T be the weighted shift with

 

weight sequence as below.

i) w = 1 if n 2.0

n
--> _.1. -. n 0
ii wn — 2 i <
Then
“T“ = su [w -nEZ]=1 and I‘T-l“-l=inf[w -neZ]=-LL
p n ' ' n ° 2
-l m '1 .
For m > 0 let fm(z) = z (2-z ) . Then by the maXimum
modulus theorem HmeA(_1_) = maXIHfm“an 'Hme-LBD]
2 2
-m -l
= max[l,2(2-—2 ) ]
_ -1
= (1-2 m 1) .
From this computation we see that lim “me 1 = 1.
m-Om A(E)

Now since “e ‘ = l we have “fm(T)“ 2_“fm(T)eO“.

0‘

We compute “fm(T)e as follows. We note first of

0“
all that

87

. -l _ mk-l _
It is easy to see that T eO — 2e_l and T e — emk-l

since mk-1 2.0 for k‘z l and m‘z 1. Thus

fm(T)eO = 8-17? ZZZ-k-le
k=1 mk—l
°° .. _ l 2 .—
Hence “f (T)e H = (1+ 24 k l) / = J13/12. Thus
m 0 k=1

there exists an MO > 0 such that “fm(T)H > “fm“ 1
A(57
if m > MO. 80 it is not possible for

||q('.r)il s “q“ for all q 6 Rat(A(%)).
M

l
3)

Proposition 6.5: Let T be the weighted

 

shift with weights as below:

i) wn = 1 if n 2.0

ii) wn = r if n < 0 where 0 < r < 1

Then “q(T)“ g_J2'“q“A(r) for all q E Rat(A(r)).

Proof: For the weighted shift above B(n) = 1
if n 2.0 and B(n) = rn if n < 0. For f E L2(B)

we have the following:

“when; = (been:

8 /~\ '1 /\
Z I(qf)(n)‘2+ 23 ‘qf(n) ‘zr2n
n=0 =-oo

. a /\ 2 ‘2 A ‘2
It is known, however, that Z)‘qf(n)‘ g_“q“a1D Z) ‘f(n)‘
n=0

=—m

-1 A ‘ c.
and 2 ‘qfihnzrznrnqniam 23 \?<n)\2r2“-

n=-m n=—w

88

Thus
2 . .2 °° A 2 .,2 °° A 22
‘lq(MZ)f“2 :1“:le nam‘fln)‘ +“qz‘ram nf__‘3__m\f<m)‘ r n
8 A
‘2

_<. 2‘\q1\§(r,nfu§ since Me): = gown)

‘

-l
+ z ‘f(n)‘2r2n.

Thus “q(MZH‘g.J2 “q“A(r) and the result follows

easily. Q.E.D.

So for the weighted shift T in Example 6.1,
A(%) is a (J2-spectral set for T. The result of
Proposition 6.5 says this type of idea will not be
helpful in getting an example where sup{cr :0 < r < l]
is infinite. There is a generalization of Proposition

6.5 which we now prove.

Proposition 6.6: Let T be an invertible

 

weighted shift with “T“ = 1 and “T-l“ = r where
0 < r < 1. Also, suppose the weight sequence {wn]

of T satisfies the following two conditions:

1) [wn:nez] c: [a

1,000 ON]
ii) I g.wn g wn+1 g_1 for all n

Then “q(T)“ g_JN “q“A(r) for all q E Rat(A(r)).

89

Proof: Let 0 = nl < n2 < ... < nN—l' We

will assume without loss of generality that:

a) wn = r = 01 if n < nl
b) wn = Gk if nk-l g_n < nk k = 2,3,‘°°,N—l
c) wn = on if nN—l g_n

This implies, of course, that r g_ok g ak+1 g_1. From

the above we have the following:

, _ n
a ) B(n) - 01 if n < nl
k-l n -n n-
I _ Z 1-1 nk-l .
b) B(n) - (i=2 01 )(OIk ) if nk_lsn<nk
N-l n -n n-n
c’) B(n) = (n 0k E i‘lmN ”'1 if n 2 nN-1
i=2
Now let P1 = 1-P where P is the orthogonal

projection of L2(B) onto H2(B). For k = 2,"',N-l

let P be the orthogonal projection of L2(B) onto

k
n .

the span of {z .nk_1 g_n < nk]. Finally, let PN

be the orthogonal projection onto the closed linear

span of [2n :n 2_nN_l].

