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Article

# Relative Growth of Series in Systems of Functions and Laplace—Stieltjes-Type Integrals

Department of Mechanics and Mathematics, Ivan Franko National University of Lviv, 79000 Lviv, Ukraine
Axioms 2021, 10(2), 43; https://doi.org/10.3390/axioms10020043
Received: 10 March 2021 / Revised: 24 March 2021 / Accepted: 24 March 2021 / Published: 25 March 2021

## Abstract

:
For a regularly converging-in-$C$ series $A ( z ) = ∑ n = 1 ∞ a n f ( λ n z ) ,$ where f is an entire transcendental function, the asymptotic behavior of the function $M f − 1 ( M A ( r ) ) ,$ where $M f ( r ) = max { | f ( z ) | : | z | = r }$, is investigated. It is proven that, under certain conditions on the functions f, $α$, and the coefficients $a n$, the equality $lim r → + ∞ α ( M f − 1 ( M A ( r ) ) ) α ( r ) = 1$ is correct. A similar result is obtained for the Laplace–Stiltjes-type integral $I ( r ) = ∫ 0 ∞ a ( x ) f ( r x ) d F ( x ) .$ Unresolved problems are formulated.
MSC:
30B50; 30D10; 30D20

## 1. Introduction

Let
$f ( z ) = ∑ k = 0 ∞ f k z k$
be an entire function, $M f ( r ) = max { | f ( z ) | : | z | = r }$, and $Φ f ( r ) = ln M f ( r )$. For an entire function g with Taylor coefficients $g n$, the study of growth of the function $Φ f − 1 ( ln M g ( r ) )$ in terms of the exponential type was initiated in papers [1,2] and was continued in . As a result, it is proven that, if $| f k − 1 / f k | ↗ + ∞$ as $k → ∞$, then
$lim ¯ r → + ∞ Φ f − 1 ( ln M g ( r ) ) r = lim ¯ k → ∞ | g n | | f n | 1 / n .$
We remark that $Φ f − 1 ( x ) = M f − 1 ( e x )$ and, thus, $Φ f − 1 ( ln M g ( r ) ) = M f − 1 ( M g ( r ) )$. The order $ρ [ g ] g = lim ¯ r → + ∞ ln M f − 1 ( M g ( r ) ) ln r$ and the lower-order $λ [ g ] f = lim ̲ r → + ∞ ln M f − 1 ( M g ( r ) ) ln r$ of the function f with respect to the function g are used in Reference . Research on the relative growth of entire functions was continued by many mathematicians (an incomplete bibliography is given in ).
Let $( λ n )$ be a sequence of positive numbers increasing to $+ ∞$. Suppose that the series
$A ( z ) = ∑ n = 1 ∞ a n f ( λ n z )$
in the system $f ( λ n z )$ is regularly convergent in $C$, i.e., $∑ n = 1 ∞ | a n | M f ( r λ n ) < + ∞$ for all $r ∈ [ 0 , + ∞ )$. Many authors have studied the representation of analytic functions by series in the system $f ( λ n z )$ and the growth of such functions. Here, we specify only the monographs of A.F. Leont’ev  and B.V. Vinnitskyi , which are references to other papers on this topic.
Since series (2) is regularly convergent in $C$ and the function A is an entire function, a natural question arises about the asymptotic behavior of the function $M f − 1 ( M A ( r ) ) .$
We suppose that the function F is nonnegative, nondecreasing, unbounded, and continuous on the right on $[ 0 , + ∞ ) ;$ that f is positive, increasing, and continuous on $[ 0 , + ∞ )$; and that a positive-on-$[ 0 , + ∞ )$ function a is such that the Laplace–Stietjes-type integral
$I ( r ) = ∫ 0 ∞ a ( x ) f ( r x ) d F ( x )$
exists for every $r ∈ [ 0 , + ∞ )$. The asymptotic behavior of such integrals in the case $f ( x ) = e x$ is studied in the monograph . A question arises again about the asymptotic behavior of the function $f − 1 ( I ( r ) )$. Here, we present some results that indicate the possibility of solving these problems.

