# Thick fluid disks around binary black holes

S. V. Chernov  Astro Space Center, Lebedev Physical Institute, 84/32
Profsoyuznaya street,
117997, Moscow, Russia<EMAIL_ADDRESS>
###### Abstract

A model of a thick fluid disk around a binary black hole is considered. A
binary black hole is described by the Majumdar-Papapetrou solution. The
hydrodynamic equations in this metric are written out. Exact analytical
solutions are presented. Generalization to the case of a toroidal magnetic
field is carried out.

##  I Introduction

Accretion disks around single black holes have important astrophysical
significance. On the other hand, On September 14, 2015, the Laser
Interferometer Gravitational-Wave Observatory detected a gravitational-wave
signal from the source GW150914 - thereby confirming the existence of binary
black holes [1]. The masses of such black holes are small, only a few tens of
solar masses. In addition, binary supermassive black holes can exist in the
center of active galactic nuclei [2]. One of the main examples is the OJ 287
system. It is believed that this system is a binary black hole with masses of
18 billion and 125 million solar masses [3]. The orbital period of such a
system is 12 years and it is assumed that an accretion disk rotates around a
more massive black hole, which is penetrated by a less massive black hole.
Such systems can be formed due to the merger of two or more galaxies with
massive black holes [2]. Thus, accretion disks may also exist around binary
black holes, the study of which is of astrophysical interest.

In this paper, thick accretion disks around binary black holes are
investigated. For simplicity, it is assumed that double black holes are
described by the Majumdar-Papapetrou metric. Chapter 2 describes the metric
and its main properties, which are used in this work. In Chapter 3, the
hydrodynamic equation in this metric is written out and exact analytical
solutions are constructed. In chapter 4, a generalization is constructed for
the case of a toroidal magnetic field. Chapter 5 provides a conclusion.

In the paper, we use the geometrical units, G=c=1.

##  II Basic equations

In this paper we will consider the simplified model of a binary black hole,
which is described by the Majumdar-Papapetrou solution [4, 5]. The solution
has the form [4, 5]

$\displaystyle ds^{2}=-\Omega^{-2}dt^{2}+\Omega^{2}(dx^{2}+dy^{2}+dz^{2}),$
(1)

where $\Omega=1+\sum\limits_{i}\frac{m_{i}}{r_{i}}$, ”i” is a number of black
holes,

$r_{i}=\sqrt{(x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}},$

$x_{i},y_{i},z_{i}$ \- are the coordinates location and $m_{i}$ \- is the mass
of black holes. For the case when one black hole is considered, this metric
passes into the metric of the extreme Reissner-Nordstrom black hole. Here we
consider the case of two black hole, when $i=2$ and without loss of
generality, we will assume that black holes are located on the z axis with
coordinates (0,0,1) and (0,0,-1) respectively.

Let’s rewrite the metric (1) in a cylindrical coordinate system. To do this,
we introduce cylindrical coordinates

$\displaystyle x$ $\displaystyle=$ $\displaystyle r\cos\phi,$ $\displaystyle
y$ $\displaystyle=$ $\displaystyle r\sin\phi,$ $\displaystyle z$
$\displaystyle=$ $\displaystyle z,$ (2)

in which the metric (1) will be written as

$\displaystyle
ds^{2}=-\Omega^{-2}dt^{2}+\Omega^{2}(dr^{2}+r^{2}d\phi^{2}+dz^{2}),$ (3)

where $\Omega=1+\sum\limits_{i=1}^{2}\frac{m_{i}}{r_{i}}$,
$r_{i}=\sqrt{r^{2}+(z-z_{i})^{2}}$. This metric has two Killing vectors,
$\xi_{(t)}=\partial/\partial t$ and $\xi_{(\phi)}=\partial/\partial\phi$ which
generate time shifts and rotations around the axis of symmetry $z$.

We will also need a metric determinant that is equal to
$\sqrt{-g}=r\Omega^{2}$.

##  III solutions of hydrodynamic equations

In this section we consider a perfect fluid in a stationary, axisymmetric
binary black holes where the self-gravity of the fluid is ignored. This means
that the following conditions are satisfied: $\partial/\partial t=0$ and
$\partial/\partial\phi=0$ and all values depend on two variables, r and z. It
is also assumed that the fluid rotates only around the axis of symmetry z.
This means that the radial and z components of the four-velocities are equal
to zero, $u^{z}=u^{r}=0$ and only the temporal and azimuthal components are
different from zero, $u^{t}\neq 0$, $u^{\phi}\neq 0$. It is not difficult to
see that under such conditions the continuity equation is performed
identically. The four-velocity of the fluid satisfies the condition,
$g_{\alpha\beta}u^{\alpha}u^{\beta}=-1$, which in our case will be rewritten
as

$\displaystyle u^{t2}-r^{2}\Omega^{4}u^{\phi 2}=\Omega^{2}.$ (4)

From the conservation laws $T^{\alpha\beta}_{\,\,\,\,\,;\beta}=0$ for the
energy-momentum tensor of an ideal fluid

$\displaystyle T^{\alpha\beta}=(P+\rho)u^{\alpha}u^{\beta}+Pg^{\alpha\beta}$
(5)

only two components (r,z) will remain nonzero, which can be written as

$\displaystyle\frac{1}{p+\rho}\frac{\partial P}{\partial
a}=\frac{1}{\Omega}\frac{\partial\Omega}{\partial a}+\frac{u^{\phi
2}}{2\Omega^{2}}\frac{\partial}{\partial a}(r^{2}\Omega^{4}),$ (6)

where notation ”a” takes two values $a=r$ or $z$ and $P$ \- is the pressure,
$\rho$ \- is the energy density.

To integrate equation (6), we need to express the pressure and energy density
in terms of enthalpy, $h$. Assuming that the entropy, $s$, is constant along
the fluid flow line, then it is easy to obtain that

$\displaystyle\frac{dP}{P+\rho}=\frac{dh}{h}.$ (7)

Thus, we have a set of equations (4) and (6), for the solution of which we
need to impose additional constrain (see, [6, 7]).

###  III.1 Fishbone-Moncrief solution

One of such constrain was proposed in the work [6]. They assumed that the
value $l=u_{\phi}u^{t}$ remains constant. Then using (4) you can get

$\displaystyle u^{\phi
2}=\frac{-1+\sqrt{1+\frac{4l^{2}}{r^{2}\Omega^{4}}}}{2r^{2}\Omega^{2}}$ (8)

and rewrite the equation (6) in the form

$\displaystyle\frac{1}{h}\frac{\partial h}{\partial
a}=\frac{1}{\Omega}\frac{\partial\Omega}{\partial
a}+\frac{-1+\sqrt{1+\frac{4l^{2}}{r^{2}\Omega^{4}}}}{4r^{2}\Omega^{4}}\frac{\partial}{\partial
a}(r^{2}\Omega^{4}).$ (9)

It is easy to integrate the above equation (9). As a result, we get

$\displaystyle\ln(h)=\frac{1}{4}\ln\left(1+\frac{2l^{2}}{r^{2}\Omega^{4}}+\sqrt{1+\frac{4l^{2}}{r^{2}\Omega^{4}}}\right)+$
$\displaystyle+\ln\Omega-\frac{1}{2}\sqrt{1+\frac{4l^{2}}{r^{2}\Omega^{4}}}-\ln(h_{c}),$
(10)

where it is convenient to determine the integration constant in the plane
(z=0) perpendicular to the axis connecting the black holes in the point
($z=0$, $r=r_{b}$)

$\displaystyle\ln(h_{c})=\ln\Omega_{z=0,r=r_{b}}-\frac{1}{2}\sqrt{1+\frac{4l^{2}}{r_{b}^{2}\Omega^{4}_{z=0,r=r_{b}}}}+$
$\displaystyle+\frac{1}{4}\ln\left(1+\frac{2l^{2}}{r_{b}^{2}\Omega^{4}_{z=0,r=r_{b}}}+\sqrt{1+\frac{4l^{2}}{r_{b}^{2}\Omega^{4}_{z=0,r=r_{b}}}}\right).$
(11)

This solution (10) is a general solution of the equations (4) and (6). The
figures (1) and (2) show examples of the contours of the logarithm of enthalpy
$\ln(h)$ for a disk around two black holes. In these figures, negative radii
correspond to azimuthal angles $\phi+\pi$. The boundary conditions were
determined at the point ($z=0$, $r_{b}=0.1$). The constant value $l$ was
chosen equal to $l^{2}=0.1$.

Figure 1: Contours $\ln(h)=0;0.05;1$ are shown by solid curves. Black holes
are located at (0,-1) and (0,1). The masses of black holes are equal,
$m_{1}=1$ and $m_{2}=1$. Figure 2: Contours $\ln(h)=0;0.05;1$ are shown by
solid curves. Black holes are located at (0,-1) and (0,1). The masses of black
holes are equal, $m_{1}=1$ and $m_{2}=0.1$, respectively.

###  III.2 Kozlowski et.al. solution

Another constrain was suggested in the paper [7]. They assumed that the value
$l=-u_{\phi}/u_{t}$ remains constant. Then it follows from the equation (4)
that $\phi$-component of the four-velocity is equal to

$\displaystyle u^{\phi
2}=\frac{l^{2}}{r^{2}\Omega^{2}(r^{2}\Omega^{4}-l^{2})}.$ (12)

It follows from this expression (12) that the condition
$r^{2}\Omega^{4}>l^{2}$ must be fulfilled, which cannot be fulfilled near the
points $r\approx 0$, $z\neq\pm 1$. Therefore, in this case, the solution will
not completely cover the space-time around the binary black hole. Formally, in
this case, we can also fully integrate these equations, (6). Using the
equations (7) and (12), we can rewrite the equation (6) as

$\displaystyle\frac{1}{h}\frac{\partial h}{\partial
a}=\frac{1}{\Omega}\frac{\partial\Omega}{\partial
a}+\frac{l^{2}}{2r^{2}\Omega^{4}(r^{2}\Omega^{4}-l^{2})}\frac{\partial}{\partial
a}(r^{2}\Omega^{4}).$ (13)

After integration we get

$\displaystyle\ln(h)=\ln\Omega+\frac{1}{2}\ln\left(1-\frac{l^{2}}{r^{2}\Omega^{4}}\right)-\ln(h_{c}).$
(14)

where the integration constant is defined in the same way as in the previous
case

$\displaystyle\ln(h_{c})=\ln\Omega_{z=0,r=r_{b}}+\frac{1}{2}\ln\left(1-\frac{l^{2}}{r^{2}_{b}\Omega^{4}_{z=0,r=r_{b}}}\right).$
(15)

###  III.3 Another exact solution

By making various assumptions, other exact analytical solutions can be
obtained. For example, assuming that the value $l=u_{\phi}u^{\phi}$ remains
constant, using expressions (6) and (7) we obtain the desired equation

$\displaystyle\frac{1}{h}\frac{\partial h}{\partial
a}=\frac{1}{\Omega}\frac{\partial\Omega}{\partial
a}+\frac{l}{2r^{2}\Omega^{4}}\frac{\partial}{\partial a}(r^{2}\Omega^{4}).$
(16)

After integration, we get an exact analytical solution

$\displaystyle\ln(h)=\ln\Omega+\frac{l}{2}\ln(r^{2}\Omega^{4})-\ln(h_{c})$
(17)

with the integration constant equal to

$\displaystyle\ln(h_{c})=\ln\Omega_{z=0,r=r_{b}}+\frac{l}{2}\ln(r_{b}^{2}\Omega^{4}_{z=0,r=r_{b}}).$
(18)

The figures (3) and (4) show examples of the distribution of the contours of
the constant enthalpy $\ln(h)=-0.1;0;0.1;0.5$ for various parameters of the
problem. In the figure (3), the mass of black holes is equal to,
$m_{1}=m_{2}=1$, and in the figure (4) the mass of black holes is equal to,
$m_{1}=1$, $m_{2}=0.1$. The value $l$ was chosen equal to, $l=0.1$.

Figure 3: Contours of the logarithm of the constant enthalpy
$\ln(h)=-0.1;0;0.1;0.5$ are shown by solid curves. Black holes are located at
(0,-1) and (0,1). The mass of black holes are equal, $m_{1}=m_{2}=1$. Figure
4: Contours of the logarithm of the constant enthalpy $\ln(h)=-0.1;0;0.1;0.5$
are shown by solid curves. Black holes are located at (0,-1) and (0,1). The
mass of black holes are equal, $m_{1}=1$ and $m_{2}=0.1$.

