1 Introduction

We say that a domain \(\mathcal {G}\subset \mathbb {C}^{n}\), is complete n-circular if \(z\lambda =(z_{1}\lambda _{1},\ldots ,z_{n}\lambda _{n})\in \mathcal {G}\) for each \(z=(z_{1},\ldots ,z_{n})\in \mathcal {G}\) and every \( \lambda =(\lambda _{1},\ldots ,\lambda _{n})\in \overline{\mathcal {U}^{n}}\), where \(\mathcal {U}\) is the unit disc \(\{\zeta \in \mathbb {C}:|\zeta |<1\}\). From now by \(\mathcal {G}\) will be denoted a bounded complete n-circular domain in \(\mathbb {C}^{n},n\ge 2.\) By \(\mathcal {H}_{\mathcal {G}}\) let us denote the space of all holomorphic functions \(f:\mathcal {G} \longrightarrow \mathbb {C}\) and by \(\mathcal {H}_{\mathcal {G}}(1)\) the collection of all \(f\in \mathcal {H_{G}}\), normalized by \(f(0)=1.\)

Many authors (cf., eg., [1, 2, 5,6,7, 11, 18, 19, 23]) considered some Bavrin’s subfamilies \(\mathcal {X_{G}}\) of the family \(\mathcal {H_{G}} (1).\) In the definitions of these families \(\mathcal {X_{G}}\) the main role play the families \(\mathcal {C_{G}}(\alpha ),\alpha \in [0,1),\)

$$\begin{aligned} \mathcal {C_{G}}(\alpha )=\{f\in \mathcal {H_{G}}(1):{\text {Re}}f(z)>\alpha ,z\in \mathcal {G}\} \end{aligned}$$

and the following invertible Temljakov [24] linear operator \(\mathcal {L}:\mathcal {H}_{\mathcal {G}}\longrightarrow \mathcal {H}_{\mathcal {G}}\)

$$\begin{aligned} \mathcal {L}f(z)=f(z)+Df(z)(z),z\in \mathcal {G}, \end{aligned}$$

where Df(z) is the Fréchet derivative of f at the point z. By a Bavrin’s family \(\mathcal {X_{G}}\) we mean a collection of functions \(f\in \mathcal {H_{G}}(1)\) whose the Temljakov transform \(\mathcal {L}f\) has a functional factorization \(\mathcal {L}f=p\cdot g\), where \(p\in \mathcal {C_{G}} \equiv \mathcal {C_{G}}(0)\) and g is from a fixed subfamily of \(\mathcal { H_{G}}(1).\) Below, we recall the factorizations which define a few well known Bavrin’s families \(\mathcal {X_{G}},\) like as

$$\begin{aligned} \mathcal {V}_{\mathcal {G}}:\mathcal {L}f= & {} p\cdot 1,p\in \mathcal {C_{G}},\\ \mathcal {M}_{\mathcal {G}}:\mathcal {L}f= & {} p\cdot f,p\in \mathcal {C_{G}},\\ \mathcal {N}_{\mathcal {G}}:\mathcal {L}f= & {} p\cdot \mathcal {LL}f,p\in \mathcal { C_{G}},\\ \mathcal {R}_{\mathcal {G}}:\mathcal {L}f= & {} p\cdot \mathcal {L}\varphi ,\varphi \in \mathcal {N}_{\mathcal {G}},p\in \mathcal {C_{G}}. \end{aligned}$$

It is known that functions of these families were used to construct biholomorphic mappings in \(\mathbb {C}^{n}\) (cf., eg., [10, 13, 20]). Let us note that the above families have geometric interpretation, in particular the functions \(f\in \mathcal {M}_{\mathcal {G}}\) map biholomorphically some planar intersections \(\mathcal {S}\) of \(\mathcal {G}\) onto starlike domains in \(\mathbb {C}\),(see [1]). It is very important, because the starlikeness plays a central role in many different subjects of geometry and topology and in particular, in geometric function theory.

Let us recall also that Bavrin showed the inclusions \(\mathcal {N}_{\mathcal {G}}\varsubsetneq \mathcal {R}_{\mathcal {G}},\mathcal {V}_{\mathcal {G}}\varsubsetneq \mathcal {R}_{\mathcal {G}}\) and pointed that the first of them can be complete to the following double inclusion \(\mathcal {N}_{\mathcal {G} }\varsubsetneq \mathcal {M}_{\mathcal {G}}\varsubsetneq \mathcal {R}_{\mathcal {G}}\). Thus, it is natural to ask whether is possible to do the same in the case of the second above inclusion. In the paper [12] the authors defined a family \(\mathcal {K}_{\mathcal {G}}^{-},\) which satisfies the inclusion \(\mathcal {V}_{\mathcal {G}}\varsubsetneq \mathcal {K}_{\mathcal {G} }^{-}\varsubsetneq \mathcal {R}_{\mathcal {G}}\). An adequate definition of \(\mathcal {K}_{\mathcal {G}}^{-}\) has the form: A function \(f\in \mathcal {H_{G}}(1)\) belongs to \(\mathcal {K}_{\mathcal {G}}^{-}\) if its Temljakov transform \( \mathcal {L}f\) has the factorization

$$\begin{aligned} \mathcal {L}f(z)=p(z)\cdot h(z)\cdot h(-z),z\in \mathcal {G},h\in \mathcal {M}_{ \mathcal {G}}\left( \frac{1}{2}\right) ,p\in \mathcal {C_{G}}, \end{aligned}$$

where the family \(\mathcal {M}_{\mathcal {G}}(\alpha ),\alpha \in [0,1), \) is defined similarly as \(\mathcal {M}_{\mathcal {G}},\) but in this case \(p\in \mathcal {C_{G}}(\alpha )\).

In the present paper we consider Bavrin’s type families \(\mathcal {K}_{ \mathcal {G}}^{k},k\ge 2(\mathcal {K}_{\mathcal {G}}^{2}=\mathcal {K}_{\mathcal { G}}^{-})\) separating also the families \(\mathcal {V}_{\mathcal {G}},\mathcal {R }_{\mathcal {G}},\) i.e., satisfying the inclusions \(\mathcal {V}_{\mathcal {G}}\varsubsetneq \mathcal {K}_{\mathcal {G}}^{k}\varsubsetneq \mathcal {R}_{ \mathcal {G}},k\ge 2.\)

The formal definition of such family has the following form.

