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ANGULAR DERIVATIVES AND SEMIGROUPS OF

HOLOMORPHIC FUNCTIONS

NIKOLAOS KARAMANLIS

Abstract. A simply connected domain Ω ⊂ C is convex in the positive

direction if for every z ∈ Ω, the half-line z + t : t ≥ 0 is contained inΩ. We provide necessary and sufficient conditions for the existence of

an angular derivative at ∞ for domains convex in the positive direction

which are contained either in a horizontal half-plane or in a horizontalstrip. This class of domains arises naturally in the theory of semigroups

of holomorphic functions and the existence of an angular derivative has

interesting consequences for the semigroup.

1. Introduction.

Let Ω ⊂ C be a simply connected domain and φ be a conformal map definedon Ω. Let ζ ∈ ∂Ω. If the function φ has non-tangential limit φ(ζ) at ζ and

the function φ(z)−φ(ζ)z−ζ has non-tangential limit at ζ, then we say that φ has

angular derivative at ζ. The non-tangential convergence here means that thelimits are taken through angular domains contained in Ω and having vertexζ. If ∞ ∈ ∂Ω, then by considering suitable conformal maps it is possibleto extend the notion of the angular derivative for ζ = ∞. We will give theappropriate definitions in detail in section 2.1.

The angular derivative problem is to give geometric conditions on theboundary of a simply connected domain Ω near ζ ∈ ∂Ω which are equiva-lent to the existence of a non-zero angular derivative at ζ for the conformalmap of Ω onto the half-plane or the disk. It is clear from the definition thatthe existence of a non-zero angular derivative only depends on the behavior ofΩ near ζ. When we say that a domain has angular derivative at a boundarypoint ζ we will mean that the conformal map of Ω onto H or D has non-zeroangular angular derivative at ζ. For a related survey, we refer to [4].

The angular derivative problem dates back at least to 1930 as Ahlforsstated it in his thesis [2]. Jenkins and Oikawa [16] and independently Rodinand Warschawski [20] showed that the existence of an angular derivative can

Date: October, 2018.

2010 Mathematics Subject Classification. 30D05, 30C45, 30C35, 31A15.Key words and phrases. Semigroup of holomorphic functions, Denjoy-Wolff point,

Koenigs map, parabolic semigroup, angular derivative, Lipschitz regions, extremal length.

1

2 NIKOLAOS KARAMANLIS

be characterized in terms of an extremal distance condition (this result is alsoincluded in [14], Ch. V, as well as in [19], Ch. 11); see section 2.2.

In general, the condition found by Rodin-Warschawski and Jenkins-Oikawacan be very hard to verify in applications. However, there are specific caseswhere it can be used to determine the existence of an angular derivative. Forexample, Jenkins in [15] was able to use this result to solve a conjecture ofRodin and Warschawski regarding certain comb domains. See also [21] and[4].

Another example where this condition can be fruitful is the class of domainswhich are convex in the positive direction. This means that if a point z lies inthe domain, then the half-line z+ t : t ≥ 0 is also contained in the domain.One way these domains can arise is as images of the unit disk under a certainclass of univalent functions (see [12]):

We say that a univalent function g : D→ C with g(0) = 0 is convex in thepositive direction if for any z ∈ D and any t > 0,

(1.1) g(z) + t ∈ g(D).

It is well known (see, for example, [1]) that for such a function g, there is apoint τ ∈ ∂D such that

(1.2) g−1(g(z) + t)→ τ,

locally uniformly in D. For simplicity, we will assume that τ = 1. The classof these functions is denoted by Σ[1]. It is clear by (1.1) that the domaing(D) is convex in the positive direction for any g ∈ Σ[1]. There is an analyticcharacterization for this class and it can be found in [12], pg. 57, Lemma 3.8:A univalent function g with g(0) = 0 is in the class Σ[1] if and only if

(1.3) Re((1− z)2g′(z)

)> 0,

for all z ∈ D. For further results on the class Σ[1] we refer to [12] and referencestherein.

Let R be a simply connected domain in C satisfying the following twoconditions:(S1) There is some horizontal strip S containing R.(S2) If z ∈ R, then z + t ∈ R, for all t > 0.We will assume that the smallest strip containing R is S, where S is thestandard horizontal strip, symmetric with respect to the real line, of widthπ. By (S2), there is a minimal a ∈ R ∪ −∞ such that (a,+∞) ⊂ R. Forx > a, let Fx be the component of R ∩ z : Rez = x containing the point xand set θ(x) = length(Fx). We will prove:

Theorem 1. The simply connected domain R has an angular derivative at+∞ if and only if

(1.4)

∫ t

s

(1

θ(x)− 1

π

)dx→ 0,

ANGULAR DERIVATIVES AND SEMIGROUPS 3

as t > s→ +∞.

We also prove a similar result for domains convex in the positive directionwhich are contained in a horizontal plane:

Suppose D ⊂ C is a simply connected domain satisfying the conditions:(H1) There is some horizontal plane H containing D.(H2) If z ∈ D, then z + t ∈ D, for all t > 0.We will assume that the smallest horizontal plane containing D is the upperhalf-plane H. We will also assume that D satisfies (2.1), which can be foundin section 2.1. For r > 0 large enough, let rη(r) denote the length of thecomponent of D ∩ ∂D(0, r) containing the point ir. Under these assumptionswe prove:

Theorem 2. The simply connected domain D has an angular derivative at∞ if and only if

(1.5)

∫ R

S

(1

rη(r)− 1

rπ

)dr → 0,

as R > S → +∞.

We remark that Theorem 2 is not a restatement of Theorem 1. Indeed, ifwe map R conformally into the upper half-plane via the map τ(z) = iez, thenτ(R) is not, in general, convex in the positive direction.

Domains convex in the positive direction which are contained either in ahorizontal strip or a horizontal plane arise naturally in the theory of semi-groups of holomorphic functions. We will now give a brief overview of thistheory and discuss how these domains come into play.

