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 \(\mu_D(k) = \max\{ a_n \lambda_n^k : n \geq 1 \}\).
According to Lepson [1], an entire function \(a(z)\) is called of bounded index if there exists \(N\in{\Bbb Z}_+\) such that \(|a^{(n)}(z)|/n!\le\max\{|a^{(k)}(z)|/k!:\,0\le k\le N\}\) for all \(n\in{\Bbb Z}_+\) and all \(z\in {\Bbb C}\). Different authors (S.M. Shah, W. Hayman, G. Fricke, and others) investigated the properties of entire functions of bounded index and their applications (a survey of results can be found in [2]).
For a positive function \(l\) continuous on \([0,\,+\infty)\), Kuzyk and Sheremeta [3] introduced the concept of an entire function of bounded \(l\)-index. An entire function \(a(z)\) is called of bounded \(l\)-index if there exists \(N\in{\Bbb Z}_+\) such that \[\frac{|a^{(n)}(z)|}{n!l^n(|z|)}\le\max\left\{\frac{|a^{(k)}(z)|}{k!l^k(|z|)}:\,0\le k\le N\right\},\] for all \(n\in{\Bbb Z}_+\) and all \(z\in {\Bbb C}\). Subsequently, this definition was extended to analytic functions in any complex domain, and the results obtained are summarized in the monograph [4]. At the beginning of this century, thanks mainly to O.B. Skaskiv and A.I. Bandura, many results were obtained on the boundedness of the \(L\)-index of holomorphic functions of several complex variables (see, for example, [5] and [6]).
Let \(M_a(r)=\max\{|a(z)|:\,|z|=r\}\) and \(l\) be a positive function continuous on \([0,\,+\infty)\). The entire function \(a(z)\) is said [7] to be of bounded \(M\)-index if there exists \(N\in{\Bbb Z}_+\) such that \(M_{a^{(n)}}(r)/n!\le\max\{M_{a^{(k)}}(r)/k!:\,0\le k\le N\}\) for all \(n\in{\Bbb Z}_+\) and all \(r\in [0,+\infty)\).
By combining the definitions of bounded \(M\)-index functions and bounded \(l\)-index functions, [8] introduced the following definition. An entire function \(a(z)\) is called of bounded \(l\)–\(M\)-index if there exists \(N\in{\Bbb Z}_+\) such that \[\dfrac{M_{a^{(n)}}(r)}{n!l^n(r)}\le\max\left\{\dfrac{M_{a^{(k)}}(r)}{k!l^k(r)}:\,0\le k\le N\right\},\] for all \(n\in{\Bbb Z}_+\) and all \(r\in [0,+\infty)\).
Let \[f(z)=\sum\limits_{k=0}^{\infty}f_kz^k, \label{t1} \tag{1}\] be an entire transcendental function, \(M_f(r)=\max\{|f(z)|:\,|z|=r\}\), \((\lambda_n)\) be a sequence of positive numbers increasing to \(+\infty\), and \[A(z)=\sum\limits_{n=1}^{\infty}a_nf(\lambda_nz), \label{t2} \tag{2}\] be the series in the system \({f(\lambda_nz)}\).
Let \(R[A]\) be the radius of regular convergence of the series (2), i.e., \[\mathfrak{M}(r,A)=\sum\limits_{n=1}^{\infty} |a_n|M_f(r\lambda_n)<+\infty, \label{t3} \tag{3}\] holds for \(r<R[A]\) and does not hold for \(r>R[A]\). Denote \(\Gamma_f(r)=\dfrac{d\ln\,M_f(r)}{d\ln\,r}\) (in points where the derivative does not exist, \(\dfrac{d\ln\,M_f(r)}{d\ln\,r}\) denotes the right-hand derivative). Since the function \(\ln\, M_f(r)\) is logarithmically convex, we have \(\Gamma_f(r)\nearrow+\infty\) as \(r\to+\infty\). It is known [9] that if \(\Gamma_f(cr)\asymp \Gamma_f(r)\) as \(r\to+\infty\) for each \(c\in (0,\,+\infty)\) and \(\ln\,n=o(\Gamma_f(\lambda_n))\) as \(n\to\infty\), then \(R[A]=\varliminf\limits_{n\to\infty}\dfrac{1}{\lambda_n}M^{-1}_f\left(\dfrac{1}{|a_n|}\right)\).
