To the boundedness of the \(l-\mathfrak{M}\) -index of functions analytic in the unit disk and represented by series in systems of functions
Year: 2025
Type: article
Abstract: Let \(f(z) = \sum\limits_{k=0}^{\infty} f_k z^k\) be an entire transcendental function, and let \((\lambda_n)\) be a sequence of positive numbers increasing to \(+\infty\). Suppose that the series \(A(z) = \sum\limits_{n=1}^{\infty} a_n f(\lambda_n z)\) is regularly convergent in \(\mathbb{D} = \{ z : |z| < 1 \}\), i.e., \(\mathfrak{M}(r, A) := \sum\limits_{n=1}^{\infty} |a_n| M_f(r \lambda_n) < +\infty\) for all \(r \in [0, 1)\). For a positive function \(l\) continuous on \([0, 1)\), the function \(A\) is said to be of bounded \(l\)-\(\mathfrak{M}\)-index if there exists \(N \in \mathbb{Z}_+\) such that \[\frac{\mathfrak{M}(r, A^{(n)})}{n! \, l^n(r)} \leq \max \left\{ \frac{\mathfrak{M}(r, A^{(k)})}{k! \, l^k(r)} : 0 \leq k \leq N \right\},\] for all \(n \in \mathbb{Z}_+\) and all \(r \in [0, 1)\). The growth of bounded \(l\)-\(\mathfrak{M}\)-index functions is studied. In particular, under the conditions \(a_n \geq 0\) and \(f_k \geq 0\), it is proved that the function \(A\) is of bounded \(l\)-\(\mathfrak{M}\)-index with \(l(r) = p(1 - r)^{-(p+1)}\), \(p > 0\), if and only if \[\lim\limits_{r \uparrow 1} (1 - r)^p \ln \mathfrak{M}(r, A) < +\infty.\] This condition is satisfied if and only if \[\lim\limits_{k \to \infty} k^{-p} \left( \ln^+ (f_k \mu_D(k)) \right)^{p+1} < +\infty,\] where \(\mu_D(k) = \max\{ a_n \lambda_n^k : n \geq 1 \}\).
Author: Myroslav M. Sheremeta
Institution: Lviv University
Language: English
Subfield: Applied Mathematics
Field: Mathematics
Domain: Physical Sciences
Open Access status: diamond
OpenAlex ID: W4412661791