# Multipoint Julia theorems

Marco Abate$\dagger$ Marco Abate
Dipartimento di Matematica
Università di Pisa
Largo Pontecorvo 5, I-56127 Pisa
Italy<EMAIL_ADDRESS>To Edoardo Vesentini

###### Abstract.

Following ideas introduced by Beardon-Minda and by Baribeau-Rivard-Wegert in
the context of the Schwarz-Pick lemma, we use the iterated hyperbolic
difference quotients to prove a multipoint Julia lemma. As applications, we
give a sharp estimate from below of the angular derivative at a boundary
point, generalizing results due to Osserman, Mercer and others; and we prove a
generalization to multiple fixed points of an interesting estimate due to
Cowen and Pommerenke. These applications show that iterated hyperbolic
difference quotients and multipoint Julia lemmas can be useful tools for
exploring in a systematic way the influence of higher order derivatives on the
boundary behaviour of holomorphic self-maps of the unit disk.

$\dagger$ Partially supported by 2017 PRIN grant “Real and Complex Manifolds:
Topology, Geometry and Holomorphic Dynamics”, Ministry of University and
Research, Italy, and by 2020 PRA grant “Sistemi dinamici in logica, geometria,
fisica matematica e scienza delle costruzioni”, University of Pisa, Italy.

2020 Mathematics Subject Classification: 30C80 (primary); 30E25, 30F45, 30J10
(secondary).

Keywords: Julia lemma; hyperbolic difference quotient; boundary dilation
coefficient; angular derivative; Cowen-Pommerenke estimate

##  1\. Introduction

The classical Schwarz-Pick lemma [Schwarz, Picka, Pickb, Caratheodory] says
that every holomorphic self-map of the unit disk $\mathbb{D}\subset\mathbb{C}$
is a weak contraction for the Poincaré distance $\omega$. More precisely, for
every $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and $z$,
$w\in\mathbb{D}$ we have

$\omega\bigl{(}f(z),f(w)\bigr{)}\leq\omega(z,w)\;,$

with equality for some $z_{0}\neq w_{0}$ if and only if there is equality
everywhere if and only if $f$ is an automorphism of $\mathbb{D}$. In
particular, if $f\notin\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$ then we
have

$\omega\bigl{(}f(z),f(w)\bigr{)}<\omega(z,w)$ (1)

for all $z\neq w$.

In the century following the appearance of this result many improvements of
(1) for non automorphisms have appeared; see, e.g., [Dieudonne, Rogosinski,
Goluzin, Mercer1997, Mercer1999, Kaptanoglu, Beardon, BeardonCarne].
Surprisingly, in 2004 Beardon and Minda [BeardonMinda] found an elegant
unified way to recover all these results, and more.

Their idea is based on the _hyperbolic difference quotient_
$f^{*}\colon\mathbb{D}\times\mathbb{D}\to\mathbb{C}$ associated to a
holomorphic self-map
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$, which is defined
as follows:

$f^{*}(z,w)=\begin{cases}\frac{f(z)-f(w)}{1-\overline{f(w)}f(z)}\bigg{/}\frac{z-w}{1-\overline{w}z}&\mathrm{if}\
z\neq w\;;\\\ f^{\prime}(z)\frac{1-|z|^{2}}{1-|f(z)|^{2}}&\mathrm{if}\
z=w\;.\end{cases}$

It is clear that for every $w\in\mathbb{D}$ the map $z\mapsto f^{*}(z,w)$ is
holomorphic. Beardon and Minda observed that (1) is equivalent to saying that,
if $f$ is not an automorphism, then $f^{*}(\cdot,w)$ is a holomorphic self-map
of $\mathbb{D}$ for every $w\in\mathbb{D}$. But then one can apply the
Schwarz-Pick lemma to $f^{*}(\cdot,w)$, obtaining the following 3-point
Schwarz-Pick lemma:

######  Theorem 1.1 (Beardon-Minda, 2004).

Let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})\setminus\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$.
Then

$\omega\bigl{(}f^{*}(z,v),f^{*}(w,v)\bigr{)}\leq\omega(z,w)$

for all $z$, $v$, $w\in\mathbb{D}$. Furthermore equality holds for some
$z_{0}\neq w_{0}$ and $v_{0}$ if and only if it holds everywhere if and only
if $f$ is a Blaschke product of degree 2.

Here a _Blaschke product_ of _degree_ $d\geq 1$ is a holomorphic self-map of
$\mathbb{D}$ of the form

$B(z)=e^{i\theta}\prod_{j=1}^{d}\frac{z-a_{j}}{1-\overline{a_{j}}z}\;,$

where $\theta\in\mathbb{R}$ and $a_{1},\ldots,a_{d}\in\mathbb{D}$. In
particular, the Blaschke products of degree 1 are exactly the automorphisms of
$\mathbb{D}$.

As mentioned above, Beardon and Minda showed how the apparently innocuous
Theorem 1.1 can be used to recover many inequalities improving the original
Schwarz-Pick lemma; we refer to their beautiful paper [BeardonMinda] for
details.

A consequence of Theorem 1.1 is that if $f$ is not a Blaschke product of
degree 2 then $f^{*}(\cdot,w)$ is not an automorphism of $\mathbb{D}$;
therefore its hyperbolic difference quotient again is a holomorphic self-map
of $\mathbb{D}$, and hence we can apply the classical Schwarz-Pick lemma to
get a 4-point Schwarz-Pick lemma — and then, iterating the procedure, a
$n$-point Schwarz-Pick lemma for any $n\geq 2$.

This idea has been explored by Baribeau, Rivard and Wegert
[BaribeauRivardWegert] in the context of the Nevanlinna-Pick interpolation
problem, and by Cho, Kim and Sugawa [ChoKimSugawa] in more generality; see
also [Rivard1, Rivard2]. To state their results, we need some notations. Given
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and $z$,
$w_{1},\ldots,w_{k}\in\mathbb{D}$ we define the _iterated hyperbolic
difference quotient_ $\Delta_{w_{k},\ldots,w_{1}}f(z)$ by setting
$\Delta_{w_{1}}f(z)=f^{*}(z,w_{1})$ and
$\Delta_{w_{k},\ldots,w_{1}}f(z)=\Delta_{w_{k}}(\Delta_{w_{k-1},\ldots,w_{1}}f)(z)$.
Then we have a multi-point Schwarz-Pick lemma:

######  Theorem 1.2 (Baribeau-Rivard-Wegert, 2009).

Given $k\geq 1$, take
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ not a Blaschke
product of degree at most $k$. Then

$\omega\bigl{(}\Delta_{w_{k},\ldots,w_{1}}f(z),\Delta_{w_{k},\ldots,w_{1}}f(w))\bigr{)}\leq\omega(z,w)$

for all $z$, $w$, $w_{1},\ldots,w_{k}\in\mathbb{D}$. Furthermore equality
holds for some $z_{0}\neq w_{0}$ and $w_{1},\ldots,w_{k}$ if and only if it
holds for all $z$, $w$, $w_{1},\ldots,w_{k}\in\mathbb{D}$ if and only if $f$
is a Blaschke product of degree $k+1$.

Another way of expressing the Schwarz-Pick lemma consists in saying that
holomorphic self-maps of the unit disk send disks for the Poincaré distance
into disks for the Poincaré distance. Julia [Julia] in 1920 noticed that by
moving the centers of these disks toward the boundary one can get a boundary
version of the Schwarz-Pick lemma, nowadays known as Julia lemma:

######  Theorem 1.3 (Julia, 1920).

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$ be such that

$\beta_{f}(\sigma):=\liminf_{z\to\sigma}\frac{1-|f(z)|}{1-|z|}=\alpha<\infty\;.$

Then there exists a unique $\tau\in\partial\mathbb{D}$ such that

$\frac{|\tau-f(z)|^{2}}{1-|f(z)|^{2}}\leq\beta_{f}(\sigma)\frac{|\sigma-z|^{2}}{1-|z|^{2}}$
(2)

for every $z\in\mathbb{D}$. Moreover, equality in (2) holds at one point if
and only if it holds everywhere if and only if
$f\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$.

The number $\beta_{f}(\sigma)\in(0,+\infty]$ is the _boundary dilation
coefficient_ of $f$ at $\sigma$, and it is the absolute value of the _angular
derivative_ $f^{\prime}(\sigma)$, the non-tangential limit of $f^{\prime}$ at
$\sigma$, which is known to exist thanks to the Julia-Wolff-Carathéodory
theorem. It is well-known that $\beta_{f}(\sigma)>0$ always; furthermore, if
$\beta_{f}(\sigma)<+\infty$ then the point $\tau$ appearing in the statement
of Julia lemma is the non-tangential limit of $f$ at $\sigma$, that we will
denote by $f(\sigma)$.

The geometrical meaning of (2) is that if $\beta_{f}(\sigma)<+\infty$ then $f$
sends horocycles centered at $\sigma$ in horocycles centered at $f(\sigma)$,
where a _horocycle_ $E(\sigma,R)$ of _center_ $\sigma\in\partial\mathbb{D}$
and _radius_ $R>0$ is given by

$E(\sigma,R)=\left\\{z\in\mathbb{D}\biggm{|}\frac{|\sigma-z|^{2}}{1-|z|^{2}}<R\right\\}\;.$

Geometrically, $E(\sigma,R)$ is an Euclidean disk or radius $R/(R+1)$
internally tangent at $\partial\mathbb{D}$ in $\sigma$.

The aim of this paper is to obtain a multipoint version of Julia lemma along
the lines of Theorems 1.1 and 1.2. The paper [BeardonMinda] contains a 3-point
Julia lemma, but its statement does not involve the hyperbolic difference
quotient, and it is in a slightly different spirit. Closer to our aims is
[Mercer2000]*Proposition 4.1; but its (Euclidean) statement is quite involved
and not easy to use (see Remark 3.8).

Our idea then is to obtain a version of Theorem 1.3 involving the iterated
hyperbolic difference quotients. The main difference between the Schwarz-Pick
lemma and the Julia lemma is that the latter works only for maps with finite
boundary dilation coefficient. So the main result allowing our approach to
start is the following (see Proposition 3.4):

######  Proposition 1.4.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$ be such that $\beta_{f}(\sigma)<+\infty$. Then

$\beta_{\Delta_{w}f}(\sigma)=\beta_{f}(\sigma)\frac{1-|f(w)|^{2}}{|f(\sigma)-f(w)|^{2}}-\frac{1-|w|^{2}}{|\sigma-w|^{2}}$

for all $w\in\mathbb{D}$. In particular
$\beta_{\Delta_{w_{0}}f}(\sigma)<+\infty$ for some $w_{0}\in\mathbb{D}$ if and
only if $\beta_{\Delta_{w}f}(\sigma)<+\infty$ for all $w\in\mathbb{D}$ if and
only if $\beta_{f}(\sigma)<+\infty$.

So if the boundary dilation coefficient is finite for $f$ it remains finite
for all the iterated hyperbolic difference quotients of $f$. This allows us to
obtain a multi-point Julia lemma (see Theorem 4.2):

######  Theorem 1.5.

Given $k\geq 1$, take
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ not a Blaschke
product of degree at most $k$. Let $\sigma\in\partial\mathbb{D}$ be such that
$\beta_{f}(\sigma)<+\infty$. Then

$\frac{|\Delta_{w_{k},\ldots,w_{1}}f(\sigma)-\Delta_{w_{k},\ldots,w_{1}}f(z)|^{2}}{1-|\Delta_{w_{k},\ldots,w_{1}}f(z)|^{2}}\leq\beta_{\Delta_{w_{k},\ldots,w_{1}}f}(\sigma)\frac{|\sigma-z|^{2}}{1-|z|^{2}}$
(3)

for all $z$, $w_{1},\ldots,w_{k}\in\mathbb{D}$. Moreover, equality occurs for
some $z_{0}$, $w_{1},\ldots,w_{k}\in\mathbb{D}$ if and only if it occurs
everywhere if and only if $f$ is a Blaschke product of degree $k+1$.

We now describe two applications of this theorem. We mentioned before that the
boundary dilation coefficient is always strictly positive. In some instances
it is useful to have a more explicit bound from below, like the classical one

$\beta_{f}(\sigma)\geq\frac{1-|f(0)|}{1+|f(0)|}\;.$ (4)

In Section 4 we shall prove a much more precise estimate (see Theorem 4.3):

######  Theorem 1.6.

Given $k\geq 0$ let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ be not a Blaschke
product of degree at most $k$. Take $\sigma\in\partial\mathbb{D}$ with
$\beta_{f}(\sigma)<+\infty$. Then

$\beta_{f}(\sigma)\geq\sum_{j=0}^{k}\frac{1-|w_{j+1}|^{2}}{|\sigma-
w_{j+1}|^{2}}\prod_{h=0}^{j}\frac{|\Delta_{w_{h},\ldots,w_{1}}f(\sigma)-\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}{1-|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}$
(5)

for every $w_{1},\ldots,w_{k+1}\in\mathbb{D}$, where
$\Delta_{w_{h},\ldots,w_{1}}f=f$ when $h=0$. Furthermore we have equality in
(5) for some $w_{1},\ldots,w_{k+1}\in\mathbb{D}$ if and only if we have
equality for all $w_{1},\ldots,w_{k+1}\in\mathbb{D}$ if and only if $f$ is a
Blaschke product of degree $k+1$.

In particular we have the following corollary (see Corollary 4.7):

######  Corollary 1.7.

Given $k\geq 0$ let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ be not a Blaschke
product of degree at most $k$. Take $\sigma\in\partial\mathbb{D}$ with
$\beta_{f}(\sigma)<+\infty$. Then

$\beta_{f}(\sigma)\geq\sum_{j=0}^{k}\prod_{h=0}^{j}\frac{1-|\Delta_{\mathbf{O}_{h}}f(0)|}{1+|\Delta_{\mathbf{O}_{h}}f(0)|}\;,$

where $\mathbf{O}_{h}=(0,\ldots,0)\in\mathbb{D}^{h}$ is the origin of
$\mathbb{C}^{h}$, and $\Delta_{\mathbf{O}_{h}}f=f$ when $h=0$. Furthermore we
have equality if and only if $f$ is a Blaschke product of degree $k+1$ and
$\Delta_{\mathbf{O}_{h}}f(0)=|\Delta_{\mathbf{O}_{h}}f(0)|\Delta_{\mathbf{O}_{h}}f(\sigma)$
for all $h=0,\ldots,k$.