A one-parameter continuous semigroup of analytic functions in the unitdisk D is a family of analytic functions φt : D→ D, t ≥ 0, satisfying:(a) φ0 is the identity map in D.(b) φt+s = φt φs, for all t, s ≥ 0.(c) φt(z)→ φs(z), as t→ s, for all s ≥ 0, z ∈ D.

The theory of these semigroups is rich and has connections with variousareas of mathematics such as operator theory, stochastic processes, geometricfunction theory and dynamical systems. We refer to [1] and [12], as well asreferences therein for more details.

It is well known (see [1], Theorem 1.4.17) that for a semigroup (φt) whichdoes not consist of elliptic automorphisms of the unit disk, there exists a pointτ ∈ D such that

φt(z)→ τ,

locally uniformly in D, as t→ +∞. This result follows from the Denjoy-Wolfftheorem (see [1], Theorem 1.3.9) and the point τ is called the Denjoy-Wolffpoint of the semigroup. In this article we are only interested in semigroupsfor which τ ∈ ∂D. Moreover, we will assume for the sake of simplicity thatτ = 1.


The Denjoy-Wolff point, τ = 1, of the semigroup has the property thatfor any t ≥ 0, φt(z) has non-tangential limit 1 at z = 1 and the angularderivative φ′t(1) exists and satisfies φ′t(1) ∈ (0, 1]. See [9], [7], and referencestherein. The semigroup will be called hyperbolic if for some (equivalently, forall) t > 0, we have φ′t(1) < 1. If for some (equivalently, for all) t > 0, we haveφ′t(1) = 1, then the semigroup will be called parabolic. See [9], Theorem 2.1.

For every semigroup with Denjoy-Wolff point 1, there is a unique conformalmap h : D→ C (see [1], Theorem 1.4.22) such that h(0) = 0 and

(1.6) φt(z) = h−1(h(z) + t),

for all t ≥ 0 and z ∈ D. This map h is called the Koenigs function of thesemigroup. The domain Ω = h(D) is the associated domain and from (1.6)it is evident that it is convex in the positive direction and thus h ∈ Σ[1].Observe that since (φt) has Denjoy-Wolff point 1, we must have

(1.7) h−1(h(z) + t)→ 1,

locally uniformly in D, as t→ +∞.A theorem of Contreras and Dıaz-Madrigal, [9] Theorem 2.1, shows that we

can distinguish between hyperbolic and parabolic semigroups, having Denjoy-Wolff point 1, only by looking at Ω: The semigroup is hyperbolic if and only ifΩ is contained in a horizontal strip. One way to classify hyperbolic semigroups,having Denjoy-Wolff point 1, is in terms of the rate of convergence of φt to1, as t → +∞. Let (φt) be such a semigroup. It is known (see, for example,[6] Theorem 4.2) that there is some λ > 0 such that eλt(1− φt(z)) convergeslocally uniformly in D to a map K(z) and K is either identically 0 or it is aunivalent function. If K ≡ 0, then we say that the corresponding semigroupis of type A. Otherwise we say that it is of type B. Moreover, K is univalentif and only if the map g := ie−λh has angular derivative at 1 (Theorem 4.2in [6]), where h is the Koenigs function of (φt). Using the definition, whichwill be given in section 2.1, one can check that this is equivalent to h(D)having an angular derivative at +∞. Thus Theorem 1 provides a geometriccharacterization for hyperbolic semigroups of type B. We will now proceedwith a classification for parabolic semigroups.

Let ρD denote the hyperbolic distance in D (for the definition see [14], Ch.I, section 4 or [19], Ch. 4). Fix s > 0 and z ∈ D. By the Schwarz-Picklemma and property (b) of the semigroup, the function ρD(φt(z), φt+s(z)) isa non-increasing function of t. It follows that the limit

(1.8) λ(s, z) := limt→+∞

ρD(φt(z), φt+s(z))

exists and λ(s, z) ≥ 0. We will say that a parabolic semigroup is of zerohyperbolic step if for some s > 0 and some z ∈ D, we have λ(s, z) = 0. Ifλ(s, z) > 0 for all s, z, then the semigroup is of positive hyperbolic step. There


is an interesting connection between the hyperbolic step of a parabolic semi-group and the image of the semigroup’s Koenigs function h. Using (1.8), (1.6)and the conformal invariance of the hyperbolic distance, it is not hard to ver-ify that λ(s, z) is equal to the hyperbolic distance in

⋃t≥0 (h(D)− t) between

the points h(z) and h(φs(z)). From this it can be shown that a parabolicsemigroup is of zero hyperbolic step if and only if

⋃t≥0 (h(D)− t) = C and

it is of positive hyperbolic step if and only if⋃t≥0 (h(D)− t) is conformally

equivalent to the upper half-plane. For these facts we refer to [3] and [11].Following [5], we will now discuss a different classification of semigroups

that involves their trajectories. Let (φt) be a semigroup and z ∈ D. The curveγz : [0,+∞)→ D with

γz(t) = φt(z)

is the trajectory starting at z. Note that for semigroups with Denjoy-Wolffpoint 1, we must have that γz tends to 1 as t→ +∞.

A horodisk at 1 is a set of the form

∆ε = z ∈ D : d(z, 1) < ε,

where ε ∈ (0,+∞) and for z ∈ D,

d(z, 1) =|1− z|2

1− |z|2.

Note that ∆ε is a Euclidean disk internally tangent to the unit circle at 1.Using Julia’s lemma (see [1]), one can show that d(γz(t), 1) is a decreasingfunction of t. It follows that the limit

k(z) := limt→+∞

d(γz(t), 1)

exists and k(z) ≥ 0. A semigroup is of finite shift if and only if

limt→+∞

d(γz(t), 1) > 0,

for some z ∈ D. Otherwise, if this limit is equal to 0, we say that the semigroupis of infinite shift. These definitions are consistent because if k(z) > 0 for somez ∈ D, then k(z) > 0, for all z ∈ D. However, the value k(z) will still dependon z. Note that a semigroup is of infinite shift if and only if every trajectoryγz(t) enters eventually every horodisk centered at the Denjoy-Wolff point 1.

