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\begin{document}
\title{The convex hull of the interpolating
{B}laschke products}
\author{Michael D. O'Neill}
\address{ Department of Mathematics, UTEP\\El Paso, TX 79968}
\email{michael@math.utep.edu}
\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
\openup 3\jot
\large
\section{Introduction and Notations }
In the sequel we prove that if a Blaschke product $B$ is continuous in the closed unit disk
except on a closed set $E\subset \Bbb T$ of measure zero, then $B$ is contained
in $9K$, where $K$ denotes the closed convex hull of the interpolating Blaschke
products. Moreover, we show that a generic Blaschke product is contained in
$27K$. By the well known theorem of Marshall, this implies that the
unit ball of $H^\infty$ is contained in $27K$. The proofs employ a technical result, given in the first section,
which may be of some independent interest.
The results in the paper improve earlier work in
\cite{MS},\cite{GN} and \cite{Oy}. We refer to these papers and to the book
\cite{G} for further
background on the questions treated here.
This paper is based on part of the author's UCLA dissertation
completed in
1995
under the direction of John Garnett. I thank him for all his help
and encouragement. My thanks also go to the referee and the editor for their help in
clarifying the presentation.
\begin{description}
\item[$\Bbb D$]  Open unit disk, $\qquad\Bbb D \equiv
\{z\in \Bbb C : z <1\}$
\\
\item[$\Bbb T$]  Unit circle, $\qquad \Bbb T \equiv
\partial \Bbb D$
\\
\item[$H\rse{\infty}$]  The space of bounded analytic
functions in $\Bbb D$.
\\
\item[$L\rse{\infty}$]  The space of essentially bounded
functions on $\Bbb T$.
\\
\item[$P_z(w)$]  The Poisson kernel in $\Bbb D$,
$\qquad P_z(w)= \frac{1z^2}{1\overline w z^2}.$
\\
\item[$\rho(z,w)$]  The ``pseudo hyperbolic distance''
between $z$ and $w$ in $\Bbb D$,$$ \rho(z,w) \equiv \bigl
\frac{zw}{1 \overline{w}z} \bigr$$
\item[$d(z,w)$]  The hyperbolic distance between $z$ and
$w$ in $\Bbb D$. $$ d(z,w) \equiv \log{\frac{1+ \rho(z,w)}{1
\rho(z,w)}}$$
\item[$d(X,Y)$]  For $X\subset \Bbb D, Y \subset \Bbb
D$,
$$
\d(X,Y) \equiv \inf\{d(x,y):x\in X,y\in Y\}
$$
\item[$A_{\delta}^{\epsilon}$]  An annulus of thickness
$2\epsilon$ about the circle $\{z = \delta \}$
$$
A_{\delta}^{\epsilon} \equiv \{z\in \Bbb D : \delta  \epsilon \le z
\le
\delta + \epsilon\}$$
\item[$I$]  A half open subinterval of $\Bbb T$
$$ I = \{e^{i\theta} \in \Bbb T: \theta \sb 1 \le \theta < \theta \sb
2 \} \qquad \text{for some} \quad 0\le \theta \sb 1 < \theta \sb 2 \le
2\pi.$$
\item[$I$]  $I = \theta \sb 2  \theta \sb 1 $, the arc
length of $I$.
\\
\item[$Q=Q(I)$]  A Carleson \lq \lq square\rq\rq \ in $\overline
{\Bbb D}$.
$$Q\equiv \{re^{i\theta}: e^{i\theta}\in I, 1 I \le r \le 1 \}
\qquad \text{for some} \quad I \subset \Bbb T$$
\item[$\ell (Q(I))$]  $\ell(Q)\equiv I $
\\
\item[$\beta Q$]  For $\beta \in \Bbb R ^+ $,
$$
\beta Q \equiv \{re^{i\theta}: \frac{\theta \sb 1 + \theta \sb 2}2
 \beta(\frac{\theta \sb 2  \theta \sb 1}2) \le \theta <
\frac{\theta \sb 1 + \theta \sb 2}2+
\beta(\frac{\theta \sb 2  \theta \sb 1}2), \quad 1 \beta (\theta
\sb 2  \theta \sb 1) \le r \le 1 \}
$$
\\
\item[$T(Q)$]  $T(Q) \equiv Q\cap \{ z\in \Bbb D : z < 1
\frac{\ell (Q)}2 \}$
\\
\item[$u\sb a$]  For an inner function $u$ and $a\in \Bbb D$ the
inner function $u\sb a$ is defined by
$$
u\sb a = \frac{ua}{1\overline{a} u}.
