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This article is cited in 1 scientific paper (total in 1 paper) Sharp univalence and sharp univalent covering domains for the class of holomorphic self-maps of a disc with two fixed boundary points O. S. Kudryavtsevaabc, A. P. Solodovab a Lomonosov Moscow State University, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia c Volgograd State Technical University, Volgograd, Russia
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    § 1. IntroductionIn this paper we consider the classical problems of univalence domains and univalent covering domains for classes of holomorphic self-maps of a disc. The key role in our studies belongs to fixed points, which are well known to affect the geometric and analytic properties of such functions. Let $\mathscr B$ be the class of holomorphic functions $f$ in the unit disc $ \mathbb D=\{{z\!\in\!\mathbb C}\colon {|z|\!<\!1}\}$ that take values in $\mathbb D$. By the Schwarz–Pick theorem (see [1]) a function $f\in \mathscr B$, $f(z)\not\equiv z$, can have at most one fixed point in the interior of ${\mathbb D}$. A boundary point $a\in\mathbb T$, $\mathbb T=\{z\in\mathbb C\colon |z|=1\}$, is called a fixed point of $f\in \mathscr B$ if $\angle\lim_{z\to a}f(z)=a$. At a boundary fixed point $a$ the angular limit $\angle \lim_{z\to a}(f(z)-a)/(z-a)$ always exists (see [2]), which is called the angular derivative of $ f$ at $a$ and denoted by $f'(a)$. The finite value of the angular derivative must be positive. The set of all fixed points of $f\in \mathscr B$ is never empty. If $f$ is not linear fractional, then by the Denjoy–Wolff theorem (see [3]–[6]) there exists, and moreover, a unique point $q\in\overline{\mathbb D}$ such that the sequence of iterates $f^n=f\circ f^{n-1}$, $n=2,3,\dots$, of $f=f^1$ converges to $q$ locally uniformly in $\mathbb D$. Furthermore, if $q\in \mathbb T$, then the angular limits $\angle\lim_{z\to q}f(z)=f(q) $ and $\angle \lim_{z\to q}f'(z)=f'(q)$ exist, and we have $f(q)=q$ and $0<f'(q)\leqslant 1$. The point $q$, which is called the Denjoy–Wolff point, is an attracting fixed point of $f$. The other fixed points, if exist, lie on the boundary of $\mathbb D$ and are repelling fixed points. It is well known that if the derivative of $f$ at an interior point is distinct from zero, then $f$ is univalent in a neighbourhood of this point. Moreover, in a neighbourhood of the image of this point a holomorphic function inverse to $f$ is defined. The main problem consists in determining the size of these neighbourhoods in its dependence on the function or the class of functions. A domain $U$ is called a univalence domain of $f$ if the images $f(z_1)$ and $f(z_2)$ of any distinct points $z_1$ and $z_2$ in it are distinct. A domain $V$ is called a univalent covering domain of $f$ if it is the univalent image of some domain $U$, that is, $U$ is a univalence domain of $f$ and $f(U)=V$. The problem of finding univalence and univalent covering domains is most often stated for classes of functions defined in terms of the values of the derivative at a fixed point (or points). The first result in this direction is due to Landau [7]. He proved that for a fixed modulus of the derivative at an interior fixed point we can find a common univalence disc for all functions satisfying this condition and also a common disc of univalent covering. For example, if the fixed point is the centre of $\mathbb D$ and $|f'(0)|=\beta$, $\beta\in(0,1)$, then $f$ is univalent in the disc $|z|<\beta/(1+\sqrt{1-\beta^2})$, and the inverse function of $f$ maps the disc $ |w|<\beta^2/(1+\sqrt{1-\beta^2})^2$ conformally onto a domain in $\mathbb D$. We turn to the case of a boundary fixed point $a\in \mathbb T$. Valiron [6] showed that the finite value of the angular derivative $f'(a)$ is a condition for local univalence, and then a univalence domain is a sector with vertex $a$,
$$
\begin{equation}
\{z\in \mathbb D\colon |{\arg(a-z)}|<\varphi,\,|z-a|<r\},
\end{equation}
\tag{1.1}
$$
where $\varphi\in(0, \pi/2)$ and $r>0$. Its opening angle can be arbitrarily close to $\pi$, and the radius depends on this angle and the function itself. An interesting refinement of this result is due to Becker and Pommerenke: depending on a function $f$ in the class $\mathscr B\{a\}=\bigl\{f\in \mathscr B\colon\angle\lim_{z\to a}f(z)=a\bigr\} $ which has a finite angular derivative $f'(a)$, they found its univalence domain, which for each $\varphi\in(0, \pi/2)$ contains a sector of the form (1.1). Theorem A (Becker and Pommerenke [8]). Let $f\in \mathscr B\{a\}$ and $f'(a)<\infty$. Then $f$ is univalent in the domain Note that by the Julia–Carathéodory theorem (see Theorem E) the left-hand side of (1.2) is at least $1$. Moreover, in the interior of each Stolz sector with vertex $a$ this quantity tends to $1$ as $z\to a$, so that it is less than 2 in the part of a sufficiently small neighbourhood lying in this sector. Thus, Valiron’s result can immediately be deduced from Theorem A, and moreover, this theorem characterizes the univalence sector numerically: for each $\varphi\in(0, \pi/2)$ we can take any sector lying in the domain (1.2) as a sector (1.1) in which $f$ is univalent. We go over to the question of univalence domains for classes of functions. Without loss of generality let $a=1$. Following Landau, for any $\alpha>1$ in the class $\mathscr B\{1\}$ we distinguish the subclass $\mathscr B_{\alpha}\{1\}$ of functions satisfying a constraint on the value of the angular derivative:
$$
\begin{equation*}
\mathscr B_{\alpha}\{1\}=\{f\in \mathscr B\{1\}\colon 1<f'(1) \leqslant \alpha\}.
\end{equation*}
\notag
$$
In [9] we presented an example of a family of Blaschke products in $\mathscr B_{\alpha}\{1\}$ with zero derivative at an arbitrary point in $\mathbb D$. Thus, in the case of a boundary fixed point a condition on the angular derivative is not sufficient for the existence of common univalence domains. It is obvious that to advance in the problem of univalence domains we must consider more narrow classes than $\mathscr B_{\alpha}\{1\}$. Since for $\alpha>1$ any function in $\mathscr B_{\alpha}\{1\}$, apart from the repelling fixed point $a=1$, must also have an attracting fixed point $q\in\overline{\mathbb D}$ (the Denjoy–Wolff point), it is natural to represent $\mathscr B_{\alpha}\{1\}$ as a union of disjoint subclasses described in terms of the position of the Denjoy–Wolff point (in the interior or on the boundary of $\mathbb D$). Let $\mathscr B[q]$, $q\in \overline{\mathbb D}$, denote the set of functions $f$ in $\mathscr B$ for which $q$ is the Denjoy–Wolff point. Then this representation of $\mathscr B_{\alpha}\{1\}$ looks as follows:
$$
\begin{equation*}
\mathscr B_{\alpha}\{1\}=\bigcup_{q\in\overline{\mathbb D}} \mathscr B_{\alpha}[q,1], \quad \text{where } \mathscr B_{\alpha}[q,1]=\mathscr B_{\alpha}\{1\}\cap \mathscr B[q].
