Some New Inequalities on Laplace–Stieltjes Transforms Involving Logarithmic Growth

Abstract

This article is devoted to exploring the properties on the logarithmic growth of entire functions represented by Laplace–Stieltjes transforms of zero order. In order to describe the growth of Laplace–Stieltjes transforms more finely, we introduce some concepts of the logarithmic indexes of the maximum term and the center index of the maximum term of Laplace–Stieltjes transforms, and establish some new inequalities focusing on the above logarithmic indexes, the logarithmic order, the (lower) logarithmic type and the coefficients of Laplace–Stieltjes transforms. Moreover, we obtain two estimation forms on the (lower) logarithmic type of entire functions represented by Laplace–Stieltjes transform by applying these inequalities. One estimation is mainly by the center indexes of the maximum term, the other is by the logarithmic order, exponent and coefficients. Finally, we obtain the equivalence condition of entire functions with the perfectly logarithmic linear growth. This result shows that the two estimation forms can be equivalent to some extent.

1. Introduction and Some Basic Notations

As we all know, the following transform

$$F\left(s\right)={\int }_0^{+\infty }{e}^{sx}d\alpha \left(x\right),s=\sigma +it,$$

(1)

is usually called a Laplace–Stieltjes transform, if $\alpha \left(x\right)$ is a bounded variation on any finite interval $[0,Y](0<Y<+\infty )$, and $\sigma$ and t are real variables. Laplace–Stieltjes transform was first named after Pierre-Simon Laplace and Thomas Joannes Stieltjes, and is also an integral transform similar to the Laplace transform. Over the past 80 years or so, it has been used in many fields of mathematics, such as functional analysis, and certain areas of theoretical and applied probability.

Yu [1] in 1963 first studied the growth and convergence of Laplace–Stieltjes transforms (1) and gave the famous Valiron–Knopp–Bohr formula of the associated abscissas of bounded convergence, absolute convergence and uniform convergence of Laplace–Stieltjes transforms, and the Borel lines of entire functions represented by Laplace–Stieltjes transforms. After his wonderful results, many mathematicians had paid considerable attention focusing on the growth and the value distribution of analytic functions defined by Laplace–Stieltjes transforms convergent in the half-plane and whole complex plane, and obtained a series of classic and important results. For example, L. N. Shang, Z. S. Gao, Z. X. Xuan, etc. further investigated the value distributions of analytic functions of some kinds of growth defined by Laplace–Stieltjes transforms, and obtained some results about the singular direction and points of Laplace–Stieltjes transforms (see [2,3,4,5]); C. Singhal, G. S. Srivastava, Y. Y. Kong, S. Y. Liu and H. Y. Xu studied the properties on the approximation of entire functions represented by Laplace-stieltjes transforms, and obtained some interesting theorems on the relationship between the error and growth (see [6,7,8,9]); O. Posiko and M. M. Sheremeta [10] in 2007 explored the relationships between the growth and the maximum term of Laplace–Stieltjes transform ${\int }_0^{\infty }f\left(x\right)e^{x\sigma }dF\left(x\right)$, where $f\left(x\right)\ge 0$, M. S. Dobushovskyi, M. M. Sheremeta [11,12] in 2017 and 2021, respectively, further analyzed the convergence and relative growth of such transform; Y. J. Bi and Y. Y. Huo [13] recently considered the growth of the double Laplace–Stieltjes transforms, and obtained some foundation growth theorems; Y. Y. Kong and his co-authors studied the growth of analytic functions defined by Laplace–Stieltjes transforms which converge in the half plane and the whole plane, and gave a great number of important theorems concerning the zero order, the generalized order, the finite and infinite order, and so on (see [14,15,16,17,18,19,20,21]).

In order to study the growth of Laplace–Stieltjes transform (1), we usually take a sequence $\left\{{\lambda }_n\right\}$ satisfying

$$0\le {\lambda }_1<{\lambda }_2<\cdots <{\lambda }_n<\cdots ,{\lambda }_n\to \infty asn\to \infty ,$$

(2)

and

$$\underset{n\to +\infty }{lim\; sup}({\lambda }_{n+1}-{\lambda }_n)<+\infty .$$

(3)

And denote

$$A_n^{*}=\underset{{\lambda }_n<x\le {\lambda }_{n+1},-\infty <t<+\infty }{sup}\left|{\int }_{{\lambda }_n}^x{e}^{ity}d\alpha \left(y\right)\right|,$$

if Laplace–Stieltjes transform (1) satisfies

$$\underset{n\to +\infty }{lim\; sup}\frac{logn}{{\lambda }_n}=D<\infty ,\underset{n\to +\infty }{lim\; sup}\frac{log{A}_n^{*}}{{\lambda }_n}=-\infty ,$$

(4)

then in view of Refs. [1,22,23], we can conclude that ${\sigma }_u^F=+\infty$, i.e., $F\left(s\right)$ is analytic on the whole plane. For convenience, let $L_{\infty }$ to denote the class of all the functions $F\left(s\right)$ of the form (1) which are analytic in the half plane $\Re s<+\infty$ and the sequence $\left\{{\lambda }_n\right\}$ satisfy (2)–(4).

Let

$$M_u(\sigma ,F)=\underset{0<x<+\infty ,-\infty <t<+\infty }{sup}\left|{\int }_0^x{e}^{(\sigma +it)y}d\alpha \left(y\right)\right|,$$

and

$$\mu (\sigma ,F)=\underset{n\in \mathbb{N}}{max}\left\{A_n^{*}{e}^{{\lambda }_n\sigma }\right\}(\sigma <+\infty ),\nu (\sigma ,F)=\underset{n}{max}\left\{{\lambda }_n\right|\mu (\sigma ,F)=A_n^{*}{e}^{{\lambda }_n\sigma }\}.$$

Usually, we utilize the order and the type to estimate the growth of $F\left(s\right)$, which are defined as follows.

Definition 1

(see [19]). If $F\left(s\right)\in L_{\infty }$ and

$$\underset{\sigma \to +\infty }{lim\; sup}\frac{{log}^{+}{log}^{+}{M}_u(\sigma ,F)}{\sigma }=\rho ,$$

we call $F\left(s\right)$ is of order ρ in the whole plane; if

$$\underset{\sigma \to +\infty }{lim\; inf}\frac{{log}^{+}{log}^{+}{M}_u(\sigma ,F)}{\sigma }=\tau ,$$

we call $F\left(s\right)$ is of lower order τ in the whole plane, where ${log}^{+}x=max\{logx,0\}$.

