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HYPONORMAL TOEPL.ITZ OPERATORS
AND NEIGHTED SHIFTS
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Dr. Sheldon Axler
Date May 1, 1984

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HYPONORMAL 'IOEPLITZ OPERATORS AND WEIGHTED SHIFTS

By

John Joseph Long, Jr.

A DISSERTATION

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

DOC'IOR OF PHILOSOPHY

Department of Mathematics

1984

ABSTRACT
HYPONORMAL TOEPLITZ OPERATORS AND WEIGHTED SHIFTS
By
John Joseph Long, Jr.

Let L2 denote the set of all Lebesgue-measurable, square
integrable functions on the unit circle, SD, and let H2 be
the usual Hardy space on 3D. For f in L00 (3D). the Toeplitz
operator Tf, mapping H2 to H2, is defined by Tfh = P(fh).
where P is the orthogonal projection of L2 onto H2.

In 1970 Paul Halmos [2] raised the question, "Is every
subnormal Toeplitz operator either normal or analytic?" A
subnormal operator is one which has a normal extension. An
analytic Toeplitz operator is one whose symbol is a bounded
analytic function on the unit disk. Chapter One contains a
more detailed discussion of this problem.

In Theorem 1, we prove that the answer to Halmos' question
is no. More precisely, we show that for 0<a<1, if v is
a Riemann mapping of the unit disk onto the ellipse with
vertices :1/(1-a) and :i/(1+a), and b = v + av, then Tb is
a subnormal unilateral weighted shift which is neither normal
nor analytic. This result was originally proven by Carl Cowen;

a different proof is given here. Theorem 2 proves that for

John Joseph Long, Jr.

any inner function u, the Toeplitz operator Thou is a direct
sum of copies of Tb. This is a special case of a more general
theorem due to Cowen [1]. These results are in Chapter Two.

In Chapter Three the work of Sun Shunhua [3] is extended.
The only finite direct sums of hyponormal weighted shifts
unitarily equivalent to a Toeplitz operator are those given
in Theorem 2. This result extends to infinite direct sums

if some additional hypotheses are assumed for the weight

sequences of the shifts.

[1] C. C. Cowen, Equivalence of Toeplitz operators, J.

Operator Theory, 7(1982) , 167-172.

[2] P. R. Halmos, Ten problems in Hilbert space, Bull. Amer.
Math. Soc. 76(1970), 887-933.

[3] Sun Shunhua, Bergman shift is not unitarily equivalent

to a Toeplitz operator, Kexue Tongbao, 28(1983),
1027-1030.

ACKNOWLEDGEMENTS

I would like to thank my advisor, Sheldon Axler. His

knowledge, patience, and encouragement were invaluable.

Thanks also to Cathy Friess and Walt Peebles, who assisted

in the preparation of the manuscript.

ii

CHAPTER ONE . . .
CHAPTER TWO . . .
CHAPTER THREE . .

L I ST 0F REFERENCES

TABLE OF CONTENTS

iii

18

34

48

CHAPTER ONE

For H a Hilbert space, let B(H) be the set of all bounded
linear operators from H to H. The unit disk in the complex
plane will be denoted by D. Let L2 denote the set of complex-
valued Lebesgue measurable functions on the unit circle, D,
which are square integrable with respect to normalized
Lebesgue measure. The Hilbert space L2 has the standard
orthonormal basis [zn : n is an integer]. Each g in L2 can

be written as

as
g = Z §(n)zn,

110-”

where g(n) is the nth Fourier coefficient of g. and the
infinite sum converges in the Lz-norm. Let H2 be the closed
' subspace of L2 consisting of those L2 functions whose negative

Fourier coefficients vanish. Each g in H2 can be written as
N n
g = Z 2(n)z
11eO

and hence g can be thought of as an analytic function on the
unit disk.

Let L denote the set of essentially bounded, (with
respect to normalized Lebesgue measure) complex-valued,
Lebesgue measurable functions on D. For any f in L , the

essential range of f will be denoted by ess ran(f). This is

2

the set of all complex numbers w such that the measure of
[2: If(z) -w| (c!

is positive for all c>0. Let H00 be the intersection of L00
and H2. Each f in H00 extends to an analytic function on D
which is bounded, and by Fatou's theorem each bounded analytic
function on D has radial limits almost everywhere. With this
identification, H0‘, is isometrically isomorphic to the set of
bounded analytic functions on D. All of these spaces are
discussed in [11].

The continuous functions on EU will be denoted by C(a D)

or simply C. The algebra H” + C is defined to be the set
[g+h : h in H°°, g in C}

This is a closed subalgebra of Loo [14], p. 191.

Let f be in LCD. The multiplication operator M1, is
defined by Mfg = fg, for all g in L2. Each Mf is a bounded
linear operator on L2, in fact IIMfH = ”film. The adjoint
of Mf is given by M; = M?‘ A normal operator is one that
comutes with its adjoint. Since all multiplication operators
comute with each other, Mf is a normal operator.

An operator S in B(H) is called subnormal if it extends
to a normal operator, that is there exists a Hilbert space
K such that H is a subspace of K, and a normal operator N
in 300 such that the restriction of N to H equals S.
Alternatively, subnormal operators are restrictions of normal

operators to invariant subspaces. A general reference for

subnormal operators is [4].

Let f be in Loo, and let P denote the orthogonal
projection of L2 onto H2. The Toeplitz operator with symbol
f, denoted Tf, is multiplication by f compressed to H2, so
Tfh = P(fh) for all h in H2. The operator Tf is contained
in am”) and um -.- urn”. The adjoint of T, is 17. s... [71.
Chapter Seven for the basic properties of Toeplitz operators.

In general Toeplitz operators do not commute with each
other. If Tng = Tng, then either both f and g are in H”,
both T and E are in H”, or there exist constants °1' 02 and
c3 (not all zero) such that clf + 02g = c3. From this it
can be shown that Tf is normal if and only if f = 01 + czg,
where c1 and c2 are constants and g is real-valued [3]. p.
98.

If f is in Hm, the Toeplitz operator Tf is called
analytic. When f is in Ha), for all h in H2, fh is also in
H2. Thus the Toeplitz operator Tf, with symbol in Hm, is the
restriction of Mf (defined on L2) to H2. Analytic Toeplitz
operators are subnormal.

A unilateral weighted shift W, defined on a Hilbert space

H, is a linear operator such that

where {en : n10] is an orthonormal basis of H and [wn : n30]
is a sequence of complex numbers. The shift W is bounded if

and only if the weight sequence [wn] is bounded, in fact

4

”W” =sup[|w :nZOI.

nl

There is no loss of generality in assuming wn_>_0, since W
is unitarily equivalent to a weighted shift with weights [Iwnli
[18], p. 52. We will thus consider only weighted shifts with

nonnegative weights. The adjoint of W is given by
t t
'W e0 = 0 and W en = wn_1en_1,

for all n > 0. If w_1 is defined to be zero, then

‘
W 8n = wn-len-l’

for all n_>_0.

Let 2 denote the identity function. Since Tzzn = 2n‘..1
for all n30, this is an example of a weighted shift with
wn = 1 for all n. The operator T2 is called the (unweighted)
unilateral shift. The function 2 is analytic, so T2 is
subnormal.

