Document [original]

Weighted Spaces of holomorphic functions

on the upper halfplane

Dissertation

submitted by

Mohammad Ali Ardalani

Paderborn 2010

Dedicated to my Wife

Zusammenfassung

In dieser Dissertation studieren wir gewichtete Räume holomorpher Funktionen auf

der offenen oberen komplexen Halbebene Gfür zwei Arten von Gewichten, die wir

Typ(I)- und Typ(II)-Gewichte nennen.

Ein Typ(I)-Gewicht ist eine Gewichtsfunktion v, die nur von den Imaginärteilen

der Elemente aus Gabhängt. Ferner ist v(it)monoton aufsteigend in tund erfüllt

limt→0v(it) = 0.

Dagegen ist vein Typ(II)-Gewicht, wenn es

1. mit einem Typ(I)-Gewicht übereinstimmt auf allen ω∈Gmit |ω| ≤ 1und

2. die Symmetriebedingung v(ω) = v(−1

ω)erfüllt für alle ω∈G.

Ferner arbeiten wir mit einer Bedingung, die die Wachstumsrate dieser Gewichte kon-

trolliert. Unsere Gewichte sollen nicht zu schnell wachsen oder fallen.

Im Mittelpunkt stehen folgende Banachräume

Hv(G) := {f|f:G→Cholomorph und ||f||v<∞ } und

Hv0(G) := {f∈Hv(G)|fv verschwindet im Unendlichen }.

Dabei sei ||f||v= supω∈G|f(ω)|v(ω).

Für viele unserer Resultate verwenden wir die Möbiustransformation α:D→G

definiert durch α(z) = 1+z

1−zi. (Dist dabei die Einheitskreisscheibe.) Wenn vein

Typ(II)-Gewicht ist, so zeigt sich, dass v◦αäquivalent zu einem radialen Gewicht auf

Dist. Dies ermöglicht uns, die wohlbekannten Resultate bezüglich der isomorphen

Klassifizierung gewichteter Räume holomorpher Funktionen auf Dzu übertragen auf

Hv(G)und Hv0(G). Deshalb erhalten wir eine vollständige isomorphe Klassifizierung

für Hv(G)und Hv0(G)im Falle von Typ(II)-Gewichten v. Unter unseren Vorausset-

zungen ist dann z.B. Hv(G)immer isomorph zu l∞oder H∞(D).

Leider kann man nicht dieselbe Methode für Typ(I)-Gewichte verwenden, denn in

diesem Fall existiert limz→1(v◦α)(z)im Allgemeinen nicht und v◦αist nicht äquivalent

zu einem radialen Gewicht auf D. Deswegen beschränken wir uns bei Typ(I)-Gewichten

auf die folgenden Teilräume von Hv(G)und Hv0(G):

Uβ

±:= {f∈Hv(G)|ω2βf(ω) = ±f(−1

ω), ω ∈G}, Uβ

±,0:= Uβ

±∩Hv0(G),

H2π

v(G) := {f∈Hv(G)|fist 2π−periodisch }

und H2π

v0(G) := H2π

v(G)∩Hv0(G).Wir erhalten eine vollständige isomorphe Klassi-

fizierung dieser Räume. Wiederum gilt, dass z.B. H2π

v(G)und Uβ

±entweder isomorph

zu l∞oder H∞(D)sind. Weiterhin zeigen wir, dass Uβ

±und Uβ

±,0komplementäre

Teilräume von Hv(G)und Hv0(G)sind.

Schliesslich studieren wir die Stetigkeit von Differential-, Kompositions- und Mul-

tiplikationsoperatoren zwischen gewichteten Räumen holomorpher Funktionen auf G

und darüberhinaus zwischen gewichteten Räumen holomorpher 2π-periodischer Funk-

tionen. Wir erhalten hinreichende (und manchmal notwendige) Bedingungen für die

Stetigkeit dieser Operatoren, wenn unsere Gewichte den Typ(I) oder den Typ(II)

haben.

Introduction

The concept of weight and weighted space of holomorphic functions on the unit

disc D={z∈C:|z|<1}has been discussed by many authers, especially by Shileds

and Willams in a series of papers. [19–21]

They studied weighted spaces :

Hυ(D) = {f|f: Ω −→ Cis holomorphic and kfkυ<∞}.

and

Hυ0(D) = {f∈Hυ(D) : fυ vanishes at infinity},

(Here kfkυ:= supz∈D|f(z)|υ(|z|)).

when the weight function υsatiesfies certain properties which they called normality.

Their work opened a new field for research and many people studied these spaces from

different aspects [1–3,11,14,15]. Among these different aspects, two subjects seem to

be particulary interesting. Firstly, finding isomorphic classifications of these weighted

spaces as Banach spaces [10,12,13], secondly, studying operators such as composition,

multiplication and differentiation operators between these spaces. [5–9,17,23].

Unlike the unit disc, the case of upper halfplane G={ω∈C:Imω > 0}has not been

studied too much.

In this Ph.D thesis, we study weighted spaces of holomorphic functions for two kinds

of weights which we call type(I) and type(II) weights (see Definition 1.2.1).

We impose some conditions on our weights in order to control the rate of growth of

these weights. Our conditions are :

(∗)I,(∗)II and (∗∗). (See Definition 1.2.7).

In chapter one, firstly we collect some preliminary facts which we need in the next

chapters. Secondly, we study equivalent properties in order to characterize (∗)I,(∗)II

and (∗∗). Moreover we present some examples and counterexamples of type(I) and

type(II) weights with the above properties.

To obtain results about the Banach spaces

Hυ(G):={f|f:G−→ Cis holomorphic and kfkυ<∞} and Hυ0(G)=: {f∈Hυ(G):

fυvanishes at infinity}(when υis a weight of type(II) and has a moderate rate of

growth) we apply the Moebius transform α:D−→ Gdefined by α(z) = 1+z

1−zi.

In this case the weight υ◦αis equivalent to a radial weight on D(see Theorem

2.2.1). This enables us to transfer wellknown results about isomorphic classification

of weighted spaces of holomorphic functions on Dto Hυ(G) or Hυ0(G). Therefore we

present a complete isomorphic classification for Hυ(G) and Hυ0(G) in Theorem 2.2.3.

For example under the certain assumptions on υ,Hυ(G) is either isomorphic to `∞

or H∞(D).

In chapter three, in a similar way( by applying Theorem 2.2.1) we use the wellknown

results for differentiation, composition and multiplication operators on Hυ(D)to ob-

tain some results about operators between weighted spaces of holomorphic functions

on upper halfplane Gfor type(II) weights.

Unfortunately, for type(I) weights, we cannot use the same method, because in this

case limz→1(υ◦α)(z)does not exist and υ◦αis not equivalent to a radial weight on

D. Therefore for type(I) weights, we restrict ourselves to some special subspaces of

Hυ(G) and Hυ0(G) such as:

Uβ

±:= {f∈Hυ(G) : ω2βf(ω) = ±f(−1

ω)∀ω∈G},Uβ

±,0:= Uβ

±∩Hυ0(G),

H2π

υ(G) := {f∈Hυ(G) : fis 2π−periodic}

and

H2π

υ0(G) := {f∈Hυ0(G) : fis 2π−periodic}.

In Theorems 2.3.12 and 2.4.13 we obtain isomorphic classifications of Uβ

±&Uβ

±,0

and H2π

υ(G) & H2π

υ0(G) respectively. Again, we have, H2π

υ(G) and Uβ

±are isomorphic

to `∞or H∞(D).

In Theorem 2.5.4 we show that Uβ

±&Uβ

±,0are complemented subspaces of Hυ(G)

and Hυ0(G) respectively. Unfortunately, the isomorphic classifications of the comple-

ments of Uβ

±and Uβ

±,0are not known( see Remark 2.5.5).

Chapter four is devoted to studying operators between weighted spaces of holomorphic

functions for type(I) weights. In this chapter, we study the continuity of differenti-

ation and composition operators not only between weighted spaces of holomorphic

functions, but also between weighted spaces of 2π-periodic functions. Our results give

sufficient( and sometimes necessary) conditions for continuity of these operators.

Acknowledgements

I am heartily thankful to my supervisor, Prof. Dr. Wolfgang Lusky, whose encour-

agement, guidance, assistance and support from the initial to the final enabled me to

develop an understanding of the subject.

Last, but not least, I offer my regards and blessing to my Mother and all of those

who supported me in any respect during the completion of the project.

Mohammad Ali Ardalani

Contents

1 Preliminaries 1

2 Isomorphic classification of weighted spaces 11

3 Operators on weighted spaces for type(II) weights. 43

4 Operators on weighted spaces for type(I) weights 59

xii CONTENTS

Chapter 1

Preliminaries

Introduction to chapter one: The goal of this chapter is to collect the definitions

and lemmas which we need in the next chapters.

In section one, we define Hυ(Ω) and Hυ0(Ω) which have the main role throughout

this thesis. Also we introduce the open unit disk Dand upper halfplane Gof Cand a

Möbius transform αwhich maps Dbiholomorphically onto G.

In section two, we define two types of weights on Gand we investigate properties

concerning the rate of growth for some weights and discuss some examples.

2CHAPTER 1. PRELIMINARIES

Section one : Basic definitions and lemmas.

Definition 1.1.1 : Let Ωbe open subset of Cand f: Ω −→ Cbe a function.

a) A weight on Ωis a function υ: Ω −→ (0,+∞).

b) We define kfkυ:= supω∈Ω|f(ω)|υ(ω).

c) We define Hυ(Ω) := {f|f: Ω −→ Cis holomorphic and kfkυ<∞}.

d) We say fυ vanishes at infinity if for any  > 0there is a compact set K⊆Ωsuch

that|f(ω)|υ(ω)<  ∀ω∈Ω\K.

e) We define Hυ0(Ω) := {f∈Hυ(Ω) : fυ vanishes at infinity}.

Definition 1.1.2: a) The sets D={z∈C:|z|<1}and G={ω∈C:Imω >

0}are the unit disc and upper halfplane respectively.

b) For any δ > 0,we define Gδ:= {ω∈C:Imω ≥δ}.

c) Suppose x∈Cand r∈R(r > 0),then x+r∂Ddenotes the circle with the center

xand radius rin the complex plane.

In particular rD:= {rz :z∈D}denotes the disc with origin as center and radius rand

r∂D={z∈C:|z|=r}.

d) Suppose f: Ω ⊆C−→ Cis a complex function. we define

M∞(f, Ω) := supω∈Ω|f(ω)|

Definition 1.1.3: For any δ, δ ≥0we define Lδ:= {ω∈G:Imω =δ}.

In particular L0:= {ω∈G:Imω = 0}is the real line.

Definition 1.1.4: Define α:D−→ Cby α(z) = 1+z

1−zi.

Remark 1.1.5: Suppose z∈D,an easy computation shows that

α(z) = −2Imz

|z|2+1−2Rez +1−|z|2

|z|2+1−2Rez i.

Hence α(D)⊆G.Put β(ω) = ω−i

ω+i∀ω∈G,then we have

α◦β=id|Gand β◦α=id|D.Hence β=α−1and α(D) = G.

It is easily seen that α−1(ω) = |ω|2−1

|ω|2+1+2Imω −2Reω

|ω|2+1+2Imω i.

Lemma 1.1.6: If z∈Dthen

i) Imα(z) = 1−|z|2

|1−z|2.

ii) α(−z) = −1

α(z).

iii) α(z2) = α(z)

2−1

2α(z).

iv) Rez ≤0if and only if |α(z)|≤ 1.

v) α(δ

1+δ+1

1+δD\{1}) = Lδ∀δ > 0.

Proof: (i), (ii) and (iii) are trivial.

iv)Since |α(z)|2=|1+z

1−z|2=1+|z|2+2Rez

1+|z|2−2Rez so |α(z)|2≤1⇔Rez ≤0.

v)Put α(z) = ω. If Imα(z) = Imω =δthen we have 1−|z|2

1+|z|2−2Rez =δ⇔

1+δ=|z|2+δ

1+δ−2δ

1+δRez

⇔1

1+δ−δ

1+δ+δ2

(1+δ)2=|z−δ

1+δ|2⇔1

(1+δ)2=|z−δ

1+δ|2⇔| α−1(ω)−δ

1+δ|=1

1+δ.

So α−1maps the line Imω =δto the circle δ

1+δ+1

1+δ∂D.

Thus α−1(Lδ) = δ

1+δ+1

1+δD\{1}or equivalently α(δ

1+δ+1

1+δD\{1}) = Lδ.

Lemma 1.1.7: If f:G−→ Cis a holomorphic function, then there are αk∈Csuch

that f(ω) = P∞

k=0 αk(ω−i

ω+i)k∀ω∈G,where the series converges uniformly on compact

subsets of G.

Proof: Since the function αis holomorphic on Dso f◦αis holomorphic on Dand it

has a Taylor series representation.Hence (f◦α)(z) = P∞

k=0 αkzkfor some αkand the

series converges uniformly on compact subsets of D.Now the lemma follows from the

fact that if α(z) = ωthen z=α−1(ω) = ω−i

ω+i.

Definition 1.1.8: a) We recall that ∂D={z∈C:|z|= 1}.Let L∞(∂D)be the

space of all essentially bounded functions on ∂D,normed by the essential supremum

norm k.k∞.

b) Define H∞(D) = {g|g:D−→ Cis holomorphic and M∞(g, D)<∞}.

Theorem 1.1.9: To every g∈H∞(D)there corresponds a function g?∈L∞(∂D),defined

4CHAPTER 1. PRELIMINARIES

almost everywhere by g?(eit) = limr→1g(reit).Then equality M∞(g, D) = kg?k∞holds.

Proof: See [18] Theorem 11.3.2 .

Note that Theorem1.1.9 is also true for any translation of the unit disk D.

Lemma 1.1.10: Let τ > δ > 0and let f:G−→ Cbe holomorphic such that f|Gδ

is a bounded function. Then M∞(f, Lτ)≤M∞(f, Lδ).

Proof: Put D1=α−1(Gδ)and D2=α−1(Gτ)then ∂D1\{1}=α−1(Lδ)and

∂D2\{1}=α−1(Lτ).Put e

f(z) = (f◦α)(z).Since f|Gδis bounded so e

fis bounded

on D1.So e

f∈H∞(D1).Now using Theorem 1.1.9, there is a function e

f∗∈L∞(∂D)such

that M∞(e

f, D) = ke

f∗

|∂D1k∞.Since one point has a Lebesgue measure zero so

f∗·χ∂D1\ {1}k∞=ke

f∗

|∂D1k∞and we have

M∞(e

f, D1) = ke

f∗·χ∂D1\ {1}k∞(1).

where χis the characteristic function. With the similar argument we have

M∞(e

f, D2) = ke

f∗·χ∂D2\ {1}k∞(2).

Since D2⊆D1we have M∞(e

f, D2)≤M∞(e

f, D1).Now (1),(2) and Theorem 1.1.9

yield M∞(f, Lτ) = k˜

f∗

|∂D2k∞=M∞(e

f, D2)≤M∞(e

f, D1) = k˜

f∗

|∂D1k∞=M∞(f, Lδ).

Remark 1.1.11: Notice that in Lemma 1.1.10 the assumption fis bounded on Gδis

necessary. The lemma is not true if fis only bounded on Lδand Lτ.See the following

example.

Example 1.1.12: Define f:G−→ Cby f(ω) = eiω.Then |f|Lδ|=eδand

|f|Lτ|=eτ,but M∞(f, Lδ)< M∞(f, Lτ)whenever τ > δ > 0.

Section two : Characterization of certain types of positive weight functions.

Definition 1.2.1: Let υ be a continuous weight on G:

a) We say υis of type(I) if limr→0+υ(ir)=0and there is a constant C > 0such that

υ(ω1)≤Cυ(ω2)whenever Imω1≤Imω2.

b) We say υis of type(II) if there is a type(I) weight υ1and a constant C > 0such

that υ(ω) = υ1(ω)if |ω|≤ 1and υ(ω)

υ(−1

ω)≤Cfor any ω∈G.

Example 1.2.2: Define υ1, υ2, υ3:G−→ (0,+∞)by

υ1(ω) = (Imω)βfor some β > 0.

υ2(ω) = min((Imω)β,1) for some β > 0.

υ3(ω) = 









(1 −ln(Imω))γif Imω ≤1

1if Imω ≥1

for some γ, γ < 0.

and υk+3(ω) = υk(Imω

max(|ω|2,1) i)where k∈ {1,2,3}.

Then υ1, υ2and υ3are type(I) weights, but υk+3 are type(II) weights which are not of

type(I).

Proof: It is easy to see that υ1, υ2and υ3are type(I) weights. Indeed for all of them

the constant Cis equal to 1.

For showing that υk+3 (k∈ {1,2,3})are type(II) weights use the fact

Im(−1

ω) = Imω

|ω|2∀ω∈G.Since sup{υk+3(ω2)

υk+3(ω1):ω1, ω2∈Gand Imω1< Imω2}=∞for

k∈ {1,2,3},so υk+3 are not of type(I). For example put k= 1, ω0

n=n2iand

ωn=ni then Imωn< Imω0

nand υ4(ωn)

υ4(ω0

n)=nβ.So sup{υ4(ωn)

υ4(ω0

n):n∈N}=∞.

In the Example 1.2.2 all the weights depend only on the imaginary part. Now we pere-

sent weights which depend both on the real and imaginary parts.

Example 1.2.3: If υis a type(I)(type(II)) weight, then

υ7(ω) = υ(Imωi) arctan(|Reω |+√3) and υ8(ω) = υ(Imωi)(sin |ω|+2) are

6CHAPTER 1. PRELIMINARIES

type(I)(type(II)) weights.

Remark 1.2.4: a) Notice that a type(I) weight υis almost constant on the lines Lδ.

Since Imω1≤Imω2and Imω2≤Imω1whenever ω1, ω2∈Lδso we have

Cυ(ω1)≤υ(ω2)≤Cυ(ω1)if ω1, ω2∈Lδ.

This means that the weight υ1with υ1(ω) = υ(Imωi)satisfies

Cυ(ω)≤υ1(ω)≤Cυ(ω)for all ω∈G.

Therefore from now on we always assume a type(I) weight υsatiesfies the property

υ(ω) = υ(Imωi).By the preceding argument this is no loss of generality.

b) If υis a type(II) weight then by definition of type(II) weight there is a constant

C > 0such that 1

Cυ(ω)≤υ(−1

ω)≤Cυ(ω)for all ω∈G.

Lemma 1.2.5: Let υbe a type(I) weight on G.Define

υ1(ω0) = inf{υ(ω) : ω∈G, Imω ≥Imω0}for any ω0∈G.Then υ1(ω)is an increas-

ing function on the positive imaginary axis. Also there is a constant C > 0such that

υ1(ω)≤υ(ω)≤Cυ1(ω).

Proof: Suppose ω1, ω2∈Gare such that Imω1≤Imω2.Then

υ1(ω1) = inf{υ(ω) : Imω ≥Imω1} ≤ inf{υ(ω) : Imω ≥Imω2}=υ1(ω2).So υ1is an

increasing function on the positive imaginary axis. Clearly υ1(ω)≤υ(ω).If Imω0≤

Imω then υ(ω0)≤Cυ(ω),so 1

Cυ(ω0)is a lower bound for the set {υ(ω) : Imω ≥

Imω0},therefore 1

Cυ(ω0)≤inf{υ(ω) : Imω ≥Imω0}=υ1(ω0).So υ(ω0)≤Cυ1(ω0).

Since ω0∈Gis arbitrary we are done. 

Remark 1.2.6: From now on we assume that a type(I) weight satisfies

υ(ω) = υ(Imωi), ω ∈Gand υ(ω1)≤υ(ω2)whenever Imω1≤Imω2.

In view of Remark 1.2.4(a) and Lemma 1.2.5, this is no loss of generality.

Definition 1.2.7: A weight υon Gsatisfies

a) (∗)I(with respect to β)if there are constants C > 0, β > 0such that

υ(ω1)

υ(ω2)≤C(Imω1

Imω2)βwhenever Imω1≥Imω2.

b) (∗)II if there are constants C > 0, β > 0such that υ(ω1)

υ(ω2)≤C(Imω1

Imω2)βwhenever

Imω1≥Imω2and |ω1|,|ω2|≤ 1.

c) (∗∗)if there are constants C > 0, γ > 0such that υ(ω1)

υ(ω2)≥C(Imω1

Imω2)γwhenever

Imω1≥Imω2and |ω1|,|ω2|≤ 1.

