Zeros of holomorphic functions in the unit disk and $ρ$-trigonometrically convex functions

Proposition 1. Let $h \in \rho$ -trc⁺ be a $2\pi$ -periodic $\rho$ -trigonometrically convex positive function, and let $g : \mathbb{R}^+ \to \mathbb{R}^+$ be a convex function with g(0) = 0. We set <span id="page-6-5"></span><span id="page-6-3"></span> $$\frac{1}{2} \le r_{\rho} := \max\left\{\frac{1}{2}, 1 - \frac{1}{\rho^2}\right\} < 1. \tag{2.11}$$ Then the function <span id="page-6-1"></span> $$z := re^{i\theta} \mapsto g\left(\frac{1-r}{r}\right)h(\theta), \quad r \in (0,1), \ \theta \in \mathbb{R}, \ z \in \mathbb{D} \setminus \{0\}, \tag{2.12}$$ belongs to the class (see (2.4)) <span id="page-6-4"></span> $$sbh_0^+(\mathbb{D}\setminus \overline{D}(r_{\rho}); \leq b_{\rho}), \text{ where } b_{\rho} := g\left(\frac{1-r_{\rho}}{r_{\rho}}\right) \max_{\theta} h(\theta). \tag{2.13}$$ Proof We use the properties (i)–(vii) of $\rho$ -trigonometrically convex functions. There is a decreasing sequence of convex positive functions $g_n \underset{n \to \infty}{\searrow} g$ on $\mathbb{R}$ such that $g_n(0) = 0$ and $g_n \in C^2(\mathbb{R}^+_)$ , $n \in \mathbb{N}$ . There is also a sequence of $2\pi$ -periodic $\rho$ -trigonometrically convex positive functions $h_n \underset{n \to \infty}{\searrow} h$ ([5, Proposition 1.4], [2, Theorem 51]) such that $h_n \in C^2(\mathbb{R}^+_)$ , $n \in \mathbb{N}$ . The limit of each decreasing sequence of positive subharmonic functions is a subharmonic positive function. Therefore it suffices to prove the subharmonicity of the function (2.12) on D outward $\overline{D}(r_\rho)$ for the case $h \in C^2(\mathbb{R})$ and $g \in C^2(\mathbb{R}^+_*)$ . The calculation of the Laplace operator of the function (2.12) in polar coordinates $(r, \theta)$ gives <span id="page-6-2"></span> $$\Delta \left( g \left( \frac{1-r}{r} \right) h(\theta) \right) \stackrel{(2.12)}{=} \left( \frac{\partial}{\partial r^2} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial}{\partial \theta} \right) \left( g(1/r - 1) h(\theta) \right) \\ = \left( g''(1/r - 1) \frac{1}{r^4} + \frac{1}{r^3} g'(1/r - 1) \right) h(\theta) + \frac{1}{r^2} g(1/r - 1) h''(\theta). \quad (2.14)$$ Each convex function $g: \mathbb{R}^+ \to \mathbb{R}^+$ , $g \in C^2(\mathbb{R}^+_*)$ , with g(0) = 0 has the following properties: <span id="page-7-0"></span> $$g'' \ge 0$$ , $g'(x) \ge \frac{g(x)}{x}$ for all $x \in \mathbb{R}_*^+$ , and $g \in C(\mathbb{R}^+)$ is increasing. (2.15) It follows from (2.14), (2.15), (1.4) that, for $h \in \rho$ -trc⁺ $\cap C^2(\mathbb{R})$ , <span id="page-7-1"></span> $$\Delta \left( g \left( \frac{1-r}{r} \right) h(\theta) \right) \\ \stackrel{(2.15)}{\geq} \left( g''(1/r-1) \frac{1}{r^4} + \frac{1}{r^2(1-r)} g(1/r-1) \right) h(\theta) + \frac{1}{r^2} g(1/r-1) h''(\theta) \\ \stackrel{(2.15),(1.4)}{\geq} \frac{1}{r^2} \left( \frac{1}{1-r} - \rho^2 \right) g(1/r-1) h(\theta) \quad \text{for all } r \in \mathbb{R}_*^+, \ \theta \in \mathbb{R}. \quad (2.16)$$ If $r \ge r_\rho$ , then the right-hand side of the inequalities (2.16) is positive. Therefore the function (2.12) is subharmonic on $\mathbb{D} \setminus \overline{D}(r_\rho)$ . Obviously, the function (2.12) is positive, since the functions $h \in \rho\text{-trc}^+$ , $g \colon \mathbb{R}^+ \to \mathbb{R}^+$ are positive, and <span id="page-7-2"></span> $$g(0) = 0 \implies \lim_{0 < x \to 0} g(x) \stackrel{\text{(2.15)}}{=} 0 \implies \lim_{1 > r \to 1} g\left(\frac{1-r}{r}\right) h(\theta) \stackrel{\text{(2.12)}}{=} 0.$$ (2.17) Besides, in view of (2.15), we have $$g\Big(\frac{1-r}{r}\Big)\max_{\theta}h(\theta)\overset{(2.11)}{\leq} g\Big(\frac{1-r_{\rho}}{r_{\rho}}\Big)\max_{\theta}h(\theta)\overset{(2.13)}{=} b_{\rho}\quad\text{for all }r\in(r_{\rho},1).$$ So, by Definition 2, in view of (2.17), the function (2.12) belongs to the class (2.13).

