Lower Bounds for Absolutely Convergent Dirichlet Series and Pits Property

Abstract

This article investigates the lower bounds of analytic functions defined by absolutely convergent Dirichlet series in the left half-plane and establishes conditions under which such functions exhibit the pits property. The study extends classical results for entire Dirichlet series and lacunary power series by refining assumptions on the sequence of exponents and resolving gaps in earlier proofs found in the literature. Central to the analysis is the introduction of a modified function $k\left(\sigma \right)$, whose behavior determines the size of exceptional sets surrounding the zeros of the series. Under suitable growth conditions on the exponents, the authors prove that outside small disks centered at the zeros, the modulus of the Dirichlet series admits explicit lower bounds involving its maximum term. Several auxiliary lemmas provide sharp estimates for the maximum modulus, the distribution of zeros, and the behavior of truncated Dirichlet polynomials. The main theorem demonstrates that the pits property holds uniformly in vertical strips approaching the imaginary axis. The paper concludes with a discussion of how the imposed step-size condition on exponents might be weakened, formulating a conjecture regarding a more general condensation index.

1. Introduction

Analytic functions that are large outside small neighborhoods of the zeros are said to have the pits effect or pits property. The paper by Littlewood and Offord [1,2] states that certain random entire functions exhibit the pits effect. And for the various approaches to the concept of a random entire function, different authors have investigated the pits effect [3,4,5] and probability of zero holes [6,7]. Offord [8,9] showed that certain gap series also exhibit the pits effect. Recently, a similar effect was discovered [10] for entire functions of the exponential type.

It is well known that for entire functions f represented by lacunary power series of the form

$$f\left(z\right)=a_0+{\displaystyle \sum _{k=1}^{+\infty }}{a}_k{z}^{n_k},$$

(1)

restrictions on the sequence of powers $\left(n_k\right)$ provide a certain regularity in the asymptotic behavior of the function f and in the distribution of its values.

In 1999, O.B. Skaskiv and M.G. Pivkach [11] (see also [12]) proved some results for the entire Dirichlet series

$$F\left(z\right)={\displaystyle \sum _{n=0}^{+\infty }}{a}_n{e}^{{\lambda }_{n}z},$$

(2)

where $z=\sigma +i\tau$, $a_n\in \mathbb{C}$ and a sequence $\lambda =\left({\lambda }_n\right)$ is such that $0\le {\lambda }_0<{\lambda }_n\uparrow +\infty$$(1\le n\to +\infty )$,

$${\lambda }_{n+1}-{\lambda }_n\ge d>0.$$

(3)

and

$$\underset{n\to +\infty }{\overline{lim}}{\displaystyle \frac{lnn}{ln{\lambda }_n}}=\alpha <1.$$

(4)

Those results are similar to results from [8] for entire functions defined by lacunary Taylor series of form (1). We will the use main notations from Dirichlet series theory (for example, see [13]):

$$\mu (\sigma ,F)=max\{|a_n|e^{{\lambda }_n\sigma }:n\ge 0\},$$

$$M(\sigma ,F)=sup\left\{\right|F(\sigma +i\tau )|:\tau \in \mathbb{R}\},$$

$$\nu (\sigma ,F)=max\{n:\mu (\sigma ,F)=|a_n|e^{{\lambda }_n\sigma }\}$$

and

$$\mathsf{\Lambda }\left(\sigma \right)=\mathsf{\Lambda }(\sigma ,F)={\lambda }_{\nu (\sigma ,F)}.$$

The functions $\mu (\sigma ,F),M(\sigma ,F),\nu (\sigma ,F)$ and $\mathsf{\Lambda }(\sigma ,F)$ are called the maximum term, maximum modulus, central index and central exponent of the maximum term, respectively. The functions $\mathsf{\Lambda }\left(\sigma \right)$ and $\nu (\sigma ,F)$ are the non-decreasing step function. The functions $ln\mu (\sigma ,F)$ and $lnM(\sigma ,F)$ are the non-decreasing convex functions on $(-\infty ,0)$.

Definition 1

([11]). Let ψ be some positive decreasing function on $(0,+\infty )$ such that

$$\underset{x\to +0}{\underline{lim}}{\displaystyle \frac{|ln\psi (x\left)\right|}{|lnx|}}>{\displaystyle \frac{1}{1-\alpha }}(0<\alpha <1).$$

And for every function $\epsilon \left(x\right)\to -0$$(x\to +0)$, one has

$$\psi \left(x\right(1+\epsilon \left(x\right)\left)\right)\le C\psi \left(x\right)\mathit{for}x\in (0,+\infty ).$$

For $\sigma \in \mathbb{R}$, we also define

$$k\left(\sigma \right)=\underset{\theta >0}{inf}max\{\mathsf{\Lambda }(\sigma +\theta ),\psi \left(\theta \right)\}.$$

Theorem 1

([11]). Suppose that an entire Dirichlet series of the form (2) satisfies (3) and (4). Then, for any $\epsilon \in (0,1-\alpha )$, there exists ${\sigma }_0={\sigma }_0\left(\epsilon \right)$ such that for each zero a of the function F with $Rea=\sigma >{\sigma }_0$, one has

$$\left|F\left(z\right)\right|>e^{-k^{\alpha +\epsilon }\left(\sigma \right)}\mu (\sigma ,F).$$

(5)

for all $|z-a|=r$ and some radius r chosen from inequality $r=r\left(a\right)\le e^{-k^{\delta }\left(a\right)}$ with $\delta <\epsilon$.

For z outside the union of the disks with centers on zeros of F (i.e., $z\notin {\displaystyle \bigcup _{a \mathit{is} \mathit{zero} \mathit{of} F}}\{z:|z-a|<r\left(a\right)\})$ and $Rez=\sigma$, inequality (5) also holds.

A similar theorem was presented in 2009 by Shang Lina and Gao Zongsheng ([14]) with a different definition of the function k and condition

$$\underset{n\to +\infty }{\underline{lim}}{\displaystyle \frac{ln{\lambda }_n}{lnn}}=\alpha >2$$

(6)

instead of condition (4).

Definition 2

([14], Definition 1). For $\sigma \in \mathbb{R}$, define

$$k_0\left(\sigma \right)=\underset{\theta >0}{inf}max\left\{\mathsf{\Lambda }(ln(e^{\sigma }+\theta )),{\displaystyle \frac{e^{\sigma }}{\theta }}\right\}.$$

Theorem 2

([14], Theorem 1). If the entire Dirichlet series (2) satisfies (6), associated with any number β satisfying ${\displaystyle \frac{1}{\alpha -1}}<\beta <1$, then for s with $Res=\sigma >{\sigma }_0$, there is a circle with center s on which

$$\left|F\left(s\right)\right|>e^{-k_0^{\beta }\left(\sigma \right)}\mu (\sigma ,F).$$

(7)

and ${\sigma }_0$ depends only on F and β. The radius r of this circle does not exceed $e^{-k_0^{\delta }\left(\sigma \right)},$ where δ satisfies $0<\delta <\beta -{\displaystyle \frac{1}{\alpha -1}}$. For s outside the pits of F and $Res=\sigma$, (7) holds.

