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\begin{document}

\title{McMillan type theorems on boundary behavior of conformal mappings.}
\author{Michael D. O'Neill }



\address{ Department of Mathematics, Claremont McKenna College, Claremont CA, 91711}




\email{moneill@mckenna.edu}
\begin{abstract}
 This paper is an exposition of the McMillan Twist point theorem and \lq \lq Area" theorem
as well as some work of O'Neill and Thurman. In particular we make a
 simplified presentation of McMillan's original proof of the twist point theorem.
 The paper is based on talks given at 
Institut Mittag Leffler and UCLA. The  point is to show that the main ideas of the 
theorems are geometric and potential theoretic.


\end{abstract}




\newtheorem{dfn}{Definition}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{claim}{Claim}
\newtheorem{theorem}{Theorem}
\newtheorem*{TheoremA}{Theorem A}
\newtheorem*{TheoremB}{Theorem B}
\newtheorem*{Case I}{Case I}
\newtheorem*{Case II}{Case II}
\newtheorem{ex}{Example}
\newtheorem{rmk}{Remark}
\newtheorem{prop}{Proposition}[section]
\def\pfclm{\noindent {\em Proof of claim}\quad}
\def\proof{\noindent{\em Proof.}\quad}
\def\pflm{\noindent {\em Proof of lemma.}\quad}
\def\pfthm{\noindent {\em Proof of theorem.}\quad}
\def\rse#1{\mbox{$^{\raisebox{.3ex}{$\scriptstyle{#1}$}}$}}

\maketitle
  
\begin{center}
2000 Mathematics subject classification: 30C85,31A20
\end{center}

\section{Introduction}
In what follows we discuss two theorems of J.E. McMillan and one due in joint work to
 R. Thurman and the author on boundary behavior of conformal mappings. 
Our purpose is to show that all three results are rather intuitive
 from the point of view of geometry and Brownian motion. This paper is part of a research
 program to answer the questions that
 we will pose in the final section. In particular, it is an introduction for 
\cite{O'Neill2}
where we will give a proof of the twist point theorem which is based only on geometry and
properties of Brownian motion. And throughout the paper we will be seeking to
 simplify arguments to the point where 
they may be useful in answering  higher dimensional versions of the same questions.
Hopefully the paper may also serve as an introduction for graduate students to some 
questions on 
boundary behavior of
 conformal mappings and harmonic functions. The paper is mostly self contained with some 
references to basic classical results.
See the end of the introduction for some convenient sources.

McMillan's twist point theorem, proved in \cite{McMillan2}, says roughly that
a typical point on the boundary of a simply connected domain is either an
interior tangent point or one about which the domain wraps arbitrarily far 
in either direction. Conformal mappings enter the picture when we clarify the 
notion of a typical boundary point. We will use harmonic measure to give a
 precise statement of the theorem and 
through out the paper. We give two equivalent 
definitions. Let $\Omega$ be a simply connected domain in $\mathbb C$ and 
$E\subset \partial \Omega$ a Borel set. Let $ z_0 \in \Omega$ denote a fixed basepoint.

\begin{dfn}
$$
\omega_{z_0}(E,\Omega)= P\{X_{\tau}\in E\}
$$
where $X_t$ is Brownian motion started at $z_0$, 
$\tau = \inf\{t>0:X_t \notin \Omega\}$
and $P$ is the Wiener measure on paths.
\end{dfn}
Thus the harmonic measure $\omega _ {z_0} (E)$ (suppressing the reference to $\Omega$)
of a Borel 
subset $E$ of $\partial \Omega$
is the probability that a Brownian traveler started at $z_0$ exits $\Omega$ first
 through $E$.

Now let $f$ denote the Riemann mapping of $\mathbb D$ onto $\Omega$ such that
$f(0)= z_0$ and (e.g.) $f'(0) >0$. Then 
\begin{dfn}
$$
\omega_{z_0}(E,\Omega)= \frac{|f^{-1}(E)|}{2\pi}
$$
where $|f^{-1}(E)|$ denotes the linear measure of $f^{-1}(E) \subset \partial \mathbb D$.
\end{dfn}
The equivalence of the above two definitions is a theorem of Kakutani
from \cite{Kakutani}. See \cite{Garnett-Marshall} for a proof. We can now state the 
twist point 
theorem. It is convenient to regard $\omega _{z_0}$ as a probability measure on the 
prime-end compactification of $\Omega$. For example, if $\Omega$ is $\mathbb D \setminus 
[0,1]$
then the positive real boundary points each represent two distinct prime ends.


Let $\Omega$ and $z_0 \in \Omega$ be as before.
\begin{theorem}
With repsect to $\omega_{z_0}$, almost every prime end $\zeta$ is either an interior 
tangent point or
has both 
$$
\liminf\limits _{z\to \zeta}\arg(z-\zeta) = -\infty
$$
and
$$
\limsup\limits _{z\to \zeta}\arg(z-\zeta) = +\infty
$$
where the limits are taken as $z\to \zeta$ inside of $\Omega$ and $\arg(z-\zeta)$
denotes a continuous branch of the argument defined in $\Omega$.
\end{theorem}

We will give a detailed proof of this result along the lines in McMillan's 
original paper but with as much simplification as possible. In particular
we will see that the original argument is quite natural from the point of view of 
Definition 2 and that in this sense the proof boils down to a combination of 
Plessner's theorem,  intuition with Brownian motion and figure 1.

