On the growth of series in system of functions and Laplace-Stieltjes integrals

Анотація

For a regularly convergent in ${\Bbb C}$ series $A(z)=\sum\nolimits_{n=1}^{\infty}a_nf(\lambda_nz)$ in the system ${f(\lambda_nz)}$, where
$f(z)=\sum\nolimits_{k=0}^{\infty}f_kz^k$ is an entire transcendental function and $(\lambda_n)$
is a sequence of positive numbers increasing to $+\infty$, it is
investigated the relationship between the growth of functions $A$ and $f$ in terms of a generalized order. It is proved that if
$a_n\ge 0$ for all $n\ge n_0$,

$\ln \lambda_n=o\big(\beta^{-1}\big(c\alpha(\frac{1}{\ln \lambda_n}\ln \frac{1}{a_n})\big)\big)$

for each $c\in (0, +\infty)$ and $\ln n=O(\Gamma_f(\lambda_n))$ as $n\to\infty$ then

$\displaystyle\varlimsup\limits_{r\to+\infty}\frac{\alpha(\ln M_A(r))}{\beta(\ln r)}=\varlimsup\limits_{r\to+\infty}\frac{\alpha(\ln M_f(r))}{\beta(\ln r)},$

where $M_f(r)=\max\{|f(z)|\colon |z|=r\}$, $\Gamma_f(r):=\frac{d\ln M_f(r)}{d\ln r}$ and positive continuous on $(x_0, +\infty)$ functions $\alpha$
and $\beta$ are such that $\beta((1+o(1))x)=(1+o(1))\beta(x)$, $\alpha(c x)=(1+o(1))\alpha(x)$ and
$\frac{d\beta^{-1}(c\alpha(x))}{d\ln x}=O(1)$ as $x\to+\infty$ for each $c\in(0, +\infty)$.\
A similar result is obtained for the Laplace-Stieltjes type integral $I(r) = \int\limits_{0}^{\infty}a(x)f(rx) dF(x)$.