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COMMUTANTS OF TOEPLITZ OPERATORS ON THE BERGMAN SPACE
presented by
V V V ,
Zeljko Cuckovic

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l 1
Date February 8, 99

 

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MSU Is An Affirmative Action/Equal Opportunity Institution
c:\citc\duedue.pm3-o.[

COMMUTANT S OF TOEPLITZ OPERATORS ON THE BERGMAN SPACE

By

Zeljko Outkovié

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
DOCTOR OF PHILOSOPHY

Department of Mathematics

1991

 

."x
,I

Lji.

ABSTRACT
COMMUTANTS OF TOEPLITZ OPERATORS ON THE BERGMAN SPACE
By
Zeljko éuzskovié

In the first chapter we show that on the Bergman space L2, two Toeplitz operators

with harmonic symbols commute only in the obvious cases. The main tool is a
characterization of harmonic functions by a conformally invariant mean value property.

The next chapter describes the commutants of certain analytic Toeplitz operators.
To underline the difference between the Bergman and Hardy spaces, we first prove that on

the Bergman space the only isometric Toeplitz operators with harmonic symbols are scalar
multiples of the identity. If "(denotes the closed subalgebra of 13(ch1) generated by all

Toeplitz operators, we show that for each positive integer n, [Tzn}'n “I is the set of all
analytic Toeplitz operators. Here [Tzn}' denotes the commutant of Tzn. Ifn > 1, then
[Tzn}' is strictly larger than the set of all analytic Toeplitz operators, but we show that

{T

zn+az

}' is precisely the set of all analytic Toeplitz operators if a at 0. Using the same

2 + + anz", where at 2 O for i = 2, ..., n,

technique, we prove that if p(z) = z + azz
and ifp(z) -p(1) has n distinct zeros, then {Tp}' = {TV : w e H“). These results are
valid for both the Bergman and Hardy space.

The dissertation concludes with descriptions of the von Neumann algebras
generated by operators Tzn on the Bergman and Hardy space. For fixed n > 1, these von
Neumann algebras have a different structure on the Bergman space than on the Hardy
space. This problem is related to the problem of finding commutants of T In and its adjoint,

and we use some results proved in the second chapter.

"- I‘Es

To my parents

ACKNOWLEDGEMENTS

I would like to thank my advisor, Professor Sheldon Axler, for all his help,

encouragement and advice. His knowledge and enthusiasm were invaluable.

iv

Introduction

Chapter 1

Chapter 2
Chapter 3

TABLE OF CONTENTS

Commuting Toeplitz operators with harmonic symbols
1. The invariant mean value property .

2. Toeplitz operators .

The commutants of certain Toeplitz operators

Von Neumann algebras generated by Tzn
1. Tzn acts on the Bergman space

2. T Zn acts on the Hardy space

Bibliography .

11
11
18
28

54
54

62

69

INTRODUCTION

The purpose of this introductory chapter is to supply basic definitions, establish
most of the notation, and list the major results of this thesis. The main goal of our study is
to investigate some commutativity properties of Toeplitz operators acting on the Bergman
space. Since this necessarily leads to a comparison with classical Toeplitz operators on the
Hardy space, we will deal with both spaces throughout the thesis. Much more is known
about the Hardy space, but techniques used there do not apply to the Bergman space.
Therefore we had to find new approaches and techniques to obtain most of our results.

We start with definitions of these function spaces. Let D = { z e C : Izl < 1 }. Let
dA denote the usual area measure on D. The complex space L2(D,dA) is a Hilbert space

with the inner product

<fl g> = ijr em.

The Bergman space L3 is the set of those functions in L2(D,dA) that are analytic on D (the
a in L3 stands for “analytic”). The Bergman space L3, is a closed subspace of L2(D,dA),
and so there is an orthogonal projection P from L2(D,dA) onto Lg. Fix w e D. Point
evaluation f -—)f(w) is a bounded linear functional on L3, (see J. B. Conway [11], Chapter

III, Corollary 10.3) so there exists a function kw e L?! such that

f(w) = <fl kw>

for all f 6 L2. This function is called the reproducing kernel for the point w. In

[5], Proposition 1.7 it is shown that

2

kw(z) = 11: “1(1 - Tvz) '2.
Using this, we get an explicit formula for Pf:

Pf(W) = <Pfi kw> = <f. kw> = 1! ’IJDflZ) (1 - w7)‘2dA(Z) (1)

For (p e L°°(D,dA), the Toeplitz operator with symbol tp, denoted Tqy is the operator
from L3, to L3, defined by w = P((pf).

Let m be normalized Lebesgue measure on the circle 8D, i.e., dm = 52%. The

complex space L2(8D) is a Hilbert space with the inner product

- d:
<, >= —.
fg Iang 2,,

For each integer n, let en denote the function eu(z) = z" for Izl = 1. Then {en} is an
orthonormal basis for L2 (8D) (see, for example [25], Theorem 4.25) and the Hardy space
H2(8D) is, by definition, the subspace span {en : n 2 0}“. For (p e L°°(6D), the Toeplitz
operator with symbol (p, denoted again by Ti” on H2(BD) is defined by TJ = Q(tpf),
where Q is the orthogonal projection from L2 (80) onto H260). It will be clear from the
context which Toeplitz operator we consider.

Although these two Toeplitz operators on different spaces differ in many ways,

they do have the same basic algebraic properties: For (pay 6 L°°(D) or L°°(BD),
7'0!va = “TWBTV

mp)" = r,

3

TvT<p=Tw if‘lf E H”,

where H°° denotes the set of all bounded analytic functions on D, or the set of all their
boundary values. Standard references for the classical Toeplitz operators are [16] and [18].

The question of when two Toeplitz operators on H2(3D) commute was answered
by Arlen Brown and Paul Halmos ([9], Theorem 9), where they proved the following

result:

Theorem 1. Suppose that (p and \l’ belong to L°°(8D). Then
TvTv = TvTv
if and only if
(2) q) andw are in H”,
or (3) OandtTlare in H”,
or (4) there exist constants a, b e C, not both 0, such that atp + but is constant an

OD.

Brown and Halmos proved their result by examining the matrix (with respect to the
orthonormal basis {en}) of products of Toeplitz operators. It is well known that Toeplitz
Operators on H2(3D) have matrices that are constant on diagonals. It is natural to ask if

Theorem 1 holds for the Bergman space Toeplitz operators. On L2 , Toeplitz operators do

n+1

 

not have nice matrices in the standard orthonormal basis e,,(z) = z", for Izl < 1
and n e N, and the techniques used by Brown and Halmos do not seem to work in this
context. In Chapter 1 we prove the analogous theorem, using function theory, rather than

matrix manipulations, assuming that symbols are harmonic functions. By a harmonic

4
function we mean a complex-valued function on D whose Laplacian is identically 0. Here

is our theorem:

Theorem 2. Suppose that (p and \V are bounded harmonic functions on D.

Then

if and only if
(5) (p and w are both analytic on D,
or (6) E and \TI are both analytic on D,
or (7) there exist constants a, b e C, not both 0, such that atp + but is constant on

D.

A special case of Theorem 1 was proved by Sheldon Axler and Pamela Gorkin ([7],
Theorem 7), using function theory techniques quite different from those that we use here.
Dechao Zheng ([33], Theorem 5) also proved a special case of Theorem 1; our proof makes
use Of some of his ideas.

There is a strong relationship between symbols of Toeplitz operators in these

theorems. Of basic importance is the Poisson integral.

Definition 3. For 0 s r < 1 and 9 real , the function

00

pr(9) = 2 rlnl cine

n=—eo

is called the Poisson kernel. IfL1(aD) denotes the space of Lebesgue integrable functions
on 8D, and f e L1(8D), then the function

5

1t
F(re"9) = IP46 - t) f(ei9)%:-

-fl

is called the Poisson integral of f. We shall write F = PU]. It is well known thath is a
harmonic function on D for every f e L1(8D).

Functions in L°°(8D) correspond, via the Poisson'integral, to bounded harmonic
functions on D, so perhaps the restriction in Theorem 1 to consideration only of Toeplitz
operators with harmonic symbols (as Opposed to Toeplitz operators with symbol in
L°°(D,dA)) is natural. More importantly, Theorem 1 does not hold if “harmonic” is
replaced by “measurable”. Paul Bourdon pointed out the following example: A function f
is called radial if f(2) = f(|zl). Suppose that (p and \it are any two radial functions in
L°°(D,dA). Then,using (1), we obtain (for fixed n e N)

(T, znxw) = Pap mm = j (p(r) 2" —1—— mo =
D (1 - wrz)

 

 

(ei9 - wr)2
'1‘

1t 1 1t
- i(n+1)6 ,
= I f(pO‘) r" glue 1 4'0 2 rd, d6 :17 I(p(r) rn'l'l f e d(e19) dr =
o 1: - wre ) 0

l
= 21:0: +1) w» j (p(r) r2n+1 dr.
0

Now, by the same reasoning as above, we Obtain

1 l
Tqu, 2"(w) = 41:2(n-i-1)2 ( j (p(r) r2n+1 dr ) . ( jmr) r2n+1 dr.)w".
0 0

 

’IESIS

6
From this, it is clear that T¢TV = Tqu, on the set of polynomials, which is dense in L2 , so

that T (9T W = T WT (p. This example suggests that the following open problem may be hard:
Find conditions on functions (p and w in L°°(D,dA) that are necessary and sufficient for Tq,
to commute with T ‘1’-

Chapter 1 concludes with a corollary that describes the normal Toeplitz operators
with harmonic symbol. For a Hilbert space H, .L'(H) denotes the algebra of all bounded
linear operators on H. An operator T e £(I-I) is called normal if T" T = T '1“. An
operator T e L(H) is called hyponormal if T" T 2 T 7". Further research could go
in the direction of describing the hyponormal Toeplitz operators with harmonic symbol.
C. Cowen [14] characterized hyponormal Toeplitz operators on the Hardy space, and it
would be interesting to do so for the Bergman space Toeplitz operators.

IfS c .801) then S '= [B e £(H): AB =BA for all A e S] is the commutant of
S. There is quite a bit of literature about commutants of different Operators. In Chapter 2
we will study commutants of certain analytic Toeplitz operators on the Bergman space.
Not much is known about this. It is well known that {T 21' = {Tq, : (p e H”}. Much
work has been done in studying commutants of analytic Toeplitz operators on H2630). In
their paper [28], Shields and Wallen studied commutants of weighted shift operators.
Deddens and Wong [15] studied the problem using operator theory and raised six questions
that stimulated much of the further work on the problem. Abrahamse [1] answered some
of Deddens-Wong questions negatively. Baker, Deddens and Ullman [8] found {Tf}' in
case that f is an entire function. In a series of papers [29] - [32], Thomson used function
theoretic methods to find commutants or intersection of commutants of certain analytic
Toeplitz operators. Finally, C. Cowen continued their work in [13] and found the
commutant of Toeplitz operators whose symbol is a finite Blaschke product or a covering

map.

