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presented by
Xiangf ei Zen;

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MSU Is An Affirmative Action/Equal Opportunity Institution
c:\clrd\datedue. pm3-p.1

TOEPLITZ OPERATORS ON BERGMAN SPACES

Xiangfei Zeng

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements

for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1992

x?

{f/‘i 5‘?

ABSTRACT
TOEPLI’I‘Z OPERATORS ON BERGMAN SPACES
By

Xiangfei Zeng

In this thesis, We study Toeplitz operators TI on the Bergman space LZ(D),
where D is the open unit disc and f is in C(D). We give a necessary and sufficient
condition for T, to be compact. We also prove that the commutator Tng — TgT;
is compact, and Show that the commutator ideal of the closed algebra generated by
{T1 :
abelianization of {T} : f E C(D)} is isomorphic to C(BD), and then use that result
to determine the essential spectrum of TI- Although the above results were known

for p = 2, the proofs in that case depend heavily on Hilbert space properties that do

f E C(D)} is equal to the ideal of compact operators. We show that the

not work for other values of p. Thus a number of new techniques must be introduced

for1<p<oo.

To my mother.

ACKNOWLEDGMENTS

I would like to express my deepest gratitude to my dissertation adviser, Professor
Sheldon Axler, who inspired me and guided me through each part of the dissertation.
I would also like to thank my dissertation committee members Professor William
Sledd, Professor Joel Shapiro, Professor Wade Ramey, and Professor Salehi Habib for

their time.

Contents

1 Introduction

2 Preliminary Results

3 Compact Toeplitz Operators
4 Hausdorff-Young Theorem

5 Compact Commutators

6 Commutator Ideal

7 Spectral Properties

10

14

17

iii

 

1 Introduction

Let D be the open unit disc of the complex plane C. Let dA be the normalized
two dimensional (Lebesgue) area measure on D. We will consider the Banach space
LP(D) (1 < p < 00) that consists of the complex—valued, measurable functions defined
on D that satisfies

/D W dA < 00.

For 1 < p < 00, the Bergman space L§(D) is the set of functions in LP(D) that
are analytic on D. It is known that the projection P with kernel (1 — 1172)"2 from
LP(D,dA) to LZ(D,dA) is bounded (See [1]). When p = 2, P is the orthogonal
projection. For f E L°°(D,dA), define the Toeplitz operator Tf by Tf(h) = P(fh).

In this thesis, we study the Toeplitz operator T; with the symbol f in C(D). The-
orem 3.4 gives the necessary and sufficient condition for TI to be compact. Theorem
4.1 describes a condition on the Taylor coefficients of the function f in L§(D), which
is used to prove that the commutator Tng — TgT; is compact (Proposition 5.3).

Let B be the set of bounded linear operators on L§(D). Let [C be the set of
compact operators in B. Let T denote the norm closed subalgebra of 8 generated by
{T1 : f E C(D)}. The commutator ideal J of T is defined to be the smallest norm
closed two sided ideal containing {TS — ST : T, S E T}. Proposition 6.1 shows that
J = K.

The last part of the thesis discusses the spectral properties of T!- An operator
T E B is called Fredholm if the kernel of T has finite dimension and the range of
T has finite codimension. The essential spectrum of T, denoted by 08(T), is defined
to be the set of complex numbers A such that T — /\ is not Fredholm. Theorem 7.9
gives a Banach algebra isomorphism between C(BD) and T/IC. This result is used

to prove that Ue(Tf) = f(aD) (Theorem 7.10).
Although the above results were known for p = 2 ([2]), the proofs in that case
depend heavily on Hilbert space properties that do not work for other values of p.

Thus a number of new techniques must be introduced for 1 < p < oo.

2 Preliminary Results

For w 6 D, define the reproducing kernel by

1
kw = —.
(z) (1 — 7.712;)2
The following Proposition and Theorem are proved in [1].
Proposition 2.1 Let 1 S p < 00, f E L§(D), and w E D. Then
_ ft?)
f(w) — [D (1 Wag/1(2). (1)

Proof. See Lemma 1.7, [1].

Corollary 2.2 Let 1 S p < 00. Then the point evaluation : f —> f(w) is bounded
on L§(D) for w E D and uniformly bounded on compact subsets of D.

Corollary 2.3 [fl S p < 00, then L§(D) is a Banach space.

For functions f, g measurable such that fg E L1(D), < f,g > is defined by

< f,g>= [Dad/1. (2)
Definition 2.4 Let] S p < 00, f E L§(D). Define Pf by
(Pom) =< M. >= [D (1%“(zi (3)

Theorem 2.5 Let 1 < p < 00. Then P is a bounded projection ofL”(D) onto LZ(D).

Proof. See Theorem 1.10, [1].

