Posts by Eduard Grigoriciuc

Abstract

In this paper we define new subclasses of univalent mappings in the case of several complex variables. We will focus our attention on a particular class, denoted $E_{1}^{\ast}$, and observe that in the case of one complex variable, $E_{1}^{\ast}(U)$\ coincides with the class of convex functions $K$ on the unit disc. However, if $n\geq2$, then $E_{1}^{\ast (\mathbb{B}^{n})$ is different from the class of convex mappings $K(\mathbb{B}^{n})$ on the Euclidean unit ball $\mathbb{B}^{n}$ in $\mathbb{C}^{n}$. Along with this, we will study other properties of the class $E_{1}^{\ast}$ on the unit polydisc, respectively on the Euclidean unit ball in $\mathbb{C}^{n}$. In the second part of the paper we discuss the Graham–Kohr extension operator n,\U{3b1} (defined by Graham and Kohr in Complex Variab. Theory Appl. 47:59-72, 2002). They proved that the extension operator $\Psi_{n,\U{3b1} }$ does not preserve convexity for $n\geq2$ for all $\U{3b1} \in \lbrack0,1]$. However, in this paper we prove that $\Psi_{n,0}(K)$ and $\Psi_{n,1}(K)$ are subsets of the class $E_{1}^{\ast}\left( \mathbb{B}^{n}\right)$ which is different from the class $K(\mathbb{B}^{n})$ for the Euclidean case.

Authors

Eduard Ştefan Grigoriciuc
Department of Mathematics, Faculty of Mathematics and Computer Science, Babeş-Bolyai University, Cluj-Napoca, Romania

Keywords

Biholomorphic mapping; Convex mapping; Starlike mapping; Extension operator.

Paper coordinates

E.S. Grigoriciuc, New subclasses of univalent mappings in several complex variables: extension operators and applications, Computational Methods and Function Theory, 23 (2023), pp. 533–555, https://doi.org/10.1007/s40315-022-00467-z

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https://doi.org/10.1007/s40315-022-00467-z

About this paper

Journal

Computational Methods and Function Theory

Publisher Name

Springer

DOI

https://doi.org/10.1007/s40315-022-00467-z

Print ISSN

1617-9447

Online ISSN

2195-3724

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New Subclasses of Univalent Mappings in Several Complex Variables. Extension Operators and Applications

\jyear

2021

[1]\fnmEduard Ştefan \surGrigoriciuc

[1]\orgdivDepartment of Mathematics, \orgnameFaculty of Mathematics and Computer Science, Babeş-Bolyai University, \orgaddress\street1 M. Kogălniceanu Str., \cityCluj-Napoca, \postcode400084, \stateCluj, \countryRomania

New Subclasses of Univalent Mappings in Several Complex Variables. Extension Operators and Applications

eduard.grigoriciuc@ubbcluj.ro *

Abstract

In this paper we define new subclasses of univalent mappings in the case of several complex variables. We will focus our attention on a particular class, denoted $E^{*}_{1}$ , and observe that in the case of one complex variable, $E^{*}_{1}(U)$ coincides with the class of convex functions $K$ on the unit disc. However, if $n\geq{2}$ , then $E^{*}_{1}(\mathbb{B}^{n})$ is different from the class of convex mappings $K(\mathbb{B}^{n})$ on the Euclidean unit ball $\mathbb{B}^{n}$ in $\mathbb{C}^{n}$ . Along with this, we will study other properties of the class $E^{*}_{1}$ on the unit polydisc, respectively on the Euclidean unit ball in $\mathbb{C}^{n}$ . In the second part of the paper we discuss about Graham-Kohr extension operator $\Psi_{n,\alpha}$ (defined by I. Graham and G. Kohr in Complex Variables Theory Appl. 47 (2002), 59–72). They proved that the extension operator $\Psi_{n,\alpha}$ does not preserve convexity for $n\geq{2}$ for all $\alpha\in[0,1]$ . However, in this paper we prove that $\Psi_{n,0}(K)$ and $\Psi_{n,1}(K)$ are subsets of the class $E^{*}_{1}(\mathbb{B}^{n})$ which is different from the class $K(\mathbb{B}^{n})$ for the Euclidean case.

keywords:

Biholomorphic mapping, Convex mapping, Starlike mapping, Extension operator

pacs:

[

MSC Classification]32H02, 30C45

1 Introduction

Let $\mathbb{C}^{n}$ denote the space of $n$ complex variables $z=(z_{1},...,z_{n})$ equipped with an arbitrary norm $\|\cdot\|_{*}$ . Let $B^{n}=\{z\in\mathbb{C}^{n}:\|z\|_{*}<1\}$ be the unit ball in $\mathbb{C}^{n}$ with respect to the arbitrary norm $\|\cdot\|_{*}$ . In the case of one complex variable, $B^{1}$ is denoted by $U$ .

Let $H(B^{n})$ denote the set of all holomorphic mappings from $B^{n}$ into $\mathbb{C}^{n}$ . If $f\in H(B^{n})$ , we say that $f$ is normalized if $f(0)=0$ and $Df(0)=I_{n}$ , where $Df(z)$ is the complex Jacobian matrix of $f$ at $z$ and $I_{n}$ is the identity operator in $\mathbb{C}^{n}$ . Let $S(B^{n})$ be the set of all normalized univalent mappings in $H(B^{n})$ . In the case of one complex variable we denote the class of normalized univalent functions by $S$ .

A mapping $f\in S(B^{n})$ is called convex (starlike) if its image is a convex (starlike with respect to the origin) set in $\mathbb{C}^{n}$ . The class of normalized convex mappings is denoted by

K(B^{n})=\big{\{}f\in S(B^{n}):f(B^{n})\text{ {is a convex set in} }\mathbb{C}% ^{n}\big{\}}

and the class of normalized starlike mappings is denoted by

S^{*}(B^{n})=\big{\{}f\in S(B^{n}):f(B^{n})\text{ {is a starlike set with % respect to zero in} }\mathbb{C}^{n}\big{\}}.

In the case of one complex variable, the sets $K(U)$ and $S^{*}(U)$ are denoted by $K$ and $S^{*}$ . For details, one may consult Dur , Goo , GrKo1 or Ko1 . In Rob Robertson introduced the class of starlike (respectively convex) functions of order $\alpha$ on the unit disc $U$ , where $\alpha\in[0,1)$ . Hence, we denote by

K(\alpha)=\bigg{\{}f\in S:\text{Re}\bigg{[}1+\dfrac{\zeta f^{\prime\prime}(% \zeta)}{f^{\prime}(\zeta)}\bigg{]}>\alpha,\quad\zeta\in U\bigg{\}}

the class of convex functions of order $\alpha$ on $U$ and by

S^{*}(\alpha)=\bigg{\{}f\in S:\text{Re}\bigg{[}\dfrac{\zeta f^{\prime}(\zeta)}% {f(\zeta)}\bigg{]}>\alpha,\quad\zeta\in U\bigg{\}}

the class of starlike functions of order $\alpha$ on $U$ . There are also generalizations of these classes in several complex variables (see Curt Cu1 , Kohr Ko2 or Ko3 , Liu Liu1 ). When we refer to these classes in the context of several complex variables, then $K(B^{n};\alpha)$ denotes the class of all normalized biholomorphic convex mappings of order $\alpha$ on $B^{n}$ and $S^{*}(B^{n};\alpha)$ denotes the class of all normalized biholomorphic starlike mappings of order $\alpha$ on $B^{n}$ .

Remark 1.

Notice that, throughout this paper, we denote by $\mathbb{Z}_{+}=\mathbb{N}\cup\{0\}$ the set of non-negative integers and by $\mathbb{Z}^{*}_{+}=\mathbb{N}=\{1,2,3,...\}$ the set of natural numbers (positive integers).

Remark 2.

During this paper we are working on different domains (unit balls in $\mathbb{C}^{n}$ with respect to different norms), as follows:

•

$\mathbb{B}^{n}$ – the Euclidean unit ball in $\mathbb{C}^{n}$ with respect to the Euclidean norm $\|z\|=\sqrt{\sum_{j=1}^{n}|z_{j}|^{2}}$ , for all $z=(z_{1},...,z_{n})\in\mathbb{C}^{n}$ .

For the Euclidean case, we denote the Euclidean norm simply $\|\cdot\|$ without any index. Hence, in this paper, when we use the notation $\|\cdot\|$ , we automatically refer to the Euclidean norm.

•

$\mathbb{U}^{n}$ – the unit polydisc in $\mathbb{C}^{n}$ with respect to the maximum norm $\|z\|_{\infty}=\max\{|z_{j}|:j=\overline{1,n}\}$ , for all $z=(z_{1},...,z_{n})\in\mathbb{C}^{n}$ .
•

$B^{n}_{p}$ – the unit ball in $\mathbb{C}^{n}$ with respect to the $p$ -norm $\|z\|_{p}=\bigg{[}\sum_{j=1}^{n}|z_{j}|^{p}\bigg{]}^{1/p}$ , for all $z=(z_{1},...,z_{n})\in\mathbb{C}^{n}$ and $p\in[1,\infty)$ .

In the case of one complex variable, each of the sets $\mathbb{B}^{1}$ , $\mathbb{U}^{1}$ and $B^{1}_{p}$ coincides with $U$ . Mention that when we work with an arbitrary norm, it will be denoted $\|\cdot\|_{*}$ . But, when the domains are those described above, we use the particular notations for unit balls and norms presented for every case. For details one may consult GrKo1 or Ko1 .

Another important result that we will refer to during this paper is Alexander’s duality theorem which says that $f\in K$ if and only if $\zeta f^{\prime}(\zeta)\in S^{*}$ , $\zeta\in U$ (see (Dur, , Theorem 2.12), (Goo, , Chapter 8, Theorem 5) or (GrKo1, , Theorem 2.2.6)).

