1 Introduction
Laplace–Stieltjes transforms
where \(\alpha(x)\) is a bounded variation on any finite interval \([0,Y]\) (\(0< Y<+\infty\)), and σ and t are real variables, as we know, if \(\alpha(t)\) is absolutely continuous, then \(F(s)\) becomes the classical Laplace integral of the form
If \(\alpha(t)\) is a step-function and satisfies
where the sequence \(\{\lambda_{n}\}_{0}^{\infty}\) satisfies
thus \(F(s)\) becomes a Dirichlet series
(σ, t are real variables), \(a_{n}\) are nonzero complex numbers. Obviously, if \(\alpha(t)\) is an increasing continuous function which is not absolutely continuous, then the integral (1) defines a class of functions \(F(s)\) which cannot be expressed either in the form (2) or (4).
$$ F(s)= \int_{0}^{+\infty}e^{sx}\,d\alpha(x),\quad s= \sigma+it, $$
(1)
$$ F(s)= \int_{0}^{\infty}e^{st}g(t)\,dt. $$
(2)
$$\alpha(x)= \textstyle\begin{cases} a_{1}+a_{2}+\cdots+a_{n}, &\lambda_{n}< x< \lambda_{n+1};\\ 0, &0\leq x< \lambda_{1};\\ \frac{\alpha(x+)+\alpha(x-)}{2},&x>0, \end{cases} $$
$$ 0=\lambda_{1}< \lambda_{2}< \lambda_{3}< \cdots< \lambda_{n}\uparrow+\infty, $$
(3)
$$ F(s)=\sum_{n=1}^{\infty}a_{n}e^{\lambda_{n}s},\quad s=\sigma+it. $$
(4)
Let a sequence \(\{\lambda_{n}\}_{n=1}^{\infty}\) satisfy (3), and
$$ \limsup_{n\rightarrow+\infty}(\lambda_{n+1}- \lambda_{n})=h< +\infty ,\qquad \limsup_{n\rightarrow+\infty} \frac{ n}{\lambda_{n}}=D< \infty. $$
(5)
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Set
if
it is easy to get \(\sigma_{u}^{F}=0\), that is, \(F(s)\) is analytic in the left half plane; if
it follows that \(\sigma_{u}^{F}=+\infty \), that is, \(F(s)\) is analytic in the whole plane. For convenience, we use \(\overline{L}_{\beta}\) to be a class of all the functions \(F(s)\) of the form (1) which are analytic in the half plane \(\Re s<\beta\) (\(-\infty<\beta<\infty\)) and the sequence \(\{\lambda_{n}\}\) satisfies (3) and (5); \(L_{0}\) to be the class of all the functions \(F(s)\) of the form (1) which are analytic in the half plane \(\Re s<0\) and the sequence \(\{\lambda_{n}\}\) satisfies (3), (5), and (6); and \(L_{\infty}\) to be the class of all the functions \(F(s)\) of the form (1) which are analytic in the whole plane \(\Re s<+\infty\) and the sequence \(\{ \lambda_{n}\}\) satisfies (3), (5), and (7). Thus, if \(-\infty<\beta<0\) and \(F(s)\in \overline{L}_{\beta}\), then \(F(s)\in L_{0}\).
$$A_{n}^{*}=\sup_{\lambda_{n}< x\leq \lambda_{n+1},-\infty< t< +\infty} \biggl\vert \int_{\lambda _{n}}^{x}e^{ity}\,d\alpha(y) \biggr\vert , $$
$$ \limsup_{n\rightarrow+\infty}\frac{\log A_{n}^{\ast}}{\lambda_{n}}=0, $$
(6)
$$ \limsup_{n\rightarrow+\infty}\frac{\log A_{n}^{\ast}}{\lambda _{n}}=-\infty , $$
(7)
In 1963, Yu [26] first proved the Valiron–Knopp–Bohr formula of the associated abscissas of bounded convergence, absolute convergence, and uniform convergence of Laplace–Stieltjes transform. Moreover, Yu [26] also estimated the growth of the maximal molecule \(M_{u}(\sigma,F)\), the maximal term \(\mu(\sigma ,F)\), by introducing the concepts of the order of \(F(s)\), and investigated the singular direction–Borel line of entire functions represented by Laplace–Stieltjes transforms converging in the whole complex plane. After his wonderful works, considerable attention has been paid to the value distribution and the growth of analytic functions represented by Laplace–Stieltjes transforms converging in the whole plane or the half plane (see [1, 3, 4, 6‐8, 11‐15, 18‐25, 27]).
Set
For \(F(s)\in L_{0}\), in view of \(M_{u}(\sigma,F)\rightarrow+\infty\) as \(\sigma\rightarrow0^{-}\), the concepts of order and type can be usually used in estimating the growth of \(F(s)\) precisely.
$$\begin{aligned} &\mu(\sigma,F)=\max_{n\in \mathbb{N}}\bigl\{ A_{n}^{*}e^{\lambda_{n}\sigma} \bigr\} \quad(\sigma< 0) , \\ & M_{u}(\sigma,F)=\sup _{0< x< +\infty,-\infty< t< +\infty} \biggl\vert \int _{0}^{x}e^{(\sigma+it)y}\,d\alpha(y) \biggr\vert \quad(\sigma< 0). \end{aligned} $$
Definition 1.1
If Laplace–Stieltjes transform (1) satisfies \(\sigma_{u}^{F}=0\) and
we call \(F(s)\) of order ρ in the left half plane, where \(\log^{+}x=\max\{\log x,0\}\). Furthermore, if \(\rho\in (0,+\infty)\), the type of \(F(s)\) is defined by
$$\limsup_{\sigma\rightarrow0^{-}}\frac{\log^{+}\log^{+}M_{u}(\sigma ,F)}{-\log(-\sigma)}=\rho, \quad0\leq\rho\leq+ \infty, $$
$$\limsup_{\sigma\rightarrow0^{-}}\frac{\log^{+}M_{u}(\sigma,F)}{(-\frac {1}{\sigma})^{\rho}}=T, \quad0\leq T\leq+\infty. $$
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Remark 1.1
However, if \(\rho=0\) and \(\rho=+\infty\), we cannot estimate the growth of such functions precisely by using the concept of type.
In 2012 and 2014, Luo and Kong [9, 10] investigated the growth of Laplace–Stieltjes transform converging on the whole plane and obtained the following.