##  IV solutions with a toroidal magnetic field

Following the work of [8], we can generalize the above solutions to the case
of the presence of a toroidal magnetic field. A set of ideal relativistic MHD
equations are

$\displaystyle g_{\alpha\beta}u^{\alpha}u^{\beta}=-1,$ $\displaystyle
T^{\alpha\beta}_{\,\,\,\,\,;\beta}=0,$
$\displaystyle(nu^{\alpha})_{;\alpha}=0,$ $\displaystyle
F_{\mu\nu,\lambda}+F_{\nu\lambda,\mu}+F_{\lambda\mu,\nu}=0,$ (19)

where $n$ is the baryon number density, $F_{\mu\nu}$ is the electromagnetic
field tensor and

$\displaystyle
T^{\alpha\beta}=(P+\rho+b^{2})u^{\alpha}u^{\beta}+(P+\frac{b^{2}}{2})g^{\alpha\beta}-b^{\alpha}b^{\beta},$
(20)

where $b^{\alpha}$ is the four-vector of magnetic field [8]. Also it’s suppose
that the flow is stationary, $\partial/\partial t$, and axisymmetric,
$\partial/\partial\phi$, the velocity of fluid and magnetic field have only
toroidal component, $u^{r}=u^{\theta}=0$, $b^{r}=b^{\theta}=0$. Then the third
and forth equations (19) are satisfied indentically. The r and z component of
second eqution (19) are rewritting in the form

$\displaystyle\frac{1}{\sqrt{-g}}\frac{\partial}{\partial
a}\left(\sqrt{-g}\left(p+\frac{1}{2}b^{2}\right)\right)=\frac{1}{2}g_{\gamma\delta,a}T^{\gamma\delta}.$
(21)

If we substitute metric coefficients (3) and the energy-momentum tensor (20)
into the equation (21), then after tedious transformations we can get an
expression of the form

$\displaystyle\frac{\partial}{\partial
a}(p+\frac{b^{2}}{2})=\frac{(p+\rho+b^{2})u^{t2}-b^{t2}}{\Omega^{3}}\frac{\partial\Omega}{\partial
a}+$ $\displaystyle+\frac{(p+\rho+b^{2})u^{\phi 2}-b^{\phi
2}}{2}\frac{\partial}{\partial a}(r^{2}\Omega^{2}).$ (22)

In order to integrate this equation, it is necessary to have an equation of
state $p=k\rho^{\gamma}$ and to associate enthalpy $P+\rho$ with magnetic
pressure $b^{2}$. Following the work [8], we assume that these quantities are
connected in a linear way of the form

$\displaystyle\beta=\frac{P+\rho}{b^{2}}.$ (23)

For simplicity, let’s assume that the parameter, $\beta$, is a constant. Using
the parameters $\beta$, we obtain a final equation that describes the
distribution of a fluid with a toroidal magnetic field around binary black
holes.

$\displaystyle\frac{\beta}{1+\beta}\frac{1}{p+\rho}\frac{\partial p}{\partial
a}+\frac{1}{2(1+\beta)b^{2}}\frac{\partial b^{2}}{\partial
a}=\frac{1}{\Omega}\frac{\partial\Omega}{\partial a}+$
$\displaystyle+\frac{u^{\phi
2}}{2\Omega^{2}}\frac{\beta}{1+\beta}\frac{\partial}{\partial
a}(r^{2}\Omega^{4})-\frac{1}{2}\frac{1}{1+\beta}\frac{1}{r^{2}\Omega^{2}}\frac{\partial}{\partial
a}(r^{2}\Omega^{2}).$ (24)

We explicitly integrate this equation below by applying various constrains to
the four-velocity of the fluid.

###  IV.1 Fishbone-Moncrief solution

Assuming that the constraints of the form, $l=u_{\phi}u^{t}$, as in the work
[6] are fulfilled and using the equation (8) from equation (24), we obtain an
exact analytical solution for the distribution of fluid in a toroidal magnetic
field around a double black hole.

$\displaystyle\frac{\beta}{1+\beta}\frac{\gamma}{\gamma-1}\ln(1+k\rho^{\gamma-1})+\frac{1}{2(1+\beta)}\ln(b^{2})=$
$\displaystyle=\ln(\Omega)+\frac{\beta}{4(1+\beta)}\ln(1+\frac{2l^{2}}{r^{2}\Omega^{4}}+\sqrt{1+\frac{4l^{2}}{r^{2}\Omega^{4}}})-$
$\displaystyle-\frac{\beta}{2(1+\beta)}\sqrt{1+\frac{4l^{2}}{r^{2}\Omega^{4}}}-\frac{\ln(r^{2}\Omega^{2})}{2(1+\beta)}-\ln(h_{c}).$
(25)

The integration constant is defined at a point ($z=0$, $r=r_{b}$) in the same
way as in the chapter (III.1). In the limiting case, when the parameter
$\beta\rightarrow\infty$ tends to infinity, we get the solution (10). The
figure (5) show examples of contours of the left side of the equation (25) for
values $-0.2;-0.1;0.25$ for the following parameters: $\beta=2$,
$m_{1}=m_{2}=1$, $l^{2}=0.1$, $r_{b}=0.1$.

If we take the mass of the second black hole equal $m_{2}=0.1$, then the
contours of the left side of the equation for the same parameters of the
problem, $m_{1}=1$, $l^{2}=0.1$, $r_{b}=0.1$ will change and take the form as
in the figure (6).

Figure 5: Contours of the left side of the equation (25) for values
$0;0.25;1.0$ are shown by solid curves. Black holes are located at (0,-1) and
(0,1). The mass of black holes are equal, $m_{1}=m_{2}=1.0$. Figure 6:
Contours of the left side of the equation (25) for values $-0.2;-0.1;0.25$ are
shown by solid curves. Black holes are located at (0,-1) and (0,1). The mass
of black holes are equal, $m_{1}=1.0$ and $m_{2}=0.1$.

Comparing Figures (1) and (5) or (2) and (6), it can be seen that the toroidal
magnetic field can greatly change the behavior of the contours and the
qualitative picture of the distribution of fluid near binary black holes.

###  IV.2 Another exact solution

Finally, consider the case when the constrain is described by the relation
$l=u_{\phi}u^{\phi}$, see (III.3). Then the solution of the equation (24) is
easy to obtain. As a result, we get

$\displaystyle\frac{\beta}{1+\beta}\frac{\gamma}{\gamma-1}\ln(1+k\rho^{\gamma-1})+\frac{1}{2(1+\beta)}\ln(b^{2})=$
$\displaystyle=\ln(\Omega)+\frac{\beta
l}{1+\beta}\ln(r\Omega^{2})-\frac{\ln(r\Omega)}{1+\beta}-\ln(h_{c}),$ (26)

where the integration constant, $\ln(h_{c})$, is defined similarly to the
previous cases.

In the figures (7) and (8) you can see the contours of the left side of the
equation (26) for the following parameters: $m_{1}=m_{2}=1$, $l=0.1$ $\beta=3$
\- for fig. (7) and $m_{1}=1$, $m_{2}=0.1$, $l=0.1$ $\beta=2$ for fig. (8).
Comparing Figures (3) (4) with Figures (7) (8), one can see a strong
difference associated with the absence of a thick disk between black holes in
the presence of a strong toroidal magnetic field.

Figure 7: Contours of the left side of the equation (26) for values
$0;0.1;0.5$ are shown by solid curves. Black holes are located at (-1,0) and
(1,0). The mass of black holes are equal, $m_{1}=m_{2}=1.0$. $\beta=3$ Figure
8: Contours of the left side of the equation (26) for values $0;0.2;0.5$ are
shown by solid curves. Black holes are located at (-1,0) and (1,0). The mass
of black holes are equal, $m_{1}=1.0$ and $m_{2}=0.1$. $\beta=2$

##  V Conclusions

In this paper, we generalized the relativistic theory of thick accretion disks
to the case of binary black holes. Black holes were described by the Majumdar-
Papapetrou solution, in which the mass of a black hole is compensated by an
electric charge. A method for constructing a thick accretion disk in the
hydrodynamic case and with the presence of a toroidal magnetic field was
described. Exact analytical solutions are written out. These solutions can be
used in numerical MHD calculations of the evolution of thick disks around
binary black holes as the initial conditions of the problem.

## References

  * [1] B.P. Abbott et al., Phys.Rev.Lett., 116, 241102 (2016).
  * [2] H. Sudou, S. Iguchi, Y. Murata, Y. Taniguchi, Science, 300, 1263 (2003).
  * [3] L. Titarchuk, E. Seifina, C. Shrader, Astron.Astrophys. 671, A159 (2023).
  * [4] S.D. Majumdar, Phys.Rev. 72, 390 (1947).
  * [5] A. Papapetrou, Proc. Roy. Irish Acad. 51, 191 (1947).
  * [6] L.G. Fishbone, V. Moncrief, Astrophys.J., 207, 962 (1976).
  * [7] M. Kozlowski, M. Jaroszynski, M.A. Abramowicz, Astron.Astrophys., 63, 209 (1978).
  * [8] S.S. Komissarov, MNRAS, 368, 993 (2006)


url]https://users.auth.gr/zarkonath

# CHARACTERIZATIONS OF CONVERGENCE BY A GIVEN SET OF ANGLES IN SIMPLY
CONNECTED DOMAINS

Konstantinos Zarvalis  Department of Mathematics, Aristotle University of
Thessaloniki, 54124, Thessaloniki, Greece<EMAIL_ADDRESS>[

###### Abstract

Let $\Delta$ be a simply connected domain and $f:\mathbb{D}\to\Delta$, where
$\mathbb{D}$ is the unit disk, be a corresponding Riemann map. Let
$\\{z_{n}\\}\subset\Delta$ be a sequence with no accumulation points inside
$\Delta$. In the present article, we give necessary and sufficient conditions
in terms of hyperbolic geometry which certify that $\\{f^{-1}(z_{n})\\}$
converges to a point of $\partial\mathbb{D}$ by a certain angle $\theta$ or by
a certain set of angles $[\theta_{1},\theta_{2}]$.

###### keywords:

Non-tangential convergence , Simply connected domain , Riemann map , Harmonic
Measure , Semigroup of holomorphic functions

###### MSC:

[2020] Primary 30C35, 30D40 , Secondary 37F44, 31A15

##  1 Introduction

Let $\Delta$ be a simply connected domain other than $\mathbb{C}$. By the
Riemann mapping theorem, this domain is conformally equivalent to the unit
disk. Let $f:\mathbb{D}\to\Delta$ be a Riemann map and $\\{z_{n}\\}$ be a
sequence in $\Delta$ with no accumulation points in $\Delta$. Then,
$\\{f^{-1}(z_{n})\\}$ is a sequence of the unit disk that has at least one
limit point lying on $\partial\mathbb{D}$ and therefore, we can talk about the
angle of the convergence. In this way, we are able to transcend the question
of convergence by angle to arbitrary simply connected domains.

A natural idea would be to examine the simply connected domain and its
geometric properties to provide an answer about the angle of the convergence
of the preimage sequence in the unit disc. However, the domain $\Delta$ can be
extremely complicated, thus making such a procedure very difficult and at
times impossible.

In [5], Bracci, Contreras, Díaz-Madrigal and Gaussier introduced an
interesting technique to tackle the above problem. Instead of looking at the
given simply connected domain $\Delta$, we find another one that contains
$\Delta$, has better geometric properties and is easier to work with. More
precisely, we find a simply connected domain $U$ such that $\Delta\subset U$
and $\Delta$ contains a suitable horodisk of $U$.

To be more concrete, we explain first the notion of a horodisk. Let
$\sigma\in\partial\mathbb{D}$ and $R>0$. Then, the horodisk of $\mathbb{D}$ of
radius $R$ centered at $\sigma$ is the set

$E_{\mathbb{D}}(\sigma,R)=\\{z\in\mathbb{D}:|\sigma-z|^{2}<R(1-|z|^{2})\\}.$

Again, we need to generalize this definition. Let $U\subsetneq\mathbb{C}$ be a
simply connected domain, $\partial_{C}U$ the set of its prime ends and
$f:\mathbb{D}\to U$ a Riemann map. Suppose that $\xi\in\partial_{C}U$ and that
the prime end $\xi$ corresponds through $f$ to the point
$\sigma\in\partial\mathbb{D}$. Then, the horodisk of $U$ of radius $R$
centered at $\xi$ is the set

$E_{U}(\xi,R)=f(E_{\mathbb{D}}(\sigma,R)).$

The “suitable horodisk” is one that is centered at the limit point of the
sequence. Setting $\xi$ the limit point and using the above terminology, we
have $E_{U}(\xi,R)\subset\Delta\subseteq U$ (the equality is allowed for the
case when $\Delta$ is already simple). Therefore, the given simply connected
domain $\Delta$ is restricted between $U$ and its horodisk and as the sequence
converges to $\xi$ the geometry of $\Delta$ seems more and more similar to
that of the simpler domain $U$.

So, the central idea is the use of a domain with simpler geometry and its
horodisks in order to find a necessary and sufficient condition for
convergence by a certain angle. In [5], the authors deal only with the case of
orthogonal convergence, when the angle of convergence is $\frac{\pi}{2}$, and
provide the corresponding theorem. In the present article, we will utilise
their method to produce an analogue result for any angle-set
$[\theta_{1},\theta_{2}]\subset(0,\pi)$.

