Results on analytic functions defined by Laplace-Stieltjes transforms with perfect ϕ-type

Simin Liu; Hongyan Xu; Yongqin Cui; Pan Gong

doi:10.1515/math-2020-0118

Article Open Access

Results on analytic functions defined by Laplace-Stieltjes transforms with perfect ϕ-type

Simin Liu , Hongyan Xu , Yongqin Cui and Pan Gong

Published/Copyright: December 31, 2020

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$Open Mathematics$

From the journal Open Mathematics Volume 18 Issue 1

Abstract

In this paper, we introduce the concept of the perfect ϕ -type to describe the growth of the maximal molecule of Laplace-Stieltjes transform by using the more general function than the usual. Based on this concept, we investigate the approximation and growth of analytic functions F ( s ) defined by Laplace-Stieltjes transforms convergent in the half plane and obtain some results about the necessary and sufficient conditions on analytic functions F ( s ) defined by Laplace-Stieltjes transforms with perfect ϕ -type, which are some generalizations and improvements of the previous results given by Kong [On generalized orders and types of Laplace-Stieltjes transforms analytic in the right half-plane, Acta Math. Sin. 59A (2016), 91–98], Singhal and Srivastava [On the approximation of an analytic function represented by Laplace-Stieltjes transformations, Anal. Theory and Appl. 31 (2015), 407–420].

Keywords: perfect ϕ-type; approximation; Laplace-Stieltjes transform; half plane

MSC 2010: 44A10; 30D15

1 Introduction

Let L ( s , F , α ) be a class of Laplace-Stieltjes transforms

(1.1) F ( s ) = ∫ 0 + ∞ e s x d α ( x ) , s = σ + i t ,

where α ( x ) is a bounded variation on any finite interval [ 0 , Y ] ( 0 < Y < + ∞ ) , and σ and t are real variables. If we choose a suitable α ( t ) , then F ( s ) can be represented as a Dirichlet series F ( s ) = ∑ n = 1 ∞ a n e λ n s , where a n ( n = 1 , 2 , … ) are nonzero complex numbers, and the sequence { λ n } n =1 ∞ satisfies

(1.2) 0 = λ 1 < λ 2 < λ 3 < ⋯ < λ n ↑ + ∞ .

Denote

A n ⁎ = sup λ n < x ≤ λ n + 1 , − ∞ < t < + ∞ ∫ λ n x e i t y d α ( y ) ,

(1.3) lim sup n → + ∞ log A n ∗ λ n = 0

and

(1.4) lim sup n → + ∞ ( λ n + 1 − λ n ) = h < + ∞ , lim sup n → + ∞ n λ n = D < ∞ ,

then in view of the Valiron-Knopp-Bohr formula [1], it leads to σ u F = 0 , that is, F ( s ) is analytic in the left half plane, where σ u F is the abscissa of uniform convergence of F ( s ) . If (1.3) is replaced by

(1.5) lim sup n → + ∞ log A n ∗ λ n = − ∞ ,

then it yields σ u F = + ∞ , that is, F ( s ) is called as an entire function. For convenience, for − ∞ < β < ∞ , we denote

L ¯ β ≔ { F ( s ) ∈ L ( s , F , α ) | σ u F = β , and { λ n } satisfy ( 1.2 ) , ( 1.4 ) } , L 0 ≔ { F ( s ) ∈ L ( s , F , α ) | A n ⁎ , { λ n } satisfy ( 1.2 ) – ( 1.4 ) }

and

L ∞ ≔ { F ( s ) ∈ L ( s , F , α ) | A n ⁎ , { λ n } satisfy ( 1.2 ) , ( 1.4 ) , ( 1.5 ) } .

Thus, if F ( s ) ∈ L 0 and − ∞ < β < 0 , then F ( s ) ∈ L ¯ β .

In order to estimate the growth of F ( s ) , Yu [1] introduced the concepts of the maximal term μ ( σ , F ) , the maximal molecule M u ( σ , F ) and the order of F ( s ) and also studied that the value distribution of entire functions defined by Laplace-Stieltjes transforms converge in the complex plane. After his wonderful works, many scholars studied the value distribution and the growth of analytic functions represented by Laplace-Stieltjes transforms converge in the whole plane or the half plane, and obtained a large number of important and interesting results (see [2,3,4,5,6,7,8,9,10,11,12,13,14]).

Define

μ ( σ , F ) = max n ∈ ℕ { A n ⁎ e λ n σ } , M u ( σ , F ) = sup 0 < x < + ∞ , − ∞ < t < + ∞ ∫ 0 x e ( σ + i t ) y d α ( y ) , ( σ < 0 ) .

For F ( s ) ∈ L 0 , in view of M u ( σ , F ) → + ∞ as σ → 0 − , the concepts of order and type can be usually used in estimating the growth of F ( s ) precisely.

Definition 1.1

[15, Definition 1.1] If Laplace-Stieltjes transform F ( s ) ∈ L 0 satisfies, we define the order and the lower order of F ( s ) as follows:

ρ = lim sup σ → 0 − log + log + M u ( σ , F ) − log ( − σ ) , μ = lim sup σ → 0 − log + log + M u ( σ , F ) − log ( − σ ) , 0 ≤ μ ≤ ρ ≤ + ∞ ,

where log + x = max { log x , 0 } . Furthermore, if ρ ∈ (0, + ∞ ) , the type and the lower type of F ( s ) are defined by

T = lim sup σ → 0 − log + M u ( σ , F ) − 1 σ ρ , t = lim inf σ → 0 − log + M u ( σ , F ) − 1 σ ρ , 0 ≤ t ≤ T ≤ + ∞ .

Remark 1.1

However, if ρ = 0 and ρ = + ∞ , we cannot estimate the growth of such functions precisely by using the concept of type.

Remark 1.2

We say that F ( s ) is of regular growth, if ρ = μ ; furthermore, F ( s ) is of perfect regular growth, if T = t , that is,

T = lim σ → 0 − log + M u ( σ , F ) − 1 σ ρ .