Theorem 1.1
(see [10])
If the L-S transform
\(F(s)\in L_{\infty}\)
and is of order
ρ (\(0<\rho <\infty\)), then
$$\rho=\limsup_{n\rightarrow+\infty}\frac{\lambda_{n}\log\lambda _{n}}{-\log A_{n}^{*}}. $$
Theorem 1.2
(see [9])
If the L-S transform
\(F(s)\in L_{\infty}\), then for
\(p=1\), we have
and for
\(p=2,3,\ldots\) , we have
where
\(h(x)\)
satisfies the following conditions:
$$\limsup_{\sigma\rightarrow+\infty}\frac{h(\log M_{u}(\sigma ,F))}{h(\sigma)}-1= \limsup_{n\rightarrow+\infty} \frac{h(\lambda_{n})}{h (-\frac {1}{\lambda_{n}}\log A_{n}^{*} )}, $$
$$\begin{aligned} \limsup_{n\rightarrow+\infty}\frac{h(\lambda_{n})}{h (-\frac {1}{\lambda_{n}} \log A_{n}^{*} )}&\leq\limsup _{\sigma\rightarrow+\infty}\frac {h(\log M_{u}(\sigma,F))}{h(\sigma)} \\ &\leq\limsup_{n\rightarrow+\infty}\frac{h(\lambda_{n})}{h (-\frac{1}{\lambda_{n}}\log A_{n}^{*} )}+1, \end{aligned}$$
(i)
\(h(x)\)
is defined on
\([a,+\infty)\)
and is positive, strictly increasing, differentiable and tends to +∞ as
\(x\rightarrow+\infty\);
(ii)
\(\lim_{x\rightarrow+\infty}\frac {d(h(x))}{d(\log^{[p]}x)}=k\in(0,+\infty)\), \(p\geq1\), \(p\in\mathbb {N}^{+}\), where
\(\log^{[0]}x=x\), \(\log^{[1]}x=\log x\), and
\(\log ^{[p]}x=\log(\log^{[p-1]}x)\).
In this paper, the first aim is to investigate the growth of analytic functions represented by Laplace–Stieltjes transforms with generalized order converging in the half plane, and we obtain some theorems about the generalized order \(A_{n}^{*}\) and \(\lambda_{n}\), which are improvements of the previous results given by Luo and Kong [9, 10]. To state our results, we first introduce the following notations and definitions.
Let Γ be a class of continuous increasing functions \(\mathscr {A}\) such that \(\mathscr{A}(x)\geq0\) for \(x\geq x_{0}\), \(\mathscr {A}(x)=\mathscr{A}(x_{0})\) for \(x\leq x_{0}\) and on \([x_{0},+\infty)\) the function \(\mathscr{A}\) increases to +∞; and \(\Gamma^{0}\) be a class such that \(\Gamma^{0}\subset\Gamma\) and \(\mathscr {A}(x(1+o(1))=(1+o(1))\mathscr{A}(x)\) as \(x\rightarrow+\infty\), for \(\mathscr{A}\in\Gamma^{0}\); further, \(\mathscr{A}\in\Gamma^{0i}\) if \(\mathscr{A}\in\Gamma\) and for any \(\eta>0\), \(\mathscr{A}(\eta x)=(1+o(1))\mathscr{A}(x)\) as \(x\rightarrow+\infty\). Obviously, it follows \(\Gamma^{0i}\subset\Gamma^{0}\) and \(h(x)\in\Gamma\).
Definition 1.2
Let \(F(s)\in L_{0}\) and \(\mathscr{A}\in\Gamma\), \(\mathscr{B}\in \Gamma\). If
then \(\rho_{\mathscr{A}\mathscr{B}}\) is called generalized order of \(F(s)\).
$$\rho_{\mathscr{A}\mathscr{B}}(F)=\limsup_{\sigma\rightarrow 0^{-}}\frac{\mathscr{A}(\log M_{u}(\sigma,F))}{\mathscr{B}(-\frac {1}{\sigma})}, $$
Remark 1.2
Let \(\mathscr{A}(x)=\log x\) and \(\mathscr{B}=\log x\), then \(\rho _{\mathscr{A}\mathscr{B}}(F)=\rho\).
Remark 1.3
Let \(\mathscr{A}(x)=\log_{p} x\) and \(\mathscr{B}=\log_{q} x\), then \(\rho_{\mathscr{A}\mathscr{B}}(F)=\rho(p,q)(F)\), where \(\rho (p,q)(F)\) is the \((p,q)\)-order of \(F(s)\) (see [2]).
Remark 1.4
Let \(\mathscr{A}(x)=\log x\) and \(\mathscr{B}=\log\log x\), then \(\rho _{\mathscr{A}\mathscr{B}}(F)=\rho_{l}(F)\), where \(\rho_{l}(F)\) is the logarithmic order of \(F(s)\).
2 Results and discussion
For generalized order of Laplace–Stieltjes transform (1), we obtain the following.
Theorem 2.1
Let
\(F(s)\in L_{0}\), \(\mathscr{A}\in\Gamma^{0i}\)
and
\(\mathscr{B}\in \Gamma^{0i}\)
be continuously differentiable, and the function
\(\mathscr{B}\)
increase more rapidly than
\(\mathscr{A}\)
such that, for any constant
\(\eta\in(0,+\infty )\),
and
If
and
then
$$ \frac{x}{\mathscr{B}^{-1}(\eta\mathscr{A}(x))}\rightarrow+\infty \quad(x_{0}\leq x \rightarrow+\infty ), $$
(8)
$$ \mathscr{A} \biggl(\frac{x}{\mathscr{B}^{-1}(\eta\mathscr {A}(x))} \biggr)=\bigl(1+o(1)\bigr) \mathscr{A}(x) \quad(x\rightarrow+\infty ). $$
(9)
$$ \lim_{\sigma\rightarrow0^{-}}\frac{\log M_{u}(\sigma,F)}{-\log (-\sigma)}=+\infty $$
(10)
$$\rho_{\mathscr{A}\mathscr{B}}(F)= \limsup_{\sigma\rightarrow 0^{-}}\frac{\mathscr{A}(\log M_{u}(\sigma,F))}{\mathscr{B}(-\frac {1}{\sigma})}, \quad 0\leq\rho_{\mathscr{A}\mathscr{B}}(F)\leq +\infty , $$
$$\rho_{\mathscr{A}\mathscr{B}}(F)=\limsup_{n\rightarrow+\infty }\frac{\mathscr{A}(\lambda_{n})}{\mathscr{B} (\frac{\lambda _{n}}{\log A_{n}^{*}} )}. $$
Theorem 2.2
Let
\(F(s)\in L_{0}\), \(\mathscr{A}\in\Gamma^{0i}\), and
\(\mathscr{B}\in \Gamma^{0i}\)
be continuously differentiable, and the function
\(\mathscr{A}\)
increase more rapidly than
\(\mathscr{B}\)
such that, for any constant
\(\eta\in(0,+\infty )\),
and
If
\(F(s)\)
satisfies (10) and
then
$$ \frac{x}{\mathscr{A}^{-1}(\eta\mathscr{B}(x))}\uparrow+\infty \quad(x_{0}\leq x \rightarrow+\infty ), $$
(11)
$$ \mathscr{B} \biggl(\frac{x}{\mathscr{A}^{-1}(\eta\mathscr {B}(x))} \biggr)=\bigl(1+o(1)\bigr) \mathscr{B}(x) \quad(x\rightarrow+\infty ). $$
(12)
$$\rho_{\mathscr{A}\mathscr{B}}(F)= \limsup_{\sigma\rightarrow 0^{-}}\frac{\mathscr{A}(\log M_{u}(\sigma,F))}{\mathscr{B}(-\frac {1}{\sigma})}, \quad 0\leq\rho_{\mathscr{A}\mathscr{B}}(F)\leq +\infty , $$
$$\rho_{\mathscr{A}\mathscr{B}}(F)=\limsup_{n\rightarrow+\infty }\frac{\mathscr{A}(\log A_{n}^{*})}{\mathscr{B} (\lambda_{n} )}. $$
If Laplace–Stieltjes transform (1) satisfies \(A^{*}_{n}=0\) for \(n\geq k+1\) and \(A^{*}_{k}\neq0\), then \(F(s)\) will be said to be an exponential polynomial of degree k usually denoted by \(p_{k}\), i.e., \(p_{k}(s)=\int^{\lambda_{k}}_{0}\exp(sy)\,d\alpha(y)\). If 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, \(\sum_{i=1}^{k}b_{i}\exp(s\lambda_{i})\). We denote \(\Pi_{k}\) to be the class of all exponential polynomials of degree almost k, that is,
$$\Pi_{k}= \Biggl\{ \sum_{i=1}^{k} b_{i}\exp(s\lambda_{i}): (b_{1},b_{2}, \ldots ,b_{k})\in \mathbb{C}^{k} \Biggr\} . $$
For \(F(s)\in\overline{L}_{\beta}\), \(-\infty<\beta<0\), 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
where
$$E_{n}(F,\beta)=\inf_{p\in\Pi_{n}}\Vert F-p \Vert_{\beta}, \quad n=1,2,\ldots, $$
$$\Vert F-p\Vert_{\beta}=\max_{-\infty< t< +\infty} \bigl\vert F(\beta +it)-p(\beta+it) \bigr\vert . $$
In 2017, Singhal and Srivastava [17] studied the approximation of Laplace–Stieltjes transforms of finite order converging on the whole plane and obtained the following theorem.