Theorem 2.6 Let 1 < p < 00. Then the dual ofLP(D) can be identified with LP'(D) .
More precisely, every bounded linear functional on L”(D) is of the form

f ——>/Df§dA

for some unique 9 E LP'(D). Furthermore, the norm of the linear functional on

LP(D) induced by g E L”'(D) is equivalent to ||g||,,; .
Proof. See Theorem 1.16, [1].

Definition 2.7 For f E L°°(D), the Toeplitz operator with symbolf is the operator
from L§(D) to L§(D) defined by

T1(g) = P(fg)- (4)
We denote the adjoint of T; by T}, which is defined such that
< T,h,g >=< )1,ng >, (5)
where h E L§(D), g E L§'(D).
Proposition 2.8 Let f, f1, f2 6 L°°(D). Then
Tf1+f2 = Tn + T1: (6)

and

T; = T}. (7)

Proof. (6) is obvious.

Let h E L§(D) and g E L§'(D). Then
" = Th = Th ‘dA 8
<h,T,g> < x,g> /D(;)g ()

3

.4.

 

/g( (w)(/[(%— )lsz(z))dA(w)

= /f(z ))h(z/—(1:_dz4w( )dA(z) (9)
=/ nah (z)(—d AM (10)
/DhTngA. (11)

The equation (9) is from Fubini’s Theorem, (10) follows from Proposition (2.1), (11)
is the same argument as (8) through (10).
Therefore ng = ng. Hence T] = Tf' QED

3 Compact Toeplitz Operators

Lemma 3.1 Let K be a compact subset ofD. Let f E L°°(D,dA) be such that f E 0
on D \ K. Then T, is a compact operator on L§(D)(1 < p < oo) .

Proof. Let {gn} be a bounded sequence in L§(D). By Corollary (2.2) {gn} is bounded
on each compact subset of D. So {gn} is a normal family. Hence there exists an ana-
lytic function g on D such that some subsequence {gnJ} of {gn} converges uniformly

on K to g. Thus {gnjf} converges in L§(D) to gf. Hence
T1(gn,) = P(fgn,) -’ P(fg)-

Therefore TI is compact. QED

IIZ

Lemma 3.2 Let 1 < p < 00. Then [[lcwllp 1 — lez)‘2/P', that is, there exist

positive constants C}, C2, such that
01(1-lwlz)‘2/p' S llkwllp S 02(1-|w|2)_2/p'

for every 11) E D.

Proof. See Lemma 3.10, [1]. QED
Lemma 3.3 law/“kw”p converges to 0 weakly in L§(D) as [w] ——-> 1.

Proof. If f E L§'(D), then by the reproducing property of km and the estimate of
"kw”p (See Lemma 3.1, [1]), we have

< fakw/llkwllp >3(1—lw|2)2/"'f(w)-

Thus if f is a bounded function in LZ'(D), then < f, lag/Ilka,”p >——» 0 as [w] ——+ 1.
Since polynomials are dense in L§’(D), this implies that < f, lay/”lea,”p >——» 0 as
[w] ——r 1 for all f E L§'(D), which means that kw/kallp converges to 0 weakly in
Lg(D,dA) as [w] ——. 1. QED

Theorem 3.4 Let f E C(D). Then T, is a compact operator on L§(D)(1 < p < 00)
If and only ifflap : 0.

Proof. <= Suppose f [30 = 0. Then f can be uniformly approximated by functions
with compact support in D. By the Lemma, T; is compact.

=> Suppose that T] is compact on L§(D). Let we 6 8D. By Lemma (3.3),
leg/”hm”p converges to 0 weakly in L§(D) as w —> wo. Hence

k kw

<Tf_‘”____>__.0 as w——>w0.
llkwllp’ llkwllp'

On the other hand, by Lemma 3.2 we have

kw k... flkwl2
|<T—,—>l = I —dA|
’llkwllp Ilkwllp' D Ilkwllpllkwllpt

I/D from —- lw|2)"’/"'”"’dA|
| [D ilkwrtl - IdeAI
| /D (fo «Md/1| —» |f(wo)|.

H2

as w ——» too.

This implies that f(wo) = 0. Therefore flap = 0.

5

4 Hausdorff-Young Theorem

The following theorem gives a necessary condition on the Taylor series of f in L§(D).

Theorem 4.1 (Hausdorff-Young Theorem) Forl < p S 2, let f 6 L§(D) have
Taylor expansion f(2) = 22:.0 akz". Then
00 pl l/p’

{gt —(k|:"1])p,_, ] s llfllp. (12)

Proof. Without loss of generality we assume that M f “p = 1.

We first suppose that f is a polynomial with degree ii. Let u denote the measure
on the set of non-negative integers such that u(k) = k + 1. Let l”(,u) denote the usual
space of sequences.