7

Our approach to the Bergman space problem will be function theoretic, based on
methods introduced by Shields and Wallen. We will not try to describe commutants of
different analytic Toeplitz operators. Instead, we will try to find those Toeplitz operators
whose commutant is equal to the set of all analytic Toeplitz operators. Methods used by
Deddens and Wong cannot be applied for the Bergman Toeplitz Operators, so that we will
not pursue the course of study they suggested.

Function f e H” is said to be an inner function if lf"'(ei9)l = 1 almost everywhere
on the unit circle. Here f *(eie) = limf(rei9) as r -) 1. One of the basic facts about the
Hardy space is that Tf“ is an isometry if and only if f is inner. To underline the difference
between the Hardy and Bergman space Toeplitz operators, we first prove the following

theorem:

Theorem 4. Suppose that he L°°(D,dA) is harmonic and that T h is an

isometry. Then h is a constant function of modulus 1.

The proof of this theorem uses, among other results, some basic aspects of
commutative Banach algebras. HA is a commutative Banach algebra with identity, the set
of all multiplicative linear functionals on A is called the maximal ideal space and is usually
denoted by M. It is easily seen that each multiplicative linear functional is continuous so
that M c A", where A" denotes the dual space of A. Given the relative weak-* topology
Of A, M is a compact Hausdorff space.

For each a e A, the function 8 :M -> C defined by 6((p) = (p(a), for (p e M , is
called the Gelfand transform of a and it is a continuous function. The map A : A —-> C(M)
defined by a —> a is a homomorphism and is called the Gelfand transform of A. We will

be particularly interested in the Banach algebras H” and L°°(BD), which are crucial for

8
understanding the McDonald-Sundberg symbol calculus for the Bergman space Toeplitz
Operators (see [23]).
After this, we will prove the first result about commutants. Let T be the closed

subalgebra of L(L(21) generated by all Toeplitz operators. We have the following theorem:

Theorem 5. Let Se Tcommute with Tzn,n e N. Then S = TV, for some

we H°°.

From the proof it is clear that operators in {Tzn}' have complicated structure. On

the other hand, it turns out that S e {T2n }' must be an analytic Toeplitz operator in the

+02

case a aé O. This is our next result:

Theorem 6. Suppose h(z) = z" + az, a =15 0, n e N, n > 1 and let S e Lug)

commutes with Th. Then S = TV for some w e H”.

At this point we use the Shields-Wallen ideas introduced in [28]. They proved that

if S commutes with T}, ,where h e H”, then S" k)‘ is an eigenvector for Th", for every

A e D. Because
Ker Tim) = (Range Tit-hm) i

we try to characterize the range of Th-h(l)~ The following theorem is based on the same

technique:

Theorem 7. Suppose p(z) = z + a222 + + anzn, where ai 2 O, for i = 2, ...,n.

pr(z) - p(l) has n distinct zeros, then {Tp}' = [TV : w e H”).

9

For Toeplitz operators whose symbols are polynomials with complex coefficients,
we cannot say much about their commutants, except in a very special case.

Chapter concludes with several examples.

In Chapter 3 we study von Neumann algebras generated by T2". Their definition is
related to the double commutants. If S C £(H) , then the double commutant of S is the set
S " = [S '}'. A von Neumann algebra A is a C*-subalgebra of £(H) such that A = A".
If T e £(I-I) , the von Neuman algebra generated by T will be denoted by W*(T). From
Corollary 7.6, p. 119 in [24] it follows that W*(T) = { T, 7")". Therefore, in order to
determine W*(Tz,,) for some fixed n e N, we have to determine {T,", T 2;)". We will
study the operator T2,, acting on the Bergman space and determine W"'(Tz,,) in this case first
and then we will consider T2,, as an operator on the Hardy space and find the von Neumann
algebra generated by this operator. We will show that they have different structures
if n > 1. The difference comes from coefficients in the Bergman space basis vectors. In
the case n = 1 the situation is simple: W*(Tz) is the set of all bounded linear operators,
either on L3, or H2(8D). The proof uses the injectivity of the map (p —> T4,. In the
Hardy space case, T q, = 0 means that all matrix elements of T¢(e,,) are zero. Since matrix
elements are Fourier coefficients of (p, it follows that (p = O. For the Bergman
space, the proof can be found in [6]. Now, if S e {T,, T,’ }', then S and S"
commute with T2. Therefore S = Tq, and S" = TV: By the previous comment, (p = {[1,
so that (p is a constant function. This means that [T 2, T z'}' = {cI : c e C}, so that

{T z, T,*}" is equal to the set of all bounded linear Operators.

The case n > 1 will use results in Chapter 2 about the commutant of T 2". For fixed

n > 1 we'll show the following:

10
W:

0
T1

W*(Tz,, ) = { T 6 mg) : T = , where L3, = X0 69 X1 69 e X“,

Tn-l

T,- :X,- —) Xi for i = 0,1,m, n-l }.

To

M TOM-

W"'(Tz,, ) = { T e L(H2(ao)) : T = z z , where

Mzn-ITOM‘Z‘n-l
H2(BD) = X0 63 X1 O O Xn-1, To : X0 -> X0, M2 and M;, are multiplication

operators} .

"- I'Esu

 

 

CHAPTER 1
COMMUTING TOEPLITZ OPERATORS WITH HARMONIC SYMBOLS

Studying special classes of Operators, like the class Of Toeplitz Operators, usually
involves the question of when two arbitrary operators from this class commute. In this
chapter we restrict our study to the class Of Bergman space Toeplitz operators with symbols
that are harmonic functions. In the first half of the chapter we examine harmonic functions
more closely from the perspective of an invariant mean value property. Then, in the second
half, we apply these results to our Operator theory problem. This chapter, once again,
shows close relationship between function theory and operator theory. It also shows how

much harder the Bergman space case is than the Hardy space case.

1. The Invariant Mean Value Property

A continuous function on the disk D is harmonic if and only if it has the mean value
property (see for example [10], Chapter X, Mean Value Theorem 1.4 and Theorem 2.11).
In this section we characterize harmonic functions in terms of an invariant mean value
property. ,

Let Aut(D) denote the set of analytic, one-to-one maps of D onto D (“Aut” stands
for “Automorphism”). A function h on D is in Aut(D) if and only if there exist a 5 8D

and [3 e D such that

 

for all 26 D.

11

1:
E

12

A function u e C(D) is said to have the invariant mean value property if

I” u(h<rei°>) are = mm»
0 211?

for every h e Aut(D) and every r 6 [0,1). Here “invariant” refers to conformal

invariance, meaning invariance under composition with elements of Aut(D). If u is

harmonic on D, then so is u-h for every h e Aut(D); thus harmonic functions have the

invariant mean value property. The converse is also true (see [26], Corollary 2 to Theorem
4.2.4): if a function u e C(D) has the invariant mean value property, then u is harmonic

on D.

The invariant mean value property concerns averages over circles with respect to arc

length measure. Because we are dealing with the Bergman space L2, we need an invariant

condition stated in terms of an area average over D. Thus we say that a function

u e C(D) n Ll(D,dA) has the area version of the invariant mean value property if

I u°h m = u(h(0))
D 1!

for every h e Aut(D). If u is in C(D) n L1(D,dA), then so is u-h for every h e Aut(D),

so the left-hand side of the above equation makes sense. Note that the area version of the
invariant mean value property deals with integrals over all of D, as opposed to integrals

over rD for r s (0,1).

 

13

If u is harmonic on D and in L1(D,dA), then so is u°h for every h e Aut(D).
Thus, by the mean value property, harmonic functions have the area version of the
invariant mean value property. Whether or not the converse is true is an open question. In
other words, if u e C(D) n L1(D,dA) has the area version of the invariant mean value
property, must it be harmonic? This question has an affirmative answer if we replace the
hypothesis that u is in C(D) n L1(D,dA) with the stronger hypothesis that u is in C(D); see
[26], Proposition 13.4.4 or [3], Proposition 10.2.

We need to consider functions that are not necessarily continuous on the closed disk
D, so the result mentioned in the last sentence will not suffice. However, our functions do
have the property that their radializations (defined below) are continuous on the closed
disk, and we will prove that this property, along with the area version of the invariant mean
value property, is enough to imply harmonicity; see Lemma 1.1.

If u e C(D), then the radialization of u, denoted 92(u), is the function on D

defined by

91(u)(W) = I02“ u(wei°) €32. .

In the following lemma, the statement SR(u-h) e C(D) means that {Nu-h) can be extended

to a continuous complex-valued function on D.

The following lemma will be a key tool in what follows.

14

Lemma 1.1. Suppose that u e C(D) n L1(D,dA). Then u is harmonic on D if
and only if

I w}: M = u(h(0)) (1.1)
D 1!
and

mot-h) e C(D) (1.2)

for every h e Aut(D).

PROOF: We first prove the easy direction (the other direction will be the one that
we need in the proof of Theorem 1). Suppose that u is harmonic on D. Let h e Aut(D).

As we discussed earlier, u-h is harmonic and so (1.1) holds. The mean value property

implies that 92(u-h) is a constant function on D, with value u(h(0)), so (1.2) also holds.

To prove the other direction, suppose that (1.1) and (1.2) hold. Let h e Aut(D),

and let

v = SR(u°h).

By (1.2), v e C(D).

We want to show that v has the area version of the invariant mean value property.

To do this, fix g e Aut(D). Then

I mm = j ma-oommam
D 1! D 113

= I I2“ u(h(g(w)eie))d_Qd_A_(_w_). (1.3)
D 0 2113 113

15

To check that interchanging the order of integration in the last integral is valid, for
each 0 e [0.21:] define f9 6 Aut(D) by

f9(w) = h(g<w)ei°).

The inverse (under composition) fe'l offe is also an analytic automorphism of D, so there

exist a e 30 and B e D such that

 

f9'1(Z)=a L -_z for allze D.
l-Bz
Thus
1.13:2
l '1’z|= _
V“ H) I1-[3zl2
sl+—'B' forallze D.
1.13:

Note that [3 =f9(0) = h(g(0)ei°); then IBI s sup {lh(z)l: Izl S.lg(0)l}. We are drinking of h
and g as fixed, so the above inequality shows there is a constant K (depending only on It

and g) such that

l(fe'1)’(z)l s K for all z e D and all e e [0,2n].

