Sharp Bounds of Fifth Coefficient and Hermitian-Toeplitz determinants for Sakaguchi Classes

Lemma 2.1. [16] If the functions $1 + \sum_{n=1}^{\infty} p_n z^n$ and $1 + \sum_{n=1}^{\infty} q_n z^n$ are members of $\mathcal{P}$ , then the same is true of the function $$1 + \sum_{n=1}^{\infty} \frac{p_n q_n}{2} z^n.$$

Lemma 2.2. [16] Let $h(z) = 1 + \beta_1 z + \beta_2 z^2 + \cdots$ and $1 + H(z) = 1 + b_1 z + b_2 z^2 + \cdots$ be functions in $\mathcal{P}$ , and set $$\gamma_n = \frac{1}{2^n} \left[ 1 + \frac{1}{2} \sum_{\nu=1}^n \binom{n}{\nu} \beta_{\nu} \right], \quad \gamma_0 = 1.$$ If $A_n$ is defined by $$\sum_{n=1}^{\infty} (-1)^{n+1} \gamma_{n-1} H^n(z) = \sum_{n=1}^{\infty} A_n z^n,$$ then $|A_n| \leq 2$ .

Lemma 2.3. [3] If $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathcal{P}$ , then for some $\xi_1, \xi_2, \xi_3 \in \overline{\mathbb{D}}$ , $$p_1 = 2\xi_1, \quad p_2 = 2\xi_1^2 + 2(1 - |\xi_1|^2)\xi_2, p_3 = 2\xi_1^3 + 4(1 - |\xi_1|^2)\xi_1\xi_2 - 2(1 - |\xi_1|^2)\overline{\xi_1}\xi_2^2 + 2(1 - |\xi_1|^2)(1 - |\xi_2|^2)\xi_3.$$ (2.1) Further, for $\xi_1, \xi_2 \in \mathbb{D}$ and $\xi_3 \in \mathbb{T}$ , there is a unique function $p(z) = (1 + \omega(z))/(1 - \omega(z)) \in \mathcal{P}$ with $p_1, p_2$ and $p_3$ as in (2.1), where <span id="page-2-0"></span> $$\omega(z) = z\Psi_{-\xi_1}(z\Psi_{-\xi_2}(\xi_3 z)), \tag{2.2}$$ that is $$p(z) = \frac{1 + (\overline{\xi_2}\xi_3 + \overline{\xi_1}\xi_2 + \xi_1)z + (\overline{\xi_1}\xi_3 + \xi_1\overline{\xi_2}\xi_3 + \xi_2)z^2 + \xi_3z^3}{1 + (\overline{\xi_2}\xi_3 + \overline{\xi_1}\xi_2 - \xi_1)z + (\overline{\xi_1}\xi_3 - \xi_1\overline{\xi_2}\xi_3 - \xi_2)z^2 - \xi_3z^3}.$$ Conversely, for given $\xi_1, \xi_2 \in \mathbb{D}$ and $\xi_3 \in \overline{\mathbb{D}}$ , we can construct a (unique) function $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathcal{P}$ such that $p_1, p_2$ and $p_3$ satisfy the identities in (2.1). For this, we define $$\omega(z) = \omega_{\xi_1, \xi_2, \xi_3}(z) = z\Psi_{-\xi_1}(z\Psi_{-\xi_2}(\xi_3 z)). \tag{2.3}$$ Moreover, if we define $p(z) = (1 + \omega(z))/(1 - \omega(z))$ , then $p_1$ , $p_2$ and $p_3$ satisfy the identities in (2.1) (see the proof of [3, Lemma 2.4]). <span id="page-2-3"></span>Assumption 2.1. Let $\varphi(z)$ be given by (1.1). The following conditions on coefficients of $\varphi$ helps us to prove the result. C1: $$|B_1^3 - 2B_1B_2 + 2B_2^2| < |2B_1^2 - B_1^3 - 2B_1B_2|$$ , C2: $|B_1^3 - B_1^2B_2 + 3B_2^2 - 3B_1B_3| < 3|B_1^3 - B_1^2 + B_2^2|$ , C3: $|B_1^7 - B_1^6(8B_2 + 3) - 6B_1^4(B_2(3B_2 + 2B_3 + 2) - 6B_3 + 9B_4) + B_1^5(7B_2(B_2 + 4) - 24B_3 + 18B_4) + 6B_1^3(B_2^3 - 2B_2^2 + 8B_2B_3 - 3B_3^2 + 6(B_2 + 1)B_4) - 6B_1B_2(3B_2^3 - 6B_3^2 + B_2^2(4B_3 - 6) + 6B_2(B_4 - 2B_3)) + 18B_2^2(-2B_3^2 + B_2((B_2 - 2)B_2 + 2B_4)) + B_1^2B_2(B_2(B_2(5B_2 + 6) - 24B_3 + 18B_4) - 36(2B_3 + B_4))| < 2|((B_1 - 2)B_1 + 2B_2)(B_1(2B_1 + B_2 - 3) + 3B_3)(4B_1^3 + 6B_2^2 - B_1^2(B_2 + 3) - 3B_1B_3)|$ , C4: $0 < (2B_1 - B_1^2 - 2B_2)/(2(B_1 - B_2)) < 1$ .

