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dissertation entitled ,
MULTI PLIERS 0F BERGMAN SPACES
presented by
Kapila Rohan Attele
has been accepted towards fulfillment
of the requirements for
Ph.D. degree in Mathematics
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/ y/ "" 81/30

MULTIPLIERS OF BERGMAN SPACES

BY

Kapila Rohan Attele

A DISSERTATION

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

DOCTOR OF PHILOSOPHY

Department of Mathematics

1983

ABSTRACT

MULTIPLIERS OF BERGMAN SPACES

BY

Kapila Rohan Attele

Let W be a plane region and Lp(W) be the usual
Lebesgue p-space relative to Lebesgue area measure dm.
The space of analytic functions in LP(W), denoted by
L§(W), is called the Bergman p-space and its 'norm' is

given by

nmp klflpm o<p<~

sup lf(z)] if p=c°
zEW

A function w on W’ is said to be a multiplier of
L§(W) to Lq(W) if ¢L§(W) C Lq(W). In Chapter 1
multipliers on Bergman p-spaces are considered and the
following theorem is proved. Here D denotes the unit

disc in the complex plane.

Theorem 1.7. Let § be a positive rational number and

f be analytic on D. Let 0 < q < p < a and suppose

Kapila Rohan Attele

lflg L§(D) c Lq(D). Then ‘f‘g e Lr(D), where
1 l l

P r q
In chapter I, §9 we will show that a Toeplitz
operator with symbol f} where f is in L:(D), is
bounded as a map from L:(D) to L:(D) if and only if
f multiplies the Bloch space to itself: a characterization

of such multipliers is also obtained.

Let B be the space of Bloch functions and let
30: {f68lf’(rele)(l-r) 4 O as r 4 1]. There are
bounded analytic functions not in 60. (see Chapter 2. §3).

However we will prove the following theorem.

Theorem 2.3. Let f be a bounded analytic function on

 

the unit disc. Then f(re19)(l —r) 4 O for almost all

6 as r 4 1.

The class of all analytic functions f on the unit
disc for which the Hankel operator Hf’ is bounded from
L:(D) to (L:(D))‘ is considered. It is shown to
contain all BMOA functions and is proved to be

contained in the space of Bloch functions.

To my parents

Attelege and Juliet Mahendrasena

ii

ACKNOWLEDGEMENTS

I would like to express my deep gratitude to my
advisor, Sheldon Axler, for his guidance,encouragement,
and patience. I'm very grateful to Daniel Luecking and
William Sledd for many valuable conversations and to
Edward Ingraham for his advice and encouragement during

my career as a graduate student.

I wish to express my deep appreciation of the
friendship of Danke Li, Mahmoud Mohseni, Ali Haj-Jafar

and Riazi-Kermani.

Finally, I would like to thank Tammy Hatfield for

typing this manuscript.

iii

Preliminaries
Chapter 1
Chapter 2

Bibliography

TABLE OF

iv

CONTENTS
Page
. 1
. 11
41
62

PRELIMINARIES

Let W be a nonempty region in the complex plane
and let LP(W) be the usual Lebesgue p-space of complex
functions with domain W, relative to the Lebesgue two
dimensional area measure dm. For 0 < p g_w, let the
Bergman p-space be defined by L§(W) = LP(W) n H(W),
where H(W) is the space of analytic functions on W.

For fEL§(W) let

E
p

if 0<p<°

anp (IW If Pam)

sup |f(z)l if p=°°
26W

The class L:(W) of bounded analytic functions on
W is usually denoted by H°(W). Let 0 < p g_~ and let
{fnl be a cauchy sequence in L§(W). Then by using a
theorem of Hardy and Littlewood ([6] Chapter 3, Lemma 3.7),
one deduces the existence of f in H(W) such that
fn a f uniformly on compact sets. It follows that if

p 2_l then L§(W) is a Banach space, and that if

0 < p < 1 then L§(W) is an F-space.

L:(W) is a Hilbert space, with the inner product

<f.g>=j‘ fEdm
W

For each wéEW there exists a unique Kw in L:(W)

such that
f(w) = I f i. dm for each f in L2(W)
w w a

This Kw is called the reproducing kernel associated
with w. Let D denote the unit disc. When W = D, we

have Ka(z) = 1 for zFD and (16D.

(1 -5a)
Easy computations show that [zn./ 5-1—1- In 2 0}
forms an orthonormal basis for L:(D). Thus for f in

O
L2(D). if f(z) = Zia zn then
a O n

 

°° lanlz P
”flip. = ("E n+1)

Let P be the orthogonal projection from L2(W) onto

L:(W), so that

P(f)(G) = [W f Ea dm

Taking this as the definition of P(f) for each f in
LP(D), Zaharjuta and Judovic [ZJ] proved that P projects
Lp(D) onto L§(D) continuously for 1 < p < 0. Later
Burbea [Bu] extended this result by replacing the disc
with multiply connected regions whose boundary satisfy
certain smoothness conditions. An immediate consequence

of the quoted theorem of Zaharjuta and Judovic is that the

dual of L§(D) can be identified with L§(D) where

l < p < a and -+ = 1.

UN"
i-Qll"

The map P does not project L1(D) to L:(D)
continuously. However L1(D) can be continuously
projected on to L:(D) (Bers [Be]). In fact, it is

not hard to see that

 

R(f)(o) '21er (1'15‘2 )3 f(z) dm(z) for fCL1(D), oCD
D(1-o z)

continuously projects L1(D) onto L:(D). Consequently
if v is a harmonic function on D, then its harmonic
conjugate also belongs to L1(D). a fact used in

Chapter 1, Proposition 5.1.

Let f€FL§(D) and let {2 ] be any subset of

con n21
the zeros of f. In general ;‘(1-—Izn I) does not
converge, but Horowitz [H2] proved that 23(1 -Izn llz

does converge, from which it follows that0

9 Z-Zn Z-Zn
H(Z) = H (—-—=—) (2 -——=—)
1 1-—znz l-znz

converges uniformly on compact subsets of D, and so H

. . p
1s 1n La(D)

is analytic. Horowitz further proved that g

and that

1/p 1/p
(f lf/HIP dm) gcq Iflp dm) ,
D D

where c is a constant that depends only on p.

Let 0 < p < a. A Blaschke product B is said to
be a universal divisor of L§(D) if whenever geEL§(D)
and g is analytic, then EGFL:(D). McDonald and
Sundberg [MS] and Horowitz [H] proved that if B is a
universal divisor for L§(D) for some p > 0, then it
is a universal divisor for every p > O and that B is
a finite product of interpolating Blaschke products. The
product H need not be bounded on D, indeed it is not
possible to factor an arbitary L:(D) function into two
L:(D) functions where one of the L:(D) functions is
required to be nonvanishing [H2]. However Horowitz [H2]

proved the following factorization theorem.

. 1_ 1 1
1161'. P, P1,...pn > 0 With 5 - 51-4- °" 51:. If

f<EL§(D). then there exists f 'fn such that each

1..

Pi 1“
f.€L (D),f= nf., and
1 a i=1 1.

n pi p
iglnfinpi S C "f”p '

where the constant c depends only on p, p1,---pn and
n.

A positive measure u on D is said to be a Carleson
(p-q) measure on D if

1 1
q 2p and (jD Iflqdu)q geQ‘D lflpdm)p for all f€L§(D)

Carleson measures also arise in the study of Hp spaces

as well as Bergman spaces; see for example [G] . Carleson

measures in various contexts were studied by Hastings

[Ha], Stegenga [Sl], Oleinik-Pavlov [OP]. Cima and Wogen
[CW], and Luecking [Lul]. A version of a characterization
of Carleson measures is as follows [Lul]: Let Db be a
hyperbolic disc with center a and radius %, let q > p,
and let u be a positive measure on D. Then

1 1
(j‘D lflqdu)q g cl(j‘D Iflpdm)p for all f EL§(D)

if and only if

£9.
Luna) 3 czu -la12) P

Here c ,c are constants.

