# Growth and Approximation of Laplace-Stieltjes Transform withp,q-Proximate Order Converges on the Whole Plane

Authors: Yong Qin Cui; Hong Yan Xu
Journal: Journal of Function Spaces
 (2020)

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## Abstract

One purpose of this paper is to study the growth of entire functions defined by Laplace-Stieltjes transform converges on the whole complex plane, by introducing the concept ofp,q-proximate order, and one equivalence theorem of the p,q-proximate order of Laplace-Stieltjes transforms is obtained. Besides, the second purpose of this paper is to investigate the approximation of entire functions defined by Laplace-Stieltjes transforms with p,q-proximate order, and some results about the p,q-proximate order, the error, and the coefficients of Laplace-Stieltjes transforms are obtained, which are generalization and improvement of the previous theorems given by Luo and Kong, Singhal, and Srivastava.


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## Body

## 1. Introduction

The main purpose of this paper is to investigate the growth and approximation of entire functions represented by Laplace-Stieltjes transforms which convergent on the whole complex plane. For Laplace-Stieltjes transforms,(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. In fact, if αt is absolutely continuous, then Gs becomes the classical Laplace integral form:(2)Gs=∫0∞e−stφtdt.Ifαt is a step function, we can choose a sequence λn0∞ satisfying(3)0≤λ1<λ2<⋯<λn<⋯,λn⟶∞,asn⟶∞,and let(4)αx=a1+a2+⋯+an,λn≤x<λn+1;0,0≤x<λ1;αx++αx−2,x>0.Thus, Gs becomes a Dirichlet series:(5)Gs=∑n=1∞aneλns,s=σ+it,where σandt are real variables and an are nonzero complex numbers; if αt is an increasing continuous function which is not absolutely continuous, then integral (1) defines a class of functions Gs which cannot be expressed either in form (2) or (6).As we know, the Laplace-Stieltjes transform, named for Pierre-Simon Laplace and Thomas Joannes Stieltjes, is an integral transform similar to the Laplace transform. For real-valued functions, it is the Laplace transform of a Stieltjes measure; however, it is often defined for functions with values in a Banach space. It can be used in many fields of mathematics, such as functional analysis, and certain areas of theoretical and applied probability.In 1963, Yu [1] first proved the Valiron-Knopp-Bohr formula of the associated abscissas of bounded convergence, absolute convergence, and uniform convergence of the LaplaceStieltjes transform.Theorem 1 (see [1]).
If the sequenceλn satisfy (3) and(6)lim supn⟶+∞lognλn=D<+∞,lim supn⟶+∞λn+1−λn=h<+∞,then LaplaceStieltjes transforms (1) satisfy(7)limsupn⟶+∞logBn∗λn≤σuG≤limsupn⟶+∞logBn∗λn+limupn⟶+∞lognλn,where σuG is called the abscissa of uniformly convergent of Fs and(8)Bn∗=supλn<x≤λn+1,−∞<t<+∞∫λnxe−itydαy.Moreover, Yu [1] introduced the concepts of the order of Gs and estimated the growth of the maximal molecule Muσ,G, the maximal term μσ,G, the Borel line, and the order of entire functions represented by LaplaceStieltjes transform convergent in the whole complex plane. After his works, considerable attention has been paid to the growth and the value distribution of the functions represented by LaplaceStieltjes transform convergent in the half plane or whole complex plane in the field of complex analysis (see [2–16]).In 2012, Luo and Kong [17] studied the following form of the LaplaceStieltjes transform:(9)Fs=∫0+∞esxdαx,s=σ+it,where αx is stated as in (1) and λn satisfies (3) and (6), which is a different form from (1). Set(10)An∗=supλn<x≤λn+1,−∞<t<+∞∫λnxeitydαy.By using the same argument as in [1], we can get the similar result about the abscissa of uniformly convergent of Fs easily. If(11)lim  supn⟶+∞logAn∗λn=−∞,by (3), (6), and Theorem 1, we have σuF=+∞; thus, it can be said that Fs is an entire function in the whole plane.Set(12)μσ,F=maxn∈NAn∗eλnσσ<+∞,Mσ,F=sup−∞<t<+∞Fσ+it,Muσ,F=sup0<x<+∞,−∞<t<+∞∫0xeσ+itydαy.SinceMσ,F and Muσ,F trend to +∞ as σ⟶+∞, in order to estimate the growth of Fs precisely, the concepts of order and type is introduced as follows.Definition 1.
If the LaplaceStieltjes transform (11) satisfies σuF=+∞ (the sequence λn satisfy (3), (9), and (20)) and(13)limsupσ⟶+∞log+log+Muσ,Fσ=ρ,0≤ρ≤+∞,we call Fs is of order ρ in the whole plane, where log+x=maxlogx,0. Furthermore, if ρ∈0,+∞, the type of Fs is defined by(14)lim  supσ⟶+∞log+Muσ,Feσρ=T,0≤T≤+∞.Remark 1.
