Toeplitz determinants of Logarithmic coefficients for Starlike and Convex functions

Lemma 1.1. [21] If $\omega(z) = \sum_{n=1}^{\infty} c_n z^n \in \Omega$ and $(\mu, \nu) \in \bigcup_{i=1}^{3} D_i$ , then $$|c_3 + \mu c_1 c_2 + \nu c_1^3| \le |\nu|,$$ where $$D_1 = \left\{ (\mu, \nu) : |\mu| \le 2, \ \nu \ge 1 \right\}, \ D_2 = \left\{ (\mu, \nu) : 2 \le |\mu| \le 4, \ \nu \ge \frac{1}{12} (\mu^2 + 8) \right\},$$ and $$D_3 = \left\{ (\mu, \nu) : |\mu| \ge 4, \ \nu \ge \frac{2}{3} (|\mu| - 1) \right\}.$$

Lemma 1.2. [9, Theorem 1] Let $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathcal{P}$ and $\mu \in \mathbb{C}$ . Then $$|p_n - \mu p_k p_{n-k}| \le 2 \max\{1, |2\mu - 1|\}, \quad 1 \le k \le n - 1.$$ The inequality is sharp for the function p(z) = (1+z)/(1-z) or its rotation when $|2\mu - 1| \ge 1$ . In case of $|2\mu - 1| < 1$ , the inequality is sharp for $p(z) = (1+z^n)/(1-z^n)$ or its rotations.

Theorem 2.1. Let $\varphi(z) = 1 + B_1 z + B_2 z^2 + B_3 z^3 + \cdots$ and $f \in S^*(\varphi)$ . If $|B_2| \ge B_1$ , then $$|\gamma_1^2 - \gamma_2^2| \le \frac{B_1^2}{4} + \frac{B_2^2}{16}.$$ The estimate is sharp. Proof. Let $f \in \mathcal{S}^*(\varphi)$ be of the form (1.1). Then there exists a Schwarz function, say $\omega(z) = \sum_{n=1}^{\infty} c_n z^n$ such that $$\frac{zf'(z)}{f(z)} = \varphi(\omega(z)), \quad z \in \mathbb{D}. \tag{2.1}$$ From the Taylor series expansions of f and $\varphi$ , we obtain $$\frac{zf'(z)}{f(z)} = 1 + a_2 z + (-a_2^2 + 2a_3)z^2 + (a_2^3 - 3a_2 a_3 + 3a_4)z^3 + \cdots$$ (2.2) and $$\varphi(\omega(z)) = 1 + B_1 c_1 z + (B_2 c_1^2 + B_1 c_2) z^2 + (B_3 c_1^3 + 2B_2 c_1 c_2 + B_1 c_3) z^3 + \cdots$$ (2.3) By comparing the same powers in (2.1) using (2.2) and (2.3), coefficients $a_2$ , $a_3$ and $a_4$ can be expressed as $$a_2 = B_1 c_1, \ a_3 = \frac{1}{2} (B_1^2 c_1^2 + B_2 c_1^2 + B_1 c_2)$$ (2.4) and $$a_4 = \frac{1}{48}((8B_1^3 + 24B_1B_2 + 16B_3)c_1^3 + (24B_1^2 + 32B_2)c_1c_2 + 16B_1c_3). \tag{2.5}$$ Further, applying $|c_n| \leq 1$ , we get $$|a_2| \le B_1. \tag{2.6}$$ Ali et al. [4, Theorem 1] established the bound of Fekete-Szegö functional for p-valent functions, which for p = 1 gives $$|a_3 - \lambda a_2^2| \le \begin{cases} \frac{1}{2} (B_1^2 + B_2 - 2\lambda B_1^2), & \text{if } 2\lambda B_1^2 \le B_1^2 + B_2 - B_1; \\ \frac{1}{2} B_1, & \text{if } B_1^2 + B_2 - B_1 \le 2\lambda B_1^2 \le B_1^2 + B_2 + B_1; \\ \frac{1}{2} (-B_1^2 - B_2 + 2\lambda B_1^2), & \text{if } 2\lambda B_1^2 \ge B_1^2 + B_2 + B_1. \end{cases}$$ Since $|B_2| \geq B_1$ , hence the above inequality directly yields $$|a_3 - \frac{1}{2}a_2^2| \le \frac{|B_2|}{2}. (2.7)$$ From (1.5), we obtain $$\left|\gamma_1^2 - \gamma_2^2\right| = \left|\frac{1}{4}\left(a_2^2 - \left(a_3 - \frac{a_2^2}{2}\right)^2\right)\right| \le \frac{1}{4}\left(|a_2|^2 + \left|a_3 - \frac{a_2^2}{2}\right|^2\right). \tag{2.8}$$ The required bound follows from (2.8) by using the bounds of $|a_2|$ and $|a_3 - (a_2^2)/2|$ from (2.6) and (2.7) respectively. To show the sharpness of the bound, consider the analytic function $k_{\varphi}: \mathbb{D} \to \mathbb{C}$ given by $$k_{\varphi}(z) = z \exp \int_{0}^{z} \frac{\varphi(it) - 1}{t} dt = z + iB_{1}z^{2} - \frac{1}{2}(B_{1}^{2} + B_{2})z^{3} + \cdots$$ (2.9) Clearly, $k_{\varphi} \in \mathcal{S}^*(\varphi)$ and for this function, a simple computation gives $$|\gamma_1^2 - \gamma_2^2| = \frac{4B_1^2 + B_2^2}{16}.$$ which shows that the bound is sharp.