## 2. Relative Growth of Series in Systems of Functions

As in , by L, we denote a class of continuous nonnegative-on-$( − ∞ , + ∞ )$ functions $α$ such that $α ( x ) = α ( x 0 ) ≥ 0$ for $x ≤ x 0$ and $α ( x ) ↑ + ∞$ as $x 0 ≤ x → + ∞$. We say that $α ∈ L 0 ,$ if $α ∈ L$ and $α ( ( 1 + o ( 1 ) ) x ) = ( 1 + o ( 1 ) ) α ( x )$ as $x → + ∞$. Finally, $α ∈ L s i ,$ if $α ∈ L$ and $α ( c x ) = ( 1 + o ( 1 ) ) α ( x )$ as $x → + ∞$ for each $c ∈ ( 0 , + ∞ ) ,$ i.e., $α$ is a slowly increasing function. Clearly, $L s i ⊂ L 0$. We need the following lemma .
Lemma 1.
If $β ∈ L$ and $B ( δ ) = lim ¯ x → + ∞ β ( ( 1 + δ ) x ) β ( x ) ,$ $δ > 0 ,$ then in order for $β ∈ L 0 ,$ it is necessary and sufficient that $B ( δ ) → 1$ as $δ → + 0$.
We need also some well-known (see, for example, ) properties of the function $ln M f ( r )$.
Lemma 2.
If a function f is transcendental, then the function $ln M f ( r )$ is logarithmically convex and, thus,
$Γ f ( r ) : = d ln M f ( r ) d ln r ↗ + ∞ , r → + ∞ ,$
(at the points where the derivative does not exist, where $d ln M f ( r ) d ln r$ means the right-hand derivative).
For $α ∈ L ,$$β ∈ L$, and entire functions f and g, we define the generalized $( α , β )$-order $ρ α , β [ g ] f$ and the generalized lower $( α , β )$-order $λ α , β [ g ] f$ of g with respect to f as follows:
$ρ α , β [ g ] f = lim ¯ r → + ∞ α ( M f − 1 ( M g ( r ) ) ) β ( r ) , λ α , β [ g ] f = lim ̲ r → + ∞ α ( M f − 1 ( M g ( r ) ) ) β ( r ) .$
Suppose that $a n ≥ 0$ for all $n ≥ 1$. Since
$A ( z ) = ∑ n = 1 ∞ a n ∑ k = 0 ∞ f k ( z λ n ) k = ∑ k = 0 ∞ f k ∑ n = 1 ∞ a n λ n k z k ,$
in view of the Cauchy inequality, we have
$M A ( r ) ≥ | f k | ∑ n = 1 ∞ a n λ n k r k ≥ a n | f k | ( λ n r ) k$
for all $n ≥ 1 ,$$k ≥ 0$ and $r ∈ [ 0 , + ∞ )$. We also remark that, if $μ f ( r ) = max { | f k | r k : k ≥ 0 }$ is the maximal term of series (1), then
$M f ( r ) ≤ ∑ k = 0 ∞ | f k | r k = ∑ k = 0 ∞ | f k | ( 2 r ) k 2 − k ≤ 2 μ f ( 2 r ) .$
We choose $n 0 ≥ 1$ such that $a n 0 > 0$ and $λ n 0 ≥ 2$. Then, from (4) and (5), we get
$M A ( r ) ≥ max { a n 0 | f k | ( λ n 0 r ) k : k ≥ 0 } ≥ a n 0 μ f ( 2 r ) ≥ a n 0 2 M f ( r ) ,$
where $M f − 1 2 d n 0 M A ( r ) ≥ r$. By Lemma 2, $d ln M f − 1 ( x ) d ln x ↘ 0$ as $x → + ∞$ and, thus, for every $c > 1$
$ln M f − 1 ( c x ) − ln M f − 1 ( x ) = ∫ x c x d ln M f − 1 ( t ) d ln t d ln t ≤ d ln M f − 1 ( x ) d ln x → 0 , x → + ∞ ,$
i.e., the function $M f − 1$ is slowly increasing. Therefore,
$M f − 1 ( M A ( r ) ) ≥ ( 1 + o ( 1 ) ) r , r → + ∞ .$