Lemma 1. ([3], [5]–[7]) Let f be a continuous function on $(r,1) \subset (0,1)$ , $\mu \in \text{Meas } \mathbb{D}$ . Under the conditions before (1.8) we have the equality <span id="page-7-4"></span> $$\int_{\mathbb{D}\setminus\overline{D}(r)} f(t)k(\theta) \,\mathrm{d}\mu(te^{i\theta}) \stackrel{(1.10)}{=} \int_{r}^{1} f(t) \,\mathrm{d}\mu^{\mathrm{rad}}(t;k). \tag{3.2}$$ By Lemma 1, we get from (3.1) the conclusion (1.12) of Main Theorem for $\rho \leq \sqrt{2}$ . Consider now the case $\rho > \sqrt{2}$ , i.e., $r_{\rho} \stackrel{(2.11)}{=} 1 - 1/\rho^2 > 1/2$ . By Proposition 1, the function (2.12) belongs to the class (2.13), and <span id="page-8-0"></span> $$\operatorname{sbh}_0^+(\mathbb{D}\setminus\overline{D}(r_\rho);\leq b_\rho)\subset \operatorname{sbh}_0^+(\mathbb{D}\setminus\overline{D}(r_\rho);\leq 1)$$ since $g(1) \le 1$ and $\max_{\theta} h(\theta) \le 1$ for [g]–[h], and $$b_{\rho} \stackrel{(2.13)}{\leq} g\left(\frac{1-(1-\rho^{-2})}{1-\rho^{-2}}\right) \max_{\theta} h(\theta) \stackrel{(2.15)}{\leq} g(1) \max_{\theta} h(\theta) \leq 1, \text{ when } \rho > \sqrt{2}.$$ Hence, by Theorem B, there is a constant $C' = \operatorname{const}_{S,M,u}^+ = \operatorname{const}_{\rho,M,u}^+$ for $S := \overline{D}(r_\rho)$ such that the inequality (2.9) holds for any function v of the form (2.12). So, we get $$\int_{\mathbb{D}\backslash\overline{D}(r_{\rho})} g\left(\frac{1-t}{t}\right) h(\theta) d\mu_{u}(te^{i\theta})$$ $$\stackrel{(2.9)}{\leq} \int_{\mathbb{D}\backslash\overline{D}(r_{\rho})} g\left(\frac{1-t}{t}\right) h(\theta) d\mu_{M}(te^{i\theta}) + C' \quad \text{for any [g]-[h].} \quad (3.3)$$ It is easy to see that there are constants $C'' := \operatorname{const}_{o,u}^+, C''' := \operatorname{const}_{o,M}^+$ such that $$\int_{D(r_{\rho})\setminus \overline{D}(1/2)} g\left(\frac{1-t}{t}\right) h(\theta) d\mu_{u}(te^{i\theta}) \leq \mu_{u}\left(D(r_{\rho})\setminus \overline{D}(1/2)\right) \leq C'',$$ $$\left|\int_{D(r_{\rho})\setminus \overline{D}(1/2)} g\left(\frac{1-t}{t}\right) h(\theta) d\mu_{M}(te^{i\theta})\right| \leq |\mu_{M}|\left(D(r_{\rho})\setminus \overline{D}(1/2)\right) \leq C''',$$ Hence, in view of (3.3), we obtain (3.1) with $C := C' + C'' + C''' = \operatorname{const}_{\rho,M,u}$ for any [g]-[h]. By Lemma 1, we again obtain from (3.1) the