1 2

In chapter I we consider the multipliers of Bergman

Spaces and we prove the following theorem.

Theorem 1. Let 0 < q g_p < Q. Then f L§(D) ciLg(D)

. . r l 1 _ 1
1f and only if feELa(D) where 54-; — 5°

Let b be an analytic function on D and let
"bHB = sup Ib'(z)] (1-lzl2)4-lb(0)l. Let
26W
6 = [b €H(D) l nan < a]. The Banach space a is called

the space of Bloch functions.

The Bergman projection P :L°(D) 4 B is continuous
on L"(n) and the dual of L:(D) can be identified with

B.

In chapter I, §9, we will show that a Toeplitz
operator with symbol f' where f is in L:(D) is
bounded as a map from L:(D) to itself if and only if
f is a multiplier of the Bloch space; a characterization

of such multipliers is also obtained.

We now introduce the space of BMOA functions. Some
properties of Bloch and BMOA functions are presented in
an attempt to draw the reader's attention to some analogies

between the two spaces.

Let I be a subarc of rm. the unit circle. and let

1 . 1 i9 d6
f be an L CE) function. Let I(f) = f(e )
TIT II 2?

where ‘1‘ denotes the normalized Lebesgue measure of
I. A function in LIKE) is said to be in BMO if

sup I(lf-I(f)l) < a. See Garnett [G] or Sarason [Sa]
fir properties of BMO functions. A function f in
BMO is said to belong to vmo if I(lf—I(f)l) tends
to zero as [1‘ 4 O. A theorem of Fefferman [F] says
that if f is in BMO. then f is the sum of a
function in LQUE) and the harmonic conjugate of a
function in Lwflm). The subspace of functions in H2
whose boundary values are in BMO (respectively in

VMO) is denoted by BMOA (respectively VMOA).

Alternatively

BMOA

{fEHZ focp -f(o) = f-f(O) VoED}
Ill 0, ”HZ n ”Hz

{f €H2 I sup llf emu-Ha)” 2 < ”i
den H

_, zt-o ,
Here ¢a(z) — 1+d'z for zGD, oED. Now the Rlesz

projection maps L°('1r) to BMO [St] and BMOA is the

dual of H1.

Now let us take an alternative View of the space of

Bloch functions. Given b in B and a in D,
the ova wanna = no -b(O)H5 .

Moreover b belongs to B if and only if the family
[b«oma-b(o)] is a normal family [P]. It is not hard
to see that a family of bounded norm in L:(D) is a

normal family. Thus

5: [bEL:(D) I323 ”beta-mmnz < e}

Indeed, Bloch functions can be viewed as a space of
functions of "Bounded Mean Oscillation". To be more
precise let a be in D and 80 be a disc in D
with center a. Let f be in L1(D) and let

J‘Sfdm

o
So(f) =

m(Sa)

Write BMO(D) for the set of functions f on D such

that
sup sa(lf-sa(f)l) < on.

where the supremum is taken over all a in D and all

So whose radius is less than (l-—]ol). The space of

Bloch functions equals the analytic functions in BMO(D)
[MS]. It may be noted here that there is a Bloch function

in all Hp classes but not in BMOA. [CCS]

A Carleson square is a set of the form

Sh={relel1-h<r<l and '9-90‘<}2'1']

for some 0 < h < l and 60 ER . A well known theorem
of Feffermann [F] characterizes BMOA as analytic

functions on D which satisfy

2
I lf'(2)l (1-1212)dm(2) = 0(h)
Sh
We now present a similar characterization of Bloch
functions. An analytic function f is a Bloch function

if and only if

f lf’(z)lz(l-lzl2)dm(z) = 0(1—lo12)

Do

As defined earlier Do is a hyperbolic disc of center a

and radius %.

We sketch a proof here. Through a coordinate trans-

formation notice that I [f’(z)]2(l-—Izlz) dm(z) is
D

comparable to o

(1 - lull Ilwlgf Ig’tw)|2 dm(W) where g(W) =f MPGW) -f(a) .

Now since Ig’lz is subharmonic,

l9’(w)l2 dm(w) zglg'mflz, and hence if

i

legt

ID lf’(z) [2(1 - lzlzldm(z) 3 0(1 - [(112)
a

then f 65.

a
If bet? and b(z) =Zanzn, then by [AS]
0

“ 2
l ‘b’(2)'2d‘“(z’ = 7’ 2%: lanl2 in
We 0 2
s c nbn; .

Thus

l If’(z) 12(1 - [212)dm(z) g c(1 - [<11sz «>ch 4(a)”;

D
(I

= c(1 -la12>nfu;§

and the proof is complete.

In Chapter 2 Hankel operators on Bergman spaces are
studied. In analogy with a well known theorem of Nehari
on bounded Hankel operators on Hardy spaces and with a
theorem of Coifman, Rochberg and Weiss [CRW Theorem VIII’].
we conjecture that for f in L:(D). Hf-:L§(D) a (L§(D))4
is bounded if and only if f<EBMOA. For an explanation
of the notation see Chapter 2, §O.
i6

Let Eb = {£6566 :f(re ) (l-r) 4 O as r 4 1]. It

is well known that H‘9 c B 'but HQ ¢ Eb [ACP]. However

10
the following theorem is also proved in Chapter 2.

a 19
Theorem 2. Let féEH . Then f(re )(l-r) 4 O for

 

almost every 9 as r 4 l.

A final word about notation. The integral I f dm
W
will be usually written as just J'f. Also c will denote
a constant, not necessarily the same each time, and n

will always denote a positive integer.

CHAPTER 1

Let W be a plane region and let v be a function
on W. we say that v is a multiplier of L§(W) to
Lq (W) if VLE (W) c Lq (W). Then the multiplication
operator MV :1}: (W) 4 Lq (W) defined by Mv(f) = vf,
for f(EL§(W), is bounded. For completeness sake we
include a proof of this fact. Since L§(W) and Lq(W)
are F-spaces we may employ the Closed Graph Theorem to
prove the assertion. So let {fn} be a sequence in L§(W]
and let fn d f in L§(W), and suppose vfn 4 g in
Lq(W). We will prove that vf = 9 almost everywhere. By
a theorem of Hardy and Littlewood ([G] Chapter 3, Lemma 3.7),
there exists a constant Kp depending only on p such
that

l
]h(§) l gxpgfilfi) [A lhlpdm)P for all h ELEM) and all g ew,

where A is an open disc in ‘W with center g. When p;;l
we can take Kp= l as a simple consequence of the subhar-
monicity of Ihlp on A. The depth of the result lies
when p < l. Replacing h by fn-f in the above

inequality, we have

11

12

*Und

I(fn-fH’s’)‘ gxpgmlfi IA Ifn-flpdm)

l
1 p _
ng(m J‘W ‘fn-fl dm)p

Letting n a a, one deduces that fn a f pointwise on
W. On the other hand [vfn} has a subsequence that
converges pointwise a.e. It follows that vf = g a.e.
Thus the continuity of Mv is established. It follows
that there exists c such that

l l
(j‘ IVIqlgiq)q 3 MI ‘9‘p)p for all g 6 LEM)

§2. This section is devoted to some examples.

Example 1. Let D denote the unit disc. It is known
that there are unbounded multipliers from L:(D) to

L2(D). For example. let
s = [rele logr < 1. [9| < (1-r)2}

and let

 

69(2) -

l
- x (2) .
(1-12M S

Q
We will prove that COL:(D) CL2 (D) . Pick f(z) = 2 anzn
n=0
from L:(D). Then

13

J‘ W lflz am
D

1
SI 11—1—5. (I 2 Z) Ia I IamIrn+md9)rdr
O ‘-r I9I<(l—r) n.m29 n

1
g 2 Z Ia I Ia I I (l -r)rn+mdr
n.mZO n m o

Ianl Iaml
(n+m-I- 2) (n+m+1)

=22
n.m29

 

Ian! 1am!