Theorem 2.3
(see [17])
If Laplace–Stieltjes 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
and
$$\rho=\limsup_{n\rightarrow+\infty}\frac{\lambda_{n}\log\lambda _{n}}{-\log E_{n-1}(F,\beta)\exp(-\beta\lambda_{n})}=\limsup _{n\rightarrow+\infty}\frac{\lambda_{n}\log\lambda_{n}}{-\log E_{n-1}(F,\beta)} $$
$$T=\limsup_{n\rightarrow+\infty}\frac{\lambda_{n}}{\rho e}\bigl(E_{n-1}(F, \beta)\exp(-\beta\lambda_{n})\bigr)^{\frac{\rho}{\lambda_{n}}} =\limsup _{n\rightarrow+\infty}\frac{\lambda_{n}}{\rho\exp(\rho \beta+1)}\bigl(E_{n-1}(F,\beta) \bigr)^{\frac{\rho}{\lambda_{n}}}. $$
In the same year, the author and Kong [20] investigated the approximation of Laplace–Stieltjes transform \(F(s)\in L_{0}\) with infinite order and obtained the following.
Theorem 2.4
(see [20, Theorem 2.5])
If the Laplace–Stieltjes transform
\(F(s)\in L_{0}\)
and is of infinite order, if
\(\lambda_{n}\sim\lambda_{n+1}\), then for any real number
\(-\infty<\beta<+\infty\), then for any fixed real number
\(-\infty<\alpha<0\), we have
where
\(0<\rho^{*}<\infty\), \(X(\cdot)\)-order can be seen in [20].
$$ \limsup_{\sigma\rightarrow0^{-}}\frac{X(\log^{+}M_{u}(\sigma,F))}{\log (-\frac{1}{\sigma})}=\rho^{*} \quad \Longleftrightarrow\quad \limsup_{n\rightarrow\infty}\frac{X(\lambda_{n})}{\log^{+}\frac {\lambda_{n}}{\log^{+} [E_{n-1}(F,\alpha)\exp\{-\alpha\lambda _{n}\} ]}}=\rho^{*}, $$
(13)
The second purpose of this paper is to study the approximation of Laplace–Stieltjes transform \(F(s)\in L_{0}\) with generalized order, and our results are listed as follows.
Theorem 2.5
Let
\(F(s)\in L_{0}\), \(\mathscr{A}\in\Gamma^{0i}\), and
\(\mathscr{B}\in \Gamma^{0i}\)
be continuously differentiable satisfying (8) and (9), and let the function
\(\mathscr{B}\)
increase more rapidly than
\(\mathscr{A}\). If
\(F(s)\)
satisfies (10) and
then for any real number
\(-\infty <\beta<0\), we have
$$\rho_{\mathscr{A}\mathscr{B}}(F)= \limsup_{\sigma\rightarrow 0^{-}}\frac{\mathscr{A}(\log M_{u}(\sigma,F))}{\mathscr{B}(-\frac {1}{\sigma})}, \quad 0\leq\rho_{\mathscr{A}\mathscr{B}}(F)\leq +\infty , $$
$$\rho_{\mathscr{A}\mathscr{B}}(F)=\limsup_{n\rightarrow+\infty }\frac{\mathscr{A}(\lambda_{n})}{\mathscr{B} (\frac{\lambda _{n}}{\log [E_{n-1}(F,\beta)\exp\{-\beta\lambda_{n}\} ]} )}. $$
Theorem 2.6
Under the assumptions of Theorem
2.2, then for any real number
\(-\infty <\beta<0\), we have
$$\rho_{\mathscr{A}\mathscr{B}}(F)=\limsup_{n\rightarrow+\infty }\frac{\mathscr{A}(\log E_{n-1}(F,\beta))}{\mathscr{B} (\lambda_{n} )}. $$
3 Conclusions
Regarding Theorems 2.1 and 2.2, the generalized order of Laplace–Stieltjes transforms are discussed by using the more abstract functions, and some related theorems among \(\lambda_{n}\), \(A_{n}^{*}\) and the generalized order are obtained. Moreover, we also investigate some properties of approximation on analytic functions defined by Laplace–Stieltjes transforms of generalized order. For the topic of the growth and approximation of Laplace–Stieltjes transforms of generalized order, it seems that this topic has never been treated before. Our theorems are generalization and improvement of the previous results given by Luo and Kong [9, 10], Singhal and Srivastava [17].
4 Methods
To prove our results, we also need to give the following lemmas (see [16]).
Let \(\Xi_{0}\) denote the set of positive unbounded functions ϕ on \((-\infty,0)\) such that the derivative \(\phi'\) is positive, continuous, and increasing to +∞ on \((-\infty,0)\). Thus, if \(\phi\in\Xi_{0}\), then \(\phi(x)\rightarrow\zeta\geq0\) and \(\phi '(x)\rightarrow0\) as \(x\rightarrow-\infty\). Let φ be the inverse function of \(\phi'\), then φ is continuous on \((0,+\infty)\) and increases to 0. Set \(\phi\in\Xi_{0}\) and \(\psi (x)=x-\frac{\phi(x)}{\phi'(x)}\). For \(-\infty< x< x+\iota<0\), since \(\phi'\) is increasing on \((-\infty,0)\), we have
that is,
Thus, it means that ψ is an increasing function on \((-\infty,0)\).
