Boundary Behaviour of Analytic Curves · 2007. 1. 4. · Int. Journal of Math. Analysis, Vol. 1,...

Int. Journal of Math. Analysis, Vol. 1, 2007, no. 3, 133 - 157

Boundary Behaviour of Analytic Curves

Yuksel Soykan

Zonguldak Karaelmas UniversityDepartment of Mathematics

Zonguldak, Turkey

Abstract

On the Hardy spaces of the unit disc and its boundary three proper-ties can be highlighted: i) nontangential limit functions exist, ii) Parse-val’s identity is satisfied and iii) Cauchy’s Integral Formula is valid. Ouraim in this paper is to show that, for some certain domains, analoguesof these properties make sense and can be verified.

Mathematics Subject Classification: Primary 42B30,31A20, 32A40;Secondary 30E20

Keywords: Cauchy’s Integral Formula, Hardy Spaces, Smirnov Spaces,Closed Analytic Curves

1 Introduction

C will be used to denote the sphere or extended plane C ∪ (∞) and as usual,T will be used to denote the boundary of open unit disc Δ. Whenever we referto a simply-connected domain D ⊆ C it will always be assumed that D hasan infinite complement, so that D can be mapped conformally onto Δ that is,there is a function ϕ which is analytic on D and maps D one-to-one onto Δ.In general, a function θ on a simply connected domain D′ ⊆ C will be calledconformal if it is meromorphic and injective on D′. Such a function will haveat most one pole on D′, which must be simple. If γ is a σ−rectifiable arc(see page 4 for definition) in C and f is integrable with respect to arc-lengthmeasure the notations ∫

γ

f(z)dz,

∫γ

f(z) |dz|

will be used to denote, respectively, the complex line integral of f along γand the integral of f with respect to arc-length measure. In the first case we

134 Yuksel Soykan

assume γ has an orientation. Arc-length measures will usually be normalized

with the factor1

2π. This will certainly be done in the case of T, so that T has

total measure 1. The notation Lp(γ) will denote the Lp space of normalizedarc length measure on γ.

Suppose D′ ⊆C is a simply connected domain. Then there is a canonical

Hilbert Space E2(D′) of analytic functions onD′. These spaces are discussed indetail in chapter 10 of [3] and the precise definition will be recalled in section2; so these spaces will be taken for granted for the moment. Also classicalH2(Δ) and H2(T) Hardy spaces will be taken for granted for the moment (seesection 2).

Three properties of H2(Δ) and H2(T) can be highlighted as follows:

I Existence of nontangential limit functions:

Every f ∈ H2(Δ) has a nontangential limit function f ∈ L2(T).

II Parseval’s identity:

The map f → f (H2(Δ) → L2(T)) is isometric. The image of H2(Δ)under this map is a suitable definition of E2(T) = H2(T) and

||f ||2H2(Δ) = ||f ||2L2(�) =1

2π

∫�

|f(z)|2|dz| (f ∈ H2(Δ)).

III Cauchy’s Integral Formula is valid:

For every f ∈ H2(Δ) and every z ∈ Δ

f(z) =1

2πi

∫�

f(ζ)

ζ − zdζ.

In this paper, the following analogous results will be proved in detail.

Theorem 1 Let C = (C1, C2, C3, ..., CN) be an N-tuple of closed distinctcurves on the sphere C and suppose that for each i, 1 ≤ i ≤ N, Ci is acircle, a line ∪{∞}, an ellipse, a parabola ∪{∞} or a branch of a hyperbola∪{∞} (5N possibilities in all). Let D be a complementary domain of ∪Ni=1Ciand suppose that D is simply connected. If Di is the complementary domainof Ci which contains D then

o For a ∈ C\D,ψ′

(ψ − a)2∈ H1(Δ), when ψ : Δ → D is a conformal equiva-

lence;

Boundary Behaviour of Analytic Curves 135

i ∂D is a σ−rectifiable closed curve and every f ∈ E2(D) has a nontangential

limit function f ∈ L2(∂D);

ii The map f → f (E2(D) → L2(∂D)) is an isometric isomorphism onto aclosed subspace E2(∂D) of L2(∂D), so

||f ||2E2(D) = ||f ||2L2(∂D) =1

2π

∫∂D

|f(z)|2|dz| (f ∈ E2(D));

iii For all f ∈ E2(D) and z ∈ D

f(z) =1

2πi

∫∂D

f (ζ)

ζ − zdζ.

Recall that a complementary domain of a closed F ⊆C is a maximal con-

nected subset of C − F, which must be a domain.

The boundary ∂D of D can be decomposed naturally as ∂D = ∂1 ∪ ∂2 ∪... ∪ ∂N where ∂i ⊆ Ci. Cauchy’s Integral Formula

f(z) =N∑j=1

1

2πi

∫∂j

f (ζ)

ζ − zdζ (f ∈ E2(D), z ∈ D)

will be validated. The function f is the “non-tangential limit function” off . These functions will be discussed in detail in the next sections. Item ois crucial in verifying i,ii,iii. In the case D = Δ it is trivial, for a conformalequivalence ψ : Δ → Δ is of the form

ψ(z) = ωz + λ

λz + 1, |ω| = 1, |λ| < 1.

Thus ψ is analytic in a neighbourhood of Δ and, hence,ψ′

(ψ − a)2is as well.

Our aim in the rest of this paper is to show that, for the domains in Theorem1, analogues of I,II,III make sense and can be verified. These properties donot make sense for a general simply connected domain D, but there is a classicexample where analogues of I,II,III are well known. Suppose that D is theinner domain of a rectifiable Jordan Curve γ, so that L2(∂D) is the L2 spaceof normalized arc length measure on γ. It is well known that γ has a tangentat almost all points a ∈ γ, so it makes sense to speak of the nontangentiallimit of a function f in E2(D) at such a point a : f(a) can be defined as thelimit, when it exists, of f(z) as z tends to a through any Stoltz angle in Dwith vertex at a. The following Theorem is proved in Duren [3].

136 Yuksel Soykan

Theorem 2 LetD be inner domain of a rectifiable Jordan curve γ and supposeψ : Δ → D is a conformal equivalence then

a ψ′ ∈ H1(Δ);

b Every f ∈ E2(D) has a nontangential limit function f ∈ L2(∂D);

c The map f → f (E2(D) → L2(∂D)) is an isometric isomorphism onto aclosed subspace E2(∂D) of L2(∂D), so

||f ||2E2(D) = ||f ||2L2(∂D) =1

2π

∫∂D

|f(z)|2|dz| (f ∈ E2(D));

d For all f ∈ E2(D) and z ∈ D

f(z) =1

2πi

∫∂D

f (ζ)

ζ − zdζ.

Proof. See [3, p.169-170].Item a here is a consequence of the celebrated F. and M. Riesz theorem. As

to b,c,d, we wish to stress that they are consequences of a only. The hypothesisthat ∂D be a rectifiable Jordan curve is a sufficient condition for a. But it isby no means necessary. Thus, in generalizing Theorem 2 in Section 6 and 7,our main effort will go into verifying the analogue o of a. Verification of i,ii,iiiwill then be a trivial modification of the proofs of b,c,d as in [3].