Definition 2

(see [19]). If $F\left(s\right)\in L_{\infty }$, and is of order $\rho (0<\rho <\infty )$, then we define the type and the lower type of Laplace–Stieltjes transform $F\left(s\right)$, respectively,

$$\underset{\sigma \to +\infty }{lim\; sup}\frac{{log}^{+}{M}_u(\sigma ,F)}{e^{\sigma \rho }}=T,\underset{\sigma \to +\infty }{lim\; inf}\frac{{log}^{+}{M}_u(\sigma ,F)}{e^{\sigma \rho }}=t.$$

For $0<\rho \left(F\right)<\infty$, Luo and Kong [19] in 2012 discussed the properties on entire functions represented by a Laplace–Stieltjes transform of finite order, and obtained.

Theorem 1

(see [19,20]). If $F\left(s\right)\in L_{\infty }$, and is of order $\rho (0<\rho <\infty )$ and of type T, then

$$\rho =\underset{n\to +\infty }{lim\; sup}\frac{{\lambda }_{n}log{\lambda }_n}{-log{A}_n^{*}},T=\underset{n\to +\infty }{lim\; sup}\frac{{\lambda }_n}{\rho e}{\left(A_n^{*}\right)}^{\frac{\rho }{{\lambda }_n}};$$

Furthermore, if ${\lambda }_n\sim {\lambda }_{n-1}$ and

$$\psi \left(n\right)=\frac{log{A}_n^{*}-log{A}_{n+1}^{*}}{{\lambda }_{n+1}-{\lambda }_n}$$

form a non-decreasing function of n, then

$$t=\underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_n}{\rho e}{\left(A_n^{*}\right)}^{\frac{\rho }{{\lambda }_n}}.$$

Remark 1.

For $\rho \left(F\right)=0$, we can see that the (lower) order and the (lower) type cannot better characterize the growth of the maximum module $M_u(\sigma ,F)$ of (1).

In view of Remark 1, Xu and Liu [9] in 2019 investigated the growth of Laplace–Stieltjes transforms for the case $\rho \left(F\right)=0$, by using the concepts of the logarithmic order and the logarithmic type below.

Definition 3

(see [9]). If $F\left(s\right)\in L_{\infty }$, and is of order $\rho =0$, and

$$\underset{\sigma \to +\infty }{lim\; sup}\frac{{log}^{+}{log}^{+}{M}_u(\sigma ,F)}{log\sigma }={\rho }_l,1\le {\rho }_l\le +\infty ,$$

then ${\rho }_l$ is called the logarithmic order of $F\left(s\right)$ of zero order. Furthermore, if $1\le {\rho }_l<+\infty$, we define the logarithmic type $T_l$ and the lower logarithmic type $t_l$ of $F\left(s\right)$, respectively,

$$\underset{\sigma \to +\infty }{lim\; sup}\frac{{log}^{+}{M}_u(\sigma ,F)}{{\sigma }^{{\rho }_l}}=T_l,\underset{\sigma \to +\infty }{lim\; inf}\frac{{log}^{+}{M}_u(\sigma ,F)}{{\sigma }^{{\rho }_l}}=t_l.$$

Remark 2.

We say that $F\left(s\right)$ is of perfectly logarithmic linear growth if and only if $0<t_l=T_l<\infty$ and $1<{\rho }_l<\infty$. Obviously, $T_l=\infty$ as ${\rho }_l=1$.

Theorem 2

(see ([9], Theorem 1.5)). If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of zero order and of logarithmic order ${\rho }_l$, then

$${\rho }_l=\underset{\sigma \to +\infty }{lim\; sup}\frac{{log}^{+}{log}^{+}{M}_u(\sigma ,F)}{log\sigma }=\underset{\sigma \to +\infty }{lim\; sup}\frac{{log}^{+}{log}^{+}\mu (\sigma ,F)}{log\sigma }.$$

(5)

Furthermore, we have

$${\rho }_l=\frac{1}{1-\underset{n\to +\infty }{lim\; sup}\frac{log{\lambda }_n}{loglog{\left(A_n^{*}\right)}^{-1}}}.$$

(6)

Theorem 3

(see ([9], Theorem 1.6)). If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of zero order and of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$ and logarithmic type $T_l$, then

$$T_l=\frac{{({\rho }_l-1)}^{{\rho }_l-1}}{{\rho }_l^{{\rho }_l}}\underset{n\to +\infty }{lim\; sup}\frac{log\frac{1}{A_n^{*}}}{{\left(\frac{1}{{\lambda }_n}log\frac{1}{A_n^{*}}\right)}^{{\rho }_l}}.$$

Remark 3.

In fact, in view of Theorem 3 and Lemma 1, we have

$$\underset{\sigma \to +\infty }{lim\; sup}\frac{{log}^{+}\mu (\sigma ,F)}{{\sigma }^{{\rho }_l}}=T_l,\underset{\sigma \to +\infty }{lim\; inf}\frac{{log}^{+}\mu (\sigma ,F)}{{\sigma }^{{\rho }_l}}=t_l,$$

and

$$\underset{n\to +\infty }{lim\; sup}\frac{{\lambda }_n}{{\rho }_l{\left[\frac{1}{{\rho }_l-1}log{\left(A_n^{*}\right)}^{-\frac{{\rho }_l}{{\lambda }_n}}\right]}^{{\rho }_l-1}}=T_l.$$

Motivated by Theorems 2 and 3, one may ask the following questions.

Question 1.

What will happened to the parameters ${\rho }_l,{\lambda }_n,A_n^{*}$, if $F\left(s\right)$ is of the lower logarithmic type $t_l$, or $F\left(s\right)$ is of perfectly logarithmic linear growth?

Question 2.

What can be said about the correlation between the logarithmic growth and the center index $\nu (\sigma ,F)$ of the maximum term $\mu (\sigma ,F)$ of Laplace–Stieltjes transform with zero order?