These results when $k=0$ recover (4) and when $k\geq 1$ improve previous
estimates due to Ünkelbach [Unkelbach1, Unkelbach2], Komatu [Komatu], Frovlova
et al. [Frovlova], Osserman [Osserman] and Mercer [Mercer2018].

When $f(0)=0$ the estimate (4) implies that $\beta_{f}(\sigma)\geq 1$. In
1982, Cowen and Pommerenke [CowenPommerenke] proved that if moreover
$f(\sigma)=\sigma$ this estimate can be improved to

$\beta_{f}(\sigma)\geq
1+\frac{|1-f^{\prime}(0)|^{2}}{1-|f^{\prime}(0)|^{2}}\;.$

More precisely, they obtained a sharp estimate valid when $f$ has a fixed
point inside and $n$ fixed points on the boundary:

######  Theorem 1.8 (Cowen-Pommerenke, 1982).

Let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})\setminus\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$
be such that $f(z_{0})=z_{0}$ for some $z_{0}\in\mathbb{D}$. Assume there
exist $\sigma_{1},\ldots,\sigma_{n}\in\partial\mathbb{D}$ distinct points with
$\beta_{f}(\sigma_{j})<+\infty$ and $f(\sigma_{j})=\sigma_{j}$ for
$j=1,\ldots,n$. Then

$\sum_{j=1}^{n}\frac{1}{\beta_{f}(\sigma_{j})-1}\leq\frac{1-|f^{\prime}(z_{0})|^{2}}{|1-f^{\prime}(z_{0})|^{2}}\;.$
(6)

Furthermore, equality holds if and only if $f$ is a Blaschke product of degree
$n+1$.

In Section 3 we shall show (see Proposition 3.11) how to obtain this result as
a consequence of our Theorem 1.5 for $k=1$. More interestingly, in Section 4
we shall generalize the estimate (6) to the case of multiple fixed points (see
Theorem 4.12):

######  Theorem 1.9.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$. Given $k\geq
1$, assume that $f$ is not a Blaschke product of degree at most $k\geq 1$ and
that there exists $z_{0}\in\mathbb{D}$ such that $f(z_{0})=z_{0}$ and
$f^{\prime}(z_{0})=\ldots=f^{(k-1)}(z_{0})=0$. Take
$\sigma_{1},\ldots,\sigma_{n}\in\partial\mathbb{D}$ distinct points such that
$\beta_{f}(\sigma_{j})<+\infty$ and

$f(\sigma_{j})=\frac{\left(\frac{\sigma_{j}-z_{0}}{1-\overline{z_{0}}\sigma_{j}}\right)^{k}+z_{0}}{1+\overline{z_{0}}\left(\frac{\sigma_{j}-z_{0}}{1-\overline{z_{0}}\sigma_{j}}\right)^{k}}\;,$

for $j=1,\ldots,n$. Then

$\sum_{j=1}^{n}\frac{1}{\left(1+2\mathop{\mathrm{Re}}\nolimits\frac{\left(f(\sigma_{j})-\sigma_{j}\right)\overline{z_{0}}}{|f(\sigma_{j})-z_{0}|^{2}}\right)\beta_{f}(\sigma_{j})-k}\leq\frac{1-\left|\frac{f^{(k)}(z_{0})}{k!}(1-|z_{0}|^{2})^{k-1}\right|^{2}}{\left|1-\frac{f^{(k)}(z_{0})}{k!}(1-|z_{0}|^{2})^{k-1}\right|^{2}}\;,$

with equality if and only if $f$ is a Blaschke product of degree $n+k$.

The proof in the general case is a bit delicate and requires the full force of
our multipoint Julia lemma. However, the case $z_{0}=0$ has a simpler
statement, and actually a much easier proof (see Corollary 4.13):

######  Corollary 1.10.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$. Given $k\geq
1$, assume that $f$ is not a Blaschke product of degree at most $k$ and that

$f(0)=\cdots=f^{(k-1)}(0)=0\;.$

Take $\sigma_{1},\ldots,\sigma_{n}\in\partial\mathbb{D}$ distinct points such
that $\beta_{f}(\sigma_{j})<+\infty$ and $f(\sigma_{j})=\sigma_{j}^{k}$ for
$j=1,\ldots,n$. Then

$\sum_{j=1}^{n}\frac{1}{\beta_{f}(\sigma_{j})-k}\leq\frac{1-\left|\frac{f^{(k)}(0)}{k!}\right|^{2}}{\left|1-\frac{f^{(k)}(0)}{k!}\right|^{2}}\;,$

with equality if and only if $f$ is a Blaschke product of degree $n+k$.

Summing up, these applications show that iterated hyperbolic difference
quotients and multipoint Julia lemmas can be an useful tool for exploring in a
systematic way the influence of higher order derivatives on the boundary
behaviour of holomorphic self-maps of the unit disk.

This paper is organized as follows. In Section 2 we shall collect a number of
preliminary definitions and results that we shall need later on. In Section 3
we shall discuss 2-point Julia lemmas, proving in particular Proposition 1.4.
Finally, in Section 4 we shall introduce our general multipoint Julia lemma
and its applications, proving in particular Theorems 1.5, 1.6, 1.9 and
Corollaries 1.7 and 1.10.

_Ackowledgments._ This paper is respectfully dedicated to the memory of my
advisor, Edoardo Vesentini, who, among (many) other things, showed me how
beautiful and elegant complex analysis can be, in one, several and infinitely
many variables.

##  2\. Preliminaries

In this section we collect a few known results that shall be useful later on.

###  2.1. Blaschke products

######  Definition 2.1.

A (finite) _Blaschke product_
$B\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ is a holomorphic
self-map of $\mathbb{D}$ continuous up to the boundary with
$B(\partial\mathbb{D})\subseteq\partial\mathbb{D}$. Since a Blaschke product
$B$ cannot vanish in a neighbourhood of $\partial\mathbb{D}$ it must have a
finite number $d\geq 0$ of zeroes in $\mathbb{D}$, counted with respect to
their multiplicity. The number $d$ is the _degree_ of $B$. We shall denote by
$\mathcal{B}_{d}$ the set of Blaschke products of degree $d\geq 1$, and by
$\mathcal{B}_{0}$ the set of constant functions of modulus 1.

######  Lemma 2.2.

A function $B\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ is a
Blaschke product of degree $d\geq 0$ if and only if there are
$\theta\in\mathbb{R}$ and $a_{1},\ldots,a_{d}\in\mathbb{D}$ such that

$B(z)=e^{i\theta}\prod_{j=1}^{d}\frac{z-a_{j}}{1-\bar{a_{j}}z}\;.$ (7)

In particular, if $\gamma\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$ then
$B\circ\gamma$ and $\gamma\circ B$ are still Blaschke products of the same
degree $d$.

###### Proof.

Since

$\frac{\sigma-a}{1-\overline{a}\sigma}=\overline{\sigma}\frac{\sigma-a}{\overline{\sigma}-\overline{a}}\in\partial\mathbb{D}$

for all $\sigma\in\partial\mathbb{D}$ and $a\in\mathbb{D}$, it is clear that
all maps of the form (7) are Blaschke products of degree $d$, with zeroes in
$a_{1},\ldots,a_{d}$.

Conversely, assume that $B$ is a Blaschke product of degree $d\geq 0$. If
$d=0$ then the maximum principle applied to $1/B$ implies that $|B|\equiv 1$,
and hence $B\equiv e^{i\theta}$ for a suitable $\theta\in\mathbb{R}$.

Assume $d\geq 1$, let $a_{1},\ldots,a_{d}\in\mathbb{D}$ be the zeroes of $B$,
listed accordingly to their multiplicities, and put

$B_{0}(z)=\prod_{j=1}^{d}\frac{z-a_{j}}{1-\bar{a_{j}}z}\;.$

Then $B/B_{0}$ is holomorphic without zeroes in $\mathbb{D}$ and
$|B/B_{0}|=|B_{0}/B|\equiv 1$ on $\partial D$. By the maximum principle we get
$|B/B_{0}|$, $|B_{0}/B|\leq 1$, and thus $B/B_{0}$ is a constant of modulus 1,
as required.

If $\gamma\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$ then $B\circ\gamma$
obviously is a Blaschke product of the same degree. On the other hand, clearly
$\gamma\circ B$ is still a Blaschke product. Moreover, if $a=\gamma^{-1}(0)$
then $(\gamma\circ B)(z)=0$ if and only if $B(z)=a$ if and only if

$e^{i\theta}\prod_{j=1}^{d}(z-a_{j})=a\prod_{j=1}^{d}(1-\bar{a_{j}}z)\;.$

This is a polynomial equation of degree exactly $d$; thus $\gamma\circ B$ has
exactly $d$ zeroes, counted with multiplicities, and we are done. ∎

In particular, the Blaschke products of degree 1 are exactly the automorphisms
of $\mathbb{D}$, that is
$\mathcal{B}_{1}=\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$. If
$w\in\mathbb{D}$ we shall denote by $\gamma_{w}$ the automorphism

$\gamma_{w}(z)=\frac{z-w}{1-\bar{w}z}\;.$

Later on we shall need the following

######  Lemma 2.3.

Let $\sigma_{1},\ldots,\sigma_{n}\in\partial\mathbb{D}$ be distinct points,
$a_{1},\ldots,a_{n}\in\mathbb{R}^{+}$ and $B\in\mathcal{B}_{d}$ for some
$n\geq 1$ and $d\geq 0$. If $B\not\equiv 1$ define
$h\colon\mathbb{D}\to\mathbb{C}$ by

$\frac{1+h(z)}{1-h(z)}=\sum_{j=1}^{n}a_{j}\frac{\sigma_{j}+z}{\sigma_{j}-z}+\frac{1+B(z)}{1-B(z)}\;.$

Then $h\in\mathcal{B}_{n+d}$.

###### Proof.

A quick computation yields

$h(z)=\frac{2B(z)+\bigl{(}1-B(z)\bigr{)}S(z)}{2+\bigl{(}1-B(z)\bigr{)}S(z)}\;,$

where

$S(z)=\sum_{j=1}^{n}a_{j}\frac{\sigma_{j}+z}{\sigma_{j}-z}\;;$

in particular $h$ is a rational function of degree $n+d$ because numerator and
denominator have no common factors.

When $z\in\mathbb{D}$ we have

$\mathop{\mathrm{Re}}\nolimits\frac{1+h(z)}{1-h(z)}=\sum_{j=1}^{n}a_{j}\mathop{\mathrm{Re}}\nolimits\frac{\sigma_{j}+z}{\sigma_{j}-z}+\mathop{\mathrm{Re}}\nolimits\frac{1+B(z)}{1-B(z)}=\sum_{j=1}^{n}a_{j}\frac{1-|z|^{2}}{|\sigma_{j}-z|^{2}}+\frac{1-|B(z)|^{2}}{|1-B(z)|^{2}}>0\;;$

this yields $h(z)\in\mathbb{D}$, and hence $h(\mathbb{D})\subseteq\mathbb{D}$.

If $\sigma\in\partial\mathbb{D}$ is different from $\sigma_{j}$ we have

$\frac{\sigma_{j}+\sigma}{\sigma_{j}-\sigma}=2i\frac{\mathop{\mathrm{Im}}\nolimits(\sigma\overline{\sigma_{j}})}{|\sigma_{j}-\sigma|^{2}}\in
i\mathbb{R}\;;$

thus if $\sigma\neq\sigma_{1},\ldots\sigma_{n}$ we have $S(\sigma)=ia\in
i\mathbb{R}$ and hence setting $B(\sigma)=e^{i\phi}\in\partial\mathbb{D}$ we
have

$h(\sigma)=\frac{2e^{i\phi}+(1-e^{i\phi})ia}{2+(1-e^{i\phi})ia}\in\partial\mathbb{D}\;.$

To deal with $\sigma=\sigma_{j}$ we write

$h(z)=\frac{a_{j}(\sigma_{j}+z)\bigl{(}1-B(z)\bigr{)}+(\sigma_{j}-z)\bigl{(}S_{j}(z)+2B(z)\bigr{)}}{a_{j}(\sigma_{j}+z)\bigl{(}1-B(z)\bigr{)}+(\sigma_{j}-z)\bigl{(}S_{j}(z)+2\bigr{)}}$

where $S_{j}(\sigma_{j})=0$; therefore we get $h(\sigma_{j})=1$, also when
$B(\sigma_{j})=1$.

Summing up, we have proved that $h$ is a Blaschke product; being a rational
function of degree $n+d$ we get $h\in\mathcal{B}_{n+d}$, and we are done. ∎

###  2.2. The hyperbolic difference quotient

######  Definition 2.4.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ be a
holomorphic self-map of the unit disk. The _hyperbolic derivative_
$f^{h}\colon\mathbb{D}\to\mathbb{C}$ of $f$ is given by

$f^{h}(z)=\frac{f^{\prime}(z)}{1-|\,f(z)|^{2}}\\!\bigg{/}\\!\frac{1}{1-|z|^{2}}=f^{\prime}(z)\frac{1-|z|^{2}}{1-|\,f(z)|^{2}}\;.$

The _hyperbolic difference quotient_
$f^{*}\colon\mathbb{D}\times\mathbb{D}\to\mathbb{C}$ is given by

$f^{*}(z,w)=\begin{cases}\frac{f(z)-f(w)}{1-\overline{f(w)}f(z)}\bigg{/}\frac{z-w}{1-\overline{w}z}&\mathrm{if}\
z\neq w\;;\\\ f^{h}(z)&\mathrm{if}\ z=w\;.\end{cases}$

It is easy to check that for every $w\in\mathbb{D}$ the function $z\mapsto
f^{*}(z,w)$ is holomorphic. Furthermore, the Schwarz-Pick lemma implies that
$|f^{*}(z,w)|\leq 1$ always, and that there exists
$(z_{0},w_{0})\in\mathbb{D}\times\mathbb{D}$ such that
$|f^{*}(z_{0},w_{0})|=1$ if and only if $|f^{*}|\equiv 1$ if and only if
$f\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$. In particular, if
$\gamma\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$ is given by

$\gamma(z)=e^{i\theta}\frac{z-a}{1-\bar{a}z}$

then it is easy to check that

$\gamma^{*}(z,w)=e^{i\theta}\frac{1-\bar{a}w}{1-a\bar{w}}\;.$

This can be seen as a particular case of the following result:

######  Proposition 2.5.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and $d\geq 1$.
Then $f\in\mathcal{B}_{d}$ if and only if $f^{*}(\cdot,w)\in\mathcal{B}_{d-1}$
for all $w\in\mathbb{D}$ if and only if
$f^{*}(\cdot,w_{0})\in\mathcal{B}_{d-1}$ for some $w_{0}\in\mathbb{D}$.