N ow

2" i9 =2“ I ll‘l’le
jo jblu(h(g(w>c may?! [0 L) u(z) (f, )(z) 479%

 

16

s K2 I lu(z)| m)
D 112

Thus we can apply Fubini’s Theorem to (1.3), getting

j V°gd_A = j“ j u(h(g(W)eie))dAM)d_Q
D 1t 0 D 11? 21C

21:
-I. Ip<u°fe><w>eeee
21c
=j0 u(fe(0)) (21% (by (1.1))

.... J’ 2" u(h(g(0)ei°) ea = m<u~h><g<0>>
o 211:

= V(g(0)).

Thus v is a continuous function on D that has the area version of the invariant mean value

property. Hence (see [26], Proposition 13.4.4 and Remark 4.1.4 or [3], Proposition 10.2,

combined with a change of variables) v is harmonic on D. Because v is also a radial

function, the mean value property implies that v is a constant function on D, with value

Recall that v = 92(wh), so

 

17
j“ (u-thci") 412 = u(h(o»
0 Zn

for every r 6 [0,1) and for each h e Aut(D). In other words, u has the invariant mean
value property (the usual version, not the area version). Thus (see [26], Corollary 2 to
Theorem 4.2.4 and Remark 4.1.4) u is harmonic on D, and the proof of Lemma 2 is

complete. I

 

18

2. Toeplitz Operators

In the previous section we obtained an invariant mean value property of harmonic
functions. In addition to this, we need some other facts about these functions. For a real-
valued harmonic function u on D, any v defined on D such that u + iv is analytic is called a
harmonic conjugate of u. Given the real harmonic function u, there is a unique real
harmonic function a that is conjugate to u with 22(0) = 0. If f = u + :22, then u = iv + 7').
In case that u is a complex-valued harmonic function, the above procedure gives u =f1 +f-2
where f1 and f2 are analytic.

Suppose now that u is a bounded harmonic complex-valued function on D. What
more can we say about f1 and f2 ? To answer this question we have to introduce another
space of functions on 8D. For a function f e L1(BD) and an are I in 8D, let f1 denote the

average of f over I:
=m(I)I f ‘1’"
For a function f e L1(BD), let

llfllBM0=sup {hi—5i; If-fll dm :1 is an arc in an}.

A function f e L1(3D) for which ll f "8M0 < oo is said to be of bounded mean
oscillation. The set of all these functions is denoted by BMO. The class BMO was first

introduced in [21]. It satisfies

L2(3D) D BMO :> L°°(3D) (1.4)

19

By assumption on u, Re u = u1 is bounded and Fatou‘s theorem guarantees the existence of
the radial limit of an almost everywhere on 6D. Also u1* e L°°(8D). The function u'i has
radial limits almost everywhere on 3D (see [22], Chapter III, Lemma 1.3) and by [17],
Chapter VI, Theorem 1.5, u]. is in BMO. Formula (1.4) implies that u1*+ iu"1* is in
L2(BD) so that 141 + id] = P[u1*+ iu'i*] is an analytic function on D whose Taylor
coefficients are square-summable. The set of all analytic functions on D with square-
summable Taylor coefficients is denoted by H2(D) and is called the Hardy space, since it is
isometrically isomorphic to H2(3D). Hence, if ul is bounded and harmonic, it is the real
part of a 112(D) function and therefore u =f1+7§, where f1, f2 6 H 2(D).

Now, we are ready to state the main theorem of this chapter.

Theorem 1.2. Suppose that (p and \V are bounded harmonic functions on D.
Then

TvTv TvT¢

if and only if
(1.5) (p and w are both analytic on D,
or (1.6) 6 and \TI are both analytic on D,

or (1.7) there exist constants a, b e C, not both 0, such that atp + but is constant

on D.

Before beginning the proof of Theorem 1.2, we need the following two lemmas.
For h e Aut(D), define an operator U h on L?! by

Uhf= (f°h)h’.

Lemma 1.3 is certainly well known.

 

20

Lemma 1.3. Let h e Aut(D). Then U h is a unitary operator from L3 onto Lg.

PROOF: When making a change of variables 2 = h(w) in an area integral, we must

make the substitution dA(z) = |h’(w)|2 dA(w). Thus

I I/(h(w))I2 Ih'(w)l2 dA(w) =J' lf(z)l2 dA(z)
D D

for every f e L2 , and hence Uh is an isometry of L?! into L3,. Clearly U ha is an inverse

for U It An invertible isometry is a unitary operator, so we are done. I

Lemma 1.4. Let h e Aut(D) and let (p e L°°(D,dA). Then

*
UhTth = TM .

moor: Define VhiLZ(D,dA) —> L2(D,dA) by Vhf= (f-h)h’. As in the proof of
Lemma 6, VI: is a unitary operator from L2(D,dA) onto L2(D,dA). Obviously VhlLfi = U h°

Thus Vh maps Lg onto Lg, so
PV,l = VhP. (1.8)

Iffe L2,then

Tthf= T,.,.(<f-h)h') = P((<P°h)(f°h)h’) = P(V,.(<po)

= Vh(P(<i>f)) (by (1.8))

21
= UhTOf'
Thus T‘p. hU h = U thv and because U h is unitary (Lemma 1.3), this implies the desired
result. I

We have now assembled all the ingredients needed to prove Theorem 1.2.
PROOF: We begin with the easy direction. First suppose that (1.5) holds, so that (p

and w are analytic on D, which means that T(p and T W are, respectively, the operators on L2
of multiplication by (p and w. Thus T¢TW = Tqur
Now suppose that (1.6) holds, so that (T) and \Tt are analytic on D. By the paragraph

above, TTT‘V = TWTTP' Take the adjoint of both sides of this equation, and then use the
. . . I! *
identities T¢ = Tq) and T1? = TV to conclude that T¢TW = Tthp‘

Finally (for the easy direction) suppose that (1.7) holds, so there exist constants a,
b e C, not both 0, such that atp + but is constant on D. Ifa at 0, then there exist
constants B, y e C such that q) = [3w + 7 on D, which means that Tq, = BTW + 71 (here 1

denotes the identity operator on L2), which clearly implies that T¢TW = Tqur If b at 0, we
conclude in a similar fashion that T q, and TW commute.

To prove the other direction of Theorem 1.2, suppose now that (p and \I’ are
bounded harmonic functions on D such that T <PTW = Tthp' Then there exist functions f1,

f2. g1. and g2 in 112(0) such that
(p=f1 +172 and w=g1+g3 onD. I (1.9)

The Hardy space H2(D) is a subspace of L2. In particular, f1, f2, g1, and g2 are all in L3,.

Let 1 denote the constant function 1 on D. Then

T‘PTWI = T¢(P\|I) = T¢(P(gl + 8—5)) = T‘p(gl + m)

22
= P([f1 +f2-llgl +m1)

=f181+mfl+P$81)+mm-

Thus
<T¢TW1, 1> = <f1gl + g2(0)fl +7581 +f2(0) g2(0), 1>

= (Dag, + iron, on. «some .n

 

= nlf1(0>g1<0)+f1(0>g2(0> +f2(0)gz(0)l +jD f_2g1dA. (1.10)

A similar formula (interchanging the f’ s and the g’s) can be obtained for <TWT¢1, 1>.

Because T‘PTW = T WTIP’ we can set the right-hand side of (1.10) equal to the

corresponding formula for <TVT¢1, 1>, getting
[D 7231 415% = male) from. (1.11)

Let h e Aut(D). Multiplying both sides of the equation T (PTW = Tqu) by Uh on
the left and by Uh“ on the right, and recalling (Lemma 1.3) that Uh is unitary (so Uh*Uh =

I), we get

UhT¢U,,*U,,TWU,,* = UhTWUh*UhTQUh*.

23

Lemma 1.4 now shows that

TQ'hT 'h = TW°th>-h‘ (1.12)

Composing both sides of the equations in (1.9) with h expresses each of the
bounded harmonic functions qrh and \V'h as the sum of an analytic function and a
conjugate analytic function:

tp-h =f1-h +54: and \V-h = g1°h + gg-h on D. (1.13)

Equation (1.11) was derived under the assumption that T¢TW = T T thus (1.12),

v «9‘
combined with (1.13), says that (1.11) is still valid when we replace each function in it by

its composition with h. In other words,
jD (73g, -f1§23°h 41.3 = 501(0)) glow» 4102(0)) 5M0».
Letting
“ = 72-8171!th

the above equation becomes

I u-h {in = u(h(0)).
D it

In other words, u has the area version of the invariant mean value property.

24

We want to show that u is harmonic on D. By the above equation and Lemma 1.1,
we need only show that 92(u'h) e C(D). To do this, represent the analytic functions f2°h

and gl-h as Taylor series:

(fz'hXZ) = E anz" and (gloh)(z) = E an" for all z e D.
n=0 n=0

Because (W: and w-h are bounded harmonic functions on D, (13) implies that f2°h and g1°h

are in H2(D); in other words,

2 lot":2 <.. and 2113"? < ..., (1.14)
n=0 n=0

Now for z e D we have

_ 2 _ . i
92((f2g1>°h)(z) = [0 " (fa-me”) (gl-the 9) gig
= £0 65,13,129".

Inequalities (1.14) imply that 2 laanl < co, so the above formula for 92((7gg1)‘h) shows
"=0

that m(('f2'gl)-h) 6 C(13); similarly, we get 92((f1g—2)-h) 6 C(13). Thus chat-h) 6 C(13),
as desired. Thus at this stage of the proof we know that u is harmonic.

Let 3/62 and 3/8? denote the usual operators defined (on smooth functions on D) by

25

If f is analytic, then the Cauchy-Riemann equations show that

a_,§[_ §Z_ a‘_—e
az‘f' a; ‘0' 32-0’ and 32"“

It is easy to check that 8/62 and 8/3? obey the usual addition and multiplication formulas for

derivatives and that

32 32 a a
$2+'a—yz=4 'a—Ea—z.

Thus, because u is harmonic, we have

a an 3 8175314155) 3 — . ._
0:4 a—-z—'(§;)=4 57[ az =4 3—;- (fzgl ”f1 82)

=4 (Far-ha?) onD.

