# The Approximation of Laplace-Stieltjes Transforms Concerning Sun’s Type Function

Authors: Xia Shen; Hong Yan Xu
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The main aim of this paper is to establish some theorems concerning the errorEnF,β, the Sun’s type function Ur, and Muσ,F of entire functions defined by Laplace-Stieltjes transforms with infinite order converge in the whole complex plane. Our results exhibit the growth of Laplace-Stieltjes transforms from the point of view of approximation.


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## Body

## 1. Introduction and Main Results

In 1946, Widder [1] considered the convergence of the following form
(1)Gs=∫0+∞e−sxdαx,s=σ+it,where αx is a bounded variation on any finite interval 0,Y0<Y<+∞, and σ and t are real variables and obtained the following theorem.Theorem 1 (see ([1], Theorem 1, Page 36)).
If(2)sup0≤u<∞∫0ue−s0tdαt=M<∞,then (1) converges for every s for which σ>σ0, and
(3)∫0+∞e−sxdαx=s−s0∫0+∞e−s−s0tβtdt,where βu=∫0ue−s0tdαt,u≥0.
As we know, (1) can be called as Laplace-Stieltjes transform, which is an integral transform similar to the Laplace transform, named for Pierre-Simon Laplace and Thomas Joannes Stieltjes. Moreover, it can be used in many fields of mathematics, such as functional analysis, and certain areas of theoretical and applied probability.
In view of Ref. [1], Gs can become the classical Laplace integral form
(4)Gs=∫0∞e−stφtdt,when αt is absolutely continuous. Moreover, if αt is a step function, choosing a sequence λn0∞ such that
(5)0≤λ1<λ2<⋯<λn<⋯,λn⟶∞asn⟶∞,(6)αx=a1+a2+⋯+an,λn≤x<λn+1,0,x=0,αx++αx−2,x>0,then we can conclude from Theorem 1 that Gs becomes a Dirichlet series
(7)Gs=fs≔∑n=1∞aneλns,s=σ+it,where σ,t is real variables, an is nonzero complex numbers. For Dirichlet series (7), it can become a Taylor series fz=∑n=0∞anzn if λn=n and es=z, and it further can also become a classical Dirichlet series fs=∑n=0∞an/ns if eλn=1/n, which is important in the fields of number theory. Hence, we can say that Laplace-Stieltjes transform is a general form of Dirichlet series. Under some conditions related to an, λn, and n, the series (7) can converge in the whole plane or the half plane; that is, fs is analytic in the whole plane or the half plane.
In the past several decades, the problem on the growth and value distribution of analytic functions has been an important and interesting subject in the fields of complex analysis. Moreover, considerable attention has been paid to the growth and the value distribution of analytic functions defined by Dirichlet series and Laplace-Stieltjes transforms, and a great deal of interesting results focusing on the growth and value distribution of such functions can be found in (see [2–17]). For example, Yu [18] in 1963 first proved a series of theorems about the Valiron-Knopp-Bohr formula of the associated abscissas of bounded convergence, absolute convergence and uniform convergence of Laplace-Stieltjes transforms, the maximal molecule Muσ,G, the maximal term μσ,G, the Borel line and the order of entire functions represented by Laplace-Stieltjes transforms convergent in the complex plane. Batty, Sheremeta, Kong, and Sun investigated the growth of analytic functions with kinds of order defined by Laplace-Stieltjes transforms (see [19–25]), and Shang, Gao, Zhang, and Xu investigated the value distribution of such functions (see [26–28]).
