Bounds for the Second Hankel Determinant of Certain Univalent Functions

Lemma 1.1. [8] If the function $p \in \mathcal{P}$ is given by the series $$(1.2) p(z) = 1 + c_1 z + c_2 z^2 + c_3 z^3 + \cdots,$$ then the following sharp estimate holds: $$|c_n| \le 2 \quad (n = 1, 2, \dots).$$

Lemma 1.2. [11] If the function $p \in \mathcal{P}$ is given by the series (1.2), then (1.4) $$2c_2 = c_1^2 + x(4 - c_1^2),$$ (1.5) $$4c_3 = c_1^3 + 2(4 - c_1^2)c_1x - c_1(4 - c_1^2)x^2 + 2(4 - c_1^2)(1 - |x|^2)z,$$ for some x, z with |x| < 1 and |z| < 1.

Theorem 2.1. Let the function $f \in \mathcal{S}^*(\varphi)$ be given by (1.1). (1) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$|B_2| \le B_1$$ , $4B_1^4 - 16B_1|B_3| + 12B_2^2 - 6B_1|B_2| + 9B_1^2 \ge 0$ , then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{B_1^2}{4}.$$ (2) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$|B_2| > B_1$$ , $4B_1^4 - 16B_1|B_3| + 12B_2^2 - 2B_1|B_2| + 5B_1^2 < 0$ , or the conditions $$|B_2| \le B_1$$ , $4B_1^4 - 16B_1|B_3| + 12B_2^2 - 6B_1|B_2| + 9B_1^2 \le 0$ , then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{1}{48}(-4B_1^4 + 16B_1|B_3| - 12B_2^2 + 6B_1|B_2| + 3B_1^2).$$ (3) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$|B_2| > B_1$$ , $4B_1^4 - 16B_1|B_3| + 12B_2^2 - 2B_1|B_2| + 5B_1^2 \ge 0$ then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{B_1^2}{12} \left( \frac{12B_1^4 - 48B_1|B_3| + 40B_2^2 - 2B_1|B_2| + 7B_1^2}{4B_1^4 - 16B_1|B_3| + 12B_2^2 + 2B_1|B_2| + B_1^2} \right).$$ PROOF. Since $f \in \mathscr{S}^*(\varphi)$ , there exists an analytic function w with w(0) = 0 and |w(z)| < 1 in $\mathbb D$ such that (2.2) $$\frac{zf'(z)}{f(z)} = \varphi(w(z)).$$ Define the functions $p_1$ by $$p_1(z) := \frac{1 + w(z)}{1 - w(z)} = 1 + c_1 z + c_2 z^2 + \cdots$$ or equivalently, (2.3) $$w(z) = \frac{p_1(z) - 1}{p_1(z) + 1} = \frac{1}{2} \left( c_1 z + \left( c_2 - \frac{c_1^2}{2} \right) z^2 + \cdots \right).$$ Then $p_1$ is analytic in $\mathbb{D}$ with $p_1(0) = 1$ and has positive real part in $\mathbb{D}$ . By using (2.3) together with (2.1), it is evident that (2.4) $$\varphi\left(\frac{p_1(z)-1}{p_1(z)+1}\right) = 1 + \frac{1}{2}B_1c_1z + \left(\frac{1}{2}B_1\left(c_2 - \frac{c_1^2}{2}\right) + \frac{1}{4}B_2c_1^2\right)z^2 + \cdots$$ Since (2.5) $$\frac{zf'(z)}{f(z)} = 1 + a_2z + (-a_2^2 + 2a_3)z^2 + (3a_4 - 3a_2a_3 + a_2^3)z^3 + \cdots,$$ it follows by (2.2), (2.4) and (2.5) that $$a_2 = \frac{B_1 c_1}{2},$$ $$a_3 = \frac{1}{8} \left[ (B_1^2 - B_1 + B_2) c_1^2 + 2B_1 c_2 \right],$$ $$a_4 = \frac{1}{48} \left[ (-4B_2 + 2B_1 + B_1^3 - 3B_1^2 + 3B_1 B_2 + 2B_3) c_1^3 + 2(3B_1^2 - 4B_1 + 4B_2) c_1 c_2 + 8B_1 c_3 \right].$$ Therefore $$a_2a_4 - a_3^2 = \frac{B_1}{96} \left[ c_1^4 \left( -\frac{B_1^3}{2} + \frac{B_1}{2} - B_2 + 2B_3 - \frac{3B_2^2}{2B_1} \right) + 2c_2c_1^2(B_2 - B_1) + 8B_1c_1c_3 - 6B_1c_2^2 \right].$$ Let $$d_1 = 8B_1, \quad d_2 = 2(B_2 - B_1),$$ (2.6) $$d_3 = -6B_1, \quad d_4 = -\frac{B_1^3}{2} + \frac{B_1}{2} - B_2 + 2B_3 - \frac{3B_2^2}{2B_1},$$ $$T = \frac{B_1}{96}.$$ Then $$|a_2a_4 - a_3^2| = T|d_1c_1c_3 + d_2c_1^2c_2 + d_3c_2^2 + d_4c_1^4|.