2 Hp-Spaces and Ep-Spaces

For 1 ≤ p ≤ ∞, Hp is a Banach space with two concrete realizations: a spaceof analytic functions on Δ and a subspace of Lp(T). In this work these twowill always be distinguished and we shall take more care than is usual withnotation. The space H∞(Δ) is just the set of all bounded analytic function onΔ with the uniform norm. For 1 ≤ p <∞, Hp(Δ) is the set of all functions fanalytic on Δ such that

sup0<r<1

1

2π

∫ 2π

0

∣∣f(reiθ)∣∣p dθ <∞. (1)

The p-th root of the left hand side of (1) here defines a complete norm on

Hp(Δ). For 1 ≤ p ≤ ∞, Hp(T) is defined as the set of all f ∈ Lp(T) such that

1

2π

∫ 2π

0

f(eiθ)eniθdθ = 0, n = 1, 2, ... (2)


so that Hp(T) is a closed subspace of Lp(T) and H2(T) is a Hilbert space withinner-product

〈f , g〉 =1

2π

∫�

f(z)g(z) |dz| =1

2π

∫ 2π

0

f (eiθ)g(eiθ)dθ (f , g ∈ H2(T)). (3)

Equivalent notations for (2) are

1

2π

∫�

f(z)zn |dz| = 0, n = 1, 2, ... (4)

and

1

2πi

∫�

f(z)zmdz = 0, m = 0, 1, 2, .... (5)

If f ∈ Hp(Δ) (1 ≤ p ≤ ∞) there is a (unique) f ∈ Hp(T) such that

limr→1

f(rω) = f(ω) for almost all ω ∈ T. (6)

Moreover, the map f → f (Hp(Δ) → Hp(T)) is an isometric isomorphismonto Hp(T).

We can say more than (6). Suppose that f ∈ Hp(Δ) and that ω ∈ T isa point where the radial limit exists as in (6); then if S is a Stoltz angle atω, that is S is an open sector of a circle with vertex at ω and S lies entirelyin Δ, then f(z) tends to f (ω) whenever z tends to ω inside S, either along a

sequence or an arc. For this reason f is called the non-tangential limit functionof f . The inverse map

f → f (Hp(T) → Hp(Δ))

can be realized by Cauchy’s integral formula

f(z) =1

2πi

∫�

f(ζ)

ζ − zdζ (f ∈ Hp(Δ), z ∈ Δ 1 ≤ p ≤ ∞) (7)

where T is positively oriented. In this work the term “Parseval’s Identity” willbe used to denote the identity

‖f‖2H2(Δ) =

1

2π

∫�

∣∣∣f (z)∣∣∣2 |dz| (f ∈ H2(Δ)). (8)

From now on we shall be interested almost exclusively in Hilbert spaces, but

we shall occasionally need to consider H1(Δ) and H1(T). What we require iscovered by the following observations. First of all

f, g ∈ H2(Δ) ⇒ fg ∈ H1(Δ) ⇒ f g = fg ∈ H1(T).

138 Yuksel Soykan

The first implication follows from Schwarz’s inequality and Equation (4) and

the second follows from Equation (6) since f g is the non-tangential limit offg. Clearly

f ∈ H2(Δ) ⇒ f2 ∈ H1(Δ) ⇒ f2 ∈ H1(T). (9)

There is a partial converse to (9). If g ∈ H1(Δ) and g is nowhere 0 in Δ then g

has an analytic square root g1/2. In this case g1/2 ∈ H2(Δ) and g1/2 ∈ H2(T).The following Lemma will be important in our studies of Cauchy’s IntegralFormula.

Lemma 3 Let ψ be a function which is conformal (i.e., meromorphic andinjective) on Δ. Suppose a ∈C − ψ(Δ) and that the function

h(z) =ψ′(z)

(ψ(z)− a)2(z ∈ Δ)

belongs to H1(Δ). Then, for all f ∈ H2(Δ) and all z ∈ Δ, and provided

ψ(z) �= ∞,

f(z) =1

2πi

∫�

f(ζ)k(ζ)ψ′(z)1/2dζ (10)

where k(ζ) is the nontangential limit of

ψ′(ζ)1/2

ψ(ζ) − ψ(z).

Proof. First note thatψ′1/2

ψ − a∈ H2(Δ). Keep z ∈ Δ fixed throughout and let

g(ζ) =ψ′(ζ)1/2ψ′(z)1/2

ψ(ζ) − ψ(z)− 1

ζ − z(ζ ∈ Δ). (11)

It will be shown first that g ∈ H2(Δ). Because ψ is conformal the first termon the right of equation 11 has a simple pole at ζ = z with residue 1, so

singularity of g at z is removable, hence g is analytic on Δ. If s =1

2(1 + |z|)

then

sup0<r≤s

1

2π

∫ 2π

0

∣∣g(reiθ)∣∣2 dθ <∞.


Now on the annulus s < |ζ| < 1 the functions ψ(ζ) −aψ(ζ) −ψ(z)

, 1ζ−zare bounded. So

sups<r<1

1

2π

∫ 2π

0

∣∣ψ′(reiθ)∣∣ |ψ′(z)|

|ψ(reiθ) − ψ(z)|2dθ <∞

and also

sups<r<1

1

2π

∫ 2π

0

1

|reiθ − z|2dθ <∞.

It follows that

sups<r<1

1

2π

∫ 2π

0

∣∣g(reiθ)∣∣2 dθ <∞,

so that g ∈ H2(Δ). Now suppose f ∈ H2(Δ) is given. Then fg ∈ H1(Δ). So

f g ∈ H1(T) and hence, by Equation (5) with m = 0,

1

2πi

∫�

f(ζ)g(ζ)dζ = 0.

Therefore the right hand side of Equation (10) is just

1

2πi

∫�

f(ζ)

ζ − zdζ = f(z).

When D is a simply connected domain contained entirely in C the followingintrinsic definition of Ep(D) is given in [3, Section 10.1].

Definition 1 Let D ⊆ C be a simply connected domain. A function g analyticin D is said to be of class Ep(D) (1 ≤ p <∞) if there exists a sequence (γn) ofrectifiable Jordan curves in D, tending to the boundary ∂D of D in the sensethat each compact subset of D is eventually in the inner domain of γn suchthat

supn≥0

∫γn

|g(z)|p |dz| <∞. (12)

It is not clear from this definition that E2(D) is a vector space, but

Lemma 4 Let D ⊆ C be a simply connected domain. Suppose that ϕ : D → Δ

is a conformal equivalence of D onto Δ and that ϕ′ 12 is an analytic branch ofthe square root of ϕ′. Then

g ∈ E2(D) ⇔ g = (f ◦ ϕ)ϕ′1/2 for some f ∈ H2(Δ). (13)

140 Yuksel Soykan

Proof. The proof can be found in e.g. Duren [3, p.169].Thus 13 provides an alternative definition of the vector space E2(D) as all

functions (f ◦ϕ)ϕ′1/2 with f ∈ H2(Δ). This definition does not depend on thechoice of Riemann mapping function ϕ. Clearly E2(Δ) = H2(Δ).

