# The Singular Points of Analytic Functions with FiniteX-Order Defined by Laplace-Stieltjes Transformations

Authors: Hong-Yan Xu; Zu-Xing Xuan
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We study the singular points of analytic functions defined by Laplace-Stieltjes transformations which converge on the right half plane, by introducing the concept ofX-order functions. We also confirm the existence of the finite X-order Borel points of such functions and obtained the extension of the finite X-order Borel point of two analytic functions defined by two Laplace-Stieltjes transformations convergent on the right half plane. The main results of this paper are improvement of some theorems given by Shang and Gao.


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## Body

## 1. Introduction

For Laplace-Stieltjes transforms(1)Fs=∫0+∞‍e-sxdαx,s=σ+it,
where α(x) is a bounded variation on any interval [0,Y](0<Y<+∞) and σ and t are real variables. We choose a sequence {λn}n=1∞,
(2)0=λ1<λ2<λ3<⋯<λn↑+∞,
which satisfies the following conditions:
(3)limsup⁡n→+∞λn+1-λn<+∞,limsup⁡n→+∞nλn=D<∞,(4)limsup⁡n→+∞log⁡An*λn=0,
where
(5)An*=sup⁡λn<x≤λn+1,-∞<t<+∞∫λnx‍e-itydα(y).Remark 1.
Dirichlet series was regarded as a special example of Laplace-Stieltjes transformations; a number of articles have focused on the growth and the value distribution of analytic functions defined by Dirichlet series; see [1–3] for some recent results.In 1963, Yu [4] proved the Valiron-Knopp-Bohr formula of the associated abscissas of bounded convergence, absolute convergence, and uniform convergence of Laplace-Stieltjes.Theorem A.
Suppose that Laplace-Stieltjes transformations (1) satisfy the first formula of (3) and limsup⁡n→+∞log⁡n/λn<+∞; then
(6)limsup⁡n→+∞log⁡An*λn≤σuF≤limsup⁡n→+∞log⁡An*λn+limsup⁡n→+∞log⁡nλn,
where σuF is called the abscissa of uniform convergence of F(s).By (3), (4), and Theorem A, one can get that σuF=0; that is, F(s) is analytic in the right half plane. Set(7)μσ,F=max⁡n∈NAn*e-λnσ(σ>0),Mσ,F=sup⁡-∞<t<+∞F(σ+it),Muσ,F=sup⁡0<x<+∞,-∞<t<+∞∫0x‍e-σ+itydαy,(σ>0).Remark 2.
From (4), for any σ>0, we have
(8)limsup⁡n→+∞log⁡An*-λnσλn=-σ<0,orlimsup⁡n→+∞log⁡An*e-λnσ=-∞.
This shows that μ(σ,F) exists.In the past few decades, many people studied some problems of analytic functions defined by Laplace-Stieltjes transformations and obtained a number of interesting and important results (including [5–11]). In those papers, there are about two methods to control the growth of the maximum modulus Mu(σ,F) or the maximum term μ(σ,F): one method is to replace the denominator in the definition of growth order by using the technique of type function U(x) (see [4, 12–14]) and the other method is to take multiple logarithm to Mu(σ,F) or μ(σ,F) in the definition of growth order (see [15, 16]). For the second method, as the logarithm function is a special function, a question rises naturally: whether we can find a relatively general function to replace the logarithm function to control the maximum growth rate.In this paper, we investigate the above question and give a positive answer to this question. Moreover, we confirm the existence of singular points for these functions by applying the main results of our paper. To do this, we introduce a completely new technique based on the concept ofX(x) which is different with the type function U(x) and more general than logarithm function and obtain the main theorems as follows.Theorem 3.
If the Laplace-Stieltjes transformationF(s) of infinite order has finite X-order and the sequence (2) satisfies (3) and (4), then one has
(9)limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ=ρ*⟺limsup⁡σ→0+X(log⁡+μ(σ,F))log⁡1/σ=ρ*.Theorem 4.
If the Laplace-Stieltjes transformationF(s) of infinite order and the sequence (2) satisfy (3) and (4), then one has(10)limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ=ρ*⟺limsup⁡n→∞X(λn)log⁡+λn/log⁡+An*=ρ*,
where 0<ρ*<∞.Remark 5.
The definitions ofX-order and the function in Theorems 3 and 4  X(x) will be introduced in Section 2.From Theorem4, we further investigate the value distribution of analytic functions with finite X-order represented by Laplace-Stieltjes transformations convergent on the right half plane and obtain the following theorems.Theorem 6.
Suppose that the sequence (2) satisfies (3) and (4) and Laplace-Stieltjes transformation F(s) is of infinite order. Let α(x)=α1(x)+iα2(x), where α1(x) is a creasing function, and, for any positive number K>0 and δ, α2(x) satisfies
(11)α2x+δ-α2x≤Kα1x+δ-α1x,hhhhhhhhhhhhhhhhih0≤x,x+δ<+∞,limsup⁡n→∞X(λn)log⁡+λn/log⁡+An*=ρ*,(0<ρ*<∞),
and then s=0 is the X-point of F(s) with X-order ϱ≥ρ*; that is, for any η>0, the inequality
(12)limsup⁡σ→0+X(n¯(σ,0,η,F=a))log⁡1/σ=ϱ≥ρ*
holds for any a∈C, except for one exception, where n¯(σ,0,η,F=a) is the counting function of distinct zero of the function F(s)-a in the strip {s:R(s)>σ,|I(s)|<η}.Theorem 7.
Suppose that the sequence (2) satisfies (3) and (4) and Laplace-Stieltjes transformation F(s) is of infinite order. Let α(x)=∫0x‍r(y)eit0ydy, where r(y) is continuous function on y∈[0,+∞), r(y)≥0, t0 is a positive real number, and
(13)limsup⁡n→∞X(λn)log⁡+λn/log⁡+An*=ρ*,0<ρ*<∞;
then s=it0 is the X-point of F(s) with X-order ϱ≥ρ*; that is, for any η>0, the inequality
(14)limsup⁡σ→0+X(n¯(σ,it0,η,F=a))log⁡1/σ=ϱ≥ρ*
holds for any a∈C, except for one exception, where n¯(σ,it0,η,F=a) is the counting function of distinct zeros of the function F(s)-a in the strip {s:R(s)>σ,|I(s)-t0|<η}.The structure of this paper is as follows. In Section2, we introduce the concept of X-order and give the proofs of Theorems 3 and 4. Section 3 is devoted to proving Theorems 6 and 7.Remark 8.
From Theorems3–7, we assume that the X-order ρ* of F(s) is finite; that is, 0<ρ<∞. For ρ*=+∞, we have studied the value distribution of analytic functions defined by Laplace-Stieltjes transformations which converge on the right half plane and obtained some results (see [17]).