conclusion (1.12) of Main Theorem already for the case $\rho > \sqrt{2}$ . Let Z be a sequence from (1.5) with $0 := \mathbb{D}$ , and let $f \in \operatorname{Hol}_*(\mathbb{D})$ be a function that vanishes on the sequence $Z \subset \operatorname{Zero}_f$ and satisfies the inequality $u := \log |f| \le M$ . By the conclusion (1.12) of Main Theorem, there exists a constant $C := \operatorname{const}_{\rho,M,f}^+$ such that we have (1.12) for $u := \log |f|$ . Here the choice of the function f is predetermined solely by the sequence Z and function f. So, f = constf and, in view of the (in)equalities $$\begin{split} \sum_{1/2 < r_k < 1} g\left(\frac{1 - r_k}{r_k}\right) h(\theta_k) &\stackrel{\text{(1.9)}}{=} \int_{1/2}^1 g\left(\frac{1 - t}{t}\right) \mathrm{d} n_{\mathsf{Z}}^{\mathsf{rad}}(t; h) \\ &\leq \int_{1/2}^1 g\left(\frac{1 - t}{t}\right) \mathrm{d} n_{\mathsf{Zero}_f}^{\mathsf{rad}}(t; h) &\stackrel{\text{(1.7)}}{=} \int_{1/2}^1 g\left(\frac{1 - t}{t}\right) \mathrm{d} \mu_u^{\mathsf{rad}}(t; h) \quad \text{for } u := \log|f|, \end{split}$$ the inequality (1.13) follows from (1.12). Proof (of Uniqueness Theorem) Without loss of generality, we can assume that $h \not\equiv 0$ , i. e., $h_0 := \max_{\theta} h(\theta) > 0$ , and g(1) > 0. If $f \in \operatorname{Hol}_*(\mathbb{D})$ , i. e., $f \neq 0$ , $f(\mathsf{Z}) = 0$ , and $|f| \leq \exp M$ on $\mathbb{D}$ , then, by Main Theorem, we have $$\frac{1}{g(1)h_0} \sum_{1/2 < r_k < 1} g(1 - r_k) h(\theta_k) \le \sum_{1/2 < r_k < 1} \frac{1}{g(1)} g\left(\frac{1 - r_k}{r_k}\right) \frac{1}{h_0} h(\theta_k) \stackrel{(1.13)}{\le} \int_{1/2}^1 \frac{1}{g(1)} g\left(\frac{1 - t}{t}\right) d\mu_M^{\text{rad}}(t; h/h_0) \le \frac{1}{g(1)h_0} \int_{1/2}^1 g\left(2(1 - t)\right) d\mu_M^{\text{rad}}(t; h) \stackrel{(1.11M)}{<} + \infty.$$ So, if $f \neq 0$ , then the latter contradicts the condition (1.11Z). Acknowledgements The authors thank the organizers of International Conferences "Complex Analysis and Related Topics 2018" (April 23–27, 2018, Euler International Mathematical Institute, St. Petersburg, Russia) and "XXVII St.Petersburg Summer Meeting in Mathematical Analysis" (August 6–11, 2018, St. Petersburg, Russia) for the invitation and for the opportunity to report the results related to the content of this article.

