Relative Growth of Series in Systems of Functions and Laplace—Stieltjes-Type Integrals

Abstract

For a regularly converging-in-$\mathbb{C}$ series $A\left(z\right)={\sum }_{n=1}^{\infty }{a}_{n}f\left({\lambda }_{n}z\right),$ where f is an entire transcendental function, the asymptotic behavior of the function $M_f^{-1}\left(M_A\left(r\right)\right),$ where $M_f\left(r\right)=max\left\{\right|f\left(z\right)|:|z|=r\}$, is investigated. It is proven that, under certain conditions on the functions f, $\alpha$, and the coefficients $a_n$, the equality ${lim}_{r\to +\infty }\frac{\alpha \left(M_f^{-1}\left(M_A\left(r\right)\right)\right)}{\alpha \left(r\right)}=1$ is correct. A similar result is obtained for the Laplace–Stiltjes-type integral $I\left(r\right)={\int }_0^{\infty }a\left(x\right)f\left(rx\right)dF\left(x\right).$ Unresolved problems are formulated.

1. Introduction

Let

$$f\left(z\right)=\sum _{k=0}^{\infty }{f}_k{z}^k$$

(1)

be an entire function, $M_f\left(r\right)=max\left\{\right|f\left(z\right)|:|z|=r\}$, and ${\mathsf{\Phi }}_f\left(r\right)=ln{M}_f\left(r\right)$. For an entire function g with Taylor coefficients $g_n$, the study of growth of the function ${\mathsf{\Phi }}_f^{-1}(ln{M}_g\left(r\right))$ in terms of the exponential type was initiated in papers [1,2] and was continued in [3]. As a result, it is proven that, if $|f_{k-1}/f_k|\nearrow +\infty$ as $k\to \infty$, then

$$\underset{r\to +\infty }{\overline{lim}}\frac{{\mathsf{\Phi }}_f^{-1}(ln{M}_g\left(r\right))}{r}=\underset{k\to \infty }{\overline{lim}}{\left(\frac{|g_n|}{|f_n|}\right)}^{1/n}.$$

We remark that ${\mathsf{\Phi }}_f^{-1}\left(x\right)=M_f^{-1}\left(e^x\right)$ and, thus, ${\mathsf{\Phi }}_f^{-1}(ln{M}_g\left(r\right))=M_f^{-1}\left(M_g\left(r\right)\right)$. The order $\rho {\left[g\right]}_g={\overline{lim}}_{r\to +\infty }\frac{ln{M}_f^{-1}\left(M_g\left(r\right)\right)}{lnr}$ and the lower-order $\lambda {\left[g\right]}_f={\underline{lim}}_{r\to +\infty }\frac{ln{M}_f^{-1}\left(M_g\left(r\right)\right)}{lnr}$ of the function f with respect to the function g are used in Reference [4]. Research on the relative growth of entire functions was continued by many mathematicians (an incomplete bibliography is given in [5]).

Let $\left({\lambda }_n\right)$ be a sequence of positive numbers increasing to $+\infty$. Suppose that the series

$$A\left(z\right)=\sum _{n=1}^{\infty }{a}_{n}f\left({\lambda }_{n}z\right)$$

(2)

in the system $f\left({\lambda }_{n}z\right)$ is regularly convergent in $\mathbb{C}$, i.e., ${\sum }_{n=1}^{\infty }\left|a_n\right|M_f\left(r{\lambda }_n\right)<+\infty$ for all $r\in [0,+\infty )$. Many authors have studied the representation of analytic functions by series in the system $f\left({\lambda }_{n}z\right)$ and the growth of such functions. Here, we specify only the monographs of A.F. Leont’ev [6] and B.V. Vinnitskyi [3], which are references to other papers on this topic.

Since series (2) is regularly convergent in $\mathbb{C}$ and the function A is an entire function, a natural question arises about the asymptotic behavior of the function $M_f^{-1}\left(M_A\left(r\right)\right).$

We suppose that the function F is nonnegative, nondecreasing, unbounded, and continuous on the right on $[0,+\infty );$ that f is positive, increasing, and continuous on $[0,+\infty )$; and that a positive-on-$[0,+\infty )$ function a is such that the Laplace–Stietjes-type integral

$$I\left(r\right)={\int }_0^{\infty }a\left(x\right)f\left(rx\right)dF\left(x\right)$$

(3)

exists for every $r\in [0,+\infty )$. The asymptotic behavior of such integrals in the case $f\left(x\right)=e^x$ is studied in the monograph [7]. A question arises again about the asymptotic behavior of the function $f^{-1}\left(I\left(r\right)\right)$. Here, we present some results that indicate the possibility of solving these problems.