Theorem 2.2. Let $\varphi(z) = 1 + B_1 z + B_2 z^2 + B_3 z^3 + \cdots$ and $f \in \mathcal{C}(\varphi)$ . If $|B_2 + \frac{1}{4}B_1^2| \ge B_1$ , then $$|\gamma_1^2 - \gamma_2^2| \le \frac{B_1^2}{16} + \frac{1}{144} \left(B_2 + \frac{B_1^2}{4}\right)^2.$$ (2.10) The estimate is sharp. Proof. Suppose $f \in \mathcal{C}(\varphi)$ be of the form (1.1). Then there exists a Schwarz function $\omega(z) = \sum_{n=1}^{\infty} c_n z^n$ such that $$1 + \frac{zf''(z)}{f'(z)} = \varphi(\omega(z)), \quad z \in \mathbb{D}.$$ After comparing the coefficients of identical powers of z with the Taylor series expansion of f, $\varphi$ and $\omega$ in the above equation, the coefficients $a_2$ and $a_3$ can be expressed as $$a_2 = \frac{B_1 c_1}{2}, \quad a_3 = \frac{1}{6} (B_1^2 c_1^2 + B_2 c_1^2 + B_1 c_2)$$ (2.11) and $$a_4 = \frac{1}{12}((4B_1^3 + 3B_1B_2 + B_3)c_1^3 + (3B_1^2 + 2B_2)c_1c_2 + B_1c_3). \tag{2.12}$$ Applying the bound $|c_n| \leq 1$ , we obtain $$|a_2| \le \frac{B_1}{2}.\tag{2.13}$$ For $f \in \mathcal{C}(\varphi)$ , Ma and Minda [17, Theorem 3] established the following bound $$|a_3 - \lambda a_2^2| \le \begin{cases} \frac{1}{6} (B_2 - \frac{3}{2}\lambda B_1^2 + B_1^2), & \text{if } 3\lambda B_1^2 \le 2(B_1^2 + B_2 - B_1); \\ \frac{1}{6} B_1, & \text{if } 2(B_1^2 + B_2 - B_1) \le 3\lambda B_1^2 \le 2(B_1^2 + B_2 + B_1); \\ \frac{1}{6} (-B_2 + \frac{3}{2}\lambda B_1^2 - B_1^2), & \text{if } 2(B_1^2 + B_2 + B_1) \le 3\lambda B_1^2. \end{cases}$$ Since $|B_2 + \frac{1}{4}B_1^2| \ge B_1$ holds, the above inequality directly gives $$|a_3 - \frac{1}{2}a_2^2| \le \frac{1}{6}|B_2 + \frac{1}{4}B_1^2|. \tag{2.14}$$ Using the bounds of $|a_2|$ and $|a_3-(a_2^2)/2|$ for $f \in \mathcal{C}(\varphi)$ given in (2.13) and (2.14), respectively, we obtain $$|\gamma_1^2 - \gamma_2^2| \le \frac{1}{4} \left( |a_2|^2 + \left| a_3 - \frac{a_2^2}{2} \right|^2 \right) \le \frac{B_1^2}{16} + \frac{1}{144} \left( B_2 + \frac{B_1^2}{4} \right)^2.$$ The equality case in (2.10) holds for the function $h_{\varphi}$ given by $$1 + \frac{zh_{\varphi}''(z)}{h_{\varphi}'(z)} = \varphi(iz). \tag{2.15}$$ Clearly, $h_{\varphi} \in \mathcal{C}(\varphi)$ and for this function, we have $$\gamma_1 = \frac{iB_1}{4}$$ and $\gamma_2 = -\frac{1}{12}(B_2 + \frac{B_1^2}{4}),$ which shows that the bound in (2.10) is sharp.