Definition 1. ([16, 8.1], [2]) Let $\rho \in \mathbb{R}_*^+$ . A $2\pi$ -periodic function $h: \mathbb{R} \to \mathbb{R}$ is called a $\rho$ -trigonometrically convex function if $$h(\theta) \leq \frac{\sin \rho(\theta_2 - \theta)}{\sin \rho(\theta_2 - \theta_1)} h(\theta_1) + \frac{\sin \rho(\theta - \theta_1)}{\sin \rho(\theta_2 - \theta_1)} h(\theta_2) \quad \text{for all } \theta \in (\theta_1, \theta_2)$$ (1.2) and for all $\theta_1, \theta_2 \in \mathbb{R}$ such that $0 < \theta_2 - \theta_1 < \pi/\rho$ . A function $h: \mathbb{R} \to \mathbb{R}$ is a 0-trigonometrically convex function if $h \equiv \text{const} \in \mathbb{R}$ . Further, the class of all $2\pi$ -periodic $\rho$ -trigonometrically convex function on $\mathbb{R}$ is denoted as $\rho$ -trc, <span id="page-1-2"></span> $$\rho\text{-trc}^+ := \{h \in \rho\text{-trc} \colon h \ge 0 \text{ on } \mathbb{R}\}, \text{ trc} := \bigcup_{\rho \in \mathbb{R}^+} \rho\text{-trc}, \text{ trc}^+ := \bigcup_{\rho \in \mathbb{R}^+} \rho\text{-trc}^+. \quad (1.3)$$ We recall some properties of $2\pi$ -periodic $\rho$ -trigonometrically convex functions that can be found in the works [15], [16], [14], [2], [17], [1]. - <span id="page-1-3"></span>(i) If $h \in \text{trc}$ , then $h \in C(\mathbb{R})$ . - (ii) If $h \in \rho$ -trc, then $h^+ := \max\{0, h\} \in \rho$ -trc⁺. - (iii) Let $h \in C^2(\mathbb{R})$ be a $2\pi$ -periodic function. $h \in \rho$ -trc if and only if <span id="page-1-5"></span> $$h''(\theta) + \rho^2 h(\theta) \ge 0$$ for all $\theta \in \mathbb{R}$ . (1.4) - (iv) A $2\pi$ -periodic continuous function h belongs to the class $\rho$ -trc if and only if $h'' + \rho^2 h \ge 0$ in the sense of the distribution theory. - (v) A $2\pi$ -periodic continuous function h belongs to the class $\rho$ -trc if and only if the function $z = re^{i\theta} \mapsto h(\theta)r^{\rho}$ is subharmonic on $\mathbb{C}$ . - (vi) If $h \in \rho$ -trc⁺ and $\rho \leq \rho' \in \mathbb{R}^+$ , then $h \in \rho'$ -trc⁺, i.,e., $\rho$ -trc⁺ $\subset \rho'$ -trc⁺. - <span id="page-1-4"></span>(vii) If a sequence of functions $h_n \in \rho$ -trc⁺, $n \in \mathbb{N}$ is decreasing, then the function $h := \lim_{n \to \infty} h_n$ belongs to the same class $\rho$ -trc⁺. Example 1 Let $\rho \in \mathbb{R}^+$ . The $2\pi$ -periodic continuation of the function $$h(\theta) := \begin{cases} \cos \rho \, \theta, & \text{if } |\theta| < \frac{\pi}{2\rho}, \ 0, & \text{if } |\theta| \geq \frac{\pi}{2\rho}, \end{cases} \qquad \theta \in (-\pi,\pi],$$ belong to the class $\rho$ -trc⁺. Example 2 Let $S \subset \mathbb{C}$ be a bounded subset. Then the support function $k_S(\theta) := \sup_{s \in S} \operatorname{Re}(se^{-i\theta})$ of S belongs to the class 1-trc. If $0 \in S$ , then $k_S \in 1$ -trc⁺. For $\rho \in \mathbb{R}^+$ , the $\rho$ -support function of $\rho$ -convex domain $S \subset \mathbb{C}$ belongs to the class $\rho$ -trc, and this $\rho$ -support function belongs to the class $\rho$ -trc⁺, when $0 \in S$ [1, Ch. VI, 2.2], [14, Ch. II, § 3], [17, § 9]. Example 3 Let u be a subharmonic function on $\mathbb{C}$ , and $\limsup_{z\to\infty}\frac{u(z)}{|z|^{\rho}}<+\infty$ . Then its $\rho$ -indicator function $$h(\theta) := \limsup_{r \to +\infty} \frac{u(re^{i\theta})}{r^{\rho}}, \quad \theta \in \mathbb{R},$$ belongs to the class $\rho$ -trc. See [15], [16], [14], [2], [17] for $u := \log |f|$ with an entire function f on $\mathbb{C}$ . Let $\mathcal{O} \subset \mathbb{C}$ be an open subset, and let <span id="page-2-1"></span> $$\mathsf{Z} := \{\mathsf{z}_k\}_{k=1,2,\dots}, \quad \mathsf{z}_k := r_k e^{i\theta_k} \in \mathfrak{O}, \quad r_k := |\mathsf{z}_k| \in \mathbb{R}^+, \ \theta_k \in \arg \mathsf{z}_k \subset \mathbb{R}, \quad (1.5)$$ be a sequence on O without limit points in O. Some points $z_k$ can repeat. It is also possible that $Z = \emptyset$ is empty. We associate with each sequence Z the integer-valued positive counting measure $n_Z$ on O by setting <span id="page-2-0"></span> $$n_{\mathsf{Z}}(S) := \sum_{\mathsf{z}_k \in S} 1, \quad S \subset \mathcal{O}; \tag{1.6}$$ $n_{\mathsf{Z}}(S)$ is the number of points $\mathsf{z}_k$ lying in S. We denote by the same symbol as the sequence $\mathsf{Z}$ the function $\mathsf{Z} \colon z \mapsto n_{\mathsf{Z}}(\{z\}), z \in \mathcal{O}$ , the divisor of the sequence $\mathsf{Z}$ . In particular, we have supp $\mathsf{Z} := \operatorname{supp} n_{\mathsf{Z}}$ for the support supp; $\mathsf{Z} \subset D$ means that $\operatorname{supp} \mathsf{Z} \subset D$ ; $z \in \mathsf{Z}$ (resp., $z \notin \mathsf{Z}$ ) means the same as $z \in \operatorname{supp} \mathsf{Z}$ (resp., as $z \notin \operatorname{supp} \mathsf{Z}$ ). Departing from the usual treatment of a sequence as a function of an integer or a positive integer variable we say that a sequence Z coincides with a sequence Z' or that they are equal (we write Z = Z') if for the associated divisors we have $Z(z) \equiv Z'(z)$ for all $z \in \mathcal{O}$ . In other words, we regard a point sequence as a representative of the equivalence class containing the sequences in $\mathcal{O}$ with equal divisors. An embedding $Z \subset Z'$ means that Z(z) < Z'(z) for all $z \in \mathcal{O}$ . See [7] in detail. We denote by $\operatorname{Zero}_f$ the zero sequence of the function $f \in \operatorname{Hol}_*(\mathcal{O})$ in $\mathcal{O}$ numbered with multiplicities taken into account. Then [18, Theorem 3.7.8] <span id="page-2-2"></span> $$n_{\text{Zero}_f} \stackrel{\text{(1.6)}}{=} \frac{1}{2\pi} \Delta \log|f| \tag{1.7}$$ is the Riesz measure of function $\log |f| \in \mathrm{sbh}_*(\mathcal{O})$ . A function $f \in \operatorname{Hol}_*(\mathcal{O})$ vanishes on Z if $Z \subset \operatorname{Zero}_f$ (we write f(Z) = 0). The function $0 \in \operatorname{Hol}(\mathcal{O})$ vanishes on any sequence $Z \subset \mathcal{O}$ . For $r \in \mathbb{R}^+$ and $z \in \mathbb{C}$ , we set $D(z,r) := \{z' \in \mathbb{C} : |z'-z| < r\}$ (i.e., D(z,r) is an open disk of radius r centered at z), D(r) := D(0,r); $\overline{D}(z,r) := \{z' \in \mathbb{C} : |z'-z| \le r\}$ (i.e., D(z,r) is a closed disk of radius r centered at z), $\overline{D}(r) := \overline{D}(0,r)$ , $\overline{D}(0) := \{0\}$ . The class of all Borel real measures, i.e., charges, on a Borel subset $S \subset \mathbb{C}$ is denoted by Meas(S), and Meas⁺(S) $\subset$ Meas(S) is the subclass of all positive measures. For a charge $\mu \in \text{Meas}(S)$ , we let $\mu^+$ , $\mu^-$ and $|\mu| := \mu^+ + \mu^-$ resp. denote its upper, lower, and total variations. Let $h: \mathbb{R} \to \mathbb{R}$ be a bounded $2\pi$ -periodic Borel function on $\mathbb{R}$ ; $\mu \in \text{Meas}(\mathbb{D})$ . We define the radial counting function $\mu^{\text{rad}}(\cdot;h)$ of charge $\mu$ with weight h on [0,1) as [5,(3.1)],[6,(0.2)] <span id="page-3-0"></span> $$\mu^{\text{rad}}(r;h) := \int_{\overline{D}(r)} h(\arg z) \, \mathrm{d}\mu(z), \quad r \in [0,1). \tag{1.8}$$ In particular, the function $\mu^{\text{rad}}(r) := \mu^{\text{rad}}(r; 1)$ with weight $h \equiv 1$ is the classical radial counting function of $\mu$ . If Z is a sequence in $\emptyset \stackrel{(1.5)}{:=} \mathbb{D}$ then [3], [5, (0.4)], [6, (0.2)] <span id="page-3-4"></span> $$n_{\mathsf{Z}}^{\mathrm{rad}}(r;h) \stackrel{(1.6)}{:=} \sum_{|\mathsf{z}_k| \le r} h(\arg \mathsf{z}_k), \quad r \in [0,1).$$ (1.9) Here and below, a reference mark over a symbol of (in)equality, inclusion, or more general binary relation, etc. means that this relation is somehow related to this reference. For $-\infty \le r < R \le +\infty$ always <span id="page-3-3"></span><span id="page-3-1"></span> $$\int_{r}^{R} \dots := \int_{(r,R)} \dots \tag{1.10}$$ A particular result of our investigation is the following Uniqueness Theorem Let $M \in \mathrm{sbh}_*(\mathbb{D})$ be a subharmonic function with Riesz measure $\mu_M := \frac{1}{2\pi} \Delta M \in \mathrm{Meas}^+(\mathbb{D})$ , and let $Z \stackrel{(1.5)}{=} \{r_k e^{i\theta_k}\}_{k=1,2,\dots} \subset \mathbb{D} \stackrel{(1.5)}{=} : 0$ be a sequence, $h \in \mathrm{trc}^+$ , and $g : \mathbb{R}^+ \to \mathbb{R}^+$ be a convex function with g(0) = 0. If a function $f \in \mathrm{Hol}(\mathbb{D})$ vanishes on Z, satisfies the inequality $|f| \leq \exp M$ on $\mathbb{D}$ , and $$\int_{1/2}^{1} g(2(1-t)) d\mu_{M}^{\text{rad}}(t;h) \stackrel{(1.8)}{<} +\infty, \tag{1.11M}$$ but <span id="page-3-5"></span><span id="page-3-2"></span> $$\sum_{1/2 < r_k < 1} g(1 - r_k) h(\theta_k) \stackrel{\text{(1.5)}}{=} + \infty, \tag{1.11Z}$$ then f is the zero function, i. e., $f \equiv 0$ on $\mathbb{D}$ . In the case M=0 with $\mu_M=0$ , $g(x)\equiv x$ , $x\in\mathbb{R}^+$ , and $h\equiv 1\in 0$ -trc⁺, the condition (1.11Z) contradicts the classical Blaschke condition $\sum_k (1-r_k)<+\infty$ . So, the Nevanlinna theorem on the distribution of zeros of bounded holomorphic functions shows that our Uniqueness Theorem is accurate in this case. By $\text{const}_{a_1,a_2,...} \in \mathbb{R}$ we denote constants that, in general, depend on $a_1,a_2,...