$$
\item[$B\sb U$]  If $B$ is a given Blaschke product with
zeros $\{z\sb {\nu}\}$ and $U$
a subset of the open unit disk, then
$$
B\sb U \equiv \prod \limits \sb {z\sb {\nu} \in U} \frac {\overline{ z \sb \nu}}
{z\sb \nu}\cdot \left(\frac {zz\sb \nu}{1\overline{ z\sb \nu} z
}\right)
$$
\item[$\mathcal I$]  The set of interpolating Blaschke products.
\\
\item[$K$]  The closed convex hull of $\mathcal I$.
\\
\item[$\mathcal F$]  The set of finite products of interpolating
Blaschke
products.
\end{description}
\section{Preliminaries}
This section includes some background on the
results presented in the paper and on two interesting open questions.
Several results recorded here will be referred to later.
A holomorphic function $u\in H\rse{\infty}(\Bbb D)$ is called an
inner function if its radial limits (which exist a.e. on $\Bbb T$ by
Fatou's theorem) satisfy $u(e^{i\theta}) =1
$ a.e on $ \Bbb T$.
An example of an inner function is
$$
u(z) =e^{i \alpha} z^m \prod \limits \sb {\nu=1} ^{\infty} \frac {\overline{ z \sb \nu}}
{z\sb \nu}\cdot \left(\frac {zz\sb \nu}{1\overline{ z\sb \nu} z
}\right)
$$
where the convergence of the infinite product is
assured by requiring that $$\sum \limits \sb {\nu} (1z\sb {\nu}) <
+\infty,$$ where each zero $z_{\nu}$ is counted with its multiplicity.
An inner function of this type is called a Blaschke product.
There are nonconstant inner functions with no zeros in $\Bbb D$
called singular functions. They are
$$
u(z) = \exp\left(\int \frac{e^{i\theta}+z}{e^{i\theta}z}\,d\lambda (e^{i\theta})
\right)
$$
where $d\lambda$ denotes a positive measure on $\Bbb T$ which is
singular with respect to the Lebesgue measure.
Now, by the canonical factorization theorem for $H\rse{\infty}$, the
only
other examples of inner functions in $\Bbb D$ are constant multiples
of products of the
above two types. The following theorem
shows that the Blaschke products are uniformly dense in the inner functions.
See \cite{G} for details.
\begin{thm}[Frostman]
Let $v(z)$ be a nonconstant inner function on the unit disc. Then for all
$\zeta$ with $\zeta<1$, except possibly for a set of logarithmic capacity
zero, the function
$$ v\sb \zeta (z) = \frac{v(z)\zeta}{1\overline{\zeta}v(z)}
$$
is a Blaschke product.
\end{thm}
This paper is concerned with those Blaschke
products
which have simple zeros lying in an interpolating sequence,
that is, a sequence of points in $\Bbb D$ such that each
interpolation problem
$$
f(z\sb {\nu}) = w\sb {\nu} \qquad \nu = 1,2,\dots \quad \{w\sb
{\nu}\}\in \ell\rse{\infty}
$$
has a solution $f\in H\rse{\infty}$. The basic facts on interpolating
sequences
are contained in the following theorem of Carleson.
\begin{thm}[Carleson] If $\{z\sb j \} $ is a sequence in the unit disc,
then the following conditions are equivalent:
\begin{enumerate}
\item The sequence is an interpolating sequence.