Around the same time, in \cite{McMillan1}, McMillan also proved another nice
  result (referred to above as the \lq \lq Area Theorem") on the almost sure character of
 boundary points of simply 
connected domains. To describe it we will need some notation. See figure 1.
Fix $z_0 \in \Omega$ and for each ideal accessible boundary point $\zeta \in \partial 
\Omega$
and $\rho < |z_0 -\zeta|$ let $\gamma(\zeta, \rho) \subset \{z:|z-\zeta|=\rho\}$ be
 the circular crosscut of $\Omega$ separating $\zeta$ from $z_0$ that can be 
joined to $\zeta$ by a Jordan arc contained in $\Omega \cap \{z:|z-\zeta| < \rho\}$.
Let $L(\zeta,\rho)$ denote the Euclidean length of $\gamma(\zeta, \rho)$ and let 
$$
A(\zeta, r) = \int \limits _ 0 ^ r L(\zeta, \rho)\, d\rho.
$$



\begin{figure}
%\centerline{\psfig{figure=figure1.ps,height=4in}}
\includegraphics*{Larfigure.eps}
\caption{An area swept out by the arcs in the definition of $L(\zeta,\rho)$}
\label{fig1}
\end{figure}


 McMillan proved the following
\begin{theorem}
With respect to the harmonic measure, almost every ideal accessible boundary point $\zeta$
satisfies
$$
\limsup \limits _ {r\to 0} \frac{A(\zeta,r)}{\pi r^2} \ge \frac12.
$$

\end{theorem}
In words, the theorem states loosely that at almost every boundary point
there is some sequence of scales at which the domain looks locally as fat as
a semi-disk. At the end of \cite{McMillan1} the question of whether 
$$
\liminf \limits _ {r\to 0} \frac{A(\zeta,r)}{\pi r^2} \le \frac12 \qquad \mbox{a.e.-}\omega
$$
is posed. The positive answer (Theorem 3) was found in \cite{OT1} and \cite{OT2}.
Here we will only outline the ideas involved with 
the aim of showing that the proofs of theorems 2 and 3 boil  down to Brownian motion 
and respectively, figures 7 and 8.
Also included in \cite{McMillan1} is an example of a domain for which 
$$
\limsup \limits _ {r\to 0} \frac{A(\zeta,r)}{\pi r^2} = 1 \qquad \mbox{a.e.-}\omega
$$
and
$$
\liminf \limits _ {r\to 0} \frac{A(\zeta,r)}{\pi r^2} =0 \qquad \mbox{a.e.-}\omega
$$

In  section 2 we will prove two lemmas of McMillan which contain
the intuition about Brownian motion that we will need. In section 3 
we will give a short discussion and definitions  of cone points, tangent points and 
twist points.
Sections 4, 5 and 6 contain the discussions of theorems 1, 2 and 3 respectively.
In the final section we record some related open questions in two and 
higher dimensions.

For background on prime ends, Plessner's theorem and a nice analytical proof of the twist
 point theorem
see \cite{Pommerenke1} and \cite{Pommerenke2}. For
 Fatou's theorem see \cite{Garnett} and for 
the Beurling projection theorem and much more about harmonic measure
see \cite{Ahlfors} and \cite{Garnett-Marshall}. For the connection between 
Brownian motion and 
harmonic measure see
 \cite{Garnett-Marshall} and \cite{Bass} and the paper \cite{oksendal}.

\section{Zero harmonic measure and escape routes for Brownian motion}

The   lemmas in this section are the main technical tool employed in 
\cite{McMillan2} and \cite{McMillan1}. The purpose of the discussion here is to show
that they are quite natural from the Brownian motion point of view. Loosely speaking
they state that in order to prove that a subset  $E$ of the boundary of a simply connected
 domain has zero harmonic measure, it is enough
 to show that  almost any Brownian traveler on its way to $E$ within the domain has
 infinitely many opportunities 
to escape to $\partial\Omega \setminus E $ with some positive probability.

The simplest version of this principal was used in \cite{McMillan1}.
Let $\Omega$ denote a simply connected domain in the plane and let $z_0 \in \Omega$ be a 
fixed
basepoint. If $S_1$ and $S_2$ are subsets of $\overline{\Omega}$ let 
$\mbox{dist}_{\Omega} (S_1,S_2) = \inf(\mbox{diam } \gamma)$ where the infimum is over 
Jordan arcs $\gamma$ which lie in $\Omega$ and join $S_1$ and $S_2$.

\begin{lemma}
Let $E\subset \partial \Omega$ consist of accessible boundary points and suppose that for 
each $a\in E$
there exists a sequence $\{c_n\}$ of crosscuts of $\Omega$ each of which separates $a$ 
from  $z_0$ and for which both $\mbox{\em{diam} } c_n \to 0$ and 
$$
\sup \limits_n \frac{\mbox{\em{diam} } c_n}{\mbox{\em{dist}}_{\Omega} (c_n,E)} < \infty.
$$
Then $\omega_{z_0} (E,\Omega) = 0$.
\end{lemma}

\begin{proof} 
We may assume that
$$
\sup \limits_n \frac{\mbox{diam } c_n}{\mbox{dist}_{\Omega} (c_n,E)} < K <+ \infty.
$$
If $z \in c_n$, Harnack's inequality  shows that $\omega _z (E^c \cap \partial \Omega, 
\Omega) > c(K) > 0$. Then by conformal invariance and lemma 2.3 we are done.
 We will \lq \lq sketch" a proof which does not rely 
on conformal invariance.

Assume as before that 
$$
\sup \limits_n \frac{\mbox{diam } c_n}{\mbox{dist}_{\Omega}
 (c_n,E)} < K <+ \infty.
$$
The set of boundary points of $\Omega$ which are not accessible by a rectifiable 
arc within $\Omega$ has harmonic measure zero. This is a purely potential 
theoretic fact and is true in higher dimensions as well. See \cite{BrelotChoquet}.
(It is also easy to see by considering the area of $\Omega$ as an integral of $|f'|^2$ 
where $f$ is 
the conformal mapping.) So a compact $K\subset \partial \Omega$ is the countable union of 
sets which are 
compact in the $\mbox{dist}_{\Omega}$ topology on $\overline{\Omega}$ and a set 
of zero harmonic measure.

Assuming that the set $E$ is compact in the $\mbox{dist}_{\Omega}$ topology we can
 cover $E$ by \lq \lq generations"  of the curves $c_n$ as in figure 2. Note that the
 width of each escape 
route is comparable to its diameter. Harnack's principle and the strong Markov 
property then finish the proof. (See \cite{O'Neill1} for the details.)