A parabolic semigroup with Denjoy-Wolff point 1 is of positive hyperbolicstep if and only if its associated domain Ω is contained in a horizontal half-plane (see for example Theorem 1 in [5], [3], and [11]). In that case, thesemigroup is of finite shift if and only if its Koenigs map h is conformal at 1 (seeTheorem 3 in [5] and Theorem 4.1 in [10]). Moreover, it follows by Proposition3.3 in [10] that a parabolic semigroup of zero hyperbolic step always hasinfinite shift. Thus Theorem 2 provides a geometric characterization for aparabolic semigroup of positive hyperbolic step being of finite shift.


2. Preliminaries.

2.1. Angular derivatives. Let D be a simply connected domain in C andζ ∈ ∂D. We will say that ∂D has an inner tangent at ζ if there is an angle θ0

such that for every α ∈ (0, π/2) there exists an ε > 0 so that

Γα,ε(ζ) := z : | arg(z − ζ)− θ0| < α, 0 < |z − ζ| < ε ⊂ D.If ∂D has an inner tangent at ζ ∈ ∂D and if φ is a conformal map defined onD, then we say that φ has angular derivative φ′(ζ) at ζ (or simply is conformalat ζ) if φ has a non-tangential limit

φ(ζ) := limΓα,ε(ζ)3z→ζ

φ(z),

for all α ∈ (0, π/2) and if

φ′(ζ) := limΓα,ε(ζ)3z→ζ

φ(z)− φ(ζ)

z − ζexists for all α ∈ (0, π/2). For a detailed presentation of the theory on angularderivatives, we refer to [14] and [19].

Now suppose that D is a simply connected domain with ∞ ∈ ∂D and sothat for any ε ∈ (0, π/2), there exists r > 0 such that

(2.1) Γα,r(∞) := z : |z| > r, 0 < arg z < π − ε ⊂ D.If φ is a conformal map on D, then we say that φ has angular derivative σ at∞ if

(2.2) limΓα,r(∞)3z→∞

φ(z) =∞

and

(2.3) limΓα,r(∞)3z→∞

z

φ(z)= σ,

for all α ∈ (0, π/2) and some r > 0. We say that D has an angular derivativeat ∞ if D satisfies (2.1) and there is a conformal map φ : D → H possessinga finite and non-zero angular derivative at ∞.

Let R be a simply connected domain in C satisfying: For any δ ∈ (0, π/2),there exists xδ > 0 so that

(2.4) Rδ := z : |Imz| < π/2− δ, Rez > xδ ⊂ R.Let S denote the standard strip of width π and suppose that f : R → S is aconformal map such that Ref(z)→ +∞, as Rez → +∞. We will say that fhas angular derivative C (−∞ < C < +∞) at +∞ if

(2.5) f(z)− z → C,

as Rez → +∞, z ∈ Rδ, for any δ ∈ (0, π/2). Note that if f has angularderivative C at +∞, then the inverse map f−1 : S → R also has an angularderivative at +∞, i.e. it satisfies Ref−1(z) → +∞ and f−1(z) − z → −C,


as Rez → +∞, z ∈ Sδ, for each δ ∈ (0, π/2), where Sδ = z : |Imz| <π/2 − δ. To see this fix δ ∈ (0, π/2), let zn ∈ Sδ be a sequence whose realpart tends to +∞ and choose 0 < ε < δ/2. Because the domains in (2.4) arecontained in R and by (2.5), we have that −ε < Imf(z) − Imz < ε, for allz ∈ yδ/2 +Rδ/2 =: Mδ for some sufficiently large yδ/2 > 0. These inequalities

imply that zn ∈ f(Mδ), for large n and thus wn = f−1(zn) ∈Mδ, for n large.Since f is conformal, it follows that the only limit point of wn is +∞. Thisshows that Ref−1(z)→ +∞ and f−1(z)− z → −C, as Rez → +∞, z ∈ Sδ.

We will say that the domain R has an angular derivative at +∞ if Rsatisfies (2.4) and there is a conformal map f : R → S possessing a finiteangular derivative at +∞. This definition does not depend on the choice ofthe map f because for any two such maps, the composition of one with theinverse of the other is an automorphism of S fixing +∞ from which it easilyfollows that one of them possesses a finite angular derivative at +∞ if andonly if the other one does.

Let τ(z) = iez. Note that τ is a conformal map of S onto H such thatτ(+∞) = ∞ and τ(−∞) = 0. Observe that half-strips of the form (2.4) aretransformed, under τ , into sets of the form (2.1). Using the map τ−1, one canalso see that condition (2.5) is really condition (2.3) for a suitable conformalmap φ. The details of the switch between definition (2.3) and (2.5) can befound in [18].

2.2. Extremal length. Given a domain Ω and a collection of rectifiablecurves Γ in Ω, the extremal length of Γ is the quantity

λΩ(Γ) = supρ≥0

(infγ∈Γ

∫γρ|dz|

)2∫Ωρ2dA

,

where |dz| denotes arc length measure, dA denotes Lebesgue area measureand where the supremum is taken over all non-negative Borel functions ρsatisfying 0 <

∫Ωρ2dA < ∞. Any such function ρ will be called a metric.

If Ω is a domain, E,F ⊂ Ω and Γ is the family of rectifiable curves in Ωconnecting E to F , we will call λΩ(Γ) the extremal distance in Ω betweenE and F and denote it by dΩ(E,F ). The main reason for the importanceof extremal length in function theory is that it is conformally invariant: iff : Ω → Ω′ is a conformal map that is continuous on Ω (for instance this isthe case when we are dealing with Jordan domains), then

dΩ(E,F ) = dΩ′(f(E), f(F )).

For more information on extremal length we refer to [14]. We can now statethe result of Rodin-Warschawski and Jenkins-Oikawa on the characterizationof the existence of an angular derivative in terms of extremal length.


