# The Growth of Entire Functions Defined by Laplace–Stieltjes Transforms Related to Proximate Order and Approximation

Authors: Wen Ju Tang; Jian Chen; Hong Yan Xu
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

In this article, we discuss the growth of entire functions represented by Laplace–Stieltjes transform converges on the whole complex plane and obtain some equivalence conditions about proximate growth of Laplace–Stieltjes transforms with finite order and infinite order. In addition, we also investigate the approximation of Laplace–Stieltjes transform with the proximate order and obtain some results containing the proximate growth order, the error,An∗, and λn, which are the extension and improvement of the previous theorems given by Luo and Kong and Singhal and Srivastava.


---

## Body

## 1. Introduction

Our main aim of this paper is to investigate some problems about the growth and approximation of entire functions represented by Laplace–Stieltjes transforms which converge on the whole complex plane. Consider Laplace–Stieltjes transforms(1)Fs=∫0+∞esxdαx,s=σ+it,σ,t∈R,where αx is a bounded variation on any finite interval 0,Y0<Y<+∞. For Laplace–Stieltjes transform (1), set(2)An∗=supλn<x≤λn+1,−∞<t<+∞∫λnxeitydαy,where the sequence λn0∞ satisfies(3)0≤λ1<λ2<⋯<λn<⋯,λn⟶∞asn⟶∞if(4)limsupn⟶+∞logAn∗λn=−∞,(5)limsupn⟶+∞lognλn=D<+∞,limsupn⟶+∞λn+1−λn=h<+∞,similar to the method in [1], and by using the Valiron–Knopp–Bohr formula, then it yields σuF=+∞, that is, Fs is an entire function in the whole plane. We denote L¯β to be a class of all the functions Fs of form (1) which are analytic in the half plane ℜs<β−∞<β<∞, and the sequence λn satisfies (3) and (5), and denote L∞ to be the class of all the functions Fs of form (1) which are analytic in the whole plane ℜs<+∞, and the sequence λn satisfies (3), (5), and (4). Thus, if −∞<β<+∞ and Fs∈L∞, then Fs∈L¯β.By Widder [2], if αt is absolutely continuous, then Fs is the classical Laplace integral form:(6)Fs=∫0∞estgtdtif αt is a step-function and a sequence λn0∞ satisfies (3), and(7)αx=a1+a2+⋯+an,λn≤x<λn+1;0,x=0,then Fs becomes a Dirichlet series:(8)Fs=∑n=1∞aneλns,s=σ+it,σ,t∈R,where 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 Fs which cannot be expressed either in form (6) or (8) (see [2]).In 1963, Yu [1] first proved the Valiron–Knopp–Bohr formula of the associated abscissas of bounded convergence, absolute convergence, and uniform convergence of Laplace–Stieltjes transform when s=−z.Theorem 1.
If the sequenceλn satisfies (3) and (5), then Laplace–Stieltjes transform (1) satisfies(9)limsupn⟶+∞logBn∗λn≤σuF≤limsupn⟶+∞logBn∗λn+limsupn⟶+∞lognλn,where σuF is called the abscissa of uniformly convergence of Fs, and(10)Bn∗=supλn<x≤λn+1,−∞<t<+∞∫λnxe−itydαy.In addition, Yu [1] introduced the concept of the order of Gz=F−s to estimate the growth of the maximal molecule Muσ,G and the maximal term μσ,G and gave the relation between the Borel line and the order of entire functions represented by Laplace–Stieltjes transform converge in the whole complex plane. After his wonderful works, considerable attention has been paid to the growth, and the value distribution of the functions represented by Laplace–Stieltjes transform or Dirichlet series converges in the half plane or the whole complex plane in the field of complex analysis (see [3–24]).Set(11)μσ,F=maxn∈NAn∗eλnσσ<+∞,Muσ,F=sup0<x<+∞,−∞<t<+∞∫0xeσ+itydαy.In view ofMuσ,F⟶+∞ as σ⟶+∞, in order to estimate the growth of Fs precisely, we usual use the concepts of order and type as follows.Definition 1.
IfFs∈L∞ and(12)limsupσ⟶+∞log+log+Muσ,Fσ=ρ,0≤ρ≤+∞,then it is said that 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(13)limsupσ⟶+∞log+Muσ,Feσρ=T,0≤T≤+∞.
LetFs be of order ρ in the whole plane and 0<ρ<+∞, and let ρσσ>σ0 be a nonnegative, continuous, and monotonous function, and it has a left-hand derivative and right-hand derivative in every σ>σ0 such that(14)limσ⟶+∞ρσ=ρ,limσ⟶+∞σρ′σ=0,then ρσ is called the proximate order; furthermore, if(15)limsupσ⟶+∞log+Muσ,Feσρσ=T,then T will be called the proximate type of Fs with respect to proximate order ρσ.