Remark 3.

It is important to notice that the Alexander’s duality theorem is no longer true in the case of several complex variables (for details one may consult (GrKo1, , Remark 6.3.15) and examples given in (GrKo1, , Problem 6.2.5), (GrKo1, , Problem 6.3.2) and (Suf1, , Example 3)).

When we refer to the starlikeness of a mapping of several complex variables, we can use the following analytical characterization result given by Matsuno (see (GrKo1, , Theorem 6.2.2) or Mat ):

Theorem 1.

Let $f:\mathbb{B}^{n}\rightarrow\mathbb{C}^{n}$ be a locally biholomorphic mapping such that $f(0)=0$ . Then $f$ is starlike if and only if

\emph{Re}\big{\langle}[Df(z)]^{-1}f(z),z\big{\rangle}>0,\quad z\in\mathbb{B}^{% n}\setminus\{0\}.

(1)

A very useful tool in the geometric function theory of one and several complex variables are the Loewner chains. Next we present the definition of the Loewner chains in $\mathbb{C}^{n}$ , as well as their connection with univalent mappings (for details, one may consult (ArBrHaKo , GrKo1 or GrKoKo1 ).

Definition 1.

(see ArBrHaKo , GrKo1 or GrKoKo1 ) A mapping $L=L(z,t):\mathbb{B}^{n}\times[0,\infty)\rightarrow\mathbb{C}^{n}$ is called a Loewner chain (normalized univalent subordination chain) if the following conditions hold:

1.

$e^{-t}L(\cdot,t)\in S(\mathbb{B}^{n})$ , for all $t\in[0,\infty)$ ;
2.

$L(\mathbb{B}^{n},s)\subseteq L(\mathbb{B}^{n},t)$ , for all $0\leq s\leq t<\infty$ .

Remark 4.

In (Pom1, , Theorem 6.1) (see also GrKo1 ) we can find a very important result in the case of one complex variable which says that for every $f\in S$ there exists a Loewner chain $L=L(\zeta,t)$ such that $f(\zeta)=L(\zeta,0)$ , for all $\zeta\in U$ .

In order to extend the previous result in the case of several complex variables, I. Graham, H. Hamada and G. Kohr (see GrHaKo1 or GrKo1 ) defined the family $S^{0}(\mathbb{B}^{n})$ of normalized univalent mappings which have parametric representation as follows

$S^{0}(\mathbb{B}^{n})=\bigg{\{}f\in S(\mathbb{B}^{n}):\exists\ L(z,t)\text{ {a% Loewner chain such that the family} }$

$\{e^{-t}L(\cdot,t)\}_{t\geq{0}}\text{ {is a normal family on} }\mathbb{B}^{n}% \text{ {and} }f=L(\cdot,0)\bigg{\}}$

Remark 5.

Recall that a family $\mathcal{F}$ is called a normal family if each sequence in $\mathcal{F}$ either has a subsequence that converges locally uniformly or else has a subsequence that is compactly divergent (see (Ko1, , Definition 1.3.4)).

Remark 6.

It is clear that if $n=1$ , then $S^{0}(\mathbb{B}^{1})=S$ . However, if $n\geq{2}$ , then $S^{0}(\mathbb{B}^{n})\subsetneqq S(\mathbb{B}^{n})$ (for details, one may consult GrHaKo1 , GrKo1 or GrKoKo1 ). Another important results regarding to the family of normalized univalent mappings which have parametric representation $S^{0}(\mathbb{B}^{n})$ can be found in GrHaKoKo1 , GrKoKo1 and GrKoPf1 .

The problem of constructing examples of starlike (respectively convex) mappings in $\mathbb{C}^{n}$ is partially solved in terms of extension operators. Two of the most important extension operators are defined by K. Roper and T.J. Suffridge (in 1995, see RoSuf0 ), respectively by I. Graham and G. Kohr (in 2002, see GrKo2 ). Such operators extend classes of univalent functions to some classes of univalent mappings in $\mathbb{C}^{n}$ . For more details about extension operators and properties of them one may consult GrHaKoSuf1 , GrKo3 and GrKoPf1 .

Let $\alpha\in[0,1]$ . In 2002, I. Graham and G. Kohr (see GrKo2 ) introduced the extension operator $\Psi_{n,\alpha}$ defined for normalized locally univalent functions on $U$ by

\Psi_{n,\alpha}(f)(z)=\bigg{(}f(z_{1}),\bigg{[}\dfrac{f(z_{1})}{z_{1}}\bigg{]}% ^{\alpha}z^{*}\bigg{)},\quad z=(z_{1},z^{*})\in\mathbb{B}^{n},

(2)

where $f(z_{1})\not=0$ , for $z_{1}\in U\setminus\{0\}$ and the branch of the power function is chosen such that $\bigg{(}\dfrac{f(z_{1})}{z_{1}}\bigg{)}^{\alpha}\bigg{|}_{z_{1}=0}=1$ .

Remark 7.

In (GrKo2, , Corollary 3.3) Graham and Kohr proved that $\Psi_{n,\alpha}(S^{*})\subseteq S^{*}(\mathbb{B}^{n})$ , for all $\alpha\in[0,1]$ . Also in (GrKo2, , Remark 3.6) the authors shows that $\Psi_{n,\alpha}$ does not preserve convexity for $n\geq{2}$ , for all $\alpha\in[0,1]$ .

In the second section of our paper, we introduce two new subclasses $E_{k}$ and $E^{*}_{k}$ of univalent mappings of one and several complex variables. We will give the definitions in a general setting (on the unit ball $B^{n}$ in $\mathbb{C}^{n}$ with respect to an arbitrary norm) and some particular examples on the Euclidean unit ball $\mathbb{B}^{n}$ .

The third section contains some general properties of the previously defined subclasses of univalent mappings. We will highlight the fact that in the case of one complex variable, $E^{*}_{1}(U)$ coincides with the class of convex functions $K$ on the unit disc $U$ , but this result is not longer true in the case of several complex variables.

In the last part of this paper, we discuss about the preservation of some new subclasses of univalent mappings by the Graham-Kohr extension operator $\Psi_{n,\alpha}$ defined in (2). It is interesting that although the operator $\Psi_{n,\alpha}$ does not keep the convexity, still in some particular cases, the class $K$ is taken to a class that we will define in the next section.

2 New Subclasses of Univalent Mappings

This section is dedicated to the study of new subclasses of univalent mappings in several complex variables. We will give the definitions in a general setting (on the unit ball $B^{n}$ in $\mathbb{C}^{n}$ with respect to an arbitrary norm $\|\cdot\|_{*}$ ), but a few properties will be presented on some particular domains, especially, on the Euclidean unit ball $\mathbb{B}^{n}$ .

Definition 2.

Let $k\in\mathbb{Z}_{+}$ and let $f:B^{n}\rightarrow\mathbb{C}^{n}$ be a holomorphic mapping such that $f(0)=0$ and $Df(0)=I_{n}$ . Also, let $f(z)=z+\sum_{m=2}^{\infty}P_{m}(z)$ be the power series expansion of $f$ on the unit ball $B^{n}$ , where

P_{m}(z)=\dfrac{1}{m!}D^{m}f(0)(z^{m}),\quad z\in B^{n}.

Then we define

\big{(}G_{k}f\big{)}(z)=\begin{cases}D^{k}f(z)(z^{k})+z+\displaystyle\sum_{m=2% }^{k-1}P_{m}(z),&k\geq{3}\\ \\ D^{2}f(z)(z^{2})+z,&k=2\\ \\ Df(z)(z),&k=1\\ \\ f(z),&k=0,\end{cases}

(3)

for all $z\in B^{n}$ .

Definition 3.

Let $k\in\mathbb{Z}_{+}$ . We define the following subclasses of normalized univalent mappings on $B^{n}$ :

E^{*}_{k}(B^{n})=\big{\{}f\in S(B^{n}):G_{k}f\in S^{*}(B^{n})\big{\}}

(4)

and

E_{k}(B^{n})=\big{\{}f\in S(B^{n}):G_{k}f\in K(B^{n})\big{\}}.

(5)

Remark 8.

According to the previous definition, it is clear than we obtain the following particular cases:

1.

If $k=0$ , then

$E^{*}_{0}(B^{n})=S^{*}(B^{n})\quad\text{and}\quad E_{0}(B^{n})=K(B^{n}).$
2.

If $k=1$ , then

$E^{*}_{1}(B^{n})=\big{\{}f\in S(B^{n}):G_{1}f\in S^{*}(B^{n})\big{\}}$

and

$E_{1}(B^{n})=\big{\{}f\in S(B^{n}):G_{1}f\in K(B^{n})\big{\}},$

where $G_{1}f(z)=Df(z)(z)$ , for $z\in B^{n}$ .

An important remark on the previous subclasses of biholomorphic mappings in $\mathbb{C}^{n}$ is based on the fact that Alexander’s duality theorem is no longer true in higher dimensions (see Remark 3). According to this, we obtain the following remarks:

Remark 9.

If $n\geq{2}$ , then

K(B^{n}_{1})\subsetneq E^{*}_{1}(B^{n}_{1})\quad\text{and}\quad K(\mathbb{U}^{% n})\subsetneq E^{*}_{1}(\mathbb{U}^{n}),

(6)

where $B^{n}_{1}$ is the unit ball in $\mathbb{C}^{n}$ with respect to the $1$ -norm, respectively $\mathbb{U}^{n}$ is the unit polydisc in $\mathbb{C}^{n}$ .

Proof.

For the sake of brevity, let us consider $n=2$ . Notice that the arguments for the case $n\geq{2}$ are similar.