2. Relative Growth of Series in Systems of Functions

As in [8], by L, we denote a class of continuous nonnegative-on-$(-\infty ,+\infty )$ functions $\alpha$ such that $\alpha \left(x\right)=\alpha \left(x_0\right)\ge 0$ for $x\le x_0$ and $\alpha \left(x\right)\uparrow +\infty$ as $x_0\le x\to +\infty$. We say that $\alpha \in L^0,$ if $\alpha \in L$ and $\alpha \left(\right(1+o\left(1\right)\left)x\right)=(1+o(1\left)\right)\alpha \left(x\right)$ as $x\to +\infty$. Finally, $\alpha \in L_{si},$ if $\alpha \in L$ and $\alpha \left(cx\right)=(1+o(1\left)\right)\alpha \left(x\right)$ as $x\to +\infty$ for each $c\in (0,+\infty ),$ i.e., $\alpha$ is a slowly increasing function. Clearly, $L_{si}\subset L^0$. We need the following lemma [9].

Lemma 1.

If $\beta \in L$ and $B\left(\delta \right)={\overline{lim}}_{x\to +\infty }\frac{\beta \left(\right(1+\delta \left)x\right)}{\beta \left(x\right)},$ $\delta >0,$ then in order for $\beta \in L^0,$ it is necessary and sufficient that $B\left(\delta \right)\to 1$ as $\delta \to +0$.

We need also some well-known (see, for example, [10]) properties of the function $ln{M}_f\left(r\right)$.

Lemma 2.

If a function f is transcendental, then the function $ln{M}_f\left(r\right)$ is logarithmically convex and, thus,

$${\mathsf{\Gamma }}_f\left(r\right):=\frac{dln{M}_f\left(r\right)}{dlnr}\nearrow +\infty ,r\to +\infty ,$$

(at the points where the derivative does not exist, where $\frac{dln{M}_f\left(r\right)}{dlnr}$ means the right-hand derivative).

For $\alpha \in L,$$\beta \in L$, and entire functions f and g, we define the generalized $(\alpha ,\beta )$-order ${\rho }_{\alpha ,\beta }{\left[g\right]}_f$ and the generalized lower $(\alpha ,\beta )$-order ${\lambda }_{\alpha ,\beta }{\left[g\right]}_f$ of g with respect to f as follows:

$${\rho }_{\alpha ,\beta }{\left[g\right]}_f=\underset{r\to +\infty }{\overline{lim}}\frac{\alpha \left(M_f^{-1}\left(M_g\left(r\right)\right)\right)}{\beta \left(r\right)},{\lambda }_{\alpha ,\beta }{\left[g\right]}_f=\underset{r\to +\infty }{\underline{lim}}\frac{\alpha \left(M_f^{-1}\left(M_g\left(r\right)\right)\right)}{\beta \left(r\right)}.$$

Suppose that $a_n\ge 0$ for all $n\ge 1$. Since

$$A\left(z\right)=\sum _{n=1}^{\infty }{a}_n\sum _{k=0}^{\infty }{f}_k{\left(z{\lambda }_n\right)}^k=\sum _{k=0}^{\infty }{f}_k\left(\sum _{n=1}^{\infty }{a}_n{\lambda }_n^k\right)z^k,$$

in view of the Cauchy inequality, we have

$$M_A\left(r\right)\ge \left|f_k\right|\left(\sum _{n=1}^{\infty }{a}_n{\lambda }_n^k\right)r^k\ge a_n\left|f_k\right|{\left({\lambda }_{n}r\right)}^k$$

(4)

for all $n\ge 1,$$k\ge 0$ and $r\in [0,+\infty )$. We also remark that, if ${\mu }_f\left(r\right)=max\left\{\right|f_k|r^k:k\ge 0\}$ is the maximal term of series (1), then

$$M_f\left(r\right)\le \sum _{k=0}^{\infty }|f_k|r^k=\sum _{k=0}^{\infty }\left|f_k\right|{\left(2r\right)}^k{2}^{-k}\le 2{\mu }_f\left(2r\right).$$

(5)

We choose $n_0\ge 1$ such that $a_{n_0}>0$ and ${\lambda }_{n_0}\ge 2$. Then, from (4) and (5), we get

$$M_A\left(r\right)\ge max\{a_{n_0}\left|f_k\right|{\left({\lambda }_{n_0}r\right)}^k:k\ge 0\}\ge a_{n_0}{\mu }_f\left(2r\right)\ge \frac{a_{n_0}}{2}{M}_f\left(r\right),$$

where $M_f^{-1}\left(\frac{2}{d_{n_0}}{M}_A\left(r\right)\right)\ge r$. By Lemma 2, $\frac{dln{M}_f^{-1}\left(x\right)}{dlnx}\searrow 0$ as $x\to +\infty$ and, thus, for every $c>1$

$$ln{M}_f^{-1}\left(cx\right)-ln{M}_f^{-1}\left(x\right)={\int }_x^{cx}\frac{dln{M}_f^{-1}\left(t\right)}{dlnt}dlnt\le \frac{dln{M}_f^{-1}\left(x\right)}{dlnx}\to 0,x\to +\infty ,$$

i.e., the function $M_f^{-1}$ is slowly increasing. Therefore,

$$M_f^{-1}\left(M_A\left(r\right)\right)\ge (1+o\left(1\right))r,r\to +\infty .$$