Z \/2(n+l) \/2(m+l)
n.m20 (n+m+1Y

 

 

g’2

2 E Ianlz
g V
n;0 2(n4-15

If”: . a.e.n.

The last inequality is due to Hilbert [D]. Thus

Mcp :L:(D)l4 L2(D) is continuous. In fact one can

verify. through a computation quite similar to the one
above. that Mm is Hilbert-Schmidt and therefore compact.

The following example was pointed out to me by Sheldon Axler.

Let I = x5. Clearly M :L2(D) » L2 (D) is bounded.
V/1-IzI
Therefore TI :L:(D) a L:(D) defined by TI =

P M MW IL:(D) is compact, where P is the
V/l-IzI

orthogonal projection from L2(D) onto L:(D). On the
other hand the cluster set of w on 6D is {0.1}.

Compare this with Proposition 5 in [MS].

14

Example 2. Let D' = [265C I O < IzI < l} and let

v(z) = log IzI for zeFD'. We claim that v L:(D')<:L2(D’).

Since L:(D') = L:(D) [ACM], given f in L:(D'), for
'6 a .9 e Ia I2
some {a I f(re1 ) = Z a rnenl and Z n <
n O n O rrtl

Also pick M such that Ilog rI2 r g_M for all r in

(0.1). Then,

1 2w i9 2
I I 1109 r f(re )1 r dr de
0 o

1 a
= 2w I Z) (log r)2 IanI2 r2n+1 dr
0 n=0

. la:2

la
2 2n n
gMZ-rrj‘o‘g‘lanl r drgM2F§m<a,

whence v L:(D') c L2(D)

Moreover it is not hard to show that Tv :L:(D') a L:(D'),

defined by TV = PMV IL:(D') is compact.

Thus we have an example of an unbounded harmonic
function which multiplies L:(W) to L2(W) for a
suitable region ‘W. However in the case of analytic
multipliers of L§(W) to L§(W) we do get the

expected result. [Also see DRS]

§3. Proposition 3.1. Let W be a region of finite area,

p > O. and suppose f L§(W) c LEM) . Then f EL:(W) .

15

Proof: We may assume that p < a. By the continuity of

M :L§(W) 4L§(W) we have

f

j‘w Ipr Ing dm g c I‘W Ing dm for all g e LEM)

Note that for every positive integer n, we have

in e LEM) and so
I‘ IfInp dm g of 1f1(n‘1)p dm .
w w

Thus,

80

(I Iprn dm)fi g c n (I‘ Ipr cam);l .
W W
1

Letting n 4 o completes the proof with "f”. g cp-
Note that W may be unbounded. When the region W is
the disc we will show that the harmonic multipliers of

L§(W) to LP(W) must be bounded.

Proposition 3.2. Let v be harmonic, O < p < a, and

suppose v L:(D) C LP(D). Then v 6 L°(D).

Proof: We have that

1 3a
(I Ipr IVIP amp g c (f Ipr amp for all f e L§(D)
D D

16

Hence IvIp dm is a Carleson measure on the disc [Lul].

Thus given a in D,

V P
IDHdm

 

c
mffigY g'c '
~where Do is a hyperbolic disc of radius % with center
a. Therefore again by the theorem of Hardy and Littlewood
[G] we have that v is bounded. One notes that the above
proof would have gone through had v been subharmonic

(instead of being harmonic) provided p 2_l.

§4. In order to classify the harmonic multipliers of
L:(D’) to L2(D’) we need the following lemma which

surely must be known: nevertheless a proof is included.

Lemma 4.0. Let u be a real-valued harmonic function

 

on D’ = {2 IO < IzI < 1}. Then there exists a in ii
such that u(z)-—c log IzI is the real part of an

analytic function on D.

Proof: Let

U
ll

D«-[z E C I z $.0}

D-{z 6 C I z 2_0}

U
M
II

w
ll

1 [Z 6 D I Im z > 0]

21
ll

{2 E D I Im Z < 0}

17

Now since D1 and D2 are simply connected, for
i = 1.2 there exist analytic maps hi on Di such
that Re h1 = u on Di' Hence there exist constants

ci such that

Imh -Imh

ll
0

1 2

and Im h1-Im h2 c2 on R2

Pick b on {2 6 D Iz g_O} and let y be a curve

that lies in D’ and joins b to b while passing

a_u_
an

independent of Y. Pick bi E Y n Ri' i = 1.2. Then

once around the origin. Let us prove that I is
Y

MEI @344 a“
Y on b to b on Y on b to b on y 55

Im h (b2)-Im h

1 (b1) +Im h

(b1)-Im h2 (b2)

1 2

and thus the asserted independence is proven.

Incidently note that we also proved that I -— = 0

provided that Y does not enclose the origin.

Hence, if Y is a curve that lies in DI and if
either Y does not enclose the origin or goes only once

around it, there exists a in nz such that

Bu 6 _
SE-a IV 55- log IzI — O

18

[Notice that I 5% log IzI = 2V or 0]. Pick zéED'.
Y

Let Y be a simple curve that joins b to 2. Then

there exists a single valued function f :D’ a C such

that
Re f(z) = u(z)-a log IzI
_ a
and Im f(z) — - I 33 (Re f)
Y
Let us show that f is analytic on D’. Since

u(z)-a log [2] is harmonic on D2. there exists an

analytic function h2 on D2 such that
Re h2(z) = u(z)-a log IzI

Let Y be a curve lying in D which joins b to z.

2
Then

Im h2(z)-Im h2 (b) = — [Y 5% (u(z))-a log IzI)

Since the definition of f does not depend on Y,

f = h +-Im h

2 2 (b) on D

2

and thus f is analytic on D Similarily, f is

2.
analytic on D-[iy'Iy‘ IO} and thus the analyticity

of f on D’ is established.

19

Proposition 4.1. Let D' = [265C] 0 < IzI < 1}. Suppose

2 2
a

 

v is a real valued, harmonic function and v L (D’)c:L (D').

Then there exists a such that v-c log I IEELQ(D’).

Proof: In view of Lemma 4.0 and since log IzI L:(D') c L2(D')
[Example 2, 92] we may assume that v is the real part of
an analytic function f on D' and prove that v€EL~(D).

The Laurent expansion of f gives

8
U

. 0 . .
f(rele) = Z a rn en18+ Z‘ _I_1 e‘nle
n n
n=0 n=1 r
Hence,
° F °° b .
2": f+f= Z (3 rn+—§) enle+ Z (a r +3) ‘n19
n n n
+ a0 + a0
2 I
so
1 ° n '5'; 2
I 2 Ianr+-I-1Irdr<co,
0 n=1 r
and so for every n,
[1! n b—nlz
a r 4--— rdr < a .
O n rn

Hence for all n > 1, we have bn = O which shows that
v extends to be a harmonic function on D. Also

L:(D’) = L:(D) [ACM], whence v L:(D)CL2(D) and now

20

the proposition follows from Proposition 3.2.

§5. Suppose v is a harmonic function on D. If
\IEIFID), then unlike the case of the circle its
harmonic conjugate is also in L1(D). (See the intro-

duction and also [SW].) We will use this fact below.

Proposition 5.1. Let 1 g_pig,w, let v be harmonic

 

on D, and suppose VL§(D)CL1(D). Then v€Lr(D)

where

"OH-4
+
HID-4
ll
H

Proof: In view of Proposition 3.1 we may assume that
1 < p, and to avoid a triviality, let p < m. We have

1
I IvI IfI g_c (I Ipr)p for all fesL§(D)

Hence, we may assume that v is real-valued, and it

also follows that
f 4 I vf for all f e L§(D)

is a continuous linear functional on L§(D). Thus by
[ZJ] for some 9 6 L:(D) we have, 2 I vf = I 3f for

all f e L§(D).

