Second Hankel determinant of logarithmic coefficients of certain analytic functions

Lemma 2.1. [15, 16] If $p \in \mathcal{P}$ is of the form (2.5) with $c_1 \geq 0$ , then <span id="page-2-1"></span> $$(2.2) c_1 = 2p_1,$$ <span id="page-2-2"></span> $$(2.3) c_2 = 2p_1^2 + 2(1 - p_1^2)p_2,$$ and <span id="page-2-3"></span> $$(2.4) c_3 = 2p_1^3 + 4(1 - p_1^2)p_1p_2 - 2(1 - p_1^2)p_1p_2^2 + 2(1 - p_1^2)(1 - |p_2|^2)p_3$$ for some $p_1 \in [0,1]$ and $p_2, p_3 \in \overline{\mathbb{D}} := \{z \in \mathbb{C} : |z| \le 1\}.$ For $p_1 \in \mathbb{T} := \{z \in \mathbb{C} : |z| = 1\}$ , there is a unique function $p \in \mathcal{P}$ with $c_1$ as in (2.2), namely <span id="page-2-0"></span> $$p(z) = \frac{1 + p_1 z}{1 - p_1 z}, \quad z \in \mathbb{D}.$$ For $p_1 \in \mathbb{D}$ and $p_2 \in \mathbb{T}$ , there is a unique function $p \in \mathcal{P}$ with $c_1$ and $c_2$ as in (2.2) and (2.3), namely (2.5) $$p(z) = \frac{1 + (p_1 + \overline{p_1}p_2)z + p_2z^2}{1 - (p_1 - \overline{p_1}p_2)z - p_2z^2}.$$ For $p_1, p_2 \in \mathbb{D}$ and $p_3 \in \mathbb{T}$ , there is unique function $p \in \mathcal{P}$ with $c_1, c_2,$ and $c_3$ as in (2.2)-(2.4), namely, $$p(z) = \frac{1 + (\overline{p_2}p_3 + \overline{p_1}p_2 + p_1)z + (\overline{p_1}p_3 + p_1\overline{p_2}p_3 + p_2)z^2 + p_3z^3}{1 + (\overline{p_2}p_3 + \overline{p_1}p_2 - p_1)z + (\overline{p_1}p_3 - p_1\overline{p_2}p_3 - p_2)z^2 - p_3z^3} \quad z \in \mathbb{D}.$$ Next we recall the following well-known result due to Choi et al. [6]. Lemma 2.2 plays an important role in the proof of our main results.

Lemma 2.2. [6] Let A, B, C be real numbers and $$Y(A, B, C) := \max_{z \in \overline{\mathbb{D}}} (|A + Bz + Cz^{2}| + 1 - |z|^{2}).$$ (i) If AC > 0, then $$Y(A, B, C) = \begin{cases} |A| + |B| + |C|, & \text{for } |B| \ge 2(1 - |C|), \\ 1 + |A| + \frac{B^2}{4(1 - |C|)}, & \text{for } |B| < 2(1 - |C|). \end{cases}$$ (ii) If AC < 0, then $$Y(A,B,C) = \begin{cases} 1 - |A| + \frac{B^2}{4(1-|C|)}, & -4AC(C^{-2}-1) \le B^2 \land |B| < 2(1-|C|), \\ 1 + |A| + \frac{B^2}{4(1+|C|)}, & B^2 < \min\{4(1+|C|)^2, -4AC(C^{-2}-1)\}, \\ R(A,B,C), & otherwise, \end{cases}$$ where $$R(A, B, C) = \begin{cases} |A| + |B| + |C|, & |C|(|B| + 4|A|) \le |AB|, \\ -|A| + |B| + |C|, & |AB| \le |C|(|B| - 4|A|), \\ (|A| + |C|)\sqrt{1 - \frac{B^2}{4AC}}, & otherwise. \end{cases}$$

Theorem 3.1. Let $-\pi/2 < \beta < \pi/2$ and $0 \le \alpha < 1$ . For every $f \in \mathcal{S}_{\beta}(\alpha)$ of the form (1.1), we have <span id="page-3-0"></span>(3.1) $$|H_{2,1}(F_f/2)| \le \frac{(1-\alpha)^2 \cos^2 \beta}{4}.$$ Equality in (3.1) holds only for the rotation of the function $$f_1(z) = \frac{z}{(1-z^2)^{(1-\alpha)\cos\beta e^{i\beta}}}$$

Theorem 3.4. Let $0 < \nu \le 1$ . If $f \in \mathcal{G}(\nu)$ given by (1.1), then $$|H_{2,1}(F_f/2)| \le \frac{\nu^2(\nu^2 + 12\nu - 44)}{192(\nu^2 + 8\nu - 32)}.$$ The inequality is sharp.

Theorem 3.5. Let $f \in \mathcal{F}_0(\lambda)$ , for $1/2 \le \lambda \le 1$ , given by (1.1). Then $$|H_{2,1}(F_f/2)| \le \frac{(2\lambda+1)^2(12\lambda^2-60\lambda-165)}{576(4\lambda^2-12\lambda-39)}$$ The inequality is sharp.