Theorem A([16], [20]). Let D be a simply connected domain satisfying (2.1)and φ : D → H a conformal map satisfying (2.2). A necessary and sufficientcondition for φ to have a finite non-zero angular derivative at ∞ (σ 6= 0,∞in (2.3)) is that D has the property

(2.6) dD(As, Ar)−1

πlog

r

s→ 0,

as r > s→∞.

Here Ar is the component of D ∩ |z| = r containing the point ir anddD(Ar, As) is the extremal length of the family of rectifiable curves connectingAs to Ar in D. We also state a version of Theorem A in the context ofdefinition (2.5):

Theorem B([16], [20]). Let R be a simply connected domain satisfying (2.4)and f : R→ S a conformal map satisfying Ref(z)→ +∞, as Rez → +∞. Anecessary and sufficient condition for f to have a finite angular derivative at+∞ is that R has the property

(2.7) dR(Fs, Ft)−|s− t|π

→ 0,

as t > s→ +∞.

Here Fx is the component of Rez = x ∩ R containing the point x anddR(Fs, Ft) is the extremal length of the family of rectifiable curves connectingFs to Ft in R.

A classical example for extremal length is the case of a rectangle. Suppose

P = z : |Imz| < H, |Rez| < L

is a rectangle and let E and F denote its vertical sides. Then the extremaldistance satisfies

dP (E,F ) = L/H.

Note that condition (2.7) of Theorem B says that a necessary and sufficientcondition for the simply connected domain R to have an angular derivativeat +∞ is that, near +∞, R is approximately the strip S. Of course, theexistence of the angular derivative depends on how well R approximates Sand this usually involves various estimates for extremal distance. Accordingto the definition of extremal distance, every choice of a metric ρ gives a lowerbound for extremal distance. It is therefore important to be able to writedown good metrics for a given domain.

Suppose R is a simply connected domain containing the real line R. LetFx denote the component of R ∩ z : Rez = x containing x. Set θ(x) =length(Fx) and fix s < t. For z = x+ iy ∈ R, define a metric ρ1 by

ρ1(z) =

1θ(x) , if s < x < t and z ∈ Fx0, elsewhere in R


For any curve γ connecting Fs to Ft in R, we have∫γ

ρ1(z)|dz| ≥∫ t

s

1

θ(x)dx.

Moreover, ∫ ∫R

ρ21dydx =

∫ t

s

1

θ(x)dx

and thus

(2.8) dR(Fs, Ft) ≥∫ t

s

1

θ(x)dx.

This estimate is due to Ahlfors. For the motivation behind the choice of themetric ρ1 see [18].

Now consider a region of the form

U = (x, y) : |y −m(x)| < θ(x)/2, s < x < t,having width θ(x) and mid-linem(x), for some continuous functions θ(x),m(x).Observe that the extremal distance between the two horizontal sides of a rec-tangle P is equal to the reciprocal of the extremal distance between its verticalsides. This fact is also true for general quadrilaterals and thus for regions likeU (see [14], Ch. IV, Theorem 4.1). Any metric gives a lower bound forthe extremal distance in U between the two components of ∂U \ (Fs ∪ Ft)and therefore it also gives an upper bound for dU (Fs, Ft). If m and θ aresufficiently smooth, then we can consider the following metric ρ2 in U :

ρ2(x+ iy) =

∣∣∣∣∇(y −m(x)

θ(x)

)∣∣∣∣=

√1

θ2(x)+

((y −m(x))θ′(x) + θ(x)m′(x)

θ2(x)

)2

.

This metric is due to Beurling (for a reference see [18]). If γ is any curve inU connecting the curve y = m(x) + θ(x)/2 to the curve y = m(x) − θ(x)/2,then ∫

γ

ρ2(z)|dz| ≥∣∣∣∣∫γ

∇(y −m(x)

θ(x)

)· dz∣∣∣∣ = 1.

Additionally,∫ ∫U

ρ22dA =

∫ t

s

dx

θ(x)+

∫ t

s

m′(x)2 + θ′(x)2/12

θ(x)dx.

Hence,

(2.9) dU (Fs, Ft) ≤∫ t

s

dx

θ(x)+

∫ t

s

m′(x)2 + θ′(x)2/12

θ(x)dx.

This inequality was discovered by Warschawski [23].


Figure 1. The dotted segments are part of ∂RM \ ∂R.

2.3. Lipschitz domains. In order to apply the extremal distance estimatesof the previous section for a domain Ω, it must be that ∂Ω is sufficientlysmooth. One way to take care of this issue is to approximate Ω by regionswhose boundary consists of graphs of Lipschitz functions. Let M > 0. Areal-valued function h defined on an interval I is called M -Lipschitz if

|h(x)− h(y)| ≤M |x− y|,for all x, y ∈ I.

Let R be a simply connected domain in C having the property that thesmallest horizontal strip containing R is S. Assume also that R satisfies (S2).As we observed before Theorem 1 in section 1, there is a minimal a ∈ R∪−∞such that (a,+∞) ⊂ R. By (S2) and because S is the smallest horizontalstrip containing R, we have

(2.10) dist(z, ∂S)→ 0,

as Rez → +∞, z ∈ ∂R. Note that (2.10) implies that R satisfies condition(2.4). Therefore, by Theorem B, R has an angular derivative at +∞ if andonly if (2.7) holds.

For M > 0 let TM denote the collection of all isosceles triangles containedin R with sides of slope ±M and base on (a,+∞). Here, the slope of a linecontained in C is the tangent of the angle between the positive real axis andthe line, measured counter clockwise. Set

(2.11) RM =⋃T : T ∈ TM.

The domain RM is the largest subregion of R which is bounded by thegraphs of two M -Lipschitz functions fM , gM , defined on (a,+∞) and satisfy-ing −π/2 ≤ gM < 0 < fM ≤ π/2. See Figure 1. Moreover, by (S2) togetherwith (2.10), it follows that fM must be non-decreasing and fM (x)→ π/2, asx→ +∞ while gM must be non-increasing and gM (x)→ −π/2, as x→ +∞.