$ and, unless otherwise specified, only on them; $\text{const}^+ \geq 0$ . Main Theorem Let $M \in \delta$ -sbh_\( $\mathbb{D}$ ) and $u \in \text{sbh}_(\mathbb{D})$ are functions, resp., with Riesz charge $\mu_M := \frac{1}{2\pi} \Delta M \in \text{Meas}(\mathbb{D})$ and with Riesz measure $\mu_u := \frac{1}{2\pi} \Delta u \in \text{Meas}^+(\mathbb{D})$ . Let $\rho \in \mathbb{R}^+$ . If $u \leq M$ on $\mathbb{D}$ , then there exists a constant $C := \text{const}_{\rho,M,u}^+ \geq 0$ such that the inequality <span id="page-4-0"></span> $$\int_{1/2}^{1} g\left(\frac{1-t}{t}\right) d\mu_u^{\text{rad}}(t;h) \stackrel{(1.8)}{\leq} \int_{1/2}^{1} g\left(\frac{1-t}{t}\right) d\mu_M^{\text{rad}}(t;h) + C \tag{1.12}$$ holds for any - [g] convex function $g: \mathbb{R}^+ \to \mathbb{R}^+$ with g(0) = 0 and $g(1) \leq 1$ , - [h] $2\pi$ -periodic $\rho$ -trigonometrically convex function $h: \mathbb{R} \to [0,1]$ . In particular, if Z is a sequence from (1.5) with $0 := \mathbb{D}$ , and there exists a function $f \in \operatorname{Hol}_*(\mathbb{D})$ , f(Z) = 0, satisfying the inequality $|f| \le \exp M$ on $\mathbb{D}$ , then there is a constant $C := \operatorname{const}_{D,M,Z}$ such that <span id="page-4-1"></span> $$\sum_{1/2 < r_k < 1} g\left(\frac{1 - r_k}{r_k}\right) h(\theta_k) \le \int_{1/2}^1 g\left(\frac{1 - t}{t}\right) \mathrm{d}\mu_M^{\mathrm{rad}}(t; h) + C \quad \textit{for any } [g]-[h]. \quad (1.13)$$ The cases u = M and $M = \log |f|$ , $Z = Zero_f$ , show that the inequalities (1.12) and (1.13) uniform with respect to [h]-[g] are optimal up to an additive constant C.

Definition 2. ([9, Definition 1]) We say that a function $v \in \operatorname{sbh}_0^+(D \setminus S)$ is a sub-harmonic test function on D outward S if the function v is bounded in $D \setminus S$ . The class of such functions w will be denoted by $\operatorname{sbh}_0^+(D \setminus S; <+\infty)$ . For $b \in \mathbb{R}^+$ , put <span id="page-5-0"></span> $$\operatorname{sbh}_{0}^{+}(D \setminus S; \leq b) \stackrel{(2.3+)}{:=} \left\{ v \in \operatorname{sbh}_{0}^{+}(D \setminus S; <+\infty) : \sup_{D \setminus S} v \leq b \right\}. \tag{2.4}$$ Thus. $$\mathrm{sbh}_0^+(D\setminus S;<+\infty)=\bigcup_{b\in\mathbb{R}^+}\mathrm{sbh}_0^+(D\setminus S;\leq b).$$ The main role will be played by the following Theorem A ([9, Main Theorem] for $\mathbb{C}$ , see also [10, Main Theorem], [11]–[13]) Let $M \in \delta$ -sbh_\(D) be a* $\delta$ -subharmonic function with Riesz charge $\mu_M = \frac{1}{2\pi}\Delta M$ , and <span id="page-5-2"></span> $$\emptyset \neq \operatorname{int} S \subset S = \operatorname{clos} S \stackrel{(2.1)}{\in} D \subset \mathbb{C}_{\infty} \neq D.$$ (2.5) Then for any point $z_0 \in \operatorname{int} S$ with $M(z_0) \in \mathbb{R}$ , any number $b \in \mathbb{R}^+$ , any regular for the Dirichlet Problem [18, 4] domain $\widetilde{D} \subset \mathbb{C}_{\infty}$ with the Green function $g_{\widetilde{D}}(\cdot, z_0)$ with a pole at $z_0$ which satisfies the conditions $S \subseteq \widetilde{D} \subset D$ and $\mathbb{C}_{\infty} \setminus \operatorname{clos} \widetilde{D} \neq \emptyset$ , any subharmonic function $u \in \operatorname{sbh}_*(D)$ satisfying the inequality $u \leq M$ on D, and any subharmonic test function $v \in \operatorname{sbh}_0^+(D \setminus S; \leq