On the other hand, since series (2) is regularly convergent in $C$, for each $r ∈ [ 0 , + ∞ )$, there exists $μ A ( r ) = max { | a n | M f ( r λ n ) : n ≥ 1 }$ and, for every $r ∈ [ 0 , + ∞ )$ and $τ > 0$, we have
$M A ( r ) ≤ ∑ n = 1 ∞ | a n | M f ( r λ n ) ≤ μ F ( ( 1 + τ ) r ) ∑ n = 1 ∞ M f ( r λ n ) M f ( ( 1 + τ ) r λ n ) .$
Then, by Lemma 2, for $r ≥ 1$, we have
$ln M f ( ( 1 + τ ) r λ n ) − ln M f ( r λ n ) = ∫ r λ n ( 1 + τ ) r λ n d ln M f ( x ) d ln x d ln x = ∫ r λ n ( 1 + τ ) r λ n Γ f ( x ) d ln x ≥ ≥ Γ f ( r λ n ) ln ( 1 + τ ) ≥ Γ f ( λ n ) ln ( 1 + τ ) .$
Therefore, if $ln n ≤ q Γ f ( λ n )$ for all $n ≥ n 0$ and $ln ( 1 + τ ) > q$, then
$∑ n = n 0 ∞ M f ( r λ n ) M f ( ( 1 + τ ) r λ n ) ≤ ∑ n = n 0 ∞ exp { − Γ f ( λ n ) ln ( 1 + τ ) } ≤ ∑ n = n 0 ∞ exp − ln ( 1 + τ ) q ln n < + ∞$
and (7) implies, for $r ≥ 1$,
$M A ( r ) ≤ T μ A ( ( 1 + τ ) r ) , T = const > 0 .$
$μ A ( r ) ≤ max | a n | ∑ k = 0 ∞ | f k | ( r λ n ) k : n ≥ 1 ≤ ≤ ∑ k = 0 ∞ max { | a n | λ n k : n ≥ 1 } | f k | r k = ∑ k = 0 ∞ μ D ( k ) | f k | r k ,$
where $μ D ( σ ) = max { | a n | exp { σ ln λ n } : n ≥ 1 }$ is the maximal term of Dirichlet series
$D ( σ ) = ∑ n = 1 ∞ | a n | exp { σ ln λ n } .$
Using estimates (6), (8), and (9), we prove the following theorem.
Theorem 1.
Let f be an entire transcendental function, $a n ≥ 0$ for all $n ≥ 1$, and series (2) be regularly convergent in $C$. Suppose that $ln n ≤ q Γ f ( λ n )$ for some $q > 0$ and all $n ≥ n 0$ and that $lim ¯ σ → + ∞ ln μ D ( σ ) σ ln M f − 1 ( e σ ) = γ .$
If $γ < 1$, then $λ α , α [ F ] f = ρ α , α [ F ] f = 1$ for every function α such that $α ( e x ) ∈ L s i$. If $γ = 0$, then $λ α , α [ F ] f = ρ α , α [ F ] f = 1$ for every function α such that $α ( e x ) ∈ L 0$.
Proof.
Since $α ∈ L 0 ,$ from (6), we get
$λ α , α [ F ] f = lim ̲ r → + ∞ α ( M f − 1 ( M F ( r ) ) ) α ( r ) ≥ lim ̲ r → + ∞ α ( ( 1 + o ( 1 ) ) r ) α ( r ) = 1 .$
On the other hand, in view of the Cauchy inequality, we have $ln | f k | ≤ ln M f ( r ) − k ln r$ for all r and k. We choose $r = r k = M f − 1 ( e k ) .$ Then, $ln | f k | ≤ k − k ln M f − 1 ( e k )$, i.e., $− ln | f k | ≥ k ( ln M f − 1 ( e k ) − 1 )$. Therefore,
$lim ¯ k → ∞ ln μ D ( k ) − ln f k ≤ lim ¯ k → ∞ ln μ D ( k ) k ( ln M f − 1 ( e k ) − 1 ) ≤ lim ¯ σ → + ∞ ln μ D ( σ ) σ ln M f − 1 ( e σ ) = σ .$
If $γ < 1$, then in view of (10), $ln μ D ( k ) − ln | f k | ≤ p$ for each $p ∈ ( γ , 1 )$ and all $k ≥ k 0$ and, thus, $μ D ( k ) ≤ | f k | − p$ for all $k ≥ k 0$. Therefore, in view of (9) and (5),
