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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 ).

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  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  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 ).

In 2012, Luo and Kong  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 , 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 ), and Shang, Gao, and Sun investigated the value distribution of such functions (see ). 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 .

## 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 , 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 , 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 .

### 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 $$\log^{}x=x, \log^{}x=\log x$$ 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.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

## 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

## References

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