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

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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Computational Methods and Function Theory

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Springer

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1617-9447
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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

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 E1, and observe that in the case of one complex variable, E1(U) coincides with the class of convex functions K on the unit disc. However, if n2, then E1(𝔹n) is different from the class of convex mappings K(𝔹n) on the Euclidean unit ball 𝔹n in n. Along with this, we will study other properties of the class E1 on the unit polydisc, respectively on the Euclidean unit ball in n. In the second part of the paper we discuss about Graham-Kohr extension operator Ψn,α (defined by I. Graham and G. Kohr in Complex Variables Theory Appl. 47 (2002), 59–72). They proved that the extension operator Ψn,α does not preserve convexity for n2 for all α[0,1]. However, in this paper we prove that Ψn,0(K) and Ψn,1(K) are subsets of the class E1(𝔹n) which is different from the class K(𝔹n) for the Euclidean case.

keywords:
Biholomorphic mapping, Convex mapping, Starlike mapping, Extension operator
pacs:
[

MSC Classification]32H02, 30C45

1 Introduction

Let n denote the space of n complex variables z=(z1,,zn) equipped with an arbitrary norm . Let Bn={zn:z<1} be the unit ball in n with respect to the arbitrary norm . In the case of one complex variable, B1 is denoted by U.

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

A mapping fS(Bn) is called convex (starlike) if its image is a convex (starlike with respect to the origin) set in n. The class of normalized convex mappings is denoted by

K(Bn)={fS(Bn):f(Bn) is a convex set in n}

and the class of normalized starlike mappings is denoted by

S(Bn)={fS(Bn):f(Bn) is a starlike set with respect to zero in n}.

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 α on the unit disc U, where α[0,1). Hence, we denote by

K(α)={fS:Re[1+ζf′′(ζ)f(ζ)]>α,ζU}

the class of convex functions of order α on U and by

S(α)={fS:Re[ζf(ζ)f(ζ)]>α,ζU}

the class of starlike functions of order α 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(Bn;α) denotes the class of all normalized biholomorphic convex mappings of order α on Bn and S(Bn;α) denotes the class of all normalized biholomorphic starlike mappings of order α on Bn.

Remark 1.

Notice that, throughout this paper, we denote by +={0} the set of non-negative integers and by +=={1,2,3,} the set of natural numbers (positive integers).

Remark 2.

During this paper we are working on different domains (unit balls in n with respect to different norms), as follows:

  • 𝔹n – the Euclidean unit ball in n with respect to the Euclidean norm z=j=1n|zj|2, for all z=(z1,,zn)n.

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

  • 𝕌n – the unit polydisc in n with respect to the maximum norm z=max{|zj|:j=1,n¯}, for all z=(z1,,zn)n.

  • Bpn – the unit ball in n with respect to the p-norm zp=[j=1n|zj|p]1/p, for all z=(z1,,zn)n and p[1,).

In the case of one complex variable, each of the sets 𝔹1, 𝕌1 and Bp1 coincides with U. Mention that when we work with an arbitrary norm, it will be denoted . 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 fK if and only if ζf(ζ)S, ζU (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:𝔹nn be a locally biholomorphic mapping such that f(0)=0. Then f is starlike if and only if

Re[Df(z)]1f(z),z>0,z𝔹n{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 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):𝔹n×[0,)n is called a Loewner chain (normalized univalent subordination chain) if the following conditions hold:

  1. 1.

    etL(,t)S(𝔹n), for all t[0,);

  2. 2.

    L(𝔹n,s)L(𝔹n,t), for all 0st<.

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 fS there exists a Loewner chain L=L(ζ,t) such that f(ζ)=L(ζ,0), for all ζU.

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 S0(𝔹n) of normalized univalent mappings which have parametric representation as follows

S0(𝔹n)={fS(𝔹n):L(z,t) a Loewner chain such that the family

{etL(,t)}t0 is a normal family on 𝔹n and f=L(,0)}

Remark 5.

Recall that a family is called a normal family if each sequence in 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 S0(𝔹1)=S. However, if n2, then S0(𝔹n)S(𝔹n) (for details, one may consult GrHaKo1 , GrKo1 or GrKoKo1 ). Another important results regarding to the family of normalized univalent mappings which have parametric representation S0(𝔹n) can be found in GrHaKoKo1 , GrKoKo1 and GrKoPf1 .

