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\begin{document}

\title{A problem of McMillan on  conformal mappings.}
\author{Michael D. O'Neill and Robert E. Thurman}



\address{ Department of Mathematics, UTEP\\El Paso, TX 79968}




\email{michael@math.utep.edu, bob@math.utep.edu}
\begin{abstract}
We answer one of two questions asked by McMillan in 1970 concerning
distortion at the boundary by conformal mappings of the disk.

\end{abstract}



\newtheorem{dfn}{Definition}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{claim}{Claim}
\newtheorem{theorem}{Theorem}
\newtheorem{Theorem}{Theorem 1}
\newtheorem{ex}{Example}
\newtheorem{rmk}{Remark}
\renewcommand{\theTheorem}{} 
\renewcommand{\thetheorem}{}
\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}
1991 Mathematics subject classification: 30C25, 30C85
\end{center}

\section{Introduction}




The purpose of this note is to answer a question of 
J.E McMillan  concerning boundary behavior of conformal mappings
which was raised in the paper \cite{McMillan1}. In that paper,
McMillan gave a sufficient geometric  condition for a subset of the
boundary
of a domain to have harmonic measure zero and used it to prove a
result which we will describe below. A similar geometric lemma was the
key to the original proof of the twist point theorem in \cite{McMillan2}.
The reader can refer to both of McMillan's papers and to
\cite{O'Neill1}
for background on these problems and more generally to \cite{Ahlfors},
\cite{Garnett-Marshall} and \cite{Pommerenke2} for the ideas used in
this paper.

 We will use $\omega(z_0,F,\Omega)$ to
denote the harmonic measure of the set $F$ in the domain $\Omega$ 
from the point $z_0$.
Let $\mathbb D$ denote the unit disk in the complex plane and
let $f: \mathbb D \to \Omega$ be a conformal map. Let $A$ denote the
set of all  ideal accessible boundary points  $f(e^{i\theta})$ of $\Omega$ when $f$ has the nontangential limit
$f(e^{i\theta})$ at $e^{i\theta}$. Note that points of $A$ are prime
ends of $\Omega$ so that a single complex coordinate may represent
more than one point of $A$.












Let $D(a,r)$ denote a disk with center $a$ and radius $r$.
Choose $r\sb 0 < d(f(0), A)$ where $d$ denotes Euclidean distance.
For each $a \in  A$ and for each  $r < r\sb 0 $ let $\gamma (a, r) \subset 
\partial D(a,r)$ be the crosscut of $\Omega $ seperating
$a$ from $f(0)$ which can be joined to $a$ by
a Jordan arc in $\Omega \cap D(a,r)$. Let $L (a,r)$ denote the Euclidean length of $\gamma
(a,r)$
and let $U(a,r) = \bigcup \limits \sb {r' < r} \gamma (a,r')$.  

Let
$$
A(a,r) = \int \limits \sb 0 ^ r L(a,\rho)\, d\rho
$$
denote the Lebesgue measure of $U(a,r)$. 

McMillan proved
\begin{thm}
The set of $a\in A$ such that
$$
\limsup \limits \sb {r\to 0} \frac{A(a,r)}{\pi r ^2} < \frac12
$$
has harmonic measure zero.
\end{thm}


Notice that this theorem implies that the set of $a \in A$ such that
$$
\limsup \limits_{r\to 0} \frac{L(a,r)}{2 \pi r} < \frac12
$$
has harmonic measure zero. 



McMillan also gave an example of a domain for which both
$$
\limsup \limits_{r\to 0} \frac{A(a,r)}{ \pi r^2} =1 \quad \omega \quad a.e.
$$
and
$$
\liminf \limits_{r\to 0} \frac{A(a,r)}{ \pi r^2} =0 \quad \omega \quad a.e.
$$
(implying the corresponding limits for $\frac{L(a,r)}{ 2 \pi r}$)
and  conjectured that

$$
E_1=\{a\in A:\liminf \limits_{r\to 0} \frac{A(a,r)}{ \pi r^2} > \frac12\}
$$
and
$$
E_2=\{a\in A:\liminf \limits_{r\to 0} \frac{L(a,r)}{2 \pi r} > \frac12\}
$$
must be  sets of harmonic measure zero.