Theorem 2.2. If $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots \in \mathcal{S}_s^*(\varphi)$ and coefficients of $\varphi(z)$ satisfy the conditions C1, C2, C3 and C4, then $$|a_5| \le \frac{B_1}{4}.$$ The bound is sharp. Proof. Let $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}_s^*(\varphi)$ , then there exist a Schwarz function $\omega(z)$ such that $$\frac{zf'(z)}{f(z) - f(-z)} = \varphi(\omega(z)).$$ By the one-to-one correspondence between the class of Schwarz functions and the class $\mathcal{P}$ , we obtain <span id="page-3-7"></span><span id="page-3-6"></span> $$\frac{zf'(z)}{f(z) - f(-z)} = \varphi\left(\frac{p(z) - 1}{p(z) + 1}\right) \tag{2.4}$$ for some $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathcal{P}$ . On the comparison of the same powers of z with the series expansions of functions f(z), $\varphi(z)$ and p(z), the above equation yields $$a_5 = \frac{B_1}{8} (\Upsilon_1 p_1^4 + \Upsilon_2 p_1^2 p_2 + \Upsilon_3 p_1 p_3 + \Upsilon_4 p_1^2 p_2 + p_4), \tag{2.5}$$ <span id="page-3-1"></span>where $$\Upsilon_{1} = \frac{B_{1}^{2} - 2B_{1} + 6B_{2} - 2B_{1}B_{2} + B_{2}^{2} - 6B_{3} + 2B_{4}}{16B_{1}},$$ $$\Upsilon_{2} = \frac{3B_{1} - B_{1}^{2} - 6B_{2} + B_{1}B_{2} + 3B_{3}}{4B_{1}}, \quad \Upsilon_{3} = \frac{B_{2} - B_{1}}{B_{1}},$$ $$\Upsilon_{4} = \frac{B_{1}^{2} - 2B_{1} + 2B_{2}}{4B_{1}}.$$ (2.6) Let us consider that $q(z) = 1 + \sum_{n=1}^{\infty} \kappa_n z^n$ and $h(z) = 1 + \sum_{n=2}^{\infty} \nu_n z^n$ are the members of $\mathcal{P}$ , then by Lemma 2.1 for $p \in \mathcal{P}$ , we have <span id="page-3-0"></span> $$1 + H(z) := 1 + \sum_{n=1}^{\infty} \frac{p_n \kappa_n}{2} z^n \in \mathcal{P}.$$ (2.7) For $h \in \mathcal{P}$ and the function 1 + H(z) given in (2.7), Lemma 2.2 gives $$A_4 = \frac{1}{2}\gamma_0\kappa_4 p_4 - \frac{1}{4}\gamma_1\kappa_2^2 p_2^2 - \frac{1}{2}\gamma_1\kappa_1\kappa_3 p_1 p_3 + \frac{3}{8}\gamma_2\kappa_1^2\kappa_2 p_1^2 p_2 - \frac{1}{16}\gamma_3\kappa_1^4 p_1^4, \tag{2.8}$$ where $\gamma_0 = 1$ , <span id="page-3-3"></span> $$\gamma_1 = \frac{1}{2} \left( 1 + \frac{1}{2} \nu_1 \right), \quad \gamma_2 = \frac{1}{4} \left( 1 + \nu_1 + \frac{1}{2} \nu_2 \right), \quad \gamma_3 = \frac{1}{8} \left( 1 + \frac{3}{2} \nu_1 + \frac{3}{2} \nu_2 + \frac{1}{2} \nu_3 \right)$$ (2.9) and <span id="page-3-5"></span><span id="page-3-2"></span> $$|A_4| \le 2. \tag{2.10}$$ Now, in order to establish the required bound, we construct functions h(z) and q(z) such that $$A_4 = \Upsilon_1 p_1^4 + \Upsilon_2 p_1^2 p_2 + \Upsilon_3 p_1 p_3 + \Upsilon_4 p_1^2 p_2 + p_4, \tag{2.11}$$ where $\Upsilon$ 's and $A_4$ are given in (2.6) and (2.8) respectively. For $0 < \tau < 1$ , define $$q(z) = \frac{1 + 2\tau z + 2\tau^2 z^2 + 2\tau z^3 + z^4}{1 - z^4},$$ <span id="page-3-4"></span>which yields $$\kappa_1 = \kappa_3 = 2\tau, \quad \kappa_2 = 2\tau^2 \quad \text{and} \quad \kappa_4 = 2.$$ (2.12) From [2, Theorem 1], we have $q \in \mathcal{P}$ . To construct function h(z), using Lemma 2.3, let $$h(z) = \frac{1 + \omega_1(z)}{1 - \omega_1(z)}$$ such that $$\omega_1(z) = z\Psi_{-\varepsilon_1}(z\Psi_{-\varepsilon_2}(\varepsilon_3 z)) \tag{2.13}$$ where $\varepsilon_1, \varepsilon_2 \in \mathbb{D}$ and $\varepsilon_3 \in \overline{\mathbb{D}}$ . Thus, we have $$\nu_1 = 2\varepsilon_1, \quad \nu_2 = 2\varepsilon_1^2 + 2(1 - |\varepsilon_1|^2)\varepsilon_2,$$ $$\nu_3 = 2\varepsilon_1^3 + 4(1 - |\varepsilon_1|^2)\varepsilon_1\varepsilon_2 - 2(1 - |\varepsilon_1|^2)\overline{\varepsilon_1}\varepsilon_2^2 + 2(1 - |\varepsilon_1|^2)(1 - |\varepsilon_2|^2)\varepsilon_3.