Theorem 2.3. Let $\varphi(z) = 1 + B_1 z + B_2 z^2 + B_3 z^3 + \cdots$ and $f \in \mathcal{S}^*(\varphi)$ . If $|B_2| \ge B_1$ and $(\mu_1, \nu_1) \in \bigcup_{i=1}^3 D_i$ hold, then $$|\gamma_2^2 - \gamma_3^2| \le \frac{1}{144} (9B_2^2 + 4B_3^2),$$ where $\mu_1 = 2B_2/B_1$ and $\nu_1 = B_3/B_1$ . The bound is sharp. Proof. Suppose $f \in \mathcal{S}^*(\varphi)$ be of the form (1.1). Then from (1.5), we have $$|\gamma_2^2 - \gamma_3^2| = \frac{1}{4} \left| \left( a_3 - \frac{a_2^2}{2} \right)^2 - \left( \frac{a_2^3}{3} - a_2 a_3 + a_4 \right)^2 \right|$$ $$\leq \frac{1}{4} \left( \left| a_3 - \frac{a_2^2}{2} \right|^2 + \left| \frac{a_2^3}{3} - a_2 a_3 + a_4 \right|^2 \right).$$ (2.16) From (2.4) and (2.5) for $f \in \mathcal{S}^*(\varphi)$ , using the values of $a_2$ , $a_3$ and $a_4$ , we obtain $$\left| \frac{a_2^3}{3} - a_2 a_3 + a_4 \right| = \frac{B_1}{3} |c_3 + \mu_1 c_1 c_2 + \nu_1 c_1^3|,$$ where $\mu_1 = 2B_2/B_1$ and $\nu_1 = B_3/B_1$ . Since $|B_2| \ge B_1$ holds, therefore $(\mu_1, \nu_1)$ is a member of either $D_1$ , $D_2$ or $D_3$ . Thus, from Lemma 1.1, we get $$\left| \frac{a_2^3}{3} - a_2 a_3 + a_4 \right| \le \frac{|B_3|}{3}. \tag{2.17}$$ Using the bounds from (2.7) and (2.17) in the inequality (2.16), the required bound is obtained The sharpness of the bound can be seen by the function $k_{\varphi}$ given by (2.9). As for this function, we have $\gamma_2 = -B_2/4$ , $\gamma_3 = -iB_3/6$ and $$\gamma_2^2 - \gamma_3^2 = \frac{1}{144} (9B_2^2 + 4B_3^2),$$ which proves the sharpness.

Theorem 2.4. Let $\varphi(z) = 1 + B_1 z + B_2 z^2 + B_3 z^3 + \cdots$ and $f \in \mathcal{C}(\varphi)$ . If $|B_2 + \frac{1}{4} B_1^2| \ge B_1$ and $(\mu_2, \nu_2) \in \bigcup_{i=1}^3 D_i$ holds, then $$|\gamma_2^2 - \gamma_3^2| \le \frac{B_1^4 + 8B_1^2 B_2 + 16B_2^2 + B_1^2 B_2^2 + 4B_1 B_2 B_3 + 4B_3^2}{2304}$$ where $\mu_2 = (B_1^2 + 4B_2)/(2B_1)$ and $\nu_2 = (B_1B_2 + 2B_3)/(2B_1)$ . The bound is sharp. Proof. In view of the equations (2.11) and (2.12) for $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}(\varphi)$ , we have $$\left| \frac{a_2^3}{3} - a_2 a_3 + a_4 \right| = \frac{B_1}{12} \left| c_3 + \mu_2 c_1 c_2 + \nu_2 c_1^3 \right|.$$ As by the hypothesis $|B_2 + \frac{1}{4}B_1^2| \ge B_1$ holds, therefore $(\mu_2, \nu_2)$ belongs to either $D_1$ , $D_2$ or $D_3$ . Hence, from Lemma 1.1, we obtain $$\left| \frac{a_2^3}{3} - a_2 a_3 + a_4 \right| \le \frac{|B_1 B_2 + 2B_3|}{24}. \tag{2.18}$$ Applying the bound from (2.14) and (2.18) in the inequality (2.16), we get $$|\gamma_2^2 - \gamma_3^2| \le \frac{B_1^4 + 8B_1^2B_2 + 16B_2^2 + B_1^2B_2^2 + 4B_1B_2B_3 + 4B_3^2}{2304}.$$ It is a simple exercise to check that the equality case holds for the function $h_{\varphi} \in \mathcal{C}(\varphi)$ given by (2.15).