If Laplace-Stieltjes transform (1.1) satisfies A n ⁎ = 0 for n ≥ k + 1 , and A k ⁎ ≠ 0 , then F ( s ) may be called as an exponential polynomial of degree k usually denoted by p k , i.e., p k ( s ) = ∫ 0 λ k exp ( s y ) d α ( y ) . If we choose a suitable function α ( y ) , the function p k ( s ) may be reduced to a polynomial in terms of exp ( s λ i ) , that is, ∑ i = 1 k b i exp ( s λ i ) . We denote Π k to be the class of all exponential polynomials of degree almost k, that is,

∏ k = ∑ i = 1 k b i exp ( s λ i ) : ( b 1 , b 2 , … , b k ) ∈ ℂ k .

For F ( s ) ∈ L ¯ β , − ∞ < β < 0 , we denote by E n ( F , β ) the error in approximating the function F ( s ) by exponential polynomials of degree n in uniform norm as

E n ( F , β ) = inf p ∈ Π n ∥ F − p ∥ β , n = 1 , 2 , … ,

where

∥ F − p ∥ β = max − ∞ < t < + ∞ | F ( β + i t ) − p ( β + i t ) | .

For convenience, the error E n − 1 ( F , β ) is always supposed to be not null.

For F ( s ) ∈ L 0 , Singhal and Srivastava [16] in 2015 studied the approximation of F ( s ) with finite order, and obtained as follows.

Theorem 1.1

[16] If Laplace-Stieltjes transform F ( s ) ∈ L 0 , and is of order ρ ( 0 < ρ < ∞ ) and of type T, then for any real number − ∞ < β < 0 , we have

ρ = lim sup n → + ∞ log + log + [ E n − 1 ( F , β ) exp { − β λ n } ] log λ n − log + log + [ E n − 1 ( F , β ) exp { − β λ n } ]

and

(1.6) T = lim sup n → + ∞ ( log + [ E n − 1 ( F , β ) exp { − β λ n } ] ) ρ + 1 ρ + 1 ρ ρ + 1 ρ ( λ n ) ρ .

From Remark 1.2 and Theorem 1.1, the following question will be suggested naturally:

Question 1.1.

What will happen when F ( s ) is of perfect regular growth, that is, T = t in Theorem 1.1?

To answer Question 1.1, we first introduce the following definition of ϕ -type of Laplace-Stieltjes transform F ( s ) , which can estimate the growth of M u ( σ , F ) or μ ( σ , F ) more widely by utilizing more general function than ( − 1 σ ) ρ in Definition 1.1.

Definition 1.2

Let Ξ 0 be the set of positive unbounded function ϕ on ( − ∞ ,0) such that the derivative ϕ ′ is positive, continuous and increasing to + ∞ on ( − ∞ ,0) . If ϕ ∈ Ξ 0 and Laplace-Stieltjes transform F ( s ) ∈ L 0 satisfies

T ϕ = lim sup σ → 0 − log + M u ( σ , F ) ϕ ( σ ) , 0 ≤ T ϕ ≤ + ∞ ,

then T is called the ϕ -type of F ( s ) . Similarly, the lower ϕ -type of F ( s ) is defined by

t ϕ = lim inf σ → 0 − log + M u ( σ , F ) ϕ ( σ ) , 0 ≤ t ϕ ≤ + ∞ .

Remark 1.3

Obviously, 0 ≤ t ϕ ≤ T ϕ ≤ + ∞ . Besides, if t ϕ = T ϕ , then we say that F ( s ) is of perfect ϕ -type, that is,

lim σ → 0 − log + M u ( σ , F ) ϕ ( σ ) = T ϕ .

Let φ be the inverse function of ϕ ′ , then φ is continuous on (0, + ∞ ) and increases to 0, and let ϕ ∈ Ξ 0 and ψ ( x ) = x − ϕ ( x ) ϕ ′ ( x ) , then in view of [17], ψ is an increasing function on ( − ∞ ,0) , and ψ ( x ) → 0 as x → 0 . Besides, let ψ − 1 be the inverse function of ψ . Then ψ − 1 is an increasing function on ( − ∞ ,0) and ϕ ′ ( ψ − 1 ( σ ) ) increases to + ∞ on ( − ∞ ,0) . For 0 < a < b < + ∞ and q > 0 , let

G 1 ( a , b , q , ϕ ) = a b b − a ∫ a b ϕ ( φ ( q t ) ) t 2 d t

and

G 2 ( a , b , q , ϕ ) = ϕ 1 b − a ∫ a b φ ( q t ) d t .

Now, we list our main results below to show the relations among the perfect ϕ -type, the error E n ( F , β ) , λ n and A n ⁎ for Laplace-Stieltjes transforms F ( s ) with the perfect ϕ -type.

Theorem 1.2

If Laplace-Stieltjes transform F ( s ) ∈ L 0 satisfies

(1.7) lim sup σ → 0 − − log ( − σ ) ϕ ( σ ) = 0 ,

then for any real number − ∞ < β < 0 and 0 < T ϕ < + ∞ , we have

lim σ → 0 − log + M u ( σ , F ) ϕ ( σ ) = T ϕ ⇔ ( i ) lim sup n → + ∞ λ n Ω n ( F , β , λ n , ϕ ′ , ψ − 1 ) = T ϕ ;

(ii) There exists a non-decreasing positive integer sequence { n v } satisfying

(1.8) lim v → + ∞ λ n v Ω n v ( F , β , λ n v , ϕ ′ , ψ − 1 ) = T ϕ

and

(1.9) lim v → + ∞ G 1 ( λ n v , λ n v + 1 , T ϕ , ϕ ) G 2 ( λ n v , λ n v + 1 , T ϕ , ϕ ) = 1 ,

where

Ω n ( F , β , λ n , ϕ ′ , ψ − 1 ) = ϕ ′ ψ − 1 1 λ n log 1 E n − 1 ( F , β ) exp { − β λ n }

and

Ω n v ( F , β , λ n v , ϕ ′ , ψ − 1 ) = ϕ ′ ψ − 1 1 λ n v log 1 E n v − 1 ( F , β ) exp { − β λ n v } .

