The Schwarz Type Inequality for Harmonic Mappings of the Unit Disc with Boundary Normalization

Abstract

Let $\mathcal H$ be the class of all complex-valued harmonic functions $F$ of the unit disk $\mathbb D$ into itself such that for every $k\in \{0,1,2\}$ and almost every $z\in T_k:=\{\mathrm {e}^{\mathrm {i}\theta }:2k\pi /3\le \theta \le 2(k+1)\pi /3\}$ the radial limit of $F$ at $z$ belongs to the angular sector determined by the convex hull spanned by the origin and arc $T_k$. The sharp estimation of the modulus $|F(z)|$ for $z\in \mathbb D$ in the class $\mathcal H$ is obtained and the extremal functions are determined.

1 Introduction

Let $\mathbb D(a,r):=\{z\in \mathbb C:|z-a|<r\}$ and $\mathbb T(a,r):=\{z\in \mathbb C:|z-a|=r\}$ for $a\in \mathbb C$ and $r>0$. In particular $\mathbb D:=\mathbb D(0,1)$ and $\mathbb T:=\mathbb T(0,1)$ are the unit disk and unit circle, respectively. We denote by $\mathrm {Har}(\mathbb D)$ the class of all complex-valued harmonic functions in $\mathbb D$, i.e., the class of twice continuously differentiable functions $F$ in $\mathbb D$ satisfying the Laplace equation

$$\begin{aligned} \frac{\partial ^2}{\partial x^2}F(z)+\frac{\partial ^2}{\partial y^2}F(z)=0,\quad z=x+\mathrm {i} y\in \mathbb D. \end{aligned}$$

In 1959 E. Heinz proved in [5] that for every $F\in \mathrm {Har}(\mathbb D)$, if $F(\mathbb D)\subset \mathbb D$ and $F(0)=0$, then

$$\begin{aligned} |F(z)|\le \frac{4}{\pi }\arctan |z|,\quad z\in \mathbb D ; \end{aligned}$$

(1.1)

cf. also [4, p. 77] and [8]. Moreover, the above estimate is sharp. The equality in (1.1) holds if

$$\begin{aligned} F(\zeta )=\mathrm {Re}\biggl [\frac{2\mathrm {i}}{\pi }\log \frac{1+\zeta }{1-\zeta }\biggr ],\quad \zeta \in \mathbb D , \end{aligned}$$

(1.2)

and if $z\in \mathbb D\cap \mathbb R$. This result can be treated as a counterpart of the well known Schwarz lemma where holomorphic mappings are replaced by the harmonic ones. The estimation (1.1) can be essentially improved under the additional assumption that $F$ is a quasiconformal mapping of $\mathbb D$ onto itself; cf. [8, Thm. 3.3] where the bound of $|F(z)|$ depends on the maximal dilatation $K$ of $F$.

In this paper we intend to find a sort of the estimation (1.1) for $F\in \mathrm {Har}(\mathbb D)$ where the normalization condition $F(0)=0$ is replaced by a certain boundary one described in Corollary 2.2. Its proof appeals to Theorem 2.1 which gives a Schwarz type estimation for $F$ determined by the Poisson integral of a certain function $f:\mathbb T\rightarrow \mathrm {cl}(\mathbb D)$. Here and subsequently, the symbol $\mathrm {cl}(A)$ stands for the closure of a set $A\subset \mathbb C$ in the Euclidian topology. From Corollary 2.2 we infer Corollary 2.4, which deals with harmonic injective mappings of $\mathbb D$ onto itself. In particular, Corollary 2.4 leads to the following result: If $F$ is a continuous function in the closed disk $\mathrm {cl}(\mathbb D)$ onto itself, harmonic and injective in $\mathbb D$, and normalized by

$$\begin{aligned} F(e_k)=e_k:=\mathrm {e}^{2\pi \mathrm {i} k/3} ,\quad k\in \{0,1,2\} , \end{aligned}$$

(1.3)

then

$$\begin{aligned} |F(z)|\le \frac{2}{\pi }\arctan \biggl (\sqrt{3}\frac{1+|z|}{1-|z|}\biggr ),\quad z\in \mathbb D. \end{aligned}$$

(1.4)

This is a direct consequence of Corollary 2.4. The right hand side of the inequality in (1.4) can be replaced by the right hand side of the equality in (2.35) or (2.36); cf. Remark 2.5. The estimations (2.5) and (2.29) are sharp. The extremal functions are described in the last section; cf. Theorem 3.3 and Corollary 3.4.

2 The Schwarz Type Inequalities

Given an integrable function $f:\mathbb T\rightarrow \mathbb C$ we denote by $\mathrm {P}[f](z)$ the Poisson integral of $f$ at $z\in \mathbb D$, i.e.,

$$\begin{aligned} \mathrm {P}[f](z):=\frac{1}{2\pi }\int _{\mathbb T}f(u)\mathrm {Re}\frac{u+z}{u-z}|\mathrm {d} u| , \quad z\in \mathbb D\ . \end{aligned}$$

(2.1)

Here and in the sequel integrable means integrable in the sense of Lebesgue. The Poisson integral $\mathrm {P}[f]$ is the unique solution to the Dirichlet problem for the unit disk $\mathbb D$ provided the boundary function $f$ is continuous; cf. e.g. [7, Thm. 2.11]. This means that $\mathrm {P}[f]$ is a harmonic mapping in $\mathbb D$ which has a continuous extension to the closed disk $\mathrm {cl}(\mathbb D)$ and its boundary limiting valued function is identical with $f$. Since

$$\begin{aligned} \mathrm {Re}\frac{u+z}{u-z}&=\mathrm {Re}\frac{(u+z)\overline{(u-z)}}{|u-z|^2}=\frac{|u|^2-|z|^2}{|u|^2-2\mathrm {Re}(u\overline{z})+|z|^2} \\&=\frac{1-|z|^2}{1-2\mathrm {Re}(z\overline{u})+|z|^2},\quad u\in \mathbb T\,,\,z\in \mathbb D , \end{aligned}$$

we conclude from (2.1) that

$$\begin{aligned} \mathrm {P}[f](r\mathrm {e}^{\mathrm {i}\theta })=\int _0^{2\pi }\mathrm {P}_r(\theta -t)f(\mathrm {e}^{\mathrm {i} t})\mathrm {d} t , \quad r\mathrm {e}^{\mathrm {i}\theta }\in \mathbb D , \end{aligned}$$

(2.2)

where

$$\begin{aligned} \mathrm {P}_r(t):=\frac{1}{2\pi }\frac{1-r^2}{1-2r\cos t+r^2} , \quad r\mathrm {e}^{\mathrm {i} t}\in \mathbb D , \end{aligned}$$

Setting $I_k:=[2\pi k/3;2\pi (k+1)/3]$ for $k\in \{0,1,2\}$ we define

$$\begin{aligned} T_k:=\{\mathrm {e}^{\mathrm {i} t}:t\in I_k\}\quad \text {and}\quad D_k:=\{r\mathrm {e}^{\mathrm {i} t}:t\in I_k\,,\,r\in [0;1]\} ,\quad k\in \{0,1,2\}. \end{aligned}$$

(2.3)

Theorem 2.1

For any integrable function $f:\mathbb T\rightarrow \mathbb C$, if

$$\begin{aligned} f(z)\in D_k\quad \text {for a.e.}\ z\in T_k ,\quad k\in \{1,2,3\} , \end{aligned}$$

(2.4)

then

$$\begin{aligned} |\mathrm {P}[f](z)|\le \frac{4}{3}-\frac{2}{\pi }\arctan \biggl (\frac{\sqrt{3}}{1+2|z|}\biggr ),\quad z\in \mathbb D , \end{aligned}$$

(2.5)

and the estimation is sharp.

