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# The approximation of Laplace-Stieltjes transforms with finite order

Journal of Inequalities and Applications20172017:164

https://doi.org/10.1186/s13660-017-1441-9

• Received: 21 March 2017
• Accepted: 27 June 2017
• Published:

## Abstract

In this paper, we study the irregular growth of an entire function defined by the Laplace-Stieltjes transform of finite order convergent in the whole complex plane and obtain some results about λ-lower type. In addition, we also investigate the problem on the error in approximating entire functions defined by the Laplace-Stieltjes transforms. Some results about the irregular growth, the error, and the coefficients of Laplace-Stieltjes transforms are obtained; they are generalization and improvement of the previous conclusions given by Luo and Kong, Singhal and Srivastava.

## Keywords

• irregular growth
• Laplace-Stieltjes transform
• approximate
• error

• 44A10
• 30D15

## 1 Introduction

Dirichlet series
$$f(s)=\sum_{n=1}^{\infty }a_{n}e^{\lambda_{n}s}, \quad s=\sigma +it,$$
(1)
where
$$0\leq \lambda_{1}< \lambda_{2}< \cdots < \lambda_{n}< \cdots , \qquad \lambda _{n}\rightarrow \infty\quad \mbox{as } n\rightarrow \infty ;$$
(2)
$$s=\sigma +it$$ (σ, t are real variables), $$a_{n}$$ are nonzero complex numbers. When $$a_{n}$$, $$\lambda_{n}$$, n satisfy some conditions, the series (1) is convergent in the whole plane or the half-plane, that is, $$f(s)$$ is an analytic function or entire function in the whole plane or the half-plane. In the past few decades, many mathematicians studied the growth and value distribution of the analytic (entire) function defined by Dirichlet series and obtained lots of interesting results (see [19]).

As we know, Dirichlet series is regarded as a special example of the Laplace-Stieltjes transform. 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.

For the Laplace-Stieltjes transforms,
$$G(s)= \int_{0}^{+\infty }e^{-sx}\,d\alpha (x), \quad s= \sigma +it,$$
(3)
where $$\alpha (x)$$ is a bounded variation on any finite interval $$[0,Y]$$ ($$0< Y<+\infty$$), and σ and t are real variables. Let
$$B_{n}^{*}=\sup_{\lambda_{n}< x\leq \lambda_{n+1},-\infty < t< +\infty } \biggl\vert \int_{\lambda_{n}}^{x}e^{-ity}\,d\alpha (y)\biggr\vert ,$$
where the sequence $$\{\lambda_{n}\}_{n=1}^{\infty }$$ satisfies (2) and
$$\limsup_{n\rightarrow +\infty }(\lambda_{n+1}- \lambda_{n})=h< +\infty.$$
(4)

In 1963, Yu [10] proved the Valiron-Knopp-Bohr formula of the associated abscissas of bounded convergence, absolute convergence, and uniform convergence of Laplace-Stieltjes.

### Theorem A

Suppose that Laplace-Stieltjes transforms (3) satisfy (2), (4) and $$\limsup_{n\rightarrow +\infty }\frac{ \log n}{\lambda_{n}}<+\infty$$, then
$$\limsup_{n\rightarrow +\infty }\frac{\log B_{n}^{\ast }}{\lambda_{n}} \leq \sigma_{u}^{G} \leq \limsup_{n\rightarrow +\infty }\frac{\log B _{n}^{\ast }}{\lambda_{n}}+\limsup_{n\rightarrow +\infty } \frac{ \log n}{\lambda_{n}},$$
where $$\sigma_{u}^{F}$$ is called the abscissa of uniform convergence of $$F(s)$$.

Moreover, Yu [10] first introduced the maximal molecule $$M_{u}(\sigma ,G)$$, the maximal term $$\mu (\sigma ,G)$$ and the Borel line, and the order of analytic functions represented by Laplace-Stieltjes transforms convergent in the complex plane. After his works, considerable attention has been paid to the growth and value distribution of the functions represented by the Laplace-Stieltjes transform convergent in the half-plane or the whole complex plane in the field of complex analysis (see [1115]).

In 2012, Luo and Kong [16] studied the following form of Laplace-Stieltjes transform:
$$F(s)= \int_{0}^{+\infty }e^{sx}\,d\alpha (x), \quad s= \sigma +it,$$
(5)
where $$\alpha (x)$$ is stated as in (3), and $$\{\lambda_{n}\}$$ satisfies (2),(4). Set
$$A_{n}^{*}=\sup_{\lambda_{n}< x\leq \lambda_{n+1},-\infty < t< +\infty } \biggl\vert \int_{\lambda_{n}}^{x}e^{ity}\,d\alpha (y)\biggr\vert .$$
By using the same argument as in [10], we can get a similar result about the abscissa of uniform convergence of $$F(s)$$ easily. If
$$\limsup_{n\rightarrow +\infty }\frac{\log n}{\lambda_{n}}=D< \infty , \qquad \limsup_{n\rightarrow +\infty }\frac{\log A_{n}^{\ast }}{\lambda_{n}}=- \infty ,$$
(6)
by (2), (4) and Theorem 1, one can get that $$\sigma_{u}^{F}=+\infty$$, $$i.e$$., $$F(s)$$ is an entire function.
Set
$$M(\sigma ,F)=\sup_{-\infty < t< +\infty }\bigl\vert F(\sigma +it)\bigr\vert ,\qquad M _{u}(\sigma ,F)=\sup_{0< x< +\infty ,-\infty < t< +\infty }\biggl\vert \int_{0} ^{x}e^{(\sigma +it)y}\,d\alpha (y)\biggr\vert$$
and
$$\mu (\sigma ,F)=\max_{n\in N}\bigl\{ A_{n}^{*}e^{\lambda_{n}\sigma } \bigr\} ( \sigma < +\infty ),\quad N(\sigma ,F)=\max \bigl\{ \lambda_{n}: ~A_{n}^{*}e ^{\lambda_{n}\sigma }=\mu (\sigma ,F)\bigr\} .$$

Since $$M(\sigma ,F)$$ and $$M_{u}(\sigma ,F)$$ tend to +∞ as $$\sigma \rightarrow +\infty$$, in order to estimate the growth of $$F(s)$$ more precisely, we will adapt some concepts of order, lower order, type, lower type as follows.