The problem of constructing examples of starlike (respectively convex) mappings in 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 n. For more details about extension operators and properties of them one may consult GrHaKoSuf1 , GrKo3 and GrKoPf1 .

Let α[0,1]. In 2002, I. Graham and G. Kohr (see GrKo2 ) introduced the extension operator Ψn,α defined for normalized locally univalent functions on U by

Ψn,α(f)(z)=(f(z1),[f(z1)z1]αz),z=(z1,z)𝔹n, (2)

where f(z1)0, for z1U{0} and the branch of the power function is chosen such that (f(z1)z1)α|z1=0=1.

Remark 7.

In (GrKo2, , Corollary 3.3) Graham and Kohr proved that Ψn,α(S)S(𝔹n), for all α[0,1]. Also in (GrKo2, , Remark 3.6) the authors shows that Ψn,α does not preserve convexity for n2, for all α[0,1].

In the second section of our paper, we introduce two new subclasses Ek and Ek of univalent mappings of one and several complex variables. We will give the definitions in a general setting (on the unit ball Bn in n with respect to an arbitrary norm) and some particular examples on the Euclidean unit ball 𝔹n.

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, E1(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 Ψn,α defined in (2). It is interesting that although the operator Ψn,α 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 Bn in n with respect to an arbitrary norm ), but a few properties will be presented on some particular domains, especially, on the Euclidean unit ball 𝔹n.

Definition 2.

Let k+ and let f:Bnn be a holomorphic mapping such that f(0)=0 and Df(0)=In. Also, let f(z)=z+m=2Pm(z) be the power series expansion of f on the unit ball Bn, where

Pm(z)=1m!Dmf(0)(zm),zBn.

Then we define

(Gkf)(z)={Dkf(z)(zk)+z+m=2k1Pm(z),k3D2f(z)(z2)+z,k=2Df(z)(z),k=1f(z),k=0, (3)

for all zBn.

Definition 3.

Let k+. We define the following subclasses of normalized univalent mappings on Bn:

Ek(Bn)={fS(Bn):GkfS(Bn)} (4)

and

Ek(Bn)={fS(Bn):GkfK(Bn)}. (5)
Remark 8.

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

  1. 1.

    If k=0, then

    E0(Bn)=S(Bn)andE0(Bn)=K(Bn).
  2. 2.

    If k=1, then

    E1(Bn)={fS(Bn):G1fS(Bn)}

    and

    E1(Bn)={fS(Bn):G1fK(Bn)},

where G1f(z)=Df(z)(z), for zBn.

An important remark on the previous subclasses of biholomorphic mappings in 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 n2, then

K(B1n)E1(B1n)andK(𝕌n)E1(𝕌n), (6)

where B1n is the unit ball in n with respect to the 1-norm, respectively 𝕌n is the unit polydisc in n.

Proof.

For the sake of brevity, let us consider n=2. Notice that the arguments for the case n2 are similar.

  1. 1.

    First, let us consider the case of the unit ball B12 with respect to the 1-norm. According to (GrKo1, , Theorem 6.3.11.a) and (GrKo1, , Remark 6.3.15), we have that if fK(B12), then Df(z)z is starlike on B12. Hence, G1fS(B12) and in view of Definition 3 we obtain that fE1(B12).

    In order to prove that the inclusion K(B12)E1(B12) is strict, it is enough to consider the mapping f:B122 given by

    f(z)=(z11z1,z2),z=(z1,z2)B12.

    Then Df(z)(z)=(z1/(1z1)2,z2) is starlike on B12 (in view of (GrKo1, , Problem 6.2.5)) and hence, fE1(B12). However, f is not convex on B12 (see (GrKo1, , Problem 6.3.2)). Then, fE1(B12)K(B12) and we conclude that

    K(B12)E1(B12).
  2. 2.

    On the other hand, in view of (GrKo1, , Theorem 6.3.11.b) and (GrKo1, , Remark 6.3.15), we have that if fK(𝕌2), then Df(z)z is starlike on 𝕌2. Hence, G1fS(𝕌2) and in view of Definition 3 we obtain that fE1(𝕌2).

    Again, to prove that the inclusion K(𝕌2)E1(𝕌2) is strict, we can consider the mapping f:𝕌22 given by

    f(z)=(z1+a2z22,z2),z=(z1,z2)𝕌2.