Proof

Given a function $f:\mathbb T\rightarrow \mathbb C$ satisfying the condition (2.4) we consider the function $F:=\mathrm {P}[f]$. Fix $r\in [0;1)$. Since the function $\mathbb R\ni \theta \mapsto |F(r\mathrm {e}^{\mathrm {i}\theta })|$ is continuous and $2\pi$-periodical we have

$$\begin{aligned} |F(r\mathrm {e}^{\mathrm {i}\theta })|\le |F(r\mathrm {e}^{\mathrm {i}\theta _r})|,\quad \theta \in \mathbb R , \end{aligned}$$

(2.6)

for a certain $\theta _r\in [0;2\pi )$. On the other hand, there exists $\alpha \in [0;2\pi )$ such that

$$\begin{aligned} F(r\mathrm {e}^{\mathrm {i}\theta _r})=\mathrm {e}^{\mathrm {i}\alpha }|F(r\mathrm {e}^{\mathrm {i}\theta _r})|\ . \end{aligned}$$

(2.7)

Therefore

$$\begin{aligned} |F(r\mathrm {e}^{\mathrm {i}\theta _r})|=\mathrm {Re}\bigl (\mathrm {e}^{-\mathrm {i}\alpha }F(r\mathrm {e}^{\mathrm {i}\theta _r})\bigr )\ . \end{aligned}$$

Hence and by (2.6),

$$\begin{aligned} |F(r\mathrm {e}^{\mathrm {i}\theta })|\le \mathrm {Re}\bigl (\mathrm {e}^{-\mathrm {i}\alpha }F(r\mathrm {e}^{\mathrm {i}\theta _r})\bigr )=\mathrm {Re}\bigl (\mathrm {e}^{-\mathrm {i}\alpha }\mathrm {P}[f](r\mathrm {e}^{\mathrm {i}\theta _r})\bigr ). \end{aligned}$$

(2.8)

Setting

$$\begin{aligned} p_k:=\int _{I_k}\mathrm {P}_r(\theta _r-t)\mathrm {d} t ,\quad k\in \{0,1,2\} , \end{aligned}$$

(2.9)

we conclude from the formula (2.2) that

$$\begin{aligned} \mathrm {P}[f](r\mathrm {e}^{\mathrm {i}\theta _r})&=\int _0^{2\pi }\mathrm {P}_r(\theta _r-t)f(\mathrm {e}^{\mathrm {i} t})\mathrm {d} t \nonumber \\&=\sum _{k=0}^2p_k\int _{I_k}\frac{1}{p_k}\mathrm {P}_r(\theta _r-t)f(\mathrm {e}^{\mathrm {i} t})\mathrm {d} t. \end{aligned}$$

(2.10)

Since each sector $D_k$, $k\in \{0,1,2\}$, is closed and convex in $\mathbb C$ we conclude from the assumption (2.4) and the integral mean value theorem for complex-valued functions that there exist $\alpha _k\in I_k$ and $r_k\in [0;1]$ for $k\in \{0,1,2\}$ such that

$$\begin{aligned} r_k\mathrm {e}^{\mathrm {i}\alpha _k}=\int _{I_k}\frac{1}{p_k}\mathrm {P}_r(\theta _r-t)f(\mathrm {e}^{\mathrm {i} t})\mathrm {d} t ,\quad k\in \{0,1,2\}. \end{aligned}$$

(2.11)

Combining this with (2.8) and (2.10) we obtain

$$\begin{aligned} |F(r\mathrm {e}^{\mathrm {i}\theta })|\le \mathrm {Re}\left[ \mathrm {e}^{-\mathrm {i}\alpha }\sum _{k=0}^2p_k r_k\mathrm {e}^{\mathrm {i}\alpha _k}\right] =\sum _{k=0}^2 c_k p_k,\quad \theta \in \mathbb R , \end{aligned}$$

(2.12)

where

$$\begin{aligned} c_k:=\mathrm {Re}\bigl (\mathrm {e}^{-\mathrm {i}\alpha }r_k\mathrm {e}^{\mathrm {i}\alpha _k})\bigr )=r_k\mathrm {Re}\bigl (\mathrm {e}^{\mathrm {i}(\alpha _k-\alpha )}\bigr )=r_k\cos (\alpha _k-\alpha ),\quad k\in \{0,1,2\}. \end{aligned}$$

(2.13)

If $\min (\{c_0,c_1,c_2)\}\le 0$, then clearly

$$\begin{aligned} c_0+c_1+c_2\le 2. \end{aligned}$$

(2.14)

Therefore, we can assume that all $c_0,c_1,c_2$ are positive, and consequently

$$\begin{aligned} c_k\le \cos (\alpha _k-\alpha ) ,\quad k\in \{0,1,2\} , \end{aligned}$$

(2.15)

because $r_k\in [0;1]$ for $k\in \{0,1,2\}$. Since $\alpha _k\in I_k$ for $k\in \{0,1,2\}$, we see that

$$\begin{aligned} \frac{2\pi k}{3}-\alpha \le \alpha _k-\alpha \le \frac{2\pi (k+1)}{3}-\alpha ,\quad k\in \{0,1,2\}. \end{aligned}$$

(2.16)

If $0\le \alpha \le 2\pi /3$, then by (2.15) and (2.16),

$$\begin{aligned} c_0\le 1 ,\quad c_1\le \cos \Bigl (\frac{2\pi }{3}-\alpha \Bigr ) \quad \text {and}\quad c_2\le \cos (2\pi -\alpha )=\cos (\alpha ) , \end{aligned}$$

and so

$$\begin{aligned} c_0+c_1+c_2\le 1+\cos \Bigl (\frac{2\pi }{3}-\alpha \Bigr )+\cos (\alpha )=1+2\cos \frac{\pi }{3}\cos \Bigl (\frac{\pi }{3}-\alpha \Bigr )\le 2\ . \end{aligned}$$