It is clear that (∗)Iimplies (∗)II .In the following lemmas we characterize the properties

(∗)I,(∗)II and (∗∗).These lemmas enable us to present examples and also counterex-

amples for these properties.

Lemma 1.2.8: i) :If υis of type(I), then υsatiesfies (∗)I⇐⇒ supn∈Z

υ(2n+1i)

υ(2ni)<∞.

ii) Let υbe a weight on Gwith the property, there exists a C > 0such that

υ(ω1)≤Cυ(ω2)whenever Imω1≤Imω2≤1.Then υsatisfies

(∗)II ⇐⇒ supn∈N∪{0}

υ(2−ni)

υ(2−n−1i)<∞.

Proof: (i): =⇒:Suppose n∈Z,put ω1= 2n+1iand ω2= 2ni. Since υsatisfies (∗)Ithen

υ(2n+1i)

υ(2ni)≤C(2n+1

2n)β=C2βso supn∈N∪{0}

υ(2n+1i)

υ(2ni)<∞.

⇐=: Let ω1, ω2∈Gbe such that Imω1=t1≥Imω2=t2>0.We can find

n∈Zand k∈N∪{0}such that 2n< t2≤2n+1 & 2n+k< t1≤2n+k+1.Then

υ(ω1)

υ(ω2)=υ(t1i)

υ(t2i)≤υ(2n+k+1i)

υ(2ni)=Qk

j=0

υ(2n+j+1i)

υ(2n+ji)≤Ck+1 where C= supn∈Z

υ(2n+1i)

υ(2ni).

Now with β=ln C

ln 2 we have υ(t1i)

υ(t2i)≤Ck+1 = 2(k+1)β= 4β(2n+k

2n+1 )β≤4β(t1

t2)β.

(ii): =⇒:suppose n∈N∪{0}is arbitrary. Put ω1=1

2niand ω2=1

2n+1i,so

Imω1≥Imω2.Now since υsatisfies (∗∗)Iso there exist C > 0and β > 0such that

υ(2−ni)

υ(2−n−1i)≤C(2−n

2−n−1)β≤C2β.Therefore supn∈N∪{0}

υ(2−ni)

υ(2−n−1i)<∞.

⇐=:Let ω1, ω2∈Gwith Imω1=t1≥Imω2=t2and |ω1|,|ω2|≤ 1be given. We

can find nand k in N∪{0}such that 2−n−k−1< t2≤2−n−kand

2−n−1< t1≤2−n.Now an argument similar to what we have done in part(i)(⇐=)

completes the proof. 

8CHAPTER 1. PRELIMINARIES

Lemma 1.2.9: Let υbe a weight on Gwith the following property:

There exists a C > 0such that υ(ω1)≤Cυ(ω2)whenever Imω1≤Imω2≤1.

Then υ satisfies(∗∗)⇐⇒ infk∈Nlim supn→∞

υ(2−n−ki)

υ(2−ni)<1.

Proof: =⇒Since υsatisfies (∗∗)so there exist Cand γ(C, γ > 0) such that

υ(ω1)

υ(ω2)≥C(Imω1

Imω2)γwhenever |ω1|,|ω2|≤ 1and Imω1≥Imω2>0.Equivalently

υ(ω2)

υ(ω1)≤1

C(Imω2

Imω1)γ.Put ω1= 2−niand ω2= 2−n−k0iwhere k0has been selected such

that 1

C2−k0γ<1.So υ(2−n−k0í)

υ(2−ni)≤1

C(2−n−k0

2−n)γ=1

C2−k0γ<1.Thus

lim supn→∞

υ(2−n−k0i)

υ(2−ni)<1and therefore infk∈Nlim supn→∞

υ(2−n−ki)

υ(2−ni)<1.

⇐=: infk∈Nlim supn→∞

υ(2−n−ki)

υ(2−ni)<1implies that there are k0, n0∈Nand

q < 1with υ(2−n−k0i)

υ(2−ni)≤qfor n≥n0.

Since υ(2−n−ki)≤υ(2−n−k0i)for k≥k0we also have υ(2−n−ki)

υ(2−ni)≤q for n ≥n0,

k≥k0.Fix m0≥k0, n0.Then we obtain υ(2−m0(j+1)i)

υ(2−m0ji)≤qfor j= 0,1, ...

Now let ω1, ω2∈Gsuch that |ω1|,|ω2|≤ 1,0< Imω2≤Imω1≤1,say

2−m0(j+1) ≤Imω1≤2−m0j,2−m0(j+k+1) ≤Imω2≤2−m0(j+k)for some jand k.

If k≥2we have υ(ω1)

υ(ω2)=υ(Imω1i)

υ(Imω2i)≥υ(2−m0(j+1)i)

υ(2−m0(j+k)i)=Qj+1

l=j+k−1

υ(2−m0li)

υ(2−m0(l+1)i)≥(1

q)k−2=

q22γm0kwhere γ=ln(n−1)

m0ln 2 .

If k∈ {0,1}then there is a constant C≤1such that υ(ω1)

υ(ω2)≥1≥Cq22γm0k.

Hence υ(ω1)

υ(ω2)≥Cq22γm0k=C2−m0γq2(2−m0j

2−m0(j+k+1) )γ≥C2−m0γq2(Imω1

Imω2)γ.

Example 1.2.10: a) The weights υ1, υ2, υ3and υ4of Example 1.2.2 satisfy (∗)I,while

the weight υ7defined on Gby

υ7(ω) = 









1−ln(Imω)Imω ≤1

eImω−1Imω > 1

only satisfies (∗)II .

b) The weights υ1, υ2of Example 1.2.2 satisfy(∗∗),while the weight υ3in this example

does not satisfy (∗∗).

Proof: a) Clearly υ1and υ2satisfy(∗)I.Moreover we have

supn∈N∪{0}

υ3(2−ni)

υ3(2−n−1i)= supn∈N∪{0}

(1+nln 2)γ

(1+(n+1) ln 2)γ.Since limn→∞(1+nln 2

1+(n+1) ln 2 )γ= 1 for any

γ < 0so supn∈N∪{0}

υ3(2−ni)

υ3(2−n−1i)= 1 <∞.Also

supn∈N∪{0}

υ3(2n+1i)

υ3(2ni)= 1 by definition of υ3.Now part(ii) of the Lemma 1.2.8 implies

that υ3satisfies(∗)I.A similar proof shows that υ4satisfies (∗)I.Finally we have

υ7|Imω≤1=υ3for γ=−1so supn∈N∪{0}

υ7(2−ni)

υ7(2−n−1i)<∞.Now part(i) of the Lemma

1.2.8 implies that υ7satisfies (∗)II .But

supn∈N∪{0}

υ7(2n+1i)

υ7(2ni)= supn∈N∪{0}e2n+1−1

e2n−1= supn∈N∪{0}e2n=∞.

b) Let k∈Nbe arbitrary. υ1(2−n−ki)

υ1(2−ni)= 2−βk <1So infk∈Nlim supn→∞

υ1(2−n−ki)

υ1(2−ni)= 0 <

1.Now Lemma 1.2.9 implies that υ1satisfies (∗∗).Similary we can show that υ2satisfies(∗∗)too.

But for υ3,since υ3(2−n−ki)

υ3(2−ni)=(1+(n+k) ln 2)γ

(1+nln 2)γand limn→∞ (1+(n+k) ln 2)γ

(1+nln 2)γ= 1 ∀γ, γ < 0.we

have infk∈Nlim supn→∞

υ3(2−n−ki)

υ3(2−ni)= 1 .Thus υ3does not satisfy (∗∗).

We conclude this chapter with the following remark.

Remark 1.2.11: υ4is a type(II) weight which is not a type(I) weight, but it satisfies

(∗)I,so the assumption being of type(I) weight together with the following assumptions

supn∈N∪{0}

υ(2−ni)

υ(2−n−1i)<∞and supn∈N∪{0}

υ(2n+1i)

υ(2ni)<∞

is a sufficient condition for concluding that the weight υsatisfies (∗)I.Also note that

10 CHAPTER 1. PRELIMINARIES

only being of type(I) does not imply that the weight υsatisfies (∗)I. υ7is an example

of such a weight.

Chapter 2

Isomorphic classification of

weighted spaces

Introduction to chapter two: The aim of this chapter is to obtain isomorphic clas-

sification of weighted spaces of holomorphic functions on the upper halfplane Gand

some subspaces of them, by using the wellknown results for weighted spaces of holo-

morphic functions on the unit disc D. This chapter is divided in to five sections:

Section one includes preliminary definitions and wellknown theorems which are nec-

essary for obtaing our results in section two.

In section two we are dealing only with the weights of type(II) and at the end of this

section, we present a complete isomorphic classification of weighted spaces of holo-

morphic function, whenever the weight υhas a moderate rate of growth.

In the remainder of this chapter we study type(I) weights.

In the section three we define special subspaces of Hυ(G)and Hυ0(G)namely,Uβ

±,Uβ

±,0

(for definitions see 1.1.1 & 2.3.6) and we obtain their isomorphic classifications.

In section four we focus on the interesting subspaces of Hυ(G)and Hυ0(G),consisting

of 2π-periodic functions( see definition 2.4.3) and again we obtain their isomorphic

claasifications (see Theorem 2.4.13).

In section five, we return to the subspaces Uβ

±,Uβ

±,0and we will show that

Uβ

±,(Uβ

±,0)are complemented subspaces in Hυ(G)(Hυ0(G)).

12 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

Sections one: Wellknown results about isomorphic classification of

weighted spaces of holomorphic functions on the unit disc.

As we said before, in this section we mention the wellknown theorems which are nec-

essary for the next section.

Remark 2.1.1: We mentioned the definitions of Hυ(Ω) and Hυ0(Ω) in Definition

1.1.1. Note that in particular we are dealing with cases Ω = Gor Ω = D.

In particular Hυ(G)(Hυ(D)) is the weighted space of holomorphic functions on

the upper halfplane G(unit disc D)and Hυ0(G)(Hυ0(D)) is the subspace of Hυ(G)

(Hυ(D)) consisting of those functions which vanish at infinity.

Definition 2.1.2: Let υbe weight on D.Then

a) we say υis a radial weight on Dif υ(z) = υ(|z|)for all z∈D.

b) we say υis a standard weight on Dif υis a radial weight and it is a continuous,

non increasing function from [0,1[ into [0,+∞)such that lim|z|→1−υ(|z|)=0.

Remark 2.1.3: a) By lim|z|→1−|f(z)|υ(z)=0uniformly, we mean that

∀ > 0∃r()>0such that ∀z∈D|z|> r()|f(z)|υ(z)< .

b) It is easy to see that |f(z)|υ(z)vanishes at infinity iff

lim|z|→1−|f(z)|υ(z) = 0 (uniformly).

c) Note that f∈Hυ(D)implies that |f(z)|υ(z)is a bounded function on D.

Definition 2.1.4: a) Let I⊂Rbe an interval. A function f:I−→ [0,∞)is called

almost decreasing (resp. almost increasing) if there exsist a positive constant C, such

that for any x>y(resp. x>y)it follows that f(y)≤C f(x).

b) Let υbe a radial weight on D.We say υis almost decreasing (resp. almost increas-

ing) on D,if υ|[0,1[ is almost decreasing (resp. almost increasing) function.

Lemma 2.1.5: Let υbe a continuous, radial and almost decreasing weight on D.

Then υ1(r) := supt≥rυ(t)is a standard weight on Dand there is a constant C > 0such

that υ(r)≤υ1(r)≤Cυ(r).

Proof: Similar to Lemma 1.2.5. 

Definition 2.1.6: We define

a) `∞:= {(αk) : αk∈C& supk∈N|αk|<∞}.

b) c0:= {(αk)∈`∞: limk→∞ αk= 0}.

c) Hn:= {P|P:∂D−→ Cis a polynomial of degree≤n}(n∈N),where Hnis

endowed with the supremum norm on ∂D,that is kfk∞= supz∈∂D|f(z)|.

d) (Pn∈N⊕Hn)0:= {(Pn)|Pn∈Hn& limn→∞ kPnk= 0},where kPnk:=

supn∈NkPnk∞.

Remark 2.1.7: It is wellknown that (Pn∈N⊕Hn)∗∗

0is isomorphic to H∞(D).See [24]

Theorem 2.1.8: Let υbe a standard weight on D,which satisfies the condition

infn∈Nυ(1−2−n−1)

υ(1−2−n)>0.(∗)0.

Then

i) If υsatisfies condition (∗∗)0: infk∈Nlim supn→∞

υ(1−2−n−k)

υ(1−2−n)<1,

then Hυ0(D)is isomorphic to c0.

ii) If υdoes not satisfy condition (∗∗)0,that is infk∈Nlim supn→∞

υ(1−2−n−k)

υ(1−2−n)= 1,then Hυ0(D)

is not isomorphic to c0.

iii) If υsatisfies both conditions (∗)0and condition (∗∗)0,then Hυ(D)is isomorphic

to `∞.

iv) If υ satisfies condition (∗)0but not condition (∗∗)0,then Hυ(D)is isomorphic

to H∞(D).

Proof: For part (i) and (ii) see [14]. For part (iii) and (iv) see [13] Corollary 1.3. 

Corollary 2.1.9: Let υbe a standard weight on Dsatisfying (∗)0.Then there are

only two possibilities for isomorphism classes of Hυ(D).Either

i) Hυ(D)is isomorphic to `∞and Hυ0(D)is isomorphic to c0.

ii) Hυ(D)is isomorphic to H∞(D)and Hυ0(D)is isomorphic to (Pn∈N⊕Hn)0.

14 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

Remark 2.1.10: Note that Corollary 2.1.9 is also true without the assumption that

the weight υsatisfies (∗)0.See [12] Theorem 1.1

Section two: Isomorphic classification of weighted spaces of holomorphic

functions on the upper halfplane for type(II) weights.

We begin this section with the following theorem which is a new result and has a

main role in obtaining the isomorphic classification of weighted spaces of holomorphic

functions for type(II) weights.

Theorem 2.2.1: Let υbe a type(II) weight on Gsatisfying (∗)II .Put

˜υ(z) = υ(α(− | z|)) = υ(1−|z|

1+|z|i).Then ˜υ(z)is a radial weight on D.

Moreover the map T, defined by (Tf)(z) = f(α(z)) ∀z∈Dis an isomorphism

from Hυ(G)(Hυ0(G)) onto H˜υ(D)(H˜υ0(D)).

Proof: Clearly ˜υ(z) = ˜υ(|z|).Since υis of type(II) there is a constant C > 0such

that

Cυ(α(z)) ≤υ(α(−z)) ≤Cυ(α(z)).(1).

(See Lemma 1.1.6(ii) and Remark 1.2.4(b)).Also there is a constant C0>0such that

υ(ω1)≤C0υ(ω2)whenever |ω1|,|ω2|≤ 1and Imω1≤Imω2.

Consider a fixed z∈D.Firstly assume Rez ≤0.Since |z|≥ −Rez, so

1−|z|

1+|z|=1−|z|

1+|z|

1+|z|=1−|z|2

1+|z|2+2|z|≤1−|z|2

1+|z|2−2Rez =1−|z|2

|1−z|2=Im(α(z)) ≤1−|z|2

1+|z|2<1.Thus

˜υ(z)≤C0υ(α(z)).(2)

Since Im(α(z)) ≥Im(α(− | z|)) and υsatisfies (∗)II ,there exsist C00 and βsuch that

υ(α(z))

υ(α(−|z|)) ≤C00(

1−|z|2

|1−z|2

1−|z|

1+|z|

)β.So

υ(α(z)) ≤C0022βυ(α(− | z|)) = C0022β˜υ(z) (3)

The relations (2) and (3) give us

˜υ(z)≤C0υ(α(z)) ≤C0C0022β˜υ(z)whenever z∈Dand Rez ≤0.

Now if Rez > 0then Re(−z)<0.Using relations (1),(2) and (3) we have

˜υ(z) = ˜υ(−z)≤C0υ(α(−z)) ≤C0Cυ(α(z)) ≤C0C2υ(α(z)) ≤C0C2υ(α(−z)) ≤

C02C2C0022β˜υ(−z)≤C02C2C0022β˜υ(z).

16 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

So the weights υ◦αand ˜υare equivalent on D.Thus the map Tis welldefined

and g∈H˜υ(D)(H˜υ0(D)) if and only if g◦α−1∈Hυ(G)(Hυ0(G)).This proves the

theorem. 

We continue this section with the following theorem which yields the goal of this

section.

Remark 2.2.2: From now on the notation ∼has two meanings :

1.Between two Banach spaces means is isomorphic to.

2.Between two norms means is equivalent to.

Theorem 2.2.3: Let υbe a type(II) weight on Gwhich satisfies (∗)II .Then

a) The follwing are equivalent.

i)υsatisfies (∗∗).

ii)Hυ0(G)is isomorphic to c0.

iii)Hυ(G)is isomorphic to `∞.

b) The follwing are equivalent.

i)υdoes not satisfy (∗∗).

ii)Hυ(G)is isomorphic to H∞(D).

iii)Hυ0(G)is isomorphic to (Pn∈N⊕Hn)0.

Proof: By Theorem 2.2.1 we have

Hυ(G)∼H˜υ(D) & Hυ0(G)∼(H˜υ0(D)) (1)

where ˜υ(z) = υ(1−|z|

1+|z|i).

Since υis of type(II) there is a constant C > 0such that if |z1|<|z2| ⇒ 1−|z2|

1+|z2|<

1−|z1|

1+|z1|<1⇒˜υ(|z2|)≤C˜υ(|z1|).Thus ˜υis an almost decreasing weight on D.Now

define υ1(z) := υ1(|z|) = supt≥|z|˜υ(t).Lemma 2.1.5 implies that υ1is a decreasing

weight on Dand

υ1(z)≥˜υ(z)≥1

Cυ1(z).(2)

Relations (1) and (2) imply that

Hυ1(D)∼Hυ(G)and H(υ1)0(D)∼Hυ0(G).(3)

We have lim|z|→1−υ1(|z|) = lim|z|→1−(supt≥|z|˜υ(t)).Now use the definition of ˜υand

note that 1+|z|

1−|z|→ ∞ as |z|→ 1−,so lim|z|→1−υ1(|z|) = limt→∞ υ(ti).Since υis of

type(II) so there is a constant C0>0such that υ(ti)≤C0υ(−1

ti ) = C0υ(1

ti).

Thus limt→∞ υ(ti) = limt→∞ υ(1

ti) = 0.Therefore υ1is a standard weight in the sense

of Definition 2.1.2(b).

Since υsatisfies (∗)II ,so Lemma 1.2.8 implies supn∈N∪{0}

υ(1

2ni)

υ(1

2n+1 i)<∞.

Put a= supn∈N∪{0}

υ(1

2ni)

υ(1

2n+1 i)then ∀ ∈ N∪{0}we have υ(1

2n+1 i)

υ(1

2ni)≥1

a.Using relation (2) so

υ1(1 −1

2n+1 )≥˜υ(1 −1

2n+1 ) = υ(1

2n+2−1i)and 1

υ1(1−1

2n)≥1

Cυ(1

2n+1−1i).

Thus υ1(1−1

2n+1 )

υ1(1−1

2n)≥υ(1

2n+2−1i)

Cυ(1

2n+1−1i).Since υis of type(II) so υ(1

2n+2 i)≤Cυ(1

2n+2−1i)

and 1

υ(1

2n+1−1i)>1

Cυ(1

2ni).

Thus υ1(1−1

2n+1 )

υ1(1−1

2n)≥υ(1

2n+2−1i)

Cυ(1

2n+1−1i)≥

Cυ(1

2n+2 i)

C2υ(1

2ni)≥1

υ(1

2n+2 i)

υ(1

2n+1 i)

υ(1

2n+1 i)

υ(1

2ni)≥1

C3a2>0.

Therefore infn∈N∪{0}

υ1(1−1

2n+1 )

υ1(1−1

2n)>0.This means that υ1satisfies (∗)0in the theorem

2.1.8.

