Wiman’s Type Inequality in Multiple-Circular Domain

Abstract

In the paper we prove for the first time an analogue of the Wiman inequality in the class of analytic functions $f\in {\mathcal{A}}_0^p\left(\mathbb{G}\right)$ in an arbitrary complete Reinhard domain $\mathbb{G}\subset {\mathbb{C}}^p$, $p\in \mathbb{N}$ represented by the power series of the form $f\left(z\right)=f(z_1,\cdots ,z_p)={\sum }_{\Vert n\Vert =0}^{+\infty }{a}_n{z}^n$ with the domain of convergence $\mathbb{G}.$ We have proven the following statement: If $f\in {\mathcal{A}}^p\left(\mathbb{G}\right)$ and $h\in {\mathcal{H}}^p$, then for a given $\epsilon =({\epsilon }_1,\dots ,{\epsilon }_p)\in {\mathbb{R}}_{+}^p$ and arbitrary $\delta >0$ there exists a set $E\subset \left|G\right|$ such that ${\int }_{E\cap {\Delta }_{\epsilon }}\frac{h\left(r\right)d{r}_1\cdots d{r}_p}{r_1\cdots r_p}<+\infty$ and for all $r\in {\Delta }_{\epsilon }\setminus E$ we have $M_f\left(r\right)\le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{\frac{p+1}{2}}{ln}^{\frac{p}{2}+\delta }h\left(r\right){ln}^{\frac{p}{2}+\delta }\left\{{\mu }_f\left(r\right)h\left(r\right)\right\}{\prod }_{j=1}^p(ln\frac{e{r}_j}{{\epsilon }_j}{)}^{\frac{p-1}{2}+\delta }.$ Note, that this assertion at $p=1,$ $\mathbb{G}=\mathbb{C},$ $h\left(r\right)\equiv \mathrm{const}$ implies the classical Wiman–Valiron theorem for entire functions and at $p=1,$ the $\mathbb{G}=\mathbb{D}:=\{z\in \mathbb{C}:|z|<1\},$ $h\left(r\right)\equiv 1/(1-r)$ theorem about the Kővari-type inequality for analytic functions in the unit disc $\mathbb{D}$; $p>1$ implies some Wiman’s type inequalities for analytic functions of several variables in ${\mathbb{C}}^n\times {\mathbb{D}}^k$, $n,k\in {\mathbb{Z}}_{+},n+k\in \mathbb{N}.$

1. Introduction: Notations and Preliminaries

Let $\mathbb{C}$, $\mathbb{R}$, $\mathbb{Z}$, $\mathbb{N}$ be sets of complex numbers, real numbers, integers, and positive integers, respectively, and ${\mathbb{Z}}_{+}=\mathbb{N}\cup \left\{0\right\}$. We denote by ${\mathcal{A}}_0^p\left(\mathbb{G}\right),$ $p\in \mathbb{N},$ the class of an analytic functions f in a complete Reinhardt domain $\mathbb{G}\subset {\mathbb{C}}^p$, represented by the power series of the form

$$f\left(z\right)=f(z_1,\dots ,z_p)=\sum _{\parallel n\parallel =0}^{+\infty }{a}_n{z}^n,$$

(1)

with the domain of convergence $\mathbb{G}$, where $z^n=z_1^{n_1}\dots z_p^{n_p},z=(z_1,\dots ,z_p)\in \mathbb{G},n=(n_1,\dots ,n_p)\in {\mathbb{Z}}_{+}^p,\parallel n\parallel ={\sum }_{j=1}^p{n}_j$; ${\mathcal{E}}^p:={\mathcal{A}}_0^p\left({\mathbb{C}}^p\right)$ is the class of entire functions of several variables (i.e., analytic functions in ${\mathbb{C}}^p$); by ${\mathcal{E}}_R:={\mathcal{A}}_0^1\left({\mathbb{D}}_R\right)$ $(0<R\le +\infty )$ we denote the class of analytic functions of one complex variable in a disk ${\mathbb{D}}_R=\{z\in \mathbb{C}:|z|<R\}$. In particular, $\mathcal{E}:={\mathcal{E}}_{+\infty }={\mathcal{E}}^1$ is the class of entire functions of one complex variable.

For a function $f\in {\mathcal{A}}_0^p\left(\mathbb{G}\right)$ of form (1) with domain of convergence $\mathbb{G}$ and $r=(r_1,\cdots ,r_p)\in \left|G\right|:=\{r=(r_1,\dots ,r_p):r_j=|z_j|,z=(z_1,\dots ,z_p)\in \mathbb{G}\}$ we denote

$$\begin{array}{c}{\Delta }_{r_0}=\{t\in |G|:t_j\ge r_j^0,j\in \{1,\cdots ,p\}\},{\mu }_f\left(r\right)=max\left\{\right|a_n|r_1^{n_1}\cdots r_p^{n_p}:n\in {\mathbb{Z}}_{+}^p\},\\ M_f\left(r\right)=max\left\{\right|f\left(z\right)|:|z_1|=r_1,\cdots ,|z_p|=r_p\},{\mathfrak{M}}_f\left(r\right)=\sum _{\parallel n\parallel =0}^{+\infty }\left|a_n\right|r^n.\end{array}$$

On the one hand, it is well-known that every analytic function f in the complete Reinhardt domain $\mathbb{G}$ with a center at $z=0$ can be represented in $\mathbb{G}$ by the series of form (1). On the other hand, the domain of convergence of each series of form (1) is the logarithmically-convex complete Reinhardt domain with the center $z=0$.

We say that a domain $\mathbb{G}\subset {\mathbb{C}}^p$ is the complete Reinhardt domain if:

(a)

$z=(z_1,\cdots ,z_p)\in \mathbb{G}⟹(\forall R=(R_1,\dots ,R_p)\in {[0,1]}^p):$$Rz=(R_1{z}_1,\cdots ,R_p{z}_p)\in \mathbb{G}$ (a complete domain);

(b)

$(z_1,\dots ,z_p)\in \mathbb{G}⟹(\forall ({\theta }_1,\dots ,{\theta }_p)\in {\mathbb{R}}^p):$$(z_1{e}^{i{\theta }_1},\dots ,z_p{e}^{i{\theta }_p})\in \mathbb{G}$ (a multiple-circular domain).