## 2. Proofs of Theorems3 and 4

We first introduce the concept ofX-order of such functions as follows.Definition 9 (see [18]).
If the Laplace-Stieltjes transformF(s) satisfies σuF=0 (the sequence (2) satisfies (3) and (4)) and
(15)limsup⁡σ→0+log⁡+log⁡+Mu(σ,F)log⁡1/σ=∞,
then F(s) is called a Laplace-Stieltjes transform of infinite order.To control the growth of the moleculeMu(σ,F) or μ(σ,F) in the definition of order, many mathematicians proposed the type functions U(x) to enlarge the growth of the denominator log⁡1/σ or -σ (see [4, 6, 11, 13, 14]). In this paper, we will investigate the growth of Laplace-Stieltjes transform of infinite order by using a class of functions to reduce the growth of Mu(σ,F) or μ(σ,F) which is different with the previous form. Thus, we should give the definition of the new function as follows.LetF be the class of all functions X which satisfies the following conditions:(i)
X(x) is defined on [a,+∞),a>0, is positive, strictly increasing, and differential, and tends to +∞ as x→+∞;(ii)
xX′(x)=o(1) as x→+∞.Definition 10.
If the Laplace-Stieltjes transformationF(s) of infinite order satisfies
(16)limsup⁡σ→0+X(log⁡Mu(σ,F))log⁡1/σ=ρ*,
where X(x)∈F, then ρ* is called the X-order of the Laplace-Stieltjes transform F(s).Remark 11.
In particular, if we takeX(x)=log⁡px,p≥2,p∈N+, where log⁡1x=log⁡x and log⁡px=log⁡(log⁡p-1x), X-order is p-order of Laplace-Stieltjes transformations with infinite order.Remark 12.
In addition,X-order is more precise than p-order. In fact, for p(≥2) being a positive integer, we can find out function X(x)∈F and a positive real function M(x) satisfying
(17)limsup⁡x→∞X(M(x))log⁡x=A,(0<A<∞),limsup⁡x→∞log⁡pM(x)log⁡x=∞,limsup⁡x→∞log⁡p+1M(x)log⁡x=0.
For example, letting M(x)=exp⁡p{(tlog⁡x)1/d},X(x)=(log⁡px)d, where t is a finite positive real constant and 0<d<1.The following lemma is very important to study the growth of analytic functions represented by Laplace-Stieltjes transforms convergent on the right half plane, which show the relation betweenMu(σ,F) and μ(σ,F) of such functions.Lemma 13 (see [7, 11]).
If the abscissaσuF=0 of uniform convergence of Laplace-Stieltjes transformation and the sequence (2) satisfy (3), then, for any given ɛ∈(0,1) and for σ(>0) sufficiently reaching 0, one has
(18)13μ(σ,F)≤Mu(σ,F)≤K(ɛ)μ((1-ɛ)σ,F)1σ,
where K(ɛ) is a constant depending on ɛ, (3), and
(19)log⁡+x=log⁡xx≥10x<1.


### 2.1. The Proof of Theorem3

Firstly, for any constantK1, we will prove that
(20)limsup⁡σ→0+X(K1log⁡μ(σ,F))log⁡1/σ=limsup⁡σ→0+X(log⁡μ(σ,F))log⁡1/σ.
Without loss of generality, we suppose that K1>1. From the Laplace-Stieltjes transformation F(s) with infinite order and Lemma 13, we have limσ→0+μ(σ,F)=∞. Then, from the Cauchy mean value theorem, there exists η(log⁡μ(σ,F)<η<K1log⁡μ(σ,F)) satisfying
(21)X(K1log⁡μ(σ,F))-X(log⁡μ(σ,F))log⁡(K1log⁡μ(σ,F))-loglogμ(σ,F)=X′(η)log⁡ξ′=ηX′η;
that is,
(22)X(K1log⁡μ(σ,F))=X(log⁡μ(σ,F))+log⁡K1ηX′(η).
From xX′(x)=o(1) as x→+∞ and the above equality, we can easily get (20).Thus, from (20) and Lemma 13, we can prove the conclusion of Theorem 3 easily.