A function \(f\in \mathcal {H_{G}}(1)\) belongs to \(\mathcal {K}_{\mathcal {G}}^{k}\) if there exist a function \(p\in \mathcal {C_{G}}\) and a function \(h\in \mathcal {M}_{\mathcal {G}}(\frac{k-1}{k})\) such that the Temljakov transform \(\mathcal {L}f\) of f has the factorization

$$\begin{aligned} \mathcal {L}f(z)=p(z)\cdot \prod \limits _{l=0}^{k-1}h(\varepsilon ^{l}z),z\in \mathcal {G}, \end{aligned}$$
(1.1)

where \(\varepsilon =\varepsilon _{k}=\exp \frac{2\pi i}{k}\) is a generator of the cyclic group of kth roots of unity.

Let us observe that \(\mathcal {K}_{\mathcal {G}}^{k},k\ge 2\) are nonempty families. Indeed, the function \(f=1\) belongs to \(\mathcal {K}_{\mathcal {G}}^{k},\) because it satisfies the factorization (1.1) with \(p=1\in \mathcal {C_{G}}\) and \(h=1\in \mathcal {M}_{\mathcal {G}}(\frac{k-1}{k)}).\)

In the future, we will use a characterization of the family \(\mathcal {K}_{ \mathcal {G}}^{k}\) by a notion of \(\left( j,k\right) \)-symmetry, which is connected with a functional decomposition with respect to the above group.

Let us observe that bounded complete n-circular domains \(\mathcal {G}\) are k-symmetric sets for \(k=2,3,\ldots ,\) that is \(\varepsilon \mathcal {G=G}\). For \( j=0,1,\ldots ,k-1\) we define the collections \(\mathcal {F}_{j,k}(\mathcal {G})\) of functions \(\left( j,k\right) \)-symmetrical, i.e., all functions \(f:\mathcal {G}\rightarrow \mathbb {C}\) such that

$$\begin{aligned} f\left( \varepsilon z\right) =\varepsilon ^{j}f\left( z\right) ,\;z\in \mathcal {G}. \end{aligned}$$

If \(n=1\) and \(\mathcal {G}=\mathcal {U},\) then we write \(\mathcal {F}_{j,k}( \mathcal {U})\).

The mentioned functional decomposition appears in the following result from [14].

Theorem A

For every function \(f:\mathcal {G}\rightarrow \mathbb {C}\) there exists exactly one sequence of functions \(f_{j,k}\in \mathcal {F}_{j,k}(\mathcal {G} ), \) \(j=0,1,\ldots ,k-1,\) such that

$$\begin{aligned} f=\sum _{j=0}^{k-1}f_{j,k}. \end{aligned}$$

Moreover,

$$\begin{aligned} f_{j,k}\left( z\right) =\frac{1}{k}\sum _{l=0}^{k-1}\varepsilon ^{-jl}f\left( \varepsilon ^{l}z\right) ,\;z\in \mathcal {G}. \end{aligned}$$

The functions \(f_{j,k},\) which are uniquely determined by the above decomposition, will be called \(\left( j,k\right) \)-symmetrical components of the function f. Some interesting applications of the above partition may also be found in [15, 16] and [17].

2 Results

Now we can present a characterization of \(f\in \mathcal {K}_{\mathcal {G}}^{k}, \) simpler than (1.1).

Theorem 1

A function \(f\in \mathcal {H_{G}}(1)\) belongs to the family \( \mathcal {K}_{\mathcal {G}}^{k},k\ge 2\) if and only if there exists a function \(g\in \mathcal {M}_{\mathcal {G}}\cap \mathcal {F}_{0,k}(\mathcal {G})\) and a function \(p\in \mathcal {C_{G}}\) such that

$$\begin{aligned} \mathcal {L}f=p\cdot g. \end{aligned}$$
(2.1)

Proof

Let \(f\in \) \(\mathcal {K}_{\mathcal {G}}^{k}.\) Then there exists \(p\in \mathcal {C_{G}}\) and \(h\in \mathcal {M}_{\mathcal {G}}(\frac{k-1}{k})\) such that

$$\begin{aligned} \mathcal {L}f(z)=p(z)\cdot g\left( z\right) ,z\in \mathcal {G}, \end{aligned}$$

where

$$\begin{aligned} g(z)=\prod \limits _{l=0}^{k-1}h(\varepsilon ^{l}z),z\in \mathcal {G}. \end{aligned}$$

It is obvious that \(g\in \mathcal {F}_{0,k}(\mathcal {G})\). We show that \(g\in \mathcal {M}_{\mathcal {G}}.\) To do it, using the differentiation product rule and the form of the operator \(\mathcal {L},\) we have at \(z\in \mathcal {G}\)

$$\begin{aligned} \frac{\mathcal {L}g(z)}{g(z)}=1+\frac{Dg(z)(z)}{g(z)}=1+\sum \limits _{l=0}^{k-1}\frac{Dh(\varepsilon ^{l}z)(\varepsilon ^{l}z)}{ h(\varepsilon ^{l}z)}=1-k+\sum \limits _{l=0}^{k-1}\frac{\mathcal {L} h(\varepsilon ^{l}z)}{h(\varepsilon ^{l}z)}. \end{aligned}$$

Hence and by the fact that \(h\in \mathcal {M}_{\mathcal {G}}(\frac{k-1}{k}),\) we obtain that \({\text {Re}}\frac{\mathcal {L}g(z)}{g(z)}>1-k+k\frac{k-1}{k}=0.\) Thus \(g\in \mathcal {M}_{\mathcal {G}}.\)

Now, let us suppose that f satisfies the equality (2.1), with a \( p\in \mathcal {C_{G}}\) and a \(g\in \mathcal {M}_{\mathcal {G}}\cap \mathcal {F} _{0,k}(\mathcal {G}).\) Let us put \(h(z)=\left( g(z)\right) ^{\frac{1}{k}},z\in \mathcal {G},\) with the power function taking the value 1 at the point 1. Since \(g(z)\ne 0\) (see [1]), the function h is holomorphic. It remains to show that \(h\in \mathcal {M}_{\mathcal {G}}(\frac{ k-1}{k})\) and the equality (1.1) is fulfilled. To this end we compute step by step

$$\begin{aligned} {\text {Re}}\frac{\mathcal {L}h(z)}{h(z)}= & {} {\text {Re}}\frac{\mathcal {L}\left( g(z)\right) ^{\frac{1}{k}}}{\left( g(z)\right) ^{\frac{1}{k}}}=1+\frac{1}{k} {\text {Re}}\frac{\left( g(z)\right) ^{\frac{1}{k}-1}Dg(z)(z)}{\left( g(z)\right) ^{\frac{1}{k}}} \\= & {} 1+\frac{1}{k}{\text {Re}}\frac{Dg(z)(z)}{g(z)}=\frac{k-1}{k}+\frac{1}{k} {\text {Re}}\frac{\mathcal {L}g(z)}{g(z)}>\frac{k-1}{k}. \end{aligned}$$