Duren [3] restricts his attention to domains D ⊆ C. But we will need toconsider more general simply connected domains D in the extended plane C,for instance D might be the exterior of a circle together with ∞. If ∞ ∈ D,definition 1 can be modified so as to make lemma 4 true in this case as well.

Definition 2 Suppose that ∞ ∈ D. A function g analytic in D is said to beof class Ep(D) if g(∞) = 0 and there exists a sequence of rectifiable Jordancurves γ1, γ2, ... in D, tending to the boundary of D in the sense that eachclosed subset of D is eventually in the outer domain of γn, such that

supn

∫γn

|g(z)|p |dz| <∞. (14)

To prove the analogue of lemma 4 the following lemma is needed

Lemma 5 Let D1, D2 ⊆ C be simply connected domains and suppose thatϕ : D1 → D2 is a conformal equivalence. Then there exists a meromorphicbranch ϕ′1/2 of the square root of ϕ′ on D1. If ∞ ∈ D1 and ϕ(∞) ∈ C then ϕ′1/2

has a simple zero at ∞. If ∞ ∈ D2 and ϕ−1(∞) ∈ C then ϕ′1/2 has a simplepole at ϕ−1(∞). At all other points of D1, ϕ

′1/2 is analytic and non-zero.

Proof. The proof is straight forward and will be omitted.Now we are ready to give the analogue of lemma 4

Lemma 6 Let D ⊆ C be a simply connected domain with ∞ ∈ D. Supposethat ϕ : D → Δ is a conformal equivalence and that ϕ′1/2 is an analytic branchof the square root of ϕ′. Then

g ∈ E2(D) ⇔ g = (f ◦ ϕ)ϕ′1/2 for some f ∈ H2(Δ). (15)

Proof. Using Lemma 5, the result follows.We now know that if D ⊆ C is any simply connected domain and ϕ :

D → Δ is any conformal equivalence then E2(D) is the set of all functions(f ◦ ϕ)ϕ′1/2 with f ∈ H2(Δ). An inner product can be defined on E2(D) bythe formula

〈g1, g2〉ϕ = 〈f1, f2〉H2(Δ) (gj = (fj ◦ ϕ)ϕ′1/2, fj ∈ H2(Δ), j = 1, 2). (16)

The inner product on E2(D) defined by (16) is independent of the choice of ϕ.So we shall now write 〈, 〉E2(D) or simply 〈, 〉 for the inner product on E2(D).

When D and the Di’s are as in Theorem 1; the boundaries of D andthe Di’s are “reasonable” and it will make sense for us to discuss to notionof a nontangential limit function and spaces E2(∂D), E2(∂Di) analogous toH2(Δ), H2(T).


3 Boundary Behaviour

From now on, we shall fix the notation as in Theorem 1. Thus C = (C1, C2, C3,..., CN) is an N -tuple of distinct Jordan curves in C. Each Ci is either a circle,an ellipse, a line, a parabola, or a single component of a hyperbola: in thelast three cases we adjoin ∞ to the finite part of Ci. Assume that D is acomplementary domain of ∪Ni=1Ci, that D is simply connected. We shall verifythat the boundary ∂D of D is a closed σ−rectifiable curve and we make senseof the non-tangential limit function f of an element f ∈ E2(D) and analoguesof I., II., III. in section 1 will be proved.

Intuition suggests that ∂D is a closed curve. This will be verified by showingthat if ψ : Δ → D is a conformal equivalence then ψ can be extended toa continuous function on Δ which maps T onto ∂D. We shall also need toconsider analytic continuations of ψ across various subarcs of T; this will bedone using a generalized versions of Schwarz’s Reflection Principle. We statethe well-known classical version first and then discuss generalizations.

Theorem 7 Schwarz’s Reflection Principle (classical): Let U, V ⊆ C be do-mains which are symmetric about the real line R and let

U+ = {z ∈ U : Im z > 0}, U− = {z ∈ U : Im z < 0}V + = {z ∈ V : Im z > 0}, V − = {z ∈ V : Im z < 0}.

Suppose that the function f is continuous on U+ ∪ (U ∩ R) and analytic onU+, that f maps U+ conformally onto V + and that f maps U ∩ R injectivelyonto V ∩R. Then f can be continued analytically into the whole of U so as todefine a conformal map of U onto V with f(U−) = V −.

Proof. See, for instance [1]. The continuation is defined explicitly by

f(z) = f(z) for z ∈ U−.

The injectivity of the extension is obvious. The only difficulty is to show thatthe singularities of f on U ∩ R are removable.

Extensions of Schwarz’s Reflection Principle involve analytic arcs and curvesso we recall the following definition.

Definition 3

i An open analytic arc γ is an arc of the form γ = h(J), where J is an opensubinterval of R (possibly R itself) and h is a function which is analyticand conformal in a neighbourhood U of J .

ii A closed analytic curve is a curve γ = k(T) where k is analytic and con-formal in a neighbourhood U of T. If γ is simple it is called an analyticJordan curve.

142 Yuksel Soykan

Clearly a circle and an ellipse is an analytic Jordan curve and a line, aparabola and a hyperbola component is an open analytic arc.

Theorem 8 Schwarz’s Reflection Principle (Analytic Arcs): Let G,G1 ∈ C

be simply connected domains and suppose that ∂ ⊆ ∂G ∩ C and ∂1 ⊆ ∂G1 ∩ C

are open analytic arcs. Let U/U1 be a domain which is symmetric about R

and suppose that J = U ∩ R /J1 = U1 ∩ R is an open interval and that h/h1

is a function which is analytic and conformal on U/U1 and satisfies h(J) =∂/h1(J1) = ∂1. Suppose further that

h(U+) ⊆ G,h(U−) ∩ G = ∅, h1(U+1 ) ⊆ G1, h1(U

−1 ) ∩ G1 = ∅.

If the function f satisfies

i f is continuous and injective on G ∪ ∂;

ii f is analytic on G;

iii f(G) = G1;

iv f(∂) = ∂1

then f can be continued analytically into G ∪ h(U) so as to define a con-formal map of G ∪ h(U) onto G1 ∪ h1(U1) with f(h(U−)) = h1(U

−1 ).

Proof. It is sufficient to prove that f can be continued analytically fromh(U+) into h(U) with f(h(U−)) = h1(U

−1 ). But this is straightforward. The

function F = h−11 ◦ f ◦ h is continuous and injective on U+ ∪ J, it maps U+

conformally onto U+1 and J onto J1. So by Theorem 7 F can be continued

analytically to a conformal map of U onto U1 with F (U−) = U−1 . So f can be

continued as required because f = h1 ◦ F ◦ h−1.Now we consider continuations across Jordan curves. If we wish to ana-

lytically continue a conformal map across a proper arc of an analytic Jordancurve Theorem 8 can be used. So we only need a theorem about continuingover the whole of a Jordan curve.

