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\begin{document}
\title[More Relative Angular Derivatives]{More Relative Angular Derivatives}
\author{Jonathan E. Shapiro}
\address{Department of Mathematics, California Polytechnic State University, San Luis
Obispo, CA 93407}
\email{jshapiro@calpoly.edu}
\date{}
\subjclass{Primary 46E22, 46E30}
\keywords{angular derivative, Hardy space, Aleksandrov measure}
\maketitle
\begin{abstract}
The notion of the angular derivative of a holomorphic self-map $b$ of the unit
disk has been generalized to that of an angular derivative of $b$ relative to
an inner function $u$. In this paper, we provide three more conditions
equivalent to the existence of an angular derivative of $b$ relative to $u$,
and use these notions to provide a generalization of Julia's lemma.
\end{abstract}
\section{Introduction}
Relative angular derivatives came into being as a generalization of the notion
of the angular derivative of a holomorphic function on the unit disk. In this
paper, we will further analyze the notion of an angular derivative of a
holomorphic self-map of the unit disk relative to a nonconstant inner function.
Let $b$ be a holomorphic self-map of the unit disk, that is, an analytic
function on the unit disk $\mathbb{D}$ of the complex plane with $|b|<1$ on
$\mathbb{D}$. We will take $u$ to be our nonconstant inner function --- a
holomorphic function on $\mathbb{D}$ with $|u|=1$ almost everywhere on
${\partial\mathbb{D}}$. This notation will remain fixed.
The analysis and definition of relative angular derivatives comes primarily
from the viewpoint of the Aleksandrov measures $\mu_{\lambda}$ and
$\nu_{\lambda}$ ($\lambda\in\partial\mathbb{D}$), which we derive from our
functions $b$ and $u$, respectively. These measures are defined and discussed
in Section \ref{aleksandrovsec}. Throughout this paper, we will use $m$ to
denote the usual normalized Lebesgue measure on the unit circle. We will also
use the notation $\mu^{a.c.}$ and $\mu^{s}$ to denote the absolutely
continuous and singular parts of the measure $\mu$ ($=\mu_{1}$) with respect
to $m$ (similarly for $\mu_{\lambda}^{a.c.}$ and $\mu_{\lambda}^{s}$). For any
function $f$ on the unit disk $\mathbb{D}$, $f_{r} $ will denote the function
on the boundary ${\partial\mathbb{D}}$ such that $f_{r}(e^{i\theta
})=f(re^{i\theta})$ for $r<1$.
The relationship between Aleksandrov measures and angular derivatives has
recently been developed by many people. The most direct connection comes
essentially from \cite{RN}, and is developed more in \cite[Ch. 7]{SDF} and
\cite[VI-7]{DS},
\begin{theorem}
The function $b$ has angular derivative at a point $z_{0}\in\partial
\mathbb{D}$ (where $\left| b(z_{0})\right| =$ $1$) exactly where its
corresponding Aleksandrov measure $\mu_{\lambda}$ has an atom, and
$\mu_{\lambda}(\left\{ z_{0}\right\} )=1/\left| b^{\prime}(z_{0})\right| .$
\end{theorem}
\noindent Others can be found in \cite{JES1}, in which the author has used
Aleksandrov measures to introduce the notion of a relative angular derivative,
\cite{CM}, in which Aleksandrov measures are used to give a very nice way to
describe the essential norm of a composition operator, and \cite{JES}, in
which the author has also used Aleksandrov measures to generalize other
theorems which give properties of composition operators and composition
operator differences in terms of angular derivatives.
In this paper we aim to develop further this type of useful generalization of
angular derivatives by studying the more broad category of relative angular derivatives.
Section \ref{aleksandrovsec} contains some background material about the
Aleksandrov measures. In Section \ref{gensec}, we are reminded how the angular
derivative was generalized to become the relative angular derivative. It is in
this section that we present the theorem from \cite{JES1} which lists six
conditions, all of which are equivalent, any one of which can be used as a
definition of a relative angular derivative.
The first goal of this paper, in Section \ref{familysec}, is to use the
properties of the family of Aleksandrov measures, $\left\{ \mu_{\lambda
}\right\} _{\lambda\in{\partial\mathbb{D}}}$, to prove that three other
conditions are each equivalent to the six which define a relative angular
derivative. These new conditions arise by consideration of the behavior of the
boundary values of the generalized difference quotient relative to the
measures ${\mu_{\lambda}}$ for $\lambda\in{\partial\mathbb{D}}$.
In Section \ref{juliasec}, we see how other results, such as Julia's lemma,
can be generalized by methods similar to those used to create relative angular derivatives.