$μ A ( r ) ≤ ∑ k = 0 k 0 − 1 + ∑ k = k 0 ∞ μ D ( k ) | f k | r k ≤ O ( r k 0 − 1 ) + ∑ k = k 0 ∞ | f k | 1 − p r k ≤ ≤ O ( r k 0 − 1 ) + 2 max { f k 1 − p ( 2 r ) k : k ≥ 0 } = = O ( r k 0 − 1 ) + 2 max { ( | f k | ( 2 r ) k / ( 1 − p ) ) 1 − p : k ≥ 0 } = = O ( r k 0 − 1 ) + 2 ( μ f ( ( 2 r ) 1 / ( 1 − p ) ) ) 1 − p ≤ μ f ( ( 2 r ) 1 / ( 1 − p ) ) , r ≥ r 0 ,$
because $ln r = o ( ln μ f ( r ) )$ as $r → + ∞$ for every entire transcendental function f and $1 − p < 1$. Therefore, from (8) and (11), we get
$M A ( r ) ≤ T μ A ( ( 1 + τ ) r ) ≤ T μ f ( ( 2 ( 1 + τ ) r ) 1 / ( 1 − p ) ) ≤ T M f ( ( 2 ( 1 + τ ) r ) 1 / ( 1 − p ) )$
and, thus, $M f − 1 ( M A ( r ) ) ≤ ( 1 + o ( 1 ) ) ( 2 ( 1 + τ ) r ) 1 / ( 1 − p )$ as $r → + ∞$. If $α ∈ L s i$, then we obtain
$lim ¯ r → + ∞ α ( M f − 1 ( M A ( r ) ) ) α ( r 1 / ( 1 − p ) ) ≤ 1 .$
Suppose that $α ( e x ) ∈ L s i .$ Then,
$α ( r 1 / ( 1 − p ) ) = α ( exp 1 1 − p ln r ) = ( 1 + o ( 1 ) ) α ( exp { ln r } ) = ( 1 + o ( 1 ) ) α ( r )$
as $r → + ∞$. Therefore, (12) implies the inequality $ρ α , α [ A ] f ≤ 1 ,$ where in view of the inequality $λ α , α [ A ] f ≥ 1$, we get $λ α , α [ A ] f = ρ α , α [ A ] f = 1 .$
If $γ = 0$, then (12) holds for every $p ∈ ( 0 , 1 )$ and all $r ≥ r 0 ( p )$. If we put $1 1 − p = 1 + δ$, then $δ → + 0$ as $p → + 0$, and in view of the condition $α ( e x ) ∈ L 0$, by Lemma 1, we have
$lim ¯ r → + ∞ α ( r 1 / ( 1 − p ) ) α ( r ) = lim ¯ r → + ∞ α ( exp { ( 1 + δ ) ln r } ) α ( exp { ln r } ) = B ( δ ) → 1 , δ → 1 .$
Therefore,
$1 ≥ lim ¯ r → + ∞ α ( M f − 1 ( M A ( r ) ) ) α ( r 1 / ( 1 − p ) ) = lim ¯ r → + ∞ α ( M f − 1 ( M A ( r ) ) ) α ( r ) · α ( r ) α ( r 1 + δ ) ≥ ≥ lim ¯ r → + ∞ α ( M f − 1 ( M A ( r ) ) ) α ( r ) lim ̲ r → + ∞ α ( r ) α ( r 1 + δ ) = ρ α , α [ F ] f B ( δ ) .$
In view of the arbitrariness of $δ$, we get $ρ α , α [ A ] f ≤ 1$, and again, $λ α , α [ A ] f = ρ α , α [ A ] f = 1$. Theorem 1 is proven. □
We remark that, if $f k ≥ 0$ for all $k ≥ 0$, then $M f ( r ) = f ( r )$. Therefore, from Theorem 1, we obtain the following statement.
Corollary 1.
Let f be an entire transcendental function, $f k ≥ 0$ for all $k ≥ 0 ,$$a n ≥ 0$ for all $n ≥ 1$, and series (2) be regularly convergent in $C$. Suppose that $f ′ ( r ) / f ( r ) ≥ h > 0$ for all $r ≥ r 0 ,$$ln n = O ( λ n )$ as $n → ∞$ and $lim ¯ σ → + ∞ ln μ D ( σ ) σ ln f − 1 ( e σ ) = γ .$
If $γ < 1$, then $λ α , α [ A ] f = ρ α , α [ A ] f = 1$ for every function α such that $α ( e x ) ∈ L s i$.
If $γ = 0$, then $λ α , α [ A ] f = ρ α , α [ A ] f = 1$ for every function α such that $α ( e x ) ∈ L 0$.