Suppose that b = (b0, b1, . . . , b", 0,0, . . .) is such that

[lblli’pm = 2 “help“? +1): 1.
k=0

We will show that

 

g 1. (13)

 

 

n ak TI

Zbk- -,1(k)[= Zbkak

k=0 k +1 k=0
Put F = |f|”, Bk = [bk|”, k = 0,1,2,...,n. There exist a function ()5 and

complex numbers ,80, B1,,62,. . . ,fln such that

f F <1», 1461 . /D < )
and
bk=B;/Pn. |flk|=1, Eamon <15)
Ic=0

Let wk(w)=wk. Then
a.=(t+1)/Df.,tde=(k+1)/DF1/p¢¢‘,.dA.

So
2 the, = Z Bat/Peak + 1) / Fl/Pddde.
k:0 D

k=0

6

Replace 1/p by z, and define
on) = z Bgdka +1)/D F’qfiz/ide. (16)
k=0
for any complex number 2. Then (I) is a entire function that is bounded on {2 : a S
Re(z) S b} for any finite a and b. We shall take a = % and b = 1, shall estimate
(1) on the edges of this strip, and then apply Phragmen-Lindeléf Theorem (See page
256, [6]) to estimate <l>(1/p).

For ——00 < y < 00, define
any) = [D FlflF‘wndA.

Since {dim/k + 1} is a orthonormal set in L2(D), by Bessel’s inequality we have

n

 

n > — 2
Zl6k(y)l2(k+1) = z / Wm ka—k+1dA
k=0 k=0 D
< lfl iy 2
_ /D|F F ¢| dA
= /FdA=1
D

and then the Schwarz inequality shows that
2 Brandi +1)c.(y)]
k=0

XnXBl/zBiyfltx/k +1)(Vk + 161410)]

k=0

n n lfl
[ZBk(k+1)-ZICt(y)I"(k+1)} s 1. (17)
k=0 k=0

1<I><§+iy>l =

 

 

|/\

The estimate

I‘P(1 +211)!

i BkBiyfldk + 1) / FF‘W/ide]
k=0 D

 

g ZBk(k+1)=1,—oo<y<oo. (18)
k=0

follows trivially from (14), (15), and (16).

We now conclude from (17), (18) and Phragmen-Lindelbf Theorem that
|<I>(x+iy)lsl (—3x<1,—oo<y3oo).
Let :c =1/p and y = 0 . We have

1
14%;» = 51.

 

n “I:
:bk-—-p(k)[ =
k=0 k+1

 

n
E bkak
k=0

 

Thus for each polynomial f = 22:0 akz" with ]]f||p = 1, (13) holds for every
b 6 [P(u) with ”b” = 1.

By the Hahn-Banach Theorem,
I /P' ' l/p'
n [aklp }1 { n [aklp }
— . k = —— < 1.
{Pour M) gourd -
As the Taylor series of any f converges in L§(D) ([8]), we have
I /p’
°° WI” 1
{g (k + ,,p,_. s 1111..

for all f E L§(D). QED

Corollary 4.2 The operator T on L§(D) defined by

00

 

at
(Tf)(z) = .2 1. +1.2", (19)
where
f(Z) = :0ka 6 MW)
k=0
is compact.

Proof. We first assume that l < p S 2. Define Tn by

 

II
[v]:
Q
,.
N
,.

(Tnf)(2)

oo

 

Let g(z)= Z2210 bkz" be a function of L” (D) with “9“,. = 1. Then

[/ (Tf)§dA] < f: Iakbkl <{i [akIP’ }1/p’{i [bklp }1/p
D n - =o(k+1)2" ,_o(k+1)r' t=o(k+1)”

l/P [bk [2 00 1/P
{gk+l+ "Pk: 0(lit-i-l)1"}

1 i,
s {i (22's,. }.,,. {191% + on”?
{ }
}

 

 

 

 

l/p

 

{/quv dA+1) +0]

°° lakl",
k=0

where 0,, Mp are constants.

Since (L:)"‘ E L§'(D) (Theorem 1.16, [1]) and g is an arbitrary unit vector in
L§'(D), the Hahn-Banach Theorem gives

00 pl l/P'
IITf||p_M{Z_:(klc_l:|1),} ,

k-O

 

for each n.

By Fatou’s Lemma, we have

00

. p, up,
IITfIIPSMp[Z(,J:'1),.} . <20)

k=0

 

Tn is a finite rank operator for each n and by (20) applied to

n
f — Z (1ka ,
k=0

we have

”(T T1111 < M i —'“—"'— W
n P — P k=n+1 (k+1)pl

0° , l/p'
MP 2 [ale]?
(12 + 1)‘/P' is.“ (k +1)?”

Mr
S WWII,» (21)

where (21) follows from Theorem 4.1.

Thus

M?
”(T- Tn)” S W ——* 0.

Therefore T is compact.

For 2 < p < oo, consider T“ on L§'(D). We have

0° akb-k
< ,T“ >=<T, >=
f 9 f9 gUHl)?

 

for all f E L§(D), g E L§'(D) . Hence T" = T.
By the above proof, T" is compact on L§'(D), which implies that T is compact.
QED

5 Compact Commutators

Definition 5.1 Let f E C(D). An operator Hf : L§(D) ——t LP(D) is called the
Hankel operator with symbol f if

Hm = (1— P)(fh)
for all h E L§(D).