But as remarked at the beginning of the section, v = Re h

for some h 6 L:(D). Thus,

21

I hf+I hr = I St for all f6H°°(D)

Let b(z) = Z3 anzn for IzI < 1
n=0
and 9(2) = Z3 bnzn for IzI < 1
n=0
a
m m

— m—
Put f — z for m‘z 1 to get 5171 _ 5171 for all m.2 1.

It follows that g-g(O) = h-h(O), whence hEL:(D). Thus

vELr(D).

 

Proposition 5.2. Suppose v is harmonic, l'g q < p.

P q r .1..1._1
and v La(D) c L (D). Then v€EL (D) where p4'r — a.

I

Proof: We may assume that q > 1. Pick q’, r such

that

1 1_

and
("Ht--l

Then by Horowitz [H1],

I

r (D)

p q’ _
La(D)La (D) — La

But then
p q’ q q’
v La(D)La (D)CL (D)La (D)
I

Hence, v L: (D)CZL1(D) .

Now envoke PrOposition 5.1 to complete the proof.

22

Proposition 5.3. Let v be a harmonic function on

 

D, q > p > O, and suppose v L§(D)c:Lq(D). Then v E O.
Sheldon Axler [A] has proved that for regions of finite

areas, if q > p and f L§(W)C:Lg(W), then f E O.

The proof for the case of the disc is quite simple.

Proof: The continuity of M.V :L§(D) 4 Lq(D) implies
the existence of c such that

1

(I lviq 1qu cam)q g c(f Ing dm)
D

for all g€Lp(D)
D a

Hence, by Luecking [Lul] for each a in D

2q

I (VIP dmg c (1- (of)?

D
G

where Do is a hyperbolic disc of radius % with center
a and c is some constant. Now there exist constants

and c such that

Cl 2

 

Moreover

P
V
[DO I l
CIVIp(0) g W [G]
a

(See [G], Chapter 3, Lemma 3.7). Hence

2'22)
lvlp<a> gc (14042) P

Therefore v(c) 4 O as IaI a l, and so v E O on D.

23

§6. We now want to show that if 0 < q < p g_l and if

f L§(D) c L§(D), then f 6 L:(D) where

DIP
+
HIP
II
QIH

The proof is harder than for the previous cases, and will
be accomplished through a series of lemmas. Be reminded
that c will denote a constant, perhaps not the same in
each occurence and n will always denote a positive

integer.

Lemma 6.1. Let O < q < p and suppose f L§(D) C Lg(D).

 

Then

1
IfIn L:p(D) C an(D) for every n

Proof:

1 l
(I Iflq Iglq)q SC (I 191p)p for every 96L§(D)

Let n > 0 be an integer. Then given 9 E L2p(D), we have

9n 6 L§(D), so
1 l l
(I lflq lying“?! sea (I leflnp)55 :
l
i.e. If)“ L:p(D) c anm)

Proposition 6.2. Suppose f has finitely many zeros,

 

o < q < p, and f L§(D) c L§(D). Then f e L:(D), where

li.l
p r

l
q.

24

Proof: Let b be the Blaschke product formed by all the
zeros of f. Then since b is a finite Blaschke product.

. q q
Mb .La(D) 4 La(D) ,

the multiplication operator by b has closed range [MS],
and so there exists c > 0 such that for each 9 in

Lg(D) for which % is analytic

1 1
(I 1%,qu g c (I lslq)q

Hence

1 l
(I If qu)q g C (I IghIq)q for every 9 6 L§(D) ,

but then
1 l

(1‘ mg 191%“? go (I IgIP)P

SO

l l
(I If glqfi) g c (I [919)]? for every 9 e L§(D) ,
whence

f/b L§(D) c L§(D)

Now pick n such that np, nq > 1. By the previous lemma

and since f/b has no zeros,

25

1
f n np nq
(S) La (D) c La (D)

Now the proposition follows from Proposition 5.2.

Lemma 6.3. Let f E L:(D) and suppose

 

r n-l r

n/r
(I IfgIr/n) gc(I (915:) r for all gELfim)

for some n > 1. Then anr g_C. Here c denotes the

same constant.

Proof: Since f E L:(D), we have fn-l E L:/n-1

(D).

Whence,

n n-l
r

(I ‘fIr/n ‘fl(n-l)r/n) SC(I 'f‘r)T

or

n—l

n/r
(I lflr) gc(I Iflr) r ;

l/r
i.e. (I IfIr) S.C - Q.E.D.

r r
Lemma 6.4. Suppose f Lg-I(D) c L:(D) for some r > O

and n > 1. Then f E L:(D).

Proof: Let Ile S.I22I g_... be the zeros of f.
Let
2'2]: z"zk
bk(Z) - (1—-_Zk-_Z) (2 1 -2kz)

and

26
B = I] b.
k .
32k 3
By Horowitz [H2] there exists c such that given 9 in

Lr/n(D) with i is analytic,
a Bk

n/r

r/n n/r
51 c r/n
(I 13k) ) g. (I lg! )

The constant c does not depend on k. Whence for some

constants independent of k.

r/n n/r n/r
(I 13:! ’ g_c (I ifgIr/“) .

(*) so
n/r n-l
r/n __.
(I IEEI ) g’c ( I IgIr/n-l) r for every g(=L:/n-1(D)
So
/ E
r n-l r
f/Bk La (D) c La (D)
But then gi has only finitely many zeros, also
-1 -l
(fl—ET) +r-1 = (Ir-1)

Hence by the corollary to Lemma 6.1, éi E L:(D). Now by

Lemma 6.2 IIE-IIr g c where c is the constant occuring in

Bk

(*). Apply Fatou's Lemma to deduce that

IIfIIr S C - Q.E.D.

27

Lemma 6.5. Let D L§(D) C Lq(D) for some nonnegative

 

function m. Then for all n > O we have

can LEI/IND) c: Lq/n (D).

Proof: Fix 9 in LE/n(D). Then by Horowitz [H1] there

exist c and 91,...,gn in L§(D) such that

n
g = H g. and

n p P/n
El (gin g c1 (sup/n .

Here c1 is independent of g, and depends only on p

and n. Since Mw :L§(D) » Lq(D) is continuous, for all

q/p
i we have I IfIq Igin g_Cq (I Igin) , so

n n q/p
I IfIq (,2 lgilq) seq 2 (I 19119)
l=1 i=1

Now the arithmetic-geometric inequality gives

n

23 (9in 2n (glq/n .
i=1
and so
q/p
n )9 lp
DI IfIq lglq/ngcqn E I n1
i=1

n q/p
q n p
re n—7—q p (Ely (91) )

Thus

28

n n
l
H

n/ c c
(I (f(q (gig/n) q _<. m

Lemma 6.6. Let 0 < q < p and suppose f L§(D) C L§(D).

 

Then

r
f L§:I(D) C L (D)

W UIH

.QIH
VIP

for some integer n > 1, where % =

Proof: By Lemma 6.1, for all n 2.1,
IfIl/n L2p(D) c an(D)

Pick n > 1 so that np, nq > 1 and such that there

exists positive R with

”H4
HIP

l;+-
nq
Then
[ill/n Lgpm) L:(D) c anm) LR(D) c Lr(D)

Now

+
+
NH“

3h“
.3h“

:3
HP”
4.
HIH
ll

1
r
(l-§)

Hence, by Horowitz [H1]

ng/n - Q.E.D.

29

np R _
La (D) La(D) — La (D) .

SO
r
l {I
If!“ La 1“(m c Lr(D)
Thus by Lemma 6.4,
r
If) 1.3%) c Lr/“(m .

and so

r/n-l r/n
f La (D) C La (D)

Theorem 6.7. Let g > 0 be a rational number, lj>p:>q:>O

and suppose IfIg L§(D) C Lq(D) for some analytic function

f. Then f 6 L:g(D), where

'Uud
+
HIH
H
(Qua

Proof: Apply Lemma 6.5 for the obvious integer and then

Lemma 6.6 and Lemma 6.5.

