On some properties of logarithmic coefficients of inverse of univalent functions

Theorem 1. Let $f \in \mathcal{S}$ and let $\Gamma_1, \Gamma_2, \Gamma_3$ be initial logarithmic coefficients of its inverse function $f^{-1}$ given by (5). Then the following sharp inequalities hold - (i) $|\Gamma_1| \leq 1$ ; - $\begin{array}{c|c} (ii) & |\Gamma_2| \leq \frac{3}{2}; \\ (iii) & |\Gamma_3| \leq \frac{10}{3}. \end{array}$

Theorem 2. Let $f \in \mathcal{S}$ be given by (1), and let $\Gamma_1$ and $\Gamma_2$ be the first two initial logarithmic coefficients of the function $f^{-1}$ . Then $$-\frac{\sqrt{2}}{2} \le |\Gamma_2| - |\Gamma_1| \le \frac{1}{2}.$$ The upper bound is sharp.

Theorem 3. Let $f \in \mathcal{S}$ be given by (1), and let $\Gamma_2$ and $\Gamma_3$ be the logarithmic coefficients of its inverse function $f^{-1}$ . Then $$|\Gamma_3| - |\Gamma_2| \le \frac{11}{6}.$$ The bound is sharp. Proof. From the proof of parts (ii) and (iii) of Theorem 1 we have $$\Gamma_2 = \frac{1}{2} \left( A_3 - \frac{1}{2} A_2^2 \right) = \frac{1}{2} \left( -a_3 + \frac{3}{2} a_2^2 \right)$$ and $$\Gamma_3 = \frac{1}{2} \left( A_4 - A_2 A_3 + \frac{1}{3} A_2^3 \right) = \frac{1}{2} \left( -a_4 + 4a_2 a_3 - \frac{10}{3} a_2^3 \right).$$ Since $|a_2|/2 < 1$ , $$|\Gamma_3| - |\Gamma_2| \le |\Gamma_3| - \frac{|a_2|}{2} |\Gamma_2| \le \left| \Gamma_3 - \frac{a_2}{2} \Gamma_2 \right|$$ $$= \frac{1}{2} \left| -a_4 + \frac{7}{2} a_2 a_3 - \frac{31}{12} a_2^3 \right|.$$ Using the relations (10), after some computations, we get $$|\Gamma_3| - |\Gamma_2| \leq \left|\omega_{33} - 3\omega_{11}\omega_{13} + \frac{3}{2}\omega_{11}^3\right| \leq |\omega_{33}| + \frac{3}{2}|\omega_{11}| \cdot |2\omega_{13} - \omega_{11}^2|.$$ Hence, applying (9) and (12), $$|\Gamma_3| - |\Gamma_2| \leq \frac{1}{3} \sqrt{1 - 3|\omega_{13}|^2} + \frac{3}{2} |\omega_{11}| \leq \frac{11}{6} \ .$$ Note that for Koebe function we have $|\Gamma_2| = \frac{3}{2}$ and $|\Gamma_3| = \frac{10}{3}$ , so in this case $$|\Gamma_3| - |\Gamma_2| = \frac{11}{6}$$ showing the sharpness of the estimate.

Lemma 1. [9] For $f \in \mathcal{C}$ and $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots$ , we have $$|a_3 - a_2^2| \le \frac{1}{3}(1 - |a_2|^2).$$ The inequality is sharp with extremal function $$f_{\lambda}(z) = \int_{0}^{z} \left(\frac{1+t}{1-t}\right)^{\lambda} \frac{1}{1-t^{2}} dt = z + \lambda z^{2} + \frac{1}{3} (2\lambda^{2} + 1)z^{3} + \frac{1}{3} (\lambda^{3} + 2\lambda)z^{4} + \cdots,$$ where $0 < \lambda < 1$ .