If $2\pi /3\le \alpha \le 4\pi /3$, then by (2.15) and (2.16),

$$\begin{aligned} c_0\le \cos \Bigl (\frac{2\pi }{3}-\alpha \Bigr ) ,\quad c_1\le 1 \quad \text {and}\quad c_2\le \cos \Bigl (\frac{4\pi }{3}-\alpha \Bigr ) , \end{aligned}$$

and so

$$\begin{aligned} c_0+c_1+c_2\le 1+\cos \Bigl (\frac{2\pi }{3}-\alpha \Bigr )+\cos \Bigl (\frac{4\pi }{3}-\alpha \Bigr )=1+2\cos \frac{\pi }{3}\cos (\pi -\alpha )\le 2\ . \end{aligned}$$

If $4\pi /3\le \alpha \le 2\pi$, then by (2.15) and (2.16),

$$\begin{aligned} c_0\le \cos (\alpha ) ,\quad c_1\le \cos \Bigl (\frac{4\pi }{3}-\alpha \Bigr ) \quad \text {and}\quad c_2\le 1 , \end{aligned}$$

and so

$$\begin{aligned} c_0+c_1+c_2\le 1+\cos (\alpha )+\cos \Bigl (\frac{4\pi }{3}-\alpha \Bigr )=1+2\cos \frac{2\pi }{3}\cos \Bigl (\frac{2\pi }{3}-\alpha \Bigr )\le 2\ . \end{aligned}$$

Thus in all the cases the inequality (2.14) holds. In order to estimate the last sum in (2.12) we consider the function

$$\begin{aligned} \mathbb R\ni \theta \mapsto Q_r(\theta ):=\int _{2\pi /3}^{4\pi /3}\mathrm {P}_r(\theta -t)\mathrm {d} t=\mathrm {P}[{\chi }_{T_1}](r\mathrm {e}^{\mathrm {i}\theta }), \end{aligned}$$

(2.17)

where ${\chi }_A$ is the characteristic function of a set $A\subset \mathbb T$, i.e., ${\chi }_A(t):=1$ for $t\in A$ and ${\chi }_A(t):=0$ for $t\in \mathbb T{\setminus }A$. Throughout the paper we understand the function $\log$ as the inverse function to $\exp _{|\Omega }$, where $\Omega :=\{z\in \mathbb C:|\mathrm {Im} z|<\pi \}$. Given $z\in \mathbb D$ we see that $T_1\subset \Omega _z:=\mathbb C\setminus \{z+t:t>0\}$, the function $\Omega _z\ni \zeta \mapsto \log (z-\zeta )$ is holomorphic and

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d} t}\log (z-\mathrm {e}^{\mathrm {i} t})=\frac{\mathrm {i}\mathrm {e}^{\mathrm {i} t}}{\mathrm {e}^{\mathrm {i} t}-z} ,\quad t\in I_1. \end{aligned}$$

Hence and by the formula (2.1) we see that

$$\begin{aligned} \mathrm {P}[{\chi }_{T_1}](z)&=\frac{1}{2\pi }\int _{T_1}{\chi }_{T_1}(u)\mathrm {Re}\frac{u+z}{u-z}|\mathrm {d} u| \\&=\frac{1}{2\pi }\int _{2\pi /3}^{4\pi /3}\mathrm {Re}\frac{\mathrm {e}^{\mathrm {i} t}+z}{\mathrm {e}^{\mathrm {i} t}-z}\mathrm {d} t \\&=\frac{1}{2\pi }\int _{2\pi /3}^{4\pi /3}\mathrm {Re}\Bigl (\frac{2\mathrm {e}^{\mathrm {i} t}}{\mathrm {e}^{\mathrm {i} t}-z}-1\Bigr )\mathrm {d} t \\&=\frac{1}{\pi }\int _{2\pi /3}^{4\pi /3}\mathrm {Im}\Bigl (\frac{\mathrm {i}\mathrm {e}^{\mathrm {i} t}}{\mathrm {e}^{\mathrm {i} t}-z}\Bigr )\mathrm {d} t-\frac{1}{3} \\&=\frac{1}{\pi }\int _{2\pi /3}^{4\pi /3}\mathrm {Im}\frac{\mathrm {d}}{\mathrm {d} t}\log (z-\mathrm {e}^{\mathrm {i} t})\mathrm {d} t-\frac{1}{3} \\&=\frac{1}{\pi }\mathrm {Im}\bigl [\log (z-e_2)-\log (z-e_1)\bigr ]-\frac{1}{3}. \end{aligned}$$

Hence and by (2.17),

$$\begin{aligned} Q_r(\theta )=\frac{1}{\pi }\mathrm {Im}\bigl [\log (r\mathrm {e}^{\mathrm {i}\theta }-e_2)-\log (r\mathrm {e}^{\mathrm {i}\theta }-e_1)\bigr ]-\frac{1}{3},\quad \theta \in \mathbb R , \end{aligned}$$

(2.18)

and consequently,

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}\theta }Q_r(\theta )&=\frac{1}{\pi }\mathrm {Im}\Bigl [\frac{\mathrm {i} r\mathrm {e}^{\mathrm {i}\theta }}{r\mathrm {e}^{\mathrm {i}\theta }-e_2}-\frac{\mathrm {i} r\mathrm {e}^{\mathrm {i}\theta }}{r\mathrm {e}^{\mathrm {i}\theta }-e_1}\Bigr ]\nonumber \\ \nonumber&=\frac{r}{\pi }\mathrm {Im}\Bigl [\frac{\mathrm {i}\mathrm {e}^{\mathrm {i}\theta }(e_2-e_1)}{(r\mathrm {e}^{\mathrm {i}\theta }-e_2)(r\mathrm {e}^{\mathrm {i}\theta }-e_1)}\Bigr ] \\ \nonumber&=\frac{r\sqrt{3}}{\pi }\frac{\mathrm {Im}[\mathrm {e}^{\mathrm {i}\theta }(r\mathrm {e}^{-\mathrm {i}\theta }-e_1)(r\mathrm {e}^{-\mathrm {i}\theta }-e_2)}{|r\mathrm {e}^{\mathrm {i}\theta }-e_2|^2|r\mathrm {e}^{\mathrm {i}\theta }-e_1|^2} \\ \nonumber&=\frac{r\sqrt{3}}{\pi }\frac{\mathrm {Im}[r^2\mathrm {e}^{-\mathrm {i}\theta }+r+\mathrm {e}^{\mathrm {i}\theta }]}{|r\mathrm {e}^{\mathrm {i}\theta }-e_2|^2|r\mathrm {e}^{\mathrm {i}\theta }-e_1|^2} \\&=\frac{r\sqrt{3}(1-r^2)\sin \theta }{\pi |r\mathrm {e}^{\mathrm {i}\theta }-e_2|^2|r\mathrm {e}^{\mathrm {i}\theta }-e_1|^2},\quad \theta \in \mathbb R. \end{aligned}$$

(2.19)

Combining this with (2.18) we obtain

$$\begin{aligned} Q_r(\theta )\ge Q_r(0)&=\frac{1}{\pi }\mathrm {Im}\bigl [\log (r-e_2)-\log (r-e_1)\bigr ]-\frac{1}{3} \nonumber \\&=\frac{2}{\pi }\arctan \frac{\sqrt{3}}{1+2r}-\frac{1}{3},\quad \theta \in \mathbb R. \end{aligned}$$