(6)

On the other hand, since series (2) is regularly convergent in $\mathbb{C}$, for each $r\in [0,+\infty )$, there exists ${\mu }_A\left(r\right)=max\left\{\right|an|M_f\left(r{\lambda }_n\right):n\ge 1\}$ and, for every $r\in [0,+\infty )$ and $\tau >0$, we have

$$M_A\left(r\right)\le \sum _{n=1}^{\infty }\left|a_n\right|M_f\left(r{\lambda }_n\right)\le {\mu }_F\left((1+\tau )r\right)\sum _{n=1}^{\infty }\frac{M_f\left(r{\lambda }_n\right)}{M_f\left((1+\tau )r{\lambda }_n\right)}.$$

(7)

Then, by Lemma 2, for $r\ge 1$, we have

$$\begin{array}{c}ln{M}_f\left((1+\tau )r{\lambda }_n\right)-ln{M}_f\left(r{\lambda }_n\right)={\int }_{r{\lambda }_n}^{(1+\tau )r{\lambda }_n}\frac{dln{M}_f\left(x\right)}{dlnx}dlnx={\int }_{r{\lambda }_n}^{(1+\tau )r{\lambda }_n}{\mathsf{\Gamma }}_f\left(x\right)dlnx\ge \\ \ge {\mathsf{\Gamma }}_f\left(r{\lambda }_n\right)ln(1+\tau )\ge {\mathsf{\Gamma }}_f\left({\lambda }_n\right)ln(1+\tau ).\end{array}$$

Therefore, if $lnn\le q{\mathsf{\Gamma }}_f\left({\lambda }_n\right)$ for all $n\ge n_0$ and $ln(1+\tau )>q$, then

$$\sum _{n=n_0}^{\infty }\frac{M_f\left(r{\lambda }_n\right)}{M_f\left((1+\tau )r{\lambda }_n\right)}\le \sum _{n=n_0}^{\infty }exp\{-{\mathsf{\Gamma }}_f\left({\lambda }_n\right)ln(1+\tau )\}\le \sum _{n=n_0}^{\infty }exp\left\{-\frac{ln(1+\tau )}{q}lnn\right\}<+\infty$$

and (7) implies, for $r\ge 1$,

$$M_A\left(r\right)\le T{\mu }_A\left((1+\tau )r\right),T=\mathrm{const}>0.$$

(8)

Additionally, we have

$$\begin{array}{c}{\mu }_A\left(r\right)\le max\left\{|a_n|\sum _{k=0}^{\infty }\left|f_k\right|{\left(r{\lambda }_n\right)}^k:n\ge 1\right\}\le \\ \le \sum _{k=0}^{\infty }max\{\left|a_n\right|{\lambda }_n^k:n\ge 1\left\}\right|f_k|r^k=\sum _{k=0}^{\infty }{\mu }_D\left(k\right)\left|f_k\right|r^k,\end{array}$$

(9)

where ${\mu }_D\left(\sigma \right)=max\left\{\right|a_n|exp\{\sigma ln{\lambda }_n\}:n\ge 1\}$ is the maximal term of Dirichlet series

$$D\left(\sigma \right)=\sum _{n=1}^{\infty }\left|a_n\right|exp\{\sigma ln{\lambda }_n\}.$$

Using estimates (6), (8), and (9), we prove the following theorem.

Theorem 1.

Let f be an entire transcendental function, $a_n\ge 0$ for all $n\ge 1$, and series (2) be regularly convergent in $\mathbb{C}$. Suppose that $lnn\le q{\mathsf{\Gamma }}_f\left({\lambda }_n\right)$ for some $q>0$ and all $n\ge n_0$ and that ${\overline{lim}}_{\sigma \to +\infty }\frac{ln{\mu }_D\left(\sigma \right)}{\sigma ln{M}_f^{-1}\left(e^{\sigma }\right)}=\gamma .$

If $\gamma <1$, then ${\lambda }_{\alpha ,\alpha }{\left[F\right]}_f={\rho }_{\alpha ,\alpha }{\left[F\right]}_f=1$ for every function α such that $\alpha \left(e^x\right)\in L_{si}$. If $\gamma =0$, then ${\lambda }_{\alpha ,\alpha }{\left[F\right]}_f={\rho }_{\alpha ,\alpha }{\left[F\right]}_f=1$ for every function α such that $\alpha \left(e^x\right)\in L^0$.

Proof. 