In view of the above questions, we will study the properties of logarithmic growth of entire functions defined by Laplace–Stieltjes transforms convergent in the whole plane, including the lower logarithmic type $t_l$, and the relations about the logarithmic type $T_l$, the lower logarithmic type $t_l$, $\nu (\sigma ,F)$, ${\lambda }_n$ and $A_n^{*}$. As far as we know, it appears that the study of the logarithmic growth of Laplace–Stieltjes transforms has seldom been involved in the literature before now. The paper is organized as follows. In Section 2, we will discuss the lower logarithmic type $t_l$ of entire functions defined by Laplace–Stieltjes transforms. In Section 3, we will study the relation among the logarithmic order ${\rho }_l$, logarithmic type $T_l$, lower logarithmic type $t_l$ and the center index $\nu (\sigma ,F)$ of the maximum term. In Section 4, we will establish the expression of the (lower) logarithmic type by the logarithmic order ${\rho }_l,{\lambda }_n,A_n^{*}$, and also obtain some equivalence conditions between the (lower) logarithmic type $T_l\left(t_l\right)$ and $\nu (\sigma ,F)$. Finally, the conclusions of this paper will be presented in Section 5.

2. The Lower Logarithmic Type of Laplace–Stieltjes Transform

We first give the following lemma, which is used to prove our two main theorems.

Lemma 1

(see [19], Lemma 2.1)). If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, for any $\sigma (-\infty <\sigma <+\infty )$ and $\epsilon (>0)$, we have

$$\frac{1}{2}\mu (\sigma ,F)\le M_u(\sigma ,F)\le C\mu ((1+2\epsilon )\sigma ,F),$$

where C is a constant.

In fact, we obtain the main result about the lower logarithmic type of Laplace–Stieltjes transform $F\left(s\right)$ in the case $\rho \left(F\right)=0$ as follows.

Theorem 4.

If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$, and of lower logarithmic type $t_l$, and if ${\lambda }_n\sim {\lambda }_{n+1}$ and the function

$$\psi \left(n\right)=\frac{log{A}_n^{*}-log{A}_{n+1}^{*}}{{\lambda }_{n+1}-{\lambda }_n}$$

(7)

is a non-decreasing function of n, then

$$\underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_n}{{\rho }_l{\left[\frac{1}{{\rho }_l-1}log{\left(A_n^{*}\right)}^{-\frac{{\rho }_l}{{\lambda }_n}}\right]}^{{\rho }_l-1}}=t_l.$$

Remark 4.

Obviously, Theorem 4 is a good supplement of Theorems 2 and 3.

In order to prove Theorem 4, we only give the proof of Theorems 5 and 6 below.

Theorem 5.

If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$ and lower logarithmic type $t_l$, and if ${\lambda }_n\sim {\lambda }_{n+1}$, then

$$\underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_n}{{\rho }_l{\left[\frac{1}{{\rho }_l-1}log{\left(A_n^{*}\right)}^{-\frac{{\rho }_l}{{\lambda }_n}}\right]}^{{\rho }_l-1}}\le t_l.$$

Proof. 

Set

$$\theta =\underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_{n-1}}{{\rho }_l{\left[\frac{1}{{\rho }_l-1}log{\left(A_n^{*}\right)}^{-\frac{{\rho }_l}{{\lambda }_n}}\right]}^{{\rho }_l-1}}.$$

(8)

Assume that $0<\theta <\infty$, for any given $\epsilon$ such that $0<\epsilon <\theta$, we have from (8) that there exists a positive integer $n_0\left(\epsilon \right)$ such that for all $n>n_0\left(\epsilon \right)$,

$$log{A}_n^{*}>-\frac{{\rho }_l-1}{{\rho }_l}{\left[\frac{{\lambda }_{n-1}}{{\rho }_l(\theta -\epsilon )}\right]}^{\frac{1}{{\rho }_l-1}}{\lambda }_n.$$

(9)

Thus, it follows by Lemma 1 and (9) that

$$log{M}_u(\sigma ,F)>-\frac{{\rho }_l-1}{{\rho }_l}{\left[\frac{{\lambda }_{n-1}}{{\rho }_l(\theta -\epsilon )}\right]}^{\frac{1}{{\rho }_l-1}}{\lambda }_n+{\lambda }_n\sigma -log2.$$

(10)

Taking $\sigma ={\left[\frac{{\lambda }_{n-1}}{{\rho }_l(\theta -\epsilon )}\right]}^{\frac{1}{{\rho }_l-1}}$, we have from (10) that

$$log{M}_u(\sigma ,F)>(1+o\left(1\right))(\theta -\epsilon ){\sigma }^{{\rho }_l},forn>n_0\left(\epsilon \right).$$

(11)

In view of (11), and combining the definition of lower logarithmic type $t_l$, we have $t_l\ge \theta$. If $\theta =0$, the conclusion holds obviously. In the case $\theta =\infty$, similar to the above argument, we can also obtain the inequality when we replace $\theta -\epsilon$ by an arbitrarily large number.

Therefore, this completes the proof of Theorem 5. □

Theorem 6.

If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$ and lower logarithmic type $t_l$, and if the function (7) form a non-decreasing function of n, then

$$\underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_n}{{\rho }_l{\left[\frac{1}{{\rho }_l-1}log{\left(A_n^{*}\right)}^{-\frac{{\rho }_l}{{\lambda }_n}}\right]}^{{\rho }_l-1}}\ge t_l.$$

Proof. 