The Reinhardt domain $\mathbb{G}$ is called logarithmically-convex if the image of the set $G^{*}=\{z\in \mathbb{G}:z_1·\dots ·z_p\ne 0\}$ under the mapping $Ln:z\to Ln\left(z\right)=(ln|z_1|,\dots ,ln|z_p\left|\right)$ is a convex set in the space ${\mathbb{R}}^p$. In one complex variable ($p=1$), a logarithmically-convex Reinhardt domain is a disc. The following complete Reinhardt domains ($p\ge 2$) are considered most frequently:

$$\begin{array}{c}{C}_p\left(R\right):=\{z\in {\mathbb{C}}^p:|z_1|<R_1,\dots ,|z_p|<R_p\},R=(R_1,\dots ,R_p)\in {(0,+\infty )}^p,\left(\mathrm{polydisk}\right),\\ {\mathbb{B}}_p\left(r\right):=\{z\in {\mathbb{C}}^p:\left|z\right|:=\sqrt{|z_1{|}^2+\dots +{\left|z_p\right|}^2}<r\}\left(\mathrm{ball}\right),\\ {\Pi }_p\left(r\right):=\{z\in {\mathbb{C}}^p:|z_1|+\dots +|z_p|<r\},r>0.\end{array}$$

Note, that $C_p\left(R\right)\subset \mathbb{G}$ for every $w=(w_1,\dots ,w_p)\in \mathbb{G}$ and $R=\left(\right|w_1|,\dots ,|w_p\left|\right)$. The domains $C_p\left(r{e}_1\right)$, $e_1=(1,\dots ,1)\in {\mathbb{R}}^p$, ${\mathbb{B}}_p\left(r\right)$, ${\Pi }_p\left(r\right)$ $(r>0)$ are the logarithmically-convex complete Reinhardt domains. However, for example, the complete Reinhardt domain ${\displaystyle G_{1,2}=\{z=(z_1,z_2):|z_1|<1,|z_2|<2\}\cup \{z=(z_1,z_2):|z_1|<2,|z_2|<1\}}$ is not a logarithmically-convex domain.

2. Wiman’s Type Inequality for Analytic Functions of One Variable

In article [1] the following statement is proved.

Theorem 1

([1]). Let a nondecreasing function $h:[0,R)\to [10,\infty )$ such that ${\int }_{r_0}^{R}h\left(r\right)dlnr=+\infty$ for some $r_0\in (0,R)$. If $f\in {\mathcal{E}}_R$, $R\in (0,+\infty ]$ is an analytic function represented by a power series of the form ${\displaystyle f\left(z\right)=\sum _{n=0}^{+\infty }{a}_n{z}^n,}$ then $(\forall \delta >0)(\exists E(\delta ,f,h)=E\subset (0,R))(\exists r_0\in (0,R))$ $(\forall r\in (r_0,R)\setminus E)$

$$M_f\left(r\right)\le h\left(r\right){\mu }_f\left(r\right){\{lnh\left(r\right)ln\left(h\left(r\right){\mu }_f\left(r\right)\right)\}}^{1/2+\delta }and{\int }_{E\cap (r_0,R)}\frac{h\left(r\right)}{r}dr<+\infty ,$$

where $M_f\left(r\right)=max\left\{\right|f\left(z\right)|:|z|=r\}$ is the maximum modulus and ${\mu }_f\left(r\right)=max\left\{\right|a_n|r^n:n\ge 0\}$ is the maximal term of power series.

For nonconstant entire functions $f\in \mathcal{E}$ we can choose $h\left(r\right)=10$ and $\delta =\epsilon /2$ for an arbitrarily given $\epsilon >0.$ Then, from Theorem 1 we obtain the assertion of the classical Wiman–Valiron theorem on Wiman’s inequality (for example see [2], [3] (p. 9), [4,5], [6] (p. 28), [7,8,9,10]), i.e., that for all $r\in (r_0,+\infty )\setminus E$, ${\int }_{E}dlnr<+\infty$, we have

$$\begin{array}{c}{M}_f\left(r\right)\le 10{\mu }_f\left(r\right){\{ln10ln\left(10{\mu }_f\left(r\right)\right)\}}^{1/2+\delta }\le {\mu }_f\left(r\right){ln}^{1/2+\epsilon }{\mu }_f\left(r\right).\end{array}$$

(2)

For analytic functions $f\in {\mathcal{E}}_1$ in the unit disk ${\mathbb{D}}_1$ we can choose $h\left(r\right)=\frac{r}{1-r}.$ Then,

$$\begin{array}{c}{M}_f\left(r\right)\le \frac{r{\mu }_f\left(r\right)}{1-r}\{ln\frac{r}{1-r}ln(\frac{r{\mu }_f\left(r\right)}{1-r}){\}}^{1/2+\delta }\le \\ \le \frac{{\mu }_f\left(r\right)}{{(1-r)}^{1+\delta }}{ln}^{1/2+\delta }\frac{{\mu }_f\left(r\right)}{1-r},\mathrm{as} r\to 1-0,r\notin E,{\int }_E\frac{dr}{1-r}<+\infty ,\end{array}$$

i.e., the theorem about the Kővari-type inequality for analytic functions in the unit disc $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$ ([11,12]).

Regarding the statement about the Wiman inequality (2), Prof. I.V. Ostrovskii in 1995 formulated the following problem: What is the best possible description of the value of an exceptional set E? In article [7], the authors found, in a sense, the best possible description of the magnitude of the exceptional set E in inequality (2) for entire functions of one complex variable. In fact, we obtain, in a sense, the best possible description for each entire function f for $h\left(r\right)=ln{\mu }_f\left(r\right)$.

The same issue was considered in a number of articles (for example, see [13,14,15]) in relation to many other relations obtained in the Wiman–Valiron theory.

Note, that for analytic functions $f\in {\mathcal{E}}_1$ such a problem is still open. Theorem 1 contains a new description of the exceptional set in the inequality (2) for analytic functions $f\in {\mathcal{E}}_1$. Perhaps the best possible description of an exceptional set is also obtained with $h\left(r\right)=ln{\mu }_f\left(r\right).$

3. Wiman’s Type Inequality for Analytic Functions of Several Variables

Some analogues of Wiman’s inequality for entire functions of several complex variables can be found in [16,17,18,19,20,21,22], and for analytic functions in the polydisc ${\mathbb{D}}^p,p\ge 2,$ in [23,24].