Clearly H j is a bounded linear operator from L§(D) to LP(D).
Suppose that ¢(z) = Z. We will show that H4, is a compact operator. We need

the following calculation first. We assume that
:2,“ *e Lf;(D
Then

P(¢h)(w) = [D 2h(z)(1—w§)‘2dA

= lim 2h(z)Z(k+1)(w§)kd/1(z)

t—vl Dr k=0

10

= 11m: k+1)w [Wilton/1(2)

 

 

 

 

 

k=0 ‘
= l k 1 1‘ aka k+2 e-i(k+l)sd (11‘
img(+)w//h (goal. he)r s
°° k+1
= ék+20k+1wk
— oo -1 oo alc+l
_. gakw" - I; k-i— 2w". (22)
Therefore
(Heh)(2) = (l-P)(¢h)(z )= 511(2 )- P(¢h)(2)
:: 11(2) — Z aka—l +2116 ak+122
__ E M °° 2'11.
- zh(2)+ z +I§ik+k22k
: ]Z[2_1h(z) ligljki ak+lz
z k_0k+22k
= (Qh)(2) + (Th)(:)a (23)
where
(Qh)(z) = lzlzz'1h(z)+£9).
_ 0° ak+1
(Th)(z) _ gk+22k.

By the proof of Corollary 4.2, T is a compact operator on L§(D). Hence Q =
H¢ - T is a bounded linear operator from L§(D) to LP(D).

Proposition 5.2 Let ¢(z) = 2", then H), is compact.

Proof. By (23) we only need to show that Q is compact. It is sufficient to show that
if {hk} is a sequence of L§(D) and hn ——+ 0 weakly, then

“thllp ——’ 0'

11

Since hn —-> 0 weakly in L§(D), by Cauchy’s formula hn ———+ 0 uniformly on

compact subsets of D. Let c > 0 and

22—
f<z>=L'——1.
Z

We can pick 9 E C(D \ 0) such that g = 0 on a neighborhood of 6D, 9 = f on a
neighborhood 0 of O, and

||f - glioo < e. (24)

Let 0, be a small open disc in O centered at 0 and

 

K: {2 ED:g(z) #0}\o..
Then K is compact and bounded away from 0.

“(0)11. s no — at”. + 11(gh.><z> + W)

Z Z

 

||(th)(z)llp = ||(fhn)(2) +

 

“10- (25)

The Principle of Uniform Boundedness gives us that {hn} is norm bounded by
some M. So by (24)

”(f — g)hnlip S ”f — gllooilhnlip < 6M (26)

Now we want to Show

hn(0)

Z

 

||(ghn)(2) + llp -* 0- (27)

The partition of D gives

hi0) ”g = [D
h..(0)

= A [(ghnxz) + ‘2;—

+ h [(ghnxz) + £591

+/
D\(KUO)

An‘i’Bn‘l’Cn-

P

dA(z)

 

 

||(ghn)(z) + (912.1(2) + “2(0)

 

 

lDdA(2:)

 

P

dA(z)

 

P

dA(z)

 

(ghn)(z) + MO)

2

 

 

12

Since hn —+ 0 uniformly on compact subsets of D, it is clear that

 

 

 

 

 

13,, ——> 0,
and p
0.. = [BMW 112(0) dA(z) ———. o.
For r sufficiently small,
2 _ p
A. = [0. ELZ—l-hno) + Eli—0) dA(z)
= 2 + h"(z) ; IMO) pal/1(2)

 

 

|/\
o\~b\-

[lht(0)l + 0020]” dA(Z) < 6

7‘

since {h:,(0)} is bounded.

Thus we proved (27). By (25), (26) and (27) we have ”thllp ——) 0. Therefore Q
is a compact operator. This completes the proof that H4, 2 Q + T is compact. QED

Proposition 5.3 Letf andg be functions in C(D). Then ng—Tng and Tng—Tng

are compact operators on L§(D).

Proof. For f E C(D), define
H} , S; :LP(D) ——+ L”(D)
by
HM = P(fh). 5,1. =(1- P)(fh),
where P is the bounded projection from LP(D) onto Lf,’(D).
The operators defined above are bounded linear operators. Straightforward cal-
culations show that
T19 ’ TJTQ = Hng (28)
Hfg=Sng + Hng. (29)

13

Let
B = {f E C(D) : H] is compact}.
Clearly B is a closed subspace of C(D). (28) and (29) shows that B is a closed
subalgebra of C (D)
We know that H1 = Hz = 0, and by Proposition 5.2, 2 E B. Thus 1, z, 2 E B.
By Stone-Weierstrass Theorem, B = C(D). Therefore H I is compact for every

f e C(D).
If f,g E C(D), then T19 — TITg = H}Hg is compact. Consequently,
TITg - Tng = (TITy — ng) + (T19 - Tng)

is also compact. QED

6 Commutator Ideal

In this section, we will Show that the commutator ideal J of 7 (defined in Introduc-

tion) is equal to the set of compact operators K.
Proposition 6.1 J = K.