$$\begin{aligned} \phi'(x)\phi(x+\iota)-\phi'(x+\iota)\phi(x)&< \phi'(x)\bigl[\phi (x+\iota)-\phi(x)\bigr] =\phi'(x) \int_{x}^{x+\iota}\phi'(t)\,dt \\ &< (x+\iota-x)\phi'(x)\phi'(x+\iota), \end{aligned}$$
$$\psi(x)=x-\frac{\phi(x)}{\phi'(x)}< x+\iota-\frac{\phi(x+\iota )}{\phi'(x+\iota)}=\psi(x+\iota). $$
Next, we will prove that \(\psi(x)\rightarrow0\) as \(x\rightarrow0\), that is, there is no constant \(\eta<0\) such that \(\psi(x)\leq\eta\) for all \(x\in(-\infty,0)\). Assume that there exist two constants η, \(K_{1}\) such that \(\psi(x)\leq\eta\) for all \(x\in(-\infty,0)\) and \(\eta< K_{1}<0\). Since ψ is an increasing function and \(\psi (x)< x<0\), then it follows \(\frac{\phi'(x)}{\phi(x)}\leq\frac{1}{x-\eta}\) for \(K_{1}\leq x<0\). Thus, it follows
Hence \(\phi(x)\leq\phi(K_{1})\frac{x-\eta}{K_{1}-\eta}\). In view of \(\phi'(x)\rightarrow+\infty\) (\(x\rightarrow0\)), we get a contradiction. Thus, it follows \(\psi(x)\rightarrow0\) as \(x\rightarrow0\).
$$\begin{aligned} \log\phi(x)&=\log\phi(K_{1})+ \int_{K_{1}}^{x}\frac{\phi'(t)}{\phi (t)}\,dt\leq\log \phi(K_{1})+ \int_{K_{1}}^{x}\frac{1}{t-\eta}\,dt \\ &=\log\phi(K_{1})+\log\frac{x-\eta}{K_{1}-\eta}. \end{aligned}$$
Besides, let \(\psi^{-1}\) be the inverse function of ψ. Then \(\psi^{-1}\) is an increasing function on \((-\infty, 0)\) and \(\phi '(\psi^{-1}(\sigma))\) increases to +∞ on \((-\infty,0)\).
Lemma 4.1
Let
\(\phi\in\Xi_{0}\), then the conclusion that
\(\log\mu(\sigma ,F)\leq\phi(\sigma)\)
for any
\(\sigma\in(-\infty,0)\)
holds if and only if
\(\log A_{n}^{*}\leq-\lambda_{n}\psi(\varphi(\lambda_{n}))\)
for all
\(n\geq0\).
Proof
Suppose that \(\log\mu(\sigma,F)\leq\phi(\sigma)\) for any \(\sigma \in(-\infty,0)\), then \(\log A_{n}^{*}\leq\phi(\sigma)-\sigma\lambda _{n}\) for all \(n>0\) and \(\sigma\in(-\infty,0)\). Thus, take \(\sigma =\varphi(\lambda_{n})\), it follows for all \(n\geq0\) that
On the contrary, assume that \(\log A_{n}^{*}\leq-\lambda_{n}\psi(\varphi (\lambda_{n}))\) for all \(n\geq0\). Since, for any \(\sigma<0\) and \(x<0\),
then it follows
$$\begin{aligned} \log A_{n}^{*}&\leq\phi\bigl(\varphi(\lambda_{n})\bigr)- \lambda_{n}\varphi(\lambda_{n}) =-\lambda_{n} \biggl(\varphi(\lambda_{n})-\frac{\phi(\varphi(\lambda _{n}))}{\phi'(\varphi(\lambda_{n}))} \biggr) =- \lambda_{n}\psi\bigl(\varphi(\lambda_{n})\bigr). \end{aligned}$$
$$(\sigma-x)\phi'(x)\leq \int_{x}^{\sigma}\phi'(t)\,dt=\phi(\sigma)- \phi(x), $$
$$\begin{aligned} \log\mu(\sigma,F)&\leq\max\bigl\{ -\lambda_{n}\psi\bigl(\varphi( \lambda _{n})\bigr)+\lambda_{n}\sigma: n\geq0\bigr\} \leq \max\bigl\{ -t\psi\bigl(\varphi(t)\bigr)+t\sigma: t\geq0\bigr\} \\ &=\max\bigl\{ -\phi'(x)\psi(x)+\sigma+\phi'(x): x>- \infty\bigr\} \\ &=\max\bigl\{ (\sigma-x)\phi'(x)+\phi(x): x>-\infty\bigr\} =\phi( \sigma). \end{aligned}$$
Therefore, this completes the proof of Lemma 4.1. □
Lemma 4.2
If the L-S transform
\(F(s)\in L_{\infty}\), then for any
σ (\(-\infty <\sigma<0\)) and
ε (>0), we have
where
\(p>2\)
and
C (≠0) are constants.
$$\frac{1}{p}\mu(\sigma,F)\leq M_{u}(\sigma,F)\leq C\mu\bigl((1- \varepsilon )\sigma,F\bigr)\frac{1}{-\sigma}, $$
Proof
We will adapt the method as in Yu [26] and Kong and Hong [5]. Set
In view of (5), there exists a positive number ξ satisfying \(0<\lambda_{n+1}-\lambda_{n}\leq\xi\) (\(n=1,2,3,\ldots\)). Thus, it yields \(e^{-\xi\sigma}<\frac{p}{2}\) for σ sufficiently close to 0−, where \(p>2\) is a constant. For \(x>\lambda_{n}\), it follows
Then, for any \(\sigma<0\) and any \(x\in(\lambda_{n},\lambda_{n+1}]\), it yields
which implies
$$I(x;\sigma+it)= \int_{0}^{x}\exp\bigl\{ (\sigma+it)y\bigr\} \,d \alpha(y). $$
$$\begin{aligned} \int_{\lambda_{n}}^{x}\exp\{ity\}\,d\alpha(y)&= \int_{\lambda_{n}}^{x}\exp\{ -\sigma y\}d_{y}I(y; \sigma+it) \\ &=I(y;\sigma+it)\exp\{-\sigma y\} \vert_{\lambda_{n}}^{x}+\sigma \int _{\lambda_{n}}^{x}\exp\{-\sigma y\}I(y;\sigma+it)\,dy. \end{aligned}$$
$$\begin{aligned} \biggl\vert \int_{\lambda_{n}}^{x}\exp\{ity\}\,d\alpha(y) \biggr\vert & \leq M_{u}(\sigma,F)\bigl[\exp\{-x\sigma\}+\exp\{-\sigma \lambda_{n}\}+ \bigl\vert \exp\{-x\sigma\}-\exp\{-\sigma \lambda_{n}\} \bigr\vert \bigr] \\ &\leq2 M_{u}(\sigma,F)\exp\bigl\{ -(\lambda_{n}+\xi)\sigma \bigr\} \leq pM_{u}(\sigma,F)\exp\{-\lambda_{n}\sigma\}, \end{aligned}$$
$$ \frac{1}{p}\mu(\sigma,F)\leq M_{u}(\sigma,F). $$
(14)
On the other hand, for any \(x>0\), it follows that there exists a positive integer \(n\in\mathbb{N}_{+}\) such that \(\lambda_{n}< x\leq \lambda_{n+1}\). Thus, it follows
