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Article

# Wiman’s Type Inequality in Multiple-Circular Domain

by *,† and
Department of Mechanics and Mathematics, Ivan Franko National University of Lviv, 79000 Lviv, Ukraine
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Axioms 2021, 10(4), 348; https://doi.org/10.3390/axioms10040348
Received: 31 October 2021 / Revised: 10 December 2021 / Accepted: 13 December 2021 / Published: 17 December 2021

## Abstract

:
In the paper we prove for the first time an analogue of the Wiman inequality in the class of analytic functions $f ∈ A 0 p ( G )$ in an arbitrary complete Reinhard domain $G ⊂ C p$, $p ∈ N$ represented by the power series of the form $f ( z ) = f ( z 1 , ⋯ , z p ) = ∑ ‖ n ‖ = 0 + ∞ a n z n$ with the domain of convergence $G .$ We have proven the following statement: If $f ∈ A p ( G )$ and $h ∈ H p$, then for a given $ε = ( ε 1 , … , ε p ) ∈ R + p$ and arbitrary $δ > 0$ there exists a set $E ⊂ | G |$ such that $∫ E ∩ Δ ε h ( r ) d r 1 ⋯ d r p r 1 ⋯ r p < + ∞$ and for all $r ∈ Δ ε ∖ E$ we have $M f ( r ) ≤ μ f ( r ) ( h ( r ) ) p + 1 2 ln p 2 + δ h ( r ) ln p 2 + δ { μ f ( r ) h ( r ) } ∏ j = 1 p ( ln e r j ε j ) p − 1 2 + δ .$ Note, that this assertion at $p = 1 ,$ $G = C ,$ $h ( r ) ≡ const$ implies the classical Wiman–Valiron theorem for entire functions and at $p = 1 ,$ the $G = D : = { z ∈ C : | z | < 1 } ,$ $h ( r ) ≡ 1 / ( 1 − r )$ theorem about the Kővari-type inequality for analytic functions in the unit disc $D$; $p > 1$ implies some Wiman’s type inequalities for analytic functions of several variables in $C n × D k$, $n , k ∈ Z + , n + k ∈ N .$
MSC:
32A10; 32A17; 32A37; 30H99; 30A05