Let L: be the set of all analytic functions on D which
are square integrable with respect to area measure. This
Hilbert space is called the Bergman space. The Bergman shift
8 is the restriction of multiplication by 2 on L2(D,area) to

L3, so S is subnormal. Clearly S shifts the orthogonal basis

[2“

n30} and normalizing shows that S is a weighted shift
with weight sequence [[(n+1)/(n+2)]1/21.
An inner function u is an H00 function such that |u| = 1

almost everywhere on 3D. A basic reference for inner functions

is [13], Chapter Seventeen. For example, given a complex

number w in D, the function
u(z) = (z - w)/(1 - Wz)

is an inner function. This is called a Blaschke factor. It
can be shown that a product of Blaschke factors b, (a Blaschke

product) is an inner function if and only if
1 - < co,
§ ( lznl)

where [2“] is the set of zeros of b counted according to
multiplicity [13], p. 333. The order of an inner function
u, ord(u), is defined to be the number of zeros of u (counted
according to multiplicity) if u is a finite Blaschke product,
otherwise the order of u is infinite.

Given a collection of Hilbert spaces [Hi : j in I} for

some index set I, the direct sum of these spaces is

H=$lHj : j inll =52 x]. : x]. innj and Z lej||2<ool.
#1 jer

If Tj is in B(Hj) for each j, then the direct sum of these

operators, denoted T = MT]- : j in I], is defined on H by:

for all x = Z x]. in H. It is easy to see that T is in

B(H) if and only if HTJ- || _<_M, for all j. In this case

”Tn = supxn'rju = ,- in n.

6

For example, Tn is a direct sum of unilateral shifts.

2
For Oijin-I, let Hj = Cl spanlznk+j

k30], where C] span
indicates the closure of the linear span of the set. Then
H2 =19[Hj : 0_<_j_<_n-1] and Tzn shifts each Hj-basis. Since
Tzn leaves each Hj and its orthogonal complement invariant,
Tzn is a direct sum of unilateral shifts.

This example can be generalized. Note that zn is an
inner function and ord(zn) is n. For any inner function u,
the Toeplitz operator Tu is a direct sum of ord(u) unilateral
shifts [5], Theorem 1. In Chapter Three we show that this
is the only example of a direct sum of weighted shifts
unitarily equivalent to an analytic Toeplitz operator.

An operator T is hyponormal if T‘T - TI" is positive.
Every subnormal operator is hyponormal. Hyponormal weighted
shifts are easily characterized. A weighted shift is
hyponormal if and only if the weight sequence [wnl is
increasing. Lemma 1.1 gives a method for constructing
hyponormal weighted shifts which are not subnormal.

In 1970 Paul Halmos asked the question "Is every
subnormal Toeplitz operator either normal or analytic?" (See
[9], p. 906 and [10], p. 537.) The surprising answer is no,
as was first shown by Carl Cowen. Later the writer of this
paper gave another proof of Cowen's result (Theorem 1). This
will be published jointly with Cowen [6].

Abrahamse [1] considered Halmos' problem and found a

large (dense in L”) class of functions for which the answer

is yes. If f or f is a function of bounded type (a ratio

7

of two H°° functions) then Tf subnormal implies Tf is normal
or analytic. Abrahamse also asked if this were true of
Toeplitz operators unitarily equivalent to weighted shifts.
In particular, is the Bergman shift unitarily equivalent to
a Toeplitz operator? If so, then this would be an example
of a non-normal, non-analytic, subnormal Toeplitz operator.
(If S were unitarily equivalent to an analytic Toeplitz
operator Tu' an analysis of the spectrum of S would imply
that u was inner and all the weights would have to be one.)

But Sun Shunhua [17] proved that the Bergman shift is
not unitarily equivalent to a Toeplitz operator. Moreover he
showed that any weighted shift W with a strictly increasing
sequence of weights that is unitarily equivalent to a Toeplitz

operator Tb' must satisfy the following conditions.

i) There exists a constant a such that 0< |a| <1 and

b - ab is analytic.

ii) The weight sequence {wnl is given by
"n = (1 _ |a|2n+2)1/2.

Steven Power and this author independently showed that
a weighted shift satisfying condition ii) is subnormal (Lenma
1.2). Theorem 1 proves the existence of a function b
satisfying condition i) and shows that Tb is a weighted shift
such that ii) holds.

Theorem 2 is a special case of a result due to Carl
Cowen [5]. A direct sum of copies of the shift in Theorem
1 is unitarily equivalent to a Toeplitz operator. These

results are in Chapter Two.

In Chapter Three the following question is considered.
Can a Toeplitz operator be unitarily equivalent to a direct
sum of distinct hyponormal weighted shifts? In many cases,
the answer is no. The only hyponormal weighted shift unitarily
equivalent to a non-analytic Toeplitz operator is the one given
in Theorem 1. The only example of a finite direct sum of
hyponormal weighted shifts that is unitarily equivalent to
a Toeplitz operator is the example in Theorem 2. This result
extends to infinite direct sums of hyponormal weighted shifts,
but some additional hypotheses are required for the weight
sequences.

Before proving these results, we list some of the facts
that will be used in the proofs.

The kernel of an.operator T in B(H) will -be denoted
by ker T. A useful fact about the kernels of Toeplitz operators
is Coburn's proposition [7], p. 185, which states that for
any nonzero Toeplitz operator Tg, either ker T8 = [0], or
ker T2 = [0].

Consider a weighted shift W with nonnegative weights

[Wu] and basis [eni. Then

kerW = Cl spanien : wn = 0}.

Suppose W is unitarily equivalent to a Toeplitz operator.
Since e0 is in ker W', Coburn's Proposition implies that ker W
is trivial. Thus W is injective and all of the weights [wn]
are positive. This result also holds for direct sums of

weighted shifts unitarily equivalent to Toeplitz operators.

9

For all the weighted shifts under consideration we may assume
that all of the weights are positive.

The spectrum of an operator T in B(H) is
[w : (w-T) is invertible]

and will be denoted by sp(T). Let K(I-I) be the set of compact
operators in B(H). This set is a closed ideal in B(H) and
the quotient space B(H)/K(H) is called the Calkin algebra;
see [7]. Chapter Five. The essential spectrum of T, denoted
spe(T), is the spectrum of T as an element of the Calkin
algebra. If T is invertible in the Calkin algebra, then T
is called a Fredholm operator. Atkinson's Theorem states that
an operator T in B(H) is Fredholm if and only if the range
of T is closed, the dimension of ker T is finite, and the
dimension of ker T. is finite [7], p. 129. The index of a

Fredholm operator is
i(T) = dim ker T - dim ker T*

The index is a continuous mapping from the set of Fredholm

Operators to the integers which satisfies
i(ST + K) = i(S) + i(T).

where S and T are Fredholm operators and K is compact [7].
p. 138.

The spectrum of an operator T in B(H) can be divided
into two parts, the left spectrum and the right spectrum. The

left spectrum of T, denoted sp1(’l‘), is the set of points w

10
such that w-T is not left invertible; the right spectrum of

T, denoted spr(T), is the set of points w such that w-T is
not right invertible. It is not hard to show that the
complement of sp1(T) consists of those numbers w such that
w-T is one to one and has closed range. ‘This last condition
is equivalent to wa being bounded away from zero, that is

there exists a constant c > 0 such that
ll <w-T>x|l :cllxll.

for all x in.IL. An operator T is right invertible if and
only if T is onto.

The left and right essential spectra of an operator T
in B(H), denoted sple(T) and spre(T) respectively, are defined
to be the left and right spectra of T as an element of the
Calkin algebra. Left and right invertible elements in the
Calkin algebra are called left and right Fredholm operators
respectively. If an operator T does not have closed range,
then T is neither left nor right Fredholm [4] p. 40. ‘Note
that sple(T) is a subset of sp1(T) and spre(T) is a subset
of spr(T).