###### Proof.

By definition we have

$\gamma_{w}(z)f^{*}(z,w)=\gamma_{f(w)}\bigl{(}f(z)\bigr{)}$ (8)

for all $z$, $w\in\mathbb{D}$. If $f^{*}(\cdot,w_{0})\in\mathcal{B}_{d-1}$ for
some $w_{0}\in\mathbb{D}$ then $B=\gamma_{w_{0}}f^{*}(\cdot,w_{0})$ is a
Blaschke product of degree $d$, and thus Lemma 2.2 implies that
$f=\gamma^{-1}_{f(w_{0})}\circ B$ is a Blaschke product of degree $d$.

Conversely, if $f$ is a Blaschke product of degree $d$, then for any
$w\in\mathbb{D}$ we have that $\gamma_{f(w)}\circ f$ is a Blaschke product of
degree $d$ vanishing at $w$. Therefore $\gamma_{w}$ is a factor of
$\gamma_{f(w)}\circ f$, and (8) implies that $f^{*}(\cdot,w)$ is a Blaschke
product of degree $d-1$. ∎

###  2.3. The classical Julia lemma

######  Definition 2.6.

The _horocycle_ $E(\sigma,R)\subset\mathbb{D}$ of _center_
$\sigma\in\partial\mathbb{D}$ and _radius_ $R>0$ is given by

$E(\sigma,R)=\biggl{\\{}z\in\mathbb{D}\biggm{|}\frac{|\sigma-z|^{2}}{1-|z|^{2}}<R\biggr{\\}}\;.$
(9)

Geometrically, $E(\sigma,R)$ is the euclidean disk of radius $R/(R+1)$
internally tangent to $\partial\mathbb{D}$ in $\sigma$.

######  Definition 2.7.

Given $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma$, $\tau\in\partial\mathbb{D}$, set

$\beta_{f}(\sigma,\tau)=\sup_{z\in\mathbb{D}}\biggl{\\{}\frac{|\tau-f(z)|^{2}}{1-|f(z)|^{2}}\biggm{/}\frac{|\sigma-z|^{2}}{1-|z|^{2}}\biggr{\\}}\;.$
(10)

The _boundary dilation coefficient_ of $f$ at $\sigma$ is given by

$\beta_{f}(\sigma)=\inf_{\tau\in\partial\mathbb{D}}\beta_{f}(\sigma,\tau)\in[0,+\infty]\;.$
(11)

######  Remark 2.8.

By definition

$\hbox{}\hfilneg\hbox to0.0pt{\hss$\forall R>0\quad\qquad$}\hfil
f\bigl{(}E(\sigma,R)\bigr{)}\subseteq
E\bigl{(}\tau,\beta_{f}(\sigma,\tau)R\bigr{)}\;.$

In particular, for every
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$ there is at most one point
$\tau\in\partial\mathbb{D}$ such that $\beta_{f}(\sigma,\tau)$ is finite.
Indeed, if we had $\beta_{f}(\sigma,\tau_{j})<+\infty$ for two distinct points
$\tau_{1}$, $\tau_{2}\in\partial\mathbb{D}$ we would get a contradiction
choosing $R$ so small that

$E(\tau_{1},\beta R)\cap E(\tau_{2},\beta
R)=\mathord{\mathchoice{\ooalign{$\hfil\displaystyle/\hfil$\crcr$\displaystyle\mathrel{\hbox{\kern
0.0pt\raise
0.8pt\hbox{$\scriptstyle\bigcirc$}}}$}}{\ooalign{$\hfil\textstyle/\hfil$\crcr$\textstyle\mathrel{\hbox{\kern
0.0pt\raise
0.8pt\hbox{$\scriptstyle\bigcirc$}}}$}}{\ooalign{$\hfil\scriptstyle/\hfil$\crcr$\scriptstyle\mathrel{\hbox{\kern
0.0pt\raise
0.8pt\hbox{$\scriptstyle\bigcirc$}}}$}}{\ooalign{$\hfil\scriptscriptstyle/\hfil$\crcr$\scriptscriptstyle\mathrel{\hbox{\kern
0.0pt\raise 0.8pt\hbox{$\scriptstyle\bigcirc$}}}$}}}\;,$

where $\beta=\max\\{\beta_{f}(\sigma,\tau_{1}),\beta_{f}(\sigma,\tau_{2})\\}$.

The following well-known result gives us an alternative way to compute the
boundary dilation coefficient (for a proof see, e.g., [Abate]*Proposition
1.2.6):

######  Proposition 2.9.

Take $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$. Then

$\beta_{f}(\sigma)=\liminf_{z\to\sigma}\frac{1-|f(z)|}{1-|z|}\;.$

Furthermore (see, e.g., [Abate]*Lemma 1.2.4):

######  Lemma 2.10.

Let $f\colon\mathbb{D}\to\mathbb{D}$ be holomorphic. Then

$\hbox{}\hfilneg\hbox to0.0pt{\hss$\forall
z\in\mathbb{D}\quad\qquad$}\hfil\frac{1-|f(z)|}{1-|z|}\geq\frac{1-|f(0)|}{1+|f(0)|}>0\;.$
(12)

In particular, for all $\sigma\in\partial\mathbb{D}$ we have

$\beta_{f}(\sigma)\geq\frac{1-|f(0)|}{1+|f(0)|}>0\;.$ (13)

Moreover, equality in (12) holds at one point $z_{0}\neq 0$ (and hence
everywhere) if and only if $f(z)=e^{i\theta}z$ for a suitable
$\theta\in\mathbb{R}$.

We can now state the classical Julia lemma [Julia]:

######  Theorem 2.11 (Julia lemma).

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$, and choose
$\sigma\in\partial\mathbb{D}$ so that $\beta_{f}(\sigma)<+\infty$. Let
$\tau\in\partial\mathbb{D}$ be the unique point of $\partial\mathbb{D}$ such
that $\beta_{f}(\sigma,\tau)<+\infty$. Then

$\frac{|\tau-f(z)|^{2}}{1-|f(z)|^{2}}\leq\beta_{f}(\sigma)\frac{|\sigma-z|^{2}}{1-|z|^{2}}\;,$
(14)

that is

$\hbox{}\hfilneg\hbox to0.0pt{\hss$\forall R>0\quad\qquad$}\hfil
f\bigl{(}E(\sigma,R)\bigr{)}\subseteq E(\tau,\beta_{f}(\sigma)R)\;.$ (15)

Moreover, equality in (14) holds at one point (and hence everywhere) if and
only if $f\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$.

As a consequence we have another way for computing the boundary dilation
coefficient:

######  Corollary 2.12.

Take $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$. Then

$\beta_{f}(\sigma)=\lim_{r\to 1^{-}}\frac{1-|f(r\sigma)|}{1-r}=\lim_{r\to
1^{-}}\frac{1-|f(r\sigma)|^{2}}{1-r^{2}}\;.$

###### Proof.

By Proposition 2.9 we have

$\beta_{f}(\sigma)\leq\liminf_{r\to 1^{-}}\frac{1-|f(r\sigma)|}{1-r}\;;$

in particular the first equality is proven when $\beta_{f}(\sigma)=+\infty$.
Assume then that $\beta_{f}(\sigma)<+\infty$. An easy computation shows that
$r\sigma\in\partial E\bigl{(}\sigma,\frac{1-r}{1+r}\bigr{)}$; therefore
Theorem 2.11 yields
$f(r\sigma)\in\overline{E\bigl{(}\tau,\beta_{f}(\sigma)\frac{1-r}{1+r}\bigr{)}}$
for a suitable $\tau\in\partial\mathbb{D}$. Since $E(\tau,R)$ is an Euclidean
disk of radius $R/(R+1)$ internally tangent to $\partial\mathbb{D}$ in $\tau$
it follows that

$1-|f(r\sigma)|\leq|\tau-f(r\sigma)|\leq
2\frac{\beta_{f}(\sigma)\frac{1-r}{1+r}}{1+\beta_{f}(\sigma)\frac{1-r}{1+r}}\;.$

Therefore

$\limsup_{r\to 1^{-}}\frac{1-|f(r\sigma)|}{1-r}\leq\limsup_{r\to
1^{-}}\frac{2\beta_{f}(\sigma)}{1+r+\beta_{f}(\sigma)(1-r)}=\beta_{f}(\sigma)$

and the first equality is proved.

To prove the second equality, first of all notice that

$\frac{2}{1+r}\frac{1-|f(r\sigma)|}{1-r}\geq\frac{1+|f(r\sigma)|}{1+r}\frac{1-|f(r\sigma)|}{1-r}=\frac{1-|f(r\sigma)|^{2}}{1-r^{2}}\geq\frac{1}{1+r}\frac{1-|f(r\sigma)|}{1-r}\;.$
(16)

From this the second equality immediately follows when
$\beta_{f}(\sigma)=+\infty$. If $\beta_{f}(\sigma)<+\infty$ then
$|f(r\sigma)|\to 1$ as $r\to 1^{-}$, and thus the assertion follows again from
(16). ∎

######  Definition 2.13.

Given $\tau\in\partial\mathbb{D}$ and $M>0$, the _Stolz region_ $K(\tau,M)$ of
_vertex_ $\tau$ and _amplitude_ $M$ is

$K(\tau,M)=\biggl{\\{}z\in\mathbb{D}\biggm{|}\frac{|\tau-z|}{1-|z|}<M\biggl{\\}}\;.$
(17)

Note that
$K(\tau,M)=\mathord{\mathchoice{\ooalign{$\hfil\displaystyle/\hfil$\crcr$\displaystyle\mathrel{\hbox{\kern
0.0pt\raise
0.8pt\hbox{$\scriptstyle\bigcirc$}}}$}}{\ooalign{$\hfil\textstyle/\hfil$\crcr$\textstyle\mathrel{\hbox{\kern
0.0pt\raise
0.8pt\hbox{$\scriptstyle\bigcirc$}}}$}}{\ooalign{$\hfil\scriptstyle/\hfil$\crcr$\scriptstyle\mathrel{\hbox{\kern
0.0pt\raise
0.8pt\hbox{$\scriptstyle\bigcirc$}}}$}}{\ooalign{$\hfil\scriptscriptstyle/\hfil$\crcr$\scriptscriptstyle\mathrel{\hbox{\kern
0.0pt\raise 0.8pt\hbox{$\scriptstyle\bigcirc$}}}$}}}$ if $M\leq 1$, for
$|\tau-z|\geq 1-|z|$.

######  Definition 2.14.

We say that a function $f\colon\mathbb{D}\to\widehat{\mathbb{C}}$ has _non-
tangential_ (or _angular_) limit $c\in\widehat{\mathbb{C}}$ at
$\sigma\in\partial\mathbb{D}$ if $f(z)\to c$ as $z$ tends to $\sigma$ within
$K(\sigma,M)$ for any $M>1$. When this happen we shall write

$\mathop{\hbox{$K$-$\lim$}}\limits_{z\to\sigma}f(z)=c$

and denote $c$ by $f(\sigma)$.

We end this preliminary section recalling the famous Julia-Wolff-Carathéodory
theorem [Wolff, Caratheodorya]; for a proof see, e.g., [Abate]*Theorem 1.2.7.

######  Theorem 2.15 (Julia-Wolff-Carathéodory; 1926).

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and $\tau$,
$\sigma\in\partial\mathbb{D}$. Then

$\mathop{\hbox{$K$-$\lim$}}\limits_{z\to\sigma}\frac{\tau-f(z)}{\sigma-z}=\tau\bar{\sigma}\beta_{f}(\sigma,\tau)\;.$
(18)

If this non-tangential limit is finite then
$\beta_{f}(\sigma)=\beta_{f}(\sigma,\tau)<+\infty$, the function $f$ has non-
tangential limit $\tau$ at $\sigma$ and

$\mathop{\hbox{$K$-$\lim$}}\limits_{z\to\sigma}f^{\prime}(z)=\tau\bar{\sigma}\beta_{f}(\sigma)\;.$
(19)

##  3\. 2-point Julia lemma

As anticipated in the introduction, our 2-point Julia lemma will be obtained
by applying the classical Julia lemma to the function $f^{*}(\cdot,w)$. To do
so we need to compute the boundary dilation coefficient of the hyperbolic
difference quotient; this is done in the next two results.

######  Lemma 3.1.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$ be such that $f$ has non-tangential limit
$f(\sigma)\in\partial\mathbb{D}$ at $\sigma$. Then

$\mathop{\hbox{$K$-$\lim$}}\limits_{z\to\sigma}f^{*}(z,w)=\frac{f(\sigma)-f(w)}{1-\overline{f(w)}f(\sigma)}\bigg{/}\frac{\sigma-w}{1-\overline{w}\sigma}=\overline{f(\sigma)}\sigma\,\frac{f(\sigma)-f(w)}{\overline{f(\sigma)}-\overline{f(w)}}\,\frac{\overline{\sigma}-\overline{w}}{\sigma-w}\in\partial\mathbb{D}$

for all $w\in\mathbb{D}$.

###### Proof.

It follows immediately from the definition of $f^{*}$. ∎

######  Definition 3.2.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$ be such that $f$ has non-tangential limit
$f(\sigma)\in\partial\mathbb{D}$ at $\sigma$. Given $w\in\mathbb{D}$ we set

$f^{*}(\sigma,w)=\mathop{\hbox{$K$-$\lim$}}\limits_{z\to\sigma}f^{*}(z,w)\;;$

in particular, $|f^{*}(\sigma,\cdot)|\equiv 1$.