### Definition 1.1

If Laplace-Stieltjes transform (5) satisfies $$\sigma_{u}^{F}=+\infty$$ (the sequence $$\{\lambda_{n}\}$$ satisfies (2), (4), and (6)) and
$$\limsup_{\sigma \rightarrow +\infty }\frac{\log^{+}\log^{+}M_{u}( \sigma ,F)}{\sigma }=\rho ,$$
we call $$F(s)$$ of order ρ in the whole plane, where $$\log^{+}x= \max \{\log x,0\}$$. If $$\rho \in (0,+\infty )$$, we say that $$F(s)$$ is an entire function of finite order in the whole plane. Moreover, the lower order of $$F(s)$$ is defined by
$$\liminf_{\sigma \rightarrow +\infty }\frac{\log^{+}\log^{+}M_{u}( \sigma ,F)}{\sigma }=\lambda .$$

### Remark 1.1

We say that $$F(s)$$ is of the regular growth, when $$\rho =\lambda$$, and $$F(s)$$ is of the irregular growth, when $$\rho \neq \lambda$$.

### Definition 1.2

If Laplace-Stieltjes transform (5) satisfies $$\sigma_{u}^{F}=+\infty$$ (the sequence $$\{\lambda_{n}\}$$ satisfies (2), (4), and (6)) and is of order ρ ($$0<\rho <\infty$$), then we define the type and lower type of L-S transform $$F(s)$$ as follows:
$$\limsup_{\sigma \rightarrow +\infty }\frac{\log^{+}M_{u}(\sigma ,F)}{e ^{\sigma \rho }}=T,\qquad \liminf _{\sigma \rightarrow +\infty }\frac{ \log^{+}M_{u}(\sigma ,F)}{e^{\sigma \rho }}=\tau .$$

### Remark 1.2

The purpose of the definition of type is to compare the growth of class functions which all have the same order. For example, let $$f(s)=e^{e ^{s}}$$, $$g(s)=e^{e^{2s}}$$, by a simple computation, we have $$\rho (f)=1= \rho (g)$$, but $$T(f)=1$$ and $$T(g)=\infty$$. Thus, we can see that the growth of $$g(s)$$ is faster than $$f(s)$$ as $$\vert s \vert \rightarrow +\infty$$.

## 2 Results and discussion

Recently, many people studied some problems on analytic functions defined by the Laplace-Stieltjes transforms and obtained a number of interesting results. Kong, Sun, Huo and Xu investigated the growth of analytic functions with kinds of order defined by the Laplace-Stieltjes transforms (see [1622]), and Shang, Gao, and Sun investigated the value distribution of such functions (see [2326]). From these references, we get the following results.

### Theorem 2.1

If Laplace-Stieltjes transform (5) satisfies $$\sigma_{u}^{F}=+\infty$$ (the sequence $$\{\lambda_{n}\}$$ satisfies (2), (4), and (6)), and is of order ρ ($$0<\rho <\infty$$) and of type T, then
$$\rho =\limsup_{n\rightarrow +\infty }\frac{\lambda_{n}\log \lambda _{n}}{-\log A_{n}^{*}}, \qquad T=\limsup _{n\rightarrow +\infty }\frac{ \lambda_{n}}{\rho e}\bigl(A_{n}^{*} \bigr)^{\frac{\rho }{\lambda_{n}}}.$$
Furthermore, if $$F(s)$$ is of the lower order λ and the lower type τ, and $$\lambda_{n}\sim \lambda_{n+1}$$ and the function
$$\psi (n)=\frac{\log A^{*}_{n}-\log A^{*}_{n+1}}{\lambda_{n+1}-\lambda _{n}}$$
forms a non-decreasing function of n for $$n>n_{0}$$, then we have
$$\lambda =\liminf_{n\rightarrow +\infty }\frac{\lambda_{n}\log \lambda _{n}}{-\log A_{n}^{*}}, \qquad \tau = \liminf _{n\rightarrow +\infty }\frac{ \lambda_{n}}{\rho e}\bigl(A_{n}^{*} \bigr)^{\frac{\rho }{\lambda_{n}}}.$$

From Definition 1.2, a natural question to ask is: What happened if $$e^{\sigma \rho }$$ is replaced by $$e^{\lambda \sigma }$$ in the definition of lower type when $$\rho \neq \lambda$$? We are going to consider this question.

### Definition 2.1

If Laplace-Stieltjes transform (5) satisfies $$\sigma_{u}^{F}=+\infty$$ (the sequence $$\{\lambda_{n}\}$$ satisfies (2), (4), and (6)), and is of order ρ ($$0<\rho <\infty$$) and of the lower order λ ($$0<\lambda <\infty$$), if $$\lambda \neq \rho$$, and
$$\liminf_{\sigma \rightarrow +\infty }\frac{\log^{+}M_{u}(\sigma ,F)}{e ^{\sigma \lambda }}=\tau_{\lambda },$$
we say that $$\tau_{\lambda }$$ is the λ-type of $$F(s)$$.

### Remark 2.1

Obviously, $$\tau_{\lambda }\geq \tau$$ and $$\tau_{\lambda }=\tau$$ as $$\rho =\lambda$$. But we cannot confirm whether $$\tau_{\lambda } \geq T$$ or $$\tau_{\lambda }\leq T$$.

The following results are the main theorems of this paper.