1.

First, let us consider the case of the unit ball $B^{2}_{1}$ with respect to the $1$ -norm. According to (GrKo1, , Theorem 6.3.11.a) and (GrKo1, , Remark 6.3.15), we have that if $f\in K(B^{2}_{1})$ , then $Df(z)z$ is starlike on $B^{2}_{1}$ . Hence, $G_{1}f\in S^{*}(B^{2}_{1})$ and in view of Definition 3 we obtain that $f\in E^{*}_{1}(B^{2}_{1})$ .

In order to prove that the inclusion $K(B^{2}_{1})\subsetneq E^{*}_{1}(B^{2}_{1})$ is strict, it is enough to consider the mapping $f:B^{2}_{1}\rightarrow\mathbb{C}^{2}$ given by

$f(z)=\bigg{(}\dfrac{z_{1}}{1-z_{1}},z_{2}\bigg{)},\quad z=(z_{1},z_{2})\in B^{% 2}_{1}.$

Then $Df(z)(z)=(z_{1}/(1-z_{1})^{2},z_{2})$ is starlike on $B^{2}_{1}$ (in view of (GrKo1, , Problem 6.2.5)) and hence, $f\in E^{*}_{1}(B^{2}_{1})$ . However, $f$ is not convex on $B^{2}_{1}$ (see (GrKo1, , Problem 6.3.2)). Then, $f\in E^{*}_{1}(B^{2}_{1})\setminus K(B^{2}_{1})$ and we conclude that

$K(B^{2}_{1})\subsetneq E^{*}_{1}(B^{2}_{1}).$
2.

On the other hand, in view of (GrKo1, , Theorem 6.3.11.b) and (GrKo1, , Remark 6.3.15), we have that if $f\in K(\mathbb{U}^{2})$ , then $Df(z)z$ is starlike on $\mathbb{U}^{2}$ . Hence, $G_{1}f\in S^{*}(\mathbb{U}^{2})$ and in view of Definition 3 we obtain that $f\in E^{*}_{1}(\mathbb{U}^{2})$ .

Again, to prove that the inclusion $K(\mathbb{U}^{2})\subsetneq E^{*}_{1}(\mathbb{U}^{2})$ is strict, we can consider the mapping $f:\mathbb{U}^{2}\rightarrow\mathbb{C}^{2}$ given by

$f(z)=\bigg{(}z_{1}+\frac{a}{2}z_{2}^{2},z_{2}\bigg{)},\quad z=(z_{1},z_{2})\in% \mathbb{U}^{2}.$

Then $Df(z)(z)=(z_{1}+az_{2}^{2},z_{2})$ is starlike on $\mathbb{U}^{2}$ for $|a|\leq 1$ (see (Suf1, , Example 3)). In view of Definition 3, we deduce that $f\in E^{*}_{1}(\mathbb{U}^{2})$ . However, $f$ is not convex for $a\not=0$ (see (Suf1, , Example 3)). Hence, there exists a mapping $f\in E^{*}_{1}(\mathbb{U}^{2})\setminus K(\mathbb{U}^{2})$ . Then

$K(\mathbb{U}^{2})\subsetneq E^{*}_{1}(\mathbb{U}^{2})$

and this completes the proof.

∎

Remark 10.

In addition to Remark 9, we can also prove (see Theorem 2) that

E^{*}_{1}(\mathbb{B}^{n})\not=K(\mathbb{B}^{n}),\quad n\geq{2},

(7)

where $\mathbb{B}^{n}$ is the Euclidean unit ball in $\mathbb{C}^{n}$ . Hence, it is not trivial to define the subclasses $E^{*}_{k}$ and $E_{k}$ of univalent mappings in the case of several complex variables. Even the simplest case $k=1$ is interesting for the Euclidean unit ball $\mathbb{B}^{n}$ in view of the difference provided by relation (7).

Next we present two examples of mappings which belongs to the class $E^{*}_{1}(B^{n}_{p})$ for the general case of the unit ball $B^{n}_{p}$ with respect to a $p$ -norm. These examples was also considered in (BrGrHaKo, , Example 3.2), (HaKo1, , Remark 3.3), (RoSuf1, , Example 5) and (RoSuf1, , Example 6).

Example 1.

Let $f:B^{2}_{p}\subset\mathbb{C}^{2}\rightarrow\mathbb{C}^{2}$ be given by

f(z)=\big{(}z_{1}+az_{2}^{2},z_{2}\big{)},\quad z=(z_{1},z_{2})\in B^{2}_{p}.

Then $f\in E^{*}_{1}(B^{2}_{p})$ if and only if

|a|\leq\dfrac{1}{2}\bigg{(}\dfrac{p^{2}-1}{4}\bigg{)}^{1/p}\bigg{(}\dfrac{p+1}% {p-1}\bigg{)},\quad p>1.

(8)

Proof.

We have that $f(0)=0$ and $Df(0)=I_{n}$ , where

Df(z)=\begin{pmatrix}1&2az_{2}\\ \\ 0&1\end{pmatrix},\quad z=(z_{1},z_{2})\in B^{2}_{p}.

Moreover,

g(z)=Df(z)(z)=\big{(}z_{1}+2az_{2}^{2},z_{2}\big{)},\quad z=(z_{1},z_{2})\in B% ^{2}_{p}.

In view of (RoSuf1, , Example 5) we know that $g=G_{1}f\in S^{*}(B^{2}_{p})$ if and only if

|2a|\leq\bigg{(}\dfrac{p^{2}-1}{4}\bigg{)}^{1/p}\bigg{(}\dfrac{p+1}{p-1}\bigg{% )},\quad p>1.

Hence, $f\in E^{*}_{1}(B^{2}_{p})$ if and only if condition (8) holds and this completes the proof. ∎

Remark 11.

It is clear that in view of the previous example, we can construct another example on the Euclidean unit ball $\mathbb{B}^{2}$ . Indeed, for the particular case $p=2$ , we obtain that $f\in E^{*}_{1}(\mathbb{B}^{2})$ if and only if $|a|\leq\frac{3\sqrt{3}}{4}$ .

Example 2.

Let $f:B^{2}_{p}\subset\mathbb{C}^{2}\rightarrow\mathbb{C}^{2}$ be given by

f(z)=\big{(}z_{1}+az_{1}z_{2},z_{2}\big{)},\quad z=(z_{1},z_{2})\in B^{2}_{p}.

Then $f\in E^{*}_{1}(B^{2}_{p})$ if and only if $|a|\leq 1/2$ , for all $1\leq p\leq\infty$ .

Proof.

We have that $f(0)=0$ and $Df(0)=I_{n}$ , where

Df(z)=\begin{pmatrix}1+az_{2}&az_{1}\\ \\ 0&1\end{pmatrix},\quad z=(z_{1},z_{2})\in B^{2}_{p}.

Moreover,

g(z)=Df(z)(z)=\big{(}z_{1}+2az_{1}z_{2},z_{2}\big{)},\quad z=(z_{1},z_{2})\in B% ^{2}_{p}.

In view of (RoSuf1, , Example 6) we know that $g=G_{1}f\in S^{*}(B^{2}_{p})$ if and only if $|2a|\leq 1$ , for all $1\leq p\leq\infty$ . Hence, $f\in E^{*}_{1}(B^{2}_{p})$ if and only if $|a|\leq\frac{1}{2}$ , and this completes the proof. ∎

Similarly, we can construct examples of mappings in the class $E_{1}(\mathbb{B}^{2})$ for the case of the Euclidean unit ball. We present these very simple examples without proofs, but for details, one may consult (RoSuf1, , Examples 7 and 8) (cf. BrGrHaKo , (HaKo1, , Remark 3.3)).

Example 3.

Let $f:\mathbb{B}^{2}\rightarrow\mathbb{C}^{2}$ be given by $f(z)=\big{(}z_{1}+az_{2}^{2},z_{2})$ for all $z\in\mathbb{B}^{2}$ . Then $f\in E_{1}(\mathbb{B}^{2})$ if and only if $|a|\leq\frac{1}{4}$ .

Example 4.

Let $f:\mathbb{B}^{2}\rightarrow\mathbb{C}^{2}$ be given by $f(z)=\big{(}z_{1}+az_{1}z_{2},z_{2})$ for all $z\in\mathbb{B}^{2}$ . Then $f\in E_{1}(\mathbb{B}^{2})$ if and only if $|a|\leq\frac{\sqrt{2}}{4}$ .

3 General Properties

In the third section we present some general properties of the previously defined subclasses of univalent mappings. We will highlight the connection between subclasses $E_{1}^{*}$ (respectively $E_{1}$ ) on the unit disc $U$ and the class of convex mappings $K(B^{n}_{p})$ in $\mathbb{C}^{n}$ . It is very interesting that results from the case $n=1$ are not longer true in the case of several complex variables.

Theorem 2.

Regarding to the class $E^{*}_{1}$ , the following statements are true:

1.

If $n=1$ , then $E_{1}^{*}(U)=K(U)=K$ .
2.

If $n\geq{2}$ , then $E^{*}_{1}(\mathbb{B}^{n})\cap K(\mathbb{B}^{n})\not=\emptyset$ and $E_{1}^{*}(\mathbb{B}^{n})\not=K(\mathbb{B}^{n})$ .

Proof.

1.