Since $\alpha \in L^0,$ from (6), we get

$${\lambda }_{\alpha ,\alpha }{\left[F\right]}_f=\underset{r\to +\infty }{\underline{lim}}\frac{\alpha \left(M_f^{-1}\left(M_F\left(r\right)\right)\right)}{\alpha \left(r\right)}\ge \underset{r\to +\infty }{\underline{lim}}\frac{\alpha \left(\right(1+o\left(1\right)\left)r\right)}{\alpha \left(r\right)}=1.$$

On the other hand, in view of the Cauchy inequality, we have $ln|f_k|\le ln{M}_f\left(r\right)-klnr$ for all r and k. We choose $r=r_k=M_f^{-1}\left(e^k\right).$ Then, $ln|f_k|\le k-kln{M}_f^{-1}\left(e^k\right)$, i.e., $-ln|f_k|\ge k(ln{M}_f^{-1}\left(e^k\right)-1)$. Therefore,

$$\underset{k\to \infty }{\overline{lim}}\frac{ln{\mu }_D\left(k\right)}{-ln{f}_k}\le \underset{k\to \infty }{\overline{lim}}\frac{ln{\mu }_D\left(k\right)}{k(ln{M}_f^{-1}\left(e^k\right)-1)}\le \underset{\sigma \to +\infty }{\overline{lim}}\frac{ln{\mu }_D\left(\sigma \right)}{\sigma ln{M}_f^{-1}\left(e^{\sigma }\right)}=\sigma .$$

(10)

If $\gamma <1$, then in view of (10), $\frac{ln{\mu }_D\left(k\right)}{-ln|f_k|}\le p$ for each $p\in (\gamma ,1)$ and all $k\ge k_0$ and, thus, ${\mu }_D\left(k\right)\le {\left|f_k\right|}^{-p}$ for all $k\ge k_0$. Therefore, in view of (9) and (5),

$$\begin{array}{c}{\mu }_A\left(r\right)\le \left(\sum _{k=0}^{k_0-1}+\sum _{k=k_0}^{\infty }\right){\mu }_D\left(k\right)\left|f_k\right|r^k\le O\left(r^{k_0-1}\right)+\sum _{k=k_0}^{\infty }{\left|f_k\right|}^{1-p}{r}^k\le \\ \le O\left(r^{k_0-1}\right)+2max\{f_k^{1-p}{\left(2r\right)}^k:k\ge 0\}=\\ =O\left(r^{k_0-1}\right)+2max\left\{\right(\left|f_k\right|{\left(2r\right)}^{k/(1-p)}{)}^{1-p}:k\ge 0\}=\\ =O\left(r^{k_0-1}\right)+2{\left({\mu }_f\left({\left(2r\right)}^{1/(1-p)}\right)\right)}^{1-p}\le {\mu }_f\left({\left(2r\right)}^{1/(1-p)}\right),r\ge r_0,\end{array}$$

(11)

because $lnr=o(ln{\mu }_f\left(r\right))$ as $r\to +\infty$ for every entire transcendental function f and $1-p<1$. Therefore, from (8) and (11), we get

$$M_A\left(r\right)\le T{\mu }_A\left((1+\tau )r\right)\le T{\mu }_f\left({\left(2(1+\tau )r\right)}^{1/(1-p)}\right)\le T{M}_f\left({\left(2(1+\tau )r\right)}^{1/(1-p)}\right)$$

and, thus, $M_f^{-1}\left(M_A\left(r\right)\right)\le (1+o\left(1\right)){\left(2(1+\tau )r\right)}^{1/(1-p)}$ as $r\to +\infty$. If $\alpha \in L_{si}$, then we obtain

$$\underset{r\to +\infty }{\overline{lim}}\frac{\alpha \left(M_f^{-1}\left(M_A\left(r\right)\right)\right)}{\alpha \left(r^{1/(1-p)}\right)}\le 1.$$

(12)

Suppose that $\alpha \left(e^x\right)\in L_{si}.$ Then,

$$\alpha \left(r^{1/(1-p)}\right)=\alpha (exp\left\{\frac{1}{1-p}lnr\right\})=(1+o\left(1\right))\alpha (exp\{lnr\})=(1+o\left(1\right))\alpha \left(r\right)$$

as $r\to +\infty$. Therefore, (12) implies the inequality ${\rho }_{\alpha ,\alpha }{\left[A\right]}_f\le 1,$ where in view of the inequality ${\lambda }_{\alpha ,\alpha }{\left[A\right]}_f\ge 1$, we get ${\lambda }_{\alpha ,\alpha }{\left[A\right]}_f={\rho }_{\alpha ,\alpha }{\left[A\right]}_f=1.$

If $\gamma =0$, then (12) holds for every $p\in (0,1)$ and all $r\ge r_0\left(p\right)$. If we put $\frac{1}{1-p}=1+\delta$, then $\delta \to +0$ as $p\to +0$, and in view of the condition $\alpha \left(e^x\right)\in L^0$, by Lemma 1, we have

$$\underset{r\to +\infty }{\overline{lim}}\frac{\alpha \left(r^{1/(1-p)}\right)}{\alpha \left(r\right)}=\underset{r\to +\infty }{\overline{lim}}\frac{\alpha (exp\{(1+\delta )lnr\left\}\right)}{\alpha (exp\{lnr\left\}\right)}=B\left(\delta \right)\to 1,\delta \to 1.$$