### Theorem 2.2

If Laplace-Stieltjes transform (5) satisfies $$\sigma_{u}^{F}=+\infty$$ (the sequence $$\{\lambda_{n}\}$$ satisfies (2), (4), and (6)), and is of order ρ and of the lower order λ, $$0\leq \lambda \neq \rho <\infty$$, then we have
$$\liminf_{\sigma \rightarrow \infty }\frac{\log M(\sigma ,F)}{e^{ \rho \sigma }}= \liminf _{\sigma \rightarrow \infty }\frac{\log \mu ( \sigma ,F)}{e^{\rho \sigma }}=0,$$
(7)
and
$$\liminf_{\sigma \rightarrow \infty }\frac{N(\sigma ,F)}{e^{\rho \sigma }}=0.$$
(8)

### Theorem 2.3

If Laplace-Stieltjes transform (5) satisfies $$\sigma_{u}^{F}=+\infty$$ (the sequence $$\{\lambda_{n}\}$$ satisfies (2), (4), and (6)), and is of order ρ and of the lower order λ, $$0< \lambda \neq \rho <\infty$$, type T, λ-type $$\tau_{\lambda }$$,
$$\limsup_{\sigma \rightarrow +\infty }\frac{N(\sigma ,F)}{e^{\rho \sigma }}=H, \qquad \liminf _{\sigma \rightarrow +\infty }\frac{N(\sigma ,F)}{e^{\rho \sigma }}=h,$$
and let
$$T_{\rho }(\sigma ,F)=\frac{\log \mu (\sigma ,F)}{\exp (\rho \sigma )},\qquad T_{\lambda }(\sigma ,F)= \frac{\log \mu (\sigma ,F)}{\exp (\lambda \sigma )},$$
then we have
\begin{aligned}& H-\rho T\leq \limsup_{\sigma \rightarrow +\infty }T'_{\rho }( \sigma ,F) \leq H, \end{aligned}
(9)
\begin{aligned}& -\infty \leq \liminf_{\sigma \rightarrow +\infty }T'_{\lambda }( \sigma ,F)\leq h-\lambda \tau_{\lambda } \end{aligned}
(10)
for almost all values of $$\sigma >\sigma_{0}$$, where $$T'_{\rho }( \sigma )$$ and $$T'_{\lambda }(\sigma )$$ are the derivatives of $$T_{\rho }(\sigma )$$ and $$T_{\lambda }(\sigma )$$ with respect to σ.

### Theorem 2.4

If Laplace-Stieltjes transform (5) satisfies $$\sigma_{u}^{F}=+\infty$$ (the sequence $$\{\lambda_{n}\}$$ satisfies (2), (4), and (6)), and is of the lower order λ ($$0\leq \lambda \neq \rho <\infty$$), if $$\lambda_{n}\sim \lambda_{n+1}$$, then
$$\tau_{\lambda }\geq \liminf_{n\rightarrow \infty } \biggl( \frac{\lambda _{n}}{e\lambda } \biggr) \bigl(A^{*}_{n} \bigr)^{\frac{\lambda }{\lambda_{n}}} \quad (0\leq \tau_{\lambda }\leq \infty ).$$
(11)
Furthermore, there exists a positive integer $$n_{0}$$ such that
$$\psi (n)=\frac{\log A^{*}_{n}-\log A^{*}_{n+1}}{\lambda_{n+1}-\lambda _{n}}$$
forms a non-decreasing function of n for $$n>n_{0}$$, then we have
$$\tau_{\lambda }= \liminf_{n\rightarrow \infty } \biggl( \frac{\lambda _{n}}{e\lambda } \biggr) \bigl(A^{*}_{n} \bigr)^{\frac{\lambda }{\lambda_{n}}} \quad (0\leq \tau_{\lambda }\leq \infty ).$$
(12)

We denote by $$\overline{L}_{\beta }$$ the class of all the functions $$F(s)$$ of the form (5) which are analytic in the half-plane $$\Re s<\beta$$ ($$-\infty <\beta <\infty$$) and the sequence $$\{\lambda _{n}\}$$ satisfies (2) and (4); and we denote by $$L_{\infty }$$ the class of all the functions $$F(s)$$ of the form (5) which are analytic in the half-plane $$\Re s<+\infty$$ and the sequence $$\{\lambda_{n}\}$$ satisfies (2), (4), and (6). Thus, if $$-\infty <\beta <+\infty$$ and $$F(s) \in \overline{L}_{\beta }$$, then $$F(s)\in L_{\infty }$$. If Laplace-Stieltjes transform (5) $$A^{*}_{n}=0$$ for $$n\geq k+1$$ and $$A^{*}_{n}\neq 0$$, then $$F(s)$$ will be called an exponential polynomial of degree k usually denoted by $$p_{k}$$, $$i.e$$., $$p_{k}(s)= \int^{\lambda_{k}}_{0}\exp (sy)\,d\alpha (y)$$. When we choose a suitable function $$\alpha (y)$$, the function $$p_{k}(s)$$ may be reduced to a polynomial in terms of $$\exp (s\lambda_{i})$$, that is, $$\sum_{i=1} ^{k}b_{i}\exp (s\lambda_{i})$$.

For $$F(s)\in \overline{L}_{\beta }$$, $$-\infty <\beta <+\infty$$, we denote by $$E_{n}(F,\beta )$$ the error in approximating the function $$F(s)$$ by exponential polynomials of degree n in uniform norm as
$$E_{n}(F,\beta )=\inf_{p\in \Pi_{n}}\Vert F-p \Vert _{\beta }, \quad n=1,2,\ldots ,$$
where
$$\Vert F-p \Vert _{\beta }=\max_{-\infty < t< +\infty }\bigl\vert F( \beta +it)-p(\beta +it) \bigr\vert .$$

In this paper, we will further investigate the relation between $$E_{n}(F,\beta )$$ and the growth of an entire function defined by the L-S transform with irregular growth. It seems that this problem has never been treated before. Our main result is as follows.

### Theorem 2.5

If the Laplace-Stieltjes transform $$F(s)\in L_{\infty }$$ and is of lower order λ ($$0\leq \lambda \neq \rho <\infty$$), if $$\lambda_{n}\sim \lambda_{n+1}$$, then for any real number $$-\infty < \beta <+\infty$$, we have
$$\tau_{\lambda }\geq \liminf_{n\rightarrow \infty } \biggl( \frac{\lambda _{n}}{e\lambda } \biggr) \bigl(E_{n-1}(F,\beta )\exp (-\beta \lambda_{n})\bigr)^{\frac{ \lambda }{\lambda_{n}}}\quad (0\leq \tau_{\lambda }\leq \infty ).$$
(13)
Furthermore, there exists a positive integer $$n_{0}$$ such that
$$\psi_{1}(n)=\frac{\log A^{*}_{n}-\log A^{*}_{n+1}}{\lambda_{n+1}- \lambda_{n}}$$
forms a non-decreasing function of n for $$n>n_{0}$$, then we have
$$\tau_{\lambda }= \liminf_{n\rightarrow \infty } \biggl( \frac{\lambda _{n}}{e\lambda } \biggr) \bigl(E_{n-1}(F,\beta )\exp (-\beta \lambda_{n}) \bigr)^{\frac{ \lambda }{\lambda_{n}}}\quad (0\leq \tau_{\lambda }\leq \infty ),$$
i.e.,
$$\exp (\beta \lambda )e\lambda \tau_{\lambda }= \liminf _{n\rightarrow \infty }\lambda_{n}\bigl(E_{n-1}(F,\beta ) \bigr)^{\frac{ \lambda }{\lambda_{n}}}.$$
(14)