The concept of type is used to compare the growth of two LaplaceStieltjes transforms of the same nonzero finite order. For example, letfs=eesandgs=ee2s, by simple computation, and we have ρf=1=ρg, but Tf=1 and Tg=∞. Thus, we can see that the growth of gs is faster than fs as s⟶+∞.Remark 2.
Valiron [18] used intermediate comparison functions called the proximate order to refine these growth scales, which make it unnecessary to consider functions of minimal or maximal type.Remark 3.
However, ifρ=0 and ρ=+∞, we cannot estimate the growth of such functions precisely by using the concept of type.
Forρ=+∞, Kong [19] studied the growth of such functions by using the concepts of the Sun’s type function Ur=rexp1/ρr, where r=e−σ, ρr is the decreasing function ↓0, ρ′r↑, and limr⟶∞rρ′rlogrlog logr=0. In fact, one can also estimate the growth of LaplaceStieltjes transforms by using the concepts of p-order were given by Sato [20]. If(15)lim supσ⟶+∞logp+1Muσ,Fσ=ρp,where log0x=x,loglogpx=logp+1x, and p∈ℕ+, then Fs is said to be of index p if ρp−1=∞ and ρp<+∞. If 0<ρp<+∞, we also compare their growth by using the concepts of p-type. However, there are two shortcomings in the concepts of p-order and p-type as follows.S1: one can hardly compare the growth of such functions of zero order by using the idea ofp-order.S2: the growth of such functions ofρp−1=∞ and ρp=0 is incomparable by using the idea of p-order.
For the first shortcoming, there were some concepts such as logarithmic order,h-order, and generalize order (see [21–24]). However, to cover these shortcomings simultaneously, the concepts of p,q-order and p,q-type were good ideas and were used to estimate the growth of a class of entire functions represented by LaplaceStieltjes transforms more precisely, which were given by Juneja, Kapor, and Bajpai [25–27].Definition 2.
If the LaplaceStieltjes transform (11) satisfies σuF=+∞ (the sequence λn satisfy (3), (9), and (20)) and(16)ρ=ρp,q=lim  supσ⟶+∞logp+1Muσ,Flogqσ,0≤ρp,q≤+∞,then we say that Fs is of p,q-order ρp,q in the whole plane, where p,q are two integers such that p≥q≥0 and p≠0, and exp0x=log0x=x, expp+1x=expexppx, and log−1x=expx, exp−1x=logx.Remark 4.
Throughout this paper, the conditionsp≥q≥0 and p≠0 are always supposed to hold.Remark 5.
It can be seen that0≤ρp,q≤+∞ if p>q and 1≤ρp,q≤+∞ if p=q. Furthermore, if 0≤ρp,q≤+∞, then ρp−n,q=∞ for n<p, ρp,q−n=0 for n<q, and ρp+n,q+n=1 for n=1,2,….Definition 3.
It is said that the LaplaceStieltjes transformFs is of index-pair p,q, p≥q≥0, and p≠0 if b<ρp,q<∞, and ρp−1,q−1 is not a finite nonzero number, where b=1 if p=q and b=0 if p>q. If Fs is of index-pair p,q, then ρ≡ρp,q is called its p,q-order.
The concept ofp,q-type is introduced below, which is used to compare the growth of such functions of the same p,q-order.Definition 4.
The LaplaceStieltjes transform (11) of p,q-order ρp,q∈0,+∞ is said to be of the p,q-type if(17)T=Tp,q=lim  supσ⟶+∞logpMuσ,Flogq−1σρ,0≤T≤+∞.Remark 6.
From the above definitions, it can be seen easily that the concepts ofp,q-order and p,q-type can cover the above shortcomings (S1) and (S2) in some sense.
However, these concepts are inadequate for comparing the growth of LaplaceStieltjes transforms of the samep,q-orders but of infinite p,q-types. Inspired by this question, one purpose of this paper is to improve the above inadequacy and investigate the growth of LaplaceStieltjes transforms.
The structure of this paper is listed as following. Section2 is to investigate the growth of LaplaceStieltjes transforms with the p,q-proximate order, and an equivalence theorem on the p,q-proximate order is obtained among the p,q-order ρp,q, p,q-type Tp,q, λn, and An∗. Section 3 is to deal with a study of the approximation on LaplaceStieltjes transforms of the p,q-proximate order and gives a relation theorem about the error EnF,β, p,q-order ρp,q, p,q-type Tp,q, and p,q-proximate order. The topic of growth and approximation of LaplaceStieltjes transforms of p,q-proximate order, it seems that this topic has never been treated before. Our results in this paper are some improvements of the previous theorems given by Kong, Luo, and Sun.