(2.20)

Combining (2.9) with (2.17) we have

$$\begin{aligned} \begin{aligned} p_0&=\int _{I_0}\mathrm {P}_r(\theta _r-t)\mathrm {d} t=\int _{I_1}\mathrm {P}_r(\theta _r+2\pi /3-t)\mathrm {d} t=Q_r(\theta _r+2\pi /3) , \\ p_1&=\int _{I_1}\mathrm {P}_r(\theta _r-t)\mathrm {d} t=Q_r(\theta _r), \\ p_2&=\int _{I_2}\mathrm {P}_r(\theta _r-t)\mathrm {d} t=\int _{I_1}\mathrm {P}_r(\theta _r-2\pi /3-t)\mathrm {d} t=Q_r(\theta _r-2\pi /3). \end{aligned} \end{aligned}$$

(2.21)

Hence and by (2.20),

$$\begin{aligned} \min (\{p_0,p_1,p_2\})\ge Q_r(0)=\frac{2}{\pi }\arctan \frac{\sqrt{3}}{1+2r}-\frac{1}{3}. \end{aligned}$$

(2.22)

From (2.9) it follows that

$$\begin{aligned} p_0+p_1+p_2=\sum _{k=0}^2\int _{I_k}\mathrm {P}_r(\theta _r-t)\mathrm {d} t=\int _0^{2\pi }\mathrm {P}_r(\theta _r-t)\mathrm {d} t=1. \end{aligned}$$

(2.23)

This together with (2.22), (2.14) and (2.15) leads to

$$\begin{aligned} \sum _{k=0}^2 c_k p_k&=\sum _{k=0}^2(p_k-Q_r(0))c_k+\sum _{k=0}^2 Q_r(0)c_k \nonumber \\ \nonumber&\le \sum _{k=0}^2(p_k-Q_r(0))+Q_r(0)\sum _{k=0}^2 c_k \\ \nonumber&\le \sum _{k=0}^2 p_k-\sum _{k=0}^2 Q_r(0)+2Q_r(0) \\&=1-Q_r(0). \end{aligned}$$

(2.24)

Combining (2.12) with (2.24) and (2.20) we infer the estimation (2.5). It is clear that the function

$$\begin{aligned} f_0:={\chi }_{T_0\cup T_2} \end{aligned}$$

(2.25)

satisfies the condition (2.4) with $f:=f_0$. From the formula (2.17) it follows that for every $z=r\mathrm {e}^{\mathrm {i}\theta }\in \mathbb D$,

$$\begin{aligned} \mathrm {P}[f_0](z)=\mathrm {P}[{\chi }_{\mathbb T{\setminus }T_1}](z)=\mathrm {P}[{\chi }_{\mathbb T}-{\chi }_{T_1}](z)=1-\mathrm {P}[{\chi }_{T_1}](z)=1-Q_r(\theta )\ . \end{aligned}$$

(2.26)

Hence and by the equalities in (2.20) we conclude that

$$\begin{aligned} \mathrm {P}[f_0](r)=1-Q_r(0)=\frac{4}{3}-\frac{2}{\pi }\arctan \frac{\sqrt{3}}{1+2r},\quad r\in [0;1). \end{aligned}$$

(2.27)

This means that $f_0$ is an extremal function in Theorem 2.1, and so the estimation (2.5) is sharp, which completes the proof. $\square$

Now we are ready to prove our main result.

Corollary 2.2

Let $F\in \mathrm {Har}(\mathbb D)$ satisfy $F(\mathbb D)\subset \mathbb D$ and for every $k\in \{0,1,2\}$,

$$\begin{aligned} \lim _{r\rightarrow 1^-}F(rz)\in D_k\quad \text {for a.e.}\ z\in T_k\cap \Lambda (F) , \end{aligned}$$

(2.28)

where $\Lambda (F)$ is the set of all $z\in \mathbb T$ such that the limit exists. Then

$$\begin{aligned} |F(z)|\le \frac{4}{3}-\frac{2}{\pi }\arctan \biggl (\frac{\sqrt{3}}{1+2|z|}\biggr ) ,\quad z\in \mathbb D , \end{aligned}$$

(2.29)

and the estimation is sharp.

Proof

Fix a harmonic function $F:\mathbb D\rightarrow \mathbb D$ satisfying the assumptions of the theorem. Then $\mathrm {Re} F$ and $\mathrm {Im} F$ are real-valued and bounded harmonic functions in $\mathbb D$, and so they have a radial limit a.e. in $\mathbb T$; cf. [3, Cor. 1, Sect. 1.2]. Therefore, there exists the limit in (2.28) for a.e. $z\in \mathbb T$. Setting

$$\begin{aligned} \mathbb T\ni z\mapsto f(z):={\left\{ \begin{array}{ll} \lim \limits _{r\rightarrow 1^-}F(rz) &{}\text {for}\ z\in \Lambda (F) ,\\ 0 &{}\text {for}\ z\in \mathbb T\setminus \Lambda (F) , \end{array}\right. } \end{aligned}$$

we see that the function $f$ is measurable and bounded. Since $F(\mathbb D)\subset \mathbb D$, we can apply the dominated convergence theorem to show that $F=\mathrm {P}[f]$. From the condition (2.28) it follows that $f(z)\in D_k$ for a.e. $z\in T_k$ and every $k\in \{0,1,2\}$. Theorem 2.1 now implies the estimation (2.29).

From the properties of the Poisson integral it follows that the function

$$\begin{aligned} \mathbb D\ni z\mapsto F_0(z):=\mathrm {P}[{\chi }_{T_0\cup T_2}](z) \end{aligned}$$

(2.30)

is harmonic in $\mathbb D$, $F_0(\mathbb D)\subset \mathbb D$ and

$$\begin{aligned} \lim _{r\rightarrow 1^-}F_0(rz)=\mathrm {P}[{\chi }_{T_0\cup T_2}](z) ,\quad z\in \mathbb T\setminus \{e_1,e_2\}. \end{aligned}$$

Therefore the function $F_0$ satisfies the condition (2.28) with $F$ replaced by $F_0$. Combining (2.30) with (2.25) and (2.27) we see that

$$\begin{aligned} F_0(r)=\frac{4}{3}-\frac{2}{\pi }\arctan \frac{\sqrt{3}}{1+2r} ,\quad r\in [0;1). \end{aligned}$$

This means that $F_0$ is an extremal function in Corollary 2.2, and so the estimation (2.29) is sharp, which completes the proof. $\square$

Remark 2.3

Let $\mathcal H$ be the class of all $F\in \mathrm {Har}(\mathbb D)$ satisfying the assumption of Corollary 2.2. Applying (2.29) for $z:=0$ we obtain the following useful estimation

$$\begin{aligned} |F(0)|\le \frac{2}{3} ,\quad F\in \mathcal H , \end{aligned}$$

which is best possible. In other words, the variability region of the origin for the class $\mathcal H$ satisfies the inclusion

$$\begin{aligned} \{F(0):F\in \mathcal H\}\subset \mathrm {cl}(\mathbb D(0,2/3)) \end{aligned}$$

(2.31)

and $2/3$ is the smallest radius of the inclusion.