## 3. Relative Growth of Laplace–Stieltjes-Type Integrals

Suppose again that f is an entire transcendental function, $f k ≥ 0$ for all $k ≥ 0$, and $x 0 > 1$ is such that $∫ 1 x 0 a ( x ) d F ( x ) ≥ > 0 .$ Then,
$I ( r ) ≥ ∫ 1 x 0 a ( x ) f ( r x ) d F ( x ) ≥ f ( r ) c ,$
i.e., as above, $f − 1 ( I ( r ) ) ≥ ( 1 + o ( 1 ) ) r$ as $r → + ∞ ,$ where for $α ∈ L 0$,
$λ α , α [ I ] f = lim ̲ r → + ∞ α ( f − 1 ( I ( r ) ) ) α ( r ) ≥ 1 .$
On the other hand, if $τ ≥ e − 1$, then as above, for $r ≥ 1$, we have
$ln f ( ( 1 + τ ) r x ) − ln f ( r x ) = ∫ r x ( 1 + τ ) r x d ln f ( x ) d ln x d ln x = ∫ r x ( 1 + τ ) r x Γ f ( x ) d ln x ≥ ≥ Γ f ( x ) ln ( 1 + x ) ,$
i.e., $f ( r x ) f ( ( 1 + τ ) r x ) ≤ e − Γ f ( x ) ln ( 1 + τ ) .$ Therefore, if $μ I ( r ) = max { a ( x ) f ( r x ) : x ≥ 0 }$ is the maximum of the integrand and $ln F ( x ) ≤ q Γ f ( x )$ for some $q > 0$ and all $x ≥ x 0$, then for $ln ( 1 + τ ) > q$ (for simplicity assuming $x 0 = 0$), we get
$I ( r ) = ∫ 0 ∞ a ( x ) f ( ( 1 + τ ) r x ) f ( r x ) f ( ( 1 + τ ) r x ) d F ( x ) ≤ μ I ( ( 1 + τ ) r ) ∫ 0 ∞ f ( r x ) f ( ( 1 + τ ) r x ) d F ( x ) ≤ ≤ μ I ( ( 1 + τ ) r ) ∫ 0 ∞ e − Γ f ( x ) ln ( 1 + τ ) d F ( x ) ≤ ≤ μ I ( ( 1 + τ ) r ) ln ( 1 + τ ) ∫ 0 ∞ e − Γ f ( x ) ln ( 1 + τ ) + ln F ( x ) d Γ f ( x ) ≤ ≤ μ I ( ( 1 + τ ) r ) ln ( 1 + τ ) ∫ 0 ∞ e − Γ f ( x ) ( ln ( 1 + τ ) − q ) d Γ f ( x ) = μ I ( ( 1 + τ ) r ) ln ( 1 + τ ) ln ( 1 + τ ) − q = = T μ I ( ( 1 + τ ) r ) .$
$μ I ( r ) = max a ( x ) ∑ k = 0 ∞ f k ( x r ) k : x ≥ 0 ≤ ≤ ∑ k = 0 ∞ max { a ( x ) x k : x ≥ 0 } f k r k = ∑ k = 0 ∞ μ J ( k ) f k r k ,$
where $μ J ( σ ) = max { a ( x ) e σ ln x : x ≥ 0 } = max { a ( x ) x ln x : x ≥ 0 }$ is the maximum of the integrand for the Laplace integral
$J ( σ ) = ∫ 0 ∞ a ( x ) e σ ln x d F ( x ) .$
Using estimates (13) and (14), and $λ α , α [ I ] f ≥ 1$, we prove the following analog of Theorem 1.
Theorem 2.
Let $ln F ( x ) ≤ q Γ f ( x )$ for some $q > 0$ and all $x ≥ x 0$, and $lim ¯ σ → + ∞ ln μ J ( σ ) γ ln f − 1 ( e σ ) = γ .$
If $γ < 1$, then $λ α , α [ I ] f = ρ α , α [ I ] f = 1$ for every function α such that $α ( e x ) ∈ L s i$.
If $γ = 0$, then $λ α , α [ I ] f = ρ α , α [ I ] f = 1$ for every function α such that $α ( e x ) ∈ L 0 .$
Proof.
As in the proof of Theorem 1, we obtain $− ln | f k | ≥ k ( ln f − 1 ( e k ) − 1 )$ and $lim ¯ k → ∞ ln μ J ( k ) − ln f k ≤ γ .$ Therefore, if $γ < 1$, then $μ D ( k ) ≤ | f k | − p$ for each $p ∈ ( γ , 1 )$ and all $k ≥ k 0$, and in view of (14) and (5), as in the proof of Theorem 1, we get $μ I ( r ) ≤ μ f ( ( 2 r ) 1 / ( 1 − p ) )$ for $r ≥ r 0 .$ Therefore, in view of (13), we get
$I ( r ) ≤ T μ I ( ( 1 + τ ) r ) ≤ T f ( ( 2 ( 1 + τ ) r ) 1 / ( 1 − p ) ) ,$
where $f − 1 ( I ( r ) ) ≤ ( 1 + o ( 1 ) ) ( 2 ( 1 + τ ) r ) 1 / ( 1 − p )$ as $r → + ∞$. If $α ∈ L s i$, then we obtain
$lim ̲ r → + ∞ α ( f − 1 ( I ( r ) ) ) α ( r 1 / ( 1 − p ) ) ≤ 1 .$
Further proof of Theorem 2 is the same as that of Theorem 1. □
Theorem 2 implies the following statement.
Corollary 2.
Let $f ′ ( x ) / f ( x ) ≥ h ,$$h > 0 ,$ $ln F ( x ) ≤ q x$ for some $q > 0$ and all $x ≥ 0 ,$ and $lim ¯ r → + ∞ ln μ J ( σ ) σ f − 1 ( e σ ) = γ .$
If $γ < 1$, then $λ α , α [ I ] f = ρ α , α [ I ] f = 1$ for every function α such that $α ( e x ) ∈ L s i .$
If $γ = 0$, then $λ α , α [ I ] f = ρ α , α [ I ] f = 1$ for every function α such that $α ( e x ) ∈ L 0$.