Theorem 1.3

If Laplace-Stieltjes transform F ( s ) ∈ L 0 satisfies (1.7), then

lim σ → 0 − log + M u ( σ , F ) ϕ ( σ ) = T ϕ ⇔ ( i ) lim sup n → + ∞ λ n Ω n ( F , A n ⁎ , λ n , ϕ ′ , ψ − 1 ) = T ϕ ;

(ii) There exists a non-decreasing positive integer sequence { n v } satisfying (1.9) and

(1.10) lim v → + ∞ λ n Ω n v ( F , A n ⁎ , λ n v , ϕ ′ , ψ − 1 ) = T ϕ ,

where

Ω n ( F , A n ⁎ , λ n , ϕ ′ , ψ − 1 ) = ϕ ′ ψ − 1 1 λ n log 1 A n ⁎

and

Ω n v ( F , A n v ⁎ , λ n v , ϕ ′ , ψ − 1 ) = ϕ ′ ψ − 1 1 λ n v log 1 A n v ⁎ .

Particularly, if F ( s ) ∈ L 0 is of finite order ρ , let ϕ ( σ ) = − 1 σ ρ , in view of Theorems 1.2 and 1.3, we can obtain the following corollaries.

Corollary 1.1

If F ( s ) ∈ L 0 is of finite order ρ ( 0 < ρ < ∞ ) , then for any real number − ∞ < β < 0 , we have

lim σ → 0 − log + M u ( σ , F ) − 1 σ ρ = T ⇔ ( i ) lim sup n → + ∞ log + [ E n − 1 ( F , β ) exp { − β λ n } ] λ n log + E n − 1 ( F , β ) exp { − β λ n } ρ = ( ρ + 1 ) ρ + 1 ρ ρ T ;

(ii) There exists a non-decreasing positive integer sequence { n v } satisfying

(1.11) lim sup v → + ∞ log + E n v − 1 ( F , β ) exp { − β λ n v } λ n v log + E n v − 1 ( F , β ) exp { − β λ n v } ρ = ( ρ + 1 ) ρ + 1 ρ ρ T , lim v → ∞ λ n v λ n v − 1 = 1 .

Corollary 1.2

If F ( s ) ∈ L 0 is of finite order ρ ( 0 < ρ < ∞ ) , then

lim σ → 0 − log + M u ( σ , F ) − 1 σ ρ = T ⇔ ( i ) lim sup n → + ∞ log + A n ⁎ λ n log + A n ⁎ ρ = ( ρ + 1 ) ρ + 1 ρ ρ T ;

(ii) There exists a non-decreasing positive integer sequence { n v } satisfying

lim sup v → + ∞ log + A n v ⁎ λ n v log + A n v ⁎ ρ = ( ρ + 1 ) ρ + 1 ρ ρ T , lim v → ∞ λ n v λ n v + 1 = 1 .

Remark 1.4

In view of Corollaries 1.1 and 1.2, this shows that our results are some generalizations and improvements of Theorem 1.1.

2 Some lemmas

To prove our results, we also need to give the following lemmas.

Lemma 2.1

[17, Lemma 2.2] If Laplace-Stieltjes transform F ( s ) ∈ L 0 , for any σ ( − ∞ < σ < 0 ) and ε ( > 0 ) , we have

1 p μ ( σ , F ) ≤ M u ( σ , F ) ≤ C μ ( ( 1 − ε ) σ , F ) − 1 σ ,

where p > 2 and C ( ≠ 0) are constants.

Lemma 2.2

Let ϕ ∈ Ξ 0 and 0 < T ϕ < + ∞ , then the conclusion that log μ ( σ , F ) ≤ T ϕ ϕ ( σ ) for any σ ∈ ( − ∞ , 0) holds if and only if log A n ⁎ ≤ − λ n ψ ( φ ( λ n T ϕ ) ) for all n ≥ 0 .

Proof

Suppose that log A n ⁎ ≤ − λ n ψ ( φ ( λ n T ϕ ) ) for all n ≥ 0 . Thus, for any σ < 0 and x < 0 , we have

( σ − x ) ϕ ′ ( x ) ≤ ∫ x σ ϕ ′ ( t ) d t = ϕ ( σ ) − ϕ ( x ) .

Hence, it follows

log μ ( σ , F ) ≤ max − λ n ψ φ λ n T ϕ + λ n σ : n ≥ 0 ≤ max { T ϕ ( − t ψ ( φ ( t ) ) + t σ ) : t ≥ 0 } = T ϕ max { − ϕ ′ ( x ) ψ ( x ) + σ + ϕ ′ ( x ) : x > − ∞ } = T ϕ max { ( σ − x ) ϕ ′ ( x ) + ϕ ( x ) : x > − ∞ } = T ϕ ϕ ( σ ) .

On the other hand, assume that log μ ( σ , F ) ≤ T ϕ ϕ ( σ ) for any σ ∈ ( − ∞ , 0 ) , then it yields log A n ⁎ ≤ T ϕ ϕ ( σ ) − σ λ n for all n > 0 and σ ∈ ( − ∞ , 0 ) . Let σ = φ ( λ n T ϕ ) , and by considering with φ being the inverse function of ϕ ′ , thus for all n ≥ 0 , it leads to

log A n ⁎ ≤ T ϕ ϕ φ λ n T ϕ − λ n φ λ n T ϕ = − λ n φ λ n T ϕ − ϕ φ λ n T ϕ ϕ ′ φ λ n T ϕ = − λ n ψ φ λ n T ϕ .

Therefore, we complete the proof of Lemma 2.2.□

Lemma 2.3

For 0 < a < b < + ∞ and q > 0 , we have

G 1 ( a , b , q , ϕ ) < G 2 ( a , b , q , ϕ ) .

Proof

First, denote G ( x ) = G 1 ( a , x , q , ϕ ) − G 2 ( a , x , q , ϕ ) , for x ∈ ( a , + ∞ ) . Thus, it is easy to get that G ( x ) → 0 as x → a + . Here, we only prove that G ( x ) is a decreasing function on ( a , + ∞ ) . In view of the definitions G 1 and G 2 , it follows

(2.1) d d x G 1 ( a , x , q , ϕ ) = a ( x − a ) 2 ϕ ( φ ( q x ) ) − a x ϕ ( φ ( q x ) ) − a ∫ a x ϕ ( φ ( q t ) ) t 2 d t = a ( x − a ) 2 ϕ ( φ ( q x ) ) − a x ϕ ( φ ( q x ) ) + a ϕ ( φ ( ( q x ) ) 1 x − ϕ ( φ ( q x ) ) 1 a − ∫ a x q ϕ ′ ( φ ( q t ) ) φ ′ ( q t ) t d t = a q ( x − a ) 2 ∫ a x ( t − a ) d φ ( q t ) = a q ( x − a ) 2 ( x − a ) φ ( q x ) − ∫ a x φ ( q t ) d t

and

(2.2) d d x G 2 ( a , x , q , ϕ ) = ϕ ′ 1 x − a ∫ a x φ ( q t ) d t − 1 ( x − a ) 2 ∫ a x φ ( q t ) d t + 1 x − a φ ( q x ) = ϕ ′ 1 x − a ∫ a x φ ( q t ) d t 1 ( x − a ) 2 ( x − a ) φ ( q x ) − ∫ a x φ ( q t ) d t .