Hence

131? = Kgl' on D. (1.15)

26
We finish the proof by showing that the above equation implies that (1.5), (1.6), or
(1.7) holds. If g1 ’ is identically 0 on D, then (1.15) shows that either g2’ is identically 0 on
D (so tit would be constant on D and (1.7) would hold) or f1 ’ is identically 0 on D (so both
(T) and \TJ would be analytic on D and (1.6) would hold). Similarly, if g2’ is identically 0 on

D, then (1.15) shows that either (1.7) or (1.5) would hold. Thus we may assume that
neither g1 ’ nor gz’ is identically 0 on D, and so (1.15) shows that

g 1 82

at all points of D except the countable set consisting of the zeroes of g1 ’gz’. The left-hand
side of the above equation'is an analytic function (on D with the zeroes of gl’gz’ deleted),
and the right-hand side is the complex conjugate of an analytic function on the same
domain, and so both sides must equal a constant c e C. Thust ’ = cgl ’ and f2’ = Egz’ on
D. Hence f1 - cgl andf—2 - cg—2 are constant on D, and so their sum, which equals (p - cw, is

constant on D; in other words, (1.7) holds and the proof of Theorem 1.2 is complete: I

Recall that an operator is called normal if it commutes with its adjoint. If

(p e L°°(D,dA), then T; = T3,, so we can use Theorem 1.2 to prove the following
corollary, which states that for (p a bounded harmonic function on D, the Toeplitz operator

T(p is normal only in the obvious case.

Corollary 1.5. Suppose that (p is a bounded harmonic function on D. Then Tq,

is a normal operator if and only if (p(D) lies on some line in C.

27

PROOF: First suppose that (p(D) lies on some line in C. Then there exist constants
a, [3 e C, with a at 0, such that one + B is real valued on D. Thus Tao + l3 is a self-
adjoint operator, and hence T‘P’ which equals CERT“, + l3 - BI), is a normal operator.

To prove the other direction, suppose now that Tq) is a normal operator. Thus
TQTE = T¢T§’ and so Theorem 1.2 implies that (p and O are both analytic on D (in which
case (p is constant, so we are done) or there are constants a, b e C, not both 0, such that
atp + b6 is constant on D. The latter condition implies that (p(D) lies on a line, completing

the proof. I

We conclude with a question. Does Theorem 1.2 remain true if we replace the disk
by an arbitrary connected region in the plane? (So the Toeplitz operators now act on the
Bergman space of this new region.) The statement of Theorem 1.2 certainly makes sense
in this context and is plausible. However, the proof of Theorem 1.2 made extensive use of
the set of analytic automorphisms of the disk. This approach will not work in general,
because nonsimply-connected regions have too few analytic automorphisms to provide any

useful information.

CHAPTER 2

THE COMMUTANTS OF CERTAIN TOEPLITZ OPERATORS

In this chapter we will be mostly interested in studying the commutants of some

Toeplitz operators on the Bergman space whose symbols are polynomials. Only a little is

known about this subject. The Bergman shift Tz has commutant that is equal to the set of
all analytic Toeplitz operators. How about T,» ?

It is well known that the Hardy space Toeplitz operator Tf is an isometry if and only
if f is inner. If f is a nonconstant inner function then Tf is a pure isometry and is unitarily
equivalent to a unilateral shift, whose commutant can be characterized matricially. Much
work about the commutants of the Hardy space Toeplitz operators is based on this fact. On
the Bergman space, a Toeplitz operator whose symbol is a nonconstant inner function is
not an isometry. We will prove even more: The only Toeplitz operator with harmonic
symbol that is an isometry is a scalar multiple of the identity. Before we prove this, we
need some facts about bounded analytic functions. Good references are Hoffman [19] and
Garnett [17].

By E” we denote the Banach algebra of bounded analytic functions on D, with the

supnorm
llfll=sup{lf(z)l:zeD]

The set of all multiplicative linear functionals on H °° is denoted by M. The obvious

elements of M are the point evaluations

28

29
(92((f) =fl 7»)

where it is a point in D. Map A —) ‘97). is a homeomorphism from D into M, so that we
regard D as a subset of M. D is an Open subset of M and by the Corona Theorem, D is
dense in M.

The Gelfand transform A : H‘” —-) C(M) is defined by f(tp) = (pm, for (p e M. The
Gelfand transform is an isometry from H” —) C(M), so that we can identify H” with the
uniformly closed subalgebra of C(M). Hoffman ([20], Lema 4.4) has proved that the
following algebras are identical:

0) C (M);

(ii) The sup norm closure of the algebra generated by H“ and 171-” (the latter set
denotes the complex conjugates of functions in H");

(iii) The sup norm closure of the algebra generated by the bounded harmonic
functions.

If m1 and m2 are in M, the pseudohyperbolic distance between m1 and m2 is

p(mt. M2) = sup 1 Iftma) sze m, "f" s 1. four) = o }

Clearly p(ml, m2) S 1. The relation m1... m2 if and only if p(ml, m2) < 1 is an

equivalence relation on M. The corresponding equivalence classes are called the Gleason

parts of M. An element m e M is said to be an one-point part if its equivalence class

consists of itself alone. The set of one-point parts in M will be denoted by M1. Let

J={(pe C(M):q)=OonM1}.

30
Let 'I(C(M)) be the closed subalgebra of 130.3,) generated by {T, : o e C(M) ) and let G be

the commutator ideal of rI(C(M)). McDonald and Sundberg [23] has proved that the

sequence
(D
O —>J —-> C(M) -) 'C(C(M))/C —)0
is exact, where

<1>((p) = Tq, + C

for (p e C(M). This implies that C(M)/J is isomorphic to 'T(C(M))/C with isomorphism

‘I’(tp+J)=T¢+C

for (p e C(M). Let H : ‘I(C(M)) —) 'T(C(M))/C be the quotient map.

Another Banach algebra we need is L°°(8D). The maximal ideal space of L°°(8D),
denoted by M(L°°), plays an important role here. If we identify each function f e H” with
its boundary values, we may regard H” as a closed subalgebra of L°°(8D), which gives a

natural map I : M (L°°) —) M. The map I is defined by restricting each complex
homomorphism of L°°(aD) to H”. It is, in fact, a homeomorphism of M (L°°) into M, so
we may think of M(L°°) as a subset of M. It turns out that M(L°°) is a subset of M1.

Now, we can prove our theorem.

Theorem 2.1. Suppose that h e L°°(D,dA) is harmonic and that T), is an

isometry. Then h is a constant function of modulus 1.

31
PROOF: Suppose that T). is an isometry, i.e., T}, T), = I. Then

H( from Th ) = II( I) (2.1)

Since h is harmonic, Hoffman's result shows that h and h e C(M). Applying ‘P'1 to
both sides of (2.1), we obtain

(h+J)-(h+J)=1+J.

This shows that

h-h-le].

By the definition of J,

h'h-1=0 oan.

As we said before, M (L°°) is a subset of M1, so that above equality means

<P(h-7)=1

for every (p e M(L°°). Since (p(h -h) = I (p(h) l2 we can conclude that

I (p(h) I = 1 (2.2)

32
for every (p e M (L°°). Since h e L°°(D,dA) is harmonic, we can identify it with its
boundary value functiOn, denoted again by h. The Gelfand transform of L°°(aD) maps
L°°(8D) isometrically and isomorphically onto C(M(L°°)) (see Hoffman, [19], p. 170), so

that we have
=sup [ |(p(h) | : (pe M(L°°) }.

Hence (2.2) implies sup { lh(z) l : z e D } = 1. Ifl h(z) l = 1 for some 2 e D, then h is
constant by the Maximum Principle. If I h(z) I < 1 for all z e D, then

1t=||1||2=||Th1ll2=llPhl|25|lh||2= JD|h(z)I2dA<JD dA=rt,

a contradiction. Here ll - II denotes the L2(D,dA) - norm. Therefore h is a constant

function, and since T h is an isometry, l h I = 1. I

We can slightly generalize this result and get the following:

Corollary 2.2. Suppose that T), is an isometry, where h = f - g", where f is
inner, g e L°°(D,dA) is harmonic, and n e N. Then h is a constant function of

modulus 1.

33
PROOF: Since f, g e C(M), it is clear that h e C(M) so that we can apply the
same reasoning as in Theorem 2.2 and thus obtain l (p(h) l = 1 for every (p e M(L°°). Since
|(p(f) | = 1 if (p e M(L°°), it follows that

1=|¢(g")|=|<p(g)ln
what gives I (p(g) l = 1 for every (p e M(L°°). Because g is bounded harmonic,
sup { Ig(z)|:ze D}=1
As before, if I g(z) | < 1 for every 2 e D, then I h(z) l < 1 for every 2 e D, and we obtain

the same contradiction. Thus I g(z) I = 1 for some 2 e D, and then g = constant, i.e.,

h = cf, 06 C. Then TAT). =1 implies Tlhl2=li i.e., l h I = 1 and we are done. I

J. Thukral asked for which harmonic h is T). a partial isometry. If T), is a partial
isometry, then by Halmos [18], Problem 98 , T), = T}, 7’}, Th. Applying ‘I"'1 II on both
sides gives

h-hZTe J.

As before this means
<P(h)i 1-|<P(h)|21= 0

for every (p e M(L°°). Then, either (p(h) = 0 or I (p(h) | = 1, so that sup { I h(z) l : z e D }
= 0 or 1. If(p(h) = 0 for every (p e M(L°°), then h = O. Ifq>(h) :16 0 for some (p e M(L°°),

34
then sup { lh(z)l:ze D } =1. If|h(z)|<1foreachze D, then takinganyuin
(Ker T}, )'L we get

ll u “2: || Thu "2: llP(hu) II2 5 II hu "2: ID I h(z) I2 I u(z) I2dA <JD |u(z) MA = II u “2.

This is again a contradiction. Hence h must be a constant function. Thus we have proved

the following theorem:

Theorem 2.3. Suppose that h e L°°(D,dA) is harmonic and that T), is a

partial isometry. Then h is either a constant function of modulus 1 or h is identically

0.

Now, we are going to consider our main problem - the commutants of some

analytic Toeplitz operators. At first we will be interested in finding the commutant of Tzn,

for arbitrary positive integer n.
Let "(be the closed subalgebra of £(L‘21) generated by all Toeplitz operators. For

(p e L°°(D,dA), the Hankel operator with symbol at, denoted HQ, is the operator from
L3 to (L2? defined by wa =(I - P )((pf). 9( will denote the ideal of all compact opertors

acting on L3. Before we state our next result, we need to define some function spaces.

For an analytic function f on D we set
llfllB=sup { (l -|z|2)lf’(z)|:ze D}

The Bloch space B is the set of all analytic functions f on D for which ll f IIB < co. The
quantity If (0) I + Ii f IIB defines a norm on B, and B equipped with this norm is a Banach

 

35
space. Contained in the Bloch space is the little Bloch space Bo, which is by definition

the set of all analytic functions f on D for which

(1-lzl2)f'(z)—>0 aslzl—>1.