## 2. Thep,q-Proximate Order of LaplaceStieltjes Transforms

Next, we introduce the definition ofp,q-proximate order of LaplaceStieltjes transforms.Definition 5.
A positive functionρσ defined on σ0,+∞, σ0>0, is a proximate order of LaplaceStieltjes transforms (11) with index-pair p,q if ρσ satisfies the following conditions:(i)
ρσ⟶ρ≔ρp,q as σ⟶+∞, b<ρ≔ρp,q<+∞, where b=1 if p=q and b=0 if p>q.(ii)
Λqσρ′σ⟶0 as σ⟶+∞, where Λqσ=∏i=0qlogiσ.(iii)
if(18)T∗=T∗p,q=lim  supσ⟶+∞logpMuσ,Flogq−1σρσ,0≤T≤+∞,then ρσ is said to be the p,q-proximate order of the LaplaceStieltjes transform Fs if 0<T∗<+∞.
Let(19)x=logq−1σρσ−ϱ,where ϱ=1 if q=1 and zero otherwise, by a simple computation, we can get that logq−1σρσ−ϱ is a monotone increase function of σ for σ>σ0. Thus, we can define a real function Φx of x>x0 such that(20)x=logq−1σρσ−ϱ⟺Φx=logq−1σ.
Here, one main theorem of this paper is as follows.Theorem 2.
If the LaplaceStieltjes transform (9) satisfies σuF=+∞ (the sequence λn satisfy (3), (6), and (11)) and is of p,q-order ρ0<ρ<∞, then(21)T∗≔T∗p,q=lim  supσ⟶+∞logpMuσ,Flogq−1σρσ⇔lim  supn⟶+∞HΦlogp−1λnlogqAn∗−1/λnρ−ϱ=T∗,where ϱ=1 if q=1 and zero otherwise, and(22)H=1eρ,ifp,q=1,0;ρ−1ρ−1ρρ,ifp,q=1,1;1,ifp≥q≥2.To prove this theorem, we require the following lemmas.Lemma 1.
For the functionΦx defined in (20) and any positive real number κ0<κ<+∞, we have(23)limx⟶+∞ΦκxΦx=κ1/ρ−ϱ,where ϱ is stated as in Theorem 2.Proof.
Here, two cases will be considered as follows.Case 1.
Ifq=0, then ϱ=0 and Φx=expσ, i.e., x=expσρσ. Thus, it follows that σ⟶+∞ as x⟶+∞ and κx=κexpσρσ, and Φκx=κ1/ρσexpσ. Since limσ⟶+∞ρσ=ρ, it yields(24)limx⟶+∞ΦκxΦx=limx⟶+∞κ1/ρσexpσexpσ=κ1/ρ.Case 2.
Ifq>0. From Definition 4, we conclude(25)limx⟶+∞dlogΦxdlogx=limx⟶+∞1ρσ−ϱ+Λqσρ′σ=1ρ−ϱ.
Thus, for any givenε>0 and κ>1, from (28) it follows(26)∫xκx1ρ−ϱ−εdlogt<∫xκxdlogΦt<∫xκx1ρ−ϱ+εdlogt,that is,(27)1ρ−ϱ−εlogκ<logΦκxΦx<1ρ−ϱ+εlogκ.
Hence,(28)κ1/ρ−ϱ−ε<ΦκxΦx<κ1/ρ−ϱ+ε.
Forκ=1, it is obviously. For 0<κ<1, by a similar argument as in (28), we can deduce(29)∫κxx1ρ−ϱ−εdlogt<∫κxxdlogΦt<∫κxx1ρ−ϱ+εdlogt,that is,(30)−1ρ−ϱ−εlogκ<logΦxΦκx<−1ρ−ϱ+εlogκ.
Hence,(31)κ−1/ρ−ϱ−ε<ΦxΦκx<κ−1/ρ−ϱ+ε.
Thus, this completes the proof of Lemma 1 from Cases 1 and 2.Lemma 2 (see [10], Lemma 2.1).
If the L-S transformFs∈L∞, for sufficiently larger σ⟶+∞ and ε>0, we have(32)12μσ,F≤Muσ,F≤Cμ1+2εσ,F,where C is a constant.Remark 7.
From Lemma2, we can easily get the following result:
If the LaplaceStieltjes transform (11) satisfies σuF=+∞ (the sequence λn satisfy (3), (9), and (20)) and is of p,q-order ρ0<ρ<∞, then(33)T∗≔T∗p,q=lim  supσ⟶+∞logpMuσ,Flogq−1σρσ=lim supσ⟶+∞logpμσ,Flogq−1σρσ.