The formula (1.1) follows from the definition of the function h. Indeed,

$$\begin{aligned} g(z)=\left( h(z)\right) ^{k}=\prod \limits _{l=0}^{k-1}h(\varepsilon ^{l}z), \quad z\in \mathcal {G}, \end{aligned}$$

because \(h\in \mathcal {F}_{0,k}(\mathcal {G}).\)

The proof is complete. \(\square \)

Now we consider an extremal problem for \(f\in \mathcal {K}_{\mathcal {G}}^{k}.\) More precisely, we look for some estimates for \(\mathcal {G}\)-balances of m -homogeneous polynomias \(Q_{f,m}\) of its unique power series expansion

$$\begin{aligned} f(z)=1+\sum _{m=1}^{\infty }Q_{f,m}(z),z\in \mathcal {G}. \end{aligned}$$
(2.2)

In our considerations the Minkowski function

$$\begin{aligned} \mu _{\mathcal {G}}(z)=inf\{t>0:\frac{1}{t}z\in \mathcal {G}\},z\in \mathbb {C}^{n}, \end{aligned}$$

will be very useful. This function gives a possibility to redefine the domain \(\mathcal {G}\) and its boundary \(\partial \mathcal {G}\) as follows:

$$\begin{aligned} \mathcal {G}=\{z\in \mathbb {C}^{n}:\mu _{\mathcal {G}}(z)<1\},\partial \mathcal {G}=\{z\in \mathbb {C}^{n}:\mu _{\mathcal {G}}(z)=1\}. \end{aligned}$$

The notion of \(\mathcal {G}\)-balance of m-homogeneous polynomial \(Q_{m}: \mathbb {C}^{n}\rightarrow \mathbb {C},\) \(m\in \mathbb {N\cup }\{0\},\) was defined in [3] as the quantity

$$\begin{aligned} \mu _{\mathcal {G}}(Q_{m})=\sup _{w\in \mathbb {C}^{n}{\setminus } \{0\}}\frac{ \left| Q_{m}(w)\right| }{(\mu _{\mathcal {G}}(w))^{m}}=\sup _{v\in \partial \mathcal {G}}\left| Q_{m}(v)\right| =\sup _{u\in \mathcal {G}}\left| Q_{m}(\mathcal {U})\right| . \end{aligned}$$

The \(\mathcal {G}\)-balance \(\mu _{\mathcal {G}}(Q_{m})\) generalizes the norm \(\left\| Q_{m}\right\| \) of the polynomial \(Q_{m}\) and if \(\mathcal {G}\) is convex ,  then \(\mu _{\mathcal {G}}(Q_{m})\) reduces to \(\left\| Q_{m}\right\| ,\) because

$$\begin{aligned} \left| Q_{m}(w)\right| \le \mu _{\mathcal {G}}(Q_{m})(\mu _{\mathcal {G}}(w))^{m},w\in \mathbb {C}^{n} \end{aligned}$$

and for bounded convex complete n-circular domains \(\mathcal {G}\) also \(\mu _{\mathcal {G}}(w)=||w||\) (see, e.g., [21]).

We present the announced estimates of \(\mathcal {G}\)-balances \(\mu _{\mathcal { G}}(Q_{f,m})\) of m-homogeneous polynomials \(Q_{f,m}\) from the Taylor series of \(f\in \mathcal {M}_{\mathcal {G}}^{k}\) in the following theorem.

Theorem 2

If the expansion of the function \(f\in \mathcal {K}_{\mathcal {G} }^{k}\), \(k\ge 2,\) into a series of m-homogenous polynomials \( Q_{f,m}\) has the form (2.2), then for the \(\mathcal {G}\)-balances \(\mu _\mathcal {G}(Q_{f,m})\) of polynomials \(Q_{f,m}\) the following sharp estimate hold:

$$\begin{aligned} \mu _{\mathcal {G}}(Q_{f,m})\,\le \left\{ \begin{array}{l} \frac{2}{m}\prod \limits _{p=1}^{\frac{m}{k}-1}\left( 1+\frac{2}{pk}\right) \text { for }m=k,2k,3k,\ldots \\ \frac{2}{m+1}\prod \limits _{p=1}^{\left\lfloor \frac{m}{k}\right\rfloor }\left( 1+\frac{2}{pk}\right) \text { for remaining }m\in \mathbb {N} \end{array} \right. , \end{aligned}$$

where \(\left\lfloor q\right\rfloor \) means the integral part of the number q. We use a standard convention that the product \(\prod \limits _{l=l_{1}}^{l_{2}}a_{l}\) is equal to 1 for \(l_{2}<l_{1}\).

Proof

Let \(f\in \) \(\mathcal {K}_{\mathcal {G}}^{k}\) be arbitrarily fixed. Then, by Theorem 1, the factorization (2.1) holds with a function \(p\in \mathcal {C_{G}}\) of the form

$$\begin{aligned} p(z)=1+\sum _{\nu =1}^{\infty }Q_{p,\nu }(z),\,\,z\in \mathcal {G} \end{aligned}$$

and a function \(g\in \mathcal {M}_{\mathcal {G}}\cap \mathcal {F}_{0,k}( \mathcal {G})\) of the form

$$\begin{aligned} g(z)=1+\sum _{\nu =1}^{\infty }Q_{g,k\nu }(z),\,\, z\in \mathcal {G}. \end{aligned}$$
(2.3)

From the above, by the series expansion of \(\mathcal {L}f\)

$$\begin{aligned} \mathcal {L}f(z)=1+\sum _{m=1}^{\infty }Q_{\mathcal {L}f,m}(z)=1+\sum _{m=1}^{ \infty }(m+1)Q_{f,m}(z),\,\, z\in \mathcal {G} \end{aligned}$$

and by the equalities \(Q_{f,0}=Q_{p,0}=Q_{g,0}=1,\) we obtain the recursive formula for \(m\in \mathbb {N}\)

$$\begin{aligned} (m+1)Q_{f,m}(z)=\sum _{l=0}^{\left\lfloor \frac{m}{k}\right\rfloor }Q_{g,kl}(z)Q_{p,m-kl}(z),\,\, z\in \mathcal {G}. \end{aligned}$$