Corollary 2.4

Let $F\in \mathrm {Har}(\mathbb D)$ be an injective mapping in $\mathbb D$ such that $F(\mathbb D)=\mathbb D$ and

$$\begin{aligned} \lim _{r\rightarrow 1^-}F(re_k)=e_k ,\quad k\in \{0,1,2\}. \end{aligned}$$

(2.32)

Then the estimation (2.29) holds.

Proof

Given a mapping $F\in \mathrm {Har}(\mathbb D)$ suppose that $F$ is injective in $\mathbb D$ and $F(\mathbb D)=\mathbb D$. Then, as shown by Choquet in [2], there exists a continuous mapping $f:\mathbb T\rightarrow \mathbb T$ such that $F=\mathrm {P}[f]$, $f(\mathbb T)=\mathbb T$ and for each $w\in \mathbb T$, $f^{-1}(\{w\})$ is a connected subset of $\mathbb T$. Therefore, $f^{-1}(\{w\})$ is a closed arc of $\mathbb T$ for $w\in \mathbb T$. This property can be also deduced from a more general result; cf. [6] or [4, Sect. 3.3]. Moreover, from the boundary normalization condition (2.32) it follows, that

$$\begin{aligned} f(e_k)=\lim _{r\rightarrow 1^-}\mathrm {P}[f](re_k)=\lim _{r\rightarrow 1^-}F(re_k)=e_k ,\quad k\in \{0,1,2\}. \end{aligned}$$

(2.33)

Suppose that $f^{-1}(\{e_2\})\cap T_0\ne \emptyset$. Since $f^{-1}(\{e_2\})$ is a closed arc of $\mathbb T$, we see that $e_0\in f^{-1}(\{e_2\})$ or $e_1\in f^{-1}(\{e_2\})$. Hence $f(e_0)=e_2$ or $f(e_1)=e_2$, which contradicts the equalities (2.33). Thus $e_2\notin f(T_0)$. Moreover, by the continuity of $f$ and by (2.33) we see that $f(T_0)$ is a closed arc of $\mathbb T$ containing the points $e_0$ and $e_1$. Consequently,

$$\begin{aligned} T_0\subset f(T_0)\subset \mathbb T\setminus \{e_2\}. \end{aligned}$$

(2.34)

Suppose that $f(T_0){\setminus }T_0\ne \emptyset$. Since

$$\begin{aligned} f(T_0{\setminus }f^{-1}(\{e_0,e_1\}))= f(T_0)\setminus \{e_0,e_1\} , \end{aligned}$$

we conclude from (2.34) that $f(T_0{\setminus }f^{-1}(\{e_0,e_1\}))$ is not a connected subset of $\mathbb T$. This is impossible, because $T_0{\setminus }f^{-1}(\{e_0,e_1\})$ is an open arc of $\mathbb T$, and so is the set $f(T_0{\setminus }f^{-1}(\{e_0,e_1\}))$. Therefore, $f(T_0)\subset T_0$. The same conclusion can be drawn for the arcs $T_1$ and $T_2$ in place of $T_0$, which means that

$$\begin{aligned} f(T_k)\subset T_k ,\quad k\in \{0,1,2\}. \end{aligned}$$

It follows that for every $k\in \{0,1,2\}$,

$$\begin{aligned} \lim _{r\rightarrow 1^-}F(rz)=f(z)\in D_k ,\quad z\in T_k , \end{aligned}$$

and consequently the condition (2.28) holds. Then Corollary 2.2 yields the estimation (2.29), which proves the corollary. $\square$

Remark 2.5

Using the trigonometric identity

$$\begin{aligned} \tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta } ,\quad \alpha ,\beta ,\alpha -\beta \in \mathbb R\setminus \{n\pi +\pi /2:n\in \mathbb Z\} , \end{aligned}$$

we can transform the right hand side of the inequality in (2.29) as follows

$$\begin{aligned} \frac{4}{3}-\frac{2}{\pi }\arctan \biggl (\frac{\sqrt{3}}{1+2|z|}\biggr )=\frac{2}{\pi }\arctan \biggl (\sqrt{3}\frac{1+|z|}{1-|z|}\biggr ) ,\quad z\in \mathbb D. \end{aligned}$$

(2.35)

Let us recall that the hyperbolic metric $\rho _h$ in $\mathbb D$ is given by the formula

$$\begin{aligned} \rho _h(z,w):=\frac{1}{2}\log \frac{|1-z\overline{w}|+|z-w|}{|1-z\overline{w}|-|z-w|} ,\quad z,w\in \mathbb D ; \end{aligned}$$

cf. e.g. [1]. Hence

$$\begin{aligned} \mathrm {e}^{2\rho _h(z,0)}=\frac{1+|z|}{1-|z|} ,\quad z\in \mathbb D , \end{aligned}$$

which together with (2.35) leads to

$$\begin{aligned} \frac{4}{3}-\frac{2}{\pi }\arctan \biggl (\frac{\sqrt{3}}{1+2|z|}\biggr )=\frac{2}{\pi }\arctan \Bigl (\sqrt{3}\mathrm {e}^{2\rho _h(z,0)}\Bigr ) ,\quad z\in \mathbb D. \end{aligned}$$

(2.36)

Therefore the right hand side of the inequalities in (2.5) and (2.29) can be replaced by the right hand side of the equality in (2.35) or (2.36).

3 The Extremal Functions

As shown in the proof of Theorem 2.1 the function $f_0$, given by the formula (2.25), yields the equality in (2.5). Consequently, the function $F_0$, given by the formula (2.30), is extremal in Corollary 2.2. Since for every $z=r\mathrm {e}^{\mathrm {i}\theta }\in \mathbb D$ the equalities (2.26) hold, we conclude from (2.18) that

$$\begin{aligned} F_0(z)=1-Q_r(\theta )=\frac{4}{3}-\frac{1}{\pi }\mathrm {Im}\bigl [\log (z-e_2)-\log (z-e_1)\bigr ] ,\quad z=r\mathrm {e}^{\mathrm {i}\theta }\in \mathbb D .\qquad \end{aligned}$$

(3.1)

We want to find all extremal functions in Theorem 2.1 and Corollary 2.2. To this end we need the following lemma.

Lemma 3.1

Given a closed arc $T\subset \mathbb T$ of length less than $\pi$, $a\in \mathbb D$, $b\in T\cup \{0\}$ and an integrable function $f:\mathbb T\rightarrow \mathrm {cl}(\mathbb D)$ suppose that

$$\begin{aligned} f(z)\in D:=\{r\zeta :\zeta \in T\,,\,r\in [0;1]\}\quad \text {for a.e.}\ z\in T \end{aligned}$$

(3.2)

as well as

$$\begin{aligned} \mathrm {P}[f\cdot {\chi }_T](a)=b\mathrm {P}[{\chi }_T](a). \end{aligned}$$

(3.3)

Then $f(z)=b$ for a.e. $z\in T$.