Since φ is an increasing function, then for x > a , it follows

(2.3) ( x − a ) φ ( q x ) − ∫ a x φ ( q t ) d t > 0 , ϕ ′ 1 x − a ∫ a x φ ( q t ) d t > q a .

In view of (2.1)–(2.3), it yields that

(2.4) d d x G ( x ) = d d x G 1 ( a , x , q , ϕ ) − d d x G 2 ( a , x , q , ϕ ) = 1 ( x − a ) 2 ( x − a ) φ ( q x ) − ∫ a x φ ( q t ) d t a q − ϕ ′ 1 x − a ∫ a x φ ( q t ) d t < 0 ,

which means that G ( x ) is a decreasing function on ( a , + ∞ ) . Thus, G ( x ) < G ( a ) → 0 as x → a + , that is, G ( x ) < 0 for all x > a . Hence, it follows that G 1 ( a , b , q , ϕ ) < G 2 ( a , b , q , ϕ ) . Therefore, this completes the proof of Lemma 2.3.□

Lemma 2.4

Suppose that F ( s ) ∈ L 0 , T ϕ ∈ ( 0 , + ∞ ) , and if for some positive integer sequence { n k } 1 ∞ increasing to + ∞ ,

(2.5) log A n k ⁎ ≥ − λ n k ψ φ λ n k T ϕ ,

then for all k ≥ k 0 and all σ ∈ [ φ ( λ n k T ϕ ) , φ ( λ n k + 1 T ϕ ) ] , we have

(2.6) log μ ( σ , F ) ≥ T ϕ ϕ ( σ ) G 1 ( λ n k , λ n k + 1 , T ϕ − 1 , ϕ ) G 2 ( λ n k , λ n k + 1 , T ϕ − 1 , ϕ ) .

Proof

Set

H ( σ ) = sup − λ n k ψ φ λ n k T ϕ + σ λ n k ; k ≥ 1

and

η k = λ n k + 1 ψ φ λ n k + 1 T ϕ − λ n k ψ φ λ n k T λ n k + 1 − λ n k .

Then, in view of (2.5), it follows that H ( σ ) ≤ log μ ( σ , F ) for all σ < 0 . Since ( t ψ ( φ ( t ) ) ) ′ = ( t φ ( t ) − ϕ ( φ ( t ) ) ) ′ = φ ( t ) , it leads to

(2.7) η k = 1 λ n k + 1 − λ n k ∫ λ n k λ n k + 1 φ t T ϕ d t .

Since φ is continuous on ( 0 , + ∞ ) and increases to 0, thus in view of (2.6), it yields that η k → 0 as k → + ∞ and φ ( λ n k T ϕ ) < η k < φ ( λ n k + 1 T ϕ ) .

For η k − 1 ≤ σ ≤ η k , if j < k , it follows that

(2.8) − λ n k ψ φ λ n k T ϕ + σ λ n k − − λ n j ψ φ λ n j T + σ λ n j = − ∑ p = j + 1 k λ n p ψ φ λ n p T ϕ − λ n p − 1 ψ φ λ n p − 1 T ϕ + σ ( λ n k − λ n j ) = − ∑ p = j + 1 k η p − 1 ( λ n p − λ n p − 1 ) + σ ( λ n k − λ n j ) = − η k − 1 ( λ n k − λ n j ) + σ ( λ n k − λ n j ) ≥ 0 .

Similarly, if j > k , we get

(2.9) − λ n k ψ φ λ n k T ϕ + σ λ n k − − λ n j ψ φ λ n j T ϕ + σ λ n j = − ∑ p = k + 1 j λ n p ψ φ λ n p T ϕ − λ n p − 1 ψ φ λ n p − 1 T ϕ + σ ( λ n j − λ n k ) = − ∑ p = k + 1 j η p − 1 ( λ n p − λ n p − 1 ) + σ ( λ n j − λ n k ) = − η k ( λ n j − λ n k ) + σ ( λ n j − λ n k ) ≥ 0 .

Hence, for η k − 1 ≤ σ ≤ η k , it yields

(2.10) H ( σ ) = − λ n k ψ φ λ n k T ϕ + σ λ n k .

Thus, for η k − 1 ≤ φ ( λ n k T ϕ ) ≤ σ ≤ η k , we can deduce

(2.11) H ( σ ) ϕ ( σ ) ′ = λ n k ϕ ( σ ) − ϕ ′ ( σ ) σ λ n k − λ n k ψ φ λ n k T ϕ ϕ 2 ( σ ) = λ n k ϕ ′ ( σ ) ψ φ λ n k T ϕ − ψ ( σ ) ϕ 2 ( σ ) ≤ 0 ,

which implies

(2.12) H ( σ ) ϕ ( σ ) ≥ H ( η k ) ϕ ( η k ) = − λ n k ψ φ λ n k T ϕ + η k λ n k G 2 ( λ n k , λ n k + 1 , T ϕ − 1 , ϕ ) = λ n k λ n k + 1 ψ φ λ n k + 1 T ϕ − λ n k ψ φ λ n k T ϕ λ n k + 1 − λ n k − ψ φ λ n k T ϕ G 2 ( λ n k , λ n k + 1 , T ϕ − 1 , ϕ ) = λ n k λ n k + 1 λ n k + 1 − λ n k ∫ λ n k λ n k + 1 T ϕ − 1 ψ φ t T ϕ ′ d t G 2 ( λ n k , λ n k + 1 , T ϕ − 1 , ϕ ) = T ϕ λ n k λ n k + 1 λ n k + 1 − λ n k ∫ λ n k λ n k + 1 ϕ φ t T ϕ t 2 d t G 2 ( λ n k , λ n k + 1 , T ϕ − 1 , ϕ ) = T ϕ G 1 ( λ n k , λ n k + 1 , T ϕ − 1 , ϕ ) G 2 ( λ n k , λ n k + 1 , T ϕ − 1 , ϕ ) .