For a positive function \(l\) continuous on \([0,+\infty)\), the entire function (2) (i.e., \(R[A]=+\infty\)) is said [10] to be of bounded \(l\)–\(\mathfrak{M}\)-index if there exists \(N\in{\Bbb Z}_+\) such that \[\frac{\mathfrak{M}(r,A^{(n)})}{n!l^n(r)}\le\max\left\{\frac{\mathfrak{M}(r,A^{(k)})}{k!l^k(r)}:\,0\le k\le N\right\}, \label{t4} \tag{4}\] for all \(n\in{\Bbb Z}_+\) and all \(r\in [0,+\infty)\). The least such integer \(N\) is called the \(l\)–\(\mathfrak{M}\)-index of \(A\) and is denoted by \(N(A; l,\mathfrak{M})\). The growth of entire functions of bounded \(l\)–\(\mathfrak{M}\)-index and the behavior of the coefficients \(a_n\) were studied.
Here, we will consider the case \(R[A]=1\) and assume that a positive continuous function \(l\) on \([0, \,1)\) satisfies the condition \(l(r)(1-r)\ge q>1\) for all \(r\in [r_0, \,1)\). We will call the function \(A\) of bounded \(l\)–\(\mathfrak{M}\)-index if (4) is satisfied for all \(n\in{\Bbb Z}_+\) and all \(r\in [0,\,1)\).
We need the following lemmas.
Lemma 1.[4, p. 76] Let \(g_1,\dots,g_n\) be positive absolutely continuous functions on \([a,\,b]\) and \(g(x):=\max\{g_k(x):\, 1\le k\le n\}\). Suppose that \(g'_k(x)\le \varphi(x)g(x)\) almost everywhere on \([a,\,b]\) for all \(1\le k\le n\), where \(\varphi\) is a positive continuous function on \([a,\,b]\). Then for all \(x\in [a,\,b]\) \[\ln\,g(x)\le \ln\,g(a)+\int\limits_{a}^{x}\varphi(t)\,dt. \label{t5}\]
Lemma 2. [4, p. 76] At the points where the derivative \(M'_f(r)\) exists, the inequality \(M'_f(r)\le M_{f'}(r)\) holds.
Using these lemmas, we first prove the following statement.
Proposition 1. Let \(l\) be a positive, continuously differentiable, and non-decreasing function on \([0,1)\) such that \(l(r)(1-r)\ge q>1\) for all \(r\in [r_0, \,1)\) and \[\varlimsup\limits_{r\to+\infty}\frac{(-l'(r))^+}{l^2(r)}=\eta<+\infty, \quad (x^+=\max\{x,\,0\}). \label{t6} \tag{6}\]
If \(N(A; l,\mathfrak{M})=N\), then \[\varlimsup\limits_{r\uparrow 1}\frac{\ln\,\mathfrak{M}(r,A)}{L(r)}\le (1+\eta)(N+1), \quad \left(L(r)=\int\limits_{0}^{r}l(t)\,dt\right). \label{t7} \tag{7}\]
Proof. First, we note that \[L(r) \ge q \int\limits_{r_0}^{r} \frac{dt}{1-t} = q\left(\ln\,\frac{1}{1-r} – \ln\,\frac{1}{1 – r_0}\right) \to +\infty \quad (r \uparrow 1).\]
Since \(A'(z) = \sum\limits_{n=1}^{\infty} a_n \lambda_n f'(\lambda_n z)\), by Lemma 2, at points where the derivative \(M'_f(r)\) exists, we have \[\mathfrak{M}(r, A') = \sum\limits_{n=1}^{\infty} |a_n| \lambda_n M_{f'}(r \lambda_n) \ge \sum\limits_{n=1}^{\infty} |a_n| \lambda_n M'_f(r \lambda_n) = \mathfrak{M}'(r, A).\]
Let \(g_k(r) = \dfrac{\mathfrak{M}(r, A^{(k)})}{k!l^k(r)}\) for \(0 \le k \le N\) and \(g(r) = \max\{g_k(r):\, 0 \le k \le N\}\). Then, almost everywhere on \([r_0,\,1)\), \[\begin{aligned} g'_k(r) &= \dfrac{\mathfrak{M}'(r, A^{(k)})}{k!l^k(r)} – \dfrac{\mathfrak{M}(r, A^{(k)})}{k!l^{k+1}(r)} k l'(r) \\ &\le \dfrac{\mathfrak{M}(r, A^{(k+1)})}{(k+1)!l^{k+1}(r)} (k+1)l(r) + \dfrac{\mathfrak{M}(r, A^{(k)})}{k!l^k(r)} k \frac{(-l'(r))^+}{l(r)} \\ &\le g(r) \left((k+1)l(r) + k \frac{(-l'(r))^+}{l(r)}\right) \le g(r)(N+1)l(r)\left(1 + \frac{(-l'(r))^+}{l^2(r)}\right). \end{aligned}\]