Hence

$$\begin{aligned} (m+1)\left| Q_{f,m}(z)\right| \le \sum _{l=0}^{\left\lfloor \frac{m}{ k}\right\rfloor }\left| Q_{g,kl}(z)\right| \left| Q_{p,m-kl}(z)\right| ,\,\, z\in \mathcal {G}. \end{aligned}$$
(2.4)

Since

$$\begin{aligned} \left| Q_{p,\nu }(z)\right| \le 2,\quad \nu \in \mathbb {N},\,\, z\in \mathcal {G}, \end{aligned}$$
(2.5)

(see [1]) we need some bounds for \(\left| Q_{g,k\mu }(z)\right| \). We show that for \(g\in \mathcal {M}_{\mathcal {G}}\cap \mathcal {F}_{0,k}(\mathcal {G})\) and \(\mu \in \mathbb {N}\) there hold the inequalities

$$\begin{aligned} \left| Q_{g,k\mu }(z)\right| \le \frac{2}{k\mu }\prod \limits _{\nu =1}^{\mu -1}\left( 1+\frac{2}{k\nu }\right) ,z\in \mathcal {G}. \end{aligned}$$
(2.6)

For this purpose let us observe that for each \(z\in \mathcal {G},\) the function

$$\begin{aligned} G(\zeta )=\zeta g(\zeta z),\zeta \in \mathcal {U} \end{aligned}$$

belongs to the family \(\mathcal {S}^{*}\cap \mathcal {F}_{1,k}(\mathcal {U} ) \) of (1, k)-symmetric univalent starlike mappings (in the unit disc \( \mathcal {U)}\) and its Taylor series has the form

$$\begin{aligned} G(\zeta )=\zeta +\sum \limits _{\mu =1}^{\infty }b_{k\mu +1}\zeta ^{k\mu +1}=1+\sum _{\mu =1}^{\infty }Q_{g,k\mu }(z)\zeta ^{k\mu +1},\zeta \in \mathcal {U}. \end{aligned}$$

Thus, in view of the estimates [25] of the coefficients of functions from \(\mathcal {S}^{*}\cap \mathcal {F}_{1,k}(\mathcal {U})\) we get the announced bounds (2.6).

In two next parts of the proof we use also the fact [4] that for every k,  \(s\in \mathbb {N{\setminus } }\left\{ 1\right\} \) there holds the identity:

$$\begin{aligned} 1+\frac{2}{k}+\sum _{l=2}^{s}\frac{2}{lk}\prod \limits _{\nu =1}^{l-1}\left( 1+ \frac{2}{\nu k}\right) =\prod \limits _{\nu =1}^{s}\left( 1+\frac{2}{\nu k} \right) . \end{aligned}$$
(2.7)

Now, we will estimate the quantities \(\left| Q_{f,m}(z)\right| ,z\in \mathcal {G},\) using all the conditions (2.4)-(2.7).

First let us assume that \(m=ks,\) where \(s\in \mathbb {N}.\) Since \( Q_{p,m-kl}(z)=1\) for \(l=s,\) we get from (2.4) that

$$\begin{aligned} (m+1)\left| Q_{f,m}(z)\right| \le \left| Q_{g,ks}(z)\right| +2\sum _{l=0}^{s-1}\left| Q_{g,kl}(z)\right| ,z\in \mathcal {G}. \end{aligned}$$

Thus for \(z\in \mathcal {G}\), in view of (2.6) and (2.7),

$$\begin{aligned} (m+1)\left| Q_{f,m}(z)\right|\le & {} \frac{2}{sk}\prod \limits _{\nu =1}^{s-1}\left( 1+\frac{2}{\nu k}\right) +2\left[ 1+\frac{2}{k} +\sum _{l=2}^{s-1}\frac{2}{lk}\prod _{\nu =1}^{l-1}\left( 1+\frac{2}{\nu k}\right) \right] \\= & {} \frac{-2}{sk}\prod \limits _{\nu =1}^{s-1}\left( 1+\frac{2}{\nu k}\right) +2\left[ 1+\frac{2}{k}+\sum _{l=2}^{s}\frac{2}{lk}\prod _{\nu =1}^{l-1}\left( 1+\frac{2}{\nu k}\right) \right] \\\le & {} \frac{-2}{sk}\prod \limits _{\nu =1}^{s-1}\left( 1+\frac{2}{\nu k} \right) +2\prod \limits _{\nu =1}^{s}\left( 1+\frac{2}{\nu k}\right) \\= & {} 2\frac{ (sk+1)}{sk}\prod \limits _{\nu =1}^{s-1}\left( 1+\frac{2}{\nu k}\right) . \end{aligned}$$

Hence, for \(m=k,2k,3k,\ldots \)

$$\begin{aligned} \left| Q_{f,m}(z)\right| \le \frac{2}{m}\prod \limits _{\nu =1}^{ \frac{m}{k}-1}\left( 1+\frac{2}{\nu k}\right) ,z\in \mathcal {G}. \end{aligned}$$

Now let us consider the case \(m=ks+r,\) where \(s\in \mathbb {N}\cup \{0\}\) and \(r\in \{1,2,\ldots ,k-1\}.\) In this case we apply in (2.4) the inequality \( \left| Q_{p,m-kl}(z)\right| \le 2,l=0,\ldots ,s=\left\lfloor \frac{m}{k} \right\rfloor ,\) which follows from estimates (2.5), because \(m-kl>0\). Thus, in view of (2.6) and (2.7) we get step by step

$$\begin{aligned} (m+1)\left| Q_{f,m}(z)\right|\le & {} 2\sum _{l=0}^{\left\lfloor \frac{ m}{k}\right\rfloor }\left| Q_{g,kl}(z)\right| \le 2\left[ 1+\frac{2 }{k}+\sum _{l=2}^{\left\lfloor \frac{m}{k}\right\rfloor }\frac{2}{lk} \prod _{\nu =1}^{l-1}\left( 1+\frac{2}{\nu k}\right) \right] \\\le & {} 2\prod \limits _{\nu =1}^{\left\lfloor \frac{m}{k}\right\rfloor }\left( 1+\frac{2}{\nu k}\right) . \end{aligned}$$