Proof

Let $T$, $a$, $b$ and $f$ satisfy the assumptions. Then $T=\{\mathrm {e}^{\mathrm {i} t}:t\in [\alpha _0-\alpha ;\alpha _0+\alpha ]\}$ for some $\alpha _0\in \mathbb R$ and $\alpha \in [0;\pi /2)$, and $D$ is a convex and closed subset of $\mathrm {cl}(\mathbb D)$. From (3.2) it follows that

$$\begin{aligned} \lambda _b\circ f(z)\in \lambda _b(D)\subset \{0\}\cup \{\zeta \in \mathbb C:\mathrm {Re}\zeta >0\}\quad \text {for a.e.}\ z\in T , \end{aligned}$$

(3.4)

where $\mathbb C\ni \zeta \mapsto \lambda _0(\zeta ):=\mathrm {e}^{-\mathrm {i}\alpha _0}\zeta$ and $\mathbb C\ni \zeta \mapsto \lambda _p(\zeta ):=1-\zeta /p$ for $p\in \mathbb T$.

From (3.4) it follows that $\mathrm {Re}\lambda _b(f(z))\ge 0$ for a.e. $z\in T$. By the equality (3.3) we have

$$\begin{aligned} \frac{\mathrm {P}[\mathrm {Re}((\lambda _b\circ f)\cdot {\chi }_T)](a)}{\mathrm {P}[{\chi }_T](a)} =\mathrm {Re}\left( \lambda _b\left( \frac{\mathrm {P}[f\cdot {\chi }_T](a)}{\mathrm {P}[{\chi }_T](a)}\right) \right) =\mathrm {Re}\lambda _b(b)=0. \end{aligned}$$

Hence the real-valued harmonic function $\mathrm {P}[\mathrm {Re}((\lambda _b\circ f)\cdot {\chi }_T)]$ attains minimum at $a$, and so, by the minimum principle for real-valued harmonic functions,

$$\begin{aligned} \mathrm {P}[\mathrm {Re}((\lambda _b\circ f)\cdot {\chi }_T)](z)=0 ,\quad z\in \mathbb D. \end{aligned}$$

Then for a.e. $z\in \mathbb T$,

$$\begin{aligned} \mathrm {Re}\lambda _b(f(z))=\mathrm {Re}((\lambda _b\circ f)\cdot {\chi }_T)(z)=\lim _{r\rightarrow 1^-}\mathrm {P}[\mathrm {Re}((\lambda _b\circ f)\cdot {\chi }_T)](rz)=0 ; \end{aligned}$$

cf. [3, Cor. 2, Sect. 1.2]. This together with (3.4) leads to $\lambda _0(f(z))=0$ for a.e. $z\in T$. Hence $f(z)=b$ for a.e. $z\in T$, which is the desired conclusion. $\square$

Remark 3.2

Setting

$$\begin{aligned} \mathbb C\ni z\mapsto S_k(z):=e_k z ,\quad k\in \{0,1,2\} , \end{aligned}$$

(3.5)

we see that the set $\{S_0,S_1,S_2\}$ is closed with respect to the composition operation of mappings $\circ$, i.e., $(\{S_0,S_1,S_2\},\circ )$ is a group. Let $\mathcal F$ be the class of all integrable functions $f:\mathbb T\rightarrow \mathbb C$ satisfying the condition (2.4). It is easily seen that

$$\begin{aligned} S_k\circ f\circ S_k^{-1}\in \mathcal F ,\quad f\in \mathcal F\,,\,k\in \{0,1,2\}. \end{aligned}$$

(3.6)

By the definition of the class $\mathcal H$ (see Remark 2.3) we also have

$$\begin{aligned} S_k\circ F\circ S_k^{-1}\in \mathcal H ,\quad F\in \mathcal H\,,\,k\in \{0,1,2\}. \end{aligned}$$

(3.7)

From the properties of the Poisson integral it may be also concluded that

$$\begin{aligned} \mathrm {P}[S_k\circ f\circ S_k^{-1}]=S_k\circ \mathrm {P}[f]\circ S_k^{-1} ,\quad f\in \mathcal F\,,\,k\in \{0,1,2\}. \end{aligned}$$

(3.8)

Theorem 3.3

Let $f:\mathbb T\rightarrow \mathbb C$ be an integrable function which satisfies the condition (2.4). Then

$$\begin{aligned} |\mathrm {P}[f](z)|=\frac{4}{3}-\frac{2}{\pi }\arctan \biggl (\frac{\sqrt{3}}{1+2|z|}\biggr ), \quad z\in \mathbb D, \end{aligned}$$

(3.9)

if and only if there exists $k\in \{0,1,2\}$ such that one of the two following conditions holds:

• (i) $z\ne 0$ and

  $$\begin{aligned} z=S_k(|z|) \quad \text {and}\quad f(u)=S_k\circ f_0\circ S_k^{-1}(u)\quad \text {for a.e.}\ u\in \mathbb T , \end{aligned}$$

  (3.10)

  where $f_0$ is the function given by the formula (2.25);

• (ii) $z=0$ and

  $$\begin{aligned} f(u)=S_k\circ f_0\circ S_k^{-1}(u) \quad \text {or}\quad f(u)=S_k\circ f_0^*\circ S_k^{-1}(u) \quad \text {for a.e.}\ u\in \mathbb T , \end{aligned}$$

  (3.11)

  where

  $$\begin{aligned} f_0^*:=\mathrm {e}^{\pi \mathrm {i}/3}\cdot {\chi }_{T_0}+e_1\cdot {\chi }_{T_1}+e_0\cdot {\chi }_{T_2}. \end{aligned}$$

  (3.12)

Proof

Fix $z\in \mathbb D$ and $f\in \mathcal F$. As shown in the final part of the proof of Theorem 2.1 the function $f_0$ satisfies (2.27). By the formula (3.12) we have

$$\begin{aligned} |\mathrm {P}[f_0^*](0)|&=\bigl |\mathrm {e}^{\pi \mathrm {i}/3}\mathrm {P}[{\chi }_{T_0}](0)+e_1\mathrm {P}[{\chi }_{T_1}](0)+e_0\mathrm {P}[{\chi }_{T_1}](0)\bigr | \\&=\frac{1}{3}\bigl |\mathrm {e}^{\pi \mathrm {i}/3}+e_1+e_0\bigr |=\frac{1}{3}\bigl |1+\mathrm {e}^{\pi \mathrm {i}/3}+\mathrm {e}^{-\pi \mathrm {i}/3}\bigr | =\frac{2}{3}=|\mathrm {P}[f_0](0)|. \end{aligned}$$

Applying now the property (3.8) we deduce the equality (3.9) from the condition (i), provided $z\ne 0$, and from the condition (ii), provided $z=0$.