Setting \(\varphi(r) = (N+1)l(r)\left(1 + \dfrac{(-l'(r))^+}{l^2(r)}\right)\), by Lemma 1 and using (5), we get \[\ln\,g(r) \le \ln\,g(r_0) + (N+1)\int\limits_{r_0}^{r} l(r)(1 + \eta + o(1))\,dr, \quad r \uparrow 1,\] and since \(\ln\,\mathfrak{M}(r, A) \le \ln\,g(r)\), in view of (6) we obtain (7). ◻
To obtain the inverse statement, we denote by \(\Omega(0)\) a class of positive functions unbounded on \((-\infty, 0)\), where each function \(\Phi\) has a positive, continuously differentiable, and strictly increasing derivative \(\Phi'\) on \((-\infty, 0)\). Let \(\Psi(x) = x – \Phi(x)/\Phi'(x)\) be the function associated with \(\Phi\) in the sense of Newton. Then \(\Psi\) is [4, p.75] continuously differentiable and increasing to \(+\infty\) on \((-\infty, 0)\). Theorem 4.3 from [4p. 79] implies the following statement.
Lemma 3. Let \(f\) be an analytic function in \({\Bbb D} = \{z:\,|z|<1\}\), and let the function \(\Phi \in \Omega(0)\) satisfy the conditions \(x + \beta(x) := x + \dfrac{\ln\,\Phi'(\Psi^{-1}(x))}{\Phi'(\Psi^{-1}(x))} < 0\) for \(x \in [x_0,\,0)\) and \(\Phi'(\Psi^{-1}(x + \beta(x))) = O(\Phi'(x))\) as \(x \uparrow 0\). Suppose that \(\Phi'(x) > \alpha/|x|\) for \(x < 0\) and \(\Phi'(x + \alpha/\Phi'(x)) = O(\Phi'(x))\) as \(x \uparrow 0\) for some \(\alpha > 1\). If \[\varlimsup\limits_{r \uparrow 1} \frac{\ln\,M_f(r)}{\Phi(\ln\,r)} < +\infty, \label{t8} \tag{8}\] then for every \(r_0 \in (0,\,1)\) there exists \(n_0 = n_0(r_0) \in {\Bbb Z}_{+}\) such that \[\frac{M_{f^{(n)}}(r)}{n!M_f(r)} \le \left(\frac{\Phi'(\ln\,r)}{r}\right)^n \label{t9} \tag{9}\] for all \(n \ge n_0\) and all \(r \in [r_0,\,1)\).
Using Lemma 3, we prove the following statement.
Proposition 2. Let \(R[A] = 1\) and suppose that the function \(\Phi\) satisfies the conditions of Lemma 3. Assume that \(f_k \ge 0\) and \(a_n \ge 0\) for all \(k\) and \(n\). If \[\varlimsup\limits_{r \uparrow 1} \dfrac{\ln\, \mathfrak{M}(r,A)}{\Phi(\ln\,r)} < +\infty, \label{t10} \tag{10}\] then the function (2) is of bounded \(l\)–\(\mathfrak{M}\)-index with \(l(r) = \dfrac{\Phi'(\ln\,r)}{r}\) for \(r \in [r_0,\,1)\).
Proof. It is clear that if \(M_A(r) = \max\{|A(z)|:\,|z| = r\}\), then \(M_A(r) \le \mathfrak{M}(r,A)\), and if \(f_k \ge 0\) and \(a_n \ge 0\) for all \(k\) and \(n\), then \(M_A(r) = \mathfrak{M}(r,A)\). Therefore, (10) implies (8), and by Lemma 3, for every \(r_0 > 0\) there exists \(n_0 = n_0(r_0) \in {\Bbb Z}_0\) such that for all \(n \ge n_0\) and \(r \in [r_0,\,1)\), \[\frac{\mathfrak{M}(r,A^{(n)})}{n!\mathfrak{M}(r,A)} \le \left(\frac{\Phi'(\ln\,r)}{r}\right)^n.\]
Setting \(l(r) = \Phi'(\ln\,r)/r\) for \(r \ge r_0\), we obtain \[\dfrac{\mathfrak{M}(r,A^{(n)})}{n!l^n(r)} \le \mathfrak{M}(r,A),\] and thus, the function \(A\) is of bounded \(l\)–\(\mathfrak{M}\)-index. ◻
Combining Propositions 1 and 2, we arrive at the following theorem.