## 3 Conclusions

From Theorems 2.2-2.5, we can see that the growth of Laplace-Stieltjes transforms is investigated under the assumption $$\rho \neq \lambda$$, and that some theorems about the λ-lower type $$\tau_{\lambda }$$, $$\lambda_{n}$$, $$A_{n}^{*}$$, and λ are obtained. In addition, we also study the problem on the error in approximating entire functions defined by the Laplace-Stieltjes transforms. This project is a new issue of Laplace-Stieltjes transforms in the field of complex analysis. Our results are generalization and improvement of the previous conclusions given by Luo and Kong [16, 27], Singhal and Srivastava [28].

## 4 Methods

### 4.1 Proofs of Theorems 2.2 and 2.3

To prove the above theorems, we require the following lemmas.

### Lemma 4.1

see [27], Lemma 2.1

If the L-S transform $$F(s)\in L_{ \infty }$$, for any $$\sigma (-\infty <\sigma <+\infty )$$ and $$\varepsilon (>0)$$, we have
$$\frac{1}{2}\mu (\sigma ,F)\leq M_{u}(\sigma ,F)\leq C\mu \bigl((1+2\varepsilon )\sigma ,F\bigr),$$
where C is a constant.

### Lemma 4.2

see [16], Lemma 2.2

If the L-S transform $$F(s)\in L_{ \infty }$$, then we have
$$\log \mu (\sigma ,F)=\log \mu (\sigma_{0},F)+ \int_{\sigma_{0}}^{ \sigma }N(t,F)\,dt$$
for $$\sigma_{0}>0$$.

#### 4.1.1 The proof of Theorem 2.2

Since $$\rho >\lambda >0$$ and $$F(s)$$ is of the lower order λ, that is,
$$\lambda =\liminf_{\sigma \rightarrow +\infty }\frac{\log \log M_{u}( \sigma ,F)}{\sigma },$$
(15)
for any small $$\varepsilon (0<\varepsilon <\rho -\lambda )$$, it follows from (15) that there exists a constant $$\sigma_{0}$$ such that, for $$\sigma >\sigma_{0}$$,
$$\log M_{u}(\sigma ,F)>\exp \bigl\{ (\lambda -\varepsilon )\sigma \bigr\} ,$$
(16)
and there exists a sequence $$\{\sigma_{k}\}$$ tending to +∞ such that
$$\log M_{u}(\sigma_{k},F)< \exp \bigl\{ ( \lambda +\varepsilon )\sigma_{k}\bigr\} .$$
(17)
Since $$0<\varepsilon <\rho -\lambda$$, it follows from (16) and (17) that
$$\liminf_{\sigma \rightarrow +\infty }\frac{\log M_{u}(\sigma ,F)}{ \exp (\rho \sigma )}=0.$$
(18)
From Lemmas 4.1 and 4.2, we have
$$\rho =\limsup_{\sigma \rightarrow +\infty }\frac{\log \log M_{u}( \sigma ,F)}{\sigma } =\limsup _{\sigma \rightarrow +\infty }\frac{ \log \log \mu (\sigma ,F)}{\sigma } =\limsup_{\sigma \rightarrow + \infty } \frac{\log N(\sigma ,F)}{\sigma }$$
and
$$\lambda =\liminf_{\sigma \rightarrow +\infty }\frac{\log \log M_{u}( \sigma ,F)}{\sigma } =\liminf _{\sigma \rightarrow +\infty }\frac{ \log \log \mu (\sigma ,F)}{\sigma } =\liminf_{\sigma \rightarrow + \infty } \frac{\log N(\sigma ,F)}{\sigma }.$$
Thus, similar to the process of (18), we can easily prove
$$\liminf_{\sigma \rightarrow +\infty }\frac{\log \mu (\sigma ,F)}{ \exp (\rho \sigma )}= \liminf_{\sigma \rightarrow +\infty } \frac{N( \sigma ,F)}{\exp (\rho \sigma )}=0.$$

Hence, this completes the proof of Theorem 2.2.

#### 4.1.2 The proof of Theorem 2.3

From Lemma 4.2, it follows that
$$\limsup_{\sigma \rightarrow +\infty }\frac{\int_{\sigma_{0}}^{\sigma }N(t,F)\,dt}{e^{\rho \sigma }} =\limsup _{\sigma \rightarrow +\infty }\frac{ \log \mu (\sigma ,F)}{e^{\rho \sigma }}= \limsup_{\sigma \rightarrow +\infty }T_{\rho }( \sigma ,F)=T$$
(19)
and
$$\liminf_{\sigma \rightarrow +\infty }\frac{\int_{\sigma_{0}}^{\sigma }N(t,F)\,dt}{e^{\lambda \sigma }} =\liminf _{\sigma \rightarrow +\infty }\frac{\log \mu (\sigma ,F)}{e^{\lambda \sigma }}= \liminf_{\sigma \rightarrow +\infty }T_{\lambda }( \sigma ,F)= \tau_{\lambda }.$$
(20)
Dividing two sides of the equality in Lemma 4.2 by $$e^{\rho \sigma }$$ and differentiating it with respect to σ, for almost all values $$\sigma >\sigma_{0}$$, we have
$$T'_{\rho }(\sigma ,F)=-\rho \frac{\log \mu (\sigma_{0},F)}{e^{\rho \sigma }}-\frac{\rho }{e^{\rho \sigma }} \int_{\sigma_{0}}^{\sigma }N(t,F)\,dt+\frac{N( \sigma ,F)}{e^{\rho \sigma }}.$$
(21)
On the basis of the assumptions of Theorem 2.3, taking lim sup in (21) when $$\sigma \rightarrow +\infty$$, from Theorem 2.2 and (19), we get (9) easily.
Similarly, dividing two sides of the equality in Lemma 4.2 by $$e^{\lambda \sigma }$$ and differentiating it with respect to σ, for almost all values $$\sigma >\sigma_{0}$$,
$$T'_{\lambda }(\sigma ,F)=-\lambda \frac{\log \mu (\sigma_{0},F)}{e ^{\lambda \sigma }}-\frac{\lambda }{e^{\lambda \sigma }} \int_{\sigma _{0}}^{\sigma }N(t,F)\,dt+\frac{N(\sigma ,F)}{e^{\lambda \sigma }}.$$
(22)
On the basis of the assumptions of Theorem 2.3, taking lim inf in (22) when $$\sigma \rightarrow +\infty$$, from Theorem 2.1 and (20), we get (10) easily.