######  Remark 3.3.

For the sake of completeness, we remark that

$\mathop{\hbox{$K$-$\lim$}}\limits_{w\to\sigma}f^{*}(z,w)=\frac{f(z)-f(\sigma)}{1-\overline{f(\sigma)}f(z)}\bigg{/}\frac{z-\sigma}{1-\overline{\sigma}z}\equiv
f(\sigma)\overline{\sigma}$

for all $z\in\mathbb{D}$. Moreover, (18) yields

$\mathop{\hbox{$K$-$\lim$}}\limits_{w\to\sigma}f^{*}(\sigma,w)=\overline{f(\sigma)}\sigma\frac{f(\sigma)\overline{\sigma}\beta_{f}(\sigma)}{\overline{f(\sigma)}\sigma\beta_{f}(\sigma)}=f(\sigma)\overline{\sigma}\;.$

######  Proposition 3.4.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$ be such that $\beta_{f}(\sigma)<+\infty$. Denote
by $f(\sigma)\in\partial\mathbb{D}$ the non-tangential limit of $f$ at
$\sigma$. Then

$\displaystyle\liminf_{z\to\sigma}\frac{1-|f^{*}(z,w)|}{1-|z|}$
$\displaystyle=\beta_{f}(\sigma)\frac{1-|f(w)|^{2}}{|f(\sigma)-f(w)|^{2}}-\frac{1-|w|^{2}}{|\sigma-w|^{2}}$
(20)
$\displaystyle=\frac{1-|f(w)|^{2}}{|f(\sigma)-f(w)|^{2}}\left[\beta_{f}(\sigma)-\frac{|f(\sigma)-f(w)|^{2}}{1-|f(w)|^{2}}\bigg{/}\frac{|\sigma-w|^{2}}{1-|w|^{2}}\right]$

for all $w\in\mathbb{D}$. Moreover, the left-hand side vanishes for some
$w_{0}\in\mathbb{D}$ if and only if it vanishes for all $w\in\mathbb{D}$ if
and only if $f\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$.

###### Proof.

First of all we have

$\displaystyle 1-|f^{*}(z,w)|^{2}$
$\displaystyle=1-\frac{\left|\frac{f(z)-f(w)}{1-\overline{f(w)}f(z)}\right|^{2}}{\left|\frac{z-w}{1-\overline{w}z}\right|^{2}}=\frac{\left(1-\left|\frac{f(z)-f(w)}{1-\overline{f(w)}f(z)}\right|^{2}\right)-\left(1-\left|\frac{z-w}{1-\overline{w}z}\right|^{2}\right)}{\left|\frac{z-w}{1-\overline{w}z}\right|^{2}}$
$\displaystyle=\frac{\frac{(1-|f(z)|^{2})(1-|f(w)|^{2})}{|1-\overline{f(w)}f(z)|^{2}}-\frac{(1-|z|^{2})(1-|w|^{2})}{|1-\overline{w}z|^{2}}}{\left|\frac{z-w}{1-\overline{w}z}\right|^{2}}$
$\displaystyle=(1-|z|^{2})\frac{\frac{1-|f(w)|^{2}}{|1-\overline{f(w)}f(z)|^{2}}\frac{1-|f(z)|^{2}}{1-|z|^{2}}-\frac{1-|w|^{2}}{|1-\overline{w}z|^{2}}}{\left|\frac{z-w}{1-\overline{w}z}\right|^{2}}$

Recalling Corollary 2.12 and the fact that $f$ has non-tangential limit
$f(\sigma)\in\partial\mathbb{D}$ at $\sigma$ we obtain

$\displaystyle\liminf_{z\to\sigma}\frac{1-|f^{*}(z,w)|}{1-|z|}$
$\displaystyle=\liminf_{r\to
1^{-}}\frac{1-|f^{*}(r\sigma,w)^{2}|}{1-r^{2}}=\frac{\frac{1-|f(w)|^{2}}{|1-\overline{f(w)}f(\sigma)|^{2}}\beta_{f}(\sigma)-\frac{1-|w|^{2}}{|1-\overline{w}\sigma|^{2}}}{\left|\frac{\sigma-w}{1-\overline{w}\sigma}\right|^{2}}$
$\displaystyle=\beta_{f}(\sigma)\frac{1-|f(w)|^{2}}{|f(\sigma)-f(w)|^{2}}-\frac{1-|w|^{2}}{|\sigma-w|^{2}}\;,$

and (20) is proved.

Finally, by Theorem 2.11 the right-hand side in (20) vanishes at one point if
and only if it vanishes identically if and only if
$f\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$, and we are done. ∎

######  Definition 3.5.

Given $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$ we put

$\beta^{*}_{f}(\sigma;w)=\beta_{f}(\sigma)\frac{1-|f(w)|^{2}}{|f(\sigma)-f(w)|^{2}}-\frac{1-|w|^{2}}{|\sigma-w|^{2}}\in[0,+\infty]$

for all $w\in\mathbb{D}$. In particular, $\beta^{*}_{f}(\sigma,w_{0})=+\infty$
for some $w_{0}\in\mathbb{D}$ if and only if
$\beta^{*}_{f}(\sigma;\cdot)\equiv+\infty$ if and only if
$\beta_{f}(\sigma)=+\infty$, and $\beta^{*}_{f}(\sigma,w_{0})=0$ for some
$w_{0}\in\mathbb{D}$ if and only if $\beta^{*}_{f}(\sigma;\cdot)\equiv 0$ if
and only if $f\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$.

As a first hint of how it is possible to use this kind of results we show how
to improve (13):

######  Corollary 3.6.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$ be such that $\beta_{f}(\sigma)<+\infty$. Then

$\beta_{f}(\sigma)\frac{1-|f(w)|^{2}}{|f(\sigma)-f(w)|^{2}}\geq\frac{|w|-\left|\frac{f(w)-f(0)}{1-\overline{f(w)}f(0)}\right|}{|w|+\left|\frac{f(w)-f(0)}{1-\overline{f(w)}f(0)}\right|}+\frac{1-|w|^{2}}{|\sigma-w|^{2}}$

for all $w\in\mathbb{D}\setminus\\{0\\}$ and

$\beta_{f}(\sigma)\geq\frac{2}{1+|f^{h}(0)|}\frac{|f(\sigma)-f(0)|^{2}}{1-|f(0)|^{2}}\geq\frac{2}{1+|f^{h}(0)|}\frac{1-|f(0)|}{1+|f(0)|}\;.$
(21)

###### Proof.

If we apply (13) to $f^{*}(\cdot,w)$ we get

$\beta^{*}_{f}(\sigma,w)\geq\frac{1-|f^{*}(0,w)|}{1+|f^{*}(0,w)|}$

for all $w\in\mathbb{D}$. Since

$f^{*}(0,w)=\begin{cases}\frac{f(w)-f(0)}{w}\frac{1}{1-\overline{f(w)}f(0)}&\mathrm{if}\
w\neq 0\;,\\\ f^{h}(0)&\mathrm{if}\ w=0\;,\end{cases}$

we immediately get the first inequality for $w\neq 0$. When $w=0$ we get

$\beta_{f}(\sigma)\frac{1-|f(0)|^{2}}{|f(\sigma)-f(0)|^{2}}\geq\frac{1-|f^{h}(0)|}{1+|f^{h}(0)|}+1=\frac{2}{1+|f^{h}(0)|}$
(22)

and we are done. ∎

Notice that when $f\notin\mathop{\mathrm{Aut}}\nolimits(D)$ we have
$|f^{h}(0)|<1$ and thus (21) is strictly stronger than (13).

The inequality (21) was already known (see, e.g., [Osserman]); however, in the
next section we shall substantially improve it (see Theorem 4.3 and its
corollaries).

We can now state and prove our 2-point Julia lemma:

######  Theorem 3.7.

Let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})\setminus\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$
and $\sigma\in\partial\mathbb{D}$ be such that $\beta_{f}(\sigma)<+\infty$.
Then

$\frac{|f^{*}(\sigma,w)-f^{*}(z,w)|^{2}}{1-|f^{*}(z,w)|^{2}}\leq\beta^{*}_{f}(\sigma;w)\frac{|\sigma-z|^{2}}{1-|z|^{2}}$
(23)

for all $z$, $w\in\mathbb{D}$. Moreover, equality in (23) occurs for some
$(z_{0},w_{0})\in\mathbb{D}\times\mathbb{D}$ if and only if it occurs
everywhere if and only if $f$ is a Blaschke product of degree 2.

###### Proof.

The inequality (23) follows from Theorem 2.11 applied to $f^{*}(\cdot,w)$. If
we have equality in (23) for some $(z_{0},w_{0})\in\mathbb{D}\times\mathbb{D}$
again Theorem 2.11 implies that
$f^{*}(\cdot,w_{0})\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$, and then
Proposition 2.5 implies that $f\in\mathcal{B}_{2}$. Conversely,
$f\in\mathcal{B}_{2}$ implies that
$f^{*}(\cdot,w)\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$ for all
$w\in\mathbb{D}$, and thus we have equality in (23) for all $z$,
$w\in\mathbb{D}$. ∎

######  Remark 3.8.

It turns out that (23) implies a (not very illuminating) Euclidean statement,
originally proved by Mercer [Mercer2000], that in our notations can be
expressed as follows: let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$ be such that $\beta_{f}(\sigma)<+\infty$. Take
$w\in\mathbb{D}$ and set $\Lambda=\frac{1-|z|^{2}}{|\sigma-z|^{2}}$,
$\hat{\beta}=\beta_{f}^{*}(\sigma;w)$,
$\phi_{w}(z)=\frac{w-z}{1-\overline{w}z}$ and

$L=\frac{1-|\overline{f(w)}\phi_{w}(z)|^{2}}{|1-\overline{f(w)}f^{*}(\sigma,w)\phi_{w}(z)|^{2}}\;.$

Then for all $z\in\mathbb{D}$ we have $|f(z)-c_{w}(z)|<r_{w}(z)$, where

$\displaystyle c_{w}(z)$
$\displaystyle=\overline{f(\sigma)}f^{*}(\sigma,w)\phi_{w}(z)\frac{1-|f(w)|^{2}}{\bigl{(}1-\overline{f(w)}f^{*}(\sigma,w)\phi_{w}(z)\bigr{)}^{2}}\frac{\hat{\beta}}{\hat{\beta}L+\Lambda}$
$\displaystyle\quad+\overline{f(\sigma)}\phi_{f(w)}\bigl{(}f^{*}(\sigma,w)\phi_{w}(z)\bigr{)}\;,$
$\displaystyle r_{w}(z)$
$\displaystyle=|\phi_{w}(z)|\frac{1-|f(w)|^{2}}{\bigl{|}1-\overline{f(w)}f^{*}(\sigma,w)\phi_{w}(z)\bigr{|}^{2}}\frac{\hat{\beta}}{\hat{\beta}L+\Lambda}\;.$

This can be recovered as follows. Theorem 3.7 says that $f^{*}(z,w)$ belongs
to the horocycle of center $f^{*}(\sigma,w)$ and radius $\beta^{*}/\Lambda$,
which is an Euclidean disk of center
$\frac{\Lambda}{\beta^{*}+\Lambda}f^{*}(\sigma,w)$ and radius
$\frac{\beta^{*}}{\beta^{*}+\Lambda}$. Since
$\phi_{w}(z)f^{*}(z,w)=\phi_{f(w)}\bigl{(}f(z)\bigr{)}$, it follows that
$\phi_{f(w)}\bigl{(}f(z)\bigr{)}$ belongs to the Euclidean disk $D$ of center
$\frac{\Lambda}{\beta^{*}+\Lambda}f^{*}(\sigma,w)\phi_{w}(z)$ and radius
$\frac{\beta^{*}}{\beta^{*}+\Lambda}|\phi_{w}(z)|$. Using the fact that
$\phi_{f(w)}^{-1}=\phi_{f(w)}$ we get that $f(z)$ belongs to the Euclidean
disk $\phi_{f(w)}(D)$; computing center and radius of this latter disk we get
the assertion.

Applying Theorem 2.15 to $f^{*}(\cdot,w)$ we get the next corollary:

######  Corollary 3.9.

Let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})\setminus\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$
and $\sigma\in\partial\mathbb{D}$ be such that $\beta_{f}(\sigma)<+\infty$.
Then

$\mathop{\hbox{$K$-$\lim$}}\limits_{z\to\sigma}\frac{f^{*}(\sigma,w)-f^{*}(z,w)}{\sigma-z}=f^{*}(\sigma,w)\overline{\sigma}\beta^{*}_{f}(\sigma;w)$

and

$\mathop{\hbox{$K$-$\lim$}}\limits_{z\to\sigma}\frac{\partial f^{*}}{\partial
z}(z,w)=f^{*}(\sigma,w)\overline{\sigma}\beta^{*}_{f}(\sigma;w)\;.$

As mentioned in the introduction, specializing Theorem 3.7 to the case
$f(z_{0})=z_{0}$ we can recover an estimate due to Cowen and Pommerenke. The
main step is contained in the following

######  Corollary 3.10.

Let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})\setminus\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$
and $\sigma\in\partial\mathbb{D}$ be such that $\beta_{f}(\sigma)<+\infty$.
Assume that $f(0)=0$. Then

$\frac{\left|\frac{f(\sigma)}{\sigma}-\frac{f(z)}{z}\right|^{2}}{1-\left|\frac{f(z)}{z}\right|^{2}}\leq\bigl{(}\beta_{f}(\sigma)-1\bigr{)}\frac{|\sigma-z|^{2}}{1-|z|^{2}}$
(24)

if $z\neq 0$ and

$\frac{\left|\frac{f(\sigma)}{\sigma}-f^{\prime}(0)\right|^{2}}{1-|f^{\prime}(0)|^{2}}\leq\beta_{f}(\sigma)-1\;.$
(25)

In particular, if furthermore $f(\sigma)=\sigma$ we get

$\frac{|1-f^{\prime}(0)|^{2}}{1-|f^{\prime}(0)|^{2}}\leq\beta_{f}(\sigma)-1\;.$
(26)

Moreover, equality occurs in (24) at some $z_{0}\in\mathbb{D}\setminus\\{0\\}$
or in (25) if and only if it always occurs if and only $f$ is a Blaschke
product of degree 2.

###### Proof.

If $f(0)=0$ then

$f^{*}(z,0)=\begin{cases}\frac{f(z)}{z}&\mathrm{if}\ z\neq 0\;,\\\
f^{\prime}(0)&\mathrm{if}\ z=0\;;\end{cases}$ (27)

moreover, $f^{*}(\sigma,0)=f(\sigma)/\sigma$ and
$\beta^{*}_{f}(\sigma;0)=\beta_{f}(\sigma)-1$. The assertions then follow from
Theorem 3.7. ∎

When $f(\sigma)=\sigma$ (26) can be restated as

$\frac{1}{\beta_{f}(\sigma)-1}\leq\frac{1-|f^{\prime}(0)|^{2}}{|1-f^{\prime}(0)|^{2}}=\mathop{\mathrm{Re}}\nolimits\frac{1+f^{\prime}(0)}{1-f^{\prime}(0)}\;.$

Recalling that $f^{\prime}(\sigma)=\beta_{f}(\sigma)$ when $f(\sigma)=\sigma$,
where $f^{\prime}(\sigma)$ is the non-tangential limit of $f^{\prime}$ at
$\sigma$ (see Theorem 2.15), using (24) we can now recover a result due to
Cowen and Pommerenke [CowenPommerenke], saying that a similar estimate still
holds when there are several fixed points in the boundary:

######  Proposition 3.11 (Cowen-Pommerenke, 1982).

Let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})\setminus\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$
be such that $f(z_{0})=z_{0}$ for some $z_{0}\in\mathbb{D}$. Assume there
exist $\sigma_{1},\ldots,\sigma_{n}\in\partial\mathbb{D}$ distinct points with
$\beta_{f}(\sigma_{j})<+\infty$ and $f(\sigma_{j})=\sigma_{j}$ for
$j=1,\ldots,n$. Then

$\sum_{j=1}^{n}\frac{1}{\beta_{f}(\sigma_{j})-1}\leq\frac{1-|f^{\prime}(z_{0})|^{2}}{|1-f^{\prime}(z_{0})|^{2}}\;.$
(28)

Furthermore, equality holds if and only if $f$ is a Blaschke product of degree
$n+1$.