Theorem 1. Let the function \(\Phi \in \Omega(0)\), and let the coefficients \(f_k\) and \(a_n\) satisfy the conditions of Proposition 2. Then the function (2) is of bounded \(l\)–\(\mathfrak{M}\)-index with \(l(r) = \Phi'(\ln\,r)/r\) for \(r \ge r_0\) and \(l(r) = \Phi'(\ln\,r_0)/r_0\) for \(0 \le r \le r_0\) if and only if (10) holds.
Indeed, if \(l(r) = \Phi'(\ln\,r)/r\), then \(L(r) = (1 + o(1)) \Phi(\ln\,r)\) as \(r \to +\infty\), \[l(r) \ge \frac{\alpha}{r|\ln\,r|} = (1 + o(1)) \frac{\alpha}{1 – r}, \quad r \uparrow 1,\] for some \(\alpha > 1\), i.e., \(l(r)(1 – r) \ge q > 1\) for some \(q > 1\) and all \(r \in [r_0,\,1)\). Thus, \[\varlimsup\limits_{r \to +\infty} \frac{(-l'(r))^+}{l^2(r)} = \varlimsup\limits_{r \to +\infty} \frac{(\Phi'(\ln\,r) – \Phi''(\ln\,r))^+}{(\Phi'(\ln\,r))^2 r} \le \varlimsup\limits_{r \to +\infty} \frac{\Phi'(\ln\,r)^+}{(\Phi'(\ln\,r))^2 r} = 0,\] because \(\Phi''(x) \ge 0\) for all \(x\). Therefore, if the function (2) is of bounded \(l\)–\(\mathfrak{M}\)-index, then by Proposition 1, (10) holds. The converse follows from Proposition 2.
Let \(p > 0\) and \(\Phi(x) = |x|^{-p}\) for \(x < 0\). Then \(\Phi \in \Omega(0)\), \(\Phi'(x) = p|x|^{-p-1}\), \(\Psi(x) = \dfrac{p+1}{p}x\) and \(\Psi^{-1}(x) = \dfrac{p}{p+1}x\) for \(x < 0\). Therefore, \[\Phi'(\Psi^{-1}(x)) = \frac{p}{|\Psi^{-1}(x)|^{p+1}} = \frac{A_p}{|x|^{p+1}}, \quad A_p = \frac{(p+1)^{p+1}}{p^p} > 1,\] i.e., \[x + \beta(x) = \dfrac{\ln\,\Phi'(\Psi^{-1}(x))}{\Phi'(\Psi^{-1}(x))} + x = \frac{|x|^{p+1} \ln\,({A_p}{|x|^{p+1}})}{A_p} – |x| = -|x|\left(1 – \frac{|x|^p \ln\,(A_p |x|^{p+1})}{A_p}\right) < 0,\] for \(x \in [x_0, 0)\), and \[\Phi'(\Psi^{-1}(x + \beta(x))) = \frac{A_p}{|x|^{p+1}} \left(1 – \frac{|x|^p \ln\,({A_p}{|x|^{p+1}})}{A_p}\right)^{-p-1} = \frac{A_p(1 + o(1))}{|x|^{p+1}} = O(\Phi'(x)),\] as \(x \uparrow 0\). Similarly, \(\Phi'(x + \alpha/\Phi'(x)) = O(\Phi'(x))\) as \(x \uparrow 0\) for each \(\alpha > 1\). Therefore, Theorem 1 implies the following corollary.
