On the difference of the moduli of the two initial logarithmic coefficients

Theorem 1. Let $f \in \mathcal{U}(\lambda)$ for some $0 < \lambda \leq 1$ . Then (i) $$-\frac{2\lambda+1}{4} \le |\gamma_2| - |\gamma_1| \le \frac{\lambda}{2} \text{ if } 0 \le \lambda \le \frac{1}{2};$$ (ii) $-\frac{\sqrt{2\lambda}}{2} \le |\gamma_2| - |\gamma_1| \le \frac{\lambda}{2} \text{ if } \frac{1}{2} \le \lambda \le 1.$ All estimates are sharp. Before proceeding with the proof let note that for $\lambda = 1$ we receive the same result as in Theorem A for the class S. That is expected since the extremal functions, $$f_1(z) = \frac{z}{1 - \sqrt{2}e^{i\theta}z + e^{2i\theta}z^2}$$ and $f_2(z) = \frac{z}{1 + e^{i\theta}z^2}$ belong to the class $\mathcal{U} \equiv \mathcal{U}(1)$ . On the other hand, that does not hold for other values of $\lambda \in (0,1)$ .

Theorem 2. For the classes $\mathcal{M}(\alpha)$ the next relations are true: - (i) $\mathcal{M}(\alpha) \subset \mathcal{S}$ for all real $\alpha$ ; - (ii) $\mathcal{M}(\alpha) \subset \mathcal{M}(\beta) \subset \mathcal{M}(0) = \mathcal{S}^*$ for $0 < \frac{\alpha}{\beta} < 1$ ; - (iii) $\mathcal{M}(\alpha) \subset \mathcal{M}(1) = \mathcal{K} \text{ for } \alpha > 1.$ Here $S^* \equiv \mathcal{M}(0)$ and $\mathcal{K} \equiv \mathcal{M}(1)$ are the well-known classes of starlike and convex functions, mapping the unit disk onto a starlike or convex domain, respectively. For the class of $\alpha$ -convex functions we also have the following

Theorem 3. For $\alpha \geq 0$ and $f \in \mathcal{A}$ , we have <span id="page-3-0"></span> $$f \in \mathcal{M}(\alpha) \quad \Leftrightarrow \quad F \in \mathcal{S}^*,$$ where (6) $$F(z) = f(z) \left[ \frac{zf'(z)}{f(z)} \right]^{\alpha}.$$ From the equation (6) we easily derive that for $\alpha > 0$ and $f \in \mathcal{M}(\alpha)$ there exists $F \in \mathcal{S}^*$ , such that (7) $$f(z) = \left(\frac{1}{\alpha} \int_0^z \frac{F^{\frac{1}{\alpha}}(t)}{t} dt\right)^{\alpha},$$ and f(z) = F(z) for $\alpha = 0$ . For example, for $F(z) = \frac{z}{(1 - e^{i\theta}z)^2} = k_{\theta}(z)$ (the Koebe function), from (7) we obtain <span id="page-3-1"></span> $$k_{\theta}(z,\alpha) = \left(\frac{1}{\alpha} \int_0^z t^{\frac{1}{\alpha} - 1} (1 - e^{i\theta} t)^{-\frac{2}{\alpha}} dt\right)^{\alpha} \quad (\alpha > 0),$$ and $k_{\theta}(z, \alpha) = k_{\theta}(z)$ for $\alpha = 0$ . We note that the function $k_{\theta}(z, \alpha)$ is extremal in many problems related with the class $\mathcal{M}(\alpha)$ , $\alpha \geq 0$ , in the same way as the Koebe function is for the class $\mathcal{S}$ . Further, from the definition of the class $\mathcal{M}(\alpha)$ we have $$(1-\alpha)\frac{zf'(z)}{f(z)} + \alpha \left[1 + \frac{zf''(z)}{f'(z)}\right] = \frac{1+\omega(z)}{1-\omega(z)},$$ where $\omega$ is a Schwartz function $(\omega(0) = 0 \text{ and } |\omega(z)| < 1 \text{ for all } z \in \mathbb{D})$ . Using $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots$ , and $\omega(z) = c_1 z + c_2 z^2 + \cdots$ , after comparing the coefficients, we receive $(1 + \alpha)a_2 = 2c_1$ and since $|c_1| \leq 1$ , $$|a_2| \le \frac{2}{1+\alpha} \quad (\alpha \ge 0).$$ The previous result is the best possible as the functions $f(z) = k_0(z, \alpha)$ , when $\alpha > 0$ , and $f(z) = \frac{z}{(1-z)^2}$ for $\alpha = 0$ , show. Now we give estimates of the difference $|\gamma_2| - |\gamma_1|$ for the class of $\alpha$ -convex functions for $\alpha$ being any non-negative real number. The obtained upper bound is sharp.