###### Proof.

First of all, let $\phi_{z_{0}}(z)=(z_{0}-z)/(1-\overline{z_{0}}z)$. Then
$\tilde{f}=\phi_{z_{0}}\circ f\circ\phi_{z_{0}}$ satisfies $\tilde{f}(0)=0$
and $\tilde{f}^{\prime}(0)=f^{\prime}(z_{0})$. Moreover if we put
$\tilde{\sigma}_{j}=\phi_{z_{0}}(\sigma_{j})$ then we have
$\tilde{f}(\tilde{\sigma}_{j})=\tilde{\sigma}_{j}$ and
$\beta_{\tilde{f}}(\tilde{\sigma}_{j})=\beta_{f}(\sigma_{j})$. Therefore in
the proof without loss of generality we can assume $z_{0}=0$.

For $j=1,\ldots,n$ set $\beta_{j}=\beta_{f}(\sigma_{j})$. We would like to
prove, by induction on $n$, that

$\displaystyle\frac{1-\left|f(z)/z\right|^{2}}{\left|1-f(z)/z\right|^{2}}$
$\displaystyle=\mathop{\mathrm{Re}}\nolimits\left(\frac{1+f(z)/z}{1-f(z)/z}\right)$
(29)
$\displaystyle\geq\sum_{j=1}^{n}\frac{1}{\beta_{j}-1}\mathop{\mathrm{Re}}\nolimits\left(\frac{\sigma_{j}+z}{\sigma_{j}-z}\right)=\sum_{j=1}^{n}\frac{1}{\beta_{j}-1}\frac{1-|z|^{2}}{|\sigma_{j}-z|^{2}}$

for all $z\in\mathbb{D}$, with equality at one point (and hence everywhere) if
and only if $f\in\mathcal{B}_{n+1}$. Clearly, when $z=0$ the expression
$f(z)/z$ is replaced by $f^{\prime}(0)$, and thus the theorem follows taking
$z=0$.

For $n=1$ (29) follows from Corollary 3.10. Assume it is true for $n-1$. In
particular we have

$\mathop{\mathrm{Re}}\nolimits\left[\frac{1+f(z)/z}{1-f(z)/z}-\sum_{j=1}^{n-1}\frac{1}{\beta_{j}-1}\frac{\sigma_{j}+z}{\sigma_{j}-z}\right]\geq
0\;,$ (30)

with equality at one point (and hence everywhere) if and only if
$f\in\mathcal{B}_{n}$. Therefore we can find
$h\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{C})$ with
$h(\mathbb{D})\subset\overline{\mathbb{D}}$ so that

$\frac{1+f(z)/z}{1-f(z)/z}-\sum_{j=1}^{n-1}\frac{1}{\beta_{j}-1}\frac{\sigma_{j}+z}{\sigma_{j}-z}=\frac{1+h(z)}{1-h(z)}\;.$
(31)

Notice that $h(\mathbb{D})\subseteq\mathbb{D}$ unless in (30) we have equality
at one point (and hence everywhere); in that case $h\equiv e^{i\theta}$ for a
suitable $\theta\in\mathbb{R}$.

If $h\equiv e^{i\theta}$, Lemma 2.3 shows that $f(z)=zB(z)$, where
$B\in\mathcal{B}_{n-1}$. But then $f$, being rational of degree $n$, can have
at most $n$ fixed points, whereas we are assuming that it has $n+1$ fixed
points, contradiction. Thus $h$ cannot be a constant, and we have the strict
inequality in (30).

A quick computation shows that

$h(z)=\frac{2\frac{f(z)}{z}-\left(1-\frac{f(z)}{z}\right)S(z)}{2-\left(1-\frac{f(z)}{z}\right)S(z)}\;,$

where

$S(z)=\sum_{j=1}^{n-1}\frac{1}{\beta_{j}-1}\frac{\sigma_{j}+z}{\sigma_{j}-z}\;,$

with the usual convention of replacing $f(z)/z$ by $f^{\prime}(0)$ when $z=0$.

Put $g(z)=zh(z)$. Then
$g\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$, $g(0)=0$ and
$g(\sigma_{n})=\sigma_{n}$, because $h(\sigma_{n})=1$. Furthermore we have

$h^{\prime}(z)=\frac{1}{\left(2-\left(1-\frac{f(z)}{z}\right)S(z)\right)^{2}}\left[\frac{4}{z}\left(f^{\prime}(z)-\frac{f(z)}{z}\right)+O\left(1-\frac{f(z)}{z}\right)\right]\;,$

and thus

$\mathop{\hbox{$K$-$\lim$}}\limits_{z\to\sigma_{n}}h^{\prime}(z)=\overline{\sigma_{n}}(\beta_{n}-1)\;.$

Since $g^{\prime}(z)=h(z)+zh^{\prime}(z)$ we get
$g^{\prime}(\sigma_{n})=\beta_{n}$. Since $h$ is not a constant we can apply
Corollary 3.10 to $g$ obtaining

$\mathop{\mathrm{Re}}\nolimits\left(\frac{1+h(z)}{1-h(z)}\right)\geq\frac{1}{\beta_{n}-1}\mathop{\mathrm{Re}}\nolimits\left(\frac{\sigma_{n}+z}{\sigma_{n}-z}\right)$
(32)

which recalling the definition of $h$ gives exactly (29).

If we have equality in one point in (29) we must have equality in one point in
(32), and this happens if and only if $g$ is a Blaschke product of degree 2,
again by Corollary 3.10. But this occurs if and only if
$h\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$; putting this in (31) we get
that $f\in\mathcal{B}_{n+1}$ by Lemma 2.3.

To prove the converse, assume that $f\in\mathcal{B}_{n+1}$ with $f(0)=0$. Then
$f(z)=zB(z)$, where $B\in\mathcal{B}_{n}$, and
$\sigma_{1},\ldots,\sigma_{n}\in\partial\mathbb{D}$ are the $n$ distinct
solutions of $B(z)=1$. Let $F\colon\mathbb{C}\to\widehat{\mathbb{C}}$ be
defined by

$F(z)=\frac{1+B(z)}{1-B(z)}-\sum_{j=1}^{n}\frac{1}{\beta_{j}-1}\frac{\sigma_{j}+z}{\sigma_{j}-z}\;.$

Then $\mathop{\mathrm{Re}}\nolimits F|_{\partial\mathbb{D}}\equiv 0$; this
implies that $\mathop{\mathrm{Re}}\nolimits F(0)=0$, which is exactly

$\frac{1-|f^{\prime}(0)|^{2}}{|1-f^{\prime}(0)|^{2}}-\sum_{j=1}^{n}\frac{1}{\beta_{j}-1}=0\;,$

and we are done. ∎

##  4\. Multipoint Julia lemmas

Our 2-point Julia lemma has been obtained by applying the classical Julia
lemma to the hyperbolic difference quotient $f^{*}(\cdot,w)$, which is a
holomorphic self-map of $\mathbb{D}$ as soon as $f$ is not an automorphism of
$\mathbb{D}$. But if we also assume that $f$ is not a Blaschke product of
degree 2 then by Proposition 2.5 $f^{*}(\cdot,w)$ is not an automorphism of
$\mathbb{D}$, and so _its_ hyperbolic difference quotient is a holomorphic
self-map of $\mathbb{D}$ to which we may apply the classical Julia lemma,
obtaining a 3-point Julia lemma.

Clearly this procedure can be iterated; to do so let us introduce some
notations.

######  Definition 4.1.

Given $k\geq 1$ and $w_{1},\ldots,w_{k}\in\mathbb{D}$ the _hyperbolic $k$-th
difference quotient_ $\Delta_{w_{k},\ldots,w_{1}}f$ of
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ is defined by
induction by setting $\Delta_{w_{1}}f(z)=f^{*}(z,w_{1})$ and

$\Delta_{w_{k},\ldots,w_{1}}f(z)=\Delta_{w_{k}}(\Delta_{w_{k-1},\ldots,w_{1}}f)(z)$

for $k\geq 2$.

Proposition 2.5 ensures that
$\Delta_{w_{k},\ldots,w_{1}}f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$
as soon as $f$ is not a Blaschke product of degree at most $k$. Moreover, if
$\sigma\in\partial\mathbb{D}$ is such that $\beta_{f}(\sigma)<+\infty$ by
applying repeatedly Proposition 3.4 we see that
$\beta_{\Delta_{w_{k},\ldots,w_{1}}f}(\sigma)$ is finite. More precisely,
$\beta_{\Delta_{w_{k},\ldots,w_{1}}f}(\sigma)$ can be recursively computed by

$\displaystyle\beta_{\Delta_{w_{1},\ldots,w_{k}}f}$ $\displaystyle(\sigma)$
$\displaystyle\\!\\!\\!=\beta_{\Delta_{w_{k-1},\ldots,w_{1}}f}(\sigma)\frac{1-|\Delta_{w_{k-1},\ldots,w_{1}}f(w_{k})|^{2}}{|\Delta_{w_{k-1},\ldots,w_{1}}f(\sigma)-\Delta_{w_{k-1},\ldots,w_{1}}f(w_{k})|^{2}}-\frac{1-|w_{k}|^{2}}{|\sigma-
w_{k}|^{2}}\;,$

and the non-tangential limit $\Delta_{w_{k},\ldots,w_{1}}f(\sigma)$ is
inductively given by

$\Delta_{w_{k},\ldots,w_{1}}f(\sigma)=\overline{\Delta_{w_{k-1},\ldots,w_{1}}f(\sigma)}\sigma\frac{\Delta_{w_{k-1},\ldots,w_{1}}f(\sigma)-\Delta_{w_{k-1},\ldots,w_{1}}f(w_{k})}{\overline{\Delta_{w_{k-1},\ldots,w_{1}}f(\sigma)}-\overline{\Delta_{w_{k-1},\ldots,w_{1}}f(w_{k})}}\frac{\overline{\sigma}-\overline{w_{k}}}{\sigma-
w_{k}}\;.$

In particular we have a multipoint Julia lemma:

######  Theorem 4.2.

Given $k\geq 1$, take
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ not a Blaschke
product of degree at most $k$. Let $\sigma\in\partial\mathbb{D}$ be such that
$\beta_{f}(\sigma)<+\infty$. Then

$\frac{|\Delta_{w_{k},\ldots,w_{1}}f(\sigma)-\Delta_{w_{k},\ldots,w_{1}}f(z)|^{2}}{1-|\Delta_{w_{k},\ldots,w_{1}}f(z)|^{2}}\leq\beta_{\Delta_{w_{k},\ldots,w_{1}}f}(\sigma)\frac{|\sigma-z|^{2}}{1-|z|^{2}}$
(33)

for all $z$, $w_{1},\ldots,w_{k}\in\mathbb{D}$. Moreover, equality occurs for
some $z_{0}$, $w_{1},\ldots,w_{k}\in\mathbb{D}$ if and only if it occurs
everywhere if and only if $f$ is a Blaschke product of degree $k+1$.

###### Proof.

The inequality (33) follows from Theorem 2.11 applied to
$\Delta_{w_{k},\ldots,w_{1}}f$. If we have equality in (33) for some $z_{0}$,
$w_{1},\ldots,w_{k}\in\mathbb{D}$ again Theorem 2.11 implies that
$\Delta_{w_{k},\ldots,w_{1}}f\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$,
and then Proposition 2.5 implies that $f\in\mathcal{B}_{k+1}$. Conversely,
$f\in\mathcal{B}_{k+1}$ implies that
$\Delta_{w_{k},\ldots,w_{1}}f\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$
for all $w_{1},\ldots,w_{k}\in\mathbb{D}$, and thus we have equality in (33)
for all $z$, $w_{1},\ldots,w_{k}\in\mathbb{D}$. ∎

The idea is that we can use this multipoint approach to improve known
estimates by involving higher order derivatives. We shall show two examples of
this: a strengthened version of Corollary 3.6 and a generalization of
Proposition 3.11.

We begin with a reformulation of Theorem 4.2 which gives a far-reaching
generalization of Corollary 3.6.

######  Theorem 4.3.

Given $k\geq 0$, let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ be not a Blaschke
product of degree at most $k$. Take $\sigma\in\partial\mathbb{D}$ with
$\beta_{f}(\sigma)<+\infty$. Then

$\beta_{f}(\sigma)\geq\sum_{j=0}^{k}\frac{1-|w_{j+1}|^{2}}{|\sigma-
w_{j+1}|^{2}}\prod_{h=0}^{j}\frac{|\Delta_{w_{h},\ldots,w_{1}}f(\sigma)-\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}{1-|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}$
(34)

for every $w_{1},\ldots,w_{k+1}\in\mathbb{D}$, where
$\Delta_{w_{h},\ldots,w_{1}}f=f$ when $h=0$. Furthermore we have equality in
(34) for some $w_{1},\ldots,w_{k+1}\in\mathbb{D}$ if and only if we have
equality for all $w_{1},\ldots,w_{k+1}\in\mathbb{D}$ if and only if $f$ is a
Blaschke product of degree $k+1$.

###### Proof.

One way to prove the assertion is to obtain by induction a formula for
$\beta_{\Delta_{w_{k},\ldots,w_{1}}f}(\sigma)$ applying repeatedly Proposition
3.4, and then to show that, with this formula, (34) is equivalent to (33). For
the sake of variety we shall describe a different proof, relying on the
classical Julia lemma.

We proceed by induction on $k$. The case $k=0$ is

$\beta_{f}(\sigma)\geq\frac{1-|w_{1}|^{2}}{|\sigma-
w_{1}|^{2}}\frac{|f(\sigma)-f(w_{1})|^{2}}{1-|f(w_{1})|^{2}}$ (35)

which is exactly the classical Julia inequality (14). In particular, we have
equality for some $w_{1}\in\mathbb{D}$ (and hence for all
$w_{1}\in\mathbb{D}$) if and only if
$f\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$.