    Then Df(z)(z)=(z1+az22,z2) is starlike on 𝕌2 for |a|1 (see (Suf1, , Example 3)). In view of Definition 3, we deduce that fE1(𝕌2). However, f is not convex for a0 (see (Suf1, , Example 3)). Hence, there exists a mapping fE1(𝕌2)K(𝕌2). Then

    K(𝕌2)E1(𝕌2)

    and this completes the proof.

Remark 10.

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

E1(𝔹n)K(𝔹n),n2, (7)

where 𝔹n is the Euclidean unit ball in n. Hence, it is not trivial to define the subclasses Ek and Ek of univalent mappings in the case of several complex variables. Even the simplest case k=1 is interesting for the Euclidean unit ball 𝔹n in view of the difference provided by relation (7).

Next we present two examples of mappings which belongs to the class E1(Bpn) for the general case of the unit ball Bpn 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:Bp222 be given by

f(z)=(z1+az22,z2),z=(z1,z2)Bp2.

Then fE1(Bp2) if and only if

|a|12(p214)1/p(p+1p1),p>1. (8)
Proof.

We have that f(0)=0 and Df(0)=In, where

Df(z)=(12az201),z=(z1,z2)Bp2.

Moreover,

g(z)=Df(z)(z)=(z1+2az22,z2),z=(z1,z2)Bp2.

In view of (RoSuf1, , Example 5) we know that g=G1fS(Bp2) if and only if

|2a|(p214)1/p(p+1p1),p>1.

Hence, fE1(Bp2) 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 𝔹2. Indeed, for the particular case p=2, we obtain that fE1(𝔹2) if and only if |a|334.

Example 2.

Let f:Bp222 be given by

f(z)=(z1+az1z2,z2),z=(z1,z2)Bp2.

Then fE1(Bp2) if and only if |a|1/2, for all 1p.

Proof.

We have that f(0)=0 and Df(0)=In, where

Df(z)=(1+az2az101),z=(z1,z2)Bp2.

Moreover,

g(z)=Df(z)(z)=(z1+2az1z2,z2),z=(z1,z2)Bp2.

In view of (RoSuf1, , Example 6) we know that g=G1fS(Bp2) if and only if |2a|1, for all 1p. Hence, fE1(Bp2) if and only if |a|12, and this completes the proof. ∎

Similarly, we can construct examples of mappings in the class E1(𝔹2) 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:𝔹22 be given by f(z)=(z1+az22,z2) for all z𝔹2. Then fE1(𝔹2) if and only if |a|14.

Example 4.

Let f:𝔹22 be given by f(z)=(z1+az1z2,z2) for all z𝔹2. Then fE1(𝔹2) if and only if |a|24.

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 E1 (respectively E1) on the unit disc U and the class of convex mappings K(Bpn) in 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 E1, the following statements are true:

  1. 1.

    If n=1, then E1(U)=K(U)=K.

  2. 2.

    If n2, then E1(𝔹n)K(𝔹n) and E1(𝔹n)K(𝔹n).

Proof.
  1. 1.

    Let us first consider the case n=1. Then we can rewrite the definition of the class E1 in the following way

    E1(𝔹1)=E1(U)={fS(𝔹1):G1fS(𝔹1)},

    where S(𝔹1)=S, S(𝔹1)=S and G1f(ζ)=ζf(ζ), for all ζ𝔹1=U. Hence,

    E1(U)={fS:Re[ζ(G1f)(ζ)G1f(ζ)]>0,ζU}

    which means that

    E1(U)={fS:Re[f(ζ)+ζf′′(ζ)f(ζ)]>0,ζU}.

    Now, it is clearly that

    E1(U)={fS:Re[1+ζf′′(ζ)f(ζ)]>0,ζU}=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

    E1(U)={fS:ζf(ζ)S,ζU}=K.
  2. 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, E1(𝔹n)K(𝔹n) and E1(𝔹n)K(𝔹n). For simplicity, let us consider the case n=2, but notice that for n2 the arguments are similar. Indeed, according to

    • Example 1 for p=2 and a=12, we can construct f:𝔹22 given by

      f(z)=(z1+az22,z2),z=(z1,z2)𝔹2.