Figure 2. The dotted segments are part of ∂DM \ ∂D.

Now let D be a simply connected domain having the property that thesmallest horizontal half-plane containing D is the upper half-plane H. Also,suppose that D satisfies conditions (H2) and (2.1). From our assumptionson D, it follows that

Imz → 0, as Rez → +∞, z ∈ ∂D.Let ε > 0. By (2.1) we can find s > 0 such that if z ∈ ∂D with Rez < −s,then z must satisfy

0 < − Imz

Rez≤ tan ε.

It follows that

(2.12) tan(π − arg z)→ 0,

as Rez → −∞. Again, we will need to approximate D using Lipschitz func-tions. Let M > 0 and consider the set

DM =⋃z + ΓM : z + ΓM ⊂ D,

where ΓM = z : Imz > M |Rez| is the angular domain with vertex at 0having sides of slope ±M . If DM is not empty, then it is bounded by thegraph of an M -Lipschitz function hM . See Figure 2. If DM is empty, sethM = +∞. The function hM is called the smallest M -Lipschitz majorant of∂D, for if the graph of an M -Lipschitz function h is contained in D, thenh ≥ hM .

Notice that by (H2) the function hM must be non-increasing. Morover,the discussion above and (2.12) show that hM (x) → 0, as x → +∞ andhM (x)x → 0, as x→ −∞.The concept of Lipschitz majorants was introduced by Burdzy [8]. Using

these smaller domains whose boundary consists of graphs of Lipschitz func-tions he was able to give a conditional necessary and sufficient condition for


the existence of an angular derivative at ∞ for domains containing the realline or the positive imaginary axis. See [14], Ch. V, Theorem 6.1 and [14],Ch. V, Exercise 9. Also, see [18].

2.4. Harmonic measure and reduced extremal distance. We will needsome basic estimates for harmonic measure. Let Ω be a Jordan domain and Ea Borel subset of ∂Ω. The harmonic measure of E relative to Ω is the solutionof the Dirichlet problem in Ω with boundary function equal to 1 on E and0 on ∂Ω \ E. We denote it by ω(z, E,Ω). For a detailed presentation of thetheory of harmonic measure we refer to [14].

Let E be a Jordan arc in D with 0 6∈ E. Let E be an arc on the unit circlewhich has the same diameter as E. It was proved in [13] and [22] that

(2.13) ω(0, E,D \ E) ≥ ω(0, E,D).

We will also need the concept of the reduced extremal distance which wenow define. Let Ω be a finitely connected Jordan domain. Fix z0 ∈ Ω andlet E be a finite union of closed subarcs of ∂Ω. Denote by Dε := D(z0, ε) thedisk centered at z0 of radius ε. The reduced extremal distance, δΩ(z0, E), isdefined as follows

δΩ(z0, E) := limε→0

(dΩ\Dε(∂Dε, E)− dΩ\Dε(∂Dε, ∂Ω)

).

For the details on why this limit exists and further applications of the reducedextremal distance we refer to [14].

We will need two facts about the reduced extremal distance:Fact 1 (see [14], Ch. V, Cor. 3.3). Let Ω be a Jordan domain, z0 ∈ Ω and Ea finite union of closed arcs on ∂Ω. Then

(2.14) ω(z0, E,Ω) ≤ e−πδΩ(z0,E).

Fact 2 (see [14], Ch. V, Theorem 3.5). Let Ω be a Jordan domain, z0 ∈ Ωand E ⊂ z ∈ ∂Ω : distΩ(0, z) ≥ R, where

distΩ(w, z) = inf∫

γ

ds : γ is a curve in Ω and z, w ∈ γ

is the euclidean distance in Ω from z to w. Then

(2.15) δΩ(0, E) ≥ 1

2πlogR.

We are now ready to begin with several lemmas needed in order to proveTheorems 1 and 2. Throughout the rest of this paper, R will denote a simplyconnected domain satisfying (S2) and having the property that the smallesthorizontal strip containing R is S. Also, D will denote a simply connecteddomain satisfying (H2), (2.1) and having the property that the smallest hor-izontal plane containing D is H.


Lemma 1. Let M > 8π. Then R has an angular derivative at +∞ if andonly if RM has an angular derivative at +∞.

Proof. Let M > 8π. Construct RM and suppose that it has an angularderivative at +∞. Note that RM ⊂ R ⊂ S. It follows that

0 ≤ dR(Fs, Ft)−|s− t|π

≤ dRM (Fs, Ft)−|s− t|π

.

By (2.7), we have

dRM (Fs, Ft)−|s− t|π

→ 0,

as t, s → +∞. By the estimates above and (2.7) again, this implies that Rhas an angular derivative at +∞.

For the converse, assume that R has an angular derivative at +∞. ClearlyArea(RM \ S) < ∞, since RM ⊂ S. Set Sa = S ∩ Rez > a, where a isthe smallest real number (it could be −∞) such that (a,+∞) ⊂ R. Thenby Theorem 6.1(b), Ch. V in [14], it follows that Area(Sa \ RM ) < ∞.Observe that (RM )M = RM . By Theorem 6.1(a), Ch. V in [14] and becauseArea(Sa \ (RM )M ) = Area(Sa \ RM ) < ∞, we have that RM has an angularderivative at +∞ if and only if Area((RM )M \ Sa) <∞. Since RM ⊂ Sa, weconclude that this area is 0 and thus RM has an angular derivative at +∞.

Remark 1. Note that Theorem 6.1, Ch. V in [14] is stated using the quantitiesArea(RM \ S) and Area(S \ RM ), under the normalization ∂R ∩ z : Rez <0 = ∂S ∩ z : Rez < 0. In our case, we may replace S with Sa because theexistence of the angular derivative depends on the behavior of R near +∞and the conditions Area(RM \ S) <∞ and Area(S \RM ) <∞ are equivalentto Area(RM \ S)s,t → 0 and Area(S \RM )s,t → 0, as s, t→ +∞ respectively.Here for a set U , we put Us,t = U ∩ z : t < Rez < s. Another thing to keepin mind here is that the proof of Lemma 1 does not rely on property (S2) ofR. We will use this fact in the proof of Theorem 2.