## 1. Introduction: Notations and Preliminaries

Let $C$, $R$, $Z$, $N$ be sets of complex numbers, real numbers, integers, and positive integers, respectively, and $Z + = N ∪ { 0 }$. We denote by $A 0 p ( G ) ,$ $p ∈ N ,$ the class of an analytic functions f in a complete Reinhardt domain $G ⊂ C p$, represented by the power series of the form
$f ( z ) = f ( z 1 , … , z p ) = ∑ ∥ n ∥ = 0 + ∞ a n z n ,$
with the domain of convergence $G$, where $z n = z 1 n 1 … z p n p , z = ( z 1 , … , z p ) ∈ G , n = ( n 1 , … , n p ) ∈ Z + p , ∥ n ∥ = ∑ j = 1 p n j$; $E p : = A 0 p ( C p )$ is the class of entire functions of several variables (i.e., analytic functions in $C p$); by $E R : = A 0 1 ( D R )$ $( 0 < R ≤ + ∞ )$ we denote the class of analytic functions of one complex variable in a disk $D R = { z ∈ C : | z | < R }$. In particular, $E : = E + ∞ = E 1$ is the class of entire functions of one complex variable.
For a function $f ∈ A 0 p ( G )$ of form (1) with domain of convergence $G$ and $r = ( r 1 , ⋯ , r p ) ∈ | G | : = { r = ( r 1 , … , r p ) : r j = | z j | , z = ( z 1 , … , z p ) ∈ G }$ we denote
$Δ r 0 = { t ∈ | G | : t j ≥ r j 0 , j ∈ { 1 , ⋯ , p } } , μ f ( r ) = max { | a n | r 1 n 1 ⋯ r p n p : n ∈ Z + p } , M f ( r ) = max { | f ( z ) | : | z 1 | = r 1 , ⋯ , | z p | = r p } , M f ( r ) = ∑ ∥ n ∥ = 0 + ∞ | a n | r n .$
On the one hand, it is well-known that every analytic function f in the complete Reinhardt domain $G$ with a center at $z = 0$ can be represented in $G$ 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 $G ⊂ C p$ is the complete Reinhardt domain if:
(a)
$R z = ( R 1 z 1 , ⋯ , R p z p ) ∈ G$ (a complete domain);
(b)
$( z 1 e i θ 1 , … , z p e i θ p ) ∈ G$ (a multiple-circular domain).
The Reinhardt domain $G$ is called logarithmically-convex if the image of the set $G * = { z ∈ G : z 1 · … · z p ≠ 0 }$ under the mapping $L n : z → L n ( z ) = ( ln | z 1 | , … , ln | z p | )$ is a convex set in the space $R p$. In one complex variable ($p = 1$), a logarithmically-convex Reinhardt domain is a disc. The following complete Reinhardt domains ($p ≥ 2$) are considered most frequently:
$C p ( R ) : = { z ∈ C p : | z 1 | < R 1 , … , | z p | < R p } , R = ( R 1 , … , R p ) ∈ ( 0 , + ∞ ) p , ( polydisk ) , B p ( r ) : = { z ∈ C p : | z | : = | z 1 | 2 + … + | z p | 2 < r } ( ball ) , Π p ( r ) : = { z ∈ C p : | z 1 | + … + | z p | < r } , r > 0 .$
Note, that $C p ( R ) ⊂ G$ for every $w = ( w 1 , … , w p ) ∈ G$ and $R = ( | w 1 | , … , | w p | )$. The domains $C p ( r e 1 )$, $e 1 = ( 1 , … , 1 ) ∈ R p$, $B p ( r )$, $Π p ( r )$ $( r > 0 )$ are the logarithmically-convex complete Reinhardt domains. However, for example, the complete Reinhardt domain $G 1 , 2 = { z = ( z 1 , z 2 ) : | z 1 | < 1 , | z 2 | < 2 } ∪ { z = ( z 1 , z 2 ) : | z 1 | < 2 , | z 2 | < 1 }$ 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 $h : [ 0 , R ) → [ 10 , ∞ )$ such that $∫ r 0 R h ( r ) d ln r = + ∞$ for some $r 0 ∈ ( 0 , R )$. If $f ∈ E R$, $R ∈ ( 0 , + ∞ ]$ is an analytic function represented by a power series of the form $f ( z ) = ∑ n = 0 + ∞ a n z n ,$ then $( ∀ δ > 0 ) ( ∃ E ( δ , f , h ) = E ⊂ ( 0 , R ) ) ( ∃ r 0 ∈ ( 0 , R ) )$ $( ∀ r ∈ ( r 0 , R ) ∖ E )$
$M f ( r ) ≤ h ( r ) μ f ( r ) { ln h ( r ) ln ( h ( r ) μ f ( r ) ) } 1 / 2 + δ a n d ∫ E ∩ ( r 0 , R ) h ( r ) r d r < + ∞ ,$
where $M f ( r ) = max { | f ( z ) | : | z | = r }$ is the maximum modulus and $μ f ( r ) = max { | a n | r n : n ≥ 0 }$ is the maximal term of power series.
For nonconstant entire functions $f ∈ E$ we can choose $h ( r ) = 10$ and $δ = ε / 2$ for an arbitrarily given $ε > 0 .$ 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 $r ∈ ( r 0 , + ∞ ) ∖ E$, $∫ E d ln r < + ∞$, we have
$M f ( r ) ≤ 10 μ f ( r ) { ln 10 ln ( 10 μ f ( r ) ) } 1 / 2 + δ ≤ μ f ( r ) ln 1 / 2 + ε μ f ( r ) .$
For analytic functions $f ∈ E 1$ in the unit disk $D 1$ we can choose $h ( r ) = r 1 − r .$ Then,
i.e., the theorem about the Kővari-type inequality for analytic functions in the unit disc $D = { z ∈ C : | z | < 1 }$ ([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 $h ( r ) = ln μ f ( r )$.
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 $f ∈ E 1$ such a problem is still open. Theorem 1 contains a new description of the exceptional set in the inequality (2) for analytic functions $f ∈ E 1$. Perhaps the best possible description of an exceptional set is also obtained with $h ( r ) = ln μ f ( r ) .$