      Because |a|=12<334 we obtain that fE1(𝔹2). Moreover, in view of (RoSuf1, , Example 7), we have that fK(𝔹2) because |a|12.

    Hence, we obtain that E1(𝔹n)K(𝔹n). On the other hand, in view of

    • Example 1 for p=2 and b=334, we can construct g:𝔹22 given by

      g(z)=(z1+bz22,z2),z=(z1,z2)𝔹2.

      Because |b|=334334 we obtain that gE1(𝔹2). But, according to (RoSuf1, , Example 7), we have that gK(𝔹2) because 334>12.

    Hence, we have that there exists gE1(𝔹2)K(𝔹2), i.e. g belongs to class E1(𝔹2), but g is not a convex mapping on 𝔹2. Finally, in view of

    • Example 2 for p=2 and c=22, we can construct h:𝔹22 given by

      h(z)=(z1+cz1z2,z2),z=(z1,z2)𝔹2.

      Because |c|=2212 we obtain that hK(𝔹2) (according to (RoSuf1, , Example 8)). However, hE1(𝔹2) because G1hS(𝔹2), where

      G1h(z)=Dh(z)(z)=(z1+2z1z2,z2),z=(z1,z2)𝔹2.

      Indeed, in view of (RoSuf1, , Example 6) it is easy to prove that G1fS(𝔹2).

    Hence, we obtain that hK(𝔹2)E1(𝔹2), i.e. h is a convex mapping on 𝔹2, but h does not belong to class E1(𝔹2). In view of all arguments presented above, we conclude that E1(𝔹n)K(𝔹n) and this completes the proof.

Remark 12.

According to Theorem 2, it is clear that if n=1, then K(U)=E1(U). However, if n2, then K(B1n)E1(B1n) and K(𝕌n)E1(𝕌n) in view of Remark 9 and K(𝔹n)E1(𝔹n) 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 α on the unit ball Bp2 (with respect to a p-norm, where p2). We will use this example given by Liu to prove that E1(𝔹n)K(𝔹n;1/2) and K(𝔹n;1/2)E1(𝔹n).

Example 5.

(see (Liu1, , Example 1)) Let α[0,1), p2 and k+={1,2,3,} such that k<pk+1. Also, let g:Bp22 be given by

g(z)=(z1+az2k+1,z2),z=(z1,z2)Bp2, (9)

where Bp2={z2:zp=(|z1|p+|z2|p)1/p}. If a satisfies the inequality |a|1αk(k+1), then gK(Bp2,α).

Theorem 3.

Regarding to the class E1, the following statements are true:

  1. 1.

    If n=1, then E1(U)K(1/2).

  2. 2.

    If n2, then E1(𝔹n)K(𝔹n;1/2) and K(𝔹n;1/2)E1(𝔹n), i.e. there exists also convex mappings of order 1/2 on 𝔹n which does not belong to class E1(𝔹n).

Proof.
  1. 1.

    Let us first consider the case n=1.

    • Let fE1(U). Then, in view of Definition 3, we have that G1fK. According to a result given by Sheil-Small and Suffridge (see (GrKo1, , Theorem 2.2.4)) we know that

      Re[2ζ(G1f)(ζ)(G1f)(ζ)(G1f)(ζ0)ζ+ζ0ζζ0]0,ζ,ζ0U.

      For ζ0=0, we obtain that

      Re[ζ(G1f)(ζ)(G1f)(ζ)]12,ζU.

      But, (G1f)(ζ)=ζf(ζ), for all ζU and using the minimum principle for harmonic functions we have that

      Re[ζ(G1f)(ζ)(G1f)(ζ)]=Re[ζ2f′′(ζ)+ζf(ζ)ζf(ζ)]=Re[1+ζf′′(ζ)f(ζ)]>12,ζU.

      In view of the definition of the class K(α) (see (Goo, , Chapter 9) or Rob ), we obtain that fK(1/2). Hence, E1(U)K(1/2).