Corollary 2.5. Let $f \in \mathcal{S}^*[A, B]$ be of the form (1.1), where $-1 \leq B < A \leq 1$ . (i) If $$|B^2 - AB| \ge A - B$$ , then $$|\gamma_1^2 - \gamma_2^2| \le \frac{(A-B)^2(4+B^2)}{16}.$$ (ii) If $$|B^2 - AB| \ge A - B$$ , and $(\mu_1, \nu_1) \in \bigcup_{i=1}^3 D_i$ , then $$|\gamma_2^2 - \gamma_3^2| \le \frac{(A-B)^2 B^2 (4B^2 + 9)}{144},$$ where $\mu_1 = -2B$ and $\nu_1 = B^2$ .

Corollary 2.6. Let $f \in C[A, B]$ be of the form (1.1), where $-1 \le B < A \le 1$ . (i) If $$|A^2 - 6AB + 5B^2| \ge 4(A - B)$$ , then $$|\gamma_1^2 - \gamma_2^2| \le \frac{(A-B)^2(A^2 + 25B^2 - 10AB + 144)}{2304}.$$ (ii) If $$|A^2 - 6AB + 5B^2| \ge 4(A - B)$$ and $(\mu_2, \nu_2) \in \bigcup_{i=1}^3 D_i$ , then $$|\gamma_2^2 - \gamma_3^2| \le \frac{(A-B)^2(A^2(B^2+1) + B^2(9B^2+25) - 2AB(3B^2+5))}{2304},$$ where $\mu_2 = (A - 5B)/2$ and $\nu_2 = (B(3B - A))/2$ . By taking A = 1 − 2α, 0 ≤ α < 1 and B = −1, the following results follow from Corollary [2.5](#page-5-2) and Corollary [2.6.](#page-5-3)

Corollary 2.7. If f ∈ S^∗ (α), 0 ≤ α < 1, then $$|\gamma_1^2 - \gamma_2^2| \leq \frac{5}{16}(2 - 2\alpha)^2 \quad and \quad |\gamma_2^2 - \gamma_3^2| \leq \frac{13}{144}(2 - 2\alpha)^2.$$

Corollary 2.8. If f ∈ C(α), 0 ≤ α < 1, then $$|\gamma_1^2 - \gamma_2^2| \leq \frac{(\alpha - 1)^2(\alpha^2 - 6\alpha + 45)}{144} \ \ \text{and} \ \ |\gamma_2^2 - \gamma_3^2| \leq \frac{(\alpha - 1)^2(2\alpha^2 - 10\alpha + 13)}{144}.$$ In particular, for α = 0, Corollary [2.7](#page-6-0) and Corollary [2.8](#page-6-1) give the bounds for the classes S ^∗ and C respectively.

Corollary 2.9. If f ∈ S^∗ , then $$|\gamma_1^2 - \gamma_2^2| \le \frac{5}{4}$$ and $|\gamma_2^2 - \gamma_3^2| \le \frac{13}{36}$ .

Corollary 2.10. If f ∈ C, then $$|\gamma_1^2 - \gamma_2^2| \le \frac{5}{16} \quad and \quad |\gamma_2^2 - \gamma_3^2| \le \frac{13}{144}.$$

Lemma 3.1. [\[4,](#page-11-12) Theorem 1] Let ϕ(z) = 1 + B1z + B2z ² + B3z ³ + · · · , and $$q_1 = \frac{3B_1^2 + 4B_2}{2B_1}, \quad q_2 = \frac{B_1^3 + 3B_1B_2 + 2B_3}{2B_1}.$$ If f ∈ S^∗ (ϕ) is of the form [\(1.1\)](#page-0-1) such that (q1, q2) ∈ ∪³ ⁱ=1Di, then $$|a_4| \leq \frac{B_1^3 + 3B_1B_2 + 2B_3}{6}.$$ The bound is sharp.