Proof of (a): i)⇒ii) : If υsatisfies (∗∗)then Lemma 1.2.9 implies that

infk∈Nlim supn→∞

υ(1

2n+ki)

υ(1

2ni)<1.With an argument similar to the above argument we

can conclude that

infk∈Nlim supn→∞

υ1(1−2−n−k)

υ1(1−2−n)<1.(∗∗)0

Part (i) of the Theorem 2.1.8 implies H(υ1)0(D)∼c0.Now relation (3) implies

Hυ0(G)∼c0.

ii⇒iii:Assume Hυ0(G)∼c0.Since Hυ0(G)∼H˜υ0(D)so H˜υ0(D)∼c0.Now using

Corollary 2.1.9 we have H˜υ(D)∼`∞.By Theorem 2.2.1 H˜υ(D)∼Hυ(G)so

Hυ(G)∼`∞.

iii⇒i:Assume Hυ(G)∼`∞.Using (3) we obtain Hυ1(D)∼`∞.Now use part (iv) of

Theorem 2.1.8 to see that υ1must satisfy (∗∗)0and it can be easily seen that υmust

satisfy (∗∗).

Proof of (b): i⇒ii:Since υdoes not satisfy (∗∗)so υ1does not satisfy (∗∗)0.

So using (iv) of Theorem 2.1.8 yields Hυ1(D)∼H∞(D).Again use (3) to show

18 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

Hυ(G)∼H∞(D).

ii⇒iii:Assume Hυ(G)∼H∞(D).By relation (3) we have Hυ1(D)∼H∞(D).Now

using part (ii) of the Corollary 2.1.9 we see that H(υ1)0(D)∼(Pn∈N⊕Hn)0.

Relation (3) gives us Hυ0(G)∼(Pn∈N⊕Hn)0.

iii⇒i:If Hυ0(G)∼(Pn∈N⊕Hn)0then by realation (3) we have H(υ1)0(D)∼(Pn∈N⊕Hn)0.

Now using part (i) of the Corollary 2.1.9 we obtain Hυ1(D)∼H∞(D).The weight

υsatisfies (∗).Assume υalso satisfies (∗∗).Then υ1satisfies both (∗)0and (∗∗)0and in

this case Theorem 2.1.8 (iii) implies Hυ1(D)∼`∞,which is a contradiction , so υdoes

not satisfy (∗∗).

Corollary 2.2.4: Let υbe a type(II) weight on Gwhich satisfies (∗)II .

Then

i) Hυ(G)∼Hυ0(G)∗∗.

ii) If υsatisfies (∗)Ithen Hυ0(G)has a Schauder basis.

Proof: i) By Theorem 2.2.3 Hυ(G)∼`∞or Hυ(G)∼H∞(D).

If Hυ(G)∼`∞then by Theorem 2.2.1 H˜υ(D)∼`∞.Now Corollary 2.1.9 implies

Hυ0(D)∼c0.Again by Theorem 2.2.1 we have Hυ0(G)∼c0.

Therefore Hυ0(G)∗∗ ∼c∗∗

0=`∞.So Hυ(G)∼Hυ0(G)∗∗.

If Hυ(G)∼H∞(D)again by Theorem 2.2.1 we have H˜υ(D)∼H∞(D).Now

Corollary 2.1.9(ii) implies H˜υ0(D)∼(Pn∈N⊕Hn)0.Again by Theorem 2.2.1

Hυ0(G)∼(Pn∈N⊕Hn)0.Therfore Hυ0(G)∗∗ ∼(Pn∈N⊕Hn)∗∗

0∼H∞(D).(See Re-

mark 2.1.7)

ii): Theorem 2.2.3 implies Hυ0(G)∼c0or Hυ0(G)∼(Pn∈N⊕Hn)0.it is known that

c0and (Pn∈N⊕Hn)0have Schauder basis.See [4] 

Corollary 2.2.5: Let υ be a standard weight satisfying (∗)0.Then

a) The following are equivalent.

i):υsatisfies (∗∗)0.

ii): Hυ(D)∼`∞.

iii): Hυ0(D)∼c0.

b) The following are equivalent.

i):υdoes not satisfy (∗∗)0.

ii): Hυ(D)∼H∞(D).

iii): Hυ0(D)∼(Pn∈N⊕Hn)0.

Proof: Use Theorem 2.1.8, Corollary 2.1.9 and an argument similar to what has been

done in the proof of the Theorem 2.2.3. 

20 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

Section three: Special subspaces of Hυ(G)and their isomorphism

classification for type(I) weights.

In this section we introduce special subspaces of Hυ(G)and Hυ0(G)(for type(I)

weights).

We will show under certain conditions that these subspaces are isomorphic to one of

the following spaces:H∞(D), `∞, c0or (Pn∈N⊕Hn)0.

Before we begin study of subspaces of Hυ(G)and Hυ0(G)we show that for the type(I)

weight υ(ω) = (Imω)β(for some β > 0) Hυ(G)and Hυ0(G)are isomorphic to `∞and c0

respectively.

Lemma 2.3.1: Let υ be a weight on G.Then the maps T0:Hυ(G)−→ Hυ◦α(D)or

1:Hυ0(G)−→ H(υ◦α)0(D)defined by T0(f) = f◦αand T0

1(f1) = f1◦αare onto

isometries.

Proof: If f∈Hυ(G)then f◦αis holomorphic on D.Since αis a one-to-one and onto

map on Dso

kT0(f)kυ◦α= supz∈D|f(α(z)) |υ(α(z)) = supω∈G|f(ω)|υ(ω) = kfkυwhere α(z) =

ω∀z∈D.Thus T0is a welldefined map.

Conversely if g∈Hυ◦α(D)then g◦α−1is holomorphic on Gand we have

kg◦α−1kυ= supω∈G|g(α−1(ω)) |υ(ω) = supz∈D|g(z)|υ(α(z)) = kgkυ.Therefore T0is

an onto map and T0(g◦α−1) = g.

Proof for the map T0

1is the same.We have only to show that if fυ vanishes at

infinity on G,then (f◦α)υ◦αvanishes at infinity on Dand if gvanishes at infin-

ity on Dthen g◦α−1vanishes at infinity on G.But the above assertions are true

since αand α−1are continuous maps. 

Lemma 2.3.2: Define υ(ω)=(Imω)β(for some β > 0). υ is of type(I) weight and

for this weight we have Hυ(G)∼`∞and Hυ0(G)∼c0.

Proof: Clearly υis of type(I). By definition of υwe have υ(α(z)) = (1−|z|2

|1−z|2)β.Now

define T:Hυ◦α(D)−→ H˜υ(D)and T1: (Hυ◦α)0(D)−→ H˜υ0(D)by T(f) = f

(1−z)2β

and T1(f1) = f1

(1−z)2βwhere ˜υ(z) = (1− | z|2)β

Claim: The map T(T1)is an onto isometry.

Now Lemma 2.3.1 and our claim imply that Hυ(G)∼H˜υ(D)and Hυ0(G)∼H˜υ0(D).

But it can be easily seen that ˜υsatisfies (∗)0and (∗∗)0therefore by Theorem 2.1.8 we

are done.

Proof of claim: Let f∈Hυ◦α(D)then f

(1−z)2βis a holomorphic function on D.Now

we have kTfk˜υ=kf

(1−z)2βk= supz∈D

f(z)

|1−z|2β(1− | z|2)β= supz∈D|f(z)|(1−|z|2

|1−z|2)β=

supz∈D|f(z)|υ(α(z)) = kfkυ◦α.So Tis welldefined and isometry map.

Suppose g∈H˜υ(D).Put f1(z) = (1 −z)2βg(z)then f1is a holomorphic function

on Dand clearly kf1kυ◦α=kgk˜υ.Therefore Tis an onto map. The proof for T1is

strightforward. 

Lemma 2.3.3: Let υ be a type(I) weight on G.Suppose a∈Ris arbitrary. Then the

translation operators Ta, Ta:Hυ(G)−→ Hυ(G), Ta:Hυ0(G)−→ Hυ0(G)defined

by (Taf)(ω) = f(ω+a)for all ω∈G,are uniformly bounded on Hυ(G)and Hυ0(G).

Proof: Clearly (Taf)is a holomorphic function on G.Since υis a type(I) weight and

Imω =Im(ω+a)∀ω∈G∀a∈R,so from part (a) of Remark 1.2.4 there exists a

constant C > 0such that 1

Cυ(ω)≤υ(ω+a)≤Cυ(ω).Thus

C|f(ω+a)|υ(ω)≤ | f(ω+a)|υ(ω+a)

≤C|f(ω+a)|υ(ω).

Therefore

kTafkυ= sup

ω∈G|f(ω+a)|υ(ω)

≤Csup

ω∈G|f(ω+a)|υ(ω+a)

=Csup

ω0∈G|f(ω0)|υ(ω0)

=Ckfkυ.

So it remains to prove that if f∈Hυ0(G)then Taf∈Hυ0(G).We must show that

for given  > 0there is a compact set K, K ⊂Gsuch that |f(ω0+a)|υ(ω0)<  for

22 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

all ω0∈G\K. Since f∈Hυ0(G)for given  > 0,there is a compact set K0∈Gsuch

that |f(ω)|υ(ω)<  ∀ω∈G\K0.

Define K=K0−a={ω−a:ω∈K0}.Clearly K⊂Gand Kis compact.We claim

|f(ω0+a)|υ(ω0)<  ∀ω0∈G\K. If there exist a ω0∈G\(K0−a)such that

|f(ω0+a)|υ(ω0)>  then, since ω0/∈(K0−a),we have ω06=ω−a∀ω∈K0or

ω0+a6=ω∀ω∈K0.Thus ω0+a∈G\K0,therefore |f(ω0+a)|υ(ω0+a)< . But

by using Remark 1.2.6 we have

 <|f(ω0+a)|υ(ω0)≤| f(ω0+a)|υ(ω0+a)<  which is a contradiction. 

Lemma 2.3.4: Let υ be a type(I) weight on G:

a) Then for any δ0>0there is a constant C > 0 (depending on δ0)such that for

any f∈Hυ(G)we have

Csup{M∞(f, Lδ)υ(δi):0< δ ≤δ0or δ ≥1

δ0}≤kfkυ≤Csup{M∞(f, Lδ)υ(δi):0<

δ≤δ0or δ ≥1

δ0}(1)

b) If υis bounded then for any δ0>0,there is a universal constant d > 0such that

for any f∈Hυ(G)

dsup{M∞(f, Lδ)υ(δi):0< δ ≤δ0}≤kfkυ≤dsup{M∞(f, Lδ)υ(δi):0< δ ≤δ0}.

Proof: a) Note that since υis of type(I), there is a constant C0>0such that

C0υ(ω1)≤υ(ω2)≤C0υ(ω1)whenever ω1, ω2∈Lδ(δ > 0).Since

kfkυ= supω∈G|f(ω)|υ(ω)we have

C0sup{M∞(f, Lδ)υ(δi) : δ > 0}≤kfkυ≤C0sup{M∞(f, Lδ)υ(δi) : δ > 0}.(2)

Thus, clearly for any δ0>0,

C0sup{M∞(f, Lδ)υ(δi):0< δ ≤δ0or δ≥1

δ0} ≤ kfkυ.If δ0≥1,then clearly

kfkυ≤C0sup{M∞(f, Lδ)υ(δi):0< δ ≤δ0or δ≥1

δ0}.

Now let 0< δ0<1be given. Since δ0<1we can divide the upper halfplane Gas

follows in to three parts.

Put G1:= {ω∈G: 0 < Imω ≤δ0}, G2:= {ω∈G:δ0≤Imω ≤1

δ0}and

G3:= {ω∈G:Imω ≥1

δ0}then

kfkυ= max{ kf|G1kυ,kf|G2kυ,kf|G3kυ}.(3)

Suppose δ0≤δ≤1

δ0.We want to estimate M∞(f, Lδ)υ(δi).Note since υ(δi)

υ(δ0i)and υ(δi)

υ(1

δ0i)

are continus functions on the compact set [δ0,1

δ0] (with respect to δ)there exist

constants C1, C2>0such that

υ(δi)≤C1υ(δ0i)and υ(δi)≤C2υ(1

δ0i).(4)

By the Phragmen-Lindelöf Theorem(See [18] Theorem 12.8) we have

M∞(f, Lδ)≤M∞(f, Lδ0)

δ0−δ

δ0−δ0M∞(f, L 1

δ0

)

δ−δ0

δ0−δ0

≤max(M∞(f, Lδ0), M∞(f, L 1

δ0

))

δ0−δ

δ0−δ0max(M∞(f, Lδ0), M∞(f, L 1

δ0

))

δ−δ0

δ0−δ0

≤max(M∞(f, Lδ0), M∞(f, L 1

δ0

)).(5)

If max(M∞(f, Lδ0), M∞(f, L 1

δ0

)) = M∞(f, Lδ0),then relations (4) & (5) imply that

M∞(f, Lδ)υ(δi)≤C1M∞(f, Lδ0)υ(δ0i).(6)

If max(M∞(f, Lδ0), M∞(f, L 1

δ0

)) = M∞(f, L 1

δ0

),then relations (4) & (5) imply that

M∞(f, Lδ)υ(δi)≤C2M∞(f, L 1

δ0

)υ(1

δ0i).(7)

Now relations (6) & (7) imply that

M∞(f, Lδ)υ(δi)≤C3sup{M∞(f, Lτ)υ(τi):0< τ ≤δ0or τ≥1

δ0}.(8)

where C3= max{C1, C2}.Now consider relations (3) & (8) and put C= max{C0, C3}we

are done.

Proof of b): Let δ0>0be given. By relation (1) we have

Csup{M∞(f, Lδ)υ(δi) : 0 < δ ≤δ0} ≤ kfkυ

Since f∈Hυ(G)and υis bounded so f|Gδis bounded. Now Lemma 1.1.10 implies

M∞(f, Lδ)≤M∞(f, Lδ0).Thus M∞(f, Lδ)υ(δi)≤M∞(f, Lδ0)υ(δ0i)υ(δi)

υ(δ0i)

Since υis bounded and it is almost increasing on the imaginary axis, there is a

constant C0>0such that υ(δi)

υ(δ0i)≤C0∀δ > δ0.So

sup{M∞(f, Lδ)υ(δi):0< δ < δ0} ≤ C0M∞(f, Lδ0)υ(δ0i).Put d= max{C, C0}.We

are done. 

Proposition 2.3.5: Suppose υis a bounded type(I) weight. Put

24 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

˜υ(z) = υ(α(− | z|)) = υ(1−|z|

1+|z|i).Then there are constants C1>0 & C2>0such that

C1kfkυ≤supa∈Rk(Taf)◦αk˜υ≤C2kfkυwhere Tais as in Proposition 2.3.3.

Proof: By part(b) of Lemma 2.3.4, there is a constant d > 0such that

dsup0<δ<1M∞(f, Lδ)υ(δi)≤ kfkυ≤dsup0<δ<1M∞(f, Lδ)υ(δi) (1)

Consider an arbitrarily fixed ω∈Gwith δ:= Imω < 1.Put a=Reω. Then since υis

a type(I) weight so there is a universal constant d1>0such that

|f(ω)|υ(ω)≤d1|(Taf)(iδ)|υ(δi)

Put z:= α−1(iδ) = δ−1

δ+1.Then |z|=−Rez =1−δ

δ+1 (since δ < 1) so δ=1−|z|

1+|z|and

|f(ω)|υ(ω)≤d1|(Taf)(1−|z|

1+|z|i)|υ(1−|z|

1+|z|i).Therefore

|f(ω)|υ(ω)≤d1|(Taf)(α(− | z|)) |υ(α(− | z|)).Since − | z|=Rez =zthe

previous relation implies

|f(ω)|υ(ω)≤d1supz∈D|(Taf)◦α(z)|˜υ(z) = d1kTaf◦αk˜υ

Therefore sup{| f(ω)|υ(ω) : ω∈G, Imω < 1} ≤ d1supa∈Rk(Taf)◦αk˜υ(2)

The relations (1) and (2) imply that there is a C1>0such that

C1kfkυ≤supa∈Rk(Taf)◦αk˜υ.Now consider a fixed z∈Dand a fixed a∈R.We

recall that 1−|z|

1+|z|≤1−|z|2

|1−z|2=Imα(z).Since υis of type(I) there is a constant d2>0such

that

υ(1−|z|

1+|z|i)≤d2υ(1−|z|2

|1−z|2i).So |(Taf)◦α(z)|˜υ(z)≤d2|(Taf)◦α(z)|υ(Im(α(z))i).

Put α(z) = ω. Then |(Taf)◦α(z)|˜υ(z)≤d2|(Taf)(ω)|υ(ω)≤d2kTafkυ≤d3kfkυ

The last inequality is a consequence of Propositon 2.3.3 so

supa∈Rsupz∈D|(Taf)◦α(z)|˜υ(z) = supa∈Rk(Taf)◦α(z)k˜υ≤d3kfkυ.

Definition 2.3.6: Let υbe a type(I) weight on G.Asumme that βis an even integer.

We define

Uβ

±:= {f∈Hυ(G) : ω2βf(ω) = ±f(−1

ω)∀ω∈G}&Uβ

±,0:= Uβ

±∩Hυ0(G).

Remark 2.3.7: a) Let f∈Hυ(G)(Hυ0(G)).Since (f◦α)(z)

(1−z)2βis a holomorphic function

on Dso there are γ0

k∈Csuch that (f◦α)(z)

(1−z)2β=P∞

k=0 γ0

kzk.Put α(z) = ω(so z=α−1(ω)).

Therefore f(ω) = P∞

k=0 γk1

(ω+i)2β(ω−i

ω+i)kwhere γk= 22βγ0

b) Note that in general Uβ

±and Uβ

±,0are not necessarily nontrivial subspaces of Hυ(G)

and Hυ0(G)respectively. But we will show that if the weight υsatisfies (∗)I,then

Uβ

±and Uβ

±,0(for certain βs)are infinite dimensional subspaces of Hυ(G)and Hυ0(G)

respectively.

Lemma 2.3.8: (i): For any f∈Uβ

+,there are γk∈Csuch that

f(ω) = P∞

k=0 γk(ω+i)−2β(ω−i

ω+i)2k.

ii) For any f∈Uβ

−,there are γk∈Csuch that f(ω) = P∞

k=0 γk(ω+i)−2β(ω−i

ω+i)2k+1.

Proof: we prove part (i). The proof of part (ii) is similar.

Suppose f∈Uβ

+.Put α(z) = 1+z

1−zi=ω, so α(−z) = −1

ω.

Since βis an even integer we obtain ω2β=(1+z)2β

(1−z)2β.Thus f(α(z))

(1−z)2β=f(α(−z))

(1+z)2β.Put

g(z) := f(α(z))

(1−z)2β.Then gis holomorphic on D(but not necessarily bounded)so there

are γ0

k∈Csuch that g(z) = P∞

k=0 γ0

kzk.Now suppose n∈Nis arbitrary and fixed.

Define hn(z) = 1

2πR2π

og(zeiφ)e−inφdφ. Since P∞

k=0 γ0

kzkis convergent on D,we have

hn(z) = 1

2πP∞

k=0 γ0

kzkR2π

oei(k−n)φdφ =γ0

nzn.Since nis fixed and g(z) = g(−z)we

obtain hn(z) = hn(−z).That is γ0

nzn=γ0

n(−z)n∀z∈D.This implies γ0

n= 0 if nis

odd. Thus g(z) = P∞

k=0 γ0

kz2k.So we have f(α(z))

(1−z)2β=P∞

k=0 γ0

kz2k.But

1−z= 1 −α−1(ω) = 2i

ω+i,therefore f(ω) = P∞

k=0 γk(ω+i)−2β(ω−i

ω+i)2k,where

γk= 2βγ0

k.

Remark 2.3.9: If the weight υon Gsatisfies (∗)I,we can choose a new β(perhaps

by increasing β)which is an even integer. From now on we always assume that βis

an even integer whenever υsatisfies (∗)Iwith respect to β.

Lemma 2.3.10: Let υbe a type(I) weight on Gsatisfying (∗)Iwith respect to β.

Then Uβ

±and Uβ

±,0are infinite dimensional subspaces of Hυ(G)and Hυ0(G)respectively.