Then for q E Rat(A(r)) and f E L2(B) we have

.2 2 N 2
(*) “q(Mz)fH2 = HquH = k§l\‘Pk(qf)“2
Now
2 '-l /\
“P1(qf)“2 = nELm‘qf(n)‘2012n

2 w
A
S’andlal) E) ‘f(n)‘2012n as in Proposition 6.5

n--:::

 

9O

2 co
g.HQHA(r) Z) ‘%(n)‘2B2(n) since

n=-m

B2(n)2r2n for all n

g ‘Iq‘|§(r)“fii§

‘ “2 “2'1 /\ 2 2n
P (qf) = Z qf(n) a
‘ 2 '2 n=0 ] 1 2
m ‘A 2n
gl‘q‘lz z (f(nnzez
023D n=—m
g.Hq“§(r)“f“: since 0:“ g B2(n) for all n .
For k = 3,4,5,-o-,N-l
nk-l
.2 /\
HPk<qf)h = Z ‘qf(n)‘2B2(n)
2 nenk_1
k—l n -n nk—l /\
S.( n a) z £-1)20;2(nk_1) Z) ‘qf(n)‘2q:n
L=2 n=nk-l
k-l n -n 2 -2( )
2 2-1 nk-l 2
S'(LE2 at ) Gk ‘q“akaI) .

Z} ‘f(n)lzafin
n=-m
k-l n —n 2 2n-2
However [ n a L z-I] Gk nk-l g_BZ(n) for all n.
L=2 1
2 2 “2 _ ...
Hence “Pk(qf)“ g_“q“A(r)“f“2 for k - 1,2,3, ,N-l.
Similarly, one can show “PN(qf)“2 g_“q“§(r)“f“§. Now,
going back to formula (*) we get “q(Mz)f“2 g_JN Hq“A(r)Hf“
for all f 6 L2(B). Hence “q(T)“ g.JN Hq“A(r) for

all q 6 Rat(A(r)). Q.E.D.

91

Again, it may not be that JN is the best

possible constant. In fact, if we pick N large

 

enough so that JNI> 24-J(1+r)/(l—r), then JN is
not the best constant by Theorem 6.2. However, JN

does give a better estimate in certain cases.

One open question in the area of c-spectral sets
is whether every operator whose spectrum is a c-spectral
set is similar to an operator whose spectrum is a
spectral set. It is possible that Example 6.1 might
answer this question in the negative. However, the
following example may indicate that it might not answer

the question.

Example 6.2: Let T be the weighted shift with

 

weights as given below:

i) wn = l for n # -1
ii) w =-£ for n = -1
n 2 '
. . . -1 -1
For this weighted shift r(T) = “T“ = l = r(T )

- -l
and “T l“ = %3 Using the same functions as in Example 6.1,

we see that o(T) = 61) is not a spectral set for T.
(Note: the part that makes the computation work is

that Tgl = 2e_l). However, by verifying the requirements

0
of Theorem 2 in Shields [23], it is easy to see that
T is similar to the unweighted shift. The unweighted

shift is normal and hence its spectrum is a spectral

set. We also note that since T is similar to a normal

Operator, its spectrum is a c-spectral set.

 

92

Suppose we could find a collection of weighted

shifts Tr for 0 < r < l satisfying:

1) “Tr“ = l and “Tr
ii) o(Tr) = A(r)

iii) sup{“6> :0 < r < 1] = M.< m where 9

‘l
r1 r
is the "analytic" projection on Lm(Br).

N A

Then if q(z) = Z) q(k)zk we would have
k=-N
N A k
“ Z)q(k)z Hal) S.“9rq” m (the spectral mapping theorem)
k=O L (B )
r
g_“0r““q“ m (assumption)
Br)
H ,
S-MCqu“A(r) (Theorem 6.2)
We note that liT_ “q”A(r) = “9“513- Thus if sup{cr =0 < r < 1)
r4

was finite, the "analytic" projection on C(51)) would
be bounded. This is not true, however, as stated
before. Hence, it would not be possible for Or to

be bounded as a function of r. Unfortunately, it is
not possible to find weighted shifts Tr satisfying
the condition above. The proposition below verifies

this.

