Entire Dirichlet series with monotonous coefficients and logarithmic h-measure

Theorem 1.1. Let (µn) be a sequence such that condition (4) holds, h ∈ L+, Φ ∈ L and F ∈ Da(Φ). If $$(\forall b > 0): \qquad \sum_{n=n_0}^{+\infty} h\left(\varphi(\lambda_n) \cdot \left(1 + \frac{b}{\mu_{n+1} - \mu_n}\right)\right) \frac{1}{\mu_{n+1} - \mu_n} < +\infty, \qquad (8)$$ then the relation (3) holds as x → +∞ outside some set E of finite logarithmic h-measure uniformly in y ∈ R. The method of proof of Theorem 1.1 differs from the method of proofs corresponding statements in [1, 3] and is close to the methods of proofs from papers [4, 5, 6].

Lemma 2.1. For all $n \ge 0$ and $k \ge 1$ inequality $$\frac{\alpha_n}{\alpha_k} e^{\tau_k(\mu_n - \mu_k)} \le e^{-\delta|n-k|} \tag{11}$$ holds, where $$\tau_k := t_k + \frac{\delta}{\mu_k - \mu_{k-1}}, \qquad t_k := \frac{\Delta_{k-1} - \Delta_k}{\mu_k - \mu_{k-1}}.$$ Proof of Lemma 2.1. We remark that $$t_k = -\delta \cdot \sum_{m=k}^{+\infty} \left( \frac{1}{\mu_m - \mu_{m-1}} + \frac{1}{\mu_{m+1} - \mu_m} \right), \tag{12}$$ $$\tau_k = -2\delta \cdot \sum_{m=k+1}^{+\infty} \frac{1}{\mu_m - \mu_{m-1}},\tag{13}$$ $$\tau_{k+1} - \tau_k = \frac{2\delta}{\mu_{k+1} - \mu_k} \quad (k \ge 1). \tag{14}$$ Since $$\ln \alpha_n - \ln \alpha_{n-1} = \Delta_n - \Delta_{n-1} = -t_n(\mu_n - \mu_{n-1}),$$ for n ≥ k + 1 we have $$\ln \frac{\alpha_n}{\alpha_k} + \tau_k(\mu_n - \mu_k) = -\sum_{j=k+1}^n t_j(\mu_j - \mu_{j-1}) + \tau_k \sum_{j=k+1}^n (\mu_j - \mu_{j-1}) =$$ $$= -\sum_{j=k+1}^n (t_j - \tau_k) (\mu_j - \mu_{j-1}) \le -\sum_{j=k+1}^n (t_j - \tau_{j-1}) (\mu_j - \mu_{j-1}) =$$ $$= -\sum_{j=k+1}^n \delta = -(n-k) \cdot \delta.$$ Similarly, for n ≤ k − 1 we obtain $$\ln \frac{\alpha_n}{\alpha_k} + \tau_k(\mu_n - \mu_k) = -\ln \frac{\alpha_k}{\alpha_n} - \tau_k(\mu_k - \mu_n) =$$ $$= \sum_{j=n+1}^k t_j(\mu_j - \mu_{j-1}) - \tau_k \sum_{j=n+1}^k (\mu_j - \mu_{j-1}) = -\sum_{j=n+1}^k (\tau_k - t_j) (\mu_j - \mu_{j-1}) \le$$ $$\le -\sum_{j=n+1}^k (\tau_j - t_j) (\mu_j - \mu_{j-1}) = -\sum_{j=n+1}^k \delta = -(k-n) \cdot \delta$$ and Lemma 1 is proved. We remark that from definitions of τ^k and t^k (see (13), (12)) and the condition + P∞ k=0 1/(µk+1 − µk) < +∞ it follows that there exists k0(δ) such that $$\tau_k \ge -1 \quad (k \ge k_0(\delta)), \qquad \tau_k < 0 \quad (k \ge 1).$$ Let J be a set of the values of the central index ν(σ, f ^∗ ), i.e. $$J = \{k \in \mathbb{N} \colon (\exists \ \sigma < 0)[\nu(\sigma, f^*) = k]\}.$$ Denote by $(R_k)$ a sequence of the points of the springs of $\nu(\sigma, f^)$ , enumerate such that $\nu(\sigma, f^) = k$ for $\sigma \in [R_k, R_{k+1})$ in the case $R_k < R_{k+1}$ . Then for $\sigma \in [R_k, R_{k+1})$ , $k \in J$ from Lemma 2.1 we have $$b_n e^{\sigma \mu_n} \le b_k e^{\sigma \mu_k} \iff \frac{b_n e^{\sigma \mu_n}}{b_k e^{\sigma \mu_k}} \le \frac{\alpha_n^{|\sigma|}}{\alpha_k^{|\sigma|}} \le e^{-|\sigma|\tau_k(\mu_n - \mu_k)} e^{-|\sigma||n - k| \cdot \delta}$$ for all $n \geq 0$ . Hence, $$\frac{b_n e^{(\sigma + |\sigma|\tau_k)\mu_n}}{b_k e^{(\sigma + |\sigma|\tau_k)\mu_k}} \le e^{-|\sigma||n - k| \cdot \delta}$$ i.e., for all $k \in J$ and for every $\sigma^* \in [R_k(1+|\tau_k|), R_{k+1}(1+|\tau_k|))$ $$\frac{b_n e^{\sigma^ \mu_n}}{b_k e^{\sigma^ \mu_k}} \le \exp\left\{-\frac{|\sigma^*||n-k| \cdot \delta}{1+|\tau_k|}\right\}.