Thus, this completes the proof of Theorem 2.3.

### 4.2 The proof of Theorem 2.4

Let
$$\vartheta =\liminf_{n\rightarrow +\infty }\frac{\lambda_{n}}{e\lambda } \bigl(A_{n}^{*}\bigr)^{\frac{\lambda }{\lambda_{n}}}\quad (0< \vartheta < + \infty ).$$
Thus, for any $$\varepsilon >0$$, there exists an integer $$n_{0}(\varepsilon )$$ such that, for $$n>n_{0}(\varepsilon )$$,
$$\lambda_{n}\bigl(A_{n}^{*} \bigr)^{\frac{\lambda }{\lambda_{n}}}>(\vartheta - \varepsilon )e\lambda .$$
(23)
By Lemma 4.1, it follows from (23) that for $$n>n_{0}(\varepsilon )$$
\begin{aligned} \frac{\log M_{u}(\sigma ,F)}{e^{\lambda \sigma }} &\geq \frac{\log A _{n}^{*}+\lambda_{n}\sigma -\log 2}{e^{\lambda \sigma }} \\ & >e^{-\lambda \sigma } \biggl( \lambda_{n}\sigma +\frac{\lambda_{n}}{ \lambda }\log \bigl[(\vartheta -\varepsilon )e\lambda \bigr] -\frac{\lambda_{n}}{ \lambda }\log \lambda_{n}-\log 2 \biggr) . \end{aligned}
(24)
Let
$$\biggl( \frac{\lambda_{n}}{\lambda \vartheta } \biggr) ^{\frac{1}{\lambda}}\leq e^{\sigma }< \biggl( \frac{\lambda_{n+1}}{\lambda \vartheta } \biggr) ^{\frac{1}{\lambda }},$$
and take
$$\sigma =\frac{1}{\lambda }\log \biggl(\frac{\lambda_{n}}{\lambda \vartheta }\biggr)+o\biggl( \frac{1}{\lambda_{n}}\biggr).$$
Then from (24) it follows
$$\frac{\log M_{u}(\sigma ,F)}{e^{\lambda \sigma }}\geq \frac{\lambda \vartheta }{\lambda_{n+1}} \biggl( \frac{\lambda_{n}}{\lambda }\log \frac{1}{ \lambda \vartheta } +\frac{\lambda_{n}}{\lambda }\log \bigl((\vartheta - \varepsilon )e\lambda \bigr)-\log 2+o(1) \biggr) .$$
(25)
Since $$\lambda_{n}\sim \lambda_{n+1}$$ and $$\lambda_{n}\rightarrow + \infty$$ as $$n\rightarrow +\infty$$, thus by a simple computation, from (25) we have $$\tau_{\lambda }\geq \vartheta$$. When $$\vartheta =0$$, $$\tau_{\lambda }\geq \vartheta$$ is obvious; if $$\vartheta =\infty$$, we also prove that $$\tau_{\lambda }\geq \vartheta$$ by using the same argument as above. Hence we prove that (11) holds.
Let $$\mu (\sigma ,F)$$ denote the maximum term for $$\Re s=\sigma$$, $$- \infty < t<+\infty$$. Since
$$\psi (n)=\frac{\log A^{*}_{n}-\log A^{*}_{n+1}}{\lambda_{n+1}-\lambda _{n}}$$
forms a non-decreasing function of n for $$n>n_{0}$$, then for $$\psi (n-1)\leq \sigma <\psi (n)$$
$$\log \mu (\sigma ,F)=\log A_{n}^{*}+\lambda_{n} \sigma .$$
Since $$\tau_{\lambda }<\infty$$, for any small $$\varepsilon >0$$, it follows from (20) that
$$\log \mu (\sigma ,F)=\log A_{n}^{*}+ \lambda_{n}\sigma \geq ( \tau_{\lambda }-\varepsilon )\exp (\lambda \sigma )$$
(26)
for $$\sigma >\sigma_{0}$$ and all n such that $$\psi (n-1)\leq \sigma <\psi (n)$$.
Let $$\Re s=\sigma >\sigma_{0}$$ and $$A_{n_{1}}^{*}\exp (\sigma \lambda_{n_{1}})$$ and $$A_{n_{2}}^{*}\exp (\sigma \lambda_{n_{2}})$$ ($$n_{1}>n_{0}, \psi (n-1)>\sigma_{0}$$) be two consecutive maximum terms such that $$n_{2}-1\geq n_{1}$$, it follows from (26) that
$$\log A_{n_{2}}^{*}+\lambda_{n_{2}}\sigma \geq ( \tau_{\lambda }-\varepsilon )\exp (\lambda \sigma )$$
for all $$\sigma >\sigma_{0}$$ satisfying $$\psi (n_{2}-1)\leq \sigma < \psi (n_{2})$$. Let $$n_{1}\leq n\leq n_{2}-1$$, then
$$\psi (n_{1})=\psi (n_{1}+1)=\cdots =\psi (n)=\cdots =\psi (n_{2}-1)$$
and $$A_{n}^{*}\exp (\lambda_{n}\sigma )=A_{n_{2}}^{*}\exp (\lambda _{n_{2}}\sigma )$$ for $$\sigma =\psi (n)$$. Then there exists a positive integer $$n_{1}$$ such that, for $$n>n_{1}$$ and $$\sigma >\sigma_{0}$$,
$$\log A_{n}^{*}>(\tau_{\lambda }-\varepsilon )e^{\lambda \sigma }- \lambda_{n}\sigma .$$
Since $$e^{x}\geq ex$$ for any x, so it follows
$$\lambda_{n}\bigl(A_{n}^{*} \bigr)^{\frac{\lambda }{\lambda_{n}}}> \frac{\lambda _{n}}{e^{\lambda \sigma }} \exp \biggl\{ \frac{\lambda (\tau_{\lambda }- \varepsilon )}{\lambda_{n}}e^{\lambda \sigma } \biggr\} >\frac{\lambda _{n}}{e^{\lambda \sigma }}\frac{e(\tau_{\lambda }-\varepsilon )\lambda }{\lambda_{n}}e^{\lambda \sigma } =e( \tau_{\lambda }-\varepsilon ) \lambda .$$
(27)
Thus, for $$\varepsilon \rightarrow 0$$ and $$n\rightarrow +\infty$$, from (27) it follows
$$\vartheta =\liminf_{n\rightarrow +\infty }\frac{\lambda_{n}}{e\lambda } \bigl(A_{n}^{*}\bigr)^{\frac{\lambda }{\lambda_{n}}}\geq \tau_{\lambda }.$$
(28)