Proof. By Proposition 5.3, Tng — TgT; E K for all f, g E C(D). Hence J C K.

We need to show that K C J. Since the Taylor series of functions in L§(D)
converge in norm (see [8]), L§(D) has Schauder basis. Therefore the set of finite rank
operators is dense in K by the similar argument to the proof in Hilbert space case
(Theorem 4.4, Chapter 2, [3]). It is sufficient to Show that J contains the set of finite
rank operators.

The following three lemmas will finish the proof of Proposition 6.1.
Lemma 6.2 The operator A0 defined by Aof = a0 , where

f(2) = iakzk E L§(D).

k=0

isinJ.

14

Proof. By the calculation in (22), we have

 

(Tszf)(w) = iakwk— 0° leak

k=0 k=l

 

(Tszf)(w) = Zakwk— kak
k=1 k=0

Hence
(TiT— TMTM (=w) éao+z(k+l)(k+2)w wk. (30)

Let A = 2(T.T. - my). Then A e .7. By (30),

n _ 2nd]c
(A mm a°+Z( ((k+1)"k+2)" wk (31)

We want to show that ”AD - An“ ——> 0. Assume 1 < p S 2. Let

Z) = Z (1ka E LEW), llfllp =1

k=0

and
z) = Z W 6 L270), llgllp1=1.

k=0

By the proof of Corollary 4.2 and (31),

 

- °° [01¢ka 2"
A —A" (1A < .
'h‘ ° ”9 ' - EH1 (k+1)"(k+2)n

 

(3- ) (gift- tit)

< (g) (1+C)——>0.

|/\

By the Hahn-Banach Theorem,

2 n
||A0_ A"|| 5 (1+ C) (5) __. 0.

Since J is a closed ideal, A0 6 J for l < p S 2. For p > 2, we have the same
QED

result by interchanging f and g in above proof.

15

Lemma 6.3 Let 4),,(2) = z", n = 0,1,2, - - -. Then the operator

an

n+1,

 

Anf = 2/016. AA = f(2) = f: e Lam)
k=0
is in J.
Proof. We use induction on n.
For n = 0, A0 6 J by Lemma 6.2.
For n = 1, we will show that A1 = AOT; E J. By (22),

 

(Tzf)(w) = Zakwk-l — Z ah“ wk. (32)
k=l k=0 k + 2
Hence
Aomn = 5’21 = Alf.

Assume An 6 J, we want to show that An“ 6 J. By (32), we have

AnTZf : an+1 = An+1f-

L
n + 2
Thus

An+l : Aan E J

by the assumption. QED

Corollary 6.4 Let [1,, be a polynomial of degree n. Then the operator defined by
13.x = /D In M. f 6 15(1))

is in J.

Proof. Since J is an ideal and B,, is a linear combination of operators in Lemma

6.3, B" E J by Lemma 6.3. QED

Lemma 6.5 J contains all finite rank operators.

 

Proof. We first show that J contains rank one operators.
Let T be a rank one operator on LZ(D). Then there exist 9 E L§(D) and d) E
L§'(D) such that
T1: ¢(f)g. 41(1) =/Df¢3dA
Since the set of polynomials is dense in L§'(D), there exists a polynomial sequence

{pn} convergent to 45 in Liz/(D). Let

¢n(f) = /D find/1, m = ¢.(f)g.

Then T,l = T93” E J as a consequence of (in E J by Corollary 6.4.

We want to show that Tn ——1 T in norm.

IIT - Tnll = SUP llg(¢ — ¢n)(f)||
11111.51

S ||91|p||¢ — ¢n||p'

= Hgllp ||¢ - Pnllp' -> 0-

Thus T E J. So J contains all rank one operators. Therefore J contains all
finite rank operators since a finite rank operator is a linear combination of rank one
operators. QED

Now Proposition 6.1 follows easily from Lemma 6.5.

7 Spectral Properties

Consider the map a : C(D) ——1 T/K defined by a(f) = T; + K . By Proposition
5.3, a is a homomorphism, and hence its range is a subalgebra of T/K . The definition
of 7 implies that a(C(D)) is dense in T/K .

Let Z denote the set of functions in C(D) that are zero on 6D. By Theorem 3.4,
the kernel of a is precisely Z. Thus there is a homomorphism 61 from C(D)/Z into
T/K defined by

ci(f+Z) =Tf+K:.

17

*5...

Now 61 is an injective homomorphism. We will show that it is bounded below.

Theorem 7.1 The map it : C(D)/Z ——1 T/K is bounded below. That is, there
exists a C > 0 such that
lld(f+ Z)“ = ”T! +’C|| 2 C ||f+ 3” = CllfllaD, (33)
for every f E C(D).
The proof of Theorem 7.1 will require several lemmas.