### 2.1. The Proof of Theorem2

Here, we first prove the necessity of Theorem2.⟹
: We will divide into two steps to prove the necessity.Step one. We first prove that(34)limsupn⟶+∞HΦlogp−1λnlogqAn∗−1/λnρ−ϱ≤T∗.Since(35)T∗≔T∗p,q=lim supσ⟶+∞logpMuσ,Flogq−1σρσ, for any given ε>0 and σ>σ0, it follows(36)logpMuσ,F≤T∗+εlogq−1σρσ.LetB=T∗+ε, then(37)logMuσ,F≤expp−1Blogq−1σρσ.It thus follows from Lemma2 that(38)logAn∗≤expp−1Blogq−1σρσ−λnσ+log2.Now, we will consider three cases as follows.Case 3.
Suppose thatp=1 and q=0, then ϱ=0. From (39), it follows(39)logAn∗≤Bexpσρσ−λnσ+log2.
Takeσ such that(40)expσρσ=λnBρ,that is,(41)eσ=λnBρ1/ρσ.
In view of (40) and (43), it yields(42)logAn∗≤λnρ−λnσ+log2≤−1+o1λnσ−1ρ,orAn∗−1/λn≥exp−1ρ1+o1λnBρ1/ρσ,that is,(43)λn1/ρσAn∗−1/λn≤1exp−1/ρ1+o11/Bρ1/ρσ.
Sinceq=0, thus x=expσρσ, that is, Φx=eσ. From (40), we have σ⟶+∞ and ρσ⟶ρ as n⟶+∞. Hence, from (43), and combining with ε is arbitrarily, we can deduce(44)limsupn⟶+∞ΦλnAn∗−1/λnρ≤T∗eρ.Case 4.
Suppose thatp=1 and q=1, then ϱ=1. From (39), it follows(45)logAn∗≤Bσρσ−λnσ+log2.
Takeσ such that(46)σρσ−1=λnBρ.
From (46), it follows σ=Φλn/Bρ. Thus, we can conclude from (45) that(47)logAn∗≤−λnσρ−1ρ1+o1=−λnρ−1ρ1+o1ΦλnBρ,that is,(48)Φλn−1/λnlogAn∗≤Φλnρ−1/ρ1+o1Φλn/Bρ.
From (46), it yields σ⟶+∞ and ρσ⟶ρ as n⟶+∞. Hence, from (48), and combining with ε is arbitrarily, we conclude(49)lim supn⟶+∞Φλn−1/λnlogAn∗ρ−1≤T∗ρρρ−1ρ−1.Case 5.
Suppose thatp≥q≥2, then ϱ=0. Take σ such that(50)logq−1σρσ=1Blogp−1λnρ,that is,(51)logq−1σ=Φ1Blogp−1λnρ.
Substituting (50) into (38), it yields(52)logAn∗≤−λnσ−1ρ+log2=−λnσ−1ρ1+o1.
Thus, it follows(53)−1λnlogAn∗≥σ−1ρ1+o1=expq−1Φ1Blogp−1λnρ−1ρ1+o1≥expq−1Φ1Blogp−1λnρ1+o1,that is,(54)logq−1−1λnlogAn∗≥Φ1Blogp−1λnρ1+o1.
In view of (54), we can deduce(55)Φlogp−1λnlogq−1−1/λnlogAn∗≤Φlogp−1λnΦ1/Blogp−1λn/ρ1+o1.
Sincelogp−1λn/ρ∼logp−1λn as n⟶+∞ and p>1, by Lemma 1, let ε⟶0 and n⟶+∞ in (56), and it yields(56)lim supn⟶+∞Φlogp−1λnlogq−1−1/λnlogAn∗ρ≤T∗.
Therefore, we can conclude from (44), (46), and (56) that(57)lim supn⟶+∞HΦlogp−1λnlogq−1−1/λnlogAn∗ρ≤T∗,where H is stated as in Theorem 2.
Step Two. Next, we will prove that(58)lim supn⟶+∞HΦlogp−1λnlogq−1−1/λnlogAn∗ρ<T∗,cannot hold. From (58), let B′ to be a real constant such that B′<T∗ and(59)lim supn⟶+∞HΦlogp−1λnlogq−1−1/λnlogAn∗ρ=B′<T∗.
Thus, for any smallε such that 0<ε<1/3T∗−B′ and k0 such that k>k0, it follows(60)Φlogp−1λnk≤B′+εH1/ρ−ϱlogq−1−1λnklogAnk∗.
Here, three cases will be considered as follows.Case 6.
Suppose thatp=1 and q=0, then ϱ=0 and H=1/eρ, and (60) becomes(61)Φλnk≤eρB′+ε1/ρexp−1λnklogAnk∗.
Thus, it follows(62)Ank∗≤exp−λnklogeρB′+ε−1/ρΦλnk,that is,(63)logAnk∗eλnkσ≤supk≥0−λnklogeρB′+ε−1/ρΦλnk+λnkσ.
Takeσ such that(64)σ=logΦλnkρB′+ε,that is,(65)eσρσ=λnkρB′+ε.
Then, in view of Lemma2 and (63), we can deduce(66)logAnk∗eλnkσ≤supk≥0λnkσ−λnklogeρB′+ε−1/ρρB′+ε1/ρ1+o1eσ≤supk≥0λnkρ1+o1≤B′+ε1+o1eσρσ≤B′+2εeσρσ.
Thus, from Lemma2 and (66) and since B′+3ε<T∗, it means(67)T∗=lim supσ⟶+∞logMuσ,Feσρσ≤B′+3ε<T∗,a contradiction.Case 7.
Suppose thatp=1 and q=1, then ϱ=1 and H=ρ−1ρ−1/ρρ. From (60), it follows(68)logAnk∗eλnkσ≤supk≥0−λnkHB′+ε1/ρ−1Φλnk+λnkσ.
Takeσ such that(69)σ=ΦλnkρB′+ε,that is,(70)σρσ−1=λnkρB′+ε.
Then, in view of Lemma1 and (68), it follows(71)logAnk∗eλnkσ≤supk≥0λnkσ−λnkHρ1/ρ−1σ1+o1≤supk≥0λnkσρ1+o1≤supk≥0B′+ε1+o1σρσ.
Thus, from Lemma2 and (71) and since B′+3ε<T∗, we conclude(72)T∗=limsupσ⟶+∞logMuσ,Fσρσ≤B′+3ε<T∗,a contradiction.Case 8.
Suppose thatp≥q≥2, then ϱ=0 and H=1. Then, it follows from (60) that(73)logAnk∗eλnkσ≤supk≥0λnkσ−λnkexpq−11B′+ε1/ρΦlogp−1λnk.
Takeσ such that(74)logq−1σ−1=Φlogp−1λnkB′+ε,that is,(75)λnk=expp−1B′+εlogq−1σ−1ρσ.
Then, from Lemma1 and (73), it yields(76)logAnk∗eλnkσ≤supk≥0λnkσ−λnkσ−11+o1≤supk≥0λnk1+o1≤supk≥0expp−1B′+εlogq−1σ−1ρσ1+o1.
Thus, from Lemma2 and (76) and since B′+3ε<T∗ and q≥2, we conclude(77)T∗=lim supσ⟶+∞logpMuσ,Flogq−1σρσ≤B′+3ε<T∗,a contradiction.
Therefore, the necessity of Theorem2 is proved from Steps one and two.
In fact, by using the same argument as in the processing of the necessity of Theorem2, we can easily prove the sufficiency of Theorem 2. Hence, this completes the proof of Theorem 2.