Conversely, assume that the equality (3.9) holds. Analyzing respective parts in the proof of Theorem 2.1 we conclude from (2.12), (2.24) and the second equality in (2.27) that

$$\begin{aligned} \sum _{k=0}^2(p_k-Q_r(0))c_k+\sum _{k=0}^2 Q_r(0)c_k=\sum _{k=0}^2(p_k-Q_r(0))+2Q_r(0) , \end{aligned}$$

and consequently

$$\begin{aligned} \sum _{k=0}^2(p_k-Q_r(0))(1-c_k)+Q_r(0)\left( 2-\sum _{k=0}^2 c_k\right) =0. \end{aligned}$$

Combining this with (2.14), (2.15) and (2.22) we see that

$$\begin{aligned} c_0+c_1+c_2=2 \end{aligned}$$

(3.13)

and

$$\begin{aligned} (p_k-Q_r(0))(1-c_k)=0 ,\quad k\in \{0,1,2\}. \end{aligned}$$

(3.14)

Since $c_k\le 1$ for $k\in \{0,1,2\}$, the equality (3.13) is impossible provided $c:=\min (\{c_0,c_1,c_2\})<0$. Therefore $0\le c\le 1$. Suppose that $c=0$. Then $c_0=0$ or $c_1=0$ or $c_2=0$.

If $c_0=0$, then by (3.13), $c_1=c_2=1$, which leads, by (2.13), to

$$\begin{aligned} r_0=0 ,\ r_1=1 ,\ \alpha _1=\frac{4\pi }{3} ,\ r_2=1 ,\ \alpha _2=\frac{4\pi }{3}. \end{aligned}$$

(3.15)

If $c_1=0$, then by (3.13), $c_0=c_2=1$, which leads, by (2.13), to

$$\begin{aligned} r_1=0 ,\ r_0=1 ,\ \alpha _0=0 ,\ r_2=1 ,\ \alpha _2=2\pi . \end{aligned}$$

(3.16)

If $c_2=0$, then by (3.13), $c_0=c_1=1$, which leads, by (2.13), to

$$\begin{aligned} r_2=0 ,\ r_0=1 ,\ \alpha _0=\frac{2\pi }{3} ,\ r_1=1 ,\ \alpha _1=\frac{2\pi }{3}. \end{aligned}$$

(3.17)

In all the cases (3.15), (3.16) and (3.17), $r_k\mathrm {e}^{\mathrm {i}\alpha _k}\in T_k\cup \{0\}$ for $k\in \{0,1,2\}$. Then (2.11) implies, by Lemma 3.1, that

$$\begin{aligned} f=\sum _{k=0}^2 r_k\mathrm {e}^{\mathrm {i}\alpha _k}\cdot {\chi }_{T_k}\quad \text {a.e. on}\ \mathbb T. \end{aligned}$$

(3.18)

In particular, (3.18) together with (3.15), (2.25) and (3.5) yields

$$\begin{aligned} f=0\cdot {\chi }_{T_0}+e_2\cdot {\chi }_{T_1}+e_2\cdot {\chi }_{T_2}=e_2\cdot {\chi }_{T_1\cup T_2}=S_2\circ f_0\circ S_2^{-1}\quad \text {a.e. on}\ \mathbb T. \end{aligned}$$

From (3.18), (3.16), (2.25) and (3.5) we see that

$$\begin{aligned} f=0\cdot {\chi }_{T_1}+e_0\cdot {\chi }_{T_0}+e_0\cdot {\chi }_{T_2}=e_0\cdot {\chi }_{T_0\cup T_2}=S_0\circ f_0\circ S_0^{-1}\quad \text {a.e. on}\ \mathbb T. \end{aligned}$$

Finally, (3.18), (3.17), (2.25) and (3.5) lead to

$$\begin{aligned} f=0\cdot {\chi }_{T_2}+e_1\cdot {\chi }_{T_0}+e_1\cdot {\chi }_{T_1}=e_1\cdot {\chi }_{T_0\cup T_1}=S_1\circ f_0\circ S_1^{-1}\quad \text {a.e. on}\ \mathbb T. \end{aligned}$$

Therefore all these equalities yield the second part of the conjunction in (3.10), provided $z\ne 0$, and the first part of the alternative in (3.11), provided $z=0$. Assuming now that $r=|z|>0$, we conclude from (2.19) that

$$\begin{aligned} Q_r(\theta )>Q_r(0) ,\quad \theta \in \mathbb R\setminus \{2n\pi :n\in \mathbb Z\}. \end{aligned}$$

(3.19)

For each $k\in \{0,1,2\}$ we conclude from (3.14) that if $c_k=0$, then $p_k=Q_r(0)$. Therefore, from (3.19), (2.21) and (3.5) it follows that:

If $c_0=0$, then $\theta _r=4\pi /3$, and so $z=S_2(|z|)$.

If $c_1=0$, then $\theta _r=0$, and so $z=S_0(|z|)$.

If $c_2=0$, then $\theta _r=2\pi /3$, and so $z=S_1(|z|)$.

These equalities yield the first part of the conjunction in (3.10), provided $z\ne 0$.

It remains to examine the case where $0<c\le 1$. Then all $c_0$, $c_1$, $c_2$ are positive. Analyzing the part of the proof of Theorem 2.1 from (2.15) until (2.17) we get the following reasoning: From (3.13) we deduce that for a given $\alpha \in I_0$,

$$\begin{aligned} 2=c_0+c_1+c_2\le 1+2\cos \frac{\pi }{3}\cos \Bigl (\frac{\pi }{3}-\alpha \Bigr ) , \end{aligned}$$

whence $\alpha =\pi /3$. Then $c_0\le 1$, $c_1\le 1/2$ and $c_2\le 1/2$. Applying (3.13) once more we get $c_0=1$, $c_1=1/2$ and $c_2=1/2$. Since

$$\begin{aligned} c_k=r_k\cos \Bigl (\alpha _k-\frac{\pi }{3}\Bigr ) ,\quad 0\le r_k\le 1 \quad \text {and}\quad \alpha _k\in I_k ,\quad k\in \{0,1,2\} , \end{aligned}$$

we see that

$$\begin{aligned} r_0=1 ,\ \alpha _0=\frac{\pi }{3} ,\ r_1=1 ,\ \alpha _1=\frac{2\pi }{3} ,\ r_2=1 ,\ \alpha _2=2\pi . \end{aligned}$$

(3.20)

In the similar manner we show that:

If $\alpha \in I_1$, then $\alpha =\pi$, $c_0=1/2$, $c_1=1$, $c_2=1/2$, and consequently

$$\begin{aligned} r_0=1 ,\ \alpha _0=\frac{2\pi }{3} ,\ r_1=1 ,\ \alpha _1=\pi ,\ r_2=1 ,\ \alpha _2=\frac{4\pi }{3}. \end{aligned}$$

(3.21)

If $\alpha \in I_2$, then $\alpha =5\pi /3$, $c_0=1/2$, $c_1=1/2$, $c_2=1$, and consequently

$$\begin{aligned} r_0=1 ,\ \alpha _0=0 ,\ r_1=1 ,\ \alpha _1=\frac{4\pi }{3} ,\ r_2=1 ,\ \alpha _2=\frac{5\pi }{3}. \end{aligned}$$