§7. We now return to the question of determining the
subharmonic multipliers of L:(D) to L1(D). Must they

be in L2(D)? A partial solution to this question is

provided by the following proposition.

I

30

Proposition 7.1. Let v > 0 be subharmonic on D and

2 1 2W '9
v La(D) c L1(D). Let M(v,p) = 2? IO v(pel )de. Then

 

M(V.p) \/1 -p = 0(1)

Proof: Since Mv :L:(D) a L1(D) is continuous, we have

1/2
I IvI Iandm g_c(I Izandm) for all n 2_o
D D
So
1 2V .
I rn+1 (%% I v(rele)d9)dr
o o

1 1/2
S C(I r2n+1dr)
0

Since v is subharmonic, M(v,r) increases with r.

Hence,

1'_ n+2 1
M(VoP)—n—% S Cm, f0]? 0 < p < 1

hence

M(v,p) gc Aim—n for all n21 and for all p in (0,1) .
1--P

Put n - [f1:51. the greatest integer less than or equal to
f%55. Then l--pn is bounded away from zero as p 4 1,
Thus,

C

 

M(V.p) _<_

 

Jl-P

31

As a corollary we have that v E Lp(D) for all p in (0,2)

Composing with increasing convex functions preserves

subharmonicity. Thus

2 q ——————— for z E D
l.- 2

is subharmonic. Hence Proposition 7.1 leads to the question

"Does —-1—- L: (D) C L1 (D) " ? Note that '1

7W fiTl’zT

Coifman and Rochberg [CR] has proved a decomposition

[L2(D).

theorem, a restricted version of which is as follows.
(See their paper for the definition of terminology used

here.)

Theorem (Coifman-Rochberg). There exists n-lattice [oi]

in D such that given f 6 L:(D) there exists [xi]

such that
a K(z,ci)
(1) 15(2): 2 Ki? 0
1:1 K(aioai)
‘° 2
(ii) Z) ‘in ,< o
i=1
Here K(z,a) = ———%:——2 is the reproducing kernel for
(1 -az)
L:(D). One wonders could (ii) be strengthened to read

(ii') .2 (lit/14%| <~

l=1

I have been unable to prove or disprove this. However, if

(ii') holds, then one can easily deduce that

32

L: (D) c L1 (D)

~/1--IzI
Let v be a real positive subharmonic multiplier of

L:(D) to L1(D). Then Pv defined by

Pv(a) = I v(z) ———lfi:—§ dm(z) for a E D

is well defined and it is easy to see that Pv is analytic

on D.

Proposition 7.2. Let P and v be as above. Then
2
Pv 6 La(D)

Proof: To begin with we will show that Pv 6 L1(D).

By Fubini's theorem it would be enough to show that

1 .
( ——2_ dm(a)) V(z)dm(2) < °°
J‘D ID I1 -azI

Indeed

rdr by Proposition 7.1.

The last integral is of course finite and so Pv E L1(D).
Again an application of Fubini's theorem shows that if

g e H°(D). then

33
I Pv'g = I vg

Since 9 4 I vg for g E L:(D) is a continuous linear
functional on L:(D), it follows that there exists h in

L:(D) such that
_ — a
I PV 9 = I hg for all g e H (D)

Now as in the proof of Proposition 5.1 one deduces that

Pv = h, and so Pv 6 L:(D).

Let P be the orthogonal projection from L2(D) to
l
(1«-az)
The Toeplitz Operator Th :L:(D) 4‘L:(D) is defined by

and let h be bounded.

L:(D). Let x(z,a')

Th = PMh IL:(D).

§8. In Proposition 3.1 we showed that Mf :L:(D)(4L:(D)
is bounded, then f is bounded. Now we intend to improve
this result. Let v be a real valued harmonic function
in L2(D). Then Tv can be defined on the dense subset
Hé(D) of L:(D). Suppose Tv extends continuously to

a bounded linear map from L:(D) to L:(D).

 

Proposition 8.1. Let Tv' defined as above, be bounded

from L:(D) 4 L:(D). Then v is bounded.

Proof: We have <TV (K( .E)). K( .5)> g,c <K( .5). K( .5)>.
Noting that K( ,5) is bounded on D for each a in D,

we have

34

<P(V K( 03)) I K( 13)) S C K(ala-) 7

i.e. (V K( ,5), K( ,E)> g c K(c,5)
Now v = Ref for some f in L:(D). Now
f(a)K(a.E) = <f K( .6). x( .3»

I f IK( ,6)12 .

Hence,

V(0() Mona)

IvIK( ,a)(2

<v K( .6). K( .3»

Thus

v g_c

But if TV is bounded, T v is also bounded, and so we

may conclude that v is bounded.

§9. In this section we determine when T__:L:(D) 4 L:(D)
f
is bounded. Also see the work of Stegenga [82]. Here

f is in L2(D).
a
Let B = {b Ib is analytic and sup Ib’(z)I(l-—IzI2)<(a}.
zED
Traditionally B has been called the space of Bloch

functions and is a Banach space when normed by

35
(1b) = lb(0)l+Hb’(z)(14212)”,

In this section H H will always denote the above norm.

Let b E 8. Then clearly

(1) (b(z)! g (b(0)1+((bll Mega-(21))

and thus B’C 0 L§(D), a fact we will use below.
1£p<~ -
*
The dual of L:(D), denoted by (L:(D)) , can be

identified with B. More precisely,
A e (L:(D))* if and only if there exists h e a
such that
A(f) = I Bf for all f 6 Han
Furthermore, there exists c independent of A such that
,1; ”b” 3 HA” 1 , g c IIbII
(La(D))

Hence T :La (D) 4 L: (D) is bounded if and only if

a»
T_:

f

"f"
416 is a bounded Operator, where

(t

* _
<T_(g),h> = <g,T_(h)> for all g(?H° and all h E B .
f f
Now fix h in B and let g(z) = zn. Then

_ *
<P(fzn) A» = <z“.T_(h)>
f

36

Now since 5 C L:(D), we have <P(fzn),h> = <fzn,h>,

whence

<zn,fh> = <z".T:(h)> .
f

*

and thus fh = T_h. Since h was arbitary we have shown
f

that f5 C 6. Conversly suppose f5 C B. It is a routine

:6’4 6 is bounded. Hence
*

f

matter to verify that Mf

there exists a bounded operator M :L:(D) 4 L:(D) such

that
*
«3,11f (h)> = <Mf(g) ,h> for all 9 6H“ and all h 68 .

Hence

_ *
<fgvh> = <Mf(g),h>

So

<P(Es).h> = <M;(f).h>

Hence T__ and N; agree on H°(D), a dense subset of
f
L:(D), and so T__ can be uniquely extended to a bounded
f
Operator from L:(D) 4 L:(D). Thus we proved the

following proposition.
. . 2 l l
PrOpOSltlon 9.1. Let f E La(D). Then T__:La(D) 4 La(D)

f
is bounded if and only if fB'c B.

37

In the next proposition we characterize the multipliers

of B to B.

proposition 9.2. f8 :2 a if and only if f e H"0 (D) and

 

sup If’(z)I(lI-IzI2)Ilog (l-IzI)I < n.
26D

Proof: Suppose f8 C B. Let us first prove that f 6 Hm(D).

Since M.f :B 4 B is continuous there exists c such that
Hf 9H s CH9“ for all g e B .

so fn 6 B for all n 2.1 and also

(inn .<. cnfn‘ln

So

“‘1an

ann g c
C 1 50 n C n = C n 0
How 5 La (D) ”f II 2 lIIf "L1 lIIfIILn

Hence

(If) n g awful/n
L

Let n 4 a to deduce that f is bounded. Now let

9 C B be arbitary. Then fg G B, so

Hf’(z) (1 - (212)9(2) +g’(z) (1 - 1212mm)!