\end{equation*}
\notag
$$
That nonempty univalence domains exist for such subclasses of $\mathscr B_{\alpha}\{1\}$ for some values of $\alpha$ was shown by Goryainov [10]. He observed that for $\alpha\in(1,2)$ $\mathscr B_{\alpha}[0,1]$ lies in the class of functions $f$ such that $f(0)=0$ and $|f'(0)|=(2-\alpha)/\alpha$, that is, in a class considered by Landau. This inclusion and Landau’s results ensure the existence of common univalence and univalent covering domains for each class $\mathscr B_{\alpha}[0, 1]$, $\alpha\in(1,2)$. As such domains we can take the discs $|z|<(1-\sqrt{\alpha-1})/(1+\sqrt{\alpha-1})$ and $|w|<(1-\sqrt{\alpha-1})^2/(1+\sqrt{\alpha-1})^2$, respectively. Of course, they are not optimal univalence or univalent covering domains. In [10] we found a univalent domain larger than this disc. As regards the case of a boundary Denjoy–Wolff point, the first result in this direction is also due to Goryainov. In [11] he found univalence domains for a wider class than $\mathscr B_{\alpha}[-1, 1]$, the class $\mathscr B_{\alpha}\{-1,1\}$ of functions satisfying a certain constraint on the value of the product of angular derivatives at the two boundary fixed points:
$$
\begin{equation*}
\mathscr B\{-1,1\}=\mathscr B\{-1\}\cap\mathscr B\{1\}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\mathscr B_\alpha\{-1,1\}=\{f\in \mathscr B\{-1,1\}\colon f'(1)f'(-1)\leqslant \alpha\}.
\end{equation*}
\notag
$$
Theorem B (Goryainov; see [11]). Let $f\in \mathscr B_{\alpha}\{-1,1\}$, where $\alpha\in(1,9)$. Then $f$ is univalent in the domain where $y^*(\alpha)$ is the unique root of the equation in the interval $0<y<\pi/2$. All these interesting results prompted investigations related to the search for maximal univalence and univalent covering domains for classes of functions with two fixed points, depending on the position of the Denjoy–Wolff point. In the case of an interior Denjoy–Wolff point the following results are valid. Theorem C ([12]). Let $\alpha\in (1,4)$ and $q\in \mathbb D$. If $f\in \mathscr B_{\alpha}[q, 1]$, then $f$ is univalent in the domain For any domain $\mathscr{V}$ such that $\mathscr D(\alpha,q)\subset\mathscr{V}\subset\mathbb D$ and $\mathscr V\neq \mathscr D(\alpha,q)$ there exists a function $f\in \mathscr B_{\alpha}[q, 1]$ that is not univalent in $\mathscr{V}$. Theorem D ([13]). Let $f\in \mathscr B_{\alpha}[q, 1]$, where $\alpha\in (1,2)$ and $q\in \mathbb D$. Then there exists a function inverse to $f$ and mapping the domain conformally onto a domain $\mathscr X\subset \mathbb D$. For any domain $\mathscr V$ such that $\mathscr Y(\alpha, q)\subset \mathscr V \subset \mathbb D$ and $\mathscr V\neq \mathscr Y(\alpha, q)$ there exists a function $f\in \mathscr B_{\alpha}[q, 1]$ that has no inverse function in $\mathscr V$. Finally, for $\alpha\geqslant 2$ there are no nonempty univalent covering domains for the class $\mathscr B_{\alpha}[q, 1]$. The aim of this paper is to find maximal univalence and univalent covering domains in the case of a boundary Denjoy–Wolff point. In combination with the results already known, this will allow us to answer all fundamental questions on univalence and univalent covering domains for classes of holomorphic self-maps of a disc with repelling boundary fixed point and a prescribed position of an attracting fixed point. To meet this objective we follow a scheme based on the (conceptually close) approaches of Landau and Becker–Pommerenke, which was successfully implemented in the proofs of Theorems C and D. The first step (see § 4) is to find a sharp inequality in the case when some value of a function in the class under consideration has at least two preimages. At the second step (see § 5) we look for a maximal domain such that for points in it two distinct preimages and their common value cannot satisfy the inequality derived at the first step. The domain occurring in this way is a sharp univalence domain for the class under consideration. The third step (see § 6) consists in looking for a sharp univalent covering domain; a key to finding it is the explicit form of the sharp univalence domain and a certain inequality relating the behaviour of a function in the interior of $\mathbb D$ to information about its fixed points (in our case this is an analogue of the Julia–Carathéodory theorem for several fixed boundary points; it is established in § 2). In § 3 we present auxiliary statements required to show that both the univalence domain and univalent covering domain are sharp. In § 7 we extend the results from § 5 and § 6 to the case of an arbitrary position of the boundary Denjoy–Wolff point. The results in § 5 and § 6 were announced in [14] and [15], respectively. § 2. An analogue of the Julia–Carathéodory theorem for several boundary fixed pointsThe classical Julia–Carathéodory theorem, relating the behaviour of a function inside the disc with the value of the angular derivative at a boundary fixed point is important for the geometric theory of functions of a complex variable. In this section we generalize this fundamental result to the case of several boundary fixed points and deduce some consequences used in our paper. Theorem E (Julia–Carathéodory). Let $f\in\mathscr B\{a\}$. Then the following inequality holds for each $z\in\mathbb D$: Inequality (2.1) is sharp and equality is attained at a linear fractional transformations of $\mathbb D$. In the case of a repelling boundary fixed point Theorem E can be formulated as follows. Lemma 1. Let $f\in \mathscr B\{a\}$, $f'(a)>1$, be distinct from a linear fractional transformation of $\mathbb D$. Then the function belongs to the class $\mathscr B$. Proof. Let $g$ be the function defined by (2.2). Consider the composition $h\circ g(z)$, where $h(z)=(a+z)/(a-z)$ is a linear fractional map of $\mathbb D$ onto the right half-plane $\mathbb H=\{\zeta\in\mathbb C\colon \operatorname{Re} \zeta >0\}$. We estimate the real part of this map:
$$
\begin{equation*}
\begin{aligned} \, \operatorname{Re} (h\circ g(z)) &=\operatorname{Re} \frac{a+g(z)}{a-g(z)}=\frac{1}{f'(a)-1}\operatorname{Re} \biggl(f'(a)\frac{a+f(z)}{a-f(z)}-\frac{a+z}{a-z}\biggr) \\ &=\frac 1{f'(a)-1}\biggl(f'(a)\frac{1-|f(z)|^2}{|a-f(z)|^2}-\frac{1-|z|^2}{|a-z|^2}\biggr). \end{aligned}
\end{equation*}
\notag
$$
It follows from (2.1) and the condition $f'(a)>1$ that $\operatorname{Re} (h\circ g(z))>0$ for all ${z\,{\in}\,\mathbb D}$. Thus, $g$ is a function in $\mathscr B$. The proof is complete. Now we generalize the Julia–Carathéodory theorem to the case when the function $f$ has several boundary fixed points. Let $\mathscr B\{a_1,\dots, a_k\}$ denote the set of functions $f$ in $\mathscr B$ that fix the distinct boundary points $a_j\in \mathbb T$, $j=1,\dots, k$:
$$
\begin{equation*}
\mathscr B\{a_1,\dots, a_k\}=\Bigl\{f\in \mathscr B\colon \angle\lim_{z\to a_j}f(z)=a_j\Bigr\}.
\end{equation*}
\notag
$$
Lemma 2. Let $f\in \mathscr B\{a_1,\dots, a_k\}$ be distinct from a linear fractional transformation of $\mathbb D$, and let $f'(a_k)>1$. Then the function belongs to the class $\mathscr B\{a_1,\dots, a_{k-1}\}$ and Proof. That $g$ (see (2.3)) belongs to $\mathscr B$ follows from Lemma 1. A direct verification shows that $g$ fixes the points $z=a_j$, $j=1,\dots, k-1$. In addition, we have formulae (2.4) for the angular derivatives of $g$ at boundary fixed points.