Let η k ≤ σ ≤ φ ( λ n k + 1 T ϕ ) , then it follows that H ( σ ) = − λ n k + 1 ψ φ λ n k + 1 T ϕ + σ λ n k + 1 and

(2.13) H ( σ ) ϕ ( σ ) ′ = λ n k + 1 ϕ ( σ ) − ϕ ′ ( σ ) σ λ n k + 1 − λ n k + 1 ψ φ λ n k + 1 T ϕ 2 ( σ ) = λ n k + 1 ϕ ′ ( σ ) ψ φ λ n k + 1 T ϕ − ψ ( σ ) ϕ 2 ( σ ) ≥ 0 ,

which shows that H ( σ ) ϕ ( σ ) is nondecreasing on [ η k , φ ( λ n k + 1 T ϕ ) ] . By combining with (2.12), for all k ≥ k 0 and all σ ∈ [ φ ( λ n k T ϕ ) , φ ( λ n k + 1 T ϕ ) ] , it yields that

H ( σ ) ϕ ( σ ) ≥ H ( η k ) ϕ ( η k ) = T ϕ G 1 ( λ n k , λ n k + 1 , T ϕ − 1 , ϕ ) G 2 ( λ n k , λ n k + 1 , T ϕ − 1 , ϕ ) .

Hence, it is easy to get (2.6).□

Lemma 2.5

[15, Lemma 2.6]. If F ( s ) ∈ L 0 and γ is any real number, then for σ ( < 0 ) sufficiently reaching 0, we have μ ( σ , F ) ≤ 4 M u ( σ , F ) and

∫ λ k ∞ exp { ( γ + i t ) y } d α ( y ) ≤ 2 ∑ n = k + ∞ A n ⁎ exp { γ λ n + 1 } .

3 Proofs of Theorems 1.2 and 1.3

3.1 The proof of Theorem 1.2

First of all, we prove the necessity of Theorem 1.2. Suppose that

(3.1) lim σ → 0 − log + M u ( σ , F ) ϕ ( σ ) = T ϕ ,

then for any sufficiently small ε ( > 0 ) , there exists σ 0 < 0 such that

( T − ε ) ϕ ( σ ) ≤ log + M u ( σ , F ) ≤ ( T ϕ + ε ) ϕ ( σ ) , as σ 0 < σ < 0 .

In view of (1.7) and Lemma 2.1, it follows that

(3.2) ( T ϕ − ε ) ϕ ( σ ) ≤ log + μ ( σ , F ) ≤ log + M u ( σ , F ) ≤ ( T ϕ + ε ) ϕ ( σ ) , as σ 0 < σ < 0 .

Since F ( s ) ∈ L 0 , thus it follows F ( s ) ∈ L ¯ β for any β ( − ∞ < β < 0 ) . And for β < σ < 0 , we have

(3.3) E n ( F , β ) ≤ ∥ F − p n ∥ β ≤ | F ( β + i t ) − p n ( β + i t ) | ≤ ∫ 0 + ∞ exp { ( β + i t ) y } d α ( y ) − ∫ 0 λ n exp { ( β + i t ) y } d α ( y ) = ∫ λ n ∞ exp { ( β + i t ) y } d α ( y ) .

In view of Lemma 2.6, it yields that A n ⁎ ≤ 4 M u ( σ , F ) e − σ λ n for any σ < 0 . By combining with this and (3.3), we obtain that

(3.4) E n ( F , β ) ≤ 2 ∑ k = n + 1 ∞ A k − 1 ⁎ exp { β λ k } ≤ 8 M u ( σ , F ) ∑ k = n + 1 ∞ exp { ( β − σ ) λ k } .

In view of (1.4), we can choose h ′ ( 0 < h ′ < h ) such that ( λ n + 1 − λ n ) ≥ h ′ for n ≥ 0 . Then for σ ≥ β 2 , in view of (3.4), we can deduce that

E n ( F , β ) ≤ 8 M u ( σ , F ) exp { λ n + 1 ( β − σ ) } ∑ k = n + 1 ∞ exp { ( λ k − λ n + 1 ) ( β − σ ) } ≤ 8 M u ( σ , F ) exp { λ n + 1 ( β − σ ) } exp − β 2 h ′ ( n + 1 ) ∑ k = n + 1 ∞ exp { α 2 h ′ k } = 8 M u ( σ , F ) exp { λ n + 1 ( β − σ ) } 1 − exp β 2 h ′ − 1 ,

that is,

(3.5) E n − 1 ( F , β ) ≤ K M u ( σ , F ) exp { λ n ( β − σ ) } ,

where K is a constant and only depends on h and β .

Thus, from (3.2) and (3.5), it follows that

(3.6) log E n − 1 ( F , β ) exp { − β λ n } ≤ ( T ϕ + ε ) ϕ ( σ ) − σ λ n − log K ,

for all n > 0 and σ 0 < σ < 0 . Let σ = φ ( λ n T ϕ + ε ) , and by using the same argument as in Lemma 2.2, it yields

log E n − 1 ( F , β ) exp { − β λ n } ≤ − ( 1 + o ( 1 ) λ n ψ φ λ n T ϕ + ε ,

which implies

(3.7) lim sup n → + ∞ λ n Ω n ( F , β , λ n , ϕ ′ , ψ − 1 ) ≤ T ϕ .

Now, we prove that the inequality

(3.8) lim sup n → + ∞ λ n Ω n ( F , β , λ n , ϕ ′ , ψ − 1 ) = T ′ < T ϕ

does not hold. In view of (3.8), for any sufficiently small ε ( 0 < ε < T ϕ − T ′ 3 ) , there exists a positive integer n 1 such that

(3.9) log E n − 1 ( F , β ) exp { − β λ n } ≤ − λ n ψ φ λ n T ′ + ε , n > n 1 .