For the basic properties of the Bloch space see [2].
Let n e N be fixed.

Theorem 2.4. Let S e ‘T commute with T2". Then S = TV for some I]! e H°°.

PROOF: The equation S T 2;: = Tzn S gives us the following:

Letgo=S 1. Then
Szn=STzn1=Tanl=zngo.
S22" =S T2» 2" = Tan z": zzngo

S 2"" = z’mgo , k = 0,1,2 .

Letg1=Sz.Then
Szn+1=STznz=Tanz=zng1

S 22"‘1'1 = S Tzn Z’H'l = Tan z’H’l = 22381

S 2’01”: zkflgl , k = 0,1,2 .

Continuing in this way, we finally get

 

36
gn-l =3 2'”
S 22’"1 = S Tzn z"'1 = Tan zn'1= z"g,,-1

S z""“”"1 = angn-1 , k = 0,1,2 .

Let X0 = span { ck”: k = O,1,2,... } ,

X1: Span [ elm-+1: k = 091929 "0 } 9

Xn-l = span { ckn+(n-l)3 k = 0.1.2, }

Then L3, =Xo e X16 e X,.-1, i.e., each fe L3, can be written as
f=f0 +fl + +ffl'19

f) e Xi, i = 0,1,2, ,n-l. Each f0 6 X0 has its Fourier series expansion

f0 = Z (fo.e1ta)ekn
It = o

m
i.e., f0 = lim sm, where sm = 2 (fo,ek,,)e1m. Since the point evaluations are bounded on
k = 0

L2, we have mm = lim sm(z), for each 2 e D, so that

(fo 'go)(2) = lim (Sm -go)(2) (2.3)

for each 2 e D. Since S 2"" = z’mgo . k = 0,1,2 it follows that S sm = sm ogo , for

every m e N. By continuity of S, we have

 

 

37
(Sfo) = lim 5 s,,, = lim s,,, -go

so that
(S fo )(2) = lim (Sm 'go)(2) (2.4)

for each 2 e D. Comparing (2.3) and (2.4) we conclude that

S fo = gofo.

for each f0 6 X0. Repeating the above reasoning, we get that S f1 = 821 f1 , for f1 6 X1 and

so on. Thus operator S can be described as

Sf=gofo+ film fife + ---+§—;‘,%f.-i. (2.5)

Let's observe another property of S, being an element of ‘T.
m: STz-TzSe 9C.
At first, assume S = Tq, , (p e L°°(D,dA). Then

T¢Tz-TZT¢=TZ¢-TZT¢=H2*H¢.

Because z 6 Bo, the operator Hf is compact (see [4]), so that Tq, T z - T, Tq, is compact.
If 11 : 130.3,) —9 Lab/9‘ denotes the natural projection, then

H<T.p)rI<T.)=H(T.)rI(Ta) (2.6)

III IA

 

38
for every (p e L°°(D,dA). If S = Tm Tq," then, because of (2.6), I'I(S T, - T, S) = 0

so that S T, = T, S e EX . An arbitrary operator S in 'I is the limit of sums of operators of
the form T01 - . -T%. In that case S = lim Sn , each S, is of the form 2 T<P1 ~ . - Twn, and

so S T, - T, S = lim (S, T, - T, Sn) 6 SK. Hence claim is proved.

Now, let's express S T, - T, S in terms of (2.5).
S Tsz = S(Zfo) = 82-1 (2fo) = glfo . because 2fo 6 X1.

S T,f1= S(zf1) = 5% (zf1) = 8221f], because zfl 6 X2, and so on. Finally we get
s Tzf=81f0 + 3,113+ gfz + + 2 goat.

From this and (2.5) we obtain

(STz'TzS)f=f0(gl'280)+f1(&zz ~31) +f2(§% - gzl)+'“+fn-l(7-80' 3%.

By the claim, (S T, - T, S)|Xo = M 81 - zgo : X0 —> L2,, is compact (M31- 180 is a multiplication
operator). Let (p = g1 - zgo.

Qaim : (p = 0.

Let w e D be fixed and let Rw 6 X0 be the reproducing kernel for X0. Recall that
M :1; : chz -—> X0 and take any g 6 X0 to obtain the following:

<g, M; kw) = <M¢g, kw> = (p(w) -g(w) = (p(w) <g, R, > = <g, (p(_w)Rw > .

W

III —" It! kw ~— R ,
It follows that M, kw = tp(w)Rw, so that M(p-m = (p(w) II kw II‘ It is well known

 

 

k , t , t k
that ll k: ll converges weakly to 0 as l w I —) 1. Since Mq, ls compact, Mq, m —) O

innormaslwl—)1. Thus

39

|
ltp(w) |——|”I:W II" —>0 as | w | —> 1 (2.7)
W

We'll calculate the numerator first.

 

Rw(z) = 2 In: 1 (Rw.Zl"')z"" = % Z (kn+l) 7"" Z”,
k=0 k=0

so that

II R, II2 = Z (kn+l) lwl2k".
k=0

alt-i

Let A = lwl2". Then we have the following:

 

IIRWII2= l2(kn+l)hk=12khk+ 121k: Li Aky + _ =
nk=0 “11:1 nk=0 1; “(11) o

= nh + 1 _l+(n-1)h
n(1-h)2 “(l-7») n(1-).)2 '

 

 

Therefore

IIRW II2 _ 1+(n-l)lw|2".
“kw “2 (l—lez")2

 

 

l 2 2
1-II22= II2n+l-IIn
( w ) [1+1w|2+lwl4+ '--+lwl2("'1)] [nw w ]

.1. 2n
anwl .

 

40
II R II
m—Wl—I 2 —1- > o, as le —> 1. Therefore by (2.7),1iml (p(w) | = 0, as le —> 1.
W

n

Thus lim

Since (I) is analytic, q) = 0, and claim is proved. This means that go = 821 .

Similarly, (S T, - T, S)|x1 = M82]: - 81 :X1 —) L?z is compact. Thus Mgz/z - 81

MZIXO = M82 - 281: X0 -—) L3 is compact. By the previous claim, g2 - zg1 = 0 so that g1 = g}

and therefore go = 5%. If we continue this way, (2.5) shows that

Sf= 80 °f= Tgof

for every f 6 L2. The function g0 must be an H” function as a multiplier of L2 (see

[28]). If we let )4! = go, the theorem is proved. I

We can extend this result. Let u e Aut(D) and define an operator V : L% —> L3, by
Vf =f-u'1. Clearly, V is a bounded linear operator, with the inverse operator V 'lf = f-u.
Observe that T,Vf= z - Vf= z ~(f-u'1) and VTuf= (u ~f)-u'1 = z ~(f-u'1) which means

that
T, V = V T, ,
i.e., T, is similar to T,. Thus

T,. V = V Tun _ (2.8)

for every n e N.

Suppose now that S e ‘T and S Tun = TunS for some n. Formula (2.8) implies

41
SVJTfV=V4T,VS.

After multiplying both sides of this equation by V from left and by V ‘1 from right, we

conclude that

VSV-16{Tzn}'.

cm: The operator B = V S V '1 has the property that B T, - T,B is in ER.
B T, — T,B = VSV'lT, - T, VS V'1= V(S V'lT,V- V‘1 T, VS)V'l
= V(S Tu- Tu SW ‘1.
The operator S Tu - Tu S is compact because u 6 Bo and the claim is proved.
Now, by the proof of Theorem 2.4, B = T (p, for some (p e H°°. This means that
s = V'qu, V, i.e., Sf= V 'qu, Vf= V ‘1(o -(f°u’1))=((p°u)~f, forfe L3,. If we let

w = (pm , we have proved the following corollary:

Corollary 2.5. Let S e ‘T commute with Tun, for some n e N. Then S = TV for

some ‘1’ e H°°.

Now, knowing that, for example, { T 22 ]' r) '1': { TV : \V e H” } leads

us to the question of finding { T }' r) '1'. The answer is rather surprising: the

22 +2

commutant { T }' itself is equal to { TV : w e H°° }, as we will see. One of the

22+ 2

crucial tools in what follows is this well-known lemma:

Lemma 2.6. If h e H” and S e £(L2) commutes with Th, then S" k) is an

eigenvector for Th" for every A e D, with eigenvalue hat) .

42
For the sake of simplicity, let h(z) = z2 + 2. By the Lemma 2.6,

5" kit 6 Ker Tit-hm = (Range Th-h(A)) i

so that we have to find Range Th4“).

Suppose that g 6 Range Th-h()(). Then g = [ h - h(A)]f for some f 6 L3,. This

implies
g(Z)=(22+z-Az-Alf=(z-A)~(Z+A+11f.
Let
Do={AeD:-A-1ED}

Suppose that A e D \D(); then 2 = - A - 1 is not a zero of g.

Claim : Range Tit-ho.) = I g 6 L3, : g0») = 0 l. ,

Clearly, Range Til-hot) is a subset of { g 6 L2 :g(A) = 0 }. Conversely, suppose
that v e { g 6 L2 : g(A) = 0 }. Then v(z) = ( z - A )u(z), for some analytic function u. It
is easy to see that u 6 L31: First choose 0 < e < 1 big enough so that A E D,, where
D, = [ z e D : Izl < e }. Since it is analytic on D, it is bounded on 5—,, let's say, by a
constant C. For 2 e D \D, , Iz - Al > k = dist ( ADD, ) so that we have the following:

JDlu(z)l dA- [Demon dA+ IDweiuml dASCZIt-I-I —dA

D\Dg|z-A|2

II v I2

5 C211: +
[(2

 

43

sothatue Lg. Wecanwritev(z)=(z-A)-(z+A+1)fiwhereflz)=-%—l—.

Recall that A e D \Do, so that z + A + l is bounded away from zero. Then f 5 L2,
because u 6 L3,, and we proved that v 6 Range Tum.) and the claim is proved.
This claim shows that Range Th-h(_),) = {kn-L for A e D \Do. Then for

these A's, S *k) = f(A)k),, where f is some function defined on D \D(). For fixed it 6 L31

we have
(s u)(A) = <5 u, hp = <u, s*k;(> = <u,f(T)k,> =f(A) - u(A)

for each A e D \Do. For such A, [( ST, - T,S )(u)](A) = [ S(zu) ](A) - [zSu ](A) = 0.
Because ( ST, - T,S )(u) is an analytic function, that is equal to 0 on an open set, it must be

identically 0 on D. The function u was arbitrary, so that ST, = T,S and therefore S = TV ,
we H°°. Hence we showed that { T22+2 }'= { Twzwe H°° }.