Corollary 1. Let \(R[A] = 1\) and \(a > p > 0\). Suppose that \(f_k \ge 0\) and \(a_n \ge 0\) for all \(k\) and \(n\). Then the function (2) is of bounded \(l\)–\(\mathfrak{M}\)-index with \(l(r) = \dfrac{p}{(1 – r)^{p+1}}\) if and only if \[\varlimsup\limits_{r \uparrow 1}(1 – r)^p \ln\,\mathfrak{M}(r, A) = \varlimsup\limits_{r \uparrow 1}(1 – r)^p \ln\,M_A(r) < +\infty. \label{t11} \tag{11}\]
Indeed, by Theorem (1), the function (2) is of bounded \(l\)–\(\mathfrak{M}\)-index with \(l(r)=\dfrac{p}{r|\ln\,r|^{p+1}}\) for \(r \in [r_0, 1)\) if and only if \(\varlimsup\limits_{r\uparrow 1}\dfrac{\ln\,\mathfrak{M}(r,A)}{|\ln\,r|^{-p}}<+\infty\). Since \(r|\ln\,r|^{p+1}=(1+o(1))(1-r)^{p+1}\) as \(r\uparrow 1\) and \(\dfrac{p}{(1-r)^{p+1}}\ge \dfrac{a}{1-r}\) for every \(a>1\) and all \(r \in [r_0(a),\,1)\), we get the required result.
Let’s find out under what conditions on \(f_k \ge 0\) and \(a_n \ge 0\) equality (11) holds.
For a function \(g(z)=\sum\limits_{k=0}^{\infty}g_k z^k\)
analytic in the unit disk, let \(M_g(r)=\max\{|g(z)|:\,|z|=r<1\}\) and
\(\mu_g(r)=\max\{|g_k|r^k:\,k\ge 0\}\)
be the maximal term. The order of \(g\)
is determined by the formula \(p=\varlimsup\limits_{r\uparrow
1}\dfrac{\ln\,\ln\,M_g(r)}{\ln\,(1/(1-r))}\) and the type is
determined by the formula
\(T=\varlimsup\limits_{r\uparrow
1}(1-r)^p\ln\,M_g(r)\) (see, for example, [11]). It is known [11] that \[T=
\varlimsup\limits_{r\uparrow 1}(1-r)^p\ln\,\mu_g(r)=
\frac{p^p}{(p+1)^{p+1}}\varlimsup\limits_{k \to
\infty}\frac{(\ln^+|g_k|)^{p+1}}{k^p}.
\label{t12} \tag{12}\]
Let’s return to the function \(A\). Put \(\mu_D(k)=\max\{a_n\lambda^k_n:\,n\ge 1\}\) and \(\mu_G(r)=\max\{|f_k|\mu_D(k)r^k:\,k\ge 0\}\). Since \[A(z)=\sum\limits_{n=1}^{\infty}a_n\sum\limits_{k=0}^{\infty}f_k(z\lambda_n)^k= \sum\limits_{k=0}^{\infty}f_k\left(\sum\limits_{n=1}^{\infty}a_n\lambda_n^k\right)z^k,\] we have \[M_A(r)\ge f_k\left(\sum\limits_{n=1}^{\infty}a_n\lambda_n^k\right)r^k\ge a_nf_k(\lambda_nr)^k,\] for all \(n\ge 1\), \(k\ge 0\) and \(r\in[0,1)\), whence it follows that \(M_A(r)\ge f_k\mu_D(k)r^k\), that is \(\mu_G(r)\le M_A(r)\).
On the other hand, since series (2) is regularly convergent in \({\Bbb D}\), for every \(r\in [0,1)\) we have \[M_A(r)\le \sum\limits_{n=1}^{\infty} |a_n|M_f(r\lambda_n)\le \mu_A\left(\frac{1+r}{2}\right)\sum\limits_{n=1}^{\infty}\frac{M_f(r\lambda_n)}{M_f\left(\dfrac{r+1}{2}\lambda_n\right)}, \label{t13} \tag{13}\] where \(\mu_A(r)=\max\{|a_n|M_f(r\lambda_n):\,n\ge 1\}\). Suppose that \(\ln\,\ln\,n=o(\ln\,\Gamma_f(c\lambda_n))\) as \(n\to\infty\) for some \(0<c<1\). Then \(\Gamma_f(c\lambda_n)\ge (\ln\,n)^{1/\varepsilon}=(\ln\,n)^{1/\varepsilon-1}\ln\,n\) for every \(\varepsilon \in (0,\,1)\) and all \(n\ge n^*(\varepsilon)\). We put \(n_0(r)=\left[\exp\left\{\left(\dfrac{8}{1-r}\right)^{\varepsilon/(1-\varepsilon)}\right\}\right]+1\). Then \(n_0(r)\ge n^*(\varepsilon)\) for \(r\in [r_0(\varepsilon),\,1)\) and for \(n\ge n_0(r)\) we get \[\frac{(1-r)\Gamma_f(c\lambda_n)}{4}\ge\frac{(1-r)(\ln\,n)^{1/\varepsilon-1}\ln\,n}{4}\ge \frac{(1-r)(\ln\,n_0(r))^{1/\varepsilon-1}\ln\,n}{4}=2\ln\,n.\]