$$ The above set of equations may be satisfied by many $\varepsilon$ 's. For our purpose, we impose some restriction on $\varepsilon$ 's and take all $\varepsilon$ 's as real numbers. Therefore, <span id="page-4-0"></span> $$\nu_{1} = 2\varepsilon_{1}, \quad \nu_{2} = 2\varepsilon_{1}^{2} + 2(1 - \varepsilon_{1}^{2})\varepsilon_{2}, \nu_{3} = 2\varepsilon_{1}^{3} + 4(1 - \varepsilon_{1}^{2})\varepsilon_{1}\varepsilon_{2} - 2(1 - \varepsilon_{1}^{2})\varepsilon_{1}\varepsilon_{2}^{2} + 2(1 - \varepsilon_{1}^{2})(1 - \varepsilon_{2}^{2})\varepsilon_{3}.$$ (2.14) In addition, if we define $$\varepsilon_{1} = \frac{B_{1}^{3} - 2B_{1}B_{2} + 2B_{2}^{2}}{2B_{1}^{2} - B_{1}^{3} - 2B_{1}B_{2}}, \quad \varepsilon_{2} = \frac{B_{1}^{3} - B_{1}^{2}B_{2} + 3B_{2}^{2} - 3B_{1}B_{3}}{3(-B_{1}^{2} + B_{1}^{3} + B_{2}^{2})},$$ $$\varepsilon_{3} = \left(B_{1}^{7} - B_{1}^{6}(8B_{2} + 3) - 6B_{1}^{4}(B_{2}(3B_{2} + 2B_{3} + 2) - 6B_{3} + 9B_{4}) + B_{1}^{5}(7B_{2}(B_{2} + 4) - 24B_{3} + 18B_{4}) + 6B_{1}^{3}(B_{2}^{3} - 2B_{2}^{2} + 8B_{2}B_{3} - 3B_{3}^{2} + 6(B_{2} + 1)B_{4}) - 6B_{1}B_{2}(3B_{2}^{3} - 6B_{3}^{2} + B_{2}^{2}(4B_{3} - 6) - 6B_{2}(2B_{3} - B_{4})) + 18B_{2}^{2}(-2B_{3}^{2} + B_{2}((B_{2} - 2)B_{2} + 2B_{4})) + B_{1}^{2}B_{2}(-36(2B_{3} + B_{4}) + B_{2}(B_{2}(6 + 5B_{2}) - 24B_{3} + 18B_{4}))\right) / \left(2((B_{1} - 2)B_{1} + 2B_{2})(B_{1}(2B_{1} + B_{2} - 3) + 3B_{3})(4B_{1}^{3} + 6B_{2}^{2} - B_{1}^{2}(3 + B_{2}) - 3B_{1}B_{3})\right)$$ and <span id="page-4-1"></span> $$\tau = \sqrt{\frac{2B_1 - B_1^2 - 2B_2}{2(B_1 - B_2)}},$$ then by the hypthesis 2.1, we have $|\varepsilon_1| < 1$ , $|\varepsilon_2| < 1$ , $|\varepsilon_3| < 1$ and $0 < \tau < 1$ . Putting these defined $\varepsilon$ 's in (2.14), we obtain $\nu_i$ 's, which in turn together with (2.9) yields $$\gamma_{1} = -\frac{(B_{1} - B_{2})^{2}}{B_{1}(B_{1}^{2} - 2B_{1} + 2B_{2})},$$ $$\gamma_{2} = -\frac{(B_{1} - B_{2})^{2}(B_{1}^{2} + 6B_{2} - B_{1}(3 + B_{2}) - 3B_{3})}{3B_{1}(B_{1}^{2} - 2B_{1} + 2B_{2})^{2}},$$ $$\gamma_{3} = -\frac{(B_{1} - B_{2})^{2}(B_{1}^{2} + 6B_{2} + B_{2}^{2} - 2B_{1}(1 + B_{2}) - 6B_{3} + 2B_{4})}{4B_{1}(B_{1}^{2} - 2B_{1} + 2B_{2})^{2}}.$$ (2.15) On putting the values of $\kappa_i's$ and $\gamma_i's$ from (2.12) and (2.15) respectively in (2.8), we get (2.11), which together with (2.5). Using the bound $|A_4| \leq 2$ in (2.11), we get $$|\Upsilon_1 p_1^4 + \Upsilon_2 p_1^2 p_2 + \Upsilon_3 p_1 p_3 + \Upsilon_4 p_1^2 p_2 + p_4| < 2.$$ which together with (2.5) gives the desired bound of $|a_5|$ . Consider the function $\tilde{f}_5(z) = z + \sum_{n=2}^{\infty} \tilde{a}_n z^n$ in the unit disk satisfying $$\frac{z\tilde{f}_5'(z)}{\tilde{f}_5(z) - \tilde{f}_5(-z)} = \varphi(z^4),$$ where $\varphi(z)$ is given by (1.1). Clearly, $\tilde{f}_5 \in \mathcal{S}_s^*(\varphi)$ . Equating the coefficients in the above equation, we obtain $\tilde{a}_2 = \tilde{a}_3 = \tilde{a}_4 = 0$ and $\tilde{a}_5 = B_1/4$ , that demonstrates the sharpness of the bound. For $-1 \le B < A \le 1$ , Consider the classes $\mathcal{S}_s^[A, B] := \mathcal{S}^((1 + Az)/(1 + Bz))$ and $\mathcal{S}_{s,SG}^ := \mathcal{S}^(2/(1 + e^{-z}))$ . These classes are analogues to the corresponding classes of starlike functions introduced and studied in [7, 8]. Theorem 2.2 directly gives the following result for these classes. <span id="page-5-0"></span>Corollary 2.3. If $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}_s^*[A, B]$ such that A and B satisfy the following conditions $$\mathbf{C1}: |(A-B)^{2}(A+B+2B^{2})| < |(A-3B-2)(A-B)^{2}|, \mathbf{C2}: |(A-B)^{3}(B+1)| < 3|(A-B)^{2}(A-1+(B-1)B)|, \mathbf{C3}: |(A-B)^{5}(B+1)(A^{2}(7B+1)+B(B(38+(12-17B)B) +15) + A(B(B(5B-31)-27)-3))| < 2|(A-B)^{4}(A -3B-2)(A(B-2)-4B^{2}+2B+3)(A(B+4)+2B(B -2)-3)|, \mathbf{C4}: 0 < (3B-A+2)/(2B+2) < 1,$$ (2.16) then $$|a_5| \le (A - B)/4.