\end{proof}
\begin{figure}
%\centerline{\psfig{figure=figure1.ps,height=4in}}
\includegraphics*{escaperoutesfigure.eps}
\caption{Two generations of escape routes for Brownian motion  from lemma 2.1}
\label{fig1}
\end{figure}



The stronger version of the lemma which was used in \cite{McMillan2}
requires Stolz cones. Fix $0<r<1$. Given $e^{i \theta} \in \mathbb D$, let 
$\Gamma_{\alpha}(e^{i \theta} )$ be  the convex hull of $\{z:|z|=r\}$ and 
$\{e^{i \theta}\}$
where
$$
\alpha(r) = 2 \arctan \frac{r}{\sqrt{1-r^2}}
$$
is the angle at the vertex $e^{i\theta}$. Suppressing the reference to $r$
we have defined our Stolz cones $\Gamma_{\alpha}(e^{i \theta} )$.
Let $f:\mathbb D \to \Omega$ be a conformal mapping and let 
$E\subset \partial \Omega$ consist of accessible boundary points.


\begin{lemma}
Suppose that for each $e^{i \theta} \in f^{-1}(E)$ there is a sequence of points 
$z_n \to e^{i\theta}$ in $\mathbb D$ such that
the $z_n$ are contained in some open triangle in $\mathbb D$ and that for each $z_n$
there is a Jordan arc $A_n \subset \Omega$ with one endpoint $f(z_n)$ and the other on
 $\partial \Omega$ such that
$$
\sup \limits_n \frac{\mbox{\em{diam} } A_n}{\mbox{\em{dist}}_{\Omega} (A_n,E)} < \infty.
$$
Then $\omega(E,\Omega)=0$.
\end{lemma}
\begin{proof}
Dividing $E$ into countably many pieces we may assume 
that the points $z_n$ lie in the union of 
all the cones $\Gamma_{\alpha}(e^{i \theta})$, $e^{i\theta} \in f^{-1}(E)$ for some
fixed $\alpha$. In the same way we may assume that
$$
\sup \limits_n \frac{\mbox{diam } A_n}{\mbox{dist}_{\Omega} (A_n,E)} <  K < \infty.
$$
By Harnack's principle there is $c(K)>0$ such that
$$
\omega_{z_n} (E^c \cap \partial \Omega,\Omega)> c(K) > 0
$$
at each $z_n$. By conformal invariance and lemma 2.3 we are done.

\end{proof}

Let $E \subset \partial \mathbb D$, $0<\alpha < \pi$ and $\epsilon >0$.
\begin{lemma}
Suppose that for each $e^{i\theta} \in E$ there exists a sequence $z_n \to e^{i\theta}$
with $\{z_n\} \subset \Gamma_{\alpha} (e^{i\theta})$ such that
$$
\omega_{z_n} (E, \mathbb D) < 1-\epsilon.
$$
Then $\omega (E) = 0$.
\end{lemma}
\begin{proof}
By Fatou's theorem $u(z) = \omega_z(E, \mathbb D)$ has the nontangential limit $1$
at almost every point of $E$. Thus the conditions of the theorem imply that
$|E|=0$.
\end{proof}

The Brownian motion proof of lemma 2.3 would run along the following lines. 
For each $\alpha$
and for almost every Brownian path $\eta$ which hits $\partial \mathbb D$ first
 in $E$ at time
$\tau(\eta)= \inf \{X_t(\eta) \notin\mathbb D\}$ there is $t_0(\eta) < \tau(\eta)$ 
such that

$$
X_t(\eta) \in \bigcup \limits _ {e^{i\theta} \in E}
\Gamma _ {\alpha} (e^{i\theta}), \qquad t_0 \le t < \tau.
$$
This follows from a stopping time argument on crossings between $\cup \Gamma _{\alpha}$
and $(\cup \Gamma _{2\alpha})^c$ and the strong Markov property. Thus almost every such 
Brownian path passes near infinitely many of the $z_n$, experiencing at each pass a 
probability greater than $c(K)>0$ of exiting $\mathbb D$ through 
$E^c \cap \partial \mathbb D$. 
By the Strong Markov property again, 
$P\{X_{\tau}\in E\} = 0$.

\begin{figure}
%\centerline{\psfig{figure=figure1.ps,height=4in}}
\includegraphics*{greenconefigure.eps}
\caption{The point $z$ is contained in $\Gamma _{\alpha} (\zeta)$, the cone of aperture
 $\alpha$ at $\zeta$}
\label{fig1}
\end{figure}



To prove lemma 2.2 without relying on conformal invariance we can, for example, 
use the ideas in the paper of Brelot
and Choquet, \cite{BrelotChoquet}, to define the analog of cones in $\Omega$ via 
Green lines.  See figure 3. Fix $z_0 \in \Omega$
and let $\zeta \in \partial \Omega$. Given $z\in \Omega$ let $\ell_0$ denote the 
Green's line from $z_0$ to $z$
(neglecting countably many critical points). Consider the Green lines starting at 
$z$ whose tangents at $z$ sweep out an angle $\alpha$
 centered at the tangent of $\ell _0$ at $z$. If $\zeta$ is one of the endpoints on 
$\partial \Omega$ of such a Green line
then put $z\in \Gamma_{\alpha}(\zeta)$. By the work of Brelot and Choquet, the set 
of such endpoints of Green lines has 
$\omega_z = \frac{\alpha}{2\pi}$. The same Brownian motion argument outlined above for 
the disk can now be made directly in $\Omega$
with  the $\Gamma_{\alpha} (\zeta)$ playing the role of the cones.