Setr=eσ, and let t=ψr=rρlogr=eσρσ and r=φt be two reciprocally inverse functions. Then, we obtain the following theorem.Theorem 2.
If Laplace–Stieltjes transformFs∈L∞ and is of proximate order ρσ and order ρ0<ρ<+∞, then(16)τ≤T≤eDρτ,where(17)τ=limsupn⟶+∞1ρeφλnAn∗1/λnρ.
Furthermore, if the sequenceλn satisfies(18)liminfn⟶+∞λn+1−λn=h′>0,then(19)T=τ=limsupn⟶+∞1ρeφλnAn∗1/λnρ.In order to estimate the growth of Laplace–Stieltjes transform of orderρ=+∞ precisely, we will use some concepts of the p-order as follows.Definition 2.
IfFs∈L∞, p≥2 is a positive integer, and(20)limsupσ⟶+∞logp+1Muσ,Fσ=ρp,0≤ρp≤+∞,then we call Fs is of p-order ρp in the whole plane, where logp+1x=logplogx for p∈ℕ+.
LetFs be of p-order ρp in the whole plane and 0<ρp<+∞, similar to the proximate type of Fs with respect to proximate order ρσ; we can give the proximate type of Fs with respect to proximate order of p-order ρp as follows. Let ρpσσ>σ0 be a nonnegative, continuous, and monotonous function, and it has a left-hand derivative and right-hand derivative in every σ>σ0 such that(21)limσ⟶+∞ρpσ=ρp,limσ⟶+∞σρp′σ=0,then ρpσ is called the p-proximate order; furthermore, if(22)limsupσ⟶+∞logpMuσ,Feσρpσ=Tp,then Tp will be called the proximate type of Fs with respect to p-proximate order ρpσ.
Setr=eσ, and let t=ψpr=rρplogr=eσρpσ and r=φpt be two reciprocally inverse functions. Then, we obtain the following theorem.Theorem 3.
If Laplace–Stieltjes transformFs∈L∞ and is of p-proximate order ρpσ and ρp0<ρp<+∞, p≥2, then(23)τp≤Tp≤eDρpτp,where(24)τp=limsupn⟶+∞φplogp−1λnAn∗1/λnρp.
Furthermore, if the sequenceλn satisfies (18), then(25)Tp=τp=limsupn⟶+∞φplogp−1λnAn∗1/λnρp.The other purpose of this article is to study the approximation of the entire function represented by Laplace–Stieltjes transform converges in the whole plane. When Laplace–Stieltjes transform (1) satisfies Ak∗≠0 and An∗=0 for n≥k+1, 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. We also use Πk to denote the class of all exponential polynomial of degree almost k, that is,(26)Πk=∑i=1kbiexpsλi:b1,b2,…,bk∈ℂk.ForFs∈L∞,−∞<β<+∞, we denote by EnF,β the error in approximating the function Fs by exponential polynomials of degree n in uniform norm as(27)EnF,β=infp∈ΠnF−pβ,n=1,2,…,where(28)F−pβ=max−∞<t<+∞Fβ+it−pβ+it.Recently, Singhal and Srivastava [15] and Xu and Liu [25] studied the approximation on Laplace–Stieltjes transforms of finite order and obtained the following theorems.Theorem 4 (see [15]).
If Laplace–Stieltjes transformFs∈L∞ and is of order ρ0<ρ<∞ and of type T, then for any real number −∞<β<+∞, we have(29)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.Theorem 5 (see [25], Theorem 2.5).
If the Laplace–Stieltjes transformFs∈L∞ and is of the lower order λ0≤λ≠ρ<∞ and if λn∼λn+1, then for any real number −∞<β<+∞, we have(30)τλ≥liminfn⟶∞λneλEn−1F,βexp−βλnλ/λn,0≤τλ≤∞.Furthermore, there exists a positive integern0 such that(31)ξn=logAn∗−logAn+1∗λn+1−λnforms a nondecreasing function of n for n>n0, and then we have(32)τλ=liminfn⟶∞λneλEn−1F,βexp−βλnλ/λn0≤τλ≤∞,i.e.,(33)expβλeλτλ=liminfn⟶∞λnEn−1F,βλ/λn.Corresponding to Theorems2 and 3, we obtain the following results concerning the approximation.Theorem 6.
If Laplace–Stieltjes transformFs∈L∞ and is of proximate order ρσ and order ρ0<ρ<+∞, then for any real number −∞<β<+∞, we have(34)τa≤T≤eDρτa,where(35)τa=limsupn⟶+∞1ρexpβρ+1φλnEn−1F,β1/λnρ.
Furthermore, if the sequenceλn satisfies (18), then(36)T=τa=limsupn⟶+∞1ρexpβρ+1φλnEn−1F,β1/λnρ.Theorem 7.
If Laplace–Stieltjes transformFs∈L∞ and is of p-proximate order ρpσ and ρp0<ρp<+∞, p≥2, then for any real number −∞<β<+∞, we have(37)τpa≤Tp≤eDρpτpa,where(38)τpa=limsupn⟶+∞1expβρpφplogp−1λnEn−1F,β1/λnρp.
Furthermore, if the sequenceλn satisfies (18), then(39)Tp=τpa=limsupn⟶+∞1expβρpφplogp−1λnEn−1F,β1/λnρp.