Assume that $0<t_l<+\infty$. From the assumption of Theorem 6, and in view of Definition 3 and Lemma 1, for any given small number $\epsilon (0<\epsilon <t_l)$, there exists a fixed ${\sigma }_0>0$ such that for all $\sigma >{\sigma }_0$,

$$log\mu (\sigma ,F)>(t_l-\epsilon ){\sigma }^{{\rho }_l},$$

that is,

$$log{A}_n^{*}+{\lambda }_n\sigma >(t_l-\epsilon ){\sigma }^{{\rho }_l}.$$

(12)

Let $\sigma >{\sigma }_0$ and let $n_1$ and $n_2(n_2-1)$ be two consecutive maximum terms, then

$$log{A}_{n_2}^{*}+{\lambda }_{n_2}\sigma >(t_l-\epsilon ){\sigma }^{{\rho }_l}$$

(13)

for all $\sigma$ satisfying $\psi (n_2-1)\le \sigma <\psi \left(n_2\right)$. Let $n_1\le n<n_2-1$, we have

$$\psi \left(n_1\right)=\psi (n_1+1)=\cdots =\psi \left(n\right)=\cdots =\psi (n_2-1),$$

(14)

and

$$A_n^{*}{e}^{{\lambda }_n\sigma }=A_{n_2}^{*}{e}^{{\lambda }_{n_2}\sigma },for\sigma =\psi \left(n\right).$$

(15)

Thus, it follows from (12)–(15) that

$$\frac{{\lambda }_n}{{\rho }_l{\left[\frac{1}{{\rho }_l-1}log{\left(A_n^{*}\right)}^{-\frac{{\rho }_l}{{\lambda }_n}}\right]}^{{\rho }_l-1}}>\frac{{\lambda }_n}{\frac{{\rho }_l^{{\rho }_l}}{{({\rho }_l-1)}^{{\rho }_l-1}}{\left[-\frac{1}{{\lambda }_n}(t_l-\epsilon ){\sigma }^{{\rho }_l}+\sigma \right]}^{{\rho }_l-1}}.$$

(16)

Let $\sigma ={\left(\frac{{\lambda }_n}{{\rho }_l(t_l-\epsilon )}\right)}^{{({\rho }_l-1)}^{-1}}$, and let $n\to +\infty$, it follows from (16) that

$$\underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_n}{{\rho }_l{\left[\frac{1}{{\rho }_l-1}log{\left(A_n^{*}\right)}^{-\frac{{\rho }_l}{{\lambda }_n}}\right]}^{{\rho }_l-1}}\ge t_l.$$

Besides, the conclusion holds obviously if $t_l=0$. By using the same argument as in the above, we can also prove the inequality in the case $t_l=\infty$ when we replace $t_l-\epsilon$ by an arbitrarily large number.

Therefore, this completes the proof of Theorem 6. □

3. Some Inequalities on the Maximum Term Index

In order to further explore the properties of logarithmic growth of Laplace–Stieltjes transform $F\left(s\right)$, we first introduce the following indicators. Let $F\left(s\right)\in L_{\infty }$ be of logarithmic order ${\rho }_l$. Here and below, unless otherwise specified, we always assume $1<{\rho }_l<+\infty$. Thus, we define

$$V=\underset{\sigma \to +\infty }{lim\; sup}\frac{\nu (\sigma ,F)}{{\sigma }^{{\rho }_l-1}},v=\underset{\sigma \to +\infty }{lim\; inf}\frac{\nu (\sigma ,F)}{{\sigma }^{{\rho }_l-1}},$$

and

$$H=\underset{\sigma \to +\infty }{lim\; sup}\frac{log\mu (\sigma ,F)}{\sigma \nu (\sigma ,F)},h=\underset{\sigma \to +\infty }{lim\; inf}\frac{log\mu (\sigma ,F)}{\sigma \nu (\sigma ,F)}.$$

Obviously, we have $v\le V$ and $h\le H$. As for the further relationship between them, we have

Theorem 7.

If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$, logarithmic type $T_l$ and lower logarithmic type $t_l$. Then we have

$$v\le {\rho }_l{t}_l\le v\left[{\rho }_l-({\rho }_l-1){\left(\frac{v}{V}\right)}^{\frac{1}{{\rho }_l-1}}\right]\le V,$$

(17)

and

$$v\le V{\left[\frac{{\rho }_{l}V-V}{{\rho }_{l}V-v}\right]}^{{\rho }_l-1}\le {\rho }_l{T}_l\le V.$$

(18)

Remark 5.

In view of $\frac{{\rho }_{l}V-V}{{\rho }_{l}V-v}\ge 1$, by combining with (17) and (18), we have

$$v\le {\rho }_l{t}_l\le {\rho }_l{T}_l\le V.$$

To prove this result, we require the following lemma.

Lemma 2

(see [20], Lemma 2.2)). If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, then we have

$$log\mu (\sigma ,F)=log\mu ({\sigma }_0,F)+{\int }_{{\sigma }_0}^{\sigma }\nu (t,F)dt,$$

for ${\sigma }_0>0$.

Proof of Theorem 7.

In view of $v\le V$ and ${\rho }_l>1$, it follows that $0\le \frac{v}{V}\le 1$ and $\frac{{\rho }_{l}V-V}{{\rho }_{l}V-v}\ge 1$. Thus,

$$v\le V{\left[\frac{{\rho }_{l}V-V}{{\rho }_{l}V-v}\right]}^{{\rho }_l-1}$$

(19)

holds obviously. Define $f\left(x\right)={\rho }_{l}x-({\rho }_l-1)x^{\frac{{\rho }_l}{{\rho }_l-1}}-1$, $0\le x\le 1,{\rho }_l>1$. Since $f^{\prime }\left(x\right)={\rho }_l\left(1-x^{\frac{1}{{\rho }_l-1}}\right)\ge 0$, then $f\left(x\right)$ is a increasing function in $0\le {\rho }_l\le 1$. Thus, $f\left(x\right)\le f\left(1\right)=0$. Replaced x by $\frac{v}{V}$, we can easily prove that

$$v\left[{\rho }_l-({\rho }_l-1){\left(\frac{v}{V}\right)}^{\frac{1}{{\rho }_l-1}}\right]\le V.$$

(20)

In view of the definitions of v and V, we have that for any $\epsilon >0$

$$(v-\epsilon ){\sigma }^{{\rho }_l-1}<\nu (\sigma ,F)<(V+\epsilon ){\sigma }^{{\rho }_l-1},for\sigma >{\sigma }_0\left(\epsilon \right).$$

(21)

By Lemma 2, for any $\sigma \ge {\sigma }_0>0$ and $\eta \ge 1$, it follows that

$$\begin{array}{cc}\hfill log\mu (\sigma {\eta }^{\frac{1}{{\rho }_l}},F)& =log\mu ({\sigma }_0,F)+{\int }_{{\sigma }_0}^{\sigma {\eta }^{\frac{1}{{\rho }_l}}}\nu (t,F)dt\hfill \\ \hfill & =O\left(1\right)+{\int }_{{\sigma }_0}^{\sigma }\nu (t,F)dt+{\int }_{\sigma }^{\sigma {\eta }^{\frac{1}{{\rho }_l}}}\nu (t,F)dt\hfill \end{array}$$