In 2012,Luo and Kong [29] studied the following form, is differ from (1), of Laplace-Stieltjes transform
(8)Fs=∫0+∞esxdαx,s=σ+it,where αx is stated as in (1), and λn satisfies (5) and
(9)limsupn→+∞λn+1−λn=h<+∞,(10)limsupn⟶+∞lognλn=D<∞.
Set(11)An∗=supλn<x≤λn+1,−∞<t<+∞∫λnxeitydαy.
By using the same argument as in [18], we can get the similar result about the abscissa of uniformly convergent of Fs easily. If
(12)limsupn⟶+∞logAn∗λn=−∞,by (5), (9)-(12), and Ref.[18], one can get that σuF=+∞, i.e., Fs is entire in the whole plane.
Set(13)μσ,F=maxn∈NAn∗eλnσσ<+∞,Mσ,F=sup−∞<t<+∞Fσ+it,Muσ,F=sup0<x<+∞,−∞<t<+∞∫0xeσ+itydαy.Definition 2 (see [30]).
If Laplace-Stieltjes transform (8) satisfies σuF=+∞ (the sequence λn satisfy (5) and (9)-(12)), we define the order and the lower order of Fs by
(14)limsupσ⟶+∞log+log+Muσ,Fσ=ρ,liminfσ⟶+∞log+log+Muσ,Fσ=λ,respectively, where log+x=maxlogx,0.Remark 3.
Ifρ=0, ρ∈0,+∞ and ρ=∞, we say that Fs is an entire function of zero order, finite order, and infinite order in the whole plane, respectively.Definition 4.
If Laplace-Stieltjes transform (8) satisfies σuF=+∞ (the sequence λn satisfy (5) and (9)-(12)) and is of order ρ0<ρ<∞, then we define
(15)T=limsupσ⟶+∞log+Muσ,Feσρ,which is called the type of Laplace-Stieltjes transform Fs.
In 2012 and 2014, Luo and Kong [29, 30] studied the growth of Laplace-Stieltjes transform of finite order and obtained the following theorem.Theorem 5 (see [29, 30]).
If Laplace-Stieltjes transform (8) satisfies σuF=+∞ (the sequence λn satisfy (5) and (9)-(12)), and is of order ρ0<ρ<∞ and of type T, then
(16)ρ=limsupn⟶+∞λnlogλn−logAn∗,T=limsupn⟶+∞λnρeAn∗ρ/λn.
In order to state our main results of this paper, we also introduce some definitions and notations below. We denote byL¯β the set of all the functions Fs of the form (8) which are analytic in the half plane Res<β−∞<β<∞ and the sequence λn satisfy (5), (9), and (10) and by L∞ the set of all the functions Fs of the form (8) which are analytic in the half plane Rs<+∞ and the sequence λn satisfy (5) and (9)-(12). Obviously, if −∞<β<+∞ and Fs∈L¯β, then Fs∈L∞. If (8) satisfies An∗=0 for n≥k+1, and Ak∗≠0, then we say that Fs is an exponential polynomial of degree k usually denoted by pk, i.e., pks=∫0λkexpsydαy. If we choose a suitable function αy, the function pks may be reduced to a polynomial in terms of expsλi, that is, ∑i=1kbiexpsλi. We denote Πn to be the class of all exponential polynomial of degree almost n, that is,
(17)Πn=∑j=1nbjeλjs:b1,b2,⋯,bn∈ℂn.
ForFs∈L¯β,−∞<β<+∞, we use EnF,β to denote the error in approximating the function Fs by exponential polynomials of degree n in uniform norm as
(18)EnF,β=infp∈ΠnF−pβ,n=1,2,⋯,where
(19)F−pβ=max−∞<t<+∞Fβ+it−pβ+it.