Proof: Define fk(ω) = 1

(ω+i)2β(ω−i

ω+i)2kfor any k∈N∪{0},clearly

26 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

ω2βfk(ω) = fk(−1

ω)for any k∈N∪{0}.

kfkkυ= sup

ω∈G

|ω+i|2β|ω−i

ω+i|2kυ(ω)

≤sup

ω∈G

υ(ω)

|ω+i|2β

= max(sup{υ(ω)

|ω+i|2β:ω∈Gand Imω ≤1},

sup{υ(ω)

|ω+i|2β:ω∈Gand Imω > 1}).

But sup{υ(ω)

|ω+i|2β:ω∈Gand Imω ≤1} ≤ sup{υ(ω) : ω∈Gand Imω ≤1} ≤ υ(i).

Also sup{υ(ω)

|ω+i|2β:ω∈Gand Imω > 1}= sup{υ(ω)

υ(i)

|ω+i|2β:ω∈Gand Imω > 1}.

Now (∗)Iimplies that there is a constant C > 0such that

sup{υ(ω)

|ω+i|2β:ω∈Gand Imω > 1} ≤ C(Imω)β

|ω+i|2βυ(i)≤Cυ(i).

Therefore for any k∈N∪{0}we have

kfkkυ≤max(υ(i), Cυ(i)) <∞.(1)

Relation (1) implies fk∈Hυ(G)for any k∈N∪{0}.

Since |ω−i

ω+i|→ 1as |ω|→ 1so 1

(ω+i)2β(ω−i

ω+i)2k(1−(ω−i

ω+i)2) = fk(ω)−fk+1(ω)∈Uβ

+,0for

any k∈N∪{0}.

Similarly define gk(ω) := (ω−i

ω+i)fk(ω)for any k∈N∪{0},then gk(ω)∈Uβ

−and

gk(ω)−gk+1(ω)∈Uβ

−,0for any k∈N∪{0}.

Proposition 2.3.11: Let υbe a type(I) weight on Gsatisfying (∗)Iwith respect to β.

Put ˜υ(z) = υ(α(− | z|)) = υ(1−|z|

1+|z|i)on D.Then Uβ

±(Uβ

±, o)are isomorphic to H˜υ(D)

(H˜υ0(D)).

Proof: Firstly, note that without loss of generality we assume that the constants(C)

which appear in the definition of type(I) weight and relation (∗)Iare the same.

Since υsatisfies (∗)Iand Im(−1

ω) = Imω

|ω|2so there is a constant C > 0such that

υ(ω)

υ(−1

ω)≤









C|ω|2βif |ω|≥ 1

Cif |ω|≤ 1

(1)

Consider f∈Uβ

±and ω∈Gwith |ω|≥ 1.Then (1) implies that

|f(ω)|υ(ω) =|f(−1

ω)|υ(−1

ω)1

|ω|2β

υ(ω)

υ(−1

ω)≤C|f(−1

ω)|υ(−1

ω).(2)

If |ω|≥ 1then | −1

ω|≤ 1.So (2) implies that

sup{| f(ω)|υ(ω) : ω∈G,|ω|≥ 1}

≤Csup{| f(ω0)|υ(ω0) : ω0∈G,|ω0|≤ 1}.(3)

Clearly

sup{| f(ω)|υ(ω) : ω∈G,|ω|≥ 1}≤kfkυ=

max(sup{| f(ω)|υ(ω) : ω∈G,|ω|≤ 1},

sup{| f(ω)|υ(ω) : ω∈G,|ω|≥ 1}) (4)

Put this in (3). We obtain

sup{| f(ω)|υ(ω) : ω∈G,|ω|≤ 1}≤kfkυ≤Csup{| f(ω)|υ(ω) : ω∈G,|ω|≤

1}(5)

We recall that with z:= α−1(ω)we have Rez ≤0iff |ω|≤ 1.Also if z∈Dand

Rez ≤0then 1−|z|

1+|z|≤1−|z|2

|1−z|2=Im(α(z)) ≤1−|z|2

1+|z|2. υ is of type(I) hence

˜υ(z)≤Cυ(α(z)) ≤C2υ(1−|z|2

1+|z|2i)for some constant C > 0.Put ω1=1−|z|2

1+|z|2iand

ω2=1−|z|

1+|z|i.

Since we satisfies (∗)Iwe have

˜υ(z)≤Cυ(α(z)) ≤C2υ(1−|z|2

1+|z|2i)≤C322βυ(1−|z|

1+|z|i) = C322β˜υ(z).(6)

The above relation is true ∀z∈Dwith Rez ≤0.

For an arbitrary z∈Dwe have 1−|z|

1+|z|≤1−|z|2

1+|z|2=Im(α(z)) which yields

C1kf◦αk˜υ≤ kf◦αkυ◦αif f∈U±βwhere C1=1

C.Clearly kf◦αkυ◦α=kfkυ.

Now using relation (5) we have

C1kf◦αk˜υ≤ kf◦αkυ◦α=kfkυ≤Csup{| f(ω)|υ(ω) : ω∈G,|ω|≤ 1}.So

C1kf◦αk˜υ≤ kf◦αkυ◦α≤Csup{| (f◦α)(z)|υ◦α(z) : z∈Dand Rez ≤0}.(7)

Using (6) we have υ(α(z)) ≤C222β˜υ(z)∀z∈DRez ≤0.So

sup{| (f◦α)(z)|υ◦α(z) : z∈D&Rez ≤0} ≤ C2sup{| (f◦α)(z)|˜υ(z) : z∈

D&Rez ≤0} ≤ C2kf◦αk˜υ,(8)

28 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

where C2=C222β.Relations (7) and (8) yield

C1kf◦αk˜υ≤ kf◦αkυ◦α

≤sup{| f(α(z)) |υ(α(z)) : z∈D&Rez ≤0}

≤C2kf◦αk˜υ.(9)

By Lemma 2.3.8 there are γk∈Csuch that f◦α(z)

(1−z)2β=P∞

k=0 γkz2kif f∈Uβ

+and

f◦α(z)

(1−z)2β=P∞

k=0 γkz2k+1 if f∈Uβ

−(βis an even integer)

Put g(z) = f◦α(z)

(1−z)2β.Then |g(z)|=|g(−z)| ∀z∈D(∀f∈Uβ

±).Since

Re(−z) = −Rez we obtain

kgk˜υ= sup{| g(z)|˜υ(z) : z∈D&Rez ≤0}

= sup{| (f◦α)(z)|˜υ(z)1

|1−z|2β:z∈D&Rez ≤0}

≤sup{| (f◦α)(z)|˜υ(z) : z∈D&Rez ≤0}

≤ kf◦αk˜υ

≤Ckf◦αkυ◦α(By(9))

≤C C2kf◦αk˜υ

≤C3C2sup{| f◦α(z)|υ◦α(z) : z∈D&Rez ≤0}(By(7))

≤C724βsup{| (f◦α)(z)|˜υ(z) : z∈D&Rez ≤0}(By(6))

≤C724βsup{| g(z)|˜υ(z)|1−z|2β:z∈D&Rez ≤0}

≤C726βsup{| g(z)|˜υ(z) : z∈D&Rez ≤0}

=C726βkgk˜υ.

(Note that in the above computation we have used the relation g(z) = ±g(−z)).

So we have proved kgk˜υ≤ kf◦αk˜υ≤Ckfkυ≤C726βkgk˜υ.(10)

But (10) implies that the maps

T:Uβ

±−→ {g∈H˜υ(D) : g(z) = ±g(−z)∀z∈D}

T:Uβ

±,0−→ {g∈H˜υ0(D) : g(z) = ±g(−z)∀z∈D}

defined by T(f) = f◦α

(1−z)2βare isomorphisms. These maps are onto isomorphisms too.

For example let

g∈ {g∈H˜υ(D) : g(z) = g(−z)∀z∈D}be given. Put f(ω)=22βg◦α−1(ω)

(ω+i)2β.Since

kfkυ=kf◦αkυ◦αso kfkυ≤22βsupz∈D|g(z)|υ(α(z)).

Again use the relation (6) and facts g(z) = g(−z) & Re(z) = Re(−z)for all

z∈D,there exsists a constant d > 0such that kfkυ≤dkgk˜υ.Since kgk˜υ<∞

we obtain f∈Hυ(G).Also we have ω2βf(ω) = f(−1

ω)which implies f∈Uβ

Finally by definition we have T(f) = g. So for completing the proof we must show

that

A±={g∈H˜υ(D) : g(z) = ±g(−z)∀z∈D}is isomorphic to H˜υ(D)and

B±={g∈H˜υ0(D) : g(z) = ±g(−z)∀z∈D}is isomorphic to H˜υ0(D).Since υsatisfies

(∗)ILemma 1.2.8 (ii) implies that supn∈N∪{0}

υ(2−ni)

υ(2−n−1i)<∞.As in the proof of the

Theorem 2.2.3 we can conclude supn∈N∪{0}

˜υ(1−2−n)

˜υ(1−2−n−1)<∞.But this relation implies

that supr<1

˜υ(r2)

˜υ(r)<∞.Put

a= supr<1

˜υ(r2)

˜υ(r)<∞.So for all rwith 0≤r < 1we have ˜υ(r2)< a ˜υ(r).(11)

Define S+:H˜υ(D)−→ A+={g∈H˜υ(D) : g(z) = g(−z)∀z∈D}and

S+:H˜υ0(D)−→ B+={g∈H˜υ0(D) : g(z) = g(−z)∀z∈D}by

(S+g)(z) = g(z2).And define

S−:H˜υ(D)−→ A−={g∈H˜υ(D) : g(z) = −g(−z)∀z∈D}

S−:H˜υ0(D)−→ B−={g∈H˜υ0(D) : g(z) = −g(−z)∀z∈D}by (S−g)(z) = zg(z2)

If we prove that S+, S−are isomorphisms then

S−1

+T:Uβ

+(Uβ

+,0)−→ H˜υ(D)(H˜υ0(D)) and S−1

−T:Uβ

−(Uβ

−,0)−→ H˜υ(D)(H˜υ0(D)) are

isomorphisms and proof will be complete.

30 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

S+is a welldefined map and

k(S+g)k˜υ= sup

z∈D|g(z2)|˜υ(z)

= sup

z∈D|g(z2)|˜υ(z2)˜υ(z)

˜υ(z2)

≤Csup

z∈D|g(z2)|˜υ(z2)

≤Ckgk˜υ.

So S+is bounded.(Note that ˜υis almost decreasing)

Define (S−1

+g)(z) = g(z1

2)then (S−1

+S+)(g) = gand

k(S−1

+g)k˜υ= supz∈D|g(z1

2)|˜υ(z) = supz∈D|g(z1

2)|˜υ(z1

2)˜υ(z)

˜υ(z1

2)≤akgk˜υby(11)

So S+is an onto isomorphism. Clearly S−is a welldefined map.

k(S−g)k˜υ= supz∈D|z| | g(z2)|˜υ(z)≤supz∈D|g(z2)|˜υ(z2)˜υ(z)

˜υ(z2)≤Ckgk˜υ

So S−is bounded. If g∈A−or g∈B−then there are αk∈Csuch that

g(z) = P∞

k=0 αkz2k+1 =zP∞

k=0 αkz2k.Put f(z) = P∞

k=0 αkzk.Then

(S−1

−g)(z) = f(z) = z−1

2g(z1

S−1

−is bounded because

k(S−1

−g)k˜υ= sup

z∈D|(S−1

−g)(z)|˜υ(z)

= max( sup

|z|≤ 1

2|(S−1

−g)(z)|˜υ(z),sup

|z|≥ 1

2|(S−1

−g)(z)|˜υ(z)).

But by using maximum principle and the fact ˜υis almost decreasing we have

sup

|z|≤ 1

2|(S−1

−g)(z)|˜υ(z)≤Csup

|z|=1

2|(S−1

−)g(z)|˜υ(1

2)˜υ(0)

˜υ(1

≤C1sup

|z|=1

2|(S−1

−g)(z)|˜υ(1

2).

So k(S−1

−g)k˜υ≤C1sup|z|≥ 1

2|(S−1

−g)(z)|˜υ(z)≤C2kgk˜υ

Thus S−1

−is an onto isomorphism and we are done. 

Theorem 2.3.12: Let υbe a type(I) weight on Gsatisfying (∗)Iwith respect to β.

a) The following are equivalent.

i)Uβ

+, Uβ

−are isomorphic to `∞.

ii)Uβ

+,0, Uβ

−,0are isomorphic to c0.

iii)υsatisfies (∗∗).

b) The following are equivalent.

i)Uβ

+, Uβ

−are isomorphic to H∞(D).

ii)Uβ

+,0, Uβ

−,0are isomorphic to (Pn∈N⊕Hn)0.

iii)υdoes not satisfy (∗∗).

Proof:(a): i)⇒ii) : By Proposition 2.3.11 we have Uβ

+∼H˜υ(D)so H˜υ(D)∼

`∞.Using Corollary 2.1.9(i) we conclude that H˜υ0(D)∼c0.Now Proposition 2.3.11

implies that Uβ

+,0∼c0and Uβ

−,0∼c0.

ii)⇒iii) : Our assumption and Proposition 2.3.11 imply H˜υ0(D)∼c0.Now Theorem

2.1.8(ii) implies ˜υsatisfies (∗∗)0,so υsatisfies (∗∗).

iii)⇒i) : υsatisfies (∗∗)so ˜υsatisfies (∗∗)0.Now Theorem 2.1.8(i) implies that H˜υ0(D)∼

c0.Now Corollary 2.1.9 implies that H˜υ(D)∼`∞.Using Proposition 2.3.11 we are

done.

Proof of (b): i)⇒ii)By Proposition 2.3.11 we have H˜υ(D)∼H∞(D).Now Corol-

lary 2.1.9(ii) implies that H˜υ0(D)∼(Pn∈N⊕Hn)0.But Proposition 2.3.11 implies that

Uβ

+,0∼H˜υ0(D)∼(Pn∈N⊕Hn)0and Uβ

−,0∼(Pn∈N⊕Hn)0.

ii)⇒iii) : By Proposition 2.3.11 we have H˜υ0(D)∼(Pn∈N⊕Hn)0.If υsatisfies (∗∗)

then ˜υsatisfies (∗∗)0.Now Theorem 2.1.8(i) imples H˜υ0(D)∼c0which is a contradic-

tion, so υdoes not satisfy (∗∗).

iii)⇒i) : υdoes not satisfy (∗∗),so ˜υdoes not satisfy (∗∗)0.Thus by Theorem 2.1.8(ii)

we have H˜υ0(D)is not isomorphic to c0.Now Corollary 2.1.9 implies that

H˜υ(D)∼H∞(D)and Proposition 2.3.11 gives us the result. 

32 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

Section four: Isomorphism classification of weighted spaces of 2π-periodic

holomorphic functions as subspace of Hυ(G)or Hυ0(G)

Definition 2.4.1: A function f:G−→ Cis called r−periodic for some r > 0

if f(ω+r) = f(ω)for all ω∈G.

Proposition 2.4.2: Suppose a > 0is arbitrary and fixed. Put fa(ω) = f(aω).If fis

r- periodic then fais r

aperiodic.

Proof:fa(ω+r

a) = f(aω +r) = f(aω) = fa(ω)∀ω∈G.

Definition 2.4.3: For any r∈R, r > 0we define

υ(G) := {f∈Hυ(G) : fis r−periodic}

and

υ0(G) := {f∈Hυ0(G) : fis r−periodic}.

Lemma 2.4.4: Let υbe a type(I) weight on Gsatisfying (∗)I.Then the operator Ua

defiend by Uaf=fa, f ∈Hr

υ(G)(f∈Hr

υ0(G)),is an isomorphism between

υ(G)(Hr

υ0(G)) and H

υ(G)(H

υ0(G)).

Proof: We have |(Uaf)(ω)|υ(ω) =|f(aω)|υ(aω)υ(ω)

υ(aω)≤ kfkυυ(ω)

υ(aω).

If a≥1then, since υis of a type(I), there exists a constant C1>0such that υ(ω)≤

C1υ(aω).

If a < 1then since υis satisfies (∗)Iso there exist C2>0and β > 0such that υ(ω)

υ(aω)≤

C2(1

a)β.Put C= max{C1, C2},so we have kUak ≤ Cmax(1,(1

a)β)for any a >

0.Clearly U−1

a=U1

aand similarly we have kU−1

ak=U1

a≤Cmax(1, aβ).So Uais a

bounded operator with bounded inverse U1

aand we are done. 

Remark 2.4.5: (a): Lemma 2.4.4 shows that without loss of generality we can always

assume r= 2π.

b) By Remark 1.2.4(a) we know that a type(I) weight υdepends only on the imaginary

part of ω(up to a universal constant).Now by a result of Stanev either Hυ(G) = {0}or

there is an integer b0>0such that υ(ω)≤eb0Imω for all ω∈G.See [22]

For avoiding the triviality we assume Hυ(G)6={0}.Hence there is a smallest integer bwith

υ(ω)≤ebImω for all ω∈G.Thus |eibω |υ(ω) = e−bImωυ(ω)is bounded but

|eikω |υ(ω)is unbounded for any integer k < b. Since υis of type(I), υ(ti)is almost

increasing for t > 0.This implies that b≥0.

Proposition 2.4.6: Let υbe a type(I) weight on G.Then for each f∈H2π

υ(G)there

exsist γk∈Csuch that f(ω) = P∞

k=bγkeikω.Here the series converges uniformly on

the compact subsets of Gand bis as in Remark 2.4.5(b).

Proof: Define τ:G−→ Gby τ(ω) = 1+eiω

1−eiω i. Indeed τ(ω) = α(eiω).Also define

η:G\ {i} −→ Gby η(ω) = −ilog(ω−i

ω+i).Then (τ◦η)(ω) = ωfor all ω∈

G\{i}and (η◦τ)(ω) = ωif −π < Reω < π. Indeed

η(τ(ω)) = (−i) log(τ(ω)−i

τ(ω)+i)and τ(ω)−i

τ(ω)+i=eiω,so η(τ(ω)) = (−i) log(τ(ω)−i

τ(ω)+i).Since

arg eiω =Reω and −π < Reω < π we obtain for the main branch of the logarithm

η(τ(ω)) = ω.

Claim: Suppose f∈H2π

υ(G).Then f◦ηis holomorphic on G\{i}.

Proof of the claim : Put g(z) = f◦η◦α(z) = f((−i) log z)for all z∈D\{0}.So g

is holomorphic on D\{z∈D:Imz = 0 & Rez ≤0}.Now consider another branch

of the logarithm, say,g

log z, g

log z= log |z|+i]arg zwhere 0<]arg z < 2π. For this

branch of the logarithm, define ˜η(ω)=(−i) log ω−i

ω+iso we have

log z=









log zif 0<arg z < π

log z+ 2πif −π < arg z < 0

Since f∈H2π

υ(G),hence f((−i) log z) = f((−i)g

log z).Therefore we have

f◦η◦α=f◦˜η◦α. (1)

But log zis holomorphic on D\{z∈D:Imz = 0 & Rez ≤0}and g

log zis holo-

morphic on D\{z∈D:Imz = 0 & Rez ≥0}.So using (1) we see that f◦η◦αis

holomorphic on D\{0}.Now since αis holomorphic on Dso f◦ηis holomorphic on

G\{i}.Since f◦ηis holomorphic onG\{i}function gis holomorphic on D\{0}so

34 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

there exist γk∈Csuch that

g(z) = P+∞

k=−∞ γkzk∀z∈D\{0}.

Thus f(ω) = f((η◦τ)(ω)) = P+∞

k=−∞ γk(τ(ω)−i

τ(ω)+i)k=P+∞

k=−∞ γkeikω if −π < Reω < π.

Since fis 2π−periodic so f(ω) = P+∞

k=−∞ γkeikω ∀ω∈G.

Using orthogonality we have 1

2πR2π

0f(ω+λ)e−ikλdx =γke−ikω.So

|γkeikω |υ(ω) = 1

2π|R2π

0f(ω+λ)e−ikλdx |υ(ω)≤sup0<λ<2π|f(ω+λ)|υ(ω)≤

Csup0<λ<2π|f(ω+λ)|υ(ω+λ)≤Ckfkυ<∞.

(Here Cis the universal constant which appears in the definition of type(I) weight).So

γkeikω ∈Hυ(G)∀k∈Z.

If k < b then |eikω |υ(ω)is unbounded by the choice of b. If there is k < b such

that γk6= 0 then γkeikω /∈Hυ(G)which is a contradiction. So for all k < b we

have γk= 0.Therefore f(ω) = P+∞

k=bγkeikω.

