On the Boundary Dieudonné–Pick Lemma

Abstract

The class of holomorphic self-maps of a disk with a boundary fixed point is studied. For this class of functions, the famous Julia–Carathéodory theorem gives a sharp estimate of the angular derivative at the boundary fixed point in terms of the image of the interior point. In the case when additional information about the value of the derivative at the interior point is known, a sharp estimate of the angular derivative at the boundary fixed point is obtained. As a consequence, the sharpness of the boundary Dieudonné–Pick lemma is established and the class of the extremal functions is identified. An unimprovable strengthening of the Osserman general boundary lemma is also obtained.

1. Introduction and Preliminaries

Let $\mathcal{B}$ be the set of holomorphic functions from the unit disk $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$ into itself. Investigations of the properties of a map $f\in \mathcal{B}$ are closely connected with the analysis of its fixed points.

An interior point $z_0$, $|z_0|<1$, is a fixed point of f if $f\left(z_0\right)=z_0$. A boundary point $\zeta$, $\left|\zeta \right|=1$, is a fixed point of f if $\angle {lim}_{z\to \zeta }f\left(z\right)=\zeta$.

Note that, if $f\in \mathcal{B}$ and $f\left(z\right)\not\equiv z$, then f can have at most one fixed point in the interior of $\mathbb{D}$. In the general case, $f\in \mathcal{B}$ need not have fixed points in $\mathbb{D}$. However, a classical result known as the Denjoy–Wolff theorem (see [1]) states that if $f\in \mathcal{B}$ is not Möbius transformation, then there exists a unique point d, $\left|d\right|⩽1$, such that the sequence of iterates $f^n=f\circ f^{n-1}$ of $f=f^1$, $n=2,3,\dots$, converges to d locally uniformly in $\mathbb{D}$. Moreover, if $d\in \mathbb{D}$, then $f\left(d\right)=d$. If d is a boundary point, then the angular limits

$$\angle \underset{z\to d}{lim}f\left(z\right)=f\left(d\right)\mathrm{a}\mathrm{n}\mathrm{d}\angle \underset{z\to d}{lim}{f}^{\prime }\left(z\right)=\angle \underset{z\to d}{lim}\frac{f\left(z\right)-d}{z-d}=f^{\prime }\left(d\right)$$

exist at d; in addition, $f\left(d\right)=d$ and the angular derivative $f^{\prime }\left(d\right)\in (0,1]$. In the literature, d is called the Denjoy–Wolff point of f.

Let us define the subclasses of functions considered in our research.

In $\mathcal{B}$, we single out the subclass $\mathcal{B}[a\to b]$ of functions that map a point a, $\left|a\right|<1$, to a point b, $\left|b\right|<1$:

$$\mathcal{B}[a\to b]=\left\{f\in \mathcal{B}:f\left(a\right)=b\right\}.$$

For brevity, the standard class $\mathcal{B}[0\to 0]$ of functions with fixed point $z_0=0$ we denote by $\mathcal{B}\left[0\right]$.

Let $\mathcal{B}\left[1\right]$ denote the set of functions f in $\mathcal{B}$ that fix the boundary point $\zeta =1$:

$$\mathcal{B}\left[1\right]=\left\{f\in \mathcal{B}:\angle \underset{z\to 1}{lim}f\left(z\right)=1\right\}.$$

The fundamental role of the Schwarz lemma and the Julia–Carathéodory theorem in complex analysis is well known (see [2,3]). These classical results characterize the behavior of the derivative on classes $\mathcal{B}\left[0\right]$ and $\mathcal{B}\left[1\right]$, respectively. The interest in developing these results stems from their applications in many directions, e.g., geometric function theory, hyperbolic geometry, complex dynamics, infinite-dimensional analysis, and the theory of composition operators (see [4,5,6,7,8]).

The purpose of this paper is to solve the following extremal problem in the spirit of the Julia–Carathéodory theorem: in the class $\mathcal{B}[a\to b,1]=\mathcal{B}[a\to b]\cap \mathcal{B}\left[1\right]$ of functions with a given value of $f^{\prime }\left(a\right)$ find the infinum of the angular derivative $f^{\prime }\left(1\right)$. The solution of this problem allows us to find the extremal functions in the boundary Dieudonné–Pick lemma and obtain an unimprovable strengthening of the Osserman general boundary lemma.