For a Toeplitz operator T , the essential range of g
is contained in both the left and right essential spectra of
Tb. First suppose that Tg is left invertible. Then Tg is
bounded away from zero. The proof of Proposition 7.6 in
Douglas' book [7] , shows that Tg bounded away from zero implies

that g is bounded away from zero. “Thus ess ran(g) lies in

sp1(Tk). If Tg is right invertible, apply the above reasoning

11

to T , which is left invertible. Thus ess ran(g) is also

contained in the right spectrum of T . Finally, suppose the

g
range of TE is closed. If T8 is also one to one, then 'I‘g
is left invertible and we are done. If T is not one to one,

8
then by Coburn's Proposition, TB is one to one, which implies

that T: is left invertible and hence g is bounded away from
zero. Thus the essential range. of g lies in the intersection
of 391e(Tg) and spre(Tg).

If f is continuous, the spectrum of Tf is easily
described [7], Theorem 7.26. The essential spectrum of Tf
is the range of the function f. For complex numbers w not
in the range of f, w-Tf is Fredholm with index equal to minus
the winding number of f about w. By Coburn's proposition,
w-Tf is invertible if its index is zero. Thus the spectrum
of Tf is the union of the range of f, together with those
components of the complement of the range where the winding
number of f about that component is not zero. For example, let
u be a finite Blaschke product. Then u- is continuous and sp(Tu)
is the unit disk. Furthermore, the essential spectrum of Tu
is the unit circle and i(w-Tu) = -ord(u), for all w in D.

This result extends to functions g in H" + C [7] . Theorem
7.36. For g in H00 + C, let g be the Poisson extension of
g to D. Suppose there exists a c>0 and r0<1, such that
for |z| = 1, we have |§(rz)| >c, for all r, r0<r<1. Define
the winding number of g about zero to be that of §(rz) for
any r>r0. Then T is Fredholm, and the index of T is the

g g
negative of the winding number of g about zero.

12

For any bounded measurable function f, the spectrum of
Tf is still related to the function f. As was noted earlier,
the essential range of f is contained in the essential spectrum
of Tf. The spectrum of Tf is contained in the closed convex
hull of the essential range of f. A deep theorem due to Widom
and Douglas states that both the spectrum and the essential
spectrum of a Toeplitz operator are connected. See [7],
Theorem 7.45 and Corollary 7.46.

Let W be an injective hyponormal weighted shift with
weights [wn] and ”W“ = 1. Then the spectrum of W is the
unit disk; the essential spectrum of W is the unit circle
[16], p. 77. Moreover i(z-W) = -1, for all z in D.

Suppose that W = $[Wj : lijSN]. where each Wj is an

injective hyponormal weighted shift and N is finite. Assume
||w|| =max1||wj|| : 1_<_j_<_N1 = 1.

In this case sp(W) is also the unit disk and spe(W) is the

union of the circles
:2 = lzl = uwjuz

for 1§j_<_N.
Now let W be an infinite direct sum of hyponormal

weighted shifts W for j 3 1. Let

it
lIWII = supzuwju = 1:11 = 1-

Then sp(W) = D, but the essential spectrum is much larger

than in the previous cases. Let lim sup! IIWJ- II} = a.

13

Proposition 1. The essential spectrum of W is the union of

the disk [2 : Izlia] and the circles [z : |z| = ”Will! for
531.
Proof: If |z| = ”will for some j, then 2 is in spe(Wj). which.

is contained in spe(W). If |z| <a, then dim ker (z-Wj) = -1
for infinitely many j's, thus 2 is in spe(W). Since essential
spectra are closed sets, [2 : Izl is} is contained in spe(W).

Suppose then that |z| >a and Izl :9 “Will for all j.
Write W = SI 0 82, where 82 is the (finite) direct sum of
[Wj : ”ij > Izl} and $1 is W 9 82. Since Izl >a, we have
”51” < |z| and thus (2-81) is invertible. Clearly, (2-82) is
Fredholm, and therefore z is not contained in the essential
spectrum of W.

Some additional algebraic properties of Toeplitz
operators will be needed in the proofs that follow. Let g

be in H” and let f be in Loo. Then for all h in H2,

Tngh = TfP(gh) = ngh = P(fgh) = ngh.

Thus Tng = ng whenever g is analytic. Taking the adjoint
of this equation shows that Tng = ng whenever f is conjugate

analytic. Together these results imply that
TquTu = Tf ,

for all f in L00 and for all inner functions u. In particular,

we have

14

'rz'rf'rz = T, .

This relation can be used to define Toeplitz operators. Given

any T in B(Hz), if
TzTTz = 'r,

then there exists an f in Lao such that T = Tf; see [3]. P.
95. Using the standard basis for H2, each T in B(H) has a
matrix representation T“ (ai,j)' where 3i,j = (sz,zi). For

-), we

a Toeplitz operator Tf with matrix representation (ai J

have

a.

. . . .+ .+
1,j = (szj.zl) = (TszTzzlml) = (szJ 1.z1 1) = ai+1,j+1°

Thus the matrix of a Toeplitz operator is constant on the
diagonals.
For g in Loo, the Hankel operator Hg from H2 to H2‘L is

defined by
th = (1—P)(gh).

where 1-P is the projection of L2 onto Hzl. Two general
references for Hankel operators are [12] and [15], Chapter

Nine. The adjoint of Hg maps H21 to H2 and is given by
It .-
th = P(gh).
where h is in HZ'L. The map g—-Hg is linear, that is

Hag + bh = ans ” th

15

for all g and h in Lm, and constants a and b. If g is analytic,
then Hg = 0. Furthermore, Nehari's Theorem [15], p. 100,

states that
III-lg” = dist(g,H°°).

The essential norm of Hg (its norm as an element of the Calkin

algebra) denoted ”Hglle- is
Illiglle = dist(g,H°°+C)

[15]. p. 101. In particular, if g is in H00 + C, then Hg
is compact.
The key connection between Hankel and Toeplitz operators

is given by the equation:

.
Tgf " '1'ng = [‘1ng
for all f and g in Loo.

Since H00 is weak-* closed in Lm, it is not hard to show

that for any f in Loo, there exists a g in H00 such that

Hf =Hg and ”Hf“ = “Hg” = ”8”,, = di8t(g.H)

Such a g is a best H00 approximation for the coset f 4- H00.
If Hg attains its norm on the unit ball of H2, then the best
approximation to HClo is unique and unimodular. More precisely,

if there exists an h in H2 such that

llhllz = 1 and IIthllz = lngll = Hall...

16

then g is the unique function of smallest norm in g + Hm.
Moreover, gh is in H21 and g is unimodular [15], p. 104. This
fact will be used in the proof of Theorem 3.

.Matrix representations of Hankel operators are similar
to the representations of Toeplitz operators; they are
constant on the cross diagonals. Let g be in L00 and suppose

Hg " (ai,j)' where ai,j = (ngj,z'1'i) for i,j:0. Then
ai'j = 2(-1-i-j)

[15] p. 100.
For H and K Hilbert spaces and an operator T mapping
H to K, the Hilbert-Schmidt norm of T, denoted ||T||2, is

defined to be
”Tllz = I g3 Il'Jl‘enll2 11/2.

where [en] is any orthonormal basis for H. This definition
is independent of the choice of basis; see [4], p. 9. It
is easy to show that

2
“Tllz = 2. [31,142.
‘1

where (ai,j) is any matrix representation of T. The Hilbert-
Schmidt norms of Hankel operators with conjugate analytic

symbol have an especially nice characterization.