Around 2017, Singhal and Srivastava [31, 32] studied the approximation of entire functions represented by Laplace-Stieltjes transforms (8) of finite order and obtained the following result.Theorem 6 (see [32]).
If Laplace-Stieltjes transformFs∈L∞ and is of order ρ0<ρ<∞ and of type T, then for any real number −∞<β<+∞, we have
(20)ρ=limsupn⟶+∞λnlogλn−logEn−1F,βexp−βλn=limsupn⟶+∞λnlogλn−logEn−1F,β,T=limsupn⟶+∞λnρeEn−1F,βexp−βλnρ/λn=limsupn⟶+∞λnρexpρβ+1En−1F,βρ/λn.
In the same year, the authors [33] further the approximation on the entire function represented by Laplace-Stieltjes transforms with irregular growth and obtained.Theorem 7 (see ([33], Theorem 6)).
If the Laplace-Stieltjes transformFs∈L∞ and is of the lower order λ0≤λ≠ρ<∞, if λn~λn+1, then for any real number −∞<β<+∞, we have
(21)τλ≥liminfn⟶∞λneλEn−1F,βexp−βλnλ/λn,0≤τλ≤∞.
Furthermore, there exists a positive integern0 such that
(22)ψ1n=logAn∗−logAn+1∗λn+1−λnforms a nondecreasing function of n for n>n0, and then we have
(23)τλ=liminfn⟶∞λneλEn−1F,βexp−βλnλ/λn,0≤τλ≤∞.
As far as we know, there are few papers focusing on the approximation of Laplace-Stieltjes transform of infinite order. Inspired by this issue, our main purpose of this paper is to deal with the approximation of Laplace-Stieltjes transforms of infinite orderρF=∞ with the help of the type function given by Sun. In 1986, Sun [34] studied the existence of type function of the complex function of infinite order and established a new type function which is more precise than Xiong’s.Theorem 8 (see [34]).
IfSr is a continuous function in a,+∞ and
(24)limsupr⟶+∞log+Srlogr=+∞,then we say that Ur is the type function of Sr, if there exist two continuous and differential functions ρr and Ur satisfying
ρr monotonous, decreasing and trend to 0, ρ′r monotonous, increasing(ii)
limr⟶+∞rρ′rlogrloglogr=0(iii)
For sufficient larger, Sr≪Ur≔rexp1/ρr(iv)
UR<1+o1Urwhere R=r+rlogr/logUrlog2logUr and Sr≪Ur mean that Sr≤Ur and ∃ a sequence rn↑+∞ such that Srn=Urn.Remark 9.
IfFs∈L∞ and is of infinite order ρ=+∞, then in view of Theorem 7, there exists a type function Ur such that
(25)limsupσ⟶+∞SrUr=1,where Sr≔logμσ,F and r=eσ.The main theorems of this article are listed as follows.Theorem 10.
LetFs be of infinite order, and the sequence λn satisfies (5), (9), (12) and
(26)limsupn⟶+∞loglognlogλn=d<1,and then for any real number −∞<β<+∞, we have
(27)limsupσ⟶+∞logMuσ,FUr=1≤Q1≔limsupσ⟶+∞−logEn−1F,βexp−βλnUψn,where ψn=En−1F,βexp−βλn−1/λn, r=eσ, and Ur are stated as in Theorem 10.Remark 11.
We can easily get (10) from (26), thus Fs∈L∞ when the sequence λn satisfies (5), (9), (12), and (26). That is to say, our condition in our theorem is better than the previous results.Theorem 12.
Under the assumptions of Theorem10, then
(28)limsupσ⟶+∞logMuσ,FUr=1≥Q2≔limsupσ⟶+∞λnU1ψn,where
(29)U1•=U•logU•log2logU•.
LetU2•=U•log2U•, and then it follows U1ψn≤U2ψn for any positive integer n and any real number β. Hence, we get the following corollary.Corollary 13.
Under the assumptions of Theorem10, then
(30)limsupσ⟶+∞logMuσ,FUr=1≥Q3≔limsupσ⟶+∞λnU2ψn.Theorem 14.
Under the assumptions of Theorem10, then we have
(31)limsupσ⟶+∞loglogMuσ,FlogUr=1⇔limsupσ⟶+∞logλnlogUψn=1.