In paper [25] some analogues of Wiman’s inequality are proven for the analytic $f\left(z\right)$ and random analytic $f(z,t)$ functions on $\mathbb{G}={\mathbb{D}}^{\ell }\times {\mathbb{C}}^{p-\ell }$, $\ell \in \mathbb{N},1\le \ell <p$, $I=\{1,\dots ,\ell \},J=\{\ell +1,\dots ,p\}$ of the form (1) and $f(z,t)={\sum }_{\parallel n\parallel =0}^{+\infty }{a}_n{Z}_n\left(t\right)z^n$, respectively. Here, $Z=\left(Z_n\right)$ is a multiplicative system of complex random variables on the Steinhaus probability space, almost surely (a.s.) uniformly bounded by the number 1. In particular, the following statements are proven:

Theorem 2

([25]). Let $f\in {\mathcal{A}}^p\left(\mathbb{G}\right)$, $\mathbb{G}={\mathbb{D}}^{\ell }\times {\mathbb{C}}^{p-\ell }$, $\ell \in \mathbb{N},1\le \ell <p$. For every $\delta >0$ there exist the sets $E_1=E_1(\delta ,f),E_2=E_2(\delta ,f)\subset {[0,1)}^l\times {(1,+\infty )}^{p-l}$ of asymptotically finite logarithmic measure (i.e., ${\int }_{{\Delta }_{\epsilon }\cap {[0,1)}^{\ell }\times {\mathbb{R}}^{p-\ell }}\frac{d{r}_1·\dots ·d{r}_{\ell }·d{r}_{\ell +1}·\dots ·d{r}_p}{(1-r_1)·\dots ·(1-r_{\ell })·r_{\ell +1}·\dots ·r_p}<+\infty$ for some $\epsilon >0$), such that the inequalities

$$\begin{array}{c}{M}_f\left(r\right)\le {\mu }_f\left(r\right)\prod _{i\in I}\frac{1}{{(1-r_i)}^{1+\delta }}{ln}^{p/2+\delta }\left({\mu }_f\left(r\right)\prod _{i\in I}\frac{1}{1-r_i}\right)(\prod _{j\in J}ln{r}_j{)}^{p+\delta },\\ M_f(r,t)\le {\mu }_f\left(r\right)\prod _{i\in I}\frac{1}{{(1-r_i)}^{1/2+\delta }}{ln}^{p/4+\delta }\left({\mu }_f\left(r\right)\prod _{i\in I}\frac{1}{1-r_i}\right)(\prod _{j\in J}ln{r}_j{)}^{p/2+\delta }.\end{array}$$

(3)

hold for all $r\in \left|G\right|\setminus E_1$ and for all $r\in \left|D\right|\setminus E_2$ a.s. in t, respectively.

The sharpness of the obtained inequalities is also proven.

The main purpose of this article is to prove analogues of Theorems 1 and 2 in the class of analytic functions $f\in {\mathcal{A}}_0^p\left(\mathbb{G}\right)$ for the arbitrary complete Reinhardt domain $\mathbb{G}$.

4. Main Result

The aim of this paper is to prove some analogues of Wiman’s inequality for the analytic functions $f\in {\mathcal{A}}_0^p\left(\mathbb{G}\right)$ represented by the series of form (1) with the arbitrary complete Reinhardt domain of convergence $\mathbb{G}$. By ${\mathcal{A}}^p\left(\mathbb{G}\right)$ we denote a subclass of functions $f\in {\mathcal{A}}_0^p\left(\mathbb{G}\right)$ such that $\frac{\partial }{\partial z_j}f(z_1,\cdots ,z_p)\not\equiv 0$ in $\mathbb{G}$ for any $j\in \{1,\dots ,p\}$.

Let ${\mathcal{H}}^p$ be the class of functions $h:\left|G\right|\to {\mathbb{R}}_{+}$ such that h is nondecreasing with respect to each variable and $h\left(r\right)>10$ for all $r\in \left|G\right|$ and

$$\underset{{\Delta }_{\epsilon }}{\int }\frac{h\left(r\right)d{r}_1\cdots d{r}_p}{r_1\cdots r_p}=+\infty$$

for every $\epsilon \in {\mathbb{R}}_{+}^p$ such that $\left|G\right|\cap {\Delta }_{\epsilon }$ is a nonempty domain in ${\mathbb{R}}_{+}^p.$

For $h\in {\mathcal{H}}^p$ we say that $E\subset \left|G\right|$ is the set of finite h-measure on$\left|G\right|$ if for some $\epsilon \in {\mathbb{R}}_{+}^p$ such that $\left|G\right|\cap {\Delta }_{\epsilon }$ is a nonempty domain in $\left|G\right|\subset {\mathbb{R}}_{+}^p$ one has

$${\nu }_h(E\cap {\Delta }_{\epsilon }):=\underset{E\cap {\Delta }_{\epsilon }}{\int }\frac{h\left(r\right)d{r}_1\cdots d{r}_p}{r_1\cdots r_p}<+\infty .$$

We denote a set of such sets by ${\mathcal{S}}_h.$

Theorem 3.

Let $f\in {\mathcal{A}}^p\left(\mathbb{G}\right).$ Then, for every $\epsilon \in {\mathbb{R}}_{+}^p,\delta >0$ there exists a set $E\in {\mathcal{S}}_h$ such that for all $r\in {\Delta }_{\epsilon }\setminus E$ the following inequality takes place:

$$\begin{array}{c}{M}_f\left(r\right)\le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{\frac{p+1}{2}}{ln}^{\frac{p}{2}+\delta }h\left(r\right){ln}^{\frac{p}{2}+\delta }\left\{{\mu }_f\left(r\right)h\left(r\right)\right\}\prod _{j=1}^p(\prod _{k=1,k\ne j}^{p}ln\frac{e{r}_k}{{\epsilon }_k}{)}^{\frac{1}{2}+\delta }.\end{array}$$

(4)

Remark 1.

Choosing $p=1$ and $\mathbb{G}={\mathbb{D}}_R$ in Theorem 3 leads to the result in Theorem 1.

5. Auxiliary Lemmas

The proof of the main result uses the probabilistic reasoning from [17,18] (see also [20]), which has already become traditional in this topic, and differs from the proofs of similar statements in [25].

Our proof actually uses a number of lemmas (Lemmas 1–4) from article [18]. But their proofs in article [18] are not written with sufficient completeness, and also contain inaccuracies in reasoning. Therefore, we present them here along with the complete proofs.

In order to prove a Wiman’s type inequality for analytic functions in $\mathbb{G}$ we need the following auxiliary results.

Let $D_f\left(r\right)=\left(D_{ij}\right)$ be a $p\times p$ matrix such that

$$D_{ij}=r_i\frac{\partial }{\partial r_i}(r_j\frac{\partial }{\partial r_j}ln{M}_f\left(r\right))={\partial }_i{\partial }_{j}ln{M}_f\left(r\right),{\partial }_i=r_i\frac{\partial }{\partial r_i},i,j\in \{1,\cdots ,p\}.$$

Let I be an identity matrix of order $p.$

For the set $E\subset {\mathbb{R}}^p$ by $\#(E\cap {\mathbb{Z}}_{+}^p)$ we denote the quantity of the elements of set $E\cap {\mathbb{Z}}_{+}^p.$

Lemma 1.