Set vM = 12 (fM + gM ) and θM = fM − gM , where fM and gM are the

M -Lipschitz functions defined on (a,+∞) whose graphs form ∂RM . Clearly,we can write

RM = x+ iy : |y − vM (x)| < θM (x)/2, x > a.Let a < s < t. Since the functions fM , gM are M -Lipschitz, it follows thatvM and θM are absolutely continuous on (a,+∞). Hence, by (2.8) and (2.9)we have the estimates∫ t

s

1

θM (x)dx ≤ dRM (Fs, Ft) ≤

∫ t

s

1

θM (x)dx+

∫ t

s

v′M (x)2 + 112θ′M (x)2

θM (x)dx.

Note that v′M (x)2 ≤ 12 (f ′M (x)2 + g′M (x)2), θ′M (x)2 ≤ 2(f ′M (x)2 + g′M (x)2), for

all x > a and θM (x) → π, as x → +∞. Also, since fM , gM are M -Lipschitz,


we have |f ′M |, |g′M | ≤ M , for a.e. x. Therefore, for s < t large enough, itfollows that∫ t

s

v′M (x)2 + 112θ′M (x)2

θM (x)dx ≤ 1

2

∫ t

s

(f ′M (x)2 + g′M (x)2

)dx

≤ M

2

∫ t

s

(f ′M (x)− g′M (x)) dx

=M

2(fM (s)− fM (t)− gM (s) + gM (t)) .

Because fM (x)→ π/2 and gM (x)→ −π/2, as x→ +∞, we have∫ t

s

v′M (x)2 + 112θ′M (x)2

θM (x)dx→ 0,

as s, t → +∞. Therefore, by (2.7), RM has an angular derivative at +∞ ifand only if ∫ t

s

1

θM (x)dx− t− s

π=

∫ t

s

(1

θM (x)− 1

π

)dx→ 0,

as s, t→ +∞. By Lemma 1, we have that R has an angular derivative at +∞if and only if there is an M > 8π such that

(2.16)

∫ t

s

(1

θM (x)− 1

π

)dx→ 0,

as s, t→ +∞.Let Fx denote the component of R ∩ Rez = x containing x. Let θ(x)

denote the length of Fx. Note that since θM ≤ θ ≤ π, for any M > 0 and allx > a, it follows that if (2.16) holds, then

(2.17)

∫ t

s

(1

θ(x)− 1

π

)dx→ 0,

as t, s → +∞. In particular, if R has an angular derivative at +∞, then(2.17) holds. Theorem 1, which will be proved in section 3, shows that theconverse is true as well.

We will now proceed with two lemmas regarding the domain D. Let M > 0and let hM be the smallest M -Lipschitz majorant of ∂D as defined in section2.3. Recall that hM is a non-increasing function such that hM (x) → 0, as

x→ +∞ and hM (x)x → 0, as x→ −∞.

Lemma 2. For R > 0 sufficiently large, the semicircle ∂D(0, R) ∩ H meetsthe graph of hM at exactly two points zlR, z

rR having real parts xlR < 0 < xrR.

Proof. Because D satisfies (2.1), we have that for R large enough, the graphGhM of hM must meet ∂D(0, R) in z : arg z ∈ (π/2, π] and in z : arg z ∈[0, π/2). The part of GhM ∩ ∂D(0, R) which lies in z : arg z ∈ (π/2, π]must consist of a single point because it is the intersection of the graphs of


a non-increasing function and an increasing function. Suppose now that thepart of GhM ∩ ∂D(0, R) which lies in z : arg z ∈ [0, π/2) consists of at leasttwo points zR, wR. Since hM is M -Lipschitz, we have

|ImwR − ImzR| ≤M |RewR − RezR|.Write zR = Reiθz and wR = Reiθw . Then the inequality above becomes

| sin θw − sin θz|| cos θw − cos θz|

≤M.

Since hM (x) → 0, as x → +∞, it follows that θw, θz → 0, as R → +∞.Using the mean value theorem, this implies that the left hand side of the lastinequality tends to +∞, as R→ +∞ which is a contradiction. The conclusionfollows.

By (2.1), for r > 0 large enough, the half-line it : t > r is contained inD. Let Rη(R) denote the length of the subarc of ∂D(0, R) having endpointszlR, z

rR and containing the point iR. Let xlR, x

rR denote the real parts of zlR, z

rR

respectively. Note that

(2.18) R =√

(xlR)2 + hM (xlR)2 =√

(xrR)2 + hM (xrR)2.

Moreover, Rη(R) is the length of the arc

Reit : a1(R) < t < a2(R),

where a1(R) = arctan(hM (xrR)xrR

)and a2(R) = π + arctan

(hM (xlR)

xlR

).

Lemma 3. There is some x0 > 0 such that the functions a1(ex) and a2(ex)are Lipschitz in the interval (x0,+∞).

Proof. We will prove the statement for a1(ex). The proof for a2 is similar. Let0 < R1 < R2 with R1, R2 sufficiently large so that hM (x)/x ≤ 1, for x > xrR1

.Using (2.18) and the fundamental theorem of calculus, we have

2(logR2 − logR1) =

∫ xrR2

xrR1

d(log(x2 + hM (x)2

)dx

dx

=

∫ xrR2

xrR1

2x+ h′M (x)hM (x)

x2 + hM (x)2dx

= 2

∫ xrR2

xrR1

1 + hM (x)h′M (x)/x

x(1 + hM (x)2

x2 )dx

≥ 1

2(log xrR2

− log xrR1),

where we used the fact that hM (x)/x ≤ 1, for x > xrR1. Note that since hM

is M -Lipschitz, we get that |h′M (x)| ≤M , for a.e. x. Using this fact together


with the mean value theorem and the estimate hM (x)/x ≤ 1, for x > xrR1, we

find

|a1(R2)− a1(R1)| =∣∣∣∣arctan

(hM (xrR2

)

xrR2

)− arctan

(hM (xrR1

)

xrR1

)∣∣∣∣≤∣∣∣∣hM (xrR2

)

xrR2

−hM (xrR1

)

xrR1

∣∣∣∣≤∫ xrR2

xrR1

|h′M (x)|+ hM (x)/x

xdx

≤ (M + 1)(log xrR2− log xrR1

).