$$ The bound is sharp. Example 2.4. For A = 0 and B = -1/2, all conditions in 2.3 are satisfied. Thus, if $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}_s^*[0, -1/2]$ , then $|a_5| \le 1/8$ . Corollary 2.5. If $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}_{s,SG}^*$ , then $|a_5| \leq 1/8$ and the bound is sharp. In case of the classes $S_{s,L}$ and $S_{s,RL}$ , the coefficients of corresponding $\varphi$ satisfy the conditions C1, C2, C3 and C4. Theorem 2.2 yields the following result for these classes: Remark 2.1. If $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}_{s,L}^*$ , then $|a_5| \le 1/8$ [9, Theorem 5(a)]. Remark 2.2. If $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}_{s,RL}^*$ , then $|a_5| \le (5 - 3\sqrt{2})/8$ [9, Theorem 5(b)].

Theorem 2.6. If $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots \in C_s(\varphi)$ and coefficients of $\varphi(z)$ satisfy the conditions C1, C2, C3 and C4, then $$|a_5| \le \frac{B_1}{20}.$$ The bound is sharp. Proof. Let $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_s(\varphi)$ , then there exists a Schwarz function $\omega(z)$ such that $$\frac{(2zf'(z))'}{(f(z) - f(-z))'} = \varphi(\omega(z)).$$ Corresponding to the Schwarz function $\omega(z)$ , let there is a function $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathcal{P}$ satisfying $p(z) = (1 + \omega(z))/(1 - \omega(z))$ . Thus, we obtain <span id="page-5-2"></span> $$\frac{(2zf'(z))'}{(f(z) - f(-z))'} = \varphi\left(\frac{1 - p(z)}{1 + p(z)}\right). \tag{2.17}$$ Comparing the coefficients of the same powers of z after applying the series expansion of f(z), $\varphi(z)$ and p(z) leads to $$a_5 = \frac{B_1}{20} (\Upsilon_1 p_1^4 + \Upsilon_2 p_1^2 p_2 + \Upsilon_3 p_1 p_3 + \Upsilon_4 p_1^2 p_2 + p_4),$$ where $\Upsilon_i$ 's are given in (2.6). Since, $\Upsilon_i$ 's are the same as in the case of $\mathcal{S}_s^*(\varphi)$ , therefore following the same methodology as in Theorem 1, we get the bound of $|a_5|$ . To see the sharpness, consider the function $\tilde{g}_5(z) = z + \sum_{n=2}^{\infty} \tilde{a}_n z^n$ in $\mathbb{D}$ such that $$\frac{(2z\tilde{g}'_{5}(z))'}{(\tilde{g}_{5}(z) - \tilde{g}_{5}(-z))'} = \varphi(z^{4}).$$ Comparison of coefficients of same powers yield $\tilde{a}_2 = \tilde{a}_3 = \tilde{a}_4 = 0$ and $\tilde{a}_5 = B_1/20$ , which proves the sharpness of the bound. We can define the classes $C_s[A, B]$ , $C_{s,e}$ , $C_{s,SG}$ , $C_{s,L}$ and $C_{s,RL}$ in a similar manner as $S_s^[A, B]$ , $S_{s,e}$ , $S_{s,SG}$ , $S_{s,L}$ and $S_{s,RL}^*$ respectively. For these classes, Theorem 2.6 yields the following: Corollary 2.7. (i) If $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in C_s[A, B]$ such that A and B satisfy the conditions given in (2.3), then $|a_5| \leq (A - B)/20$ . (ii) If $$f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in C_{s,e}$$ , then $|a_5| \le 1/20$ . (iii) If $$f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in C_{s,L}$$ , then $|a_5| \le 1/40$ . (iv) If $$f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in C_{s,RL}$$ , then $|a_5| \le (5 - 3\sqrt{2})/40$ . (v) If $$f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in C_{s,SG}$$ , then $|a_5| \le 1/40$ . All these bounds are sharp.