$$ Since the function $p(e^{i\theta}z)$ $(\theta \in \mathbb{R})$ is in the class $\mathscr{P}$ for any $p \in \mathscr{P}$ , there is no loss of generality in assuming $c_1 > 0$ . Write $c_1 = c$ , $c \in [0,2]$ . Substituting the values of $c_2$ and $c_3$ respectively from (1.4) and (1.5) in (2.7), it follows that $$|a_2a_4 - a_3^2| = \frac{T}{4} |c^4(d_1 + 2d_2 + d_3 + 4d_4) + 2xc^2(4 - c^2)(d_1 + d_2 + d_3)$$ $$+ (4 - c^2)x^2(-d_1c^2 + d_3(4 - c^2)) + 2d_1c(4 - c^2)(1 - |x|^2)z|.$$ Replacing |x| by $\mu$ and substituting the values of $d_1$ , $d_2$ , $d_3$ and $d_4$ from (2.6), yield $$|a_{2}a_{4} - a_{3}^{2}| \leq \frac{T}{4} \left[ c^{4} \left( -2B_{1}^{3} + 8|B_{3}| - 6\frac{B_{2}^{2}}{B_{1}} \right) + 4|B_{2}|\mu c^{2}(4 - c^{2}) + \mu^{2}(4 - c^{2})(2B_{1}c^{2} + 24B_{1}) + 16B_{1}c(4 - c^{2})(1 - \mu^{2}) \right]$$ $$= T \left[ \frac{c^{4}}{4} \left( -2B_{1}^{3} + 8|B_{3}| - 6\frac{B_{2}^{2}}{B_{1}} \right) + 4B_{1}c(4 - c^{2}) + |B_{2}|(4 - c^{2})\mu c^{2} + \frac{B_{1}}{2}\mu^{2}(4 - c^{2})(c - 6)(c - 2) \right]$$ $$\equiv F(c, \mu).$$ Note that for $(c, \mu) \in [0, 2] \times [0, 1]$ , differentiating $F(c, \mu)$ in (2.8) partially with respect to $\mu$ yields (2.9) $$\frac{\partial F}{\partial \mu} = T \left[ |B_2|(4-c^2) + B_1 \mu (4-c^2)(c-2)(c-6) \right].$$ Then for $0 < \mu < 1$ and for any fixed c with 0 < c < 2, it is clear from (2.9) that $\frac{\partial F}{\partial \mu} > 0$ , that is, $F(c,\mu)$ is an increasing function of $\mu$ . Hence for fixed $c \in [0,2]$ , the maximum of $F(c,\mu)$ occurs at $\mu = 1$ , and $$\max F(c, \mu) = F(c, 1) \equiv G(c).$$ Also note that $$G(c) = \frac{B_1}{96} \left[ \frac{c^4}{4} \left( -2B_1^3 + 8|B_3| - 6\frac{B_2^2}{B_1} - |B_2| - \frac{B_1}{2} \right) + 4c^2(|B_2| - B_1) + 24B_1 \right].$$ Let (2.10) $$P = \frac{1}{4} \left( -2B_1^3 + 8|B_3| - 6\frac{B_2^2}{B_1} - |B_2| - \frac{B_1}{2} \right),$$ $$Q = 4(|B_2| - B_1),$$ $$R = 24B_1.$$ Since (2.11) $$\max_{0 \le t \le 4} (Pt^2 + Qt + R) = \begin{cases} R, & Q \le 0, P \le -\frac{Q}{4}; \\ 16P + 4Q + R, & Q \ge 0, P \ge -\frac{Q}{8} \text{ or } Q \le 0, P \ge -\frac{Q}{4}; \\ \frac{4PR - Q^2}{4P}, & Q > 0, P \le -\frac{Q}{8}, \end{cases}$$ we have $$|a_2a_4 - a_3^2| \le \frac{B_1}{96} \begin{cases} R, & Q \le 0, P \le -\frac{Q}{4}; \\ 16P + 4Q + R, & Q \ge 0, P \ge -\frac{Q}{8} \text{ or } Q \le 0, P \ge -\frac{Q}{4}; \\ \frac{4PR - Q^2}{4P}, & Q > 0, P \le -\frac{Q}{8} \end{cases}$$ where P, Q, R are given by (2.10) REMARK 2.1. When $B_1 = B_2 = B_3 = 2$ , Theorem 2.1 reduces to [16, Theorem 3.1]. COROLLARY 2.1. - (1) If $f \in \mathcal{S}^*(\alpha)$ , for $0 < \alpha \le 3/4$ , $|a_2a_4 a_3^2| \le (1 \alpha)^2$ . And for $3/4 \le \alpha \le 1$ , $|a_2a_4 a_3^2| \le (1 \alpha)^2$ . $|a_3| < (1-\alpha)^2 [13-16(1-\alpha)^2]/12.$ - (2) If $f \in \mathscr{S}_{L}^{}$ , then $|a_{2}a_{4} a_{3}^{2}| \leq 1/16 = 0.0625$ . (3) If $f \in \mathscr{S}_{P}^{}$ , then $|a_{2}a_{4} a_{3}^{2}| \leq 16/\pi^{4} \approx 0.164255$ . (4) If $f \in \mathscr{S}_{\beta}^{*}$ , then $|a_{2}a_{4} a_{3}^{2}| \leq \beta^{2}$ .