\section{Cone points, tangent points, conformality.}


In preparation for our discussion of the twist point theorem in Section 4 we 
make the following definitions. 


\begin{dfn}
A  point $\zeta \in \partial \Omega$ is an inner tangent to $\partial \Omega$
if there exists exactly one $0\le \phi_0< 2 \pi$ such that
 for each $0< \epsilon < \frac{\pi}2$ there is a $\delta >0$ so that
$$
\{\zeta + \rho e^{i\phi} : 0 < \rho < \delta, | \phi - \phi_0| < \frac{\pi}2 - \epsilon\}
\subset \Omega.
$$
\end{dfn}

\begin{dfn}
The boundary point $\zeta$ is called a cone point if it is the vertex of some open
 triangle
contained in $\Omega$.
\end{dfn}

\begin{dfn}
The conformal mapping $f$ from $\mathbb D$ onto $\Omega$ is said to be conformal 
at the boundary point $e^{i\theta}$ if the non-tangential limit of $f'(z)$ at 
$e^{i\theta}$ exists and is not $0$ or $\infty$.
\end{dfn}

With respect to harmonic measure on $\partial\Omega$, almost every cone point is an inner 
tangent point.
This is yet another result from \cite{McMillan2}. We note that a proof depends only on the 
construction of 
the correct Lipschitz sub-domains of $\Omega$ and Harnack's inequality.
 (See \cite{Garnett-Marshall}.)
Moving to the disk, let
 $$
\mbox{Sect}(f) = f^{-1}(\{\mbox{Cone Points}\}).
$$
 Then $f$ is conformal at almost
every point of $\mbox{Sect(f)}$. (See \cite{Pommerenke2} chapter 4.)

It is also convenient to pull the notion of twisting back to the disk.
\begin{dfn}
The conformal mapping $f$ from $\mathbb D$ onto $\Omega$ is twisting at $e^{i\theta}$
if as $z$ approaches $e^{i\theta}$ within a Stolz cone at $e^{i\theta}$,
$$
\liminf\limits _{z\to e^{i\theta}}\arg(f(z)-f(e^{i\theta})) = -\infty
$$
and
$$
\limsup\limits _{z\to e^{i\theta}}\arg(f(z)-f(e^{i\theta})) = +\infty.
$$
\end{dfn}
The twist point theorem can now be stated as follows. At almost every $e^{i\theta}\in
\mathbb D$ either $f$ is conformal or $f$ is twisting.

We now recall Plessner's theorem. (See \cite{Pommerenke2} chapter 6.) It is 
a consequence of the
 theorem of Fatou used in lemma 2.3 together with another construction of
Lipschitz subdomains and the  conformal invariance of harmonic measure.
Let $g$ be a meromorphic function in $\mathbb D$.
\begin{thm}
For almost every $e^{i\theta}$ either $g$ has a finite angular limit at $e^{i\theta}$
or  for any $0<\alpha < \pi$, $g(\Gamma_{\alpha}(e^{i\theta}))$ is a dense subset of 
the plane.
\end{thm}
An immediate consequence of Plessner's theorem is the Privalov uniqueness theorem.
\begin{thm}If the meromorphic function $g$ is not identically zero then it has 
the angular limit
zero almost nowhere on $\partial \mathbb D$.
\end{thm}

Applying Plessner's theorem to the function $\log f'$ we see that the failure
 of conformality at $e^{i\theta}$
is almost everywhere equivalent to the unboundedness above and below  of 
$\arg f'$ in each Stolz cone $\Gamma _\alpha (e^{i\theta})$. Thus to prove the twist 
point theorem it is enough to show
that  unboundedness from above and below of $\arg f'$ is almost everywhere 
equivalent to the 
condition  that $f$ is twisting.

In the next section we will show that the set $E$ where $\arg f'$ is unbounded 
from above and below in Stolz cones and at the same time 
$$
\limsup\limits _{z\to e^{i\theta}}\arg(f(z)-f(e^{i\theta})) < +\infty,
$$
has harmonic measure zero. Since the same reasoning applies with $\limsup$ replaced
by $\liminf$, that will prove the theorem.


\section{The twist point theorem}
We will now give a proof of the twist point theorem following McMillan's 
original argument.
Our purpose is to show that the argument is physically intuitive and to simplify matters
 wherever possible. 
The general strategy is depicted in figure 4.
\begin{figure}
%\centerline{\psfig{figure=figure1.ps,height=4in}}
\includegraphics*{twistfigure.eps}
\caption{A large increase in $\arg f'$ implies that a path winds around many boundary
 points.
Some possible escape routes are shown in gray. The point $w^*$ is on the 
curve $\sigma$ near 
the center of the spiral.}
\label{fig1}
\end{figure}
 A curve along which $\arg f'$ grows 
without bound is forced to wind 
many times around many boundary points, thus excluding them from membership in the set $E$
and making them good escape destinations as in  lemma 2.2.
In \cite{O'Neill2} we  will give another proof of the twist 
point theorem which does not use any conformal mapping. Then the 
main task is to make the same argument directly using 
the failure of the cone point condition, i.e. to show that as Brownian motion approaches 
a set of positive harmonic measure which fails the cone condition
 it is forced to see
large  (in the sense of harmonic measure) parts of the boundary about which
 the domain winds
by a large amount. 

The proof below will be done in four steps. In the first step we will reduce the set
$E$ by chopping it into countably many pieces. In step 2 we will see that the reduction
in step 1 allows us to 
show that the curve  $\sigma$ shown in figure 4 starting at $w(0)$ is forced to wind
 around $w^*$, a later point on the curve $\sigma$.

In the third step we will see that the point $w^*$ can be chosen so that a path 
from $w^*$ to the boundary of $\Omega$ can be found which lies in a half plane 
who's boundary contains $w^*$. This already makes it evident, as in figure 4, 
that there must be
boundary within the wound up parts of $\sigma$ but in step 4 we will make this precise by 
 showing how step 3 lets us apply lemmas 2.2 and 2.3 to finish the proof. 

We come to the details.