Remark 2.4.7: a) Notice that in Proposition 2.4.6 we did not require the property (∗)I.

we only assumed that υis a type(I) weight.

b) Take bas in Remark 2.4.5(b) and put υb(ω) = e−bImωυ(ω) =|eibω |υ(ω)∀ω∈

G.We recall that υbis a bounded weight on G.

Proposition 2.4.8: Define the operator T, by (Tf)(ω) = eibωf(ω).Then Tmaps

Hυ(G),(Hυ0(G)) isometrically onto Hυb(G),(H(υb)0(G)). T also maps

H2π

υ(G),(H2π

υ0(G)) isometrically onto H2π

υb(G),(H2π

(υb)0(G)).

Proof: Clearly if fis 2π−periodic then e−ibωf(ω)is 2π−periodic and if

f∈Hυ0(G)then e−ibωf(ω)∈H(υb)0(G).Also we have

kTfkυb= supω∈G|e−ibω || f(ω)|| eibω |υ(ω) = supω∈G|f(ω)|υ(ω) = kfkυ.

Definition 2.4.9: Define ˜υ(z) := υb(ln( 1

|z|)i) =|z|bυ(ln( 1

|z|)i)for all z∈D\{0}

and put ˜υ(0) = 0.Clearly ˜υ(z) = ˜υ(|z|)and lim|z|→1−˜υ(|z|) = 0.

Remark 2.4.10: We should notice that the weight ˜υin Definition 2.4.9 may be non-

continuous at zero. But we replace it by an equivalent weight ˜υ1which is continuous

on [0,1].Define

˜υ1(z) := 









˜υ(z)if |z|≥ 1

˜υ(1

2)if |z|≤ 1

Put C=sup|z|≤1

2˜υ(z)

˜υ(2−1).If h∈H˜υ(D)and z0∈Dsuch that |z0|≤ 1

2then

˜υ(z0)

˜υ(2−1)≤C. So ˜υ(z0)≤C˜υ(1

2).Also by maximum modulus principle we have

|h(z0)|˜υ(z0)≤Csup|z|=1

2|h(z)|˜υ(1

2).Therefore we obtain

khk˜υ1≤ khk˜υ≤Ckhk˜υ1.So k.k˜υand k.k˜υ1are equivalent.

Theorem 2.4.11: The operator S, defined by (Sf)(z) = f(−ilog z)∀z∈D\{0}maps

H2π

υb(G)(H2π

(υb)0(G)) isometrically to H˜υ(D)(H˜υ0(D)).

Proof: υis a type(I) weight, so kfkυis equivalent to supω∈G|f(ω)|υ(Imωi)and

kfkυbis equivalent to supω∈G|f(ω)eibω |υ(Imωi) = supω∈G|f(ω)|e−bImωυ(Imωi).

For a f∈H2π

υb(G)put g(ω) = eibωf(ω).Hence

kgkυ= supω∈G|f(ω)|| eibω |υ(ω) = kfkυb<∞.So g∈H2π

υ(G).Then by Propo-

sition 2.4.6 there are γk∈Csuch that g(ω) = P∞

k=bγkeikω and therefore f(ω) =

P∞

k=bγkei(k−b)ω.We have

(Sf)(z) = f(−ilog z) = P∞

k=bγkei(k−b)(−i) log z=P∞

k=bγkzk−b.Put k−b=jso (Sf)(z) =

P∞

j=0 γj+bzj.

kfkυb= sup

ω∈G|f(ω)|υb(Imωi)

= sup

ω∈G|

∞

k=b

γkei(k−b)ω|| eibω |υ(Imωi)

= sup

ω∈G|

∞

k=b

γkeikω |υ(Imωi).

Put eiω =zso ω= ln( 1

|z|)i+ arg z. Therefore υ(Imωi) = υ(ln( 1

|z|)i).So

kfkυb= sup

z∈D|

∞

k=b

γkzk| | z−b| | zb|υ(ln( 1

|z|)i)

= sup

z∈D|

∞

k=b

γkzk−b|˜υ(z) = sup

z∈D|(Sf)(z)|˜υ(z).

Therefore we have

kfkυb=k(Sf)k˜υ.(1)

36 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

Relation (1) implies that S is an isometry from H2π

υb(G)to H˜υ(D).A similar argument

shows that Sis also an isometry from H2π

(υb)0(G)in to H˜υ0(D).

Define S−1:H˜υ(D)−→ H2π

υb(G)or S−1:H˜υ0(D)−→ H2π

(υb)0(G)by (S−1h)(ω) =

h(eiω).Then (S−1S)(f)(z) = f(z)and (SS−1)(h)(ω) = h(ω).Relation (1) also implies

that S−1is bounded on H˜υ(D)(H˜υ0(D)) and we have k(S−1h)(ω)kυ=khk˜υ.this proves

the assertion of the theorem. 

Proposition 2.4.12: Let ˜υbe as in Definition 2.4.9.Then ˜υis almost decreasing on

the set {z∈D:|z|≥ 1

2}.

Proof: We have ˜υ(z) =|z|bυ(ln( 1

|z|)i)∀z∈D\{0}.Suppose 1

2≤t≤s≤1.Then

ln 1

t≥ln 1

s.Since υis of type(I) there is constant C > 0such that

Cυ(ln(1

t)i)≥υ(ln(1

s)i),so

Ctbυ(ln(1

t)i)≥sbυ(ln(1

s)i)( t

s)b

≥(1

2)bsbυ(ln(1

s)i).

Thus ˜υ(s)≤C2b˜υ(t)and we are done. 

Theorem 2.4.13: Let υbe a type(I) weight on Gsatisfying (∗)II .Then

i) H2π

υ(G)(H2π

υ0(G)) is isomorphic to `∞(c0)⇔υsatisfies (∗∗).

ii) H2π

υ(G)(H2π

υ0(G)) is isomorphic to H∞(D)((Pn∈N⊕Hn)0)⇔υdoes not satisfy (∗∗).

Proof: Using Proposition 2.4.8, Theorem 2.4.11 and Remark 2.4.10 we have

H2π

υ(G)∼H˜υ1(D).(1)

and

H2π

υ0(G)∼H(˜υ1)0(D).(2)

By Proposition 2.4.12 ˜υ1is almost decreasing. So using Lemma 2.1.5, there is a de-

creasing weight which is equivalent to ˜υ1.We call it again ˜υ1(Note that ˜υ1is a standard

weight). υ satisfies (∗)II ,so Lemma 1.2.8 implies that supn∈N∪{0}

υ(2n+1i)

υ(2ni)<∞.

An argument similar to what has been done in the Theorem 2.2.3 shows that

infn∈N∪{0}

˜υ1((1−1

2n+1 ))

˜υ1((1−1

2n)) >0.(∗)0

If υsatisfies (∗∗),then Lemma 1.2.9 implies that infk∈Nlim supn→∞

υ(2−n−ki)

υ(2−ni)<1.

So infk∈Nlim supn→∞

˜υ1((1−2−n−k))

˜υ1((1−2−n)) <1.(∗∗)0

Proof of (i): ⇒:If H2π

υ(G)∼`∞then by relation (1) H˜υ1(D)∼`∞.Now, by part

(iii) and (iv) of Theorem 2.1.8, ˜υ1satisfies (∗∗)0.So υsatisfies (∗∗).If H2π

υ0(G)∼

c0then, by relation (2),H(˜υ1)0(D)∼c0.Now part (ii) of Theorem 2.1.8 imples

that ˜υ1satisfies (∗∗)0.So υsatisfies (∗∗).

⇐:If υsatisfies (∗∗)then ˜υ1satisfies (∗∗)0.Since ˜υ1satisfies both (∗)0and (∗∗)0so The-

orem 2.1.8(iii) implies that H˜υ1(D)∼`∞.By relation (1),we have H2π

υ(G)∼`∞.Also

Theorem 2.1.8(i) implies H(˜υ1)0(D)∼c0.Relation (2) implies H2π

υ0(G)∼c0.

Proof of (ii): ⇒:We have H2π

υ(G)∼H∞(D),so, by relation (1) H˜υ1(D)∼H∞(D).

Now Theorem 2.1.8(iii) implies that ˜υ1does not satisfy (∗∗)0,so υdoes not satisfy (∗∗).

If H2π

υ0(G)∼(Pn∈N⊕Hn)0then, by realation(2), H( ˜υ1)0(D)∼(Pn∈N⊕Hn)0.

Corollary 2.1.9(i) implies that H˜υ1(D)∼H∞(D).Now Theorem 2.1.8 implies that ˜υdoes

not satisfy (∗∗)0,therefore υdoes not satisfy (∗∗).

⇐:If υsatisfies (∗)but not (∗∗),then ˜υ1satisfies (∗)0but not (∗∗)0.Now part

(iii) of Theorem 2.1.8 implies that H˜υ1(D)∼H∞(D).By using relation (1) we

obtain H2π

υ(G)∼H∞(D).Also Corollary 2.1.9(ii) implies that

H( ˜υ1)0(D)∼(Pn∈N⊕Hn)0.Now using relation (2) we are done.

Theorem 2.4.13 implies:

Corollary 2.4.14: Let υbe a type(I) weight on Gsatisfying (∗)II .Then

either H2π

υ(G)is isomorphic to `∞and H2π

υ0(G)is isomorphic to c0

H2π

υ(G)is isomorphic to H∞(D)and H2π

υ0(G)is isomorphic to (Pn∈N⊕Hn)0.

38 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

Section five: Some complemented subspaces of Hυ(G)and Hυ0(G).

In this section we will prove Uβ

±, Uβ

±,0are complemented subspaces of Hυ(G)and

Hυ0(G)respectively for a type(I) weight υ.

Before we come to the main result of this section we recall some elementary facts from

the geometry of Banach spaces.

Lemma 2.5.1: Suppose Xis a vector space and Aand Yare subspaces of X. If Yis

finite codimensional in X, then A∩Yis finite codimensional in A.

Proof: We have dim X

Y=n < ∞,so there are x1, ..., xn∈Xsuch that

{x1+Y, ..., xn+Y}is a basis for X

Y.Consider an arbitrary x∈X. Then there exist

unique α1, ..., αn∈Csuch that

x+Y=α1(x1+Y)+...+αn(xn+Y)=(α1x1+...+αnxn)+Y, so x−(α1x1+...+αnxn)∈

Y. Now define φ1, ..., φn:X−→ Cby φk(x) = αkfor 1≤k≤n, so

Y=Tn

k=1 ker φk.Thus Y∩A=Tn

k=1 ker(φk|A).Therefore dim A

A∩Y<∞.

Lemma 2.5.2: Suppose Aand Bare closed subspaces of the Banach space Xand

Bis subspace of A(B⊆A).If Bis a complemented subspace in Xand dim A

∞,then Ais a complemented subspace in X, too.

Proof: Since Bis a complemented subspace in X, there is a bounded projection

P:X−→ Bsuch that X=B⊕ker P. Now the assumption B⊆Aimplies

A=B⊕ker(P|A).Since dim A

B=n < ∞there is a finite dimensional subspace

of A, say Nsuch that A=B⊕Nand N=h{a1, ..., an}i where {a1+B, ..., an+B}is

a basis for A

B.Comparing the relations A=B⊕Nand A=B⊕ker(P|A),we

conclude that ker(P|A)is a finite dimensional subspace of X, so there is a bounded

projection Q:X−→ ker(P|A).

Now we claim P+Q(id −P) : X−→ Ais a bounded projection. Hence Ais comple-

mented in X. Since Pand Qare bounded operators so P+Q(id −P)is a bounded

operator. We obtain

(P+Q(id −P))(P+Q(id −P)) = P2+PQ(id −P) + Q(id −P)P

+Q(id −P)Q(id −P)

=P+0+0+Q(id −P)

so P+Q(id −P)is a projection. We have P(X) = B⊆Aand

Q(id −P)(X)⊆ker(P|A)so

P(X) + Q(id −P)(X)⊂A. (1)

Suppose a∈A. P(a)+(Q(id −P))(a) = P(a)+(id −P)(a) = a. So

(P+Q(id −P))|A=id|A(2)

Relations (1) and (2) imply that P+Q(id −P)maps Xonto A. 

Lemma 2.5.3: Suppose Aand Yare closed subspaces of the Banach space X. Also

assume that dim X

Y<∞and P:Y−→ Ais a bounded linear operator with P|A∩Y=

id. Then A∩Yand Aare complemented subspaces of X.

Proof: Since dim X

Y<∞,by Lemma 2.5.1 we have dim( A

A∩Y)<∞.Thus there is a

bounded projection Q1:A−→ A∩Y. Since

(Q1PQ1P)(y) = (Q1P)[(Q1P)(y)]

=Q1(P(y))

= (Q1P)(y)

we obtain that (Q1P) : Y−→ A∩Yis a bounded projection.

dim X

Y<∞implies that Yis complemented in X. So there is a bounded projection

Q2:X−→ Y:= rangeQ2.Note that

(Q1PQ2)(Q1PQ2)(x) = (Q1PQ2)(Q1P(Q2(x)))

= (Q1P)(Q1P)Q2(x)

= (Q1PQ2)(x)

Thus Q1PQ2:X−→ A∩Yis a bounded projection. Therefore A∩Yis a comple-

mented subspace of X. Now Lemma 2.4.2 implies Ais also a complemented subspace

40 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

of X. 

Theorem 2.5.4: Let υbe a type(I) weight on Gsatisfying (∗)Iwith respect to β.(Where

βis a positive even intiger).

Then Uβ

±(Uβ

±,0)are complemented subspaces of Hυ(G)(Hυ0(G)).

Proof: We prove the theorem for Uβ

+.The proofs for Uβ

−, Uβ

±,0are similar.

ω2β+1 has 2βzeros of the form cos( 1

2β(π+2πk))+isin( 1

2β(π+2πk)) for 0≤k≤2β−1.

βof this zeros, say ω1, ..., ωβare in G.Put h(ω) = ω2β+ 1 then h(ω)and 1

ω2β+ 1

have the same zeros in G.Also, if ωkis a zero of h(ω),then −1

ωkis a zero of h(ω)too.Define

X:= {f∈Hυ(G) : f(ωk)=0, k = 1,2, ..., β}.

Claim 1: Xis closed and finite codimensional in Hυ(G).

Proof of Claim 1: For each k, 1≤k≤β, define the evaluvation functional

δωk:Hυ(G)−→ Cby δωk(f) = f(ωk).We have

|δωk(f)|=|f(ωk)|υ(ωk)1

υ(ωk)≤1

υ(ωk)kfkυ.So for 1≤k≤β, δωkis a bounded linear

functional. In fact kδωkk ≤ 1

υ(ωk).Now since X=Tβ

k=1 ker δωk,so Xis closed and finite

codimensional in Hυ(G).

Define P:X−→ Hυ(G)by (Pf)(ω) = 1

ω2β+1(f(ω) + f(−1

ω)).

Claim 2: Pis welldefined.

Proof of Claim 2: Firstly note that for any x∈Cand any r > 0by B(x, r)we

mean the set {ω∈C:|ω−x|< r}.Also B(x, r)is the closure of the set B(x, r).

We must prove Pf ∈Hυ(G)∀f∈X. Suppose f∈X, then to any ωk(1 ≤k≤

β)there coresponds a rk>0such that f(ω) = P∞

j=1 αkj(ω−ωk)jin B(ωk, rk).Put

r=1

3min{rk: 1 ≤k≤β},so B(ωk, r)∩B(ωk0, r) = ∅whenever 1≤k, k0≤βand k6=

k0.Since −1

ωk∈ {ω1, ..., ωβ}(1 ≤k≤β)we have −1

ωk∈Gand f(−1

ωk)is holomorphic

on G.So to any ωk(1 ≤k≤β)there coresponds a r0

k>0such that

f(−1

ωk) = P∞

j=1 γkj(ω−ωk)jin B(ωk, r0

k)for suitable γkj∈C.

Put r0=1

3min{r0

k: 1 ≤k≤β},so B(ωk, r0)∩B(ωk0, r0) = ∅whenever 1≤k, k0≤

βand

k6=k0.Put r00 =1

3min{r, r0}and Uk=¯

B(ωk, r00)for 1≤k≤β. Clearly, for any

1≤k≤β, Ukis compact, Uk∩Uk0=∅whenever 1≤k, k0≤βand k6=k0and the Tay-

lor series of f(ω)and f(−1

ω)are convergent in Uk.Put U=Sβ

k=1 Uk.If ω∈G\U, then

|(Pf)(ω)|υ(ω)≤| f(ω)|υ(ω)

|ω2β+1|+|f(−1

ω)|υ(−1

ω)1

|ω2β+1|

υ(ω)

υ(−1

ω)

Put d= infω∈G\U|ω2β+ 1 |>0,so 1

|ω2β+1|<1

d∀ω∈G\U. Since υis a type(I)

weight which satisfies (∗)I,there exsist constants C1, C0such that

supω∈G\U1

|ω2β+1|

υ(ω)

υ(−1

ω)≤supω∈G\U











C1|ω|2β

|ω2β+1|if |ω|≥ 1

|ω2β+1|if |ω|≤ 1

If |ω|is sufficently large then |ω|2β

|ω2β+1|<1,if 1<|ω|<2then

|ω|2β

|ω2β+1|<22β

d.So |ω|2β

|ω2β+1|<max(1,22β

d)for all ω∈G\U, |ω|≥ 1.

Therefore there is a universal constant C > 0such that

supω∈G\U|(Pf)(ω)|υ(ω)≤Ckfkυ.(1).

Now let ω∈U, say ω∈Uk(Since Uk∩Uk0=∅if k6=k0,so we have that ωis only an

element of one Uk).

Since ω2β+ 1 = (ω−ω1)(ω−ω2), ..., (ω−ω2β),where ωk(1 ≤k≤2β)are the zeros

of ω2β+ 1 we have

f(ω)

ω2β+1 =1

πj6=k(ω−ωj)P∞

j=1 αkj(ω−ωk)j−1and f(−1

ω)

ω2β+1 =1

πj6=k(ω−ωj)P∞

j=1 γkj(ω−ωk)j−1.

Thus f(ω)

ω2β+1 and f(−1

ω)

ω2β+1 are bounded on Uk.So

supω∈U|(Pf)(ω)|υ(ω)<∞.(2).

Relations (1) and (2) imply supω∈G|(Pf)(ω)|υ(ω) = kPfkυ<∞.So Pf ∈Hυ(G).

Clearly ω2β(Pf)(ω) = (Pf)(−1

ω),so Pf ∈Uβ

+.Also if f∈X∩Uβ

+,then

(Pf)(ω) = 1

ω2β+1(f(ω) + ω2βf(ω)) = f(ω).So P|X∩Uβ

+=id.

Claim 3: Pis a bounded map on X.

Proof of Claim 3: We use the closed graph theorem. Suppose (f, g)∈graph(P)closure,

then there exists a sequence (fn, Pfn)∈graph(P)such that

(fn, Pfn)−→ (f, g)so 









fn−→ f

Pfn−→ g

(pointwise).If Pf =gthen graph(P)is closed

42 CHAPTER 2. ISOMORPHIC CLASSIFICATION OF WEIGHTED SPACES

and Pis bounded. Xis a closed subspase of Hυ(G)so fn−→ f(pointwise)implies

f∈Xand Pf ∈Hυ(G).Also fn−→ f(pointwise)implies 









fn(ω)−→ f(ω)

fn(−1

ω)−→ f(−1

ω)

(pointwise).Thus

g= lim

n→∞(Pfn)(ω)

= lim

n→∞

ω2β+ 1(fn(ω) + fn(−1

ω))

ω2β+ 1(f(ω) + f(−1

ω))

=Pf.

We have Uβ

+and Xare closed subspace of Hυ(G),dim Hυ(G)

X<∞,

P:X−→ Uβ

+⊆Hυ(G)is bounded and P|Uβ

+∩X=id. Now use Lemma 2.5.3

with A=Uβ

+and Y=X, we are done. 

Remark 2.5.5: Looking more closely, the proof of the Lemma 2.5.4 reveals that the

complement of Uβ

+in Hυ(G)is U0

−:= {f∈Hυ(G) : f(ω) = −f(−1

ω), ω ∈G}up to a

finite dimensional subspace. Unfortunately the isomorphic classification of U0

−is not

known.