\item There is a $\delta >0$ such that
$$
\prod \limits \sb {j,j\ne k}\left \frac {z\sb k z\sb j}{1\bar z \sb j z\sb k
}
\right \ge \delta, \qquad k=1,2,\dots
$$
\item The points $z \sb j $ are separated,
$$ \left \frac{z\sb j  z\sb k }{1\bar z\sb j z \sb k}
\right \ge a>0, \qquad j\ne k,
$$
and there is a constant C such that for every \lq\lq square" $Q$
%$$ Q=\{re^{i\theta} : \theta \sb 0<\theta< \theta \sb 0 + \ell (Q) ,
% 1\ell(Q) 0$
there exists an interpolating Blaschke product $B^* $ such that
$$
\sup \limits \sb {z\in \Bbb D } B(z)B^*(z) <\epsilon .
$$
\end{lemma}
The lemma shows that we may as well pose problems 1 and 2 for the
class
$\mathcal F$ instead of $\mathcal I$. A characterization of $\mathcal F$
in terms of zero sequences is given in \cite{GN}.
\begin{lemma}[Garnett, Nicolau] Let $B$ be a Blaschke product and
let $\{z\sb \nu \} $ be its zeros, counted with their multiplicities.
Then the following are equivalent:
\begin{enumerate}
\item $B=B\sb 1 \dots B\sb N $ , and each $B\sb j $ is an interpolating
Blaschke product.
\item There is a constant C such that for every \lq\lq square" $Q$
%$$ Q=\{re^{i\theta} : \theta \sb 0<\theta< \theta \sb 0 + \ell (Q) ,
% 1\ell(Q) 0.
$$
There exist $\lambda=\lambda(\delta),0<\lambda <1$, and $r=r(\delta),00$ is given and $\epsilon(\lambda) >0$ is
sufficiently small,
we have by lemma 2.3
$$
\{z\in \mathbb D : u(z)\le \epsilon \} \subset
\bigcup \limits \sb {i=1} ^N \{z\in \mathbb D : B\sb i(z) \le
\epsilon^{\frac1N}\} \subset \bigcup \limits \sb n \{z:\rho(z,z\sb n) < \lambda \}.
$$
If $\lambda >0$ is sufficiently small then
by the density condition on the set $\{z\sb n\}$ there is an $M>0$
such that any connected component of $\bigcup \limits \sb n
\{z:\rho(z,z\sb n) < \lambda \}$ is the union of at most $M$ of the
disks $\{z:\rho(z,z\sb n) < \lambda \}$. By the argument principle,
$u\sb a \in \mathcal F$ whenever $a <\epsilon(\lambda)$. This shows
that if $v$ is any inner function, then the set
$\left\{a \in \Bbb D: v\sb a \in
\cal F\right\}$
is open.}
\end{rmk}
The last (well known) lemma of this
section
gives us information on the modulus of a Blaschke product in
terms of its zero sequence. It's proof is an exercise with the
identity of Lagrange:
$$
\frac{(1z^2)(1w^2)}{1\overline{w}z^2}=1\left\frac{zw}
{1\overline{w}z} \right^2 \qquad \text{for all}\quad z,w \in \mathbb D
$$
\begin{lemma}
Let $0a >0
\qquad \text{for each}\quad \mu.
$$
Then
$$
\frac12 \sum \limits \sb {\mu}
\frac{(1z^2)(1z\sb {\mu}^2)}{1\overline{z\sb {\mu}}z^2}
\le \log{\frac1{B(z)}}
\le
\frac12 C(a)\sum \limits \sb {\mu}
\frac{(1z^2)(1z\sb {\mu}^2)}{1\overline{z\sb {\mu}}z^2}.
$$
The left hand inequality holds without any condition except $0\delta >0$ be
fixed. Let $E\sb {\delta}=\{z\in \mathbb D: u(z)= \delta\}.$
The following two statements
are equivalent.
\begin{enumerate}
\item There exists an $\epsilon>0$ such that
$$
u\sb a\in \cal F, \quad \text{for each}\quad a \in A\sb{\delta}^{\epsilon}.
$$
\item There are
$ M>0 $ and $ \eta>0$ such that for each $z\in E\sb \delta$,
there exists a square $Q$ with the following two properties:
\begin{enumerate}
\item
$d(z,T(Q))0$ and $M>0$, we
can find some
$ z= z(\eta, M)\in E\sb {\delta} $ such that
all squares $Q$ with $d(z,T(Q))0$ is sufficiently large
then
$u\sb 2(w) > \frac{\delta}2$ for all $w$ on
the
top edge of $Q\sb z$. We further require that
$$
(\frac{\delta}2)^{\alpha^N} > e^{\eta}
$$
where $\alpha $ is the constant from lemma 3.1
and that
$$
\int \limits \sb {\mathbb T \setminus NQ\sb z}
P\sb {z'}(e^{i\theta})\,d\theta \le \frac {\eta}{10\log{\frac 1{\delta}}}
\qquad \text{for each} \quad z' \in S(z,N).