## 2. Some Lemmas

Lemma 1.
For the functionsψ,φ, and r=eσ, we have(40)limσ⟶+∞rψ′rψr=ρ,limt⟶+∞tφ′tφt=1ρ,(41)limσ⟶+∞ψkrψr=kρ,limt⟶+∞φktφt=k1/ρ,k>0.Proof.
Sincetφ′t/φt=ψr/rψ′r and(42)limσ⟶+∞rψ′rψr=limσ⟶+∞eσeσρσρσ+σρ′σe−σeσρσ=ρ,then (40) follows. Moreover,(43)limσ⟶+∞ψkrψr=limσ⟶+∞kρσψrψr=kρ.
Next, we prove the last equality. Sinceφt=eσ and t=eσρσ, thus, it follows σ⟶+∞ as t⟶+∞ and kt=keσρσ, and φkt=k1/ρσeσ. Since limσ⟶+∞ρσ=ρ, then it yields(44)limt⟶+∞φktφt=limσ⟶+∞k1/ρσeσeσ=κ1/ρ.
Therefore, this completes the proof of Lemma1.Remark 1.
Obviously, the functionsψp and φp have the same conclusions as ψ and φ in Lemma 1.Lemma 2.
Leta,b>0; then, the function(45)hr=aψr−blogrcan obtain the minimum(46)b1ρ1+o1−logφbaρ1+o1inr=φbaρ1+o1.Proof.
Let(47)h′r=aψ′r−br=1rarψ′r−b=0.
Then, it meansarψ′r−b=0. In view of the first equality in (40), we deduce b=aρ1+o1ψr, that is,(48)r=φbaρ1+o1,asb⟶+∞.
We can see that, as the value ofr increases to the given value above, the value of hr changes from a negative value to a positive value. Thus, hr can obtain minimum(49)b1ρ1+o1−logφbaρ1+o1,inr=φbaρ1+o1.
Therefore, this completes the proof of Lemma2.Lemma 3.
Leta>0; then, the function(50)gt=at−tlogφtcan obtain the maximum(51)1ρ1+o1ψeae1+o1/ρ,int=ψeae1+o1/ρa⟶+∞.Proof.
Let(52)g′t=a−logφt−tφ′tφt=0.
Then, it follows from (40) that a−logφt−1/ρ1+o1=0. In view of the first equality in (40), we deduce φt=ea/e1+o1/ρ, that is,(53)t=ψeae1+o1/ρ,asa⟶+∞.
It is easy to see thatgt obtain the maximum(54)ψeae1+o1/ρa−a+1ρ1+o1=1ρ1+o1ψeae1+o1/ρ,a⟶+∞,int=ψeae1+o1/ρ.
Therefore, this completes the proof of Lemma3.Lemma 4.
If Laplace–Stieltjes transformFs∈L∞, for any σ0<σ<+∞ and ε>0, we have(55)12μσ,F≤Muσ,F≤CeKσμσ+D+ε,F,where C and K are constants.Proof.
We will adapt the method as in Yu [1] and Kong and Hong [7]. Set(56)Ix;σ+it=∫0xexpσ+itydαy.
Forx>λn, it follows(57)∫λnxexpitydαy=∫λnxexp−σydyIy;σ+it=Iy;σ+itexp−σyλnx+σ∫λnxexp−σyIy;σ+itdy.
Then, for anyσ>0 and any x∈λn,λn+1, it yields(58)∫λnxexpitydαy≤2Muσ,Fexp−σλn−exp−xσ≤2Muσ,Fexp−λnσ,which implies(59)12μσ,F≤Muσ,F.
On the contrary, for anyx>0, it follows that there exists a positive integer n∈ℕ+ such that λn<x≤λn+1. Thus, it follows(60)∫0xexpσ+itydαy=∑k=1n−1∫λkλk+1expσ+itydαy+∫λnxexpσ+itydαy.
SetIkx;it=∫λnxexpitydαy,λk<x≤λk+1; then, for any real number t and σ>0, it follows Ikx;it≤Ak∗≤μσ,Fe−λkσ. Thus, for any ε∈0,1 and σ>0, it yields Ikx;it≤μσ+D+ε,Fe−λkσ+D+ε and(61)∫0xexpσ+itydαy=∑k=1n−1expλk+1σIkλk+1;it−σ∫λkλk+1expσyIky;itdy+expxσInx;it−σ∫λnxexpσyIny;itdy.
In view of (5), there exists a constant K such that λn+1≤λn+K for n≥1. So, we can deduce(62)∫0xexpσ+itydαy≤∑k=1n−1μσ+D+ε,Fexp−λkσ+D+εexpλk+1σ+expλk+1σ−expλkσ+μσ+D+ε,Fexp−λnσ+D+εexpxσ+expxσ−expλnσ≤2μσ+D+ε,FeKσ∑k=1nexp−λkD+ε=2μσ+D+ε,FeKσ∑k=1nexp−λkD+12εD+εD+1/2ε.
In view of (5), for the above ε and sufficiently large n, we have logn<λnD+1/2ε,(63)exp−λkD+12εD+εD+1/2ε<1nD+ε/D+1/2ε,and ∑k=1+∞1/nD+ε/D+1/2ε is convergent. Therefore, this completes the proof of Lemma 4.