## 2. The Proof of Theorem 6

To prove Theorem10, we require the following lemma.Lemma 15 (see ([1], Theorem 6b)).
Iffx and φx are continuous and αx is of bounded variation in a≤x≤b, and if
(32)γx=∫cxφxdαt,a≤x≤b,a≤c≤b,then
(33)∫abfxdγx=∫abfxφxdαx.Lemma 16.
If Laplace-stieltjes transformFs is of infinite order, and the sequence λn satisfies (5), (9), (12), and (26), then
(34)limsupσ⟶+∞logMuσ,FUr=1⟺limsupσ⟶+∞logμσ,FUr=1,where r=eσ, Sr=logμσ,F, and Ur are stated as in Theorem 7.Proof.
The idea of the proof of this lemma come from Ref. [22]. Next, we will show the completely details.
Set(35)Inx;σ+it=∫λnxexpσ+itydαy.
From (9), there exists η>0 satisfying 0<λn+1−λn≤ηn=1,2,3,⋯, and then it follows e−ησ<1, as σ>0. Thus, for x>λn and by Theorem 1 and Lemma 15, we deduce
(36)∫λnxexpitydαy=∫λnxe−σydyIny;σ+it=Iy;σ+ite−σyλnx+σ∫λnxe−σyIny;σ+itdy;that is,
(37)∫λnxexpitydαy≤Muσ,Fe−σx+e−σλn+e−σx−e−σλn≤2Muσ,Fe−σx.
Hence, for anyσ>0 and any x∈λn,λn+1, it follows
(38)∫λnxexpitydαy≤2Muσ,Fe−σλne−ση≤2Muσ,Fe−σλn;that is,
(39)μσ,F≤2Muσ,F.
Therefore, we can conclude from (39) that
(40)limsupσ⟶+∞logMuσ,FUr≥limsupσ⟶+∞logμσ,FUr.
On the other hand, assume that(41)limsupσ⟶+∞logμσ,FUr=1,then for any fixed ε∈0,1/2 and sufficiently large σ, it follows
(42)logμσ,F≤1+εUr.
For any positive real numberx, in view of (5), there exists a positive integer n such that λn<x≤λn+1; thus, it yields
(43)∫0xexpσ+itydαy=∑k=1n−1∫λkλk+1expσ+itydαy+∫λnxexpσ+itydαy=∑k=1n−1∫λkλk+1expσydyIky;it+∫λnxexpσydyIny;it=∑k=1n−1expλk+1σIkλk+1;it−σ∫λkλk+1expσyIky;itdy+expxσInx;it−σ∫λnxexpσyIny;itdy,where Ikx;it=∫λkxexpitydαy. Similar to the argument as in (38), it follows
(44)∫0xexpσ+itydαy≤2∑k=1nAk∗eλkσ.
SetQ=σ+log1+logr/logUrlog2logUr=logR, where r=eσ, and then from (44), we have
(45)Muσ,F≤2∑n=1+∞An∗expλnQ−λnlog1+logrlogUrlog2logUr≤2μQ,F∑n=1+∞e−λnE,where
(46)E≔log1+logrlogUrlog2logUr.
Thus, for any real numberδ∈0,1−d, in view of (26), there exists a positive integer N∈ℕ+ such that
(47)λn>logn1/d+δ>logn,asn>N.
Hence, we can conclude from (45) and (47) that
(48)Muσ,F≤2μQ,F∑n=1+∞e−λnE≤2μQ,FN+∑n=N+∞e−Elogn1/d+δ.
SetT=exp2/Ed+δ/1−d−δ+1, where x is an integral function. Then, it follows Elogn1−d−δ/d+δ>2 as n>T. So, from (48), we can deduce
(49)Muσ,F≤2μQ,FN+∑n=NTe−Elogn+∑n=T+1+∞e−2logn≤2μQ,FN+∫1Tt−Edt=2μQ,FN+11−ET1−E.
Hence, from (39), (49), and by Theorem 8, it becomes
(50)logMuσ,F≤logμQ,F+log2+logN+log11−E+1−ElogT≤1+εUR+K1+2Ed+δ/1−d−δ≤1+2εUr,where K1 is a finite constant. Since ε is arbitrary and δ∈0,1−d, then we conclude
(51)limsupσ⟶+∞logMuσ,FUr≤limsupσ⟶+∞logμσ,FUr=1.
Therefore, this completes the proof of Lemma 2 from (40) and (51).
Proof of Theorem 10. In view of(52)Q1≔limsupσ⟶+∞−logEn−1F,βexp−βλnUψn,and then it is obvious that the conclusion of Theorem 10 holds as Q1=+∞. Next, we will prove that the conclusion of Theorem 10 holds for Q1∈0,+∞.
IfQ1∈0,+∞, then for any fixed real number ε∈0,1/3, there exists a positive integer N∈ℕ+ such that
(53)−logEn−1F,βexp−βλn<Q1+εUψn,forn>N.
LetV≔Vσ=Ueσ/σ and W be the inverse function of V, and then we know in view of Theorem 8 that V is an increasing function for σσ→+∞. Thus, from the above inequality, we can deduce
(54)λnQ1+ε≤λn−logEn−1exp−βλnUψn;that is,
(55)logEn−1expσ−βλn≤−λnWλnQ1+ε−σ.
For sufficiently largeσ, set
(56)H≔Hσ=Q1+εUr+rlogrlogUrlog2logUr=Q1+εUR,and then it follows
(57)HQ1+εQ=URQ,WHQ1+εQ=X.
Ifλnσ≤H, it yields from (55) and (57) that
(58)logEn−1expσ−βλn≤−λnWλnQ1+ε−σ<λnσ≤H≤Q1+2εUr.
Ifλnσ>H, that is, λn>H/σ, it yields from (55) and (57) that
(59)logEn−1expσ−βλn<−λnWλnQ1+ε−σ<−λnWHQ1+εσ−σ=0.
Hence, we can conclude from (58) and (59) that
(60)logEn−1expσ−βλn≤Q1+2εUr,forn>N.
For anyβ<+∞, it follows
(61)An∗expβλn=supλn<x≤λn+1,−∞<t<+∞∫λnxexpitydαyexpβλn≤supλn<x≤λn+1,−∞<t<+∞∫λnxexpβ+itydαy≤sup−∞<t<+∞∫λn∞expβ+itydαy,and thus for any p∈Πn−1, we can deduce by combining the above inequalities that
(62)An∗expβλn≤Fβ+it−pβ+it≤F−pβ.
On the other hand, there existsp1∈Πn−1 such that
(63)F−p1≤2En−1F,β.
Hence, from (60)–(63) and for any β<+∞, it follows
(64)logAn∗expσλn≤log2En−1expσ−βλn≤Q1+3εUr,forn>N;that is,
logμσ,F≤Q1+3εUr,forσ⟶+∞.
Sinceε is arbitrary and by Lemma 16, we get
limsupσ⟶+∞logMuσ,F/Ur=1≤Q1.
We therefore completes the Proof of Theorem 6.