Theorem 2.2. Let the function $f \in \mathcal{C}(\varphi)$ be given by (1.1). (1) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$B_1^2+4|B_2|-2B_1\leq 0,\quad B_1^4-B_1^2|B_2|-6B_1|B_3|+4B_2^2+4B_1^2\geq 0,$$ then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{B_1^2}{36}.$$ (2) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$B_1^2 + 4|B_2| - 2B_1 \ge 0$$ , $2B_1^4 - 2B_1^2|B_2| - 12B_1|B_3| + 8B_2^2 + 4B_1|B_2| + B_1^3 + 6B_1^2 \le 0$ , or the conditions $$B_1^2 + 4|B_2| - 2B_1 \le 0$$ , $B_1^4 - B_1^2|B_2| - 6B_1|B_3| + 4B_2^2 + 4B_1^2 \le 0$ , then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{1}{144}(-B_1^4 + B_1^2|B_2| + 6B_1|B_3| - 4B_2^2).$$ П (3) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$B_1^2 + 4|B_2| - 2B_1 > 0$$ , $2B_1^4 - 2B_1^2|B_2| - 12B_1|B_3| + 8B_2^2 + 4B_1|B_2| + B_1^3 + 6B_1^2 \ge 0$ , then the second Hankel determinant satisfies $$|a_2a_4-a_3^2| \leq \frac{B_1^2}{576} \left( \frac{17B_1^4 - 8B_1^2|B_2| - 96B_1|B_3| + 80B_2^2 + 12B_1^3 + 48B_1|B_2| + 36B_1^2}{B_1^4 - B_1^2|B_2| - 6B_1|B_3| + 4B_2^2 + B_1^3 + 4B_1|B_2| + 2B_1^2} \right).$$ PROOF. Since $f \in \mathcal{C}(\varphi)$ , there exists an analytic function w with w(0) = 0 and |w(z)| < 1 in $\mathbb{D}$ such that (2.12) $$1 + \frac{zf''(z)}{f'(z)} = \varphi(w(z)).$$ Since $$(2.13) 1 + \frac{zf''(z)}{f'(z)} = 1 + 2a_2z + (-4a_2^2 + 6a_3)z^2 + (8a_2^3 - 18a_2a_3 + 12a_4)z^3 + \cdots,$$ equations (2.4), (2.12) and (2.13) yield $$\begin{aligned} a_2 &= \frac{B_1 c_1}{4}, \\ a_3 &= \frac{1}{24} \left[ (B_1^2 - B_1 + B_2) c_1^2 + 2B_1 c_2 \right], \\ a_4 &= \frac{1}{192} \left[ (-4B_2 + 2B_1 + B_1^3 - 3B_1^2 + 3B_1 B_2 + 2B_3) c_1^3 + 2(3B_1^2 - 4B_1 + 4B_2) c_1 c_2 + 8B_1 c_3 \right]. \end{aligned}$$ Therefore $$a_{2}a_{4} - a_{3}^{2} = \frac{B_{1}}{768} \left[ c_{1}^{4} \left( -\frac{4}{3}B_{2} + \frac{2}{3}B_{1} - \frac{1}{3}B_{1}^{3} - \frac{1}{3}B_{1}^{2} + \frac{1}{3}B_{1}B_{2} + 2B_{3} - \frac{4}{3}\frac{B_{2}^{2}}{B_{1}} \right) + \frac{2}{3}c_{2}c_{1}^{2}(B_{1}^{2} - 4B_{1} + 4B_{2}) + 8B_{1}c_{1}c_{3} - \frac{16}{3}B_{1}c_{2}^{2} \right].$$ By writing $$d_1 = 8B_1$$ , $d_2 = \frac{2}{3}(B_1^2 - 4B_1 + 4B_2)$ , (2.14) $$d_{3} = -\frac{16}{3}B_{1}, \quad d_{4} = -\frac{4}{3}B_{2} + \frac{2}{3}B_{1} - \frac{1}{3}B_{1}^{3} - \frac{1}{3}B_{1}^{2} + \frac{1}{3}B_{1}B_{2} + 2B_{3} - \frac{4}{3}\frac{B_{2}^{2}}{B_{1}},$$ $$T = \frac{B_{1}}{768},$$ we have $$(2.15) |a_2a_4 - a_3^2| = T|d_1c_1c_3 + d_2c_1^2c_2 + d_3c_2^2 + d_4c_1^4|.$$ Similar as in Theorems 2.1, it follows from (1.4) and (1.5) that $$|a_2a_4 - a_3^2| = \frac{T}{4} |c^4(d_1 + 2d_2 + d_3 + 4d_4) + 2xc^2(4 - c^2)(d_1 + d_2 + d_3) + (4 - c^2)x^2(-d_1c^2 + d_3(4 - c^2)) + 2d_1c(4 - c^2)(1 - |x|^2)z|$$ Replacing |x| by $\mu$ and then substituting the values of $d_1$ , $d_2$ , $d_3$ and $d_4$ from (2.14) yield $$|a_{2}a_{4} - a_{3}^{2}| \leq \frac{T}{4} \left[ c^{4} \left( -\frac{4}{3}B_{1}^{3} + \frac{4}{3}B_{1}B_{2} + 8B_{3} - \frac{16B_{2}^{2}}{3B_{1}} \right) + 2\mu c^{2}(4 - c^{2}) \left( \frac{2}{3}B_{1}^{2} + \frac{8}{3}B_{2} \right) \right.