Moreover, when we speak of the geometry of these simply connected domains, we
mostly talk in terms of hyperbolic geometry. So, before mentioning the theorem
in [5], we first give some useful definitions to render its statement clearer.
To begin with, the hyperbolic distance in $\mathbb{D}$ is given by (see e.g.
[4, Section 1.3])

$k_{\mathbb{D}}(z,w)=\frac{1}{2}\log\frac{1+\left|\frac{z-w}{1-\bar{z}w}\right|}{1-\left|\frac{z-w}{1-\bar{z}w}\right|},\;\;z,w\in\mathbb{D}.$

Now, let $\Delta\subsetneq\mathbb{C}$ be a simply connected domain and
$f:\Delta\to\mathbb{D}$ a corresponding Riemann map. We denote by $k_{\Delta}$
the hyperbolic distance in $\Delta$ given by

$k_{\Delta}(z,w)=k_{\mathbb{D}}(f(z),f(w)),\;\;z,w\in\Delta.$

It can be verified that this definition is indeed independent of the choice of
the conformal map $f$. Moreover, if $T\subset\mathbb{R}$ is a segment (open,
closed, semi-open) and $\gamma:T\to\Delta$ is a smooth curve, then we denote

$k_{\Delta}(z,\gamma)=\inf\limits_{t\in
T}k_{\Delta}(z,\gamma(t)),\;z\in\Delta.$

Now, we are ready for the theorem.

######  Theorem ([5]).

Let $\Delta\subsetneq\mathbb{C}$ be a simply connected domain and
$f:\mathbb{D}\to\Delta$ a Riemann map. Let $\\{z_{n}\\}\subset\Delta$ be a
sequence with no accumulation points in $\Delta$. Then, there exists a
$\sigma\in\partial\mathbb{D}$ such that $\\{f^{-1}(z_{n})\\}$ converges
orthogonally to $\sigma$ if and only if there exist a simply connected domain
$U\subsetneq\mathbb{C},\;\xi\in\partial_{C}U$ and $R>0$ such that

  1. (i)

$E_{U}(\xi,R)\subset\Delta\subseteq U$,

  2. (ii)

$\lim\limits_{n\to+\infty}k_{U}(z_{n},\gamma)=0$, where $\gamma:[0,+\infty)\to
U$ is any geodesic for the hyperbolic distance in $U$ such that
$\lim\limits_{t\to+\infty}\gamma(t)=\xi$ in the Carathéodory topology of $U$.

We will generalize this theorem to give a result for convergence by any angle
$\theta\in(0,\pi)$, by providing a corollary to our main theorem which will be
an even more general result concerning the convergence by any angle-set
$[\theta_{1},\theta_{2}]\subset(0,\pi)$.

In order to obtain the proof, we will mostly utilize techniques of harmonic
measure; techniques upon which we are going to elaborate further in the next
section. The corresponding proof given in [5] for orthogonal convergence
relies on properties of the hyperbolic distance.

We need to provide a rigid definition of convergence by angle.

######  Definition 1.1.

A sequence $\\{z_{n}\\}\subset\mathbb{D}$ is said to converge by angle
$\theta$, $\theta\in[0,\pi],$ to a point $\sigma\in\partial\mathbb{D}$,
provided $\lim\limits_{n\to+\infty}z_{n}=\sigma$ and
$\lim\limits_{n\to+\infty}\arg(1-\bar{\sigma}z_{n})=\frac{\pi}{2}-\theta$. In
a similar manner, a continuous curve $\gamma:[0,+\infty)\to\mathbb{D}$
converges by angle $\theta$, $\theta\in[0,\pi]$, to a point
$\sigma\in\partial\mathbb{D}$, provided
$\lim\limits_{t\to+\infty}\gamma(t)=\sigma$ and
$\lim\limits_{t\to+\infty}\arg(1-\bar{\sigma}\gamma(t))=\frac{\pi}{2}-\theta$.

In the extreme cases when $\theta=0$ or $\theta=\pi$, the convergence is
characterized as tangential. Nevertheless, a sequence in the unit disk may
oscillate while converging to its limit on the unit circle and as a result
this sequence can have various (and even infinite) subsequences that converge
to the limit by different angles. In this case, we cannot say that the
sequence converges by a certain angle. We are in need of a new, more inclusive
definition.

######  Definition 1.2.

A sequence $\\{z_{n}\\}\subset\mathbb{D}$ is said to converge by angle-set
$[\theta_{1},\theta_{2}]\subset[0,\pi]$ to a point
$\sigma\in\partial\mathbb{D}$, provided
$\lim\limits_{n\to+\infty}z_{n}=\sigma$ and the cluster set of
$\arg(1-\bar{\sigma}z_{n})$ is
$\left[\frac{\pi}{2}-\theta_{2},\frac{\pi}{2}-\theta_{1}\right]$.

In this article, we will only entertain the possibility when
$[\theta_{1},\theta_{2}]\subset(0,\pi)$. Clearly, as was the case previously,
the definition can be extended to a continuous curve
$\gamma:[0,+\infty)\to\mathbb{D}$.

It is a well known fact that the cluster set of
$\arg(1-\bar{\sigma}\gamma(t))$ for a continuous curve $\gamma$ is a compact,
connected subset of $[0,\pi]$. So, this cluster set must be either a singleton
$\\{\theta\\}$ or a continuum $[\theta_{1},\theta_{2}]$. Consequently, a
continuous curve falls into one of the two categories described by Definitions
1.1 and 1.2. However, for a sequence $\\{z_{n}\\}$, things can get much more
complicated. For example, the cluster set of $\arg(1-\bar{\sigma}z_{n})$ could
be a finite set of angles, or even a sequence of angles. In particular, if the
cluster set of $\arg(1-\bar{\sigma}z_{n})$ is a subset of $(0,\pi)$, the
convergence is called non-tangential. For the purposes of this article, we
only examine the first two categories.

The goal of our article is to provide a necessary and sufficient condition for
convergence by angle-set. To this direction, we first need the notion of
hyperbolic sectors. Let $\Delta$ be a simply connected domain and
$\gamma:[\alpha,+\infty)\to\Delta,\;\alpha\geq-\infty$, a geodesic for the
hyperbolic distance in $\Delta$ with the added property
$\lim\limits_{t\to+\infty}k_{\Delta}(\gamma(t),\gamma(t_{0}))=+\infty$, for
some $t_{o}\in(a,+\infty)$. For $R>0$, we call the set

$S_{\Delta}(\gamma,R)=\\{z\in\Delta:k_{\Delta}(z,\gamma)<R\\}$

a hyperbolic sector around $\gamma$ of amplitude $R$.

In [6], again Bracci, Contreras, Díaz-Madrigal and Gaussier proved the
following theorem that characterizes non-tangential convergence:

######  Theorem ([6]).

Let $\Delta\subsetneq\mathbb{C}$ be a simply connected domain and let
$f:\mathbb{D}\to\Delta$ be a Riemann map. Let $\\{z_{n}\\}\subset\Delta$ be a
sequence with no accumulation points in $\Delta$. Then, $\\{f^{-1}(z_{n})\\}$
converges non-tangentially to a point $\sigma\in\partial\mathbb{D}$ if and
only if there exist a simply connected domain $U\subsetneq\mathbb{C}$, a
geodesic $\gamma:[0,+\infty)\to U$ of $U$ such that
$\lim\limits_{t\to+\infty}k_{U}(\gamma(t),\gamma(0))=+\infty$ and $R>R_{0}>0$
such that

  1. (i)

$S_{U}(\gamma,R)\subset\Delta\subseteq U$,

  2. (ii)

there exists $n_{0}\geq 0$ such that $z_{n}\in S_{U}(\gamma,R_{0})$ for all
$n\geq n_{0}$.

From their theorem, we borrow the central idea of our characterization of
convergence by angle-set. This idea is the use of hyperbolic sectors and we
combine this idea again with the use of a simple, in terms of geometry, simply
connected domain and its horodisks. However, the statement of the theorem
demands certain new definitions and notations which will be dealt with in
detail in Section 3.

Briefly, let $\\{z_{n}\\}\subset\Delta$ be a sequence with no accumulation
points in $\Delta$, where $\Delta$ is a simply connected domain, and
$\lim\limits_{n\to+\infty}z_{n}=\xi\in\partial_{C}\Delta$. We will introduce
the sets $A_{\Delta}(\gamma,\theta_{1},\theta_{2})$, where
$\gamma:[0,+\infty)\to\Delta$ is a geodesic for the hyperbolic distance in
$\Delta$ with $\lim\limits_{t\to+\infty}\gamma(t)=\xi$ and
$0<\theta_{1}<\theta_{2}<\pi$ are the angles that define the angle-set
$[\theta_{1},\theta_{2}]$ of convergence. The definition of these sets relies
exclusively on the geodesic $\gamma$ and its hyperbolic sectors with
amplitudes depending on the angles. We will say that the sequence
$\\{z_{n}\\}$ exhausts the set $A_{\Delta}(\gamma,\theta_{1},\theta_{2})$ if
it is eventually contained in every set
$A_{\Delta}(\gamma,\omega_{1},\omega_{2}),\;\omega_{1}<\theta_{1},\omega_{2}>\theta_{2}$
and, in a way that will be explained later, eventually “fills” the set
$A_{\Delta}(\gamma,\theta_{1},\theta_{2})$. Our theorem is the following.

######  Theorem 1.1.

Let $\Delta\subsetneq\mathbb{C}$ be a simply connected domain and
$f:\mathbb{D}\to\Delta$ be a Riemann map. Let $\\{z_{n}\\}\subset\Delta$ be a
sequence with no accumulation points in $\Delta$. Then, there exists a
$\sigma\in\partial\mathbb{D}$ such that $\\{f^{-1}(z_{n})\\}$ converges by
angle-set $[\theta_{1},\theta_{2}]\subset(0,\pi)$ to $\sigma$ if and only if
there exist a simply connected domain $U\subsetneq\mathbb{C}$,
$\xi\in\partial_{C}U$, $R>0$ and a geodesic $\gamma:[0,+\infty)\to U$ for the
hyperbolic distance in $U$ with $\lim\limits_{t\to+\infty}\gamma(t)=\xi$ in
the Carathéodory topology of $U$ such that

  1. (i)

$E_{U}(\xi,R)\subset\Delta\subseteq U,$

  2. (ii)

$\gamma(0)\in\Delta,$

  3. (iii)

$\\{z_{n}\\}$ exhausts the set $A_{U}(\gamma,\theta_{1},\theta_{2}).$

Again, our proof relies heavily on harmonic measure and the conformally
invariant nature of hyperbolic distance and harmonic measure. In [6], the
authors rely on Gromov’s hyperbolicity theory and quasi-geodesics. Our proof
thus shows that Gromov’s theory is not necessary. However, Gromov’s theory and
the notion of quasi-geodesics are certainly very useful for a profound
understanding of the problem.

Finally, in Section 4, we will give an application on continuous semigroups of
holomorphic functions of the unit disk.

##  2 Preliminaries

###  2.1 Hyperbolic Distance

Before moving on to the main body of the article, we must first review some
basic facts and properties concerning two conformal invariants: hyperbolic
distance and harmonic measure.

Starting with the hyperbolic distance, its property of conformal invariance is
directly implied by its definition with respect to an arbitrary simply
connected domain, other than the complex plane. In addition, hyperbolic
distance has a monotonicity property: let $\Delta_{1},\Delta_{2}$ be two
simply connected domains satisfying $\Delta_{1}\subset\Delta_{2}$. Then, for
all $z,w\in\Delta_{1}$ we have that

$k_{\Delta_{2}}(z,w)\leq k_{\Delta_{1}}(z,w).$

###  2.2 Harmonic Measure

Next, we turn to harmonic measure, which is conformally invariant as well. A
comprehensive presentation of its theory can be found in [18]. Let $\Delta$ be
a domain in $\mathbb{C}$ with non-polar boundary. Let $B$ be a Borel subset of
$\partial\Delta$. Then, the harmonic measure of $B$ with respect to $\Delta$
is exactly the solution of the generalized Dirichlet problem for the Laplacian
in $\Delta$ with boundary function equal to $1$ on $B$ and to $0$ on
$\partial\Delta\setminus B$. For the harmonic measure of $B$ with respect to
$\Delta$ and for $z\in\Delta$ we use the notation $\omega(z,B,\Delta)$. It is
known that for a fixed $z\in\Delta$, $\omega(z,\cdot,\Delta)$ is a Borel
probability measure on $\partial\Delta$.