\noindent {\bf Step 1}\quad
Let $E \subset \mathbb D$ denote the set of $e^{i\theta}$ where
$\arg f'$ is unbounded above on $\Gamma_{\alpha} (e^{i\theta})$
but  $\limsup \limits _{z\in \mathbb D} \arg(f(z) - f(e^{i\theta})) < +\infty$.
We want to show that $\omega (E)=0$. Let $D$ denote a disk with rational center and radius
and let $\Omega _D$ be one of the countably many components of $D\cap \Omega$.
Dividing $E$ into countably many pieces, we may assume
\begin{enumerate}
\item[(i)] $f(z) \in \Omega _D$ if $z\in \Gamma _{\alpha} (e^{i\theta})$ is sufficiently
close to $e^{i\theta} \in E$.

\item[(ii)] $f(D_{r(\alpha)}(0)) \cap D = \emptyset$

\item[(iii)] $\mbox{Arg}(w-f(e^{i\theta})) - \mbox{Arg}(w_0-f(e^{i\theta}))\le M < +\infty$
for each $w\in \Omega _D$, each $w_0 \in \partial \Omega_D \cap \Omega$ and  
each $e^{i\theta} \in E$.


\end{enumerate}

For $(iii)$, note that $\Omega_D$ is a  simply connected subset of the disk $D$ so that 
all values of $\mbox{Arg}(w_0-f(e^{i\theta}))$ for $w_0 \in \partial \Omega_D \cap \Omega$
lie in some interval of length $2\pi$.


\noindent {\bf Step 2}\quad

Choose $z_0$ such that $f(z_0) \in \partial \Omega_ D \cap \Omega$ and 
fix $e^{i\theta} \in E$. Let $z^* \in \Gamma_{\alpha}(e^{i\theta})$ be a point such that
$$
\arg f'(z^*) - \arg f'(z_0) > 2 M.
$$
Let $z(t) = (1-t) z_0 + t z^*$,  $w(t) = f(z(t))$ and $\sigma = \{w(t): 0\le t\le 1\}$.
Moving $z_0$  and $z^*$ if necessary, we may assume that $\sigma \subset \Omega _D$.

The function $g(\tau,t)=w(t)- w(\tau)$ is continuous  and nowhere zero on 
$T= \{(\tau,t):0<t\le 1, 0\le \tau < t\}$. So we may determine a branch of the argument 
so that
$\phi(\tau,t)= \arg(w(t) - w(\tau))$ is continuous on $T$. Because
 $w(0) \in \partial \Omega _D 
\cap \Omega$ and $\sigma \subset \Omega _D$, we can determine the
 branch so that 
$$
0\le \phi(0,t)\le 2\pi
$$
 for sufficiently small $t$ and therefore so that
$$
-\pi \le \phi(0,t) \le 3\pi
$$
 for all $0\le t \le 1$. Since $w'(t)$ is continuous
and never zero
$$
\phi(t_0) = \lim \limits_ {(\tau,t) \to (t_0,t_0)} \phi(\tau,t)
$$
exists for each $0\le t_0 \le 1$. Thus $\phi (t)$ is continuous
and  $\phi(t) - \arg f'(z(t))$ is constant. It follows that
$$
\phi(1) - \phi(0) = \arg f'(z^*) - \arg f'(z_0).
$$ 
Since both
$$
-\pi \le \phi(0,1) \le 3\pi
$$
and
$$
-\pi \le \phi(0) \le 3\pi
$$
it follows that 
$$
\phi(1) - \phi(0,1) \ge \arg f'(z^*) - \arg f'(z_0) - 4 \pi.
$$
But $\phi(1) - \phi(0,1) $ is the change in $\phi(\tau,1)$ as $\tau$ increases from
$0$ to $1$. So the curve $\sigma$ winds around $w^* = f(z^*)$ by an angle of at 
least $2M - 4\pi$.

\noindent {\bf Step 3}\quad

Let $S$ denote the component of the level curve of $\arg f'$ which contains $z^*$.
We may assume that $S$ contains no critical point of $\arg f'$. By applying the maximum
 principle to $\log|f'|$
we see that $S$ has exactly two endpoints on $\partial \mathbb D$ which 
we label $a$ and $b$.
By the Privalov uniqueness theorem and  
by increasing $M$ if necessary, we may assume that $S$ is contained in a  disk of
 radius $1/2$ centered at $e^{i\theta}$. Then we may also assume that in moving 
along the level curve from $a$ to 
$b$ that  $z^*$ is the last point on $S$ contained in
$\overline{\Gamma_{\alpha}(e^{i\theta})}$.
Let $A$ denote the piece of $S$ running from $z^*$ to $b$. (See figure 5.)
\begin{figure}
%\centerline{\psfig{figure=figure1.ps,height=4in}}
\includegraphics*{step3.eps}
\caption{Step 3 }
\label{fig1}
\end{figure}
 We claim that
$f(A)$ is contained in a half plane whose boundary contains $w^* = f(z^*)$.
To see this let $z(t)$, $0\le t\le 1$ be a parameterization of $A$ with $z(0) = z^*$ and
$z(1) = b$. The function $\arg f'(z(t))$ is constant and we may assume 
that $\log |f'(z(t))|$
is decreasing. Let $H$ denote the half plane containing $b$ whose
 boundary contains $z^*$ and 
$e^{i\theta}$.

Move $H$ to the upper half plane by $\zeta = \eta z + \nu$ with $\zeta (z^*) = 0$ and
 $|\eta| = 1$.
Let $F(\zeta) = f(\frac{\zeta - b}{a} )$ and $\zeta(t) = u(t) + i v(t) = \eta z(t) + \nu$.
Now $\arg F'(\zeta(t)) = \lambda $ is constant and $|F'(\zeta (t))|$ is decreasing
in $0\le t\le 1$.

We have
$$
F(\zeta(t)) - F(0) = e^{i\lambda}\left(\int \limits _0 ^t |F'(\zeta(t))|\, du(t) + 
i \int \limits _0 ^t |F'(\zeta(t))|\, dv(t) \right ).
$$
Because $v(t)$ is non-negative and $|F'(\zeta(t))|$ is decreasing we have
$$
\int \limits _0 ^t |F'(\zeta(t))|\, dv(t) = |F'(\zeta(t))| v(t) - 
\int \limits _0 ^t v(t) \, d|F'(\zeta(t))| \ge 0
$$
and the claim is proved.