Proposition 2. Let g be in H”. Then

 

”Hgllz = [(area g(D))/Tr 11/2,

17

where the area of g(D) is counted according to multiplicity.
(If g is an N to one mapping, then the area is counted N

times.)

an
Proof: Let g = Z cnzn. If Hg "' (ai j)' then
n-o '

“1.1: °i+i+1'

for i,j3_0. Thus

2 °° 2
lng||2= Z nlcnl -

11']

I/WfdA

g(D)

1”] ls'lsz

D

Now

(area g(D) )/1r

1 2n .
2 f(1/21T) f Ig'(re1t)|2dt rdr
0 0

I
n
2 f '2: nzlcnIZrzn"1 dr
0

at

1:4

1
w
2 Z) nzlcnl2 ] r2n-1 dr
0

‘° 2
Z nlcnl .

11-!

This proposition will be used in the proof of Theorem 1.

17

where the area of g(D) is counted according to multiplicity.
(If g is an N to one mapping, then the area is counted N

times.)
an

Proof: Let g = 2 cnzn. If Hg "' (ai,j)' then
use

ai,j = °i+j+1'

for i,j_>_0. Thus

2 °° 2
“Hg”2= 2 "Ion! ‘

1U]

llfijdA

g(D)

ll“)r lar'l2 dA

D

Now

(area g(D) )/7r

1 2n .
2 f (1/2“) I Ig'(re1t)|2dt rdr
0 0

n2lcn| 2 1,211.]. dl‘

u
N
3“”;
3E4:

u
N
M8
:3
N
o

1
nlz Jf r2n-1 dr
1|" 0

55 l I2
n C o
11-! n

This proposition will be used in the proof of Theorem 1.

CHAPTER TWO

The main result of this chapter is Theorem 1.

Theorem 1. For 0<a< 1, let v be the Riemann mapping of D
onto the interior of the ellipse E with vertices :1/(1+a) and
:i/(l-a), and let b = v+av. Then Tb is a subnormal weighted
shift which is neither normal nor analytic.

In Leanna 1.2 two proofs of subnormality for the weighted
shift in Theorem 1 are given. (Leanna 1.2 actually proves
subnormality for a larger class of weighted shifts.) In
general the problem of determining whether a hyponormal
operator is subnormal is quite difficult. One criterion
equivalent to subnormality is due to Bram and Halmos [4], p.
117. An operator S in B(H) is subnormal if and only if for

all x0,x1, . . . ,x inH, we have

n

n . .
(SIX-,Slx.))0.
13:10 3 1 -

When 8 is a weighted shift, the Bram-Halmos condition holds
whenever a certain infinite collection of matrices has
positive determinant. This is Lemma 1.1 and is the basis for

the first proof of Lenma 1.2.

18

 

 

19
Another way to determine whether a weighted shift is

subnormal is due to Berger, Gellar and Wallen; see [4] p.
159 and [8]. Aweighted shift Wwith weights [wn] is subnormal
if and only if there exists a probability measure v defined

on [0,1] , with 1 in the support of v, such that
1
0

for all n31. This is the basis of the second proof of Lemma

1.2.

Lenlna 1.1. Let W be a weighted shift on a Hilbert space H

 

with orthonormal basis [en] and positive weights [wn] for n_>_0.

For all i and k_>_0, define will by

[i] __
Wk “wk"kd-l - - “kn-1°

Let B(n,k) be the (n+1) by (n+1) matrix with (i,j)-th entry

wgli "£111; for 01i,j_<_n. If the determinant of B(n,k) is

positive for all nonnegative n and k, then W is subnormal.

Proof: It suffices to show that the Halmos-Bram condition
holds for any xo,x1, . . . ,xn lying in a dense subset of
H. Assume then that each xj is a finite linear combination

of basis elements.

[i]

Fix n30 and N30. Define e- and wj to be zero for

J
j < 0, and let wlgo] equal one. For 0:j_<_n, let

X'=a

J j,-nej-n + aj,-n-¢-lej-n+1 + ' ° ' 1' aj,Nej+N’

20

where each aj,k is a complex number. Note that Wiek = winehk

and (Wiek,Wjem) = win will] (ei+k'ej+m)'
Recall that a matrix B = (bid) is positive semidefinite

if and only if

1\
5.1) .a. >0
LED 1 is] J '-

for all complex scalars a0,a1, . . . ,an.

With this notation we have

n . ,
(4:0

N i j
Li-O tun g4 aJ'.l!lai,k(w ej+m:w 914*)

[i] [i]

= the 1.4» figs-s aI.mai.kw1+m"’i+k .(eii'it‘m’ehhk)

N ['1 I 1
2 Z ai,m“"j,mwj+1-mwilm

szO mm
=....;:. +
= 31...; aimwiirinwilnii aj,m

+ 12:: .%:51+m,-mWij+m] WEi+mlaj+m,-m-

Let B(n,k,m) be the matrix with (i,j)-th entry

[j+m] [i+m]
i+k wj+k '

21

for 0_<_i,j_<_n. The first term in the sum is nonnegative for
all ai,j if and only if the matrix B(n,m,0) ispositive
semidefinite, and the second term is nonnegative if and only
if B(n-m,0,m) is positive semidefinite.

In fact, for all n, k and m_>_0, B(n,k,m) is positive
definite. Any self-adjoint matrix is positive definite if all
its principal (upper left) submatrices have positive
determinant. Since the principal submatrices of B(n,k,m) are:
B(0,k,m), B(1,k,m), . . . ,B(n-1,k,m), it suffices to prove
that each B(n,k,m) has positive determinant. The (i,j)-th

entry of B(n,k,m) can be written:

[j+m]w[i+m] [m] [j] [m] [i]

wi+k j+k =wi+k wi+k+mwj+k wj4-k4»m

The first factor is constant in each row and the third factor
is constant in each column, and both are positive. Thus
B(n,k,m) has positive determinant precisely when
B(n,k+m,0) = B(n,k+m) also has positive determinant.

Lema 1.1 can be used to construct examples of hyponormal
operators which are not subnormal. The proof of the lema
shows that W is subnormal if and only if the matrix B(n,k)
is positive semidefinite for all n and k_>_0. But if the
determinant of B(n,k) is negative for any values of n and
k, then B(n,k) is not positive semidefinite. Hyponormality
alone implies that B(n,k) has nonnegative determinant for all
k_>_0, and n_<_ 2. For example, any hyponormal shift with initial

weights satisfying

22

"o = 1; *1 = 21”; w, = 2; w3 = (9/2)1/2; w, > (14/3)1/2

is not subnormal since B(3,0) has negative determinant.

Gellar and Wallen [8] prove a much stronger version of
Leanna 1.1. They show that W is subnormal if and only if the
matrices B(n,0) and B(n,1) are positive semidefinite for all

n30.

Lemma 1.2. Let W be a weighted shift with weights [wn]

 

satisfying
wn = (1 - sue)“2
for 0 < a < 1 and 0 < c < 1. ThenW is subnormal.

First Proof: Consider the matrices B(n,k) as defined in Lenlna

 

1.1. We will show by induction on n that each matrix has
positive determinant.

For n = 0, we have B(0,k) = 1.