Lemma 7.2 Iffor every f E C(aD) there«is a continuous extension off to D that
satisfies (33), then Theorem 7.1 holds.

Proof. For each g E C(D), consider the function glaD on 3D. Let g be the
continuous extension of glaD to D such that (33) holds.

Since g —§ 6 Z. Thus g = g in C(D)/Z.

Therefore that (33) holds for g implies that (33) holds for g. QED

Let {wn} C D such that wn —-) wo E (9D. Let k,, = kw", and

kn
an = —-
llknllp

Then ||gn||p = 1 and g,, ——1 0 weakly in L§(D) by Lemma 3.3.

Lemma 7.3 Let C be as in (33). Iffor every f E C(D), there exists a sequence
{wn} ofD with w,, ———i we 6 0D, such that

HTjgnllp Z CllfllaDa (34)
then Theorem 7.1 holds.

Proof. For every K E K, we have

 

 

IITJ+K Z l|(Tx+1\')gn||p
Z llTjgnllp— l|1\'9n||p
Z Cllf||60-||1\'gn|lp-

18

 

 

Since K is compact, ||Kg,,|]p —--> 0. Thus
“T; + K“ 2 Cllfllao-
So we have (33). QED
Lemma 7.4 Let C be as in (33). Iffor every f E C(D), there exists a sequence

{wn} ofD with w, ——+ we 6 6D, such that
C ”New

(1‘ lwnlzl2 ,

1 LMMI 2

D [l — wnil“ (35)

then Theorem 7.1 holds.

Proof. By Lemma 3.2 and (35), we have

 

 

 

 

k k
” f "’ ’11kn11. ||k,,||,,,

—1-—|<Tk k >|
111i”. “1,11,, ’ '“ "

s (1 — 1w.1212/P+2/P’ 1(T1k.1<w.11

= (1—lwn12)2|P(fkn)l

__ _ 22 f(2)kn(2)

— (1 1w.1) ]/D—(1—w,.2)2dA(z)

_ _ 22 L2)

_ (1 1111.1) /D|1_wn§|4dA(z)

2 01111....

Hence (34) holds. By Lemma 7.3, Theorem 7.1 follows. QED
For w E D, let <15“, be the function defined by

 

For w and z in D, the pseudo-hyperbolic distance d(w,z) between w and z is
defined by
(100.2) = l¢w(2)l-

19

 

 

For w E D and 0 < r < 1, the pseudo—hyperbolic disk D(w, r) with center w and
radius r is defined by

D(w,r) = {z E D: d(w,z) < r}.

Since of”, is a fractional linear transformation, the pseudo-hyperbolic disk D(w,r) is
also a Euclidean disk.

Simple calculation shows that (15“, preserves pseudo-hyperbolic distance
d(/\.Z) = d(¢w(x\), ¢w(2))
forall/\, 26D.

Lemma 7.5 Let w E D and let 0 < r < 1. Then

(w,r) I1 — quI‘1 — (1 — |w|2)2'

/ dA(z) r2
D

Proof. See Lemma 4.7, [1].

Lemma 7.6 Let u be a harmonic function in D such that ”all S 1 and u(0) = 0.
Then

[11(2)] S i arctan [2].
7r
Proof. See Lemma in [5].

Corollary 7.7 Let w and z in D. Let A > 0. Then there exists a 6 > 0 such that

for any f harmonic in D, we have

|f(w) — f(2)l S AllfllaD

whenever d(w, z) < 6 .

20

Proof. Suppose that B = (1531(2). Then by the harmonicity of f o qiw and Lemma
7.6 ,

|f(w)-f(z)1 = |f(0)—f(¢w(fl))l
s ”jeans;arctannm
= 1111130§arctand(w,z)
s 11111aoc.d(w.z).

where 01 > 0.
Let 6 = /\/C1. The proof is completed. QED

Lemma 7.8 There exists a constant C > 0 such that iffl E C(8D), then there is a

continuous extension f of f; to D that satisfies

f(z)
[D—dA(z) 2

[l- wnfl‘I

0 ”film

(1- 1w.12)2 (36)

 

for some sequence {wn} ofD with w, ———» wo 6 (9D.

Proof . Let f1 6 C(aD). We first assume that fl is real valued and f1(1) =
llflllaD-

We first consider the harmonic extension of f1 to D, still denoted by f1. Then ‘

|f1| S ||f1||aD in D. Since f1(1) > 0, there exists an a > 0 such that f1 2 0 in the

region B, shown in the Figure.

 

21

Let {wn} C (0,1) {'1 B, be such that w" —> 1, and

11m.) — 11015111111130
for each n.

By Corollary 7.7, there is a 6 > 0 ( independent of f1) such that

1
MW) - f1(wn)l S - ||f1||aD
for each w E D(wn, 6), where D(wn, 6) is the pseudo-hyperbolic disk centered at w,
with radius 6.