## 4. Examples

Here, we consider the case when $f ( z ) = E ρ ( z ) ,$ where
$E ρ ( z ) = ∑ k = 0 ∞ z k Γ ( 1 + k ρ ) , 0 < ρ < + ∞ ,$
is the Mittag–Leffler function. The properties of this function have been used in many problems in the theory of entire functions. We only need the following property of the Mittag–Leffler function: if $0 < ρ < + ∞$, then ( p. 85)
$M E ρ ( r ) = E ρ ( r ) = ( 1 + o ( 1 ) ) ρ e r ρ , r → + ∞$
and, if $1 / 2 < ρ < + ∞$, then 
$E ρ ′ ( r ) / E ρ ( r ) = ρ r ρ − 1 + O ( r ρ − 2 e − r ρ ) , r → + ∞ .$
From (15), it follows that $E ρ − 1 ( x ) = ( 1 + o ( 1 ) ) ln 1 / ρ x$ as $x → + ∞$. Therefore, for $f ( x ) = E ρ ( x )$, we have $σ ln f − 1 ( e σ ) = 1 + o ( 1 ) ρ σ ln σ$ as $σ → + ∞$. Since in (16), $Γ E ρ ( r ) = ρ r ρ + o ( 1 )$ as $r → + ∞ ,$ then if $ln F ( x ) ≤ q ρ x ρ$ for some $q > 0$ and all $x ≥ x 0$, and
$lim ¯ σ → + ∞ ln μ J ( σ ) σ ln σ = 0 ,$
then for $α ( x ) = ln x$$( x ≥ e )$, by Theorem 2, we get
$lim r → + ∞ ln E ρ − 1 ( I ρ ( r ) ) ln r = 1 , I ρ ( r ) = ∫ 0 ∞ a ( x ) E ρ ( r x ) d F ( x ) .$
Let us now find out under what conditions (17) holds on $a ( x )$. For this, as in ( p. 29), by $Ω$, we denote a class of positive unbounded functions $Φ$ on $( − ∞ , + ∞ )$ such that the derivative $Φ 0$ is positive, continuously differentiable, and increasing to $+ ∞$ on $( − ∞ , + ∞ ) .$ For $Φ ∈ Ω$, let $φ$ be the inverse function to $Φ ′$ and $Ψ ( σ ) = σ − Φ ( σ ) Φ ′ ( σ )$ be the function associated with $Φ$ in the sense of Newton.
By Theorem 2.2.1 from ( p. 30), $ln max { a ( x ) e σ x : x ≥ 0 } ≤ Φ ( σ ) ∈ Ω$ for all $σ ≥ σ 0$ if and only if $ln a ( x ) ≤ − x Ψ ( φ ( x ) )$ for all $x ≥ x 0$. Choosing $Φ ( σ ) = ϵ σ ln σ$ for $σ ≥ σ 0 ,$ we obtain $Φ ′ ( σ ) = ϵ ( ln σ + 1 ) ,$$φ ( x ) = exp { x / ϵ − 1 }$ and $x Ψ ( φ ( x ) ) = x φ ( x ) − Φ ( φ ( x ) ) = ϵ exp { x / ϵ − 1 }$ for $x ≥ x 0$. Therefore, $ln μ J ( σ ) ≤ ε σ ln σ$ for all $σ ≥ σ 0$ if and only if $ln a ( x ) ≤ − ε exp { ln x / ε − 1 }$ for $x ≥ x 0$. Hence, it follows that, if $ln x = o ( ln ln ( 1 / a ( x ) ) )$ as $x → + ∞$, then (17) holds. Thus, the following statement is true.