(22)

holds for any fixed positive number ${\sigma }_0>0$. Since $\nu (\sigma ,F)$ is an increasing function of $\sigma$, we have from (21) and (22) that

$$log\mu (\sigma {\eta }^{\frac{1}{{\rho }_l}},F)<O\left(1\right)+\frac{V+\epsilon }{{\rho }_l}{\sigma }^{{\rho }_l}+\nu (\sigma {\eta }^{\frac{1}{{\rho }_l}},F)({\eta }^{\frac{1}{{\rho }_l}}-1)\sigma ,$$

(23)

for all $\sigma >{\sigma }_0$. In view of Remark 1.2, and let $\sigma \to +\infty$, it follows from (23) that

$$\begin{array}{cc}\hfill & T_l\le \frac{V}{\eta {\rho }_l}+\frac{V({\eta }^{\frac{1}{{\rho }_l}}-1)}{{\eta }^{\frac{1}{{\rho }_l}}},\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill & t_l\le \frac{V}{\eta {\rho }_l}+\frac{v({\eta }^{\frac{1}{{\rho }_l}}-1)}{{\eta }^{\frac{1}{{\rho }_l}}}.\hfill \end{array}$$

(25)

Thus, let $\eta =1$ in (24), and let $\eta ={\left(\frac{V}{v}\right)}^{\frac{{\rho }_l}{{\rho }_l-1}}$ in (), we have

$${\rho }_l{T}_l\le V,{\rho }_l{t}_l\le v\left[{\rho }_l-({\rho }_l-1){\left(\frac{v}{V}\right)}^{\frac{1}{{\rho }_l-1}}\right].$$

(26)

By combining with the first inequality and (22), we also obtain that

$$\begin{array}{cc}\hfill & T_l\ge \frac{v}{\eta {\rho }_l}+\frac{V({\eta }^{\frac{1}{{\rho }_l}}-1)}{\eta },\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill & t_l\ge \frac{v}{\eta {\rho }_l}+\frac{v({\eta }^{\frac{1}{{\rho }_l}}-1)}{\eta }.\hfill \end{array}$$

(28)

Thus, let $\eta =1$ in (), and let $\eta ={\left(\frac{{\rho }_{l}V-V}{{\rho }_{l}V-v}\right)}^{{\rho }_l}$ in (27), we have

$${\rho }_l{t}_l\ge v,{\rho }_l{T}_l\ge V{\left(\frac{{\rho }_{l}V-V}{{\rho }_{l}V-v}\right)}^{{\rho }_l-1}.$$

(29)

By combining with (19), (20), (26) and (29), we can prove the conclusions of Theorem 7 easily.

Therefore, this completes the proof of Theorem 7. □

Next, the following results show the relations among the quotas $v,V,h$ and H.

Theorem 8.

If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$. Then we have

$$\frac{v}{{\rho }_{l}V}\le h\le H\le \frac{V}{{\rho }_{l}v}.$$

(30)

Proof. 

By making use of Lemma 2 and (21), and combining with the definitions of h and H, we can prove the conclusions of Theorem 8 easily. □

Theorem 9.

If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$, logarithmic type $T_l$ and lower logarithmic type $t_l$, $(0<t_l\le T_l<+\infty )$. Then we have

(i)

$$\underset{\sigma \to +\infty }{lim}\frac{log{M}_u(\sigma ,F)}{{\sigma }^{{\rho }_l}}=T_l⟺\underset{\sigma \to +\infty }{lim}\frac{\nu (\sigma ,F)}{{\sigma }^{{\rho }_l-1}}={\rho }_l{T}_l;$$

(ii)

$$0<t_l\le T_l<+\infty ⟺0<v\le V<+\infty .$$

Remark 6.

From Theorem 9 (i), we can see that $t_l=T_l⟺v=V={\rho }_l{T}_l$.

Proof. 

(i) $⟹:$ If

$$\underset{\sigma \to +\infty }{lim}\frac{log{M}_u(\sigma ,F)}{{\sigma }^{{\rho }_l}}=T_l,$$

in view of Lemma 1, it follows $t_l=T_l$ and

$$\underset{\sigma \to +\infty }{lim}\frac{log\mu (\sigma ,F)}{{\sigma }^{{\rho }_l}}=T_l.$$

By Lemma 2, for $\eta =1+\vartheta$, $\vartheta >0$, we have

$$\begin{array}{cc}\hfill T_l\left[{\left(\sigma {\eta }^{\frac{1}{{\rho }_l}}\right)}^{{\rho }_l}-{\sigma }^{{\rho }_l}+o\left({\sigma }^{{\rho }_l}\right)\right]& =log\mu \left(\sigma {\eta }^{\frac{1}{{\rho }_l}},F\right)-log\mu (\sigma ,F)\hfill \\ \hfill & ={\int }_{\sigma }^{\sigma {\eta }^{\frac{1}{{\rho }_l}}}\nu (t,F)dt<\nu \left(\sigma {\eta }^{\frac{1}{{\rho }_l}},F\right)\sigma ({\eta }^{\frac{1}{{\rho }_l}}-1).\hfill \end{array}$$

(31)

Dividing ${\left(\sigma {\eta }^{\frac{1}{{\rho }_l}}\right)}^{{\rho }_l-1}$ into two side of (31), and let $\sigma \to +\infty$, we have

$$\underset{\sigma \to +\infty }{lim\; inf}\frac{\nu \left(\sigma {\eta }^{\frac{1}{{\rho }_l}},F\right)}{{\left(\sigma {\eta }^{\frac{1}{{\rho }_l}}\right)}^{{\rho }_l-1}}\ge \frac{T_l(\eta -1)}{({\eta }^{\frac{1}{{\rho }_l}}-1){\eta }^{\frac{{\rho }_l-1}{{\rho }_l}}}=\frac{\vartheta T_l}{\left[(1+\vartheta )-{(1+\vartheta )}^{\frac{{\rho }_l-1}{{\rho }_l}}\right]}.$$

(32)