    • However, we can prove that the previous inclusion is strict, i.e. there is a function fK(1/2) which does not belog to class E1(U). Indeed, if we consider f:U given by f(ζ)=ζ+aζ2 for all ζU with a=16, then according to (MeRoSco, , Theorem 2) (for α=1/2) we have that

      n=2(n12)n|an|=(212)2|a|=1212

      and then fK(1/2) by Alexander’s duality theorem. On the other hand, let us consider

      g(ζ)=G1f(ζ)=ζf(ζ)=ζ+2aζ2,ζU,a=1/6 (10)

      and h:U given by

      h(ζ)=1+ζg′′(ζ)g(ζ)=4ζ+32ζ+3,ζU. (11)

      Then h is a Möbius transformation on U such that h(3/2)=, h(0)=1, h(1)=75 and h(i)=1713+613i. Then

      h(U)=𝒰(1/5,6/5)={x+iy:(x0.2)2+y2<1.44},

      where 𝒰(1/5,6/5) is the open disc of center w0=1/5 and radius r=6/5, as we can observe in Figure 1.

      Refer to caption
      Figure 1: The open disc 𝒰(1/5,6/5) of center w0=1/5 and radius r=6/5

      Moreover, for every point w𝒰(1/5,6/5){x+iy:x<0}, we have that Rew<0, i.e. ζ0U such that Re[h(ζ0)]<0. For example, if ζ0=910, then ζ0U and simple computations show that

      Re[h(ζ0)]=Re[1+ζ0g′′(ζ0)g(ζ0)]=Re[4ζ0+32ζ0+3]=12<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=G1fK, and then fE1(U). Finally, we obtain that fK(1/2)E1(U).

  2. 2.

    For the second part of the proof, let us consider n=2. However, notice that for n2, the arguments are similar.

    • First, we can use the example given by Liu (see (Liu1, , Example 1)) in the particular case n=2 and α=12. For this, let us consider f:𝔹22 given by

      f(z)=(z1+az22,z2),z=(z1,z2)𝔹2.

      According to the result given by Liu and Example 3, we obtain that

      fE1(𝔹2)K(𝔹2;1/2)

      for |a|14. Hence,

      E1(𝔹n)K(𝔹n;1/2).
    • In order to prove that fK(𝔹n;1/2)E1(𝔹n), i.e. there is a convex mapping of order 1/2 on 𝔹n such that fE1(𝔹n), 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 α=12 and a=16. Also, let

      f(z)=z+az,u2u,z=(z1,z2)𝔹2,u=(1,0)2.

      Then, according to (LiuZhu, , Example 2), we have that fK(𝔹2;1/2). On the other hand, if

      g(z)=G1f(z)=Df(z)(z)=(z1+2az12,z2)=z+2az,u2u,

      for z𝔹2 and u=(1,0)2, then |2a|=13>14. In view of (LiuZhu, , Example 2), we deduce that g=G1fK(𝔹2) and hence fE1(𝔹2). Finally, we conclude that fK(𝔹n;1/2)E1(𝔹n) and this completes the proof.

Remark 13.

Let us consider f(z)=(z1+az2k+1,z2) with |a|12k(k+1) and k+. In view of the Example 5 (for α=1/2) we obtain that fK(Bp2,1/2). On the other hand,

Df(z)(z)=(z1+(k+1)az2k+1,z2),z=(z1,z2)Bp2

and

|(k+1)a|=(k+1)|a|(k+1)12k(k+1)=12k.

To prove that fE1(Bp2), we must obtain some conditions for G1f to be convex on Bp2. According to Example 5 (for α=0) we can impose condition 12k1k(k+1) which leads us to k1. Hence

fE1(Bp2)K(Bp2;1/2),

for k+ with k1. 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 ).

This remark together with Example 5 shows us that a function defined by relation (9) belongs to the both classes E1(Bpn) and K(Bpn;1/2) if k=1, p=2 and |a|14, as we saw in the proof of Theorem 3.

Definition 4.

Let k+ and α[0,1). In view of Definition 3, we define

Ek(Bn;α)={fS(Bn):GkfS(Bn;α)}, (12)

where S(Bn,α) is the family of starlike mappings of order α in 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 Ek and Ek.

Theorem 4.

Let k+. Then

Ek(Bn)Ek(Bn;1/2)Ek(Bn), (13)

where Bn is the unit ball of n with respect to an arbitrary norm .

Proof.

Let fEk(Bn). In view of Definition 3, we have that Gkf is a convex mapping on Bn. In view of (GrKo1, , Theorem 6.3.18 and Theorem 6.3.19) we obtain that Gkf is starlike of order 1/2 and this means that fEk(Bn;1/2). It is obvious that Ek(Bn;1/2)Ek(Bn) and this completes the proof. ∎

Corollary 5.