We also consider the class of functions with interior and boundary fixed points,

$$\mathcal{B}[0,1]=\mathcal{B}\left[0\right]\cap \mathcal{B}\left[1\right].$$

Note that classes of functions with several fixed points are studied in connection with various problems, e.g., the problem of fractional iteration of holomorphic function [9], the problem of finding domains of univalence [10,11,12], the problem of describing the Taylor coefficients [13,14], as well as estimating the angular derivatives [15] and the Schwarzian derivatives [16].

The paper is organized as follows. In Section 2 and Section 3, we consider the extremal problems that correspond to various versions of the Schwarz lemma and Julia–Carathéodory theorem. In Section 4, we give a solution of the above-mentioned extremal problem. As a consequence, we identify the class of the extremal functions in the boundary Dieudonné–Pick lemma and formulate an unimprovable strengthening of the Osserman general boundary lemma. In Section 5, we give proofs of the main results.

2. Schwarz–Pick and Dieudonné –Pick Lemmas

In the class $\mathcal{B}\left[0\right]$, the Schwarz lemma is a result with deep sense and important consequences. In particular, it describes the character of the fixed point of the function $f\in \mathcal{B}\left[0\right]$: namely, excluding the case of the rotations of the unit disk, in the class $\mathcal{B}\left[0\right]$, the point $z_0=0$ is attractive.

Theorem 1

(Schwarz Lemma). Let $f\in \mathcal{B}\left[0\right]$. Then

$$\begin{array}{cc}\hfill & \left|f\right(z\left)\right|⩽\left|z\right|,\forall z\in \mathbb{D},\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill & |f^{\prime }\left(0\right)|⩽1.\hfill \end{array}$$

(2)

If the equality in (2) is attained or the equality in (1) is attained for at least one $z\in \mathbb{D}$, $z\ne 0$, then $f\left(z\right)\equiv e^{i\theta }z$, $\theta \in \mathbb{R}$. In this case, the equality in (1) is attained for all $z\in \mathbb{D}$.

Inequality (2) can be interpreted as a solution of the following extremal problem: in the class $\mathcal{B}\left[0\right]$, find the supremum of $|f^{\prime }\left(0\right)|$. The solution of this problem, as follows from Theorem 1, has the form

$$\underset{f\in \mathcal{B}\left[0\right]}{sup}\left|f^{\prime }\left(0\right)\right|=1.$$

Moreover, the extremal functions are the rotations of the unit disk.

The Schwarz lemma has an invariant form established by Pick. Let us give the necessary part of the formulation of the Pick lemma.

Theorem 2

(Pick Lemma). Let $f\in \mathcal{B}$. Then for all $z\in \mathbb{D}$

$$\begin{array}{c}\hfill |f^{\prime }\left(z\right)|⩽\frac{1-{\left|f\left(z\right)\right|}^2}{1-{\left|z\right|}^2}.\end{array}$$

(3)

If the equality in (3) is attained for at least one $z\in \mathbb{D}$, then $f\left(z\right)\equiv e^{i\theta }(z-q)/(1-\overline{q}z)$, $q\in \mathbb{D}$, $\theta \in \mathbb{R}$. In this case, the equality in (3) is attained for all $z\in \mathbb{D}$.

The Pick lemma allows one to solve the following extremal problem: in the class $\mathcal{B}[a\to b]$, find the supremum of $|f^{\prime }\left(a\right)|$. As can be seen from Theorem 2, the solution of this problem is as follows:

$$\underset{f\in \mathcal{B}[a\to b]}{sup}\left|f^{\prime }\left(a\right)\right|=\frac{1-{\left|b\right|}^2}{1-{\left|a\right|}^2}.$$

Moreover, the extremal functions are Möbius transformations with $f\left(a\right)=b$.

A further restriction of the class $\mathcal{B}[a\to b]$ leads to a refinement of the estimate of the derivative at the interior point. The following result is called the Dieudonné lemma (see [17]).