The reasoning in papers [14,15] contains gaps. We will only point out that in the proof in [14], the authors, for example, implicitly used the condition ${\lambda }_{n+1}-{\lambda }_n>1$ ($n\ge 0$), which is not present in their formulation of Theorem 2. Proof of Theorem 2 depends on a chain of lemmas. Among them there is Lemma 3 in [14] (see also proof of Lemma 2 in [15]). In its proof, such an estimate was used:

$${\displaystyle \sum _{{\lambda }_n\ge k+1}^{\infty }}{\lambda }_n^q{e}^{{\lambda }_n\sigma -G_{{\lambda }_n}}\le {\displaystyle \sum _{\nu =k+1}^{\infty }}{\nu }^q{e}^{\sigma \nu -G_{\nu }}.$$

The authors replaced ${\lambda }_n$ by $\nu$. But if at least two consecutive ${\lambda }_n$ and ${\lambda }_{n+1}$ belong to $[m,m+1)$ for some $m\ge k+1$, then such a replacement is incorrect. Moreover, in the proof of Lemma 5 in [14] (see also proof of Lemma 4 [15]), such an inequality was used:

$$|\prod _{{\displaystyle \genfrac{}{}{0pt}{}{{\lambda }_j\in I}{{\lambda }_j\ne \mathsf{\Lambda }\left(\sigma \right)}}}(\mathsf{\Lambda }\left(\sigma \right)-{\lambda }_j)|\ge {\left([q/2]!\right)}^2,$$

(8)

where $I=\{{\lambda }_j:j\le q\left(\sigma \right)\}\setminus \left\{\mathsf{\Lambda }\left(\sigma \right)\right\}$, and the number $q=q\left(\sigma \right)$, in particular, is such that $\mathsf{\Lambda }\left(\sigma \right)<{\lambda }_q.$ Obviously, this inequality is incorrect in general (for an arbitrary sequence of exponents). According to the formulated statement, inequality (8) must hold for an arbitrary sequence of exponents $\left({\lambda }_j\right)$. However, we assume that for some sequence ${\sigma }_j$, the previous largest exponent $\lambda ={\lambda }_{\nu \left({\sigma }_j\right)-1}$ can be directed to the central exponent $\lambda \to \mathsf{\Lambda }\left({\sigma }_j\right)-0$. But ${\lambda }_{\nu \left({\sigma }_j\right)-1}$ is smaller than the central exponent $\mathsf{\Lambda }\left({\sigma }_j\right)$ and we fix values of the other parameters included in inequality (8). Then, the left-hand side of inequality (8) will tend to zero, while the right-hand side of inequality (8) will remain unchanged and positive. This is a contradiction.

As we will see from our further considerations, under condition (3), the statement of Theorem 2 is likely to be correct in the form in which we have formulated it. However, the same cannot be said about the proofs presented in these papers. We also note that the reason for this is that the authors repeat step by step the corresponding proof for lacunary power series from [8], where the exponents of the powers ${\lambda }_n$ are natural numbers, and thus condition (3) is satisfied. The results of [8] are correct. However, there are some inaccuracies in the reasoning in their proofs that do not affect the claims themselves. Some of these inaccuracies in the reasoning, as well as the above-mentioned lack of condition (3), have also been carried over into the cited papers [14,15].

In this article, we limit ourselves to the case of absolutely convergent Dirichlet series in the half-plane $\{z:\mathrm{Re}z<0\}$. It is easy to see from the proofs in this article that the main auxiliary statements remain valid in the case of entire Dirichlet series.

The goal of this article is to study the Dirichlet series, which have form (2), and their convergence domain is the half-plane $\{z:\mathrm{Re}z<0\}$.

2. Main Result

Furthermore, suppose that the series (2) satisfies

$$\underset{\sigma \to 0_{-}}{\underline{lim}}{(-\sigma )}^{\gamma }\mathsf{\Lambda }\left(\sigma \right)>1,$$

(9)

where $\gamma >{\displaystyle \frac{1}{1-\alpha }}$, $\alpha \in (0,1)$ is defined in (4).

For $\sigma <0$ and $\theta \in (\sigma ,0)$, we put

$$\varkappa (\sigma ,\theta )={\varkappa }_{\gamma }(\sigma ,\theta )=max\left\{\mathsf{\Lambda }\left(\theta \right),{\displaystyle \frac{1}{{(\theta -\sigma )}^{\gamma }}}\right\},$$

$$k\left(\sigma \right)=\underset{\theta \in (\sigma ;0)}{inf}\varkappa (\sigma ,\theta ).$$

(10)

The properties of function $k\left(\sigma \right)$ are given in Lemma 1 below. In particular, it is a right-continuous non-decreasing function of $\sigma <0$.

Theorem 3.

If Dirichlet series (2) is absolutely convergent in the half-plane $\{z:\mathrm{Re}z<0\}$ and it satisfies (3), (4) and (9), i.e., $F\left(z\right)={\sum }_{n=0}^{+\infty }{a}_n{e}^{{\lambda }_{n}z},$ ${\lambda }_{n+1}-{\lambda }_n\ge d>0,$${\overline{lim}}_{n\to +\infty }{\displaystyle \frac{lnn}{ln{\lambda }_n}}=\alpha <1$, then for any number $\epsilon \in (0,1-\alpha )$, there exists ${\sigma }_0={\sigma }_0\left(\epsilon \right)\in (-\infty ,0)$ such that for each zero a of the function F with $Rea=\sigma \in ({\sigma }_0,0)$, the following estimate is valid for any points from the circle with center a and radius r:

$$\left|F\left(z\right)\right|>e^{-k^{\alpha +\epsilon }\left(\sigma \right)}\mu (\sigma ,F),$$

(11)

where $k\left(\sigma \right)$ is defined in (10). The radius r of this circle can be chosen from the condition $r=r\left(a\right)\le e^{-k^{\delta }\left(a\right)}$ with $\delta <\epsilon$.

For z outside the union of the disks with centers on zeros of F (i.e., $z\notin {\displaystyle \bigcup _{a \mathit{is} \mathit{zero} \mathit{of} F}}\{z:|z-a|<r\left(a\right)\})$ and $Rez=\sigma \in ({\sigma }_0,0)$, inequality (11) also holds.

3. Preliminary Lemmas and Proof of the Main Theorem

Lemma 1.

The above-defined function $k\left(\sigma \right)$ has the following properties:

(1) 

$k\left(\sigma \right)$ is a non-decreasing function on the set $\sigma \in (-\infty ,0);$

(2) 

$k\left(\sigma \right)$ is a right-continuous function of $\sigma \in (-\infty ,0);$

(3) 

$\mathsf{\Lambda }\left(\sigma \right)\le k\left(\sigma \right)\le \mathsf{\Lambda }(\sigma +{\mathsf{\Lambda }}^{-\frac{1}{\gamma }}\left(\sigma \right))$ for $\sigma \in ({\sigma }_0,0)$, where γ is a constant from condition (9);

(4) 

For any $\epsilon >0$ and $\eta \in (\frac{1}{\gamma },1-\alpha )$, there exists ${\sigma }_0$ such that for $\sigma \in ({\sigma }_0,0)$,

$$k\left(\sigma +{\displaystyle \frac{1}{k^{\eta }\left(\sigma \right)}}\right)\le (1+\epsilon )k\left(\sigma \right);$$

(5) 

For any $\epsilon >0$, there exists ${\sigma }_0$ such that for $\sigma \in ({\sigma }_0,0),$

$$k\left(\sigma +{\displaystyle \frac{1}{k\left(\sigma \right)}}\right)\le (1+\epsilon )k\left(\sigma \right).$$

Proof. 