b)$ the following inequality holds: <span id="page-5-4"></span> $$\widetilde{C}u(z_0) + \int_{D \setminus S} v \, d\mu_u \le \int_{D \setminus S} v \, d\mu_M + \int_{\widetilde{D} \setminus S} v \, d\mu_M^- + \widetilde{C} \, \overline{C}_M, \tag{2.6}$$ where $\mu_u := \frac{1}{2\pi} \Delta u$ is the Riesz measure of the function u, <span id="page-5-3"></span> $$\widetilde{C} := \operatorname{const}_{z_0, S, \widetilde{D}, b}^+ := \frac{b}{\inf_{z \in \partial S} g_{\widetilde{D}}(z, z_0)} > 0, \tag{2.7}$$ and the value $+\infty$ is possible for the constant <span id="page-5-1"></span> $$\overline{C}_M := \int_{\widetilde{D}\setminus\{z_0\}} g_{\widetilde{D}}(\cdot, z_0) \,\mathrm{d}\mu_M + \int_{\widetilde{D}\setminus S} g_{\widetilde{D}}(\cdot, z_0) \,\mathrm{d}\mu_M^- + M^+(z_0), \tag{2.8}$$ but for $\widetilde{D} \subseteq D$ this is a certain constant $\overline{C}_M \stackrel{(2.8)}{=} \operatorname{const}^+_{z_0, S, \widetilde{D}, M, D} < +\infty$ . We use the following simplified version of Theorem A. Theorem B Under the agreements (2.1), and (2.5), let $M \in \delta$ -sbh_\(D) be a function with Riesz charge $\mu_M \in \text{Meas}(D)$ . Then, for any function $u \in \text{sbh}_(D)$ with Riesz measure $\mu_u$ satisfying the inequality $u \leq M$ on D, we have the inequality <span id="page-5-5"></span> $$\int_{D\setminus S} v \, \mathrm{d}\mu_u \le \int_{D\setminus S} v \, \mathrm{d}\mu_M + C \quad \text{for all } v \overset{(2.4)}{\in} \mathrm{sbh}_0^+(D\setminus S; \le 1), \tag{2.9}$$ where a constant $C := \operatorname{const}_{D,S,u,M}^+ \in \mathbb{R}^+$ is independent of $v \overset{(2.4)}{\in} \operatorname{sbh}_0^+(D \setminus S; \leq 1)$ . <span id="page-6-0"></span>Proof There exists always a point $z_0 \in \text{int } S$ and $r_0 \in \mathbb{R}^+_*$ such that [9, 3.1] $$D(z_0, r_0) \equiv \text{int } S, \quad u(z_0) \neq -\infty, \quad M(z_0) \neq \pm \infty,$$ $$\left| \int_{D(z_0, r_0)} \log|z - z_0| \, \mathrm{d}\mu_M \right| < +\infty.$$ (2.10) There is always a regular for the Dirichlet Problem domain $\widetilde{D}$ such that $\operatorname{int} S \subseteq \widetilde{D} \subseteq D$ [18, 4]. The choice of such point $z_0$ and such domain $\widetilde{D}$ is predetermined solely by sets S,D. We choice b:=1. Thus, $\widetilde{C} \stackrel{(2,7)}{=} \operatorname{const}_{D,S}^+ \in \mathbb{R}^+$ is a constant depending only on S and D. In view of (2.10), by the definition (2.8), the constant $\overline{C}_M \stackrel{(2.8),(2.10)}{=} \operatorname{const}_{D,S,M}^+ \in \mathbb{R}^+$ is depending only on D,S,M. Hence the constant $$C \stackrel{(2.6)}{:=} |\widetilde{C}u(z_0)| + |\mu_M|(\widetilde{D} \setminus S) + \widetilde{C}\overline{C}_M \ge -\widetilde{C}u(z_0) + \int_{\widetilde{D} \setminus S} v \, d\mu_M^- + \widetilde{C}\overline{C}_M,$$ depends only on D, S, u, M, i. e., $C = \text{const}_{D.S.u.M}^+ \in \mathbb{R}^+$ . So, (2.9) follows from (2.6). A method of constructing subharmonic test functions on $\mathbb{D}$ outward D(r) by means of $\rho$ -trigonometrically convex positive functions is given by the following