\section{The Aleksandrov Measures\label{aleksandrovsec}}
For $\lambda\in{\partial\mathbb{D}}$, the function Re$(\frac{\lambda
+b}{\lambda-b})$ is positive, and, as the real part of an analytic function,
harmonic (on the disk, $\mathbb{D}$). It is thus the Poisson integral of a
positive measure on ${\partial\mathbb{D}}$, which we will call ${\mu_{\lambda
}} $. We have, then,
\[
\text{Re}\left( \frac{\lambda+b(z)}{\lambda-b(z)}\right) =\int
_{\partial\mathbb{D}}P(\theta,z)d{\mu_{\lambda}}(e^{i\theta})=P{\mu_{\lambda}%
}(z)
\]
and the Herglotz integral representation,
\[
\frac{\lambda+b(z)}{\lambda-b(z)}=\int_{\partial\mathbb{D}}H(\theta
,z)d{\mu_{\lambda}}(e^{i\theta})+i\,\text{Im}\frac{\lambda+b(0)}{\lambda
-b(0)}.
\]
Note that for $z\in\mathbb{D}$, the Poisson kernel, $P(\theta,z)=\frac
{1-|z|^{2}}{|e^{i\theta}-z|^{2}}$, is the real part of the Herglotz kernel,
$H(\theta,z)=\frac{e^{i\theta}+z}{e^{i\theta}-z}$. The measure $\mu_{1}$ we
shall simply call $\mu$. The measure $\nu$ is similarly defined to correspond
with the inner function $u$.
The following are some properties of the Aleksandrov measures defined above:
\begin{itemize}
\item All positive Borel measures on ${\partial\mathbb{D}}$ are associated
with functions in this way.
\item The absolutely continuous part of $\mu$ is given by $\frac{1-|b|^{2}%
}{|1-b|^{2}}$ times the normalized Lebesgue measure (on ${\partial\mathbb{D}}$).
\item The measure $\mu$ is singular if and only if $b$ is an inner function,
i.e., $|b|=1$ almost everywhere on ${\partial\mathbb{D}}$.
\item For $\mu_{\lambda}^{s}$-a.e. $\xi\in{\partial\mathbb{D}}$ we have
$P{\mu_{\lambda}}(\xi)=\infty$ and thus $b(\xi)=\lambda$.
\end{itemize}
\section{Relative Angular Derivatives\label{gensec}}
In \cite{JES1}, the relative angular derivatives were defined by extending the
notion of the angular derivative of the function $b$ by replacing the identity
function $z$ by an arbitrary (nonconstant) inner function $u$ in the
denominator of the standard difference quotient, $\frac{b(z)-b(z_{0})}%
{z-z_{0}}$. The behavior of this generalized difference quotient,
$\frac{1-b(z)}{1-u(z)}$, was then studied.
The main theorem from \cite{JES1} which provided the basis for the definition
of the relative angular derivative was:
\begin{theorem}
\label{mainthm} The following conditions are equivalent:
\begin{enumerate}
\item $\nu\ll\mu$ and $\frac{d\nu}{d\mu}\in L^{2}(\mu)$;\label{measurecond}
\item $\frac{1-b}{1-u}k_{0}^{u}\in\mathcal{H}(b)$ \ (the de Branges-Rovnyak space);\label{quotconda}
\item $\frac{1-b}{1-u}k_{w}^{u}\in\mathcal{H}(b)$\ \ for all $w\in\mathbb{D}$;\label{quotcondb}
\item $\int_{\partial\mathbb{D}}\left| \frac{1-b_{r}}{1-u_{r}}\right| d\nu$
stays bounded as $r\nearrow1$;\label{intcond}
\item $\frac{1-b}{1-u}\in H^{2}$ and $\frac{1-b}{1-u}\in H^{2}(\mu^{a.c.})$;\label{h2accond}
\item $\frac{1-b}{1-u}\in H^{2}$ and $\frac{1-b}{1-u}\in H^{2}(\mu)$;\label{h2cond}
\end{enumerate}
\end{theorem}
If any of the above hold, then we say that $b$ has an angular derivative
relative to $u$.