Let B be parallelepiped in ${\mathbb{R}}^p$ with edges of the lengths $l_1,l_2,\dots ,l_p$ so that there exists an isometry $H:{\mathbb{R}}^p\to {\mathbb{R}}^p$ such that

$$H:B\to \{x\in {\mathbb{R}}^p:|x_j|\le l_j/2,j\in \{1,2,\dots p\}\}.$$

Then,

$$\#(B\cap {\mathbb{Z}}^p)\le {\lambda }_p\prod _{j=1}^p(l_j+1),$$

where ${\lambda }_p$ is the inverse value to the volume of a sphere with the radius $\frac{1}{2}$ in ${\mathbb{R}}^p,$ i.e., ${\lambda }_p=\frac{2^n·\mathrm{\Gamma }\left(\frac{n+2}{2}\right)}{{\pi }^{n/2}}$.

Proof. 

Denote

$$\begin{array}{c}{B}^{\prime }=\{x\in {\mathbb{R}}^p:\left|x_j\right|\le \frac{l_j}{2},j\in \{1,\dots ,p\}\},\\ B^{*}=\{x\in {\mathbb{R}}^p:\left|x_j\right|\le \frac{l_j+1}{2},j\in \{1,\dots ,p\}\}\supset B.\end{array}$$

Let $S\left(n\right)$ be an open sphere with a center at $n\in {\mathbb{Z}}_{+}^p$ with radius $\frac{1}{2}.$ Note that

$$\bigcup _{n\in {\mathbb{Z}}_{+}^p\cap B}S\left(n\right)\subseteq B^{*}.$$

By the monotony of the Lebesgue measure $\mu$ in ${\mathbb{R}}^p$ we obtain

$$\mu \left(\bigcup _{n\in {\mathbb{Z}}_{+}^p\cap B}S\left(n\right)\right)\le \mu \left(B^{*}\right).$$

Finally, by the additivity of this measure, we obtain

$$\begin{array}{c}\mu \left(S(1/2)\right)·\#\{B\cap {\mathbb{Z}}_{+}^p\}\le \prod _{j=1}^p(l_j+1),\#\{B\cap {\mathbb{Z}}_{+}^p\}\le \frac{1}{\mu \left(S\right(1/2\left)\right)}\prod _{j=1}^p(l_j+1).\end{array}$$

 □

By $A^{+}$ we denote the Moore–Penrose inverse matrix of A ([18,26]), i.e.,

$$A^{+}=\underset{\delta \to 0}{lim}{A}^T{(A{A}^T+\delta I)}^{-1}.$$

Lemma 2.

Let $\alpha \in {\mathbb{R}}^p,C>0$, A be a $p\times p$ nonnegative matrix $0<m=\mathrm{rank}A\le p$ and

$$E=\{x\in {\mathbb{R}}^m:(x-\alpha )A^{+}{(x-\alpha )}^T\le C\}.$$

There exists a constant $\delta =\delta (C,p)>0$ that does not depend on A and α such that

$$\#\{E\cap {\mathbb{Z}}_{+}^m\}\le \delta {(det(A+I))}^{1/2}.$$

Proof. 

Let $0<{\lambda }_1\le {\lambda }_2\le \dots \le {\lambda }_m$ be positive eigenvalues of the matrix $A.$ Then, $\frac{1}{{\lambda }_1},\frac{1}{{\lambda }_2},\dots ,\frac{1}{{\lambda }_m}$ are eigenvalues of the matrix $A^{+}.$ Thus, there exists an isometry $H:{\mathbb{R}}^m\to {\mathbb{R}}^m,$ such that

$$\begin{array}{c}H:E\to \{x\in {\mathbb{R}}^m:\sum _{j=1}^m\frac{x_j^2}{{\lambda }_j}\le C\},E\subset H^{-1}\{x\in {\mathbb{R}}^m:x\le \sqrt{C{\lambda }_j},j\in \{1,2,\dots ,m\}\}.\end{array}$$

By Lemma 1 there exists a constant ${\delta }^{\prime }>0$ such that

$$\#\{E\cap {\mathbb{Z}}_{+}^m\}\le {\delta }^{\prime }\prod _{j=1}^m(2{\left(C{\lambda }_j\right)}^{1/2}+1)={\delta }^{\prime }\prod _{j=1}^p(2{\left(C{\lambda }_j\right)}^{1/2}+1).$$

It remains to remark that

$$\begin{array}{c}\prod _{j=1}^p({\lambda }_j+1)=det(A+I),\#\{E\cap {\mathbb{Z}}_{+}^p\}\le {\delta }^{\prime }{\left(\sqrt{2C}\right)}^p(\prod _{j=1}^p({\lambda }_j+1){)}^{1/2}\le {\delta }^{″}{(det(A+I))}^{1/2}.\end{array}$$

 □

Lemma 3.

Let $\xi ={({\xi }_1,{\xi }_2,\dots ,{\xi }_p)}^T$ be a random vector, $\alpha =M\xi ={(M{\xi }_1,M{\xi }_2,\dots ,M{\xi }_p)}^T,$ A covariance matrix of ξ, $\delta >0,0<m=\mathrm{rank}A\le p.$ Then,

$$P\{\omega :(\xi \left(\omega \right)-\alpha )A^{+}{(\xi \left(\omega \right)-\alpha )}^T\le \delta \}\ge 1-\frac{m}{\delta }.$$

Proof. 

Let us consider the random variable

$$Z\left(\omega \right)={(\xi \left(\omega \right)-\alpha )}^T{A}^{+}(\xi \left(\omega \right)-\alpha ).$$

As A is non-negative, then $\forall \omega \in \mathrm{\Omega }:Z\left(\omega \right)\ge 0.$ Moreover, as A is also symmetric, there exists an orthogonal matrix G such that $G{G}^T=G^{T}G=I$ and $G^{T}AG=Q.$ Here, I is the identity matrix of order p and $Q=\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m,0,\dots ,0)$ is the diagonal matrix with the ordered eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \dots \ge {\lambda }_m>0,$ $0<m=\mathrm{rank}A\le p.$ Then (see, for example [26,27]),

$$\begin{array}{c}{G}^{T}AG=Q,G{G}^{T}AG{G}^T=GQ{G}^T⟹A=GQ{G}^T,\\ A^{+}={\left(GQ{G}^T\right)}^{+}={\left(G^T\right)}^{+}{Q}^{+}{G}^{+}={\left(G^T\right)}^{-1}{Q}^{+}{G}^{-1}=G{Q}^{+}{G}^T,\end{array}$$