Chapter 3

Operators on weighted spaces for

type(II) weights.

Introduction to chapter three: There are many wellknown results about differen-

tiation, composition and multiplication operators on the weighted spaces of holomor-

phic functions on the unit disc D. In this chapter we want to obtain similar results for

weighted spaces of holomorphic functions on the upper halfplane, where our weights

are of type(II) and satisfy certain properties. For reaching this aim Theorem 2.2.1 is

a very useful tool. We divide this chapter in to three sections:

In section one we are dealing with the differentiation operator between two weighted

spaces of holomorphic functions and we study boundedness and surjectivity of this

operator.

In section two we obtain necessary and sufficient conditions for continuity and com-

pactness of composition operators between two weighted spaces of holomorphic func-

tions.

In section three, we study multiplication operators on a weighted space of holomorphic

functions into itself. We want to know under which conditions such a multiplication

operator is:

a) a Fredholm operator

44CHAPTER 3. OPERATORS ON WEIGHTED SPACES FOR TYPE(II) WEIGHTS.

b) a continuous operator

c) an isomorphism.

Section one: Differentiation operator between weighted spaces of holomor-

phic functions on the upper halfplane for type(II) weights.

We recall that a standard weight u on Dsatisfies:

(*)0if infn∈Nu(1−2−n−1)

u(1−2−n)>0

(∗∗)0if infk∈Nlim supn→∞

u(1−2−n−k)

u(1−2−n)<1.

For arriving at the main theorem of this section we need the following wellknown result

on the unit disc.

Definition 3.1.1: For any open set O⊆Cand any holomorphic function f:O−→ C

we put Df =f0.

Theorem 3.1.2: Suppose uis a standard weight on Dwhich satisfies (∗)0.Then

i) The differentiation operator D, D :Hu(D)−→ Hu1(D),is bounded where

u1(t) = (1 −t)u(t)

ii) Let usatisfy (∗∗)0in addition. Then Dis a surjective map.

iii) If udoes not satisfy (∗∗)0,then Dis not a surjective map.

Proof: See [16] Theorem 3.1.

Lemma 3.1.3: Theorem 3.1.2remains valid if uis an almost decreasing radial weight

(i.e u(z) = u(|z|)is an almost decreasing function) on the unit disc D.

Proof: Let ube an almost decreasing weight on D,then (1 −t)u(t)is also an almost

decreasing weight on D.By Lemma 2.1.5 u∗(z) = supt≥|z|u(t)is a decreasing weight

on D.Now consider the following diagram

D:Hu(D)−→ H(1−t)u(t)(D)

↓id ↓id1

D1:Hu∗(D)−→ H(1−t)u∗(t)(D)

where Dand D1are differentiation operators and id1again is the identity map.

Clearly id1◦D=D1◦id. Now we have if D1is bounded (surjective,nonsurjective),

then Dis bounded (surjective,nonsurjective), respectively.Now using Theorem 3.1.2

46CHAPTER 3. OPERATORS ON WEIGHTED SPACES FOR TYPE(II) WEIGHTS.

we are done. 

Proposition 3.1.4: Let υbe a type(II) weight on Gsatisfying (∗)II .Define

υ2:G−→ (0,+∞)by

υ2(ω) = 









Imω υ(Imωi)if |ω|≤ 1

Im(−1

ω)υ(Im(−1

ω)i)if |ω|≥ 1

Then υ2is a type(II) weight on Gwhich satisfies (∗)II .

Proof: Since limδ→0+υ(x+iδ) = 0 for all x∈R,so limδ→0+υ2(x+iδ) = 0 for

all x∈R.Let ω0∈Gwith |ω0|= 1 be given. We have

limω→ω0|ω|≤1υ2(ω) = limω→ω0|ω|≤1(Imω υ(Imω i)) and

lim

ω→ω0|ω|≥1υ2(ω) = lim

ω→ω0|ω|≥1(Im(−1

ω)υ(Im(−1

ω)i))

= lim

ω→ω0|ω|≤1(Imω υ(Imω i)).

So limω→ω0|ω|≤1υ2(ω) = limω→ω0|ω|≥1υ2(ω).Since υis continuous so υ2is continuous.

If |ω|≤ 1then υ2(ω) = Imω υ(Imωi).Since υis a type(II) weight so there is a

type(I) weight υ1such that υ2(ω) = Imω υ1(Imωi)whenever |ω|≤ 1.

Now suppose |ω1|,|ω2|≤ 1and Imω1≤Imω2,thus there exists a constant Csuch

that υ2(ω1)≤Cυ2(ω2).Also by definition of υ2we have υ2(ω)

υ2(−1

ω)= 1 ∀ω∈G.So υ2is

a type(II) weight. Suppose |ω1|,|ω2|≤ 1and Imω1≥Imω2.Since

υsatisfies (∗)II ,there exist C > 0and β > 0such that

υ2(ω1)

υ2(ω2)=Imω1υ(Imω1i)

Imω2υ(Imω2i)≤CImω1

Imω2(Imω1

Imω2)β=C(Imω1

Imω2)β+1.

So υ2satisfies (∗)II .

Proposition 3.1.5: Let υbe a type(II) weight on Gwhich satisfies (∗)II .Then the

weights ˜υ(z) = υ(α(− | z|)) = υ(1−|z|

1+|z|i)and ˜υ3(z) = υ((1− | z|)i)are equivalent.

Proof: For any z∈D,we have 1−|z|

1+|z|≤1− | z|≤ 1.Since υis of type(II) there is a

constant C > 0such that

υ(1−|z|

1+|z|i)≤Cυ((1− | z|)i).Since υsatisfies (∗)II ,there exist C1>0and β > 0

such that υ((1−|z|)i)

υ(1−|z|

1+|z|i)≤C12β.Thus we are done. 

Remark 3.1.6: Theorem 2.2.1 and Proposition 3.1.5 imply that the map

T:Hυ(G)−→ H˜υ3(D)defined by T(f) = f◦αis an isomorphism.

Lemma 3.1.7: Let υbe a type(II) weight on Gwhich satisfies (∗)II .Define

δ:Hυ(G)−→ Hυ2(G)by δ(f) = P∞

k=1 kαk(ω−i

ω+i)k−1where f(ω) = P∞

k=0 αk(ω−i

ω+i)kand

υ2is as in Proposition 3.1.4.Then δis a bounded (welldefined) map.

Proof: Define ˜

δ:H˜υ3(D)−→ H˜υ2(D)by ˜

δ(˜

f) = δ(f)◦αwhere

˜υ3(z) = υ((1− | z|)i),˜υ2(z) = υ2((1− | z|)i)and ˜

f=f◦αfor all f∈Hυ(G).

Clearly δ(f)◦α=D(˜

f).Also

˜υ2(z) = υ2((1− | z|)i) = (1− | z|)υ((1− | z|)i) = (1− | z|)˜υ3(z).Now Lemma 3.1.3

and Theorem 3.1.2 imply that there is a constant C0>0such that

k˜

δ(˜

f)k˜υ2≤C0k˜

fk˜υ3.(1)

But by Proposition 3.1.4, Proposition 3.1.5, Remark 3.1.6 and Theorem 2.2.1 we have

Hυ(G)∼H˜υ3(D)and Hυ2(G)∼H˜υ2(D).Now the relation equivalent with relation

(1) on Gis kδ(f)kυ2≤Ckfkυfor some universal constant C. 

Lemma 3.1.8: Let υbe as in the Lemma 3.1.7.If υsatisfies (∗∗),then

δ:Hυ(G)−→ Hυ2(G)is a surjective map, where υ2is as in Proposition 3.1.4.

Proof: Let ˜

δ:H˜υ3(D)−→ H˜υ2(D)be as in Lemma 3.1.7. Since υsatisfies (∗)II and (∗∗)

so ˜υ3satisfies (∗)0and (∗∗)0.Now Theorem 3.1.2 (ii) implies that ˜

δis surjective.

Suppose g∈Hυ2(G),then ˜g=g◦α∈H˜υ2(D).(Use Theorem 2.2.1 and Proposition

3.1.5 for υ2).Since ˜

δis surjective there exists a

˜g1∈H˜υ3(D)such that ˜

δ(˜g1) = ˜gwhere g1∈Hυ(G)and ˜g1=g◦α.

So δ(g1)◦α=g◦αor equivalently δ(g1) = g. Thus δis a surjective map.

Lemma 3.1.9: Define T:Hυ2(G)−→ H:= {2i

(ω+i)2h|h∈Hυ2(G)}by

T(h) = 2i

(ω+i)2h. Then H⊆Hυ2(G)and Tis bounded.

Proof: Firstly note that:

If ω∈Gand ω=x+iy then |ω+i|2=x2+ (y+ 1)2.Since y > 0we have 1

|ω+i|2<1

48CHAPTER 3. OPERATORS ON WEIGHTED SPACES FOR TYPE(II) WEIGHTS.

for all ω∈G.Suppose g∈H,then there is a h∈Hυ2(G)such that g=2i

(ω+i)2h.

kgkυ2=k2i

(ω+i)2hkυ2

= sup

ω∈G|2i

(ω+i)2h(ω)|υ2(ω)≤sup

ω∈G|2i

(ω+i)2|sup

ω∈G|h(ω)|υ2(ω)

≤2khkυ2.

Since h∈Hυ2(G)we obtain kgkυ2≤2khkυ2<∞.Hence H⊆Hυ2(G)and

kT(h)kυ2≤2khkυ2.So kTk ≤ 2.

Corollary 3.1.10: Let υbe a type(II) weight on Gwhich satisfies (∗)II .Then the

differentiation operator D:Hυ(G)−→ Hυ2(G)is a bounded operator.

Proof: Consider the composition map T◦δ:Hυ(G)−→ H.

(T◦δ)(f) = T(δ(f)) = 2i

(ω+i)2δ(f) = D(f)for any f∈Hυ(G)so D=T◦δ. Now using

Lemma 3.1.8 and Lemma 3.1.9 we deduce Dis bounded. 

Lemma 3.1.11: Let υbe a type(II) weight on Gwhich satisfies (∗)II .Then the dif-

ferentiation operator D:Hυ(G)−→ His a surjective map.

Proof: Clearly Tis a surjective map. Lemma 3.1.8 implies, δis a surjective map.

Therefore T◦δ=Dis a surjective map. 

Lemma 3.1.12: If the differentiation operator D, D :Hυ(G)−→ H,is a surjective

map, then υsatisfies (∗∗).

Proof: If υdoes not satisfy (∗∗)then ˜υ3does not satisfy (∗∗)0.Since Dis surjective, D =

T◦δand Tis surjective so δis surjective.

Consider ˜g=g◦α∈H˜υ2(D),then g∈Hυ2(G).Since δis surjective there exists

af∈Hυ(G)such that δ(f) = g. Let ˜

D:H˜υ3(D)−→ H˜υ2(D)be the differentiation

operator. We have δ(f)◦α=g◦αor equivalently

D(˜

f) = ˜g, which means ˜

Dis surjective. Since ˜υdoes not satisfy (∗∗)0,surjectivity

of ˜

Dis a contradiction with the Theorem 3.1.2(iii).Therefore υmust satisfy (∗∗).

Now we summarize what we have obtained in the following theorem.

Theorem 3.1.13: Let υbe a type(II) weight on Gwhich satisfies (∗)II .

i) The differentiation operator D:Hυ(G)−→ Hυ2(G)is a bounded operator, where υ2is

as in Proposition 3.1.4.

ii) υsatisfies (∗∗)⇔D:Hυ(G)−→ His a surjective map.

Proof:i: Corollary 3.1.10.

ii: ⇒:Lemma 3.1.11.

⇐:Lemma 3.1.12.

50CHAPTER 3. OPERATORS ON WEIGHTED SPACES FOR TYPE(II) WEIGHTS.

Section two: Composition operators between two weighted spaces of

holomorphic functions on the upper halfplane for type(II) weights

Remark 3.2.1: Throughout this section we assume that u:D−→ [0,+∞)and

η:D−→ [0,+∞)are standard weights on D.

Firstly we state wellknown results for composition operators between two weighted

spaces of holomorphic functions on the unit disc.

Definition 3.2.2: Suppose Ωis an open subset of C

a) We define H(Ω) := {f|f: Ω −→ Cis holomorphic}

b) Suppose ϕ: Ω −→ Ωis a holomorphic function. We define the composition operator

Cϕ:H(Ω) −→ H(Ω) by Cϕ(f) = f◦ϕ.

In the following we are dealing with the cases Ω = Dor Ω = G.

Theorem 3.2.3: The following conditions are equivalent for the composition operator

Cϕ:Hu(D)−→ Hη(D).

i) The composition operator Cϕis continuous.

ii) supn∈N

kϕ(z)nkη

kznku<∞.

In this case kCϕk ≤ supn∈Nkϕ(z)nkη

kznku<∞.

Proof: See Proposition 2.1of [7] and Proposition 6of [6].

Theorem 3.2.4: The following conditions are equivalent for the composition operator

Cϕ:Hu(D)−→ Hη(D).

i) The composition operator Cϕis compact.

ii) limn→∞ kϕ(z)nkη

kznku= 0.

Proof: See Theorem 3.3of [7].

Now we want to obtain results similar to Theorem 3.2.3 and Theorem 3.2.4 for the

composition operators between two weighted spaces of holomorphic functions on the

upper halfplane.

Remark 3.2.5: Consider the composition operator Cϕ:H(G)−→ H(G) (Put Ω =

Gin Definition 3.2.2).In the following we also deal with Cα−1◦ϕ◦α:H(D)−→ H(D).

Note that Cα−1◦ϕ◦α(˜

f) = f◦ϕ◦α, where ˜

f=f◦α.

Remark 3.2.6: In the remainder of this section we assume υand υ0are type(II)

weights on Gwhich satisfy (∗)II and put ˜υ(z) = υ(α(− | z|)) &

˜υ0(z) = υ0(α(− | z|)).

Now Theorem 2.2.1 implies that

Hυ(G)∼H˜υ(D) (1)

and

Hυ0(G)∼H˜υ0(D) (2)

Relations (1) and (2) are the main tools for proving our results.

Remark 3.2.7: Assume that Cϕmaps Hυ(G)into Hυ0(G).Then Cα−1◦ϕ◦αmaps H˜υ(D)

into H˜υ0(D)and vice versa.

Proof: If ˜

f=f◦α∈H˜υ(D)then relation (1) implies that f∈Hυ(G),so Cϕ(f) = f◦ϕ

belongs to Hυ0(G)and now relation (2) implies that f◦ϕ◦α=Cα−1◦ϕ◦α(˜

f)∈

H˜υ0(D).The proof for the converse is similar. 

Lemma 3.2.8: The composition operator Cϕ:Hυ(G)−→ Hυ0(G)is continuous iff

the composition operator Cα−1◦ϕ◦α:H˜υ(D)−→ H˜υ0(D)is continuous.

Proof: It is clear because of relations (1),(2) and continuity of maps α, α−1and ϕ. 

Theorem 3.2.9: The following conditions are equivalent for the composition operator

Cϕ:Hυ(G)−→ Hυ0(G).

i) The operator Cϕis continuous.

ii) supn∈N

k(ϕ(ω)−i

ϕ(ω)+i)nkυ0

k(ω−i

ω+i)nkυ<∞.

In this case there is a constant Cυ0depending on the weight υ0such that kCϕk ≤

Cυ0supn∈N

k(ϕ(ω)−i

ϕ(ω)+i)nkυ0

k(ω−i

ω+i)nkυ

Proof: i ⇒ii :Using Lemma 3.2.8, we see that Cα−1◦ϕ◦αis continuous. Now using

52CHAPTER 3. OPERATORS ON WEIGHTED SPACES FOR TYPE(II) WEIGHTS.

Theorem 3.2.3 for Cα−1◦ϕ◦αwe have supn∈N

k[(α−1◦ϕ◦α)(z)]nk˜υ0

kznk˜υ<∞.But because of the

relations (1) and (2) (see Remark 3.2.6),we have

C1k[(f◦α−1)(ω)]nkυ≤ kznk˜υ≤C2k[(foα−1)(ω)]nkυ

and

C3k[(ϕ1◦α−1)(ω)]nkυ0≤ kϕ1(z)nk˜υ0≤C4k[(ϕ1◦α−1)(ω)]nkυ0.

Here C1, C2, C3and C4are universal constants, f(z) = zand ϕ1=α−1◦ϕ◦α. Since

k[(ϕ1◦α−1)(ω)]nkυ0=k(ϕ(ω)−i

ϕ(ω)+i)nkυ0and k[(f◦α−1)(ω)]nkυ=k(ω−i

ω+i)nkυso

supn∈N

kϕ1(z)nk˜υ0

kznk˜υ<∞ ⇔ supn∈N

k(ϕ(ω)−i

ϕ(ω)+i)nkυ0

k(ω−i

ω+i)nkυ<∞.(1)

ii ⇒i: Do the above proof in the reverse direction and again use Lemma 3.2.8.

For proving the last assertion of the theorem firstly note that by definition we have

kCα−1◦ϕ◦αk= sup{kCα−1◦ϕ◦α(f◦α)k˜

υ0:kf◦αk˜υ≤1}&

kCϕk= sup{kCϕ(f)kυ0:kfkυ≤1}

Now our assumptions imply that k.kυ∼ k.k˜υ&k.kυ0∼ k.k˜

υ0.

Using relation (1),the above facts and Theorem 3.2.3 there is a constant Cυ0such that

kCϕk ≤ Cυ0supn∈N

k(ϕ(ω)−i

ϕ(ω)+i)nkυ0

k(ω−i

ω+i)nkυ



Consider the following diagram where ϕ1=α−1◦ϕ◦αand T1&T2are the maps of

Theorem 2.2.1.

Cϕ:Hυ(G)−→ Hυ0(G).

T1↓ ↓ T2(1)

Cϕ1:H˜υ(D)−→ H˜υ0(D)

Lemma 3.2.10: The diagram (1) is commutative, that is

Cϕ1=T2◦Cϕ◦T−1

1:H˜υ(D)−→ H˜υ0(D).

Proof: Firstly note that ˜

f=f◦α. So

(T2◦Cϕ◦T−1

1)( ˜

f) = (T2◦Cϕ)(T−1

1(˜

f))

= (T2◦Cϕ)(f)

=T2(f◦ϕ)

=f◦ϕ◦α

=Cϕ1(˜

f).

Corollary 3.2.11: The composition operator Cϕis compact iff the composition operator

Cα−1◦ϕ◦αis compact.

Proof: Theorem 2.2.1 implies that the maps T1and T2in Lemma 3.2.10 are isomor-

phisms, so we are done. 

Theorem 3.2.12: The following conditions are equivalent for the composition operator

Cϕ:Hυ(G)−→ Hυ0(G).

i) The composition operator Cϕis compact.

ii) limn→∞

k(ϕ(ω)−i

ϕ(ω)+i)nkυ0

k(ω−i

ω+i)nkυ= 0.

Proof: Suppose (i) holds. Corollary 3.2.11 implies that the composition operator

Cα−1◦ϕ◦α:H˜υ(D)−→ H˜υ0(D)is compact. If we use Theorem 3.2.4 for Cα−1◦ϕ◦αand

the argument of the Theorem 3.2.9[i⇒ii],then the proof is complete. 

Remark 3.2.13: Theorem 1 of [6] states that the compactness and weak compactness

of the composition operator Cϕ:Hγ(G1)−→ Hδ(G2)are equivalent if C∗\G1has no

one point component, where C∗is the extended complex plane, G1and G2are open

conected domains in C, γ :G1−→ (0,+∞)and δ:G2−→ (0,+∞)are continuous

bounded strictly positive weights.

Now since the sets Dand Gsatisfy the assumption of theorem so compactness and

weak compactness of the composition operator Cϕ(see Theorem 3.2.12) for bounded

type(II) weights are the same.

We finish this section with the following example.

54CHAPTER 3. OPERATORS ON WEIGHTED SPACES FOR TYPE(II) WEIGHTS.

Example 3.2.14: If ϕ:D−→ Dis an automorphism on D,then

α◦ϕ◦α−1:G−→ Gis an automorphism on G.