Lemma 3.1. If $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots \in S_s^*(\varphi)$ and $B_1 \leq |B_2|$ , then $$|a_3| \le \frac{|B_2|}{2}.$$

Lemma 3.2. If $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots \in C_s(\varphi)$ and $B_1 \leq |B_2|$ , then $$|a_3| \le \frac{|B_2|}{6}.$$

Theorem 3.1. If $f \in \mathcal{S}_{s}^{*}(\varphi)$ and $B_{1} \leq |B_{2}|$ , then $$T_{3,1}(f) < 1$$ . The bound is sharp. Proof. Let $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}_s^*(\varphi)$ , then $$T_{3,1}(f) = 1 + 2|a_2|^2 - |a_3|^2 + 2\operatorname{Re}(a_2^2\bar{a}_3).$$ (3.1) Applying the inequality $2\operatorname{Re}(a_2^2\bar{a}_3) \leq 2|a_2^2||a_3|$ in the last equation, we obtain $$T_{3,1}(f) \le 1 + 2|a_2|^2 - |a_3|^2 + 2|a_2|^2 |a_3| =: g(x),$$ where $g(x) = 1 + 2|a_2|^2 - x^2 + 2|a_2^2|x$ with $x = |a_3|$ . For $f \in \mathcal{S}_s^*(\varphi)$ , we have $|a_2| \leq B_1/2$ and from Lemma 3.1, $|a_3| \leq |B_2|/2$ . Thus $|a_2| \in [0,1]$ and $x = |a_3| \in [0,1]$ . As g'(x) = 0 at $x = |a_2|^2$ and g''(x) < 0 for all $x \in [0,1]$ . Consequently, we have $$T_{3,1}(f) \le \max g(x)$$ = $g(|a_2|^2) = (|a_2|^2 - 1)^2 \le 1$ . Since the identity function f(z) = z is a member of the class $S_s^*(\varphi)$ and for this function, we have $a_2 = 0$ , $a_3 = 0$ and $T_{3,1}(f) = 1$ , which shows that the bound is sharp.

Theorem 3.2. If $f \in C_c(\varphi)$ and $B_1 \leq |B_2|$ , then $$T_{3,1}(f) \leq 1.$$ The result is sharp. Proof. Let $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_s(\varphi)$ , then using the inequality $\operatorname{Re}(a_2^2 \bar{a}_3) \leq |a_2|^2 |a_3|$ in (??) for $f \in \mathcal{C}_s(\varphi)$ , we obtain $$T_{3,1}(f) \le 1 + 2|a_2|^2 - |a_3|^2 + 2|a_2|^2|a_3| =: g(x),$$ where $g(x) = 1 + 2|a_2|^2 - x^2 + 2|a_2|^2x$ . Since $|a_2| \le B_1/4$ and from Lemma 3.2, we have $|a_3| \le |B_2|/6$ , therefore $|a_2| \in [0, 1/2]$ and $|a_3| \in [0, 1/3]$ . Also, note that g(x) attains its maximum value at $x = |a_2|^2$ . Hence $$T_{3,1}(f) \le \max g(x)$$ = $g(|a_2|^2) = (|a_2|^2 - 1)^2 \le 1$ . The equality case holds for f(z) = z.

Theorem 3.3. If $f \in \mathcal{S}_s^*(\varphi)$ such that $B_1^2 > 2B_2$ , then the following estimates hold: $$T_{3,1}(f) \ge \begin{cases} \min\left\{1 - \frac{B_1^2}{4}, 1 - \frac{B_1^2}{2} + \frac{B_1^2 B_2}{4} - \frac{B_2^2}{4}\right\}, & \sigma_1 \notin [0, 4], \\ 1 - \frac{B_1^2}{2} + \frac{B_1^2 B_2}{4} - \frac{B_2^2}{4}, & \sigma_1 = 4, \\ 1 - \frac{B_1^3 (B_1^3 + 4B_1^2 - 4B_1 - 8B_2)}{16(B_1^3 + B_1^2 (B_2 - 1) - 2B_1 B_2 - B_2^2)}, & \sigma_1 \in (0, 4), \end{cases}$$ where $$\sigma_1 = \frac{2B_1(B_1^2 - 2B_2)}{(B_1^2 - B_1 - B_2)(B_1 + B_2)}.$$ First two inequalities are sharp. Proof. Let $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}_s^*(\varphi)$ , then from (2.4), we obtain $$a_2 = \frac{B_1 p_1}{4}$$ and $a_3 = \frac{1}{8}(-B_1 p_1^2 + B_2 p_1^2 + 2B_1 p_2).$ Since the class $\mathcal{S}_s^*(\varphi)$ and the class $\mathcal{P}$ is rotationally invariant, therefore we can take $p_1 = p \in [0, 2]$ . Moreover, Libera et al. [14] showed that $2p_2 = p_1^2 + (4 - p_1^2)\zeta$ , $\zeta \in \mathbb{D}$ for $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathcal{P}$ . Thus, we have $$-|a_3|^2 = -\frac{1}{64} \left( B_2^2 p_1^4 + B_1^2 (4 - p_1^2)^2 |\zeta|^2 + 2B_1 B_2 p_1^2 (4 - p_1^2) \operatorname{Re} \bar{\zeta} \right),$$ $$2 \operatorname{Re}(a_2^2 \bar{a}_3) = \frac{1}{64} B_1^2 p_1^2 \left( (B_2 - B_1) p_1^2 + B_1 (p_1^2 + (4 - p_1^2) \operatorname{Re} \bar{\zeta}) \right).