\noindent {\bf Step 4}\quad
It remains to check that the curve $\sigma$ has a winding angle bigger than $M$ 
around a set 
of boundary points 
which has large $\omega _{w^*}$ measure. This is practically evident from step 3. We will
 give a proof mostly via figure 6. The following list of notations 
should be checked off 
on the figure.
\begin{figure}
%\centerline{\psfig{figure=figure1.ps,height=4in}}
\includegraphics*{step4.eps}
\caption{Step 4}
\label{fig1}
\end{figure}

Let $z(t)$ be the parameterization of $A$ from step 3 and 
$w(t) = f(z(t))$. Let $d= \mbox{dist} (w^*, \partial \Omega)$,
$$
D_{w^*}=\{w:|w-w^*|< d\},
$$
$$
t^* = \sup \{0\le t\le 1:w(t)\in D_{w^*}\},
$$
$$
p=w(t^*).
$$
Let $\beta$ be the component of $\partial D_{w^*}\cap \Omega$ which contains $p$ and let 
$q$ be the nearer endpoint of $\beta$ to $p$. Let $\beta^*$ be the part of $\beta$ 
connecting 
$p$ and $q$. Let $D_q(d/4)$ denote the disk of radius $d/4$ centered at $q$. Then
$\partial D_q(d/4) \cap \Omega$ has exactly one connected component arc which 
intersects $D_{w^*}$.
Denote it by $\Theta$. Let $F$ be the set of prime ends separated from $w^*$ by $\Theta$
and let $q^*$ be some point of $F$. We claim that the curve $\sigma$ winds around $q^*$
nearly as much as it does around $w^*$.

The winding of $\sigma$ around $w^*$ is within $4\pi$ of its winding around any 
other point of 
$f(A) \cup \beta^* \cup F$ by step 3. For the winding angle of $\sigma$ about $w^*$ to 
exceed the 
winding around
$q^*$ by more than $8\pi$, $\sigma$ would have to cross $\beta^*$ twice. By deleting 
appropriate 
parts of $\sigma$ and replacing
them with arcs of $\beta^*$ we could then construct a closed curve in $\Omega$ whose 
interior 
contains $f(A)$.
This is a contradiction because $\Omega$ is simply connected. The argument is simpler if
$f(b) \in \partial D_{w^*}$.

By the Beurling projection theorem,(e.g. See \cite{Ahlfors} or \cite{Garnett-Marshall}), 
$\omega_{w^*} (F) > c > 0$ where $c$ is some universal constant.
So if $M$ is sufficiently large, then $F\subset E^c$. An appeal to lemma 2.3 now 
completes McMillan's proof of the twist point theorem.









\section{McMillan's Area theorem}

The paper \cite{O'Neill1} contains a  shortened but detailed presentation of the proof of
 McMillan's 
area theorem from \cite{McMillan1}. Here we will only give the main ideas  of the 
proof together with
figure X which illustrates the construction of the required escape routes for
 Brownian motion. We note that the proof needs no mention of conformal mappings.


Let $w_0\in \Omega$ be a fixed base point and 
let $E$ denote the set of accessible boundary points $a$ for which
$$
\limsup \limits _ {r\to 0} \frac{A(a,r)}{\pi r^2} < \frac12.
$$

\noindent {\bf Theorem 2}\quad

$$
\omega(E)=0
$$



\begin{figure}
%\centerline{\psfig{figure=figure1.ps,height=4in}}
\includegraphics*{area1.eps}
\caption{The proof of theorem 2.}
\label{fig1}
\end{figure}

\begin{proof}

Let $w_0\in \Omega$ be a fixed base point and let $C_1,C_2,\dots$ denote 
positive constants which are either numerical or depend only on $m$ and $k$.
If we use the phrase \lq \lq bounded below" or \lq\lq bounded away from"
it will also  mean by a constant depending only on $m$ and $k$. As in the proof of the
 twist point
 theorem
we gain control by breaking $E$ into countably many pieces $E= \cup E_{m,k}$ where
$$
E_{m,k}= \{a: \frac{A(a,r)}{\pi r^2 } < \frac12 -\frac1m, \quad \forall r< \frac1k\}
$$
As mentioned in the proof of lemma 2.1, it is enough to then show that if $F\subset
E_{m,k}$ is compact in the $\mbox{dist}_{\Omega}$ topology then $\omega(F)=0$. 
Given such a set $F$ we will show that for any $a\in F$ and $\epsilon >0$ there is a 
crosscut $c$ of 
$\Omega$ with diameter less than $C_1 \epsilon$ separating $a$ from the base point $w_0$ 
such that
$$
\mbox{diam } c < C_2 \mbox{dist}_{\Omega} (c,F).
$$
By lemma 2.1 or 2.2 then, we will be done.

There is a $1> C_3 >0$  such that for any $r< \frac1k$ and $a\in E_{m,k}$ the set 
$\{ r' \in (C_3 r,r): L(a,r') < (\frac12 - \frac 2m ) 2\pi r' \}$ has positive Lebesgue
 measure.
To any point $a\in F$  and any $\frac 1k> \epsilon >0$ we associate a single line segment 
which is a crosscut of $\Omega$
as pictured in figure 7a) and  7b). In the figure, $C_3 \epsilon < r' < \epsilon$,
$L(a,r') < (\frac12 - \frac 2m ) 2\pi r'$ and 
$d > C_4 \epsilon$. The segment is chosen so that it can be joined to $a$ by a
 Jordan curve in $\Omega$ which does not touch any other 
part of the line $L$ or the circle $\{z:|a-z|= r'\}$. We will refer to such a 
segment below as a basic segment at $a$.

 We will now construct the required escape route for a given point $a\in F$ at a given 
scale $\epsilon > 0$.