### 2.2. The Proof of Theorem4

(
⇐
): Suppose that
(23)limsup⁡n→∞X(λn)log⁡+λn/log⁡An*=ρ*.We consider two steps in this case as follows.Step 1.
We will prove thatlimsup⁡σ→0+X(log⁡+Mu(σ,F))/log⁡1/σ≥ρ*. From (23) and 0<ρ*<∞, there exists a positive sequence {nk}; for any ɛ(0<ɛ<ρ*),
(24)log⁡Ank*>λnkexp⁡-1ρ*-ɛX(λnk),k=1,2,3,….
From (18) and (23), we have
(25)log⁡Muσ,F≥log⁡Ank*-λnkσ+O(1)>λnkexp⁡-1ρ*-ɛXλnk-σλnk+O(1).
Setting σ=σk and taking σk=1/2exp⁡-1/ρ*-ɛX(λnk), (k=1,2,3,….), then, from (24) and (25), we have
(26)log⁡Muσk,F>12λnkexp⁡-1ρ*-ɛXλnk1+o1;
that is,
(27)(1+o(1))1σklog⁡Mu(σk,F)>λnk.
From σk=1/2exp⁡-1/ρ*-ɛX(λnk), we have
(28)X(λnk)=ρ*-ɛlog⁡12σk.
Thus, from (27), (28) and X(x) being strictly increasing function, we have
(29)X(1+o(1))1σklog⁡Mu(σk,F)>(ρ*-ɛ)log⁡12σk.
Since σ>0, from the Cauchy mean value theorem, there exists ζ(log⁡Muσk,F<ζ<(1+o(1))1/σklog⁡Mu(σk,F)) satisfying
(30)X(1+o(1))1/σklog⁡Mu(σk,F)-X(log⁡Mu(σk,F))log⁡(1+o(1))1/σklog⁡Mu(σk,F)-loglogMu(σk,F)=X′ζlog⁡ζ′=ζX′ζ;
that is,
(31)X(1+o(1))1σklog⁡Mu(σk,F)=X(log⁡Mu(σk,F))+log⁡1+o11σkζX′(ζ).
Since xX′(x)=o(1) as x→+∞, from (23) and (31), we can get
(32)limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ≥ρ*.Step 2.
We will prove thatlimsup⁡σ→0+X(log⁡+Mu(σ,F))/log⁡1/σ≤ρ*.
From0<ρ*<∞ and (23), there exists a positive integer n0; for any ɛ(>0), we have
(33)log⁡An*<λnexp⁡-1ρ*+ɛX(λn),n>n0.
From (3), for any ɛ′(>0) and the above ɛ, there exists a positive integer N1(>n0) satisfying
(34)N1≤(D+ɛ′)W(ρ*+ɛ)log⁡2σ<N1+1,kkkkkkkkkkhhhkkkkkkkikkasσ⟶0+,
where W(x) and X(x) are two reciprocally inverse functions. Set λn=W(ρ*+ɛ)log⁡(2/σ), and then
(35)∑n=N1+1∞‍exp⁡λnexp⁡-1ρ*+ɛX(λn)-λnσ<exp⁡-λnσ2<∑n=N1+1∞‍exp⁡-nσ2D+ɛ′=exp⁡-σ(N1+1)/2(D+ɛ′)1-exp⁡-σ/2(D+ɛ′).
From the above inequality and (34), we have
(36)log⁡exp⁡-σ(N1+1)/2(D+ɛ′)1-exp⁡-σ/2(D+ɛ′)=-σ(N1+1)2(D+ɛ′)+log⁡2σ+O(1)<-σ2Wρ*+ɛlog⁡2σ+log⁡2σ+O1,σ⟶0+.
Next, we will prove that(37)log⁡exp⁡-σ(N1+1)/2(D+ɛ′)1-exp⁡-σ/2(D+ɛ′)⟶-∞,ggggggggggggggghhhhjgggggggggσ⟶0+.
By using the same argument as in (31), we can get that
(38)X2σ2log⁡2σ=Xlog⁡2σ+o(1)log⁡2σ2,kkkkkkkkkkkkhhkgggfikkkkkkkkkasσ⟶0+.
Since F(s) is infinite order and finite X-order, from the definition of X(x), we can get the following equality easily:
(39)limsup⁡x→∞X(x)x=0.
From (38), (39) and 0<ρ*<∞, we have
(40)ρ*+ɛlog⁡2σ>Xlog⁡2σ+o(1)log⁡2σ2=X2σ2log⁡2σ.
Thus, from (40), we can deduce
(41)limsup⁡σ→0+-σ/2W(ρ*+ɛ)log⁡2/σlog⁡2/σ=limsup⁡σ→0+-2σ⟶-∞.
From (36), we can get that
(42)log⁡exp⁡-σ(N1+1)/2(D+ɛ′)1-exp⁡-σ/2(D+ɛ′)⟶-∞,hhhhhhhhjjhhhhhkhhhhhhhhhhhhhhhσ⟶0+.
That is,
(43)exp⁡-σ(N1+1)/2(D+ɛ′)1-exp⁡-σ/2(D+ɛ′)⟶0,σ⟶0+.
Hence, we have
(44)∑n=N1+1∞‍exp⁡λnexp⁡-1ρ*+ɛX(λn)-λnσ⟶0,hhhhhhhhhhhhhhhhhhhhhihhhhhhhhhhhσ⟶0+.
SinceX∈F and xβ′(x)→0 as x→∞, then there exists a positive integer N2(N2>n0) satisfying
(45)12<1-1ρ*+ɛλnX′(λn)<2,n>N2.
Take
(46)λn=W-(ρ*+ɛ)log⁡σ1-1ρ*+ɛλnX′λn-1,jjjjjjjjjjjjjjjjijjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjn>N2.
From (45) and W(x) being a strictly increasing function, we can get that N2<N1. Then, from (33), (45), (47), and σ>0, we have
(47)max⁡N2≤n≤N1An*exp⁡(-σλn)≤exp⁡λnexp⁡-1ρ*+ɛX(λn)-σλn≤exp⁡λnσ1-1ρ*+ɛλnX′(λn)-1-σ≤exp⁡-λnσ1-1-1ρ*+ɛλnX′(λn)-1≤exp⁡{σλn}≤exp⁡σW(ρ*+ɛ)log⁡2σ.
For any σ>0 and any t∈R, we have
(48)∫0x‍e-σ+itydαy≤∑k=1n-1‍Ak*e-λk+1σ+e-λk+1σ-e-λkσ+An*e-σx+e-σx-e-λnσ≤∑k=1n‍Ak*e-λkσ.
Thus, from (33), (34), (45)–(47), and the above inequality, we have
(49)Muσ,F≤∑n=1∞‍An*e-λnσ≤∑n=1N2‍+∑n=N2+1N1‍+∑n=N1+1∞‍An*exp⁡(-σλn)≤H(N2)+(D+ɛ′)W(ρ*+ɛ)log⁡2σ×exp⁡σWρ*+ɛlog⁡2σ+o(1),fffffffffffffffjjfffffffffffffffffffffσ⟶0+,
where H(N2) is the sum of finite items of the series ∑n=1∞‍An*e-λnσ. Hence, from the above inequality, we have
(50)log⁡Mu(σ,F)≤σWρ*+ɛlog⁡2σ1+o1,hhhhhhhhhhhhhhhhgghhhhhhhhhhhhhσ⟶0+;
that is,
(51)O11σlog⁡Muσ,F≤Wρ*+ɛlog⁡2σ,σ⟶0+.
Then, by using the same argument as in (31), we can deduce
(52)limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ≤ρ*.
From Steps 1 and 2, the sufficiency of the theorem is completed.By using the same argument as in the above proof of the theorem, we can prove the necessity of the theorem easily.Thus, we complete the proof of Theorem4.