Proposition 6.7: Suppose that [Tr :0 < r < l]
:is a collection of weighted shifts with “Tr“ = l and

HT -l“-1 = r. Then sup{“9r“ : 0 < r < l] = no.

r

93

Proof: Suppose sup{“ér‘ :0 < r < 1] = M < m,

N

 

A
Then for q(z) = Z) q(k)zk we have
k=-N
N A k N A
“ Z) q(k)z “ = “ Z)q(k)T “ (von Neumann's theorem)
k=O 51D k=O
- 1| ‘
- ngq“ a
L (Br)
= Miiq',‘ ...
L (B )
r
However, I claim 1im “g“ m = “q“ . To see this,
r41” L (Br) BI)
let T be the unweighted shift. Then “q(T)“ = “q“aI)
since T is normal and o(T) = BI).
Now

k

" N A . k .
l|q(Tr)-q(T)“ g 23 ‘q(k)“‘Tr ..T “
k=-N

N N A _
g Z\%(k)(<1-rk)+ Z) \q(-k)((r k-1) .
k=1 k=1

Since the sums above are finite 1im “q(Tr)-—q(T)“ = 0.
r41-

Thus 1im_“q“ = 1im “q(T )“ = “q“ . The claim

r41 L°°(Br) r41” I‘ an
is now verified. Using this, we have
N A k
I 1'
“RE—:0 q(k)z (‘51) g Mllqi‘aD '

As before, this inequality says the "analytic" projection
on C(51)) is bounded. This is false and so

sup{“é ::0 < r < l] = w. Q.E.D.

r“

94

This is all we will say about weighted shifts.

We will end this chapter by considering linear operators
. - -1 .

T on ¢2 with “T“ = l and “T l“ = r. Any linear

Operator T on C2 has a matrix representation

C1 0

ll 12
T = where aij 6 ¢. By unitary equivalence,
G21 G22
11 G
we may assume that T has the form with
0 12

respect to some orthonormal basis for ¢2. The norm of
a 2)(2 matrix can be computed easily by noting that
2 * . * . . . . .
“T“ = “T T“. Since T T is selfadjOint, its norm is
* _ ...
sup{‘x‘ :x e o(T T)]. By similar reasoning, “T 1“
*
will be inf[‘x‘ 3X E o(T T)]. One can find the
*
eigenvalues of T T by finding the solutions to the
*

quadratic equation det(tI-—T T) = 0. If one does this for

1 a

l _1 -

T = with “T“ = 1, “T H = r, we get the

0 12

following restrictions on 11,12, and a.

i) ‘A1‘1A2) = r

ii) ‘a‘2=(1-‘A1‘2)(l'])‘2‘2)

We will use these, but first we will find out what

q(T) is for q e Rat(A(r)).

‘ o
‘ O
t
A 1
l
.--._..

 

95

X1 a
Proposition 6.8: Let T = and let
0 12

q be analytic on a neighborhood of o(T) = [11,12]. Then

q(ll) a[q().l)-q().2)]/().1-)2)

q(T)
o q(lz)

Note: if 11 = 12. then (g(Al)‘q(A2))/(A1‘A2) is

replaced by q’(xl) in the above formula.

Proof: First suppose )1 = 12 = x and let q be

analytic on U = {2: ‘z-x‘ < e]. Let T = [z :‘z-x‘ = e/2]

and q(z) = 23(q(n)(x)/h1)(z-x)n for z 6 U. Then by the
n=0

Riesz functional calculus

 

1 r
2W1 Jr

Q(T) = q(z)(z-T)-1dz

Thus (q(T)ei,ej) = 3%; ‘r q(T)((z-T)-lei,ej)dz for

i,j = 1,2. One easily computes that
(z-l)-1 a(z->.)'2

(z-T)-1 = . Thus (q(T)ei,ei) =
O (z-).)'1

5%3-‘r q(z)((z—T)-lei.ei)dz

g(X) by the Cauchy integral

1. P
2W1 J

 

Q(Z)a(z-))-2dz =

formula. Also (q(T)e ,e ) =
l 2 r

aq’(x) by the residue theorem.