Proposition 1.
If $ρ > 1 / 2 ,$$ln F ( x ) = O ( x ρ )$ and $ln x = o ( ln ln ( 1 / a ( x ) ) )$ as $x → + ∞$, then (18) holds.
Remark 1.
If $ρ = 1$, then $E ρ ( r ) = E 1 ( r ) = e r$, and we have a usual Laplace–Stieltjes integral $I 1 ( r ) = ∫ 0 ∞ a ( x ) e r x d F ( x )$. Therefore, if $ln F ( x ) = O ( x )$ and $ln x = o ( ln ln ( 1 / a ( x ) ) )$ as $x → + ∞$, then $p R [ I 1 ] : = lim r → + ∞ ln ln I 1 ( r ) ln r = 1$. On the other hand, the quantity $p R [ I 1 ]$ is called the logarithmic R-order of $I 1$, and in ( p. 83), it is proven that, if $ln F ( x ) = O ( x )$ as $x → + ∞$, then $p R [ I 1 ] = lim ¯ x → + ∞ ln x ln ( 1 x ln 1 a ( x ) ) = 1 ,$ i.e., if $ln F ( x ) = O ( x )$ and $ln x = o ( ln ln ( 1 / a ( x ) ) )$ as $x → + ∞$, then $p R [ I 1 ] = 1 .$
Similarly, we can prove the following statement.
Proposition 2.
Let $ρ ≥ 1 / 2$, $ln n = O ( λ n ρ )$ as $n → ∞ ,$ $a n ≥ 0$ for all $n ≥ 1$ and series $A ρ ( z ) = ∑ n = 1 ∞ a n E ρ ( λ n z )$ be regularly convergent in $C$. If $ln n = o ( ln ln ( 1 / a n ) )$ as $n → ∞$, then $lim r → + ∞ ln E ρ − 1 ( M A ρ ( r ) ) ln r = 1 .$
Remark 2.
If $ρ = 1$, then we have a Dirichlet series $A 1 ( z ) = ∑ n = 1 ∞ a n e λ n z$. Therefore, if this Dirichlet series is absolutely convergent in $C$, $a n ≥ 0$ for all $n ≥ 1$, $ln n = O ( λ n )$, and $ln n = o ( ln ln ( 1 / a n ) )$ as $n → ∞$, then $p R [ A 1 ] : = lim r → + ∞ ln ln M A 1 ( r ) ln r = 1$. On the other hand, the quantity $p R [ A 1 ]$ is called the logarithmic R-order of $A 1$ and $p R [ A 1 ] = lim ¯ n → + ∞ ln λ n ln ( 1 λ n ln 1 a n ) = 1$ provided $ln n = O ( λ n )$ as $n → ∞$ , i.e., if $ln n = O ( λ n )$ and $ln λ n = o ( ln ln ( 1 / a n ) )$ as $n → ∞$, then $p R [ A 1 ] = 1 .$