Theorem 3

(Dieudonné Lemma). Let $f\in \mathcal{B}\left[0\right]$. Then for all $z\in \mathbb{D}$, $z\ne 0$,

$$\left|f^{\prime }\left(z\right)-\frac{f\left(z\right)}{z}\right|⩽\frac{{\left|z\right|}^2-{\left|f\left(z\right)\right|}^2}{\left|z\right|(1-|z{|}^2)}.$$

Note that Theorem 3 is a simple corollary of the sharpened version of the Schwarz lemma established by Mercer [18].

Let the mapping $T_q$ be defined by

$$T_q\left(z\right)=\frac{q-z}{1-\overline{q}z},q\in \mathbb{D}.$$

It takes the unit disk $\mathbb{D}$ onto itself with $T_q\left(q\right)=0$ and $T_q\left(0\right)=q$. An application of Pick’s ideas to Theorem 3 allows one to describe for the class $\mathcal{B}$ the set of the values of the derivative in terms of two points and their images under the mapping $T_q$ (see also [19]).

Theorem 4

(Dieudonné–Pick Lemma). Let $f\in \mathcal{B}$ with $f\left(z\right)=w$ and $f\left(z_1\right)=w_1$. Then

$$\left|f^{\prime }\left(z\right)-c\right|⩽r,$$

where

$$\begin{array}{cc}\hfill c& =\frac{T_w\left(w_1\right)}{T_z\left(z_1\right)}\frac{1-|T_z\left(z_1\right){|}^2}{1-|T_w\left(w_1\right){|}^2}\frac{1-{\left|w\right|}^2}{1-{\left|z\right|}^2},r=\frac{|T_z\left(z_1\right){|}^2-{\left|T_w\left(w_1\right)\right|}^2}{|T_z\left(z_1\right){|}^2(1-|T_w\left(w_1\right){|}^2)}\frac{1-{\left|w\right|}^2}{1-{\left|z\right|}^2}.\hfill \end{array}$$

3. Julia–Carathéodory Theorem and Boundary Schwarz Lemma

To formulate the Julia–Carathéodory theorem, consider the linear fractional transformation

$$L_q\left(z\right)=\frac{1-\overline{q}}{1-q}\frac{z-q}{1-\overline{q}z},q\in \mathbb{D}.$$

It takes the unit disk $\mathbb{D}$ onto itself and satisfies the conditions $L_q\left(q\right)=0$ and $L_q\left(1\right)=1$.

Theorem 5

(Julia–Carathéodory Theorem). Let $f\in \mathcal{B}\left[1\right]$. Then for all $z\in \mathbb{D}$

$$f^{\prime }\left(1\right)⩾\frac{{|1-f\left(z\right)|}^2}{1-{\left|f\left(z\right)\right|}^2}\frac{1-{\left|z\right|}^2}{{|1-z|}^2}.$$

(4)

If the equality in (4) is attained for at least one $z\in \mathbb{D}$, then $f\left(z\right)\equiv L_q\left(z\right)$, $q\in \mathbb{D}$. In this case, the equality in (4) is attained for all $z\in \mathbb{D}$.

For proof, see [3]. The Julia–Carathéodory theorem can be interpreted as a solution of the following extremal problem: in the class $\mathcal{B}[a\to b,1]$, find the infinum of the angular derivative $f^{\prime }\left(1\right)$. The solution of this problem follows from Theorem 5:

$$\underset{f\in \mathcal{B}[a\to b,1]}{inf}{f}^{\prime }\left(1\right)=\frac{{|1-b|}^2}{1-{\left|b\right|}^2}\frac{1-{\left|a\right|}^2}{{|1-a|}^2}.$$

(5)

Moreover, the extremal function is Möbius transformation with $f\left(a\right)=b$ and $f\left(1\right)=1$.

In the class $\mathcal{B}[0,1]$ the ratio (5) takes the form

$$\underset{f\in \mathcal{B}[0,1]}{inf}{f}^{\prime }\left(1\right)=1.$$

(6)

Moreover, the equality in (6) is attained only if $f\left(z\right)\equiv z$. The estimate (6) means that on the class $\mathcal{B}[0,1]$ the point $\zeta =1$ is repulsive or neutral.

It is known that the estimate (6) can be strengthened if there is an additional information about $f^{\prime }\left(0\right)$. Unkelbach [20] obtained the following boundary Schwarz lemma.