Therefore,

$$\begin{array}{c}1\ge \underset{r\to +\infty }{\overline{lim}}\frac{\alpha \left(M_f^{-1}\left(M_A\left(r\right)\right)\right)}{\alpha \left(r^{1/(1-p)}\right)}=\underset{r\to +\infty }{\overline{lim}}\left(\frac{\alpha \left(M_f^{-1}\left(M_A\left(r\right)\right)\right)}{\alpha \left(r\right)}·\frac{\alpha \left(r\right)}{\alpha \left(r^{1+\delta }\right)}\right)\ge \\ \ge \underset{r\to +\infty }{\overline{lim}}\frac{\alpha \left(M_f^{-1}\left(M_A\left(r\right)\right)\right)}{\alpha \left(r\right)}\underset{r\to +\infty }{\underline{lim}}\frac{\alpha \left(r\right)}{\alpha \left(r^{1+\delta }\right)}=\frac{{\rho }_{\alpha ,\alpha }{\left[F\right]}_f}{B\left(\delta \right)}.\end{array}$$

In view of the arbitrariness of $\delta$, we get ${\rho }_{\alpha ,\alpha }{\left[A\right]}_f\le 1$, and again, ${\lambda }_{\alpha ,\alpha }{\left[A\right]}_f={\rho }_{\alpha ,\alpha }{\left[A\right]}_f=1$. Theorem 1 is proven. □

We remark that, if $f_k\ge 0$ for all $k\ge 0$, then $M_f\left(r\right)=f\left(r\right)$. Therefore, from Theorem 1, we obtain the following statement.

Corollary 1.

Let f be an entire transcendental function, $f_k\ge 0$ for all $k\ge 0,$$a_n\ge 0$ for all $n\ge 1$, and series (2) be regularly convergent in $\mathbb{C}$. Suppose that $f^{\prime }\left(r\right)/f\left(r\right)\ge h>0$ for all $r\ge r_0,$$lnn=O\left({\lambda }_n\right)$ as $n\to \infty$ and ${\overline{lim}}_{\sigma \to +\infty }\frac{ln{\mu }_D\left(\sigma \right)}{\sigma ln{f}^{-1}\left(e^{\sigma }\right)}=\gamma .$

If $\gamma <1$, then ${\lambda }_{\alpha ,\alpha }{\left[A\right]}_f={\rho }_{\alpha ,\alpha }\left[A\right]f=1$ for every function α such that $\alpha \left(e^x\right)\in L_{si}$.

If $\gamma =0$, then ${\lambda }_{\alpha ,\alpha }{\left[A\right]}_f={\rho }_{\alpha ,\alpha }{\left[A\right]}_f=1$ for every function α such that $\alpha \left(e^x\right)\in L^0$.

3. Relative Growth of Laplace–Stieltjes-Type Integrals

Suppose again that f is an entire transcendental function, $f_k\ge 0$ for all $k\ge 0$, and $x_0>1$ is such that ${\int }_1^{x_0}a\left(x\right)dF\left(x\right)\ge >0.$ Then,

$$I\left(r\right)\ge {\int }_1^{x_0}a\left(x\right)f\left(rx\right)dF\left(x\right)\ge f\left(r\right)c,$$

i.e., as above, $f^{-1}\left(I\left(r\right)\right)\ge (1+o\left(1\right))r$ as $r\to +\infty ,$ where for $\alpha \in L^0$,

$${\lambda }_{\alpha ,\alpha }{\left[I\right]}_f=\underset{r\to +\infty }{\underline{lim}}\frac{\alpha \left(f^{-1}\left(I\left(r\right)\right)\right)}{\alpha \left(r\right)}\ge 1.$$

On the other hand, if $\tau \ge e-1$, then as above, for $r\ge 1$, we have

$$\begin{array}{c}lnf\left((1+\tau )rx\right)-lnf\left(rx\right)={\int }_{rx}^{(1+\tau )rx}\frac{dlnf\left(x\right)}{dlnx}dlnx={\int }_{rx}^{(1+\tau )rx}{\mathsf{\Gamma }}_f\left(x\right)dlnx\ge \\ \ge {\mathsf{\Gamma }}_f\left(x\right)ln(1+x),\end{array}$$

i.e., $\frac{f\left(rx\right)}{f\left(\right(1+\tau \left)rx\right)}\le e^{-{\mathsf{\Gamma }}_f\left(x\right)ln(1+\tau )}.$ Therefore, if ${\mu }_I\left(r\right)=max\{a\left(x\right)f\left(rx\right):x\ge 0\}$ is the maximum of the integrand and $lnF\left(x\right)\le q{\mathsf{\Gamma }}_f\left(x\right)$ for some $q>0$ and all $x\ge x_0$, then for $ln(1+\tau )>q$ (for simplicity assuming $x_0=0$), we get