Assume that (34) holds for $k-1$, and take $w_{1}\in\mathbb{D}$. Since $f$ is
not a Blaschke product of degree at most $k$, by Proposition 2.5
$\Delta_{w_{1}}f$ is not a Blaschke product of degree at most $k-1$. So by the
inductive hypothesis we have

$\displaystyle\beta_{\Delta_{w_{1}}f}(\sigma)$
$\displaystyle\geq\sum_{j=1}^{k}\frac{1-|w_{j+1}|^{2}}{|\sigma-
w_{j+1}|^{2}}\prod_{h=1}^{j}\frac{|\Delta_{w_{h},\ldots,w_{2}}(\Delta_{w_{1}}f)(\sigma)-\Delta_{w_{h},\ldots,w_{2}}(\Delta_{w_{1}}f)(w_{h+1})|^{2}}{1-|\Delta_{w_{h},\ldots,w_{2}}(\Delta_{w_{1}}f)(w_{h+1})|^{2}}$
$\displaystyle=\sum_{j=1}^{k}\frac{1-|w_{j+1}|^{2}}{|\sigma-
w_{j+1}|^{2}}\prod_{h=1}^{j}\frac{|\Delta_{w_{h},\ldots,w_{1}}f(\sigma)-\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}{1-|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}$

for all $w_{2},\ldots,w_{k+1}\in\mathbb{D}$, with equality for some (and hence
all) $w_{2},\ldots,w_{k+1}\in\mathbb{D}$ if and only if $\Delta_{w_{1}}f$ is a
Blaschke product of degree $k$, that is, by Proposition 2.5, if and only if
$f$ is a Blaschke product of degree $k+1$.

Now Proposition 3.4 yields

$\beta_{f}(\sigma)=\frac{|f(\sigma)-f(w_{1})|^{2}}{1-|f(w_{1})|^{2}}\left[\frac{1-|w_{1}|^{2}}{|\sigma-
w_{1}|^{2}}+\beta_{\Delta_{w_{1}}f}(\sigma)\right]\;;$

therefore

$\displaystyle\beta_{f}(\sigma)$
$\displaystyle\geq\frac{|f(\sigma)-f(w_{1})|^{2}}{1-|f(w_{1})|^{2}}$
$\displaystyle\quad\times\\!\Biggl{[}\frac{1-|w_{1}|^{2}}{|\sigma-
w_{1}|^{2}}+\sum_{j=1}^{k}\frac{1-|w_{j+1}|^{2}}{|\sigma-
w_{j+1}|^{2}}\prod_{h=1}^{j}\frac{|\Delta_{w_{h},\ldots,w_{1}}f(\sigma)-\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}{1-|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}\Biggr{]}$
$\displaystyle=\sum_{j=0}^{k}\frac{1-|w_{j+1}|^{2}}{|\sigma-
w_{j+1}|^{2}}\prod_{h=0}^{j}\frac{|\Delta_{w_{h},\ldots,w_{1}}f(\sigma)-\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}{1-|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}\;,$

with equality for some (and hence all) $w_{1},\ldots,w_{k+1}\in\mathbb{D}$ if
and only if $f$ is a Blaschke product of degree $k+1$, and we are done. ∎

######  Remark 4.4.

It is easy to see that the estimate (34) becomes better and better as $k$
increases.

Theorem 4.3 has a number of corollaries that it is worthwhile to state.

######  Corollary 4.5.

Given $k\geq 0$, let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ be not a Blaschke
product of degree at most $k$. Take $\sigma\in\partial\mathbb{D}$ with
$\beta_{f}(\sigma)<+\infty$. Then

$\beta_{f}(\sigma)\geq\sum_{j=0}^{k}\frac{1-|w_{j+1}|^{2}}{|\sigma-
w_{j+1}|^{2}}\prod_{h=0}^{j}\frac{1-|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|}{1+|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|}$
(36)

for every $w_{1},\ldots,w_{k+1}\in\mathbb{D}$, where
$\Delta_{w_{h},\ldots,w_{1}}f=f$ when $h=0$. Furthermore we have equality in
(36) if and only if $f$ is a Blaschke product of degree $k+1$ and
$w_{1},\ldots,w_{k+1}\in\mathbb{D}$ are such that

$\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})=|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|\Delta_{w_{h},\ldots,w_{1}}f(\sigma)$

for all $h=0,\ldots,k$.

###### Proof.

It follows from (34) using the standard estimate $|\tau-z|\geq 1-|z|$ valid
for all $\tau\in\partial\mathbb{D}$ and $z\in\mathbb{D}$, with equality if and
only if $z=|z|\tau$. ∎

######  Remark 4.6.

If take $k=1$ and $w_{1}=w_{2}=z$ and we assume $\sigma=f(\sigma)=1$ then (36)
becomes exactly [Mercer2018]*Theorem 2.1.

######  Corollary 4.7.

Given $k\geq 0$ let
$f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ be not a Blaschke
product of degree at most $k$. Take $\sigma\in\partial\mathbb{D}$ with
$\beta_{f}(\sigma)<+\infty$. Then

$\displaystyle\beta_{f}(\sigma)$
$\displaystyle\geq\sum_{j=0}^{k}\prod_{h=0}^{j}\frac{|\Delta_{\mathbf{O}_{h}}f(\sigma)-\Delta_{\mathbf{O}_{h}}f(0)|^{2}}{1-|\Delta_{\mathbf{O}_{h}}f(0)|^{2}}$
$\displaystyle\geq\sum_{j=0}^{k}\prod_{h=0}^{j}\frac{1-|\Delta_{\mathbf{O}_{h}}f(0)|}{1+|\Delta_{\mathbf{O}_{h}}f(0)|}$
(37)

where $\mathbf{O}_{h}=(0,\ldots,0)\in\mathbb{D}^{h}$ is the origin of
$\mathbb{C}^{h}$, and $\Delta_{\mathbf{O}_{h}}f=f$ when $h=0$. Furthermore we
have equality on the left of (37) if and only if $f$ is a Blaschke product of
degree $k+1$, and on the right if and only if

$\Delta_{\mathbf{O}_{h}}f(0)=|\Delta_{\mathbf{O}_{h}}f(0)|\Delta_{\mathbf{O}_{h}}f(\sigma)$

for all $h=0,\ldots,k$.

###### Proof.

It follows from Theorem 4.3 taking $w_{h}=0$ for $h=1,\ldots,k+1$. ∎

######  Remark 4.8.

(37) for $k=0$ is exactly (13), while for $k=1$ it yields (21), because
$\Delta_{0}f(0)=f^{h}(0)$.

######  Corollary 4.9.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ be not a
Blaschke product, and $\sigma\in\partial\mathbb{D}$ with
$\beta_{f}(\sigma)<+\infty$. Then

$\displaystyle\beta_{f}(\sigma)$
$\displaystyle\geq\sum_{j=0}^{\infty}\frac{1-|w_{j+1}|^{2}}{|\sigma-
w_{j+1}|^{2}}\prod_{h=0}^{j}\frac{|\Delta_{w_{h},\ldots,w_{1}}f(\sigma)-\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}{1-|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|^{2}}$
(38) $\displaystyle\geq\sum_{j=0}^{\infty}\frac{1-|w_{j+1}|^{2}}{|\sigma-
w_{j+1}|^{2}}\prod_{h=0}^{j}\frac{1-|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|}{1+|\Delta_{w_{h},\ldots,w_{1}}f(w_{h+1})|}$

for any sequence $\\{w_{h}\\}\subset\mathbb{D}$, where
$\Delta_{w_{h},\ldots,w_{1}}f=f$ when $h=0$ as usual. In particular,

$\displaystyle\beta_{f}(\sigma)$
$\displaystyle\geq\sum_{j=0}^{\infty}\prod_{h=0}^{j}\frac{|\Delta_{\mathbf{O}_{h}}f(\sigma)-\Delta_{\mathbf{O}_{h}}f(0)|^{2}}{1-|\Delta_{\mathbf{O}_{h}}f(0)|^{2}}$
$\displaystyle\geq\sum_{j=0}^{\infty}\prod_{h=0}^{j}\frac{1-|\Delta_{\mathbf{O}_{h}}f(0)|}{1+|\Delta_{\mathbf{O}_{h}}f(0)|}\;.$

###### Proof.

It follows from Theorem 4.3, Corollary 4.7 and Remark 4.4. ∎

A natural question, that we leave open, is whether the first inequality in
(38) actually is an equality, at least when $f$ is an infinite Blaschke
product.

To give an idea of the actual content of (34), let us reformulate it for small
values of $k$ and particular values of $w_{1},\ldots,w_{k+1}$.

For $k=0$ we get (35), that we already noticed to be equivalent to the
classical Julia lemma.

For $k=1$ we get

$\beta_{f}(\sigma)\geq\frac{|f(\sigma)-f(w_{1})|^{2}}{1-|f(w_{1})|^{2}}\left[\frac{1-|w_{1}|^{2}}{|\sigma-
w_{1}|^{2}}+\frac{1-|w_{2}|^{2}}{|\sigma-
w_{2}|^{2}}\frac{|\Delta_{w_{1}}f(\sigma)-\Delta_{w_{1}}f(w_{2})|^{2}}{1-|\Delta_{w_{1}}f(w_{2})|^{2}}\right]\;.$

Since $\Delta_{0}f(0)=f^{h}(0)$ and

$\Delta_{0}f(\sigma)=\overline{f(\sigma)}\overline{\sigma}\frac{f(\sigma)-f(0)}{\overline{f(\sigma)}-\overline{f(0)}}\;,$
(39)

putting $w_{1}=w_{2}=0$ we obtain

$\beta_{f}(\sigma)\geq\frac{|f(\sigma)-f(0)|^{2}}{1-|f(0)|^{2}}\left[1+\frac{\left|\overline{f(\sigma)}\overline{\sigma}\frac{f(\sigma)-f(0)}{\overline{f(\sigma)}-\overline{f(0)}}-f^{h}(0)\right|^{2}}{1-|f^{h}(0)|^{2}}\right]\;,$

which is a slightly more precise version of (21). If moreover $f(0)=0$ we find
again (25).

The case $k=2$ with $w_{1}=w_{2}=w_{3}=0$ yields

$\displaystyle\beta_{f}(\sigma)$
$\displaystyle\geq\frac{|f(\sigma)-f(0)|^{2}}{1-|f(0)|^{2}}$ (40)
$\displaystyle\quad\times\left[1+\frac{\left|\overline{f(\sigma)}\overline{\sigma}\frac{f(\sigma)-f(0)}{\overline{f(\sigma)}-\overline{f(0)}}-f^{h}(0)\right|^{2}}{1-|f^{h}(0)|^{2}}\left(1+\frac{|\Delta_{0,0}f(\sigma)-(\Delta_{0}f)^{h}(0)|^{2}}{1-|(\Delta_{0}f)^{h}(0)|^{2}}\right)\right]\;,$

where we have used the equality $\Delta_{0,0}f(0)=(\Delta_{0}f)^{h}(0)$. To
compute $(\Delta_{0}f)^{h}(0)$ first all we notice that

$\displaystyle(\Delta_{w_{0}}f)^{\prime}(z)$
$\displaystyle=\frac{f^{\prime}(z)(z-w_{0})-\bigl{(}f(z)-f(w_{0})\bigr{)}}{(z-w_{0})^{2}}\cdot\frac{1-\overline{w_{0}}z}{1-\overline{f(w_{0})}f(z)}$
$\displaystyle\quad+\frac{f(z)-f(w_{0})}{z-w_{0}}\cdot\frac{-\overline{w_{0}}\bigl{(}1-\overline{f(w_{0})}f(z)\bigr{)}+(1-\overline{w_{0}}z)\overline{f(w_{0})}f^{\prime}(z)}{\bigl{(}1-\overline{f(w_{0})}f(z)\bigr{)}^{2}}\;,$

and so

$\displaystyle(\Delta_{w_{0}}f)^{h}(0)=\frac{1}{1-|\Delta_{w_{0}}f(0)|^{2}}\biggl{[}$
$\displaystyle\frac{f(w_{0})-f(0)-f^{\prime}(0)w_{0}}{w_{0}^{2}}\cdot\frac{1}{1-\overline{f(w_{0})}f(0)}$
$\displaystyle+\frac{f(w_{0})-f(0)}{w_{0}}\frac{\overline{f(w_{0})}f^{\prime}(0)-\overline{w_{0}}\bigl{(}1-\overline{f(w_{0})}f(0)\bigr{)}}{\bigl{(}1-\overline{f(w_{0})}f(0)\bigr{)}^{2}}\biggr{]}.$

In particular putting $w_{0}=0$ we get

$\displaystyle(\Delta_{0}f)^{h}(0)$
$\displaystyle=\frac{1}{1-|f^{h}(0)|^{2}}\biggl{[}\frac{f^{\prime\prime}(0)}{2\bigl{(}1-|f(0)|^{2}\bigr{)}}+\frac{\overline{f(0)}f^{\prime}(0)^{2}}{\bigl{(}1-|f(0)|^{2}\bigr{)}^{2}}\biggr{]}$
(41)
$\displaystyle=\frac{1}{1-|f^{h}(0)|^{2}}\biggl{[}\frac{f^{\prime\prime}(0)}{2\bigl{(}1-|f(0)|^{2}\bigr{)}}+\overline{f(0)}f^{h}(0)^{2}\biggr{]}\;.$

Applying (39) to $\Delta_{0}f$ we also get

$\Delta_{0,0}f(\sigma)=f(\sigma)\frac{\overline{f(\sigma)}-\overline{f(0)}}{f(\sigma)-f(0)}\frac{\Delta_{0}f(\sigma)-f^{h}(0)}{\overline{\Delta_{0}f(\sigma)}-\overline{f^{h}(0)}}\;,$

and we have all the terms appearing in (40). In particular,

$\beta_{f}(\sigma)\geq\frac{1-|f(0)|}{1+|f(0)|}\left[1+\frac{1-|f^{h}(0)|}{1+|f^{h}(0)|}\frac{2}{1+|(\Delta_{0}f)^{h}(0)|}\right]\;,$

and thus if $f(0)=0$ we obtain

$\beta_{f}(\sigma)\geq
1+\frac{2(1-|f^{\prime}(0)|)}{1+|f^{\prime}(0)|+\frac{|f^{\prime\prime}(0)|}{2(1-|f^{\prime}(0)|)}}$

that improves (25). If moreover $f^{\prime}(0)=0$ we also get

$\beta_{f}(\sigma)\geq 1+\frac{2}{1+\frac{1}{2}|f^{\prime\prime}(0)|}\geq 2$

because $\frac{1}{2}|f^{\prime\prime}(0)|=|\Delta_{0,0}f(0)|\leq 1$.