For fixed n>0, let R = (ri,j) be the (n+1) by (n+1)

matrix with (i,j)-th entry ri,j defined by:

1.1.] =1for0§j§n, and
rj+1,j = ’wk+j for 0_<_j_<_n-1.
Set r- o = 0 elsewhere. Then RB(n,k) has first column

1,]
[1,0,0 . . .Olt, and foriandj>0 the (i,j)-th entry is

[5-1] [i-1l 2 2
wk+i Wk+j (Wk+i+j-1‘Wk+i-1)°

23

The formula for the wn implies that
2 2 _ ' k+i-1
(wk+i+j-1 - “kn-1)- (1-a3)a c.

Thus, for i and j > 0, the (i,j)-th entry of RB(n,k) is:

cak+i-1(1-aj)wl[(1;1] “£41311.

k+i-1

Now ca is constant for each row and (l-aj) is

constant for each column. Thus the determinant of B(n,k) is
[1-1] [i-l]

positive if the matrix with (i,j)-th entry wk” wk+j for

1:i,jin has positive determinant. But this is the matrix

B(n-1,k+1), and the proof follows by induction.

Second Proof: This proof is due to Steven Power. The use
of the q-binomial theorem was suggested by Richard Askey.

The q-binomial theorem [2]. p. 350, formula (3.3) is
‘° k
1 + k2: (r;q)kx /(q;q)k = (rx;q)m/(x;q)m,
-l

where (r;q)k = (1-_r)(l-rq) . . .(l-qu‘l) and (r;q)°°is the
infinite product. Let r = 0, x = anc and q = a. Let p0 = 1,
pi = (c;a)j for j> 0 and let pm be the corresponding infinite
product. This gives

1 4» kg! ckank/pk =[(1-anc)(1-an+lc) . . . 1‘1 = pn/pm.

Then

(wowl . . . wn__1)2

ll
'U
:3

24

1
= ft“ dv(t).
o

where v is the discrete measure on [0,1] with mass poo at 1
and mass pmck/pk at ak/z for k: 1. The conclusion now follows

from the Berger-Gellar-Wallen condition [4], p. 159.

Theorem 1. For 0<a< 1, let v be the Riemann mapping of D
onto the interior of the ellipse E with vertices _4_-_1/(l+a) and
_+_i/(1-a), and let b = v+av. Then Tb is a subnormal weighted

shift which is neither normal nor analytic.

Proof: We will show that Tb is a weighted shift with respect
to some orthonormal basis [en : n_>_0] with weights [wn] which
satisfy the hypothesis of Leuma 1.2. If Tb is to be a weighted

shift, then we must have
2
(1 - TBTb)en = (1 - wn)en.

Thus the basis which Tb shifts can be characterized as a basis
of eigenvectors for the compact operator (1 - TBTb)°

Since v is a Riemann mapping, v is continuous on 3D
and maps 8D onto 3E. The map z—~z + a2 sends the boundary
of E back onto the boundary of D and preserves the winding
number of curves in E about zero. Thus b is continuous and
unimodular. Furthermore, ker T5 has dimension one.

Let K = 1 - TBTb' Then K = HgHb, which is compact.

Note that b - ab = (1-a2)v, which is an analytic function.

This gives

25

- * _ 2* - 2
K -anb- a H5115 - a (1 - TbTB).

Then
KTb = a2(1 - TbTB)Tb
= asz(1 - TBTb)
= aszK.
Let e0 be in ker T5 with Ileoll = 1. For n30, define
en+1 = Then, "Tben" ,
and let wn = ”Then I. Then Then = wnen+1 for all n30 and

 

each en is of norm one.
The next step is to show by induction on n, that each
en is an eigenvector for K with corresponding eigenvalue a2n+2.

For n = 0, we have

Keo = a2(l - Tb'l‘5)e0 = azeo.

= a2k+2e

Assume that Kek k' This gives

Kek+1 = KTbek/Wk

aszKeklwk

= a2k+4Tbeklwk

2k+4
a ek+1 ,

which is the claim for n = k+1.
Thus [a2n+2 : n30] are eigenvalues of K, and since K

is self-adjoint, is n30] is an orthonormal set. This set

n

will be a basis for H2 if each a2n+2 has multiplicity one

26

and there are no other eigenvalues.

The trace of Hgl-Ib is the sum of its eigenvalues [4],

p. 16. Thus

‘
Z a2n+2£ tr(H;Hb)

mo
2
= ”Hb [I 2.
where II ”2 is the Hilbert-Schmidt norm. Now
2 2
”Kb” 2 = ”Hv + 83v” 2
2
= azllflv“ 2

a2[area v(D)] / Tl'

a2/(1 - a2).

‘
But “20 a2n+2 = a2/(1 - a2), so [azm’2 : n_>_0] is a complete

set of non-zero eigenvalues and each has multiplicity one.
Now ker K = ker HgHb = ker Hb = qu, where u is an
inner function or u = 0. Since KTD = aszK, if f is in ker K,

then be is also in ker K. Apply these observations to u:

since u is in ker K, Tbu = bu - Hbu = bu is in ker K, and
thus bu = ug for some g in H2. But b is not in H2, so u
must be zero. Thus ker K = [0] and 0 is not an eigenvalue

of K. Therefore [en : n30] is a basis for H2 and Tb is a

weighted shift.

For each n: 0,

w: = ”when“2
= (TBTbensen)
= (en - Ken,en)

27

1 - (Ken, en)

Thus wn = (1 - 82n+2)1/2 and this is the formula for the weights
in Lemma 1.2 (with a and c replaced by a2). Then Tb is
subnormal, b is clearly not analytic, and the range of b is
not contained in a line segment so Tb is not normal. This
proves the theorem.

Unfortunately, this result does not give a general
description of subnormal Toeplitz operators. If Tb is a
subnormal Toeplitz operator, what can be said about the symbol
b? In the above example, b is continuous, but this is not

a necessary condition. This follows from Carl Cowen's result

[5];

Theorem Suppose f is in L00 and u is an inner function of
order n (where n is a positive integer or °°). Then Tfou is
a direct sum of copies of Tf. The number of copies is the
order of 11.

Let u be an inner function of infinite order and let
b be as in Theorem 1. Then Tbou is a direct sum of subnormal
operators, hence Thou is subnormal, but b-u is not continuous,

or even in H°° + C. (An easy way to see this is to note that

.
K = 1 ' TbouTbou = Hboquou

is not compact if the order of u is infinite.)

28

In view of this result, perhaps the right question to
consider is, "What are the irreducible subnormal Toeplitz
operators?" Do the symbols of such operators lie in H00 + C?
Are the symbols either in H00 or in C?

Theorem 2 is a direct proof of the above theorem for
the weighted shift in Theorem 1. The proof follows the outline

of Theorem 1 .

Theorem 2. Let g = v + a9, where v is as in Theorem 1, and
let u be an inner function. Then Tgou is unitarily equivalent
to the direct sum of copies of T . The number of direct

g
summands is the order of 11.

Proof: Let b = gou, and let N = ord(u). We will show that

H2 has a basis [e : n30, 1_<_j§N] such that for each fixed

n,j

j, we have Tb°n,j = wnen+1,j' where wn = (1 - a2n+2)1/2. Then

Tb is a direct sum of weighted shifts of the type in Theorem
1.

As in Theorem 1, b is unimodular and

b - ab- = (1 - a2)vou,

which is analytic. Let K = 1 - TBTb' Then
x = 11311,, = 323335 and KTb = aZTbK

as before.
The first step is to show that the dimension of ker T5
equals N. If N is finite, then u is continuous which implies

that b is continuous. Then dim ker T5 = winding number of

29

b about zero, which equals the winding number you about zero,
and this is N since v is a Riemann mapping. Conversely, suppose
dim ker T5 is finite. Now ”K” = a<1, so 1 - K = TBTb is
invertible. Thus T5 is right invertible and must be Fredholm.