## 3. Proofs of Theorems2 and 3

### 3.1. The Proof of Theorem2

We firstly proveτ≤T. Assume that T<+∞. From(64)T=limsupσ⟶+∞log+Muσ,Feσρσ,then for any positive number ε and sufficiently large σ, it yields(65)log+Muσ,F<eσρσT+ε=ψrT+ε.By Lemmas2 and 4, it follows(66)logAn∗<ψrT+ε−λnlogr+log2≤λn1ρ1+o1−logφλnT+ερ1+o1,n⟶∞.Hence, in view of Lemma1, we can deduce(67)An∗1/λn≤exp1/ρ1+o1φλn/T+ερ1+o1=exp1/ρ1+o1ρT+ε1+o11/ρφλn,n⟶∞,that is,(68)φλnAn∗1/λnρ≤exp1+o1ρT+ε1+o1,n⟶∞.Sinceε is arbitrary, we conclude from the above inequality that(69)τ=limsupn⟶+∞1ρeφλnAn∗1/λnρ≤T,and τ<+∞ as T<+∞.Next, we proveT≤eDρτ. Assume that 0≤τ<+∞; then, for any ε>0, there exists a constant K1 such that(70)An∗<K1ρeτ+ε1/ρφλnλn,n≥1.Thus, from Lemma3, Lemma 4, and (70), it yields(71)logMuσ,F≤logC+logK1+Kσ+maxnlogeσ+D+ερeτ+ε1/ρφλnλnlogMuσ,F≤logC+logK1+Kσ+maxnlogeσ+D+ερeτ+ε1/ρφλnλn≤logC+logK1+Kσ+1ρ1+o1ψeD+ε−ρ−11+o1ρeτ+ε1/ρeσ=logC+logK1+Kσ+1ρ1+o1eρD+ε+o1ρτ+εψeσ=logC+logK1+Kσ+eρD+ε+o1τ+ε+o1ψeσ,σ⟶+∞,which implies(72)limsupσ⟶+∞logMuσ,Fψeσ≤eρD+ετ+ε.Sinceε is arbitrary, let ε⟶0 in (72); it follows that T≤eDρτ and T<+∞ as τ<+∞.Therefore, (16) is proved.Furthermore, if the sequenceλn satisfies (18), then for any ε∈0,h′, there exists a positive integer n0 such that λn+1−λn>h′−ε for n≥n0. Thus, for any positive integer k, it follows(73)λk+n0−λn0=λk+n0−λk−1+n0+⋯+λ1+n0−λn0>kh′−ε.Hence,(74)k+n0λk+n0<λk+n0−λn0λk+n01h′−ε+n0λk+n0.Sinceε is arbitrary, then in view of (74), we conclude(75)D=limsupn⟶+∞lognλn≤limsupn⟶+∞lognnlimsupn⟶+∞nλn≤limsupn⟶+∞lognnlimsupk⟶+∞k+n0λk+n0≤1h′limsupn⟶+∞lognn,that is, D=0. Thus, in view of (16), it follows T=τ=limsupn⟶+∞1/ρeφλnAn∗1/λnρ.Therefore, this completes the proof of Theorem2.