## 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 $D p , p ≥ 2 ,$ in [23,24].
In paper [25] some analogues of Wiman’s inequality are proven for the analytic $f ( z )$ and random analytic $f ( z , t )$ functions on $G = D ℓ × C p − ℓ$, $ℓ ∈ N , 1 ≤ ℓ < p$, $I = { 1 , … , ℓ } , J = { ℓ + 1 , … , p }$ of the form (1) and $f ( z , t ) = ∑ ∥ n ∥ = 0 + ∞ a n Z n ( t ) z n$, respectively. Here, $Z = ( Z n )$ 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 $f ∈ A p ( G )$, $G = D ℓ × C p − ℓ$, $ℓ ∈ N , 1 ≤ ℓ < p$. For every $δ > 0$ there exist the sets $E 1 = E 1 ( δ , f ) , E 2 = E 2 ( δ , f ) ⊂ [ 0 , 1 ) l × ( 1 , + ∞ ) p − l$ of asymptotically finite logarithmic measure (i.e., $∫ Δ ε ∩ [ 0 , 1 ) ℓ × R p − ℓ d r 1 · … · d r ℓ · d r ℓ + 1 · … · d r p ( 1 − r 1 ) · … · ( 1 − r ℓ ) · r ℓ + 1 · … · r p < + ∞$ for some $ε > 0$), such that the inequalities
$M f ( r ) ≤ μ f ( r ) ∏ i ∈ I 1 ( 1 − r i ) 1 + δ ln p / 2 + δ ( μ f ( r ) ∏ i ∈ I 1 1 − r i ) ( ∏ j ∈ J ln r j ) p + δ , M f ( r , t ) ≤ μ f ( r ) ∏ i ∈ I 1 ( 1 − r i ) 1 / 2 + δ ln p / 4 + δ ( μ f ( r ) ∏ i ∈ I 1 1 − r i ) ( ∏ j ∈ J ln r j ) p / 2 + δ .$
hold for all $r ∈ | G | ∖ E 1$ and for all $r ∈ | D | ∖ E 2$ 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 $f ∈ A 0 p ( G )$ for the arbitrary complete Reinhardt domain $G$.