By applying L’Hospital’s rule, and let $\vartheta \to 0^{+}$, it is easy to obtain

$$\underset{\vartheta \to 0+}{lim}\frac{\vartheta T_l}{\left[(1+\vartheta )-{(1+\vartheta )}^{\frac{{\rho }_l-1}{{\rho }_l}}\right]}={\rho }_l{T}_l.$$

(33)

Thus, in view of (32) and (33), we have $v\ge {\rho }_l{T}_l$. Similarly, let $\eta =1-\vartheta ,0<\vartheta <1$, we have

$$\begin{array}{cc}\hfill T_l\left[{\sigma }^{{\rho }_l}-{\left(\sigma {\eta }^{\frac{1}{{\rho }_l}}\right)}^{{\rho }_l}+o\left({\sigma }^{{\rho }_l}\right)\right]& =log\mu (\sigma ,F)-log\mu \left(\sigma {\eta }^{\frac{1}{{\rho }_l}},F\right)\hfill \\ \hfill & ={\int }_{\sigma {\eta }^{\frac{1}{{\rho }_l}}}^{\sigma }\nu (t,F)dt>\nu \left(\sigma {\eta }^{\frac{1}{{\rho }_l}},F\right)\sigma (1-{\eta }^{\frac{1}{{\rho }_l}}),\hfill \end{array}$$

and

$$V=\underset{\sigma \to +\infty }{lim\; sup}\frac{\nu \left(\sigma {\eta }^{\frac{1}{{\rho }_l}},F\right)}{{\left(\sigma {\eta }^{\frac{1}{{\rho }_l}}\right)}^{{\rho }_l-1}}\le {\rho }_l{T}_l.$$

By combining with $v\le V$, we have

$$v=V=\underset{\sigma \to +\infty }{lim}\frac{\nu (\sigma ,F)}{{\sigma }^{{\rho }_l-1}}={\rho }_l{T}_l.$$

Now, we will prove the sufficiency of Theorem 9 (i). Let $\underset{\sigma \to +\infty }{lim}\frac{\nu (\sigma ,F)}{{\sigma }^{{\rho }_l-1}}={\rho }_l{T}_l$, in view of the definitions of v and V, we have $v=V={\rho }_l{T}_l$. By combining with Remark 5, we obtain that $t_l=T_l$, that is,

$$\underset{\sigma \to +\infty }{lim}\frac{log{M}_u(\sigma ,F)}{{\sigma }^{{\rho }_l}}=T_l.$$

Therefore, this completes the conclusion (i) of Theorem 9.

(ii) We first prove the sufficiency of Theorem 9 (ii). Let $0<v\le V<+\infty$. In view of Remark 5 and $0<{\rho }_l<+\infty$, it follows that $T_l<+\infty$ and $t_l>0$. Furthermore, in view of Theorem 7 (i), we can obtain that $t_l=T_l$ if $v=V$. Thus, the sufficiency of Theorem 9 (ii) is proved.

Next, we will prove the necessity of Theorem 9 (ii). Let $0<t_l\le T_l<+\infty$. Then it follows that $v>0$ and $V<+\infty$. Otherwise, if $v=0$, then we have from () that $T_l\ge \eta {\rho }_l{t}_l$. This is a contradiction since $T_l<+\infty$ and $\eta$ is arbitrary. Similarly, if $V=+\infty$, then we have from (27) that $T_l\ge \frac{V({\eta }^{\frac{1}{{\rho }_l}}-1)}{\eta }$. This is a contradiction since $T_l<+\infty$ and $\eta >1$. Besides, in view of Theorem 7 (i), we can obtain that $v=V$ if $t_l=T_l$. Thus, the necessity of Theorem 9 (ii) is proved.

Therefore, we complete the proof of Theorem 9. □

4. Applications

In this section, we will establish some results to reveal the relationship between the logarithm order ${\rho }_l$, the logarithm type $T_l$, the lower logarithm type $t_l$, the form exponent ${\lambda }_n$ and the form coefficients $A_n^{*}$ of Laplace–Stieltjes transformation of small growth, by applying the inequalities given in Section 1 and Section 2. Denote

$$w=\underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_n}{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}},W=\underset{n\to +\infty }{lim\; sup}\frac{{\lambda }_n}{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}}.$$

Theorem 10.

If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$, logarithmic type $T_l$ and lower logarithmic type $t_l$. If ${\lambda }_n\sim {\lambda }_{n+1}$ and

$$\sum _{m=n_0}^{n-1}{\lambda }_m^k({\lambda }_{m+1}-{\lambda }_m)\sim \frac{{\lambda }_n^{k+1}}{k+1},(k\ge 0,n_0\ge 1),$$

(34)

then we have

$$w\le {\rho }_l{t}_l\le {\rho }_l{T}_l\le W.$$

(35)

The following example shows that the inequalities in (35) are best possible to some extent.

Example 1.

Let ${\lambda }_n=n$ and $\alpha \left(x\right)$ satisfy

$$\alpha \left(x\right)=1+e^{-1}+e^{-2^2}+\cdots +e^{-{(n-1)}^2},(n-1<x<n,n=1,2,\dots ,).$$

Then (1) can be expressed as the form

$$F\left(s\right)=\sum _{n=1}^{\infty }{e}^{-n^2}{e}^{sn}.$$

In view of Theorems 2-4, by simple calculation, we have ${\rho }_l\left(F\right)=2$, $T_l\left(F\right)=t_l\left(F\right)=\frac{1}{4}$ and $w=W=\frac{1}{2}$. Thus, this shows that the equal sign situation in (35) can be attained.

Proof. 