The proof is complete. Using Lemma 2 we can solve a number of extremal problems on classes of holomorphic self-maps of a disc with several boundary fixed points. In particular, it underlies the search for sharp domains of mutual variation of Taylor coefficients for these classes of functions (see [16]). In this paper we use the construction (2.3) to find sharp univalence and univalent covering domains. Namely, we require some corollaries to Lemma 2. Proof. If $f$ is a linear fractional self-map of the disc $\mathbb D$, that is, $ f(z)\!=\!{(z\!-\!s)}/{(1\!-\!s z)}$, where $s\in(-1,1)$, then it is easy to check that $f'(-1)f'(1)=1$. If $f$ is not linear fractional, then one of the points $z=-1$ and $z=1$ is repelling. For definiteness let $z=1$ be a repelling point. Then by Lemma 2 the function is in the class $\mathscr B\{-1\}$ and Since $g'(-1)>0$ and $f'(1)>1$, from this we obtain $f'(-1)f'(1)>1$.
The corollary is proved. Note that the above estimate is well known in the literature and can be proved in various ways. Corollary 2. Let $f\!\in\! \mathscr B_\alpha[-1,1]$ be different from a linear fractional self-map of $\mathbb D$. Then the function is in the class $\mathscr B[-1]$. Proof. By Lemma 2 the function $g$ belongs to $\mathscr B\{-1\}$, and we have Hence from the inequality $f'(-1)\leqslant 1$ we deduce that $g'(-1)\leqslant 1$, that is, $z=-1$ is the Denjoy–Wolff point of $g$, and therefore $ g\in \mathscr B[-1]$.
The proof is complete. Corollary 3. Let $f\!\in\!\mathscr B_\alpha[-1,1]$. Then the following inequality holds for each ${z\!\in\!\mathbb D}$: Proof. If $f$ is a linear fractional transformation of $\mathbb D$, that is, $f(z)={(z\!-\!s)}/{(1\!-\!s z)}$, where $s=(\alpha-1)/(\alpha+1)$, then it is easy to verify that we have equality in (2.6). If $f$ is not linear fractional, then by Corollary 2 and Theorem E the function $g$ defined by (2.5) satisfies
$$
\begin{equation*}
\frac{|1+g(z)|^2}{1-|g(z)|^2}\leqslant \frac{|1+z|^2}{1-|z|^2}.
\end{equation*}
\notag
$$
Taking (2.5) into account this inequality can be transformed into
By Theorem E the second factor on the left in (2.7) is positive, which proves (2.6). Note that in the case of a single fixed point we can only say that the right-hand side of (2.6) is nonnegative (see Theorem E). Thus, (2.6) is a refinement of (2.1) in the case of two boundary fixed points, one of which is attracting. § 3. Auxiliary statementsFor an upper estimate of univalence and univalent covering domains we need the following auxiliary statements. Lemma 3. Let $\alpha$ and $\beta$ be positive numbers, and let $z_0\in \mathbb D$. If a function ${f\in\mathscr B\{-1,1\}}$ satisfies $f'(-1)=\beta$, $f'(1)=\alpha$ and $f'(z_0)=0$, then for any point $\zeta\in \mathbb D$ such that there exists a function $ g\in\mathscr B\{-1,1\}$ satisfying $g'(-1)=\beta$, $g'(1)=\alpha$ and $g'(\zeta)=0$. If, in addition to the above, the function $f$ has no other fixed points, then $g$ can also be chosen to have no other fixed points. Proof. Assume that we have found a required function $f$ for $\alpha$, $\beta$ and $z_0$ as in the hypotheses. For each $s\in(-1,1)$ consider the linear fractional map which satisfies $T_s(\pm 1)=\pm 1$. Since $f\in\mathscr B\{-1,1\}$, the function $g_s=T^{-1}_s\circ f\circ T_s$ also belongs to $\mathscr B\{-1,1\}$. It is easy to check that $g'_s(-1)=\beta$ and $g'_s(1)=\alpha$. In addition, $g'_s(\zeta(s))=0$, where $\zeta(s)=T^{-1}_s(z_0)$.
It is easy to see that each point on the curve $\zeta(s)$, $s\in(-1,1)$, satisfies (3.1). Since $\zeta(s)$ is continuous, $\zeta(1-0)=1$ and $\zeta(-1+0)=-1$, this curve is one of the two mirror symmetric (relative to the real diameter) arcs of circles forming the locus of points satisfying (3.1). The second arc $\overline{\zeta(s)}$, $s\in(-1,1)$, consists of the zeros of the derivatives of the functions $h_s(z)=\overline{g_s(\overline{z}})$, each of which belongs to $\mathscr B\{-1,1\}$ and satisfies $h'_s(-1)=\beta$ and $h'_s(1)=\alpha$. The second part of the lemma follows directly from the following fact: $a\in \overline{\mathbb D}$ is a fixed point of $g_s$ (of $h_s$) if and only if the point $T_s(a)$ (the point $\overline{T_s(a)}$) is a fixed point of $f$. The lemma is proved. Lemma 4. Let $\alpha$ and $\beta$ be positive numbers, and let $w_0\in \mathbb D$. If a second-order Blaschke product $f\in\mathscr B\{-1,1\}$ satisfies $f'(-1)=\beta$ and $f'(1)=\alpha$, and if $w_0$ is a branch point of the inverse function of $f$, then for any point $w\in \mathbb D$ satisfying there exists a second-order Blaschke product $g\in\mathscr B\{-1,1\}$ such that $g'(-1)=\beta$, $g'(1)=\alpha$ and $w$ is a branch point of the inverse function of $g$. Proof. Assume that we have found a required function $f$ for $\alpha$, $\beta$ and $w_0$ under consideration. Since $f\in\mathscr B\{-1,1\}$, the composition $g_s=T^{-1}_s\circ f\circ T_s$, where $T_s$ is given by (3.2), belongs to the class $\mathscr B\{-1,1\}$, and we have $g'_s(-1)=\beta$ and $g'_s(1)=\alpha$. In addition, $w(s)=T_s^{-1}(w_0)$ is a branch point of $g^{-1}_s$.
It is easy to verify that each point on the curve $w(s)$, $s\in(-1,1)$, satisfies (3.3). Since $w(s)$ is continuous, $w(1-0)=1$ and $w(-1+0)=-1$, this curve is one of the two mirror symmetric (relative to the real diameter) arcs of circles forming the locus of points satisfying condition (3.3). The second arc $\overline{w(s)}$, $s\in(-1,1)$, consists of the branch points of the functions $h_s(z)=\overline{g_s(\overline{z}})$, all of which belong to $\mathscr B\{-1,1\}$ and satisfy $h'_s(-1)=\beta$ and $h'_s(1)=\alpha$. The lemma is proved. § 4. Improving the Becker–Pommerenke inequalityAs mentioned in the introduction, the first step in the solution of the problems under consideration consists in the derivation of a sharp inequality estimating the common value of a function at two distinct points. In particular, to prove Theorem A, Becker and Pommerenke established the following inequality. Lemma A (Becker and Pommerenke [8]). Let $f\in \mathscr B\{1\}$, and let $a, b\in \mathbb D$, $a\neq b$, be points such that $f(a)=f(b)=c$. Then We will prove a similar inequality for functions in $\mathscr B[-1,1]$. Applying Lemma A to a function $f\in\mathscr B[-1,1]$ that takes two distinct points $a$ and $b$ to a point $c$, we obtain that the quantities
$$
\begin{equation*}
f'(1)\frac{1-|c|^2}{|1-c|^2}-\frac{1-|a|^2}{|1-a|^2}-\frac{1-|b|^2}{|1-b|^2} \qquad \text{and}\qquad \frac{1-|c|^2}{|1+c|^2}-\frac{1-|a|^2}{|1+a|^2}-\frac{1-|b|^2}{|1+b|^2}
\end{equation*}
\notag
$$
are nonnegative. In fact, a stronger result holds. Lemma 5. If $f\in\mathscr B[-1,1]$ and $a,b\in \mathbb D$, $a\neq b$, are points such that $f(a)\,{=}\,f(b)\,{=}\,c$, then To prove Lemma 5 (and also the results in § 7) we use the linear fractional transformation
$$
\begin{equation}
S_q(z)=\frac{(3q-1)z-(1+q)}{(1+q)z-(3-q)}, \qquad q\in \mathbb T.