The same reasoning would show that [ Th }' = [ TV : \l’ e H°° }, in the case that
h(z) = (z - a)(z - b), where a,b e C and b if: - a. Of course, when b = - a, h(z) = 22 - a2
and { Th }' = [T22 }'. The difference between {T22 }' and {T22+z}' leads us to the

question of characterizing the commutant of T1,», 02 , a :19 0, and n e N. For that reason,

let's fix it e N. The case lal > 1 is easy:

Suppose S e {T,»+ az}'. Denote h(z) = z" + az and we will find Range Th-h(;(). If

g 6 Range Til-MA): then

g(z) = [(z" - A") + a(z - A)]f= (z - A)[ z"'1 + Az"'2 + + A”1 + a ]f. (2.9)

Let p(z) = z"'1 + Az'l'2 + ---+ N”. Since lal > 1, there exists 8 > 0 such that l - é > e,

 

i.e., < lal. For this 8, let D, = { z e D : Izl < e }, and suppose A e De. Then

 

44
IP(Z)I<1+8+82+...+en-1<

 

<|a|
l-e

so that p(z) + a =16 0 for every 2 e D. In fact p(z) + a is bounded away from 0 if
A. 6 DE 2

 

lp(z)+a|.>_|al-|p(z)I>|a|-11 >0.

-8

From (2.9) we conclude that Range T,-),(r, = { g e L3, : g(A) = o } if A e 1),. As in the
example above, it will follow that (S u)(A) =f(A) - u(A), for some function f and A 6 D,.

This will again give
(ST: - T25 )0!) = 0

for u 6 L3,, since (ST, - T ,S )(u) is an analytic function that is equal to 0 on an open set
D,. Therefore S e {T,}', so that S = TV , for some ‘1’ e H°°.
Now we want to show that the same holds for any a e C \ {0]. This is the

content of the next theorem:

Theorem 2.7. Suppose h(z) = z" + az, where a =16 0, n e N, and n > 1. If

S e £(Lg) commutes with T,., then S = TV! for some \i’ e H°°.

PROOF: The case lal > 1 has already been discussed. Then we may assume

0 < lal S 1. By Lemma 2.6, S" k) is an eigenvector for Th", for every A e D. Again,
suppose g 6 Range Th-h(;()_ Then g(z) = ( z" + az - A" - aA )f.
GOAL: We want to find an open subset U of D such that z = A is the only solution

ofz" + az-A"-aA=OinD,ifAe U, and furthermore z=Ais simple.

 

45

So, we shall look at the equation

2" + az = A" + aA (2.10)

For explanatory reasons, we'll consider two cases.

Easel: a>0.

For such a, choose A = l and look at the equation

zn+az=1+a (2.11)

ForzeD,lz"+az|$|znl+alz|<1+a,sothereisnosolutionof(2.11)in

D. Ifz 6 8D satisfies (2.11), we have

l+a=|zn+az|S|znl+alzl=1+a.

Equality in the triangle inequality-occurs only in the case that all terms are linearly
dependent. Thus arg ( z" + az ) = arg az + 2klt , k e Z. But 2" + az e R so that
arg z = arg az = 2m: , for some m e Z, i.e., z = 1. Thus, (2.11) has only one root in D :
z = l = A.

Since(z"+az-1-a )'(1)=(nz"'1+a)(1)=n+a;‘-0 (becausea>0),z= 1
is a simple zero.

Let F : C xC —> C be defined by F(A,z) = z" + az - An- aA. By the previous
arguments, F (1,2) has n - l zeros outside D: 21, 22, ---, 2,4. The other zero is z = 1. Each
zero 2; is simple, because ( z" + az -1 - a )'(z,-) = n z,“‘1 + a 4: O, for a S 1.

Let's choose 8 > 0 small enough such that K(zi, e) n D = 0 for every i, and

K(z,-,8)r)K(z,-,e)=0,foriséj. HereK(z,-,e)={ze D:lz-z,-|<e}.

 

45

We know that F(1,21) = 0. Taking partial derivatives with respect to 2 gives g);- =

n z":1 + a and-g; (1,21) = my“1 «I» a :16 0, because 0 < lal S 1. By the Implicit Function
Theorem, there exists W1, an open neighborhood of 1, and (p1 : W1 -> C continuous such
that (p1(1) = 21 and F(A,(p1(A)) = 0 for all A 6 W1. By continuity of cm, there exists V1,
an open subset of W1, such that 1 6 V1 and K(z1, e) D (p1(V1).

Similarly, F (1,z2) = 0 and-g; (1,22) =1e 0. By the Implicit Function Theorem, there
exists W2, an open neighborhood of 1, and (p2 : W2 -> C continuous such that (p2(1) =22
and F (A,(p2(A)) = 0 for all A 6 W2. By continuity of (p2, there exists V2, an open subset
of W2, such that l 6 V2 and K (22, e) 3 (p2(V2). Let's continue in this way. Then we get
V1, V2, ,Vn-1 as open neighborhoods of 1. Let V = n V). Then V is also an open
neighborhood of 1 and U = V n D is a nonempty open subset of D. For A e U fixed,
A e V,- for every i, and thus (A,tp1(A)), (A,q>2(A)), ..., (A,(p,,-1(A)) are zeros of F. This
means that (p1(A), (p2(A), ,(pn-1(A) are roots of (2.10). Since (1).-(A) e K (2,, e), we have
(pi(A) e E. In addition to that, (not) ¥= (pj(A), ifi aé j, so that we have exactly n - 1 roots
outside 5. The only other root is z = A. Thus we found a nonempty open set U of D such

that foreachAe U,z"+az-A"-aA=0 has exactly onerootinD:z=A.

Gent-431m: a=rei9,0<r<1.
For such a, choose A0 = exp(i 53;). Then A()"'1 = em, and hence

|A0"‘1+al=l+lal

sothat le"+aAol=le(Ao"'l+a)l=1+|a|.

We will investigate the equation

2" + az = A0" + (1 A0 (2.12)

47

IfzeD,then lzn+az|<1+|al=len+aAol,sothereisnosolutionof
(2.12)inD.

If 2 6 8D satisfies the above equation, then
1+la|=|Ao”+aAol=|zn+azlS|znl+la Ilz|=1+|a I.

This means that equality holds in the triangle inequality so that

arg(z"+az)=argaz+2k1t=9+argz+2k7t,ke Z.

On the other hand,

arg(z"+az)=arg (Ao"+aAo)=arg(exp(ii%)+rexp(in%+i6))=ief.

Comparing these two quantities , we get

6+argz+2k1t=£fel

and therefore org 2 = ’fi - 2k1t , which means that z = A0. Hence, we have proved that

z = A0 is the only solution of (2.12) in 13'.

Since (2" + az - Aon-aAo)'(Ao) = nAon‘l + a, and lal S 1 it follows that
(2" + az - A0" - aAo)'(Ao) at: 0, so that z = A0 is a simple zero. Again, by the same
reasoning as above, the Implicit Function Theorem gives a nonempty open set U in D such
that for each A e U , z" + az - A" - aA=O has exactly one root in D.

Let's fix one A e U. If g 6 Range Th-h(;() it is clear that g( A) = 0. On the other
hand, if g 6 L2 and g(A) = 0 , then g(z) = (z - A)u(z) , where u 6 L3. We can write g as

 

48

_ _ _ , u(Z)
g(z)—(z"+az A" 0A) zn‘1+Az"'2+---+A"'1+a .

 

By the previous discussion, the polynomial Z":1 + Az’i'2 + + A"'1 + a has no zeros in D,
u(z)

so it is bounded away from 0 in D. Hence,
2’“1 + Az"‘2 + + An'1+ a

2
6 La, so

that g 6 Range T h-h(7~)~ In other words, we have proved that Range T h-MA) = { g 6 L31 :

g(A) = 0 } if A e U. Thus for A e U, S' k) =fa)k),, where f is some function defined
on U. This leads to the conclusion that S e { T, }' so that S = TV, for some

WEE”.-

After proving this theorem one may ask if this method can help us describing the
commutant of a Toeplitz operator whose symbol is a polynomial. It turns out that the same
method works in this situation, under some additional conditions. The nicest result can be
obtained under the assumption that all polynomial coefficients are nonnegative. In order to
apply the same reasoning, we should look at the proof of the previous theorem more
carefully. The crucial point in the proof was observing that the equation z" + az = 1 + a ,
with 0 < a < 1, has n distinct zeros. If we let p(z) = z" + az, we needed that p(z) - p(l)
has n distinct zeros. Then we could apply the Implicit Function Theorem. In light of this,

we can understand the following theorem:

Theorem 2.8. Suppose p(z) = z + agz2 + + anz", where ai 20, for

i = 2,...,n. pr(z) - p(l) has n distinct zeros, then [Tpl' = {TV : ‘V e H°°}.
PROOF: Consider the equation p(z) - p(l) = 0 , i.e.,

z + 0222 + +anz" = 1 + a2 + + an_ (2.13)

 

49

For 2 e D , I z + a2z2 + +a,,z I < 1 + a2 + + an . and consequently there is

no solution of (2.13) in D.
Ifz e 30 satisfies (2.13), then

1+a2+m+an=lz+a222+ ---+a,,z"|$|z|+lazilzzI+--~+|a,,||z"|
s 1 +a2+--- +a,,.

Therefore, 2, 22, ..., z" are linearly dependent. Hence arg (z + 0222 + +a,,z" ) =
= org 2 + 2km

On the other hand, 2 + 0222 + +a,,z is positive so that arg z = 2m1t, showing
that z = 1. Therefore 2 = 1 is the only solution of 2.13 in D. Other zeros 21, 22, ---, 2,-1 are
outside D. Let's choose 6 > 0 small enough such that K (z,, e) n D = 0 for every i, and
K(zi, e) n K(zj-, 8) = (0, for i #j. Define a function F: C xC -) C by

F (A,z) = z + a2z2 + + (1,2 - A - a2A2- - anA".

Then z —) F (1,2) has n - l zeros outside 5. After taking partial derivatives of F with

respect to 2, we get g;- = p'(z) , so that for every i

1.21) =P'(Zi) = [19(2) - 19(1) 100 r 0,

31%

because of the assumption that all zeros ofp(z) - p(l) are simple. By the Implicit Function

Theorem, we can find a nonempty open set U in D such that for each A e U, p(z) - p(A) =0
has exactly one root in D. By the same argument as before, S = TV, for some

we H°°.-

 

 

50
Corollary 2.9. Suppose p(z) = z + azz2 + + anz", where a,- 2 O, for

i= 2,...,n. pr'(z) has no zeros outside D, then {T,,}' = ”W : w e H°°}.