As an special case consider ϕ:D−→ Ddefined by ϕ(z) = iz, then

(α◦ϕ◦α−1)(ω) = (α◦ϕ)(ω−i

ω+i) = α(ω−i

ω+ii) = (i−1)ω+i−1

(1−i)ω+i−1or equivalently

(α◦ϕ◦α−1)(ω) =

√2

2ω+√2

−√2

2ω+√2

The above relation is also an automorphism on Gwhich is obtained by rotation 90◦

counterclockwise on D.

Section three: Pointwise multiplication operators on weighted spaces of

holomorphic functions on the upper halfplane for type(II) weights

Definition: 3.3.1 Let Ωbe an open subset of C.Suppose ϕ∈H(Ω) and ϕ6=

0 (i.e ∃ω∈Ωsuch that ϕ(ω)6= 0).We define the pointwise multiplication operator

Mϕ:H(Ω) −→ H(Ω) by Mϕ(f) = fϕ.

Note that in the following we deal with the cases Ω = Dor Ω = G.

Suppose u:D−→ (0,+∞)is a continuous and strictly positive weight function. The

following is wellknown :

Theorem 3.3.2: i) Mϕ(Hu(D)) ⊆Hu(D)iff Mϕis continuous iff Mϕ∈H∞(D)and

in this case

kϕk∞=kMϕk,where k.k∞is the supremum norm on D.

ii) Mϕis injective.

If ϕ∈H∞(D),then

iii) Mϕis an isomorphism on Hu(D)iff 1

ϕ∈H∞(D).

iv) Mϕis a Fredholm operator iff there is a  > 0such that |ϕ(z)|≥ 

for all z with 1−≤| (z)|<1.

Proof: See [5,8] 

Definition 3.3.3: Suppose ϕ∈H(G)and ϕ6= 0 (i.e ∃ω∈Gsuch that ϕ(ω)6= 0).

Consider the pointwise multiplication operator Mϕfrom H(G)into H(G).Corresponding

to Mϕ,define ˜ϕ=ϕ◦α∈H(D).Then we have

M˜ϕ:H(D)−→ H(D)where M˜ϕ(˜

f) = ˜

f˜ϕ= (f◦α)(ϕ◦α).

We call ˜

Mϕthe pointwise multiplication operator corresponding to Mϕ.

Remark 3.3.4: Throughout this section we assume υis a type(II) weight on Gwhich

satisfies (∗)II .So by Theorem 2.2.1 we have that Hυ(G)is isomorphic to H˜υ(D),where

˜υ(z) = υ(α(− | z|)).

From the definition of Mϕand M˜ϕthe following proposition is clear:

56CHAPTER 3. OPERATORS ON WEIGHTED SPACES FOR TYPE(II) WEIGHTS.

Proposition 3.3.5: The pointwise multiplication operator Mϕon Hυ(G)is continu-

ous iff the pointwise multiplication operator M˜ϕon H˜υ(G)is continuous.

Theorem 3.3.6: Let the pointwise multiplication operator Mϕ:H(G)−→ H(G)be

given.Then

i) Mϕ(Hυ(G)) ⊆Hυ(G)iff Mϕis continuous on (Hυ(G)) iff ϕ∈H∞(G),and in this

case kϕk∞=kMϕk.

ii) Mϕis injective.

If ϕ∈H∞(G),then

iii) Mϕis an isomorphism on Hυ(G)iff 1

ϕ∈H∞(G).

iv) Mϕis a Fredholm operator iff there exists  > 0such that

|ϕ(ω)|≥ for all ω∈Gsuch that 1−≤| ω−i

ω+i|<1.

Proof: i) Since Hυ(G)is isomorphic to H˜υ(D)so Mϕ(Hυ(G)) ⊆Hυ(G)

iff Mϕ(H˜υ(D)) ⊆H˜υ(D).Using Theorem 3.3.2, we obtain M˜ϕ(H˜υ(D)) ⊆H˜υ(D)iff M˜ϕis

continuous on (H˜υ(D)) iff ˜ϕ∈H∞(D)and in this case k˜ϕk∞=kM˜ϕk.

Now using Proposition 3.3.5, we conclude M˜ϕis continuous iff Mϕis continuous. If

˜ϕ= (ϕ◦α)∈H∞(D),since α−1∈H∞(G),so ˜ϕ◦α−1=ϕ∈H∞(G)and if

ϕ∈H∞(G)then ˜ϕ= (ϕ◦α)∈H∞(D).

ii) Since ϕis holomorphic on Gand ϕ6= 0 we are done.

iii) From the definition of Mϕand M˜ϕ,part(ii) of this theorem and part(ii) of The-

orem 3.3.2, it is clear that Mϕand M˜ϕare one-to-one linear maps. Also it is easy to

see that Mϕis onto iff M˜ϕis onto. Since Hυ(G)is isomorphic to H˜υ(D),so Mϕis an

isomorphism on (Hυ(G)) iff M˜ϕis an isomorphism on H˜υ(D)and (by Theorem 3.3.2

(ii))iff 1

˜ϕ∈H∞(D).But

˜ϕ◦α−1=1

ϕ.Since α−1∈H∞(G),so 1

ϕ∈H∞(G).If 1

ϕ∈H∞(G),then

ϕ◦α=1

˜ϕ∈H∞(D).

iv) Mϕ(Hυ(G)) is closed and is finite codimensional in Hυ(G)iff M˜ϕ(H˜υ(D)) is closed

and is finite codimensional in H˜υ(D).Also ker Mϕ= ker M˜ϕ={0}.So Mϕis a Fred-

holm operator iff M˜ϕis a Fredholm operator (now using Theorem 3.3.2(iv))iff there

is a  > 0such that |˜ϕ(z)|≥ for all z∈Dwith 1−≤| z|(and transfering this

to G)iff there is a  > 0such that |ϕ(ω)|≥ for all ω∈Gwith

1−≤| ω−i

ω+i|<1.

58CHAPTER 3. OPERATORS ON WEIGHTED SPACES FOR TYPE(II) WEIGHTS.

Chapter 4

Operators on weighted spaces for

type(I) weights

Introduction to chapter four: In this chapter we investigate differentiation and

composition operators on weighted spaces of holomorhpic functions for type(I) weights.

We begin section one by studying the differentiation operator. We continue this section

by obtaining some various sufficient conditions for continuity of composition operators

where at least one of the weights are bounded.

In section two, by using Theorem 3.2.9, we will obtain another sufficient condition

for the continuity of the composition operator when our weights are not necessarily

bounded.

In section three we find a sufficient condition for the continuity of the composition

operator Cϕ(see Definition 3.2.2) between two weighted spaces of 2π−periodic func-

tions, where ϕ:G−→ Gis a 2π−periodic and holomorphic function.

In section four we study the differentiation operator between two weighted spaces

of 2π−periodic functions. We obtain necessary and sufficient condition for contiuity

of the differentiation operator.

Also we want to know when this operator will be surjective.

60CHAPTER 4. OPERATORS ON WEIGHTED SPACES FOR TYPE(I) WEIGHTS

Section one: Differentiation and composition operators between weighted

spaces of holomorphic functions on the upper halfplane for type(I) weights.

We begin this section by studing the differentiation operator. For this aim we need

the following.

Note that throughout this section the weight υis of type(I).

Remark 4.1.1: Using Lemma 1.2.8 (i), if υsatisfies(∗)I,then

infn∈N˜υ(1−2−n+1)

˜υ(1−2−n)>0

where ˜υ(z) = υ(α(− | z|)) = υ((1−|z|

1+|z|)i).

As a consequence of Theorem 3.1.2 we obtain:

Corollary 4.1.2: Let υbe a type(I) weight satisfying (∗)I.Consider the standard weight

˜υ(z) = υ((1−|z|

1+|z|)i).Then the differentiation operator D:H˜υ(z)(D)−→ H˜υ(z)(1−|z|)(D)is

a bounded operator.

Proof: Since ˜υsatisfies (∗)0(see begining of chapter three) and it is a standard weight

so this follows from Theorem 3.1.2.

Theorem 4.1.3: Let υbe a type(I) weight on G.If υis bounded and satisfies (∗)I(for

a bounded weight (∗)Iand (∗)II are equivalent),then the differentiation operator

D:Hυ(G)−→ Hυ1(G)is a bounded operator where υ1(ω) = min(Imω, 1)υ(Imωi).

Proof: Firstly note that it is easy to see that υ1is a bounded weight of type(I).Now

Corollary 4.1.2 and Proposition 2.3.5 imply that there exists a constant C00 >0such

that

supa∈RkDz((Taf)◦α)k˜υ(z)(1−|z|)≤C00kfkυ,(1)

where Dzdenotes the differentiation operator on the unit disc Dand Tais as in

Lemma 2.3.3.We have

kDz((Taf)◦α)k˜υ(z)(1−|z|)=k(Taf0)◦α·α0k˜υ(z)(1−|z|)=kTaf0◦αk|α0(z)|˜υ(z)(1−|z|)

and υ1(ω) = min(Imω, 1)υ(Imωi).If Imω ≤1,then υ1(ω) = Imωυ(Imωi) =

υ1(Imωi).Suppose ω∈Gis such that Imω ≤1.So

a=Reω. Put z∈Dsuch that

α−1(Imωi) = z. So Imω−1

Imω+1 =z:= −r. Thus |z|=r, |z|=−zand z=− | z|.

|f0(ω)|υ1(ω) = |Df(ω)|υ1(Imωi)

=|(Taf0)(Imωi)|υ1(Imωi)

=|(Taf0)(Imωi)|Imωυ(Imωi)

=|(Taf0)◦α(−r)|Imα(−r)υ(α(z))

=|(Taf0)◦α(z)|Imα(z)υ(α(z))

=|(Taf)0◦α(−r)|Imα(− | z|)υ(α(− | z|)).

Since Imα(− | z|) = 1

2|α0(− | z|)|(1− | z|2)so

|Df(ω)|υ1(Imωi) = 1

2|(Taf0)◦α(z)|| α0(z)|˜υ(z)(1− | z|2).(2)

By Lemma 2.3.4 (b)there is a constant d > 0such that

kDfkυ1≤dsupω∈G,Imω≤1|Df(ω)|υ1(Imωi).(3)

By relation (2) we have

sup

ω∈G,Imω≤1|Df(ω)|υ1(Imωi)≤1

2sup

a∈Rsup

z∈DRez≤0Imz=0 |(Taf0)◦α(z)|| α0(z)|˜υ(z)(1− | z|2)

≤sup

a∈Rk(Taf0)◦α(z)·α0(z)k˜υ(z)(1−|z|)

≤sup

a∈RkDz((Taf)◦α)k˜υ(z)(1−|z|).

Therefore

supω∈G,Imω≤1|Df(ω)|υ1(Imωi)≤supa∈RkDz((Taf)◦α)k˜υ(z)(1−|z|).(4)

Now relations (3),(4) and (1) imply that there exists a constant C > 0such that

kDfkυ1≤Ckfkυand the proof is complete. .

We continue this section with studying composition operators (See Remark 3.2.2).

62CHAPTER 4. OPERATORS ON WEIGHTED SPACES FOR TYPE(I) WEIGHTS

We present sufficient conditions for boundedness of composition operators.

Proposition 4.1.4: Let υand ˆυbe type(I) weights on Gand assume that ˆυis bounded.

Moreover, put ˜υ(z) = υ(α(− | z|)) and ˜

ˆυ(z) = ˆυ(α(− | z|)), z ∈D.Let ϕ:G−→

Gbe a holomorphic function such that

supa∈Rsupn∈N

k(ϕ(α(z)+a)−i

ϕ(α(z)+a)+i)nk˜

ˆυ

kznk˜υ<∞.Then the composition operator

Cϕ:Hυ(G)−→ Hˆυ(G)is bounded.

Proof: Using Proposition 2.3.5 there are constants C1, C2>0such that for all f∈

Hˆυ(G)we have

C1kCϕfkˆυ≤supa∈RkTa(Cϕf)◦α)k˜

ˆυ≤C2kCϕfkˆυ.(1)

Moreover,(Ta(Cϕf))(α(z)) = (Cϕf)(α(z) + a) = f(ϕ(α(z) + a)).Put

ϕa(z) = α−1(ϕ(α(z) + a)).Then ϕais a holomorphic map from Dinto Dand we

obtain, with ˜

f=f◦α,

(Cϕa˜

f)(z)=(TaCϕf)(α(z)) (2)

By our assumption we have

supa∈Rsupn∈N

k(ϕa(z))nk˜

ˆυ

kznk˜υ<∞.

The above relation implies that the operators Cϕa(a∈R)from H˜υ(D)into H˜

ˆυ(D)are

uniformly bounded.(see Theorem 3.2.3). Therefore C0:= supa∈RkCϕak<∞.

Now using relations (1)& (2) and the fact that Imα(− | z|)≤Imα(z)we have

C1kCϕfkˆυ≤sup

a∈RkTa(Cϕf)◦α)k˜

ˆυ

= sup

a∈RkCϕa˜

fk˜

ˆυ

≤C0k˜

fk˜υ

=C0sup

z∈D|f(α(z)) |υ(α(− | z|))

≤C0sup

z∈D|f(α(z)) |υ(α(z))

=C0kfkυ

Therefore there is a universal constant Csuch that for any f∈Hυ(G)

kCϕfkˆυ≤Ckfkυ.

Corollary 4.1.5: Let υand ˆυbe type(I) weights on Gsuch that ˆυis bounded and υsatisfies (∗)I.

Put υ1(ω) = υ(2Imω

|ω|2+1i).Assume that ϕ:G−→ Gis a holomorphic function and

satisfies

supa∈Rsupn∈N

k(ϕ(ω+a)−i

ϕ(ω+a)+i)nkˆυ

k(ω−i

ω+i)nkυ1

<∞.Then Cϕ:Hυ(G)−→ Hˆυ(G)is bounded.

Proof: Let ω=α(z), z ∈D.Since ˜

ˆυ(z) = ˆυ(α(− | z|)) ≤ˆυ(α(z)) we obtain

k(ϕ(α(z)+a)−i

ϕ(α(z)+a)+i)nk˜

ˆυ≤ k(ϕ(ω+a)−i

ϕ(ω+a)+i)nkˆυ(1)

Since α(− | z|) = 1−|z|

1+|z|≤1−|z|2

1+|z|2and υsatisfies (∗)Iwe obtain

υ(1−|z|2

1+|z|2i)≤Cυ(α(− | z|)) = C˜υ(z)for some universal constant C > 0.But

1−| ω−i

ω+i|2

1+|ω−i

ω+i|2=|ω+i|2−|ω−i|2

|ω+i|2+|ω−i|2=2Imω

|ω|2+1.

Therefore ˜υ(z)≥1

Cυ1(ω).This implies

kznk˜υ≥1

Ck(ω−i

ω+i)nkυ1(2)

Our assumption,(1) and (2) yield

supa∈Rsupn∈N

k(ϕ(α(z)+a)−i

ϕ(α(z)+a)+i)nk˜

ˆυ

kznk˜υ<∞.

Now Proposition 4.1.4 concludes the proof.

64CHAPTER 4. OPERATORS ON WEIGHTED SPACES FOR TYPE(I) WEIGHTS

Section two: Continuity of the composition operators for type(I) weights

which satisfy(∗)I

In Proposition 4.1.4we presented a sufficient condition for boundedness of the com-

position operator Cϕ:Hυ(G)−→ Hˆυ(G)where υand ˆυare type(I) weights and ˆυis

also bounded.

Now we want to find a suficient condition for boundedness of the composition operator

Cϕ,when our weights are not necessarily bounded, but satisfy condition (∗)I.

For proving our result we define special type(II) weights and we use Lemma 3.2.9.

Definition 4.2.1: Let υbe a type(I) weight on G.For any n∈Nwe define υn(ω) =

υ(nω)

Lemma 4.2.2: υnis a type(I) weight on Gfor all n∈N.

Proof: υis of type(I). So by definition there exists a C > 0such that

υ(ω1)≤Cυ(ω2)whenever Imω1≤Imω2.Note that Imnω =nImω ∀ω∈G.

If Imω1≤Imω2⇒nImω1=Im(nω1)≤nImω2=Im(nω2).Therefore

υ(nω1) = υn(ω1)≤Cυ(nω2) = υn(ω2).

Remark 4.2.3: Note that the weights υnare of type(I) with the same constant which

appears in the definition of type(I) weight υ.

Definition 4.2.4: For all ω∈Gdefine ˆυ(ω) = min(υ(ω), Cυ(−1

ω)) and for all n∈

Nand all ω∈Gdefine ˆυn(ω) = min(υn(ω), Cυn(−1

ω)).

Here Cis the constant which appears in the definition of type(I) weights υn.

Lemma 4.2.5: ˆυand ˆυn(n∈N)are type(II) weights on G.

Proof: We prove the lemma for ˆυ. The proof for ˆυn(n∈N)is similar.

If |ω|≤ 1then Im(−1

ω) = Imω

|ω|2≥Imω. Therefore

υ(ω)≤Cυ(−1

ω).(1)

Relation (1) implies that ˆυ(ω) = υ(ω)whenever |ω|≤ 1.Now we prove that there

exists a d > 0such that ˆυ(ω)

ˆυ(−1

ω)≤d∀ω∈G.By definition of ˆυ,

ˆυ(−1

ω) = min(υ(−1

ω), Cυ(ω)).

Using relation (1), if |ω|≤ 1then

ˆυ(ω)

ˆυ(−1

ω)≤max( υ(ω)

υ(−1

ω),υ(ω)

Cυ(ω))≤max(C, 1

C).

If |ω|>1then Im(−1

ω) = Imω

|ω|2≤Imω. Therefore

υ(−1

ω)≤Cυ(ω).(2)

Now considering that ˆυ(ω) = υ(ω)⇔ ≤ Cυ(−1

ω)⇔| ω|≤ 1and relation (2),if

|ω|>1,then

ˆυ(ω)

ˆυ(−1

ω)≤max( υ(ω)

υ(−1

ω),Cυ(−1

ω)

υ(−1

ω))≤C.

Now put d= max{C, 1

C}.So we are done. 

Definition 4.2.6: For any n∈Nwe define Tn:Hυ(G)−→ Hˆυn(G)by

Tn(f)(ω) = f(nω)for all f∈Hυ(G).

Lemma 4.2.7: For all n∈N, Tnis a contractive map.

Proof:

k(Tnf)kˆυn= sup

ω∈G|(Tnf)(ω)|ˆυn(ω)

= sup

ω∈G|f(nω)|min(υn(ω), Cυn(−1

ω))

≤sup

ω∈G|f(nω)|υn(ω)

= sup

ω∈G|f(nω

n)|υ(nω

=kfkυ.

Lemma 4.2.8: If the weight υsatisfies (∗)I,then ˆυn(n∈N)satisfies (∗)II .

Proof: Since υsatisfies (∗)I,so ∃C > 0,∃β > 0such that

υ(ω1)

υ(ω2)≤C(Imω1

Imω2)βwhenever Imω1≥Imω2.Let n∈Nbe arbitrary.

If Imω1≥Imω2then Im(nω1)≥Im(nω2).So υn(ω1)

υn(ω2)=υ(nω1)

υ(nω2)≤C(Imω1

Imω2)β.

So we have proved that υnsatisfies (∗)Ifor all n∈N.Now since for all n∈N

ˆυn(ω) = υn(ω)whenever |ω|≤ 1 (see Lemma 4.2.5) so ˆυnsatisfies (∗)II for all n∈

66CHAPTER 4. OPERATORS ON WEIGHTED SPACES FOR TYPE(I) WEIGHTS

N.

Lemma 4.2.9: Let f:G−→ Cbe a holomorphic function. If supn∈NkTnfkˆυn<

∞then f∈Hυ(G)and kfkυ= supn∈NkTnfkˆυn.

Proof:

kTnfkˆυn= sup

ω∈G|(Tnf)(ω)|ˆυn(ω)

= sup

ω∈G|(Tnf)(ω

n)|ˆυn(ω

= sup

ω∈G|f(ω)|min(υn(ω

n), Cυn(−n

ω))

= sup

ω∈G|f(ω)|min(υ(ω), Cυ(−n2

ω)).(1)

So we have kTnfkˆυn≤supω∈G|f(ω)|υ(ω) = kfkυ.Since n∈Nis arbitrary

so supn∈NkTnfkˆυn≤ kfkυ(2)

Consider an arbitrary and fixed ω0∈G.Since Im(−n2

ω0) = n2Imω0

|ω0|2so for large

enough n∈Nwe have Im(−n2

ω0)≥Imω0.Therefore υ(ω0)≤Cυ(−n2

ω0)for large

enough n∈N.So

|f(ω0)|υ(ω0) =|f(ω0)|min(υ(ω0), Cυ(−n2

ω0)) ≤supω∈G|f(ω)|min(υ(ω), Cυ(−n2

ω))

Now using relation (1) we obtain |f(ω0)|υ(ω0)≤ kTnfkˆυnfor large enough

n∈N.Thus

kfkυ= supω0∈G|f(ω0)|υ(ω0)≤supn∈NkTnfkˆυn.(3)

Now relations (2),(3) and our assumption imply that kfkυ= supn∈NkTnfkˆυn<

∞.Therefore we are done. 