## 4. Main Result

The aim of this paper is to prove some analogues of Wiman’s inequality for the analytic functions $f ∈ A 0 p ( G )$ represented by the series of form (1) with the arbitrary complete Reinhardt domain of convergence $G$. By $A p ( G )$ we denote a subclass of functions $f ∈ A 0 p ( G )$ such that $∂ ∂ z j f ( z 1 , ⋯ , z p ) ≢ 0$ in $G$ for any $j ∈ { 1 , … , p }$.
Let $H p$ be the class of functions $h : | G | → R +$ such that h is nondecreasing with respect to each variable and $h ( r ) > 10$ for all $r ∈ | G |$ and
$∫ Δ ε h ( r ) d r 1 ⋯ d r p r 1 ⋯ r p = + ∞$
for every $ε ∈ R + p$ such that $| G | ∩ Δ ε$ is a nonempty domain in $R + p .$
For $h ∈ H p$ we say that $E ⊂ | G |$ is the set of finite h-measure on$| G |$ if for some $ε ∈ R + p$ such that $| G | ∩ Δ ε$ is a nonempty domain in $| G | ⊂ R + p$ one has
$ν h ( E ∩ Δ ε ) : = ∫ E ∩ Δ ε h ( r ) d r 1 ⋯ d r p r 1 ⋯ r p < + ∞ .$
We denote a set of such sets by $S h .$
Theorem 3.
Let $f ∈ A p ( G ) .$ Then, for every $ε ∈ R + p , δ > 0$ there exists a set $E ∈ S h$ such that for all $r ∈ Δ ε ∖ E$ the following inequality takes place:
$M f ( r ) ≤ μ f ( r ) ( h ( r ) ) p + 1 2 ln p 2 + δ h ( r ) ln p 2 + δ { μ f ( r ) h ( r ) } ∏ j = 1 p ( ∏ k = 1 , k ≠ j p ln e r k ε k ) 1 2 + δ .$
Remark 1.
Choosing $p = 1$ and $G = D R$ 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 $G$ we need the following auxiliary results.
Let $D f ( r ) = ( D i j )$ be a $p × p$ matrix such that
$D i j = r i ∂ ∂ r i ( r j ∂ ∂ r j ln M f ( r ) ) = ∂ i ∂ j ln M f ( r ) , ∂ i = r i ∂ ∂ r i , i , j ∈ { 1 , ⋯ , p } .$
Let I be an identity matrix of order $p .$
For the set $E ⊂ R p$ by $# ( E ∩ Z + p )$ we denote the quantity of the elements of set $E ∩ Z + p .$
Lemma 1.
Let B be parallelepiped in $R p$ with edges of the lengths $l 1 , l 2 , … , l p$ so that there exists an isometry $H : R p → R p$ such that
$H : B → { x ∈ R p : | x j | ≤ l j / 2 , j ∈ { 1 , 2 , … p } } .$
Then,
$# ( B ∩ Z p ) ≤ λ p ∏ j = 1 p ( l j + 1 ) ,$
where $λ p$ is the inverse value to the volume of a sphere with the radius $1 2$ in $R p ,$ i.e., $λ p = 2 n · Γ ( n + 2 2 ) π n / 2$.
Proof.
Denote
$B ′ = { x ∈ R p : | x j | ≤ l j 2 , j ∈ { 1 , … , p } } , B * = { x ∈ R p : | x j | ≤ l j + 1 2 , j ∈ { 1 , … , p } } ⊃ B .$
Let $S ( n )$ be an open sphere with a center at $n ∈ Z + p$ with radius $1 2 .$ Note that
$⋃ n ∈ Z + p ∩ B S ( n ) ⊆ B * .$
By the monotony of the Lebesgue measure $μ$ in $R p$ we obtain
$μ ⋃ n ∈ Z + p ∩ B S ( n ) ≤ μ ( B * ) .$
Finally, by the additivity of this measure, we obtain
$μ ( S ( 1 / 2 ) ) · # { B ∩ Z + p } ≤ ∏ j = 1 p ( l j + 1 ) , # { B ∩ Z + p } ≤ 1 μ ( S ( 1 / 2 ) ) ∏ j = 1 p ( l j + 1 ) .$
□
By $A +$ we denote the Moore–Penrose inverse matrix of A ([18,26]), i.e.,
$A + = lim δ → 0 A T ( A A T + δ I ) − 1 .$
Lemma 2.
Let $α ∈ R p , C > 0$, A be a $p × p$ nonnegative matrix $0 < m = rank A ≤ p$ and
$E = { x ∈ R m : ( x − α ) A + ( x − α ) T ≤ C } .$
There exists a constant $δ = δ ( C , p ) > 0$ that does not depend on A and α such that
$# { E ∩ Z + m } ≤ δ ( det ( A + I ) ) 1 / 2 .$
