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. <span id="page-2-0"></span>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 <span id="page-2-1"></span> $$\frac{zf'(z)}{f(z)} = \varphi(\omega(z)), \quad z \in \mathbb{D}. \tag{2.1}$$ 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) <span id="page-2-2"></span>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) <span id="page-2-5"></span>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 <span id="page-2-4"></span><span id="page-2-3"></span> $$|a_2| \le B_1. \tag{2.6}$$ 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 <span id="page-3-1"></span> $$|a_3 - \frac{1}{2}a_2^2| \le \frac{|B_2|}{2}. (2.7)$$ From (1.5), we obtain <span id="page-3-0"></span> $$\left|\gamma_1^2 - \gamma_2^2\right| = \left|\frac{1}{4}\left(a_2^2 - \left(a_3 - \frac{a_2^2}{2}\right)^2\right)\right| \le \frac{1}{4}\left(|a_2|^2 + \left|a_3 - \frac{a_2^2}{2}\right|^2\right). \tag{2.8}$$ 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 <span id="page-3-4"></span><span id="page-3-3"></span> $$|\gamma_1^2 - \gamma_2^2| = \frac{4B_1^2 + B_2^2}{16}.$$ 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) <span id="page-3-6"></span>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 <span id="page-3-5"></span><span id="page-3-2"></span> $$|a_2| \le \frac{B_1}{2}.\tag{2.13}$$ 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 <span id="page-4-0"></span> $$|a_3 - \frac{1}{2}a_2^2| \le \frac{1}{6}|B_2 + \frac{1}{4}B_1^2|. \tag{2.14}$$ 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 <span id="page-4-3"></span> $$1 + \frac{zh_{\varphi}''(z)}{h_{\varphi}'(z)} = \varphi(iz). \tag{2.15}$$ 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 <span id="page-4-2"></span> $$|\gamma_2^2 - \gamma_3^2| \le \frac{1}{144} (9B_2^2 + 4B_3^2),$$ 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 <span id="page-4-1"></span> $$\left| \frac{a_2^3}{3} - a_2 a_3 + a_4 \right| \le \frac{|B_3|}{3}. \tag{2.17}$$ 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 <span id="page-5-0"></span> $$\left| \frac{a_2^3}{3} - a_2 a_3 + a_4 \right| = \frac{B_1}{12} \left| c_3 + \mu_2 c_1 c_2 + \nu_2 c_1^3 \right|.$$ 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).

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 <span id="page-8-4"></span> $$q_1 = \frac{3B_1^2 + 4B_2}{2B_1}, \quad q_2 = \frac{B_1^3 + 3B_1B_2 + 2B_3}{2B_1}.$$ 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 <span id="page-8-0"></span> $$a_4 = \frac{1}{192} \left( (B_1^3 - 3B_1^2 + 2B_1 - 4B_2 + 3B_1B_2 + 2B_3)p_1^3 + (6B_1^2 + 8B_2 - 8B_1)p_1p_2 + 8B_1p_3 \right). \tag{3.4}$$ 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 <span id="page-8-2"></span> $$a_4 = \frac{1}{12}B_1(c_3 + q_1c_1c_2 + q_2c_1^3),$$ 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 <span id="page-8-1"></span> $$|a_4| \le \frac{B_1^3 + 3B_1B_2 + 2B_3}{24}. (3.6)$$ Employing the bounds of $|a_2|$ and $|a_4|$ from (2.13) and (3.6) respectively, we get <span id="page-9-0"></span> $$|a_2 - a_4| \le |a_2| + |a_4| \le \frac{B_1}{2} + \frac{B_1^3 + 3B_1B_2 + 2B_3}{24}.$$ (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.