Hence, this proves that (12) holds.

### 4.3 The proof of Theorem 2.5

To prove this theorem, we require the following lemma.

### Lemma 4.3

If the abscissa $$\sigma_{u}^{F}=+\infty$$ of uniform convergence of the Laplace-Stieltjes transformation $$F(s)$$ and sequence (2) satisfies (4), (6), then for any real number β, we have
$$\biggl\vert \int_{\lambda_{k}}^{\infty }\exp \bigl\{ (\beta +it)y\bigr\} \,d\alpha (y)\biggr\vert \leq 2\sum_{n=k}^{+\infty }A_{n}^{*} \exp \{\beta \lambda_{n+1}\},$$
where
$$A_{n}^{*}=\sup_{\lambda_{n}< x\leq \lambda_{n+1},-\infty < t< +\infty } \biggl\vert \int_{\lambda_{n}}^{x}e^{ity}\,d\alpha (y)\biggr\vert .$$

### Proof

Set
$$I(x;it)= \int_{0}^{x}\exp \{ity\}\,d\alpha (y).$$
For any real number β, since
$$\biggl\vert \int_{\lambda_{k}}^{\infty }\exp \bigl\{ (\beta +it)y\bigr\} \,d\alpha (y)\biggr\vert =\lim_{b\rightarrow +\infty }\biggl\vert \int_{\lambda_{k}}^{b} \exp \bigl\{ (\beta +it)y\bigr\} \,d \alpha (y)\biggr\vert .$$
Set $$I_{j+k}(b;it)=\int_{\lambda_{j+k}}^{b}\exp \{ity\}\,d\alpha (y)$$, $$(\lambda_{j+k}< b\leq \lambda_{j+k+1})$$, then we have $$\vert I_{j+k}(b;it) \vert \leq A_{j+k}^{*}$$. Thus, it follows
\begin{aligned}& \biggl\vert \int_{\lambda_{k}}^{b}\exp \bigl\{ (\beta +it)y\bigr\} \,d \alpha (y)\biggr\vert \\& \quad = \Biggl\vert \sum_{j=k}^{n+k-1} \int_{\lambda_{j}}^{\lambda_{j+1}}\exp \{ \beta y\}d_{y}I_{j}(y;it)+ \int_{\lambda_{n+k}}^{b}\exp \{\beta y\}d _{y}I_{n+k}(y;it) \Biggr\vert \\& \quad = \Biggl\vert \Biggl[ \sum_{j=k}^{n+k-1}e^{\lambda_{j+1}\beta }I_{j}( \lambda_{j+1};it)-\beta \int_{\lambda_{j}}^{\lambda_{j+1}}e^{\beta y}I_{j}(y;it) \,dy\Biggr] \\& \quad \quad {} +e^{\beta b}I_{n+k}(b;it)-\beta \int_{\lambda_{n+k}}^{b}e^{\beta y}I_{j}(y;it) \,dy\Biggr\vert \\& \quad \leq \sum_{j=k}^{n+k-1}\bigl[ A_{j}^{*}e^{\lambda_{j+1}\beta }+A_{j} ^{*} \bigl(e^{\lambda_{j+1}\beta }-e^{\lambda_{j}\beta }\bigr)\bigr] +2e^{\beta \lambda_{n+k+1}}A_{n+k}^{*}-e^{\beta \lambda_{n+k}}A_{n+k}^{*} \\& \quad \leq 2\sum_{j=k}^{n+k} A_{n}^{*}e^{\lambda_{n+1}\beta }. \end{aligned}
When $$n\rightarrow +\infty$$, we have $$b\rightarrow +\infty$$, thus we have
$$\biggl\vert \int_{\lambda_{k}}^{\infty }\exp \bigl\{ (\beta +it)y\bigr\} \,d\alpha (y)\biggr\vert \leq 2\sum_{n=k}^{+\infty }A_{n}^{*} \exp \{\beta \lambda_{n+1}\}.$$
□

Now, we are going to prove Theorem 2.5.