3 CIIQII -

38

But f e H"(D). So

sup Ig’(z)(1—IzI2)f(z)I gens" -
ZED

Fix a e D with a ¥ 0. Let ga(z) = log(1- a 2). Then
a
"9a" 3.2 and 9a(a) = 109(1-IGI). It follows that

If’(c)I(l-IcI2)Ilog(l-IaI)I gc for all a e D,

completing the proof in one direction.

To prove the converse we need

0
(2) H c 6 .
This is well known. See for example [ACP]. Perhaps the
following proof of (2) which is patterned after Luecking's

[Lul] Lemma 2.1 has not appeared elsewhere.

NIH

For a E D let Do be the hyperbolic disc of radius
with center a. Luecking [Lu2] has gone through the

computations to show that

(a) Il-—5'wI 2_C(1-IGI2) whenever w 6 DO

and (b) m(Da) g C(1 - IC1I2)2

Now since 6 C L:(D), given f 6,6, a normal family

argument shows that If’(O)I g_c I IfIdm.
Do
z+—a

Let ”0(2) = (£1355)

39

 

 

 

Then
2 2
I(f ova)’(o)( g c I If(w)I(1-1_<_II)2r dm(w)
Dc Il-a wI
SO
(f’(a>l(1-Ia(2> gc (f(w)) 1 mm
J‘13“ (l —Io(12)2
m(D )
S CIIfII, a
(l-lmlz)2
S CHfII, - Q.E.D.

Once armed with (1) (mentioned above) and (2), verifying

the converse is easy and will be omitted.

Now if fBCB then f 6H”, so limf(re19). denoted

* i9 r41
by f (e ), exists for almost every 6.

 

Corollary. Suppose f6 C.B. Then f* 6 H” n Hé-tC.

Therefore for example an infinite Blaschke product

will not multiply B in to B.

 

Since H" n VMO = H” n H°+c [Sa], it would be
*
enough to show that if fB’C B, then f 6 VMO. This

will be proved by using a theorem of Feffermann-Stein [F].

Proof: A Carleson square Sh is a set of the form

 

sh= [re19 Il-h < r < 1, 19-90] <2)

40

C

Let fB’C 5. Then I f’(z)I g. 2 .
(1 - IZI )Ilog(1 - IZIHI

Hence

I (grad 121% -IZI2)dm
S

1
5h (1 -|z)2)(1og(1 - Izhf2

h
1 90+“
3 c I I h

2
l-h
9 2'

g_c I dm

p d9 d9
(1-p2)(lo<_)(1-p))2
O

h [ '1 11 = c -——-17- .

Hence, % I Igrad fI2(l-IzI2)dm 4 0 as h 4 O and so
. Sh
f is a VMO function.

CHAPTER 2

§(0). Let (L:(D))'L be the orthogonal complement of
L:(D) in L2(D), let P be the orthogonal projection
of L2(D) onto L:(D), and let Q be the orthogonal
projection of L2(D) onto (L:(D))L. Let f be in

L°(D). Write M for the multiplication Operator on

f
L2(D) defined by

Mf(g) = fg for every 9 6 L2 (D) .

and let T :L:(D) 4 L:(D) be the compression

f
PMf I L:(D). Define

2 2 .L _ 2
H :La(D) 4 (La(D)) by H —QMfILa(D)

f f

Traditionally T is called a Toeplitz operator, and

f

H is called a Hankel operator. We can express Mf as

f
an operator-valued matrix with respect to the decomposition

13(1)) = L:(D) o (1.:(Dn‘L by

”T 11*-

f E
Mf=

H K

_f fd

 

 

41

42

‘*
To see why the first row second column slot is H__

f
consider the matrix for .M_ and.compare that with the

0

111

one for M* Here K -(L2(D))L 4 (L2(D))'L is some
f' f T a a

operator. Let f,g E Lc(D). Then

—

 

 

TT+H*H TH"'+H"PKfl
£9 E 9 f9 7,—- 9
MfMg = .
*
+
beTg 1(ng HfH_+Kng
g .-
Comparison with Mfg gives the following useful identities:

T

*
TT+H H ..
f9 £9 E 9

H H T 4-K H
9

£9 f g f
For later reference we also note that

— -+l n-m .
1r(nn+m1 )a if ogngm
T2116”) (a) =
0 if m > n 2_O

2 2" — ie
Let S = (g E L (D)I I f g(re )d6 = O for almost every r
O

and for all f in L:(D)]. Then clearly S is a subspace
O(D) =

[f I f E L:(D), f(0) = O] and the subspace A

 

of (L:(D))*. Hence for example L:

. 2
l in L (D)

spanned by

[f xA I f 6 La O(D), A is an annulus in D]

43

lie in (L:(D))*.

§l. Let f E L:(D). Then H__ can be defined on the
f
dense subspace H'(D) of L§(D). We try to find

necessary and sufficient conditions on f for H__ to

 

be continuous, more precisely, for H__ to be extgnded
to a bounded operator from L:(D) tof (L:(D))*.
Proposition 1.1. Suppose f 6 BMOA. Then
H__:L:(D) 4 (L:(D))‘ is continuous.

f
Egoogz Note that H__:L:(D) 4 (L:l(D))'L is bounded if
and only if f

)<?e.h>l g.cHgH2HhH2 for every 9 e H°(D) .

for all h e (L‘.§(D))‘L

Now, since f 6 BMOA a theorem of Fefferman [F] says

that f = f -+f where f1, f2 6 BMOA and

l 2
G 0
Re f1 6 L (D), Im f2 6 L (D). Since H__ is continuous
f
if and only if H _, is continuous, we may now assume
if
that Re f 6 L” (D) . But then H_ = H_ = H2 f and thus
f f+f Re

the proof is complete.

Notice that the proof shows that

(a?) g c((anMOA

 

Corollary 1.2. Let f E VMOA. Then H__ is compact.
f

44

Proof: We first show that H__ is compact. Consider

N

n n4—1
w

 

Let gn = 2 Then {gn I n 2_O] is an
orthonormal basis for L:(D) and for each n 2.0

*
H_ H__ (g
z 2

[see §O].

 

_ w
n) — (n+1) (n+2) g’n

*
Hence H__H__ is Hilbert-Schmidt [also see [BS], Theorem 7]
z z
and so compact. Thus H__ is compact. From §O
2

H = H_ T_+ K_ H_
zn+1 Zn 2 Zn 2

 

So, by an inductive argument H___ is compact for each
n

z
n 2_O, and so H__ is compact for any analytic polynomial

P
p. It follows that H___ is compact where fr(z) = f(rz).
f
r
But ”HF-HE” g cIIfr-fIIBMOA. since f e VMOA.
r
”fr"f”BMOA 4 0 and thus HE. lS compact.

Later (Proposition 3.3) we will produce a noncompact

H__ for some f E H°(D). Hence HH__;—H_H in general
f

does not tend to zero as r 4 l. r

Proposition 1.3. Let f 6 BMOA. Then H__ 4 H__ S.0.T.:
fr f2
in other words H__(g) 4 H_(g) for every 9 6 La(D).
f f
r

Proof: Now since H”(D) is dense in L:(D) and since

"H__” is bounded, it suffices to show that

fr

45

H_(g) 4H_(g) for each 9 6 H°(D) ,
f

fr

Q
so fix 9 6 H°(D) and let f(z) = Zlanzn. Then
0

Ile_(9) -H.f_(9)ll g H (fr 4599112
r

S IIgIIQIIfr " f” 2

"9","? an(1 -r )z "2

 

a I3 '2 I_rn 2
((gn../(w§ “ (1 ))

n+-l

B t th 5 Ian‘z IIfIIz < I d l L b

u en T = a an so app y e esgue
(D n+-I 2

Dominated Convergence Theorem to finish the proof.