Theorem 3.1. Let ϕ(z) = 1 + B1z + B2z ² + B3z ³ + · · · such that $$6B_1^2 \le B_1(3B_1^2 + 2B_2) \le B_1^2 + 2B_1^4 + 3B_1^2B_2 + 3B_2^2 - 2B_1B_3,$$ and $$q_1 = \frac{3B_1^2 + 4B_2}{2B_1}, \quad q_2 = \frac{B_1^3 + 3B_1B_2 + 2B_3}{2B_1}.$$ If f ∈ S^∗ (ϕ) and (q1, q2) ∈ ∪³ ⁱ=1Di, then $$|T_{3,2}(f)| \le \left(B_1 + \frac{B_1^3 + 3B_1B_2 + 2B_3}{6}\right) \left(B_1^2 + \frac{B_1^4}{3} + \frac{B_1^2B_2}{2} + \frac{B_2^2}{2} - \frac{B_1B_3}{3}\right).$$ The bound is sharp. Proof. Let $f \in \mathcal{S}^*(\varphi)$ be of the form (1.1). Then from (2.1), we have $$zf'(z) = f(z)\varphi(\omega(z)), \quad z \in \mathbb{D}.$$ Corresponding to the Schwarz function $\omega$ , there exists $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathcal{P}$ such that w(z) = (p(z) - 1)/(p(z) + 1). The comparison of identical powers of z using the power series expansions of f, $\varphi$ and p yield $$a_2 = \frac{B_1 p_1}{2}, \ a_3 = \frac{1}{8} (B_1^2 - B_1 + B_2) p_1^2 + 2B_1 p_2)$$ and $$a_4 = \frac{1}{48} \left( (B_1^3 - 3B_1^2 + 2B_1 - 4B_2 + 3B_1B_2 + 2B_3)p_1^3 + (6B_1^2 - 8B_1 + 8B_2)p_1p_2 + 8B_1p_3 \right).$$ Using these values of $a_2$ , $a_3$ and $a_4$ in terms of $p_1$ , $p_2$ and $p_3$ , it follows that $$|a_2^2 - 2a_3^2 + a_2 a_4| = \left| \frac{B_1^2 p_1^2}{4} - \frac{(B_1^2 - 3B_1^3 + 2B_1^4 - 2B_1 B_2 + 3B_1^2 B_2 + 3B_2^2 - 2B_1 B_3) p_1^4}{96} - \frac{B_1 (3B_1^2 - 2B_1 + 2B_2) p_1^2 p_2}{48} - \frac{B_1^2}{8} p_2^2 + \frac{B_1^2}{12} p_1 p_3 \right|.$$ Keeping in mind that $B_1^2 + 2B_1^4 + 3B_1^2B_2 + 3B_2^2 - 2B_1B_3 - B_1(3B_1^2 + 2B_2) \ge 0$ and by applying the bound $|p_n| \le 2$ , $n \in \mathbb{N}$ (see [8, Page- 41]), we get $$\begin{split} |a_2^2 - 2a_3^2 + a_2 a_4| & \leq \frac{3B_1^2}{2} + \frac{\left(B_1^2 - 3B_1^3 + 2B_1^4 - 2B_1B_2 + 3B_1^2B_2 + 3B_2^2 - 2B_1B_3\right)}{6} \\ & + \frac{B_1^2}{6} \bigg| p_3 - \bigg(\frac{3B_1^2 - 2B_1 + 2B_2}{4B_1}\bigg) p_1 p_2 \bigg|. \end{split}$$ Since $3B_1^2 + 6B_2 \ge 6B_1$ , therefore from Lemma 1.2, we obtain $$|a_2^2 - 2a_3^2 + a_2 a_4| \le B_1^2 + \frac{B_1^4}{3} + \frac{B_1^2 B_2}{2} + \frac{B_2^2}{2} - \frac{B_1 B_3}{3}.$$ (3.1) Further, we have $|a_2 - a_4| \le |a_2| + |a_4|$ . Using the bounds of $|a_2|$ and $|a_4|$ from (2.6) and Lemma 3.1 respectively, we get $$|a_2 - a_4| \le B_1 + \frac{B_1^3 + 3B_1B_2 + 2B_3}{6}$$ From (1.3), a simple computation reveals that $$|T_{3,2}(f)| = |(a_2 - a_4)(a_2^2 - 2a_3^2 + a_2 a_4)|. (3.2)$$ The required estimated is determined by putting the bounds given in (3.1) and (3.2) in the above equation. The function $k_{\varphi}$ defined by (2.9) plays the role of extremal functions. As for this function, we have $$a_2 = iB_1$$ , $a_3 = -\frac{1}{2}(B_1^2 + B_2)$ , $a_4 = -\frac{i}{6}(B_1^3 + 3B_1B_2 + 2B_3)$ and $$|T_{3,2}(k_{\phi})| = \left(B_1 + \frac{B_1^3 + 3B_1B_2 + 2B_3}{6}\right) \left(B_1^2 + \frac{B_1^4}{3} + \frac{B_1^2B_2}{2} + \frac{B_2^2}{2} - \frac{B_1B_3}{3}\right)$$ proving the sharpness.