i.e., $A=GQ{G}^T,A^{+}=G{Q}^{+}{G}^T.$ Therefore,

$$\begin{array}{c}Z={(\xi -\alpha )}^T{A}^{+}(\xi -\alpha )={(\xi -\alpha )}^{T}G{Q}^{+}{G}^T(\xi -\alpha )=\\ ={(\xi -\alpha )}^{T}G{Q}^{-1/2}{Q}^{-1/2}{G}^T(\xi -\alpha )={\left(Q^{-1/2}{G}^T(\xi -\alpha )\right)}^T\left(Q^{-1/2}{G}^T(\xi -\alpha )\right)=Y^{T}Y,\end{array}$$

where $Y=Q^{-1/2}{G}^T(\xi -\alpha ),Q^{-1/2}=\mathrm{diag}({\lambda }_1^{-1/2},{\lambda }_2^{-1/2},\dots ,{\lambda }_m^{-1/2},0,\dots ,0).$ The expected value and covariance of the random vector Y satisfy the equations

$$\begin{array}{c}MY=M\left(Q^{-1/2}{G}^T(\xi -\alpha )\right)=Q^{-1/2}{G}^{T}M(\xi -\alpha )=0,\\ \mathrm{cov}Y=\mathrm{cov}\left(Q^{-1/2}{G}^T(\xi -\alpha )\right)=\mathrm{cov}\left(Q^{-1/2}{G}^T\xi \right)=Q^{-1/2}{G}^T\mathrm{cov}\left(\xi \right){\left(Q^{-1/2}{G}^T\right)}^T=\\ =Q^{-1/2}{G}^{T}AG{Q}^{-1/2}=Q^{-1/2}Q{Q}^{-1/2}=\mathrm{diag}(\underset{m \mathrm{times}}{\underbrace{1,\dots ,1}},\underset{p-m \mathrm{times}}{\underbrace{0,\dots ,0}}).\end{array}$$

Therefore,

$$MZ=M\left(Y^{T}Y\right)=M\left(\sum _{j=1}^p{Y}_j^2\right)=\sum _{j=1}^{p}M\left(Y_i^2\right)=\sum _{j=1}^{p}D\left(Y_j\right)=m.$$

Finally, using Markov’s inequality we obtain

$$\begin{array}{c}P\{\omega :Z\left(\omega \right)\ge \delta \}=P\{\omega :(\xi \left(\omega \right)-\alpha )A^{+}{(\xi \left(\omega \right)-\alpha )}^T\le \delta \}\le \frac{MZ}{\delta }=\frac{m}{\delta }.\\ P\{\omega :(\xi \left(\omega \right)-\alpha )A^{+}{(\xi \left(\omega \right)-\alpha )}^T\le \delta \}\ge 1-\frac{m}{\delta }.\end{array}$$

 □

Lemma 4

(Theorem 3.1, [18]). Let $f\in {\mathcal{A}}^p.$ There exists a constant $C_0\left(p\right)$ such that

$${\mathfrak{M}}_f\left(r\right)\le C_0\left(p\right){\mu }_f\left(r\right){(det(D_f\left(r\right)+I))}^{1/2},$$

where I is the identity $p\times p$ matrix.

Proof. 

Let us consider random vector $X\left(\omega \right)=(X_1\left(\omega \right),X_2\left(\omega \right),\dots ,X_p\left(\omega \right))$ such that

$$\begin{array}{c}P\{\omega :X_j\left(\omega \right)=n_j,j\in \{1,\dots ,p\}\}=\frac{1}{{\mathfrak{M}}_f\left(r\right)}\left|a_{n_1\dots n_p}\right|r_1^{n_1}\dots r_p^{n_p},k\in {\mathbb{Z}}_{+}.\end{array}$$

Then for $j\in \{1,2,\dots ,p\}$ we obtain

$$\begin{array}{c}M{X}_j=\frac{1}{{\mathfrak{M}}_f\left(r\right)}\sum _{\parallel n\parallel =0}^{+\infty }{n}_j\left|a_n\right|r^n=r_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right).\end{array}$$

$D_f\left(r\right)$ is covariance matrix of random vector $X\left(\omega \right)$.

One can choose $\delta =2p$ in Lemma 3. We then obtain

$$\begin{array}{c}\frac{1}{2}\le 1-\frac{m}{2p}\le P\{\omega :(x-\alpha )D_f^{+}{(x-\alpha )}^T\le 2p\}\le \\ \le \frac{{\mu }_f\left(r\right)}{{\mathfrak{M}}_f\left(r\right)}·\#\{x\in {\mathbb{R}}_{+}^p:(x-\alpha )D_f^{+}{(x-\alpha )}^T\le 2p\}\le 2p\frac{{\mu }_f\left(r\right)}{{\mathfrak{M}}_f\left(r\right)}{\left(\det (D_f\left(r\right)+I)\right)}^{1/2},\\ {\mathfrak{M}}_f\left(r\right)\le 4p{\mu }_f\left(r\right){(det(D_f\left(r\right)+I))}^{1/2}.\end{array}$$

 □

Lemma 5.

Let $f\in {\mathcal{A}}^p.$ Then for $\epsilon \in {\mathbb{R}}_{+}^p,\delta >0$ there exists a set $E\in {\mathcal{S}}_h$ such that for all $r\in {\Delta }_{\epsilon }\setminus E$ the inequalities

$$\begin{array}{c}det(D_f\left(r\right)+I)\le \end{array}$$

$$\begin{array}{c}\le h\left(r\right)\prod _{j=1}^p(r_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right)+ln(\frac{e{r}_j}{{\epsilon }_j}\left)\right)\prod _{j=1}^p{ln}^{1+\delta }(r_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right)+ln(\frac{e{r}_j}{{\epsilon }_j}\left)\right),\end{array}$$

(5)

$$\begin{array}{c}{r}_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right)\le h\left(r\right){ln}^{1+\delta }{\mathfrak{M}}_f\left(r\right)\prod _{k=1,k\ne j}^p{ln}^{1+\delta }\left(\frac{e{r}_k}{{\epsilon }_k}\right),j\in \{1,\cdots ,p\}\end{array}$$

(6)

hold.

Proof. 