Here, we will verify McMillan's conjecture that the set $E_2$ must always have zero harmonic
measure.
\section{There are no points of density in  $f^{-1}(E_2)$.}
With the notations and definitions of the introduction we prove
\begin{thm}\label{lengththeorem}
The harmonic measure of the set $E_2$ is zero.
\end{thm}
\begin{proof}
For any positive integers $m$ and $k$, let
$$
E_{m,k}= \{a\in A|\, L(a,r) > (\frac12 + \frac{1}{m}) 2 \pi r\quad
\forall r<\frac{1}{k}\}.
$$
Since $E_2$ is the countable union of sets $E_{m,k}$, it suffices to
show that each $E_{m,k}$ has harmonic measure zero.

We will require the following lemma (see \cite{Pommerenke2}, pg 142)
which is a consequence of results of Beurling, \cite{Beurling}.
\begin{lemma}\label{crowdlemma}
Let $f$ map $\mathbb D$ conformally  into $\mathbb C$ and let $0<
\delta
< 1$. If $z\in \mathbb D$ and $I$ is an arc of $\mathbb T$ with 
$\omega(z,I)\ge \alpha >0$ then there exists a Borel set $B\subset I$
with $\omega(z,B) > (1 - \delta) \omega(z,I)$ such that 
$$
| f(\xi) - f(z)| \le \Lambda(f(S)) < K(\delta,\alpha) d_f(z) \qquad \mbox{
for } \xi \in B
$$
where $\Lambda$ denotes linear measure, $d_f(z)$ is the euclidean
distance from $f(z)$ to the boundary of $f(\mathbb D)$, $S$ is the non-euclidean segment from $z$ to $\xi$ and where
$K(\delta,\alpha)$ depends only on $\delta$ and $\alpha$.

\end{lemma}
 The basic idea of the proof of theorem \ref{lengththeorem}
is that since  points of $E_{m,k}$ are separated from $f(0)$ by
circular arcs of wide angle and large radius, if $f^{-1}(E_{m,k})$
has a point of density then lemma \ref{crowdlemma} will provide
enough wide angled circular arcs of a fixed radius to wrap around
on themselves and disconnect the domain $\Omega$.

Suppose then that $\eta \in \mathbb T$ is a point of density of
$f^{-1}(E_{m,k})$ and let $I$ denote an arc of $\mathbb T$ centered at
$\eta$.



Given $\delta_1 >0$ we can choose $I$ such that
\begin{equation}\label{eq1}
\frac{|f^{-1}(E_{m,k})\cap I|}{|I|} > (1-\delta_1).
\end{equation}

Given $\delta _2 >0$ we can find $0< r(I,\delta_2)<1$ such that
$$
\omega((1-r(I,\delta_2)) \eta, I ,\mathbb D) = 1- \delta_2
$$
and this determines the point $z_I = (1-r(I,\delta_2)) \eta$.
 
If we are given $\delta_3>0$ then  if $\delta_1$ is sufficiently
small,
(\ref{eq1}) implies that
$$
\omega(f^{-1}(E_{m,k}),z_I,\mathbb D)> (1-\delta_3).
$$
By lemma \ref{crowdlemma}, if we are given $\delta_4 >0$ then there is
a Borel set $B\subset I$ such that
$$
\omega(z_I,B,\mathbb D)> (1-\delta_4)(1-\delta_2)
$$
and such that
\begin{equation}\label{eq2}
|f(\xi) - f(z_I)| < K(\delta_4,(1-\delta_2))d_f(z_I) \quad \forall
\xi \in B.
\end{equation}
It follows that
\begin{equation}\label{eq3}
\omega(f^{-1}(E_{m,k}) \cap B, z_I, \mathbb D) > 1- (\delta_2 + \delta_3 +
\delta_4 - \delta_2 \delta_4)
\end{equation}
and that (\ref{eq2}) holds for all $\xi \in f^{-1}(E_{m,k})\cap B$.
Notice that the constant $K$ only depends on $\delta_2$ and $\delta_4$.