Proof.
Let $0 < λ 1 ≤ λ 2 ≤ … ≤ λ m$ be positive eigenvalues of the matrix $A .$ Then, $1 λ 1 , 1 λ 2 , … , 1 λ m$ are eigenvalues of the matrix $A + .$ Thus, there exists an isometry $H : R m → R m ,$ such that
$H : E → { x ∈ R m : ∑ j = 1 m x j 2 λ j ≤ C } , E ⊂ H − 1 { x ∈ R m : x ≤ C λ j , j ∈ { 1 , 2 , … , m } } .$
By Lemma 1 there exists a constant $δ ′ > 0$ such that
$# { E ∩ Z + m } ≤ δ ′ ∏ j = 1 m ( 2 ( C λ j ) 1 / 2 + 1 ) = δ ′ ∏ j = 1 p ( 2 ( C λ j ) 1 / 2 + 1 ) .$
It remains to remark that
$∏ j = 1 p ( λ j + 1 ) = det ( A + I ) , # { E ∩ Z + p } ≤ δ ′ ( 2 C ) p ( ∏ j = 1 p ( λ j + 1 ) ) 1 / 2 ≤ δ ″ ( det ( A + I ) ) 1 / 2 .$
□
Lemma 3.
Let $ξ = ( ξ 1 , ξ 2 , … , ξ p ) T$ be a random vector, $α = M ξ = ( M ξ 1 , M ξ 2 , … , M ξ p ) T ,$ A covariance matrix of ξ, $δ > 0 , 0 < m = rank A ≤ p .$ Then,
$P { ω : ( ξ ( ω ) − α ) A + ( ξ ( ω ) − α ) T ≤ δ } ≥ 1 − m δ .$
Proof.
Let us consider the random variable
$Z ( ω ) = ( ξ ( ω ) − α ) T A + ( ξ ( ω ) − α ) .$
As A is non-negative, then $∀ ω ∈ Ω : Z ( ω ) ≥ 0 .$ Moreover, as A is also symmetric, there exists an orthogonal matrix G such that $G G T = G T G = I$ and $G T A G = Q .$ Here, I is the identity matrix of order p and $Q = diag ( λ 1 , λ 2 , … , λ m , 0 , … , 0 )$ is the diagonal matrix with the ordered eigenvalues $λ 1 ≥ λ 2 ≥ … ≥ λ m > 0 ,$ $0 < m = rank A ≤ p .$ Then (see, for example [26,27]),
$G T A G = Q , G G T A G G T = G Q G T ⟹ A = G Q G T , A + = ( G Q G T ) + = ( G T ) + Q + G + = ( G T ) − 1 Q + G − 1 = G Q + G T ,$
i.e., $A = G Q G T , A + = G Q + G T .$ Therefore,
$Z = ( ξ − α ) T A + ( ξ − α ) = ( ξ − α ) T G Q + G T ( ξ − α ) = = ( ξ − α ) T G Q − 1 / 2 Q − 1 / 2 G T ( ξ − α ) = ( Q − 1 / 2 G T ( ξ − α ) ) T ( Q − 1 / 2 G T ( ξ − α ) ) = Y T Y ,$
where $Y = Q − 1 / 2 G T ( ξ − α ) , Q − 1 / 2 = diag ( λ 1 − 1 / 2 , λ 2 − 1 / 2 , … , λ m − 1 / 2 , 0 , … , 0 ) .$ The expected value and covariance of the random vector Y satisfy the equations
Therefore,
$M Z = M ( Y T Y ) = M ( ∑ j = 1 p Y j 2 ) = ∑ j = 1 p M ( Y i 2 ) = ∑ j = 1 p D ( Y j ) = m .$
Finally, using Markov’s inequality we obtain
$P { ω : Z ( ω ) ≥ δ } = P { ω : ( ξ ( ω ) − α ) A + ( ξ ( ω ) − α ) T ≤ δ } ≤ M Z δ = m δ . P { ω : ( ξ ( ω ) − α ) A + ( ξ ( ω ) − α ) T ≤ δ } ≥ 1 − m δ .$
□
Lemma 4
(Theorem 3.1, [18]). Let $f ∈ A p .$ There exists a constant $C 0 ( p )$ such that
$M f ( r ) ≤ C 0 ( p ) μ f ( r ) ( det ( D f ( r ) + I ) ) 1 / 2 ,$
where I is the identity $p × p$ matrix.
Proof.
Let us consider random vector $X ( ω ) = ( X 1 ( ω ) , X 2 ( ω ) , … , X p ( ω ) )$ such that
$P { ω : X j ( ω ) = n j , j ∈ { 1 , … , p } } = 1 M f ( r ) | a n 1 … n p | r 1 n 1 … r p n p , k ∈ Z + .$
Then for $j ∈ { 1 , 2 , … , p }$ we obtain
$M X j = 1 M f ( r ) ∑ ∥ n ∥ = 0 + ∞ n j | a n | r n = r j ∂ ∂ r j ln M f ( r ) .$
$D f ( r )$ is covariance matrix of random vector $X ( ω )$.
One can choose $δ = 2 p$ in Lemma 3. We then obtain
$1 2 ≤ 1 − m 2 p ≤ P { ω : ( x − α ) D f + ( x − α ) T ≤ 2 p } ≤ ≤ μ f ( r ) M f ( r ) · # { x ∈ R + p : ( x − α ) D f + ( x − α ) T ≤ 2 p } ≤ 2 p μ f ( r ) M f ( r ) ( det ( D f ( r ) + I ) ) 1 / 2 , M f ( r ) ≤ 4 p μ f ( r ) ( det ( D f ( r ) + I ) ) 1 / 2 .$