Theorem 9 Let G be the inner domain of an analytic Jordan curve γ ⊆C and let G1 be the inner domain of an analytic Jordan curve γ1 ⊆ C. Ifψ is a conformal equivalence mapping G onto G1 then ψ can be continuedanalytically across γ to define a conformal map of a neighbourhood of G ontoa neighbourhood of G1.

Proof. First of all, the well-known Osgood theorem (see, for example, [5,corollary 4.10, p.446]) tells us that ψ can be extended to a homeomorphismof G onto G1 with ψ(γ) = γ1. Although it is natural to parametrize a Jordan


curve using T it can also be parametrized using R∪(∞), because the conformalautomorphism

μ(z) =z − i

z + i

maps R ∪ (∞) onto T. So there is a domain U ⊆ C which is a neighbourhoodof R∪ (∞) and a function h which is analytic and conformal on U and satisfiesh(R∪(∞)) = γ. There is no loss of generality in assuming that U is symmetricabout R. Thus U is the complement of a compact subset of the upper half-planeand its mirror image in R.Moreover, we can assume, because of conformality onR, that h(U+) ⊆ G and h(U−)∩G = ∅.We can also define an analogous domainU1 and mapping h1 for G1 and γ1. When we set up the function F = h−1

1 ◦ψ◦hit can be continued analytically across R by the classical theorem Theorem 7.The singularity of F at ∞ will be removable because F will be analytic andbounded outside of any disc which contains the complement of U . Now useh1 ◦ F ◦ h−1 to continue ψ as required.

4 σ-rectifiable arcs and Hypothesis-R

In this section σ − rectifiable arcs and their images under conformal transfor-mations will be considered. An arc or closed curve γ is called σ− rectifiable ifand only if it is a countable union of rectifiable arcs in C, together with (∞)in the case when ∞ ∈ γ. For instance, a parabola without ∞ is σ− rectifiablearc, and a parabola with ∞ is σ − rectifiable Jordan curve.

Lemma 10 Let γ ⊆ C be a compact, rectifiable arc and suppose that γ1 =μ(γ), where μ is a function which is analytic and conformal on a neighbourhoodof γ, then

i γ1 is compact and rectifiable;

ii a set E ⊆ γ has arc length measure 0 if and only if μ(E) ⊆ γ1 has arc lengthmeasure 0.

Proof. The notation γ1 = μ(γ) is taken to mean that if h ∈ C [a, b] parametrizesγ then μ ◦ h parametrizes γ1.Only rectifiability causes any problem in i. But,using l to denote arc length measure,

l(γ1) = l(μ(γ)) =

∫γ

|μ′(z)| |dz| <∞,

144 Yuksel Soykan

because μ′ is continuous on the compact set γ. To prove ii note that if E ⊆ γhas arc length measure 0 then

l(μ(E)) =

∫E

|μ′(z)| |dz| = 0.

The reverse implication follows by symmetry using μ−1.

Lemma 11 i If γ is a simple open analytic arc then γ is σ − rectifiable;

ii Let γ and γ1 be simple, open analytic arcs. Suppose that V and V1 areopen neighbourhoods of γ and γ1, respectively, and that ψ is a conformalequivalence mapping V onto V1 with ψ(γ) = γ1. Then a set E ⊆ γ hasarc length measure 0 if and only if ψ(E) ⊆ γ1 has arc length measure 0.

Proof. Let J ⊆ R be an open interval and suppose γ = h(J) where h isanalytic and conformal in an open set U containing J . J can be written as theunion of an increasing sequence (Jn) of compact subintervals, so that γ is theunion of the sequence (h(Jn)) of compact arcs. Each of these arcs is rectifiablebecause

l(h(Jn)) =

∫Jn

|h′(t)|dt <∞,

where l is arc length measure, as before. So γ is σ− rectifiable.To prove ii putβn = h(Jn), as above. If E ⊆ γ is a set of measure 0 then

E =

∞⋃n=1

E ∩ βn,

so that each E∩βn has measure 0. Applying Lemma 10 to the compact rectifi-able arcs βn it follows that each set ψ(E∩βn) has measure 0 as well. Therefore,ψ(E) =

⋃∞n=1 ψ(E ∩ βn) has measure 0. The reverse implication follows by

symmetry.Many of the transformations we shall use will be fractional linear, that is

of the form

μ(z) =αz + β

γz + δ; Λ := αδ − βγ �= 0. (17)

Simple manipulation shows that

μ′(z) =Λ

(γz + δ)2, (18)

so it is easy to define a branch of μ′ 12 by choosing a square root Λ1/2 of Λ and

putting

μ′(z)1/2 =Λ1/2

γz + δ. (19)


Definition 4 A simple σ−rectifiable arc γ ⊆ C is said to satisfy hypothesis-Rif and only if, for some (hence all) a ∈ C − γ

∫γ

|dz||z − a|2 <∞.

Equivalently, if the function f(z) =1

z − abelongs to L2(γ).

The “some hence all” rider is easy to verify. For suppose a, b ∈ C − γ.Choose closed discs δa, δb centred on a, b respectively, both of them disjoint

from γ. Then each of the functionsz − a

z − band

z − b

z − ais bounded on γ. Hence

∫γ

|dz||z − a|2 <∞ ⇔

∫γ

|dz||z − b|2 <∞.

Lemma 12 Let γ be a simple, σ−rectifiable arc in C. Suppose that a ∈ C−γand that μ(z) =

1

z − a. Then γ satisfies hypothesis-R if and only if μ(γ) is

rectifiable.

Proof. Note that

l(μ(γ)) =

∫γ

|μ′(z)| |dz| =

∫γ

|dz||z − a|2 <∞.

So the result is clear.

Example 1 An analytic Jordan curve in C obviously satisfies hypothesis-R.It is also easy to see that if γ satisfies hypothesis-R so does μ(γ), where μ isa linear equivalence: μ(z) = αz + β. For if a ∈ C − μ(γ) then a = μ(b), forsome b ∈ C − γ, so

∫μ(γ)

|dz||z − a|2 =

∫γ

|μ′(ζ)||μ(ζ) − μ(b)|2 |dζ| =

1

|α|∫γ

|dζ||ζ − b|2 .

Obviously R satisfies hypothesis-R and therefore so does any line; and us-ing the parametrizations of the standard parabola P0 and standard hyperbolacomponents Hα, as in section 3, we see that every parabola and hyperbola com-ponent satisfies hypothesis-R.