$$ Taking these into account in (1.4), we get $$T_{3,1}(f) = \frac{1}{64} \left( (B_1^2 - B_2) B_2 p_1^4 - B_1^2 (4 - p_1^2)^2 |\zeta|^2 + B_1 (B_1^2 - 2B_2) p_1^2 (4 - p_1^2) \operatorname{Re} \bar{\zeta} \right)$$ $$- \frac{B_1^2 p_1^2}{8} + 1 =: F(p_1, |\zeta|, \operatorname{Re} \bar{\zeta}).$$ It can be seen that $F(p_1, |\zeta|, \operatorname{Re} \bar{\zeta}) \geq F(p_1, |\zeta|, -|\zeta|) =: G(x, y)$ by considering $p_1^2 = x$ and $|\zeta| = y$ , where $$G(x,y) = \frac{1}{64} \left( (B_1^2 - B_2)B_2x^2 - B_1^2(4-x)^2y^2 - B_1(B_1^2 - 2B_2)x(4-x)y \right)$$ $$-\frac{B_1^2x}{8} + 1.$$ Whenever $B_1^2 > 2B_2$ , we have $$\frac{\partial G}{\partial y} = \frac{1}{64} (-2B_1^2 (4-x)^2 y - B_1 (B_1^2 - 2B_2) x (4-x)) \le 0$$ for $x \in [0,4]$ and $y \in [0,1]$ , which means that G(x,y) is a decreasing function of y and $G(x,y) \ge G(x,1) = I(x)$ with $$I(x) = \frac{1}{64}(B_1^3 + B_1^2(B_2 - 1) - 2B_1B_2 - B_2^2)x^2 + \frac{B_1}{16}(2B_2 - B_1^2)x - \frac{B_1^2}{4} + 1.$$ An easy computation yields that I'(x) = 0 at $$x_0 = \frac{2B_1(B_1^2 - 2B_2)}{(B_1^2 - B_1 - B_2)(B_1 + B_2)}$$ and $$I''(x_0) = \frac{1}{32}(B_1^2 - B_1 - B_2)(B_1 + B_2).$$ Since $B_1^2 > 2B_2$ , therefore numerator of $x_0$ is always positive. Moreover, denominator of $x_0$ and numerator of $G''(x_0)$ are same, therefore $x_0 < 0$ (or $x_0 > 0$ ) iff $I''(x_0) < 0$ (or $I''(x_0) > 0$ ). Here we discuss the following cases: Case I: Whenever $x_0 \in (0,4)$ , then $I''(x_0) > 0$ . Thus I(x) attains its minimum value at $x_0$ , which gives $$T_{3,1}(f) \ge I(x_0)$$ $$= 1 - \frac{B_1^3(B_1^3 + 4B_1^2 - 4B_1 - 8B_2)}{16(B_1^3 + B_1^2(B_2 - 1) - 2B_1B_2 - B_2^2)}$$ Case II: When $x_0 < 0$ or $x_0 > 4$ , which indicates that I(x) does not have any critical point, therefore $$\begin{split} T_{3,1}(f) &\geq \min\{I(0), I(4)\} \\ &= \min\bigg\{1 - \frac{B_1^2}{4}, 1 - \frac{B_1^2}{2} + \frac{B_1^2 B_2}{4} - \frac{B_2^2}{4}\bigg\}. \end{split}$$ For $x_0 = 4$ , $T_{3,1}(f) \ge I(4)$ . Function $\tilde{f}_2 \in \mathcal{S}_s^(\varphi)$ and $\tilde{f}_3 \in \mathcal{S}_s^(\varphi)$ given by $$\frac{z\tilde{f}_{2}'(z)}{\tilde{f}_{2}(z) - \tilde{f}_{2}(-z)} = \varphi(z), \quad \frac{z\tilde{f}_{3}'(z)}{\tilde{f}_{3}(z) - \tilde{f}_{3}(-z)} = \varphi(z^{2})$$ shows that these bounds are sharp as $$T_{3,1}(\tilde{f}_2) = 1 - \frac{B_1^2}{2} + \frac{B_1^2 B_2}{4} - \frac{B_2^2}{4}$$ and $T_{3,1}(\tilde{f}_3) = 1 - \frac{B_1^2}{4}$ , which completes the