$$\begin{array}{c}I\left(r\right)={\int }_0^{\infty }a\left(x\right)f\left((1+\tau )rx\right)\frac{f\left(rx\right)}{f\left(\right(1+\tau \left)rx\right)}dF\left(x\right)\le {\mu }_I\left((1+\tau )r\right){\int }_0^{\infty }\frac{f\left(rx\right)}{f\left(\right(1+\tau \left)rx\right)}dF\left(x\right)\le \\ \le {\mu }_I\left((1+\tau )r\right){\int }_0^{\infty }{e}^{-{\mathsf{\Gamma }}_f\left(x\right)ln(1+\tau )}dF\left(x\right)\le \\ \le {\mu }_I\left((1+\tau )r\right)ln(1+\tau ){\int }_0^{\infty }{e}^{-{\mathsf{\Gamma }}_f\left(x\right)ln(1+\tau )+lnF\left(x\right)}d{\mathsf{\Gamma }}_f\left(x\right)\le \\ \le {\mu }_I\left((1+\tau )r\right)ln(1+\tau ){\int }_0^{\infty }{e}^{-{\mathsf{\Gamma }}_f\left(x\right)(ln(1+\tau )-q)}d{\mathsf{\Gamma }}_f\left(x\right)={\mu }_I\left((1+\tau )r\right)\frac{ln(1+\tau )}{ln(1+\tau )-q}=\\ =T{\mu }_I\left((1+\tau )r\right).\end{array}$$

(13)

Additionally, as above, we have

$$\begin{array}{c}{\mu }_I\left(r\right)=max\left\{a\left(x\right)\sum _{k=0}^{\infty }{f}_k{\left(xr\right)}^k:x\ge 0\right\}\le \\ \le \sum _{k=0}^{\infty }max\{a\left(x\right)x^k:x\ge 0\}{f}_k{r}^k=\sum _{k=0}^{\infty }{\mu }_J\left(k\right)f_k{r}^k,\end{array}$$

(14)

where ${\mu }_J\left(\sigma \right)=max\{a\left(x\right)e^{\sigma lnx}:x\ge 0\}=max\{a\left(x\right)x^{lnx}:x\ge 0\}$ is the maximum of the integrand for the Laplace integral

$$J\left(\sigma \right)={\int }_0^{\infty }a\left(x\right)e^{\sigma lnx}dF\left(x\right).$$

Using estimates (13) and (14), and ${\lambda }_{\alpha ,\alpha }{\left[I\right]}_f\ge 1$, we prove the following analog of Theorem 1.

Theorem 2.

Let $lnF\left(x\right)\le q{\mathsf{\Gamma }}_f\left(x\right)$ for some $q>0$ and all $x\ge x_0$, and ${\overline{lim}}_{\sigma \to +\infty }\frac{ln{\mu }_J\left(\sigma \right)}{\gamma ln{f}^{-1}\left(e^{\sigma }\right)}=\gamma .$

If $\gamma <1$, then ${\lambda }_{\alpha ,\alpha }{\left[I\right]}_f={\rho }_{\alpha ,\alpha }{\left[I\right]}_f=1$ for every function α such that $\alpha \left(e^x\right)\in L_{si}$.

If $\gamma =0$, then ${\lambda }_{\alpha ,\alpha }{\left[I\right]}_f={\rho }_{\alpha ,\alpha }{\left[I\right]}_f=1$ for every function α such that $\alpha \left(e^x\right)\in L^0.$

Proof. 

As in the proof of Theorem 1, we obtain $-ln|f_k|\ge k(ln{f}^{-1}\left(e^k\right)-1)$ and ${\overline{lim}}_{k\to \infty }\frac{ln{\mu }_J\left(k\right)}{-ln{f}_k}\le \gamma .$ Therefore, if $\gamma <1$, then ${\mu }_D\left(k\right)\le {\left|f_k\right|}^{-p}$ for each $p\in (\gamma ,1)$ and all $k\ge k_0$, and in view of (14) and (5), as in the proof of Theorem 1, we get ${\mu }_I\left(r\right)\le {\mu }_f\left({\left(2r\right)}^{1/(1-p)}\right)$ for $r\ge r_0.$ Therefore, in view of (13), we get

$$I\left(r\right)\le T{\mu }_I\left((1+\tau )r\right)\le Tf\left({\left(2(1+\tau )r\right)}^{1/(1-p)}\right),$$

where $f^{-1}\left(I\left(r\right)\right)\le (1+o\left(1\right)){\left(2(1+\tau )r\right)}^{1/(1-p)}$ as $r\to +\infty$. If $\alpha \in L_{si}$, then we obtain

$$\underset{r\to +\infty }{\underline{lim}}\frac{\alpha \left(f^{-1}\left(I\left(r\right)\right)\right)}{\alpha \left(r^{1/(1-p)}\right)}\le 1.$$

Further proof of Theorem 2 is the same as that of Theorem 1. □

Theorem 2 implies the following statement.

Corollary 2.