Theorem 3.2. Let $\varphi(z) = 1 + B_1 z + B_2 z^2 + B_3 z^3 + \cdots$ such that $$16B_1^2 - 4B_1B_2 \le 7B_1^3 \le 5B_1^4 + 2B_1^2 - 4B_1B_2 + 7B_1^2B_2 + 8B_2^2 - 6B_1B_3, \tag{3.3}$$ and $$q_1 = \frac{3B_1^2 + 4B_2}{2B_1}, \quad q_2 = \frac{B_1^3 + 3B_1B_2 + 2B_3}{2B_1}.$$ If $f \in \mathcal{C}(\varphi)$ and $(q_1, q_2) \in \bigcup_{i=1}^3 D_i$ , then $$|T_{3,2}(f)| \le \frac{1}{144} \left( \frac{B_1}{2} + \frac{B_1^3 + 3B_1B_2 + 2B_3}{24} \right) (5B_1^4 + 36B_1^2 + 7B_1^2B_2 + 8B_2^2 - 6B_1B_3).$$ The bound is sharp. Proof. Suppose $f \in \mathcal{C}(\varphi)$ be of the form (1.1), then we have $$f'(z) + zf''(z) = f'(z)\varphi(\omega(z)).$$ Corresponding to the Schwarz function $\omega(z) = \sum_{n=1}^{\infty} c_n z^n$ , there exists $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathcal{P}$ such that w(z) = (p(z) - 1)/(p(z) + 1). The comparison of same powers of z in the above equation after the series expansions yield that $$a_2 = \frac{B_1 p_1}{4}, \quad a_3 = \frac{1}{24}((B_1^2 - B_1 + B_2)p_1^2 + 2B_1 p_2)$$ and $$a_4 = \frac{1}{192} \left( (B_1^3 - 3B_1^2 + 2B_1 - 4B_2 + 3B_1B_2 + 2B_3)p_1^3 + (6B_1^2 + 8B_2 - 8B_1)p_1p_2 + 8B_1p_3 \right). \tag{3.4}$$ Using these expressions for $a_2$ , $a_3$ and $a_4$ in terms of the coefficients $p_1$ , $p_2$ and $p_3$ , a simple computation gives $$|a_2^2 - 2a_3^2 + a_2 a_4| = \left| \frac{1}{2304} \left( (2B_1^2 - 7B_1^3 + 5B_1^4 - 4B_1B_2 + 7B_1^2B_2 + 8B_2^2 - 6B_1B_3) p_1^4 + 32B_1^2 p_2^2 - 144B_1^2 p_1^2 - 24B_1^2 p_1 \left( p_3 - \frac{(14B_1^3 - 8B_1^2 + 8B_1B_2)}{24B_1^2} p_1 p_2 \right) \right) \right|.$$ In view of the hypothesis $2B_1^2 + 5B_1^4 - 4B_1B_2 + 7B_1^2B_2 + 8B_2^2 - 6B_1B_3 \ge 7B_1^3$ and by applying the bound $|p_n| \le 2 \ (n \in \mathbb{N})$ , we get $$|a_2^2 - 2a_3^2 + a_2 a_4| \le \frac{1}{2304} \left( 16(2B_1^2 - 7B_1^3 + 5B_1^4 - 4B_1B_2 + 7B_1^2B_2 + 8B_2^2 - 6B_1B_3) + 128B_1^2 + 576B_1^2 + 48B_1^2 \left( \left| p_3 - \frac{(14B_1^3 - 8B_1^2 + 8B_1B_2)}{24B_1^2} p_1 p_2 \right| \right) \right).$$ Since $7B_1^2 + 4B_2 \ge 16B_1$ holds, therefore from Lemma 3.1, it follows that $$|a_2^2 - 2a_3^2 + a_2 a_4| \le \frac{1}{144} (36B_1^2 + 5B_1^4 + 7B_1^2 B_2 + 8B_2^2 - 6B_1 B_3). \tag{3.5}$$ Now, we only need to maximize $|a_2 - a_4|$ for $f \in \mathcal{C}(\varphi)$ . By the one to one correspondence between the class $\mathcal{P}$ and the class of Schwarz functions, the coefficients $a_4$ in (3.4) can be expressed as $$a_4 = \frac{1}{12}B_1(c_3 + q_1c_1c_2 + q_2c_1^3),$$ where $q_1 = (3B_1^2 + 4B_2)/(2B_1)$ and $q_2 = (B_1^3 + 3B_1B_2 + 2B_3)/(2B_1)$ . As by the hypothesis $(q_1, q_2) \in \bigcup_{i=1}^3 D_i$ , from Lemma 1.1, we obtain $$|a_4| \le \frac{B_1^3 + 3B_1B_2 + 2B_3}{24}. (3.6)$$ Employing the bounds of $|a_2|$ and $|a_4|$ from (2.13) and (3.6) respectively, we get $$|a_2 - a_4| \le |a_2| + |a_4| \le \frac{B_1}{2} + \frac{B_1^3 + 3B_1B_2 + 2B_3}{24}.$$ (3.7) Thus, applying the bounds of $|a_2^2 - 2a_3^2 + a_2a_4|$ and $|a_2 - a_4|$ from (3.5) and (3.7) respectively in (3.2), we get the desired result. The result is sharp for the function $h_{\varphi}$ defined in (2.15). As for this function, we have $a_2=iB_1/2$ , $a_3=-(B_1^2+B_2)/6$ , $a_4=-i(B_1^3+3B_1B_2+2B_3)/24$ and $$|T_{3,2}(f)| = \frac{1}{144} \left( \frac{B_1}{2} + \frac{B_1^3 + 3B_1B_2 + 2B_3}{24} \right) \left( 5B_1^4 + 36B_1^2 + 7B_1^2B_2 + 8B_2^2 - 6B_1B_3 \right)$$ proving the sharpness of the bound.