### 3.2. The Proof of Theorem3

We firstly proveτp≤Tp. Assume that Tp<+∞. From(76)Tp=limsupσ⟶+∞logpMuσ,Feσρσ,then for any positive number ε and sufficiently large σ, it yields(77)logpMuσ,F<eσρpσTp+ε=ψprTp+ε,where r=eσ. By Lemmas 2 and 4, for sufficiently large σ, it follows(78)logAn∗≤logMuσ,F−λnσ+log2<expp−1ψprTp+ε−λnlogr+log2,where exp0x=x,exppx=expexpp−1x,p≥1. In view of Lemma 2, it is easy to see that expp−1ψprTp+ε−λnlogr+log2 can attain the minimum when r satisfies the following equation:(79)rψp′r∏j=1p−1expjTp+εψpr=λnTp+ε.By Lemma1, we have rψp′r=ψprρp1+o1 as σ⟶+∞. Thus, it follows from (79) that(80)∏j=1p−1expjTp+εψpr=λnρp1+o1,n⟶+∞.Hence, the functionexpp−1ψprTp+ε−λnlogr+log2 can obtain the minimum(81)−1+o1λnlogφp1Tp+εlogp−1λnρp1+o1,n⟶+∞,as r satisfies equation (80). From p≥2 and Lemma 1, it yields(82)φp1Tp+εlogp−1λnρp1+o1=φp1+o1Tp+εlogp−1λn=1+o1Tp+ε1/ρpφplogp−1λn,n⟶+∞.So, we can deduce from (78), (81), and (82) that(83)logAn∗≤−1+o1λnlogφp1Tp+εlogp−1λnρp1+o1≤−1+o1λnlog1+o1Tp+ε1/ρpφplogp−1λn,n⟶+∞,that is,(84)φplogp−1λnAn∗1/λnρp≤1+o1Tp+ε,n⟶∞.Sinceε is arbitrary, we conclude from the above inequality that(85)τp=limsupn⟶+∞φplogp−1λnAn∗1/λnρp≤Tp,and τp<+∞ as Tp<+∞.Next, we proveTp≤eDρpτp. Assume that 0≤τp<+∞; then, for any ε>0, there exists a constant K2 such that(86)An∗<K2τp+ε1/ρpφplogp−1λnλn,n≥1.Thus, from Lemma3, Lemma 4, and (86), it yields(87)logMuσ,F≤logC+logK2+Kσ+maxnlogeσ+D+ετp+ε1/ρpφplogp−1λnλn.By using the same argument as in Lemma3, we can conclude that the function(88)logeσ+D+ετp+ε1/ρpφplogp−1λnλncan attain the maximum(89)1+o1ρp∏j=0p−2logjλnexpp−1ψp1+o1expσ+D+ετp+ε1/ρp,n⟶+∞,as λn satisfies(90)logexpσ+D+εTp+ε1/ρp−logφplogp−1λn−1+o1ρp∏j=0p−2logjλn=0,n⟶+∞.Thus, by Lemma1, we can deduce from (87) and (89) that(91)limsupσ⟶+∞logpMuσ,Fψpeσ≤eρpD+ετp+ε.Sinceε is arbitrary, let ε⟶0 in (89); it follows that Tp≤eDρpτp and Tp<+∞ as τp<+∞. Hence, we prove (23).Furthermore, if the sequenceλn satisfies (18), then we see D=0 from the proof of Theorem 2. Thus, it is easy to prove that Tp=τp from (23).Therefore, this completes the proof of Theorem3.