Set \(I_{k}(x;it)=\int_{\lambda_{n}}^{x}\exp\{ity\}\,d\alpha(y)\) (\(\lambda _{k}< x\leq\lambda_{k+1}\)), then for any real number t and \(\sigma <0\), it follows \(|I_{k}(x;it)|\leq A_{k}^{*}\leq\mu(\sigma,F)e^{-\lambda _{k}\sigma}\). Thus, for any \(\varepsilon\in(0,1)\) and \(\sigma<0\), it yields \(|I_{k}(x;it)|\leq\mu((1-\varepsilon)\sigma,F)e^{-\lambda _{k}(1-\varepsilon)\sigma}\) and
Hence, we can deduce
In view of (5), for the above ε, there exists \(N_{1}\in N_{+}\) such that, for any \(n>N_{1}\), we have \(\lambda_{n} >\frac {n}{D+\varepsilon}\). Hence it follows for \(\sigma\rightarrow0^{-}\) that
where C is a constant on ε and (5). Therefore, this lemma is proved from (14) and (15). □
$$\int_{0}^{x}\exp\bigl\{ (\sigma+it)y\bigr\} \,d \alpha(y)=\sum_{k=1}^{n-1} \int _{\lambda_{k}}^{\lambda_{k+1}}\exp\bigl\{ (\sigma+it)y\bigr\} \,d \alpha(y) + \int_{\lambda_{n}}^{x}\exp\bigl\{ (\sigma+it)y\bigr\} \,d \alpha(y). $$
$$\begin{aligned} \int_{0}^{x}\exp\bigl\{ (\sigma+it)y\bigr\} \,d \alpha(y)&=\sum_{k=1}^{n-1}\biggl[\exp\{ \lambda_{k+1}\sigma\}I_{k}(\lambda_{k+1};it)-\sigma \int_{\lambda _{k}}^{\lambda_{k+1}}\exp\{\sigma y\}I_{k}(y;it) \,dy\biggr] \\ &\quad{}+\exp\{x\sigma\}I_{n}(x;it)-\sigma \int_{\lambda_{n}}^{x}\exp\{\sigma y\}I_{n}(y;it) \,dy. \end{aligned}$$
$$\begin{aligned} & \biggl\vert \int_{0}^{x}\exp\bigl\{ (\sigma+it)y\bigr\} \,d \alpha(y) \biggr\vert \\ &\quad \leq \sum_{k=1}^{n-1}\mu\bigl((1- \varepsilon)\sigma,F\bigr)\exp\bigl\{ -\lambda _{k}(1-\varepsilon)\sigma \bigr\} \bigl(\exp\{\lambda_{k+1}\sigma\}+ \bigl\vert \exp\{\lambda _{k+1}\sigma\}-\exp\{\lambda_{k}\sigma\} \bigr\vert \bigr) \\ &\qquad {} +\mu\bigl((1-\varepsilon)\sigma,F\bigr)\exp\bigl\{ -\lambda_{n}(1- \varepsilon )\sigma\bigr\} \bigl(\exp\{x\sigma\}+ \bigl\vert \exp\{x\sigma\}- \exp\{\lambda _{n}\sigma\} \bigr\vert \bigr) \\ &\quad =\mu\bigl((1-\varepsilon)\sigma,F\bigr)\sum_{k=1}^{+\infty } \exp\{\lambda _{k}\varepsilon\sigma\}. \end{aligned}$$
$$ \sum_{k=1}^{+\infty}\exp\{ \lambda_{k}\varepsilon\sigma\}\leq\sum_{k=1}^{N_{1}} \exp\{\lambda_{k}\varepsilon\sigma\}+ \sum_{k=N_{1}+1}^{+\infty } \exp \biggl\{ k\frac{\varepsilon\sigma }{D+\varepsilon} \biggr\} < C \frac{1}{-\sigma}, $$
(15)
4.1 Proofs of Theorems 2.1 and 2.2
4.1.1 The proof of Theorem 2.1
Suppose that \(\rho:=\rho_{\mathscr{A}\mathscr{B}}(F)<+\infty \) and
In view of the definition of generalized order and Lemma 4.2, for any \(\varepsilon>0\), there exists a constant \(\sigma_{0}<0\) such that, for all \(0>\sigma>\sigma_{0}\),
that is,
Choosing
we conclude from (9) and (16) that
which implies
Since \(\mathscr{A}\in\Gamma^{0i}\), \(\mathscr{B}\in\Gamma^{0i}\) and let \(\varepsilon\rightarrow0^{+}\), we can conclude from (17) that \(\vartheta\leq\rho\).
$$\vartheta=\limsup_{n\rightarrow+\infty}\frac{\mathscr{A}(\lambda _{n})}{\mathscr{B} (\frac{\lambda_{n}}{\log A_{n}^{*}} )}. $$
$$ \log\mu(\sigma,F)\leq\mathscr{A}^{-1} \biggl((\rho+\varepsilon ) \mathscr{B} \biggl(-\frac{1}{\sigma} \biggr) \biggr)+\log p, $$
$$ \log A_{n}^{*}\leq\mathscr{A}^{-1} \biggl(( \rho+\varepsilon)\mathscr {B} \biggl(-\frac{1}{\sigma} \biggr) \biggr)- \lambda_{n}\sigma+\log p, \quad n\geq0. $$
(16)
$$-\frac{1}{\sigma}=\mathscr{B}^{-1} \biggl(\frac{1}{\rho+\varepsilon } \mathscr{A} \biggl(\frac{\lambda_{n}}{\mathscr{B}^{-1} (\frac {\mathscr{A}(\lambda_{n})}{\rho+\varepsilon} )} \biggr) \biggr), $$
$$\begin{aligned} \log A_{n}^{*}&\leq\frac{\lambda_{n}}{\mathscr{B}^{-1} (\frac{\mathscr{A}(\lambda_{n})}{\rho+\varepsilon} )}+\frac {\lambda_{n}}{\mathscr{B}^{-1} (\frac{1}{\rho+\varepsilon} \mathscr{A} (\frac{\lambda_{n}}{\mathscr{B}^{-1} (\frac {\mathscr{A}(\lambda_{n})}{\rho+\varepsilon} )} ) )}+\log p \\ &=\frac{(1+o(1))\lambda_{n}}{\mathscr{B}^{-1} ((1+o(1))\frac {\mathscr{A}(\lambda_{n})}{\rho+\varepsilon} )},\quad \mbox{as } n\rightarrow+\infty , \end{aligned}$$
$$ \mathscr{A}(\lambda_{n})\leq(\rho+\varepsilon) \mathscr{B} \biggl(\frac{(1+o(1))\lambda_{n}}{\log A_{n}^{*}} \biggr), \quad\mbox{as } n\rightarrow+\infty . $$
(17)
Assume \(\vartheta< \rho\), then we can choose a constant \(\rho_{1}\) such that \(\vartheta<\rho_{1}<\rho\). Since \(\mathscr{B}^{-1} (\frac {\mathscr{A}(x)}{\rho_{1}} )\) is an increasing function, then there exists a positive integer \(n_{0}\) such that, for \(n\geq n_{0}\),
where here and further \(K_{j}\) is a constant.