Let us first consider the case $n=1$ . Then we can rewrite the definition of the class $E_{1}$ in the following way

$E_{1}^{*}(\mathbb{B}^{1})=E_{1}^{*}(U)=\big{\{}f\in S(\mathbb{B}^{1}):G_{1}f% \in S^{*}(\mathbb{B}^{1})\big{\}},$

where $S(\mathbb{B}^{1})=S$ , $S^{*}(\mathbb{B}^{1})=S^{*}$ and $G_{1}f(\zeta)=\zeta f^{\prime}(\zeta)$ , for all $\zeta\in\mathbb{B}^{1}=U$ . Hence,

$E_{1}^{*}(U)=\bigg{\{}f\in S:\text{Re}\bigg{[}\dfrac{\zeta(G_{1}f)^{\prime}(% \zeta)}{G_{1}f(\zeta)}\bigg{]}>0,\quad\zeta\in{U}\bigg{\}}$

which means that

$E_{1}^{*}(U)=\bigg{\{}f\in S:\text{Re}\bigg{[}\dfrac{f^{\prime}(\zeta)+\zeta f% ^{\prime\prime}(\zeta)}{f^{\prime}(\zeta)}\bigg{]}>0,\quad\zeta\in{U}\bigg{\}}.$

Now, it is clearly that

$E_{1}^{*}(U)=\bigg{\{}f\in S:\text{Re}\bigg{[}1+\dfrac{\zeta f^{\prime\prime}(% \zeta)}{f^{\prime}(\zeta)}\bigg{]}>0,\quad\zeta\in{U}\bigg{\}}=K.$

Another simple argument for the equality between these two classes is given by the fact that Alexander’s Duality Theorem holds in the case of one complex variable. Then, in view of Definition 3 and Remark 8, we have that

$E^{*}_{1}(U)=\big{\{}f\in S:\zeta f^{\prime}(\zeta)\in S^{*},\quad\zeta\in U% \big{\}}=K.$
2.
For the second part of the proof it is enough to consider some examples which shows us that, in the case of several complex variables, $E_{1}^{*}(\mathbb{B}^{n})\cap K(\mathbb{B}^{n})\not=\emptyset$ and $E^{*}_{1}(\mathbb{B}^{n})\not=K(\mathbb{B}^{n})$ . For simplicity, let us consider the case $n=2$ , but notice that for $n\geq{2}$ the arguments are similar. Indeed, according to
- •
  
  Example 1 for $p=2$ and $a=\frac{1}{2}$ , we can construct $f:\mathbb{B}^{2}\rightarrow\mathbb{C}^{2}$ given by
  
  $f(z)=\big{(}z_{1}+az_{2}^{2},z_{2}\big{)},\quad z=(z_{1},z_{2})\in\mathbb{B}^{% 2}.$
  
  Because $|a|=\frac{1}{2}<\frac{3\sqrt{3}}{4}$ we obtain that $f\in E^{*}_{1}(\mathbb{B}^{2})$ . Moreover, in view of (RoSuf1, , Example 7), we have that $f\in K(\mathbb{B}^{2})$ because $|a|\leq\frac{1}{2}$ .
Hence, we obtain that $E_{1}^{*}(\mathbb{B}^{n})\cap K(\mathbb{B}^{n})\not=\emptyset$ . On the other hand, in view of
- •
  
  Example 1 for $p=2$ and $b=\frac{3\sqrt{3}}{4}$ , we can construct $g:\mathbb{B}^{2}\rightarrow\mathbb{C}^{2}$ given by
  
  $g(z)=\big{(}z_{1}+bz_{2}^{2},z_{2}\big{)},\quad z=(z_{1},z_{2})\in\mathbb{B}^{% 2}.$
  
  Because $|b|=\frac{3\sqrt{3}}{4}\leq\frac{3\sqrt{3}}{4}$ we obtain that $g\in E^{*}_{1}(\mathbb{B}^{2})$ . But, according to (RoSuf1, , Example 7), we have that $g\not\in K(\mathbb{B}^{2})$ because $\frac{3\sqrt{3}}{4}>\frac{1}{2}$ .
Hence, we have that there exists $g\in E^{*}_{1}(\mathbb{B}^{2})\setminus K(\mathbb{B}^{2})$ , i.e. $g$ belongs to class $E^{*}_{1}(\mathbb{B}^{2})$ , but $g$ is not a convex mapping on $\mathbb{B}^{2}$ . Finally, in view of
- •
  
  Example 2 for $p=2$ and $c=\frac{\sqrt{2}}{2}$ , we can construct $h:\mathbb{B}^{2}\rightarrow\mathbb{C}^{2}$ given by
  
  $h(z)=\big{(}z_{1}+cz_{1}z_{2},z_{2}\big{)},\quad z=(z_{1},z_{2})\in\mathbb{B}^% {2}.$
  
  Because $|c|=\frac{\sqrt{2}}{2}\leq\frac{1}{\sqrt{2}}$ we obtain that $h\in K(\mathbb{B}^{2})$ (according to (RoSuf1, , Example 8)). However, $h\not\in E^{*}_{1}(\mathbb{B}^{2})$ because $G_{1}h\not\in S^{*}(\mathbb{B}^{2})$ , where
  
  $G_{1}h(z)=Dh(z)(z)=\big{(}z_{1}+\sqrt{2}z_{1}z_{2},z_{2}\big{)},\quad z=(z_{1}% ,z_{2})\in\mathbb{B}^{2}.$
  
  Indeed, in view of (RoSuf1, , Example 6) it is easy to prove that $G_{1}f\not\in S^{*}(\mathbb{B}^{2})$ .
Hence, we obtain that $h\in K(\mathbb{B}^{2})\setminus E^{*}_{1}(\mathbb{B}^{2})$ , i.e. $h$ is a convex mapping on $\mathbb{B}^{2}$ , but $h$ does not belong to class $E^{*}_{1}(\mathbb{B}^{2})$ . In view of all arguments presented above, we conclude that $E^{*}_{1}(\mathbb{B}^{n})\not=K(\mathbb{B}^{n})$ and this completes the proof.

∎

Remark 12.

According to Theorem 2, it is clear that if $n=1$ , then $K(U)=E^{*}_{1}(U)$ . However, if $n\geq{2}$ , then $K(B^{n}_{1})\subsetneq E^{*}_{1}(B^{n}_{1})$ and $K(\mathbb{U}^{n})\subsetneq E^{*}_{1}(\mathbb{U}^{n})$ in view of Remark 9 and $K(\mathbb{B}^{n})\not=E^{*}_{1}(\mathbb{B}^{n})$ in view of Theorem 2.

Next, we present an example given by M.S. Liu in Liu1 . He obtained a very nice condition for a function to be convex of order $\alpha$ on the unit ball $B^{2}_{p}$ (with respect to a $p$ -norm, where $p\geq{2}$ ). We will use this example given by Liu to prove that $E_{1}(\mathbb{B}^{n})\cap K(\mathbb{B}^{n};1/2)\not=\emptyset$ and $K(\mathbb{B}^{n};1/2)\setminus E_{1}(\mathbb{B}^{n})\not=\emptyset$ .

Example 5.

(see (Liu1, , Example 1)) Let $\alpha\in[0,1)$ , $p\geq{2}$ and $k\in\mathbb{Z}^{*}_{+}=\{1,2,3,...\}$ such that $k<p\leq{k+1}$ . Also, let $g:B^{2}_{p}\rightarrow\mathbb{C}^{2}$ be given by

g(z)=\big{(}z_{1}+az_{2}^{k+1},z_{2}\big{)},\quad z=(z_{1},z_{2})\in B^{2}_{p},

(9)

where $B^{2}_{p}=\big{\{}z\in\mathbb{C}^{2}:\|z\|_{p}=(|z_{1}|^{p}+|z_{2}|^{p})^{1/p}% \big{\}}$ . If $a$ satisfies the inequality $|a|\leq\frac{1-\alpha}{k(k+1)}$ , then $g\in K(B^{2}_{p},\alpha)$ .

Theorem 3.

Regarding to the class $E_{1}$ , the following statements are true:

1.

If $n=1$ , then $E_{1}(U)\subsetneq K(1/2)$ .
2.

If $n\geq{2}$ , then $E_{1}(\mathbb{B}^{n})\cap K(\mathbb{B}^{n};1/2)\not=\emptyset$ and $K(\mathbb{B}^{n};1/2)\setminus E_{1}(\mathbb{B}^{n})\not=\emptyset$ , i.e. there exists also convex mappings of order $1/2$ on $\mathbb{B}^{n}$ which does not belong to class $E_{1}(\mathbb{B}^{n})$ .

Proof.

1.
Let us first consider the case $n=1$ .
- •
  
  Let $f\in E_{1}(U)$ . Then, in view of Definition 3, we have that $G_{1}f\in K$ . According to a result given by Sheil-Small and Suffridge (see (GrKo1, , Theorem 2.2.4)) we know that
  
  $\text{Re}\bigg{[}\dfrac{2\zeta(G_{1}f)^{\prime}(\zeta)}{(G_{1}f)(\zeta)-(G_{1}% f)(\zeta_{0})}-\dfrac{\zeta+\zeta_{0}}{\zeta-\zeta_{0}}\bigg{]}\geq{0},\quad% \forall\ \zeta,\zeta_{0}\in U.$
  
  For $\zeta_{0}=0$ , we obtain that
  
  $\text{Re}\bigg{[}\dfrac{\zeta(G_{1}f)^{\prime}(\zeta)}{(G_{1}f)(\zeta)}\bigg{]% }\geq\dfrac{1}{2},\quad\zeta\in U.$
  
  But, $(G_{1}f)(\zeta)=\zeta f^{\prime}(\zeta)$ , for all $\zeta\in U$ and using the minimum principle for harmonic functions we have that
  