As a final example, we record that a similar argument with $k=3$ and
$w_{1}=w_{2}=w_{3}=w_{4}=0$ yields

$\beta_{f}(\sigma)\geq\frac{1-|f(0)|}{1+|f(0)|}\biggl{[}\frac{1-|f^{h}(0)|}{1+|f^{h}(0)|}\left(\frac{1-|(\Delta_{0}f)^{h}(0)|}{1+|(\Delta_{0}f)^{h}(0)|}\frac{2}{1+|(\Delta_{0,0}f)^{h}(0)|}+1\right)+1\biggr{]}\;,$

where

$(\Delta_{0,0}f)^{h}(0)=\frac{1}{1-|(\Delta_{0}f)^{h}(0)|^{2}}\biggl{[}\frac{(\Delta_{0}f)^{\prime\prime}(0)}{2\bigl{(}1-|f^{h}(0)|^{2}\bigr{)}}+\overline{f^{h}(0)}(\Delta_{0}f)^{h}(0)^{2}\biggr{]}\;,$

with

$(\Delta_{0}f)^{\prime\prime}(0)=\frac{1}{1-|f(0)|^{2}}\biggl{[}\frac{1}{3}f^{\prime\prime\prime}(0)+2\overline{f(0)}f^{h}(0)f^{\prime\prime}(0)+\overline{f(0)}^{2}f^{h}(0)^{2}f^{\prime}(0)\biggr{]}\;.$

We now proceed toward the promised generalization of Proposition 3.11. Let us
start with the following reformulation of the case $k=2$ of Theorem 4.2 valid
when $f(0)=0$:

######  Proposition 4.10.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$, not a
Blaschke product of degree at most 2, and $\sigma\in\partial\mathbb{D}$ be
such that $\beta_{f}(\sigma)<+\infty$. Assume that $f(0)=0$. Then

$\frac{\left|\frac{\overline{\sigma}\bigl{(}f(\sigma)\overline{\sigma}-f^{\prime}(0)\bigr{)}}{1-\overline{f^{\prime}(0)}f(\sigma)\overline{\sigma}}-\frac{\frac{1}{z}\left(\frac{f(z)}{z}-f^{\prime}(0)\right)}{1-\overline{f^{\prime}(0)}\frac{f(z)}{z}}\right|^{2}}{1-\left|\frac{\frac{1}{z}\left(\frac{f(z)}{z}-f^{\prime}(0)\right)}{1-\overline{f^{\prime}(0)}\frac{f(z)}{z}}\right|^{2}}\leq\left[\frac{1-|f^{\prime}(0)|^{2}}{\left|\frac{f(\sigma)}{\sigma}-f^{\prime}(0)\right|^{2}}\bigl{(}\beta_{f}(\sigma)-1\bigr{)}-1\right]\\!\frac{|\sigma-z|^{2}}{1-|z|^{2}}$
(42)

for $z\neq 0$, and

$\frac{\left|\frac{\overline{\sigma}\bigl{(}f(\sigma)\overline{\sigma}-f^{\prime}(0)\bigr{)}}{1-\overline{f^{\prime}(0)}f(\sigma)\overline{\sigma}}-\frac{f^{\prime\prime}(0)}{2\bigl{(}1-|f^{\prime}(0)|^{2}\bigr{)}}\right|^{2}}{1-\left|\frac{f^{\prime\prime}(0)}{2\bigl{(}1-|f^{\prime}(0)|^{2}\bigr{)}}\right|^{2}}\leq\frac{1-|f^{\prime}(0)|^{2}}{\left|\frac{f(\sigma)}{\sigma}-f^{\prime}(0)\right|^{2}}\bigl{(}\beta_{f}(\sigma)-1\bigr{)}-1\;.$
(43)

Moreover, equality occurs in (42) at some $z_{0}\in\mathbb{D}\setminus\\{0\\}$
or in (43) if and only if it always occurs if and only $f$ is a Blaschke
product of degree 3.

###### Proof.

We would like to apply Theorem 4.2 with $k=2$ and $w_{1}=w_{2}=0$.

First of all

$\beta_{\Delta_{0,0}f}(\sigma)=\beta_{\Delta_{0}f}(\sigma)\frac{1-|\Delta_{0}f(0)|^{2}}{|\Delta_{0}f(\sigma)-\Delta_{0}f(0)|^{2}}-1=\frac{1-|f^{\prime}(0)|^{2}}{\left|\frac{f(\sigma)}{\sigma}-f^{\prime}(0)\right|^{2}}\bigl{(}\beta_{f}(\sigma)-1\bigr{)}-1\;,$

where we used $f(0)=0$ and $\Delta_{0}f(0)=f^{h}(0)=f^{\prime}(0)$. Next
recalling (27) we get

$\Delta_{0,0}f(z)=\begin{cases}\frac{\frac{f(z)}{z}-f^{\prime}(0)}{z}\frac{1}{1-\overline{f^{\prime}(0)}\frac{f(z)}{z}}&\mathrm{if}\
z\neq 0\;,\\\
\frac{f^{\prime\prime}(0)}{2(1-|f^{\prime}(0)|^{2})}&\mathrm{if}\
z=0\end{cases}$

where we used (41). The assertion then follows from Theorem 4.2. ∎

Notice that (26) is equivalent to saying that the right-hand side of (43) is
non-negative; so (43) is an improvement of (26).

If $f(0)=f^{\prime}(0)=0$ (42) and (43) simplify becoming

$\frac{\left|\frac{f(\sigma)}{\sigma^{2}}-\frac{f(z)}{z^{2}}\right|^{2}}{1-\left|\frac{f(z)}{z^{2}}\right|^{2}}\leq\bigl{(}\beta_{f}(\sigma)-2\bigr{)}\frac{|\sigma-z|^{2}}{1-|z|^{2}}$

for $z\neq 0$ and

$\frac{\left|\frac{f(\sigma)}{\sigma^{2}}-\frac{1}{2}f^{\prime\prime}(0)\right|^{2}}{1-\left|\frac{1}{2}f^{\prime\prime}(0)\right|^{2}}\leq\beta_{f}(\sigma)-2\;.$

These formulas suggest the following

######  Proposition 4.11.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$ and
$\sigma\in\partial\mathbb{D}$ be such that $\beta_{f}(\sigma)<+\infty$. Given
$k\geq 1$, assume that $f$ is not a Blaschke product of degree at most $k\geq
1$, and that $f(0)=\cdots=f^{(k-1)}(0)=0$. Then

$\frac{\left|\frac{f(\sigma)}{\sigma^{k}}-\frac{f(z)}{z^{k}}\right|^{2}}{1-\left|\frac{f(z)}{z^{k}}\right|^{2}}\leq\bigl{(}\beta_{f}(\sigma)-k\bigr{)}\frac{|\sigma-z|^{2}}{1-|z|^{2}}$
(44)

for $z\neq 0$ and

$\frac{\left|\frac{f(\sigma)}{\sigma^{k}}-\frac{1}{k!}f^{(k)}(0)\right|^{2}}{1-\left|\frac{1}{k!}f^{(k)}(0)\right|^{2}}\leq\beta_{f}(\sigma)-k\;.$
(45)

Moreover, equality occurs in (44) at some $z_{0}\in\mathbb{D}\setminus\\{0\\}$
or in (45) if and only if it always occurs if and only $f$ is a Blaschke
product of degree $k+1$.

###### Proof.

By induction it is easy to prove that

$\Delta_{\mathbf{O}_{k}}f(z)=\begin{cases}\frac{f(z)}{z^{k}}&\mathrm{if}\
z\neq 0\;,\\\ \frac{1}{k!}f^{(k)}(0)&\mathrm{if}\ z=0\;,\end{cases}$ (46)

and that $\beta_{\Delta_{\mathbf{O}_{k}}f}(\sigma)=\beta_{f}(\sigma)-k$. The
assertion then follows from Theorem 4.2. ∎

In particular, if $f(\sigma)=\sigma^{k}$ the left-hand sides of (44) and (45)
become independent of $\sigma$. This suggests that we might obtain a
generalization of Proposition 3.11 with multiple fixed points. It turns out
that this is easy when the multiple fixed point is the origin (see Corollary
4.13 below), but the statement and the proof of the general result when the
multiple fixed point is not the origin are considerably harder:

######  Theorem 4.12.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$. Given $k\geq
1$, assume that $f$ is not a Blaschke product of degree at most $k$ and that
there exists $z_{0}\in\mathbb{D}$ such that $f(z_{0})=z_{0}$ and
$f^{\prime}(z_{0})=\ldots=f^{(k-1)}(z_{0})=0$. Take
$\sigma_{1},\ldots,\sigma_{n}\in\partial\mathbb{D}$ distinct points such that
$\beta_{f}(\sigma_{j})<+\infty$ and

$f(\sigma_{j})=\frac{\left(\frac{\sigma_{j}-z_{0}}{1-\overline{z_{0}}\sigma_{j}}\right)^{k}+z_{0}}{1+\overline{z_{0}}\left(\frac{\sigma_{j}-z_{0}}{1-\overline{z_{0}}\sigma_{j}}\right)^{k}}\;,$
(47)

for $j=1,\ldots,n$. Then

$\sum_{j=1}^{n}\frac{1}{\left(1+2\mathop{\mathrm{Re}}\nolimits\frac{\left(f(\sigma_{j})-\sigma_{j}\right)\overline{z_{0}}}{|f(\sigma_{j})-z_{0}|^{2}}\right)\beta_{f}(\sigma_{j})-k}\leq\frac{1-\left|\frac{f^{(k)}(z_{0})}{k!}(1-|z_{0}|^{2})^{k-1}\right|^{2}}{\left|1-\frac{f^{(k)}(z_{0})}{k!}(1-|z_{0}|^{2})^{k-1}\right|^{2}}\;,$
(48)

with equality if and only if $f$ is a Blaschke product of degree $n+k$.

###### Proof.

The fact that $z_{0}$ is multiple fixed point of $f$ of order $k$ is
equivalent to saying that we can write

$f(z)=z_{0}+\frac{1}{k!}f^{(k)}(z_{0})(z-z_{0})^{k}+O\bigl{(}(z-z_{0})^{k+1}\bigr{)}\;.$

We claim that then

$\Delta_{(\mathbf{z_{0}})_{h}}f(z)=\frac{1}{k!}f^{(k)}(z_{0})(1-|z_{0}|^{2})^{h-1}(z-z_{0})^{k-h}+O\bigl{(}(z-z_{0})^{k-h+1}\bigr{)}$
(49)

for all $h=1,\ldots,k$, where
$(\mathbf{z_{0}})_{h}=(z_{0},\ldots,z_{0})\in\mathbb{D}^{h}$. We proceed by
induction. For $h=1$ we have

$\displaystyle\Delta_{z_{0}}f(z)$
$\displaystyle=\frac{f(z)-z_{0}}{z-z_{0}}\frac{1-\overline{z_{0}}z}{1-\overline{z_{0}}f(z)}=\frac{1}{k!}f^{(k)}(z_{0})(z-z_{0})^{k-1}\bigl{[}1+O(z-z_{0})\bigr{]}$
$\displaystyle=\frac{1}{k!}f^{(k)}(z_{0})(z-z_{0})^{k-1}+O\bigl{(}(z-z_{0})^{k}\bigr{)}\;,$

as claimed. Assume that (49) holds for $1\leq h-1\leq k-1$. Then
$\Delta_{(\mathbf{z_{0}})_{h-1}}f(z_{0})=0$ yields

$\displaystyle\Delta_{(\mathbf{z_{0}})_{h}}f(z)$
$\displaystyle=\frac{\Delta_{(\mathbf{z_{0}})_{h-1}}f(z)}{z-z_{0}}(1-\overline{z_{0}}z)$
$\displaystyle=\frac{1}{k!}f^{(k)}(z_{0})(1-|z_{0}|^{2})^{h-2}(z-z_{0})^{k-h}\bigl{[}1-|z_{0}|^{2}+O(z-z_{0})\bigr{]}$
$\displaystyle=\frac{1}{k!}f^{(k)}(z_{0})(1-|z_{0}|^{2})^{h-1}(z-z_{0})^{k-h}+O\bigl{(}(z-z_{0})^{k-h+1}\bigr{)}\;,$

as wanted.

Now we claim that

$\beta_{\Delta_{(\mathbf{z_{0}})_{h}}f}(\sigma_{j})=\frac{1-|z_{0}|^{2}}{|\sigma_{j}-z_{0}|^{2}}\left[\left(1+2\mathop{\mathrm{Re}}\nolimits\frac{\bigl{(}f(\sigma_{j})-\sigma_{j}\bigr{)}\overline{z_{0}}}{|f(\sigma_{j})-z_{0}|^{2}}\right)\beta_{f}(\sigma_{j})-h\right]$
(50)

for all $h=1,\ldots,k$ and $j=1,\ldots,n$. We again proceed by induction on
$h$. For $h=1$ Proposition 3.4 yields

$\displaystyle\beta_{\Delta_{z_{0}}f}(\sigma_{j})$
$\displaystyle=\frac{1-|z_{0}|^{2}}{|f(\sigma_{j})-z_{0}|^{2}}\beta_{f}(\sigma_{j})-\frac{1-|z_{0}|^{2}}{|\sigma_{j}-z_{0}|^{2}}$
$\displaystyle=\frac{1-|z_{0}|^{2}}{|\sigma_{j}-z_{0}|^{2}}\bigl{(}\beta_{f}(\sigma)-1\bigr{)}+\left(\frac{1-|z_{0}|^{2}}{|f(\sigma_{j})-z_{0}|^{2}}-\frac{1-|z_{0}|^{2}}{|\sigma_{j}-z_{0}|^{2}}\right)\beta_{f}(\sigma)$
$\displaystyle=\frac{1-|z_{0}|^{2}}{|\sigma_{j}-z_{0}|^{2}}\left(\beta_{f}(\sigma)-1+\frac{|\sigma_{j}-z_{0}|^{2}-|f(\sigma_{j})-z_{0}|^{2}}{|f(\sigma_{j})-z_{0}|^{2}}\beta_{f}(\sigma_{j})\right)$
$\displaystyle=\frac{1-|z_{0}|^{2}}{|\sigma_{j}-z_{0}|^{2}}\left[\left(1+2\mathop{\mathrm{Re}}\nolimits\frac{\bigl{(}f(\sigma_{j})-\sigma_{j}\bigr{)}\overline{z_{0}}}{|f(\sigma_{j})-z_{0}|^{2}}\right)\beta_{f}(\sigma_{j})-1\right]\;,$

as claimed. Assume that (50) holds for $1\leq h-1\leq k-1$. Using again the
fact that $\Delta_{(\mathbf{z_{0}})_{h-1}}f(z_{0})=0$ we get

$\displaystyle\beta_{\Delta_{(\mathbf{z_{0}})_{h}}f}(\sigma_{j})$
$\displaystyle=\beta_{\Delta_{(\mathbf{z_{0}})_{h-1}}f}(\sigma_{j})-\frac{1-|z_{0}|^{2}}{|\sigma_{j}-z_{0}|^{2}}$
$\displaystyle=\frac{1-|z_{0}|^{2}}{|\sigma_{j}-z_{0}|^{2}}\left[\left(1+2\mathop{\mathrm{Re}}\nolimits\frac{\bigl{(}f(\sigma_{j})-\sigma_{j}\bigr{)}\overline{z_{0}}}{|f(\sigma_{j})-z_{0}|^{2}}\right)\beta_{f}(\sigma_{j})-h\right]\;,$

and we are done.