Let $h=\frac{\delta}{10} \sin{(\frac 2m)}$. Cover $F$ with finitely many \lq \lq disks "
$\Delta _ {\nu}$ (in the metric $\mbox{dist}_{\Omega}$) of relative diameter less than
$h \epsilon$. Assume that $a\in \Delta _1$ and construct the basic segment $c_0$ for $a$.
If $\mbox{dist}_{\Omega}(c_0,F) \ge h\epsilon$ we are done. Otherwise the construction 
continues. Choose a disk $\Delta_{\nu}$ such that 
$\mbox{dist}_{\Omega}(c_0,\Delta _{\nu} \cap F)< h\epsilon$
and a point $a_{\nu}\in F\cap \Delta_{\nu}$ and construct a basic segment for $a_{\nu}$.
Two possibilities for this second step are shown in figure 7c) and 7d).
In c) the second stage arc is $c_1$. In d) it is the doubled arc. Note that in d)
the angle of intersection $\Psi$ is bounded away from $\pi$. The construction continues 
until
no disks are within $h\epsilon$ of the constructed polygonal arc. At the $n^{\mbox{th}}$ 
stage
a polygonal arc $c_n$ has been constructed. If there is a disk $\Delta_{\nu}$
 within $h\epsilon$ of $c_n$
in the relative distance, select one and construct the basic segment at some point 
of $F \cap \Delta _{\nu}$. There are three possibilities to consider depending 
on whether this 
new basic segment intersects $c_n$ in $0$, $1$ or more than $1$ points.

In the $0$ intersection case, the situation is as in figure 7c). The new arc $c_{n+1}$
is simply the basic segment for the offending disk at the $n^{\mbox{th}}$ stage.
 It is easy to show
that $a$ is separated from $w_0$ by this arc.

In case of more than one intersection, it takes a brief topological argument to show
that there can be at most two intersections. Then the situation is as in figure 7e).
Note that all the angles pictured are bounded by $\Psi < \pi$.

In the one intersection case denote the new basic segment by $c^*$. Then
 the new arc $c_{n+1}$
is the union of the piece of $c_n$ on the $w_0$ side of $c^*$ and the 
piece of $c^*$ on the $w_0$
side of $c_n$. Again it takes a short argument to verify that $c_{n+1}$ 
separates $a$ from $w_0$.
The control on the angles of intersection at all steps by the 
angle $\Psi < \pi$ ensures that the 
diameter of the constructed cross cut is bounded by $C_4 \epsilon$ at 
every stage. After exhausting the 
finite supply of disks
or stopping earlier because the disks are far enough away, we are done.
\end{proof}

We refer to \cite{O'Neill1} or \cite{McMillan1} for the details. (There is a 
typo in \cite{O'Neill1}
for the one intersection case. In line 6 from the bottom 
of page 233 $a^*$ should be $c^*_0$.) 




\section{Another theorem of the same type.}

In \cite{OT1} and \cite{OT2} the positive answers to the questions asked  in 
\cite{McMillan1} are given in detail. Our aim here is 
again to show that the heart of the matter  is a simple geometric idea
together with intuition about Brownian motion or Harnack's inequality.

Let $w_0\in \Omega$ be a fixed base point and  let $E$ denote the set of accessible 
boundary points such that
$$
\liminf \limits _ {r\to 0} \frac{A(a,r)}{\pi r^2} > \frac12.
$$

\noindent {\bf Theorem 3}\quad
$$
\omega (E) = 0
$$

\begin{proof}
We again break the set in question into countably many pieces.
$$
E_{m,k}= \{a: \frac{A(a,r)}{\pi r^2 } > \frac12 +\frac1m \quad \forall r< \frac1k\}
$$
but now the strategy of the proof is different. As before, let $C_1,C_2,\dots$
denote constants depending only on $m$ or $k$ or which are purely numerical.

We will show that the geometric condition describing points of $E_{m,k}$ is incompatible 
with the
 existence of a point of density of $f^{-1}(E_{m,k})$, where $f$ is a conformal mapping
from $\mathbb D$ to $\Omega$.
In fact we show that there is $\delta(m,k) > 0$ such that if
 $\omega(z_0, E_{m,k}) > 1-\delta$ for some $z_0 \in \Omega$, 
then it will be possible to construct a closed curve in $\Omega$ surrounding 
$z_1\in \partial \Omega$, 
a nearest boundary point to $z_0$. This will contradict the simple
 connectivity of $\Omega$.

Referring now to figure 8, suppose the sub-region of $\Omega$ consisting of annular 
corridors
 has been constructed as shown. It is assumed to extend, at the current stage, 
by an angle of $\theta_n$ around $z_1$ and to have a thickness bounded from below by 
$C_1 |z_0 - z_1|$. The lower bound on the thickness gives Brownian travelers 
starting at $z_0$
a probability $>> \delta$ of reaching boundary near $z_n$. Assume also that in each
 previous stage 
of construction our corridor has been 
further extended by at least the angle $\alpha (m)>0$ around $z_1$. The density condition 
$\omega(z_0,E_{m,k}) > 1-\delta$ implies that there are many points of $E_{m,k}$ 
near $z_n$.
The precise argument uses only Harnack's inequality and the Beurling projection theorem.
 We claim 
that the large harmonic measure of points of $E_{m,k}$ near $z_n$ together with
 the geometric condition
on the points allows us to proceed to the next step of the construction.

\begin{figure}
%\centerline{\psfig{figure=figure1.ps,height=4in}}
\includegraphics*{area2.eps}
\caption{The proof of theorem 3.}
\label{fig1}
\end{figure}


Notice that in figure 8  the piece of corridor labelled S is different than the ones above
 it.
The pieces above S represent the inductive step which we are about to describe. It is
 the inductive step which assures our path turns by some fixed 
additional amount, $\alpha (m) >0$, around $z_1$.
The inclusion of S is a technical neccessity which prevents the constructed corridor from
 turning the wrong way after several steps. In \cite{OT2} we simply included an 
S step in each stage of the construction and we indicate how to do this below.


We are assuming that we have a subset $J\subset E_{m,k}$ with $\omega_{z_0}$ measure 
larger than
$C_2>0$ near $z_n$ at the end of the currently constructed corridor.