96

If 11 #’xz and q is analytic on a neighborhood
of [11,12] we choose two disjoint circles T1 and

T2 in U surrounding 11 and 12 respectively. Then

 

 

 

 

_ l -l
(q(T)ei.ej) - 2W1 Ir q(2)((z-T) ei.ej)dz
l
+ l -1
'3}; [r q(z)((z-T) ei.ej)dz -
2
,
(z-).l)-1 a(z-).1)"l(z-).2)"l
-1
If 11 # 12 then (z-T) = .
-l
b 0 (2-12) 4
H < (T) e ) —-—i— < )( - )‘1dz+ 1 (z)(z- )‘1dz
ence q ei, i — 2vi I? q z 2 xi 2Wi IT q Ki
1 2

g(k ) since for 1 # j, I 9(Z)(Z-1-)-ldz
l i
T.
3
= 0

because q(z)(z—).i)-l is analytic inside of Fj. Now

(q(T)eloez) 371;; II‘ q(z)a((z->.l)'1(z-12)dz

1

._l_

+ .
2Fl

I q(2)a((z-)l)-1(z-12)dz
1‘2

a[q(>.l)/().l-).2) +q()2)/().2-).1)]

= a[(q(ll)-q(12))/(ll->.2)] as desired-
Q.E.D.

 

_

97

We end this chapter with the following proposition.
It says that for two-dimensional linear operators
sup{cr :0 < r < 1] is finite. A conjecture might
be that this is true for operators on n-dimensional
space. The constants cr would however perhaps depend

upon the dimension of the space.

Proposition 6.9: Let T e 6(¢2) satisfy

_ -1 - I
HTH = 1 and “T 1I = s when IIq<T>II g ser l/ZIlq“A(r,
for all g E Rat(A(r)).
11 a
Proof: We let T have the form and
O 12

. .. 2 2 2
thus have i) [11“12‘ = r and ii) ‘a‘ = (1-‘11‘ )(1-‘12‘ ).
We also note that if A = (aij) is any Operator on ¢2

then “A“ g_4 max[‘a.

lj‘ :i,j = 1,2]. By Proposition 6.8

we know that if *1 #'12 Then q(r) =

q(hl) a(q(xl)-q(12))/().l-).2)

Since [11,12] = o(T)c:A(r)
0 g(lz)

we have ‘q(xi)‘ g_“q“A(r) for 1 1,2. Thus we need
. . -l .

only investigate ‘a“q(xl)-q(12)“xl-xz‘ - We will

use three cases. The first case will also indicate what

to do if A1 = A2-

98

 

 

Case 1: Assume ‘xl‘ = ‘12‘ = j; and let
’q<z)-q(xl) .
2-11 1f ZEA(r). za’hl
f(z) =
Lq’ul) if z = 11

Then f is a function analytic on a neighborhood of

A(r). Thus “f“A(r) = max[“f“al),“f“ral)] by the

maximum modulus theorem. It is easy to see that

Hf!)81D g 2l‘q“A(r,/<l-J?) and IIfIIraD _<. 2IIqIIA(r,/<fi-r>

and ‘a‘2 = (1- (11‘2)(1-‘12‘2) = (1—r)2.

-1

Hence IaIIqul)-q(>.2)IIll-12"1g4r /2“qHA(r).

Case 2: Assume J? < ‘xl‘ g_3?. Then

{/r3 g ‘12‘ < r/f. In this case we let f(z) be defined

as in Case 1.

. -l _. _
Then ‘f(z)‘ g 2“q“A(r) max{(l-rl/4) ,(Jr-r) l}

and ‘a‘2 g_(l-r)(l-r2) g_2(l-r)2. Thus
‘s'I‘q<>~2)-q(>~1)HA2-A1"1 s a)? r'”2 \‘q%‘A(r) since

1/4 1/4

(l-r)/(l-r )==(1+J§)(l+r ) and (l-r)/(Jr-r)==(l+j?)/fr .

_C_§_§_e__3_: Assume 4)? < ‘11‘ g 1. Then
r g ‘12‘< 4):]. In this case we let f(z) =
(Q(Z)-Q(Al))/(Z-11) for z 6 [Z 6 ¢ = r g ‘2‘ g ff}. Then
‘Q().2)-CI().1)“A2-)(1‘-l _<_ 2“q“A(r)(4(/-l: -fr'rl and
‘a‘2 g(l-J?)(1-r2)g4(1-V’_f)2.

99

Thus ‘a“q(12)-q().1)“).2-Al‘-l g 8r'1/4HqHA‘r) _<.
Br-l/ZHqHA(r)-

Using the results of these three cases, we have

-l/2

“q(T)“ g 64r “q“Mr) for all q 6 Rat(A(r))

BIBLIOGRAPHY

[l]

[2]

[3]

[4]

[5]

[6]

[7]

[8]

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[11]

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"I7'1)‘11))1‘4‘71‘1‘1111‘7'“