Theorem 4. Let $f \in \mathcal{M}(\alpha)$ , $\alpha \geq 0$ and let $\gamma_1$ and $\gamma_2$ be its initial logarithmic coefficients. (i) If $$0 \le \alpha \le \frac{1+\sqrt{3}}{2}$$ , then $$-\frac{1}{\sqrt{2(\alpha^2 + 3\alpha + 1)}} \le |\gamma_2| - |\gamma_1| \le \frac{1}{2(1 + 2\alpha)}.$$ (ii) If $$\alpha \ge \frac{1+\sqrt{3}}{2}$$ , then $$-\frac{6\alpha^2 + 10\alpha + 3}{4(2\alpha + 1)(\alpha^2 + 3\alpha + 1)} \le |\gamma_2| - |\gamma_1| \le \frac{1}{2(1 + 2\alpha)}.$$ Given upper bounds are sharp. Proof. Let put $$J(f, \alpha; z) = (1 - \alpha) \frac{zf'(z)}{f(z)} + \alpha \left[ 1 + \frac{zf''(z)}{f'(z)} \right].$$ Then, by definition of the class $\mathcal{M}(\alpha)$ , $$H(f,\alpha;z) = \frac{1 - J(f,\alpha;z)}{1 + J(f,\alpha;z)}$$ $$= -\frac{1}{2}(1 + \alpha)a_2z - \left[ (1 + 2\alpha)a_3 - \frac{\alpha^2 + 8\alpha + 3}{4}a_2^2 \right]z^2 + \cdots$$ is a Schwartz function, and by using the inequalities $|c_1| \le 1$ and $|c_2| \le 1 - |c_1|^2$ , for $\alpha \ge 0$ we receive that $\frac{1}{2}(1+\alpha)|a_2| \le 1$ and $$\left| (1+2\alpha)a_3 - \frac{\alpha^2 + 8\alpha + 3}{4}a_2^2 \right| \le 1 - \frac{1}{4}(1+\alpha)^2 |a_2|^2.$$ From these inequalities for $\alpha \geq 0$ we obtain $|a_2| \leq \frac{2}{1+\alpha}$ and <span id="page-4-0"></span>(8) $$\left| a_3 - \frac{\alpha^2 + 8\alpha + 3}{4(1+2\alpha)} a_2^2 \right| \le \frac{1}{(1+2\alpha)} - \frac{(1+\alpha)^2}{4(1+2\alpha)} |a_2|^2.$$ Here, we can note that from (8), for $\alpha = 0$ (the starlike case) we have $$\left| a_3 - \frac{3}{4}a_2^2 \right| \le 1 - \frac{1}{4}|a_2|^2,$$ and for $\alpha = 1$ (the convex case): $$\left|a_3 - a_2^2\right| \le \frac{1}{3}(1 - |a_2|^2)$$ (Trimble [18]). For the upper bounds (right estimates) in both cases (i) and (ii), using (3) and (8), we have for $\alpha \geq 0$ : $$\begin{split} |\gamma_2| - |\gamma_1| &= \frac{1}{2} \left| a_3 - \frac{1}{2} a_2^2 \right| - \frac{1}{2} |a_2| \\ &= \frac{1}{2} \left| \left( a_3 - \frac{\alpha^2 + 8\alpha + 3}{4(1 + 2\alpha)} a_2^2 \right) + \frac{\alpha^2 + 4\alpha + 1}{4(1 + 2\alpha)} a_2^2 \right| - \frac{1}{2} |a_2| \\ &\leq \frac{1}{2} \left| a_3 - \frac{\alpha^2 + 8\alpha + 3}{4(1 + 2\alpha)} a_2^2 \right| + \frac{\alpha^2 + 4\alpha + 1}{8(1 + 2\alpha)} |a_2|^2 - \frac{1}{2} |a_2| \\ &\leq \frac{1}{2(1 + 2\alpha)} - \frac{(1 + \alpha)^2}{8(1 + 2\alpha)} |a_2|^2 + \frac{\alpha^2 + 4\alpha + 1}{8(1 + 2\alpha)} |a_2|^2 - \frac{1}{2} |a_2| \\ &= \frac{1}{2(1 + 2\alpha)} + \frac{\alpha}{4(1 + 2\alpha)} |a_2|^2 - \frac{1}{2} |a_2| \\ &\leq \frac{1}{2(1 + 2\alpha)}, \end{split}$$ since $$\frac{\alpha}{4(1+2\alpha)}|a_2|^2 - \frac{1}{2}|a_2| \le 0$$ for $0 \le |a_2| \le \frac{2}{1+\alpha}$ . The result is sharp as the functions (see (7)) defined by $$f(z) = \left(\frac{1}{\alpha} \int_0^z t^{\frac{1}{\alpha} - 1} (1 - t^2)^{-\frac{1}{\alpha}} dt\right)^{\alpha} = z + \frac{1}{1 + 2\alpha} z^3 + \cdots \quad \text{(for } \alpha > 0),$$ and $f(z) = \frac{z}{1 - z^2}$ (for $\alpha = 0$ ) show. (i) Now, for the lower bound in the first case, let $0 \le \alpha \le \frac{1+\sqrt{3}}{2}$ . Then the left hand side of inequality to be proven is equivalent to $$\frac{1}{2} \left| a_3 - \frac{1}{2} a_2^2 \right| - \frac{1}{2} |a_2| \ge -\frac{1}{\sqrt{2(\alpha^2 + 3\alpha + 1)}},$$ i.e., to $$\left| a_3 - \frac{1}{2} a_2^2 \right| \ge |a_2| - \sqrt{\frac{2}{\alpha^2 + 3\alpha + 1}}.$$ If $0 \le |a_2| < \sqrt{\frac{2}{\alpha^2 + 3\alpha + 1}}$ , then the previous inequality is true. Further, let $\sqrt{\frac{2}{\alpha^2+3\alpha+1}} \leq |a_2| \leq \frac{2}{1+\alpha}$ . Then using (3) and (8) we have <span id="page-5-0"></span> $$|\gamma_{2}| - |\gamma_{1}| = \frac{1}{2} \left| a_{3} - \frac{1}{2} a_{2}^{2} \right| - \frac{1}{2} |a_{2}|$$ $$= \frac{1}{2} \left| \left( a_{3} - \frac{\alpha^{2} + 8\alpha + 3}{4(1 + 2\alpha)} a_{2}^{2} \right) + \frac{\alpha^{2} + 4\alpha + 1}{4(1 + 2\alpha)} a_{2}^{2} \right| - \frac{1}{2} |a_{2}|$$ $$\geq \frac{\alpha^{2} + 4\alpha + 1}{8(1 + 2\alpha)} |a_{2}|^{2} - \frac{1}{2} \left| a_{3} - \frac{\alpha^{2} + 8\alpha + 3}{4(1 + 2\alpha)} a_{2}^{2} \right| - \frac{1}{2} |a_{2}|$$ $$\geq \frac{\alpha^{2} + 4\alpha + 1}{8(1 + 2\alpha)} |a_{2}|^{2} - \frac{1}{2} |a_{2}| + \frac{(1 + \alpha)^{2}}{8(1 + 2\alpha)} |a_{2}|^{2} - \frac{1}{2} |a_{2}|$$ $$= \frac{\alpha^{2} + 3\alpha + 1}{4(1 + 2\alpha)} |a_{2}|^{2} - \frac{1}{2} |a_{2}| - \frac{1}{2(1 + 2\alpha)}.