Although items (1) and (2), given geometric considerations, are obvious, for the sake of completeness, we will provide their full proofs. Remark, at first, that

$$\varkappa (\sigma ,\theta )=\left\{\begin{array}{cc}\mathsf{\Lambda }\left(\theta \right),\hfill & \sigma \le {\sigma }_1\left(\theta \right),\hfill \\ {(\theta -\sigma )}^{-\gamma },\hfill & {\sigma }_1\left(\theta \right)<\sigma <\theta ,\hfill \end{array}\right.$$

where ${\sigma }_1\left(\theta \right)=\theta -{\left(\mathsf{\Lambda }\left(\theta \right)\right)}^{-1/\gamma }.$

(1)

It is easy to see that

$$k\left(\sigma \right)=\underset{0<\theta <-\sigma }{inf}max\{\mathsf{\Lambda }(\sigma +\theta ),{\theta }^{-\gamma }\}.$$

(12)

Then, for ${\sigma }_1<{\sigma }_2$, one has

$$k\left({\sigma }_1\right)\le \underset{0<\theta <-{\sigma }_2}{inf}max\{\mathsf{\Lambda }({\sigma }_1+\theta ),{\theta }^{-\gamma }\}\le \underset{0<\theta <-{\sigma }_2}{inf}max\{\mathsf{\Lambda }({\sigma }_2+\theta ),{\theta }^{-\gamma }\}=k\left({\sigma }_2\right).$$

(2)

Consistently, we have the following for $\theta >\sigma +\delta ,\delta >0$:

$$\begin{array}{c}\varkappa (\sigma +\delta ,\theta )=max\left\{\mathsf{\Lambda }\left(\theta \right);{\displaystyle \frac{1}{{(\theta -\sigma -\delta )}^{\gamma }}}\right\}=max\left\{\mathsf{\Lambda }\left(\theta \right);{\displaystyle \frac{1}{{(\theta -\sigma )}^{\gamma }}}{\left(1+{\displaystyle \frac{\delta }{\theta -\sigma -\delta }}\right)}^{\gamma }\right\}\le \\ \le {\left(1+{\displaystyle \frac{\delta }{\theta -\sigma -\delta }}\right)}^{\gamma }max\left\{\mathsf{\Lambda }\left(\theta \right);{\displaystyle \frac{1}{{(\theta -\sigma )}^{\gamma }}}\right\}={\left(1+{\displaystyle \frac{\delta }{\theta -\sigma -\delta }}\right)}^{\gamma }\varkappa (\sigma ,\theta ).\end{array}$$

(13)

We now assume that $\theta ={\theta }_0={\theta }_0\left(\sigma \right)$ is such that

$$k\left(\sigma \right)=\underset{\theta \in (\sigma ;0)}{inf}\varkappa (\sigma ,\theta )\le \varkappa (\sigma ,{\theta }_0)<(1+{\epsilon }_1)\underset{\theta \in (\sigma ;0)}{inf}\varkappa (\sigma ,\theta )=(1+{\epsilon }_1)k\left(\sigma \right).$$

(14)

For a given $\epsilon \in (0,1)$, we now choose

$$\delta ={\epsilon }_1{\displaystyle \frac{\theta -\sigma }{2}}.$$

Here, ${\epsilon }_1>0$ is chosen such that

$${(1+{\epsilon }_1)}^{1+\gamma }=1+\epsilon .$$

Then, from (13) and (14), we get

$$\varkappa (\sigma +\delta ,{\theta }_0)\le {(1+{\epsilon }_1)}^{\gamma }max\left\{\mathsf{\Lambda }\left({\theta }_0\right);{\displaystyle \frac{1}{{({\theta }_0-\sigma )}^{\gamma }}}\right\}={(1+{\epsilon }_1)}^{\gamma }\varkappa (\sigma ,{\theta }_0).$$

Hence,

$$k(\sigma +\delta )\le \varkappa (\sigma +\delta ,{\theta }_0)\le {(1+{\epsilon }_1)}^{\gamma }\varkappa (\sigma ,{\theta }_0)={(1+{\epsilon }_1)}^{1+\gamma }k\left(\sigma \right).$$

Therefore,

$$k\left(\sigma \right)\le k(\sigma +\delta )<(1+\epsilon )k\left(\sigma \right)$$

for $\delta >0$ and

$$\underset{\delta \to +0}{lim}k(\sigma +\delta )=k\left(\sigma \right).$$

This means that $k\left(\sigma \right)$ is the right-continuous function at the given point $\sigma <0$.

(3)

Since $\mathsf{\Lambda }\left(t\right)$ is the right-continuous function of variable $t<0$, inequality (9) implies

$${\mathsf{\Lambda }\left(\sigma \right)>\left|\sigma \right|}^{-\gamma }(\sigma \in ({\sigma }_0,0)),$$

From (12), it follows that

$$k\left(\sigma \right)\ge \underset{0<\theta <-\sigma }{inf}\mathsf{\Lambda }(\sigma +\theta )=\mathsf{\Lambda }\left(\sigma \right)>{\left|\sigma \right|}^{-\gamma }(\sigma \in ({\sigma }_0,0)).$$

Therefore,

$$k^{1/\gamma }\left(\sigma \right)\ge {\mathsf{\Lambda }}^{1/\gamma }\left(\sigma \right)>{\left|\sigma \right|}^{-1}(\sigma \in ({\sigma }_0,0)),$$

and for $\theta ={\mathsf{\Lambda }}^{-1/\gamma }\left(\sigma \right)<-\sigma$ as $\sigma \in ({\sigma }_0,0)$, we obtain consistently

$$\begin{array}{c}k\left(\sigma \right)=\underset{0<\theta <-\sigma }{inf}max\{\mathsf{\Lambda }(\sigma +\theta ),{\theta }^{-\gamma }\}\le \\ \le max\left\{\mathsf{\Lambda }(\sigma +{\mathsf{\Lambda }}^{-\frac{1}{\gamma }}\left(\sigma \right)),\mathsf{\Lambda }\left(\sigma \right)\right\}=\mathsf{\Lambda }(\sigma +{\mathsf{\Lambda }}^{-\frac{1}{\gamma }}\left(\sigma \right)).\end{array}$$

(4)

If we choose $\delta ={\delta }_0:=C{\varkappa }^{-\eta }(\sigma ,\theta )$, $C>0,$ in inequality (13), then

$$\begin{array}{c}\varkappa (\sigma +{\delta }_0,\theta )=max\left\{\mathsf{\Lambda }\left(\theta \right);{\displaystyle \frac{{\varkappa }^{\eta \gamma }(\sigma ,\theta )}{{\left((\theta -\sigma ){\varkappa }^{\eta }(\sigma ,\theta )-C\right)}^{\gamma }}}\right\}.\end{array}$$