Assume that $0<w\le W<+\infty$. From the definitions of w and W, for a fixed positive integer $n_0$, then we obtain that for any $\epsilon >0$, the following inequalities

$$\frac{1}{W}-\epsilon <\frac{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}}{{\lambda }_n}<\frac{1}{w}+\epsilon$$

(36)

hold for all $n\ge n_0$. Thus, for any positive integer $m\ge n_0$, we have

$${\left[(\frac{1}{W}-\epsilon ){\lambda }_m\right]}^{\frac{1}{{\rho }_l-1}}({\lambda }_{m+1}-{\lambda }_m)<log{A}_m^{*}-log{A}_{m+1}^{*}<{\left[(\frac{1}{w}+\epsilon ){\lambda }_m\right]}^{\frac{1}{{\rho }_l-1}}({\lambda }_{m+1}-{\lambda }_m).$$

(37)

Let $m=n_0,n_0+1,\dots ,n-1$ in (37), adding them, then it follows that

$$\begin{array}{cc}\hfill {(\frac{1}{W}-\epsilon )}^{\frac{1}{{\rho }_l-1}}\sum _{m=n_0}^{n-1}{\lambda }_m^{\frac{1}{{\rho }_l-1}}({\lambda }_{m+1}-{\lambda }_m)& <log{A}_{n_0}^{*}-log{A}_n^{*}\hfill \\ \hfill & <{(\frac{1}{w}+\epsilon )}^{\frac{1}{{\rho }_l-1}}\sum _{m=n_0}^{n-1}{\lambda }_m^{\frac{1}{{\rho }_l-1}}({\lambda }_{m+1}-{\lambda }_m).\hfill \end{array}$$

(38)

In view of (34) and (38), for all $n\ge n_0$, then we obtain that

$${(\frac{1}{W}-\epsilon )}^{\frac{1}{{\rho }_l-1}}\frac{{\lambda }_n^{\frac{1}{{\rho }_l-1}+1}}{\frac{{\rho }_l}{{\rho }_l-1}}<log{A}_{n_0}^{*}-log{A}_n^{*}<{(\frac{1}{w}+\epsilon )}^{\frac{1}{{\rho }_l-1}}\frac{{\lambda }_n^{\frac{1}{{\rho }_l-1}+1}}{\frac{{\rho }_l}{{\rho }_l-1}}.$$

(39)

Thus, it follows from (39) that

$$w\le \underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_n}{{\left[\frac{1}{{\rho }_l-1}log{\left(A_n^{*}\right)}^{-\frac{{\rho }_l}{{\lambda }_n}}\right]}^{{\rho }_l-1}}\le \underset{n\to +\infty }{lim\; sup}\frac{{\lambda }_n}{{\left[\frac{1}{{\rho }_l-1}log{\left(A_n^{*}\right)}^{-\frac{{\rho }_l}{{\lambda }_n}}\right]}^{{\rho }_l-1}}\le W.$$

(40)

By combining with Remark 3 and Theorem 4, we have from (40) that

$$w\le {\rho }_l{t}_l\le {\rho }_l{T}_l\le W.$$

(41)

If $w=0$ or $W=+\infty$, the conclusions (41) are obvious. If $w=+\infty$, then $W=+\infty$. We can obtain (40) by replacing $w,W$ by an arbitrarily large number. If $W=0$, then $w=0$. We also obtain (38) by replacing $\frac{1}{W}-\epsilon$ by an arbitrarily large number. Thus, we can obtain (41) in either case.

Therefore, this completes the proof of Theorem 10. □

Theorem 11.

If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$ and logarithmic type $T_l(0<T_l<+\infty )$. If the sequence $\left\{{\lambda }_n\right\}$ satisfy (34) and $\psi \left(n\right)$ form a non-decreasing function of $n(\ge n_0)$, then we have

$${\rho }_l{T}_l\le W\le {\left(\frac{{\rho }_l}{{\rho }_l-1}\right)}^{{\rho }_l-1}{\rho }_l{T}_l<e{\rho }_l{T}_l.$$

(42)

Proof. 

From the assumptions of Theorem 11, and the definitions of logarithmic type $T_l$, for any given $\epsilon >0$, there exists a positive integer $n_0\left(\epsilon \right)$ such that for all $n>n_0\left(\epsilon \right)$, we have

$${\lambda }_n^{\frac{{\rho }_l}{{\rho }_l-1}}<-\frac{{\rho }_l}{{\rho }_l-1}{({\rho }_l{T}_l+\epsilon )}^{\frac{1}{{\rho }_l-1}}log{A}_n^{*}.$$

(43)

Thus, it follows that

$$A_n^{*}<exp\left[-\frac{{\rho }_l-1}{{\rho }_l}{({\rho }_l{T}_l+\epsilon )}^{-\frac{1}{{\rho }_l-1}}{\lambda }_n^{\frac{{\rho }_l}{{\rho }_l-1}}\right],n>n_0\left(\epsilon \right),$$

(44)

and

$$\begin{array}{cc}\hfill log{A}_m^{*}+log\frac{A_{m+1}^{*}}{A_m^{*}}& +\cdots +log\frac{A_n^{*}}{A_{n-1}^{*}}<-\frac{{\rho }_l-1}{{\rho }_l}{({\rho }_l{T}_l+\epsilon )}^{-\frac{1}{{\rho }_l-1}}{\lambda }_n^{\frac{{\rho }_l}{{\rho }_l-1}},m>n_0\left(\epsilon \right).\hfill \end{array}$$

(45)

By combining with the non-decreasing function $\psi \left(m\right)$, we obtain

$$log{A}_m^{*}-({\lambda }_n-{\lambda }_m)\psi (n-1)<-\frac{{\rho }_l-1}{{\rho }_l}{({\rho }_l{T}_l+\epsilon )}^{-\frac{1}{{\rho }_l-1}}{\lambda }_n^{\frac{{\rho }_l}{{\rho }_l-1}},$$

that is,

$$\frac{{\lambda }_n}{\psi {(n-1)}^{{\rho }_l-1}}<{\left(\frac{{\rho }_l-1}{{\rho }_l}\right)}^{{\rho }_l-1}({\rho }_l{T}_l+\epsilon ){\left(\frac{{\lambda }_n-{\lambda }_m}{{\lambda }_n}\right)}^{{\rho }_l-1}(1+o\left(1\right)).$$

(46)

In view of $\psi (n-1)\le \psi \left(n\right)$, and let $n\to +\infty$, then we obtain from (46) that

$$W=\underset{n\to +\infty }{lim\; sup}\frac{{\lambda }_n}{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}}\le {\left(\frac{{\rho }_l-1}{{\rho }_l}\right)}^{{\rho }_l-1}{\rho }_l{T}_l.$$

(47)

By combining with the fact that $e^x>1+x$ for $0<x<+\infty$, it follows from (47) that

$$W\le {\left(\frac{{\rho }_l-1}{{\rho }_l}\right)}^{{\rho }_l-1}{\rho }_l{T}_l<e{\rho }_l{T}_l.$$

(48)

Thus, we can obtain (42) from (35) and (48) immediately.