## 3. Proofs of Theorems 7 and 8

### 3.1. The Proof of Theorem 7

Here, we will adopt the reduction to absurdity to prove Theorem 7. Suppose that(65)limsupσ⟶+∞logMuσ,FUr=1<Q2.IfQ2∈1,+∞, set t∈0,Q2−1/7∩0,1. From (65), for any small t and any positive integer n, we have
(66)logMuσ,F≤1+tUr.IfFs∈L∞ and β−∞<β<+∞, we have Fs∈L¯β. Moreover, for β<σ<+∞ and pn∈Πn, it follows
(67)EnF,β≤F−pnβ≤Fβ+it−pnβ+it≤∫0+∞expβ+itydαy−∫0λnexpβ+itydαy=∫λn∞expβ+itydαy.Similar to (44), we have
(68)∫λk∞expβ+itydαy≤2∑n=k+∞An∗expβλn+1.Thus, for anyσβ<σ<+∞, we can conclude from (39), (67), and (68) that
(69)EnF,β≤2∑k=n+1∞Ak−1∗expβλk≤4Muσ,F∑k=n+1∞expβ−σλk.From (9), there exists h′0<h′<h such that λn+1−λn≥h′ for n≥0. Then, for σ≥β+1, it follows from (69) that
(70)EnF,β≤4Muσ,Fexpλn+1β−σ∑k=n+1∞expλk−λn+1β−σ≤4Muσ,Fexpλn+1β−σexph′n+1∑k=n+1∞exp−h′k=4Muσ,Fexpλn+1β−σ1−exph′−1;that is,
(71)En−1F,β≤K1Muσ,Fexpλnβ−σ,where K1 is a constant. Hence, we conclude from (66) and (71) that
(72)logEn−1expσ−βλn≤logK1Muσ,F≤1+2tUr,forσ⟶+∞.On the other hand, in view ofQ2 and t, there exists a subsequence np such that
(73)λnp>Q2−tU1ψnp≥1+6tU1ψnp.We choose a sequenceσnp such that
(74)1+2tUrnp=λnplog1+1logUψnplog2logUψnp,and then it follows from (72) and (74) that
(75)σnp≤logψnp+log1+1logUψnplog2logUψnp.Hence,(76)Urnp≤Uψnp1+1logUψnplog2logUψnp≤Uψnp1+logψnplogUψnplog2logUψnp≤1+tUψnp.In view oft∈0,Q2−1/7∩0,1, then it follows 1+2t1+t<1+5t. Thus, by combining (74) and (76), we deduce
(77)λnp=1+2tUrnplog1+1logUψnplog2logUψnp−1≤1+2t1+t1+o1U1ψnp≤1+5tU1ψnp,which is a contradiction with (73).IfQ2=+∞, we choose t=1, and by using the same argument as above, we also get a contradiction.Therefore, this completes the Proof of Theorem 7.