Lemma 2. [8] If $\omega(z) = c_1 z + c_2 z^2 + c_3 z^3 + \cdots$ is analytic in $\mathbb{D}$ , satisfies the condition $|\omega(z)| < 1$ , $z \in \mathbb{D}$ , and if $$\Psi(\omega) = |c_3 + \mu c_1 c_2 + \nu c_1^3|,$$ then the following sharp estimate $\Psi(\omega) \leq \Phi(\mu, \nu)$ holds, where $$\Phi(\mu,\nu) = \begin{cases} |\nu|, & (\mu,\nu) \in D_6, \\ \frac{2}{3}(|\mu|+1) \left(\frac{|\mu|+1}{3(|\mu|+1+\nu)}\right)^{\frac{1}{2}}, & (\mu,\nu) \in D_8 \end{cases},$$ with $$D_6 = \left\{ (\mu, \nu) : 2 \le |\mu| \le 4, \ \nu \ge \frac{1}{12} (\mu^2 + 8) \right\}$$ and $$D_8 = \left\{ (\mu, \nu) : \frac{1}{2} \le |\mu| \le 2, \ -\frac{2}{3} (|\mu| + 1) \le \nu \le \frac{4}{27} (|\mu| + 1)^3 - (|\mu| + 1) \right\}.$$ Here we used the notations from [8]. It is known (see, Ponnusamy et al, [7]) that

Theorem 4. Let $f \in \mathcal{C}$ , $f^{-1}$ be given by (4), and let $\Gamma_1$ , $\Gamma_2$ , $\Gamma_3$ be its first three logarithmic coefficients. Then - (i) $|\Gamma_1| \le \frac{1}{2}$ ; (ii) $|\Gamma_2| \le \frac{1}{4}$ ; $$(iii)$$ $|\Gamma_3| \leq \frac{1}{6}$ . All estimates are sharp. We shall find the lower and the upper bound of the differences $|\Gamma_2| - |\Gamma_1|$ and $|\Gamma_3| - |\Gamma_2|$ when $f \in \mathcal{C}$ . For the second difference we need to express coefficients of $f \in \mathcal{C}$ in terms of appropriate coefficients of a Schwarz function $\omega$ . From the definition of convex functions we have $$1 + \frac{zf''(z)}{f'(z)} = \frac{1 + \omega(z)}{1 - \omega(z)}, \qquad |\omega(z)| < 1, z \in \mathbb{D},$$ and from here $$[zf'(z)]' = \{1 + 2[\omega(z) + \omega^2(z) + \cdots]\} \cdot f'(z).$$ If we put $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots$ and $\omega(z) = c_1 z + c_2 z^2 + c_3 z^3 + \cdots$ , then from the last relation, after comparing the coefficients, we get (14) $$a_2 = c_1,$$ $$a_3 = \frac{1}{3}(c_2 + 3c_1^2),$$ $$a_4 = \frac{1}{6}(c_3 + 5c_1c_2 + 6c_1^3).$$ Now, we are ready to prove the final theorem.