(3.22)

In all the cases (3.20), (3.21) and (3.22), $r_k\mathrm {e}^{\mathrm {i}\alpha _k}\in T_k$ for $k\in \{0,1,2\}$. Then (2.11) implies, by Lemma 3.1, the equality (3.18). In particular, (3.18) together with (3.20), (3.12) and (3.5) yields

$$\begin{aligned} f=\mathrm {e}^{\pi \mathrm {i}/3}\cdot {\chi }_{T_0}+e_1\cdot {\chi }_{T_1}+e_0\cdot {\chi }_{T_2}=f_0^*=S_0\circ f_0^*\circ S_0^{-1} \quad \text {a.e. on}\ \mathbb T. \end{aligned}$$

From (3.18), (3.21), (3.12) and (3.5) we see that

$$\begin{aligned} f=e_1\cdot {\chi }_{T_0}+\mathrm {e}^{\pi \mathrm {i}}\cdot {\chi }_{T_1}+e_2\cdot {\chi }_{T_2}=S_1\circ f_0^*\circ S_1^{-1} \quad \text {a.e. on}\ \mathbb T. \end{aligned}$$

Finally, (3.18), (3.22), (3.12) and (3.5) lead to

$$\begin{aligned} f=e_0\cdot {\chi }_{T_0}+e_2\cdot {\chi }_{T_1}+\mathrm {e}^{5\pi \mathrm {i}/3}\cdot {\chi }_{T_2}=S_2\circ f_0^*\circ S_2^{-1} \quad \text {a.e. on}\ \mathbb T. \end{aligned}$$

Therefore all these equalities yield the second part of the alternative in (3.11). Moreover, as shown above, $c_i=c_j=1/2$ for some $i,j\in \{0,1,2\}$ such that $i\ne j$. Then by the condition (3.14) it follows that $p_i=p_j=Q_r(0)$. However, from (2.21) and (3.19) we see that this is impossible provided $|z|=r>0$. Therefore, $z=0$, which completes the proof. $\square$

Corollary 3.4

For any function $F\in \mathrm {Har}(\mathbb D)$ and $z\in \mathbb D$, if $F(\mathbb D)\subset \mathbb D$ and the condition (2.28) holds for every $k\in \{0,1,2\}$, then

$$\begin{aligned} |F(z)|=\frac{4}{3}-\frac{2}{\pi }\arctan \biggl (\frac{\sqrt{3}}{1+2|z|}\biggr ) \end{aligned}$$

(3.23)

iff there exists $k\in \{0,1,2\}$ such that one of the two following conditions holds:

• (i) $z\ne 0$ and

  $$\begin{aligned} z=S_k(|z|) \quad \text {and}\quad F(u)=S_k\circ F_0\circ S_k^{-1}(u) ,\quad u\in \mathbb D , \end{aligned}$$

  (3.24)

  where $F_0:=\mathrm {P}[f_0]$ with $f_0$ given by the formula (2.25);

• (ii) $z=0$ and

  $$\begin{aligned} F(u)=S_k\circ F_0\circ S_k^{-1}(u) \quad \text {or}\quad F(u)=S_k\circ F_0^*\circ S_k^{-1}(u) ,\quad u\in \mathbb D , \end{aligned}$$

  (3.25)

  where $F_0^*:=\mathrm {P}[f_0^*]$ with $f_0^*$ given by the formula (3.12).

Proof

From the definitions of the classes $\mathcal H$ and $\mathcal F$ it follows that $\mathcal H=\mathrm {P}[\mathcal F]$; cf. [3, Cor. 1 and Cor. 2, Sect. 1.2]. Thus the corollary follows directly from Theorem 3.3, the property (3.8) as well as the equalities $F_0=\mathrm {P}[f_0]$ and $F_0^*=\mathrm {P}[f_0^*]$. $\square$

Remark 3.5

The formula (3.1) gives an explicite form of the function $F_0$. From (2.30) we have

$$\begin{aligned} \mathrm {P}[{\chi }_{T_1}]=\mathrm {P}[1-{\chi }_{T_0\cup T_2}]=1-\mathrm {P}[{\chi }_{T_0\cup T_2}]=1-F_0\ . \end{aligned}$$

Hence and by (3.5) we see that

$$\begin{aligned} \mathrm {P}[{\chi }_{T_0}]&=\mathrm {P}[{\chi }_{T_1}\circ S_2^{-1}]=\mathrm {P}[{\chi }_{T_1}]\circ S_2^{-1}=1-F_0\circ S_1 , \\ \mathrm {P}[{\chi }_{T_2}]&=\mathrm {P}[{\chi }_{T_1}\circ S_1^{-1}]=\mathrm {P}[{\chi }_{T_1}]\circ S_1^{-1}=1-F_0\circ S_2. \end{aligned}$$

Using now the formula (3.12) we deduce that

$$\begin{aligned} F_0^*&=\mathrm {P}[f_0^*]=\mathrm {e}^{\pi \mathrm {i}/3}\cdot \mathrm {P}[{\chi }_{T_0}]+e_1\cdot \mathrm {P}[{\chi }_{T_1}]+e_0\cdot \mathrm {P}[{\chi }_{T_2}] \\&=\mathrm {e}^{\pi \mathrm {i}/3}\bigl ((1-F_0\circ S_1)+\mathrm {e}^{\pi \mathrm {i}/3}(1-F_0)+\mathrm {e}^{-\pi \mathrm {i}/3}(1-F_0\circ S_2)\bigr ) \\&=\mathrm {e}^{\pi \mathrm {i}/3}\bigl (2-F_0\circ S_1-\mathrm {e}^{\pi \mathrm {i}/3}F_0-\mathrm {e}^{-\pi \mathrm {i}/3}F_0\circ S_2)\bigr ). \end{aligned}$$

Thus we obtain the following explicite form of the function $F_0^*$,

$$\begin{aligned} F_0^*(z)=\mathrm {e}^{\pi \mathrm {i}/3}\Bigl (2-F_0(\mathrm {e}^{2\pi \mathrm {i}/3}z)-\mathrm {e}^{\pi \mathrm {i}/3}F_0(z)-\mathrm {e}^{-\pi \mathrm {i}/3}F_0(\mathrm {e}^{4\pi \mathrm {i}/3}z)\Bigr ),\quad z\in \mathbb D, \end{aligned}$$

which leads to the more symmetric one

$$\begin{aligned} \mathrm {e}^{-\pi \mathrm {i}/3}F_0^*(\mathrm {e}^{\pi \mathrm {i}/3}z)=2-F_0(-z)-\mathrm {e}^{\pi \mathrm {i}/3}F_0(\mathrm {e}^{\pi \mathrm {i}/3}z)-\mathrm {e}^{-\pi \mathrm {i}/3}F_0(\mathrm {e}^{-\pi \mathrm {i}/3}z),\quad z\in \mathbb D. \end{aligned}$$