  $\text{Re}\bigg{[}\dfrac{\zeta(G_{1}f)^{\prime}(\zeta)}{(G_{1}f)(\zeta)}\bigg{]% }=\text{Re}\bigg{[}\dfrac{\zeta^{2}f^{\prime\prime}(\zeta)+\zeta f^{\prime}(% \zeta)}{\zeta f^{\prime}(\zeta)}\bigg{]}=\text{Re}\bigg{[}1+\dfrac{\zeta f^{% \prime\prime}(\zeta)}{f^{\prime}(\zeta)}\bigg{]}>\dfrac{1}{2},\quad\zeta\in U.$
  
  In view of the definition of the class $K(\alpha)$ (see (Goo, , Chapter 9) or Rob ), we obtain that $f\in K(1/2)$ . Hence, $E_{1}(U)\subseteq K(1/2)$ .
- •
  
  However, we can prove that the previous inclusion is strict, i.e. there is a function $f\in K(1/2)$ which does not belog to class $E_{1}(U)$ . Indeed, if we consider $f:U\rightarrow\mathbb{C}$ given by $f(\zeta)=\zeta+a\zeta^{2}$ for all $\zeta\in U$ with $a=\frac{1}{6}$ , then according to (MeRoSco, , Theorem 2) (for $\alpha=1/2$ ) we have that
  
  $\sum_{n=2}^{\infty}\bigg{(}n-\frac{1}{2}\bigg{)}n|a_{n}|=\bigg{(}2-\frac{1}{2}% \bigg{)}\cdot 2\cdot|a|=\frac{1}{2}\leq\frac{1}{2}$
  
  and then $f\in K(1/2)$ by Alexander’s duality theorem. On the other hand, let us consider
  
  $g(\zeta)=G_{1}f(\zeta)=\zeta f^{\prime}(\zeta)=\zeta+2a\zeta^{2},\quad\zeta\in U% ,\quad a=1/6$ (10)
  
  and $h:U\rightarrow\mathbb{C}$ given by
  
  $h(\zeta)=1+\dfrac{\zeta g^{\prime\prime}(\zeta)}{g^{\prime}(\zeta)}=\dfrac{4% \zeta+3}{2\zeta+3},\quad\zeta\in U.$ (11)
  
  Then $h$ is a Möbius transformation on $U$ such that $h(-3/2)=\infty$ , $h(0)=1$ , $h(1)=\frac{7}{5}$ and $h(i)=\frac{17}{13}+\frac{6}{13}i$ . Then
  
  $h(U)=\mathcal{U}(1/5,6/5)=\big{\{}x+iy\in\mathbb{C}:(x-0.2)^{2}+y^{2}<1.44\big% {\}},$
  
  where $\mathcal{U}(1/5,6/5)$ is the open disc of center $w_{0}=1/5$ and radius $r=6/5$ , as we can observe in Figure 1.
  
  Figure 1: The open disc $\mathcal{U}(1/5,6/5)$ of center $w_{0}=1/5$ and radius $r=6/5$
  
  Moreover, for every point $w\in\mathcal{U}(1/5,6/5)\cap\{x+iy\in\mathbb{C}:x<0\}$ , we have that Re $w<0$ , i.e. $\exists\ \zeta_{0}\in U$ such that Re $[h(\zeta_{0})]<0$ . For example, if $\zeta_{0}=-\frac{9}{10}$ , then $\zeta_{0}\in U$ and simple computations show that
  
  $\text{Re}[h(\zeta_{0})]=\text{Re}\bigg{[}1+\dfrac{\zeta_{0}g^{\prime\prime}(% \zeta_{0})}{g^{\prime}(\zeta_{0})}\bigg{]}=\text{Re}\bigg{[}\dfrac{4\zeta_{0}+% 3}{2\zeta_{0}+3}\bigg{]}=-\dfrac{1}{2}<0.$
  
  Hence, according to the behavior of the function $h$ on $U$ and the analytical characterization of convexity (see for example (GrKo1, , Theorem 2.2.3)), we deduce that $g=G_{1}f\not\in K$ , and then $f\not\in E_{1}(U)$ . Finally, we obtain that $f\in K(1/2)\setminus E_{1}(U)$ .
2.
For the second part of the proof, let us consider $n=2$ . However, notice that for $n\geq{2}$ , the arguments are similar.
- •
  
  First, we can use the example given by Liu (see (Liu1, , Example 1)) in the particular case $n=2$ and $\alpha=\frac{1}{2}$ . For this, let us consider $f:\mathbb{B}^{2}\rightarrow\mathbb{C}^{2}$ given by
  
  $f(z)=(z_{1}+az_{2}^{2},z_{2}),\quad z=(z_{1},z_{2})\in\mathbb{B}^{2}.$
  
  According to the result given by Liu and Example 3, we obtain that
  
  $f\in E_{1}(\mathbb{B}^{2})\cap K(\mathbb{B}^{2};1/2)$
  
  for $|a|\leq\frac{1}{4}$ . Hence,
  
  $E_{1}(\mathbb{B}^{n})\cap K(\mathbb{B}^{n};1/2)\not=\emptyset.$
- •
  
  In order to prove that $\exists\ f\in K(\mathbb{B}^{n};1/2)\setminus E_{1}(\mathbb{B}^{n})$ , i.e. there is a convex mapping of order $1/2$ on $\mathbb{B}^{n}$ such that $f\not\in E_{1}(\mathbb{B}^{n})$ , we can use a particular form of the example considered by Liu and Zhu in the general case of complex Hilbert spaces in (LiuZhu, , Example 2). Notice that, the example considered fits better with the example used in the case $n=1$ , in the first part of the proof.
  
  Let us consider $\alpha=\frac{1}{2}$ and $a=\frac{1}{6}$ . Also, let
  
  $f(z)=z+a\langle z,u\rangle^{2}u,\quad z=(z_{1},z_{2})\in\mathbb{B}^{2},\quad u% =(1,0)\in\mathbb{C}^{2}.$
  
  Then, according to (LiuZhu, , Example 2), we have that $f\in K(\mathbb{B}^{2};1/2)$ . On the other hand, if
  
  $g(z)=G_{1}f(z)=Df(z)(z)=(z_{1}+2az_{1}^{2},z_{2})=z+2a\langle z,u\rangle^{2}u,$
  
  for $z\in\mathbb{B}^{2}$ and $u=(1,0)\in\mathbb{C}^{2}$ , then $|2a|=\frac{1}{3}>\frac{1}{4}$ . In view of (LiuZhu, , Example 2), we deduce that $g=G_{1}f\not\in K(\mathbb{B}^{2})$ and hence $f\not\in E_{1}(\mathbb{B}^{2})$ . Finally, we conclude that $f\in K(\mathbb{B}^{n};1/2)\setminus E_{1}(\mathbb{B}^{n})$ and this completes the proof.

∎

Remark 13.

Let us consider $f(z)=(z_{1}+az_{2}^{k+1},z_{2})$ with $|a|\leq\frac{1}{2k(k+1)}$ and $k\in\mathbb{Z}^{*}_{+}$ . In view of the Example 5 (for $\alpha=1/2$ ) we obtain that $f\in K(B^{2}_{p},1/2)$ . On the other hand,

Df(z)(z)=\big{(}z_{1}+(k+1)az_{2}^{k+1},z_{2}\big{)},\quad z=(z_{1},z_{2})\in B% ^{2}_{p}

and

\big{|}(k+1)a\big{|}=(k+1)|a|\leq(k+1)\dfrac{1}{2k(k+1)}=\dfrac{1}{2k}.

To prove that $f\in E_{1}(B^{2}_{p})$ , we must obtain some conditions for $G_{1}f$ to be convex on $B^{2}_{p}$ . According to Example 5 (for $\alpha=0$ ) we can impose condition $\frac{1}{2k}\leq\frac{1}{k(k+1)}$ which leads us to $k\leq{1}$ . Hence

f\in E_{1}(B^{2}_{p})\cap K(B^{2}_{p};1/2),

for $k\in\mathbb{Z}_{+}^{*}$ with $k\leq{1}$ . It is clear that $k=1$ and then the only possible value for $p$ is $p=2$ (this means that we have to use the Euclidean norm $\|\cdot\|$ ).

This remark together with Example 5 shows us that a function defined by relation (9) belongs to the both classes $E_{1}(B^{n}_{p})$ and $K(B^{n}_{p};1/2)$ if $k=1$ , $p=2$ and $|a|\leq\frac{1}{4}$ , as we saw in the proof of Theorem 3.

Definition 4.

Let $k\in\mathbb{Z}_{+}$ and $\alpha\in[0,1)$ . In view of Definition 3, we define

E_{k}^{*}(B^{n};\alpha)=\bigg{\{}f\in S(B^{n}):G_{k}f\in S^{*}(B^{n};\alpha)% \bigg{\}},

(12)

where $S^{*}(B^{n},\alpha)$ is the family of starlike mappings of order $\alpha$ in $\mathbb{C}^{n}$ (see Cu1 or Ko2 ).

Using the previous definition, we can obtain another form of the Marx-Strohhacker theorem (see (GrKo1, , Theorem 2.3.2) for $n=1$ and (GrKo1, , Theorem 6.3.19) for the case of several complex variables) in the context of classes $E_{k}$ and $E^{*}_{k}$ .

Theorem 4.

Let $k\in\mathbb{Z}_{+}$ . Then

E_{k}(B^{n})\subseteq E^{*}_{k}(B^{n};1/2)\subseteq E^{*}_{k}(B^{n}),

(13)

where $B^{n}$ is the unit ball of $\mathbb{C}^{n}$ with respect to an arbitrary norm $\|\cdot\|_{*}$ .

Proof.

Let $f\in E_{k}(B^{n})$ . In view of Definition 3, we have that $G_{k}f$ is a convex mapping on $B^{n}$ . In view of (GrKo1, , Theorem 6.3.18 and Theorem 6.3.19) we obtain that $G_{k}f$ is starlike of order 1/2 and this means that $f\in E_{k}^{*}(B^{n};1/2)$ . It is obvious that $E^{*}_{k}(B^{n};1/2)\subseteq E^{*}_{k}(B^{n})$ and this completes the proof. ∎

Corollary 5.