Theorem 3.4. If $f \in C_s(\varphi)$ and $3B_1^2 \geq 8B_2$ , then the following estimates hold: $$T_{3,1}(f) \ge \begin{cases} \min\left\{1 - \frac{B_1^2}{144}, 1 + \frac{1}{144}(3B_1^2(B_2 - 6) - 4B_2^2)\right\}, & \sigma_2 \notin [0, 4], \\ 1 + \frac{1}{144}(3B_1^2(B_2 - 6) - 4B_2^2), & \sigma_2 = 4, \\ 1 - \frac{B_1^2(9B_1^4 + 114B_1^3 + 289B_1^2 - 304B_1B_2 - 36B_1^2B_2 + 48B_2^2)}{576(3B_1^3 - 8B_1B_2 + 3B_1^2B_2 - 4B_2^2)}, & \sigma_2 \in (0, 4), \end{cases}$$ where $$\sigma_2 = \frac{2(17B_1^2 + 3B_1^3 - 8B_1B_2)}{3B_1^3 - 8B_1B_2 + 3B_1^2B_2 - 4B_2^2}.$$ First two inequalities are sharp. Proof. Let $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_s(\varphi)$ , then from (2.17), we obtain <span id="page-9-0"></span> $$a_2 = \frac{B_1 p_1}{8}, \quad a_3 = \frac{1}{24}((B_2 - B_1)p_1^2 + 2B_1 p_2).$$ (3.2) The rotationally invariant property of the classes $C_s(\varphi)$ and $\mathcal{P}$ allows to take $p_1 \in [0, 2]$ . Using the formula $2p_2 = p_1^2 + (4 - p_1^2)\zeta$ (see [14]) in (3.2), we get $$-|a_3|^2 = -\frac{1}{576} (B_2^2 p_1^4 + B_1^2 (4 - p_1^2)^2 |\zeta|^2 + 2B_1 B_2 p_1^2 (4 - p_1^2) \operatorname{Re} \bar{\zeta}),$$ $$2 \operatorname{Re}(a_2^2 \bar{a}_3) = \frac{B_1^2 p_1^2}{768} (-B_1 p_1^2 + B_2 p_1^2 + B_1 (p_1^2 + (4 - p_1^2) \operatorname{Re} \bar{\zeta})).$$ These above values together with (1.4) leads to $$T_{3,1}(f) = \left(\frac{B_1^2 B_2}{768} - \frac{B_2^2}{576}\right) p_1^4 - \frac{1}{576} B_1^2 (4 - p_1^2)^2 |\zeta|^2 + \left(\frac{3B_1^3 - 8B_1 B_2}{2304}\right) p_1^2 (4 - p_1^2) \operatorname{Re} \bar{\zeta}$$ $$- \frac{B_1^2 p_1^2}{32} + 1 =: F(p_1, |\zeta|, \operatorname{Re} \bar{\zeta}).$$ As Re $\bar{\zeta} \geq -|\zeta|$ , hence $F(p_1,|\zeta|,\operatorname{Re}\bar{\zeta}) \geq F(p_1,|\zeta|,-|\zeta|) := G(x,y)$ , where $$G(x,y) = \left(\frac{B_1^2 B_2}{768} - \frac{B_2^2}{576}\right) x^2 - \frac{1}{576} B_1^2 (4-x)^2 y^2 - \left(\frac{3B_1^3 - 8B_1 B_2}{2304}\right) x (4-x) y$$ $$-\frac{B_1^2 p_1^2}{32} + 1$$ for $x = p_1^2 \in [0.4]$ and $y = |\zeta| \in [0, 1]$ . Whenever $3B_1^2 \ge 8B_1B_2$ , we have $$\frac{\partial G(x,y)}{\partial y} = -\frac{1}{288}B_1^2(4-x)^2y^2 - \left(\frac{3B_1^3 - 8B_1B_2}{2304}\right)x(4-x)y \le 0.$$ Therefore, G(x,y) is decreasing function of y and $G(x,y) \geq G(x,1) =: I(x)$ , where $$I(x) = \frac{(3B_1^3 - 8B_1B_2 + 3B_1^2B_2 - 4B_2^2)}{2304}x^2 - \frac{B_1(17B_1 + 3B_1^2 - 8B_2)}{576}x - \frac{B_1^2}{144} + 1.$$ An elementary calculation reveals that I'(x) = 0 at $$x_0 = \frac{2(17B_1^2 + 3B_1^3 - 8B_1B_2)}{3B_1^3 - 8B_1B_2 + 3B_1^2B_2 - 4B_2^2},$$ and $$I''(x) = \frac{3B_1^3 - 8B_1B_2 + 3B_1^2B_2 - 4B_2^2}{1152}.$$ Since $3B_1^2 \ge 8B_2$ and $B_1 > 0$ , therefore numerator of $x_0$ is always positive. Also, note that, denominator $x_0$ and numerator of I''(x) is same, therefore sign of $x_0$ and I''(x) changes simultaneously. Here, two cases arise: Case I: When $0 < x_0 < 4$ . In this case I''(x) > 0, so the minimum of I(x) attains at $x_0$ , which gives $$T_{3,1}(f) \ge I(x_0)$$ $$= 1 - \frac{B_1^2(9B_1^4 + 114B_1^3 + 289B_1^2 - 304B_1B_2 - 36B_1^2B_2 + 48B_2^2)}{576(3B_1^3 - 8B_1B_2 + 3B_1^2B_2 - 4B_2^2)}.$$ Case II: When $x_0 < 0$ or $x_0 > 4$ , that means I(x) has no critical point. Thus $$T_{3,1}(f) \ge \min\{I(0), I(4)\}$$ $$=\min\bigg\{1-\frac{B_1^2}{144},1-\frac{B_1^2}{8}+\frac{B_1^2B_2}{48}-\frac{B_2^2}{36}\bigg\}.$$ For the case $x_0 = 4$ , we have $T_{3,1}(f) \ge I(4)$ . The sharpness of these bounds follows from the functions $\tilde{g}_2(z)$ and $\tilde{g}_3(z)$ defined by $$\frac{(2z\tilde{g}_{2}'(z))'}{(\tilde{g}_{2}(z)-\tilde{g}_{2}(-z))'}=\varphi(z),\quad \frac{(2z\tilde{g}_{3}'(z))'}{(\tilde{g}_{3}(z)-\tilde{g}_{3}(-z))'}=\varphi(z^{2}).$$ Since $$T_{3,1}(\tilde{g}_2) = 1 - \frac{B_1^2}{8} + \frac{B_1^2 B_2}{48} - \frac{B_2^2}{36}, \quad T_{3,1}(\tilde{g}_2) = 1 - \frac{B_1^2}{144},$$ which completes the