146 Yuksel Soykan

5 Extensions of Mapping Functions

Each Ci ∈ C has two complementary domains Di, D′i. Because D is connected

and D ⊆ Di ∪D′i it can be assumed that Di, D

′i are labelled so that D ⊆ Di,

D ∩ D′i = ∅. Assume that each Ci is oriented so that Di, hence D, is to the

left of Ci. A simple and nontrivial example is the following one. We urge thereader to draw a picture including the orientations on the circles in C whichkeep D to the left.

Example 2 Let

C1 = T, D1 = C\Δ = {z ∈ C : |z| > 1} ∪ (∞);

C2 = {z ∈ C : |z| = 2}, D2 = {z ∈ C : |z| < 2};C3 = {z ∈ C : Re z = −1} ∪ (∞), D3 = {z ∈ C : Re z > −1};C4 = {z ∈ C : |z − 3| = 1}, D4 = {z ∈ C : |z − 3| > 1} ∪ (∞);

C5 = {z ∈ C :

∣∣∣∣z − 11

2

∣∣∣∣ =1

2}, D5 = {z ∈ C :

∣∣∣∣z − 11

2

∣∣∣∣ > 1

2} ∪ (∞).

Then D = ∩5i=1Di is a connected and simply connected complementary domain

of C. Note that ∂D is not a Jordan curve.

In the simplest cases, for instance, if N = 2 and C1, C2 are circles, then

D1 ∩D2 is connected. But in general ∩Ni=1Di need not be connected.

Example 3 Suppose that C is a collection of circles all of the same radiusr > 0 and that the inner domains of the Ci’s cover the unit circle T = ∂Δ. LetD be the unbounded complementary domain of ∪Ci. Clearly Di is the outerdomain of Ci. Of course, in this case each Ci is negatively oriented. If r ≥ 1then D = ∩Di, but if r < 1 then ∩Di is disconnected.

(To return to the general case), clearly ∂D �= ∅. First note that, since D isa maximal domain in C − ∪Ni=1Ci, we must have ∂D ⊆ ∪Ni=1Ci. Next observethat ∂D contains no isolated points. A simple proof runs as follows. If δ ⊆ C

is an open disc centre a and δ− (a) ⊆ D then we must have a ∈ D, otherwisea ∈ Ci for some i, but δ ∩ Ci is infinite, contradicting δ − (a) ⊆ D. The proofthat ∞ is not an isolated point of ∂D is similar.

Definition 5 i A point a ∈ C will be called a multiple point of C if a ∈ Ci∩Cjfor two distinct values of i and j; ∞ is called a multiple point if some Ciis unbounded;

ii we denote by F ′ the set of all multiple point of C and define F = F ′ ∩ ∂D.


If Ci ∈ C and we remove the set F ′ ∩ Ci of multiple points on Ci we areleft with the disjoint union of finitely many open analytic subarcs.

Definition 6 An arc Γ ⊆ C will be called a maximal open subarc (abbreviationis m.o.s.) of C if

i for some i, 1 ≤ i ≤ N, Γ ⊆ Ci and Γ ∩ Cj = ∅ for j �= i;

ii the end points of Γ belong to F ′.

If, in addition, Γ ⊆ ∂D then Γ will be called a maximal open subarc of ∂D.The end points of an m.o.s need not be distinct and there is an extreme casewhere Ci is a circle or an ellipse and contains no multiple points (see Example2). In this case we shall call to whole of Ci a maximal open subarc. Item iiabove will be taken vacuously. Obviously 2 different m.o.s are disjoint.

Lemma 13 Let Γ be a maximal open subarc of C. Then either Γ ⊆ ∂D orΓ ∩ ∂D = ∅.

Proof. A connectedness argument will be used. Suppose Γ ⊆ C1, say, andlet us assume first that C1 is unbounded. Let h : R → C be a parametriza-tion of C1 such that lim|t|→∞ h(t) = ∞ and h is analytic and conformal in aneighbourhood of R in C.

Assume that the orientation of C1 corresponds to increasing t. Thus pointsin the upper half-plane close to R will be mapped into D1 and points in thelower half-plane close to R will be mapped into D′

1. To show that Γ ∩ ∂D isopen in Γ suppose a ∈ Γ ∩ ∂D. It can be assumed that a = h(0). Choose anopen disc δ centre 0 such that

i h is conformal in a neighbourhood of δ;

ii h(δ) ∩ Cj = ∅ for j ≥ 2.

Let δ+, δ− be the parts of δ contained in the upper, lower, open half-planesrespectively. Since D1, D

′1 are complementary domains of C1 it follows from i

that h(δ+) ⊆ D1 and h(δ−) ⊆ D′1. And since D is a complementary domain

of ∪Ni=1Ci and h(δ+) ∩D �= ∅ it follows from ii that h(δ+) ⊆ D. Clearly thenh(δ ∩ R) ⊆ ∂D and hence Γ ∩ ∂D is open in Γ. Now ∂D is a compact subsetof C so Γ − ∂D is open in Γ. Since Γ is connected we see that either Γ ⊆ ∂Dor Γ ∩ ∂D = ∅. If Γ is a circle or an ellipse we can use a similar proof with aparametric function h for Γ which is analytic and conformal on a neighborhoodof the unit circle T.

There is an important corollary to this Lemma which it will be needed inthe next section. If G is a domain and a ∈ ∂G then a is said to be accessible

148 Yuksel Soykan

from G if given any sequence (an) ⊆ G which tends to a there is an arc γ ⊆ Gwhich interpolates (an) and tends to a. More precisely, there is a continuousfunction k : [0, 1) → G and an increasing sequence (bn) ⊆ [0, 1) which tends to1 such that an = k(bn) (n ≥ 1) and limt→1 k(t) = a.

Lemma 14 If Γ is a maximal open subarc of ∂D then every point a ∈ Γ isaccessible from D.

Proof. Let h, δ, δ+, δ− be as in the proof of the Lemma 13. If (an) ⊆ D andan → a it can be assumed that (an) ⊆ h(δ+) so that (h−1(an)) is a sequence inthe convex set δ+. Obviously (h−1(an)) can be interpolated with an piecewiselinear arc γ′ which tends to 0; so we can put γ = h(γ′).

In the case Ci is a circle or an ellipse the parametric function h is conformalin a neighborhood of the circle T. This time the positive orientation of T

corresponds to the orientation of Ci so the analogue of δ+ is the intersectionof a disc with the inner domain Δ of T. This set is convex.

Another way of stating Lemma 14 is to say that each maximal open subarcof ∂D is one-sided. Now we can return to our analysis of ∂D with

Lemma 15 There exists a finite collection Γ1,Γ2, ..., Γm of distinct (hence dis-joint) maximal open subarcs of C such that

∂D = Γ1 ∪ Γ2 ∪ ... ∪ Γm.

Proof. The boundary of D consists of finitely many maximal open subarcsand finitely many multiple points. Let Γ1,Γ2, ..., Γm be the (distinct) maximalopen subarcs of ∂D. Now Γi is just Γi together with its endpoints. Since ∂Dis closed we have

Γ1 ∪ Γ2 ∪ ... ∪ Γm = Γ1 ∪ Γ2 ∪ ... ∪ Γm ⊆ ∂D.