But, $\varkappa (\sigma ,\theta )\ge {(\theta -\sigma )}^{-\gamma }$; hence,

$$(\theta -\sigma ){\varkappa }^{\eta }(\sigma ,\theta )\ge {\varkappa }^{\eta -1/\gamma }(\sigma ,\theta )\ge k^{\eta -1/\gamma }\left(\sigma \right)\to +\infty \mathrm{as}\sigma \to -0$$

and

$${\displaystyle \frac{C}{(\theta -\sigma ){\varkappa }^{\eta }(\sigma ,\theta )}}\le k^{1/\gamma -\eta }\left(\sigma \right)\le {\epsilon }_1(\sigma \in ({\sigma }_0,0))$$

for some ${\sigma }_0<0.$ Therefore, at ${\delta }_0=C{\varkappa }^{-\eta }(\sigma ,\theta )$ and $\sigma \in ({\sigma }_0,0)$, by inequality (14), one has

$$\begin{array}{c}k(\sigma +{\delta }_0)\le \varkappa (\sigma +{\delta }_0,{\theta }_0)=max\left\{\mathsf{\Lambda }\left({\theta }_0\right);{\displaystyle \frac{{\varkappa }^{\eta \gamma }(\sigma ,{\theta }_0)}{{\left(({\theta }_0-\sigma ){\varkappa }^{\eta }(\sigma ,{\theta }_0)-C\right)}^{\gamma }}}\right\}\le \\ \le {\displaystyle \frac{1}{{(1-{\epsilon }_1)}^{\gamma }}}max\left\{\mathsf{\Lambda }\left({\theta }_0\right);{\displaystyle \frac{1}{{({\theta }_0-\sigma )}^{\gamma }}}\right\}={\displaystyle \frac{1}{{(1-{\epsilon }_1)}^{\gamma }}}\varkappa (\sigma ,{\theta }_0)\le {\displaystyle \frac{1+{\epsilon }_1}{{(1-{\epsilon }_1)}^{\gamma }}}k\left(\sigma \right).\end{array}$$

Again, for a given $\epsilon >0$, we choose ${\epsilon }_1\in (0,1)$ such that

$${\displaystyle \frac{1+{\epsilon }_1}{{(1-{\epsilon }_1)}^{\gamma }}}\le 1+\epsilon .$$

Using inequality (14), we obtain the following with $C={(1+{\epsilon }_1)}^{\gamma }$:

$$k\left(\sigma +k^{-\eta }\left(\sigma \right)\right)\le k\left(\sigma +C{\varkappa }^{-\eta }(\sigma ,{\theta }_0)\right)\le (1+\epsilon )k\left(\sigma \right).$$

(5)

We note that

$$k\left(\sigma +{\displaystyle \frac{1}{k\left(\sigma \right)}}\right)\le k\left(\sigma +{\displaystyle \frac{1}{k^{\eta }\left(\sigma \right)}}\right),$$

because $\eta <1$ and k is non-decreasing.

□

Lemma 2.

For any $\epsilon >0$, there exists ${\sigma }_0$ such that for $\sigma \in ({\sigma }_0,0)$,

$$\mu \left(\sigma +{\displaystyle \frac{1}{k\left(\sigma \right)}},F\right)\le (1+\epsilon )e\mu (\sigma ,F).$$

Proof. 

Let $\nu =\nu (\sigma +\delta )$. Then,

$$\mu (\sigma +\delta ,F)=\left|a_{\nu }\right|e^{(\sigma +\delta )\mathsf{\Lambda }(\sigma +\delta )}=\left|a_{\nu }\right|e^{\sigma \mathsf{\Lambda }(\sigma +\delta )}{e}^{\delta \mathsf{\Lambda }(\sigma +\delta )}\le \mu (\sigma ,F)e^{\delta \mathsf{\Lambda }(\sigma +\delta )}.$$

If $\delta =1/k\left(\sigma \right)$, then

$$\delta \mathsf{\Lambda }(\sigma +\delta )=\delta \mathsf{\Lambda }(\sigma +1/k\left(\sigma \right))\le {\displaystyle \frac{k(\sigma +1/k(\sigma \left)\right)}{k\left(\sigma \right)}}\le (1+\epsilon )(\sigma \in ({\sigma }_0,0)).$$

So,

$$\mu (\sigma +1/k\left(\sigma \right),F)\le \mu (\sigma ,F)e^{1+\epsilon }.$$

□

Lemma 3.

If a function given by (2) satisfies (3), (4) and (9), then for any number $\epsilon >0$, there exists ${\sigma }_0$ such that for $z=\sigma +i\tau$ with $\sigma \in ({\sigma }_0,0)$ and $\zeta =x+iy$ with $|x-\sigma |\le k^{-\eta }\left(\sigma \right)$,

$$R\left(\zeta \right)=\left|{\displaystyle \sum _{n=q+1}^{+\infty }}{a}_n{e}^{{\lambda }_n\zeta }\right|\le \mu (\sigma ,F)e^{-k^{\beta }\left(\sigma \right)},$$

where

$$q=q\left(\sigma \right)=\sum _{{\lambda }_n\le 3k\left(\sigma \right)}1,\beta \in (\alpha ,1).$$

Proof. 

Let

$$F^{*}\left(z\right)={\displaystyle \sum _{n=0}^{+\infty }}{a}_n^{*}{e}^{{\lambda }_{n}z}$$

be the Newton majorant of the function $F\left(z\right)={\sum }_{n=0}^{+\infty }{a}_n{e}^{{\lambda }_{n}z}$, i.e., (see [16]) $|a_n|\le a_n^{*}$; the sequence ${\displaystyle \frac{ln{a}_{n-1}^{*}-ln{a}_n^{*}}{{\lambda }_n-{\lambda }_{n-1}}}$ is non-decreasing, $\mu (\sigma ,F)=\mu (\sigma ,F^{*}),$$\nu (\sigma ,F)=\nu (\sigma ,F^{*})$.

Then, ${\varkappa }_n:={\displaystyle \frac{ln{a}_{n-1}^{*}-ln{a}_n^{*}}{{\lambda }_n-{\lambda }_{n-1}}}\nearrow 0$$(n\to +\infty )$:

$$R\left(\zeta \right)\le {\displaystyle \sum _{n=q+1}^{+\infty }}{a}_n^{*}{e}^{{\lambda }_{n}x},\mu (x,F)=\mu (x,F^{*})=a_s^{*}{e}^{{\lambda }_{s}x}(x\in [{\varkappa }_s,{\varkappa }_{s+1}]).$$

Let $s=\nu \left(x_1\right)$, $x_1=x+k^{-\eta }\left(x\right)$, $x=\sigma +k^{-\eta }\left(\sigma \right)$ with $\eta \in (\frac{1}{\gamma },1-\alpha )$; hence,

$${\lambda }_s=\mathsf{\Lambda }\left(x_1\right)\le k\left(x_1\right)\le (1+\epsilon )k\left(x\right)\le {(1+\epsilon )}^{2}k\left(\sigma \right)\le 2k\left(\sigma \right).$$