## 2.1. The Proof of Theorem3

Firstly, for any constantK1, we will prove that
(20)limsup⁡σ→0+X(K1log⁡μ(σ,F))log⁡1/σ=limsup⁡σ→0+X(log⁡μ(σ,F))log⁡1/σ.
Without loss of generality, we suppose that K1>1. From the Laplace-Stieltjes transformation F(s) with infinite order and Lemma 13, we have limσ→0+μ(σ,F)=∞. Then, from the Cauchy mean value theorem, there exists η(log⁡μ(σ,F)<η<K1log⁡μ(σ,F)) satisfying
(21)X(K1log⁡μ(σ,F))-X(log⁡μ(σ,F))log⁡(K1log⁡μ(σ,F))-loglogμ(σ,F)=X′(η)log⁡ξ′=ηX′η;
that is,
(22)X(K1log⁡μ(σ,F))=X(log⁡μ(σ,F))+log⁡K1ηX′(η).
From xX′(x)=o(1) as x→+∞ and the above equality, we can easily get (20).Thus, from (20) and Lemma 13, we can prove the conclusion of Theorem 3 easily.


## 2.2. The Proof of Theorem4

(
⇐
): Suppose that
(23)limsup⁡n→∞X(λn)log⁡+λn/log⁡An*=ρ*.We consider two steps in this case as follows.Step 1.
We will prove thatlimsup⁡σ→0+X(log⁡+Mu(σ,F))/log⁡1/σ≥ρ*. From (23) and 0<ρ*<∞, there exists a positive sequence {nk}; for any ɛ(0<ɛ<ρ*),
(24)log⁡Ank*>λnkexp⁡-1ρ*-ɛX(λnk),k=1,2,3,….
From (18) and (23), we have
(25)log⁡Muσ,F≥log⁡Ank*-λnkσ+O(1)>λnkexp⁡-1ρ*-ɛXλnk-σλnk+O(1).
Setting σ=σk and taking σk=1/2exp⁡-1/ρ*-ɛX(λnk), (k=1,2,3,….), then, from (24) and (25), we have
(26)log⁡Muσk,F>12λnkexp⁡-1ρ*-ɛXλnk1+o1;
that is,
(27)(1+o(1))1σklog⁡Mu(σk,F)>λnk.
From σk=1/2exp⁡-1/ρ*-ɛX(λnk), we have
(28)X(λnk)=ρ*-ɛlog⁡12σk.
Thus, from (27), (28) and X(x) being strictly increasing function, we have
(29)X(1+o(1))1σklog⁡Mu(σk,F)>(ρ*-ɛ)log⁡12σk.
Since σ>0, from the Cauchy mean value theorem, there exists ζ(log⁡Muσk,F<ζ<(1+o(1))1/σklog⁡Mu(σk,F)) satisfying
(30)X(1+o(1))1/σklog⁡Mu(σk,F)-X(log⁡Mu(σk,F))log⁡(1+o(1))1/σklog⁡Mu(σk,F)-loglogMu(σk,F)=X′ζlog⁡ζ′=ζX′ζ;
that is,
(31)X(1+o(1))1σklog⁡Mu(σk,F)=X(log⁡Mu(σk,F))+log⁡1+o11σkζX′(ζ).
Since xX′(x)=o(1) as x→+∞, from (23) and (31), we can get
(32)limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ≥ρ*.Step 2.
We will prove thatlimsup⁡σ→0+X(log⁡+Mu(σ,F))/log⁡1/σ≤ρ*.
From0<ρ*<∞ and (23), there exists a positive integer n0; for any ɛ(>0), we have
(33)log⁡An*<λnexp⁡-1ρ*+ɛX(λn),n>n0.
From (3), for any ɛ′(>0) and the above ɛ, there exists a positive integer N1(>n0) satisfying
(34)N1≤(D+ɛ′)W(ρ*+ɛ)log⁡2σ<N1+1,kkkkkkkkkkhhhkkkkkkkikkasσ⟶0+,
where W(x) and X(x) are two reciprocally inverse functions. Set λn=W(ρ*+ɛ)log⁡(2/σ), and then
(35)∑n=N1+1∞‍exp⁡λnexp⁡-1ρ*+ɛX(λn)-λnσ<exp⁡-λnσ2<∑n=N1+1∞‍exp⁡-nσ2D+ɛ′=exp⁡-σ(N1+1)/2(D+ɛ′)1-exp⁡-σ/2(D+ɛ′).