Let $k\in\mathbb{Z}_{+}$ and $n=1$ . Then $E_{k}(U)\subseteq E^{*}_{k}(U;1/2)\subseteq E^{*}_{k}(U)$ .

Corollary 6.

Let us consider $k\in\{0,1\}$ .

1.

If $n=1$ , then

$K=E_{0}(U)\subseteq E^{*}_{0}(U;1/2)=S^{*}(1/2)\subseteq S^{*}$ (14)

and

$E_{1}(U)\subseteq E^{*}_{1}(U;1/2)=K(1/2)\subseteq K.$ (15)
2.

On the other hand, if $n\geq{2}$ , then

$E_{0}(B^{n}_{p})=K(B^{n}_{p})\subseteq E^{*}_{0}(B^{n}_{p};1/2)=S^{*}(B^{n}_{p% };1/2)\subseteq S^{*}(B^{n}_{p})$ (16)

and

$E_{1}(B^{n}_{p})\subseteq E_{1}^{*}(B^{n}_{p};1/2),\quad\text{for}\quad 1\leq p% <\infty,$ (17)

where $B^{n}_{p}$ is the unit ball in $\mathbb{C}^{n}$ with respect to the $p$ -norm and

$E_{1}(\mathbb{U}^{n})\subseteq E_{1}^{*}(\mathbb{U}^{n};1/2),$ (18)

where $\mathbb{U}^{n}$ is the unit polydisc in $\mathbb{C}^{n}$ .

Proof.

1.

First, according to Theorem 4 (for $n=1$ , $k=0$ , respectively $k=1$ ) we obtain immediately the inclusions from relations (14) and (15). Moreover, in view of Definition 3 and the Alexander’s duality theorem for the classes $S^{*}(\alpha)$ and $K(\alpha)$ , we have that

$E_{0}^{*}(U;1/2)=\big{\{}f\in S:G_{0}f\in S^{*}(1/2)\big{\}}=\big{\{}f\in S:f% \in S^{*}(1/2)\big{\}}=S^{*}(1/2)$

and

$E_{1}^{*}(U;1/2)=\big{\{}f\in S:G_{1}f\in S^{*}(1/2)\big{\}}=\big{\{}f\in S:f% \in K(1/2)\big{\}}=K(1/2),$

where $G_{1}f(\zeta)=\zeta f^{\prime}(\zeta)$ for all $\zeta\in U$ . Relation (14) is also proved in (GrKo1, , Theorem 2.3.2) and relation (15) follows from Theorem 3.
2.
Next, let us consider $n\geq{2}$ .
- •
  
  In view of Definitions 3 and 4, Remark 8 and Theorem 4, we obtain the inclusions and equalities from relation (16).
- •
  
  The inclusions from relations (17) and (18) are based on Theorem 4 for the unit ball $B^{n}_{p}$ in $\mathbb{C}^{n}$ with respect to the $p$ -norm, where $p\in[0,\infty)$ , respectively for the unit polydisc $\mathbb{U}^{n}$ in $\mathbb{C}^{n}$ , and this completes the proof.

∎

4 Extension Operators and Applications

In last part of this paper we consider two particular cases of the Graham-Kohr extension operator $\Psi_{n,\alpha}$ defined by relation (2) for $\alpha=0$ and $\alpha=1$ . Although the operator $\Psi_{n,\alpha}$ does not preserve convexity, we can still observe an interesting property related to the subclass $E^{*}_{1}$ in the particular cases mentioned above.

Proposition 7.

Let us consider the Graham-Kohr extension operator $\Psi_{2,0}$ defined in (2) for $n=2$ and $\alpha=0$ . Then

\Psi_{2,0}(K)=\Psi_{2,0}\big{(}E_{1}^{*}(U)\big{)}\subseteq E_{1}^{*}(\mathbb{% B}^{2})\not=K(\mathbb{B}^{2}).

(19)

Proof.

Let us consider the case $n=2$ and $f\in K$ . Also, let $F:\mathbb{B}^{2}\rightarrow\mathbb{C}^{2}$ be given by

F(z)=\Psi_{2,0}(f)(z)=\big{(}f(z_{1}),z_{2}\big{)},

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ . We want to prove that

F\in E_{1}^{*}(\mathbb{B}^{2})\quad\Leftrightarrow\quad F\in S(\mathbb{B}^{2})% \quad\text{and}\quad G_{1}F\in S^{*}(\mathbb{B}^{2}).

1.

In (GrHaKoSuf1, , Theorem 2.1) or (GrKo2, , Theorem 3.2) the authors proved that $F\in S^{0}(\mathbb{B}^{2})$ , so according to this result it is clear that $F\in S(\mathbb{B}^{2})$ .
2.

For the second part, let us consider the mapping $G:\mathbb{B}^{2}\rightarrow\mathbb{C}^{2}$ given by

$G(z)=DF(z)(z)=\begin{pmatrix}f^{\prime}(z_{1})&0\\ \\ 0&1\end{pmatrix}\begin{pmatrix}z_{1}\\ z_{2}\end{pmatrix}=\big{(}z_{1}f^{\prime}(z_{1}),z_{2}\big{)},$

for all $z\in\mathbb{B}^{2}$ . To prove that $G$ is starlike on $\mathbb{B}^{2}$ it is enough to observe that

$G(z)=\big{(}G_{1}(z_{1}),G_{2}(z_{2})\big{)},$

where $G_{j}(z_{j})\in S^{*}$ for $j\in\{1,2\}$ . Indeed, according to Alexander’s duality theorem we have that $G_{1}\in S^{*}$ if and only if $f\in K$ . Moreover, it is clear that $G_{2}$ is a starlike function on the unit disc. Hence, $G=G_{1}F\in S^{*}(\mathbb{B}^{2})$ and this completes the proof.

∎

Remark 14.

Notice that we proved the previous result in the case $n=2$ . However, the arguments in the case $n\geq{2}$ are similar.

Another proof of Proposition 7 can be given using a surprising property of the mapping $G_{k}$ . It seems that the relationship between $G_{k}$ and $\Psi_{n,\alpha}$ is commutative (in the sense presented in the following lemma).

Lemma 8.

Let us consider $n=2$ , $\alpha=0$ and $k_{0}\in\{0,1,2\}$ . Then

\Psi_{2,0}(G_{k_{0}}f)=G_{k_{0}}(\Psi_{2,0}(f)).

(20)

Proof.

1.

Let $k_{0}=0$ . In view of Definition 2 it is clear that for $k_{0}=0$ we obtain

$\Psi_{2,0}(G_{0}f)=\Psi_{2,0}(f)=G_{0}(\Psi_{2,0}(f)).$
2.

Let $k_{0}=1$ . Then

$\Psi_{2,0}(G_{1}f)(z)=\big{(}G_{1}f(z_{1}),z_{2}\big{)}=\big{(}z_{1}f^{\prime}% (z_{1}),z_{2}\big{)},$

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ and

$G_{1}\big{(}\Psi_{2,0}(f)\big{)}(z)=D\Psi_{2,0}(f)(z)=\begin{pmatrix}f^{\prime% }(z_{1})&0\\ \\ 0&1\end{pmatrix}\begin{pmatrix}z_{1}\\ z_{2}\end{pmatrix}=\big{(}z_{1}f^{\prime}(z_{1}),z_{2}\big{)},$

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ . Hence,

$\Psi_{2,0}(G_{1}f)=G_{1}(\Psi_{2,0}(f)).$
3.

For the last part of the proof, let us consider $k_{0}=2$ . Then

$\Psi_{2,0}(G_{2}f)(z)=\big{(}G_{2}f(z_{1}),z_{2}\big{)}=\big{(}z_{1}^{2}f^{% \prime\prime}(z_{1})+z_{1},z_{2}\big{)},$

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ . On the other hand,

$\displaystyle G_{2}\big{(}\Psi_{2,0}(f)\big{)}(z)$ $\displaystyle=D^{2}\Psi_{2,0}(f)(z)(z^{2})+z=\begin{pmatrix}z_{1}f^{\prime% \prime}(z_{1})&0\\ \\ 0&0\end{pmatrix}\begin{pmatrix}z_{1}\\ z_{2}\end{pmatrix}+z$

$\displaystyle=\big{(}z_{1}^{2}f^{\prime\prime}(z_{1})+z_{1},z_{2}\big{)},$

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ and then

$\Psi_{2,0}(G_{2}f)=G_{2}(\Psi_{2,0}(f)).$

Hence, we proved the commutative property of the mapping $G_{k}$ for the particular cases $k\in\{0,1,2\}$ and this completes the proof. ∎

Remark 15.

According to the previous proof we deduce that Lemma 8 is true also in the case $n\geq{2}$ , $\alpha=0$ and $k_{0}\in\{0,1,2\}$ .

Remark 16.

Using Lemma 8 we can prove Proposition 7 for the case $n\geq{2}$ in the following alternative way:

f\in K\Leftrightarrow f\in E^{*}_{1}(U)\Leftrightarrow f\in S\quad\text{and}% \quad G_{1}f\in S^{*}

which implies that

\Psi_{n,0}(f)\in S(\mathbb{B}^{n})\quad\text{and}\quad\Psi_{n,0}(G_{1}f)\in S^% {*}(\mathbb{B}^{n}),

where the last implication follows from the property that $\Psi_{n,0}$ preserves the starlikeness (see Remark 7). But, $\Psi_{n,0}(G_{1}f)=G_{1}(\Psi_{n,0}(f))$ . Then

\Psi_{n,0}(f)\in S(\mathbb{B}^{n})\quad\text{and}\quad G_{1}(\Psi_{n,0}(f))\in S% ^{*}(\mathbb{B}^{n})\Leftrightarrow\Psi_{n,0}(f)\in E^{*}_{1}(\mathbb{B}^{n}),

in view of Definition 3. Hence,

f\in K=E^{*}_{1}(U)\Rightarrow\Psi_{n,0}(f)\in E^{*}_{1}(\mathbb{B}^{n})

and this completes the proof.