Corollary 3.3. For $-1 \le B < A \le 1$ , let $$6(A-B)^2 \le (3A-5B)(A-B)^2 \le (A-B)^2(2A^2-7AB+6B^2+1)$$ and $$q_1 = \frac{3A - 7B}{2}, \quad q_2 = \frac{A^2 - 5AB + 6B^2}{2}.$$ If $f \in \mathcal{S}^*[A, B]$ and $(q_1, q_2) \in \bigcup_{i=1}^3 D_i$ , then $$|T_{3,2}(f)| \le \frac{1}{36}(A-B)^2(2A^2-7AB+6B^2+6)(A^3+6A-6B-6A^2B+11AB^2-6B^3).$$ The estimates is sharp.

Corollary 3.4. For $-1 \le B < A \le 1$ , let $$4(A-B)^{2}(4+B) \le 7(A-B)^{3} \le (A-B)^{2}(2+5A^{2}+4B-17AB+14B^{2})$$ and $$q_1 = \frac{3A - 7B}{2}, \quad q_2 = \frac{A^2 - 5AB + 6B^2}{2}.$$ If $f \in \mathcal{C}[A, B]$ and $(q_1, q_2) \in \bigcup_{i=1}^3 D_i$ , then $$|T_{3,2}(f)| \le \frac{1}{3456}(A-B)^2(5A^2-17AB+14B^2+36)(A^3+12A-12B-6A^2B+11AB^2-6B^3).$$ The estimates is sharp. When $A = 1 - 2\alpha$ and B = -1, the conditions in Corollary 3.3 and 3.4 are true and $(q_1, q_2) \in D_3$ for $\alpha \in [0, 1/7]$ . Thus, we obtain the following bounds for the classes $\mathcal{S}^*(\alpha)$ and $\mathcal{C}(\alpha)$ .

Corollary 3.5. If $f \in S^*(\alpha)$ , then $$|T_{3,2}(f)| \le \frac{4}{9}(1-\alpha)^3(16\alpha^4 - 100\alpha^3 + 268\alpha^2 - 345\alpha + 189)$$ for $\alpha \in [0, 1/7]$ . The bound is sharp.