From the above inequality and (34), we have
(36)log⁡exp⁡-σ(N1+1)/2(D+ɛ′)1-exp⁡-σ/2(D+ɛ′)=-σ(N1+1)2(D+ɛ′)+log⁡2σ+O(1)<-σ2Wρ*+ɛlog⁡2σ+log⁡2σ+O1,σ⟶0+.
Next, we will prove that(37)log⁡exp⁡-σ(N1+1)/2(D+ɛ′)1-exp⁡-σ/2(D+ɛ′)⟶-∞,ggggggggggggggghhhhjgggggggggσ⟶0+.
By using the same argument as in (31), we can get that
(38)X2σ2log⁡2σ=Xlog⁡2σ+o(1)log⁡2σ2,kkkkkkkkkkkkhhkgggfikkkkkkkkkasσ⟶0+.
Since F(s) is infinite order and finite X-order, from the definition of X(x), we can get the following equality easily:
(39)limsup⁡x→∞X(x)x=0.
From (38), (39) and 0<ρ*<∞, we have
(40)ρ*+ɛlog⁡2σ>Xlog⁡2σ+o(1)log⁡2σ2=X2σ2log⁡2σ.
Thus, from (40), we can deduce
(41)limsup⁡σ→0+-σ/2W(ρ*+ɛ)log⁡2/σlog⁡2/σ=limsup⁡σ→0+-2σ⟶-∞.
From (36), we can get that
(42)log⁡exp⁡-σ(N1+1)/2(D+ɛ′)1-exp⁡-σ/2(D+ɛ′)⟶-∞,hhhhhhhhjjhhhhhkhhhhhhhhhhhhhhhσ⟶0+.
That is,
(43)exp⁡-σ(N1+1)/2(D+ɛ′)1-exp⁡-σ/2(D+ɛ′)⟶0,σ⟶0+.
Hence, we have
(44)∑n=N1+1∞‍exp⁡λnexp⁡-1ρ*+ɛX(λn)-λnσ⟶0,hhhhhhhhhhhhhhhhhhhhhihhhhhhhhhhhσ⟶0+.
SinceX∈F and xβ′(x)→0 as x→∞, then there exists a positive integer N2(N2>n0) satisfying
(45)12<1-1ρ*+ɛλnX′(λn)<2,n>N2.
Take
(46)λn=W-(ρ*+ɛ)log⁡σ1-1ρ*+ɛλnX′λn-1,jjjjjjjjjjjjjjjjijjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjn>N2.
From (45) and W(x) being a strictly increasing function, we can get that N2<N1. Then, from (33), (45), (47), and σ>0, we have
(47)max⁡N2≤n≤N1An*exp⁡(-σλn)≤exp⁡λnexp⁡-1ρ*+ɛX(λn)-σλn≤exp⁡λnσ1-1ρ*+ɛλnX′(λn)-1-σ≤exp⁡-λnσ1-1-1ρ*+ɛλnX′(λn)-1≤exp⁡{σλn}≤exp⁡σW(ρ*+ɛ)log⁡2σ.
For any σ>0 and any t∈R, we have
(48)∫0x‍e-σ+itydαy≤∑k=1n-1‍Ak*e-λk+1σ+e-λk+1σ-e-λkσ+An*e-σx+e-σx-e-λnσ≤∑k=1n‍Ak*e-λkσ.
Thus, from (33), (34), (45)–(47), and the above inequality, we have
(49)Muσ,F≤∑n=1∞‍An*e-λnσ≤∑n=1N2‍+∑n=N2+1N1‍+∑n=N1+1∞‍An*exp⁡(-σλn)≤H(N2)+(D+ɛ′)W(ρ*+ɛ)log⁡2σ×exp⁡σWρ*+ɛlog⁡2σ+o(1),fffffffffffffffjjfffffffffffffffffffffσ⟶0+,
where H(N2) is the sum of finite items of the series ∑n=1∞‍An*e-λnσ. Hence, from the above inequality, we have
(50)log⁡Mu(σ,F)≤σWρ*+ɛlog⁡2σ1+o1,hhhhhhhhhhhhhhhhgghhhhhhhhhhhhhσ⟶0+;
that is,
(51)O11σlog⁡Muσ,F≤Wρ*+ɛlog⁡2σ,σ⟶0+.
Then, by using the same argument as in (31), we can deduce
(52)limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ≤ρ*.
From Steps 1 and 2, the sufficiency of the theorem is completed.By using the same argument as in the above proof of the theorem, we can prove the necessity of the theorem easily.Thus, we complete the proof of Theorem4.