\end{equation}
\tag{4.2}
$$
Since $|q|=1$, $\operatorname{Re} q<1$ and $|(1+q)/(3q-1)|=(1+\operatorname{Re} q)/(5-3\operatorname{Re} q)<1$, the function $S_q$ maps the unit disc $\mathbb D$ onto itself and satisfies $S_q(1)=1$ and $S_q(-1)=q$. Proof of Lemma 5. Consider the function
$$
\begin{equation*}
g(z)=\frac{1-\overline{c}}{1-c}\, \frac{f(z)-c}{1-\overline{c}f(z)}.
\end{equation*}
\notag
$$
It is easy to see that $g$ belongs to $\mathscr B\{1\}$ and $g(a)=g(b)=0$. By the Schwarz–Pick lemma $g$ has the representation
$$
\begin{equation*}
g(z)=\frac{1-\overline{a}}{1-a}\,\frac{z-a}{1-\overline{a}z}\,\frac{1-\overline{b}}{1-b}\, \frac{z-b}{1-\overline{b}z} h(z),
\end{equation*}
\notag
$$
where either $h\in \mathscr B\{1\}$ or $h(z)\equiv 1$. If $h(z)\equiv 1$, then it is straightforward to see that we have equality in (4.1).
In the nontrivial case when $h\in \mathscr B\{1\}$, for
$$
\begin{equation*}
h(z)=\frac{1-a}{1-\overline{a}}\,\frac{1-\overline{a}z}{z-a}\, \frac{1-b}{1-\overline{b}}\,\frac{1-\overline{b}z}{z-b} \, \frac{1-\overline{c}}{1-c}\,\frac{f(z)-c}{1-\overline{c}f(z)}
\end{equation*}
\notag
$$
we have the following relations:
$$
\begin{equation*}
\begin{gathered} \, h(-1)=-\varkappa, \\ h'(-1)=\varkappa \biggl(\frac{1-|c|^2}{|1+c|^2}-\frac{1-|a|^2}{|1+a|^2}-\frac{1-|b|^2}{|1+b|^2}\biggr) \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
h'(1)=f'(1)\frac{1-|c|^2}{|1-c|^2}-\frac{1-|a|^2}{|1-a|^2}-\frac{1-|b|^2}{|1-b|^2},
\end{equation*}
\notag
$$
where $\varkappa$ denotes the quantity
$$
\begin{equation*}
\varkappa=\frac{1-\overline{c}}{1-c}\,\frac{1+c}{1+\overline{c}}\, \frac{1-a}{1-\overline{a}}\,\frac{1+\overline{a}}{1+a}\, \frac{1-b}{1-\overline{b}}\,\frac{1+\overline{b}}{1+b}.
\end{equation*}
\notag
$$
Consider the linear fractional transformation $S_{-\overline{\varkappa}}$ (see (4.2)). Then the composition $u(z)=S_{-\overline{\varkappa}}\circ h(z)$ belongs to $\mathscr B$ and fixes the points $z=-1$ and $z=1$. Applying Corollary 1 to $u\in\mathscr B\{-1,1\}$ we obtain $u'(-1)u'(1)\geqslant 1$. Since
$$
\begin{equation*}
u'(1)=f'(1)\frac{1-|c|^2}{|1-c|^2}-\frac{1-|a|^2}{|1-a|^2}-\frac{1-|b|^2}{|1-b|^2}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
u'(-1)=\frac{4}{|1+\varkappa|^2} \biggl(\frac{1-|c|^2}{|1+c|^2}-\frac{1-|a|^2}{|1+a|^2}-\frac{1-|b|^2}{|1+b|^2}\biggr),
\end{equation*}
\notag
$$
this inequality is equivalent to (4.1).
The lemma is proved. § 5. Univalence domainIn this section we prove our first main result, a theorem on a sharp univalence domain for the class $\mathscr B_{\alpha}[-1,1]$. Theorem 1. Let $\alpha\in (1,4)$. If $f\in \mathscr B_{\alpha}[-1, 1]$, then $f$ is univalent in the domain For any domain $\mathscr{V}$ such that $\mathscr U(\alpha) \subset \mathscr{V}\subset\mathbb D$ and $\mathscr{V}\neq \mathscr U(\alpha)$ there exists a function $f\in \mathscr B_{\alpha}[-1, 1]$ that is not univalent in $\mathscr{V}$. Following the scheme from the introduction, before we proceed to the proof of Theorem 1, we prove an inequality opposite to (4.1) in a certain sense. Lemma 6. Let $\alpha\in (1,4)$. Then for any points $a,b\in \mathscr U(\alpha)$ and $c\in \mathbb D$ such that Proof. It is easy to see that the boundary of $\mathscr U(\alpha)$ consists of arcs of circles. The linear fractional transformation of the right half-plane $\mathbb H$ onto $\mathbb D$ takes rays in $\mathbb H$ to them. Hence for some positive $r$ and $s$ and some $\varphi$ and $\psi$ in the interval $\bigl(-\arctan({1}/{\sqrt{\alpha-1}}),\arctan({1}/{\sqrt{\alpha-1}})\bigr)$ we have
$$
\begin{equation}
a=\frac{re^{i\varphi}-1}{re^{i\varphi}+1}\quad\text{and} \quad b=\frac{se^{i\psi}-1}{se^{i\psi}+1}.
\end{equation}
\tag{5.5}
$$
Of course, any point in $\mathbb D$ is also an image of a point in $\mathbb H$ under the action of $L$. Hence for some $r>0$ and $\theta \in (-{\pi}/{2},{\pi}/{2})$ we have In view of (5.5) and (5.6) we have the equalities
$$
\begin{equation}
\begin{split} &\biggl|1+\frac{1-\overline{c}}{1-c} \, \frac{1+c}{1+\overline{c}} \, \frac{1-a}{1-\overline{a}} \, \frac{1+\overline{a}}{1+a} \, \frac{1-b}{1-\overline{b}} \, \frac{1+\overline{b}}{1+b} \biggr|^2 \\ &\qquad=4(1+\tan \varphi\tan\theta+\tan\psi\tan\theta-\tan\varphi\tan\psi)^2 \cos^2\varphi\cos^2\psi\cos^2\theta \\ &\qquad=4\cos^2(\varphi+\psi-\theta), \end{split}
\end{equation}
\tag{5.7}
$$
$$
\begin{equation}
\alpha\frac{1-|c|^2}{|1-c|^2}-\frac{1-|a|^2}{|1-a|^2}-\frac{1-|b|^2}{|1-b|^2}=\alpha t\cos \theta-r\cos\varphi-s\cos\psi,
\end{equation}
\tag{5.8}
$$
$$
\begin{equation}
\frac{1-|c|^2}{|1+c|^2}-\frac{1-|a|^2}{|1+a|^2}-\frac{1-|b|^2}{|1+b|^2} =\frac{\cos\theta}{t}-\frac{\cos\varphi}{r}-\frac{\cos\psi}{s}.
\end{equation}
\tag{5.9}
$$
Taking (5.7)–(5.9) into account we can write (5.3) as
$$
\begin{equation}
(\alpha t\cos \theta-r\cos\varphi-s\cos\psi) \biggl(\frac{\cos\theta}{t}-\frac{\cos\varphi}{r}-\frac{\cos\psi}{s}\biggr) <\cos^2(\varphi+\psi-\theta).