A useful property of harmonic measure is its domain monotonicity. In detail,
let
$\Delta_{1}\subset\Delta_{2},\;B\subset\partial\Delta_{1}\cap\partial\Delta_{2}$
Borel. Then, for all $z\in\Delta_{1}$, we have

$\omega(z,B,\Delta_{1})\leq\omega(z,B,\Delta_{2}).$

The above statement can be made more precise by means of what is called the
strong Markov property of harmonic measure (see e.g. [12, p.307]). Again, let
$\Delta_{1}\subset\Delta_{2}$ and
$B\subset\partial\Delta_{1}\cap\partial\Delta_{2}$ Borel. Then, for
$z\in\Delta_{1},$

$\omega(z,B,\Delta_{2})=\omega(z,B,\Delta_{1})+\int\limits_{\partial\Delta_{1}\setminus\partial\Delta_{2}}\omega(\zeta,B,\Delta_{2})\cdot\omega(z,d\zeta,\Delta_{1}).$

######  Remark 2.1.

Let $B$ be an arc of $\partial\mathbb{D}$ with endpoints $a,b$. Then, we know
(see e.g. [8, p.155]) that the level set

$L_{k}=\\{z\in\mathbb{D}:\omega(z,B,\mathbb{D})=k\\},\;0<k<1,$

is a circular arc (or a diameter in case $B$ is a half-circle and
$k=\frac{1}{2}$) in $\mathbb{D}$ with endpoints $a,b$ that meets
$\partial\mathbb{D}$ with angle $k\pi$. Therefore, a sequence
$\\{z_{n}\\}\subset\mathbb{D}$ that converges to $a$ or $b$, converges by
angle $\theta$ if and only if
$\lim\limits_{n\to+\infty}\omega(z_{n},B,\mathbb{D})=\frac{\theta}{\pi}$. The
analogous result is also true for any curve converging to $a$ or $b$ as well.
Certainly, inside the subdomain of the unit disk bounded by the level set
$L_{k}$ and the arc $B$, the harmonic measure $\omega(z,B,\mathbb{D})$ is
larger than $k$, while inside the complementary subdomain, the corresponding
harmonic measure is smaller than $k$.

Finally, we refer to [18] for some results about harmonic measure on certain
particular domains. Denoting by $U_{1}$ the upper half plane $\\{z:\Im
z>0\\}$, we have

$\omega(z,[a,b],U_{1})=\frac{1}{\pi}\arg\left(\frac{z-b}{z-a}\right),$

where $[a,b]\subset\mathbb{R},z\in U_{1}$. Then, for any sequence
$\\{z_{n}\\}\subset U_{1}$ with $\lim\limits_{n\to+\infty}z_{n}=\infty$, it is
easily calculated by the above formula that
$\lim\limits_{n\to+\infty}\omega(z_{n},[a,b],U_{1})=0$. Furthermore, denoting
by $U_{2}$ the set $\\{z:a<\arg z<\beta\\}$, where
$(a,\beta)\subset(-\pi,\pi)$, for $z\in U_{2}$ we get

$\omega(z,\\{\arg z=\beta\\},U_{2})=\frac{\arg z-a}{\beta-a}.$

A direct consequence of the last relation is that for the right half plane
$\mathbb{H}=\\{z:\Re z>0\\}$ we have

$\omega(t,\\{\arg z=-\frac{\pi}{2}\\},\mathbb{H})=\omega(t,\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})=\frac{1}{2},$

for all positive $t$. Of course, using conformal mappings and combining the
above equalities, we can find numerous other results.

###  2.3 Convergence by Angle

It is necessary to understand the elementary origin of Definitions 1.1 and 1.2
and comprehend what they signify. To this goal, let
$\sigma\in\partial\mathbb{D}$ and $T$ be the tangent of $\partial\mathbb{D}$
at $\sigma$. Consider $\Gamma$ to be a straight line that intersects
$\partial\mathbb{D}$ at $\sigma$. Restricting ourselves to the half-plane
determined by $T$ and $\mathbb{D}$, the lines $T$ and $\Gamma$ form two
angles. We call $\theta$ the angle formed by $\Gamma$ and the ray of $T$
extending to the right of $\sigma$ (where right and left are determinded by
considering the half-plane we chose as the lower one, as shown in Figure 1).
Let $\\{z_{n}\\}\subset\Gamma\cap\mathbb{D}$ with
$\lim\limits_{n\to+\infty}z_{n}=\sigma$. Geometrically speaking, we could say
that $\\{z_{n}\\}$ converges by angle $\theta$ to $\sigma$. To be more rigid,
for all $n\in\mathbb{N}$, we can write
$z_{n}=\sigma+r_{n}e^{i\theta_{n}},\;r_{n}>0,\;\theta_{n}\in[0,2\pi)$. Looking
at Figure 1 and after doing some elementary computations, it is easily
verified that for every term of the sequence $\\{z_{n}\\}$, we have

$\arg(1-\bar{\sigma}z_{n})=\frac{\pi}{2}-\theta,$

and of course the same is true for the limit as $n\to+\infty$. As a result,
the practical meaning of Definition 1.1 is that for a sequence $\\{z_{n}\\}$
to converge by angle $\theta$ to $\sigma$, it is necessary that there is a
$n_{0}\in\mathbb{N}$ such that for all $n>n_{0}$, all the terms $z_{n}$ are
sufficiently close to the line $\Gamma$. A similar conclusion holds in the
case of a continuous curve.

Figure 1: Origin of Definition 1.1

##  3 Convergence by angle-set

Being restricted to $(0,\pi)$, the convergence by angle-set is non-tangential.
Therefore, if a sequence $\\{z_{n}\\}\subset\mathbb{D}$ converges to a point
$\sigma\in\partial\mathbb{D}$ by a specific angle-set, it is necessary that
this sequence is eventually contained in a Stolz region of vertex $\sigma$,
that is the set

$\\{z\in\mathbb{D}:|\sigma-z|<R(1-|z|)\\},$

for some $R>1$. Of course, the inverse implication is also true. It is known
that if we take a geodesic of $\mathbb{D}$ with endpoint at $\sigma$, then
every Stolz region of vertex $\sigma$ is equivalent to a hyperbolic sector
around $\gamma$, a notion we have already introduced. Thus, we could roughly
say that convergence by angle-set might be characterized with the help of
hyperbolic sectors. Indeed, the principal role of this section will be played
by hyperbolic sectors. We will, first, need some lemmas and definitions.

We commence with a known fact about geodesics. Let
$\Delta\subsetneq\mathbb{C}$ be a simply connected domain and
$\gamma:(a,b)\to\Delta$, where $a,b$ can be $-\infty,\infty$ respectively, a
geodesic for the hyperbolic distance in $\Delta$. We suppose that
$\lim\limits_{t\to a^{+}}\gamma(t),\lim\limits_{t\to
b^{-}}\gamma(t)\in\partial\Delta$ or, in other words, that with this
parametrization we have the whole geodesic and not only part of it. Then, the
set $\Delta\setminus\gamma((a,b))$ consists of two simply connected
components.

Using the above remark, the following notation can be stated naturally. Let
$\gamma:(-1,1)\to\mathbb{D}$ be a geodesic in $\mathbb{D}$ such that
$\lim\limits_{t\to-1^{+}}\gamma(t)=\tau\in\partial\mathbb{D}$ and
$\lim\limits_{t\to 1^{-}}\gamma(t)=\sigma\in\partial\mathbb{D}$. Then,
$\mathbb{D}\setminus\gamma((-1,1))$ consists of two simply connected
components. Consider $\gamma$ to be oriented with direction towards $\sigma$.
We denote by $\mathbb{D}_{\sigma}^{+}$ the simply connected component of
$\mathbb{D}\setminus\gamma((-1,1))$ that lies to the right of $\gamma$. We
denote by $\mathbb{D}_{\sigma}^{-}$ the one that lies to the left.

Now, we need some more important notations to render everything that will
follow more legible. Let $[\theta_{1},\theta_{2}]\subset(0,\pi)$,
$\sigma\in\partial\mathbb{D}$ and $\gamma:[0,1)\to\mathbb{D}$ be a geodesic
(actually part of it) in $\mathbb{D}$ with $\lim\limits_{t\to
1^{-}}\gamma(t)=\sigma$. Of course, the geodesic $\gamma$ can be extended in a
unique way in order to reach its second endpoint on $\partial\mathbb{D}$.
Through this extension, we can naturally obtain, again, the sets
$\mathbb{D}_{\sigma}^{-}$ and $\mathbb{D}_{\sigma}^{+}$. From now on, we set

$R(\theta)=k_{\mathbb{H}}(1,e^{i\theta}),\;\theta\in(-\frac{\pi}{2},\frac{\pi}{2}).$

The importance of this number will be understood better later on. For now, we
can mention some properties. First of all, it is obvious that $R(0)=0$. In
addition, because of the symmetry of the right half-plane with respect to the
real axis, it is easily seen that $R(\theta)=R(-\theta)$. Moreover, vertical
translations and scalings are conformal automorphisms of $\mathbb{H}$.
Therefore,

$R(\theta)=k_{\mathbb{H}}(1+iy,e^{i\theta}+iy)=k_{\mathbb{H}}(r,re^{i\theta}),$

for all $y\in\mathbb{R}$ and for all $r>0$. Composing a vertical translation
and a scaling we can get new combinations. Finally, using the Cayley transform
$C:\mathbb{H}\to\mathbb{D}$, with $C(z)=\frac{z-1}{z+1}$, we get

$R(\theta)=k_{\mathbb{D}}\left(0,\frac{e^{i\theta}-1}{e^{i\theta}+1}\right)=k_{\mathbb{D}}\left(0,\left|\frac{e^{i\theta}-1}{e^{i\theta}+1}\right|\right).$

Using the formula for the hyperbolic distance in the unit disk, we can find
that $R(\theta)=\textrm{arctanh}(|\tan(\frac{\theta}{2})|)$.

Next, we mention the following lemma.

######  Lemma 3.1 ([6]).

Let $\gamma:[0,+\infty)\to\mathbb{H}$ be a geodesic for the hyperbolic
distance in $\mathbb{H}$ such that
$\gamma([0,+\infty))=[r_{0},+\infty),\;\gamma(0)=r_{0}$, for some $r_{0}>0$.
Then, for every $R>0$, there exists $\beta\in(0,\frac{\pi}{2})$, with
$k_{\mathbb{H}}(r_{0},r_{0}e^{i\beta})=R$, such that

$S_{\mathbb{H}}(\gamma,R)=\\{\rho
e^{i\theta}:\rho>r_{0},|\theta|<\beta\\}\cup\\{z\in\mathbb{H}:k_{\mathbb{H}}(r_{0},z)<R\\}.$

Figure 2: Hyperbolic sector in the right half-plane

With the help of Lemma 3.1, we can see (Figure 2) that a hyperbolic sector in
the right half-plane around a ray on the positive semi-axis is actually a
simply connected domain bounded by a circular arc that joins the two points
$r_{0}e^{i\beta},r_{0}e^{-i\beta}$ and by the two rays $\\{\rho
e^{i\beta}:\rho\geq r_{0}\\},\\{\rho e^{-i\beta}:\rho\geq r_{0}\\}$. Also, it
is symmetric with respect to the real line. Thanks to conformal invariance,
the right half-plane and the ease in picturing its hyperbolic sectors will
play a crucial role later on.

Now, we can picture a hyperbolic sector of amplitude $R(\theta)$. Let
$\gamma:[0,+\infty)\to\mathbb{H}$ be a geodesic of the right half-plane such
that its image is a ray on the positive real axis with $\gamma(0)=r_{0}$.
Then, utilising the above lemma, for
$\theta_{0}\in(-\frac{\pi}{2},\frac{\pi}{2})\setminus\\{0\\}$ we have

$S_{\mathbb{H}}(\gamma,R(\theta_{0}))=\\{\rho
e^{i\theta}:\rho>r_{0},\;|\theta|<|\theta_{0}|\\}\cup\\{z\in\mathbb{H}:k_{\mathbb{H}}(r_{0},z)<R(\theta_{0})\\}.$

So, the sector $S_{\mathbb{H}}(\gamma,R(\theta_{0}))$ eventually contains
every ray $\\{z:\arg z=\theta\\}$, for
$\theta\in(-|\theta_{0}|,|\theta_{0}|)$. Using harmonic measure, this means
that the sector $S_{\mathbb{H}}(\gamma,R(\theta_{0}))$ eventually contains
every level set $\\{z:\omega(z,\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})=\frac{\theta+\frac{\pi}{2}}{\pi}\\}$, for
$\theta\in(-|\theta_{0}|,|\theta_{0}|)$. But this gives us some useful
information for convergence by angle. Through the Cayley transform that
corresponds $\infty$ to $1$, all the above certify that this sector eventually
contains sequences whose images converge to $1$ by angles inside
$(-|\theta_{0}|+\frac{\pi}{2},|\theta_{0}|+\frac{\pi}{2})$.

Moreover, for any $y\in\mathbb{R}$ and for
$\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$ fixed, we can easily compute that

$\lim_{\begin{subarray}{c}\zeta\to\infty\\\
\arg\zeta=\theta\end{subarray}}\omega(\zeta+iy,\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})=\frac{\theta+\frac{\pi}{2}}{\pi}.$

Of course, the ray $t+iy,\;t\geq r_{0}$, where $r_{0}>0$, is a geodesic of
$\mathbb{H}$. Considering $\gamma(t)=r_{0}+t+iy$, then the hyperbolic sector
$S_{\mathbb{H}}(\gamma,R(\theta_{0}))$ again eventually contains all the rays
$\\{z:\arg z=\theta\\}+iy$, for $\theta\in(-|\theta_{0}|,|\theta_{0}|)$.
Therefore, for any
$\theta\in(-|\theta_{0}|+\frac{\pi}{2},|\theta_{0}|+\frac{\pi}{2})$, any
sequence such that its image in $\mathbb{D}$ converges to $1$ by angle
$\theta$ is contained in some sector $S_{\mathbb{H}}(\gamma,R(\theta_{0}))$,
where $\gamma:[0,+\infty)\to\mathbb{H}$ is a geodesic of $\mathbb{H}$ such
that $\lim\limits_{t\to+\infty}\gamma(t)=\infty$. Through the conformal
invariance of the hyperbolic distance, a similar thing can be stated for any
simply connected domain.