Theorem 6

(Boundary Schwarz Lemma). Let $f\in \mathcal{B}[0,1]$. Then

$$f^{\prime }\left(1\right)⩾\frac{2}{1+|f^{\prime }\left(0\right)|}.$$

(7)

Moreover, the equality in (7) holds if and only if

$$f\left(z\right)\equiv z\frac{z-u}{1-uz}$$

for some $u\in (-1,0]$.

The boundary Schwarz lemma is related to the following extremal problem: in the class $\mathcal{B}[0,1]$ of functions with a given value of $|f^{\prime }\left(0\right)|$, find the infinum of the angular derivative $f^{\prime }\left(1\right)$. The solution of this problem follows from Theorem 6:

$$\underset{f\in \mathcal{B}[0,1]:|f^{\prime }\left(0\right)|=M}{inf}{f}^{\prime }\left(1\right)=\frac{2}{1+M},M\in [0,1).$$

Moreover, the extremal function has the form $f\left(z\right)=z(z+M)/(1+Mz)$.

In [21], Cowen and Pommerenke proved the following result, which is an improvement of Theorem 6. Using another method, this result was also obtained by Goryainov [13]. We identify the specific form of the corresponding extremal functions.

Theorem 7.

Let $f\in \mathcal{B}[0,1]$, $f\left(z\right)\not\equiv z$. Then

$$f^{\prime }\left(1\right)⩾1+\frac{{\left|1-f^{\prime }\left(0\right)\right|}^2}{1-{\left|f^{\prime }\left(0\right)\right|}^2}.$$

(8)

Moreover, the equality in (8) holds if and only if $f\left(z\right)\equiv z{L}_q\left(z\right)$ for some $q\in \mathbb{D}$.

Theorem 7 allows one to solve the following extremal problem: in the class $\mathcal{B}[0,1]$ of functions with a given value of $f^{\prime }\left(0\right)$, find the infinum of the angular derivative $f^{\prime }\left(1\right)$. The solution of this problem has the form

$$\underset{f\in \mathcal{B}[0,1]:f^{\prime }\left(0\right)=c}{inf}{f}^{\prime }\left(1\right)=1+\frac{{|1-c|}^2}{1-{\left|c\right|}^2}.$$

Moreover, the extremal function has the form $f\left(z\right)=z{L}_q\left(z\right)$, $q=-c(1-\overline{c})/(1-c)$.

A horocycle at the point 1 with parameter $k>0$ is called a disk

$$\begin{array}{cc}\hfill {\mathbb{O}}_k& =\left\{z\in \mathbb{D}:\frac{{|1-z|}^2}{1-{\left|z\right|}^2}⩽k\right\}=\left\{z\in \mathbb{D}:\left|z-\frac{1}{k+1}\right|⩽1-\frac{1}{k+1}\right\}.\hfill \end{array}$$

In view of this definition, Theorem 7 can be rewritten as follows.

Theorem 8.

Let $f\in \mathcal{B}[0,1]$. Then

$$\left|f^{\prime }\left(0\right)-\frac{1}{f^{\prime }\left(1\right)}\right|⩽1-\frac{1}{f^{\prime }\left(1\right)}.$$

(9)

Moreover, the equality in (9) holds if and only if $f\left(z\right)\equiv z$ or $f\left(z\right)\equiv z{L}_q\left(z\right)$ for some $q\in \mathbb{D}$.

Theorem 8 allows one to solve another extremal problem: in the class $\mathcal{B}[0,1]$ of functions with a given value of the angular derivative $f^{\prime }\left(1\right)$, find the exact set of the values of $f^{\prime }\left(0\right)$. In view of (9), the solution of this problem is as follows: in the class $\mathcal{B}[0,1]$ of functions with $f^{\prime }\left(1\right)=\alpha$, the set of the values of $f^{\prime }\left(0\right)$ belongs to the disk

$$\left\{z\in \mathbb{C}:\left|z-\frac{1}{\alpha }\right|⩽1-\frac{1}{\alpha }\right\}.$$

Furthermore, $f^{\prime }\left(0\right)$ lies on the boundary of this disk if and only if $f\left(z\right)=z{L}_q\left(z\right)$, $q\in \partial {\mathbb{O}}_{1/(\alpha -1)}$.

Theorem 6 was improved by Osserman [22] by removing the assumption that the origin is fixed.