Corollary 9.

Let us consider the Graham-Kohr extension operator $\Psi_{n,0}$ defined in (2) for $\alpha=0$ . Then, for any $k\in\{0,1,2\}$ , we have the embedding

\Psi_{n,0}\big{(}E_{k}^{*}(U)\big{)}\subseteq E_{k}^{*}(\mathbb{B}^{n}).

Theorem 10.

Let us consider the Graham-Kohr extension operator $\Psi_{2,1}$ defined in (2) for $n=2$ and $\alpha=1$ . Then

\Psi_{2,1}(K)=\Psi_{2,1}\big{(}E_{1}^{*}(U)\big{)}\subseteq E_{1}^{*}(\mathbb{% B}^{2})\not=K(\mathbb{B}^{2}).

(21)

Proof.

Let us consider the case $n=2$ and $f\in K$ . Let $F:\mathbb{B}^{2}\rightarrow\mathbb{C}^{2}$ be given by

F(z)=\Psi_{2,1}(f)(z)=\bigg{(}f(z_{1}),z_{2}\dfrac{f(z_{1})}{z_{1}}\bigg{)},

(22)

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ . We want to prove that

F\in E_{1}^{*}(\mathbb{B}^{2})\quad\Leftrightarrow\quad F\in S(\mathbb{B}^{2})% \quad\text{and}\quad G_{1}F\in S^{*}(\mathbb{B}^{2}).

1.

In (GrHaKoSuf1, , Theorem 2.1) or (GrKo2, , Theorem 3.2) the authors proved that $F\in S^{0}(\mathbb{B}^{2})$ , so according to this result it is clear that $F\in S(\mathbb{B}^{2})$ .
2.

In order to show that $G_{1}F\in S^{*}(\mathbb{B}^{2})$ , let us consider the mapping $G:\mathbb{B}^{2}\rightarrow\mathbb{C}^{2}$ given by

$G(z)=DF(z)(z)=\begin{pmatrix}f^{\prime}(z_{1})&0\\ \\ \dfrac{z_{2}}{z_{1}}\bigg{[}f^{\prime}(z_{1})-\dfrac{f(z_{1})}{z_{1}}\bigg{]}&% \dfrac{f(z_{1})}{z_{1}}\end{pmatrix}\begin{pmatrix}z_{1}\\ z_{2}\end{pmatrix},$

where $F$ is given by (22) and $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ . Hence

$G(z)=\big{(}z_{1}f^{\prime}(z_{1}),z_{2}f^{\prime}(z_{1})\big{)},$ (23)

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ . Next, to prove that $G_{1}F=G\in S^{*}(\mathbb{B}^{2})$ , we can use the analytical characterization of starlikeness from Theorem 1.

It is clear that $G\in H(\mathbb{B}^{2})$ , $G(0)=0$ and $DG(0)=I_{2}$ , where

DG(z)=\begin{pmatrix}f^{\prime}(z_{1})+z_{1}f^{\prime\prime}(z_{1})&0\\ \\ z_{2}f^{\prime\prime}(z_{1})&f^{\prime}(z_{1})\end{pmatrix},

for $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ . Also $G$ is locally biholomorphic on $\mathbb{B}^{2}$ since

J_{G}(z)=\det[DG(z)]=f^{\prime}(z_{1})\cdot\big{[}f^{\prime}(z_{1})+z_{1}f^{% \prime\prime}(z_{1})\big{]}\not=0,\quad z_{1}\in U

in view of the fact that $f\in K$ . Moreover, we have that

	$\displaystyle[DG(z)]^{-1}$	$\displaystyle=\dfrac{1}{f^{\prime}(z_{1})[f^{\prime}(z_{1})+z_{1}f^{\prime% \prime}(z_{1})]}\begin{pmatrix}f^{\prime}(z_{1})&0\\ \\ -z_{2}f^{\prime\prime}(z_{1})&f^{\prime}(z_{1})+z_{1}f^{\prime\prime}(z_{1})% \end{pmatrix}$
		$\displaystyle=\begin{pmatrix}\dfrac{1}{f^{\prime}(z_{1})+z_{1}f^{\prime\prime}% (z_{1})}&0\\ \\ -\dfrac{z_{2}f^{\prime\prime}(z_{1})}{f^{\prime}(z_{1})[f^{\prime}(z_{1})+z_{1% }f^{\prime\prime}(z_{1})]}&\dfrac{1}{f^{\prime}(z_{1})}\end{pmatrix},\quad z% \in\mathbb{B}^{2}.$

Thus,

$\displaystyle[DG(z)]^{-1}G(z)$	$\displaystyle=\bigg{(}\dfrac{z_{1}f^{\prime}(z_{1})}{f^{\prime}(z_{1})+z_{1}f^% {\prime\prime}(z_{1})},z_{2}-\dfrac{z_{1}z_{2}f^{\prime\prime}(z_{1})}{f^{% \prime}(z_{1})+z_{1}f^{\prime\prime}(z_{1})}\bigg{)}$
	$\displaystyle=\bigg{(}\dfrac{z_{1}f^{\prime}(z_{1})}{f^{\prime}(z_{1})+z_{1}f^% {\prime\prime}(z_{1})},\dfrac{z_{2}f^{\prime}(z_{1})+z_{1}z_{2}f^{\prime\prime% }(z_{1})-z_{1}z_{2}f^{\prime\prime}(z_{1})}{f^{\prime}(z_{1})+z_{1}f^{\prime% \prime}(z_{1})}\bigg{)}$
	$\displaystyle=\bigg{(}\dfrac{z_{1}f^{\prime}(z_{1})}{f^{\prime}(z_{1})+z_{1}f^% {\prime\prime}(z_{1})},\dfrac{z_{2}f^{\prime}(z_{1})}{f^{\prime}(z_{1})+z_{1}f% ^{\prime\prime}(z_{1})}\bigg{)},\quad z\in\mathbb{B}^{2}.$	(24)

Then

	$\displaystyle\text{Re}\big{\langle}[DG(z)]^{-1}G(z),z\big{\rangle}$	$\displaystyle=\text{Re}\bigg{\langle}\bigg{(}\dfrac{z_{1}f^{\prime}(z_{1})}{f^% {\prime}(z_{1})+z_{1}f^{\prime\prime}(z_{1})},\dfrac{z_{2}f^{\prime}(z_{1})}{f% ^{\prime}(z_{1})+z_{1}f^{\prime\prime}(z_{1})}\bigg{)},(z_{1},z_{2})\bigg{\rangle}$
		$\displaystyle=\text{Re}\bigg{[}\dfrac{z_{1}\overline{z}_{1}f^{\prime}(z_{1})}{% f^{\prime}(z_{1})+z_{1}f^{\prime\prime}(z_{1})}+\dfrac{z_{2}\overline{z}_{2}f^% {\prime}(z_{1})}{f^{\prime}(z_{1})+z_{1}f^{\prime\prime}(z_{1})}\bigg{]}$
		$\displaystyle=\|z_{1}\|^{2}\cdot\text{Re}\bigg{[}\dfrac{f^{\prime}(z_{1})}{f^{% \prime}(z_{1})+z_{1}f^{\prime\prime}(z_{1})}\bigg{]}+\|z_{2}\|^{2}\cdot\text{Re}% \bigg{[}\dfrac{f^{\prime}(z_{1})}{f^{\prime}(z_{1})+z_{1}f^{\prime\prime}(z_{1% })}\bigg{]}$
		$\displaystyle=\\|z\\|^{2}\cdot\text{Re}\bigg{[}\dfrac{f^{\prime}(z_{1})}{f^{% \prime}(z_{1})+z_{1}f^{\prime\prime}(z_{1})}\bigg{]},$

for all $z\in\mathbb{B}^{2}\setminus\{0\}$ . Hence,

\text{Re}\big{\langle}[DG(z)]^{-1}G(z),z\big{\rangle}=\|z\|^{2}\cdot\text{Re}% \bigg{[}\dfrac{1}{1+z_{1}f^{\prime\prime}(z_{1})/f^{\prime}(z_{1})}\bigg{]}>0,

for all $z\in\mathbb{B}^{2}\setminus\{0\}$ according to the fact that $f\in K$ . Using the characterization of starlikeness in $\mathbb{B}^{n}$ (see Theorem 1), we obtain that $G\in S^{*}(\mathbb{B}^{2})$ , i.e. $G_{1}F\in S^{*}(\mathbb{B}^{2})$ . Hence, $F=\Psi_{2,1}(f)\in E^{*}_{1}(\mathbb{B}^{2})$ and this completes the proof. ∎

Remark 17.

Notice that Theorem 10 is true also in the case $n\geq{2}$ , but for the sake of brevity we give the proof only in the case $n=2$ . A general alternative proof of Theorem 10 can be given using the following arguments:

1.

Let $f\in K$ and $F=\Psi_{n,1}(f)$ . Then $F\in S(\mathbb{B}^{n})$ according to (GrKo2, , Theorem 3.2).
2.

If we define $h(w)=wf^{\prime}(w)$ , for all $w\in U$ , then $f\in K$ implies that $h\in S^{*}$ (in view of Alexander’s theorem).
3.