Finally, combining all the above information, we can even predict the
following: any sequence of $\mathbb{H}$ such that its image converges to $1$
by angle-set $[\theta_{1},\theta_{2}]$ must be eventually contained in a
sector $S_{\mathbb{H}}(\gamma,R(\frac{\pi}{2}-\theta_{1}))$ and not be
eventually contained in the sector
$S_{\mathbb{H}}(\gamma,R(\frac{\pi}{2}-\theta_{2}))$ or the inverse, depending
on the angles.

So, we are in a position to define the set
$A_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$, for
$[\theta_{1},\theta_{2}]\subset(0,\pi)$, where $\gamma:[0,1)\to\mathbb{D}$ is
a geodesic in $\mathbb{D}$ such that $\lim\limits_{t\to
1^{-}}\gamma(t)=\sigma$. But first, we ought to define the auxiliary sets
$\tilde{A}_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$ as follows:

  1. (i)

if $[\theta_{1},\theta_{2}]\subset(0,\frac{\pi}{2})$, then
$\tilde{A}_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})=\left[S_{\mathbb{D}}(\gamma,R(\frac{\pi}{2}-\theta_{1}))\setminus
S_{\mathbb{D}}(\gamma,R(\frac{\pi}{2}-\theta_{2}))\right]\cap\mathbb{D}_{\sigma}^{+}$,

  2. (ii)

if $[\theta_{1},\theta_{2}]\subset(\frac{\pi}{2},\pi)$, then
$\tilde{A}_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})=\left[S_{\mathbb{D}}(\gamma,R(\frac{\pi}{2}-\theta_{2}))\setminus
S_{\mathbb{D}}(\gamma,R(\frac{\pi}{2}-\theta_{1}))\right]\cap\mathbb{D}_{\sigma}^{-}$,

  3. (iii)

if $[\theta_{1},\theta_{2}]=[\theta,\pi-\theta]$, where
$\theta\in(0,\frac{\pi}{2})$, then
$\tilde{A}_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})=S_{\mathbb{D}}(\gamma,R(\frac{\pi}{2}-\theta_{1}))=S_{\mathbb{D}}(\gamma,R(\frac{\pi}{2}-\theta_{2}))$,

  4. (iv)

if $[\theta_{1},\theta_{2}]=[\theta_{1},\frac{\pi}{2}]$, then
$\tilde{A}_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})=S_{\mathbb{D}}(\gamma,R(\frac{\pi}{2}-\theta_{1}))\cap\mathbb{D}_{\sigma}^{+}$,

  5. (v)

if $[\theta_{1},\theta_{2}]=[\frac{\pi}{2},\theta_{2}]$, then
$\tilde{A}_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})=S_{\mathbb{D}}(\gamma,R(\frac{\pi}{2}-\theta_{2}))\cap\mathbb{D}_{\sigma}^{-}$,

  6. (vi)

if $[\theta_{1},\theta_{2}]$ satisfies $\theta_{1}<\frac{\pi}{2}<\theta_{2}$,
then
$\tilde{A}_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})=\left[S_{\mathbb{D}}(\gamma,R(\frac{\pi}{2}-\theta_{1}))\cap\mathbb{D}_{\sigma}^{+}\right]\cup\left[S_{\mathbb{D}}(\gamma,R(\frac{\pi}{2}-\theta_{2}))\cap\mathbb{D}_{\sigma}^{-}\right]$.

Finally, for any of the above six cases, we define

$A_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2}):=\overline{\tilde{A}_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})}\cap\mathbb{D},$

where the overline signifies the closure of the set. In other words, the set
$A_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$ is just the closure of the
auxiliary set $\tilde{A}_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$ minus the
point $\sigma$. Obviously, the definition of the sets
$\tilde{A}_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$ is not really needed,
but we try to avoid the repetitive use of the closure in all of the six cases.

|  |
---|---
|
|

(a) ${[\theta_{1},\theta_{2}]}\subset(0,\frac{\pi}{2})$

(b) ${[\theta_{1},\theta_{2}]}\subset(\frac{\pi}{2},\pi)$

(c) ${[\theta_{1},\theta_{2}]}={[\theta,\pi-\theta]}$

(d) ${[\theta_{1},\theta_{2}]}={[\theta_{1},\frac{\pi}{2}]}$

(e) ${[\theta_{1},\theta_{2}]}={[\frac{\pi}{2},\theta_{2}]}$

(f) $\theta_{1}<\frac{\pi}{2}<\theta_{2}$

Figure 3: The set $A_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$

In Figure 3, we have chosen the radius tending to $\sigma$ as the geodesic
$\gamma$ and we present all the different versions of the set
$A_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$ depending on the angles
$\theta_{1},\theta_{2}$.

We now give the corresponding general definitions. Let
$\Delta\subsetneq\mathbb{C}$ be a simply connected domain,
$f:\mathbb{D}\to\Delta$ a Riemann map and $\gamma:[0,+\infty)\to\Delta$ a
geodesic for the hyperbolic distance in $\Delta$. We can define the set
$\tilde{A}_{\Delta}(\gamma,\theta_{1},\theta_{2})$ in a similar manner, using
this time the hyperbolic distance in $\Delta$, corresponding hyperbolic
sectors and the images
$f(\mathbb{D}_{\sigma}^{+}),f(\mathbb{D}_{\sigma}^{-})$, where this time,
$\sigma$ is the point at which the preimage $f^{-1}\circ\gamma$ intersects the
unit circle or equivalently
$\lim\limits_{t\to+\infty}f^{-1}(\gamma(t))=\sigma$. For example, for
$[\theta_{1},\theta_{2}]\subset(0,\frac{\pi}{2})$, we have

$\tilde{A}_{\Delta}(\gamma,\theta_{1},\theta_{2})=\left[S_{\Delta}(\gamma,R(\frac{\pi}{2}-\theta_{1}))\setminus
S_{\Delta}(\gamma,R(\frac{\pi}{2}-\theta_{2}))\right]\cap
f(\mathbb{D}_{\sigma}^{+}),$

and then
$A_{\Delta}(\gamma,\theta_{1},\theta_{2})=\overline{\tilde{A}_{\Delta}(\gamma,\theta_{1},\theta_{2})}\cap\Delta$.

######  Definition 3.1.

Let $\Delta\subsetneq\mathbb{C}$ be a simply connected domain and
$\\{z_{n}\\}$ a sequence in $\Delta$. We say that $\\{z_{n}\\}$ exhausts the
set $A_{\Delta}(\gamma,\theta_{1},\theta_{2})$, where
$\gamma:[0,+\infty)\to\Delta$ is a geodesic for the hyperbolic distance in
$\Delta$ and $\theta_{1},\theta_{2}\in(0,\pi)$ if:

  1. (i)

all the sets $A_{\Delta}(\gamma,\omega_{1},\omega_{2})$, where
$\omega_{1}<\theta_{1},\omega_{2}>\theta_{2}$, eventually contain
$\\{z_{n}\\}$

  2. (ii)

for any $\theta\in[\theta_{1},\theta_{2}]$ and any
$\epsilon\in(0,\min\\{\theta_{1},\pi-\theta_{2}\\})$, there is a subsequence
of $\\{z_{n}\\}$ that is eventually contained in
$A_{\Delta}(\gamma,\theta-\epsilon,\theta+\epsilon)$.

This definition extends naturally to continuous curves as well.

Next, we need the following proposition. During its proof, by $[z_{1},z_{2}]$
we will mean the line segment joining the complex numbers $z_{1},z_{2}$.

######  Proposition 3.1.

Let $\Delta\subsetneq\mathbb{C}$ be a simply connected domain and
$f:\mathbb{D}\to\Delta$ a Riemann map. Suppose that
$\mathbb{H}+a\subset\Delta\subseteq\mathbb{H}$, for some $a>0$. Let
$\delta:[0,+\infty)\to\Delta$ be a continuous curve in $\Delta$ with
$\lim\limits_{t\to+\infty}\delta(t)=\infty$. Suppose that $\delta$ exhausts
$A_{\mathbb{H}}(\gamma,\theta_{1},\theta_{2})$, where
$\gamma:[0,+\infty)\to\mathbb{H}$ is a geodesic for the hyperbolic distance in
$\mathbb{H}$ such that $\gamma(0)=r_{0}>a$ and
$\lim\limits_{t\to+\infty}\gamma(t)=\infty$. Then, there exists a
$\sigma\in\partial\mathbb{D}$ such that $f^{-1}(\delta(t))$ converges by
angle-set $[\theta_{1},\theta_{2}]$ to $\sigma$, as $t\to+\infty$.

###### Proof.

Let $f$ be a Riemann map for $\Delta$. By Carathéodory’s Theorem, $f$ induces
a homeomorphism
$\tilde{f}:\mathbb{D}\cup\partial_{C}\mathbb{D}\to\Delta\cup\partial_{C}\Delta$.
Also, there is a one-to-one correspondence between prime ends of the unit disk
and points of the unit circle. Moreover, even though there might exist more
than one prime ends of $\Delta$ corresponding to the point at infinity, there
is a unique one that is determined by any chain of crosscuts that extends to
the half-plane $\mathbb{H}+a$. Suppose that this unique prime end corresponds
through $f^{-1}$ to a point $\sigma\in\partial\mathbb{D}$. Then, since
$\delta([0,+\infty))$ is eventually contained in $\mathbb{H}+a$, it is
necessary that $f^{-1}(\delta(t))\to\sigma$, as $t\to+\infty$. All that
remains is to verify that the convergence is indeed by angle-set
$[\theta_{1},\theta_{2}]$.

Since the geometry of $A_{\mathbb{H}}(\gamma,\theta_{1},\theta_{2})$ depends
firmly on the selected angles, we will provide the proof in the case when
$[\theta_{1},\theta_{2}]\subset(0,\frac{\pi}{2})$. In the rest of the cases,
the proof is almost the same, albeit with some minor modifications. Our
hypotheses directly imply that the image of the geodesic $\gamma$ is the
subset $[r_{0},+\infty)$ of the real line. Using Lemma 3.1, we find that the
boundary of $A_{\mathbb{H}}(\gamma,\theta_{1},\theta_{2})$ consists of the two
rays $\\{re^{i(\theta_{1}-\frac{\pi}{2})}:r\geq
r_{0}\\},\;\\{re^{i(\theta_{2}-\frac{\pi}{2})}:r\geq r_{0}\\}$, the two lower
halves $B_{1},B_{2}$ of the circular arcs that join the points
$r_{0}e^{i(\theta_{1}-\frac{\pi}{2})},r_{0}e^{i(\frac{\pi}{2}-\theta_{1})}$
and
$r_{0}e^{i(\theta_{2}-\frac{\pi}{2})},r_{0}e^{i(\frac{\pi}{2}-\theta_{2})}$
respectively and a segment $[r_{1},r_{2}]\subset\mathbb{R}$, with
$0<r_{1}<r_{2}<r_{0}$, where $\\{r_{1}\\}=B_{1}\cap\mathbb{R}$ and
$\\{r_{2}\\}=B_{2}\cap\mathbb{R}$.

Figure 4: The set $\Gamma$

Without loss of generality, we can suppose that $a=1$ and that
$\partial\Delta\cap\partial\mathbb{H}=\partial\Delta\cap(\partial\mathbb{H}+1)=\emptyset$
(in the case where $\Delta=\mathbb{H}$, the proof becomes much more
straightforward).

Set $x_{0}=\inf\\{x\in\mathbb{R}\cap\partial\Delta\\}$. Since
$\partial\Delta\cap\partial\mathbb{H}=\emptyset$ and therefore
$\Delta\subsetneq\mathbb{H}$, such a $x_{0}$ exists and, in fact,
$x_{0}\in(0,1)$. Moreover, there exists a connected component $L$ of
$\partial\Delta$ such that $x_{0}\in L,\inf\\{\Im z:z\in L\\}=-\infty$ and
$\sup\\{\Im z:z\in L\\}=+\infty$. If either $\inf\\{\Im z:z\in
L\\}\in(-\infty,0]$ or $\sup\\{\Im z:z\in L\\}\in[0,+\infty)$, then either
$\inf\\{x\in\mathbb{R}\cap\partial\Delta\\}<x_{0}$ or
$(-\infty,x_{0})\subset\Delta$. Both cases lead to a contradiction and
consequently there must exist such a component $L$. Therefore, there is a
prime end $\xi$ of $\Delta$ corresponding to $x_{0}$ (if there are more than
one, we just choose one randomly). It is possible that the impression $I[\xi]$
of this prime end is not a singleton and that $x_{0}$ is not accessible. In
this case, since $I[\xi]\subset L$, we slightly change the construction and
take as $x_{0}$ an accessible point belonging to $I[\xi]$ (for more on prime
ends see [17, Chapter 9]). As a result, through $f^{-1}$, the point
$x_{0}\in\partial\Delta$ corresponds to a point
$\tau\in\partial\mathbb{D}\setminus\\{\sigma\\}$.