Let $E_0\subset \left|G\right|$ be a set for which inequality (5) does not hold. Now we prove that $E_0\in {\mathcal{S}}_h.$ Since $r_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right)>0$, there for any $r\in {\Delta }_{\epsilon }$ we have

$$r_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right)+ln\left(\frac{e{r}_j}{{\epsilon }_j}\right)>1,j\in \{1,\cdots ,p\}.$$

Then,

$$\begin{array}{c}{\nu }_h(E_0\cap {\Delta }_{\epsilon })=\underset{E_0\cap {\Delta }_{\epsilon }}{\int \dots \int }\frac{h\left(r\right)d{r}_1\cdots d{r}_p}{r_1\cdots r_p}\le \\ \le \underset{E_0\cap {\Delta }_{\epsilon }}{\int \dots \int }\frac{det(D_f\left(r\right)+I)d{r}_1\cdots d{r}_p}{{\displaystyle \prod _{j=1}^p}{r}_j{\displaystyle \prod _{j=1}^p}(r_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right)+ln{r}_j){\displaystyle \prod _{j=1}^p}{ln}^{1+\delta }(r_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right)+ln{r}_j)}.\end{array}$$

Let $U:\left|G\right|\to {\mathbb{R}}_{+}^p$ be a mapping such that $U=(u_1\left(r\right),u_2\left(r\right),\cdots ,u_p\left(r\right))$ and $u_j\left(r\right)=r_j\frac{\partial }{\partial r_j}ln{M}_f\left(r\right)+ln\left(\frac{e{r}_j}{{\epsilon }_j}\right),j\in \{1,\cdots ,p\},r=(r_1,r_2,\cdots ,r_p).$ Then for $i,j\in \{1,2,\cdots ,p\}$ we obtain

$$\begin{array}{c}\frac{\partial u_i}{\partial r_i}=\frac{\partial }{\partial r_i}(r_i\frac{\partial }{\partial r_i}ln{\mathfrak{M}}_f\left(r\right)+ln(\frac{e{r}_j}{{\epsilon }_j}\left)\right)=\frac{1}{r_i}{\partial }_i{\partial }_{i}ln{\mathfrak{M}}_f\left(r\right)+{\frac{1}{r}}_i,\\ \frac{\partial u_i}{\partial r_j}=\frac{\partial }{\partial r_j}(r_i\frac{\partial }{\partial r_i}ln{\mathfrak{M}}_f\left(r\right)+ln(\frac{e{r}_j}{{\epsilon }_j}\left)\right)=\frac{1}{r_j}{\partial }_i{\partial }_{j}ln{\mathfrak{M}}_f\left(r\right),i\ne j.\end{array}$$

Hence, the Jacobian

$$J_1:=\frac{D(u_1,u_2,\cdots ,u_p)}{D(r_1,r_2,\cdots ,r_p)}=\left|\begin{array}{ccc}\frac{\partial u_1}{\partial r_1}& \cdots & \frac{\partial u_1}{\partial r_p}\\ \cdots & \ddots & \cdots \\ \frac{\partial u_p}{\partial r_1}& \cdots & \frac{\partial u_p}{\partial r_p}\end{array}\right|=\prod _{j=1}^p\frac{1}{r_j}·det(D_f\left(r\right)+I).$$

Therefore,

$$\begin{array}{c}{\nu }_h(E_0\cap {\Delta }_{\epsilon })\le \underset{U(E_0\cap {\Delta }_{\epsilon })}{\int \dots \int }\frac{d{u}_{1}d{u}_2\cdots d{u}_p}{u_1{ln}^{1+\delta }{u}_1\dots u_p{ln}^{1+\delta }{u}_p}\le \\ \le \underset{1}{\overset{+\infty }{\int }}\cdots \underset{1}{\overset{+\infty }{\int }}\frac{d{u}_{1}d{u}_2\cdots d{u}_p}{u_1{ln}^{1+\delta }{u}_1\dots u_p{ln}^{1+\delta }{u}_p}<+\infty .\end{array}$$

Let $E_1\subset \left|G\right|$ be a set for which inequality (6) does not hold for $j=1$. Now we prove that $E_1\cap {\Delta }_{\epsilon }\in {\mathcal{S}}_h.$

Then,

$$\begin{array}{c}{\nu }_h(E_1\cap {\Delta }_{\epsilon })=\underset{E_1\cap {\Delta }_{\epsilon }}{\int \dots \int }\frac{h\left(r\right)d{r}_1\cdots d{r}_p}{r_1\cdots r_p}\le \underset{E_1\cap {\Delta }_{\epsilon }}{\int \dots \int }\frac{r_1\frac{\partial }{\partial r_1}ln{\mathfrak{M}}_f\left(r\right)d{r}_1\cdots d{r}_p}{\left({\displaystyle \prod _{j=1}^p}{r}_j\right){ln}^{1+\delta }{\mathfrak{M}}_f\left(r\right){\displaystyle \prod _{j=2}^p}\left({ln}^{1+\delta }\right(\frac{e{r}_j}{{\epsilon }_j}\left)\right)}.\end{array}$$

Let $V:\left|G\right|\to {\mathbb{R}}_{+}^p$ be a mapping such that $V=(v_1\left(r\right),v_2\left(r\right),\cdots ,v_p\left(r\right))$ and $v_1\left(r\right)=ln{\mathfrak{M}}_f\left(r\right),v_j=ln\left(\frac{e{r}_j}{{\epsilon }_j}\right)j\in \{2,\cdots ,p\},r=(r_1,r_2,\cdots ,r_p).$ Therefore, the Jacobian

$$\begin{array}{c}{J}_2:=\frac{D(v_1,v_2,\cdots ,v_p)}{D(r_1,r_2,\cdots ,r_p)}=\\ =\left|\begin{array}{cccc}\frac{\partial }{\partial r_1}ln{\mathfrak{M}}_f\left(r\right)& \frac{\partial }{\partial r_2}ln{\mathfrak{M}}_f\left(r\right)& \cdots & \frac{\partial }{\partial r_p}ln{\mathfrak{M}}_f\left(r\right)\\ 0& \frac{1}{r_2}& \cdots & 0\\ \cdots & \cdots & \ddots & \cdots \\ 0& 0& \cdots & \frac{1}{r_p}\end{array}\right|=\prod _{j=1}^p\frac{1}{r_j}·r_1\frac{\partial }{\partial r_1}ln{\mathfrak{M}}_f\left(r\right).\end{array}$$

Therefore,

$${\nu }_h(E_1\cap {\Delta }_{\epsilon })\le \underset{U(E_0\cap {\Delta }_{\epsilon })}{\int \dots \int }\frac{d{u}_{1}d{u}_2\cdots d{u}_p}{{(u_1{u}_2\dots u_p)}^{1+\delta }}\le \underset{1}{\overset{+\infty }{\int }}\cdots \underset{1}{\overset{+\infty }{\int }}\frac{d{u}_{1}d{u}_2\cdots d{u}_p}{{(u_1{u}_2\dots u_p)}^{1+\delta }}<+\infty .$$

Let $E_j\subset \left|G\right|$ be a set for which inequality (6) does not hold for $j\in \{2,\dots ,p\}$. Similarly, $E_j\cap {\Delta }_{\epsilon }\in {\mathcal{S}}_h$ for $j\in \{2,\dots ,p\}$. It remains to remark that the set $E={\bigcup }_{j=0}^p{E}_j$ is also a set of finite h-measure in $\left|G\right|$. □

6. Proof of the Main Theorem

Proof of Theorem 2.