## 3.1. The Proof of Theorem2

We firstly proveτ≤T. Assume that T<+∞. From(64)T=limsupσ⟶+∞log+Muσ,Feσρσ,then for any positive number ε and sufficiently large σ, it yields(65)log+Muσ,F<eσρσT+ε=ψrT+ε.By Lemmas2 and 4, it follows(66)logAn∗<ψrT+ε−λnlogr+log2≤λn1ρ1+o1−logφλnT+ερ1+o1,n⟶∞.Hence, in view of Lemma1, we can deduce(67)An∗1/λn≤exp1/ρ1+o1φλn/T+ερ1+o1=exp1/ρ1+o1ρT+ε1+o11/ρφλn,n⟶∞,that is,(68)φλnAn∗1/λnρ≤exp1+o1ρT+ε1+o1,n⟶∞.Sinceε is arbitrary, we conclude from the above inequality that(69)τ=limsupn⟶+∞1ρeφλnAn∗1/λnρ≤T,and τ<+∞ as T<+∞.Next, we proveT≤eDρτ. Assume that 0≤τ<+∞; then, for any ε>0, there exists a constant K1 such that(70)An∗<K1ρeτ+ε1/ρφλnλn,n≥1.Thus, from Lemma3, Lemma 4, and (70), it yields(71)logMuσ,F≤logC+logK1+Kσ+maxnlogeσ+D+ερeτ+ε1/ρφλnλnlogMuσ,F≤logC+logK1+Kσ+maxnlogeσ+D+ερeτ+ε1/ρφλnλn≤logC+logK1+Kσ+1ρ1+o1ψeD+ε−ρ−11+o1ρeτ+ε1/ρeσ=logC+logK1+Kσ+1ρ1+o1eρD+ε+o1ρτ+εψeσ=logC+logK1+Kσ+eρD+ε+o1τ+ε+o1ψeσ,σ⟶+∞,which implies(72)limsupσ⟶+∞logMuσ,Fψeσ≤eρD+ετ+ε.Sinceε is arbitrary, let ε⟶0 in (72); it follows that T≤eDρτ and T<+∞ as τ<+∞.Therefore, (16) is proved.Furthermore, if the sequenceλn satisfies (18), then for any ε∈0,h′, there exists a positive integer n0 such that λn+1−λn>h′−ε for n≥n0. Thus, for any positive integer k, it follows(73)λk+n0−λn0=λk+n0−λk−1+n0+⋯+λ1+n0−λn0>kh′−ε.Hence,(74)k+n0λk+n0<λk+n0−λn0λk+n01h′−ε+n0λk+n0.Sinceε is arbitrary, then in view of (74), we conclude(75)D=limsupn⟶+∞lognλn≤limsupn⟶+∞lognnlimsupn⟶+∞nλn≤limsupn⟶+∞lognnlimsupk⟶+∞k+n0λk+n0≤1h′limsupn⟶+∞lognn,that is, D=0. Thus, in view of (16), it follows T=τ=limsupn⟶+∞1/ρeφλnAn∗1/λnρ.Therefore, this completes the proof of Theorem2.


## 3.2. The Proof of Theorem3

We firstly proveτp≤Tp. Assume that Tp<+∞. From(76)Tp=limsupσ⟶+∞logpMuσ,Feσρσ,then for any positive number ε and sufficiently large σ, it yields(77)logpMuσ,F<eσρpσTp+ε=ψprTp+ε,where r=eσ. By Lemmas 2 and 4, for sufficiently large σ, it follows(78)logAn∗≤logMuσ,F−λnσ+log2<expp−1ψprTp+ε−λnlogr+log2,where exp0x=x,exppx=expexpp−1x,p≥1. In view of Lemma 2, it is easy to see that expp−1ψprTp+ε−λnlogr+log2 can attain the minimum when r satisfies the following equation:(79)rψp′r∏j=1p−1expjTp+εψpr=λnTp+ε.By Lemma1, we have rψp′r=ψprρp1+o1 as σ⟶+∞. Thus, it follows from (79) that(80)∏j=1p−1expjTp+εψpr=λnρp1+o1,n⟶+∞.Hence, the functionexpp−1ψprTp+ε−λnlogr+log2 can obtain the minimum(81)−1+o1λnlogφp1Tp+εlogp−1λnρp1+o1,n⟶+∞,as r satisfies equation (80). From p≥2 and Lemma 1, it yields(82)φp1Tp+εlogp−1λnρp1+o1=φp1+o1Tp+εlogp−1λn=1+o1Tp+ε1/ρpφplogp−1λn,n⟶+∞.So, we can deduce from (78), (81), and (82) that(83)logAn∗≤−1+o1λnlogφp1Tp+εlogp−1λnρp1+o1≤−1+o1λnlog1+o1Tp+ε1/ρpφplogp−1λn,n⟶+∞,that is,(84)φplogp−1λnAn∗1/λnρp≤1+o1Tp+ε,n⟶∞.Sinceε is arbitrary, we conclude from the above inequality that(85)τp=limsupn⟶+∞φplogp−1λnAn∗1/λnρp≤Tp,and τp<+∞ as Tp<+∞.Next, we proveTp≤eDρpτp. Assume that 0≤τp<+∞; then, for any ε>0, there exists a constant K2 such that(86)An∗<K2τp+ε1/ρpφplogp−1λnλn,n≥1.Thus, from Lemma3, Lemma 4, and (86), it yields(87)logMuσ,F≤logC+logK2+Kσ+maxnlogeσ+D+ετp+ε1/ρpφplogp−1λnλn.By using the same argument as in Lemma3, we can conclude that the function(88)logeσ+D+ετp+ε1/ρpφplogp−1λnλncan attain the maximum(89)1+o1ρp∏j=0p−2logjλnexpp−1ψp1+o1expσ+D+ετp+ε1/ρp,n⟶+∞,as λn satisfies(90)logexpσ+D+εTp+ε1/ρp−logφplogp−1λn−1+o1ρp∏j=0p−2logjλn=0,n⟶+∞.Thus, by Lemma1, we can deduce from (87) and (89) that(91)limsupσ⟶+∞logpMuσ,Fψpeσ≤eρpD+ετp+ε.Sinceε is arbitrary, let ε⟶0 in (89); it follows that Tp≤eDρpτp and Tp<+∞ as τp<+∞. Hence, we prove (23).Furthermore, if the sequenceλn satisfies (18), then we see D=0 from the proof of Theorem 2. Thus, it is easy to prove that Tp=τp from (23).Therefore, this completes the proof of Theorem3.