$$
Let $\mathcal C_N$
be the collection of $N 2^{2N}$ disjoint Carleson squares
contained in $NQ\sb z$, with
$$
\ell (Q\sb j) = 2^{2N} \ell(Q\sb z) \qquad \text{for each} \quad Q\sb j \in
\mathcal C_N.
$$
Note that any square $S$ whose top half contains a zero has
$\frac{\mu(S)}{\ell(S)}$
differing by at least $\frac12$ from some square of exactly half the
size
contained inside of it.
So if $M$ is large enough and $\eta< \frac12$ then $\mu_{u\sb 1}$ is
supported
in $\bigcup \limits \sb {\cal C_N} Q\sb
j$.
We have
\begin{eqnarray}
\biggl\log{u(z')}& & 2\pi \frac{\mu \sb u (Q\sb z)}{\ell (Q\sb z)}
\biggr \le
\biggl\log{u(z')} +\log{u\sb 1 (z') }
\biggr \nonumber
\\{}\nonumber\\&+& \biggl\log{\frac1{u\sb 1 (z')}}  \int \limits \sb {NQ\sb z}
\frac{1z'^2}{1\overline{w}z'^2}\,d\mu\sb u (w) \biggr
\nonumber
\\{}\nonumber\\&+&\biggl\int \limits \sb {NQ\sb z}
\frac{1z'^2}{1\overline{w}z'^2}\,d\mu\sb u (w) 
\sum\limits \sb {Q\sb j \in \cal C_N}\frac
{\mu\sb u(Q\sb z)}{\ell(Q\sb z)}\int \limits \sb {Q\sb j \cap \Bbb T}
P\sb {z'}(e^{i\theta})\,d\theta \biggr
\nonumber
\\{}\nonumber\\ &+&\biggl\sum\limits \sb {Q\sb j \in \cal C_N}\frac
{\mu\sb u(Q\sb z)}{\ell(Q\sb z)}\int \limits \sb {Q\sb j \cap \Bbb T}
P\sb {z'}(e^{i\theta})\,d\theta

2\pi \frac{\mu \sb u (Q\sb z)}{\ell (Q\sb z)}
\biggr
\nonumber
\end{eqnarray}
for any $z'\in S(z,N)$.
The first and fourth terms on the right are each smaller than
$\eta$
by our choice of $N$. By application of lemma 2.4 and properties of the
Poisson
kernel respectively, the second and
third terms are each smaller than $\eta$ if $M$ is
sufficiently
large. This shows that $u(z') \delta
 < 4\eta $ for any $z'\in S(z,N)$. As $N$ can be arbitrarily large,
Schwarz's lemma and remark 2 after lemma 2.3 then show that condition
(1) of the theorem fails.
For the remaining half of the proof, let $Q\sb z$ be as
before,
the square with $z$ at the center of its top edge.
Let $I=Q\sb z \cap \Bbb T$ denote the base of $Q\sb z$ and let
$S\sb m$ denote the union of $2^m$ disjoint Carleson squares of length
$r_m\equiv 2^{m}\ell(Q\sb z)$ contained in $Q\sb z$.
Suppose now that for any given $\tau>0$ and integer $n>0$,
there exists a point $z\sb 0(\tau,n)\in \Bbb D$ such that
$u(z\sb 0)=\delta$,
but $u(z\sb 0)u(w)<\tau$ for all $w\in 2^nQ\sb {z\sb 0}\cap \{\xi
\in \mathbb D:\xi\le12^{2n}\ell(Q\sb {z\sb 0})\}$.
By lemma 2.2 and Schwarz's lemma, this is true for any inner function $u$ for which
condition (1) of the theorem fails.