Therefore, for $x\in [{\varkappa }_s,{\varkappa }_{s+1}]$,

$$\begin{array}{c}R\left(\zeta \right)\le \mu \left(x\right){\displaystyle \sum _{n=q+1}^{+\infty }}{\displaystyle \frac{a_n^{*}}{a_s^{*}}}{e}^{x({\lambda }_n-{\lambda }_s)}=\mu \left(x\right){\displaystyle \sum _{n=q+1}^{+\infty }}{\displaystyle \frac{a_n^{*}}{a_{n-1}^{*}}}{\displaystyle \frac{a_{n-1}^{*}}{a_{n-2}^{*}}}·\dots ·{\displaystyle \frac{a_{s+1}^{*}}{a_s^{*}}}{e}^{x({\lambda }_n-{\lambda }_s)}\end{array}$$

From ${\varkappa }_m={\displaystyle \frac{ln{a}_{m-1}^{*}-ln{a}_m^{*}}{{\lambda }_m-{\lambda }_{m-1}}}$, we obtain ${\displaystyle \frac{a_m^{*}}{a_{m-1}^{*}}}\le exp(-{\varkappa }_m({\lambda }_m-{\lambda }_{m-1}))$ for any $m\in \mathbb{N}.$ Substituting this equality above instead of each fraction ${\displaystyle \frac{a_m^{*}}{a_{m-1}^{*}}}$ for $m\in \{s+1,\dots ,n\},$ we deduce

$$\begin{array}{c}R\left(\zeta \right)\le \mu \left(x\right){\displaystyle \sum _{n=q+1}^{+\infty }}{e}^{-{\varkappa }_n({\lambda }_n-{\lambda }_{n-1})}·\dots ·e^{-{\varkappa }_{s+1}({\lambda }_{s+1}-{\lambda }_s)}{e}^{x({\lambda }_n-{\lambda }_s)}.\end{array}$$

Since ${\varkappa }_n\nearrow 0$ as $n\to \infty$, one has ${\varkappa }_m\ge {\varkappa }_{s+1}$ for each $m\in \{s+1,\dots ,n\}$. Substituting it in above estimate, we obtain

$$\begin{array}{c}R\left(\zeta \right)\le \mu \left(x\right){\displaystyle \sum _{n=q+1}^{+\infty }}{e}^{(x-{\varkappa }_{s+1})({\lambda }_n-{\lambda }_s)}.\end{array}$$

Let us define $Q=e^{x-{\varkappa }_{s+1}}$; then, using condition (4), we get

$$\begin{array}{c}{\displaystyle \sum _{n=q+1}^{+\infty }}{Q}^{{\lambda }_n}=Q^{{\lambda }_{q+1}}+{\displaystyle \sum _{n=q+2}^{+\infty }}{\displaystyle \frac{1}{{\lambda }_n-{\lambda }_{n-1}}}{Q}^{{\lambda }_n}({\lambda }_n-{\lambda }_{n-1})\le \\ \le Q^{{\lambda }_{q+1}}+{\displaystyle \frac{1}{d}}{\displaystyle \sum _{n=q+2}^{+\infty }}{\int }_{{\lambda }_{n-1}}^{{\lambda }_n}{Q}^{t}dt\le Q^{{\lambda }_{q+1}}+{\displaystyle \frac{1}{d}}{\int }_{{\lambda }_{q+1}}^{+\infty }{Q}^{t}dt=\\ =Q^{{\lambda }_{q+1}}\left(1+{\displaystyle \frac{1}{d|lnQ|}}\right)\le Q^{{\lambda }_{q+1}}\left(1+{\displaystyle \frac{1}{d(1-Q)}}\right).\end{array}$$

Combining with Lemma 2, we have

$$\begin{array}{c}R\left(\zeta \right)\le \mu (\sigma +1/k\left(\sigma \right),F)Q^{-{\lambda }_s}{\displaystyle \sum _{n=q+1}^{+\infty }}{Q}^{{\lambda }_n}\le \\ \le e(1+\epsilon )\mu (\sigma ,F)Q^{{\lambda }_{q+1}-{\lambda }_s}\left(1+{\displaystyle \frac{1}{d(1-Q)}}\right).\end{array}$$

(15)

Since ${\lambda }_{q+1}-{\lambda }_s\ge 3k\left(\sigma \right)-2k\left(\sigma \right)\ge k\left(\sigma \right)$ and $s=\nu \left(x_1\right)$, $x_1=x+k^{-\eta }\left(x\right)\in [{\varkappa }_s,{\varkappa }_{s+1})$, then $x-{\varkappa }_{s+1}<-k^{-\eta }\left(x\right)\le -{(1+\epsilon )}^{-\eta }{k}^{-\eta }\left(\sigma \right)$. Therefore,

$$Q=e^{-({\varkappa }_{s+1}-x)}\le e^{-{(1+\epsilon )}^{-\eta }{k}^{-\eta }\left(\sigma \right)},$$

and

$$Q^{{\lambda }_{q+1}-{\lambda }_s}\le e^{-C{k}^{1-\eta }\left(\sigma \right)}$$

with $C={(1+\epsilon )}^{-\eta }.$ Let us now replace $\eta \in (\frac{1}{\gamma },1-\alpha )$ by ${\eta }_1\in (\frac{1}{\gamma },\eta )$. Then,

$$e(1+\epsilon )Q^{{\lambda }_{q+1}-{\lambda }_s}\left(1+{\displaystyle \frac{1}{d(1-Q)}}\right)\le e^{-C{k}^{1-{\eta }_1}\left(\sigma \right)}$$

(16)

Combining (15) and (16) and choosing $\beta =1-{\eta }_1$, we deduce

$$R\left(\zeta \right)\le \mu (\sigma ,F)e^{-C{k}^{\beta }\left(\sigma \right)},$$

where $\beta \in (\alpha ,1-\frac{1}{\gamma })\subset (\alpha ,1)$, $\gamma >\frac{1}{1-\alpha }$ is defined in (9), and $\alpha \in (0,1)$ is defined in (4). Since $\epsilon >0$ is chosen arbitrarily, we can move on to the limit as $\epsilon \to 0$ and obtain $C={(1+\epsilon )}^{-\eta }\to 1$. Then,

$$R\left(\zeta \right)\le \mu (\sigma ,F)e^{-k^{\beta }\left(\sigma \right)}.$$

This completes the proof of Lemma 3. □

Note that from condition (4), it follows that

$$\underset{\sigma \to -0}{\overline{lim}}{\displaystyle \frac{lnq\left(\sigma \right)}{ln\left(3k\right(\sigma \left)\right)}}=\alpha <1.$$

Hence, for any $\epsilon >0$ and for all $\sigma \in ({\sigma }_0,0)$, one has

$$q\left(\sigma \right)<k^{\alpha +\epsilon }\left(\sigma \right).$$

(17)

Let $P\left(z\right)={\sum }_{n=0}^q{a}_n{e}^{{\lambda }_{n}z}$. We define an operator

$$D_{I}P\left(z\right)=(\prod _{{\lambda }_j\in I}{d}_{{\lambda }_j})P\left(z\right),$$

where $I=\{{\lambda }_0,\dots ,{\lambda }_q\}\setminus {\lambda }_{\nu \left(\sigma \right)}$ and

$$d_{{\lambda }_j}P\left(z\right)=e^{{\lambda }_{j}z}{\displaystyle \frac{d}{dz}}\left(e^{-{\lambda }_{j}z}P\left(z\right)\right).$$