\end{equation}
\tag{5.10}
$$
Let $p$ denote the first factor on the left. Note that by the assumptions of the lemma (see (5.2) and (5.8)) $p$ is nonnegative. In the new notation (5.10) is equivalent to the inequality
$$
\begin{equation*}
p\biggl(\frac{\alpha \cos^2{\theta}}{p+r\cos{\varphi}+s\cos\psi} -\frac{\cos{\varphi}}{r}-\frac{\cos{\psi}}{s}\biggr) <\cos^2(\varphi+\psi-\theta),
\end{equation*}
\notag
$$
which takes the following form after some simple transformations:
$$
\begin{equation}
\begin{aligned} \, \notag &\frac12 \, rs\bigl(\alpha p - (p+r \cos \varphi+s\cos\psi)\cos 2(\varphi+\psi)\bigr)\cos 2\theta \\ \notag &\qquad\qquad -\frac12\, rs (p+r \cos \varphi+s\cos\psi)\sin 2(\varphi+\psi) \sin 2\theta \\ &\qquad <-\frac12\, \alpha p r s +(p+r \cos \varphi+s\cos\psi) \biggl(\frac12\, r s +p r \cos\psi+p s \cos \varphi\biggr). \end{aligned}
\end{equation}
\tag{5.11}
$$
To prove (5.11) it is sufficient to verify that the least upper bound of (5.11), taken over all $\theta\in(-\pi/2,\pi/2)$, is less than the right-hand side of (5.11). Namely, it remains to verify the following inequality:
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl(\frac14 \, r^2 s^2\bigl(\alpha p - (p+r \cos \varphi+s\cos\psi)\cos 2(\varphi+\psi)\bigr)^2 \\ \notag &\qquad\qquad +\frac14\, r^2s^2 (p+r \cos \varphi+s\cos\psi)^2\sin^2 2(\varphi+\psi)\biggr)^{1/2} \\ &\qquad <-\frac12\, \alpha p r s +(p+r \cos \varphi+s\cos\psi) \biggl(\frac12\, r s +p r \cos\psi+p s \cos \varphi\biggr). \end{aligned}
\end{equation}
\tag{5.12}
$$
Note that the right-hand side of (5.12) is positive. In fact, as $\cos^2\varphi>(\alpha-1)/\alpha$ and $\cos^2\psi>(\alpha-1)/\alpha$, it follows that
$$
\begin{equation*}
\begin{aligned} \, &-\frac12 \, \alpha p r s +(p+r \cos \varphi+s\cos\psi) \biggl(\frac12\, r s +p r \cos\psi+p s \cos \varphi\biggr) \\ &\qquad >-\frac12\, \alpha p r s +\biggl(p+(r+s)\sqrt{\frac{\alpha-1}{\alpha}}\biggr) \biggl(\frac12 \, r s +p(r+ s) \sqrt{\frac{\alpha-1}{\alpha}}\biggr) \\ &\qquad\geqslant p rs \frac{(\alpha-1)(8-\alpha)}{\alpha}>0. \end{aligned}
\end{equation*}
\notag
$$
Squaring both sides of (5.12) and making some transformations we reduce this inequality to the following form:
$$
\begin{equation*}
\begin{aligned} \, &p (p+r \cos\varphi+s \cos\psi) \bigl(\alpha r^2 s^2 \sin^2(\varphi+\psi)+\alpha p r^2 s \cos\psi+\alpha p r s^2\cos\varphi \\ &\qquad- (p+r \cos\varphi+s \cos\psi)(r\cos\psi+s\cos\varphi) (rs+pr\cos\psi+ps\cos\varphi)\bigr) <0. \end{aligned}
\end{equation*}
\notag
$$
Clearly, it suffices to check that the third factor is negative. Removing brackets and grouping like terms we represent it as
$$
\begin{equation}
\begin{aligned} \, \notag &r^2s^2(\alpha \sin^2(\varphi+\psi)-\cos^2\varphi-\cos^2\psi)+p r^2 s((\alpha-1) \cos\psi-2\cos^2\varphi\cos\psi) \\ \notag &\qquad+prs^2((\alpha-1) \cos\varphi-2\cos\varphi\cos^2\psi) -p r^2 s\cos^3\psi-prs^2\cos^3\varphi -p^2 r^2 \cos^2\psi \\ \notag &\qquad-p^2 s^2 \cos^2\varphi -2p^2 rs\cos\varphi\cos\psi-r^3s\cos\varphi\cos\psi-rs^3\cos\varphi\cos\psi \\ &\qquad-p r^3\cos\varphi\cos^2\psi-ps^3\cos^2\varphi\cos\psi. \end{aligned}
\end{equation}
\tag{5.13}
$$
We use the estimates
$$
\begin{equation*}
\begin{gathered} \, p^2r^2\cos^2\psi+p^2s^2\cos^2\varphi \geqslant 2 p^2rs \cos\varphi\cos\psi, \\ r^3s\cos\varphi\cos\psi+rs^3\cos\varphi\cos\psi\geqslant 2 r^2 s^2 \cos\varphi\cos\psi \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, &pr^2s\cos^3\psi+p r s^2\cos^3\varphi+p r^3\cos\varphi\cos^2\psi+ps^3\cos^2\varphi\cos\psi \\ &\qquad =pr^2(r\cos\varphi+s\cos\psi)\cos^2\psi+ps^2(r\cos\varphi+s\cos\psi)\cos^2\varphi \\ &\qquad \geqslant2p r s (r\cos\varphi+s\cos\psi)\cos\varphi\cos\psi, \end{aligned}
\end{equation*}
\notag
$$
which show that the expression in (5.13) is at most
$$
\begin{equation*}
\begin{aligned} \, &-4 p^2rs \cos\varphi\cos\psi+ r^2s^2(\alpha \sin^2(\varphi+\psi)-\cos^2\varphi-\cos^2\psi-2\cos\varphi\cos\psi) \\ &\qquad+p r^2 s((\alpha-1) \cos\psi-4\cos^2\varphi\cos\psi) +prs^2((\alpha-1) \cos\varphi-4\cos\varphi\cos^2\psi), \end{aligned}
\end{equation*}
\notag
$$
which is a negative quantity because the coefficients of $p^2rs$, $r^2s^2$, $pr^2s$ and $prs^2$ are negative. It is sufficient to consider the coefficients of $r^2s^2$ and $pr^2s$ because the latter is similar to the coefficient of $prs^2$. Taking the ranges of $\varphi$, $\psi$ and $\alpha$ into account we obtain
$$
\begin{equation*}
\begin{aligned} \, &\alpha \sin^2(\varphi+\psi)-\cos^2\varphi-\cos^2\psi-2\cos\varphi\cos\psi \\ &\qquad =2\cos^2\frac{\varphi+\psi}{2}\bigl((\alpha-1)(1+\sin\varphi\sin\psi) -(\alpha+1)\cos\varphi\cos\psi\bigr) \\ &\qquad <2\cos^2\frac{\varphi+\psi}{2}\biggl((\alpha-1)\biggl(1+\frac{1}{\alpha}\biggr) -(\alpha+1)\biggl(1-\frac{1}{\alpha}\biggr) \biggr)=0, \\ &(\alpha-1) \cos\psi-4\cos^2\varphi\cos\psi=\cos\psi( (\alpha-1)-4\cos^2\varphi) \\ &\qquad <\cos\psi\biggl((\alpha-1)-\frac{4(\alpha-1)}{\alpha}\biggr) =\cos\psi\frac{(\alpha-1)(\alpha-4)}{\alpha}< 0. \end{aligned}
\end{equation*}
\notag
$$
Thus, the quantity (5.13) is negative, and therefore (5.3) holds.