$$ Thus, as $x = \frac{1}{|\sigma^*|} \in \left[-\frac{1}{R_k(1+|\tau_k|)}, -\frac{1}{R_{k+1}(1+|\tau_k|)}\right)$ we get $$\frac{|a_n|e^{x\lambda_n}}{|a_k|e^{x\lambda_k}} \le \exp\left\{-\frac{|n-k| \cdot \delta}{1+|\tau_k|}\right\}$$ for all $k \in J$ , $n \ge 0$ . Therefore, $$\nu(x,F) = k, \ \mu(x,F) = |a_k|e^{x\lambda_k}, \ x \in \left[ -\frac{1}{R_k(1+|\tau_k|)}, -\frac{1}{R_{k+1}(1+|\tau_k|)} \right]. \tag{15}$$ Denote $$E^(\delta) := \bigcup_{k=k_0(\delta)}^{+\infty} \left[ -\frac{1}{R_k(1+|\tau_k|)}, -\frac{1}{R_{k+1}(1+|\tau_k|)} \right), \ E(\delta) := [0, +\infty) \setminus E^(\delta).$$ Then for every x > 0, $x \notin E(\delta)$ $$|F(x+iy) - a_{\nu(x,F)}e^{(x+iy)\lambda_{\nu(x,F)}}| \le$$ $$\le \mu(x,F) \cdot \sum_{n \ne \nu(x,F)} \exp\left\{-\frac{\delta \cdot |n - \nu(x,F)|}{1 + |\tau_{\nu(x,F)}|}\right\} \le \frac{2e^{-\delta/2}}{1 - e^{-\delta/2}} \cdot \mu(x,F)$$ (16) because $1 + |\tau_{\nu(x,F)}| < 2$ $(x \in E^*(\delta))$ . It remains to prove that the logarithmic h-measure of a set $E(\delta)$ is finite. Using $$E(\delta) \subset [0, x_0) \bigcup \Big( \bigcup_{k=k_0(\delta)+1}^{+\infty} \Big[ -\frac{1}{R_k(1+|\tau_{k-1}|)}, -\frac{1}{R_k(1+|\tau_k|)} \Big) \Big),$$ $$x_0 = -\frac{1}{R_{k_0(\delta)}(1+|\tau_{k_0(\delta)-1}|)},$$ we obtain h-log-meas $$(E \cap [x_0, +\infty)) = \sum_{k=k_0(\delta)+1}^{+\infty} \int h(x)d\ln x \le \sum_{k=k_0(\delta)+1}^{+\infty} h\left(-\frac{1}{R_k(1+|\tau_{k-1}|)}\right) \ln \frac{1+|\tau_{k-1}|}{1+|\tau_k|} = \sum_{k=k_0(\delta)+1}^{+\infty} h\left(-\frac{1}{R_k(1+|\tau_k|)}\right) \ln \left(1+\frac{|\tau_{k-1}|-|\tau_k|}{1+|\tau_k|}\right) \le \le \sum_{k=k_0(\delta)}^{+\infty} h\left(-\frac{1}{R_k(1+|\tau_{k-1}|)}\right) (|\tau_k|-|\tau_{k+1}|),$$ Hence, using equality (14) we have h-log-meas $$(E \cap [x_0, +\infty)) \le 2\delta \cdot \sum_{k=k_0(\delta)}^{+\infty} h\left(-\frac{1}{R_{k+1}(1+|\tau_{k+1}|)}\right) \frac{1}{\mu_{k+1} - \mu_k}.$$ (17) The condition $F \in \mathcal{D}_a(\Phi)$ implies $$x\Phi(x) \le \ln \mu(x, F) = -\mu_{\nu(x-0,F)} + x\lambda_{\nu(x-0,F)} \le x\lambda_{\nu(x-0,F)} \quad (x \ge x_1 > 0),$$ therefore $$x \le \varphi(\lambda_{\nu(x-0,F)}) \quad (x \ge x_1 > 0). \tag{18}$$ Denote $\theta_k := -\frac{1}{R_{k+1}(1+|\tau_k|)}$ . By (15) we have $\nu(\theta_k - 0) = k$ , thus from (14) and (18) it follows $$-\frac{1}{R_{k+1}(1+|\tau_{k+1}|)} = \theta_k \cdot \frac{1+|\tau_k|}{1+|\tau_{k+1}|} = \theta_k \cdot \left(1+\frac{|\tau_k|-|\tau_{k+1}|}{1+|\tau_{k+1}|}\right) \le$$ $$\le \theta_k \cdot \left(1+\frac{2\delta}{\mu_{k+1}-\mu_k}\right) \le \varphi(\lambda_k) \cdot \left(1+\frac{2\delta}{\mu_{k+1}-\mu_k}\right)$$ (19) for all $k \geq k_1(\delta)$ . Using inequality (19) to inequality (17), we get $$h-\log-\max(E(\delta)\cap[x_0,+\infty)) \leq$$ $$\leq 2\delta \cdot \sum_{k=k_0(\delta)}^{k_2(\delta)-1} h\left(-\frac{1}{R_{k+1}(1+|\tau_{k+1}|)}\right) \frac{1}{\mu_{k+1}-\mu_k} +$$ $$+2\delta \cdot \sum_{k=k_2(\delta)}^{+\infty} h\left(\varphi(\lambda_k)\cdot\left(1+\frac{2\delta}{\mu_{k+1}-\mu_k}\right)\right) \frac{1}{\mu_{k+1}-\mu_k} := K(\delta) < +\infty, \quad (20)$$ where $k_2(\delta) = \max\{k_0(\delta), k_1(\delta)\}$ . Relation (20) implies that $$(\forall \delta > 0)$$ : h-log-meas $(E(\delta) \cap [x, +\infty)) = o(1) \quad (x \to +\infty).$ We put now $\delta_n = n, \varepsilon_n = 2^{-n}$ $(n \ge 1)$ . Then for every $n \ge 1$ there exists $x_n \ge x_0$ such that h-log-meas $$(E(\delta_n) \cap [x_n, +\infty)) \le \varepsilon_n$$ . Without loss of generality we may assume that $x_n < x_{n+1}$ $(n \ge 1)$ . Denote $E = \bigcup_{n=1}^{+\infty} \left( E(\delta_n) \cap [x_n; x_{n+1}) \right)$ . Define a function $\gamma \colon [x_1, +\infty) \to [0, +\infty)$ by equality $\gamma(x) = \frac{2e^{-\delta_n/2}}{1 - e^{-\delta_n/2}}$ for $x \in [x_n, x_{n+1})$ . Then from inequality (16) it follows $$|F(x+iy) - a_{\nu(x,F)}e^{(x+iy)\lambda_{\nu(x,F)}}| \le \gamma(x) \cdot \mu(x,F)$$ (21) for all $x \in [x_1, +\infty) \setminus E$ uniformly in $y \in \mathbb{R}$ . But $\gamma(x) \to 0 \ (x \to +\infty)$ and $$\text{h-log-meas}(E \cap [x_1, +\infty)) \leq \sum_{n=1}^{+\infty} \text{h-log-meas}(E(\delta_n) \cap [x_n, x_{n+1})) \leq \sum_{n=1}^{+\infty} \varepsilon_n = 1.$$ Thus, h-log-meas $(E) < +\infty$ .

Theorem 3.1. Let $(\mu_n)$ be a sequence such that condition (4) holds, $h \in \mathcal{L}_+, \Phi \in \mathcal{L}_+$ and $F \in \mathcal{D}_a^*(\Phi)$ . If $$(\forall b > 0): \qquad \sum_{n=n_0}^{+\infty} \frac{h\left(b\varphi(b\lambda_n)\right)}{\mu_{n+1} - \mu_n} < +\infty, \tag{22}$$ then relation (3) holds as $x \to +\infty$ outside some set E of finite logarithmic h-measure uniformly in $y \in \mathbb{R}$ . Remark 3.2. In the case $\Phi(x) = e^x/x$ we obtain that $\mathcal{D}_a^*(\Phi)$ is the class Dirichlet series of nonzero lower R-order $$\underline{\lim_{x \to +\infty}} \frac{\ln \ln M(x, F)}{x} := \rho_R[F] \in (0, +\infty]$$ and condition (22) from condition $$(\forall b > 0): \qquad \sum_{n=n_0}^{+\infty} \frac{h(b \ln \lambda_n)}{\mu_{n+1} - \mu_n} < +\infty$$ (23) follows. Example 3.2. Well known, if $\ln n = o(\lambda_n \ln \lambda_n)$ and $\ln \lambda_{n+1} \sim \ln \lambda_n$ $(n \to +\infty)$ then for the function $F \in D$ with coefficients $$|a_n| = \exp\left\{-\frac{1}{\rho}\lambda_n \ln \lambda_n\right\}$$ we have $\rho_R[F] = \rho$ . Thus condition (23) follows from the condition $$(\forall b > 0): \qquad \sum_{n=n_0}^{+\infty} \frac{h(b \ln \lambda_n)}{\lambda_{n+1} \ln \lambda_{n+1} - \lambda_n \ln \lambda_n} < +\infty.$$ Question 3.1. Is the description of exceptional sets in our Theorems 1.1 and 3.1 the best possible? Question 3.2. Are conditions (8) and (22) in our Theorems 1.1 and 3.1 necessary?