Let $r_1$ be as shown and recall the definition of $\gamma(\zeta,r)$ from
the introduction. We now further assume that if $\zeta \in J$
then for any $r< r_1$, the arc $\gamma(\zeta,r)$ intersects the whole previously 
constructed 
pathway. If this is not true for some $\zeta\in J$ and some $r< r_1$ 
then  we begin the whole process over starting from the other side of the initial
 annular corridor $A_0$, but working in the opposite direction around $z_1$.
A brief topological argument shows that we do not encounter the same difficulty again.

By the density condition on $E_{m,k}$, the diameter of $J$ is bounded below
 by $C_3 |z_1 - z_0|$ and we can find many pairs of points 
near $z_n$ which are a given small distance $\sim C_4|z_1-z_0|$ apart. Because
 the angles of the arcs $\gamma(\zeta,r)$ are larger than $\pi + 
\frac1 m$ for a set of $r< r_1$ with measure larger than $C_5 r_1$
we can for each $\zeta \in J$ find a set of $r$'s with measure larger than $C_6(m) r_1$
such that $\gamma(\zeta,r)$ intersects $L_1$ for each such $r$ and so that these arcs 
continue past $L_1$ by an angle
greater than $\alpha(m) >0$. For fixed $\zeta$, the
 set of intersections with $L_1$ has positive linear measure on $L_1$ larger than
$C_7(m) r_1$. A pigeonhole argument on the different points in $J$ then lets us find 
two pairs
of arcs as shown in figure 9, passing through the same two points $a$ and $b$ on 
$L_1$,
with $|a-b| \ge C_8|z_1 - z_0|$.
\begin{figure}
%\centerline{\psfig{figure=figure1.ps,height=4in}}
\includegraphics*{figure1.eps}
\caption{$S^*(a,b)$ can contain no point of $\partial \Omega$.}
\label{fig1}
\end{figure}
 The two centers of these pairs of arcs can be chosen
far enough apart in $J$ and the points $a$ and $b$ close enough together so that,
as shown in the figure, the new annular corridor traverses an angle larger than
 $\alpha(m)>0 $
past $L_1$ with out meeting $\partial \Omega$. The newly constructed annular
 corridor has thickness greater than
 $C_9(m) |z_1 - z_0|$ and once it passes $L_1$ by an angle of $\alpha (m)$,
we may do the S construction to make sure we keep things moving around $z_1$.
Since $\alpha (m) >0$ is fixed, we finish in finitely many steps, re-entering 
$D(z_0,|z_1-z_0|)$ and obtaining our contradiction.

It only remains to start the construction. This is done by choosing $r_1 \sim |z_1-z_0|$
so that $\partial D(z_0,|z_1-z_0|)$ is close to being a diameter of $D(z_1,r_1)$.
The density condition and the Beurling projection theorem provide
 many points of $E_{m,k}$ near $z_1$
so that we can construct the initial annular corridor $A_0$. There 
will be at most 
$\sim\frac{\pi}{\alpha(m)}$ steps in the construction and in each 
step the new annular corridor that is added
may be dilated within a fixed factor compared to the previous one. 
This technicality is also
easily dealt with by the initial choice for the size of $A_0$.

See \cite{OT2} for the details.


\end{proof}


\section{Remarks and questions}
\begin{enumerate}



\item  
Is there a proof of the twist point theorem which uses only geometry and 
Harnack's inequality? In \cite{O'Neill2}, we will give a proof which does not use 
conformal mapping
but requires Ito's formula and the harmonicity of the argument of the gradient 
of Green's function.
The author was asked this question around 1994 by Juan Donaire and Jose Llorente.

\item
With respect to Hausdorff dimension, the set of twist points which support
 harmonic measure is typically very small compared to the set of all twist points.
Is there a purely geometric characterization of the twist points
 which support harmonic measure? What about if the domain is of the type considered by 
Carleson in \cite{Carleson}, i.e. if the harmonic measure can be represented as 
a shift invariant measure on a symbol space?

\item
Do theorems 1,2 and 3 have natural generalizations to $\mathbb R^3$?
Regarding theorem 1, K. Burdzy asked the following question
(see \cite{Burdzy} for similar questions).

Let $\Omega$ be a domain in $\mathbb R^3$ and let $X_t$ be a Brownian motion started
 at some fixed point $x_0 \in \Omega$ with $\eta$ denoting a path. Let 
$$
\tau(\eta) = \inf \{t: X_t(\eta) \notin \Omega\}
$$
and 
$$
E(\eta) = \bigcap \limits _{\epsilon >0} 
\overline{\left\{ \frac{X_t(\eta) - X_{\tau}(\eta)}{|X_t(\eta) - X_{\tau}(\eta)|} 
 :\tau - \epsilon \le t\le \tau \right\}       }.
$$
Is $E(\eta)$ almost surely either a sphere or a hemisphere? It is clearly a hemisphere if 
$\partial \Omega$ is smooth. If $\Omega = V \times \mathbb R$ where $V$ is the Von-Koch 
snowflake domain then $E(\eta)$ is almost surely a sphere because
 almost every point of $\partial V$
is a twist point.

\item
The following conjecture is from \cite{Bishop}. Let $D(x,r)$ denote a disk of radius $r$
centered at $x$. and let $\Omega \subset \mathbb C$ be simply connected.
It is known from the work of Makarov, McMillan and Pommerenke, \cite{Mak}, \cite {Pomm3},
that for $\omega$-a.e. $x\in \partial \Omega$
$$
\lim \limits _{r\to 0} \frac{\omega (D(x,r))}{r}
$$
exists and is not $0$ or $\infty$. It is also known that
for $\omega$-a.e. twist point $x$,
$$
\limsup \limits _ {r\to 0} \frac{\omega (D(x,r))}{r}= \infty.
$$
It is conjectured that 
$$
\liminf \limits _ {r\to 0} \frac{\omega (D(x,r))}{r}=0
$$
for $\omega$-a.e. twist point $x$.

\end{enumerate}






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\end{document}
 