Since $f(\eta)\in A$ we can choose $I$ so that $ K d_f(z_I) <<
\frac{1}{k}$ where $k$ is the integer in the definition of $E_{m,k}$.
 The finite  number of steps required to get 
a contradiction in the construction to follow will only depend on the number $m$ in the definition of
$E_{m,k}$.  By
choosing a sufficiently small arc $I$, we can arrange that in each 
step of our construction, the positive number 
$$
\delta \equiv \delta_2 + \delta_3 +
\delta_4 - \delta_2 \delta_4
$$
 is small enough so that the construction
can proceed to the next step. We assume that these conditions hold  on the size of the
interval $I$.

Let $w_0 = f(0)$, $w_1 = f(z_I)$, $d_0= d_f(z_I)$ and let  $x_1$ be a
point of $\partial \Omega$ such that $|x_1 - w_1| = d_0$.
Let the letters $c_1,c_2,\dots$ denote positive constants which will
be assumed to be sufficiently small in each step of the construction
but will ultimately depend only on the number $m$ in the definition of
the set $E_{m,k}$ and not on $f$, $\Omega$, or $\delta$. Let $C_1,C_2,\dots$ denote other constants
which may be purely numerical or which may depend only on the number $m$.

First let $0 <c_0<<1$ and $c_1 << \frac{\pi}{m}c_0$.
We will see that these choices allow for rotation by a fixed positive
angle of certain separating circular arcs in consecutive steps of the
construction to follow. The arc of $\partial D(x_1,c_1d_0)$
which intersects the interior of $D(w_1,d_0)$ extends to a crosscut of
$\Omega$ and determines a unique subdomain  $U_1\subset\Omega$ not containing $w_1$.
We proceed to find a point close to $x_1$ which is contained in
$E_{m,k}$. By Harnack's inequality, 
$$
\omega(w_1,\partial U_1 \cap \partial \Omega \cap D(x_1, c_1 d_0), \Omega)\ge C_1 
\omega(w'_1,\partial U_1 \cap \partial \Omega \cap D(x_1,c_1 d_0), \Omega)
$$
where $w'_1$ is the point on the line between $w_1$ and $x_1$
such that $|x_1 - w'_1| = \frac{c_1 d_0}2$. By the comparison
principle for harmonic measure and the Beurling projection theorem, 
 (\cite{Ahlfors}, pg.43),
$$
\omega(w'_1,\partial U_1 \cap \partial \Omega \cap D(x_1,c_1d_0), \Omega)\ge C_2>0.
$$
So by lemma \ref{crowdlemma} and equation (\ref{eq3}), if $\delta$ is
sufficiently small, ($\delta << C_1C_2$), there
is a
 constant $C_3$ such that
$$
\omega(w_1,\partial U_1 \cap \partial \Omega \cap D(x_1,c_1d_0) \cap E_{m,k}, \Omega)\ge C_3>0.
$$ 
Choose a point  $x_1^*\in \partial U_1 \cap \partial \Omega \cap
D(x_1,c_1 d_0) \cap
E_{m,k}$. If $c_0$ is sufficiently small then 
 the arc of $ \{z \in \mathbb C: |x_1^* - z| = c_0 d_0\}$
which intersects $D(w_1,d_0)$ has an angle greater than
$\pi( 1 - \frac{1}{2 m})$. This arc must therefore be part of the
crosscut whose length is $L(x_1^*,c_0 d_0)>\pi(1+\frac1m)$.
Denote by $\overline{ab}$ the segment with endpoints $a\in \mathbb C$
and
 $b\in mathbb C$.
Let $w_1^*$ be the point on $\overline{x_1^* w_1}$ with $|x_1-w_1^*| =
c_0 d_0$ and  consider the annulus 
$$
R_1=\{z\in \mathbb C:(1-c_2)|x_1^* - w^*_1| < |x_1^* - z| <
(1+c_2)|x_1^* - w^*_1|\} 
$$
where $c_2 << \frac{\pi}{m}c_0$. Let $S_1$ be the component of
$R_1 \cap \Omega$ which intersects $D(w_1,d_0)$ and
let $x_2$ be a point of $\partial \Omega \cap S_1$ such that
$\overline{ x_1^* x_2}$ has minimal angle clockwise from
$\overline{x_1^* w_1^*}$.