We will
show that given $M>0$ and $\eta>0$, if $n$ is large enough and $\tau$ small
enough, then all squares $Q$ with $d(z\sb 0,T(Q))0$, we will have
$$
B\sb {N,\sigma}(z\sb \sigma)= (1z\sb {\sigma}^2)B'\sb N(z\sb
{\sigma})
> 1\epsilon \quad \text{for each}\quad \sigma
$$
if $N= N(\epsilon)$ is sufficiently large.
Altering the previous notation, we write
$$
B\sb N ^w \equiv \frac{B\sb N w}{1\overline{w}B\sb N}.
$$
By lemma 2.3, there are
positive numbers
$00$ satisfy $\frac1{2\pi}\log{\frac1{\delta}}=2^{(N+1)}$.
Then by theorem 3.1 we have
$$
(*) \qquad\frac{BB\sb N +a}{1+\overline{a}BB\sb N} \in
\mathcal F \qquad\text{for each}\quad a
\text{ with } a= \delta.
$$
Now choose $w\in \Bbb D$ with $r\sb N ^2<w1\frac{\pi}{128}$ and further, that each dyadic square
$Q$ with $\ell(Q)= \frac{\pi}{128}$ satisfies
$$
\frac{\mu \sb B(Q)}{\ell(Q)} < \epsilon \sb 0.
$$
Let $\{Q\sb j ^1\}= \mathcal G_1$ be the set of maximal dyadic squares with
$$
\frac{\mu \sb B (Q\sb j ^1)}{\ell (Q \sb j ^1)}>2^{M}\epsilon\sb 0,
$$
where $M>0$ will later be chosen as large as needed and $\epsilon \sb 0>0$ as
small as needed. We have
$$
\sum \limits \sb{\mathcal G \sb 1}\ell (Q\sb j^1)\le 2\pi2^{M}.
$$
Now let $\{V\sb k ^1\}$ be the set of dyadic squares which are
maximal among dyadic squares contained in some $Q\sb j ^1 \in \mathcal G
\sb 1$
with respect to the property that
$$\frac{\mu(V\sb k ^1)}{\ell (V\sb k ^1)}<\epsilon \sb 0\,.$$
Because $B \to 1$ nontangentially almost everywhere, we have
$$
\sum \limits\sb{V\sb k ^1 \subset Q\sb j ^1}\ell (V\sb k ^1) =
\ell (Q\sb j ^1)
\qquad \text{for each}\quad Q\sb j ^1 \in \cal G \sb 1.
$$
To see this, let $Q$ be a square such that $z\in \Bbb D$ is contained in
$T(Q)$. We then have
$$
\log {B(z)} \le C\sum \limits \sb {w\sb n \in Q}
\frac{(1z^2)(1w\sb n^2)}{1\overline{w\sb n}z^2}
$$
$$
\le C'\sum \limits \sb {w\sb n \in Q}\frac{1w\sb n}{1z}
$$
$$
\le C'\frac{\mu \sb B(Q)}{\ell(Q)}
$$
and this implies what was claimed.
Let
$$
R\sb j ^1 = Q\sb j ^1 \setminus \bigcup \limits \sb{V\sb k ^1
\subset Q\sb j ^1}
V\sb k ^1 \qquad \text{for each}\quad Q\sb j ^1 \in \cal G\sb 1.
$$
Define $\cal G\sb 2 =\{Q\sb j ^2\}$ to be the
set of dyadic squares which are
maximal among squares contained in $V\sb k^1 $ for some $k$
with respect to the property that
$$\frac{\mu(Q\sb j ^2)}{\ell (Q\sb j ^2)}\ge 2^{M}\epsilon \sb 0\,.$$
We have
$$
\sum \limits \sb{Q\sb j ^2 \subset V\sb k ^1}\ell (Q\sb j ^2)\le
2^{M}\ell (V \sb k ^1),\qquad \text{for each}\quad k.
$$
We form the set of squares $\{V\sb k ^2\}$ and regions $R\sb j ^2$ as
before
and continue to obtain $Q\sb j ^n$, $ R\sb j ^n$ and $V\sb k ^n$.
If $2^{M}\epsilon\sb 0
<\frac12$ our assumptions imply that all zeros of $B$ are contained in the
interiors of the regions $R\sb j ^n$.
Let $\sigma$ and $\nu$ index the pairs $(n,k)$.