Let i(TlB) = n and let f = vou. We have
"1 - (1-a2)'b'f||°° = Ilb - (l-a2)f||°° = ausum = a < 1.

Thus TBTf is invertible and Tf is Fredholm. The index of Tf
is -n. Now f is in H”, so the winding number of f about zero
is 11. Since f = v-u the winding number of u about zero is
also n. Therefore dim ker TB is finite if and only if N is
finite and in this case, they are equal.

Let [and : 1£j_<_N] be an orthonormal basis-for ker T5.
For n_>_0, define °n+1,j = Then,jlwn,j’ where “n,j = IITben’jll.

Then The“.j = "n,jen+1,j' and each end has norm one.

Claim 1: Each en,j is an eigenvector for K with corresponding

eigenvalue a2!”2 .

The proof is by induction on n. For n = 0,
- - - 2
Keo’j -82(1 TbTB)80,j -a eo’j.
Assume that Ke - = a2k+2e . This gives
ks] k’j

Kek+1’j = K'rbek’j/Wk’j

2
a TbKek,j/wk,j

2k+
a 4Tbek,j/wk,j

82k+4ek+1’j9

30

which is the claim for n = k+1.

Claim 2: For all n30 and 131, we have “n,j = (1-32n+2)1/2.
Let wn = “n,j for all n30. Then Them-Li = "nemj'

By Claim 1,
(l‘TBTb)en’j = a2n+2en

Thus

_ +
(1 a2“ 2)en,j TBTben’j

= wn, jTBen+1 , j '
Therefore,

[(1-a2n+2)/wn’ jlen,j = Then+1, j .

Then

[(1-a2n+2)/wn,-]

n,j'

and the claim follows. Define w_1 to be zero, so that

Themj = wn-l‘:"n-1,j

for all n30.

Claim 3: The set is :n_>_0, lijiN] is orthonormal.

n,j
In view of Claim 1, and since K is self-adjoint, we have
em] .Lem,k whenever n 1: m. It must be shown that en,j lemk

whenever j at: k. This is true by definition when n = 0. For

31

any nonnegative n,
wn(en+1,k s en+1’j ) = (Then’k a en+1’j)
= (911,1. ' TB°n+1,j)

= wn(en,k , en,j)

and the claim follows by induction on n.

The last step is to show that ”n,j : n30, 1_<_j_gN]
is a basis for H2. If N<°°, then the method used in the proof
of Theorem 1 works here. The operator K is compact and

as
2
N 2 a2n+2 :azlle.ullz

THO

= a2(area vou(D) ) / Tl

= Na2/(1 - a2).

(Recall that the area of vou(D) is counted according to
multiplicity.) Thus {82114-2 : n_>_0] is a complete set of

eigenvalues and each has multiplicity N.

Note that zero is not an eigenvalue for K. The proof

is the same as in Theorem 1.

Claim 4: Let M be the orthogonal complement of the set

”n,j :n_>_0, 1_<_j£N]. ThenTb(M) =M.

Let h be an element of M. Then for all n_>_0,
(Tbh , end) = (h , TBen’j)
= Wn_1(h , en_1’j)

=0.

32
Thus Tbh is also in M. If h is in M, then h _|_ ker T5,

which implies that h is in the range of Tb. (The range is
closed, since Tb is left invertible.) Let h = be. Then for

all 1130,

(f, Tfien+1’j)/Wn

(bes en+1’j)/Wn

0.

Thus f is also in M and therefore Tb(M) = M.

Claim 5: M=[0].
Given an h in M, there exists a sequence [fn] in H2

such that h = Tgfn. Then for all positive 11, it follows from
KTb = aszK that

n
Kh KTb fn

2n n
a Tb Kfn.

Taking the norm of this equation yields ||Kh||ia2n+2||fn

 

 

If there exists a constant C, such that Ilfnllic, for all
n_>_1, then Kh = 0 which implies h = 0. Since Tb is left
invertible, it suffices to show that IISnII _<_C, where S is a
left inverse for Tb'

Let s = (TBTb)'1TB. Now

0‘ .
(TB'rbYl = (1-K)‘1 = 23 K].
1'0

33
av .
Let Ln = Z azan], for n_>_0. Note that
i=0

co . ‘° .
||Ln||_<_ 2: .2n1||x||1_<_ :2 am”)! = 1/(1-a2nt2).
18° 180

Since KTb = aszK, we have TBK = azKTB and hence

2 azanBKj

TEL j-O

n

w

2: a'(21'1'1'2)].Kjv1\5
r°

.. 2 ... - 2
Thus S - LOTB and S - L0(TBL0)TB - LOLITB'

For any n, this implies that

n- n
S - L0 0 o o [In-1TB.

Then
n
IISnII = llLo - . - Ln-1T5||
§.||Lo||- - -||Ln-1H
i 1/[(1'82) o o o (1-82n)]

1/[(1-a2)(1-a4)(1-a6) .. .1

IA

which is a convergent infinite product since a < 1.

Therefore, M = {0] and [e :n_>_0, l_<_j_<_N] is a basis

n,j

for H2. Thus Tb is a direct sum of weighted shifts.

CHAPTER THREE

Theorem 2 gives examples of direct sums of subnormal
weighted shifts which are unitarily equivalent to Toeplitz
operators. Are there any hyponormal weighted shifts or direct
sums of such which are unitarily equivalent to Toeplitz
operators? In this chapter we prove that any Toeplitz operator
unitarily equivalent to a finite direct sum of hyponormal
weighted shifts is either an analytic Toeplitz operator or
one of the form in Theorem 2. We also obtain some similar
conclusions for infinite direct sums. These results are based

on Theorem 3 .

Theorem 3. Let [Wj] be a finite or countably infinite
collection of hyponormal weighted shifts; each W]. has weights
{wn,j} and orthonormal basis [en’j] for n30 and lijiN. Let
supiIIWjII: lijiN] = 1, and assume that not all the weights

are one. Suppose there exists b in Lao such that Tb is unitarily

equivalent to the direct sum of [Wj : l_<_j_<_N]. Assume that:
i) b is unimodular, and

ii) inflwn :1_<_j_<_N] =w0’1.

.1
Then b = c(vou + a'v'ou), where c is a constant of modulus one,
u is an inner function, and v is the Riemann mapping of Theorem

1. Thus Tb is of the form in Theorem 2.

34

35
Corollary 3.1. Let Tb be unitarily equivalent to a finite

direct sum of the shifts [Wj]. Then the conclusion of Theorem

3 holds.

Corollary 3.2. Let Tb be unitarily equivalent to an infinite

direct sum of the shifts in Theorem 3. For 1130, define

cn=inflwn :j=1,2,3...].

.1

Then the conclusion of Theorem 3 holds whenever w0,1 = on
and lim cn = 1.

The hypothesis in Theorem 3 that not all the weights
are one eliminates the analytic case. Direct sums of weighted
shifts are analytic Toeplitz operators if and only if all the
weights are equal. This is Theorem 4. The other assumptions
i) and ii) actually concern the infinite case, since these
conditions are always satisfied by finite direct sums.
Corollary 3.2 shows that the extra assumptions needed in the
infinite case can be restated in terms of the of the given
weight sequences. The constant c in the conclusion of Theorem

3 corresponds to a rotation of the ellipse E given in Theorem

1.