Let E be the exceptional set from Lemma 2. Then, using Lemma 1 we obtain for all $r\in {\Delta }_{\epsilon }\setminus E$

$$\begin{array}{c}{M}_f\left(r\right)\le {\mathfrak{M}}_f\left(r\right)\le C_0{\mu }_f\left(r\right){(det(D_f\left(r\right)+I))}^{1/2}\le C_0{\mu }_f\left(r\right)\sqrt{h\left(r\right)}\times \\ \times \prod _{j=1}^p(r_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right)+ln(\frac{e{r}_j}{{\epsilon }_j}\left){)}^{1/2}\prod _{j=1}^p{ln}^{(1+\delta )/2}\right(r_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right)+ln\left(\frac{e{r}_j}{{\epsilon }_j}\right))\le \\ \le C_0{\mu }_f\left(r\right)\sqrt{h\left(r\right)}\prod _{j=1}^p\left(h\left(r\right){ln}^{1+\delta }{\mathfrak{M}}_f\left(r\right)\prod _{k=1,k\ne j}^p{ln}^{1+\delta }\right(\frac{e{r}_k}{{\epsilon }_k})+ln(\frac{e{r}_j}{{\epsilon }_j}){)}^{1/2}\times \\ \times \prod _{j=1}^p{ln}^{(1+\delta )/2}\left(h\left(r\right){ln}^{1+\delta }{\mathfrak{M}}_f\left(r\right)\prod _{k=1,k\ne j}^p{ln}^{1+\delta }\right(\frac{e{r}_k}{{\epsilon }_k})+ln(\frac{e{r}_j}{{\epsilon }_j}\left)\right)\le \\ \le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{\frac{p+1}{2}}{ln}^{\frac{p}{2}+p\delta }h\left(r\right){ln}^{\frac{p}{2}}{\mathfrak{M}}_f\left(r\right){ln}^{\frac{p}{2}+p\delta }ln{\mathfrak{M}}_f\left(r\right)\prod _{j=1}^p(\prod _{k=1,k\ne j}^{p}ln\frac{e{r}_k}{{\epsilon }_k}{)}^{1/2+\delta }\le \\ \le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{\frac{p+1}{2}}{ln}^{\frac{p}{2}+p\delta }h\left(r\right){ln}^{\frac{p}{2}}{\mathfrak{M}}_f\left(r\right){ln}^{\frac{p}{2}+p\delta }ln{\mathfrak{M}}_f\left(r\right)\prod _{j=1}^p(ln\frac{e{r}_j}{{\epsilon }_j}{)}^{\frac{p-1}{2}(1+4\delta )},\\ ln{\mathfrak{M}}_f\left(r\right)\le \\ \le ln{\mu }_f\left(r\right)+\left(\frac{p+1}{2}+\delta \right)lnh\left(r\right)+\left(\frac{p}{2}+\delta \right)lnln{\mathfrak{M}}_f\left(r\right)+\left(\frac{p-1}{2}(1+4\delta )\right)\sum _{j=1}^p{ln}^{+}ln\frac{e{r}_j}{{\epsilon }_j}.\end{array}$$

Note that we can chose set E such that $\forall r\in ({\Delta }_{\epsilon }\cap |G\left|\right)\setminus E$

$${\mathfrak{M}}_f\left(r\right)>C_p^{*},{\mu }_f\left(r\right)>1,$$

where $C_p^{*}$ is some constant such that $C_p^{*}\ge {ln}^{2p}{C}_p^{*}.$ Then,

$$\begin{array}{c}{\mathfrak{M}}_f\left(r\right)\ge {ln}^{2p}{\mathfrak{M}}_f\left(r\right),ln{\mathfrak{M}}_f\left(r\right)\ge 2plnln{\mathfrak{M}}_f\left(r\right),\\ \frac{1}{2}ln{\mathfrak{M}}_f\left(r\right)\le ln{\mathfrak{M}}_f\left(r\right)-\left(\frac{p}{2}+\delta \right)lnln{\mathfrak{M}}_f\left(r\right)\le \\ \le ln{\mu }_f\left(r\right)+\left(\frac{p+1}{2}+\delta \right)lnh\left(r\right)+\left(\frac{p-1}{2}(1+4\delta )\right)\sum _{j=1}^p{ln}^{+}ln\frac{e{r}_j}{{\epsilon }_j},\\ ln{\mathfrak{M}}_f\left(r\right)\le (1+p+2\delta )ln\left\{{\mu }_f\left(r\right)h\left(r\right)\prod _{j=1}^{p}ln\frac{e{r}_j}{{\epsilon }_j}\right\}.\\ M_f\left(r\right)\le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{\frac{p+1}{2}}{ln}^{\frac{p}{2}+\delta }h\left(r\right){ln}^{\frac{p}{2}}\left\{{\mu }_f\left(r\right)h\left(r\right)\prod _{j=1}^{p}ln\frac{e{r}_j}{{\epsilon }_j}\right\}\times \\ \times {ln}^{1/2+\delta }ln\left\{{\mu }_f\left(r\right)h\left(r\right)\prod _{j=1}^{p}ln\frac{e{r}_j}{{\epsilon }_j}\right\}\prod _{j=1}^p(ln\frac{e{r}_j}{{\epsilon }_j}{)}^{\frac{p-1}{2}(1+5\delta )}\le \\ \le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{\frac{p+1}{2}}{ln}^{\frac{p}{2}+\delta }h\left(r\right){ln}^{\frac{p}{2}+{\delta }_1}\left\{{\mu }_f\left(r\right)h\left(r\right)\right\}\prod _{j=1}^p(ln\frac{e{r}_j}{{\epsilon }_j}{)}^{\frac{p-1}{2}+{\delta }_1},{\delta }_1=5p\delta .\end{array}$$

 □

7. Corollaries Hypotheses

Let us consider the case when domain G is bounded. Then there exists $R>0$ such that $G\subset C_p\left(R\right):=\{z\in {\mathbb{C}}^p:|z_i|<R,i\in \{1,\dots ,p\}\}$. Therefore we have for all $r\in {\Delta }_{\epsilon }\setminus E$

$$\prod _{j=1}^p(\prod _{k=1,k\ne j}^{p}ln\frac{e{r}_k}{{\epsilon }_k}{)}^{\frac{1}{2}+\delta }\le \prod _{k=1}^p(ln\frac{eR}{{\epsilon }_k}{)}^{\frac{p}{2}+p\delta }.$$