Therefore, for \(r \in [r_0(\varepsilon),\,1) \subset [c, 1)\) and \(n \ge n_0(r)\) we have \[\begin{aligned} \ln\,M_f\left(\dfrac{r+1}{2}\lambda_n\right)-\ln\,M_f(r\lambda_n)=&\int\limits_{r\lambda_n}^{(r+1)\lambda_n/2}\Gamma_f(x)d\ln\,x\\ \ge&\Gamma_f(r\lambda_n)\ln\,\frac{1+r}{2r}\\ =& \Gamma_f(r\lambda_n)\ln\,\left(1+\frac{1-r}{2r}\right)\\ \ge& \frac{(1-r)\Gamma_f(c\lambda_n)}{4}\ge 2n. \end{aligned}\]
From this, it follows that \[\begin{aligned} \sum\limits_{n=1}^{\infty}\frac{M_f(r\lambda_n)}{M_f\left(\dfrac{r+1}{2}\lambda_n\right)}\le&\left(\sum\limits_{n=1}^{n_0(r)-1}+\sum\limits_{n_0(r)}^{\infty}\right)\exp\left\{-\frac{(1-r)\Gamma_f(r\lambda_n)}{4}\right\}\\ \le&\sum\limits_{n=1}^{n_0(r)-1}1+\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\le n_0(r)+2\le \exp\left\{\left(\dfrac{8}{1-r}\right)^{\varepsilon/(1-\varepsilon)}\right\}+3. \end{aligned}\]
Thus, (13) implies \[M_A(r)\le \mu_A\left(\frac{1+r}{2}\right) \left(\exp\left\{\left(\dfrac{8}{1-r}\right)^{\varepsilon/(1-\varepsilon)}\right\}+3\right), \quad (r\in [r_0(\varepsilon),\,1)). \label{t14} \tag{14}\]
Also, we have \[\begin{aligned} \mu_A(r)\le\max\left\{a_n\sum\limits_{k=0}^{\infty}f_k(r\lambda_n)^k:\,n\ge 1\right\}\le& \sum\limits_{k=0}^{\infty}\max\{a_n\lambda^k_n:\,n\ge 1\}f_kr^k\\ =&\sum\limits_{k=0}^{\infty}\mu_D(k)f_k\left(\frac{1+r}{2}\right)^k\left(\frac{2r}{1+r}\right)^k\\ \le&\mu_G\left(\frac{1+r}{2}\right)\frac{1+r}{1-r}\le\frac{2}{1-r}\mu_G\left(\frac{1+r}{2}\right). \end{aligned}\]
Therefore, if \(\varepsilon < p/(p+1)\), then in view of (10) \[\ln\,\mu_G(r)\le\ln\,M_A(r)\le \ln\,\mu_G\left(\frac{3+r}{4}\right)+\ln\,\frac{4}{1-r}+ \left(\dfrac{8}{1-r}\right)^{\varepsilon/(1-\varepsilon)}+\ln\,6,\] whence \[\begin{aligned} \varlimsup\limits_{r\uparrow 1}(1-r)^p\ln\,\mu_G(r)\le \varlimsup\limits_{r\uparrow 1}(1-r)^p\ln\,M_A(r)\le \varlimsup\limits_{r\uparrow 1}(1-r)^p\ln\,\mu_G\left(\frac{3+r}{4}\right) =4^p\varlimsup\limits_{r\uparrow 1}(1-r)^p\ln\,\mu_G(r). \end{aligned}\]
In view of (12), we get that \(\varlimsup\limits_{r\uparrow 1}(1-r)^p\ln\,\mu_G(r)<+\infty\) if and only if \[\varlimsup\limits_{k \to \infty}k^{-p}(\ln^+(f_k\mu_D(k)))^{p+1}<+\infty. \label{t15} \tag{15}\]
Thus, the following theorem holds.
Theorem 2. If \(f_k \ge 0\) and \(a_n \ge 0\) for all \(k\) and \(n\), then the function (2) is of bounded \(l\)–\(\mathfrak{M}\)-index with \(l(r) = \dfrac{p}{(1-r)^{p+1}}\) if and only if (15) holds.
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