Theorem 9

(General Boundary Lemma). Let $f\in \mathcal{B}\left[1\right]$. Then

$$f^{\prime }\left(1\right)⩾\frac{2{(1-|f\left(0\right)\left|\right)}^2}{1+|f^{\prime }{\left(0\right)|-|f\left(0\right)|}^2}.$$

(10)

Note that in the general case estimate (10) is unattainable and sharp only if $f\left(0\right)>0$. In Section 5 we get the sharp estimate for $f^{\prime }\left(1\right)$ and find the class of the extremal functions.

4. Boundary Dieudonné–Pick Lemma and Its Consequences

The following result is the boundary Dieudonné–Pick lemma in which we identify the extremal functions.

Theorem 10.

Let $f\in \mathcal{B}\left[1\right]$. Then for all $z\in \mathbb{D}$

$$\left|f^{\prime }\left(z\right)-\frac{1}{f^{\prime }\left(1\right)}\frac{{\left(1-f\left(z\right)\right)}^2}{{(1-z)}^2}\right|⩽\frac{1-{\left|f\left(z\right)\right|}^2}{1-{\left|z\right|}^2}-\frac{1}{f^{\prime }\left(1\right)}\frac{{\left|1-f\left(z\right)\right|}^2}{{|1-z|}^2}.$$

(11)

If the equality in (11) is attained for at least one $z\in \mathbb{D}$, then $f\left(z\right)\equiv L_q\left(z\right)$, $q\in \mathbb{D}$, or $f\left(z\right)\equiv L_q\left(z\right)L_p\left(z\right)$, $q,p\in \mathbb{D}$. In this case, the equality in (11) is attained for all $z\in \mathbb{D}$.

The first part of Theorem 10 is essentially due to Yanagihara [23]. It was also obtained by Mercer [24] as a limit form of Theorem 4 in cases when $f\left(z_1\right)\to 1$ as $z_1\to 1$. However, the sharpness of inequality (11) was not discussed. In Section 5, we give another proof of inequality (11) based on the Julia–Carathéodory theorem and Pick’s ideas. We also establish its nonimprovability and describe the class of the extremal functions.

Theorem 10 allows one to solve the following extremal problem: in the class $\mathcal{B}[a\to b,1]$ of functions with a given value of the angular derivative $f^{\prime }\left(1\right)$, find the exact set of the values of $f^{\prime }\left(a\right)$. In view of (11) the solution of this problem is as follows: in the class $\mathcal{B}[a\to b,1]$ of functions with $f^{\prime }\left(1\right)=\alpha$, the set of the values of $f^{\prime }\left(a\right)$ belongs to the disk

$$\left\{z\in \mathbb{C}:\left|z-\frac{1}{\alpha }\frac{{(1-b)}^2}{{(1-a)}^2}\right|⩽\frac{1-{\left|b\right|}^2}{1-{\left|a\right|}^2}-\frac{1}{\alpha }\frac{{|1-b|}^2}{{|1-a|}^2}\right\}.$$

Furthermore, $f^{\prime }\left(a\right)$ lies on the boundary of this disk if and only if $f\left(z\right)=L_q\left(z\right)$, where $q\in \mathbb{D}$, such that $f\left(a\right)=b$, $f^{\prime }\left(1\right)=\alpha$, or $f\left(z\right)=L_q\left(z\right)L_p\left(z\right)$, where $q,p\in \mathbb{D}$ such that $f\left(a\right)=b$, $f^{\prime }\left(1\right)=\alpha$.

Theorem 10 admits a reformulation, which can be considered as a strengthening of the Julia–Carathéodory theorem (Theorem 5) in the case when for $f\in \mathcal{B}\left[1\right]$ in addition to the value of $f\left(z\right)$, $z\in \mathbb{D}$, the value of the derivative $f^{\prime }\left(z\right)$ is known.

Theorem 11.