Let $f^{\prime }\left(x\right)/f\left(x\right)\ge h,$$h>0,$ $lnF\left(x\right)\le qx$ for some $q>0$ and all $x\ge 0,$ and ${\overline{lim}}_{r\to +\infty }\frac{ln{\mu }_J\left(\sigma \right)}{\sigma f^{-1}\left(e^{\sigma }\right)}=\gamma .$

If $\gamma <1$, then ${\lambda }_{\alpha ,\alpha }{\left[I\right]}_f={\rho }_{\alpha ,\alpha }{\left[I\right]}_f=1$ for every function α such that $\alpha \left(e^x\right)\in L_{si}.$

If $\gamma =0$, then ${\lambda }_{\alpha ,\alpha }{\left[I\right]}_f={\rho }_{\alpha ,\alpha }{\left[I\right]}_f=1$ for every function α such that $\alpha \left(e^x\right)\in L^0$.

4. Examples

Here, we consider the case when $f\left(z\right)=E_{\rho }\left(z\right),$ where

$$E_{\rho }\left(z\right)=\sum _{k=0}^{\infty }\frac{z^k}{\mathsf{\Gamma }(1+\frac{k}{\rho })},0<\rho <+\infty ,$$

is the Mittag–Leffler function. The properties of this function have been used in many problems in the theory of entire functions. We only need the following property of the Mittag–Leffler function: if $0<\rho <+\infty$, then ([11] p. 85)

$$M_{E_{\rho }}\left(r\right)=E_{\rho }\left(r\right)=(1+o\left(1\right))\rho e^{r^{\rho }},r\to +\infty$$

(15)

and, if $1/2<\rho <+\infty$, then [12]

$$E_{\rho }^{\prime }\left(r\right)/E_{\rho }\left(r\right)=\rho r^{\rho -1}+O\left(r^{\rho -2}{e}^{-r^{\rho }}\right),r\to +\infty .$$

(16)

From (15), it follows that $E_{\rho }^{-1}\left(x\right)=(1+o\left(1\right)){ln}^{1/\rho }x$ as $x\to +\infty$. Therefore, for $f\left(x\right)=E_{\rho }\left(x\right)$, we have $\sigma ln{f}^{-1}\left(e^{\sigma }\right)=\frac{1+o\left(1\right)}{\rho }\sigma ln\sigma$ as $\sigma \to +\infty$. Since in (16), ${\mathsf{\Gamma }}_{E_{\rho }}\left(r\right)=\rho r^{\rho }+o\left(1\right)$ as $r\to +\infty ,$ then if $lnF\left(x\right)\le q\rho x^{\rho }$ for some $q>0$ and all $x\ge x_0$, and

$$\underset{\sigma \to +\infty }{\overline{lim}}\frac{ln{\mu }_J\left(\sigma \right)}{\sigma ln\sigma }=0,$$

(17)

then for $\alpha \left(x\right)=lnx$$(x\ge e)$, by Theorem 2, we get

$$\underset{r\to +\infty }{lim}\frac{ln{E}_{\rho }^{-1}\left(I_{\rho }\left(r\right)\right)}{lnr}=1,I_{\rho }\left(r\right)={\int }_0^{\infty }a\left(x\right)E_{\rho }\left(rx\right)dF\left(x\right).$$

(18)

Let us now find out under what conditions (17) holds on $a\left(x\right)$. For this, as in ([7] p. 29), by $\mathsf{\Omega }$, we denote a class of positive unbounded functions $\mathsf{\Phi }$ on $(-\infty ,+\infty )$ such that the derivative ${\mathsf{\Phi }}_0$ is positive, continuously differentiable, and increasing to $+\infty$ on $(-\infty ,+\infty ).$ For $\mathsf{\Phi }\in \mathsf{\Omega }$, let $\phi$ be the inverse function to ${\mathsf{\Phi }}^{\prime }$ and $\mathsf{\Psi }\left(\sigma \right)=\sigma -\frac{\mathsf{\Phi }\left(\sigma \right)}{{\mathsf{\Phi }}^{\prime }\left(\sigma \right)}$ be the function associated with $\mathsf{\Phi }$ in the sense of Newton.

By Theorem 2.2.1 from ([7] p. 30), $lnmax\{a\left(x\right)e^{\sigma x}:x\ge 0\}\le \mathsf{\Phi }\left(\sigma \right)\in \mathsf{\Omega }$ for all $\sigma \ge {\sigma }_0$ if and only if $lna\left(x\right)\le -x\mathsf{\Psi }\left(\phi \right(x\left)\right)$ for all $x\ge x_0$. Choosing $\mathsf{\Phi }\left(\sigma \right)=ϵ\sigma ln\sigma$ for $\sigma \ge {\sigma }_0,$ we obtain ${\mathsf{\Phi }}^{\prime }\left(\sigma \right)=ϵ(ln\sigma +1),$$\phi \left(x\right)=exp\{x/ϵ-1\}$ and $x\mathsf{\Psi }\left(\phi \right(x\left)\right)=x\phi \left(x\right)-\mathsf{\Phi }\left(\phi \right(x\left)\right)=ϵexp\{x/ϵ-1\}$ for $x\ge x_0$. Therefore, $ln{\mu }_J\left(\sigma \right)\le \epsilon \sigma ln\sigma$ for all $\sigma \ge {\sigma }_0$ if and only if $lna\left(x\right)\le -\epsilon exp\{lnx/\epsilon -1\}$ for $x\ge x_0$. Hence, it follows that, if $lnx=o(lnln(1/a\left(x\right)\left)\right)$ as $x\to +\infty$, then (17) holds. Thus, the following statement is true.