Any point of the complement ∂D− (Γ1 ∪Γ2 ∪ ...∪Γm) must be an isolatedmultiple point. But recall that ∂D has no isolated points. Hence ∂D =Γ1 ∪ Γ2 ∪ ... ∪ Γm.

Corollary 16 ∂D = Γ1 ∪ Γ2 ∪ ... ∪ Γm ∪ F , disjointly.

Proof. F is precisely the set of endpoints of Γ1,Γ2, ..., Γm, see Definition 5.The notations of this lemma will be kept fixed from now on. Let us choose

and fix a Riemann mapping function (abbreviation is RMF) for D, that isϕ : D → Δ is a conformal map of D onto Δ. ψ will always be used to denoteϕ−1. Thus ψ : Δ → D is conformal. Further information about ∂D can beobtained from two standard theorems: one about extending RMFs, the otherabout extending inverse RMFs. The first result is


Theorem 17 Let G ⊆ C be a simply connected domain and suppose thatθ : G → Δ is an RMF. Let A(G) be the set of all points of ∂G which areaccessible from G. Then θ extends by continuity to a (unique) continuousinjective function mapping G ∪ A(G) into Δ. The extended function mapsA(G) into T.

Proof. See [2, theorem 5.12, p.53; exercise 6, p.56] and [7, theorem 14.18,p.289].

Corollary 18 The RMF ϕ of D extends by continuity to a continuous injec-tive map of D ∪ (Γ1 ∪ Γ2 ∪ ...∪Γm) into Δ. Each point of Γ1 ∪Γ2 ∪ ...∪Γm ismapped into T by the extended function.

Proof. An immediate consequence of Lemma 14 and theorem 17.

Definition 7 i Let G be a domain, suppose a ∈ ∂G, E ⊆ G and a ∈ E. If fis a function defined on E we write

limz→aE

f(z) = α

if f(z) tends to α as z tends to a through points of E.

ii Let ϕ denote the restriction to Γ1 ∪Γ2 ∪ ...∪Γm of the continuous extensionof ϕ. Then

ϕ(a) = limz→aD

ϕ(z) (a ∈ Γ1 ∪ Γ2 ∪ ... ∪ Γm).

Before stating the theorem on inverse RMFs we need to set up some nota-tions and recall some definitions.

Definition 8

i Given a ∈ C and r > 0 let Nr(a) denote the open disc centre a, radius r.

ii A subset E ⊆ C is said to be locally connected (abbreviation is lc) at a pointa ∈ E if given ε > 0 there exist some δ > 0 such that any 2 pointsz1, z2 ∈ E∩ Nδ(a) can be joined by a connected set γ ⊆ E∩ Nε(a).

iii A subset E is called locally connected if it is locally connected at each of itspoints.

iv A subset E is called uniformly locally connected (abbreviation is ulc) if it islocally connected and if δ in ii can be chosen to be independent of a ∈ E.

150 Yuksel Soykan

Corollary 19 i A compact lc subset of C is ulc.

ii A finite union of compact (uniformly) lc sets is (uniformly) lc.

Proof. See [4, corollary 1 to theorem 13.1 in chapter VI and corollary totheorem 8.1 in chapter IV ]. Note that Pommerenke [6] takes Newman’s [4]definition of uniformly local connectedness as local connectedness.

Theorem 20 Let Ω ⊆ C be a bounded simply connected domain and supposethat χ : Δ → Ω is an inverse RMF for Ω. The following are equivalent

i χ extends by continuity to a continuous function mapping Δ into Ω.

ii ∂Ω is uniformly locally connected.

iii ∂Ω is a closed curve.

Proof. See [6, Theorem 2.1, p.20].

Corollary 21 If i, ii, iii of theorem 20 are true then ∂Ω = χ(T).

Proof. First of all ∂Ω ⊆ χ(T). Suppose b ∈ ∂Ω. Choose a sequence (an) ⊆ Δsuch that χ(an) → b. It can be assumed that (an) converges to some a ∈ Δ.Because χ is an open mapping on Δ we cannot have a ∈ Δ, otherwise b = χ(a)belongs to (the interior of ) Ω. Hence a ∈ T and b = χ(a) ⊆ χ(T).

χ(T) ⊆ ∂Ω will be proved by contradiction. Suppose a ∈ T but b = χ(a) ∈Ω. Then b has another counter image c = χ−1(b) ∈ Δ. Let U be an openneighbourhood of c such that U ⊆ Δ. Now choose a sequence (an) ⊆ Δ − Usuch that an → a, hence also χ(an) → b = χ(c). Since χ(U) is an openneighbourhood of b, an ∈ χ(U), for large n. Hence, for large n, χ(an) has twodifferent counter images: an ∈ Δ − U and χ−1(χ(an)) ∈ U. This contradictsthe conformality of χ.

6 The case where D is bounded

In this section ∂D will be elucidated when D is bounded domain in C. It willbe verified that every f ∈ E2(D) has a non-tangential limit function and thatParseval’s identity and Cauchy’s Integral formula are valid.

Lemma 22 If D is bounded then D and ψ satisfies i., ii., iii. of Theorem 20and Corollary 21 .


Proof. By lemma 15

∂D = Γ1 ∪ Γ2 ∪ ... ∪ Γm

where Γ1,Γ2, ..., Γm are the maximal open subarcs of ∂D. Each Γi is eitherhomeomorphic to a closed interval or homeomorphic to a circle. So ∂D is afinite union of compact l.c. sets and is therefore u.l.c. The result now followsfrom theorem 20.

Remark 1 The last three items show that when D is bounded ψ extends to acontinuous map of Δ onto D which maps T onto ∂D. The notation ψ will beused for the restriction of the extended map to T. Thus

ψ(z) = limζ→zΔ

ψ(ζ) (z ∈ T).

This does not conflict with our previous notation (see Equation 6) because ψ(z)as defined here is necessarily the non-tangential limit of ψ at z.

∂D can now be written as

∂D = ψ(T) = Γ1 ∪ Γ2 ∪ ... ∪ Γm = Γ1 ∪ Γ2 ∪ ... ∪ Γm ∪ F (20)

where F is the set of multiple points of ∂D. For each i we have two subsets of

T : ϕ(Γi),which must be connected, and ψ−1

(Γi),which must be open.

Lemma 23 For each i, ψ−1

(Γi) = ϕ(Γi).

Proof. To prove that ψ−1

(Γi) ⊆ ϕ(Γi) suppose a ∈ ψ−1

(Γi). Then b :=

ψ(a) = limz→aΔ

ψ(z) ∈ Γi. Choose a sequence (zn) ⊆ Δ such that zn → a. Then

b = ψ(a) = limn→∞

ψ(zn). So, by definition of ϕ,

ϕ(b) = limn→∞

ϕ(ψ(zn)) = limn→∞

zn = a.