Let $f\in \mathcal{B}\left[1\right]$ be not Möbius transformation. Then for all $z\in \mathbb{D}$

$$f^{\prime }\left(1\right)⩾\frac{{\left|1-f\left(z\right)\right|}^2}{1-{\left|f\left(z\right)\right|}^2}\frac{1-{\left|z\right|}^2}{{|1-z|}^2}\left(1+\frac{{\left|1-f^{\prime }\left(z\right)\frac{1-{\left|z\right|}^2}{1-{\left|f\left(z\right)\right|}^2}\frac{1-z}{1-\overline{z}}\frac{1-\overline{f\left(z\right)}}{1-f\left(z\right)}\right|}^2}{1-{\left|f^{\prime }\left(z\right)\right|}^2{\left(\frac{1-{\left|z\right|}^2}{1-{\left|f\left(z\right)\right|}^2}\right)}^2}\right).$$

(12)

If the equality in (12) is attained for at least one $z\in \mathbb{D}$, then $f\left(z\right)\equiv L_q\left(z\right)L_p\left(z\right)$, $q,p\in \mathbb{D}$. In this case, the equality in (12) is attained for all $z\in \mathbb{D}$.

Theorem 11 can be interpreted as a solution of the following extremal problem: in the class $\mathcal{B}[a\to b,1]$ of functions with a given value of $f^{\prime }\left(a\right)$, find the infinum of the angular derivative $f^{\prime }\left(1\right)$. Since $|f^{\prime }\left(a\right)|<(1-{\left|b\right|}^2{)/(1-\left|a\right|}^2)$, for given $a,b\in \mathbb{D}$ and c such that $\left|c\right|<(1-{\left|b\right|}^2{)/(1-\left|a\right|}^2)$ the solution of this problem has the form

$$\underset{f\in \mathcal{B}[a\to b,1]:f^{\prime }\left(a\right)=c}{inf}{f}^{\prime }\left(1\right)=\frac{{|1-b|}^2}{1-{\left|b\right|}^2}\frac{1-{\left|a\right|}^2}{{|1-a|}^2}\left(1+\frac{{\left|1-c\frac{1-{\left|a\right|}^2}{1-{\left|b\right|}^2}\frac{1-a}{1-\overline{a}}\frac{1-\overline{b}}{1-b}\right|}^2}{1-{\left|c\right|}^2{\left(\frac{1-{\left|a\right|}^2}{1-{\left|b\right|}^2}\right)}^2}\right).$$

Moreover, the extremal function is the Blaschke product $f\left(z\right)=L_q\left(z\right)L_p\left(z\right)$, where $q,p\in \mathbb{D}$ such that $f\left(a\right)=b$, $f^{\prime }\left(a\right)=c$.

A simple corollary of Theorem 10 is a strengthening of Theorem 9.

Corollary 1.

Let $f\in \mathcal{B}\left[1\right]$. Then

$$f^{\prime }\left(1\right)⩾\frac{2\frac{{|1-f\left(z\right)|}^2}{{|1-z|}^2}}{|f^{\prime }\left(z\right)|+\frac{1-{\left|f\left(z\right)\right|}^2}{1-{\left|z\right|}^2}}.$$

(13)

Although inequality (13) is less sharp than inequality (12), it has a more compact form and is a refinement of inequality (4), except for the case $f\left(z\right)\equiv L_q\left(z\right)$, $q\in \mathbb{D}$.

Corollary 2

(strengthening of Theorem 9). Let $f\in \mathcal{B}\left[1\right]$. Then

$$f^{\prime }\left(1\right)⩾\frac{{2|1-f\left(0\right)|}^2}{1+|f^{\prime }{\left(0\right)|-|f\left(0\right)|}^2}.$$

(14)

Moreover, the equality in (14) holds if and only if

$$f\left(z\right)\equiv L_q\left(z\right)L_p\left(z\right)$$

for some $q,p\in \mathbb{D}$, ${q/(1-|q|}^2{)+p/(1-\left|p\right|}^2)$ is a non-positive real number.

5. Proof of the Main Results

For completeness, we present a simple proof of Theorem 7 based on Pick’s ideas.

Proof of Theorem 7. 