For the second inequality we used the fundamental theorem of calculus. Weconclude that

|a1(R2)− a1(R1)| ≤ 4(M + 1)(logR2 − logR1)

and the proof is complete.

Set

vM (x) =a1(ex) + a2(ex)− π

2,

θM (x) = a2(ex)− a1(ex),

and

DM = x+ iy : |y − vM (x)| < θM (x)/2.

Let τ(z) = iez and observe that τ is a conformal map of S onto H with

τ(z) → ∞, as Rez → +∞. Moreover, we have that τ(DM ) = DM . Theestimates of section 2.2 yield for s < t∫ t

s

1

θM (x)dx ≤ dDM (Fs, Ft) ≤

∫ t

s

1

θM (x)dx+

∫ t

s

v′M (x)2 + 112θ′M (x)2

θM (x)dx.

Note that by our conditions on hM , we have a1(R) → 0 and a2(R) → π, asR → +∞. It follows that θM (x) → π, as x → +∞. Furthermore, using theexpressions for vM and θM , we have

v′M (x)2 +1

12θ′M (x)2 ≤ 1

2

((da1(ex)

dx

)2

+

(da2(ex)

dx

)2).

Since a1(ex), a2(ex) are Lipschitz functions, we can use an argument similarto the one after Remark 1 to show that∫ t

s

v′M (x)2 + 112θ′M (x)2

θM (x)dx→ 0,


Figure 3. The integral∫ xrj+1

xlj+1

(θ(x)− θM (x)) dx is bounded

above by the area of the right triangle with the dotted sides.

as s, t→ +∞. By (2.7) and these estimates, we conclude that DM , and thusDM , has an angular derivative at ∞ if and only if

(2.19)

∫ t

s

(1

θM (x)− 1

π

)dx→ 0,

as s, t→ +∞. By a change of variables, condition (2.19) is equivalent to∫ R

S

(1

rηM (r)− 1

rπ

)dr → 0,

as R,S → +∞. Here ηM (ex) = θM (x). For a related result see [14], Ch. V,Cor. 5.8.

3. Proof of Theorem 1.

By (2.17) and the discussion following it, we see that if R has an angularderivative at +∞, then (2.17) holds. Before we proceed with the converse,note that condition (2.17) is equivalent to

(3.1)

∫ ∞a

(1

θ(x)− 1

π

)dx <∞.


Conversely, assume that (3.1) holds. Let M > 8π and construct RM .Following part of the proof of Theorem 6.1, Ch. V in [14], we write ∂RM\∂R =∪jσj , where each σj is an open arc with endpoints zlj , z

rj ∈ ∂RM∩∂R satisfying

Rezlj < Rezrj . See Figure 3. By the construction of RM , it follows that eachσj consists of at most two line segments. We claim that every σj must be asingle line segment. To see this, assume that there is some σj which consistsof exactly two line segments that meet at a vertex v ∈ R. Then we musthave that Rezlj < Rev < Rezrj and |Imv| > max(|Imzlj |, |Imzrj |). Because Rsatisfies property (S2), we have that x+ iy : x > Rev, 0 < y < Imv ⊂ R,if Imv > 0 or x + iy : x > Rev, 0 > y > Imv ⊂ R, if Imv < 0. Both ofthese cases are absurd, since zrj is always in one of these two sets but not in

R. Therefore, each σj is a single line segment with endpoints zlj , zrj satisfying

xlj = Rezlj < Rezrj = xrj , |ylj = Imzlj | < |yrj = Imzrj | and |yrj−ylj | = M(xrj−xlj).Observe that the integral in (3.1) is comparable to∫ ∞

a

(π − θ(x)) dx <∞.

By property (S2) of R and elementary geometry (see Figure 3), we have∫ ∞a

(θ(x)− θM (x))dx ≤∑j

|yrj − ylj |(xrj − xlj) =1

M

∑j

(yrj − ylj)2.

Since the intervals (yrj , ylj) (or (ylj , y

rj )) are pairwise disjoint and all subsets of

(−π/2, π/2), we conclude that∑j(y

rj − ylj)2 <∞. Therefore,∫ ∞

a

(θ(x)− θM (x))dx <∞.

It follows that ∫ ∞a

(π − θM (x)) dx <∞

and thus

(3.2)

∫ t

s

(π − θM (x)) dx→ 0,

as t, s → +∞. By the discussion preceding Lemma 2, we have that RM hasan angular derivative at +∞ if and only if (2.16) holds. Since the integrals in(2.16) and (3.2) are comparable, we deduce that RM has an angular derivativeat +∞. The conclusion follows by Lemma 1.

4. Proof of Theorem 2.

Throughout the proof we will use the notation θ(x) = η(ex).


Suppose that D has an angular derivative at ∞. Since τ(z) = iez is a

conformal map from S onto H fixing ∞, it follows that D = τ−1(D) ⊂ S hasan angular derivative at +∞. Let N > 8π. By Lemma 1 and Remark 1, we

have that DN has an angular derivative at +∞. We conclude that τ(DN ) hasan angular derivative at ∞. By the proof of Lemma 1 in [17], we have that

τ(DN ) is the graph of an N ′-Lipschitz function near ±∞, for some constantN ′ that depends only on N . By the definition of the Lipschitz majorant, thisimplies

τ(DN ) ∩ |z| > R ⊂ DN ′ ∩ |z| > R ⊂ D,

for some R sufficiently large. Let U be any of the simply connected domains

τ(DN ), DN ′ , or D. For each r, let Ar be the component of U ∩ ∂D(0, r)containing the point ir. For R < s < r, let Us,r be the connected componentof U\(Ar∪As) containing the segment (is, ir). The extension rule for extremallength (see [14], Ch. IV, pg. 134) then yields for r, s large

1

dτ(DN )s,r(As, Ar)

≤ 1

d(DN′ )s,r (As, Ar)≤ 1

dDs,r (Ar, As).