Therefore, we complete the proof of Theorem 11. □

Theorem 12.

If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$. If ${\lambda }_n$ satisfy ${\lambda }_n\sim {\lambda }_{n+1}$ and $\psi \left(n\right)$ form a non-decreasing function of $n(\ge n_0)$, then we have

$$w=vandW=V.$$

(49)

Proof. 

From the assumptions of Theorem 12, and the definitions of v and V, for any positive number $\epsilon$, there exists ${\sigma }_0\left(\epsilon \right)$ such that for all $\sigma >{\sigma }_0\left(\epsilon \right)$,

$$v-\epsilon <\frac{\nu (\sigma ,F)}{{\sigma }^{{\rho }_l-1}}<V+\epsilon .$$

(50)

Since $\psi \left(n\right)$ is an increasing function of n, taking

$$\frac{log{A}_{n-1}^{*}-log{A}_n^{*}}{{\lambda }_n-{\lambda }_{n-1}}\le \sigma <\frac{log{A}_n^{*}-log{A}_{n+1}^{*}}{{\lambda }_{n+1}-{\lambda }_n},$$

(51)

then we have that $A_n^{*}{e}^{{\lambda }_n\sigma }$ is the maximum term for $\Re s=\sigma$, that is, ${\lambda }_n=\nu (\sigma ,F)$. In view of (50) and (51), we have

$$(v-\epsilon ){\left[\frac{log{A}_{n-1}^{*}-log{A}_n^{*}}{{\lambda }_n-{\lambda }_{n-1}}\right]}^{{\rho }_l-1}<{\lambda }_n<(V+\epsilon ){\left[\frac{log{A}_n^{*}-log{A}_{n+1}^{*}}{{\lambda }_{n+1}-{\lambda }_n}\right]}^{{\rho }_l-1},$$

(52)

for all $n>n_0$. Thus, let $n\to +\infty$, it follows from (52) that

$$V\ge \underset{n\to +\infty }{lim\; sup}\frac{{\lambda }_n}{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}},v\le \underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_{n+1}}{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}}.$$

(53)

and in view of ${\lambda }_n\sim {\lambda }_{n+1}$, the second inequality in (53) becomes

$$v\le \underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_n}{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}}.$$

(54)

Obviously, (53) and (40) hold for $v=0$ and $V=+\infty$. Besides, if $v=+\infty$, we can obtain (54) by replacing $v-\epsilon$ by an arbitrary large number in (50). Similarly, we can obtain (53) for $V=0$.

On the other hand, from the definition of V, we have $\nu (\sigma ,F)>(V-\epsilon ){\sigma }^{{\rho }_l-1}$ for a sequence of values of $\sigma ={\sigma }_1,{\sigma }_2,\dots$, tending to ∞. Thus, in view of (51), corresponding to the sequence $\left\{{\sigma }_n\right\}$, we obtain

$${\lambda }_n>(V-\epsilon ){\left[\frac{1}{{\lambda }_n-{\lambda }_{n-1}}log\frac{A_{n-1}^{*}}{A_n^{*}}\right]}^{{\rho }_l-1}.$$

In view of ${\lambda }_n\sim {\lambda }_{n+1}$, for a sequence of values of $n\to +\infty$, we have

$$\frac{{\lambda }_n}{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}}>V-\epsilon .$$

(55)

Hence, we obtain

$$\underset{n\to +\infty }{lim\; sup}\frac{{\lambda }_n}{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}}\ge V.$$

(56)

Similar to the above argument, we have

$$\underset{n\to +\infty }{lim\; inf}\frac{{\lambda }_n}{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}}\le v.$$

(57)

Thus, in view of (54)–(56), we can obtain $w=v$ and $W=V$.

Therefore, this completes the proof of Theorem 12. □

Theorem 13.

If Laplace–Stieltjes transform $F\left(s\right)\in L_{\infty }$, and is of logarithmic order ${\rho }_l(1<{\rho }_l<+\infty )$. If ${\lambda }_n$ satisfy ${\lambda }_n\sim {\lambda }_{n+1}$ and $\psi \left(n\right)$ form a non-decreasing function of $n(\ge n_0)$. We have

(i) $F\left(s\right)$ is of perfectly logarithmic linear growth if, and only if,

$$\underset{n\to +\infty }{lim}\frac{{\lambda }_n}{{\left[\frac{1}{{\lambda }_{n+1}-{\lambda }_n}log\frac{A_n^{*}}{A_{n+1}^{*}}\right]}^{{\rho }_l-1}}={\rho }_l{T}_l;$$

(ii) if $0<t_l\le T_l<\infty$, then $0<w\le W<\infty$.

Proof. 

(i) From Theorem 9 (i) and Theorem 12, we can obtain Theorem 13 (i) easily.

(ii) Similar to the argument as in the proof of Theorem 9 (ii), and combining with the conclusions of Theorem 12, we can prove Theorem 13 (ii).

Therefore, this completes the proof of Theorem 13. □

5. Conclusions

In view of Theorems 7–13, we can see that these results reveal the relationships between the logarithmic growth and some indexes of entire functions represented by Laplace–Stieltjes transforms of finite logarithmic order ${\rho }_l$. In fact, Theorems 7–11 and Remark 5 exhibit the relationships concerning some indexes including ${\rho }_l,t_l,T_l,v,V,h,H$. These theorems show that the (lower) logarithmic type $T_l\left(t_l\right)$ of Laplace–Stieltjes transform can be bounded not only by the center indexes $\nu (\sigma ,F)$ of the maximum terms (see Theorems 7 and 8), but also by the logarithmic order ${\rho }_l$, $A_n^{*}$ and ${\lambda }_n$ (see Theorems 10 and 11). Finally, Theorems 12 and 13 depict the equivalence conditions between the (lower) logarithmic type $T_l\left(t_l\right)$ and $\nu (\sigma ,F)$ of Laplace–Stieltjes transforms with certain restricts. These are very obvious differences since the growth indexes are usual estimated by $A_n^{*},{\lambda }_n$ (can be founded in Theorems 1–3).