$$ $$\left. + \mu^{2}(4 - c^{2}) \left( \frac{8}{3}B_{1}c^{2} + \frac{64}{3}B_{1} \right) + 16B_{1}c(4 - c^{2})(1 - \mu^{2}) \right]$$ $$(2.16) \qquad = T \left[ \frac{c^{4}}{3} \left( -B_{1}^{3} + B_{1}|B_{2}| + 6|B_{3}| - 4\frac{B_{2}^{2}}{B_{1}} \right) + 4B_{1}c(4 - c^{2}) + \frac{1}{3}\mu c^{2}(4 - c^{2})(B_{1}^{2} + 4|B_{2}|) \right.$$ $$\left. + \frac{2B_{1}}{3}\mu^{2}(4 - c^{2})(c - 4)(c - 2) \right]$$ $$\equiv F(c, \mu).$$ Again, differentiating $F(c, \mu)$ in (2.16) partially with respect to $\mu$ yield (2.17) $$\frac{\partial F}{\partial \mu} = T \left[ \frac{c^2}{3} (4 - c^2) (B_1^2 + 4|B_2|) + \frac{4B_1}{3} \mu (4 - c^2) (c - 4) (c - 2) \right].$$ It is clear from (2.17) that $\frac{\partial F}{\partial \mu} > 0$ . Thus $F(c, \mu)$ is an increasing function of $\mu$ for $0 < \mu < 1$ and for any fixed c with 0 < c < 2. So the maximum of $F(c, \mu)$ occurs at $\mu = 1$ and $$\max F(c,\mu) = F(c,1) \equiv G(c).$$ Note that $$G(c) = T \left[ \frac{c^4}{3} \left( -B_1^3 + B_1 |B_2| + 6|B_3| - 4\frac{B_2^2}{B_1} - B_1^2 - 4|B_2| - 2B_1 \right) + \frac{4}{3}c^2 (B_1^2 + 4|B_2| - 2B_1) + \frac{64}{3}B_1 \right].$$ Let $$P = \frac{1}{3} \left( -B_1^3 + B_1 |B_2| + 6|B_3| - 4\frac{B_2^2}{B_1} - B_1^2 - 4|B_2| - 2B_1 \right),$$ $$Q = \frac{4}{3} (B_1^2 + 4|B_2| - 2B_1),$$ $$R = \frac{64}{3} B_1,$$ By using (2.11), we have $$|a_2a_4 - a_3^2| \le \frac{B_1}{768} \begin{cases} R, & Q \le 0, P \le -\frac{Q}{4}; \\ 16P + 4Q + R, & Q \ge 0, P \ge -\frac{Q}{8} \text{ or } Q \le 0, P \ge -\frac{Q}{4}; \\ \frac{4PR - Q^2}{4P}, & Q > 0, P \le -\frac{Q}{8} \end{cases}$$ where P, Q, R are given in (2.18). REMARK 2.2. For the choice of $\varphi(z) = (1+z)/(1-z)$ , Theorem 2.2 reduces to [16, Theorem 3.2].

Theorem 3.1. Let $0 \le \gamma \le 1$ , $\tau \in \mathbb{C} \setminus \{0\}$ and the function f as in (1.1) is in the class $\mathscr{R}^{\tau}_{\nu}(\varphi)$ . Also, let $$p = \frac{8}{9} \frac{(1+\gamma)(1+3\gamma)}{(1+2\gamma)^2}.$$ (1) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$2|B_2|(1-p) + B_1(1-2p) \le 0$$ , $|B_1B_3 - pB_2^2| - pB_1^2 \le 0$ , then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{|\tau|^2 B_1^2}{9(1+2\gamma)^2}.$$ (2) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$2|B_2|(1-p)+B_1(1-2p) \ge 0$$ , $2|B_1B_3-pB_2^2|-2(1-p)B_1|B_2|-B_1 \ge 0$ , or the conditions $$2|B_2|(1-p)+B_1(1-2p) \le 0$$ , $|B_1B_3-pB_2^2|-B_1^2 \ge 0$ , then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{|\tau|^2}{8(1+\gamma)(1+3\gamma)} |B_3B_1 - pB_2^2|.$$ (3) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$2|B_2|(1-p)+B_1(1-2p)>0, \quad 2|B_1B_3-pB_2^2|-2(1-p)B_1|B_2|-B_1^2\leq 0,$$ then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \leq \frac{|\tau|^2 B_1^2}{32(1+\gamma)(1+3\gamma)} \left( \frac{4p|B_3B_1 - pB_2^2| - 4(1-p)B_1[|B_2|(3-2p) + B_1]}{-4B_2^2(1-p)^2 - B_1^2(1-2p)^2}}{|B_3B_1 - pB_2^2| - (1-p)B_1(2|B_2| + B_1)} \right).$$ PROOF. For $f \in \mathcal{R}^{\tau}_{\gamma}(\varphi)$ , there exists an analytic function w with w(0) = 0 and |w(z)| < 1 in $\mathbb{D}$ such that (3.1) $$1 + \frac{1}{\tau}(f'(z) + \gamma z f''(z) - 1) = \varphi(w(z)).