Thus for w E D(wmb), we have

111(w1—m111 s lf1(w)—f1(wn)|+|f1(wn)—f1(1)|
3 $111.11.... (37)

Since [[f1]|3D = f1(1), equation (37) gives

new) 2 1111.1131) (38)

for w E D(wmb).

By (38) we know there is some room along the border of B0, and Do (See the
Figure), so that we can define a continuous function f on D such that f = 0 on Do,
f = ft on 60. Ifl 3 111.1130 on D. f 2 0 on B... and f = ft on anD<w..61.

By the definition of f, equation (38) holds for f. The definition of f also implies
that

| Jfl—dmznz/ —f(z)——-dA(z)— flaw). (39)

D [1 — wn§|4 D(w,,,1s) |1— wnil“ A |1— wnil“
We will estimate the two terms separately.

By (38) and Lemma 7.5, the first term becomes

(w..,6) I1 — wn2|4 _ 2 D(w,..6) |1— wn2|4

1 Milan 2
5—(1—lwn]2)26 . (40)

 

 

 

 

The elementary inequality

1— 2wnrcosa + r2

wiZl—cosza, 0<r<1

gives us a boundary for the second term in the following calculation:

11(211 .1 AM S

A ll — wnzl“

|/\

|/\

where C1 > 0.
Now by (39), (40) and (42), we

 

 

[lfllaD 7r‘ /1—a /2:—°— rdtdr

[1 — w—,,re“|4

11111.02/_

|1— w rire'al“

 

rdr
2
HfHaD [.01 (1 — 2wnr cosa + r2w2)2
rdr
211x11... /_ (1—_S,—),
2oz
llfllab m
Cl ”New 3
have

f(2) f(2) |f(2)|
/o |1— wnil“ MM 2 /D(wn,6) |1— wnil“ (“(2) —./A [1— wnil“
1 f
2 562 Hfi)‘ — 0.111111...
_ llfllaD
- mééz- _H'Cl(l_lwnl))

Since C1 (1 — |w,,]2)2 ——> 0,

_1_2_ _ 22>_I_2
26 C1(1 |wn|) _36 ,

for n large.

Let C = ébz. Then (36) holds.

dA(z)

(42)

If fl is real and f1(wo) = ||f1||aD, then replacing {wn} by {wnwo} and applying
(36) to g(w) = f1(wow) gives us the desired result.

Now we consider complex function f1 = u1+iv1 E C(BD). Suppose that ”ulllap >
llflllaD/2. Then there is an extension u of u] satisfying (36) for some {wn} C D with
112,. ——» we 6 D.

Let U be an extension of v1 and let f = u + iv. We have

 

 

 

f(z) 2 u(z) 2
/D 11 _ 1.0.214 “(2) 2 h) 11 — 10.214 “(2)
> (Cllullan)2
— (1“ lwnlz)4
Czllfllép
2(1- 1w.12)4 '

If ||v1||aD > ||f1||3D/2, a similar argument shows that (43) still holds. QED
Since 6 bounded below implies that (I has closed range, we have proved that
a is an isomorphism from C(D)/Z onto T/IC. Furthermore, there is an obvious

isomorphism between C (8D) and C (D) / Z . Therefore we have the following theorem
Theorem 7.9 There is a Banach algebra isomorphism between C(BD) and T/IC.

Now we are ready to study the essential spectrum of T; for f E C(D). Let 08(Tf)

denote the essential spectrum of T]. Then
05(Tf) = {A E C : T; — )1 is not Fredholm} (43)
Theorem 7.10 Let f E C(D). Then U6(Tf) = f(0D).

Proof. If /\ ¢ f(8D), then f — A is invertible in C(BD). By Theorem 7.9,
T; — /\ + IC = T;_,\ + IC is invertible in T/IC. Thus TI — A + IC is invertible in B/IC.
By Atkinson’s Theorem )1 ¢ U6(Tf). Therefore 06(Tf) C f((9D).

If /\ E f(8D), then for some we 6 3D, f(wo) = A. We want to show that T; — /\

is not Fredholm operator.

Without loss of generality we can assume that f (1) = 0. We will show that T; is
not Fredholm operator.

Define the function sequence {gn} by

2+1
2

 

gn(z)=( )",n=1,2,3,---_

Then fgn ——-> 0 uniformly since f(1) = 0. Thus
TIT n = T19" —' 0- (44)

Suppose that T; is Fredholm. Then there exist an operator 5 E B and a compact
operator K such that
ST; = 1 + K. (45)
By (44), ST;Tgn ——-—) 0. Equation (45) gives

ST;T . = T9,, + KT," ——» 0.

So Tgn+K:——>0in T/IC.

By Theorem 7.9, there is an Banach algebra isomorphism (I) : T/IC -——-> C(BD)
such that

WT; + ’C) = flan-
Therefore
‘NTgn + ’C) = gnlaD —-* 0,

which contradicts g,,(1) = 1 . QED

In the second part of the proof above, we showed that if 0 E f(3D), then T; is
not Fredholm. We will further show that the same condition also implies that T; is
not bounded below.