Let k+ and n=1. Then Ek(U)Ek(U;1/2)Ek(U).

Corollary 6.

Let us consider k{0,1}.

  1. 1.

    If n=1, then

    K=E0(U)E0(U;1/2)=S(1/2)S (14)

    and

    E1(U)E1(U;1/2)=K(1/2)K. (15)
  2. 2.

    On the other hand, if n2, then

    E0(Bpn)=K(Bpn)E0(Bpn;1/2)=S(Bpn;1/2)S(Bpn) (16)

    and

    E1(Bpn)E1(Bpn;1/2),for1p<, (17)

    where Bpn is the unit ball in n with respect to the p-norm and

    E1(𝕌n)E1(𝕌n;1/2), (18)

    where 𝕌n is the unit polydisc in n.

Proof.
  1. 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(α) and K(α), we have that

    E0(U;1/2)={fS:G0fS(1/2)}={fS:fS(1/2)}=S(1/2)

    and

    E1(U;1/2)={fS:G1fS(1/2)}={fS:fK(1/2)}=K(1/2),

    where G1f(ζ)=ζf(ζ) for all ζU. Relation (14) is also proved in (GrKo1, , Theorem 2.3.2) and relation (15) follows from Theorem 3.

  2. 2.

    Next, let us consider n2.

    • 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 Bpn in n with respect to the p-norm, where p[0,), respectively for the unit polydisc 𝕌n in 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 Ψn,α defined by relation (2) for α=0 and α=1. Although the operator Ψn,α does not preserve convexity, we can still observe an interesting property related to the subclass E1 in the particular cases mentioned above.

Proposition 7.

Let us consider the Graham-Kohr extension operator Ψ2,0 defined in (2) for n=2 and α=0. Then

Ψ2,0(K)=Ψ2,0(E1(U))E1(𝔹2)K(𝔹2). (19)
Proof.

Let us consider the case n=2 and fK. Also, let F:𝔹22 be given by

F(z)=Ψ2,0(f)(z)=(f(z1),z2),

for all z=(z1,z2)𝔹2. We want to prove that

FE1(𝔹2)FS(𝔹2)andG1FS(𝔹2).
  1. 1.

    In (GrHaKoSuf1, , Theorem 2.1) or (GrKo2, , Theorem 3.2) the authors proved that FS0(𝔹2), so according to this result it is clear that FS(𝔹2).

  2. 2.

    For the second part, let us consider the mapping G:𝔹22 given by

    G(z)=DF(z)(z)=(f(z1)001)(z1z2)=(z1f(z1),z2),

    for all z𝔹2. To prove that G is starlike on 𝔹2 it is enough to observe that

    G(z)=(G1(z1),G2(z2)),

    where Gj(zj)S for j{1,2}. Indeed, according to Alexander’s duality theorem we have that G1S if and only if fK. Moreover, it is clear that G2 is a starlike function on the unit disc. Hence, G=G1FS(𝔹2) 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 n2 are similar.

Another proof of Proposition 7 can be given using a surprising property of the mapping Gk. It seems that the relationship between Gk and Ψn,α is commutative (in the sense presented in the following lemma).

Lemma 8.

Let us consider n=2, α=0 and k0{0,1,2}. Then

Ψ2,0(Gk0f)=Gk0(Ψ2,0(f)). (20)
Proof.
  1. 1.

    Let k0=0. In view of Definition 2 it is clear that for k0=0 we obtain

    Ψ2,0(G0f)=Ψ2,0(f)=G0(Ψ2,0(f)).
  2. 2.

    Let k0=1. Then

    Ψ2,0(G1f)(z)=(G1f(z1),z2)=(z1f(z1),z2),

    for all z=(z1,z2)𝔹2 and

    G1(Ψ2,0(f))(z)=DΨ2,0(f)(z)=(f(z1)001)(z1z2)=(z1f(z1),z2),

    for all z=(z1,z2)𝔹2. Hence,

    Ψ2,0(G1f)=G1(Ψ2,0(f)).
  3. 3.