□
Lemma 5.
Let $f ∈ A p .$ Then for $ε ∈ R + p , δ > 0$ there exists a set $E ∈ S h$ such that for all $r ∈ Δ ε ∖ E$ the inequalities
$det ( D f ( r ) + I ) ≤$
$≤ h ( r ) ∏ j = 1 p ( r j ∂ ∂ r j ln M f ( r ) + ln ( e r j ε j ) ) ∏ j = 1 p ln 1 + δ ( r j ∂ ∂ r j ln M f ( r ) + ln ( e r j ε j ) ) ,$
$r j ∂ ∂ r j ln M f ( r ) ≤ h ( r ) ln 1 + δ M f ( r ) ∏ k = 1 , k ≠ j p ln 1 + δ ( e r k ε k ) , j ∈ { 1 , ⋯ , p }$
hold.
Proof.
Let $E 0 ⊂ | G |$ be a set for which inequality (5) does not hold. Now we prove that $E 0 ∈ S h .$ Since $r j ∂ ∂ r j ln M f ( r ) > 0$, there for any $r ∈ Δ ε$ we have
$r j ∂ ∂ r j ln M f ( r ) + ln ( e r j ε j ) > 1 , j ∈ { 1 , ⋯ , p } .$
Then,
$ν h ( E 0 ∩ Δ ε ) = ∫ … ∫ E 0 ∩ Δ ε h ( r ) d r 1 ⋯ d r p r 1 ⋯ r p ≤ ≤ ∫ … ∫ E 0 ∩ Δ ε det ( D f ( r ) + I ) d r 1 ⋯ d r p ∏ j = 1 p r j ∏ j = 1 p ( r j ∂ ∂ r j ln M f ( r ) + ln r j ) ∏ j = 1 p ln 1 + δ ( r j ∂ ∂ r j ln M f ( r ) + ln r j ) .$
Let $U : | G | → R + p$ be a mapping such that $U = ( u 1 ( r ) , u 2 ( r ) , ⋯ , u p ( r ) )$ and $u j ( r ) = r j ∂ ∂ r j ln M f ( r ) + ln ( e r j ε j ) , j ∈ { 1 , ⋯ , p } , r = ( r 1 , r 2 , ⋯ , r p ) .$ Then for $i , j ∈ { 1 , 2 , ⋯ , p }$ we obtain
$∂ u i ∂ r i = ∂ ∂ r i ( r i ∂ ∂ r i ln M f ( r ) + ln ( e r j ε j ) ) = 1 r i ∂ i ∂ i ln M f ( r ) + 1 r i , ∂ u i ∂ r j = ∂ ∂ r j ( r i ∂ ∂ r i ln M f ( r ) + ln ( e r j ε j ) ) = 1 r j ∂ i ∂ j ln M f ( r ) , i ≠ j .$
Hence, the Jacobian
$J 1 : = D ( u 1 , u 2 , ⋯ , u p ) D ( r 1 , r 2 , ⋯ , r p ) = ∂ u 1 ∂ r 1 ⋯ ∂ u 1 ∂ r p ⋯ ⋱ ⋯ ∂ u p ∂ r 1 ⋯ ∂ u p ∂ r p = ∏ j = 1 p 1 r j · det ( D f ( r ) + I ) .$
Therefore,
$ν h ( E 0 ∩ Δ ε ) ≤ ∫ … ∫ U ( E 0 ∩ Δ ε ) d u 1 d u 2 ⋯ d u p u 1 ln 1 + δ u 1 … u p ln 1 + δ u p ≤ ≤ ∫ 1 + ∞ ⋯ ∫ 1 + ∞ d u 1 d u 2 ⋯ d u p u 1 ln 1 + δ u 1 … u p ln 1 + δ u p < + ∞ .$
Let $E 1 ⊂ | G |$ be a set for which inequality (6) does not hold for $j = 1$. Now we prove that $E 1 ∩ Δ ε ∈ S h .$
Then,
$ν h ( E 1 ∩ Δ ε ) = ∫ … ∫ E 1 ∩ Δ ε h ( r ) d r 1 ⋯ d r p r 1 ⋯ r p ≤ ∫ … ∫ E 1 ∩ Δ ε r 1 ∂ ∂ r 1 ln M f ( r ) d r 1 ⋯ d r p ( ∏ j = 1 p r j ) ln 1 + δ M f ( r ) ∏ j = 2 p ( ln 1 + δ ( e r j ε j ) ) .$
Let $V : | G | → R + p$ be a mapping such that $V = ( v 1 ( r ) , v 2 ( r ) , ⋯ , v p ( r ) )$ and $v 1 ( r ) = ln M f ( r ) , v j = ln ( e r j ε j ) j ∈ { 2 , ⋯ , p } , r = ( r 1 , r 2 , ⋯ , r p ) .$ Therefore, the Jacobian
$J 2 : = D ( v 1 , v 2 , ⋯ , v p ) D ( r 1 , r 2 , ⋯ , r p ) = = ∂ ∂ r 1 ln M f ( r ) ∂ ∂ r 2 ln M f ( r ) ⋯ ∂ ∂ r p ln M f ( r ) 0 1 r 2 ⋯ 0 ⋯ ⋯ ⋱ ⋯ 0 0 ⋯ 1 r p = ∏ j = 1 p 1 r j · r 1 ∂ ∂ r 1 ln M f ( r ) .$
Therefore,
$ν h ( E 1 ∩ Δ ε ) ≤ ∫ … ∫ U ( E 0 ∩ Δ ε ) d u 1 d u 2 ⋯ d u p ( u 1 u 2 … u p ) 1 + δ ≤ ∫ 1 + ∞ ⋯ ∫ 1 + ∞ d u 1 d u 2 ⋯ d u p ( u 1 u 2 … u p ) 1 + δ < + ∞ .$
Let $E j ⊂ | G |$ be a set for which inequality (6) does not hold for $j ∈ { 2 , … , p }$. Similarly, $E j ∩ Δ ε ∈ S h$ for $j ∈ { 2 , … , p }$. It remains to remark that the set $E = ⋃ j = 0 p E j$ is also a set of finite h-measure in $| G |$. □