### 3.2. The Proof of Theorem 8

From Lemma15, it is easy to get the following lemma.Lemma 17.
If Laplace-Stieltjes transformFs is of infinite order, and the sequence λn satisfied (5), (9), (12), and (26), then
(78)limsupσ⟶+∞loglogMuσ,FlogUr=1⟺limsupσ⟶+∞loglogμσ,FlogUr=1.
Proof of Theorem 8. IfQ3=+∞, then it follows
(79)limsupσ⟶+∞λnUψnlog2Uψn=+∞.
By combining Corollary13, it yields
limsupσ⟶+∞logMuσ,F/Ur=+∞,
which is a contradiction with the properties ofUr. Hence, Q3<+∞. Thus, for any fixed ε>0 and sufficiently large n, we have
(80)λn<UQ3+εψn,logV1λn1/Q3+ε≤−logEn−1F,βexp−βλnλn,where r=V1x and x=Ur are two reciprocally inverse functions; that is,
(81)logEn−1F,βexpσ−βλn≤−λnlogV1λn1/Q3+ε−σ.
For any fixed sufficiently largeσ, take
(82)I≔Iσ=σUQ3+εr+rlogrlogUrlog2logUr,and then it yields
(83)logV1Iσ1/Q3+ε=σ+log1+logrlogUrlog2logUr.
Ifλnσ≤I, from (81) and (83), it follows
(84)logEn−1F,βexpσ−βλn≤−λnlogV1λn1/Q3+ε−σ<σλn≤I≤UQ3+3εr.
Ifλnσ>I, that is, λn>I/σ, thus in view of (81) and (83), it yields
(85)logEn−1F,βexpσ−βλn≤−λnlogV1λn1/Q3+ε−σ<−λnlogV1Iσ1/Q3+ε−σ<0.
Hence, from (62), (63), (84), and (85), we deduce
(86)loglogμσ,F≤Q3+4εlogUr,forn>N,σ⟶+∞.
Letε⟶0 and by Lemma 17, and it leads to
(87)limsupσ⟶+∞loglogMuσ,FlogUr≤Q3.
Suppose that(88)1=limsupσ⟶+∞loglogMuσ,FlogUr<Q3.
Sett∈0,Q3−1/7, and then for any positive integer n and sufficiently large σ, from (71), we have
(89)logEn−1F,βexpσ−βλn≤logMuσ,F+logK1≤U1+2tr,where K1 is a constant. Since 1+6t≤Q3−t, then there exists a subsequence np such that
(90)λnp≥UQ3−tψnp>U1+6tψnp.
We choose a sequenceσnp such that
(91)U1+2trnp=λnplog1+1log2Uψnp.
Thus, it follows from (89) and (91) that
(92)eσnp≤ψnp1+1log2Uψnp.That is,
(93)Urnp≤Uψnp1+1log2Uψnp≤1+tUψnp.
We therefore can conclude from (91) that
(94)λnp=U1+2trnplog1+1log2Uψnp−1≤2U1+2trnplog2Uψnp≤21+t1+2tU1+2tψnplog2Uψnp≤U1+5tψnp,which is a contradiction with the inequality (90).
Therefore, this completes the Proof of Theorem 8.


## 3.1. The Proof of Theorem 7

Here, we will adopt the reduction to absurdity to prove Theorem 7. Suppose that(65)limsupσ⟶+∞logMuσ,FUr=1<Q2.IfQ2∈1,+∞, set t∈0,Q2−1/7∩0,1. From (65), for any small t and any positive integer n, we have
(66)logMuσ,F≤1+tUr.IfFs∈L∞ and β−∞<β<+∞, we have Fs∈L¯β. Moreover, for β<σ<+∞ and pn∈Πn, it follows
(67)EnF,β≤F−pnβ≤Fβ+it−pnβ+it≤∫0+∞expβ+itydαy−∫0λnexpβ+itydαy=∫λn∞expβ+itydαy.Similar to (44), we have
(68)∫λk∞expβ+itydαy≤2∑n=k+∞An∗expβλn+1.Thus, for anyσβ<σ<+∞, we can conclude from (39), (67), and (68) that
(69)EnF,β≤2∑k=n+1∞Ak−1∗expβλk≤4Muσ,F∑k=n+1∞expβ−σλk.From (9), there exists h′0<h′<h such that λn+1−λn≥h′ for n≥0. Then, for σ≥β+1, it follows from (69) that
(70)EnF,β≤4Muσ,Fexpλn+1β−σ∑k=n+1∞expλk−λn+1β−σ≤4Muσ,Fexpλn+1β−σexph′n+1∑k=n+1∞exp−h′k=4Muσ,Fexpλn+1β−σ1−exph′−1;that is,
(71)En−1F,β≤K1Muσ,Fexpλnβ−σ,where K1 is a constant. Hence, we conclude from (66) and (71) that
(72)logEn−1expσ−βλn≤logK1Muσ,F≤1+2tUr,forσ⟶+∞.On the other hand, in view ofQ2 and t, there exists a subsequence np such that
(73)λnp>Q2−tU1ψnp≥1+6tU1ψnp.We choose a sequenceσnp such that
(74)1+2tUrnp=λnplog1+1logUψnplog2logUψnp,and then it follows from (72) and (74) that
(75)σnp≤logψnp+log1+1logUψnplog2logUψnp.Hence,(76)Urnp≤Uψnp1+1logUψnplog2logUψnp≤Uψnp1+logψnplogUψnplog2logUψnp≤1+tUψnp.In view oft∈0,Q2−1/7∩0,1, then it follows 1+2t1+t<1+5t. Thus, by combining (74) and (76), we deduce
(77)λnp=1+2tUrnplog1+1logUψnplog2logUψnp−1≤1+2t1+t1+o1U1ψnp≤1+5tU1ψnp,which is a contradiction with (73).IfQ2=+∞, we choose t=1, and by using the same argument as above, we also get a contradiction.Therefore, this completes the Proof of Theorem 7.