Let $S_1^*$ denote the sector of $R_1$ clockwise between 
$\overline{x_1^* w_1^*}$ and $\overline{ x_1^* x_2}$.
The circular arc $\partial D(x_2, c_2 d_0) \cap S_1^*$ is part of  a
crosscut of $ \Omega$ which determines a unique subdomain
$U_2$ of $\Omega$ not containing $w_1^*$. By an argument similar to the
previous one using Harnack's inequality, the comparison principle for
harmonic measure and the Beurling projection theorem but now in the
annular sector
$S_1$, it follows that
$$
\omega(w_1, \partial \Omega \cap \partial U_2 \cap D(x_2, c_2 d_0),\Omega) > C_4>0
$$
We remark that $C_4$ depends on $c_0,c_1,c_2$ and therefore only on
$m$
and that the remaining constants $C_j$ may have similar dependence
on $m$.





A simple geometric argument shows that there is a point $x_2^{*}$ in
$D(x_2,c_2 d_0) \cap E_{m,k}$ and a constant $c_3 >0$ determined by 
the diameter of the $E_{m,k} \cap D(x_2,c_2 d_0)$ such that the set of distances
$$
\{|x_2^{*} - w|: w \in D(x^*_1, c_1 d_0) \cap \partial \Omega\}
$$
contains an interval $J_1$ of length greater than $c_3 d_0$.



Let $R_2 = \{w\in \mathbb C: |w- x_2^{*}|\in J_1\}$ and let $S_2$ be
the component of $R_2 \cap \Omega$ which intersects $S_1$. Each of the
circular arcs of $S_2$ centered at $x_2^{*}$ is a  crosscut of
$\Omega$. If there is such a crosscut $L_1 \subset S_2 \cap \Omega$
which does not separate $x_2^{*}$ from $w_0$ then we repeat the above
construction of $S_2$ but in the counterclockwise direction from
$\overline{x_1^* w_1^*}$. Then any circular arc  $L_2 \subset S_2
\cap \Omega$ centered at $x_2^{*}$ which intersects $S_1$, separates 
$x_2^{*}$ from $w_0$. For otherwise, $w_0$ is contained in both
subdomains
of $\Omega$ determined by the concave sides of $L_1$ and $L_2$. Since
  $w_0$ lies on the convex side of any circular arc
which defines $L(a,r)$ for some $a\in A$ and $r>0$ and therefore of any
arc of $S_1$,
this is impossible. If one choice of $S_2$, clockwise or
counterclockwise from $\overline{x_1^{*}w_1^*}$, fails to 
separate $x_2^{*}$ from $w_0$ we choose the other. Otherwise, the
construction can continue, as described below, in both directions until the non-separating 
 case occurs
and after that point, a  topological argument similar to the  above
allows the construction to continue in the remaining direction.
 
