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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
(This article belongs to the Special Issue Complex Analysis)

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)
z = ( z 1 , , z p ) G ( R = ( R 1 , , R p ) [ 0 , 1 ] p ) :   R z = ( R 1 z 1 , , R p z p ) G (a complete domain);
(b)
( z 1 , , z p ) G ( ( θ 1 , , θ p ) R p ) :   ( 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,
M f ( r ) r μ f ( r ) 1 r { ln r 1 r ln ( r μ f ( r ) 1 r ) } 1 / 2 + δ μ f ( r ) ( 1 r ) 1 + δ ln 1 / 2 + δ μ f ( r ) 1 r , as   r 1 0 , r E , E d r 1 r < + ,
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 × C p , N , 1 < p , I = { 1 , , } , J = { + 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 × C p , N , 1 < 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 ) × R p d r 1 · · d r · d r + 1 · · d r p ( 1 r 1 ) · · ( 1 r ) · r + 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
M Y = M ( Q 1 / 2 G T ( ξ α ) ) = Q 1 / 2 G T M ( ξ α ) = 0 , cov Y = cov ( Q 1 / 2 G T ( ξ α ) ) = cov ( Q 1 / 2 G T ξ ) = Q 1 / 2 G T cov ( ξ ) ( Q 1 / 2 G T ) T = = Q 1 / 2 G T A G Q 1 / 2 = Q 1 / 2 Q Q 1 / 2 = diag ( 1 , , 1 m   times , 0 , , 0 p m   times ) .
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 | .
If we additionaly suppose that
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.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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Kuryliak, A.; Skaskiv, O. Wiman’s Type Inequality in Multiple-Circular Domain. Axioms 2021, 10, 348. https://doi.org/10.3390/axioms10040348

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Kuryliak, Andriy, and Oleh Skaskiv. 2021. "Wiman’s Type Inequality in Multiple-Circular Domain" Axioms 10, no. 4: 348. https://doi.org/10.3390/axioms10040348

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