Proof of Corollary 3.1 Since Tb is a finite direct sum of
the shifts [Wj], the essential spectrum of Tb is a union of
the circles [z : |zI == ”WjH]. But the essential spectrum
of a Toeplitz operator is connected, thus each of the shifts
W- has norm one and spe(Tb) = 3D. ‘The essential range of b

1
lies in spe(Tb), so b is unimodular.

36
Relabel the shifts (if necessary) so that wo liw0,j

for all j, then ii) holds.

If Tb is an infinite direct sum of hyponormal weighted
shifts, then the technique used in Corollary 3.2 fails to show
that b is unimodular. The connectedness of the spe(Tb) only
implies that the lim supIIWjII is one and thus the essential
spectrum of Tb is the entire unit disk. But the essential
range of b is also contained in the left essential spectrum
of Tb' Since Tb is one to one, its left essential spectrum
is equal to its left spectrum. The conditions on the weight
sequences in Corollary 3.2 imply that the left spectrum of

Tb is contained in the unit circle.

Proof of Corollary 3.2. It must be shown that b is a
unimodular function.
Since Tb is a direct sum of weighted shifts, for any

vector f,

||T§f||

Iv

inIIWO,jwl’j o s swk-1’j:j31}”f||

Iv

 

(0001 o o .ck_1)IIf

and lim c = 1, thus

Now0<w01=c0§c1..._<_c n

n

(0001 o o OCR-1)].lk

also converges to 1. Given w in D, choose k so large that

(00 o o OCR-1) - IWlk> 00

37
Then

Hui: -w*‘>fII _>_ ||T§fll - lekllfll

Iv

Thus T5 - wk is bounded away from zero, and this together

with the equation
TL} - wk =(Tbk'1 + wrbk-Z . . .+ wkwrb - w)

implies that (Tb - w) is bounded away from zero. Thus w is
not contained in spl(Tb). Since w was an arbitrary point in
D, the left spectrum of Tb lies in 3D. Therefore b is
unimodular and the hypotheses of Theorem 3 are satisfied.
Before proving Theorem 3, we will eliminate the case
where b is analytic. If Tb is analytic and a direct sum of

weighted shifts, then these shifts are easily characterized.

Theorem 4. Suppose Tb is unitarily equivalent to a direct

 

sum of hyponormal weighted shifts, Wj, with basis ”n,j! and
weights iwn,j3° Let IITbII = 1. Then the following are

equivalent:
1) b is an inner function.
2) Tb is analytic.

3) For allnandj, =1.

Wn’j
4) Tb is hyponormal and there is a weight w ,- equal

to one.

38
1) implies 2): Obvious.

 

2) implies 3): Let f = |b|2. Since b is analytic, we have

TBTb = Tf and Tfen,j = '31,j°n,j for all n and 1. Thus the

Toeplitz operator with symbol f - “31.1 has nontrivial kernel.

But this operator is self-adjoint; by Coburn's proposition
2

f - "n,j = 0, for all n and j. Since IIbIIm = 1, all the

weights must be equal to one.

3) implies 4): Clear.

 

4) implies 1): Suppose wmj = 1. Then

 

1 = ”Them,- II = ”Hump” : llben,j|| g Hem-n = 1.

Since the last inequality is actually an equality, b is
unimodular.

Now

ker Hb [h : bh is in H2]

= {h : Tbh=bhl

[

3‘

: IITthI = IIhII] (since b is unimodular)

°w -=1].

Cl spanien’j . n’]

This last set is invariant for Tb since "n,jiwn+1,j for all
n_>_0 and j_>_1. Hypothesis 4) implies that ker Hb is non-
trivial, hence ker Hb = qu, where u is an inner function.

Nowu is in ker Hb, so

Tbu=bu -Hbu =bu

39

is also in ker Hb' Thus bu = uh for some h in H2 and b is
analytic as well as unimodular. Therefore 1) holds.

To prove Theorem 3, two lenmas are needed. These lemas
make use of the H2 inner - outer factorization, :A function
g in H2 is called an outer function if the set gH0° is dense
in H2. .An outer function g is determined (up to a constant)
by its modulus. 'Given two outer functions f and g such that
[fl = Igl, we have f = cg for some constant c of modulus
one. If h is in H2, then h can be written as h = ug, where
u is an inner function and g is an outer function [7], pp.
158-159. Another fact about H2 used in Lemma 3.1 is that no
nonzero H2 function can be zero on a set of positive measure
[7], p. 154. In the proofs that follow, let Hg denote the

space 2H2.

Lemma 3.1. Let f be in L”, with f t 0, and let h be in H2.
Suppose ker Tf # [0] and.M is a nontrivial subspace of ker Tf.

Then:

1) For all inner functions u, if uh is in ker Tf, then
h is in ker Tf. In particular, ker Tf contains an

outer function.

2) If M is invariant for T2, then f is conjugate

analytic.

3) If u is an inner function and M is invariant under

multiplication by u, then u is constant.

40
Proof of 1. Let g = uh. Then

Proof of 2. If h is in ker Tf, then P(fh) = 0 and fh is
in HZ‘L. Thus Th- is in H3, which means that fh and th are
conjugate analytic.

Choose h in M of the form

.. k k+l
h-z +°k+lz +...

for some 1:30. Apply T2 k times to h to obtain a function

in M of the form
1 + °k+lz + ck+222 + . . .
Thus we can assume that M contains an h such that
h = 1 + zg,

where g is in H2. Then Tzh = g is also in M. Since M is
contained in ker Tf' both fh and zfg are conjugate analytic.

Therefore, f = fh - zfg is conjugate analytic.

Proof of 3. Let h be in M with h t 0. Then 73 is in H3
and for all n30, f'huu is in H3. Let Th = ungn, where gn
is in H2. Divide out the cannon outer factor of Th- and gn

to obtain v0 = unv where v is the inner factor of gn for

n' n

all 1130. Then for any w such that IwI < 1, we have

Ivo(w)I i Iu(w)|n.

41

If u is nonconstant, then by the maximum modulus principle,
|u(w)| <1, for all w in D. This implies that v0 = 0. Then
fh = 0, but h cannot be zero on a set of positive measure,
so f must be zero almost everywhere - contradiction. Therefore

11 is a constant.

Lemma 3.2. Let f and g be unimodular. If [0] #- ker T and

 

8
ker T8 is contained in ker Tf, then there exists an inner
function u such that g = uf. Moreover multiplication by 11

maps ker Tg into ker Tf .

Proof: Define Afh = fzh for all h in L2. If h is in ker Tf,

then Th is in H2 and so Afh is in H2. Hence
TfAfh = P(ff'z'h) = pc‘zTi) = 0.

Therefore, Af maps ker Tf into itself. Since f is unimodular,

A: is the identity map on ker Tf and

IAth = Ifth = IhI.

This last equation implies that the outer factor of Afh is
the outer factor of h. The same properties hold for the

analogously de fined Ag.

By Lemma 3.1.1 there is an outer function h in ker T8°

Then Agh = vh for some inner function v. Thus Agvh = h. Since

h and vh are also in ker Tf, we have Afvh = uh for some inner

function u. Then

42

uA vh = uh = Afvh
or
ugzvh = fzv .

Therefore g = uf. The map Ang sends ker Tg into ker Tf,

and for all h in ker Tg,

AfA h = Af(gzh) = ‘fgh = uh.