Consider in the unit disk $\mathbb{D}$ the set
$\Gamma=(\tau,f^{-1}(1)]\cup[f^{-1}(1),\sigma)$ (see Figure 4). Obviously,
$\Gamma$ separates the unit circle $\partial\mathbb{D}$ into two connected
arcs. Passing to $\Delta$ through $f$, it is clear that $f(\Gamma)$ joins
$x_{0}$ and $1$ and separates the prime ends of $\Delta$ into two sets
$\partial_{C}\Delta^{+}$ and $\partial_{C}\Delta^{-}$. Denote as
$\partial_{C}\Delta^{+}$ the subset of $\partial_{C}\Delta$ that contains the
prime end of $\Delta$ determined by any chain of crosscuts with one end on the
upper half of $L$ (i.e. the subset of $L$ with one end on $x_{0}$ and the
other at infinity, but with imaginary parts tending to $+\infty$). Then,
$\partial_{C}\Delta^{-}=\partial_{C}\Delta\setminus\partial_{C}\Delta^{+}$. We
denote by $\partial\Delta^{+}$ and $\partial\Delta^{-}$ the sets of points of
$\partial\Delta$ that correspond to the prime ends of $\partial_{C}\Delta^{+}$
and $\partial_{C}\Delta^{-}$ respectively. By our construction,
$\partial\Delta^{+}$ and $\partial\Delta^{-}$ correspond through $f^{-1}$ to
circular arcs of $\partial\mathbb{D}$ with common ends.

Figure 5: The sets of Proposition 3.1

For the rest of the proof we denote by $D_{1}$ the simply connected domain
bounded by $\\{z:\arg z=\frac{\pi}{2}\\}\cup[0,x_{0}]\cup
f((\tau,f^{-1}(1)])\cup[\\{z:\arg z=-\frac{\pi}{2}\\}+1]$ and by $D_{2}$ the
simply connected domain bounded by $\\{z:\arg
z=-\frac{\pi}{2}\\}\cup[0,x_{0}]\cup f((\tau,f^{-1}(1)])\cup[\\{z:\arg
z=\frac{\pi}{2}\\}+1]$. Then, we set $\Delta_{1}:=D_{1}\cap\Delta$ and
$\Delta_{2}:=D_{2}\cap\Delta$ (see Figure 5). By the construction above, both
$\Delta_{1}$ and $\Delta_{2}$ are simply connected domains.

Now, we will make use of the harmonic measure to prove that the convergence is
indeed by angle-set $[\theta_{1},\theta_{2}]$. By our hypothesis, there
certainly exists a $t_{0}>0$, such that $\Re(\delta(t))>1$, for all $t>t_{0}$.
Then, by the monotonicity property of the harmonic measure, we see that

$\omega(\delta(t),\partial\Delta^{+},\Delta_{1})\leq\omega(\delta(t),\partial\Delta^{+},\Delta),$

for all $t>t_{0}$. Moreover, by the maximum principle for harmonic functions,

$\omega(\delta(t),\partial\Delta^{+},\Delta_{1})\geq\omega(\delta(t),\\{\arg
z=\frac{\pi}{2}\\},D_{1}),$

for all $t>t_{0}$. Combining the two and letting $t\to+\infty$, we find that

$\limsup\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{+},\Delta)\geq\limsup\limits_{t\to+\infty}\omega(\delta(t),\\{\arg
z=\frac{\pi}{2}\\},D_{1}).$ (1)

Then, we need to evaluate the quantity $\omega(\delta(t),\\{\arg
z=\frac{\pi}{2}\\},D_{1})$. Since $D_{1}\subset\mathbb{H}$, by the Strong
Markov Property, for all $t>t_{0}$,

$\displaystyle\omega(\delta(t),\\{\arg z=\frac{\pi}{2}\\},\mathbb{H})$
$\displaystyle=$ $\displaystyle\omega(\delta(t),\\{\arg
z=\frac{\pi}{2}\\},D_{1})+$ $\displaystyle+$
$\displaystyle\int\limits_{\partial
D_{1}\setminus\partial\mathbb{H}}\omega(s,\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})\cdot\omega(\delta(t),ds,D_{1}).$

Of course, $\partial D_{1}\setminus\partial\mathbb{H}=(0,x_{0}]\cup
f((\tau,f^{-1}(1)])\cup[\\{z:\arg z=-\frac{\pi}{2}\\}+1]$. Consequently, the
last integral can be broken into the two integrals

$I_{1}(t)=\int\limits_{(0,x_{0}]\cup f((\tau,f^{-1}(1)])}\omega(s,\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})\cdot\omega(\delta(t),ds,D_{1}),$

$I_{2}(t)=\int\limits_{[\\{\arg z=-\frac{\pi}{2}\\}+1]}\omega(s,\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})\cdot\omega(\delta(t),ds,D_{1}).$

We remind that these integrals are valid for $t>t_{0}$. Since the harmonic
measure has $1$ as an upper bound, it is directly computed that

$I_{1}(t)\leq\omega(\delta(t),(0,x_{0}]\cup f((\tau,f^{-1}(1)]),D_{1}).$

Without loss of generality, we can assume that there exists a line segment $A$
that joins $1$ with $\\{z:\arg z=\frac{\pi}{2}\\}$ such that $A\subset D_{1}$
(except, of course, for the endpoints). In case $f(\Gamma)$ reaches $1$
tangentially with respect to the vertical line $\\{z:\Re z=1\\}$ or even
intersects $\mathbb{H}+1$, we can simply consider another $a>1$ such that
$f((\tau,f^{-1}(a)])$ does not intersect $\mathbb{H}+a$ and is not tangent to
the vertical line $\\{z:\Re z=a\\}$. This is possible, because
$\tau\neq\sigma$ and the sets $f^{-1}(\mathbb{H}+a)$ are decreasing as
$a\to+\infty$, with
$\bigcap\limits_{a>1}\overline{f^{-1}(\mathbb{H}+a)}=\sigma$. Therefore, we
can find a suitable $a>1$ such that, in the unit disk $\mathbb{D}$, the line
segment $(\tau,f^{-1}(a)]$ does not intersect $f^{1}(\mathbb{H}+a)$ and is not
tangent to $f^{1}(\mathbb{H}+a)$ at the point $f^{-1}(a)$. Then, we take
$\Gamma=(\tau,f^{-1}(a)]\cup[f^{-1}(a),\sigma)$ and the construction of the
line segment $A$ becomes well defined, while the proof can be executed for
$\mathbb{H}+a\subset\Delta\subseteq\mathbb{H}$ without altering the result.

Denote by $H_{1}$ the simply connected domain bounded by $\\{z:\arg
z=\frac{\pi}{2}\\},A$ and $[\\{z:\arg z=-\frac{\pi}{2}\\}+1]$ (see Figure 6).
Obviously $H_{1}\subset D_{1}$. Then, by the maximum principle for harmonic
functions,

$\omega(\delta(t),(0,x_{0}]\cup
f((\tau,f^{-1}(1)]),D_{1})\leq\omega(\delta(t),A,H_{1}),$

for all $t>t_{0}$. We can see that

$\displaystyle\omega(\delta(t),A,H_{1})$ $\displaystyle=$
$\displaystyle\omega(\delta(t)-1,A-1,H_{1}-1)$ $\displaystyle\leq$
$\displaystyle\omega(\delta(t)-1,A-1,\\{-\frac{\pi}{2}<\arg z<\phi\\}),$

where $\phi$ is the angle that $A$ forms with the real axis. Using
consecutively the conformal mappings
$g_{1}(z)=e^{i(\frac{\pi}{4}-\frac{\phi}{2})}z$,
$g_{2}(z)=z^{\frac{2\pi}{\pi+2\phi}}$ and $g_{3}(z)=iz$, we map
$\\{z:-\frac{\pi}{2}<\arg z<\phi\\}$ to the upper half-plane. It is easily
checked that $g_{3}\circ g_{2}\circ g_{1}$ maps $A-1$ onto a line segment
$[\alpha,\beta]$ on the negative semi-axis and $\delta(t)-1$ to
$\delta_{1}(t)=i[(\delta(t)-1)e^{i(\frac{\pi}{4}-\frac{\phi}{2})}]^{\frac{2\pi}{\pi+2\phi}}$.
Then, by the conformal invariance of the harmonic measure,

$\omega(\delta(t)-1,A-1,\\{-\frac{\pi}{2}<\arg
z<\phi\\})=\omega(\delta_{1}(t),[\alpha,\beta],\\{\Im z>0\\}).$

Combining all the above, we get

$\limsup\limits_{t\to+\infty}I_{1}(t)\leq\limsup\limits_{t\to+\infty}\omega(\delta_{1}(t),[\alpha,\beta],\\{\Im
z>0\\}).$

Figure 6: Construction of $A$ and $H_{1}$

Since $\delta_{1}(t)$ is eventually contained in an angle and converges to
$\infty$, by our formula the last limit actually exists and is equal to 0,
which leads to $\lim\limits_{t\to+\infty}I_{1}(t)=0$.

Next, we move to $I_{2}(t)$. Let $s\in[\\{\arg z=-\frac{\pi}{2}\\}+1]$ or
$s=1-iy$, for some $y>0$. Again, by a known formula,

$\omega(s,\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})=\frac{\arg(1-iy)+\frac{\pi}{2}}{\pi},$

with the last quantity tending to 0 as $y\to+\infty$. Let $\epsilon>0$.
Because of the limit, there is a $y_{0}>0$ such that

$\omega(1-iy,\\{\arg z=\frac{\pi}{2}\\},\mathbb{H})<\epsilon,$

for all $y>y_{0}$. Then,

$\displaystyle I_{2}(t)$ $\displaystyle\leq$
$\displaystyle\int\limits_{[1,1-iy_{0}]}\omega(s,\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})\cdot\omega(\delta(t),ds,D_{1})+\epsilon$
$\displaystyle\leq$
$\displaystyle\omega(\delta(t),[1,1-iy_{0}],D_{1})+\epsilon.$

With the exact same technique as before, we use a conformal mapping onto the
upper half-plane to show that

$\lim\limits_{t\to+\infty}\omega(\delta(t),[1,1-iy_{0}],D_{1})=0.$

This means that $\limsup\limits_{t\to+\infty}I_{2}(t)\leq\epsilon$, for all
positive $\epsilon$, something that in turn leads to
$\lim\limits_{t\to+\infty}I_{2}(t)=0$. Returning to (2) and remembering (1),
we get

$\displaystyle\limsup\limits_{t\to+\infty}\omega(\delta(t),\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})$ $\displaystyle=$
$\displaystyle\limsup\limits_{t\to+\infty}\omega(\delta(t),\\{\arg
z=\frac{\pi}{2}\\},D_{1})$ $\displaystyle\leq$
$\displaystyle\limsup\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{+},\Delta).$

The same is obviously true for the $\liminf$. We know from the given formulas
that for every $\zeta$ on the ray $\\{\arg z=\theta\\}$, the harmonic measure
$\omega(\zeta,\\{\arg z=\frac{\pi}{2}\\},\mathbb{H})$ remains constant and
equal to $\frac{\theta+\frac{\pi}{2}}{\pi}$. Given that $\delta$ is a
continuous curve that exhausts $A_{\mathbb{H}}(\gamma,\theta_{1},\theta_{2})$
and by the shape of $A_{\mathbb{H}}(\gamma,\theta_{1},\theta_{2})$, it follows
that $\liminf\limits_{t\to+\infty}\arg(\delta(t))=\theta_{1}-\frac{\pi}{2}$
and $\limsup\limits_{t\to+\infty}\arg(\delta(t))=\theta_{2}-\frac{\pi}{2}$. As
a result,

$\displaystyle\limsup\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{+},\Delta)$
$\displaystyle\geq$
$\displaystyle\limsup\limits_{t\to+\infty}\omega(\delta(t),\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})$ $\displaystyle=$
$\displaystyle\limsup\limits_{t\to+\infty}\frac{\arg\delta(t)+\frac{\pi}{2}}{\pi}$
$\displaystyle=$ $\displaystyle\frac{\theta_{2}}{\pi}.$

Likewise,

$\liminf\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{+},\Delta)\geq\frac{\theta_{1}}{\pi}.$

Following a similar process, but this time with the use of $\\{z:\arg
z=-\frac{\pi}{2}\\}$, $\partial\Delta^{-}$, $D_{2}$, $\Delta_{2}$, we find
that

$\limsup\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{-},\Delta)\geq\frac{\pi-\theta_{1}}{\pi},$
$\liminf\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{-},\Delta)\geq\frac{\pi-\theta_{2}}{\pi}.$

However, since harmonic measure is always at most $1$, we have

$\limsup\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{+},\Delta)+\liminf\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{-},\Delta)\leq
1=\frac{\theta_{2}}{\pi}+\frac{\pi-\theta_{2}}{\pi},$
$\liminf\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{+},\Delta)+\limsup\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{-},\Delta)\leq
1=\frac{\theta_{1}}{\pi}+\frac{\pi-\theta_{1}}{\pi}.$

Consequently, all inequalities actually become equalities. Since $\delta$ is
continuous, it is necessary that the cluster set of
$\omega(\delta(t),\partial\Delta^{+},\Delta)$ is equal to
$[\frac{\theta_{1}}{\pi},\frac{\theta_{2}}{\pi}]$. Through $f^{-1}$, the set
$\partial\Delta^{+}$ corresponds to a circular arc $B$ of $\partial\mathbb{D}$
with one end at $\sigma$ and the other at $\tau$. By conformal invariance,

$\liminf\limits_{t\to+\infty}\omega(f^{-1}(\delta(t)),B,\mathbb{D})=\frac{\theta_{1}}{\pi},$
$\limsup\limits_{t\to+\infty}\omega(f^{-1}(\delta(t)),B,\mathbb{D})=\frac{\theta_{2}}{\pi}.$

We are now done, because in view of Remark 2.1, the last two equalities imply
that as $t\to+\infty$, $f^{-1}(\delta(t))$ converges to $\sigma$ by angle-set
$[\theta_{1},\theta_{2}]$. ∎

With some minor changes, a similar proof can be executed for any horizontal
geodesic of the right half-plane, since vertical translations are conformal
automorphisms of $\mathbb{H}$ and so, they do not affect harmonic measure. We
can think of it as a parallel move that does not affect the angles.