## 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 $r ∈ Δ ε ∖ E$
$M f ( r ) ≤ M f ( r ) ≤ C 0 μ f ( r ) ( det ( D f ( r ) + I ) ) 1 / 2 ≤ C 0 μ f ( r ) h ( r ) × × ∏ j = 1 p ( r j ∂ ∂ r j ln M f ( r ) + ln ( e r j ε j ) ) 1 / 2 ∏ j = 1 p ln ( 1 + δ ) / 2 ( r j ∂ ∂ r j ln M f ( r ) + ln ( e r j ε j ) ) ≤ ≤ C 0 μ f ( r ) h ( r ) ∏ j = 1 p ( h ( r ) ln 1 + δ M f ( r ) ∏ k = 1 , k ≠ j p ln 1 + δ ( e r k ε k ) + ln ( e r j ε j ) ) 1 / 2 × × ∏ j = 1 p ln ( 1 + δ ) / 2 ( h ( r ) ln 1 + δ M f ( r ) ∏ k = 1 , k ≠ j p ln 1 + δ ( e r k ε k ) + ln ( e r j ε j ) ) ≤ ≤ μ f ( r ) ( h ( r ) ) p + 1 2 ln p 2 + p δ h ( r ) ln p 2 M f ( r ) ln p 2 + p δ ln M f ( r ) ∏ j = 1 p ( ∏ k = 1 , k ≠ j p ln e r k ε k ) 1 / 2 + δ ≤ ≤ μ f ( r ) ( h ( r ) ) p + 1 2 ln p 2 + p δ h ( r ) ln p 2 M f ( r ) ln p 2 + p δ ln M f ( r ) ∏ j = 1 p ( ln e r j ε j ) p − 1 2 ( 1 + 4 δ ) , ln M f ( r ) ≤ ≤ ln μ f ( r ) + p + 1 2 + δ ln h ( r ) + p 2 + δ ln ln M f ( r ) + p − 1 2 ( 1 + 4 δ ) ∑ j = 1 p ln + ln e r j ε j .$
Note that we can chose set E such that $∀ r ∈ ( Δ ε ∩ | G | ) ∖ E$
$M f ( r ) > C p * , μ f ( r ) > 1 ,$
where $C p *$ is some constant such that $C p * ≥ ln 2 p C p * .$ Then,
$M f ( r ) ≥ ln 2 p M f ( r ) , ln M f ( r ) ≥ 2 p ln ln M f ( r ) , 1 2 ln M f ( r ) ≤ ln M f ( r ) − p 2 + δ ln ln M f ( r ) ≤ ≤ ln μ f ( r ) + p + 1 2 + δ ln h ( r ) + p − 1 2 ( 1 + 4 δ ) ∑ j = 1 p ln + ln e r j ε j , ln M f ( r ) ≤ ( 1 + p + 2 δ ) ln μ f ( r ) h ( r ) ∏ j = 1 p ln e r j ε j . M f ( r ) ≤ μ f ( r ) ( h ( r ) ) p + 1 2 ln p 2 + δ h ( r ) ln p 2 μ f ( r ) h ( r ) ∏ j = 1 p ln e r j ε j × × ln 1 / 2 + δ ln μ f ( r ) h ( r ) ∏ j = 1 p ln e r j ε j ∏ j = 1 p ( ln e r j ε j ) p − 1 2 ( 1 + 5 δ ) ≤ ≤ μ f ( r ) ( h ( r ) ) p + 1 2 ln p 2 + δ h ( r ) ln p 2 + δ 1 { μ f ( r ) h ( r ) } ∏ j = 1 p ( ln e r j ε j ) p − 1 2 + δ 1 , δ 1 = 5 p δ .$
□