### 4.4 The proof of Theorem 2.5

Let
$$\vartheta_{1}= \liminf_{n\rightarrow \infty } \biggl( \frac{\lambda_{n}}{e \lambda } \biggr) \bigl(E_{n-1}(F,\beta )\exp (-\beta \lambda_{n})\bigr)^{\frac{ \lambda }{\lambda_{n}}} \quad (0< \vartheta_{1}< + \infty ).$$
Then, for any small $$\varepsilon >0$$, there exists an integer $$n_{0}(\varepsilon )$$ such that, for any $$n>n_{0}(\varepsilon )$$,
$$\log \bigl(E_{n-1}(F,\beta )\exp (-\beta \lambda_{n})\bigr)>\frac{\lambda_{n}}{ \lambda }\log \frac{(\vartheta_{1}-\varepsilon )e\lambda }{\lambda _{n}}.$$
(29)
Since $$F(s)\in L_{\infty }$$, thus for any constant β ($$-\infty <\beta <+\infty$$), we have $$F(s)\in \overline{L}_{\beta }$$. For $$\beta <\sigma <+\infty$$. It follows from the definitions of $$E_{n}(F,\beta )$$ and $$p_{n}$$ that
\begin{aligned} E_{n}(F,\beta ) &\leq \Vert F-p_{n} \Vert _{\beta }\leq \bigl\vert F(\beta +it)-p_{n}(\beta +it) \bigr\vert \\ &\leq \biggl\vert \int_{0}^{+\infty }\exp \bigl\{ (\beta +it)y\bigr\} \,d\alpha (y)- \int_{0}^{\lambda_{n}}\exp \bigl\{ (\beta +it)y\bigr\} \,d \alpha (y)\biggr\vert \\ &=\biggl\vert \int_{\lambda_{n}}^{\infty }\exp \bigl\{ (\beta +it)y\bigr\} \,d\alpha (y)\biggr\vert . \end{aligned}
(30)
Thus, from the definition of $$A_{n}^{*}$$ and $$M_{u}(\sigma ,F)$$, and by Lemma 4.1, we have $$A_{n}^{*}\leq 2 M_{u}(\sigma ,F)e^{-\sigma \lambda _{n}}$$ for any σ ($$\beta <\sigma <+\infty$$). It follows from (30) and Lemma 4.3 that
$$E_{n}(F,\beta )\leq 2\sum_{k=n+1}^{\infty }A_{k-1}^{*} \exp \{\beta \lambda_{k}\}\leq 4M_{u}(\sigma ,F)\sum _{k=n+1}^{\infty }\exp \bigl\{ ( \beta -\sigma ) \lambda_{k}\bigr\} .$$
(31)
From (4), take $$h'$$ ($$0< h'< h$$) such that $$(\lambda_{n+1} -\lambda_{n}) \geq h'$$ for $$n\geq 0$$. Then, for $$\sigma \geq \frac{\beta }{2}$$, it follows from (31) that
\begin{aligned} E_{n}(F,\beta ) &\leq 4M_{u}(\sigma ,F)\exp \bigl\{ \lambda_{n+1}(\beta - \sigma )\bigr\} \sum_{k=n+1}^{\infty } \exp \bigl\{ (\lambda_{k}-\lambda_{n+1}) ( \beta -\sigma )\bigr\} \\ &\leq 4M_{u}(\sigma ,F)\exp \bigl\{ \lambda_{n+1}(\beta - \sigma )\bigr\} \exp \biggl\{ -\frac{ \beta }{2}h'(n+1)\biggr\} \sum _{k=n+1}^{\infty }\biggl(\exp \biggl\{ \frac{\beta }{2}h'k \biggr\} \biggr) \\ &=4M_{u}(\sigma ,F)\exp \bigl\{ \lambda_{n+1}(\beta -\sigma ) \bigr\} \biggl( 1- \exp \biggl\{ \frac{\beta }{2}h'\biggr\} \biggr) ^{-1}, \end{aligned}
that is,
$$E_{n-1}(F,\beta )\leq KM_{u}(\sigma ,F)\exp \bigl\{ \lambda_{n}(\beta - \sigma )\bigr\} ,$$
(32)
where K is a constant. Let
$$\gamma_{n}= E_{n-1}(F,\beta )\exp (-\beta\lambda_{n})\quad (n=1,2,\ldots ).$$
Thus, from (29) and (32), it follows that for $$n>n_{0}(\varepsilon )$$
\begin{aligned} \frac{\log M_{u}(\sigma ,F)}{e^{\lambda \sigma }} &\geq \frac{\log \gamma_{n}+\lambda_{n}\sigma -\log K}{e^{\lambda \sigma }} \\ & >e^{-\lambda \sigma } \biggl( \lambda_{n}\sigma +\frac{\lambda_{n}}{ \lambda }\log \bigl[(\vartheta_{1}-\varepsilon )e\lambda \bigr]-\frac{\lambda _{n}}{\lambda } \log \lambda_{n}-\log K \biggr) . \end{aligned}
(33)
By using the same argument as in Theorem 2.4, we can easily prove that $$\tau_{\lambda }\geq \vartheta_{1}$$.
From the proof of Theorem 2.4, we have that there exists a positive integer $$n_{1}$$ such that
$$\log A_{n}^{*}>(\tau_{\lambda }-\varepsilon )e^{\lambda \sigma }- \lambda_{n}\sigma$$
for $$n>n_{1}$$ and $$\sigma >\sigma_{0}$$. Since for any $$\beta <+\infty$$, from the definition of $$E_{k}(F,\beta )$$, there exists $$p_{1} \in \Pi_{n-1}$$ such that
$$\Vert F-p_{1} \Vert \leq 2E_{n-1}(F,\beta ).$$
(34)
And since
\begin{aligned} A_{n}^{*}\exp \{\beta \lambda_{n}\} & = \sup_{\lambda_{n}< x\leq \lambda_{n+1},-\infty < t< +\infty }\biggl\vert \int _{\lambda_{n}}^{x}\exp \{ity\}\,d\alpha (y)\biggr\vert \exp \{\beta \lambda _{n}\} \\ &\leq \sup_{\lambda_{n}< x\leq \lambda_{n+1},-\infty < t< + \infty }\biggl\vert \int_{\lambda_{n}}^{x}\exp \bigl\{ (\beta +it)y\bigr\} \,d\alpha (y)\biggr\vert \\ &\leq \sup_{-\infty < t< +\infty }\biggl\vert \int_{\lambda_{n}}^{ \infty }\exp \bigl\{ (\beta +it)y\bigr\} \,d \alpha (y)\biggr\vert , \end{aligned}
thus for any $$p\in \Pi_{n-1}$$, it follows
$$A_{n}^{*}\exp \{\beta \lambda_{n} \}\leq \bigl\vert F(\beta +it)-p(\beta +it) \bigr\vert \leq \Vert F-p \Vert _{\beta }.$$
(35)
Hence from (34) and (35), for any $$\beta <+\infty$$ and $$F(s)\in L _{\infty }$$, we have
$$A_{n}^{*}\exp \{\beta \lambda_{n}\}\leq 2E_{n-1}(F,\beta ).$$
Since $$e^{x}\geq ex$$ for any x, so it follows
\begin{aligned} \lambda_{n}(\gamma_{n})^{\frac{\lambda }{\lambda_{n}}}&> \frac{\lambda _{n}}{e^{\lambda \sigma }} \exp \biggl\{ \frac{\lambda (\tau_{\lambda }- \varepsilon )}{\lambda_{n}}e^{\lambda \sigma }- \frac{\lambda \log 2}{ \lambda_{n}} \biggr\} \\ &>\frac{\lambda_{n}}{e^{\lambda \sigma }} \biggl( \frac{e(\tau_{\lambda }-\varepsilon )\lambda }{\lambda_{n}}e^{\lambda \sigma }\exp \bigl\{ o(1) \bigr\} \biggr) =e(\tau_{\lambda }-\varepsilon )\lambda . \end{aligned}
(36)
Thus, for $$\varepsilon \rightarrow 0$$ and $$n\rightarrow +\infty$$, from (36) it follows
$$\vartheta_{1}=\liminf_{n\rightarrow \infty }\frac{\lambda_{n}}{e \lambda }( \gamma_{n})^{\frac{\lambda }{\lambda_{n}}}\geq \tau_{\lambda }.$$
Since $$[E_{n-1}(F,\beta )\exp (-\beta \lambda_{n})]^{\frac{\lambda }{ \lambda_{n}}}=[E_{n-1}(F,\beta )]^{\frac{\lambda }{\lambda_{n}}} \exp (-\beta \lambda )$$, then (14) follows.