Summing up the results of both cases we get

$$\begin{aligned} \left| Q_{f,m}(z)\,\right| \le \left\{ \begin{array}{l} \frac{2}{m}\prod \limits _{\nu =1}^{\frac{m}{k}-1}\left( 1+\frac{2}{\nu k} \right) \text { for }m=k,2k,3k,\ldots \\ \frac{2}{m+1}\prod \limits _{\nu =1}^{\left\lfloor \frac{m}{k}\right\rfloor }\left( 1+\frac{2}{\nu k}\right) \text { for remaining }m\in \mathbb {N} \end{array} \right. ,z\in \mathcal {G} \end{aligned}$$

and consequently

$$\begin{aligned} \sup _{z\in \mathcal {G}}\left| Q_{f,m}(z)\,\right| \,\le \left\{ \begin{array}{l} \frac{2}{m}\prod \limits _{\nu =1}^{\frac{m}{k}-1}\left( 1+\frac{2}{\nu k} \right) \text { for }m=k,2k,3k,\ldots \\ \frac{2}{m+1}\prod \limits _{\nu =1}^{\left\lfloor \frac{m}{k}\right\rfloor }\left( 1+\frac{2}{\nu k}\right) \text { for remaining }m\in \mathbb {N} \end{array} \right. . \end{aligned}$$

These inequalities and the definition of \(\mathcal {G}\)-balances \(\mu _{ \mathcal {G}}(Q_{f,m})\) of m-homogeneous polynomials imply the estimates from the statement of the theorem.

Now, we will show the sharpness of the above estimates.

For the linear functional \(I=\left( \mu _{\mathcal {G}}(J)\right) ^{-1}J,\) with

$$\begin{aligned} J(z)=\sum \limits _{l=1}^{n}z_{l},z=(z_{1},\ldots ,z_{n})\in \mathbb {C}^{n}, \end{aligned}$$

let us denote by \(\mathcal {Z}\) an analytic set \(\mathcal {\ G}\cap \, I^{-1}\{0\}\) and let \(I^{m}(z)=\left( Iz\right) ^{m},z\in \mathcal {G},m\in \mathbb {N}\cup \{0\}.\) The equalities in our estimates are achieved for the following function \(f\in \mathcal {K}_{\mathcal {G}}^{k},\,k\ge 2,\)

$$\begin{aligned} f\left( z\right) =\left\{ \begin{array}{lc} \frac{\sum \limits _{l=0}^{k-1}I^{l-1}(z)}{\left( 1-I^{k}(z)\right) ^{\frac{2 }{k}}}-\frac{1}{I(z)}-&{}\sum \limits _{l=3}^{k-1}\frac{l-2}{l}I^{l-1}(z)H(\frac{ 2}{k},\frac{l}{k},\frac{l+k}{k},I^{k}(z))\hbox { for } z\in \mathcal {G}\diagdown \mathcal {Z} \\ 1 &{}\qquad \qquad \quad {\textit{for}}\, z\in \mathcal {Z} \end{array} ,\right. \end{aligned}$$
(2.8)

where \(H(a,b,c,\zeta ):\mathcal {U}\rightarrow \mathbb {C}\) is a hypergeometric function

$$\begin{aligned} H(a,b,c,\zeta )=\sum \limits _{\nu =0}^{\infty }\frac{\left( a\right) _{\nu }\left( b\right) _{\nu }}{\left( c\right) _{\nu }}\frac{\zeta ^{\nu }}{\nu !} ,\zeta \in \mathcal {U}, \end{aligned}$$

defined by Pochhamer symbols \(\ \left( a\right) _{\nu },\left( b\right) _{\nu },\left( c\right) _{\nu }:\)

$$\begin{aligned} \left( a\right) _{\nu }=\left\{ \begin{array}{ll} a(a+1)\ldots (a+\nu -1), &{}\nu \in \mathbb {N} \\ 1, &{}\nu =0 \end{array} \right. , \end{aligned}$$

and the branch of the power function \(x^{\frac{2}{k}}\) takes the value 1 at the point \(x=1.\) In the case \(k=2,3\) we use a standard convention that the sum

$$\begin{aligned} \sum \limits _{l=3}^{k-1}\frac{l-2}{l}I^{l-1}(z)H\left( \frac{2}{k},\frac{l}{k}, \frac{l+k}{k},I^{k}(z)\right) ,\,\, z\in \mathcal {G} \end{aligned}$$

is equal to zero, if the superscript of the sum is smaller than the subscript.

In the paper [4], it was proven that the above function gives the equalities in the bounds from the statement of the theorem. It remains to show that \(f\in \mathcal {K}_{\mathcal {G}}^{k}\) for \(k\ge 2.\) To do it, let us observe that as shown in [4]

$$\begin{aligned} \mathcal {L}f(z)=\frac{1+I(z)}{1-I(z)}\frac{1}{\left( 1-I^{k}(z)\right) ^{ \frac{2}{k}}},\,\, z\in \mathcal {G}. \end{aligned}$$

This implies, in view of Theorem 1, the relation \(f\in \mathcal {K}_{\mathcal { G}}^{k},\) because the functions

$$\begin{aligned} p(z)=\frac{1+I(z)}{1-I(z)},g(z)=\frac{1}{\left( 1-I^{k}(z)\right) ^{\frac{2}{ k}}},\,\, z\in \mathcal {G} \end{aligned}$$

belong to \(\mathcal {C}_{\mathcal {G}}\) and to \(\mathcal {M}_{\mathcal {G}}\cap \mathcal {F}_{0,k}(\mathcal {G}),\) respectively. \(\square \)

We use the estimates of \(\mathcal {G}\)-balances \(\mu _{\mathcal {G}}(Q_{f,m})\) of polynomials \(Q_{f,m}\) to solve the mentioned separation problem for the families \(\mathcal {V}_{\mathcal {G}},\mathcal {K}_{\mathcal {G}}^{k},\mathcal {R} _{\mathcal {G}}\). We prove the following theorem:

Theorem 3

For every \(k\ge 2\) there holds the double inclusion

$$\begin{aligned} \mathcal {V}_{\mathcal {G}}\varsubsetneq \mathcal {K}_{\mathcal {G}}^{k}\varsubsetneq \mathcal {R}_{\mathcal {G}}. \end{aligned}$$