Let $G(z)=DF(z)(z)$ , for $z\in\mathbb{B}^{n}$ . According to relation (23) and the previous step, we have that

$G(z)=\big{(}z_{1}f^{\prime}(z_{1}),z^{*}f^{\prime}(z_{1})\big{)}=\bigg{(}h(z_{% 1}),z^{*}\dfrac{h(z_{1})}{z_{1}}\bigg{)},\quad z=(z_{1},z^{*})\in\mathbb{B}^{n}.$
4.

Now it is clear that $G(z)=\Psi_{n,1}(h)(z)$ .
5.

I. Graham and G. Kohr proved in (GrKo2, , Corollary 3.3) that $h\in S^{*}$ implies $\Psi_{n,1}(h)=G\in S^{*}(\mathbb{B}^{n})$ .
6.

Hence, $F\in S(\mathbb{B}^{n})$ and $G=G_{1}F\in S^{*}(\mathbb{B}^{n})$ . In view of Definition 3, we obtain that $F\in E_{1}^{*}(\mathbb{B}^{n})$ .

Lemma 11.

Let us consider $n=2$ , $\alpha=1$ and $k_{0}\in\{0,1,2\}$ . Then

\Psi_{2,1}(G_{k_{0}}f)=G_{k_{0}}(\Psi_{2,1}(f)).

(25)

Proof.

1.

Let $k_{0}=0$ . In view of Definition 2 it is clear that for $k_{0}=0$ we obtain

$\Psi_{2,1}(G_{0}f)=\Psi_{2,1}(f)=G_{0}(\Psi_{2,1}(f)).$

Let $k_{0}=1$ . Then

\Psi_{2,1}(G_{1}f)(z)=\bigg{(}G_{1}f(z_{1}),\dfrac{G_{1}f(z_{1})}{z_{1}}z_{2}% \bigg{)}=\big{(}z_{1}f^{\prime}(z_{1}),z_{2}f^{\prime}(z_{1})\big{)},

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ and

	$\displaystyle G_{1}\big{(}\Psi_{2,1}(f)\big{)}(z)$	$\displaystyle=D\Psi_{2,1}(f)(z)(z)=\begin{pmatrix}f^{\prime}(z_{1})&0\\ \\ z_{2}\dfrac{z_{1}f^{\prime}(z_{1})-f(z_{1})}{z_{1}^{2}}&\dfrac{f(z_{1})}{z_{1}% }\end{pmatrix}\begin{pmatrix}z_{1}\\ z_{2}\end{pmatrix}$
		$\displaystyle=\bigg{(}z_{1}f^{\prime}(z_{1}),\dfrac{z_{2}}{z_{1}}\big{[}z_{1}f% ^{\prime}(z_{1})-f(z_{1})\big{]}+\dfrac{z_{2}}{z_{1}}f(z_{1})\bigg{)}$
		$\displaystyle=\big{(}z_{1}f^{\prime}(z_{1}),z_{2}f^{\prime}(z_{1})\big{)},$

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ . Hence,

\Psi_{2,1}(G_{1}f)=G_{1}(\Psi_{2,1}(f)).

3.

For the last part of the proof, let us consider $k_{0}=2$ . Then

$\displaystyle\Psi_{2,1}(G_{2}f)(z)$ $\displaystyle=\bigg{(}G_{2}f(z_{1}),z_{2}\dfrac{G_{2}f(z_{1})}{z_{1}}\bigg{)}$

$\displaystyle=\big{(}z_{1}^{2}f^{\prime\prime}(z_{1})+z_{1},z_{2}z_{1}f^{% \prime\prime}(z_{1})+z_{2}\big{)}$

$\displaystyle=\big{[}z_{1}f^{\prime\prime}(z_{1})+1\big{]}\cdot(z_{1},z_{2}),$

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ . On the other hand,

$\displaystyle G_{2}\big{(}\Psi_{2,1}(f)\big{)}(z)$ $\displaystyle=D^{2}\Psi_{2,1}(f)(z)(z^{2})+z$

$\displaystyle=\big{(}z_{1}^{2}f^{\prime\prime}(z_{1})+z_{1},z_{2}z_{1}f^{% \prime\prime}(z_{1})+z_{2}\big{)}$

$\displaystyle=\big{[}z_{1}f^{\prime\prime}(z_{1})+1\big{]}\cdot(z_{1},z_{2}),$

where

$D^{2}\Psi_{2,1}(f)(z)(z^{2})=\begin{pmatrix}z_{1}f^{\prime\prime}(z_{1})&0\\ \\ z_{2}f^{\prime\prime}(z_{1})-\dfrac{z_{2}}{z_{1}}f^{\prime}(z_{1})+\dfrac{z_{2% }}{z_{1}^{2}}f(z_{1})&f^{\prime}(z_{1})-\dfrac{f(z_{1})}{z_{1}}\end{pmatrix}% \begin{pmatrix}z_{1}\\ z_{2}\end{pmatrix},$

for all $z=(z_{1},z_{2})\in\mathbb{B}^{2}$ . Thus,

$\Psi_{2,1}(G_{2}f)=G_{2}(\Psi_{2,1}(f)).$

Hence, we proved the commutative property given by (25) for the particular cases $k\in\{0,1,2\}$ and this completes the proof. ∎

Remark 18.

Notice that, for the sake of brevity, we considered in the previous lemma the case $n=2$ , but the arguments in the case $n\geq{2}$ are similar.

Remark 19.

Using Lemma 11 in the general case $n\geq{2}$ , Remark 7 and Definition 3, we can prove Theorem 10 in the following alternative way:

f\in K\Leftrightarrow f\in E^{*}_{1}(U)\Leftrightarrow f\in S\quad\text{and}% \quad G_{1}f\in S^{*}

which implies that

\Psi_{n,1}(f)\in S(\mathbb{B}^{n})\quad\text{and}\quad\Psi_{n,1}(G_{1}f)\in S^% {*}(\mathbb{B}^{n}).

But, $\Psi_{n,1}(G_{1}f)=G_{1}(\Psi_{n,1}(f))$ . Then

\Psi_{n,1}(f)\in S(\mathbb{B}^{n})\quad\text{and}\quad G_{1}(\Psi_{n,1}(f))\in S% ^{*}(\mathbb{B}^{n})\Leftrightarrow\Psi_{n,1}(f)\in E^{*}_{1}(\mathbb{B}^{n}).

Hence,

f\in K=E^{*}_{1}(U)\Rightarrow\Psi_{n,1}(f)\in E^{*}_{1}(\mathbb{B}^{n})

and this completes the proof.

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New Subclasses of Univalent Mappings in Several Complex Variables: Extension Operators and Applications

5 months ago

Abstract In this paper we define new subclasses of univalent mappings in the case of several complex variables. We will…

On some convex combinations of biholomorphic mappings in several complex variables

1 year ago

Abstract In this paper, our interest is devoted to study the convex combinations of the form $(1-\lambda)f+\lambda g$ where \(\lambda…

	$\displaystyle G_{2}\big{(}\Psi_{2,0}(f)\big{)}(z)$	$\displaystyle=D^{2}\Psi_{2,0}(f)(z)(z^{2})+z=\begin{pmatrix}z_{1}f^{\prime% \prime}(z_{1})&0\\ \\ 0&0\end{pmatrix}\begin{pmatrix}z_{1}\\ z_{2}\end{pmatrix}+z$
		$\displaystyle=\big{(}z_{1}^{2}f^{\prime\prime}(z_{1})+z_{1},z_{2}\big{)},$

	$\displaystyle\Psi_{2,1}(G_{2}f)(z)$	$\displaystyle=\bigg{(}G_{2}f(z_{1}),z_{2}\dfrac{G_{2}f(z_{1})}{z_{1}}\bigg{)}$
		$\displaystyle=\big{(}z_{1}^{2}f^{\prime\prime}(z_{1})+z_{1},z_{2}z_{1}f^{% \prime\prime}(z_{1})+z_{2}\big{)}$
		$\displaystyle=\big{[}z_{1}f^{\prime\prime}(z_{1})+1\big{]}\cdot(z_{1},z_{2}),$

	$\displaystyle G_{2}\big{(}\Psi_{2,1}(f)\big{)}(z)$	$\displaystyle=D^{2}\Psi_{2,1}(f)(z)(z^{2})+z$
		$\displaystyle=\big{(}z_{1}^{2}f^{\prime\prime}(z_{1})+z_{1},z_{2}z_{1}f^{% \prime\prime}(z_{1})+z_{2}\big{)}$
		$\displaystyle=\big{[}z_{1}f^{\prime\prime}(z_{1})+1\big{]}\cdot(z_{1},z_{2}),$

Posts by Eduard Grigoriciuc

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New Subclasses of Univalent Mappings in Several Complex Variables. Extension Operators and Applications

Abstract

keywords:

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1 Introduction

Remark 1.

Remark 2.

Remark 3.

Theorem 1.

Definition 1.

Remark 4.

Remark 5.

Remark 6.

Remark 7.

2 New Subclasses of Univalent Mappings

Definition 2.

Definition 3.

Remark 8.

Remark 9.

Proof.

Remark 10.

Example 1.

Proof.

Remark 11.

Example 2.

Proof.

Example 3.

Example 4.

3 General Properties

Theorem 2.

Proof.

Remark 12.

Example 5.

Theorem 3.

Proof.

Remark 13.

Definition 4.

Theorem 4.

Proof.

Corollary 5.

Corollary 6.

Proof.

4 Extension Operators and Applications

Proposition 7.

Proof.

Remark 14.

Lemma 8.

Proof.

Remark 15.

Remark 16.

Corollary 9.

Theorem 10.

Proof.

Remark 17.

Lemma 11.

Proof.

Remark 18.

Remark 19.

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