Notation: For each a in D, let

_ z+-a
(pa(z) — (fl—+32) for every 2 6 D

Lemma 1.4. Let a 6 D. For each 9 defined on D,

 

9 EL: (D) “there exists k 6L: (D) such that g= (kowahpc;

Proof of : Suppose g = (k.o$h)$é for some k 6 L:(D).

Then
I Ikowaflmglz =I (HZ .

and so 9 6 L:(D).

46

“4

Noting that 6 HQ(D), we observe that the proof yields

an":

that for each 9, if g E L:(D), then

2
g 0% G La(D)

Proof of =: Let g 6 L:(D). Since $% 6 Ha, we have
a
$% 6 L:(D) and hence as observed in the proof of the
a
direction

 

gain
a -G

9"@_a
So let k = —7————— , completing the proof.

CDC OCp_a

Corollary 1.5. Let h E (L:(D))‘. Then

 

(hscpmmo’K e (L:(D))'L for every o in D.

Proof: Given 9 E L:(D), pick k 6 L:(D) such that

_ I
g - (k.o$a)¢a. Then

*

I _ I—
<9. hoea (pop — I koqia (pa hocpch (pa

Ikfi=o. Q.E.D.

Let s = [f e L:(D) IH_:L:(D) 4 (L:(D))'L is bounded].
f
Define a norm. H HS on S by

Hst = HHEHi-If(O)I for each f E S

Let [fn] be a Cauchy sequence in S. Then [H__] is a
fn
Cauchy sequence in the Banach space of bounded operators

47

from L:(D) to L2(D), and thus there exists a bounded
operator T :L:(D) 4 L2(D) such that HH__g-TH 4 0.
Notice that there exists c such that HHEH 2_chH2

for every f E L:(D). Hence {fn} is a Cfiuchy sequence
in L:(D) and so there exists f 6 L:(D) such that
an--fH2 4 0. Moreover, clearly T maps L:(D) into

(L:(D))‘. Fix 9 6 H” and let h E (L:(D))L. Then

<H_f_(g) .h> = <?9.h>

lim <fng.h> = lim (H_(g) ,h>

n4» n49 fn
= <T(g) .h>
Thus H_(g) = T(g). Now clearly (S,H HS) is a Banach
f
space.

proposition 1.6. If f e s and c e D, then
Hf “9c --f(d)))S = If -f(0)IIS
Let GED, gEH"(D). and h6(L:(D))" .
Then (gacgangéerl’m) and (hocgmngc’l e (L§(D))* .
SO
(<f(g . (gang; (h oceaheg>l g((fHSH (9 . (2G)=2;112H(h . Tammy

Hence

48
I <f °°2a)g.h> I g Hfflsllgllzllgflz .
and thus
Hf era-f(mns _<. ((f-flmns.
Replace a by -a and f by f<>ga to get
Hf -f.(o)HS g Hf opa -f(a)lls

Later in Proposition 1.8 we will show that S c B where
B is the space of Bloch functions (§8, Chapter 1). We

propose a conjecture now.

Conjecture: Let (V. H H) be some Banach space of

 

analytic functions between BMOA and 8. Suppose for
every f in V and for each a 6 D, we have f'o¢h<EV

and

HfoCPa—fkl)” Hf—f(0)H .

Then either S = BMOA or S = B.

Let 4' be the subspace of L1(D) spanned by
[g XA I 96L:(D).g(0) =0, and A is an annulus in D]

* 1 _ 1 _ — a.
Let a n La(D) — {f e La(D) I<g.f> — I £9 for gEL rm

is a bounded linear functional on 0'}.

 

49

Proposition 1.7. Let 0' be defined as above. Then

 

4’ n L:(D) c BMOA.

*-
Proof: Suppose f E d’ D L:(D). Then

I I E9 XA I $.6H9 XAH1 for every 9 e L:(D)

and for every annulus A such that A C D. Set 9

ll
HI

and let A be an annulus such that A": D.

so (A (£12 go (A If)

Hence I IfI2 < c and
A

I IfI2 go (I (r12)%(m(n))%
A A

I (£12

So 21—— g c2
m(A)
2” 2 is
Now since I IfI (re )de increases with r, it follows
0
that
"fl! :0
H2
i6 . i6 .
Hence f(e ) = 11m f(re ) eXlsts for almost every 9
r41
and

2v

f(rele) = g% I Pr(9-t)f(eit)dt ,

0

where Pr(9 -t) is the Poisson kernel. Let g be an

analytic polynomial and O < p < l, with A = [z Ip<<IzI<(l].

50

__ 1. 2W 2W ‘_
Then IA fg= %I ‘I0 Io Pr(9 -t.)f(ei ti)g(re e)rdrdedt

-—I1f(eiH)g(r e it)rdrdt

 

 

P
IA E<1 .
8° 'IZ — )ge(1t)dt
m(A) o
1
2w . I [g(rZ el ti)‘-9(e t)]rdr
=~I halt” p ) dt
0 1
I rdr
P

Use the continuity of g on D. and the Lebesgue Dominated

Convergence Theorem to deduce that

fmfe

4 IZW )g(eit)dt as p 4 1

Hence for each analytic polynomial 9

2w ‘_ 't I.IA gI
[IO f(e1 )g(si t)dt I go Wgc ”9” H1 ,

*
so f is an analytic function in (H1) , and thus

f E BMOA.

Let QO be the orthogonal projection from L2(D)

2 2 .
to La,o(D) and let f 6 La(D). Define

 

c :H°°(D) CL2(D) 4L2
E a a,

o (D) by

C_Ig) = 00(fg) for every g(EHa(D)
f

51

Coifman, Rochberg and Weiss [CRW] have proved the following

proposition.

 

Proposition 1.8. Let fEL:(D). Then Cf can be extended

 

2

to a bounded operator from L:(D) to La 0

(D) if and only
if f E B.

We give another proof of this fact.

Proof: Let g E L:(D) and suppose f 6 5. Then
I f(z)g(z)dm(z) =I f’(z) (1 -IZI2)9(Z)dm(z)
D D

Since f E B, we have f’(z)(l(-IzI2) 6 L°(D), so

If E<91) SC (I91! 1
D L

 

Hence given 91 E H°(D) and 92 E L O(D), we have

2
I I g 91 §_2—I S C Hgl EEHI
S. c HglIIzngIIz ,

so C__ is bounded. Conversly suppose
f

<C_f_(91) .92> _<. C HQIHZHSZHZ

 

2

for every g1 6 HQ and for every 92 6 La 0

(D)

SO <Egl.92> g c (191(I2"92Hz

52

 

for every 91 6 Ho and for every 92 6 L: 0

We may assume that f(0) = 0. Pick a E D and put

1

g (2) = k (z) = _
1 a (1 ~02)

 

92(2) = kc(z)r-l .
Then
(I an 1. 4am.) (so)! 1. I’-
D U.-cz) 0,-az)
(1-10112)2
Now
I f(z)ka(z)dm(z) = f(c) ,
and so
- 6 65 ‘—r
I f(Z) —-_—-—z dm(z) = f (a)
(l-az)
Hence,

(mm (1 - Ia|2)21 g c .

SO

53

If’(c)(l-IOI2)I g c (by [D], Theorem 5.5) ,

and hence f 6 B as desired. Let g 6 L:(D). Then it is
not hard to see that g(c)(1.-IOI2) 4 0 as IcI 4 l.
k
a . 2 . .
Hence tends to zero weakly in L (D). USlng this
HkaH2 a
fact and examinig the above proof one gets that if C__ is

compact, then f’(c)(l'—IcI2) 4 O as IcI 4 1. Also see

[CRW].

 

Corollary 1.9. Let f E L:(D). Then a necessary condition
for H__:L:(D) 4 (L:(D))‘ to be compact is that

f 2
f’(c)(1-—IaI ) 4 0 as IO] 4 1.