$$ \log A_{n}^{*}\leq\frac{\lambda_{n}}{\mathscr{B}^{-1} (\frac {\mathscr{A}(\lambda_{n})}{\rho_{1}} )}\leq \int_{\lambda _{n_{0}}}^{\lambda_{n}}\frac{1}{\mathscr{B}^{-1} (\frac{\mathscr {A}(t)}{\rho_{1}} )}\,dt+K_{1}, $$
(18)
Since \(\phi\in\Xi_{0}\), and let
Then it follows
and
$$\phi(\sigma)= \int_{-\frac{1}{\sigma_{0}}}^{-\frac{1}{\sigma}}\frac {\mathscr{A}^{-1}(\rho_{1}\mathscr{B}(t))}{t^{2}}\,dt+K_{2} \quad \mbox{for } 0>\sigma\geq\sigma_{0}. $$
$$\begin{aligned} &\phi'(\sigma)=\mathscr{A}^{-1}\biggl(\rho_{1} \mathscr{B}\biggl(-\frac{1}{\sigma }\biggr)\biggr), \qquad \varphi( \lambda_{n})=-\frac{1}{\mathscr{B}^{-1} (\frac{\mathscr{A}(\lambda_{n})}{\rho_{1}} )}, \\ &\bigl[\lambda_{n}\psi\bigl(\varphi(\lambda_{n})\bigr) \bigr]'=\bigl[\lambda_{n}\varphi(\lambda _{n})-\phi\bigl(\varphi(\lambda_{n})\bigr) \bigr]'=\varphi(\lambda_{n}), \end{aligned} $$
$$ -\lambda_{n}\psi\bigl(\varphi(\lambda_{n}) \bigr)=- \int_{\lambda_{n_{0}}}^{\lambda _{n}}\varphi(t)\,dt+K_{2}= \int_{\lambda_{n_{0}}}^{\lambda_{n}}\frac {1}{\mathscr{B}^{-1} (\frac{\mathscr{A}(t)}{\rho_{1}} )}\,dt+K_{2}. $$
(19)
Thus, in view of (18) and (19), it follows
Since \(\mathscr{A}\in\Gamma^{0i}\), in view of (10), (20) and by applying Lemma 4.2, we can deduce \(\rho_{\mathscr {A}\mathscr{B}}(F)\leq\rho_{1}\), which implies a contradiction with \(\rho_{\mathscr{A}\mathscr {B}}(F)>\rho_{1}\). Hence \(\vartheta=\rho_{\mathscr{A}\mathscr{B}}(F)\).
$$\begin{aligned} \log\mu(\sigma,F)&\leq\phi(\sigma)= \int_{-\frac {1}{\sigma_{0}}}^{-\frac{1}{\sigma}}\frac{\mathscr{A}^{-1}(\rho _{1}\mathscr{B}(t))}{t^{2}}\,dt+K_{2} \\ &\leq\mathscr{A}^{-1} \biggl(\rho_{1}\mathscr{B} \biggl(-\frac {1}{\sigma} \biggr) \biggr) \int_{-\frac{1}{\sigma_{1}}}^{-\frac {1}{\sigma}}\frac{dt}{t^{2}}+K_{3} \\ &\leq-\sigma_{1} \mathscr{A}^{-1} \biggl( \rho_{1}\mathscr{B} \biggl(-\frac{1}{\sigma} \biggr) \biggr)+K_{3}. \end{aligned}$$
(20)
If \(\rho_{\mathscr{A}\mathscr{B}}(F)=+\infty \), by using the same argument as above, it is easy to prove that the conclusion is true. Therefore, this completes the proof of Theorem 2.1.
4.1.2 The proof of Theorem 2.2
Suppose that \(\rho:=\rho_{\mathscr{A}\mathscr{B}}(F)<+\infty \) and
In view of the definition of generalized order and Lemma 4.2, for any \(\varepsilon>0\), there exists a constant \(\sigma_{0}<0\) such that, for all \(0>\sigma>\sigma_{0}\),
that is,
Choosing \(-\frac{1}{\sigma}=\lambda_{n}\), we conclude from (21) that
Since \(\mathscr{A}\in\Gamma^{0i}\), \(\mathscr{B}\in\Gamma^{0i}\) and let \(\varepsilon\rightarrow0^{+}\), we can conclude from (22) that \(\vartheta_{1}\leq\rho\).
$$\vartheta_{1}=\limsup_{n\rightarrow+\infty}\frac{\mathscr{A}(\log A_{n}^{*})}{\mathscr{B} (\lambda_{n} )}. $$
$$ \log\mu(\sigma,F)\leq\mathscr{A}^{-1} \biggl((\rho+\varepsilon ) \mathscr{B} \biggl(-\frac{1}{\sigma} \biggr) \biggr)+\log p, $$
$$ \log A_{n}^{*}\leq\mathscr{A}^{-1} \biggl(( \rho+\varepsilon)\mathscr {B} \biggl(-\frac{1}{\sigma} \biggr) \biggr)- \lambda_{n}\sigma+\log p, \quad n\geq0. $$
(21)
$$\begin{aligned} \log A_{n}^{*}&\leq\mathscr{A}^{-1}\bigl(( \rho+\varepsilon)\mathscr {B}(\lambda_{n})\bigr)+1+\log p \\ &\leq \bigl(1+o(1)\bigr)\mathscr{A}^{-1}\bigl((\rho +\varepsilon)\mathscr{B}( \lambda_{n})\bigr),\quad \mbox{as } n\rightarrow +\infty . \end{aligned}$$
(22)
Assume \(\vartheta_{1}< \rho\), then we can choose a constant \(\rho_{2}\) such that \(\vartheta_{1}<\rho_{2}<\rho\). It means that there exists a positive integer \(n_{0}\) such that, for \(n\geq n_{0}\),
that is,
$$\log A_{n}^{*}\leq\mathscr{A}^{-1}\bigl(\rho_{2} \mathscr{B}(\lambda_{n})\bigr), $$
$$ \log\mu(\sigma,F)\leq\max\bigl\{ \mathscr{A}^{-1}\bigl( \rho_{2}\mathscr {B}(\lambda_{n})\bigr)+\lambda_{n} \sigma: n\geq n_{0}\bigr\} +K_{5}. $$
(23)
In view of (10), the following equation
has a unique solution \(t_{1}:=t(\sigma)\) such that \(t_{1}\uparrow+\infty \) as \(\sigma\rightarrow0^{-}\), and for \(t\geq t_{1}\) we can deduce that \(\mathscr{A}^{-1}(\rho_{2}\mathscr{B}(t))+t\sigma\leq0\). Hence, it follows
In view of
it follows from (12) that
Hence, we can deduce from (24) and (25) that
which implies \(\rho_{\mathscr{A}\mathscr{B}}(F)\leq\rho_{2}<\rho _{\mathscr{A}\mathscr{B}}(F)\) by combining Lemma 4.2 and (10), a contradiction. Therefore, \(\vartheta_{1}=\rho_{\mathscr {A}\mathscr{B}}(F)\).
$$\mathscr{A}^{-1}\bigl(\rho_{2}\mathscr{B}(t)\bigr)+t\sigma=0 $$
$$\begin{aligned} \log\mu(\sigma,F)&\leq\max\bigl\{ \mathscr{A}^{-1}\bigl(\rho _{2}\mathscr{B}(t)\bigr)+t\sigma: t_{0}\leq t\leq t_{1}\bigr\} +K_{6} \\ &\leq\mathscr{A}^{-1}\bigl(\rho_{2}\mathscr{B}(t_{1}) \bigr)+K_{6}. \end{aligned}$$
(24)
$$-\frac{1}{\sigma}=\frac{t_{1}}{\mathscr{A}^{-1}(\rho_{2}\mathscr{B}(t_{1}))}, $$
$$ \mathscr{B} \biggl(-\frac{1}{\sigma} \biggr)=\mathscr{B} \biggl(\frac {t_{1}}{\mathscr{A}^{-1}(\rho_{2}\mathscr{B}(t_{1}))} \biggr)=\bigl(1+o(1)\bigr)\mathscr{B}(t_{1}), \quad\sigma\rightarrow0^{-}. $$
(25)
$$\log\mu(\sigma,F)\leq\mathscr{A}^{-1} \biggl(\rho _{2} \bigl(1+o(1)\bigr)\mathscr{B} \biggl(-\frac{1}{\sigma} \biggr) \biggr),\quad \mbox{as } \sigma\rightarrow0^{-}, $$
If \(\rho_{\mathscr{A}\mathscr{B}}(F)=+\infty \), by using the same argument as above, it is easy to prove that the conclusion is true. Therefore, this completes the proof of Theorem 2.2.