Proof

We start with the inclusion \(\mathcal {V}_{\mathcal {G}}\subset \mathcal {K}_{ \mathcal {G}}^{k}.\) To do it, let us assume that \(f\in \mathcal {V}_{\mathcal {G }},\) then \(\mathcal {L}f\) \(\in \mathcal {C_{G}}.\) Putting \(p=\mathcal {L}f\) and \(h=1,\) we obtain the factorization (1.1) with \(p\in \mathcal {C_{G}}\) and \(g = 1 \in \mathcal {M}_{\mathcal {G}}\cap \mathcal {F}_{0,k}( \mathcal {G}).\) Hence \(f\in \mathcal {K}_{\mathcal {G}}^{k}.\) It remains to show the relation \(\mathcal {V}_{\mathcal {G}}\ne \mathcal {K}_{\mathcal {G} }^{k}.\) To do it, let us observe that for \(f\in \mathcal {V}_{\mathcal {G}}\) there hold the sharp estimates \(\mu _{\mathcal {G}}(Q_{f,m})\le \) \(\frac{2}{ m+1},m\in \mathbb {N}\) (cf., eg., [1]), while for \(f\in \,\mathcal {K}_{\mathcal {G}}^{k}\) the sharp estimates \(\mu _{\mathcal {G} }(Q_{f,m})\le \) B(m) (Theorem 2.), with the obvious bound \(B(m)>\frac{ 2}{m+1},m\in \mathbb {N}\diagdown \{1\}.\) Hence, the extremal function \(f\in \mathcal {K}_{\mathcal {G}}^{k}\) does not belong to \(\mathcal {V}_{\mathcal {G} }. \)

Now we prove that \(\mathcal {K}_{\mathcal {G}}^{k}\subset \mathcal {R}_{ \mathcal {G}}.\)To this end, let us suppose that \(f\in \mathcal {K}_{\mathcal {G} }^{k}.\) Then there exist functions \(p\in \mathcal {C_{G}},\) \(g\in \mathcal {M} _{\mathcal {G}}\cap \mathcal {F}_{0,k}(\mathcal {G})\) such that \(\mathcal {L} f=p\cdot g.\) Denoting \(\varphi =\mathcal {L}^{-1}g,\) we have that \(\varphi \in \mathcal {N}_{\mathcal {G}}\) (by the Aleksander type theorem [1]) and \(\mathcal {L}f=p\mathcal {L}\varphi .\) Thus \(f\in \mathcal {R}_{ \mathcal {G}}\). It remains to show the relation \(\mathcal {K}_{\mathcal {G} }^{k}\ne \mathcal {R}_{\mathcal {G}}.\) For this purpose, let us observe that in the above estimates \(\mu _{\mathcal {G}}(Q_{f,m})\le \) \(B(m),m\in \mathbb { N},\) we have \(B(m)\le 1,m\in \mathbb {N}\) (see below), while for \(f\in \mathcal {R}_{\mathcal {G}}\) there hold the sharp estimates \(\mu _{\mathcal {G} }(Q_{f,m})\le m+1(\)see for instance [1]). Therefore, the extremal function \(f\in \mathcal {R}_{\mathcal {G}}\) does not belong to \(\mathcal {K}_{ \mathcal {G}}^{k}.\)

To complete the proof, we show that \(B(m)\le 1,\) \(m\in \mathbb {N}.\) To do it, we consider two cases, according to the partition \(m=ks+r,\) \(r\in \{0,1,\ldots ,k-1\},\) from the proof of Theorem 2.

1. Let us suppose that \(r=0.\) Then, if \(s=\frac{m}{k}=1,\) we see that the superscript \(s-1\) of the first product in Theorem 2 is smaller than its subscript 1. Hence, we replace the referred product by 1 and consequently, we get \(\mu _{\mathcal {G}}(Q_{f,m})\le \frac{2}{m}\le 1,\) because \( m=k\ge 2\). Next, if \(s\ge 2,\) then from Theorem 2, by the inequality \(1+ \frac{2}{\nu k}\le \frac{\nu +1}{\nu },\nu \in \mathbb {N},k\in \mathbb {N} \diagdown \{1\},\) we obtain

$$\begin{aligned} \mu _{\mathcal {G}}(Q_{f,m})\le \frac{2}{m}\prod \limits _{\nu =1}^{s-1}\frac{\nu +1}{ \nu }\le \frac{2}{m}s=\frac{2}{k}\le 1. \end{aligned}$$

2. Let us suppose that \(r\in \{1,\ldots ,k-1\}.\) Then, if \( s=\left\lfloor \frac{m}{k}\right\rfloor =0,\) we see that the superscript of the second product in Theorem 2 is smaller than its subscript. Hence we replace the referred product by 1 and consequently, we get \(\mu _{G}(Q_{f,m})\le \frac{2}{m+1}\le 1,\) because \(m\le k-1.\) Next, if \( s=\left\lfloor \frac{m}{k}\right\rfloor \ge 1,\) then similarly as in step 1,  we obtain

$$\begin{aligned} \mu _{\mathcal {G}}(Q_{f,m})\le \frac{2}{m+1}\prod \limits _{\nu =1}^{s}\frac{\nu +1}{ \nu }\le \frac{2}{m+1}(s+1)\le \frac{2(s+1)}{ks+2}\le \frac{2(s+1)}{2s+2} \le 1. \end{aligned}$$

\(\square \)

Now, we give a growth theorem for \(f\in \mathcal {K}_{\mathcal {G}}^{k}\) and its Temljakov transform \(\mathcal {L}f.\)

Theorem 4

For functions \(f\in \mathcal {K}_{\mathcal {G}}^{k}\) there follow the following sharp estimates

$$\begin{aligned} \frac{1-r}{1+r}\frac{1}{\left( 1+r^{k}\right) ^{\frac{2}{k}}}\le & {} \left| \mathcal {L}f(z)\right| \le \frac{1+r}{1-r}\frac{1}{\left( 1-r^{k}\right) ^{\frac{2}{k}}},r=\mu _{\mathcal {G}}(z)\in [0,1),\nonumber \\ \end{aligned}$$
(2.9)
$$\begin{aligned} \frac{1}{r}\int \limits _{0}^{r}\frac{1-\varrho }{1+\varrho }\frac{1}{\left( 1+\varrho ^{k}\right) ^{\frac{2}{k}}}d\varrho\le & {} \left| f(z)\right| \le \frac{1}{r}\int \limits _{0}^{r}\frac{1+\varrho }{1-\varrho } \frac{1}{\left( 1-\varrho ^{k}\right) ^{\frac{2}{k}}}d\varrho ,\nonumber \\ r= & {} \mu _{\mathcal {G}}(z)\in [0,1). \end{aligned}$$
(2.10)