And since

A n ⁎ exp { β λ n } = sup λ n < x ≤ λ n + 1 , − ∞ < t < + ∞ ∫ λ n x exp { i t y } d α ( y ) exp { β λ n } ≤ sup λ n < x ≤ λ n + 1 , − ∞ < t < + ∞ ∫ λ n x exp { ( β + i t ) y } d α ( y ) ≤ sup − ∞ < t < + ∞ ∫ λ n ∞ exp { ( β + i t ) y } d α ( y ) ,

thus for any p ∈ Π n − 1 and β < 0 , it yields

(3.10) A n ⁎ exp { β λ n } ≤ | F ( β + i t ) − p ( β + i t ) | ≤ ∥ F − p ∥ β ,

and there exists p 1 ∈ Π n − 1 such that

(3.11) ∥ F − p 1 ∥ ≤ 2 E n − 1 ( F , β ) .

Hence, we can conclude in view of (3.10) and (3.11) that

(3.12) A n ⁎ exp { β λ n } ≤ 2 E n − 1 ( F , β ) ,

for any β < 0 and F ( s ) ∈ L 0 . Then from (3.9) and (3.12), it follows

(3.13) log A n ⁎ ≤ − ( 1 + o ( 1 ) λ n ψ φ λ n T ′ + ε .

By Lemma 2.2, it yields

log μ ( σ , F ) ≤ ( T ′ + ε ) ϕ ( σ ) .

Thus, in view of Lemma 2.1, (1.7) and 0 < ε < T ϕ − T ′ 3 , we obtain

lim sup σ → 0 − log + M u ( σ , F ) ϕ ( σ ) ≤ T ′ + 2 ε < T ϕ ,

which is a contradiction with (3.1). Therefore, we have

(3.14) lim sup n → + ∞ λ n Ω n ( F , β , λ n , ϕ ′ , ψ − 1 ) = T ϕ .

To prove (1.8), in view of (3.14), we only need to prove

lim inf v → + ∞ λ n v Ω n v ( F , β , λ n v , ϕ ′ , ψ − 1 ) = T ϕ .

In view of (3.2), we can conclude that there exists a positive integer subsequence { n v } such that

(3.15) log [ E n v − 1 ( F , β ) exp { − β λ n v } ] ≥ − λ n v ψ φ λ n v T ϕ − ε

and

(3.16) ( T ϕ + ε ) G 1 ⁎ ( λ n v ′ , λ n v ″ , ( T ϕ + ε ) − 1 , ϕ ) ≥ ( T ϕ − ε ) G 2 ⁎ ( λ n v ′ , λ n v ″ , ( T ϕ + ε ) − 1 , ϕ ) ,

where

G 1 ⁎ ( λ n v ′ , λ n v ″ , ( T ϕ + ε ) − 1 , ϕ ) = λ n v ′ λ n v ″ λ n v ″ − λ n v ′ ∫ λ n v ′ λ n v ″ ϕ φ t T ϕ + ε t 2 d t , G 2 ⁎ ( λ n v ′ , λ n v ″ , ( T ϕ + ε ) − 1 , ϕ ) = ϕ 1 λ n v ″ − λ n v ′ ∫ λ n v ′ λ n v ″ φ t T ϕ + ε d t .

Indeed, we suppose that such sequence { n v } does not exist, that is, there exist two sequences { n v ′ } , { n v ″ } increasing to + ∞ such that

(3.17) log [ E p − 1 ( F , β ) exp { − β λ p } ] < − λ p ψ φ λ p T ϕ − ε , for n v ′ ≤ p ≤ n v ″ ,

and

(3.18) ( T ϕ + ε ) G 1 ⁎ ( λ n v ′ , λ n v ″ , ( T ϕ + ε ) − 1 , ϕ ) < ( T ϕ − ε ) G 2 ⁎ ( λ n v ′ , λ n v ″ , ( T ϕ + ε ) − 1 , ϕ ) .

Assume that λ n v + 1 ′ > λ n v ″ . Denote

η v 1 = λ n v ″ ψ φ λ n v ″ T ϕ + ε − λ n v ′ ψ φ λ n v ′ T ϕ + ε λ n v ″ − λ n v ′ .

Thus, by Lemma 2.4, it follows that

(3.19) η v 1 = 1 λ n v ″ − λ n v ′ ∫ λ n v ′ λ n v ″ φ t T ϕ + ε d t ,

and η v 1 → 0 as k → + ∞ , φ ( λ n v ′ T ϕ + ε ) < η v 1 < φ ( λ n v ″ T ϕ + ε ) .

If λ p < λ n v ′ , then we have

− λ p ψ φ λ p T ϕ + ε + η v 1 λ p ′ = − φ λ p T ϕ + ε + η v 1 > − φ λ n v ′ T ϕ + ε + η v 1 > 0 ,

which means that

(3.20) − λ p ψ φ λ p T ϕ + ε + η v 1 λ p < − λ n v ′ ψ φ λ n v ′ T ϕ + ε + η v 1 λ n v ′ .

If λ p > λ n v ″ , then

− λ p ψ φ λ p T ϕ + ε + η v 1 λ p ′ = − φ λ p T ϕ + ε + η v 1 < − φ λ n v ″ T ϕ + ε + η v 1 < 0 ,

which means that

(3.21) − λ p ψ φ λ p T ϕ + ε + η v 1 λ p < − λ n v ″ ψ φ λ n v ″ T ϕ + ε + η v 1 λ n v ″ .

Since

(3.22) − λ n v ′ ψ φ λ n v ′ T ϕ + ε + η v 1 λ n v ′ = − λ n v ″ ψ φ λ n v ″ T ϕ + ε + η v 1 λ n v ″ ,

thus in view of (3.20)–(3.22), it yields that for any λ p ∉ [ n v ′ , n v ″ ]

(3.23) − λ p ψ φ λ p T ϕ + ε + η v 1 λ p < λ n v ′ λ n v ″ λ n v ″ − λ n v ′ ψ φ λ n v ″ T ϕ + ε − ψ φ λ n v ′ T ϕ + ε = ( T ϕ + ε ) λ n v ′ λ n v ″ λ n v ″ − λ n v ′ ∫ λ n v ′ λ n v ″ ϕ φ t T ϕ + ε t 2 d t .