4.2 Proofs of Theorems 2.5 and 2.6
4.2.1 The proof of Theorem 2.5
Suppose that \(\rho:=\rho_{\mathscr{A}\mathscr{B}}(F)<+\infty \) and
In view of the definition of generalized order and Lemma 4.2, for any \(\varepsilon>0\), there exists a constant \(\sigma_{0}<0\) such that, for all \(0>\sigma>\sigma_{0}\),
Since \(F(s)\in L_{0}\), and for any constant β (\(-\infty<\beta<0\)), then \(F(s)\in \overline{L}_{\beta}\). Hence, for \(\beta<\sigma<0\) and \(p_{k}\in\Pi _{k}\), it follows
$$\vartheta_{3}=\limsup_{n\rightarrow+\infty}\frac{\mathscr {A}(\lambda_{n})}{\mathscr{B} (\frac{\lambda_{n}}{\log [E_{n-1}(F,\beta)\exp\{-\beta\lambda_{n}\} ]} )}. $$
$$ \log M_{u}(\sigma,F)\leq\mathscr{A}^{-1} \biggl((\rho+\varepsilon )\mathscr{B} \biggl(-\frac{1}{\sigma} \biggr) \biggr). $$
(26)
$$\begin{aligned} E_{k}(F,\beta) &\leq\Vert F-p_{k} \Vert_{\beta}\leq \bigl\vert F(\beta +it)-p_{k}(\beta+it) \bigr\vert \\ &\leq \biggl\vert \int_{0}^{+\infty}\exp\{sy\}\,d\alpha(y)- \int _{0}^{\lambda_{k}}\exp\{sy\}\,d\alpha(y) \biggr\vert = \biggl\vert \int_{\lambda_{k}}^{\infty}\exp\{sy\}\,d\alpha(y) \biggr\vert . \end{aligned}$$
(27)
Let
then \(|I_{j+k}(b;it)|\leq A_{j+k}^{*}\). In view of
where \(-\infty <\beta<0\), hence
Therefore, we conclude
In view of Lemma 4.2, it follows \(A_{n}^{*}\leq p M_{u}(\sigma ,F)e^{-\sigma\lambda_{n}}\). So, for any σ (\(\beta<\sigma<0\)), it yields from (27) and (28) that
$$I_{j+k}(b;it)= \int_{\lambda_{j+k}}^{b}\exp\{ity\}\,d\alpha(y) \quad ( \lambda_{j+k}< b\leq\lambda_{j+k+1}), $$
$$\biggl\vert \int_{\lambda_{k}}^{\infty}\exp\bigl\{ (\beta+it)y\bigr\} \,d\alpha (y) \biggr\vert =\lim_{b\rightarrow+\infty} \biggl\vert \int_{\lambda _{k}}^{b}\exp\bigl\{ (\beta+it)y\bigr\} \,d \alpha(y) \biggr\vert , $$
$$\begin{aligned} & \biggl\vert \int_{\lambda_{k}}^{b}\exp\bigl\{ (\beta+it)y\bigr\} \,d\alpha (y) \biggr\vert \\ &\quad = \Biggl\vert \sum_{j=k}^{n+k-1} \int_{\lambda_{j}}^{\lambda _{j+1}}\exp\{\beta y\}d_{y}I_{j}(y;it)+ \int_{\lambda_{n+k}}^{b}\exp\{\beta y\}d_{y}I_{n+k}(y;it) \Biggr\vert \\ &\quad = \Biggl\vert \Biggl[\sum_{j=k}^{n+k-1}e^{\lambda_{j+1}\beta }I_{j}( \lambda_{j+1};it)-\beta \int_{\lambda_{j}}^{\lambda _{j+1}}e^{\beta y}I_{j}(y;it) \,dy \Biggr] \\ &\qquad{}+e^{\beta b}I_{n+k}(b;it)-\beta \int_{\lambda _{n+k}}^{b}e^{\beta y}I_{j}(y;it) \,dy \Biggr\vert \\ &\quad \leq \sum_{j=k}^{n+k-1} \bigl[A_{j}^{*}e^{\lambda_{j+1}\beta }+A_{j}^{*} \bigl(e^{\lambda_{j+1}\beta}-e^{\lambda_{j}\beta}\bigr) \bigr] +2e^{\beta\lambda_{n+k+1}}A_{n+k}^{*}-e^{\beta\lambda _{n+k}}A_{n+k}^{*} \\ &\quad \leq 2\sum_{j=k}^{n+k} A_{n}^{*}e^{\lambda_{n+1}\beta}. \end{aligned}$$
$$ \biggl\vert \int_{\lambda_{k}}^{\infty}\exp\bigl\{ (\beta+it)y\bigr\} \,d\alpha (y) \biggr\vert \leq2\sum_{n=k}^{+\infty}A_{n}^{*} \exp\{\beta\lambda _{n+1}\}, \quad \mbox{as } n\rightarrow+\infty . $$
(28)
$$ E_{n}(F,\beta)\leq2\sum_{k=n+1}^{\infty}A_{k-1}^{*}\exp\{\beta\lambda _{k}\}\leq2pM_{u}( \sigma,F)\sum_{k=n+1}^{\infty}\exp\bigl\{ (\beta- \sigma )\lambda_{k}\bigr\} . $$
(29)
In view of (5), we can choose \(h'\) (\(0< h'< h\)) such that \((\lambda_{n+1} -\lambda_{n})\geq h'\) for \(n\geq0\). Then, for \(\sigma\geq \frac{\beta}{2}\), it follows from (29) that
that is,
where K is a constant. Hence, it follows from (26) and (30) that
Let
we conclude from (9) and (31) that
which implies
Since \(\mathscr{A}\in\Gamma^{0i}\), \(\mathscr{B}\in\Gamma^{0i}\) and let \(\varepsilon\rightarrow0^{+}\), we can conclude from (32) that \(\vartheta_{3}\leq\rho\).