Denote

$$K:=\{z\in G:{ln}^{\delta }h\left(r\right)\le \prod _{k=1}^p(ln\frac{eR}{{\epsilon }_k}\left){}^{\frac{p}{2}+p\delta }\right\}.$$

In addition, ${\nu }_h(E\cap {\Delta }_{\epsilon })$ is finite when

$${\nu }_h^{*}(E\cap {\Delta }_{\epsilon })=\underset{E\cap {\Delta }_{\epsilon }}{\int }h\left(r\right)d{r}_1\cdots d{r}_p<+\infty .$$

Note that

$$\begin{array}{c}{\nu }_h^{*}(K\cap {\Delta }_{\epsilon })=\underset{K\cap {\Delta }_{\epsilon }}{\int }h\left(r\right)d{r}_1\cdots d{r}_p\le exp\left\{\prod _{k=1}^p\right(ln\frac{eR}{{\epsilon }_k}\left){}^{\frac{p}{2\delta }+p}\right\}\underset{K\cap {\Delta }_{\epsilon }}{\int }d{r}_1\cdots d{r}_p\le \\ \le exp\left\{\prod _{k=1}^p\right(ln\frac{eR}{{\epsilon }_k}\left){}^{\frac{p}{2\delta }+p}\right\}\underset{G}{\int }d{r}_1\cdots d{r}_p<+\infty .\end{array}$$

Finally, for all $r\in {\Delta }_{\epsilon }\setminus (E\cup K)$ we obtain

$$\begin{array}{c}{M}_f\left(r\right)\le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{\frac{p+1}{2}}{ln}^{\frac{p}{2}+\delta }h\left(r\right){ln}^{\frac{p}{2}+\delta }\left\{{\mu }_f\left(r\right)h\left(r\right)\right\}\prod _{j=1}^p(\prod _{k=1,k\ne j}^{p}ln\frac{e{r}_k}{{\epsilon }_k}{)}^{\frac{1}{2}+\delta }\le \\ \le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{\frac{p+1}{2}}{ln}^{\frac{p}{2}+\delta }h\left(r\right){ln}^{\frac{p}{2}+\delta }\left\{{\mu }_f\left(r\right)h\left(r\right)\right\}\prod _{k=1}^p(ln\frac{eR}{{\epsilon }_k}{)}^{\frac{p}{2}+p\delta }\le \\ \le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{\frac{p+1}{2}}{ln}^{\frac{p}{2}+2\delta }h\left(r\right){ln}^{\frac{p}{2}+\delta }\left\{{\mu }_f\left(r\right)h\left(r\right)\right\}.\end{array}$$

Thus, we prove such a statement.

Theorem 4.

Let $f\in {\mathcal{A}}^p\left(\mathbb{G}\right),$ G is bounded. Then for every $\epsilon \in {\mathbb{R}}_{+}^p,\delta >0$ there exists a set $E\in {\mathcal{S}}_h$ such that for all $r\in {\Delta }_{\epsilon }\setminus E$ we have

$$\begin{array}{c}{M}_f\left(r\right)\le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{\frac{p+1}{2}}{ln}^{\frac{p}{2}+\delta }h\left(r\right){ln}^{\frac{p}{2}+\delta }\left\{{\mu }_f\left(r\right)h\left(r\right)\right\}.\end{array}$$

(7)

In the case when

$$G={\mathbb{B}}_p\left(1\right):=\{z\in {\mathbb{C}}^p:\left|z\right|:=\sqrt{|z_1{|}^2+\dots +{\left|z_p\right|}^2}<1\}$$

one can choose $h\left(r\right)={(1-|r\left|\right)}^{-p}$, $\left|r\right|={(r_1^2+\dots +r_p^2)}^{1/2}.$

Theorem 5.

Let $f\in {\mathcal{A}}^p\left({\mathbb{B}}_p\left(1\right)\right),$ $h\left(r\right)={(1-|r\left|\right)}^{-p}$. Then, for every $\epsilon \in {\mathbb{R}}_{+}^p,\delta >0$ there exists a set $E\in {\mathcal{S}}_h$ such that for all $r\in {\Delta }_{\epsilon }\setminus E$ we have

$$\begin{array}{c}{M}_f\left(r\right)\le \frac{{\mu }_f\left(r\right)}{{(1-|r\left|\right)}^{\frac{1}{2}(p^2+p)+\delta }}{ln}^{\frac{p}{2}+\delta }\frac{{\mu }_f\left(r\right)}{1-\left|r\right|}.\end{array}$$

If we additionaly suppose that

$$h\left(r\right)=\prod _{j=1}^p{h}_j\left(r_j\right)$$

(8)

then (see [23]) inequality (6) from Lemma 5 can be replayced by

$$r_j\frac{\partial }{\partial r_j}ln{\mathfrak{M}}_f\left(r\right)\le h^{\delta }\left(r\right)h_j^{1-\delta }\left(r_j\right){ln}^{1+\delta }{\mathfrak{M}}_f\left(r\right)\prod _{k=1,k\ne j}^p{ln}^{1+\delta }\left(\frac{e{r}_k}{{\epsilon }_k}\right),j\in \{1,\cdots ,p\}.$$

We therefore have the following statement:

Theorem 6.

Let $f\in {\mathcal{A}}^p\left(\mathbb{G}\right),$ $h\in {\mathcal{H}}^p$ satisfies condition (8). Then for every $\epsilon \in {\mathbb{R}}_{+}^p,\delta >0$ there exists a set $E\in {\mathcal{S}}_h$ such that for all $r\in {\Delta }_{\epsilon }\setminus E$ we have

$$\begin{array}{c}{M}_f\left(r\right)\le {\mu }_f\left(r\right){\left(h\left(r\right)\right)}^{1+\delta }{ln}^{\frac{p}{2}+\delta }\left\{{\mu }_f\left(r\right)h\left(r\right)\right\}\prod _{j=1}^p(\prod _{k=1,k\ne j}^{p}ln\frac{e{r}_k}{{\epsilon }_k}{)}^{\frac{1}{2}+\delta }.\end{array}$$

Inequality (3) follows from this statement if we choose $h\left(r\right)={\prod }_{i\in I}\frac{1}{(1-r_i)}.$

8. Discussion

In view of the obtained results we can formulate the following conjectures:

Conjecture 1. The descriptions of exceptional sets in the Theorems 1–3 are in a sense the best possible.

Conjecture 2. For a given $h\in \mathcal{H}$, the inequality (4) is sharp in the general case.