In view of (3.5), (3.12), (3.17)–(3.19) and (3.23), we can deduce

log μ ( η v 1 , F ) ≤ sup { log [ E p − 1 ( F , β ) exp { − β λ p } ] + η v 1 λ p + O ( 1 ) ; p > 0 } = sup { sup { log [ E p − 1 ( F , β ) exp { − β λ p } ] + η v 1 λ p + O ( 1 ) : p ∉ [ n v ′ , n v ″ ] } , sup { log [ E p − 1 ( F , β ) exp { − β λ p } ] + η v 1 λ p + O ( 1 ) : p ∈ [ n v ′ , n v ″ ] } } < sup ( T ϕ + ε ) λ n v ′ λ n v ″ λ n v ″ − λ n v ′ ∫ λ n v ′ λ n v ″ ϕ φ t T ϕ + ε t 2 d t , sup { − ( 1 + o ( 1 ) ) λ p ψ φ λ p T ϕ − ε + η v 1 λ p : p ∈ [ n v ′ , n v ″ ] } ≤ sup { ( T ϕ + ε ) G 1 ⁎ ( λ n v ′ , λ n v ″ , ( T ϕ + ε ) − 1 , ϕ ) , ( T ϕ − ε ) ϕ ( η v 1 ) } ≤ ( T ϕ − ε ) ϕ ( η v 1 ) ,

which is a contradiction with (3.2). Hence, we conclude that there exists a positive integer subsequence { n v } satisfying (3.15) and (3.16). Let v → ∞ and ε → 0 + , and by combining with (3.14) and Lemma 2.3, then we can deduce (1.8) and (1.9).

Thus, the proof of the necessity for Theorem 1.2 is completed.

Now, we prove the sufficiency of Theorem 1.2. Suppose that

lim sup n → + ∞ λ n Ω n ( F , β , λ n , ϕ ′ , ψ − 1 ) = T ϕ ,

then in view of (3.12), for any sufficiently small ε ( > 0 ) , there exists a positive integer n 2 such that

(3.24) log A n ⁎ ≤ − ( 1 + o ( 1 ) λ n ψ φ λ n T ϕ + ε , for n > n 2 .

Thus, in view of (1.7), (3.24) and Lemmas 2.1 and 2.2, we can easily obtain

(3.25) lim sup σ → 0 − log + M u ( σ , F ) ϕ ( σ ) = T ϕ .

On the other hand, in view of (1.8), it follows that for any sufficiently small ε ( > 0 ) , there exists a positive integer n v such that

(3.26) log [ E n v − 1 ( F , β ) exp { − β λ n v } ] ≥ − λ n v ψ φ λ n v T ϕ − ε .

Then, from (1.7), (3.5) and by Lemma 2.1, it yields

log A n v ⁎ ≥ − ( 1 + o ( 1 ) ) λ n v ψ φ λ n v T ϕ − ε .

So, by Lemma 2.4, it follows that for all v ≥ v 0 and all σ ∈ [ φ ( λ n v T ϕ − ε ) , φ ( λ n v + 1 T ϕ − ε ) ] , we have

log μ ( σ , F ) ≥ ( T ϕ − ε ) ϕ ( σ ) G 1 ( λ n v , λ n v + 1 , ( T ϕ − ε ) − 1 , ϕ ) G 2 ( λ n v , λ n v + 1 , ( T ϕ − ε ) − 1 , ϕ ) .

Hence, by combining with (1.9), it leads to

log μ ( σ , F ) ≥ ( T ϕ − ε ) ϕ ( σ ) ,

which implies

(3.27) lim inf σ → 0 − log + μ ( σ , F ) ϕ ( σ ) ≥ T ϕ .

Thus, in view of (3.25) and (3.26), and by applying Lemma 2.1, (3.1) holds. The proof of the sufficiency for Theorem 1.2 is completed.

Therefore, this completes the proof of Theorem 1.2.

3.2 The proof of Theorem 1.3

By using the same argument as in the proof of Theorem 1.3, and combining with the relation between E n − 1 ( F , β ) exp { − β λ n } with A n ⁎ (see (3.5) and (3.12)), it is easy to prove the conclusions of Theorem 1.3.

4 Proofs of Corollaries 1.1 and 1.2

4.1 The proof of Corollary 1.1

Without loss of generalization, assume that T ϕ = 1 . Set ϕ ( σ ) = ( − σ ) − ρ , otherwise, let ϕ ( σ ) = T ( − σ ) − ρ . Thus,

ϕ ′ ( σ ) = ρ ( − σ ) − ρ − 1 , φ ( x ) = − ρ x 1 ρ + 1 , ψ ( σ ) = ρ + 1 ρ σ ,

x ψ ( φ ( x ) ) = − ( ρ + 1 ) x ρ ρ ρ + 1

and

G 1 ( a , b , 1 , ϕ ) = ( ρ + 1 ) ρ − ρ ρ + 1 a b b − a a − 1 ρ + 1 − b − 1 ρ + 1 ,

G 2 ( a , b , 1 , ϕ ) = ( ρ + 1 ) − ρ ρ ρ 2 ρ + 1 b ρ ρ + 1 − a ρ ρ + 1 b − a − ρ .

Thus, it yields that

(4.1) Ω n ( F , β , λ n , ϕ ′ , ψ − 1 ) = ϕ ′ ψ − 1 1 λ n log 1 E n − 1 ( F , β ) exp { − β λ n } = ( ρ + 1 ) ρ + 1 ρ ρ λ n log + E n − 1 ( F , β ) exp { − β λ n } ρ + 1

and

(4.2) G 1 ( λ n v , λ n v + 1 , 1 , ϕ ) G 2 ( λ n v , λ n v + 1 , 1 , ϕ ) = ( ρ + 1 ) ρ + 1 ρ ρ λ n v λ n v + 1 λ n v + 1 − λ n v 1 λ n v 1 ρ + 1 − 1 λ n v + 1 1 ρ + 1 λ n v + 1 ρ ρ + 1 − λ n v ρ ρ + 1 λ n v + 1 − λ n v ρ .