## 4. Proofs of Theorems6 and 7

### 4.1. The Proof of Theorem6

Firstly of all, we proveτa≤T. Assume that T<+∞. By using the same argument as in the proof of Theorem 2, for any positive number ε and sufficiently large σ, we have (64). Since Fs∈L∞, then we have Fs∈L¯β for any β−∞<β<+∞. Hence, from the definitions of EnF,β and pn, it follows(92)EnF,β≤F−pnβ≤Fβ+it−pnβ+it≤∫0+∞expβ+itydαy−∫0λnexpβ+itydαy=∫λn∞expβ+itydαy,β<σ<+∞.Let(93)Jx;it=∫0xexpitydαy,Jj+kb;it=∫λj+kbexpitydαy,λj+k<b≤λj+k+1.Then,Jj+kb;it≤Aj+k∗. Since(94)∫λk∞expβ+itydαy=limb⟶+∞∫λkbexpβ+itydαy,−∞<β<+∞,then we can deduce(95)∫λkbexpβ+itydαy=∑j=kn+k−1∫λjλj+1expβydyJjy;it+∫λn+kbexpβydyJn+ky;it=∑j=kn+k−1eλj+1βJjλj+1;it−β∫λjλj+1eβyJjy;itdy+eβbJn+kb;it−β∫λn+kbeβyJjy;itdy≤∑j=kn+k−1Aj∗eλj+1β+Aj∗eλj+1β−eλjβ.Hence, from the above inequalities, we conclude(96)∫λk∞expβ+itydαy≤2∑n=k+∞An∗expβλn+1,asn⟶+∞.Thus, for anyσβ<σ<+∞, from (92) and (96) and by combining An∗≤2Muσ,Fe−σλn, it yields(97)EnF,β≤2∑k=n+1∞Ak−1∗expβλk≤4Muσ,F∑k=n+1∞expβ−σλk.ChoosingH0<H<h such that λn+1−λn≥H for n≥0, thus, for σ≥β+1, we can deduce from (97) that(98)EnF,β≤4Muσ,Fexpλn+1β−σ∑k=n+1∞expλk−λn+1β−σ≤4Muσ,Fexpλn+1β−σexpHn+1∑k=n+1∞exp−Hk=4Muσ,Fexpλn+1β−σ1−expH−1,that is,(99)En−1F,β≤K3Muσ,Fexpλnβ−σ,where K3 is a constant.Thus, it follows from (65) and (99) that(100)logEn−1F,βexp−βλn<ψrT+ε−λnlogr+logK3.In view of (100) and using the same argument as in the proof of Theorem 2, we conclude from the above inequality that(101)τa=limsupn⟶+∞1ρexpβρ+1φλnEn−1F,β1/λnρ≤T,and τa<+∞ as T<+∞.Next, we proveT≤eDρτa. Assume that 0≤τa<+∞; then, for any ε>0, there exists a constant K4 such that(102)En−1F,βexp−βλn<K4ρeτa+ε1/ρφλnλn,n≥1.And since(103)An∗expβλn=supλn<x≤λn+1,−∞<t<+∞∫λnxexpitydαyexpβλn≤supλn<x≤λn+1,−∞<t<+∞∫λnxexpβ+itydαy≤sup−∞<t<+∞∫λn∞expβ+itydαy,thus, it means(104)An∗expβλn≤Fβ+it−pβ+it≤F−pβ,for any p∈Πn−1. In addition, there exists p1∈Πn−1 such that(105)F−p1≤2En−1F,β.Thus, for anyβ<+∞ and Fs∈L∞, from (105) and (104), it yields(106)An∗expβλn≤2En−1F,β.Hence, in view of (102) and (106) and by using the same argument as in the proof of Theorem 2, we conclude(107)limsupσ⟶+∞logMuσ,Fψeσ≤eρD+ετa+ε,which implies T≤eDρτa and T<+∞ as τa<+∞. Hence, (34) holds.Furthermore, if the sequenceλn satisfies (18), from the proof of Theorem 2, it follows D=0, which means T=τa.Therefore, this completes the proof of Theorem6.