Our goal now is to continue the study of the boundary values of the
generalized difference quotient, $\frac{1-b}{1-u}$, relative to the measures
${\mu_{\lambda}}$, and to show
\begin{theorem}
\label{mainthm2}The following are equivalent, and each is equivalent to the
six conditions in Theorem \ref{mainthm}:
\begin{enumerate}
\item $\frac{1-b}{1-u}\in H^{2}(\mu^{a.c.})$ and there is a function $h\in
L^{1}(m)$ such that for almost every $\lambda\in{\partial\mathbb{D}}$ we have
$\int_{\partial\mathbb{D}}\left| \frac{1-b}{1-u}\right| ^{2}d{\mu_{\lambda}%
}c$. We will then have, since the
Poisson kernel is an approximate identity, for $\mu^{s}$-a.e. $\xi$ in such a
set,
\[
\frac{P\nu(z)}{P\mu(z)}=\frac{P\frac{d\nu}{d\mu}\mu(z)}{P\mu(z)}%
\rightarrow\frac{d\nu}{d\mu}(\xi)>c
\]
as $z\rightarrow\xi$ nontangentially. Hence no constant $c$ can make the
desired inequality true for all $z\in\mathbb{D}$. This completes the proof of
the theorem. \hfill$\square$
We can look at the restriction of the above theorem in the same special case
we were studying earlier, i.e., where $u(z)=\overline{z_{0}}z$ for some
$z_{0}\in{\partial\mathbb{D}}$. Then we have
\begin{theorem}
[Julia's Lemma]\label{julialemma} For a holomorphic self-map of the disk $b$
with an angular derivative at the point $z_{0}$,
\[
\frac{|b(z)-b(z_{0})|^{2}}{1-|b(z)|^{2}}\le c\frac{|z-z_{0}|^{2}}{1-|z|^{2}}%
\]
for all $z\in\mathbb{D}$ where we can take $c=|b^{\prime}(z_{0})|$.
\end{theorem}
\noindent\emph{Proof:} This is an application of Theorem \ref{juliagen} with
the function $\overline{b(z_{0})}b$ and $u(z)=\overline{z_{0}}z$. Again, we
get $\nu=\delta_{z_{0}}$, and since $b$ has an angular derivative at $z_{0}$,
$\mu$ has an atom at $z_{0}$, and we have $\nu\ll\mu$ and $\frac{d\nu}{d\mu
}\in L^{2}(\mu)$. In fact, by a result in the proof of Theorem 15 in
\cite{JES1}, we have $\frac{d\nu}{d\mu}(z_{0})=|b^{\prime}(z_{0})|$, and is
zero at all other points, so $\left\| \frac{d\nu}{d\mu}\right\| _{\infty
}=\frac{d\nu}{d\mu}(z_{0})=|b^{\prime}(z_{0})|$, and the theorem applies. The
theorem gives us directly that for all $z\in\mathbb{D}$
\[
\frac{|1-\overline{b(z_{0})}b(z)|^{2}}{1-|\overline{b(z_{0})}b(z)|^{2}}%
\leq\left\| \frac{d\nu}{d\mu}\right\| _{\infty}\left( \frac{|1-\overline
{z_{0}}z|^{2}}{1-|\overline{z_{0}}z|^{2}}\right)
\]
and this, given that $|b(z_{0})|=1$ (since $b$ has an angular derivative at
$z_{0}$) is easily seen to be equivalent to
\[
\frac{|b(z)-b(z_{0})|^{2}}{1-|b(z)|^{2}}\leq|b^{\prime}(z_{0})|\frac
{|z-z_{0}|^{2}}{1-|z|^{2}}.
\]
This completes the proof of Julia's lemma as a special case of the Theorem
\ref{juliagen}. \hfill$\square$
Another way to look at our generalization of Julia's lemma is to see that we
showed, for appropriate $b$ and $u$,
\[
\frac{\displaystyle\frac{|1-b(z)|^{2}}{1-|b(z)|^{2}}}{\displaystyle
\frac{|1-u(z)|^{2}}{1-|u(z)|^{2}}}=\frac{P\nu(z)}{P\mu(z)}%
\]
or
\begin{equation}
\frac{\displaystyle\frac{|1-b(z)|}{1-|b(z)|}}{\displaystyle\frac
{|1-u(z)|}{1-|u(z)|}}=\frac{\left( {\displaystyle\frac{1+|b(z)|}{1+|u(z)|}%
}\right) \left( {\displaystyle\frac{P\nu(z)}{P\mu(z)}}\right)
}{\displaystyle\left| \frac{1-b(z)}{1-u(z)}\right| }.\label{jquotient}%
\end{equation}
The left side of the equality above is the ratio of two terms. The top,
$\frac{|1-b(z)|}{1-|b(z)|}$ is, as $z\rightarrow\xi$ for some $\xi\in
{\partial\mathbb{D}}$, in some sense, the angle at which $b(z)$ approaches
$b(\xi)$. (Note that $b(\xi)$ $=1$ for $\nu$-a.e. $\xi$). Similarly, the lower
term is the same for $u$. The right side of the equality approaches $1$ as
$z\rightarrow\xi$ nontangentially for $\nu$-a.e. $\xi\in{\partial\mathbb{D}}$.