Lemma 6 is proved. Proof of Theorem 1. Fix $\alpha\in (1,4)$ and assume that a function $f$ in the class $\mathscr B_{\alpha}[-1, 1]$ is not univalent in $\mathscr U(\alpha)$, that is, there exist distinct points $a, b\in \mathscr U(\alpha)$ such that $f(a)=f(b)=c$. By Lemma A we have inequalities (5.2) for $\alpha=f'(1)$, $a$, $b$ and $c$, and by Lemma 5 we have (4.1). On the other hand, by Lemma 6 we have inequality (5.3), which is opposite to (4.1). This contradiction proves the first part of the theorem.
The second part was proved in our joint paper with Goryainov [17]. Nevertheless, we present the proof for completeness. To verify the second part of the assertion of the theorem, it is sufficient to find, for each boundary point of $\mathscr U(\alpha)$, a function in the class $\mathscr B_{\alpha}[-1,1]$ with derivative vanishing at this point. Consider the function
$$
\begin{equation}
h(\zeta)=\frac{1}{\alpha}\biggl(\zeta+i(\alpha-1)+\frac{\alpha-1}{\zeta+i}\biggr).
\end{equation}
\tag{5.14}
$$
It takes the right half-plane $\mathbb H$ to itself with precisely two fixed points, $\zeta=0$ and $\zeta=\infty$. In addition, the derivative of $h$ has a zero at $\zeta_0=\sqrt{\alpha-1}-i$.
Now consider the composition $f=L\circ h\circ L^{-1}$, where $L$ is defined by (5.4). The function $f$ takes $\mathbb D$ to itself with precisely two fixed points, $z= -1$ and ${z=1}$, one of which is the Denjoy–Wolff point by the Denjoy–Wolff theorem. Easy calculations show that $f$ has the following angular derivatives at the fixed points: ${f'(-1)\!=\!1}$ and $f'(1)=\alpha$. Thus, $z=-1$ is the Denjoy–Wolff point and ${f\in\mathscr B_{\alpha}[-1,1]}$. The derivative of $f$ has a zero at
$$
\begin{equation}
z_0=L(\zeta_0)=\frac{\sqrt{\alpha-1}-1-i}{\sqrt{\alpha-1}+1-i},
\end{equation}
\tag{5.15}
$$
and we have $|1-z_0^2|/(1-|z_0|^2)=\sqrt{\alpha/(\alpha-1)}$.
Thus, $f\!\in\!\mathscr B_{\alpha}[-1,1]$ and its derivative has a zero on the boundary of $\mathscr U(\alpha)$ (at $z_0$). Then by Lemma 3, for an arbitrary point $z\in\mathbb D$ on the boundary of $\mathscr U(\alpha)$ there exists a function $g\in\mathscr B_{\alpha}[-1,1]$ with derivative vanishing at this point. Let $\mathscr V$ be a domain such that $\mathscr U(\alpha)\subset \mathscr V\subset \mathbb D$ and $\mathscr V\neq \mathscr U(\alpha)$. Then some boundary point of $\mathscr U(\alpha)$ lies in the interior of $\mathscr V$. As shown above, there exists a function in the class $\mathscr B_{\alpha}[-1,1]$ whose derivative has a zero at this point. Hence this function is not univalent in $\mathscr V$. The proof of Theorem 1 is complete. In Figure 1 we show the structure and mutual position of the univalence domains from Theorems B and 1. Remark 1. These authors are sure that both the domains $\mathscr D(\alpha, q)$ (see theorem C) and $\mathscr U(\alpha)$ (see (5.1)) are maximal univalence domains for the corresponding classes not only for $\alpha\in (1,4)$, but also for $\alpha\geqslant 4$. However, the proof of this encounters some additional technical problems and has not been obtained so far. We can explain the core difficulties in the case of the class $\mathscr B_{\alpha}[-1,1]$. As already mentioned, the proof of Theorem 1 is based on inequality (4.1), which holds for all ${\alpha=f'(1)>1}$. On the other hand the ‘opposite’ inequality (5.3) only holds for $\alpha\in (1,4)$. Thus, to prove the result of Theorem 1 for $\alpha\geqslant 4$ we must supplement estimate (4.1) with another inequality, which will significantly complicate the technical complexity. In the conclusion of the section we indicate the disc of maximum radius in which all functions in $\mathscr B_{\alpha}[-1,1]$ are univalent. Corollary 4. Let $f\in \mathscr B_{\alpha}[-1, 1]$, where $\alpha\in(1,4)$. Then $f$ is univalent in the disc For any disc $\mathscr O'\subset \mathbb D$ of radius greater than that of $\mathscr O(\alpha)$ there exists a function $f\in \mathscr B_{\alpha}[-1, 1]$ that is not univalent in $\mathscr O'$. § 6. Domain of univalent coveringIn this section we treat our second main result in this paper, a theorem on a sharp univalent covering domain for the class $\mathscr B_{\alpha}[-1,1]$. Note again that, as we will see from the proof, our key to finding a sharp domain of univalent covering is the sharp univalence domain found in § 5. Theorem 2. Let $f\in\mathscr B_{\alpha}[-1,1]$, $\alpha\in (1,2)$. Then there exists a function inverse to $f$ and mapping the domain conformally onto a domain $\mathscr X\subset \mathbb D$. For any domain $\mathscr{V}$ such that $\mathscr W(\alpha) \subset \mathscr{V}\subset\mathbb D$ and $\mathscr{V}\neq \mathscr W(\alpha)$ there exists a function $f\in \mathscr B_{\alpha}[-1, 1]$ that has no inverse function in $\mathscr{V}$. Finally, for $\alpha\geqslant 2$ there exist no nonempty domains of univalent covering for the class $\mathscr B_{\alpha}[-1,1]$. Proof. First we estimate a domain of univalent covering from above, thus proving the second part of the theorem. As in the proof of Theorem 1, consider the composition $f=L\circ h\circ L^{-1}$, where the functions $h$ and $L$ are defined by (5.14) and (5.4). We showed above that $f\in\mathscr B_{\alpha}[-1,1]$ and the derivative of $f$ has a zero at the point $z_0$ defined by (5.15). This means that the inverse function of $f$ branches at the point and we have $|1-w_0^2|/(1-|w_0|^2)=\alpha/(2\sqrt{\alpha-1})$.
Thus, $f\in\mathscr B_{\alpha}[-1,1]$ and the inverse function of $f$ has a branching point on the boundary of $\mathscr W(\alpha)$ (namely, the point $w_0$). By Lemma 3 the existence of such a function means that for an arbitrary point $w\in\mathbb D$ on the boundary of $\mathscr W(\alpha)$ there exists a function $g\in\mathscr B_{\alpha}[-1,1]$ whose inverse function branches at $w$. Let $\mathscr V$ be an arbitrary domain such that $\mathscr W(\alpha)\subset \mathscr V\subset \mathbb D$ and $\mathscr V\neq \mathscr W(\alpha)$. Then some boundary point of $\mathscr W(\alpha)$ lies in $\mathscr V$. As shown above, there exists a function in the class $\mathscr B_{\alpha}[-1,1]$ such that the inverse function branches at this point. Hence this function has no inverse in $\mathscr V$. Next we show that the domain $\mathscr W(\alpha)$ is univalently covered by all functions in $\mathscr B_{\alpha}[-1, 1]$. We take an arbitrary $f\in \mathscr B_{\alpha}[-1, 1]$. It is sufficient to check that for each $w\in\mathscr W(\alpha)$ the equation $f(z)=w$ has a unique solution in some subdomain of $\mathbb D$. Let us show that we can take $\mathscr U(\alpha)$ (see (5.1)) as such a subdomain. The idea of the proof consists in making a round along the boundary of $\mathscr U(\alpha)$ and using the argument principle. An additional technical complication is that the boundary of $\mathscr U(\alpha)$ contains corners at $z=-1$ and $z=1$, where $f$ is not assumed to be analytic. To overcome this difficulty we replace small parts of the boundary of $\mathscr U(\alpha)$ near the corners by suitable smooth curves lying fully in the interior of $\mathbb D$. We change the travelling trajectory in this way below. First we verify that if $z$ is a boundary point of $\mathscr U(\alpha)$ (other than $z=-1$ or $z=1$), then $f(z)$ cannot occur in the domain $\mathscr W(\alpha)$. The domain $\mathscr U(\alpha)$ is bounded by two arcs of circles which are mirror symmetric relative to the real diameter, namely, by the arcs
$$
\begin{equation}
z^+_r=\frac{r(\sqrt{\alpha-1}+i)-\sqrt{\alpha}}{r(\sqrt{\alpha-1}+i)+\sqrt{\alpha}}\quad\text{and} \quad z^-_r=\frac{r(\sqrt{\alpha-1}-i)-\sqrt{\alpha}}{r(\sqrt{\alpha-1}-i)+\sqrt{\alpha}}, \quad \text{where } r\in(0,\infty).