## 7. Corollaries Hypotheses

Let us consider the case when domain G is bounded. Then there exists $R > 0$ such that $G ⊂ C p ( R ) : = { z ∈ C p : | z i | < R , i ∈ { 1 , … , p } }$. Therefore we have for all $r ∈ Δ ε ∖ E$
$∏ j = 1 p ( ∏ k = 1 , k ≠ j p ln e r k ε k ) 1 2 + δ ≤ ∏ k = 1 p ( ln e R ε k ) p 2 + p δ .$
Denote
$K : = { z ∈ G : ln δ h ( r ) ≤ ∏ k = 1 p ( ln e R ε k p 2 + p δ .$
In addition, $ν h ( E ∩ Δ ε )$ is finite when
$ν h * ( E ∩ Δ ε ) = ∫ E ∩ Δ ε h ( r ) d r 1 ⋯ d r p < + ∞ .$
Note that
$ν h * ( K ∩ Δ ε ) = ∫ K ∩ Δ ε h ( r ) d r 1 ⋯ d r p ≤ exp { ∏ k = 1 p ( ln e R ε k p 2 δ + p ∫ K ∩ Δ ε d r 1 ⋯ d r p ≤ ≤ exp { ∏ k = 1 p ( ln e R ε k p 2 δ + p ∫ G d r 1 ⋯ d r p < + ∞ .$
Finally, for all $r ∈ Δ ε ∖ ( E ∪ K )$ we obtain
$M f ( r ) ≤ μ f ( r ) ( h ( r ) ) p + 1 2 ln p 2 + δ h ( r ) ln p 2 + δ { μ f ( r ) h ( r ) } ∏ j = 1 p ( ∏ k = 1 , k ≠ j p ln e r k ε k ) 1 2 + δ ≤ ≤ μ f ( r ) ( h ( r ) ) p + 1 2 ln p 2 + δ h ( r ) ln p 2 + δ { μ f ( r ) h ( r ) } ∏ k = 1 p ( ln e R ε k ) p 2 + p δ ≤ ≤ μ f ( r ) ( h ( r ) ) p + 1 2 ln p 2 + 2 δ h ( r ) ln p 2 + δ { μ f ( r ) h ( r ) } .$
Thus, we prove such a statement.
Theorem 4.
Let $f ∈ A p ( G ) ,$ G is bounded. Then for every $ε ∈ R + p , δ > 0$ there exists a set $E ∈ S h$ such that for all $r ∈ Δ ε ∖ E$ we have
$M f ( r ) ≤ μ f ( r ) ( h ( r ) ) p + 1 2 ln p 2 + δ h ( r ) ln p 2 + δ { μ f ( r ) h ( r ) } .$
In the case when
$G = B p ( 1 ) : = { z ∈ C p : | z | : = | z 1 | 2 + … + | z p | 2 < 1 }$
one can choose $h ( r ) = ( 1 − | r | ) − p$, $| r | = ( r 1 2 + … + r p 2 ) 1 / 2 .$
Theorem 5.
Let $f ∈ A p ( B p ( 1 ) ) ,$ $h ( r ) = ( 1 − | r | ) − p$. Then, for every $ε ∈ R + p , δ > 0$ there exists a set $E ∈ S h$ such that for all $r ∈ Δ ε ∖ E$ we have
$M f ( r ) ≤ μ f ( r ) ( 1 − | r | ) 1 2 ( p 2 + p ) + δ ln p 2 + δ μ f ( r ) 1 − | r | .$
$h ( r ) = ∏ j = 1 p h j ( r j )$
then (see [23]) inequality (6) from Lemma 5 can be replayced by
$r j ∂ ∂ r j ln M f ( r ) ≤ h δ ( r ) h j 1 − δ ( r j ) ln 1 + δ M f ( r ) ∏ k = 1 , k ≠ j p ln 1 + δ ( e r k ε k ) , j ∈ { 1 , ⋯ , p } .$
We therefore have the following statement:
Theorem 6.
Let $f ∈ A p ( G ) ,$ $h ∈ H p$ satisfies condition (8). Then for every $ε ∈ R + p , δ > 0$ there exists a set $E ∈ S h$ such that for all $r ∈ Δ ε ∖ E$ we have
$M f ( r ) ≤ μ f ( r ) ( h ( r ) ) 1 + δ ln p 2 + δ { μ f ( r ) h ( r ) } ∏ j = 1 p ( ∏ k = 1 , k ≠ j p ln e r k ε k ) 1 2 + δ .$
Inequality (3) follows from this statement if we choose $h ( r ) = ∏ i ∈ I 1 ( 1 − r i ) .$

## 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 $h ∈ H$, the inequality (4) is sharp in the general case.

## Author Contributions

Conceptualization, O.S.; investigation, A.K.; supervision, O.S.; writing—original draft preparation, A.K. All authors have read and agreed to the published version of the manuscript.

## Funding

This research received no external funding.

Not applicable.

Not applicable.

Not applicable.

## Conflicts of Interest

The authors declare no conflict of interest.

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