## 3.2. The Proof of Theorem 8

From Lemma15, it is easy to get the following lemma.Lemma 17.
If Laplace-Stieltjes transformFs is of infinite order, and the sequence λn satisfied (5), (9), (12), and (26), then
(78)limsupσ⟶+∞loglogMuσ,FlogUr=1⟺limsupσ⟶+∞loglogμσ,FlogUr=1.
Proof of Theorem 8. IfQ3=+∞, then it follows
(79)limsupσ⟶+∞λnUψnlog2Uψn=+∞.
By combining Corollary13, it yields
limsupσ⟶+∞logMuσ,F/Ur=+∞,
which is a contradiction with the properties ofUr. Hence, Q3<+∞. Thus, for any fixed ε>0 and sufficiently large n, we have
(80)λn<UQ3+εψn,logV1λn1/Q3+ε≤−logEn−1F,βexp−βλnλn,where r=V1x and x=Ur are two reciprocally inverse functions; that is,
(81)logEn−1F,βexpσ−βλn≤−λnlogV1λn1/Q3+ε−σ.
For any fixed sufficiently largeσ, take
(82)I≔Iσ=σUQ3+εr+rlogrlogUrlog2logUr,and then it yields
(83)logV1Iσ1/Q3+ε=σ+log1+logrlogUrlog2logUr.
Ifλnσ≤I, from (81) and (83), it follows
(84)logEn−1F,βexpσ−βλn≤−λnlogV1λn1/Q3+ε−σ<σλn≤I≤UQ3+3εr.
Ifλnσ>I, that is, λn>I/σ, thus in view of (81) and (83), it yields
(85)logEn−1F,βexpσ−βλn≤−λnlogV1λn1/Q3+ε−σ<−λnlogV1Iσ1/Q3+ε−σ<0.
Hence, from (62), (63), (84), and (85), we deduce
(86)loglogμσ,F≤Q3+4εlogUr,forn>N,σ⟶+∞.
Letε⟶0 and by Lemma 17, and it leads to
(87)limsupσ⟶+∞loglogMuσ,FlogUr≤Q3.
Suppose that(88)1=limsupσ⟶+∞loglogMuσ,FlogUr<Q3.
Sett∈0,Q3−1/7, and then for any positive integer n and sufficiently large σ, from (71), we have
(89)logEn−1F,βexpσ−βλn≤logMuσ,F+logK1≤U1+2tr,where K1 is a constant. Since 1+6t≤Q3−t, then there exists a subsequence np such that
(90)λnp≥UQ3−tψnp>U1+6tψnp.
We choose a sequenceσnp such that
(91)U1+2trnp=λnplog1+1log2Uψnp.
Thus, it follows from (89) and (91) that
(92)eσnp≤ψnp1+1log2Uψnp.That is,
(93)Urnp≤Uψnp1+1log2Uψnp≤1+tUψnp.
We therefore can conclude from (91) that
(94)λnp=U1+2trnplog1+1log2Uψnp−1≤2U1+2trnplog2Uψnp≤21+t1+2tU1+2tψnplog2Uψnp≤U1+5tψnp,which is a contradiction with the inequality (90).
Therefore, this completes the Proof of Theorem 8.



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*Source: 9970287-2021-06-18.xml*