The order of composition is less important, because $d_{{\lambda }_j}$ commutes. This is proven below by direct calculations of $d_{{\lambda }_k}\left(d_{{\lambda }_j}P\left(z\right)\right).$ It easy to see that

$$\begin{array}{c}{d}_{{\lambda }_j}P\left(z\right)=P\prime \left(z\right)-{\lambda }_{j}P\left(z\right),d_{{\lambda }_k}\left(d_{{\lambda }_j}P\left(z\right)\right)=\\ =d_{{\lambda }_k}\left(P\prime \left(z\right)-{\lambda }_{j}P\left(z\right)\right)=P″\left(z\right)-({\lambda }_k+{\lambda }_j)P\prime \left(z\right)+{\lambda }_k{\lambda }_{j}P\left(z\right),\\ d_{{\lambda }_s}\left(d_{{\lambda }_k}\left(d_{{\lambda }_j}P\left(z\right)\right)\right)=\\ =P‴\left(z\right)-({\lambda }_s+{\lambda }_k+{\lambda }_j)P″\left(z\right)+({\lambda }_k{\lambda }_j+{\lambda }_k{\lambda }_s+{\lambda }_s{\lambda }_j)P\prime \left(z\right)-{\lambda }_s{\lambda }_k{\lambda }_{j}P\left(z\right).\end{array}$$

Using the method of mathematical induction, it is easy to prove that

$$D_{I}P\left(z\right)={\displaystyle \sum _0^q}{(-1)}^{j+1}{C}_{q,j}{P}^{\left(j\right)}\left(z\right),$$

where ${\sum }_{j=0}^q{C}_{q,j}{t}^j={\prod }_{{\lambda }_j\in I}({\lambda }_j+t)$. It easily yields the following inequality, which we will formulate below in the form of a lemma.

Lemma 4.

The inequality

$$|D_{I}P\left(z\right)|\le {\displaystyle \sum _{j=0}^q}{C}_{q,j}\left|P^{\left(j\right)}\left(z\right)\right|$$

holds, where the coefficients $C_{q,j}$ are non-negative and they are determined from the following equality:

$${\displaystyle \sum _{j=0}^q}{C}_{q,j}{t}^j=\prod _{{\lambda }_j\in I}({\lambda }_j+t).$$

Lemma 5.

There exists ${\sigma }_0$ such that for each $\sigma \in ({\sigma }_0,0)$, there exists ${\xi }_1$ with $|{\xi }_1-z|={\displaystyle \frac{1}{k^2\left(\sigma \right)}}$ for which this lower estimate holds:

$$|F\left({\xi }_1\right)|\ge e^{-k^{\alpha +\epsilon }\left(\sigma \right)}\mu (\sigma ,F).$$

Proof. 

We put $m,0\le m\le q$, such that

$$|P^{\left(m\right)}\left(z\right)|=max\{|P^{\left(j\right)}\left(z\right)|:\le j\le q\}.$$

Denote

$$|P\left({\xi }_1\right)|=max\{|P\left(\xi \right)|:|\xi -z|={\displaystyle \frac{1}{k^2\left(\sigma \right)}}\}.$$

From Lemma 4, one has

$$|D_{I}P\left(z\right)|\le |P^{\left(m\right)}\left(z\right)|\prod _{{\lambda }_j\in I}({\lambda }_j+1).$$

The Cauchy integral formula gives

$$\left|P^{\left(m\right)}\left(z\right)\right|={\displaystyle \frac{m!}{2\pi }}\left|{\int }_{|z-\xi |=k^{-2}\left(\sigma \right)}{\displaystyle \frac{P\left(\xi \right)}{{(\xi -z)}^{m+1}}}d\xi \right|\le q!k^{2q}\left(\sigma \right)\left|P\left({\xi }_1\right)\right|.$$

Combining the two last inequalities, we have

$$|P\left({\xi }_1\right)|\ge {\displaystyle \frac{|D_{I}P\left(z\right)|}{q!{\prod }_{{\lambda }_j\in I}({\lambda }_j+1)k^{2q}\left(\sigma \right)}}.$$

(18)

From the definition of the operator D and condition (3), we have

$$\left|D_{I}P\left(z\right)\right|=|{\displaystyle \sum _{m=0}^q}\prod _{{\lambda }_j\in I}({\lambda }_m-{\lambda }_j)a_m{e}^{z{\lambda }_m}|=|\prod _{{\lambda }_j\in I}({\lambda }_{\nu }-{\lambda }_j)|\mu (\sigma ,F)\ge d^q{([\frac{q}{2}]!)}^2\mu (\sigma ,F).$$

Moreover, ${\lambda }_{q\left(\sigma \right)}\le 3k\left(\sigma \right)$, and thus ${\displaystyle \prod _{{\lambda }_j\in I}}({\lambda }_j+1)\le {\left(3k\left(\sigma \right)\right)}^q·C$; here, $C={\prod }_{j=1}^{+\infty }(1+1/{\lambda }_j)<+\infty$ by condition (4). Remark that ${\left(\left[\frac{q}{2}\right]!\right)}^2\ge 2^{-q}q!$. Therefore,

$$\begin{array}{c}{\displaystyle \frac{|D_{I}P\left(z\right)|}{q!{\prod }_{{\lambda }_j\in I}({\lambda }_j+1)k^{2q}\left(\sigma \right)}}\ge {\displaystyle \frac{d^q{\left(\left[\frac{q}{2}\right]!\right)}^2}{q!{\prod }_{{\lambda }_j\in I}({\lambda }_j+1)k^{2q}\left(\sigma \right)}}\mu (\sigma ,F)\ge {\displaystyle \frac{d^q\mu (\sigma ,F)}{C{\left(6k\left(\sigma \right)\right)}^q{k}^{2q}\left(\sigma \right)}}.\end{array}$$

Hence, using inequalities (17) and (18), by definition of the function $q,$ we obtain

$$|P\left({\xi }_1\right)|\ge {\displaystyle \frac{d^q\mu (\sigma ,F)}{{\left(3ek\left(\sigma \right)\right)}^q{k}^{2q}\left(\sigma \right)}}\ge e^{-k^{\alpha +\frac{\epsilon }{2}}\left(\sigma \right)}\mu (\sigma ,F)$$

for any $\epsilon >0$. Then,

$$|F\left({\xi }_1\right)|\ge |P\left({\xi }_1\right)|-|R\left({\xi }_1\right)|\ge (exp\{-k^{\alpha +\frac{\epsilon }{2}}\left(\sigma \right)\}-exp\{-k^{\beta }\left(\sigma \right)\})\mu (\sigma ,F),$$

where $\beta$ is a constant from Lemma 3. If we now choose $\beta \in (\alpha +\frac{\epsilon }{2},1)$, then we get

$$|F\left({\xi }_1\right)|\ge \mu (\sigma ,F)(1-\epsilon )exp\{-k^{\alpha +\frac{\epsilon }{2}}\left(\sigma \right)\}\ge \mu (\sigma ,F)exp\{-k^{\alpha +\epsilon }\left(\sigma \right)\}$$

for any $\epsilon >0$ and $\sigma \in ({\sigma }_0,0).$ The proof of Lemma 5 is complete. □

Lemma 6.