If p(z) is a polynomial with complex coefficients, then we need very strong

restrictions on coefficients , in order to obtain similar result. We'll omit that case.

Remark: A reader should realize that proofs of Theorem 2.8, Corollary 2.9, and
Theorem 2.10 are valid for both the Bergman and Hardy spaces.
We'll conclude this chapter with several examples, where we use the same

technique.

a-z

 

, where as D is fixed and 26 D. We

Example: Suppose h(z) = z .

l-dz

 

. . a -z . .
want to find the commutant of T,.. The function (pa(z) = is an analytic

l-a_z

automorphism of D that is inverse to itself, i.e.,

‘Pa ((Pa (2)) = Z (2.14)

for z e D . Suppose g 6 Range Th-h(;(), for some A e D. Then

g(z)=[z-a’z J..a-A ]f= (z-A)[a-z-A+azA]f

1- dz 1- EA (1- Ez)(l - 5A)

 

 

 

for some f 6 L2. This fraction is equal 0 if z = A or z = a , so that we conclude

l-EiA

51

 

(“‘7')=0I.
1-dA

In other words Ker T 54,3) = span {k1, k%(l)}_. If S commutes with Th, then

Range Th-h(),) = { g 6 L3, 2800:19

S‘k), e Ker Tim) for every A e D. Thus

s M =flMkA + 898th» .

for some functions f and g defined on D. Now
(S a)(i») =f(A) 140») + 20») u(<po 00).
Therefore,
S u =f- u + g - (u°(pa) = (Tf+ Tg C(M) )(u) , (2.15)

where C‘Pa :Lfz —) L3, is the composition operator defined by CW, = uwpa. This shows

what an operator in [ T), }' looks like. Suppose, in addition to this, that S e ”I. Let's

calculate S T, - T, S:
(STz-TzS)(u)=S(z-u)-z-(Su)=z-f-u+g-[(z-u)°<po]-z'f-u-z-g~(u°<po)=
=8' “90' Z) ' (“"90).

In other words, S T, - T, S = T3090 - ,) Cw. In Theorem 2.4, we proved that S T, - T, S

must be compact, so that T8(<Pa - ,) C,“ is compact. Note that C¢a'1 = C4,“ (because of

 

52
(2.14)) and that implies that T8(‘Pa - ,) is compact. A multiplication operator Tf( i.e.,

f e H °° ) has spectrum equal to f(D) and can be compact only if f = 0, because the
spectrum of a compact operator must be countable. Hence g - (q), - z) = 0. If a = 0, then

h(z) = - 22 and we already proved that S must be an analytic Toeplitz operator. If a :15 O,
<Pa(Z) if 2, so that g = 0. Then (2.15) gives that
S u = f . u

for u 6 L%,. In particular, f = S 1 e L?! so that S must be an analytic Toeplitz operator.

Example: For fixed w e D, let h(z) = kw(z). Then g 6 Range Tkw - kwo.)

 

 

 

implies
(z - A)(2W - W22 - WA)
(1 — wz)2(1-WA)
The numerator is equal toOif z=Aor z— 2:39)”. Inthe latter case
W
Izlz (2"'WM)> 1 >1
wl lwl

so that Range Tkw - kw(7~) = [ g 6 L3, : g(A) = 0 } , and as before, if S commutes with

Tkw, it must be an analytic Toeplitz operator.

Examplg: Let h(z) =§-:%, a IE b , lb l > 1. We want to find the commutant of

Th. If g 6 Range Th-h(),) then

53
= (z - A)(a - b)
(2 - bxx - b)

 

8(2) f.

Hence, RangeTh-h(),)=(ge Lg:g(A)=0}and { Th }'={Tw:\ye H°°}.

We conclude this chapter with a question related to Theorem 2.4. Suppose
s e Lag) is such thatS T, - T, s e :x. Then s T,. - Tan e at, for every n. If

S T zn - Tzn S = 0 for some n, then Theorem 2.4 shows that S must be an analytic Toeplitz
operator. Now, the same assumption S T, - T, S 6 IX implies that S Tp - Tp S 6 CK for
every polynomial p. If S T p - Tp S = 0 for some polynomial p (other than 2"), must S be an
analytic Toeplitz operator ?

CHAPTER 3

VON NEUMANN ALGEBRAS GENERATED BY Tzn

In the previous chapter we studied the commutant of an operator Tzn acting on L2
and we found its intersection with the algebra '1'. Now we are interested in
characterizing the von Neumann algebra generated by Tzn, denoted by W*(Tzn).
Because W“ (Tzn) = { T Zn, T} 1", our problem is again related to the problem of finding
commutants of certain operators. In this chapter we will study operators Tzn acting on the
Bergman and Hardy spaces and show that von Neumann algebras generated by them have

different structure if n > 1.

1. Tzn acts on the Bergman space

We will prove the following theorem:

Theorem 3.1. Let n e N be fixed. Then
0

T1

W*(Tz,,) = ( T e Lag): T = , where L3, = X0 e X1 69 e X,-1,

T

n-l

T,- :Xi —9X,°for i = 0,1, , n-l}.

This theorem follows easily from the following theorem:

Theorem 3.2. LetS e Lag), and let n e N be fixed. Then s e { T,., T}. 1'

ifand onlyif
n-l
Sf=aofo+alfl +-"+an-lfn-l .at' E C. f= 2ft. fi 6 Xi .
i=0

54

55

PROOF: Let s be in ( T,., T}. }'. Then s T,.. = Tan and s T}. = T}. s. This

means that S, 5* e { tht }'. From Chapter 2, Theorem 2.4 we know that

Sf=gofo+ fir. + fife +-~+fi:%f..-r

and

A A A
* -l
s f=§ofo+ firm fif2+ ---+fi.—,f.-r

for fe L3,, where gj=5 ziandg} = S*zi,j = 0, 1, , n-l. We want to show that all g,-
and g} are polynomials.

Let's start with go.
<go, zkn> = <5 1, zkn> = <1, S*z""> = <1, go zkn>
so that<go,zk">=0ifk>0,i.e.,go .1. {ekn:k>0 }. Similarly
<go, zkn+1> = <5 1, zkn+1> = <1, S‘zk"+1> = <1, £1 zkn>

so that <go, zkn+1> = 0 ifk > 0, i. e. go 1 { elm.“ : k > 0 }. Continuing this procedure

gives that go 1 { ekn+j1k > 0 ] forj = 0, 1,, n-l. Therefore

n-l
so = 2 akek
k=0

i.e., go is a polynomial of degree at most n - 1.

 

.56

By the exactly same argument, go is a polynomial of degree at most n - 1.
Now look at g1. Similarly as before we have

<g1, z""> = <5 2, Z”) = <z, S*zk"> = <z, £0 zkn>
so that <g1,zk">=0ifk>0, i. e.,gl .L {elm : k>0 }. Also
<g1, zkn+1> = <5 2, zkn+1> = <z, S*z""+1> = <z, g1 zk">

so that <g1, zkn+1> = 0 ifk > O, i. e., g1 J. { e10.“ : k > O }. Continuing this procedure

gives that gl 1 {eknfl- : k > 0 } forj = 0, 1,, n-l. Therefore

n-l
31 = 2 after
k=0

i.e., g1 is a polynomial of degree at most n - 1. By the same argument, £1 is a polynomial

of degree 5 n - 1. In this way we can show that all g; and g} are polynomials of degree at

most n -1. Hence,

Sf=pofo + fin + fife + +fi3—if.-.

and

 

A A A
* __I\ 22 Pn-l
s f—pofo+ 3291+ z,fat + z,,.,f...r . (3.1)

We will observe these polynomials more closely. Let

57

p,- = 06 + aiz + agz2 + + anflzn‘l
13,-: b6 + biz + béoz2 + + bnflz’l'l ,

for i = 0, 1, ..., n - 1. Let f0 = a0 + anz" + 112,22" + be an arbitrary function in X0.

Look at the following scalar products:

<SfOt 2""> = <f0. 5*zk"> = <fo.130 zkn> = <fo. b8 zkn> = b8 aka krill

 

<Sfo. 2""“> = sfo.S*2""+1> = <fo. 1'31 Z""> = <f0,b(1)zkn> = bi) “k" krill

* A - - 1C
<Sf0, ZION-"4) = #09 S zkn+n-l> = 40: Pn-l 2k") = bnol akn [01+].

Thus

S fo = 2 (5 f0. ek)ek = 2 (Sfo. ekn)ekn +
i=0 k=0

2 (SfOt ekn+l)ekn+l '1‘ + Z (Sf09 ekn+n-l)ekn+n-1 =
k=0 15:0

- °° -1 kn+2 -2 z kn+3
= b8 2 aknzkn + b0 ”+1 aknzkn+l '1' Do le'l aknzkn+2 + ... +
"0 i=0 k=0

 

 

(ff—122 aknzk"+"'1 (3.2)

-n-1
+ b 0 kn+l
I50

On the other hand, S f() = pofo, as was shown in (3.1). Now

58

n—l

PO= 21 (p0. elder = 21 (S 1 ek)ek= "212301 S‘Z")2"= I‘—+1—(1,pft)2"=
k=0

n-l
=2 (k+l)5’52k.
k=0

Thus

SfO = (58 + 21362 + 3b(2)22 + + nbnélzn'1)f0 =

= b8 3; aknzkn + b3 2 2a,,zkn+1 + 133 2 3aknzkn+2 + + 5'31 2 naknzk"+"'1 (3.3)
k=0 k=0 [1:0 k=0

Equating (3.2) and (3.3) , we get

 

'1kn+2 ' kn+3 - k+l n " -
b0 km on," = b}, zoom, og— km a)“: b2 3a,, , b"01£kn+—i 01k, = b"01 not)”,

for all k. The function fo was arbitrary, so that the above equalities are true for all a's.