Choose an integer $N>0$ and let
$B\sb N \in \cal I$ be the product with one zero
$z\sb \nu= z\sb {(n,k)}$
at the center of the top edge of each square $2^{N}V\sb k^n$ for all
$n$ and $k$.
We will also write $V\sb k^n= V\sb {\nu}$. Let
$$
B\sb {N,\sigma}\equiv \prod \limits \sb {\nu\ne \sigma}\frac{\overline{z\sb \nu}}
{z\sb \nu}\frac{zz\sb \nu}{1\overline{z\sb \nu}z}
$$
We claim that given $\epsilon>0$, if $N$ and $M$ are sufficiently
large, then
$$
B\sb {N,\sigma}(z\sb \sigma)\ge1\epsilon.
$$
Let $S\sb n$ denote the square $2^{(Nn)}V\sb {\sigma}$.
Then we have
$$
\log{\frac1{B\sb {N,\sigma}(z\sb{\sigma})}}\sim \sum \limits\Sb z\sb \nu \in V\sb \sigma\\
\nu\ne\sigma \endSb
\frac{(1z\sb \sigma^2)(1z\sb \nu^2)}{1\overline{z\sb \sigma}z\sb \nu^2}
+
\sum \limits \sb {n> N}\sum \limits \sb {z\sb \nu \in S\sb n\setminus
S\sb {n1}}
\frac{(1z\sb \sigma^2)(1z\sb \nu^2)}{1\overline{z\sb \sigma}z\sb \nu^2}.
$$
As in the
previous proof the sum on the right is as small as we like if $N$
is large enough. The sum on the left is
$$
\le C\sum \limits \sb {n\le N}\sum \limits \sb {z\sb \nu \in S\sb n\setminus
S\sb {n1}}2^{2n}\left(\frac{1z\sb \nu^2}{1z\sb \sigma^2}\right)
$$
$$
\le C' \frac{2^N}{\ell(V\sb \sigma)}\sum \limits \sb {n\le N}\sum \limits \sb
{z\sb \nu \in S\sb n\setminus
S\sb {n1}}2^{2n}(1z\sb \nu^2)
$$
$$
\le C'' \frac{2^N}{\ell(V\sb \sigma)}\sum \limits \sb {n\le N}2^{2n}
\cdot2^{(M+N)}\ell(V\sb \sigma)
$$
$$
\le C' \,2^{M}\sum \limits \sb {n\le N}2^{n}
$$
and therefore is as small as we like if $M$ is sufficiently large.
As in the proof of the previous theorem there are positive numbers
$01$ , $ \cal H\sb n$ is the set of maximal squares in
$$
\cal G' \sb n \setminus \bigcup \limits \sb {i=1}^{n1} \cal H\sb i.
$$
In the terminology of \cite{G} we have divided each set of squares
$\mathcal G'\sb n$ into generations $\mathcal H\sb i$.
Note that if $\Omega \sb j ^n \in \mathcal H \sb i$ then
$$
\sum \limits \sb {\{\Omega \sb k ^n \in \mathcal H\sb {i+2}:\Omega \sb k ^n
\subset \Omega \sb j ^n\}} \ell (\Omega \sb k ^n )
\le
\frac14 \ell (\Omega \sb j ^n ).
$$
Define
$$
\cal G\sb {n,0}=\bigcup \limits \sb {i=1}^{\infty}\{Q\sb j ^n \in \cal G\sb n:
\Omega \sb j ^n \in \cal H\sb {2i1}\}
$$
and
$$
\cal G\sb {n,1}=\bigcup \limits \sb {i=1}^{\infty}\{Q\sb j ^n \in \cal G\sb n:
\Omega \sb j ^n \in \cal H\sb {2i}\},
$$
and define
$$
W\sb {n,\alpha}=\bigcup \limits \sb {Q\sb j ^n \in \cal G\sb {n,\alpha}}
R\sb j ^n \qquad \alpha=0,1
$$
and
$$
U\sb \alpha = \bigcup \limits \sb {n=1}^{\infty}W\sb {n,\alpha}\qquad \alpha=
0,1.
$$
We will now use these definitions to
factor
the product $B$.