There exists ${\sigma }_0\in (-\infty ;0)$ such that for any $z=\sigma +i\tau$ with $\sigma \in ({\sigma }_0,0)$ and for all $\xi =x+iy:|\xi -z|<{\displaystyle \frac{1}{k\left(\sigma \right)}}$, the estimate

$$M(x,F)<k^{\alpha +\epsilon }\mu (\sigma ,F)$$

is true for any $\epsilon >0$.

Proof. 

By (17) and Lemmas 1–3, one has

$$\begin{array}{c}F\left(\xi \right)\le \left|P\left(\xi \right)\right|+\left|R\left(\xi \right)\right|\le q\mu \left(x\right)+\mu (\sigma ,F)exp\{-k^{\beta }\left(\sigma \right)\}\le \\ \le \left(k^{\alpha +\frac{\epsilon }{2}}\left(x\right)+exp\{-k^{\beta }\left(\sigma \right)\}\right)\mu (\sigma ,F)\le k^{\alpha +\epsilon }\left(\sigma \right)\mu (\sigma ,F).\end{array}$$

□

Lemma 7.

The number of zeros of function F in the disk

$$\overline{K}(z,{\displaystyle \frac{1}{2k\left(\sigma \right)}})=\{\xi :|\xi -z|\le {\displaystyle \frac{1}{2k\left(\sigma \right)}}\}$$

does not exceed $k^{\alpha +\epsilon }\left(\sigma \right)$.

Proof. 

Let $n({\xi }_1,t)$ be the number of zeros of the function F at the disk $K({\xi }_1,t)$, where ${\xi }_1$ is a number from Lemma 5. Let us define

$$\begin{array}{c}{\rho }_1={\displaystyle \frac{1}{2k\left(\sigma \right)}},{\rho }_2={\displaystyle \frac{1}{2k\left(\sigma \right)}}+{\displaystyle \frac{1}{k^2\left(\sigma \right)}},{\rho }_3={\displaystyle \frac{1}{k\left(\sigma \right)}}.\end{array}$$

Then,

$$\overline{K}(z,{\rho }_1)\subset \overline{K}({\xi }_1,{\rho }_2)\subset \overline{K}({\xi }_1,{\rho }_3).$$

It is obvious that

$$n({\xi }_1,{\rho }_2)\le {\displaystyle \frac{1}{ln\frac{{\rho }_3}{{\rho }_2}}}{\int }_{{\rho }_2}^{{\rho }_3}{\displaystyle \frac{n({\xi }_1,t)}{t}}dt.$$

By the Jensen formula,

$${\int }_0^{{\rho }_3}{\displaystyle \frac{n({\xi }_1,t)}{t}}dt={\displaystyle \frac{1}{2\pi }}{\int }_o^{2\pi }ln|F({\xi }_1+{\rho }_3{e}^{i\theta })|d\theta -ln|F\left({\xi }_1\right)|.$$

Then, by Lemmas 5 and 6,

$$n({\xi }_1,{\rho }_2)\le {\displaystyle \frac{(1+\epsilon )}{ln2}}{k}^{\alpha +\epsilon }\left(\sigma \right)\le k^{\alpha +\epsilon }\left(\sigma \right)$$

and

$$n(z,{\rho }_1)\le n({\xi }_1,{\rho }_2)\le k^{\alpha +\epsilon }\left(\sigma \right).$$

□

Let us consider the Blaschke product for the set $A=\{a_1,a_2,\dots ,a_n\}$ consisting of all zeros of the analytic function $F\left(z\right)$ within a disk with the center z and the radius $\rho ={\displaystyle \frac{1}{2k\left(\sigma \right)}}$:

$$B(\xi ,A)=B(\xi ,\rho ,z,A)={\displaystyle \prod _{i=1}^n}{\displaystyle \frac{\rho (\xi -a_i)}{{\rho }^2-\overline{(a_i-z)}(\xi -z)}}.$$

It is well-known that

$$\left|B\right(\xi ,A\left)\right|=1\left(\mathrm{for}\right|\xi -z|=\rho ),\left|B\right(\xi ,A\left)\right|<1\left(\mathrm{for}\right|\xi -z|<\rho ).$$

Lemma 8.

If Dirichlet series (2) is absolutely convergent in the half-plane $Rez<0$ and it satisfies (3), (4) and (9) and if $a_1,a_2,\dots ,a_n$ are the zeros of $F\left(z\right)$ in the disk $\overline{K}(z,\rho )=\{\xi :|\xi -z|\le \rho \}$, then

$$\left|F\left(z\right)\right|\ge e^{-k^{\alpha +\epsilon }\left(\sigma \right)}\mu (\sigma ,F)k^{-n}\left(\sigma \right)\left|{\displaystyle \prod _{i=1}^n}(\xi -a_i)\right|$$

for all $\xi \in \overline{K}(z,\frac{\rho }{2})=\{\xi :|\xi -z|\le \frac{\rho }{2}\}$, where $\rho ={\displaystyle \frac{1}{2k\left(\sigma \right)}}$ and $n\le k^{\alpha +\epsilon }\left(\sigma \right)$, $z=\sigma +i\tau$.

Proof. 

We write

$$F\left(z\right)=\mathsf{\Phi }\left(z\right)B(\xi ,A).$$

Then, on the circle $\{\xi :|\xi -z|=\rho \}$, one has

$$\left|\mathsf{\Phi }\right(\xi \left)\right|=\left|F\right(\xi \left)\right|.$$

So, on the disk $\{\xi :|\xi -z|\le \rho \}$, by the maximum modulus principle and Lemma 6, we get

$$\left|\mathsf{\Phi }\left(\xi \right)\right|=\left|F\left(\xi \right)\right|\le k^{\alpha +\epsilon }\left(\sigma \right)\mu (\sigma ,F).$$

(19)

By Lemma 5, we can choose ${\xi }_1$ such that

$$|\xi -{\xi }_1|=k^{-2}\left(\sigma \right)\le \frac{\rho }{2}$$

and

$$|\mathsf{\Phi }\left({\xi }_1\right)|\ge e^{-k^{\alpha +\frac{\epsilon }{2}}\left(\sigma \right)}\mu (\sigma ,F).$$

(20)

Consider the functions $F\left(z\right)$ and $\mathsf{\Phi }\left(z\right)$ at the disks

$$\overline{K}(z,\frac{\rho }{2})\subset \overline{K}({\xi }_1,\frac{5\rho }{8})\subset \overline{K}({\xi }_1,\frac{3\rho }{4})\subset \overline{K}(z,\rho ).$$

Since ${\displaystyle \frac{\mathsf{\Phi }\left(\xi \right)}{\mathsf{\Phi }\left({\xi }_1\right)}}$ has no zeros at the disk $|\xi -{\xi }_1|\le \frac{3\rho }{4}$, the function

$$G\left(\xi \right)=ln{\displaystyle \frac{\mathsf{\Phi }\left(\xi \right)}{\mathsf{\Phi }\left({\xi }_1\right)}}$$

has an analytic branch at this disk.