Figure 7: Hyperbolic sector in the unit disk

The last step, before proving the main theorem of this section, is to talk
about hyperbolic sectors in the unit disk. Specifically, we will make some
simple geometric considerations concerning hyperbolic sectors around a
diameter. We can suppose that the diameter is the segment $(-1,1)$. Then a
hyperbolic sector around this diameter (see Figure 7) is bounded by two
circular arcs that join $-1$ and $1$ and intersect the unit circle with angles
$\theta$ and $\pi-\theta$, where $\theta\in(0,\frac{\pi}{2})$. Restricting
ourselves to a part of the diameter, for example the radius tending to 1, it
is easy to see that the corresponding hyperbolic sector is bounded by two
circular arcs $B_{1},B_{2}$ that tend to $1$ and a third circular arc that
joins $B_{1}$ and $B_{2}$ and is symmetric with respect to $(-1,1)$.
Therefore, such a hyperbolic sector is eventually equivalent to a Stolz region
and contains sequences that can converge to $1$ only by angle-set that is a
subset of $(\theta,\pi-\theta)$. Keeping this in mind, we can understand how
the sets $A_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$ and the sequences that
exhaust them behave.

We are now ready to prove our main result. For the sake of convenience, we
will state the result again.

######  Theorem 1.1.

Let $\Delta\subsetneq\mathbb{C}$ be a simply connected domain and
$f:\mathbb{D}\to\Delta$ a Riemann map. Let $\\{z_{n}\\}\subset\Delta$ be a
sequence with no accumulation points in $\Delta$. Then, there exists a
$\sigma\in\partial\mathbb{D}$ such that $\\{f^{-1}(z_{n})\\}$ converges by
angle-set $[\theta_{1},\theta_{2}]\subset(0,\pi)$ to $\sigma$ if and only if
there exist a simply connected domain $U\subsetneq\mathbb{C}$, a prime end
$\xi\in\partial_{C}U$, $R>0$ and a geodesic $\gamma:[0,+\infty)\to U$ for the
hyperbolic distance in $U$ with $\lim\limits_{t\to+\infty}\gamma(t)=\xi$ in
the Carathéodory topology of $U$ such that

  1. (i)

$E_{U}(\xi,R)\subset\Delta\subseteq U,$

  2. (ii)

$\gamma(0)\in\Delta,$

  3. (iii)

$\\{z_{n}\\}$ exhausts the set $A_{U}(\gamma,\theta_{1},\theta_{2}).$

###### Proof.

“$\implies$” First, suppose that $\\{f^{-1}(z_{n})\\}$ converges by angle-set
$[\theta_{1},\theta_{2}]$ to some $\sigma\in\partial\mathbb{D}$. By conformal
invariance, we can take $\Delta=U=\mathbb{D},\xi=\sigma$ and
$f=id_{\mathbb{D}}$. Therefore, condition (i) is obviously satisfied, for all
$R>0$. Moreover, now $\\{f^{-1}(z_{n})\\}\equiv\\{z_{n}\\}$, so $\\{z_{n}\\}$
converges by angle set $[\theta_{1},\theta_{2}]$ to $\sigma$. Consider
$\gamma:[0,1)\to\mathbb{D}$ to be part of the radius that ends at $\sigma$.
Condition (ii) is satisfied as well. Finally, by the way that the numbers
$R_{\mathbb{D}}(\frac{\pi}{2}-\theta_{i}),\;i=1,2,$ were chosen for the
definition of the set $A_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$, the
equivalence between hyperbolic sectors and Stolz regions and the geometry of
the hyperbolic sectors around a radius, condition (iii) holds too.
“$\impliedby$” Conversely, assume that conditions (i),(ii) and (iii) are
satisfied. By conformal invariance, we can suppose that $U=\mathbb{H}$ and
$\xi$ is the prime end of $\infty$ determined by the positive semi-axis. Then,
there is an $a>0$ depending on $R$, such that
$E_{\mathbb{H}}(\infty,R)=\mathbb{H}+a$. Therefore, condition (i) becomes
$\mathbb{H}+a\subset\Delta\subseteq\mathbb{H}$. Also, the geodesic $\gamma$ of
$\mathbb{H}$ satisfies $\lim\limits_{t\to+\infty}\gamma(t)=\infty$, which
means that $\gamma$ must be a horizontal ray. Then, by the definition of
exhaustion and Proposition 3.1, condition (iii) ensures that
$\\{f^{-1}(z_{n})\\}$ converges by angle-set $[\theta_{1},\theta_{2}]$ to some
$\sigma\in\partial\mathbb{D}$, as $t\to+\infty$. ∎

######  Remark 3.1.

Examining the hyperbolic sectors around horizontal geodesics of $\mathbb{H}$,
we can see that the existence of a suitable geodesic in the statement of
Theorem 1.1 is equivalent to stating that the result is true for every such
geodesic.

######  Remark 3.2.

Let $\theta\in(0,\pi)$. Suppose that $\\{z_{n}\\}$ is a sequence of points in
the unit disk $\mathbb{D}$ with no accumulation points inside $\mathbb{D}$.
Let $\sigma\in\partial\mathbb{D}$ and $\gamma:[0,+\infty)\to\mathbb{D}$ be the
radius tending to $\sigma$ (or, even, any geodesic for the hyperbolic distance
in the unit disk such that $\lim\limits_{t\to+\infty}\gamma(t)=\sigma$). We
assume that $\\{z_{n}\\}$ is eventually contained in
$A_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$, for all
$\theta_{1}\in(0,\theta)$ and all $\theta_{2}\in(\theta,\pi)$. Then, with the
help of the conformal invariance of the harmonic measure and our formulas for
the right half-plane, it is clear to see that $\\{z_{n}\\}$ necessarily
converges to $\sigma$ and in fact, by angle $\theta$.

Conversely, if $\\{z_{n}\\}$ converges to $\sigma$ by angle $\theta$, then it
is easily verified that $\\{z_{n}\\}$ is eventually contained in every set
$A_{\mathbb{D}}(\gamma,\theta_{1},\theta_{2})$, for all
$\theta_{1}\in(0,\theta)$ and all $\theta_{2}\in(\theta,\pi)$.

Through a Riemann map, we can extend Remark 3.2 to any simply connected domain
$\Delta\subsetneq\mathbb{C}$ in a suitable way. Therefore, a direct corollary
of Theorem 1.1 is the following.

######  Corollary 3.1.

Let $\Delta\subsetneq\mathbb{C}$ be a simply connected domain and
$f:\mathbb{D}\to\Delta$ a Riemann map. Let $\\{z_{n}\\}\subset\Delta$ be a
sequence with no accumulation points in $\Delta$. Then, there exists a
$\sigma\in\partial\mathbb{D}$ such that $\\{f^{-1}(z_{n})\\}$ converges by
angle $\theta\in(0,\pi)$ to $\sigma$ if and only if there exist a simply
connected domain $U\subsetneq\mathbb{C}$, a prime end
$\xi\in\partial_{C}U,R>0$ and a geodesic $\gamma:[0,+\infty)\to U$ for the
hyperbolic distance in $U$ with $\lim\limits_{t\to+\infty}\gamma(t)=\xi$ in
the Carathéodory topology of $U$ such that

  1. (i)

$E_{U}(\xi,R)\subset\Delta\subseteq U$,

  2. (ii)

$\gamma(0)\in\Delta$,

  3. (iii)

$\\{z_{n}\\}$ is eventually contained in the set
$A_{U}(\gamma,\theta_{1},\theta_{2})$, for all $\theta_{1}\in(0,\theta)$ and
all $\theta_{2}\in(\theta,\pi)$.

Suppose, as above, that $\\{z_{n}\\}$ is eventually contained in the set
$A_{U}(\gamma,\theta_{1},\theta_{2})$, for all
$\theta_{1}\in(0,\frac{\pi}{2})$ and all $\theta_{2}\in(\frac{\pi}{2},\pi)$.
Going back to the definition of $A_{U}(\gamma,\theta_{1},\theta_{2})$ and
keeping in mind that $R(0)=0$, our assumption directly implies that
$\lim\limits_{n\to+\infty}k_{U}(z_{n},\gamma)=0$. On the other hand, one can
see that if $\lim\limits_{n\to+\infty}k_{U}(z_{n},\gamma)=0$, then
$\\{z_{n}\\}$ is eventually contained in every hyperbolic sector around
$\gamma$. Thus, it is also eventually contained in every set of the form
$A_{U}(\gamma,\theta_{1},\theta_{2})$, where
$0<\theta_{1}<\frac{\pi}{2}<\theta_{2}<\pi$. As a result, in the case that
$\theta=\frac{\pi}{2}$ and the convergence is orthogonal, Corollary 3.1 is
actually equivalent with Theorem 1.1 in [5] (for its statement see Section 1).
Thus, Corollary 3.1 is a generalization for the convergence of a sequence by a
certain angle, providing a necessary and sufficient condition.

######  Remark 3.3.

As we mentioned in the Introduction, the idea to tackle the problem of
convergence by angle or angle-set in a simply connected domain $\Delta$ is the
use of a larger simply connected domain $U$ that has a simpler geometry and is
easier to work with. A natural question that arises after the proof of Theorem
1.1 and Corollary 3.1 is if the behavior of the sequence $\\{z_{n}\\}$ is the
same with respect to both $\Delta$ and $U$. In particular, consider a Riemann
map $f_{\Delta}:\mathbb{D}\to\Delta$ and a Riemann map $f_{U}:\mathbb{D}\to U$
and suppose that $E_{U}(\xi,R)\subset\Delta\subseteq U$, for some prime end
$\xi\in\partial_{C}U$ and some $R>0$. Since $\Delta$ contains such a horodisk
of $U$, it follows that $\xi\in\partial_{C}\Delta$ as well. Suppose that $\xi$
corresponds through $f_{\Delta}^{-1}$ to a point
$\sigma_{\Delta}\in\partial\mathbb{D}$ and through $f_{U}^{-1}$ to a point
$\sigma_{U}\in\partial\mathbb{D}$. Then, is it true that
$\\{f_{\Delta}^{-1}(z_{n})\\}$ converges by angle-set
$[\theta_{1},\theta_{2}]$ (or by angle $\theta$) to $\sigma_{\Delta}$ if and
only if $\\{f_{U}^{-1}(z_{n})\\}$ converges by angle-set
$[\theta_{1},\theta_{2}]$ (or by angle $\theta$) to $\sigma_{U}$? Actually,
the answer to this question is positive. By conformal invariance, we can
assume that $U=\mathbb{H}$ and that $\xi$ is the unique prime end of
$\mathbb{H}$ corresponding to $\infty$. Then, following the proof of
Proposition 3.1, we remember that in the end,

$\limsup\limits_{t\to+\infty}\omega(\delta(t),\\{\arg
z=\frac{\pi}{2}\\},\mathbb{H})=\limsup\limits_{t\to+\infty}\omega(\delta(t),\partial\Delta^{+},\Delta).$

The same is true for the liminf as well. Moreover, the positive semi-axis
separates $\partial\mathbb{H}$ into the two connected components $\\{\arg
z=\frac{\pi}{2}\\}$ and $\\{\arg z=-\frac{\pi}{2}\\}$, which through
$f_{\mathbb{H}}^{-1}$ correspond to two circular arcs $B_{\mathbb{H}}^{+}$ and
$B_{\mathbb{H}}^{-}$ respectively, that meet at $\sigma_{\mathbb{H}}$.
Therefore, by the conformal invariance of the harmonic measure, we have

$\limsup\limits_{t\to+\infty}\omega(f_{\mathbb{H}}^{-1}(\delta(t)),B_{\mathbb{H}}^{+},\mathbb{D})=\limsup\limits_{t\to+\infty}\omega(f_{\Delta}^{-1}(\delta(t)),B,\mathbb{D}).$

Again, the same is true for the liminf. As a result, in view of Remark 2.1, we
get the desired result for the continuous curve $\delta$. Thus, the analogous
result is true for the sequence $\\{z_{n}\\}$ as well.