Corollary 4.1. (i) If $f \in \mathcal{S}_{s}^{*}[A, B]$ and $A - B \leq |B^{2} - AB|$ , then $T_{3,1}(f) \leq 1$ . (ii) If $$f \in \mathcal{C}_s[A, B]$$ and $A - B \leq |B^2 - AB|$ , then $T_{3,1}(f) \leq 1$ . Theorem 3.3 and 3.4 yield the following lower bound of $T_{3,1}(f)$ for these classes. Corollary 4.2. If $f \in \mathcal{S}_s^*[A, B]$ such that $A^2 - B^2 > 0$ , then the following estimates hold: 1. If $$(A-B)^2(1-A)(1-B) < 0$$ or $2(A-B)^2(A(2B-1)-B+2) > 0$ , then $$\det T_{3,1}(f) \ge \min\left\{1 - \frac{(A-B)^2}{4}, 1 - \frac{(A-B)^2AB + 2}{4}\right\}.$$ 2. If $2(A-B)^2(A(2B-1)-B+2)=0$ , then $$\det T_{3,1}(f) \ge 1 - \frac{(A-B)^2 AB + 2}{4}.$$ 3. If $0 < 2(A-B)^2(A+B) < 4(A-B)^2(1-A)(1-B)$ , then $$\det T_{3,1}(f) \ge 1 + \frac{(A-B)^2(A^2 + B^2 + 4B - 2A(B-2) - 4)}{16(1-A)(1-B)}$$ First two inequalities are sharp. Corollary 4.3. If $f \in C_c(\varphi)$ and $3A^2 + 2AB - 5B^2 \ge 0$ , then the following estimates hold: 1. If $$-(A-B)^2(3A(B-1)+B(B-5)) < 0$$ or $2(A-B)^2(2B^2-5B+A(6B-3)+17) > 0$ , then $$\det T_{3,1}(f) \ge \min\left\{1 - \frac{(A-B)^2}{144}, 1 - \frac{(A-B)^2(B^2+3AB+18)}{144}\right\}. \tag{4.1}$$ 2. If $$(A - B)^2(2B^2 - 5B + A(6B - 3) + 17) = 0$$ , then $$\det T_{3,1}(f) \ge 1 - \frac{(A-B)^2(B^2 + 3AB + 18)}{144}.$$ (4.2) 3. If $$0 < 2(A - B)(3A^2 + A(2B + 17) - B(5B + 17)) < 4(A - B)^2(3A(1 - B) + (5 - B)B)$$ , then $$\det T_{3,1}(f) \ge 1 + \frac{(A - B)^2(9A^2 + 21B^2 + 190B + 6A(3B + 19) + 289)}{576(3A(B - 1) + (B - 5)B)}.$$ (4.3) First two inequalities are sharp. For $\varphi(z) = (1+(1-2\alpha)z)/(1-z)$ and (1+z)/(1-z) in $\mathcal{S}_s^(\varphi)$ , we obtain the class $\mathcal{S}_s^(\alpha)$ and Sakaguchi's class, $\mathcal{S}_s^*$ respectively, where $\alpha \in [0,1]$ . For more detail of these classes, we refer [15, 22]. Theorem 3.1 and 3.3 yield the following sharp lower and upper bound of $T_{3,1}(f)$ for these classes, proved by Kumar and Kumar [10]. Remark 4.1. (i) If $f \in \mathcal{S}_s^*(\alpha)$ , then $(3-2\alpha)\alpha^2 \leq T_{3,1}(f) \leq 1$ [10, Theorem 2.2]. (ii) If $f \in \mathcal{S}_s^*$ , then $0 \le T_{3,1}(f) \le 1$ [10, Corollary 2.3]. For other subclasses of $S_s^*$ , the following sharp bounds follow from Theorem 3.3. Corollary 4.4. If $f \in S_{s,SG}^*$ , then $T_{3,1}(f) \geq 2009/2304$ . Remark 4.2. (i) If $f \in \mathcal{S}_{s,L}^*$ , then $T_{3,1}(f) \ge 221/256$ [10, Theorem 3.1]. (ii) If $f \in \mathcal{S}_{s,RL}^*$ , then $T_{3,1}(f) \geq (863 - 444\sqrt{2})/256$ [10, Theorem 3.3]. Theorem 3.2 and 3.4 give the following corollaries for different subclasses of $C_c$ . Corollary 4.5. (i) If $f \in C_s[A, B]$ and $A - B \le |B^2 - AB|$ , then $T_{3,1}(f) \le 1$ . - (ii) If $f \in C_s(\alpha)$ , then $T_{3,1}(f) \leq 1$ . - (iii) If $f \in C_s$ , then $T_{3,1}(f) \leq 1$ . All these bounds are sharp. Corollary 4.6. (i) If $f \in C_{s,SG}$ , then $T_{3,1}(f) \ge 40165/41472$ . - (ii) If $f \in C_{s,L}$ , then $T_{3,1}(f) \geq 4459/4608$ . - (iii) If $f \in C_{s,RL}$ , then $T_{3,1}(f) \ge (-3731 + 5835\sqrt{2})/4608$ . All these bounds are sharp.