Theorem 5. Let $f \in \mathcal{C}$ and let $f^{-1}$ be given by (4). Also, let $\Gamma_1$ , $\Gamma_2$ , $\Gamma_3$ be its initial logarithmic coefficients. Then $$\begin{array}{ll} (i) & -\frac{\sqrt{10}}{10} \leq |\Gamma_2| - |\Gamma_1| \leq \frac{1}{6}. \\ (ii) & |\Gamma_3| - |\Gamma_2| \leq \frac{2\sqrt{10}}{75}. \end{array}$$ (ii) $$|\Gamma_3| - |\Gamma_2| \le \frac{2\sqrt{10}}{75}$$ . All inequalities are sharp. Proof. (i) Using similar consideration as in Theorem 3, the result of Lemma 2, and $|a_2| \leq 1$ , for the upper bound we get $$\begin{aligned} |\Gamma_2| - |\Gamma_1| &= \frac{1}{2} \left| -a_3 + \frac{3}{2} a_2^2 \right| - \frac{1}{2} |-a_2| \\ &= \frac{1}{2} \left| (-a_3 + a_2^2) + \frac{1}{2} a_2^2 \right| - \frac{1}{2} |a_2| \\ &\leq \frac{1}{2} |a_3 - a_2^2| + \frac{1}{4} |a_2|^2 - \frac{1}{2} |a_2| \\ &\leq \frac{1}{6} (1 - |a_2|^2) + \frac{1}{4} |a_2|^2 - \frac{1}{2} |a_2| \\ &\leq \frac{1}{6}. \end{aligned}$$ The inequality is sharp for the function $f_0(z) = z + \frac{1}{3}z^3 + \cdots$ given in Lemma 1. The lower bound of the inequality is equivalent with (15) $$\left| -a_3 + \frac{3}{2}a_2^2 \right| \ge |a_2| - \sqrt{\frac{2}{5}}.$$ If $0 \le |a_2| < \sqrt{2/5}$ , then the inequality (15) obviously holds. Now, let $\sqrt{2/5} \le |a_2| \le 1$ . Then, using Lemma 2, since $$\left| -a_3 + \frac{3}{2}a_2^2 \right| \ge \frac{1}{2}|a_2|^2 - |-a_3 + a_2^2|$$ $$\ge \frac{1}{2}|a_2|^2 - \frac{1}{3}(1 - |a_2|^2)$$ $$= \frac{5}{6}|a_2|^2 - \frac{1}{3},$$ we conclude that to prove (15) it is enough to show that $$\frac{5}{6}|a_2|^2 - \frac{1}{3} \ge |a_2| - \sqrt{\frac{2}{5}}.$$ The last inequality is equivalent to $$\left(|a_2| - \sqrt{\frac{2}{5}}\right) \left(\frac{5}{6} \left(|a_2| + \sqrt{\frac{2}{5}}\right) - 1\right) \ge 0,$$ which is true since by assumption $|a_2| - \sqrt{\frac{2}{5}} \ge 0$ and $\frac{5}{6} \left( |a_2| + \sqrt{\frac{2}{5}} \right) - 1 \ge \frac{5}{6} \cdot 2\sqrt{\frac{2}{5}} - 1 = \sqrt{\frac{10}{9}} - 1 > 0$ . This proves the inequality (15) and the lower bound of (i). (ii) In the proof of Theorem 3 we obtained that $\Gamma_2 = \frac{1}{2} \left( -a_3 + \frac{3}{2} a_2^2 \right)$ and $\Gamma_3 = \frac{1}{2} \left( -a_4 + 4a_2a_3 - \frac{10}{3}a_2^3 \right)$ . Since $|a_2| \le 1$ , so also $|a_2| \le \frac{5}{4}$ and $-1 \le -\frac{4}{5}|a_2|$ . Therefore, $$|\Gamma_3| - |\Gamma_2| \le |\Gamma_3| - \frac{4}{5}|a_2||\Gamma_2|$$ $$\le \left|\Gamma_3 + \frac{4}{5}a_2\Gamma_2\right|$$ $$= \frac{1}{2}\left|a_4 - \frac{16}{5}a_2a_3 + \frac{32}{15}a_2^3\right|.$$ Using (14) and Lemma 2 (case $D_8$ ), after some calculations, we get $$|\Gamma_3| - |\Gamma_2| \le \frac{1}{12} \left| c_3 - \frac{7}{5} c_1 c_2 - \frac{2}{5} c_1^3 \right| \le \frac{1}{12} \cdot \frac{8}{5} \sqrt{\frac{2}{5}} = \frac{2\sqrt{10}}{75}.$$ For showing the sharpness, it is enough to consider the function $f_{\lambda}$ from Lemma 1 with $\lambda = \sqrt{2/5}$ , i.e., $$f_{\sqrt{2/5}}(z) = z + \sqrt{\frac{2}{5}}z^2 + \frac{3}{5}z^3 + \frac{4}{5}\sqrt{\frac{2}{5}}z^4 \cdots,$$ such that $$a_2 = \sqrt{\frac{2}{5}}$$ , $a_3 = \frac{3}{5}$ , and $a_4 = \frac{4}{5}\sqrt{\frac{2}{5}}$ , and further, $$\Gamma_1 = -\frac{\sqrt{10}}{10} \ , \quad \Gamma_2 = 0 \quad \mbox{ and } \quad \Gamma_3 = \frac{2\sqrt{10}}{75} \ .$$ The above means that all inequalities in Theorem 5 are sharp.