## 3. Proofs of Theorems6 and 7

Similar to the definition ofX-order of Laplace-Stieltjes transformations in the right half plane, we will give the definition of X-order of Laplace-Stieltjes transformations in the level half-strip as follows.Definition 14.
LetF(s) be the analytic function with infinite order represented by Laplace-Stieltjes transformations convergent on the right half plane; set S(t0,l)={σ+it:σ>0,t-t0≤l}, where t0 is a real number and l is a positive number. Let X∈F and
(53)τSX=limsup⁡σ→0+X(log⁡+MS(σ,F))log⁡1/σ,
where MS(σ,F)=sup⁡|t-t0|≤l|F(σ+it)|; then τSX is called the X-order of F(S) in the level half-strip S(t0,l).To prove Theorems6 and 7, we need some lemmas as follows.Lemma 15.
If the Laplace-Stieltjes transformationF(s) of infinite order and the sequence (2) satisfy (3) and (4) and α(x)=α1(x)+iα2(x), where α1(x) is a creasing function, and, for any positive number K>0 and δ, α2(x) satisfies
(54)α2x+δ-α2x≤Kα1x+δ-α1x,ggggggggghgghgggigggg0≤x,x+δ<+∞,
then, for any ɛ>0, one has
(55)ρ*=limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ=limsup⁡σ→0+X(log⁡+MSɛ(σ,F))log⁡1/σ=τSɛ.Proof.
We will prove this lemma by using the similar argument as in [11]. From the assumptions of Lemma 15, for any 0<x≤∞,
(56)MSɛσ,F≥∫0∞‍e-σydαy≥∫0∞‍e-σydα1(y)≥1K+1∫0∞‍e-σydα1y≥1K+1∫0x‍e-σydα1y≥1K+1∫0∞‍e-(σ+it)ydα(y).
Thus, we have
(57)M(σ,F)≥MSɛ(σ,F)≥1K+1Mu(σ,F)≥1K+1M(σ,F).
SinceF(s) is an analytic function with infinite order, from the above inequality and the definition of X-order, we can get the conclusion of Lemma 15 easily.Lemma 16 (see [11, Lemma 2.4]).
Let(58)z=1-sinh⁡s1+sinh⁡s,s∈B=s:R(s)>0,Is<π2.
Then, one has the following:(i)
this mapping maps the horizontal half-stripB to the unit disc {z:|z|<1}, and its inverse mapping is
(59)s=Ψ(z)=sinh⁡-11-z1+z;(ii)
min⁡0≤θ≤2πR[Ψ(reiθ)]≥Ψ(r),(0<r<1);(iii)
max⁡0≤θ≤π/4R[Ψ(reiθ)]≤Ψ(r2),(0<r<1);(iv)
Ψ({z:|z|<r})⊆{s:R(s)>Ψ(r),|I(s)|<π/2},(0<r<1).Definition 17 (see [19, 20]).
Letf be a meromorphic function in D and limr→1-T(r,f)=∞. Then
(60)Df:=limsup⁡r→1-Tr,f-log⁡⁡1-r
is called the (upper) index of inadmissibility of f. If D(f)<∞, f is called nonadmissible; if D(f)=∞, f is called admissible.Lemma 18 (see [20, Theorem 2]).
Letf be a meromorphic function in D and limr→1-T(r,f)=∞, let q be a positive integer, and let a1,a2,…,aq be pairwise distinct complex numbers. Then, for r→1-, r∉E,
(61)(q-2)T(r,f)≤∑j=1q‍N¯r,1f-aj+log⁡11-r+S(r,f).Remark 19.
We can see thatS(r,f)=olog⁡1/1-r holds in Lemma 18 without a possible exception set when 0<D(f)<∞.Remark 20.
The termlog⁡1/1-r in Lemma 18 can enter the remainder S(r,f)=o(T(r,f)) when the function f satisfies D(f)=∞.From Lemmas 3.3 and 3.4 in [21], we can get the following result for nonadmissible functions in the unit disc which is used in this paper.Lemma 21 (see [19, page 282 (1.8)]).
Leth be analytic in the disc z=r<1; then
(62)T(r,h)≤log⁡M(r,h)≤1+r1-rT(r,h),
where M(r,h) is the maximum modulus of h in the disc |z|=r<1.