Hence a = ϕ(b) ∈ ϕ(Γi). The proof that ϕ(Γi) ⊆ ψ−1

(Γi) is similar. Supposec ∈ ϕ(Γi), say c = ϕ(d), where d ∈ Γi. Choose a sequence (wn) ⊆ D such thatwn → d. Then c = lim

n→∞ϕ(wn) and hence

d = limn→∞

wn = limn→∞

ψ(ϕ(wn)) = ψ(c).

From now on, define

γi = ϕ(Γi) = ψ−1

(Γi) 1 ≤ i ≤ n. (21)

152 Yuksel Soykan

Each γi is an open and connected subset of T. Hence γi is either a proper

open arc or the whole of T. Use G to denote the counter image ψ−1

(F ) of themultiple points F of ∂D. Thus, in addition to equation 20,

T = γ1 ∪ γ2 ∪ ... ∪ γm ∪ G. (22)

Clearly ψ|γi is a homeomorphism onto Γi, in particular ψ is injective on

each γi. And since the Γi’s are disjoint, ψ is injective on T −G.

Lemma 24 If D is bounded then G is precisely the set of end points of thearcs γ1, γ2, ..., γm.

Proof. Recall that G is the complement of m disjoint open subarcs of T, soG can be written as

G = I1 ∪ I2 ∪ ... ∪ Imwhere each Ii is either a compact arc or a point. It will be shown that each Iiis necessarily a point.

The set F of multiple points of ∂D is finite, say F = {p1, p2, ..., pk}. Now

Ii ⊆ G = ψ−1

(F ) so Ii can be written as

Ii =k⋃j=1

ψ−1

(pj) ∩ Ii (23)

and also note that ψ(z) ≡ pj on ψ−1

(pj)∩Ii. Now D is bounded so ψ ∈ H∞(Δ)and obviously ψ is not constant. So, as is well known [3, theorem 2.2, p.17]

ψ cannot be constant on a set of positive arc-length measure. Therefore eachset on the right-hand side of 23 must have measure zero. So Ii cannot benon-trivial arc: it must be a point.

Remark 2 ψ can be used to parametrize the closed curve ∂D. Since ψ(γi) = Γiis an open analytic arc ψ can be continued analytically across γi (see Theorem

8). Hence, by conformality, the orientation which ψ induces on ∂D is thepositive orientation which keeps D to the “left” of ∂D.

Remark 3 Now that the continuations of ϕ and ψ have been noted, ψ(ζ)

will be used rather that ψ(ζ) when ζ ∈ T, and ϕ(z) rather that ϕ(z) whenz ∈ Γ1 ∪ Γ2 ∪ ... ∪ Γm.

The next result is an analogue statement of theorem 2 for the boundedcase.


Theorem 25 If D is bounded then

o ψ′ ∈ H1(Δ);

i Every f ∈ E2(D) has a nontangential limit function f ∈ L2(∂D);

ii (Parseval’s Identity) The map f → f (E2(D) → L2(∂D)) is an isometricisomorphism onto a closed subspace E2(∂D) of L2(∂D), so

||f ||2E2(D) = ||f ||2L2(∂D) =1

2π

∫∂D

|f(z)|2|dz| (f ∈ E2(D));

iii (Cauchy’s Integral Formula) For all f ∈ E2(D) and z ∈ D

f(z) =1

2πi

∫∂D

f (ζ)

ζ − zdζ.

Proof. To prove o two results are needed from Duren’s book [3] which arestated here.

Lemma 26 [3, theorem 3.10] If f ∈ H1(Δ) and f is equal almost everywhere

to a function of bounded variation on T then f is absolutely continuous on T.

Lemma 27 [3, theorem 3.11] If f is continuous on Δ and analytic in Δ then

f is absolutely continuous on T if and only if f ′ ∈ H1(Δ).

To prove o note that ψ ∈ H∞(Δ) ⊆ H1(Δ), since D is bounded. Recall

that ψ is continuous on T and injective on the complement of the finite set G.Clearly, the total variation of ψ on T is the total length of ψ(T) = ∂D. But,in the bounded case, ∂D is a finite union of rectifiable arcs, so, by lemma 26,ψ is absolutely continuous with respect to arc length measure on T. Now itfollows from lemma 27 that ψ′ ∈ H1(Δ).

To prove i note first that, by Lemma 11 (ii) applied to each γi, E ⊆ T hasmeasure 0 if and only if ψ(E) ⊆ ∂D has measure 0.

Now let f ∈ E2(D), then f = (g ◦ϕ)ϕ′1/2, for some g ∈ H2(Δ). Let E ⊆ T

be a set of measure 0 such that g(ζ) exists, for all ζ ∈ T− E. We can assumethat E ⊇ G := ψ−1(F ), where F is the set of multiple points of ∂D.

If ζ ∈ T − E then ψ is conformal at ζ. So if S is a Stolz angle in Δ withvertex ζ then ψ(S) contains a Stolz angle in D with vertex at ψ(ζ). Conversely,if S ′ is a Stolz angle in D with vertex at ψ(ζ) then ϕ(S ′) contains a Stolz anglein Δ with vertex at ζ. It follows easily from this that the non-tangential limitg ◦ ϕ(z) exists, for all z ∈ ψ(T − E), hence almost everywhere. Now ϕ′1/2 is

continuous on Γ1 ∪ Γ2 ∪ ...∪Γm so ϕ′1/2 exists almost everywhere. Hence f (z)exists for almost all z ∈ ∂D.

154 Yuksel Soykan

The proof of Parseval’s identity ii is now easy. Let f = (g ◦ ϕ)ϕ′1/2 asbefore. Then substituting z = ψ(ζ) we have

1

2π

∫∂D

|f(z)|2|dz| =1

2π

∫∂D

|g(ϕ(z))|2 |ϕ′(z)||dz|

=1

2π

∫�

|g(ζ)|2|dζ|= ||g||2H2(Δ)

= ||f ||2E2(D).

To prove iii let f = (g ◦ ϕ)ϕ′1/2 with g ∈ H2(Δ) and fix z ∈ D. Let

I(z) =1

2πi

∫∂D

f (ζ)

ζ − zdζ.

Substituting ζ = ψ(u) it follows that

I(z) =1

2πi

∫∂D

g(ϕ(ζ))ϕ′(ζ)1/2

ζ − zdζ =

1

2πi

∫�

g(u)ψ′(u)1/2

ψ(u) − ψ(ϕ(z))du.

Now, if a ∈ C − D then ψ(u) − a is bounded away from 0 on Δ. Given that

ψ′ ∈ H1(Δ) we see thatψ′

(ψ − a)2∈ H1(Δ) as well. It follows from Lemma 3

(p.138) that

I(z) = g(ϕ(z))1

ψ′(ϕ(z))1/2= g(ϕ(z))ϕ′(z)1/2 = f(z).