This is true because as $z\rightarrow\xi$ nontangentially for $\nu$-a.e.
$\xi\in{\partial\mathbb{D}}$, both $b(z)$ and $u(z)\rightarrow1$, so
$\frac{1+|b(z)|}{1+|u(z)|}\rightarrow1$. Also, we know $\frac{1-b(z)}%
{1-u(z)}\rightarrow\frac{d\nu}{d\mu}(\xi)$, which is nonzero for $\nu$-a.e.
$\xi$, and, as we have seen before, $\frac{P\nu(z)}{P\mu(z)}\rightarrow
\frac{d\nu}{d\mu}(\xi)$. Our conclusion is
\begin{theorem}
\label{angle} If $b$ has an angular derivative relative to $u$, then for $\nu
$-a.e. $\xi\in{\partial\mathbb{D}}$, as $z\rightarrow\xi$ nontangentially, the
angle at which $b(z)$ approaches 1 is the same (in the limit) as the angle at
which $u(z)$ approaches 1.
\end{theorem}
A different way to look at equation (\ref{jquotient}) is to write it as
\begin{align*}
\frac{1-|b(z)|}{1-|u(z)|} & =\frac{\displaystyle\left| \frac{1-b(z)}%
{1-u(z)}\right| ^{2}}{\left( {\displaystyle\frac{1+|b(z)|}{1+|u(z)|}%
}\right) \left( {\displaystyle\frac{P\nu(z)}{P\mu(z)}}\right) }\\
& \rightarrow{\displaystyle\frac{d\nu}{d\mu}(\xi)}%
\end{align*}
as $z\rightarrow\xi$ nontangentially, for $\nu$-a.e. $\xi$. This is true since
as $z\rightarrow\xi$ nontangentially, for $\nu$-a.e. $\xi$, $\left|
\frac{1-b(z)}{1-u(z)}\right| \rightarrow\frac{d\nu}{d\mu}(\xi)$,
$\frac{1+|b(z)|}{1+|u(z)|}\rightarrow1$, and $\frac{P\nu(z)}{P\mu
(z)}\rightarrow\frac{d\nu}{d\mu}(\xi)$. Since $\frac{d\nu}{d\mu}\in L^{1}%
(\nu)$ this gives us
\begin{theorem}
If $b$ has an angular derivative relative to $u$, then the function on
${\partial\mathbb{D}}$ whose values are the nontangential limits of
$\frac{1-|b|}{1-|u|}$ is in $L^{1}(\nu)$ and its norm is equal to $\left\|
\frac{d\nu}{d\mu}\right\| _{L^{1}(\nu)}$, or $\left\| \frac{d\nu}{d\mu
}\right\| _{L^{2}(\mu)}^{2}$.
\end{theorem}
\medskip
\noindent\emph{Remark:} Because of the above theorem, we might want to add the
condition
\[
\text{the boundary function of}\ \ \frac{1-|b|}{1-|u|}\ \ \text{is
in}\ \ L^{1}(\nu)
\]
to the list of conditions equivalent to the assertion that $b$ has an angular
derivative relative to $u$. We cannot do this, however, because the converse
of the above theorem does not hold. Pick, for example, $b(z)=-z$ and $u(z)=z$.
$\frac{1-|b|}{1-|u|}=1\in L^{1}(\nu)$, (note that again we define the boundary
values of $\frac{1-|b|}{1-|u|}$ as the nontangential limits of values in the
disk) but because $\nu=\delta_{1}$ and $\mu=\delta_{-1}$ we do not have
$\nu\ll\mu$. In this case, however, $-b$ has an angular derivative relative to
$u$. We can pick $b$ and $u$, however, so that $\bar{\xi}b$ will not have an
angular derivative relative to $u$ for any $\xi\in{\partial\mathbb{D}}$. Such
an example can be found by taking $b(z)=z^{3}$ and $u(z)=z^{2}$. Then the
boundary function of $\frac{1-|b|}{1-|u|}=\frac{1-|z^{3}|}{1-|z^{2}|}%
=\frac{(1-|z|)(1+|z|+|z|^{2})}{(1-|z|)(1+|z|)}=3/2$ (everywhere on
${\partial\mathbb{D}}$), but $\mu$ has atoms at the cube roots of unity,
whereas $\nu$ has atoms at the square roots of unity, so $\nu\not \ll\mu$,
therefore $b$ does not have an angular derivative relative to $u$, nor, it is
clear, would $\bar{\xi}b$ have an angular derivative relative to $u$ for any
$\xi\in{\partial\mathbb{D}}$.
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\end{document}