## 2.1. The Proof of Theorem2

Here, we first prove the necessity of Theorem2.⟹
: We will divide into two steps to prove the necessity.Step one. We first prove that(34)limsupn⟶+∞HΦlogp−1λnlogqAn∗−1/λnρ−ϱ≤T∗.Since(35)T∗≔T∗p,q=lim supσ⟶+∞logpMuσ,Flogq−1σρσ, for any given ε>0 and σ>σ0, it follows(36)logpMuσ,F≤T∗+εlogq−1σρσ.LetB=T∗+ε, then(37)logMuσ,F≤expp−1Blogq−1σρσ.It thus follows from Lemma2 that(38)logAn∗≤expp−1Blogq−1σρσ−λnσ+log2.Now, we will consider three cases as follows.Case 3.
Suppose thatp=1 and q=0, then ϱ=0. From (39), it follows(39)logAn∗≤Bexpσρσ−λnσ+log2.
Takeσ such that(40)expσρσ=λnBρ,that is,(41)eσ=λnBρ1/ρσ.
In view of (40) and (43), it yields(42)logAn∗≤λnρ−λnσ+log2≤−1+o1λnσ−1ρ,orAn∗−1/λn≥exp−1ρ1+o1λnBρ1/ρσ,that is,(43)λn1/ρσAn∗−1/λn≤1exp−1/ρ1+o11/Bρ1/ρσ.
Sinceq=0, thus x=expσρσ, that is, Φx=eσ. From (40), we have σ⟶+∞ and ρσ⟶ρ as n⟶+∞. Hence, from (43), and combining with ε is arbitrarily, we can deduce(44)limsupn⟶+∞ΦλnAn∗−1/λnρ≤T∗eρ.Case 4.
Suppose thatp=1 and q=1, then ϱ=1. From (39), it follows(45)logAn∗≤Bσρσ−λnσ+log2.
Takeσ such that(46)σρσ−1=λnBρ.
From (46), it follows σ=Φλn/Bρ. Thus, we can conclude from (45) that(47)logAn∗≤−λnσρ−1ρ1+o1=−λnρ−1ρ1+o1ΦλnBρ,that is,(48)Φλn−1/λnlogAn∗≤Φλnρ−1/ρ1+o1Φλn/Bρ.
From (46), it yields σ⟶+∞ and ρσ⟶ρ as n⟶+∞. Hence, from (48), and combining with ε is arbitrarily, we conclude(49)lim supn⟶+∞Φλn−1/λnlogAn∗ρ−1≤T∗ρρρ−1ρ−1.Case 5.
Suppose thatp≥q≥2, then ϱ=0. Take σ such that(50)logq−1σρσ=1Blogp−1λnρ,that is,(51)logq−1σ=Φ1Blogp−1λnρ.
Substituting (50) into (38), it yields(52)logAn∗≤−λnσ−1ρ+log2=−λnσ−1ρ1+o1.
Thus, it follows(53)−1λnlogAn∗≥σ−1ρ1+o1=expq−1Φ1Blogp−1λnρ−1ρ1+o1≥expq−1Φ1Blogp−1λnρ1+o1,that is,(54)logq−1−1λnlogAn∗≥Φ1Blogp−1λnρ1+o1.
In view of (54), we can deduce(55)Φlogp−1λnlogq−1−1/λnlogAn∗≤Φlogp−1λnΦ1/Blogp−1λn/ρ1+o1.
Sincelogp−1λn/ρ∼logp−1λn as n⟶+∞ and p>1, by Lemma 1, let ε⟶0 and n⟶+∞ in (56), and it yields(56)lim supn⟶+∞Φlogp−1λnlogq−1−1/λnlogAn∗ρ≤T∗.
Therefore, we can conclude from (44), (46), and (56) that(57)lim supn⟶+∞HΦlogp−1λnlogq−1−1/λnlogAn∗ρ≤T∗,where H is stated as in Theorem 2.