### 3.1. The Proof of Theorem6

From the sequence (2) satisfying (3) and (4), Laplace-Stieltjes transformation F(s) of infinite order, and limsup⁡n→∞X(λn)/log⁡+λn/log⁡+An*=ρ*,(0<ρ*<∞), from Theorem 4, we have
(63)limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ=ρ*,
and, from Lemma 15 and (63), for any ɛ>0, we have
(64)limsup⁡σ→0+X(log⁡+MSɛ(σ,F))log⁡1/σ=ρ*.
Thus, we can get
(65)limsup⁡σ→0+X(log⁡M(σ,F,Sɛ))log⁡1/σ=ρ*,
where Sɛ={s:R(s)>0,|I(s)|≤ɛ} and M(σ,F,Sɛ)=sup⁡{|F(s)|:R(s)≥σ,s∈Sɛ}.Setg(z)=F2ɛ/πΨ(z), where Ψ(z) is stated as in Lemma 16; then, from Lemma 16, we have that g(z) is analytic in the unit disc |z|<1 and satisfies
(66)M2ɛπΨ(r2),F,Sɛ1≤M(r,g)≤M2ɛπΨ(r),F,Sɛ,
where 0<ɛ1<ɛ. Therefore, from (66) and Lemma 16, we have
(67)limsup⁡r→1-log⁡+log⁡+M(r,g)log⁡1/1-r=∞,limsup⁡r→1-X(log⁡+M(r,g))log⁡1/1-r=ρ*.
From Lemma 21, by using the same argument as in (31), we can get
(68)X1+r1-rT(r,g)=X(T(r,g))+o(1)log⁡1+r1-r.
From (67), (68), and Lemma 21, we can get that g is an admissible function in |z|<1 and
(69)limsup⁡r→1-X(T(r,g))log⁡1/1-r=ρ*.
Then, from Lemma 18, we can get that there at most exists one except value a satisfying
(70)limsup⁡r→1-X(n¯(r,g=a))log⁡1/1-r=ϱ≥ρ*;
that is, for any η>0, the inequality
(71)limsup⁡σ→0+X(n¯(σ,0,η,F=a))log⁡1/σ=ϱ≥ρ*
holds for any a∈C, except for one exception, where n(σ,0,η,F=a) is the counting function of zeros of the function F(s)-a in the strip {s:R(s)>σ,|I(s)|<η}.Thus, we complete the proof of Theorem6.


### 3.2.  The Proof of Theorem7

Sincer(y) is a continuous function on y∈[0,+∞), then we can get that α(x)=∫0x‍r(y)eit0ydy is a function of bounded variation on x∈[0,X](0<X<∞). Set
(72)S^ɛ=s:Rs>0,Is-t0≤ɛ,MS^ɛ(σ,F)=sup⁡|t-t0|≤ɛFσ+it.
From the assumptions of Theorem 7, for any real number x(0<x≤∞), we have
(73)MS^ɛσ,F=sup⁡|t-t0|≤ɛFσ+it=sup⁡|t-t0|≤ɛ∫0∞‍e-syr(y)eit0ydy≥∫0∞‍e-σyr(y)dy≥∫0x‍e-σyr(y)dy≥∫0x‍e-syryeit0ydy=∫0∞‍e-sydαy;
that is,
(74)M(σ,F)≤Mu(σ,F)≤MS^ɛ(σ,F)≤M(σ,F).
From the assumption of Theorem 7, by (74) and Theorem 4, we can get
(75)limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ=limsup⁡σ→0+X(log⁡+MS^ɛ(σ,F))log⁡1/σ=limsup⁡σ→0+X(log⁡+M(σ,F))log⁡1/σ=ρ*.
From (75), for any ɛ>0, we have
(76)limsup⁡σ→0+X(log⁡+M(σ,F,S^ɛ))log⁡1/σ=ρ*,
where M(σ,F,S^ɛ)=sup⁡{|F(s)|:R(s)≥σ,s∈S^ɛ}.Setg(z)=F2ɛ/πΨ(z)+it0, where Ψ(z) is stated as in Lemma 16; then, from Lemma 16, we have that g(z) is analytic in the unit disc |z|<1 and satisfies
(77)M2ɛπΨ(r2),F,Sɛ2≤M(r,g)≤M2ɛπΨ(r),F,Sɛ,
where 0<ɛ2<ɛ. Therefore, from the above inequality and Lemma 16, we have
(78)limsup⁡r→1-log⁡+log⁡+M(r,g)log⁡1/1-r=∞,limsup⁡r→1-X(log⁡+M(r,g))log⁡1/1-r=ρ*.
Then, similar to the proof of Theorem 6, we can prove that, for any η>0, the inequality
(79)limsup⁡σ→0+X(n¯(σ,it0,η,F=a))log⁡1/σ=ϱ≥ρ*
holds for any a∈C, except for one exception, where n¯(σ,it0,η,F=a) is stated as in Theorem 7.Thus, we complete the proof of Theorem7.