$$\begin{aligned} E_{n}(F,\beta) &\leq M_{u}(\sigma,F)\exp\bigl\{ \lambda_{n+1}(\beta-\sigma)\bigr\} \sum_{k=n+1}^{\infty}\exp\bigl\{ (\lambda_{k}-\lambda_{n+1}) (\beta-\sigma)\bigr\} \\ &\leq M_{u}(\sigma,F)\exp\bigl\{ \lambda_{n+1}(\beta-\sigma) \bigr\} \exp \biggl\{ -\frac{\beta}{2}h'(n+1) \biggr\} \sum _{k=n+1}^{\infty}\exp \biggl\{ \frac{\beta}{2}h'k \biggr\} \\ &=M_{u}(\sigma,F)\exp\bigl\{ \lambda_{n+1}(\beta-\sigma)\bigr\} \biggl(1-\exp\biggl\{ \frac{\beta}{2}h'\biggr\} \biggr)^{-1}, \end{aligned}$$
$$ E_{n-1}(F,\beta)\leq KM_{u}(\sigma,F)\exp\bigl\{ \lambda_{n}(\beta-\sigma )\bigr\} , $$
(30)
$$ \log\bigl[E_{n-1}(F,\beta)\exp\{-\beta \lambda_{n}\}\bigr]\leq\mathscr {A}^{-1} \biggl((\rho+ \varepsilon)\mathscr{B} \biggl(-\frac{1}{\sigma } \biggr) \biggr)- \lambda_{n}\sigma+\log K, \quad n\geq0. $$
(31)
$$-\frac{1}{\sigma}=\mathscr{B}^{-1} \biggl(\frac{1}{\rho+\varepsilon } \mathscr{A} \biggl(\frac{\lambda_{n}}{\mathscr{B}^{-1} (\frac {\mathscr{A}(\lambda_{n})}{\rho+\varepsilon} )} \biggr) \biggr), $$
$$\begin{aligned} \log\bigl[E_{n-1}(F,\beta)\exp\{-\beta\lambda_{n}\} \bigr]&\leq\frac {\lambda_{n}}{\mathscr{B}^{-1} (\frac{\mathscr{A}(\lambda _{n})}{\rho+\varepsilon} )}+\frac{\lambda_{n}}{\mathscr {B}^{-1} (\frac{1}{\rho+\varepsilon} \mathscr{A} (\frac{\lambda_{n}}{\mathscr{B}^{-1} (\frac {\mathscr{A}(\lambda_{n})}{\rho+\varepsilon} )} ) )}+\log K \\ &=\frac{(1+o(1))\lambda_{n}}{\mathscr{B}^{-1} ((1+o(1))\frac {\mathscr{A}(\lambda_{n})}{\rho+\varepsilon} )}, \quad\mbox{as } n\rightarrow+\infty , \end{aligned}$$
$$ \mathscr{A}(\lambda_{n})\leq(\rho+\varepsilon) \mathscr{B} \biggl(\frac{(1+o(1))\lambda_{n}}{\log[E_{n-1}(F,\beta)\exp\{-\beta \lambda_{n}\}]} \biggr), \quad \mbox{as } n\rightarrow+ \infty . $$
(32)
Assume \(\vartheta_{3}< \rho\), then we can choose a constant \(\rho_{3}\) such that \(\vartheta_{3}<\rho_{3}<\rho\). Since \(\mathscr{B}^{-1} (\frac {\mathscr{A}(x)}{\rho_{3}} )\) is an increasing function, then there exists a positive integer \(n_{0}\) such that, for \(n\geq n_{0}\),
$$ \log\bigl[E_{n-1}(F,\beta)\exp\{-\beta \lambda_{n}\}\bigr]\leq\frac{\lambda _{n}}{\mathscr{B}^{-1} (\frac{\mathscr{A}(\lambda_{n})}{\rho _{3}} )}\leq \int_{\lambda_{n_{0}}}^{\lambda_{n}}\frac {1}{\mathscr{B}^{-1} (\frac{\mathscr{A}(t)}{\rho_{3}} )}\,dt+K_{1}. $$
(33)
For any \(\beta<0\), then there exists \(p_{1}\in\Pi_{n-1}\) such that
And since
thus, for any \(p\in\Pi_{n-1}\), it follows
Hence, for any \(\beta<0\) and \(F(s)\in L_{0}\), it follows from (34) and (35) that
Hence, (18) follows from (33) and (36).
$$ \Vert F-p_{1} \Vert \leq2E_{n-1}(F,\beta). $$
(34)
$$\begin{aligned} A_{n}^{*}\exp\{\beta\lambda_{n}\}& =\sup _{\lambda_{n}< x\leq\lambda_{n+1},-\infty< t< +\infty } \biggl\vert \int_{\lambda_{n}}^{x}\exp\{ity\}\,d\alpha(y) \biggr\vert \exp\{\beta\lambda_{n}\} \\ &\leq\sup_{\lambda_{n}< x\leq\lambda_{n+1},-\infty< t< +\infty } \biggl\vert \int_{\lambda_{n}}^{x}\exp\bigl\{ (\beta+it)y\bigr\} \,d \alpha(y) \biggr\vert \\ &\leq\sup_{-\infty< t< +\infty} \biggl\vert \int_{\lambda _{n}}^{\infty}\exp\bigl\{ (\beta+it)y\bigr\} \,d \alpha(y) \biggr\vert , \end{aligned}$$
$$ A_{n}^{*}\exp\{\beta\lambda_{n}\}\leq \bigl\vert F(\beta+it)-p(\beta +it) \bigr\vert \leq \Vert F-p \Vert _{\beta}. $$
(35)
$$ A_{n}^{*}\leq2E_{n-1}(F,\beta)\exp\{-\beta \lambda_{n}\}. $$
(36)
Since \(\phi\in\Xi_{0}\), and let
and
By using the same argument as in the proof of Theorem 2.1, we conclude
Since \(\mathscr{A}\in\Gamma^{0i}\), in view of (10), (37) and by applying Lemma 4.2, we can deduce \(\rho_{\mathscr {A}\mathscr{B}}(F)\leq\rho_{3}\), which implies a contradiction with \(\rho_{\mathscr{A}\mathscr {B}}(F)>\rho_{1}\). Hence \(\vartheta_{3}=\rho_{\mathscr{A}\mathscr{B}}(F)\).
$$\phi(\sigma)= \int_{-\frac{1}{\sigma_{0}}}^{-\frac{1}{\sigma}}\frac {\mathscr{A}^{-1}(\rho_{1}\mathscr{B}(t))}{t^{2}}\,dt+K_{2} \quad \mbox{for } 0>\sigma\geq\sigma_{0}, $$
$$\varphi(\lambda_{n})=-\frac{1}{\mathscr{B}^{-1} (\frac{\mathscr {A}(\lambda_{n})}{\rho_{1}} )}. $$
$$\begin{aligned} \log\mu(\sigma,F)&\leq\phi(\sigma)= \int_{-\frac {1}{\sigma_{0}}}^{-\frac{1}{\sigma}}\frac{\mathscr{A}^{-1}(\rho _{1}\mathscr{B}(t))}{t^{2}}\,dt+K_{2} \\ &\leq-\sigma_{1} \mathscr{A}^{-1} \biggl( \rho_{1}\mathscr{B} \biggl(-\frac{1}{\sigma} \biggr) \biggr)+K_{3}. \end{aligned}$$
(37)
If \(\rho_{\mathscr{A}\mathscr{B}}(F)=+\infty \), by using the same argument as above, it is easy to prove that the conclusion is true. Therefore, this completes the proof of Theorem 2.5.
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.
Competing interests
The authors declare that none of the authors have any competing interests in the manuscript.
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