This forces

b0: b0: =b";,1=0

so that p0 = b8 = constant. Therefore S fo = constant - fo .
Let f1 = alz + an+1zn+1 + a2,+1z2"+1 + be an arbitrary function in X1. Look at

the following scalar products:
<Sfl. zkn>= <fl S z"">= <f1 p0 Z"”>= Sfl b0 Z""+1>= blalwl— Ita+2

 

(Sflt zkn+l> = (fl, S*an+l> = <f1,$12kn> = (fl, b} an+l> = bl aka-1'1 kn+2

 

59

(Sfl zkn+n-l>_ Sf] S zk"+"'1>=<f1pn-1zk">=bnilakn+1—kn+2

Then

Sfl = 2(Sf1. ekn)ekn '1' 2(Sf1. ekn+l)ekn+1 + ‘1' 2 (Sfltekn+n-1)ekn+n-l =
[0:0 k=0

                

k=0
=b(1) -12 akn+lzkn+1 + b2 2 iii—3 alcni-lzkn'1'2 + '1‘
k=0 *0 k=0
" - k+l n
+ bnll (7532— akn+1zkn+”‘1. (3.4)
k=0

On the other hand Sf1 = fifl. so that

n-1 n-1
P1 - 2 (Pl. 6/081 = "Z (S 2 ek)elt

—l
=zk—fl(z S z")2"= 21k _(2 13102" =ijlzk

k=0

Thus

Sfl =%f1 =(—21"+-b%z+3-21-z2 +... +nTzn-1)§=

 

 

 

= D? 2 5 akn+lzkn “I" D% 2 akn+lzkn+l '1' b? 2 '5 akn+12kn+2 + ...
It=o k=0 k=0

+ 3";1 2 "5 akn+lzk"+"'1 . (3.5)
k=0

Comparing (3.4) and (3.5) we get (similarly as above)
b9=b§ = =bni1 =0

so that p1 = biz = constant -z. Therefore S f1 = constant -f1.
Contining in this way, we get pk = constant ~z", so that S f}, = constant - fk ,

where

k=0, 1, ,n - 1. Hence

Sf=aofo+a1f1+...+a,,-1f,,-1, ai e C.

Thus, we have proved that { Tzn, T :n }' is a subset of

1

,_
{S e L<L%,):Sf=aofo+arfr +... +a..-tf..-r . a. e C. f= 2f.- 1.
k=0

On the other hand, if S belongs to this set, then

5 Tznf= 5 T2" (f0 +f1 + +fn-l ) = S (2% + 2an + + z"f,,-1) =

= aoznfo + alznfl + + an-1z"fn-1 (since znfi e X,- ) = z" Sf= Tzn Sf

61

for eachfe L3,. Since S*f= Eofo + 51 f1 + + a-,-1f,,-1 , then 5* Tzn = Tzn 5* so

that S e { T Zn, T21: }'. Hence, we proved Theorem 3.2.

The second step in our study is to find the commutant of
1

,_
{S e £(LZ):Sf=aofo+alf1 +... +aa-lfn-l . 0i e C. f= zfi},1.6-,W*(T,n)-
M

It is readily seen that

0
W*(Tz,,) = { Te L(L%):T={r T1. ,whereLg =Xo O X1 O O Xn-1,
T

n-l

Tiin-9 Xifor i = 0,1, ..., n-1}

and Theorem 3.1 is proved.

62

2. Tzn acts on the Hardy space

We will prove a theorem that is analogous to the Theorem 3.1 in the Hardy space

case. Here is our result. We don't know if this theorem is already known.

Theorem 3.3. Let n e N be fixed. Then
To

M T '-
W*(Tzn) = { Te £(H2(3D)) : T = 2 0M2. , where
Mzn-ITOMgn-l
H2(8D) = X0 O X1 O O Xn-1, To: X0 -—> X0, M2 and M; are multiplication

operators}.
This theorem will follow from the following theorem:

Theorem 3.4. LetS e £(H2(BD)), and letn e N be fixed. Then
S e { Tzn, T2; }' ifand only if

Sf=pofo + fifr + fife + +fi=1‘-f.-r .

n—l
where p(), p1, , pn-1 are polynomials of degree at most n - 1, where f = 2ft. with
k=0

fie Xi.

PROOF: Suppose S e { Tzn, T} }'. By the same argument as for the Bergman

space,

Sf=pofo+ fif. + fife + atfillfnh (3.6)

63

where p0 , p1 , , p,-1 are polynomials of degree at most n - 1.
Suppose, conversely, that S is an operator of the form (3.6).
A A A

. '1' .~ - .
Claim: 5 f=Pofo+ fif. + fife + +fiT‘faa. r“).- are polynomials.

Let p,- = a6 + a‘iz + aéz2 + + anf'lzfl'l , and letfo = a0 + anz" + 012,22" + be

an arbitrary function in X0. Then

<S*fo, zkn> = <fo. s ,1...) = <fo. pozkn> = 58 an.

<5. f0. zk"""‘1> = <fo. S 2""+"'1> = <f0. tin-12""l> = 5"(31 aka

Then

Satfo = Z (S*fo. ekn)ekn + 2 (S‘fo. ekn+l)ekn+l + + 2 (S70. ekn+n-l)ekn+n-l =
k=0 k=0 It=o

= (.18 Z aknz’m + a}, 2 aknz’m+1 + 08 2 aknzkn+2 + + 0’51 2 aknzkn+"'1
k=0 1c=o ‘ i=0 k=0

- - - - _ A
= (08 + a(l)z 4» a322 + + anolzn'1)fo =pofo .

Let f1 = (112 + an+1zn+1 + a2n+1zzn+1 + be an arbitrary function in X1. Look at the

following scalar products:

<S*fr. ,1...) = <fl. s zk~> = <fr. pozkn> = a? akn+1 .

 

, _
<5 fl. 2""+"'1> = <fl. S 2""+"'1> = <fl. Pn-lzk”> = anil alcn+l
Thus

571: 2 ($71: ekn)ekn + 2 (S171, ekn+l)ekn+l ‘1' '1' z (571: ekn+n-l)ekn+n-l =
Ic=0 k=0 k=0

-0 °° _ °° _ oo - - 00
= 01 2 akn‘i’lzkn + 0% Z akn+lzkn+l + 0% 2 akn+12kn+2 '1' ... + anll z akn+lzkn+n-1
k=0 k=0 k=0 k=0

'0 .... _ co
_1 2 akn+lzkn+l + 0% 2 akn+lzkn+l

zk=0 k=0

9

+ a%z 2 akn+1zkn+1 + + a"ilz"'2 2 “’01“,an =
k=0 It=0

-0 A
a — — —
= (~z—1-I-ai +a%z + + anilz”2 )f1=ezlf1.

A
Ifwe continue in this way , we'll get S'fi= gfi , for i = 0, 1, , n-l , where 13,-

are polynomials of degree at most n - 1. Hence we have proved the claim.

Now, it is easy to see that S and 5* both commute with Tzn and we proved the

theorem.
Our goal is to determine ( T2», T} }". Suppose that T e { Tzn, T}: }". Then T

commutes with all operators S of the form

Sf=pofo + fifr + fife + + fij—ff.-. .

 

65

where p0, p1, , p,.1 are polynomials, by Theorem 3.4. In particular, T commutes with

operators of the form

Sf=aofo+a1f1+...+a,,-1f,,-1, a,- E C.

0
T1

Thus T= , where H’ZQD) =X0OX1 O OX,“ and T,- :X,- —) X).

T

n-l

Choose S such that S f = pofo , for any f e H2(8D) and such that

O 0

p0 = do ‘1‘ all + agzz + ... ‘1' an9lzn-l.

Thus
ST f = P0 (Tofo) = 08(Tofo) + 092(Tofo) + 0322(Tofo) + + 0,912"'1(Tofo).
On the other hand, this is equal to
TSf= T (pofo) = a8 (Tofo) + a? T1 (zfo) + a‘,’ T2 (2%) + + (1,9, m (zn-lfo).
After comparing these two expressions, we get

2(Tofo) = Tl (2fo).
22(Tofo) = T2 (zzfo).

2"‘1(Tofo) = Tn-l (2"'1fo).

66

or, using multiplication operators, we obtain

M, To = T1 M, lxo

Mzn-l To = Tn-1 Mzfl‘l/XO o

Letfl = alz + anslel + a2n+1z2M1 + = 2(a1 + an+lzn + a2n+122fl + ...) be an

arbitrary function inXl. Denote fi = a1 + anrlz” + 012,...122" + 6 X0. Hence,
Tlf1= T1 (21“.) = Tl M. Inf. = M. Tofi = M. To Mifl
so that
T1 = M, To M}.
Let f2 = (ltzz2 + i)i,,+2z"+2 + a2n+2zzn+2 + = 22(012 + amtzzn + 012,922" + ...) be an

arbitrary function ian. Denote fé = a1 + an+1zn + a2n+122" + 6 X0. As before,

we have
T2f2 = T2 (221%) = T2 Mz2/xofi = M.2 Tori = M.2 To M22f2.
and therefore
T2 = M,2 To Mp.

Continuing this way, we finally get

67

Tn-1=Mzn-1 To Min-1.

Thus

To
MzTOM?

Mzn- lToMgn-l

Conversely, suppose T is of the above form. Let S be as in (3.6). We want to

show that T commutes with S. First,

5 Tf= ST (fo +fl + +fn-l) = S (Tofo + Mz To Mifl + + Man-1 To Min-lfn-1)=

po(Tofo) + pl(To Mifl) + + Pn-1(T0 Mrfn-l )-

On the other hand,

TSf= T (pofo + fifr + fife + + ”"—‘,‘f...) =

2’”
_ 0 1- 2-2 n-l —,,-1
—To(a0fo+aozf1+aozf2+...+a0 z f,-1)+

+ M, To M,(a(1)zfo + a}f1+ a? 'zfg + + a"i12"'2f,,-1)+

+ M,n—l To Mgn-1(an9l zn'lfo + 0,111 zn'zfl + “’31 z"'3f2 + + affl fn.1) =
a8<Tofo> + aim) szr) + + a".3‘(ToMzn-if.-t) +
a? 2(Tofo) + aizab Mifl) + + a"i‘z(To Min-lfn-l) +

(3.7)

68

0,31 zn'1(Tofo) + anhzn'lao M, f1) + + 02:}23'100 Min-lfn_1) =

P0(T0f0) +P1(T0 Mifl) + + Pn-1(TO MEfn-l ). (3.8)
Comparing (3.7) and (3.8), we get that

ST=TS.

Hence T e { Tzn, T2; }". Thus, we proved the following:

To
* n ° - MZTOM-Z-
Te {Tzn, Tzn}1fandonlylf T=

Mzn-ITOM‘z'n-l

Therefore, we proved Theorem 3.3.

If we compare theorems 3.1 and 3.3 we see that in both cases operators in W(Tz,)
have diagonal matrices. In the Bergman space case, each diagonal entry is an arbitrary
operator T,. while on the Hardy space each diagonal element is of the form Mzi-lToM-gi-l,

for some To. The difference comes from the coefficients of the orthonormal basis in the

Bergman space.

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