Therefore, we complete the proof of Theorem 2.5.

### 4.5 Remarks

From the proof of Theorem 2.5, and combining those results of the Laplace-Stieltjes transforms in Ref. [14, 16, 27], we can obtain the following results on the approximation of Laplace-Stieltjes transforms, which can be found partly in [28].

### Theorem 4.1

If the L-S transform $$F(s)\in L_{\infty }$$ and is of order ρ ($$0<\rho <\infty$$) and of type T, then for any real number $$-\infty < \beta <+\infty$$, we have
$$\rho =\limsup_{n\rightarrow +\infty }\frac{\lambda_{n}\log \lambda _{n}}{-\log E_{n-1}(F,\beta )\exp (-\beta \lambda_{n})}= \limsup _{n\rightarrow +\infty }\frac{\lambda_{n}\log \lambda_{n}}{- \log E_{n-1}(F,\beta )}$$
and
\begin{aligned} T&=\limsup_{n\rightarrow +\infty }\frac{\lambda_{n}}{\rho e}\bigl(E_{n-1}(F, \beta )\exp (-\beta \lambda_{n})\bigr)^{\frac{\rho }{\lambda_{n}}} \\ &= \limsup _{n\rightarrow +\infty }\frac{\lambda_{n}}{\rho \exp (\rho \beta +1)}\bigl(E_{n-1}(F,\beta ) \bigr)^{\frac{\rho }{\lambda_{n}}}. \end{aligned}
Furthermore, if $$F(s)$$ is of the lower order λ and the lower type τ, and $$\lambda_{n}\sim \lambda_{n+1}$$ and the function
$$\psi (n)=\frac{\log A^{*}_{n}-\log A^{*}_{n+1}}{\lambda_{n+1}-\lambda _{n}}$$
forms a non-decreasing function of n for $$n>n_{0}$$, then we have
$$\lambda =\liminf_{n\rightarrow +\infty }\frac{\lambda_{n}\log \lambda _{n}}{-\log E_{n-1}(F,\beta )}, \qquad \tau = \liminf _{n\rightarrow + \infty }\frac{\lambda_{n}}{\rho \exp (\rho \beta +1)}\bigl(E_{n-1}(F, \beta ) \bigr)^{\frac{\rho }{\lambda_{n}}}.$$

### Theorem 4.2

If the L-S transform $$F(s)\in L_{\infty }$$, then for any real number $$-\infty <\beta <+\infty$$. For $$p=1$$, we have
$$\limsup_{\sigma \rightarrow +\infty }\frac{h(\log M_{u}(\sigma ,F))}{h( \sigma )}-1= \limsup _{n\rightarrow +\infty }\frac{h(\lambda_{n})}{h ( -\frac{1}{\lambda_{n}} \log [E_{n-1}(F,\beta )\exp (-\beta \lambda_{n})] ) },$$
and for $$p=2,3,\ldots$$ , we have
\begin{aligned} &\limsup_{n\rightarrow +\infty }\frac{h(\lambda_{n})}{h ( -\frac{1}{ \lambda_{n}} \log [E_{n-1}(F,\beta )\exp (-\beta \lambda_{n})] ) } \\ &\quad \leq \limsup _{\sigma \rightarrow +\infty }\frac{h(\log M_{u}(\sigma ,F))}{h( \sigma )} \\ &\quad \leq \limsup_{n\rightarrow +\infty }\frac{h(\lambda_{n})}{h ( -\frac{1}{ \lambda_{n}} \log [E_{n-1}(F,\beta )\exp (-\beta \lambda_{n})] ) }+1, \end{aligned}
where $$h(x)$$ satisfies the following conditions:
1. (i)

$$h(x)$$ is defined on $$[a,+\infty )$$ and is positive, strictly increasing, differentiable and tends to +∞ as $$x\rightarrow + \infty$$;

2. (ii)

$$\lim_{x\rightarrow +\infty } \frac{d(h(x))}{d(\log^{[p]}x)}=k\in (0,+\infty )$$, $$p\geq 1, p\in \mathbb{N}^{+}$$, where $$\log^{[0]}x=x, \log^{[1]}x=\log x$$ and $$\log^{[p]}x=\log (\log^{[p-1]}x)$$.

## Declarations

### Acknowledgements

We thank the referee(s) for reading the manuscript very carefully and making a number of valuable and kind comments which improved the presentation.

## Authors’ Affiliations

(1)
School of Mathematics and Statistics, Xidian University, Xi’an, Shaanxi, 710126, China
(2)
Department of Informatics and Engineering, Jingdezhen Ceramic Institute, Jingdezhen, Jiangxi, 333403, China

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