## 5. Discussion Open Problems

1. The natural problem studied was the relative growth when the domain of regular convergence of series (2) is the disk $D R = { z : | z | < R < + ∞ }$ and the function f is either entire or analytic in $D R$.
2. It is well known that the study of the growth of entire functions of many complex variables involves many options. The following problem is the simplest.
Let f be an entire function and the series $A ( z , w ) = ∑ m = 1 , n = 1 ∞ a m , n f ( λ m z + μ n w )$ be regularly convergent in $C 2$. A question arises about the asymptotic behavior of the function $M f − 1 ( M A ( r , ρ ) )$, where $M A ( r , ρ ) = max { | A ( z , w ) | : | z | ≤ r , | w | ≤ ρ }$.
3. The condition $ρ ≥ 1 / 2$ in Propositions 1 and 2 arose in connection to the application of Equation (16). Probably, it is superfluous in the above statements.

## Funding

This research received no external funding.

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Not applicable.

## Data Availability Statement

This research did not report any data.

## Conflicts of Interest

The author declares no conflict of interest.

## References

1. Nachbin, L. An extension of the notion of integral function of the finite exponential type. An. Acad. Brasil Sci. 1944, 16, 143–147. [Google Scholar]
2. Boas, R.P.; Buck, R.C. Polynomial Expansions of Analytic Functions; Springer: Berlin, Germany, 1958. [Google Scholar]
3. Vinnitsky, B.V. Some Approximation Properties of Generalized Systems of Exponentials; Dep. in UkrNIINTI 25 February 1991; Drogobych Pedagogical Institute: Drogobych, Ukraine, 1991. (In Russian) [Google Scholar]
4. Roy, C. On the relative order and lower order of an entire function. Bull. Soc. Cal. Math. Soc. 2010, 102, 17–26. [Google Scholar]
5. Mulyava, O.M.; Sheremeta, M.M. Relative growth of Dirichlet series with different abscissas of absolute convergence. Ukr. Math. J. 2020, 72, 1535–1543. [Google Scholar]
6. Leont’ev, A.F. Generalizations of Exponential Series; Nauka: Moscow, Russia, 1981. (In Russian) [Google Scholar]
7. Sheremeta, M.M. Asymptotical Behavior of Laplace-Stietjes Integrals; VNTL Publishers: Lviv, Ukraine, 2010. [Google Scholar]
8. Sheremeta, M.N. Connection between the growth of the maximum of the modulus of an entire function and the moduli of the coefficients of its power series expansion. Izv. Vyssh. Uchebn. Zaved. Mat. 1967, 2, 100–108. (In Russian) [Google Scholar]
9. Sheremeta, M.M. On two classes of positive functions and the belonging to them of main characteristics of entire functions. Mat. Stud. 2003, 19, 75–82. [Google Scholar]
10. Pólya, G.; Szegő, G. Aufgaben und Lehrsatze aus der Analysis. II; Springer: Berlin, Germany, 1964. [Google Scholar]
11. Gol’dberg, A.A.; Ostrovskii, I.V. Value Distribution of Meromorphic Functions; AMS: Providence, RI, USA, 2008; (Translated from Russian ed. Nauka: Moscow, USSR, 1970). [Google Scholar]
12. Gol’dberg, A.A. An estimate of modulus of logarithmic derivative of Mittag-Leffler function with applications. Mat. Stud. 1995, 5, 21–30. [Google Scholar]
13. Reddy, A.R. On entire Dirichlet series of zero order. Tohoku Math. J. 1966, 18, 144–155. [Google Scholar] [CrossRef]
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Sheremeta, M. Relative Growth of Series in Systems of Functions and Laplace—Stieltjes-Type Integrals. Axioms 2021, 10, 43. https://doi.org/10.3390/axioms10020043

AMA Style

Sheremeta M. Relative Growth of Series in Systems of Functions and Laplace—Stieltjes-Type Integrals. Axioms. 2021; 10(2):43. https://doi.org/10.3390/axioms10020043

Chicago/Turabian Style

Sheremeta, Myroslav. 2021. "Relative Growth of Series in Systems of Functions and Laplace—Stieltjes-Type Integrals" Axioms 10, no. 2: 43. https://doi.org/10.3390/axioms10020043

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