We need the following Borel–Carathéodory theorem (see [17]): If a function g is analytic on a closed disc $K(0,R)$ and $r<R$, then

$$\underset{\left|z\right|\le r}{max}\left|g\left(z\right)\right|\le {\displaystyle \frac{2r}{R-r}}\underset{\left|z\right|\le R}{max}\mathrm{Re}g\left(z\right)+{\displaystyle \frac{R+r}{R-r}}\left|g\left(0\right)\right|.$$

Denote

$$M=\underset{|\xi -{\xi }_1|\le \frac{5\rho }{8}}{max}\left|G\left(\xi \right)\right|.$$

By the Borel–Carathéodory inequality applied to the disks $\overline{K}({\xi }_1,\frac{5\rho }{8})$ and $\overline{K}({\xi }_1,\frac{3\rho }{4})$, using inequalities (19) and (20), we obtain

$$M\le 11\underset{|\xi -{\xi }_1|\le \frac{3\rho }{4}}{max}\left|ReG\left(\xi \right)\right|=11\underset{|\xi -{\xi }_1|\le \frac{3\rho }{4}}{max}\left|ln\left|{\displaystyle \frac{\mathsf{\Phi }\left(\xi \right)}{\mathsf{\Phi }\left({\xi }_1\right)}}\right|\right|\le C{k}^{\alpha +\frac{\epsilon }{2}}\left(\sigma \right).$$

Hence, for $|\xi -{\xi }_1|\le \frac{5\rho }{8}$, one has

$$ln\left|{\displaystyle \frac{\mathsf{\Phi }\left(\xi \right)}{\mathsf{\Phi }\left({\xi }_1\right)}}\right|\ge -M\ge -C{k}^{\alpha +\epsilon }\left(\sigma \right)$$

and so,

$$ln\left|\mathsf{\Phi }\left(\xi \right)\right|\ge ln\mu (\sigma ,F)-C{k}^{\alpha +\frac{\epsilon }{2}}\left(\sigma \right).$$

Therefore,

$$\left|F\left(\xi \right)\right|\ge e^{-C{k}^{\alpha +\frac{\epsilon }{2}}\left(\sigma \right)}\mu (\sigma ,F)\left|B(\xi ,A)\right|.$$

Finally, it is easy to see that

$$\left|B(\xi ,A)\right|\ge {\left(2\rho \right)}^n\left|{\displaystyle \prod _{i=1}^n}(\xi -a_i)\right|$$

and, therefore, the statement of the lemma is proved. □

Proof of Theorem 3. 

Clearly, the points $\xi$ such that

$${\displaystyle \prod _{i=1}^n}|\xi -a_i|\le h^n$$

can be covered at most n disks of radius h. If we now choose

$$h={\displaystyle \frac{1}{2n}}exp\{-k^{\delta }\left(\sigma \right)\},$$

then the sum of diameters of those disks is $exp\{-k^{\delta }\left(\sigma \right)\}$. Therefore, the disk centered at z and of radius $2exp\{-k^{\delta }\left(\sigma \right)\}$ must contain a circle of radius of at least $exp\{-k^{\delta }\left(\sigma \right)\}$.

On this circle, we get

$$|\xi -a_i|={\displaystyle \frac{1}{2n}}exp\{-k^{\delta }\left(\sigma \right)\}\ge {\displaystyle \frac{exp\{-k^{\delta }\left(\sigma \right)\}}{2k^{\alpha +\epsilon }\left(\sigma \right)}}\ge exp\{-k^{\delta }\left(\sigma \right)-2(\alpha +\epsilon )lnk\left(\sigma \right)\}.$$

Hence,

$$k^{-n}\left(\sigma \right){\displaystyle \prod _{i=1}^n}|\xi -a_i|\ge exp\left\{(-k^{\delta }\left(\sigma \right)-Clnk\left(\sigma \right))k^{\alpha +\frac{\epsilon }{4}}\left(\sigma \right)\right\}.$$

Finally, if we choose $\delta <\frac{\epsilon }{4}$, then

$$\left|F\left(z\right)\right|\ge \mu (\sigma ,F)exp\{-k^{\alpha +\frac{\epsilon }{2}}\left(\sigma \right)\}exp\{-k^{\alpha +\frac{\epsilon }{2}}\left(\sigma \right)\}.$$

Theorem 3 is proved. □

4. Discussion

Given the obtained results, we can pose a natural conjecture.

Conjecture 1.

In Theorem 3, condition (3) must be replaced by the condition

$$\prod _{\genfrac{}{}{0pt}{}{{\lambda }_j\le k\left(\sigma \right)}{{\lambda }_j\ne \mathsf{\Lambda }\left(\sigma \right)}}|\mathsf{\Lambda }\left(\sigma \right)-{\lambda }_j|\ge {\left(\left[q/2\right]!\right)}^2;$$

(21)

here, $q=q\left(\sigma \right)={\sum }_{{\lambda }_j\le 3k\left(\sigma \right)}1$ and $\mathsf{\Lambda }\left(\sigma \right)={\lambda }_{\nu (\sigma ,F)}$. Clearly, condition (21) is weaker than condition (3). The condition was used in the proof of Lemma 5 in [14] (see also proof of Lemma 4 [15]) and it was justified that (3) implies (21). However, it is unclear how to prove the other auxiliary lemmas and final theorems using (21).

Question 1.

Can condition (3) be rewritten as a condition on the condensation index $\delta :=\underset{n\to +\infty }{\overline{lim}}{\displaystyle \frac{-ln|L^{\prime }\left({\lambda }_n\right)|}{{\lambda }_n}}$, where $L\left(\lambda \right)={\prod }_{n=1}^{+\infty }(1-{\lambda }^2/{\lambda }_n^2)$ is an entire function of zero order? It is well-known that the conditions $n=o\left({\lambda }_n\right)(n\to +\infty )$ and ${\lambda }_{n+1}-{\lambda }_n\ge d>0$ $(n\ge 1)$ imply $\delta =0$ (for example, see [18,19]). In particular, in paper [19], by the condition $\delta =0$, an investigation is carried out on the equality on a ray and in a half-plane of generalized orders of growth of analytic functions represented by Dirichlet series which are absolutely convergent in the half-plane. In some sense, this problem is close to the problem investigated in the present article.

It is well known that there are many papers in which the results established for the Dirichlet series are generalized to functions defined by Laplace–Stieltjes integrals [20,21,22,23]. Such interest is generated by the fact that Dirichlet series can be written as the Laplace–Stieltjes integral. The last one has a more general form than the Dirichlet series. In these articles, one can find results concerning domains of the convergence of the Laplace–Stieltjes integral. Additionally, a connection was investigated between the growth of the logarithms of the Laplace–Stiltjes integral and the maximum of its integrand, as well as the logarithmic growth of entire functions represented by Laplace–Stieltjes transforms of zero order. The most significant contribution in theory of the Laplace–Stieltjes integral was made by M. Sheremeta and his co-authors over the last decade [24,25]. They investigated the following topics concerning the Laplace–Stieltjes integral: the Banach spaces, the growth of series in systems of functions, its application to the lacunary power series, and lower and upper estimates for the integral. New estimates were obtained for the abscissa of the convergence of the Laplace–Stiltjes integral and, therefore, for the Laplace integrals and Dirichlet series. It is therefore quite natural to ask whether there is an analog of our results on Dirichlet series pits for Laplace–Stieltjes integrals [26,27].