Proposition 1.

If $\rho >1/2,$$lnF\left(x\right)=O\left(x^{\rho }\right)$ and $lnx=o(lnln(1/a\left(x\right)\left)\right)$ as $x\to +\infty$, then (18) holds.

Remark 1.

If $\rho =1$, then $E_{\rho }\left(r\right)=E_1\left(r\right)=e^r$, and we have a usual Laplace–Stieltjes integral $I_1\left(r\right)={\int }_0^{\infty }a\left(x\right)e^{rx}dF\left(x\right)$. Therefore, if $lnF\left(x\right)=O\left(x\right)$ and $lnx=o(lnln(1/a\left(x\right)\left)\right)$ as $x\to +\infty$, then $p_R\left[I_1\right]:={lim}_{r\to +\infty }\frac{lnln{I}_1\left(r\right)}{lnr}=1$. On the other hand, the quantity $p_R\left[I_1\right]$ is called the logarithmic R-order of $I_1$, and in ([7] p. 83), it is proven that, if $lnF\left(x\right)=O\left(x\right)$ as $x\to +\infty$, then $p_R\left[I_1\right]={\overline{lim}}_{x\to +\infty }\frac{lnx}{ln(\frac{1}{x}ln\frac{1}{a\left(x\right)})}=1,$ i.e., if $lnF\left(x\right)=O\left(x\right)$ and $lnx=o(lnln(1/a\left(x\right)\left)\right)$ as $x\to +\infty$, then $p_R\left[I_1\right]=1.$

Similarly, we can prove the following statement.

Proposition 2.

Let $\rho \ge 1/2$, $lnn=O\left({\lambda }_n^{\rho }\right)$ as $n\to \infty ,$ $a_n\ge 0$ for all $n\ge 1$ and series $A_{\rho }\left(z\right)={\sum }_{n=1}^{\infty }{a}_n{E}_{\rho }\left({\lambda }_{n}z\right)$ be regularly convergent in $\mathbb{C}$. If $lnn=o(lnln(1/a_n))$ as $n\to \infty$, then ${lim}_{r\to +\infty }\frac{ln{E}_{\rho }^{-1}\left(M_{A_{\rho }}\left(r\right)\right)}{lnr}=1.$

Remark 2.

If $\rho =1$, then we have a Dirichlet series $A_1\left(z\right)={\sum }_{n=1}^{\infty }{a}_n{e}^{{\lambda }_{n}z}$. Therefore, if this Dirichlet series is absolutely convergent in $\mathbb{C}$, $a_n\ge 0$ for all $n\ge 1$, $lnn=O\left({\lambda }_n\right)$, and $lnn=o(lnln(1/a_n))$ as $n\to \infty$, then $p_R\left[A_1\right]:={lim}_{r\to +\infty }\frac{lnln{M}_{A_1}\left(r\right)}{lnr}=1$. On the other hand, the quantity $p_R\left[A_1\right]$ is called the logarithmic R-order of $A_1$ and $p_R\left[A_1\right]={\overline{lim}}_{n\to +\infty }\frac{ln{\lambda }_n}{ln(\frac{1}{{\lambda }_n}ln\frac{1}{a_n})}=1$ provided $lnn=O\left({\lambda }_n\right)$ as $n\to \infty$ [13], i.e., if $lnn=O\left({\lambda }_n\right)$ and $ln{\lambda }_n=o(lnln(1/a_n))$ as $n\to \infty$, then $p_R\left[A_1\right]=1.$

5. Discussion Open Problems

1. The natural problem studied was the relative growth when the domain of regular convergence of series (2) is the disk $D_R=\{z:|z|<R<+\infty \}$ and the function f is either entire or analytic in $D_R$.

2. It is well known that the study of the growth of entire functions of many complex variables involves many options. The following problem is the simplest.

Let f be an entire function and the series $A(z,w)={\sum }_{m=1,n=1}^{\infty }{a}_{m,n}f({\lambda }_{m}z+{\mu }_{n}w)$ be regularly convergent in ${\mathbb{C}}^2$. A question arises about the asymptotic behavior of the function $M_f^{-1}\left(M_A(r,\rho )\right)$, where $M_A(r,\rho )=max\left\{\right|A(z,w)|:|z|\le r,|w|\le \rho \}$.

3. The condition $\rho \ge 1/2$ in Propositions 1 and 2 arose in connection to the application of Equation (16). Probably, it is superfluous in the above statements.