Let
$$
B\sb \alpha = B\sb {U\sb \alpha}, \quad \alpha=0,1
$$
and let
$$
B\sb {N,n,\alpha}= \prod \limits \sb {\{\nu=(n,k),j:V\sb k ^n \subset
G\sb j ^n \in \mathcal G\sb {n,\alpha}\}}
\frac{\overline{z\sb \nu}}
{z\sb \nu}\left(\frac{zz\sb \nu}{1\overline{z\sb \nu}z}\right),
\qquad \alpha=0,1
$$
and
$$
B\sb {N,\alpha} = \prod \limits \sb n B\sb {N,n,\alpha}\,\,,\qquad \alpha=0,1.
$$
Then $B= B\sb 0 B\sb 1 $ and $B\sb N = B\sb {N,0}B\sb {N,1}$.
Let
$$
B\sb {n,\alpha}^* =
B\sb {\alpha}\sb {W\sb {n,\alpha}}\cdot B\sb {N,n,\alpha}
$$
and let
$$
B\sb {\alpha} ^*= \prod \limits \sb n B\sb {n,\alpha}^*
= B\sb {\alpha}B\sb {N,\alpha}.
$$
We claim that
$$
\frac{B\sb \alpha ^*\delta}{1\overline{\delta}B\sb \alpha ^*}\in \cal F,
\qquad \alpha=0,1
$$
for all $\delta$ with $\frac1{2\pi}\log{\frac1{\delta}}
=\frac78 2^{N}$.
This will follow from theorem 3.1 once we
show that every point
in $\Bbb D$ is within bounded hyperbolic distance of some dyadic square
$Q$ with
$$
\frac{\mu\sb {B\sb \alpha ^*}(Q)}{\ell(Q)}\notin[\frac342^{N},2^{N}).
$$
Let $\alpha$ be fixed.
If $Q\sb j ^n \in \cal G\sb {n,\alpha}$ then
$$
\frac{\mu \sb {B\sb {n, \alpha} ^*}(\Omega \sb j ^n)}
{\ell(\Omega \sb j ^n)} \le \left(\frac{2^{N}+2^M\epsilon\sb 0}2\right)
+\left(\frac14\right)\left(\frac{2^{N}+2^M\epsilon\sb 0}2\right)
+\left(\frac14\right)^2\left(\frac{2^{N}+2^M\epsilon\sb 0}2\right)+
\cdots
$$
$$
\le \frac46(2^{N}+2^M\epsilon\sb0).
$$
Here the factor $\frac14$ is provided by the \lq \lq oddeven\rq\rq\
splitting within each set $\mathcal G \sb n$.
Now summing over all the sets $\mathcal G\sb n$ we have
$$
\frac{\mu \sb {B\sb \alpha ^*}(\Omega \sb j ^n)}{\ell (\Omega \sb j ^n)}
\le (1+ 2^{M}+2^{2M}+\cdots)\cdot
\frac46(2^{N}+2^M\epsilon\sb 0)<\frac342^{N}
$$
if $\epsilon \sb 0$ is small.
Therefore if $z$ is not in any $R\sb j ^n$ with $Q\sb j ^N\in
\cal G\sb {n,\alpha}$
then it is hyperbolically near the top half of a dyadic square $Q$ with
$\frac {\mu \sb {B\sb {\alpha} ^*}(Q)}{\ell(Q)}<\frac342^{N}$.
This is because the point $z$ is inside some square $V\sb {\nu}$ but
outside
any $R\sb j ^n$. Moving a fixed hyperbolic distance away from the zero
$z\sb {\nu}$ if necessary, there are only squares with
$\frac{\mu}{\ell}< \frac34 2 ^{N}$ directly below.
On the other hand,
if $z\in R\sb j ^n$ with $Q\sb j ^n \in \cal G\sb {n,\alpha}$
then $z$ is contained in the top
half of some dyadic square $Q$ with $\frac {\mu \sb {B\sb {\alpha} ^*}(Q)}
{\ell(Q)}\ge 2^{N} $. The claim is proved.
Following the proof of the previous theorem, we may replace $B\sb N$
by $B\sb N ^w=\frac{B\sb N w}{1\overline{w}B\sb N}$ where
$r\sb N ^2<w