    For the last part of the proof, let us consider k0=2. Then

    Ψ2,0(G2f)(z)=(G2f(z1),z2)=(z12f′′(z1)+z1,z2),

    for all z=(z1,z2)𝔹2. On the other hand,

    G2(Ψ2,0(f))(z) =D2Ψ2,0(f)(z)(z2)+z=(z1f′′(z1)000)(z1z2)+z
    =(z12f′′(z1)+z1,z2),

    for all z=(z1,z2)𝔹2 and then

    Ψ2,0(G2f)=G2(Ψ2,0(f)).

Hence, we proved the commutative property of the mapping Gk for the particular cases k{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 n2, α=0 and k0{0,1,2}.

Remark 16.

Using Lemma 8 we can prove Proposition 7 for the case n2 in the following alternative way:

fKfE1(U)fSandG1fS

which implies that

Ψn,0(f)S(𝔹n)andΨn,0(G1f)S(𝔹n),

where the last implication follows from the property that Ψn,0 preserves the starlikeness (see Remark 7). But, Ψn,0(G1f)=G1(Ψn,0(f)). Then

Ψn,0(f)S(𝔹n)andG1(Ψn,0(f))S(𝔹n)Ψn,0(f)E1(𝔹n),

in view of Definition 3. Hence,

fK=E1(U)Ψn,0(f)E1(𝔹n)

and this completes the proof.

Corollary 9.

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

Ψn,0(Ek(U))Ek(𝔹n).
Theorem 10.

Let us consider the Graham-Kohr extension operator Ψ2,1 defined in (2) for n=2 and α=1. Then

Ψ2,1(K)=Ψ2,1(E1(U))E1(𝔹2)K(𝔹2). (21)
Proof.

Let us consider the case n=2 and fK. Let F:𝔹22 be given by

F(z)=Ψ2,1(f)(z)=(f(z1),z2f(z1)z1), (22)

for all z=(z1,z2)𝔹2. We want to prove that

FE1(𝔹2)FS(𝔹2)andG1FS(𝔹2).
  1. 1.

    In (GrHaKoSuf1, , Theorem 2.1) or (GrKo2, , Theorem 3.2) the authors proved that FS0(𝔹2), so according to this result it is clear that FS(𝔹2).

  2. 2.

    In order to show that G1FS(𝔹2), let us consider the mapping G:𝔹22 given by

    G(z)=DF(z)(z)=(f(z1)0z2z1[f(z1)f(z1)z1]f(z1)z1)(z1z2),

    where F is given by (22) and z=(z1,z2)𝔹2. Hence

    G(z)=(z1f(z1),z2f(z1)), (23)

    for all z=(z1,z2)𝔹2. Next, to prove that G1F=GS(𝔹2), we can use the analytical characterization of starlikeness from Theorem 1.

It is clear that GH(𝔹2), G(0)=0 and DG(0)=I2, where

DG(z)=(f(z1)+z1f′′(z1)0z2f′′(z1)f(z1)),

for z=(z1,z2)𝔹2. Also G is locally biholomorphic on 𝔹2 since

JG(z)=det[DG(z)]=f(z1)[f(z1)+z1f′′(z1)]0,z1U

in view of the fact that fK. Moreover, we have that

[DG(z)]1 =1f(z1)[f(z1)+z1f′′(z1)](f(z1)0z2f′′(z1)f(z1)+z1f′′(z1))
=(1f(z1)+z1f′′(z1)0z2f′′(z1)f(z1)[f(z1)+z1f′′(z1)]1f(z1)),z𝔹2.

Thus,

[DG(z)]1G(z) =(z1f(z1)f(z1)+z1f′′(z1),z2z1z2f′′(z1)f(z1)+z1f′′(z1))
=(z1f(z1)f(z1)+z1f′′(z1),z2f(z1)+z1z2f′′(z1)z1z2f′′(z1)f(z1)+z1f′′(z1))
=(z1f(z1)f(z1)+z1f′′(z1),z2f(z1)f(z1)+z1f′′(z1)),z𝔹2. (24)

Then

Re[DG(z)]1G(z),z =Re(z1f(z1)f(z1)+z1f′′(z1),z2f(z1)f(z1)+z1f′′(z1)),(z1,z2)
=Re[z1z¯1f(z1)f(z1)+z1f′′(z1)+z2z¯2f(z1)f(z1)+z1f′′(z1)]
=|z1|2Re[f(z1)f(z1)+z1f′′(z1)]+|z2|2Re[f(z1)f(z1)+z1f′′(z1)]
=z2Re[f(z1)f(z1)+z1f′′(z1)],

for all z𝔹2{0}. Hence,

Re[DG(z)]1G(z),z=z2Re[11+z1f′′(z1)/f(z1)]>0,

for all z𝔹2{0} according to the fact that fK. Using the characterization of starlikeness in 𝔹n (see Theorem 1), we obtain that GS(𝔹2), i.e. G1FS(𝔹2). Hence, F=Ψ2,1(f)E1(𝔹2) and this completes the proof. ∎

Remark 17.