Step Two. Next, we will prove that(58)lim supn⟶+∞HΦlogp−1λnlogq−1−1/λnlogAn∗ρ<T∗,cannot hold. From (58), let B′ to be a real constant such that B′<T∗ and(59)lim supn⟶+∞HΦlogp−1λnlogq−1−1/λnlogAn∗ρ=B′<T∗.
Thus, for any smallε such that 0<ε<1/3T∗−B′ and k0 such that k>k0, it follows(60)Φlogp−1λnk≤B′+εH1/ρ−ϱlogq−1−1λnklogAnk∗.
Here, three cases will be considered as follows.Case 6.
Suppose thatp=1 and q=0, then ϱ=0 and H=1/eρ, and (60) becomes(61)Φλnk≤eρB′+ε1/ρexp−1λnklogAnk∗.
Thus, it follows(62)Ank∗≤exp−λnklogeρB′+ε−1/ρΦλnk,that is,(63)logAnk∗eλnkσ≤supk≥0−λnklogeρB′+ε−1/ρΦλnk+λnkσ.
Takeσ such that(64)σ=logΦλnkρB′+ε,that is,(65)eσρσ=λnkρB′+ε.
Then, in view of Lemma2 and (63), we can deduce(66)logAnk∗eλnkσ≤supk≥0λnkσ−λnklogeρB′+ε−1/ρρB′+ε1/ρ1+o1eσ≤supk≥0λnkρ1+o1≤B′+ε1+o1eσρσ≤B′+2εeσρσ.
Thus, from Lemma2 and (66) and since B′+3ε<T∗, it means(67)T∗=lim supσ⟶+∞logMuσ,Feσρσ≤B′+3ε<T∗,a contradiction.Case 7.
Suppose thatp=1 and q=1, then ϱ=1 and H=ρ−1ρ−1/ρρ. From (60), it follows(68)logAnk∗eλnkσ≤supk≥0−λnkHB′+ε1/ρ−1Φλnk+λnkσ.
Takeσ such that(69)σ=ΦλnkρB′+ε,that is,(70)σρσ−1=λnkρB′+ε.
Then, in view of Lemma1 and (68), it follows(71)logAnk∗eλnkσ≤supk≥0λnkσ−λnkHρ1/ρ−1σ1+o1≤supk≥0λnkσρ1+o1≤supk≥0B′+ε1+o1σρσ.
Thus, from Lemma2 and (71) and since B′+3ε<T∗, we conclude(72)T∗=limsupσ⟶+∞logMuσ,Fσρσ≤B′+3ε<T∗,a contradiction.Case 8.
Suppose thatp≥q≥2, then ϱ=0 and H=1. Then, it follows from (60) that(73)logAnk∗eλnkσ≤supk≥0λnkσ−λnkexpq−11B′+ε1/ρΦlogp−1λnk.
Takeσ such that(74)logq−1σ−1=Φlogp−1λnkB′+ε,that is,(75)λnk=expp−1B′+εlogq−1σ−1ρσ.
Then, from Lemma1 and (73), it yields(76)logAnk∗eλnkσ≤supk≥0λnkσ−λnkσ−11+o1≤supk≥0λnk1+o1≤supk≥0expp−1B′+εlogq−1σ−1ρσ1+o1.
Thus, from Lemma2 and (76) and since B′+3ε<T∗ and q≥2, we conclude(77)T∗=lim supσ⟶+∞logpMuσ,Flogq−1σρσ≤B′+3ε<T∗,a contradiction.
Therefore, the necessity of Theorem2 is proved from Steps one and two.
In fact, by using the same argument as in the processing of the necessity of Theorem2, we can easily prove the sufficiency of Theorem 2. Hence, this completes the proof of Theorem 2.