## 3.1. The Proof of Theorem6

From the sequence (2) satisfying (3) and (4), Laplace-Stieltjes transformation F(s) of infinite order, and limsup⁡n→∞X(λn)/log⁡+λn/log⁡+An*=ρ*,(0<ρ*<∞), from Theorem 4, we have
(63)limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ=ρ*,
and, from Lemma 15 and (63), for any ɛ>0, we have
(64)limsup⁡σ→0+X(log⁡+MSɛ(σ,F))log⁡1/σ=ρ*.
Thus, we can get
(65)limsup⁡σ→0+X(log⁡M(σ,F,Sɛ))log⁡1/σ=ρ*,
where Sɛ={s:R(s)>0,|I(s)|≤ɛ} and M(σ,F,Sɛ)=sup⁡{|F(s)|:R(s)≥σ,s∈Sɛ}.Setg(z)=F2ɛ/πΨ(z), where Ψ(z) is stated as in Lemma 16; then, from Lemma 16, we have that g(z) is analytic in the unit disc |z|<1 and satisfies
(66)M2ɛπΨ(r2),F,Sɛ1≤M(r,g)≤M2ɛπΨ(r),F,Sɛ,
where 0<ɛ1<ɛ. Therefore, from (66) and Lemma 16, we have
(67)limsup⁡r→1-log⁡+log⁡+M(r,g)log⁡1/1-r=∞,limsup⁡r→1-X(log⁡+M(r,g))log⁡1/1-r=ρ*.
From Lemma 21, by using the same argument as in (31), we can get
(68)X1+r1-rT(r,g)=X(T(r,g))+o(1)log⁡1+r1-r.
From (67), (68), and Lemma 21, we can get that g is an admissible function in |z|<1 and
(69)limsup⁡r→1-X(T(r,g))log⁡1/1-r=ρ*.
Then, from Lemma 18, we can get that there at most exists one except value a satisfying
(70)limsup⁡r→1-X(n¯(r,g=a))log⁡1/1-r=ϱ≥ρ*;
that is, for any η>0, the inequality
(71)limsup⁡σ→0+X(n¯(σ,0,η,F=a))log⁡1/σ=ϱ≥ρ*
holds for any a∈C, except for one exception, where n(σ,0,η,F=a) is the counting function of zeros of the function F(s)-a in the strip {s:R(s)>σ,|I(s)|<η}.Thus, we complete the proof of Theorem6.


## 3.2.  The Proof of Theorem7

Sincer(y) is a continuous function on y∈[0,+∞), then we can get that α(x)=∫0x‍r(y)eit0ydy is a function of bounded variation on x∈[0,X](0<X<∞). Set
(72)S^ɛ=s:Rs>0,Is-t0≤ɛ,MS^ɛ(σ,F)=sup⁡|t-t0|≤ɛFσ+it.
From the assumptions of Theorem 7, for any real number x(0<x≤∞), we have
(73)MS^ɛσ,F=sup⁡|t-t0|≤ɛFσ+it=sup⁡|t-t0|≤ɛ∫0∞‍e-syr(y)eit0ydy≥∫0∞‍e-σyr(y)dy≥∫0x‍e-σyr(y)dy≥∫0x‍e-syryeit0ydy=∫0∞‍e-sydαy;
that is,
(74)M(σ,F)≤Mu(σ,F)≤MS^ɛ(σ,F)≤M(σ,F).
From the assumption of Theorem 7, by (74) and Theorem 4, we can get
(75)limsup⁡σ→0+X(log⁡+Mu(σ,F))log⁡1/σ=limsup⁡σ→0+X(log⁡+MS^ɛ(σ,F))log⁡1/σ=limsup⁡σ→0+X(log⁡+M(σ,F))log⁡1/σ=ρ*.
From (75), for any ɛ>0, we have
(76)limsup⁡σ→0+X(log⁡+M(σ,F,S^ɛ))log⁡1/σ=ρ*,
where M(σ,F,S^ɛ)=sup⁡{|F(s)|:R(s)≥σ,s∈S^ɛ}.Setg(z)=F2ɛ/πΨ(z)+it0, where Ψ(z) is stated as in Lemma 16; then, from Lemma 16, we have that g(z) is analytic in the unit disc |z|<1 and satisfies
(77)M2ɛπΨ(r2),F,Sɛ2≤M(r,g)≤M2ɛπΨ(r),F,Sɛ,
where 0<ɛ2<ɛ. Therefore, from the above inequality and Lemma 16, we have
(78)limsup⁡r→1-log⁡+log⁡+M(r,g)log⁡1/1-r=∞,limsup⁡r→1-X(log⁡+M(r,g))log⁡1/1-r=ρ*.
Then, similar to the proof of Theorem 6, we can prove that, for any η>0, the inequality
(79)limsup⁡σ→0+X(n¯(σ,it0,η,F=a))log⁡1/σ=ϱ≥ρ*
holds for any a∈C, except for one exception, where n¯(σ,it0,η,F=a) is stated as in Theorem 7.Thus, we complete the proof of Theorem7.



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*Source: 865069-2015-03-02.xml*