These estimates show that DN ′ has an angular derivative at ∞ and thus by(2.16) we must have ∫ t

s

(1

θN ′(x)− 1

π

)dx→ 0,

as s, t → +∞. Since θN ′ ≤ θ, we can use a change of variables to arrive at(1.5).

Conversely, assume that (1.5) holds. ForM > 0 and becauseDM ⊂ D ⊂ H,we have the inequalities

1

d(DM )s,r (As, Ar)− π

log r/s≤ 1

dDs,r (As, Ar)− π

log r/s≤ 0.

Therefore, it suffices to show that DM has an angular derivative at ∞ forsome M . By (2.16), this is equivalent to∫ t

s

(π − θM (x))dx→ 0,

as t, s→∞. The condition in (1.5), via a change of variables, is equivalent to∫ t

s

(π − θ(x))dx→ 0,

as s, t→ +∞. It is evident from these facts that it suffices to prove

(4.1)

∫ t

s

(θ(x)− θM (x))dx→ 0,


Figure 4. The union of the two dotted segments is Er.

as s, t→ +∞. At this point it is more convenient to work with η and as suchwe will make another change of variables to transform this condition into

(4.2)

∫ S

R

η(r)− ηM (r)

rdr → 0,

as R,S → +∞. Using the same argument as in the proof of Theorem 1,we can write ∂DM \ ∂D = ∪jσj , where each σj is a segment of slope −M(because hM is non-increasing) with endpoints zlj , z

rj satisfying Rezlj < Rezrj .

Note that when Rezlj > 0, we have |zrj | > |zlj | while the reverse inequalityholds when Rezrj < 0. For each r > 0, let a(r), b(r) be the endpoints ofthe arc of ∂D(0, r) ∩ D which contains ir and has length rη(r). Also, letaM (r), bM (r) be the endpoints of the arc of ∂D(0, r) ∩ DM which containsir and has length rηM (r). Because DM ⊂ D, we can assume that Rea(r) <ReaM (r) < RebM (r) < Reb(r). Then η(r) − ηM (r) = arg a(r) − arg aM (r) +arg bM (r) − arg b(r). Note that if ∂D(0, r) ∩ ∪jσj = ∅, then η(r) = ηM (r).We will now find an upper bound for the difference η(r) − ηM (r) for any rsufficiently large, satisfying ∂D(0, r) ∩ ∪jσj 6= ∅. Choose some j and r suchthat ∂D(0, r) meets σj . We assume that σj lies in arg z ∈ (π/2, π). Thecase σj ⊂ arg z ∈ (0, π/2) is treated in a similar manner.

Suppose that σj ⊂ arg z ∈ (π/2, π). Let r ∈ (|zrj |, |zlj |) and let c(r) denotethe point in arg z ∈ (π/2, π) where ∂D(0, r) meets the line y = Imzrj . Thenby property (H2) of D, arg a(r) − arg aM (r) ≤ arg c(r) − arg aM (r). Letd(r) = ReaM (r) + iImc(r). Set Er = [c(r), d(r)] ∪ [d(r), aM (r)] and let Ir =reit : arg aM (r) < t < arg c(r). See Figure 4. Observe that diam(Er) =


Figure 5. The domain Br ⊂ rD and the sets Ir,Er.

diam(Ir), because the chord of Ir is the hypotenuse of the right triangle withvertices aM (r), d(r), c(r). Let Br denote the component of rD \ Er whichcontains 0. By (2.13) and (2.14), we have

arg a(r)− arg aM (r)

r≤ arg c(r)− arg aM (r)

r

=2π

rω(0, Ir, rD)

≤ 2π

rω(0, Er, Br)

≤ 2π

re−πδBr (0,Er),

where δBr (0, Er) is the reduced extremal distance. See Figure 5. Note thatEr ⊂ z ∈ ∂Br : distBr (0, z) = |z| ≥ |d(r)|. Therefore, by (2.15) andbecause |d(r)| > |zrj |, we have the estimate

(4.3) δBr (0, Er) ≥1

2πlog |d(r)| ≥ 1

2πlog |zrj |.

Note that

|zrj ||zlj |

=|Rezrj ||Rezlj |

√1 + (Imzrj /Rezrj )2√1 + (Imzlj/Rezlj)

2.

Since σj has slope −M , it follows that

RezrjRezlj

=M + Imzlj/RezljM + Imzrj /Rezrj

.


By (2.12),

ImzljRezlj

,ImzrjRezrj

→ 0,

as Rezrj → −∞. This implies that|zrj ||zlj |→ 1, as Rezrj → −∞. Hence, when

|zrj | is large, we must have

(4.4) |zrj | ≥ |zlj |/2 ≥ r/2.

Using (4.3) together with (4.4) and the estimates preceding (4.3), we get

(4.5)arg a(r)− arg aM (r)

r≤ 2π

√2

r3/2,

for r ∈ (|zrj |, |zlj |). We conclude that

η(r)− ηM (r)

r≤ C

r3/2,

for some constant C, and r sufficiently large. Because r−3/2 is integrable near∞, it follows that (4.2) holds and thus the proof is complete.

Acknowledgment. I would like to thank Prof. Dimitrios Betsakos for all his

help with this work. I also thank the referee for his suggestions and comments.

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Department of Mathematics, Aristotle University of Thessaloniki, 54124 Thes-

saloniki, GreeceE-mail address: [email protected]

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