Let $f\in \mathcal{B}[0,1]$. By Theorem 1 the function $g\left(z\right)=f\left(z\right)/z$ takes the unit disk $\mathbb{D}$ into itself. Since $z=1$ is a fixed point of f, the function g belongs to $\mathcal{B}\left[1\right]$. Applying Theorem 5 to the function g, for all $z\in \mathbb{D}$ we get

$$\frac{{|1-g\left(z\right)|}^2}{1-{\left|g\left(z\right)\right|}^2}\frac{1-{\left|z\right|}^2}{{|1-z|}^2}⩽g^{\prime }\left(1\right),$$

which is equivalent to the inequality

$$\frac{{|1-f\left(z\right)/z|}^2}{{1-|f\left(z\right)/z|}^2}\frac{1-{\left|z\right|}^2}{{|1-z|}^2}⩽f^{\prime }\left(1\right)-1.$$

(15)

Setting $z=0$ in (15), we arrive at estimate (8), and the first part of the theorem is proved.

We turn to the proof of the second part of the theorem. Let for some function $f\in \mathcal{B}[0,1]$, $f\left(z\right)\not\equiv z$, in (8) the equality is attained. Let us rewrite it in terms of the function $g\left(z\right)=f\left(z\right)/z$ from the class $\mathcal{B}\left[1\right]$:

$$\frac{{|1-g\left(0\right)|}^2}{1-{\left|g\left(0\right)\right|}^2}=g^{\prime }\left(1\right).$$

By Theorem 5, the last equality is possible only for Möbius transformation $g\left(z\right)=L_q\left(z\right)$, $q\in \mathbb{D}$. Thus, $f\left(z\right)=z{L}_q\left(z\right)$, $q\in \mathbb{D}$.

Conversely, let $f\left(z\right)=z{L}_q\left(z\right)$ for some $q\in \mathbb{D}$. Obviously, $f\in \mathcal{B}[0,1]$ and since

$$f^{\prime }\left(0\right)=-q\frac{1-\overline{q}}{1-q},f^{\prime }\left(1\right)=\frac{2-q-\overline{q}}{(1-q)(1-\overline{q})},$$

it is easy to see that the function f turns inequality (8) into an equality. This completes the proof of the theorem. □

Now we can proceed to the proof of the main results of this paper.

Proof of Theorem 11. 

Fix the point $z_0\in \mathbb{D}$. If the function f belongs to $\mathcal{B}\left[1\right]$, then the composition

$$h\left(w\right)=L_{f\left(z_0\right)}\circ f\circ L_{z_0}^{-1}\left(w\right)$$

(16)

belongs to $\mathcal{B}[0,1]$. Moreover, its derivatives at the fixed points have the form

$$\begin{array}{c}{h}^{\prime }\left(0\right)=f^{\prime }\left(z_0\right)\frac{1-z_0}{1-\overline{z_0}}\frac{1-\overline{f\left(z_0\right)}}{1-f\left(z_0\right)}\frac{1-|z_0{|}^2}{1-|f\left(z_0\right){|}^2},\end{array}$$

(17)

$$\begin{array}{c}{h}^{\prime }\left(1\right)=f^{\prime }\left(1\right)\frac{|1-z_0{|}^2}{1-|z_0{|}^2}\frac{1-|f\left(z_0\right){|}^2}{|1-f\left(z_0\right){|}^2}.\end{array}$$

(18)

The function h satisfies the conditions of Theorem 7, and hence

$$h^{\prime }\left(1\right)⩾1+\frac{|1-h^{\prime }{\left(0\right)|}^2}{1-|h^{\prime }{\left(0\right)|}^2}.$$

(19)

Plugging (17) and (18) into (19), and taking into account the arbitrary choice of the point $z_0$, we get estimate (12). The first part of the theorem is proved.

We turn to the proof of the second part of the theorem. Let, for a function $f\in \mathcal{B}\left[1\right]$ in (12), the equality be attained at some point $z_0\in \mathbb{D}$. In terms of the function $h\in \mathcal{B}[0,1]$ defined by (16), in view of (17) and (18), this equality can be written as

$$h^{\prime }\left(1\right)=1+\frac{|1-h^{\prime }{\left(0\right)|}^2}{1-|h^{\prime }{\left(0\right)|}^2}.$$

But then, by Theorem 7, there is a point $s\in \mathbb{D}$ such that $h\left(w\right)=w{L}_s\left(w\right)$. It follows that $f\left(z\right)=L_{f\left(z_0\right)}^{-1}\circ h\circ L_{z_0}\left(z\right)$ is a Blaschke product of degree 2.