Proof

First, let us observe that the above estimates are true for \(z=0\) (in (2.10) the values at \(r=0,\) of the left and right hand sides, mean the limit if \(r\rightarrow 0^{+}).\) Thus, in the sequel we will assume that \( z\in \mathcal {G\diagdown }\{0\}.\) We start with the estimates (2.9). Since \(f\in \mathcal {K}_{\mathcal {G}}^{k},\) there exist a function \(p\in \mathcal {C_{G}}\) and a function \(g\in \mathcal {M}_{\mathcal {G}}\cap \mathcal { F}_{0,k}(\mathcal {G})\) such that the factorization (2.1) holds. Therefore, we show for such functions g the following inequalities

$$\begin{aligned} \frac{1}{\left( 1+r^{k}\right) ^{\frac{2}{k}}}\le \left| g(z)\right| \le \frac{1}{\left( 1-r^{k}\right) ^{\frac{2}{k}}},r=\mu _{ \mathcal {G}}(z)\in (0,1). \end{aligned}$$

To this aim, let us fix arbitrarily a point \(z\in \mathcal {G}\) such that \( \mu _{\mathcal {G}}(z)=r\in (0,1)\) and let us consider the function

$$\begin{aligned} G(\zeta )=\zeta g(\zeta \frac{z}{\mu _{\mathcal {G}}(z)}),\zeta \in \mathcal {U }. \end{aligned}$$

Then \(\ G\) is (1, k)-symmetric, holomorphic, normalized and satisfies the condition

$$\begin{aligned} {\text {Re}}\frac{\zeta G^{\prime }(\zeta )}{G(\zeta )}={\text {Re}}\frac{ \mathcal {L}g(\zeta \frac{z}{\mu _{\mathcal {G}}(z)})}{g(\zeta \frac{z}{\mu _{ \mathcal {G}}(z)})}>0,\zeta \in \mathcal {U}. \end{aligned}$$

Hence \(G\in \mathcal {S}^{*}\cap \mathcal {F}_{1,k}(\mathcal {U})\) and by [9, Thm. 2.2.13]

$$\begin{aligned} \frac{|\zeta |}{\left( 1+|\zeta |^{k}\right) ^{\frac{2}{k}}}\le \left| G(\zeta )\right| \le \frac{|\zeta |}{\left( 1-|\zeta |^{k}\right) ^{ \frac{2}{k}}},\zeta \in \mathcal {U}. \end{aligned}$$

Putting \(\zeta =\mu _{\mathcal {G}}(z)\) in the above we obtain, by the definition of the function G, the announced inequality.

On the other hand, there hold for \(p\in \mathcal {C_{G}}\) the following estimates [1]

$$\begin{aligned} \frac{1-r}{1+r}\le \left| p(z)\right| \le \frac{1+r}{1-r},r=\mu _{ \mathcal {G}}(z)\in (0,1), \end{aligned}$$

Using the estimates of \(\ \left| p(z)\right| \) and \(\left| g(z)\right| \) we get the estimates (2.9). The sharpness of the upper bounds (2.9) confirms the function given by (2.8). Indeed, for \(r\in (0,1)\) and function \(f\in \mathcal {K}_{\mathcal {G}}^{k}\) given by (2.8), we get

$$\begin{aligned} \mathcal {L}f(z)=\frac{1+r}{1-r}\frac{1}{\left( 1-r^{k}\right) ^{\frac{2}{k}}} \end{aligned}$$

at points \(z\in \mathcal {G},\) \(\mu _{\mathcal {G}}(z)=r\in (0,1)\) such that \(I(z)=r\) (this condition is fulfilled by the points \(z=rz^{*},\) where \( z^{*}\in \partial \mathcal {G}\) and \(I(z^{*})=1).\)

The sharpness of the lower bounds (2.9) can be proven in a similar way.

Now, we prove the estimates (2.10). To obtain the upper bound (2.10), we use the proved above upper bound (2.9) and the fact that the Temljakov operator \(\mathcal {L}\) is invertible and

$$\begin{aligned} \mathcal {L}^{-1}u(z)=\int \limits _{0}^{1}u(tz)dt,u\in \mathcal {H_{G}},z\in \mathcal {G}. \end{aligned}$$

Indeed, we have for \(f\in \mathcal {K}_{\mathcal {G}}^{k}\) and \(z\in \mathcal {G},\,\mu _{\mathcal {G}}(z)=r\in (0,1),\)

$$\begin{aligned} |f(z)|= & {} \left| \mathcal {L}^{-1}\mathcal {L}f(z)\right| =\left| \int \limits _{0}^{1}\mathcal {L}(tz)dt\right| \le \int \limits _{0}^{1} \frac{1+rt}{(1-rt)\left( 1-(rt)^{k}\right) ^{\frac{2}{k}}}dt\\= & {} \frac{1}{r} \int \limits _{0}^{r}\frac{1+\varrho }{(1-\varrho )\left( 1-\varrho ^{k}\right) ^{\frac{2}{k}}}d\varrho . \end{aligned}$$

To prove the lower bound (2.10) let us consider the function

$$\begin{aligned} F(\zeta )=\zeta f\left( \zeta \frac{z}{\mu _{\mathcal {G}}(z)}\right) ,\zeta \in \mathcal {U}, \end{aligned}$$

with arbitrarily fixed \(f\in \mathcal {K}_{\mathcal {G}}^{k}\) and \(z\in \mathcal {G},\,\mu _{\mathcal {G}}(z)=r\in (0,1).\) Since

$$\begin{aligned} F^{\prime }(\zeta )=\mathcal {L}f\left( \zeta \frac{z}{\mu _{\mathcal {G}}(z)} \right) ,\zeta \in \mathcal {U}, \end{aligned}$$

we get, by Theorem 1, that there exist functions \(g\in \mathcal {M}_{ \mathcal {G}}\cap \mathcal {F}_{0,k}(\mathcal {G})\) and \(p\in \mathcal {C_{G}}\) such that the factorization (2.1) is true. Thus

$$\begin{aligned} F^{\prime }(\zeta )=P(\zeta )\cdot G(\zeta ),\zeta \in \mathcal {U}, \end{aligned}$$

where for \(\zeta \in \mathcal {U}\)

$$\begin{aligned} G(\zeta )=\zeta g\left( \zeta \frac{z}{\mu _{\mathcal {G}}(z)}\right) ,P(\zeta )=p\left( \zeta \frac{z}{\mu _{\mathcal {G}}(z)}\right) . \end{aligned}$$

Moreover, \(G\in \mathcal {S}^{*}\cap \mathcal {F}_{1,k}(\mathcal {U})\) ( see the proof of the estimates (2.9)) and \(P:\mathcal {U}\rightarrow \mathbb {C},P(0)=1,\) is a holomorphic function with a positive real part. Therefore, \(\ F\) belongs to a subclass \(\mathcal {K}^{(k)}\) (considered in [22] and for \(k=2\) in [8]) of the class of close-to-convex functions. Hence, F is univalent in the disc \(\mathcal {U}\).