\end{equation}
\tag{6.1}
$$
By Corollary 3 the point $f(z)$ must lie in the disc
$$
\begin{equation*}
\mathscr E(z)=\biggl\{\zeta\in\mathbb D\colon \frac 1{\alpha-1}\,\frac {1-|z|^2}{|1+z|^2}\, \biggl|\alpha\frac{1+\zeta}{1-\zeta}-\frac{1+z}{1-z}\biggr|^2\leqslant \alpha\frac{1-|\zeta|^2}{|1-\zeta|^2}-\frac{1-|z|^2}{|1-z|^2} \biggr\},
\end{equation*}
\notag
$$
for any $z \in \mathbb D$, for instance, for $z$ lying on a curve in (6.1).
Supposing that $f(z)\in\mathscr W(\alpha)$ for some $z$ on the curve (6.1), we can find at least one boundary point of $\mathscr W(\alpha)$ lying in the interior of $\mathscr E(z)$. The boundary of $\mathscr W(\alpha)$ is formed by two mirror symmetric curves relative to the real diameter,
$$
\begin{equation}
\zeta^+_s=\frac{s (2\sqrt{\alpha-1}+i(2-\alpha))-\alpha}{s(2\sqrt{\alpha-1}+i(2-\alpha))+\alpha}\quad\text{and} \quad \zeta^-_s=\frac{s (2\sqrt{\alpha-1}-i(2-\alpha))-\alpha}{s(2\sqrt{\alpha-1}-i(2-\alpha))+\alpha},
\end{equation}
\tag{6.2}
$$
where $s\in(0,\infty)$. All this means that for some $r, s\in(0, \infty)$ the pair of points $z$ and $\zeta$ of the form (6.1) and (6.2), respectively, satisfies the inequality
$$
\begin{equation}
\begin{aligned} \, \frac 1{\alpha-1}\,\frac {1-|z|^2}{|1+z|^2}\, \biggl|\alpha\frac{1+\zeta}{1-\zeta}-\frac{1+z}{1-z}\biggr|^2< \alpha\frac{1-|\zeta|^2}{|1-\zeta|^2}-\frac{1-|z|^2}{|1-z|^2}. \end{aligned}
\end{equation}
\tag{6.3}
$$
By symmetry we can limit ourselves to the case when $z=z^+_r$ and $\zeta=\zeta^+_s$. Then after some transformations we can express (6.3) in the following equivalent form: $\alpha(\sqrt{\alpha}\, s-r)^2<0$.
This inequality cannot hold for any $r, s\in(0, \infty)$. Thus, the assumption that $f(z)$ lies in the interior of $\mathscr W(\alpha)$ fails. That is, we have proved that for a point $z$ on the boundary of $\mathscr U(\alpha)$ its image $f(z)$ cannot lie in $\mathscr W(\alpha)$. In Figure 2 we present an illustration of the above argument: the point $z$ travels along the boundary of $\mathscr{U}(\alpha)$, while the corresponding disc $\mathscr{E}(z)$ rolls about the boundary of $\mathscr{W}(\alpha)$. Now we fix some point $w\in\mathscr W(\alpha)$ and describe the change of trajectory mentioned above. We indicate parts of $\mathbb D$ containing the modified parts of the trajectory. We begin with a neighbourhood of $z=1$. Set
$$
\begin{equation}
\varepsilon_1=\min\biggl\{\sqrt{\frac{\alpha-1}\alpha},\frac{|1-w|}5 \biggr\}
\end{equation}
\tag{6.4}
$$
and consider the sector
$$
\begin{equation}
V_1=\biggl\{z\in\mathbb D\colon|1-z|<\varepsilon_1,\, |{\arg(1-z)}|<\arctan \frac 1{\sqrt{\alpha-1}}\biggr\}.
\end{equation}
\tag{6.5}
$$
Note that all points of the curves (6.1) occurring in the $\varepsilon_1$-neighbourhood of $z=1$ lie inside the sector $V_1$. Fix some point $z$ in $V_1$. Taking (6.4) into account we have
$$
\begin{equation}
\begin{aligned} \, \notag (\alpha-1)\frac{|1-z|^2}{1-|z|^2} &=(\alpha-1) \frac{|1-z|}{2\cos(\arg(1-z))-|1-z|}<(\alpha-1) \frac{|1-z|}{2\sqrt{(\alpha-1)/\alpha}-\varepsilon_1} \\ &<\sqrt{\alpha(\alpha-1)}|1-z| <\varepsilon_1\sqrt{\alpha(\alpha-1)}. \end{aligned}
\end{equation}
\tag{6.6}
$$
Since the disc $\mathscr E(z)$ can also be expressed as $|\zeta-a|\leqslant r$, where
$$
\begin{equation*}
\begin{aligned} \, a&=\biggl(z-\frac{\alpha-1}{4}|1-z|^2\biggr) \biggl(1+(\alpha-1)\frac{1+\operatorname{Re} z}{2}\, \frac{|1-z|^2}{1-|z|^2}\biggr)^{-1}, \\ r&=\biggl(\frac{\alpha-1}{4}\, \frac{|1-z|^2}{1-|z|^2} |1+z|^2\biggr) \biggl(1+(\alpha-1)\frac{1+\operatorname{Re} z}{2}\, \frac{|1-z|^2}{1-|z|^2} \biggr)^{-1}, \end{aligned}
\end{equation*}
\notag
$$
taking (6.4)–(6.6) into account, for each $\zeta$ in $\mathscr E(z)$ we deduce the estimate
$$
\begin{equation}
\begin{aligned} \, \notag |1-\zeta| &\leqslant |1-a|+r \leqslant|1-z|+(\alpha-1)\frac{1+\operatorname{Re} z}{2}\, \frac{|1-z|^2}{1-|z|^2} \\ \notag &\qquad +\frac{\alpha-1}{4}|1-z|^2+\frac{\alpha-1}{4}\,\frac{|1-z|^2}{1-|z|^2} |1+z|^2 \\ \notag &<|1-z|+(\alpha-1) \frac{|1-z|^2}{1-|z|^2}+\frac{\alpha-1}{4}|1-z|+(\alpha-1)\frac{|1-z|^2}{1-|z|^2} \\ &<2\varepsilon_1\sqrt{\alpha(\alpha-1)}+\frac{\alpha+3}{4}\varepsilon_1 < 5 \varepsilon_1<|1-w|. \end{aligned}
\end{equation}
\tag{6.7}
$$
For $z=-1$ set
$$
\begin{equation}
\varepsilon_{-1}=\min\biggl\{\sqrt{\frac{\alpha-1}\alpha},\frac{|1+w|}4\biggr\}
\end{equation}
\tag{6.8}
$$
and consider the sector
$$
\begin{equation}
V_{-1}=\biggl\{z\in\mathbb D\colon|1+z|<\varepsilon_{-1}, \,|{\arg(1+z)}|<\arctan \frac 1{\sqrt{\alpha-1}}\biggr\}.