Corollary 3.6. If $f \in C(\alpha)$ , then $$|T_{3,2}(f)| \le \frac{1}{108} (1-\alpha)^3 (20\alpha^4 - 124\alpha^3 + 381\alpha^2 - 576\alpha + 432)$$ for $\alpha \in [0, 1/7]$ . The bound is sharp. Remark 3.1. In particular, when $\alpha = 0$ , the following bounds for the classes $\mathcal{S}^*$ and $\mathcal{C}$ follow as special case proved in [3]. - (i) If $f \in S^*$ , then $|T_{3,2}(f)| \le 84$ [3, Theorem 2.3]. - (ii) If $f \in C$ , then $|T_{3,2}(f)| \le 4$ [3, Theorem 2.11]. In case of $\varphi(z) = ((1+z)/(1-z))^{\beta}$ , $0 < \beta \le 1$ , the classes $\mathcal{S}^(\varphi)$ and $\mathcal{C}(\varphi)$ reduce to the class of strongly starlike functions of order $\beta$ and the class of strongly convex functions of order $\beta$ , denoted by $\mathcal{SS}^(\beta)$ and $\mathcal{CC}(\beta)$ respectively (see [8]).

Corollary 3.7. If $f \in SS^*(\beta)$ , then $$|T_{3,2}(f)| \le \frac{4}{81}\beta^3(160 + 742\beta^2 + 799\beta^4)$$ for $\beta \in [3/4, 1]$ . The bound is sharp.

Corollary 3.8. If $f \in \mathcal{CC}(\beta)$ , then $$|T_{3,2}(f)| \le \frac{1}{324}\beta^3(323 + 650\beta^2 + 323\beta^4)$$ for $\beta \in [8/9, 1]$ . The bound is sharp. For $-1/2 < \lambda \le 1$ and $f \in \mathcal{A}$ such that f is a locally univalent functions, Robertson [22] considered the class $$\mathcal{F}(\lambda) = \left\{ f \in \mathcal{A} : \operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) > \frac{1}{2} - \lambda \right\}.$$ Clearly, when $1/2 \le \lambda \le 1$ , functions in $\mathcal{F}(\lambda)$ are close-to-convex [12]. For $-1/2 < \lambda \le 1/2$ , the functions in $\mathcal{F}(\lambda)$ are convex. Vasudevarao et al. [26] derived the sharp bound of $|T_{3,2}(f)|$ for $f \in \mathcal{F}(\lambda)$ when $1/2 \le \lambda \le 1$ , that is the class of Ozaki close-to-convex functions. Consider $$\varphi_{\lambda}(z) = \frac{1 + 2\lambda z}{1 - z}, \quad z \in \mathbb{D}.$$ The function $\varphi_{\lambda}$ maps the unit disk onto the right half plane for $-1/2 < \lambda \le 1/2$ such that $\operatorname{Re} \varphi_{\lambda} > (1/2 - \lambda)$ . Clearly, $\mathcal{C}(\varphi_{\lambda}) \subset \mathcal{F}(\lambda)$ for $\lambda \in (-1/2, 1]$ and $\mathcal{C}(\varphi_{\lambda}) = \mathcal{F}(\lambda)$ when $\lambda \in (-1/2, 1/2]$ . The Taylor's series expansion of $\varphi_{\lambda}$ gives $B_1 = B_2 = B_3 = (1 + 2\lambda)$ , which satisfy the condition (3.3) for $\lambda \in [5/14, 1/2]$ . Thus, from Theorem 3.2, we obtain the following sharp bound of $|T_{3,2}(f)|$ for the class $\mathcal{F}(\lambda)$ when $5/14 \le \lambda \le 1/2$ .

Corollary 3.9. If $f \in \mathcal{F}(\lambda)$ , $5/14 \le \lambda \le 1/2$ , then $$|T_{3,2}(f)| \le \frac{1}{864} (1+2\alpha)^3 (9+5\alpha+2\alpha^2)(25+17\alpha+10\alpha^2).$$ Remark 3.2. Vasudevarao et al. [26, Theorem 4.3] proved the same bound as given in Corollary 3.9 for $1/2 \le \lambda \le 1$ . Thus, Corollary 3.9 shows that the result is also true for $5/14 \le \lambda \le 1/2$ .