$$ Since f has the Maclaurin series given by (1.1), a computation shows that $$(3.2) \quad 1 + \frac{1}{\tau}(f'(z) + \gamma z f''(z) - 1) = 1 + \frac{2a_2(1+\gamma)}{\tau}z + \frac{3a_3(1+2\gamma)}{\tau}z^2 + \frac{4a_4(1+3\gamma)}{\tau}z^3 + \cdots$$ It follows from (3.1), (2.4) and (3.2) that $$\begin{split} a_2 &= \frac{\tau B_1 c_1}{4(1+\gamma)}, \\ a_3 &= \frac{\tau B_1}{12(1+2\gamma)} \left[ 2c_2 + c_1^2 \left( \frac{B_2}{B_1} - 1 \right) \right], \\ a_4 &= \frac{\tau}{32(1+3\gamma)} [B_1 (4c_3 - 4c_1c_2 + c_1^3) + 2B_2 c_1 (2c_2 - c_1^2) + B_3 c_1^3]. \end{split}$$ Therefore $$a_{2}a_{4} - a_{3}^{2} = \frac{\tau^{2}B_{1}c_{1}}{128(1+\gamma)(1+3\gamma)} \left[ B_{1}(4c_{3} - 4c_{1}c_{2} + c_{1}^{3}) + 2B_{2}c_{1}(2c_{2} - c_{1}^{2}) + B_{3}c_{1}^{3} \right]$$ $$- \frac{\tau^{2}B_{1}^{2}}{144(1+2\gamma)^{2}} \left[ 4c_{2}^{2} + c_{1}^{4} \left( \frac{B_{2}}{B_{1}} - 1 \right)^{2} + 4c_{2}c_{1}^{2} \left( \frac{B_{2}}{B_{1}} - 1 \right) \right]$$ $$= \frac{\tau^{2}B_{1}^{2}}{128(1+\gamma)(1+3\gamma)} \left\{ \left[ (4c_{1}c_{3} - 4c_{1}^{2}c_{2} + c_{1}^{4}) + \frac{2B_{2}c_{1}^{2}}{B_{1}}(2c_{2} - c_{1}^{2}) + \frac{B_{3}}{B_{1}}c_{1}^{4} \right]$$ $$- \frac{8}{9} \frac{(1+\gamma)(1+3\gamma)}{(1+2\gamma)^{2}} \left[ 4c_{2}^{2} + c_{1}^{4} \left( \frac{B_{2}}{B_{1}} - 1 \right)^{2} + 4c_{2}c_{1}^{2} \left( \frac{B_{2}}{B_{1}} - 1 \right) \right] \right\},$$ which yields $$|a_{2}a_{4} - a_{3}^{2}| = T \left| 4c_{1}c_{3} + c_{1}^{4} \left[ 1 - 2\frac{B_{2}}{B_{1}} - p\left(\frac{B_{2}}{B_{1}} - 1\right)^{2} + \frac{B_{3}}{B_{1}} \right] - 4pc_{2}^{2} - 4c_{1}^{2}c_{2} \left[ 1 - \frac{B_{2}}{B_{1}} + p\left(\frac{B_{2}}{B_{1}} - 1\right) \right] \right|,$$ (3.3) where $$T = \frac{|\tau|^2 B_1^2}{128(1+\gamma)(1+3\gamma)} \quad \text{and} \quad p = \frac{8}{9} \frac{(1+\gamma)(1+3\gamma)}{(1+2\gamma)^2}.$$ It can be easily verified that $p \in \left[\frac{64}{81}, \frac{8}{9}\right]$ for $0 \le \gamma \le 1$ . Let (3.4) $$d_1 = 4, \quad d_2 = -4 \left[ 1 - \frac{B_2}{B_1} + p \left( \frac{B_2}{B_1} - 1 \right) \right],$$ $$d_3 = -4p, \quad d_4 = 1 - 2\frac{B_2}{B_1} - p \left( \frac{B_2}{B_1} - 1 \right)^2 + \frac{B_3}{B_1}.$$ Then (3.3) becomes $$|a_2a_4 - a_3^2| = T|d_1c_1c_3 + d_2c_1^2c_2 + d_3c_2^2 + d_4c_1^4|.$$ It follows that $$|a_2a_4 - a_3^2| = \frac{T}{4} |c^4(d_1 + 2d_2 + d_3 + 4d_4) + 2xc^2(4 - c^2)(d_1 + d_2 + d_3)$$ $$+ (4 - c^2)x^2(-d_1c^2 + d_3(4 - c^2)) + 2d_1c(4 - c^2)(1 - |x|^2)z|.$$ An application of triangle inequality, replacement of |x| by $\mu$ and substituting the values of $d_1$ , $d_2$ , $d_3$ and $d_4$ from (3.4) yield $$|a_{2}a_{4} - a_{3}^{2}| \leq \frac{T}{4} \left[ 4c^{4} \left| \frac{B_{3}}{B_{1}} - p \frac{B_{2}^{2}}{B_{1}^{2}} \right| + 8 \left| \frac{B_{2}}{B_{1}} \right| \mu c^{2} (4 - c^{2}) (1 - p) \right.