$$ From the previous relation, it is enough to show that $$\frac{\alpha^2 + 3\alpha + 1}{4(1+2\alpha)}|a_2|^2 - \frac{1}{2}|a_2| - \frac{1}{2(1+2\alpha)} \ge -\frac{1}{\sqrt{2(\alpha^2 + 3\alpha + 1)}},$$ which is equivalent to $$\left[|a_2| - \sqrt{\frac{2}{\alpha^2 + 3\alpha + 1}}\right] \left[\frac{\alpha^2 + 3\alpha + 1}{2(1 + 2\alpha)} \left(|a_2| + \sqrt{\frac{2}{\alpha^2 + 3\alpha + 1}}\right) - 1\right] \ge 0.$$ The last inequality is indeed true since by assumption $\sqrt{\frac{2}{\alpha^2+3\alpha+1}} \leq |a_2| \leq \frac{2}{1+\alpha}$ and $$\frac{\alpha^2 + 3\alpha + 1}{2(1 + 2\alpha)} \left( |a_2| + \sqrt{\frac{2}{\alpha^2 + 3\alpha + 1}} \right) - 1 \ge \frac{\sqrt{2(\alpha^2 + 3\alpha + 1)}}{1 + 2\alpha} - 1 \ge 0,$$ since also $0 \le \alpha \le \frac{1+\sqrt{3}}{2}$ . Finally, let $\alpha \geq \frac{1+\sqrt{3}}{2}$ . Then, using the beginning and end of the relation (9) we get $$|\gamma_2| - |\gamma_1| \ge \frac{\alpha^2 + 3\alpha + 1}{4(1 + 2\alpha)} |a_2|^2 - \frac{1}{2} |a_2| - \frac{1}{2(1 + 2\alpha)}$$ $$\ge -\frac{6\alpha^2 + 10\alpha + 3}{4(2\alpha + 1)(\alpha^2 + 3\alpha + 1)},$$ since the last function attains its minimum for $|a_2|_0 = \frac{1+2\alpha}{\alpha^2+3\alpha+1}$ and since $$\sqrt{\frac{2}{\alpha^2 + 3\alpha + 1)}} \le |a_2|_0 < \frac{2}{1 + \alpha}$$ for $\alpha \geq \frac{1+\sqrt{3}}{2}$ . Corollary 1. For $\alpha = 0$ , i.e., for the starlike functions $\mathcal{S}^{\star}$ , as for the class $\mathcal{S}$ , we have $$-\frac{1}{\sqrt{2}} \le |\gamma_2| - |\gamma_1| \le \frac{1}{2}.$$ (see [6, 13]). Both inequalities are sharp. For $\alpha = 1$ , i.e., for the convex function from the class K, we have $$-\frac{1}{\sqrt{10}} \le |\gamma_2| - |\gamma_1| \le \frac{1}{6}.$$ (see [13]). The right hand side of the inequality is sharp.

Theorem 5. If $f \in \mathcal{G}(\alpha)$ , $0 < \alpha \le 1$ , then $$-\frac{\alpha(17-\alpha)}{12(8-\alpha)} \le |\gamma_2| - |\gamma_1| \le \frac{\alpha}{12}.$$ The right-hand side inequality is sharp.