Conversely, let $f\left(z\right)=L_q\left(z\right)L_p\left(z\right)$ for some $q,p\in \mathbb{D}$. Obviously, $f\in \mathcal{B}\left[1\right]$. Plugging the derivatives

$$\begin{array}{cc}\hfill f^{\prime }\left(z\right)& =-\frac{(1-\overline{q})(1-\overline{p})}{(1-q)(1-p)}\frac{{q(1-|p|}^2{)+p(1-\left|q\right|}^2{)-2(1-\left|q\right|}^2{\left|p\right|}^2)z+\left(\overline{q}{(1-|p|}^2)+\overline{p}{(1-|q|}^2)\right)z^2}{{(1-\overline{q}z)}^2{(1-\overline{p}z)}^2},\hfill \\ \hfill f^{\prime }\left(1\right)& =-\frac{(q+\overline{q}){(1-|p|}^2)+(p+\overline{p}){(1-|q|}^2{)-2(1-\left|q\right|}^2{\left|p\right|}^2)}{{|1-q|}^2{|1-p|}^2}\hfill \end{array}$$

into inequality (12), we obtain the equality for all $z\in \mathbb{D}$. The theorem is proved. □

Proof of Corollary 1. 

By the triangle inequality we have the lower estimate

$$\left|f^{\prime }\left(z\right)-\frac{1}{f^{\prime }\left(1\right)}\frac{{\left(1-f\left(z\right)\right)}^2}{{(1-z)}^2}\right|⩾\frac{1}{f^{\prime }\left(1\right)}\frac{{\left|1-f\left(z\right)\right|}^2}{{|1-z|}^2}-\left|f^{\prime }\left(z\right)\right|.$$

Complementing it with the upper estimate (11), we obtain inequality (13). □

Proof of Corollary 2. 

The first part of the result is Corollary 1 when $z=0$.

Now we prove the second part. Let for some function $f\in \mathcal{B}\left[1\right]$ in (14), the equality is attained. By Theorem 11, the function f has the form $f\left(z\right)=L_q\left(z\right)L_p\left(z\right)$ for some $q,p\in \mathbb{D}$. Plugging $f\left(z\right)=L_q\left(z\right)L_p\left(z\right)$ into equality (14), we find a condition satisfied by q and p. In view of the relations

$$\begin{array}{cc}\hfill f\left(0\right)& =\frac{1-\overline{q}}{1-q}\frac{1-\overline{p}}{1-p}qp,\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill |f^{\prime }\left(0\right)|& =\left|{q(1-|p|}^2{)+p(1-\left|q\right|}^2)\right|,\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill f^{\prime }\left(1\right)& =\frac{1-{\left|q\right|}^2}{{|1-q|}^2}+\frac{1-{\left|p\right|}^2}{{|1-p|}^2}\hfill \end{array}$$

(22)

we have

$$\begin{array}{c}\hfill f^{\prime }\left(1\right)-\frac{{2|1-f\left(0\right)|}^2}{1+|f^{\prime }{\left(0\right)|-|f\left(0\right)|}^2}=\frac{2\left(\frac{\Re q}{1-{\left|q\right|}^2}+\frac{\Re p}{1-{\left|p\right|}^2}+\left|\frac{q}{1-{\left|q\right|}^2}+\frac{p}{1-{\left|p\right|}^2}\right|\right){|1-qp|}^2}{{|1-q|}^2{|1-p|}^2{\left(\frac{1-{\left|q\right|}^2{\left|p\right|}^2}{{(1-|q|}^2\left)\right(1-{\left|p\right|}^2)}+\left|\frac{q}{1-{\left|q\right|}^2}+\frac{p}{1-{\left|p\right|}^2}\right|\right)}^2}.\end{array}$$

Its right-hand side vanishes if and only if ${q/(1-|q|}^2{)+p/(1-\left|p\right|}^2)⩽0$.

Conversely, let $f\left(z\right)=L_q\left(z\right)L_p\left(z\right)$ for some $q,p\in \mathbb{D}$, that satisfy the condition ${q/(1-|q|}^2{)+p/(1-\left|p\right|}^2)⩽0$. Then $f\in \mathcal{B}\left[1\right]$ and relations (20)–(22) are correct. Substituting (20)–(22) in (14) and taking into account the condition connecting q and p, we obtain the equality. □