We have now arranged that each of the circular arcs of $S_2$ centered
at $x_2^{*}$ separates $x_2^{*}$ from $w_0$ and can be joined
to $x_2^{*}$ by a Jordan arc lying inside $S_1$. Therefore, since
$x_2^* \in E_{m,k}$, each
circular
arc of $S_2$ has an
angular measure greater than $(1+ \frac{2}{m})\pi$.
Let $w_2$ be a point of $S_2 \cap S_1$ and let $x_3$ be
a point of $\partial \Omega \cap \overline{S_2}$ which minimizes the clockwise
angle from $\overline{x_2^{*}w_2}$ to $\overline{x_2^{*} x_3}$.
Let $S_2^*$ denote the sector of $R_2$ clockwise between
$\overline{x_2^{*}w_2}$ and $\overline{x_2^{*} x_3}$.
As before the circular arc $\partial D(x_3, c_{3} d_0) \cap S^*_2$
extends to a crosscut of $\Omega$ which determines a unique subdomain
of $U_3 \subset\Omega$ not containing $w_1$.
The same harmonic measure argument as before but now done in the union
of annular corridors $S_1 \cup S_2$ shows that
$$
\omega(w_1, \partial \Omega \cap \partial U_3 \cap D(x_3, c_{3}d_0),\Omega) > C_6>0.
$$
If $\delta>0$ is sufficiently small, then as before, lemma
\ref{crowdlemma} and (\ref{eq3}) imply that
$$
\omega(w_1, \partial \Omega \cap \partial U_3 \cap D(x_3, c_{3}d_0)
\cap E_{m,k},\Omega) > C_7>0
$$
and we find $x_3^{*} \in \partial \Omega 
\cap \partial U_3 \cap D(x_3, c_{3}d_0)\cap
E_{m,k}$ such that the set of distances
$$
\{|x_3^{*} - w|: w\in D(x_2^{*}, c_{3} d_0) \cap \partial \Omega\}
$$
contain an interval $J_3$ of length greater than $c_{4} d_0$, where
$c_4$ depends only on the previous $c_i$ and on $m$.
Note that since  the constants satisfy $c_i<< c_0 \frac{\pi}m$,  there is a
numerical constant $c>0$ such  that the clockwise angle from
$\overline{x_1^{*} x_2^*}$ to $\overline{x_2^{*} x_3^{*}}$ is
at least $(1 + \frac{c}{m})\pi$. The construction continues in this
way
so that having found annular corridors $S_1,\dots S_j$ with centers $x_1^*,x_2^{*},\dots x_j^{*}$
we find $x_{j+1}^{*} \in E_{m,k}$ so that there is an interval of distances
$J_{j}$ between $x_{j+1}^{*}$ and the part of $\partial \Omega$
in a disk of radius $c_{\ell} d_0$ centered at $x_j^{*}$. The
intersection of the annulus centered at $x_{j+1}^*$ determined by $J_j$ with $\Omega$ contains a
component $S_{j+1}$ which intersects $S_j$. Concentric circular
arcs of this annular piece separate $x_{j+1}^{*}$ from $w_0$
(or else the construction continues in the other direction)
and each such circular arc can be joined to $x_{j+1}^{*}$ through
the annular corridor $S_{j}$ by a Jordan arc contained in the circle.
Therefore, each such arc
has an angle  greater than $(1+ \frac2{m})\pi$. Let $w_{j+1}$
be a point of $S_{j+1} \cap S_{j}$ and find $x_{j+2}$ which minimizes
the clockwise angle between $\overline{x_{j+1}^{*} w_{j+1}}$ and
$\overline{x_{j+1}^{*} x_{j+2}}$.
 The
construction can continue if $\delta >0$ is sufficiently small
since the harmonic measure of  the end of $S_{j+1}$ near $x_{j+2}$ 
from $w_1$ in $S_1 \cup S_2 \cup \dots S_{j+1}\cup D(w_1,d_0)$ is greater than some
positive numerical constant.

But it is clear from the construction that the union of annular
corridors  $S_1\cup \dots\cup S_j$ must wrap around on
itself after a finite number of  steps which only depends on $m$. 
 The union of annular corridors thus formed, being a subset of
$\Omega$,
 would contain a closed curve in $\Omega$ whose interior component
contains the points $x_i^{*} \in \partial \Omega$. Since 
$\Omega$ is simply connected, this contradiction shows that
$f^{-1}(E_{m,k})$ does not contain a point of density and therefore
must have measure zero. Therefore $E_{m,k}$ has harmonic measure zero
in $\Omega$.



\end{proof}







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\bibitem{Beurling}
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\bibitem{Garnett-Marshall}
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\bibitem{McMillan1}
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\bibitem{McMillan2}
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\newblock Boundary behavior of a conformal mapping.
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\bibitem{O'Neill1}
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\bibitem{Pommerenke2}
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\newblock {\em Boundary Behavior of Conformal Maps}.
\newblock Springer-Verlag, Berlin, 1991.





\end{thebibliography}



\end{document}