8

Thus multiplication by u maps ker T

g into ker Tf .

Theorem 3. Let [Wj] be a finite or countably infinite
collection of hyponormal weighted shifts; each Wj- has weights
[wmj] and orthonormal basis [emj] for n_>_0 and l_<_j_gN. Let
supIIIWj II 1£j_<_N] = 1, and assume that not all the weights
are one. Suppose there exists b in Lao such that Tb is unitarily

equivalent to the direct sum of [Wj : l_<-j_<_N]. Assume that:
i) b is unimodular, and

ii) inftwn,j:1_<_j_<_N} =WO’10
Then b = c(v-u + a-v'ou), where c is a constant of modulus one,
u is an inner function, and v is the Riemann mapping of Theorem

1. Thus Tb is of the form in Theorem 2.

Proof: The key idea in the proof is as follows. Recall from

*
Theorem 2 that Hbe = 82143115 and ”HE” = 1. This means that
ab is the unique function of smallest norm in the coset of

b + H”. The technique here is to start with the function

43

of smallest norm in b + H” and argue that this is indeed ab.
Assumption ii) guarantees that this function is unimodular.
Since Tb is hyponormal, the equivalence of conditions
3) and 4) in Theorem 4 shows that no weight is equal to one.
From the proof of 4) implies l) in Theorem 4, we can conclude
that ker Hb = [0].
By Coburn's proposition and the fact that ker TB is

nontrivial, ker Tb = [0] and thus "0,1 > 0. Let

a 2’ ,1/2

j =(1’Wn j

n.

and let a = a0 1. Since b is unimodular,
l-TBTb = Tbg - 'rfi'rb = Hgnb .

Then Hbeen,j = “121.jen,j: The norm of Hb is equal- to a and
is attained by 90,1“ Now, there exists a g in Lao such that
b-ag is analytic and ”gum: 1. Thus Hb = aHg and g must
be unimodular since IIHg°o,1II2 = l = IIgIIm. , We will show
that b = czg, where c is a constant of modulus one, from which

the conclusion will follow. Now

ker Tg ker Tng

ker (1 - HEHg)

ker (a2 - Hgnb)

Cl spanien’j :wmj =w0'1l.

This last set contains e0 1 and is invariant for TB since

wnd-ierLj.

44
Claim: The kernel of T9 is contained in ker T5.

If Tb is a finite direct sum of subnormal shifts, then
this claim is easily proven. For a subnormal weighted shift,
if the first two weights are equal, then all the weights are
equal [18], Theorem 6. This cannot occur here since no weight
equals one. Thus "n,j <w1’j for all j, and the claim holds.

Let M = T5(ker TE). By the preceeding argument, M is
contained in ker Tg. We will show that T2(M) is contained
in M, and then by Lenlna 3.1.2, M = [0] or g is analytic.
Let h be in M. Then there exists an f1 in ker T , such that
TBfl = h. Let e be the outer factor of 90,1’ By Lemma 3.1.1,
e is contained in ker T5 and in ker TE' Let zf = fl - de,
where d is a constant chosen so that f1 - de is in H3. Since

e is in ker TB we have
TBzf =T5f1 = h.
Now zf is in ker Tg which implies that f is in ker Tg. Thus
Tzh = TzTBTzf = TBf,

which is in M. Therefore T2 (M) is contained in M.

If g is analytic, then g(b-ag) = bg-a is also analytic,
which implies that g is in ker Hb. But ker Hb = [0], therefore
M = [0] and the claim is proven.

Now apply Lemma 3.2 to E and b obtaining an inner
function u such that g = Bu and multiplication by 11 sends
ker Tg into ker TB' Suppose u is not constant. Then by Lenlna

3.1.3, there exists an e in ker T? such that ue is contained

45
in ker T5 but not in ker TE' Let f = ue and h = Tgf 1: 0.

Then
Tub = TuTgf = TuTEue = Tge = 0.
Thus Eh _I_ H2. Now

- - _ a - _ . 2
Tgh-Tngf-f HgHgf—f Hbef/a,

which is contained in ker T5. Let

bEh = gh =Tgh +‘z'1'E,

where k is in H2. Multiplying by 3 gives

Eh - b(Tgh) = bzk.

Since Tgh is in ker T5, we have 3(T8h) _I_ H2. Then D7]? is
orthogonal to H2, and hence bzk lies in H3. Now bk is in
H2 which implies that k is in ker Hb. Thus k = 0. But then
bfih = Tgh which is in H2, implying that bh is in H2 and h
is in ker Hb. Thus h = 0 - contradiction. Hence u = b-g' is
constant.

Let bi: c2, where Ic| = 1. Then
b-aE=b-aczb=c(3b-acb)

is analytic. Let (1-a2)f = 3b - acb, so f is analytic. Solving
for 3b yields (f + a?) = 3b. Since b is unimodular, f + a?
is also unimodular. Writing 3b as the composition of f and
the map 2 —- z + a; shows that the essential range of f

(as a function on the unit circle) lies in the boundary of

46
the ellipse E with vertices _+_-1/(1+a) and :i/(l-a). As a

function on D, f + a? is the Poisson extension of 3b, hence
|f(w) + am“ <1 for all w in D. Thus f(D) is contained
in B. Let v be the Riemann mapping of D onto E. Now v'1
is a conformal mapping of E onto D which extends to a
homeomorphism of E onto 5. Let u = v'lof. The function u
is analytic on D and since v"1 is continuous on 3E, 11 is

unimodular. Thus
b = c(vou + aVou)

as required.
The following result is a weakening of the hypotheses

of Corollary 3. 2.

Corollary 3.3. Let Tb be unitarily equivalent to an infinite
direct sum of the shifts in Theorem 3. Let c0 be defined

as in Corollary 3.2. For n30, define

i-m
Then the conclusion of Theorem 3 holds whenever w0,1 = c0,
lim dn = 1 and lle II = 1, for all j_>_1.

Proof: As in Corollary 3.2, we will show that the left
spectrum of Tb is the unit circle. Choose a sequence of

positive numbers [an] such that lim 3n = 0 and 3n<dn for

all n_>_0. Let r =d -s

n n n' Then for each fixed n, there

are only finitely many j's such that w < r . Now r0 > 0 and

n,j n

lim rn = 1, so

47

1k
(r0 0 o o rk_1) /

also converges to one. Fix w in D. Choose k such that
(to o o o rk_1) - IWIk > 00
Write Tb = V1 9 V2, where

V1 = OIWj :wn’ Zrn whenever 0_<_n_<_k-1]

J'
and V2 = Tb 6 V1. Then the proof of Corollary 3.2 applies
to V1 and thus w-Vl is bounded away from zero. The second
sunrnand is a finite direct sum of hyponormal weighted shifts,
each of which has norm one, hence w-V2 is also bounded away

from zero. Therefore, w-Tb is left invertible, and sp1(Tb)

is contained in the unit circle.

L I ST OF REFERENCES

[1]

[2]

[3]

[4]

[5]

[6]

[7]

[8]

[9]

[10]

[ll]

[12]

[13]

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J. B. Conway, Subnormal Operators, Pitman, Boston,
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C. C. Cowen, Equivalence of Toeplitz operators, J.
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R. G. Douglas, Banach Algebra Techniques in Operator
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R. Gellar and L. J. Wallen, Subnormal weighted shifts
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P. R. Halmos, Ten problems in Hilbert space, Bull. Amer.
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, Function theory on the unit circle, Lecture
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A. L. Shields, Weighted shift operators and analytic
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