$$ $$\left. + (4 - c^{2}) \mu^{2} (4c^{2} + 4p(4 - c^{2})) + 8c(4 - c^{2}) (1 - \mu^{2}) \right]$$ $$= T \left[ c^{4} \left| \frac{B_{3}}{B_{1}} - p \frac{B_{2}^{2}}{B_{1}^{2}} \right| + 2c(4 - c^{2}) + 2\mu \left| \frac{B_{2}}{B_{1}} \right| c^{2} (4 - c^{2}) (1 - p) + \mu^{2} (4 - c^{2}) (1 - p) (c - \alpha) (c - \beta) \right]$$ $$\equiv F(c, \mu)$$ where $\alpha = 2$ , $\beta = 2p/(1-p) > 2$ . Similarly as in the previous proofs, it can be shown that $F(c, \mu)$ is an increasing function of $\mu$ for $0 < \mu < 1$ . So for fixed $c \in [0, 2]$ , let $$\max F(c, \mu) = F(c, 1) \equiv G(c)$$ , which is $$G(c) = T \left\{ c^4 \left[ \left| \frac{B_3}{B_1} - p \frac{B_2^2}{B_1^2} \right| - (1 - p) \left( 2 \left| \frac{B_2}{B_1} \right| + 1 \right) \right] + 4c^2 \left[ 2 \left| \frac{B_2}{B_1} \right| (1 - p) + 1 - 2p \right] + 16p \right\}.$$ Let (3.7) $$P = \left| \frac{B_3}{B_1} - p \frac{B_2^2}{B_1^2} \right| - (1 - p) \left( 2 \left| \frac{B_2}{B_1} \right| + 1 \right),$$ $$Q = 4 \left[ 2 \left| \frac{B_2}{B_1} \right| (1 - p) + 1 - 2p \right],$$ $$R = 16p.$$ Using (2.11), we have $$|a_2a_4 - a_3^2| \le T \begin{cases} R, & Q \le 0, P \le -\frac{Q}{4}; \\ 16P + 4Q + R, & Q \ge 0, P \ge -\frac{Q}{8} \text{ or } Q \le 0, P \ge -\frac{Q}{4}; \\ \frac{4PR - Q^2}{4P}, & Q > 0, P \le -\frac{Q}{8} \end{cases}$$ where P, Q, R are given in (3.7). REMARK 3.1. For the choice $\varphi(z) := (1+Az)/(1+Bz)$ with $-1 \le B < A \le 1$ , Theorem 3.1 reduces to [6, Theorem 2.1].

Theorem 3.2. Let the function f given by (1.1) be in the class $\mathscr{G}_{\alpha}(\varphi)$ , $0 \le \alpha \le 1$ . Also, let $$p = \frac{8}{9} \frac{(1+2\alpha)}{(1+\alpha)}.$$ (1) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$B_1^2\alpha(3-2p)+2|B_2|(1+\alpha-p)+B_1(1+\alpha-2p)\leq 0,$$ $$B_1^4 \alpha (2\alpha - 1 - p\alpha) + \alpha B_1^2 |B_2| (3 - 2p) + (\alpha + 1)B_1 |B_3| - p(B_1^2 + B_2^2) \le 0,$$ then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{B_1^2}{9(1+\alpha)^2}.$$ (2) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$\begin{split} B_1^2\alpha(3-2p) + 2|B_2|(1+\alpha-p) + B_1(1+\alpha-2p) &\geq 0, \\ 2B_1^4\alpha(2\alpha-1-p\alpha) + 2\alpha B_1^2|B_2|(3-2p) - B_1^3\alpha(3-2p) \\ + 2(\alpha+1)B_1|B_3| - 2(1+\alpha-p)B_1|B_2| - (1+\alpha)B_1^2 - 2pB_2^2 &\geq 0, \end{split}$$ or $$B_1^2\alpha(3-2p)+2|B_2|(1+\alpha-p)+B_1(1+\alpha-2p)\leq 0,$$ $$B_1^4 \alpha (2\alpha - 1 - p\alpha) + \alpha B_1^2 |B_2| (3 - 2p) + (\alpha + 1)B_1 |B_3| - p(B_1^2 + B_2^2) \ge 0,$$ then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{B_1^4\alpha(2\alpha - 1 - p\alpha) + \alpha B_1^2|B_2|(3 - 2p) + (\alpha + 1)B_1|B_3| + p(B_2^2 - B_1^2)}{8(1 + \alpha)(1 + 2\alpha)}.$$ (3) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$\begin{split} B_1^2\alpha(3-2p) + 2|B_2|(1+\alpha-p) + B_1(1+\alpha-2p) &> 0, \\ 2B_1^4\alpha(2\alpha-1-p\alpha) + 2\alpha B_1^2|B_2|(3-2p) - B_1^3\alpha(3-2p) \\ + 2(\alpha+1)B_1|B_3| - 2(1+\alpha-p)B_1|B_2| - (1+\alpha)B_1^2 - 2pB_2^2 &\leq 0, \end{split}$$ then the second Hankel determinant satisfies $$|a_2a_4-a_3^2| \leq \frac{B_1^2}{32(1+\alpha)(1+2\alpha)} \left[ 4p - \frac{[B_1^2\alpha(3-2p)+2|B_2|(1+\alpha-p)+B_1(1+\alpha-2p)]^2}{B_1^4\alpha(2\alpha-1-p\alpha)+\alpha B_1^2|B_2|(3-2p)-B_1^3\alpha(3-2p)} + (\alpha+1)B_1|B_3| - (1+\alpha-p)B_1(2|B_2|+1)-pB_2^2 \right].