##  4 Applications to Semigroups

In this last part of the article, we connect the above work with the study of
continuous semigroups of holomorphic functions of the unit disk (or from now
on semigroups in $\mathbb{D}$) and examine a consequence of Proposition 3.1. A
semigroup in $\mathbb{D}$ is a family of holomorphic functions
$\phi_{t}:\mathbb{D}\to\mathbb{D},t\geq 0,$ with the properties:

  1. (i)

$\phi_{0}$ is the identity in $\mathbb{D}$,

  2. (ii)

$\phi_{t+s}=\phi_{t}\circ\phi_{s}$, for all $t,s\geq 0$,

  3. (iii)

$\lim\limits_{t\to s}\phi_{t}(z)=\phi_{s}(z)$, for all $s\geq 0$ and all
$z\in\mathbb{D}$.

An extensive presentation of the theory of semigroups can be found in the
books [1],[4],[15].

When the functions of the semigroup $(\phi_{t})$ do not have any fixed points,
then $(\phi_{t})$ is called non-elliptic. For a non-elliptic semigroup
$(\phi_{t})$ in $\mathbb{D}$, there exists a unique point
$\tau\in\partial\mathbb{D}$, namely the Denjoy-Wolff point of the semigroup,
such that for every $z\in\mathbb{D}$,

$\lim\limits_{t\to+\infty}\phi_{t}(z)=\tau.$

Usually, we assume without loss of generality, that the Denjoy-Wolff point is
$1$.

An important tool connected to the study of a semigroup $(\phi_{t})$ is the
Koenigs function $h$ of the semigroup. This function is the unique conformal
mapping $h:\mathbb{D}\to\mathbb{C}$ satisfying $h(0)=0$ and

$\phi_{t}(z)=h^{-1}(h(z)+t),$

for all $t\geq 0$ and all $z\in\mathbb{D}$. The simply connected domain
$\Omega=h(\mathbb{D})$ is called the associated planar domain of the semigroup
and plays a great role in the classification of semigroups. It is convex in
the positive direction (also known as starlike at infinity), namely
$\Omega+t\subset\Omega$, for every $t\geq 0$.

Going over to the classification of semigroups, a semigroup $(\phi_{t})$ is
called hyperbolic if its associated planar domain $\Omega$ is contained in a
horizontal strip. Otherwise, it is called parabolic. In addition, we say that
a parabolic semigroup is of zero hyperbolic step if for some (equivalently for
every) $s>0$ and some (equivalently every) $z\in\mathbb{D}$,

$\lim\limits_{t\to+\infty}k_{\mathbb{D}}(\phi_{t}(z),\phi_{t+s}(z))=0.$

If this limit is always positive, then we say that the parabolic semigroup is
of positive hyperbolic step. Consequently, we can divide non-elliptic
semigroups in three classes: hyperbolic, parabolic of zero hyperbolic step and
parabolic of positive hyperbolic step. For a study on the classification of
semigroups in terms of the geometry of $\Omega$, we refer to [2],[4, Chapter
9].

An interesting topic in the study of semigroups in $\mathbb{D}$ is the so-
called “slope” problem. In this problem, we examine the trajectories of the
semigroups (namely the sets $\\{\phi_{t}(z):t\geq 0\\},z\in\mathbb{D}$) and
most importantly the cluster set of $\arg(1-\bar{\tau}\phi_{t}(z))$, as
$t\to+\infty$, where $\tau$ is the Denjoy-Wolff point of the semigroup. It has
been proven (see [9]) that in case of a hyperbolic semigroup,
$\\{\phi_{t}(z)\\}$ always converges non-tangentially to the Denjoy-Wolff
point, as $t\to+\infty$, while in case of a parabolic semigroup of positive
hyperbolic step, the convergence is always tangential. In the third case, when
the semigroup $(\phi_{t})$ is parabolic of zero hyperbolic step, the situation
is rather chaotic. At first, it was conjectured that the slope is a singleton,
something that was proven under some additional assumptions in [13, 14].
However, in general, the convegence can be tangential or non-tangential and
the cluster set of $\arg(1-\bar{\tau}\phi_{t}(z))$, as $t\to+\infty$, can be
either a singleton $\\{\theta\\}$ or a set $[\theta_{1},\theta_{2}]$, but it
is always a compact connected subset of $[-\frac{\pi}{2},\frac{\pi}{2}]$. In
[3, 11] examples are constructed where the slope is the whole interval
$[-\frac{\pi}{2},\frac{\pi}{2}]$, while in [6, 16] examples are constructed
where the slope is a closed subinterval of $(-\frac{\pi}{2},\frac{\pi}{2})$.
Furthermore, in [5], the authors find geometric conditions that guarantee that
the slope reduces to $\\{0\\}$. Finally, in [7], the authors characterize the
type of convergence (tangential or non-tangential) of the trajectories of a
semigroup $(\phi_{t})$ to its Denjoy-Wolff point in terms of the shape of the
associated planar domain.

In the present article, we provide some geometric conditions on the associated
planar domain $\Omega$ that are sufficient for the slope to be a singleton
$\\{\frac{\pi}{2}-\theta\\},\;\theta\in(0,\pi)$. For this purpose, we will
first prove the next proposition which is very similar to Proposition 3.1, but
concerns the convergence by a certain angle and not by an angle-set. In
particular, this next proposition is a generalization of Proposition 4.1 in
[5]. We set $U_{\theta}=\\{z:-\theta<\arg z<\pi-\theta\\}$, where
$\theta\in(0,\pi)$. Actually, for $\theta=\frac{\pi}{2}$, the statements of
the two propositions are identical. However, the proofs are entirely
different. The proof in [5] is based on hyperbolic geometry, while this next
proof relies on harmonic measure. We can consider Proposition 4.1 as a subcase
of Proposition 3.1. In fact, the following proof is directly implied by the
one of Proposition 3.1.

######  Proposition 4.1.

Let $\Delta\subsetneq\mathbb{C}$ be a simply connected domain and
$f:\mathbb{D}\to\Delta$ a Riemann map. We suppose that
$U_{\theta}+a\subset\Delta\subseteq U_{\theta}$, for some $a>0$. Then, there
exists a $\sigma\in\partial\mathbb{D}$ such that $f^{-1}(t)$ converges by
angle $\theta$ to $\sigma$, as $t\to+\infty$.

###### Proof.

Using the rotation $g(z)=e^{i(-\frac{\pi}{2}+\theta)}z$ which is a conformal
mapping and, as such, keeps the harmonic measure invariant, we can assume that
$\mathbb{H}+a\subset\Delta\subseteq\mathbb{H}$. Consider the continuous curve
$\delta:[0,+\infty)\to\Delta$ with
$\delta(t)=e^{i(-\frac{\pi}{2}+\theta)}(t+a)$, which is actually a
reparametrization and then a rotation through $g$ of the positive semi-axis.
Obviously, $\lim\limits_{t\to+\infty}\delta(t)=\infty$. By the geometry of the
right half-plane and its hyperbolic sectors and by the definition of the sets
$A_{\mathbb{H}}(\gamma,\theta_{1},\theta_{2})$, it is easy to check that
$\delta([0,+\infty))$ is eventually contained in every set of the form
$A_{\mathbb{H}}(\gamma,\theta_{1},\theta_{2})$, where $\gamma$ is the geodesic
for the hyperbolic distance of the right half-plane that has as its image the
positive semi-axis and $0<\theta_{1}<\theta<\theta_{2}<\pi$. Therefore, with
the exact same proof as in Proposition 3.1, we find that $\delta(t)$ converges
to $\infty$ by angle $\theta$, as $t\to+\infty$. Rotating back through
$g^{-1}$ and returning to the unit disk through $f^{-1}$, we get the desired
result.

∎

In the following corollary, $\Omega$ will not be contained in any horizontal
strip and so the semigroup cannot be hyperbolic. In addition, we already said
that the slope will not reduce to $\\{-\frac{\pi}{2}\\}$ or
$\\{\frac{\pi}{2}\\}$ or in other words the convergence will not be
tangential. Therefore, the semigroup will not be parabolic of positive
hyperbolic step. Having restricted ourselves to non-elliptic semigroups, we
can only talk about a parabolic semigroup $(\phi_{t})$ of zero hypebolic step.
For $\alpha_{1},\alpha_{2}\in(0,\pi]$, consider the sector

$U(\alpha_{1},\alpha_{2})=\\{z:-\alpha_{1}<\arg z<\alpha_{2}\\}.$

######  Corollary 4.1.

Let $(\phi_{t})$ be a parabolic semigroup in $\mathbb{D}$ of zero hyperbolic
step, $h$ the corresponding Koenigs function, $\Omega=h(\mathbb{D})$ and
$\tau\in\partial\mathbb{D}$ the Denjoy-Wolff point of $(\phi_{t})$. If there
exist $\alpha_{1},\alpha_{2}\in(0,\pi]$ with $\alpha_{1}+\alpha_{2}\geq\pi$
and $a>0$ such that $U(\alpha_{1},\alpha_{2})+a\subset\Omega\subseteq
U(\alpha_{1},\alpha_{2})$, then
$\lim\limits_{t\to+\infty}\arg(1-\bar{\tau}\phi_{t}(z))=\frac{\pi}{2}\cdot\frac{\alpha_{2}-\alpha_{1}}{\alpha_{1}+\alpha_{2}}$,
for every $z\in\mathbb{D}$.

###### Proof.

Let $z\in\mathbb{D}$ such that $h(z)\in\mathbb{R}$. Surely, the trajectory
$\phi_{t}(z)$ converges to the Denjoy-Wolff point $\tau$, as $t\to+\infty$.
Through the Koenigs function $h$, all the trajectories correspond to
horizontal rays that converge to $\infty$ in the positive direction. So, we
have a curve $h(\phi_{t}(z))\subset\mathbb{R},t\geq 0,$ with
$\lim\limits_{t\to+\infty}h(\phi_{t}(z))=\infty$ and
$U(\alpha_{1},\alpha_{2})+a\subset\Omega\subseteq U(\alpha_{1},\alpha_{2})$.
Applying consecutively the conformal mappings
$g_{1}(z)=e^{i\frac{\alpha_{1}-\alpha_{2}}{2}}z$,
$g_{2}(z)=z^{\frac{\pi}{\alpha_{1}+\alpha_{2}}}$ and
$g_{3}(z)=e^{-i\frac{\pi}{2}\cdot\frac{\alpha_{1}-\alpha_{2}}{\alpha_{1}+\alpha_{2}}}z$,
we can map $U(\alpha_{1},\alpha_{2})$ conformally onto the set
$U=\\{z:-\frac{\pi\alpha_{1}}{\alpha_{1}+\alpha_{2}}<\arg
z<\frac{\pi\alpha_{2}}{\alpha_{1}+\alpha_{2}}\\}$, while also
$g_{t}(z)=g_{3}\circ g_{2}\circ g_{1}\circ h\circ\phi_{t}(z)\in\mathbb{R}$,
for all $t\geq 0$. We can easily see that the set $U$ is actually a half-plane
and as a result we can directly make use of Proposition 4.1 to find that
$g_{t}(z)$ converges by angle $\frac{\pi\alpha_{1}}{\alpha_{1}+\alpha_{2}}$ to
$\infty$, as $t\to+\infty$. Therefore, keeping in mind the conformal
invariance of the harmonic measure, $\phi_{t}(z)$ converges by angle
$\frac{\pi\alpha_{1}}{\alpha_{1}+\alpha_{2}}$ to $\tau$, as $t\to+\infty$. By
definition, this means that
$\lim\limits_{t\to+\infty}\arg(1-\bar{\tau}\phi_{t}(z))=\frac{\pi}{2}\cdot\frac{\alpha_{2}-\alpha_{1}}{\alpha_{1}+\alpha_{2}}$.
Finally, we know (see e.g. [4, p.492]) that the dynamic behavior of every
trajectory is the same. As a consequence, the desired result holds for every
$z\in\mathbb{D}$. ∎

######  Remark 4.1.

In the statement of this last corollary, we demand that
$\alpha_{1}+\alpha_{2}\geq\pi$. This added condition is needed because in the
case when $\alpha_{1}+\alpha_{2}<\pi$, the chain of conformal mappings we
used, does not guarantee that Proposition 4.1 can be applied.

It is worth mentioning that a more general result was proven in [10], where
the authors use functions which measure the angular displacement of the
boundary of the associated planar domain with respect to a fixed vertical
straight line. In fact, the above Corollary follows immediately from Corollary
5.11 in [10].

## Acknowledgements

I thank Professor D. Betsakos, my advisor, for his advice during the
preparation of this work. I also thank the referee for his/her very valuable
suggestions.

This research did not receive any specific grant from funding agencies in the
public, commercial, or not-for-profit sectors.
Declarations of interest: none.

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