We need one more preliminary computation. We claim that

$\Delta_{(\mathbf{z_{0}})_{h}}f(\sigma_{j})=\overline{f(\sigma_{j})}\sigma_{j}^{h}\left(\frac{\overline{\sigma_{j}}-\overline{z_{0}}}{\sigma_{j}-z_{0}}\right)^{h}\frac{f(\sigma_{j})-z_{0}}{\overline{f(\sigma_{j})}-\overline{z_{0}}}$
(51)

for $j=1,\ldots,n$ and $h=1,\ldots,k$. As always, we argue by induction on
$h$. For $h=1$ we have

$\Delta_{z_{0}}f(\sigma_{j})=\frac{f(\sigma_{j})-z_{0}}{1-\overline{z_{0}}f(\sigma_{j})}\frac{1-\overline{z_{0}}\sigma_{j}}{\sigma_{j}-z_{0}}=\overline{f(\sigma_{j})}\sigma_{j}\frac{\overline{\sigma_{j}}-\overline{z_{0}}}{\sigma_{j}-z_{0}}\frac{f(\sigma_{j})-z_{0}}{\overline{f(\sigma_{j})}-\overline{z_{0}}}$

as claimed. Assume that (51) holds for $1\leq h-1\leq k-1$. Recalling that
$\Delta_{(\mathbf{z_{0}})_{h-1}}f(z_{0})=0$ we obtain

$\Delta_{(\mathbf{z_{0}})_{h}}f(\sigma_{j})=\Delta_{(\mathbf{z_{0}})_{h-1}}f(\sigma_{j})\frac{1-\overline{z_{0}}\sigma_{j}}{\sigma_{j}-z_{0}}=\overline{f(\sigma_{j})}\sigma_{j}^{h}\left(\frac{\overline{\sigma_{j}}-\overline{z_{0}}}{\sigma_{j}-z_{0}}\right)^{h}\frac{f(\sigma_{j})-z_{0}}{\overline{f(\sigma_{j})}-\overline{z_{0}}}\;,$

and (51) is proved. In particular, we have
$\Delta_{(\mathbf{z_{0}})_{k}}f(\sigma_{j})=1$ if and only if

$\frac{1-\overline{f(\sigma_{j})}z_{0}}{\overline{f(\sigma_{j})}-\overline{z_{0}}}=\left(\frac{\sigma_{j}-z_{0}}{1-\overline{z_{0}}\sigma_{j}}\right)^{k}$

if and only if

$f(\sigma_{j})=\frac{\left(\frac{\sigma_{j}-z_{0}}{1-\overline{z_{0}}\sigma_{j}}\right)^{k}+z_{0}}{1+\overline{z_{0}}\left(\frac{\sigma_{j}-z_{0}}{1-\overline{z_{0}}\sigma_{j}}\right)^{k}}\;.$

In other words, condition (47) is just another way of writing
$\Delta_{(\mathbf{z_{0}})_{k}}f(\sigma_{j})=1$.

We can now apply Theorem 4.2. Recalling the assumption
$\Delta_{(\mathbf{z_{0}})_{k}}f(\sigma_{j})=1$ we get

$\frac{|1-\Delta_{(\mathbf{z_{0}})_{k}}f(z)|^{2}}{1-|\Delta_{(\mathbf{z_{0}})_{k}}f(z)|^{2}}\leq\frac{1-|z_{0}|^{2}}{|\sigma_{j}-z_{0}|^{2}}\left[\left(1+2\mathop{\mathrm{Re}}\nolimits\frac{\bigl{(}f(\sigma_{j})-\sigma_{j}\bigr{)}\overline{z_{0}}}{|f(\sigma_{j})-z_{0}|^{2}}\right)\beta_{f}(\sigma_{j})-k\right]\frac{|\sigma_{j}-z|^{2}}{1-|z|^{2}}$
(52)

for all $z\in\mathbb{D}$ and $j=1,\ldots,n$, where

$\Delta_{(\mathbf{z_{0}})_{k}}f(z)=\frac{1}{k!}f^{(k)}(z_{0})(1-|z_{0}|^{2})^{k-1}+O(z-z_{0})$

by (49). Furthermore, equality in (52) holds in one point (and hence
everywhere) if and only if $f$ is a Blaschke product of degree $k+1$.

We now claim that

$\displaystyle\mathop{\mathrm{Re}}\nolimits\biggl{(}$
$\displaystyle\frac{1+\Delta_{(\mathbf{z_{0}})_{k}}f(z)}{1-\Delta_{(\mathbf{z_{0}})_{k}}f(z)}\biggr{)}$
(53)
$\displaystyle\geq\sum_{j=1}^{n}\frac{|\sigma_{j}-z_{0}|^{2}}{1-|z_{0}|^{2}}\frac{1}{\left(1+2\mathop{\mathrm{Re}}\nolimits\frac{\left(f(\sigma_{j})-\sigma_{j}\right)\overline{z_{0}}}{|f(\sigma_{j})-z_{0}|^{2}}\right)\beta_{f}(\sigma_{j})-k}\mathop{\mathrm{Re}}\nolimits\left(\frac{\sigma_{j}+z}{\sigma_{j}-z}\right)\;,$

with equality in one point (and hence everywhere) if and only if $f$ is a
Blaschke product of degree $n+k$.

We argue by induction on $n$. For $n=1$ (53) is exactly equivalent to (52).
Assume that (53) holds for $n-1$. In particular we have

$\mathop{\mathrm{Re}}\nolimits\left[\frac{1+g(z)}{1-g(z)}-\sum_{j=1}^{n-1}a_{j}\frac{\sigma_{j}+z}{\sigma_{j}-z}\right]\geq
0\;,$ (54)

with equality in one point (and hence everywhere) if and only if
$f\in\mathcal{B}_{n+k-1}$, where $g=\Delta_{(\mathbf{z_{0}})_{k}}f$ and

$a_{j}=\frac{|\sigma_{j}-z_{0}|^{2}}{1-|z_{0}|^{2}}\frac{1}{\left(1+2\mathop{\mathrm{Re}}\nolimits\frac{\left(f(\sigma_{j})-\sigma_{j}\right)\overline{z_{0}}}{|f(\sigma_{j})-z_{0}|^{2}}\right)\beta_{f}(\sigma_{j})-k}=\frac{1}{\beta_{\Delta_{(\mathbf{z_{0}})_{k}}f}(\sigma_{j})}>0\;.$

Therefore we can find
$h\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{C})$ with
$h(\mathbb{D})\subset\overline{\mathbb{D}}$ so that

$\frac{1+g(z)}{1-g(z)}-\sum_{j=1}^{n-1}a_{j}\frac{\sigma_{j}+z}{\sigma_{j}-z}=\frac{1+h(z)}{1-h(z)}\;.$

Notice that either $h(\mathbb{D})\subseteq\mathbb{D}$ or $h\equiv
e^{i\theta}\in\partial\mathbb{D}$, and the latter case occurs if and only if
we have equality in (54).

If $h\equiv e^{i\theta}$ Lemma 2.3 implies that $g\in\mathcal{B}_{n-1}$. So
$g$ is a rational function of degree $n-1$; but we are assuming that the
equation $g(z)=1$ has at least $n$ distinct solutions, contradiction.

So $h\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$. Since
$g(\sigma_{n})=1$, $\beta_{g}(\sigma_{n})=\frac{1}{a_{n}}$ and
$g^{\prime}(\sigma_{n})=\overline{\sigma_{n}}/a_{n}$, where the latter
equality follows from (19), a quick computation yields $h(\sigma_{n})=1$ and
$h^{\prime}(\sigma_{n})=\overline{\sigma_{n}}/a_{n}$.

Put $\tilde{h}(z)=zh(z)$. Then we have $\tilde{h}(0)=0$,
$\tilde{h}(\sigma_{n})=\sigma_{n}$ and
$\beta_{\tilde{h}}(\sigma_{n})=\frac{1}{a_{n}}+1$. So we can apply Corollary
3.10 to $\tilde{h}$ obtaining

$\mathop{\mathrm{Re}}\nolimits\left(\frac{1+h(z)}{1-h(z)}\right)\geq
a_{n}\mathop{\mathrm{Re}}\nolimits\left(\frac{\sigma_{n}+z}{\sigma_{n}-z}\right)\;,$

with equality in one point (and hence everywhere) if and only if
$h\in\mathop{\mathrm{Aut}}\nolimits(\mathbb{D})$. Recalling (54) and Lemma 2.3
we see that we have proven (53), with equality in one point (and hence
everywhere) implying that $g$ is a Blaschke product of degree $n$, and thus
that $f$ is a Blaschke product of degree $n+k$, by Proposition 2.5.

In particular, (48) follows taking $z=z_{0}$ in (53), and equality there
implies that $f\in\mathcal{B}_{n+k}$.

To prove the converse, assume that $f\in\mathcal{B}_{n+k}$, so that
$g=\Delta_{(\mathbf{z_{0}})_{k}}f\in\mathcal{B}_{n}$, and
$\sigma_{1},\ldots,\sigma_{n}\in\partial\mathbb{D}$ are the $n$ distinct
solutions of $g(z)=1$. Let $F\colon\mathbb{C}\to\widehat{\mathbb{C}}$ be
defined by

$F(z)=\frac{1+g\circ\phi_{z_{0}}(z)}{1-g\circ\phi_{z_{0}}(z)}-\sum_{j=1}^{n}a_{j}\frac{\sigma_{j}+\phi_{z_{0}}(z)}{\sigma_{j}-\phi_{z_{0}}(z)}\;,$

where $\phi_{z_{0}}(z)=(z_{0}-z)/(1-\overline{z_{0}}z)$; notice that
$g\circ\phi_{z_{0}}$ is still a Blaschke product thanks to Lemma 2.2. Then
$\mathop{\mathrm{Re}}\nolimits F|_{\partial\mathbb{D}}\equiv 0$; this implies
$\mathop{\mathrm{Re}}\nolimits F(0)=0$, which gives exactly

$\frac{1-|g(z_{0})|^{2}}{|1-g(z_{0})|^{2}}=\sum_{j=1}^{n}a_{j}\frac{1-|z_{0}|^{2}}{|\sigma_{j}-z_{0}|^{2}}=\sum_{j=1}^{n}\frac{1}{\left(1+2\mathop{\mathrm{Re}}\nolimits\frac{\left(f(\sigma_{j})-\sigma_{j}\right)\overline{z_{0}}}{|f(\sigma_{j})-z_{0}|^{2}}\right)\beta_{f}(\sigma_{j})-k}\;,$

and we are done. ∎

######  Corollary 4.13.

Let $f\in\mathop{\mathrm{Hol}}\nolimits(\mathbb{D},\mathbb{D})$. Given $k\geq
1$, assume that $f$ is not a Blaschke product of degree at most $k$ and that
$f(0)=\cdots=f^{(k-1)}(0)=0$. Take
$\sigma_{1},\ldots,\sigma_{n}\in\partial\mathbb{D}$ distinct points such that
$f(\sigma_{j})=\sigma_{j}^{k}$ and $\beta_{f}(\sigma_{j})<+\infty$ for
$j=1,\ldots,n$. Then

$\sum_{j=1}^{n}\frac{1}{\beta_{f}(\sigma_{j})-k}\leq\frac{1-\left|\frac{f^{(k)}(0)}{k!}\right|^{2}}{\left|1-\frac{f^{(k)}(0)}{k!}\right|^{2}}\;,$

with equality if and only if $f$ is a Blaschke product of degree $n+k$.

###### Proof.

It immediately follows from Theorem 4.12 applied with $z_{0}=0$.

Alternatively, we can apply directly Proposition 3.11 to
$g=\Delta_{\mathbf{O}_{k-1}}f$. Indeed, (46) shows that $g(z)=f(z)/z^{k-1}$
for $z\neq 0$ and $g(0)=0$; in particular,
$g^{\prime}(0)=\frac{1}{k!}f^{(k)}(0)$. Moreover, $g(\sigma_{j})=\sigma_{j}$
and $g^{\prime}(\sigma_{j})=\beta_{f}(\sigma_{j})-(k-1)$ for all
$j=1,\ldots,n$; hence the assertion follows immediately from (28). ∎

Notice that when $k=1$ the condition (47) becomes

$f(\sigma_{j})=\frac{\frac{\sigma_{j}-z_{0}}{1-\overline{z_{0}}\sigma_{j}}+z_{0}}{1+\overline{z_{0}}\frac{\sigma_{j}-z_{0}}{1-\overline{z_{0}}\sigma_{j}}}=\sigma_{j}\;.$

So (48) reduces to (28), and thus Theorem 4.12 for $k=1$ recovers exactly
Proposition 3.11.

######  Remark 4.14.

We have seen that Proposition 3.11 for a generic fixed point $z_{0}$ followed
immediately from the case $z_{0}=0$, just replacing the map $f$ by the
composition $\phi_{z_{0}}\circ f\circ\phi_{z_{0}}$. Such an approach however
does not allow to easily deduce Theorem 4.12 from Corollary 4.13 because the
boundary dilation coefficient depends in a complicated way on the higher order
derivatives, and so we need the iterated hyperbolic difference quotients to
keep everything under control.

## References