$$ PROOF. For $f \in \mathscr{G}_{\alpha}(\varphi)$ , a calculation shows that $$|a_{2}a_{4} - a_{3}^{2}| = T \left| 4(1+\alpha)B_{1}c_{1}c_{3} + c_{1}^{4} \left[ -3\alpha B_{1}^{2} + \alpha(2\alpha - 1)B_{1}^{3} + B_{1}(1+\alpha) + 3\alpha B_{1}B_{2} \right] + (1+\alpha)(B_{3} - 2B_{2}) - p \frac{(\alpha B_{1}^{2} - B_{1} + B_{2})^{2}}{B_{1}} \right] - 4pB_{1}c_{2}^{2}$$ $$(3.8) + 2c_{1}^{2}c_{2} \left[ -2(1+\alpha)B_{1} + 3\alpha B_{1}^{2} + 2(1+\alpha)B_{2} - 2p(\alpha B_{1}^{2} - B_{1} + B_{2}) \right]$$ where $$T = \frac{B_1}{128(1+\alpha)(1+2\alpha)}$$ and $p = \frac{8(1+2\alpha)}{9(1+\alpha)}$ . It can be easily verified that for $0 \le \alpha \le 1$ , $p \in \left[\frac{8}{9}, \frac{4}{3}\right]$ . Let (3.9) $d_1 = 4(1+\alpha)B_1,$ $$d_1 = 4(1+\alpha)B_1$$ $$d_2 = 2 \left[ -2(1+\alpha)B_1 + 3\alpha B_1^2 + 2(1+\alpha)B_2 - 2p(\alpha B_1^2 - B_1 + B_2) \right],$$ $$d_3 = -4pB_1$$ $$d_4 = -3\alpha B_1^2 + \alpha(2\alpha - 1)B_1^3 + B_1(1 + \alpha) + 3\alpha B_1B_2 + (1 + \alpha)(B_3 - 2B_2) - p\frac{(\alpha B_1^2 - B_1 + B_2)^2}{B_1},$$ Then $$(3.10) |a_2a_4 - a_3^2| = T|d_1c_1c_3 + d_2c_1^2c_2 + d_3c_2^2 + d_4c_1^4|.$$ Similar as in earlier theorems, it follows that $$(3.11) \quad |a_{2}a_{4} - a_{3}^{2}| = \frac{T}{4} \left| c^{4} (d_{1} + 2d_{2} + d_{3} + 4d_{4}) + 2xc^{2} (4 - c^{2}) (d_{1} + d_{2} + d_{3}) \right. \\ \left. + (4 - c^{2})x^{2} (-d_{1}c^{2} + d_{3}(4 - c^{2})) + 2d_{1}c(4 - c^{2})(1 - |x|^{2})z \right| \\ \leq T \left[ c^{4} \left[ B_{1}^{3}\alpha(2\alpha - 1 - p\alpha) + \alpha B_{1}|B_{2}|(3 - 2p) + (\alpha + 1)|B_{3}| - p\frac{B_{2}^{2}}{B_{1}} \right] \right. \\ \left. + \mu c^{2} (4 - c^{2})[B_{1}^{2}\alpha(3 - 2p) + 2|B_{2}|(1 + \alpha - p)] + 2c(4 - c^{2})B_{1}(1 + \alpha) \right. \\ \left. + \mu^{2} (4 - c^{2})B_{1}(1 + \alpha - p)(c - 2) \left( c - \frac{2p}{1 + \alpha - p} \right) \right] \\ \equiv F(c, \mu),$$ and for fixed $c \in [0,2]$ , $\max F(c,\mu) = F(c,1) \equiv G(c)$ with $$\begin{split} G(c) &= T \left[ c^4 \left[ B_1^3 \alpha (2\alpha - 1 - p\alpha) + \alpha B_1 | B_2 | (3 - 2p) - B_1^2 \alpha (3 - 2p) + (\alpha + 1) | B_3 | \right. \\ &- (1 + \alpha - p) (2|B_2| + B_1) - p \frac{B_2^2}{B_1} \right] + 4c^2 [B_1^2 \alpha (3 - 2p) + 2|B_2| (1 + \alpha - p) \\ &+ B_1 (1 + \alpha - 2p)] + 16p B_1 \right]. \end{split}$$ Let $$P = B_1^3 \alpha (2\alpha - 1 - p\alpha) + \alpha B_1 |B_2| (3 - 2p) - B_1^2 \alpha (3 - 2p) + (\alpha + 1) |B_3|$$ $$- (1 + \alpha - p)(2|B_2| + B_1) - p \frac{B_2^2}{B_1}$$ $$Q = 4 \left[ B_1^2 \alpha (3 - 2p) + 2|B_2| (1 + \alpha - p) + B_1 (1 + \alpha - 2p) \right],$$ $$R = 16pB_1,$$ By using (2.11), we have $$|a_2a_4 - a_3^2| \le T \begin{cases} R, & Q \le 0, P \le -\frac{Q}{4}; \\ 16P + 4Q + R, & Q \ge 0, P \ge -\frac{Q}{8} \text{ or } Q \le 0, P \ge -\frac{Q}{4}; \\ \frac{4PR - Q^2}{4P}, & Q > 0, P \le -\frac{Q}{8} \end{cases}$$ where P, O, R are given in (3.13) REMARK 3.2. For $\alpha=0$ , Theorem 3.2 reduces to Theorem 2.2. For $0\leq\alpha<1$ , let $\varphi(z):=(1+(1-2\alpha)z)/(1-z)$ . For this function $\varphi$ , $B_1=B_2=B_3=2(1-\alpha)$ . In this case, Theorem 3.2 reduces to [39, Theorem 3.1]. П