7 The unbounded case

Suppose now that D is unbounded: either ∞ ∈ D or ∞ ∈ ∂D. Clearly C−Dis non-empty. So let us choose and fix an a ∈ C −D and define a mapping μand a domain Da by

μ(z) =1

z − a, Da = μ(D). (24)

Clearly Da is a bounded simply connected domain. Our plan is to showthat Da satisfies all the items in the last section 6. Everything in section5 is already satisfied for Da. Because μ is a homeomorphism of the spheremost topological matters can be settled quickly without much comment; forinstance, all but finitely many points of ∂Da are accessible from Da. Havingshowing that Da satisfies the key Theorem 25 we shall quickly deduce that ourunbounded D satisfies i, ii, and iii of Theorem 25 and a modified o. Let

ϕa = ϕ ◦ μ−1, ψa = ϕ−1a = μ ◦ ψ (25)


so that ϕa is an RMF for Da and ψa is an inverse RMF for Da. It is easy toshow that ϕa, ψa satisfy the hypothesis of the extension theorems Theorem 17and Theorem 20, respectively.

By Corollary 16 ∂D = Γ1 ∪Γ2 ∪ ...∪Γm ∪F, hence ∂Da = μ(Γ1)∪μ(Γ2)∪... ∪ μ(Γm) ∪ μ(F ) : both unions are disjoint. Each μ(Γi) is an open one-sidedanalytic arc with end points in μ(F ), and μ(F ) is precisely the set of endpoints of the maximal open subarcs μ(Γ1), μ(Γ2), ..., μ(Γm). So Theorem 17and Schwarz reflection principle prove

Lemma 28 ϕa can be extended to a function which is analytic and conformal(= injective) in a neighbourhood of Da ∪ μ(Γ1) ∪ μ(Γ2) ∪ ... ∪ μ(Γm) and ϕamaps μ(Γ1) ∪ μ(Γ2) ∪ ... ∪ μ(Γm) injectively into T.

Hence also, by (25), ϕ can be extended to a function which is analytic andconformal in a neighbourhood ofD∪Γ1∪Γ2∪...∪Γm and ϕmaps Γ1∪Γ2∪...∪Γminjectively into T.

As before we are not using a separate notation ϕa, ϕ the boundary valuesof ϕa on ∂Da and ϕ on ∂D, respectively.

Now let us consider ψa. Because ∂D = Γ1 ∪ Γ2 ∪ ... ∪ Γm (Lemma 15) itfollows that ∂Da = μ(Γ1) ∪ μ(Γ2) ∪ ... ∪ μ(Γm). Each μ(Γi) is homeomorphicto a closed interval or a circle. Hence ∂Da is u.l.c (see Corollary 19). So, byTheorem 20 and Corollary 21

Lemma 29 ψa can be extended uniquely to a continuous function on Δ whichmaps T onto ∂Da. Hence also ψ can be extended to a continuous function onΔ which maps T onto ∂D.

Lemma 23 is purely topological, so the argument of the proof shows that

ψ−1a (μ(Γi)) = ϕa(μ(Γi)) (1 ≤ i ≤ m).

But, by (25)

ϕa(μ(Γi)) = ϕ ◦ μ−1(μ(Γi)) = ϕ(Γi) = γi

as in (22) Thus T can be written as the disjoint union T = γ1∪γ2∪ ...∪γm∪G,where G = ψ−1

a (μ(F )) = ψ−1(F ) is as before. If G is thought as ψ−1a (μ(F ))

then, because Da is bounded, the argument in the proof of Lemma 24 showsthat G has measure zero. Hence G consists of m points: the end points of theγi’s. As before ψa and hence ψ are injective on T −G, so they can be used toparametrize ∂Da and ∂D respectively. Because each Γi is an analytic arc ψaand ψ can be continued analytically across each γi, so that they are conformalat almost all points of T.

We can now state our Theorem for unbounded D and attend to the re-maining details of its proof.

156 Yuksel Soykan

Theorem 30 Let D, a, Da, ψa, ψ be as above. Then

oψ′

(ψ − a)2∈ H1(Δ);

i Every f ∈ E2(D) has a non-tangential limit function f ∈ L2(∂D);

ii (Parseval’s identity) The map f → f is an isometric isomorphism of E2(D)onto a closed subspace E2(∂D) of L2(∂D), so

||f ||2E2(D) = ||f ||2L2(∂D) =1

2π

∫∂D

|f(z)|2|dz| (f ∈ E2(D));

iii (Cauchy’s Integral Formula) For all f ∈ E2(D) and z ∈ D

f(z) =1

2πi

∫∂D

f (ζ)

ζ − zdζ.

Proof. Because ψa = μ ◦ ψ =1

ψ − awe see that

ψ′

(ψ − a)2= −ψ′

a. Lemmas 26

and 27 show that this function is in H1(Δ) provided ∂Da is rectifiable. Butthis follows because each Γi satisfies hypothesis−R (see Example 1). Thususing l for arc-length measure, we have

l(∂Da) =m∑i=1

l(μ(Γi)) =m∑i=1

∫Γi

|μ′(ζ)| |dζ|

=m∑i=1

∫Γi

|dζ||ζ − a|2 <∞.

So o is proved. To prove i note that γi,Γi and μ(Γi) are σ-compact. Lemma 11can now be used to show that if E ⊆ γi then l(E) = 0 if and only if l(ψ(E)) = 0if and only if l(ψa(E)) = 0. Now if f ∈ E2(D), say f = (g ◦ ϕ)ϕ′1/2, whereg ∈ H2(Δ), then the conformality argument in the proof of Theorem 25 i

can be used to show that f has a non-tangential limit function f . Parseval’sidentity is verified as before by integrating by substitution. Thus

1

2π

∫∂D

|f(z)|2|dz| =1

2π

∫�

|g(ζ)|2|dζ| = ||f ||2E2(D).

If some Γi is unbounded we integrate |g|2 over a sequence of compact arcs onwhich ψ′ is absolutely continuous and take a limit. Because o is true Cauchy’sintegral formula can be verified exactly as in Theorem 25.


References

[1] L. V. Ahlfors, Complex Analysis, McGraw-Hill Book Company, New York,1966.

[2] B. J. Conway, Functions of One Complex Variable II, Springer-Verlag, NewYork, 1995.

[3] P. L. Duren, H p Spaces, Academic Press, New York, 1970.

[4] M.H.A. Newman, Elements of the Topology of Plane Sets of Points, DoverPublications, Inc., New York, 1992.

[5] P.M. Bruce, An introduction to Complex Function Theory, Springer-Verlag,New York, Berlin,1991.

[6] C. Pommerenke, Boundary behaviour of conformal maps, Springer-Verlag,Berlin, Heidelberg, 1992.

[7] W. Rudin, Real and Complex Analysis, McGraw-Hill, New York, Interna-tional Edition, 1987.

Received: September 13, 2006

Boundary Behaviour of Analytic Curves · 2007. 1. 4. · Int. Journal of Math. Analysis, Vol. 1,...

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