On the other hand, by the lower bound (2.9), we have that

$$\begin{aligned} |F^{\prime }(\zeta )|\ge \frac{1-|\zeta |}{1+|\zeta |}\frac{1}{\left( 1+|\zeta |^{k}\right) ^{\frac{2}{k}}}, \end{aligned}$$

because \(r=\mu _{\mathcal {G}}\left( \zeta \frac{z}{\mu _{\mathcal {G}}(z)} \right) =|\zeta |.\) Now we show that

$$\begin{aligned} |F(\zeta )|\ge \int \limits _{0}^{r}\frac{1-\varrho }{1+\varrho }\frac{1}{ \left( 1+\varrho ^{k}\right) ^{\frac{2}{k}}}d\varrho ,|\zeta |=r\in (0,1). \end{aligned}$$

To this aim, it is sufficient to show that it holds for the nearest point \( F(\zeta _{0})\) from zero \((\left| \zeta _{0}\right| =r\in (0,1)),\) otherwise, we have \(|F(\zeta )|\ge |F(\zeta _{0})|,\left| \zeta \right| =r.\) Since F is univalent in the disc \(\mathcal {U},\) the original image of the line segment \(\overline{0,F(\zeta _{0})}\) is a piece of arc \(F^{-1}\left( \overline{0,F(\zeta _{0})}\right) \) in the disc \(r \overline{\mathcal {U}}.\) Thus

$$\begin{aligned} |F(\zeta _{0})|= & {} \int \limits _{\overline{0,F(\zeta _{0})}}|dw|=\int \limits _{F^{-1}\left( \overline{0,F(\zeta _{0})}\right) }\left| F^{\prime }(\zeta )\right| |d\zeta \nonumber \\\ge & {} \int \limits _{0}^{r}\frac{ 1-\varrho }{1+\varrho }\frac{1}{\left( 1+\varrho ^{k}\right) ^{\frac{2}{k}}} d\varrho ,r\in (0,1)). \end{aligned}$$

Thus, by the definition of F, we get

$$\begin{aligned} \left| \zeta f\left( \zeta \frac{z}{\mu _{\mathcal {G}}(z)}\right) \right| \ge \int \limits _{0}^{r}\frac{1-\varrho }{1+\varrho }\frac{1}{ \left( 1+\varrho ^{k}\right) ^{\frac{2}{k}}}d\varrho ,|\zeta |=r\in (0,1). \end{aligned}$$

Hence, putting \(\zeta =\) \(\mu _{\mathcal {G}}\left( z\right) \) \(=r\in (0,1),\) we have the lower bound (2.10).

Finally, let us note that we obtain the equalities in the inequalities (2.10) for the function (2.8) in adequate points \(z\in \mathcal {G}.\) \(\square \)

We close the paper with a sufficient condition guaranteeing that a function \( f\in \mathcal {H_{G}}(1)\) belongs to \(\mathcal {K}_{\mathcal {G}}^{k}.\) We formulate it in the term of \(\mathcal {G}\)-balances of m-honogeous polynomials in developments of functions from \(\mathcal {H_{G}}(1).\)

Theorem 5

Let \(f\in \mathcal {H_{G}}(1)\) has the form (2.2). If there exists a function \(g\in \mathcal {M}_{\mathcal {G}}\cap {\mathcal {F}}_{0,k}( \mathcal {G})\) of the form (2.3) such that

$$\begin{aligned} \sum _{m=1}^{\infty }(m+1)\mu _{\mathcal {G}}\left( Q_{f,m}\right) \mathcal {+} \sum _{m=1}^{\infty }\mu _{\mathcal {G}}\left( Q_{g,mk}\right) \le 1, \end{aligned}$$

then \(f\in \mathcal {K}_{\mathcal {G}}^{k}.\)

Proof

Since g, as a function from \(\mathcal {M}_{\mathcal {G}}\) omits zero [1], we will prove that

$$\begin{aligned} {\text {Re}}\frac{\mathcal {L}f(z)}{g(z)}>0,z\in \mathcal {G}. \end{aligned}$$

To do it, we compute step by step

$$\begin{aligned}&|\mathcal {L}f(z)-g(z)|-|\mathcal {L}f(z)+g(z)| \\&\quad =\left| \sum _{m=1}^{\infty }(m+1)Q_{f,m}\left( z\right) \mathcal {-} \sum _{m=1}^{\infty }Q_{g,mk}\left( z\right) \right| -\left| 2+\sum _{m=1}^{\infty }(m+1)Q_{f,m}\left( z\right) \right. \\&\qquad \left. \mathcal {+} \sum _{m=1}^{\infty }Q_{g,mk}\left( z\right) \right| \\&\quad \le 2\left[ \sum _{m=1}^{\infty }(m+1)\left| Q_{g,mk}\left( z\right) \right| +\sum _{m=1}^{\infty }\left| Q_{g,mk}\left( z\right) \right| -1\right] \\&\quad \le 2\left[ \sum _{m=1}^{\infty }(m+1)\mu _{\mathcal {G}}\left( Q_{f,m}\right) +\sum _{m=1}^{\infty }\mu _{\mathcal {G}}\left( Q_{g,mk}\right) -1\right] \le 0. \end{aligned}$$

Thus

$$\begin{aligned} \left| \frac{\mathcal {L}f(z)}{g(z)}-1\right| \le \left| \frac{ \mathcal {L}f(z)}{g(z)}+1\right| ,z\in \mathcal {G} \end{aligned}$$

and hence

$$\begin{aligned} {\text {Re}}\frac{\mathcal {L}f(z)}{g(z)}\ge 0,z\in \mathcal {G}. \end{aligned}$$

This gives the mentioned inequality by a maximum principle for pluriharmonic functions of several complex variables. Putting \(p(z)=\frac{\mathcal {L}f(z)}{ g(z)},\) \(z\in \mathcal {G},\) we obtain that the transform \(\mathcal {L}f\) has the factorization (1) with \(g\in \mathcal {M}_{\mathcal {G}}\cap \mathcal {F} _{0,k}(\mathcal {G})\) and \(p\in \mathcal {C_{G}}.\ \)Consequently, \(f\in \mathcal {K}_{\mathcal {G}}^{k}\). \(\square \)