Recall that D(w,r) is the pseudo-hyperbolic disk with center w E D and radius

25

 

Lemma 7.11 Let w e D and r 6 (0,1]. Then

kadA= 1— 22‘2” 1— ‘ 2p“‘d
[D(MI 1 < 1w1> hm' wzl A(z)

Proof. Recall that

 

w — z
¢w(z) = 1— tin
Hence
, (lwl2 - 1)

Thus we have

,, _ 4 z
[D(w,r)lkwI dA — L(w,r)l1—w2|2pdA( )
<1 — 1w121-P /D(W)1¢:.(z)1P-21¢:.(z>12dA(z)
= (1—1w121-P h...) 1¢i<¢w(z))1P-2dA(z),

where the last equality uses the change of variables formula. But

(1 — 7112)2

¢iu=(¢w(z)) m,

so we have

kwpdAz 1— w22—2p/ l—szP'4dAz .
[D(w,r)ll (11) W1 1 ()

QED
Proposition 7.12 Let f 6 C(13). If 0 E f(aD), then T; is not bounded below.

Proof. Without loss of generality we can assume that f(1) = 0.
For w E D, let kw(z) = 1/(1 — wz)2. Then [cw/“kw”p has norm 1.

Since

J—p)llp=

     

 

T

it is sufficient to show that

II— II
M: Ilp ,,

26

asw—vl.

D(w,r) still denote the pseudo-hyperbolic disk D(w, r) with center w and radius
r (0 < r <1). Then by Lemma 7.11,

 

 

 

 

 

 

lfkwl” = lfkwl" / Ifkwlpd
D ”kw”; D(w,r) “kwllg D\D(w,r) ”kw”;
lkwl” llfll”
S sup f / dA+ °°/ kadA
(ma' ') 01w) 111cm”; ”an: D\D(..,.)' '
S sup lf|+ llfll’;:(/ lkwIPdA—f lkwlpdA). (46)
D(w-T) “kn/”P D 00”.")

By Lemma 3.2 and Lemma 7.11, the second term of (46) becomes

11f112. , p z _ ~ ,,
”kw“; (Alkwl dA( ) /D(w,r)|kw| dA)

_ llfllgo _w22—2p —1T)22p-4 Z— —1T)22p_4 Z
‘11:..1110") (/D11 1 dA()/D(ovr)|1 1 «m»

 

 

s 11f11:.(/011 —wz12P-4dA<z>— [D )11—w21QP-4daz».

(0,r

Let c > 0. If p Z 2, there exists a 6 6 (0,1) (independent of w) such that for
r 6 (6,1), the above quantity is less than 6. Let w ——> 1. The first term of (46) goes
to 0:

sup 11(211 —» f(1) = 0.
D(w,r)
Therefore T; is not bounded below on L§(D) for p Z 2.
For 1 < p < 2, we want to prove that
1—- ”Md/12 —/ 1—- 2”“dA 0
[DI «021 1 1 0,0,)1 sz (z) —»
uniformly with respect to w as r ——1 1.

Let 9(2) 2 (1 — z)P—2, so g E LZ(D). Suppose that l

g(z) = f: anz".

11:0

27 1

Then
11g11 = ): —'“"' < co
2 ”:0 71 +1

and

2 2n
2dA=/ 2 2dA= r2 l_“"l+’"1 ,
(010,.) 191 D If(r2)l r :—
Therefore

_. — 2P-4 __ _ - 2112—4
All wzl dA(z) ANN) I1 wz] dA(z)

[D Ig( wz) )|2 dA(z )—/D Ig( (wrz) )zlerdA( )

_ Z— lan_l_2w 2n ”2% lanlzwznrz" __, 0

uniformly as r ——> 1, QED

References

[1] Axler, Sheldon. Bergman Spaces and their Operators, Surveys of Some Recent
Results in Operator Theory, Vol. 1, Pitman Research Notes in Math. 171, 1988.

[2] Axler, Sheldon, John Conway and Gerald McDonald. Toeplitz Operators 0n
Bergman Spaces, Can. J. Math., Vol. XXXIV, No. 2, 1982, pp. 466-483.

[3] Conway, John. A Course in Functional Analysis, Springer-Verlag, 1985.
[4] Duren, Peter. Theory of H 7’ Spaces, Academic Press, 1970.

[5] Heinz, Erhard. On One- To-One Harmonic Mappings, Pacific J. of Mathematics,
9 (1959), 101-105.

[6] Rudin, Walter. Real And Complex Analysis, Third Edition, McGraw-Hill Book
Company.

[7] Zhu, Kehe. Duality of Bloch Spaces and Norm Convergence of Taylor Series,

preprint.

[8] Zhu, Kehe. Duality of Bloch Spaces and Norm Convergence of Taylor Series,
preprint.

29

 

 

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