### 4.2. The Proof of Theorem7

By using the same argument as in the proof of Theorem3 and combining (99), (102), and (106), we can prove the conclusions of Theorem 7 easily.


## 4.1. The Proof of Theorem6

Firstly of all, we proveτa≤T. Assume that T<+∞. By using the same argument as in the proof of Theorem 2, for any positive number ε and sufficiently large σ, we have (64). Since Fs∈L∞, then we have Fs∈L¯β for any β−∞<β<+∞. Hence, from the definitions of EnF,β and pn, it follows(92)EnF,β≤F−pnβ≤Fβ+it−pnβ+it≤∫0+∞expβ+itydαy−∫0λnexpβ+itydαy=∫λn∞expβ+itydαy,β<σ<+∞.Let(93)Jx;it=∫0xexpitydαy,Jj+kb;it=∫λj+kbexpitydαy,λj+k<b≤λj+k+1.Then,Jj+kb;it≤Aj+k∗. Since(94)∫λk∞expβ+itydαy=limb⟶+∞∫λkbexpβ+itydαy,−∞<β<+∞,then we can deduce(95)∫λkbexpβ+itydαy=∑j=kn+k−1∫λjλj+1expβydyJjy;it+∫λn+kbexpβydyJn+ky;it=∑j=kn+k−1eλj+1βJjλj+1;it−β∫λjλj+1eβyJjy;itdy+eβbJn+kb;it−β∫λn+kbeβyJjy;itdy≤∑j=kn+k−1Aj∗eλj+1β+Aj∗eλj+1β−eλjβ.Hence, from the above inequalities, we conclude(96)∫λk∞expβ+itydαy≤2∑n=k+∞An∗expβλn+1,asn⟶+∞.Thus, for anyσβ<σ<+∞, from (92) and (96) and by combining An∗≤2Muσ,Fe−σλn, it yields(97)EnF,β≤2∑k=n+1∞Ak−1∗expβλk≤4Muσ,F∑k=n+1∞expβ−σλk.ChoosingH0<H<h such that λn+1−λn≥H for n≥0, thus, for σ≥β+1, we can deduce from (97) that(98)EnF,β≤4Muσ,Fexpλn+1β−σ∑k=n+1∞expλk−λn+1β−σ≤4Muσ,Fexpλn+1β−σexpHn+1∑k=n+1∞exp−Hk=4Muσ,Fexpλn+1β−σ1−expH−1,that is,(99)En−1F,β≤K3Muσ,Fexpλnβ−σ,where K3 is a constant.Thus, it follows from (65) and (99) that(100)logEn−1F,βexp−βλn<ψrT+ε−λnlogr+logK3.In view of (100) and using the same argument as in the proof of Theorem 2, we conclude from the above inequality that(101)τa=limsupn⟶+∞1ρexpβρ+1φλnEn−1F,β1/λnρ≤T,and τa<+∞ as T<+∞.Next, we proveT≤eDρτa. Assume that 0≤τa<+∞; then, for any ε>0, there exists a constant K4 such that(102)En−1F,βexp−βλn<K4ρeτa+ε1/ρφλnλn,n≥1.And since(103)An∗expβλn=supλn<x≤λn+1,−∞<t<+∞∫λnxexpitydαyexpβλn≤supλn<x≤λn+1,−∞<t<+∞∫λnxexpβ+itydαy≤sup−∞<t<+∞∫λn∞expβ+itydαy,thus, it means(104)An∗expβλn≤Fβ+it−pβ+it≤F−pβ,for any p∈Πn−1. In addition, there exists p1∈Πn−1 such that(105)F−p1≤2En−1F,β.Thus, for anyβ<+∞ and Fs∈L∞, from (105) and (104), it yields(106)An∗expβλn≤2En−1F,β.Hence, in view of (102) and (106) and by using the same argument as in the proof of Theorem 2, we conclude(107)limsupσ⟶+∞logMuσ,Fψeσ≤eρD+ετa+ε,which implies T≤eDρτa and T<+∞ as τa<+∞. Hence, (34) holds.Furthermore, if the sequenceλn satisfies (18), from the proof of Theorem 2, it follows D=0, which means T=τa.Therefore, this completes the proof of Theorem6.


## 4.2. The Proof of Theorem7

By using the same argument as in the proof of Theorem3 and combining (99), (102), and (106), we can prove the conclusions of Theorem 7 easily.



---

*Source: 5809767-2020-04-09.xml*