## 3. The Approximation of LaplaceStieltjes Transforms

We denoteL¯β to be a class of all the functions Fs of form (9) which are analytic in the half plane ℜs<β−∞<β<∞ and the sequence λn satisfy (3) and (6), and denote L∞ to be the class of all the functions Fs of form (9) which are analytic in the half plane ℜs<+∞ and the sequence λn satisfy (3), (6), and (11). Thus, if −∞<β<+∞ and Fs∈L¯β, then Fs∈L∞. If the LaplaceStieltjes transform (11), An∗=0 for n≥k+1, and Ak∗≠0, then Fs will be called an exponential polynomial of degree k usually denoted by pk, i.e., pks=∫0λkexpsydαy. When we choose a suitable function αy, the function pks may be reduced to a polynomial in terms of expsλi, that is, ∑i=1kbiexpsλi.ForFs∈L¯βand−∞<β<+∞, we denote by EnF,β the error in approximating the function Fs by exponential polynomials of degree n in uniform norm as(78)EnF,β=infp∈ΠnF−pβ,n=1,2,…,where(79)F−pβ=max−∞<t<+∞Fβ+it−pβ+it.In 2017, Singhal and Srivastava [28] studied the approximation of LaplaceStieltjes transforms of finite order and obtained the following theorem.Theorem 3 (see [28]).
If the LaplaceStieltjes transformFs∈L∞ and is of order ρ0<ρ<∞ and of type T, then for any real number −∞<β<+∞, we have(80)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.Recently, the second author investigated the approximation of LaplaceStieltjes transforms of infinite order by using the concepts ofX-order and the type function and obtained some theorems as follows.Theorem 4 (see [29]).
If the LaplaceStieltjes transformFs∈L0, then for any fixed real number −∞<β<0, we have(81)limsupσ⟶0−Xlog+Muσ,Flog−1/σ=ρ∗⇔limsupn⟶∞Xλnlog+λn/log+En−1F,βexp−βλn=ρ∗,where 0<ρ∗<∞.Theorem 5 (see [29]).
If the LaplaceStieltjes transformFs∈L0, then for any fixed real number −∞<β<0, we have(82)limsupσ⟶0−Xlog+Muσ,FlogU−1/σ=1⇔limsupn⟶∞XλnlogUλn/log+En−1F,βexp−βλn=1.Remark 8.
The definitions ofX-order and the type function Ux can be found in [30, 31].
The other purpose of this paper is to investigate the approximation of LaplaceStieltjes transforms with proximate orderρσ, the result concerning the error EnF,β, and the proximate order of such function is obtained. The main theorem is listed below.Theorem 6.
If the LaplaceStieltjes transformFs∈L∞ and is of (p, q) proximate order T∗0<T∗<∞, then for any real number −∞<β<+∞, we have(83)T∗=limsupn⟶+∞HΦlogp−1λnlogqEn−1F,β−1/λnρ−ϱ,ifp,q=1,1orq≥2;limsupn⟶+∞1ρexpβρ+1ΦλnEn−1F,β−1/λnρ,ifp,q=1,0.where ρσ, ϱ, and H are stated as in Theorem 2.Proof.
Suppose that(84)T∗≔T∗p,q=limsupσ⟶+∞logpMuσ,Flogq−1σρσ,then for any given ε>0and σ>σ0, we have(85)logMuσ,F≤expp−1Blogq−1σρσ,where B=T∗+ε. Since Fs∈L∞, thus for any constant β−∞<β<+∞, we have Fs∈L¯β. For β<σ<+∞, it follows from the definitions of EnF,β and pn that(86)EnF,β≤F−pnβ≤Fβ+it−pnβ+it≤∫0+∞expβ+itydαy−∫0λnexpβ+itydαy=∫λn∞expβ+itydαy.
Set(87)Ix;it=∫0xexpitydαy,Ij+kb;it=∫λj+kbexpitydαyλj+k<b≤λj+k+1.
Then, we haveIj+kb;it≤Aj+k∗. Since(88)∫λk∞expβ+itydαy=limb⟶+∞∫λkbexpβ+itydαy,