Definition 2.1. Let $\varphi : \mathbb{D} \to \mathbb{C}$ be analytic and the Maclaurin series of $\varphi$ is given by (2.1) $$\varphi(z) = 1 + B_1 z + B_2 z^2 + B_3 z^3 + \cdots, \quad (B_1, B_2 \in \mathbb{R}, B_1 > 0).$$ The class $\mathscr{S}^(\varphi)$ of Ma-Minda starlike functions with respect to* $\varphi$ consists of functions $f \in \mathscr{A}$ satisfying the subordination $$\frac{zf'(z)}{f(z)} \prec \varphi(z).$$ For the function $\varphi$ given by $\varphi_{\alpha}(z) := (1 + (1 - 2\alpha)z)/(1 - z)$ , $0 < \alpha \le 1$ , the class $\mathscr{S}^(\alpha) := \mathscr{S}^(\varphi_{\alpha})$ is the well-known class of starlike functions of order $\alpha$ . Let $$\varphi_{PAR}(z) := 1 + \frac{2}{\pi^2} \left( \log \frac{1 + \sqrt{z}}{1 - \sqrt{z}} \right)^2.$$ Then the class $$\mathscr{S}_{P}^{} := \mathscr{S}^{}(\varphi_{PAR}) = \left\{ f \in \mathscr{A} : \operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) > \left|\frac{zf'(z)}{f(z)} - 1\right| \right\}$$ is the parabolic starlike functions introduced by Rønning [34]. For a survey of parabolic starlike functions and the related class of uniformly convex functions, see [3]. For $0 < \beta \le 1$ , the class $$\mathscr{S}_{\beta}^{} := \mathscr{S}^{}\left(\left(\frac{1+z}{1-z}\right)^{\beta}\right) = \left\{f \in \mathscr{A} : \left|\arg\left(\frac{zf'(z)}{f(z)}\right)\right| < \frac{\beta\pi}{2}\right\}$$ is the familiar class of strongly starlike functions of order $\beta$ . The class $$\mathscr{S}_L^ := \mathscr{S}^(\sqrt{1+z}) = \left\{ f \in \mathscr{A} : \left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| < 1 \right\}$$ is the class of lemniscate starlike functions studied in [37].

Definition 2.2. Let $\varphi : \mathbb{D} \to \mathbb{C}$ be analytic and $\varphi(z)$ is given as in (2.1). The class $\mathscr{C}(\varphi)$ of Ma-Minda convex functions with respect to $\varphi$ consists of functions f satisfying the subordination $$1 + \frac{zf''(z)}{f'(z)} \prec \varphi(z).$$

Definition 3.1. Let $\varphi : \mathbb{D} \to \mathbb{C}$ be analytic and $\varphi(z)$ as given in (2.1). Let $0 \le \gamma \le 1$ and $\tau \in \mathbb{C} \setminus \{0\}$ . A function $f \in \mathscr{A}$ is in the class $\mathscr{R}^{\tau}_{\gamma}(\varphi)$ if it satisfies the following subordination: $$1 + \frac{1}{\tau}(f'(z) + \gamma z f''(z) - 1) \prec \varphi(z).$$

Definition 3.2. Let $\varphi : \mathbb{D} \to \mathbb{C}$ be analytic and $\varphi(z)$ as given in (2.1). For fixed real number $\alpha$ , function $f \in \mathscr{A}$ is in the class $\mathscr{G}_{\alpha}(\varphi)$ if it satisfies the following subordination: $$(1-\alpha)f'(z) + \alpha\left(1 + \frac{zf''(z)}{f'(z)}\right) \prec \varphi(z).$$ Al-Amiri and Reade [1] introduced the class $\mathscr{G}_{\alpha} := \mathscr{G}_{\alpha}((1+z)/(1-z))$ and they have shown that $\mathscr{G}_{\alpha} \subset \mathscr{S}$ for $\alpha < 0$ . Univalence of the functions in the class $\mathscr{G}_{\alpha}$ was also investigated in [35,36]. Singh et al. also obtained the bound for the second Hankel determinant of functions in $\mathscr{G}_{\alpha}$ . The following theorem provides a bound for the second Hankel determinant of the functions in the class $\mathscr{G}_{\alpha}(\varphi)$ .