Hence, in view of (4.1), and by combining with the conclusions of Theorem 1.2, we can prove that the conclusion of Corollary 1.1(i) and the first conclusion of Corollary 1.1(ii) hold. Thus, it remains to prove that (4.2) is equivalent to the second formula of (1.11).

Set λ n v + 1 = ( 1 + δ v ) λ n v . Then (4.2) becomes the following form:

(4.3) f ( δ v ) = ( ρ + 1 ) ρ + 1 ρ ρ 1 + δ v δ v 1 − 1 ( 1 + δ v ) 1 ρ + 1 ( 1 + δ v ) ρ ρ + 1 − 1 δ v ρ .

By the aforementioned arguments and in view of Theorem 1.2, it remains to prove that f ( δ v ) → 1 as v → + ∞ if and only if δ v → 0 as v → + ∞ . Since λ n + 1 > λ n , it follows δ k > 0 . Assume that f ( δ v ) → 1 as v → + ∞ , we will prove that δ v → 0 as v → + ∞ as follows.

If lim sup v → + ∞ δ v = + ∞ , then for some increasing sequence v j , it follows

f ( δ v j ) = ( ρ + 1 ) ρ + 1 ρ ρ 1 δ v j ρ ρ + 1 → 0 , as j → + ∞ ,

which is a contradiction with f ( δ v ) → 1 as v → + ∞ .

If lim sup v → + ∞ δ v = δ > 0 , then for some increasing sequence v j , it follows

f ( δ v j ) = ( 1 + o ( 1 ) ) ( ρ + 1 ) ρ + 1 ρ ρ 1 + δ δ 1 − 1 ( 1 + δ ) 1 ρ + 1 ( 1 + δ ) ρ ρ + 1 − 1 δ ρ ,

as j → + ∞ . Set x = ( 1 + δ ) 1 ρ + 1 , then x > 1 and f ( δ v j ) = ( 1 + o ( 1 ) ) g ( x ) , where

g ( x ) = ( ρ + 1 ) ρ + 1 ρ ρ x ρ ( x − 1 ) x ρ + 1 − 1 x ρ − 1 x ρ + 1 − 1 ρ .

Let

g 1 ( x ) = ( ρ + 1 ) ρ + 1 x − 1 x ρ + 1 − 1 − ρ ρ x ρ + 1 − 1 x ρ + 1 − x ρ ,

then

d d x g 1 ( x ) = ( ( ρ + 1 ) x ρ − ρ x ρ + 1 − 1 ) ( ρ + 1 ) ρ + 1 ( x ρ + 1 − x ) ρ + 1 − ρ ρ + 1 ( x ρ + 1 − 1 ) ρ + 1 ( x ρ + 1 − 1 ) 2 ( x ρ + 1 − x ) ρ + 1 .

In view of ( ρ + 1 ) x ρ − ρ x ρ + 1 − 1 < 0 and for all x > 1 ( ρ + 1 ) ( x ρ + 1 − x ) − ρ ( x ρ + 1 − 1 ) > 0 , then we obtain d d x g 1 ( x ) < 0 for all x > 1 . Since lim x → 1 + g 1 ( x ) = ( ρ + 1 ) ρ − ( ρ + 1 ) ρ = 0 and lim x → + ∞ g 1 ( x ) = − ρ ρ < 0 , thus g 1 ( x ) < g 1 ( 1 ) = 0 for all x > 1 , that is,

( ρ + 1 ) ρ + 1 x − 1 x ρ + 1 − 1 − ρ ρ x ρ + 1 − 1 x ρ + 1 − x ρ < 0 .

Hence, for all x > 1 , we can deduce

g ( x ) = ( ρ + 1 ) ρ + 1 ρ ρ x ρ ( x − 1 ) x ρ + 1 − 1 x ρ − 1 x ρ + 1 − 1 ρ < 1 .

Thus, for all δ > 0 , it yields

( ρ + 1 ) ρ + 1 ρ ρ 1 + δ δ 1 − 1 ( 1 + δ ) 1 ρ + 1 ( 1 + δ ) ρ ρ + 1 − 1 δ ρ < 1 ,

which is a contradiction with f ( δ v ) → 1 as v → + ∞ .

If lim v → + ∞ δ v = 0 , then

f ( δ v ) = ( ρ + 1 ) ρ + 1 ρ ρ 1 + δ v δ v δ v ρ + 1 + ( 1 + o ( 1 ) ) ( ρ + 2 ) δ v 2 2 ( ρ + 1 ) 2 ρ ρ + 1 − ( 1 + o ( 1 ) ) ρ δ v 2 ( ρ + 1 ) 2 ρ = ( ρ + 1 ) ρ + 1 ρ ρ ( 1 + δ v ) 1 ρ + 1 + O ( δ v ) ρ ρ + 1 + O ( δ v ) ρ ,

and f ( δ v ) → 1 as v → + ∞ .

Therefore, f ( δ v ) → 1 as v → + ∞ if and only if δ v → 0 as v → + ∞ . Thus, (1.9) is equivalent to lim v → ∞ λ n v λ n v + 1 = 1 .

This completes the proof of Corollary 1.1.

4.2 The proof of Corollary 1.2

By using the same argument as in the proof of Corollary 1.1, it is easy to prove the conclusions of Corollary 1.2.

Acknowledgments

The authors thank the referee(s) for reading the manuscript very carefully and making a number of valuable and kind comments which improved the presentation.

Data availability: No data were used to support this study.
Conflict of interest: The authors declare that they have no competing interests.
Author contributions: Conceptualization, H. Y. Xu; writing-original draft preparation, H. Y. Xu; writing-review and editing, H. Y. Xu, S. M. Liu, Y. Q. Cui and P. Gong; funding acquisition, S. M. Liu, H. Y. Xu and P. Gong.
Funding: This work was supported by the National Natural Science Foundation of China no. 11561033, the Natural Science Foundation of Jiangxi Province in China no. 20181BAB201001 and no. 20151BAB201008 and the Foundation of Education Department of Jiangxi no. GJJ190876 and no. GJJ190895 of China.

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Received: 2020-01-10

Revised: 2020-10-21

Accepted: 2020-11-12

Published Online: 2020-12-31

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2020-0118

Keywords for this article

perfect ϕ-type; approximation; Laplace-Stieltjes transform; half plane

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