\end{equation}
\tag{6.9}
$$
Fix a point $z$ in this sector. Taking (6.8) into account we obtain
$$
\begin{equation}
\begin{aligned} \, \notag (\alpha-1)\frac{|1+z|^2}{1-|z|^2} &=(\alpha-1)\frac{|1+z|}{2\cos(\arg(1+z))-|1+z|} <(\alpha-1)\frac{|1+z|}{2\sqrt{(\alpha-1)/\alpha}-\varepsilon_{-1}} \\ &<\sqrt{\alpha(\alpha-1)}|1+z| <\varepsilon_{-1}\sqrt{\alpha(\alpha-1)}. \end{aligned}
\end{equation}
\tag{6.10}
$$
In view of (6.8)–(6.10), for each $\zeta$ in the disc $\mathscr E(z)$ we have
$$
\begin{equation}
\begin{aligned} \, \notag |1+\zeta| &\leqslant |1+a|+r \leqslant|1+z|+\biggl|(\alpha-1)\frac{1+\operatorname{Re} z}{2}\, \frac{|1-z|^2}{1-|z|^2}-\frac{\alpha-1}{4}|1-z|^2\biggr| \\ \notag &\qquad +\frac{\alpha-1}{4}\,\frac{|1+z|^2}{1-|z|^2} |1-z|^2 \\ \notag &=|1+z|+\frac{\alpha-1}{2}\,\frac{|1+z|^2}{1-|z|^2} |1-z|^2 \\ &<|1+z|+2(\alpha-1)\frac{|1+z|^2}{1-|z|^2} <\varepsilon_{-1}+2\varepsilon_{-1}\sqrt{\alpha(\alpha-1)} <4 \varepsilon_{-1}<|1+w|. \end{aligned}
\end{equation}
\tag{6.11}
$$
We travel anticlockwise around $w$ along the smooth curve formed by parts of the curve (6.1) which are closed by a smooth curve lying in $V_1$ in the $\varepsilon_1$-neighbourhood of $z=1$ and by another smooth curve lying in $V_{-1}$ in the $\varepsilon_{-1}$-neighbourhood of $z=-1$. We can cut corners in this way because all points of the curves (6.1) occurring in these neighbourhoods lie inside $V_1$ and $V_{-1}$, respectively. In travelling along the curves (6.1), as mentioned above, the image of $z$ cannot occur in $\mathscr W(\alpha)$ this completing making a rotation around the point $w$. On the other hand, in travelling along the modified pieces lying in neighbourhoods of $z=1$ and $z=-1$ $f(z)$ cannot go around $w$ because of (6.7) and (6.11). Thus, when the point $z$ makes a round the argument of $f(z)-w$ increases by $2\pi$. Since the modified pieces of curves can lie arbitrarily close to the corners, the equation $f(z)=w$ has a unique solution in the whole domain $\mathscr U(\alpha)$. Since we have proved the unique solvability of the equation $f(z)=w$ for any points $w\in\mathscr W(\alpha)$, we have proved the first part of the theorem. Finally, to prove the second part of the theorem it is sufficient to observe that the boundary of $\mathscr W(\alpha)$ deforms continuously to the circle $\mathbb T$ as $\alpha\to 1+0$ and to the interval $[-1,1]$ as $\alpha\to 2-0$. Furthermore, the function
$$
\begin{equation*}
f(z)=\biggl(\frac{2-i}{2+i}\biggr)^2 \biggl(\frac{z-(1-2i)/5}{1-z(1+2i)/5}\biggr)^2
\end{equation*}
\notag
$$
belongs to the class $\mathscr B_2[-1, 1]$, and the inverse function branches at the point $w=0$. Hence by Lemma 4 each point in $(-1,1)$ is a branch point of some function inverse to a Blaschke product in the class $\mathscr B_2[-1, 1]$. Thus, each point in $\mathbb D$ is a branch point of an inverse function of a function in $\mathscr B_2[-1, 1]$, so that even for $\mathscr B_2[-1, 1]$ there exist no nonempty domains of univalent covering. The proof of Theorem 2 is complete. Remark 2. The curve with the property that the discs $\mathscr E(z)$ corresponding to its points are disjoint from $\mathscr W(\alpha)$ is unique. Only the boundary of $\mathscr U(\alpha)$ can have this property. If $z$ does not lie on the boundary of $\mathscr U(\alpha)$, then $\mathscr E(z)$ and $\mathscr W(\alpha)$ intersect. Remark 3. Geometrically, the boundary of the sharp univalent covering domain for the class $\mathscr B_{\alpha}[-1, 1]$ can be characterized as the envelope of the family of discs $\mathscr E(z)$ corresponding to the points $z$ ranging over the boundary of the sharp univalence domain for the same class. Remark 4. It is clear from the proof that the statement of the theorem can be refined as follows: a domain $\mathscr X$ equal to the conformal image of $\mathscr W(\alpha)$ under a branch of the inverse function of $f\in \mathscr B_{\alpha}[-1, 1]$ can be taken inside $\mathscr U(\alpha)$. In the conclusion of this section we indicate a disc of maximum radius covered univalently by the values of each function in $\mathscr B_{\alpha}[-1,1]$. Corollary 5. Let $f\in \mathscr B_{\alpha}[-1, 1]$, where $\alpha\in(1,2)$. Then there exists a function inverse to $f$ that maps the disc conformally onto a domain $\mathscr X\subset \mathbb D$. For any disc $\mathscr C'\subset \mathbb D$ of radius greater than that of $\mathscr C(\alpha)$ there exists a function $f\in \mathscr B_{\alpha}[-1, 1]$ that has no inverse in $\mathscr C'$. § 7. Case of an arbitrary boundary Denjoy–Wolff pointWe can generalize Theorems 1 and 2 to the case when the Denjoy–Wolff point $q$ has an arbitrary position on the unit circle $\mathbb T$. To do this we use the map $S_q$ (see (4.2)). If $f\in\mathscr B_{\alpha}[-1,1]$, then the composition $\widetilde{f}(\xi)=S_q\circ f\circ S_q^{-1}(\xi)$ belongs to $\mathscr B_{\alpha}[q,1]$. Applying $S_q^{-1}$ to the domains $\mathscr U(\alpha)$ and $\mathscr W(\alpha)$ we obtain the following results. Theorem 3. Let $\alpha\in(1,4)$ and $q\in\mathbb T\setminus \{1\}$. If $f\in \mathscr B_{\alpha}[q, 1]$, then $f$ is univalent in the domain For any domain $\mathscr V$ such that $\mathscr U(\alpha,q)\subset \mathscr V\subset \mathbb D$ and $\mathscr V\neq \mathscr U(\alpha, q)$ there exists $f\in\mathscr B_{\alpha}[q,1]$ that is not univalent in $\mathscr V$. Theorem 4. Let $f\in \mathscr B_{\alpha}[q, 1]$, $\alpha\in(1,2)$ and $q\in\mathbb T\setminus \{1\}$. Then there exists a function inverse to $f$ that maps the domain conformally onto a domain $\mathscr X\subset \mathbb D$. For any domain $\mathscr V$ such that $\mathscr W(\alpha, q)\subset \mathscr V \subset \mathbb D$ and $\mathscr V\neq \mathscr W(\alpha, q)$ there exists a function $f\in \mathscr B_{\alpha}[q, 1]$ that has no inverse in $\mathscr V$. Finally, for $\alpha\,{\geqslant}\, 2$ the class $\mathscr B_{\alpha}[q, 1]$ has no nonempty univalent covering domains. In Figure 3 we show the boundaries of the domains $\mathscr{U}(\alpha,q)$ and $\mathscr{W}(\alpha,q)$ for an arbitrary position of the point $q$ on the unit circle $\mathbb T$. 
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\Bibitem{KudSol25} |
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