whereβ is an any real number, it follows(89)∫λkbexpβ+itydαy=∑j=kn+k−1∫λjλj+1expβydyIjy;it+∫λn+kbexpβydyIn+ky;it=∑j=kn+k−1eλj+1βIjλj+1;it−β∫λjλj+1eβyIjy;itdy+eβbIn+kb;it−β∫λn+kbeβyIjy;itdy≤∑j=kn+k−1Aj∗eλj+1β+Aj∗eλj+1β−eλjβ+2eβλn+k+1An+k∗−eβλn+kAn+k∗≤2∑j=kn+kAn∗eλn+1β.
Whenn⟶+∞, we have b⟶+∞. It yields that(90)∫λk∞expβ+itydαy≤2∑n=k+∞An∗expβλn+1.
Thus, from the definition ofAn∗ and Muσ,F and by Lemma 2, we have An∗≤2Muσ,Fe−σλn for any σβ<σ<+∞, it follows from (86) and (90) that(91)EnF,β≤2∑k=n+1∞Ak−1∗expβλk≤4Muσ,F∑k=n+1∞expβ−σλk.
From (6), take h′0<h′<h such that λn+1−λn≥h′ for n≥0.Then, for σ≥β+1, it follows from (91) that(92)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,(93)En−1F,β≤KMuσ,Fexpλnβ−σ,where K is a constant. Thus, from (85) and (93), it yields that(94)logEn−1F,βexp−βλn≤expp−1Blogq−1σρσ−λnσ+logK,where ε>0and σ>σ0.Next, three cases will be considered.Case 9.
Suppose thatp=1 and q=0, then ϱ=0 and H=1/eρ. From (92), it follows(95)logEn−1F,βexp−βλn≤Beσρσ−λnσ+logK.
Takingσ such that(96)expσ=ΦλnBρ,by using the same argument as in Case 3, we can easily get(97)limsupn⟶+∞ΦλnEn−1F,βexp−βλn−1/λnρ≤T∗eρ,that is,(98)limsupn⟶+∞1ρexpβρ+1ΦλnEn−1F,β−1/λnρ≤T∗.
Now, we need to prove that(99)limsupn⟶+∞1ρexpβρ+1ΦλnEn−1F,β−1/λnρ<T∗,does not hold. Let B′ to be a real constant such that B′<T∗ and(100)limsupn⟶+∞1ρexpβρ+1ΦλnEn−1F,β−1/λnρ=B′<T∗.
Thus, for any smallε such that 0<ε<1/3T∗−B′ and k0 such that k>k0, it follows(101)En−1F,βexp−βλn≤exp−λnklogeρB′+ε−1/ρΦλnk.
Since for anyβ<+∞, from the definition of EkF,β, there exists p1∈Πn−1 such that(102)F−p1≤2En−1F,β.
And since(103)An∗expβλn=supλn<x≤λn+1,−∞<t<+∞∫λnxexpitydαyexpβλn≤supλn<x≤λn+1,−∞<t<+∞∫λnxexpβ+itydαy≤sup−∞<t<+∞∫λn∞expβ+itydαy,thus for any p∈Πn−1, it means that(104)An∗expβλn≤Fβ+it−pβ+it≤F−pβ.
Hence, from (103) and (104), for any β<+∞ and Fs∈L∞, we can deduce(105)An∗expβλn≤2En−1F,β.
From (101) and (105), it follows(106)logAnk∗eλnkσ≤supk≥0−λnklogeρB′+ε−1/ρΦλnk+λnkσ+log2.
Takingσk such that(107)σk=logΦλnkρB′+ε,by using the same argument as in Case 6, we have(108)T∗=limsupσ⟶+∞logMuσ,Feσρσ≤B′+3ε<T∗,a contradiction.Case 10.
Suppose thatp=1and q=1; then, ϱ=1and H=ρ−1ρ−1/ρρ. Then, it follows from (101) that(109)logEn−1F,βexp−βλn≤Bσρσ−λnσ+logK.
Takingσ such that(110)σ=ΦλnBρ,by using the same argument as in Case 3, we can easily get(111)limsupn⟶+∞HΦλn−1/λnlogEn−1F,βexp−βλnρ−1≤T∗.
If(112)limsupn⟶+∞HΦλn−1/λnlogEn−1F,βexp−βλnρ−1<T∗,by combining the proof of Case 7 and Case 9, a contradiction can be easily obtained.Case 11.
Suppose thatp≥q≥2, then ϱ=0 and H=1. Take σ such that(113)logq−1σ=Φ1Blogp−1λnρ.
By using the same argument as in Case5, we can conclude from (94) that(114)limsupn⟶+∞Φlogp−1λnlogq−1−1/λnlogEn−1F,βexp−βλnρ≤T∗.
If(115)limsupn⟶+∞Φlogp−1λnlogq−1−1/λnlogEn−1F,βexp−βλnρ<T∗,by combining the proof of Case 8 and Case 9, a contradiction can be easily obtained.
Therefore, this completes the proof of Theorem6.



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*Source: 9624131-2020-02-13.xml*