Notice that Theorem 10 is true also in the case n2, 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. 1.

    Let fK and F=Ψn,1(f). Then FS(𝔹n) according to (GrKo2, , Theorem 3.2).

  2. 2.

    If we define h(w)=wf(w), for all wU, then fK implies that hS (in view of Alexander’s theorem).

  3. 3.

    Let G(z)=DF(z)(z), for z𝔹n. According to relation (23) and the previous step, we have that

    G(z)=(z1f(z1),zf(z1))=(h(z1),zh(z1)z1),z=(z1,z)𝔹n.
  4. 4.

    Now it is clear that G(z)=Ψn,1(h)(z).

  5. 5.

    I. Graham and G. Kohr proved in (GrKo2, , Corollary 3.3) that hS implies Ψn,1(h)=GS(𝔹n).

  6. 6.

    Hence, FS(𝔹n) and G=G1FS(𝔹n). In view of Definition 3, we obtain that FE1(𝔹n).

Lemma 11.

Let us consider n=2, α=1 and k0{0,1,2}. Then

Ψ2,1(Gk0f)=Gk0(Ψ2,1(f)). (25)
Proof.
  1. 1.

    Let k0=0. In view of Definition 2 it is clear that for k0=0 we obtain

    Ψ2,1(G0f)=Ψ2,1(f)=G0(Ψ2,1(f)).
  2. 2.

    Let k0=1. Then

    Ψ2,1(G1f)(z)=(G1f(z1),G1f(z1)z1z2)=(z1f(z1),z2f(z1)),

    for all z=(z1,z2)𝔹2 and

    G1(Ψ2,1(f))(z) =DΨ2,1(f)(z)(z)=(f(z1)0z2z1f(z1)f(z1)z12f(z1)z1)(z1z2)
    =(z1f(z1),z2z1[z1f(z1)f(z1)]+z2z1f(z1))
    =(z1f(z1),z2f(z1)),

    for all z=(z1,z2)𝔹2. Hence,

    Ψ2,1(G1f)=G1(Ψ2,1(f)).
  3. 3.

    For the last part of the proof, let us consider k0=2. Then

    Ψ2,1(G2f)(z) =(G2f(z1),z2G2f(z1)z1)
    =(z12f′′(z1)+z1,z2z1f′′(z1)+z2)
    =[z1f′′(z1)+1](z1,z2),

    for all z=(z1,z2)𝔹2. On the other hand,

    G2(Ψ2,1(f))(z) =D2Ψ2,1(f)(z)(z2)+z
    =(z12f′′(z1)+z1,z2z1f′′(z1)+z2)
    =[z1f′′(z1)+1](z1,z2),

    where

    D2Ψ2,1(f)(z)(z2)=(z1f′′(z1)0z2f′′(z1)z2z1f(z1)+z2z12f(z1)f(z1)f(z1)z1)(z1z2),

    for all z=(z1,z2)𝔹2. Thus,

    Ψ2,1(G2f)=G2(Ψ2,1(f)).

Hence, we proved the commutative property given by (25) for the particular cases k{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 n2 are similar.

Remark 19.

Using Lemma 11 in the general case n2, Remark 7 and Definition 3, we can prove Theorem 10 in the following alternative way:

fKfE1(U)fSandG1fS

which implies that

Ψn,1(f)S(𝔹n)andΨn,1(G1f)S(𝔹n).

But, Ψn,1(G1f)=G1(Ψn,1(f)). Then

Ψn,1(f)S(𝔹n)andG1(Ψn,1(f))S(𝔹n)Ψn,1(f)E1(𝔹n).

Hence,

fK=E1(U)Ψn,1(f)E1(𝔹n)

and this completes the proof.

2023

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