§2. In Chapter 1, §8 we proved that
O
H C B .

In this section we will show that, for each f E H”

f’(re19)(l-r) 4 O for almost every 9 as r 4 1

Lemma 2.1. Let 0 < r < l, t E II, and

 

2)3 1
2) (ll-2rcost-I-r2)2

(l-—r
(l+-r

c(9(r.t) =

 

Then

V
g% I m(r,t)dt = l for every r in (0,1)
-T

54

1 w. 1«-r2 2
fij cp(r,t)dt = T—T HPr(t)HL2

-w ( -+r ) (TU

2 0 . a .
(LLEE) (l + 2 Z rjcosjt, 1 + 2 Z rjcosjt>
l-+r 1 l

2 G .
l-r l 21
P——fi[l+4 Zr ]
l-kr 2 1

- <1——2"‘2H1+——22r2 1 = 1
l-+r l-r

Lemma 2.2. Let u be a complex measure on (E and suppose

 

Du(60) exists. Then

v
f% I-w T(r.90-t)du(t) 4 u(BO) as r 4 1-

Proof: The usual approximate identity proof for the
Poisson kernel Pr(9) works for ¢(r,6) as well.
Without loss of generality 60 = 0. Let A = Du(0)

and

i9 __ 1 V
u(re )— fif cp<r.e-t)du(t>
-w

Then

i6 _ 1 7’
u(re )-A — fij‘ @(r,t) [dm(t) -Adt]
-w

= 21? [ca(r.t)u<t) 4t]:

TT
- 717?] (u(t) -At) g? dt .

-w

55

where u(t) is the function of bounded variation associ-
ated with u. The first term of the partial integration

tends to o as r41. Fix 5>o andlet 5g]t‘gw.

Then

3
r2) 1

4 O as r 4 1
8 sin66/2

 

1§5§(r.t)1_<.(1 -

Hence for each 6 > O. u(rele)‘-A--I5 4 0, where

16 = - fij’ (u(t) -At) ft dt

lu(t) -u(-t)
*2t

II

+

H"
‘fi

-A] t [ $72] dt

Given 6 > 0, choose 5 > 0 such that

 

1L1”) LEM-t) -A1 < e for all t in (0,6) ,

so that

W
‘15] g e/ZF Io tl-—§% m(r.t)l dt

notice that QEéEéEl is negative for each t in (O,w)
and for every r in (0,1). Hence
w
6 ET
'15‘ Sfi‘rot (-61? (rut)dt

TT
W
= _ 2% [t ¢(r't)]o+2£1'r«ro cp(r.t)dt

7T

S.f% I-w T(r.t)dt = e

56

Theorem 2.3. Let f 6 H” (D) . Then f’(rele) (1 -r2) 4O

 

for almost every 6 as r 4 1-.

Proof: We have that

 

. 1r 2 .
f(rele) = éL.f (1-r ) 2 f(elt)dt
1r —Tr (l-2rcos(6-t)+r)

Differentiate with respect to r to get

ei6 f’(reie) = Il(r) -12(r)

 

 

 

where
w it
-TT (1 -2rcos(9 -t) +r)
and
1 7’ (1 -r2) (-2cos(6 -t) + 2r)f(eit)
12(r) = my 2 2 dt
-'IT (l-2rcos(9-t)+r)
Now
(1 -r)Il(r)
2r 1 ” (1-r2) it
= - T__)- f(e )dt
1+1: EI_W(1 -2rcos(9 -t) +r2)
4 - f(eie) for almost every 9 as r 4 1

1r 2 .
12(1-r) = 21? (1-r)j’ (1": )(‘zcc’sm'ung’ f(elt)dt

-1r (1 -2rcos(9 -t) +r)

 

+ f(elt)dt

 

(1 -r) I" (1 —r2) (2r —2)
Tr -1r (1 -2rcos(6 -t) + r?)

57

The second term of the expression for 12(1 -r) equals

— 2(l+r2) _1_ 7’

(1+ r)2 2” -1r

 

CD(r,9-t)f(eit)dt

4 - f(ele) for almost every 8 as r 4 l, by Lemma 2.2.

It is easy to see that the first term in 12(1 —r) tends to

 

 

zero as r 4 1—. Indeed
w 2 .
I§L(l-r) I (l -r )2r(l-—cos(6-Et;) f(elt)dtl
1r —1r (l-2rcos(6-t)+r)
w 2 .
g 2—17?(1 -r) f (1 ’r) 7 ‘f(elt) ldt

-w (1 -2rcos(9-—t)4—r )

g <1 -r)llfH. .

The proof is complete.

§3. Let 6b = {b 6 B’Ib’(z)(l-—Izl2) 4 O as ‘2] 4 1].

We give an example of a Blaschke product b i Eb.

Lemma 3.1. Let 0 < r < l.

 

Then
(i) If 0 < x < r < 1, then

-Ln(1 -x) < (l—ETS-x

 

O O 1 - X C l
(11) l-—xr decreases Wlth x on («n,f)
... l-x . . 1
(111) 1 -xr increases Wlth x on (-m,f)

 

The proofs are elementary and therefore omitted.

 

Lemma 3.2. Let on = 1-j% for n 2,1 and let
an+flfil 2
Bn = ———77——— for n 2’1. Then
a 5-
N
1_1_11II ll_af§l>o
N4a n=l

Proof: We first show that, for some constant cl

independent of N

o a 5
Z: - Ln(-2--§—) g_c < w
n=N+l 1_an BN 1

Indeed

E3 an-BN
-£n(r——)
n=N+1 'an BN
a (1 -a )(1+B )

= Z -Ln(l- n N
n=N+l l.-an BN

)

 

1 . (1 -an) (1+BN)

 

 

S (1

1 _ 'aN+1) n=N+1 (1 -an BN)

(1 + 3N)

 

l -GN+lEN

(use Lemma 3.1 ii and i)

B) E (l-cn)2

 

 

 

59

l+x . . 1
Now FEB—N increaseSW1th x on (-co, EN) .

SO

(1 -BN) (1 +an) (1 ‘31:) (1 +aN)
1 -c1n 3N S (1 “(IN BN)

 

 

 

 

 

 

Thus
N on -B
Z -Ln(-B——§-)
n=1 1 -an BN
N (1 -B )(l+a )
=Z-Ln(l- Tynan)
n=l n N
g (1 -aN BN) N (1 -BN) (1+an)
BN 'GN n=l 1 -an BN
(Lemma 2.2 iii and i)
g (1 ”an BN)(1—BN) g 2
BN-GN n=l u-anj
l “I; g n
g. _ (l -B ) 2.2
Mann “1? N 1
.1_
N
:4 21 2%.4 (2N+1-1)
2N+l
g 64 . Q.E.D.
a z-d
Let b(z) = n21 (far-1%) for ‘2] < l, where {an} was

defined in Lemma 3.2.

for each n such that lgngN .

60

Proposition 3.3. For the above Blaschke product b,

 

b’(x)(l-x) 7&0 as x41-

Proof: For suppose b'(x)(l-—x) 4 O as x 4 1.

Then given a positive 3, there exists N such that
Ib’(x)](l-—X) < e for every x in (dN,l)

le x in (aN,l) and pick n so that on < x < cn+1.

x
Now b(x) = f b’ so
a
n
n+1

1
b(x) e -——— dx = 6 Ln 2

n

Thus a contradiction to Lemma 3.2 is obtained, so

b'(x)(1-x) 7‘0 as x4l-.

We note that Proposition 3.3 combined with Corollary

1.9 shows that H__ is not compact.
b

BIBLIOGRAPHY

[ACP]

[AS]

[A]

[ACM]

[BS]

[Be]

[Bu]

[CCS]

[CW]

[CR]

[CRW]

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"I7'71((1111111))11(s