Theorem 4.2.10: Let υand υ0be type(I) weights on Gwhich satisfy (∗)I. If

a= supn∈Nsupm∈N

nϕ(nω)−i

nϕ(nω)+i)mkˆ

υ0n

k(ω−i

ω+i)mkˆυn

<∞,then the composition operator

Cϕ:Hυ(G)−→ Hυ0(G)is welldefined and bounded. Here ˆυnand ˆ

υ0nare as in

Definition 4.2.4.

Proof: Let n∈Nbe arbitrary and fixed. Consider the following diagram, where Tnis

as in Definition 4.2.6and T0

nis defined by T0

n(f) = f(nω)for all f∈Hυ0(G).

Cϕ:Hυ(G)−→ Hυ0(G)

Tn↓ ↓ T0

n◦Cϕ◦T1

n:Hˆυn(G)−→ Hˆ

υ0n(G)

It is easy to see that (T0

n◦Cϕ◦T1

n)(f) = f(1

nϕ(nω)) = Cϕn(f)where ϕn(ω) =

ϕ(nω)

n.So T0

n◦Cϕ◦T1

nis really a composition operator on Hˆυn(G).ˆυnand ˆ

υ0nare

type(II) weights which satisfy (∗)II (see Lemma 4.2.5 and Lemma 4.2.8)also by our

assumption for any n∈N

supm∈N

nϕ(nω)−i

nϕ(nω)+i)mkˆ

υ0n

k(ω−i

ω+i)mkˆυn

<∞.Therefore Theorem 3.2.9 implies that T0

n◦Cϕ◦T1

nis

bounded as an operator from Hˆυn(G)into Hˆ

υ0n(G)and in this case

kCϕnk ≤ Cˆ

υ0nsupm∈N

k(nϕ(ω

n)−i

nϕ(ω

n)+i)mkˆ

υ0n

k(ω−i

ω+i)mkˆυn

.So supn∈NkCϕnk ≤ asupn∈NCˆ

υ0n.

Here Cˆ

υ0nare the constants of Theorem 3.2.9.

Now looking at the proof of the Theorem 2.2.1 we see that the constants Cˆ

υ0ndepend

on the constants which appear in the definition of the type(II) weights ˆ

υ0n(n∈N)

and property (∗)II .

But the definition of the type(II) weights ˆ

υ0n(n∈N),Remark 4.2.3,Lemma 4.2.5and

Lemma 4.2.8show that the constants does not depend on n∈N.Therefore

supn∈NkCϕnk≤∞.This means that the family of the composition operators

{T0

n◦Cϕ◦T1

n}is uniformly bounded. Therefore by using the relation

kCϕnk= sup{kCϕn(h)kˆ

υ0n:khkˆυn≤1}we deduce that

∃C > 0such that ∀n∈N∀h∈Hˆυn(G)

k(T0

n◦Cϕ◦T1

n)(h)kˆ

υ0n≤Ckhkˆυn.(2)

From Lemma 4.2.9 we have

kfkυ= supn∈NkTnfkˆυn.∀f∈Hυ(G) (3)

and

kgkυ0= supn∈NkT0

nfkˆ

υ0n.∀g∈Hυ0(G) (4)

Let f∈Hυ(G)be arbitrary. Then

68CHAPTER 4. OPERATORS ON WEIGHTED SPACES FOR TYPE(I) WEIGHTS

supn∈NkT0

n◦Cϕfkˆ

υ0n= supn∈NkT0

n◦Cϕ◦T1

n◦(Tnf)kˆ

υ0n

Now relation (2) implies that supn∈NkT0

n◦Cϕ◦T1

n(Tnf)kˆ

υ0n≤CkTnfkˆυn

Using relation (3) we have

supn∈NkT0

n◦Cϕ(f)kˆ

υ0n≤Ckfkυ<∞(since f∈Hυ(G))

Again Lemma 4.2.9 implies that Cϕ(f)∈Hυ0(G)and kCϕfkυ0= supn∈NkT0

n◦Cϕfkˆ

υ0n.

Therefore kCϕfkυ0≤Ckfkυfor all f∈Hυ(G)

Section three: Composition operator between weighted spaces of 2π-periodic

holomorphic functions

As we said before we study in this section the continuity of the composition operator Cϕ

(see Remark 3.2.5) between two weighted spaces of 2π−periodic functions where ϕ:

G−→ Gis a 2π−periodic and holomorphic function. Before arriving at our result we

present an example of a 2π−periodic and holomorphic function from Ginto G.

Example 4.3.1: Define ϕ(ω) = eiω +ifor any ω∈G.Clearly

Im(eiω +i) = (e−Imω sin Reω + 1) >0and ϕ(ω+ 2π) = ϕ(ω).So ϕ(G)⊆Gand ϕis

2π−periodic.

Let υand υ0be type(I) weights on G.Also let b1and b2be the smallest inte-

gers such that e−b1Imωυ(ω)and e−b2Imωυ0(ω)are bounded (see Remark 2.4.5(b)).

Put b= max{b1, b2}then bis the smallest integer such that υb(ω) = e−bImωυ(ω)and

υ0

b=e−bImωυ0(ω)both are bounded weights. Also we define ˜υ(z) := υb(ln( 1

|z|)i)and

˜υ0(z) := υ0

b0(ln( 1

|z|)i).Hence ˜υ(z) =|z|bυ(−iln |z|)and ˜υ0(z) =|z|bυ0(−iln |z|).

Proposition 2.4.8implies that the maps T:H2π

υ(G)−→ H2π

υb(G)defined by T(g) =

eibωg(ω)and T1:H2π

υ0(G)−→ H2π

υ0

b(G)defined by T1(g1) = eibωg1(ω)are isometries. Also

Theorem 2.4.11 implies that the maps S:H2π

υb(G)−→ H˜υ(D)defined by S(h)(z) =

h(−ilog z)and S1:H2π

υ0

b(G)−→ H˜

υ0(D)defined by S1(h1)(z) = h1(−ilog z)are

isometries. So the maps ST :H2π

υ(G)−→ H˜υ(D)and S1T1:H2π

υ0(G)−→ H˜

υ0(D)are

isometries.

Remark 4.3.2: Note that ˜υand ˜

υ0are not necessary continuous at zero. But there

are radial weights ˜υ1and ˜

υ0

1which are continuous and k.k˜υ∼ k.k˜υ1and k.k˜

υ0∼

k.k˜

υ0

1.(see Remark 2.4.10).Using Remark 1.2.6,without loss of generality we can

assume ˜υ1and ˜

υ0

1are also decreasing weights on the unit disk D.Therefore the

maps ST :H2π

υ(G)−→ H˜υ1(D)and S1T1:H2π

υ0(G)−→ H˜

υ0

1(D)are isomorphisms.

Suppose ϕ:G−→ Gis a 2π−periodic and holomorphic function. Consider the com-

position operator Cϕand the following diagram

70CHAPTER 4. OPERATORS ON WEIGHTED SPACES FOR TYPE(I) WEIGHTS

Cϕ:H2π

υ(G)−→ H2π

υ0(G).

ST ↓ ↓ S1T1(1)

Cϕ:H˜υ1(D)−→ H˜

υ0

1(D)

Here ˜

Cϕis defined as follows ˜

Cϕ:= (S1T1)◦Cϕ◦(ST)−1.Since ST and S1T1are

isomorphisms (see Remark 4.3.2)so Cϕis continuous iff ˜

Cϕis continuous. So it is

enough to study the operator ˜

Cϕ.Suppose f∈H˜υ1(D)then T−1S−1(f) = e−ibωf(eiω)and

(˜

Cϕf)(z) = (S1T1)◦Cϕ◦(ST)−1(f) = zbe−ibϕ(−ilog z)f(eiϕ(−ilog z)).

We can consider ˜

Cϕas a weighted composition operator. There is a wellknown result

for continuity of weighted composition operators on the unit disk D.

We formulate it in the following way.

Theorem 4.3.3: Let υ1and υ2be a standard weights on the unit disk D.Also suppose

ϕ1, ψ1∈H(D)and ϕ1(D)⊆D.Then the weighted composition operator

Cϕ1,ψ1:Hυ1(D)−→ Hυ2(D)defined by Cϕ1,ψ1(f) = ψ1·(f◦ϕ1)is continuous iff

supn∈N

kϕ1(z)nψ1(z)kυ2

kznkυ1<∞.

Proof: Use Proposition 3.1 of [9] and Proposition 6 of [6] 

Now we can obtain the following necessary and sufficient condition for the continuity

of the composition operator Cϕin diagram (1).

Theorem 4.3.4: Let υand υ0be type(I) weights on G.The composition operator

Cϕ:H2π

υ(G)−→ H2π

υ0(G)defined by Cϕ(f) = f◦ϕis continuous iff

supn∈N

keinϕ(ω)kυ0

keinϕ(ω)kυ<∞.

(Here ϕ:G−→ Gis 2π−periodic and holomorphic)

Proof: Use Theorem 4.3.3for the operator ˜

Cϕin diagram (1) where υ1= ˜υ1, υ2=

υ0

1, ψ1(z) = zbe−ibϕ(−ilog z)and ϕ1(z) = eiϕ(−ilog z).

Also note that since ϕ:G−→ Gis holomophic so ψ1, ϕ1∈H(D).we also have

|ϕ1(z)|=|eiϕ(−ilog z)|=|e−Imϕ(−ilog z)|<1∀z∈Dsince Imϕ(ω)>0for

all ω∈G.Therefore

supn∈N

keinϕ(−ilog z)zbe−ibϕ(−ilog z)k˜

υ0

kznk˜υ1<∞.(1).

Use the definition of ˜υand ˜

υ0to see that relation (1) is equivalent to

supn∈N

kei(n−b)ϕ(−ilog z)z2bku0

kzn+bku<∞(2).

Here u0(z) = υ0(ln( 1

|z|)i)and u(z) = υ(ln( 1

|z|)i).

Put m=n−b. Since mis non-positive for only finitely many n∈N,(exactly n=

1,2, ..., b)so relation (2) is equivalent to

supm∈N

keimϕ(−ilog z)z2bku0

kzn+bku<∞.(3).

It is wellknown that if M∈Nthen the map TM:Hυ(D)−→ Hυ(D)defined

by TM(f) = zMf(z)is an isomorphism onto a finite codimensional subspace of Hυ(D)for

any standard weight υon the unit disc D.See [5,8]

Consider the relations T2b(1) = z2b, T2b(zn−b) = zn+band m=n−b. So relation (3) is

equivalent to supm∈N

keimϕ(−ilog z)ku0

kzmku<∞.Now put ω=−ilog zor equivalently z=

eiω we have supm∈N

keimϕ(ω)kυ0

keimωkυ<∞.Then Theorem 4.3.3 completes the proof. 

72CHAPTER 4. OPERATORS ON WEIGHTED SPACES FOR TYPE(I) WEIGHTS

Section four: Differentiation operator between weighted spaces of 2π-periodic

holomorphic functions

In this section we obtain a suficient condition for boundedness of the differentiation

operator between weighted spaces of 2π−periodic holomorphic functions.

Lemma 4.4.1: Let υbe a type(I) weight on Gsatisfying the property

supn∈N

υ(1

2ni)

υ(1

2n+1 i)<∞.Then

i) ˜υ(z)and ˜υ1(z)satisfy (∗)0.(See Theorem 2.1.8,Definition 2.4.9and Remark 2.4.10 for

definitions of (∗)0,˜υ(z)and ˜υ1(z)resp.)

ii) Put υ0(ω) = (1 −e−Imω)υ(ω).Then υ0is a type(I) weight on G.

Proof: i) ˜υ(1−2−n)

˜υ(1−2−n−1)=(1−2−n)bυ(−iln(1−2−n))

(1−2−n−1)bυ(−iln(1−2−n−1)).Clearly (1−2−n)b

(1−2−n−1)b<1.Using the

mean value theorem for the function ln xon the interval [1 −2−n,1],there exists

aC0∈(1 −2−n,1) such that ln 1 −ln(1 −2−n) = 1

C02n.Therefore

−ln(1 −2−n)≤1

(1−2−n)2n<1

2n−2for any n≥3.

Similarly there exists a C00 ∈(1 −2−n−1,1) such that −ln(1 −2−n−1) = 1

C00

2n+1 .So

−ln(1 −2−n−1)≥1

2n+1 .

Now since υis a type(I) weight there exists a C > 0such that

υ(−iln(1 −2−n)) ≤Cυ(1

2n−2i) & υ(1

2n+1 i)≤Cυ(−iln(1 −2−n−1)).

Therefore supn∈N˜υ(1−2−n)

˜υ(1−2−n−1)≤C2supn≥3

υ(2−n+2i)

υ(2−n+1i)supn≥3

υ(2−n+1i)

υ(2−ni)supn≥3

υ(2−ni)

υ(2−n−1i).

Now our assumption implies supn∈N˜υ(1−2−n)

˜υ(1−2−n−1)<∞So ˜υsatisfies (∗)0.We also have

˜υ1(z) = ˜υ(z)for all zwith |z|≥ 1

2.So ˜υ1(z)satisfies (∗)0.

ii) It is clear. 

Remark 4.4.2: Let υbe a type(I) weight on G.

i) If we define υ0(ω) = (1 −e−Imω)υ(ω),then ˜

υ0(z) = (1− | z|)˜υ(z).Also we have

k.k˜υ0

1∼ k.k˜

υ0.

ii) k.k˜υ∼ k.k˜υ1implies that H˜υ(D)∼H˜υ1(D).Similarly we have H˜υ0(D)∼H˜υ0

1(D).

Lemma 4.4.3: Let band b0be the smallest integers ksuch that

supω∈Ge−kImωυ(ω)<∞& supω∈Ge−kImωυ0(ω)<∞respectively, where υis of

type(I) and υ0is as in Lemma 4.4.1.

i) Let a > 0be given. There is a constant d > 0 (ddepends on aand υ0)such

that A:= sup{e−b0Imωυ0(ω) : ω∈G, Imω ≥a} ≤ supω∈Ge−b0Imωυ0(ω)≤dA.

ii) b=b0.

Proof: i) supω∈Ge−b0Imωυ0(ω) = max(A, sup{e−b0Imωυ0(ω) : ω∈G,0< Imω <

a}).Since υ0is of type(I) so there is a constant d0>0such that

sup{e−b0Imωυ0(ω) : ω∈G,0< Imω < a} ≤ sup{υ0(ω) : ω∈G,0< Imω < a} ≤ d0.

Therefore supω∈Ge−b0Imωυ0(ω)≤dA with d=max(d0,A)

ii) If b0> b then by definition we have

∞= supω∈Ge−bImω(1−e−Imω)υ(ω)<supω∈Ge−bImωυ(ω)<∞which is a contradiction.

If b0< b then by definition

supω∈Ge−b0Imωυ(ω) = ∞(1)

and

supω∈Ge−b0Imωυ0(ω)<∞(2)

Now relation (2) implies that

sup{e−b0Imωυ0(ω) : ω∈G, Imω ≥a}<∞.(3)

Put a= ln 2 in relation (3). Then we have

sup{e−b0Imωυ(ω) : ω∈G, Imω ≥ln 2}<∞(4)

sup{e−b0Imωυ(ω) : ω∈G,0< Imω ≤ln 2} ≤ sup{υ(ω) : ω∈G,0< Imω ≤ln 2} ≤

Cln 2 <∞.(5)

Now relations (4) and (5) contradict relation (1).

Now we prove the following theorem.

Theorem 4.4.4: Let υbe a type(I) weight on Gsatisfying (∗)II .Put υ0(ω) = (1 −

e−Imω)υ(ω).Then the differentation operator D:H2π

υ(G)−→ H2π

υ0(G)is a welldefined

and bounded map.

Proof: Consider the following diagram.

74CHAPTER 4. OPERATORS ON WEIGHTED SPACES FOR TYPE(I) WEIGHTS

D: H2π

υb(G)−→ H2π

υ0

b(G)

S↓ ↓ S1

D1:H˜υ(D)−→ H˜υ0(D)

Here Dis differentation operator, D1=S1◦D◦S−1and S, S1,˜υ, ˜υ0are defined as in

Theorem 2.4.11 and Definition 2.4.9respectively.

So S1◦D◦S−1is welldefined and bounded iff Dis welldefined and bounded. Since

(S1◦D◦S−1)(h) = izh0(z)so S1◦D◦S−1is welldefined and bounded iff the differ-

entiation operator Dz:H˜υ(D)−→ H˜υ0(D)is welldefined and bounded.

Now consider the following diagram.

D:H2π

υ(G)−→ H2π

υ0(G)

T↓ ↓ T1

D2:H2π

υb(G)−→ H2π

υ0

b(G)

Here Dis differentiation operator and

D2:= (T1◦D◦T−1) (1).

Also Tand T1are as in Proposition 2.4.8.Clearly

D2(g) := (−ibeiω(b0−b)id +eiω(b0−b)D)(g)for any g∈H2π

υb(G).Since b0=b(see Lemma

4.4.3(ii))so the above relation reduces to

D2(g) = (id(−bi) + D)(g)∀g∈H2π

υb(G).(2)

Note that since υ0

b(ω)≤υb(ω)so kgkυ0

b≤ kgkυb.This implies H2π

υb(G)⊆H2π

υ0

b(G).

Therefore relation (2) is welldefined.

Since Tand T1are isometries, relation (1) implies D2is welldefined and bounded

iff Dis welldefined and bounded from H2π

υ(G)into H2π

υ0(G).Relation (2) implies

that D2is welldefined and bounded iff Dis welldefined and bounded from H2π

υb(G)

into H2π

υ0

b(G).Therefore we have the following:

Dfrom H2π

υ(G)into H2π

υ0(G)is welldefined and bounded iff Dzfrom H˜υ(D)into H˜υ0(D)is

welldefined and bounded.

Lemma 4.4.1(i) and Theorem 3.1.2 imply that Dz:H˜υ1(z)(D)−→ H(1−|z|)˜υ1(z)(D)is

welldefined and bounded. Also Part(ii) of Remark 4.4.2,implies that

id :H˜υ1(z)(z)(D)−→ H˜υ(D)is welldefined and bounded.

The above facts and the following diagram imply that the differentiation operator Dzfrom

H˜υ(D)into H(1−|z|)˜υ(z)(D)is welldefined and bounded iff

id :H(1−|z|)˜υ1(z)(D)−→ H(1−|z|)˜υ(z)(D)is welldefined and bounded.

D:H˜υ1(z)(D)−→ H(1−|z|)˜υ1(z)(D)

id ↓ ↓ id

D:H˜υ(D)−→ H(1−|z|)˜υ(z)(D)

But we are done if we prove k.k(1−|z|)˜υ1(z)∼ k.k(1−|z|)˜υ(z).(3)

Since ˜υ1and ˜υare decreasing and bounded weights such that ˜υ1(1

2),˜υ(1

2)>0so

kfk(1−|z|)˜υ(z)∼sup|z|≥ 1

2|f(z)|(1− | z|)˜υ(z).(4)

kfk(1−|z|)˜υ1(z)∼sup|z|≥ 1

2|f(z)|(1− | z|)˜υ1(z) (5)

The definition of ˜υ1and ˜υimply ˜υ1|[2−1,1] ∼˜υ|[2−1,1] .

Therefore sup|z|≥ 1

2|f(z)|(1− | z|)˜υ1(z)∼sup|z|≥ 1

2|f(z)|(1− | z|)˜υ(z).(6)

Now relations (4),(5) and (6) give relation (3) and the proof is complete. 

76CHAPTER 4. OPERATORS ON WEIGHTED SPACES FOR TYPE(I) WEIGHTS

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