The third Hankel determinant for inverse coefficients of starlike functions of order 1/2

Lemma 1.1. [7, 10, 16] Let $p \in \mathcal{P}$ be given by (1.8) with $c_1 > 0$ . Then $$2c_{2} = c_{1}^{2} + \delta(4 - c_{1}^{2}),$$ $$4c_{3} = c_{1}^{3} + 2(4 - c_{1}^{2})c_{1}\delta - (4 - c_{1}^{2})c_{1}\delta^{2} + 2(4 - c_{1}^{2})(1 - |\delta|^{2})\eta,$$ $$8c_{4} = c_{1}^{4} + (4 - c_{1}^{2})\delta(c_{1}^{2}(\delta^{2} - 3\delta + 3) + 4\delta) - 4(4 - c_{1}^{2})(1 - |\delta|^{2})(c_{1}(\delta - 1)\eta + \overline{\delta}\eta^{2} - (1 - |\eta|^{2})\rho)$$ for some $\delta$ , $\eta$ , and $\rho$ such that $|\delta| \leq 1$ , $|\eta| \leq 1$ and $|\rho| \leq 1$ .

Theorem 2.1. Let $f \in \mathcal{S}^*(1/2)$ be defined by (1.3). Then $$|H_{3,1}(f^{-1})| \le \frac{1}{9}.$$ The result is sharp for the function $f_0 \in S^*(1/2)$ given by <span id="page-3-6"></span>(2.1) $$f_0(z) = \frac{z}{(1-z^3)^{1/3}}.$$ Remark 2.1. The conclusion of Theorem 2.1 reveals the following: - (i) The sharp bound of $|H_3(1)(f^{-1})|$ for the class $\mathcal{S}^*(1/2)$ is 1/9. Consequently, Question 1.1 is answered successfully. - (ii) The sharp bound of the functionals $|H_3(1)(f^{-1})|$ and $|H_3(1)(f^{-1})|$ are equal with 1/9. Hence, the Question 1.2 is answered successfully. Proof of Theorem 2.1. Let $f \in \mathcal{S}^*(1/2)$ . Then from (1.3) there exists a holomorphic function $p \in \mathcal{P}$ of the form (1.8) such that <span id="page-3-1"></span>(2.2) $$zf'(z) = \frac{1}{2}(p(z)+1)f(z), z \in \mathbb{D}$$ Putting the series of (1.1) and (1.8) in (2.2) and further simplification yields the following coefficients <span id="page-3-2"></span>(2.3) $$\begin{cases} a_2 = \frac{1}{2}c_1 \\ a_3 = \frac{1}{8}(2c_2 + c_1^2) \\ a_4 = \frac{1}{48}(8c_3 + 6c_1c_2 + c_1^3) \\ a_5 = \frac{1}{384}(48c_4 + 32c_1c_3 + 12c_2^2 + 12c_1^2c_2 + c_1^4). \end{cases}$$ Substituting (2.3) into (1.7), we obtain Substituting (2.3) into (1.7), we obtain <span id="page-3-3"></span> $$(2.4) A_2 = -\frac{c_1}{2},$$ (2.5) $$A_3 = \frac{3}{8}c_1^2 - \frac{1}{4}c_2,$$ (2.6) $$A_4 = -\frac{1}{3}c_1^3 + \frac{1}{2}c_1c_2 - \frac{1}{6}c_3,$$ <span id="page-3-4"></span>(2.7) $$A_5 = \frac{125}{384}c_1^4 - \frac{25}{32}c_1^2c_2 + \frac{5}{32}c_2^2 + \frac{5}{12}c_1c_3 - \frac{1}{8}c_4.$$ Then by using (2.4)-(2.7) in (1.5), a simple computation shows that <span id="page-3-5"></span> $$H_3(1)(f^{-1}) = \frac{1}{9216} (17c_1^6 - 102c_1^4c_2 + 32c_1^3c_3 + 180c_1^2c_2^2 - 144c_1^2c_4 + 192c_1c_2c_3 - 216c_2^3)$$ $$(2.8) + 288c_2c_4 - 256c_3^2)$$ Since the classes $\mathcal{P}$ , $\mathcal{S}^*(1/2)$ and the functional $H_3(1)(f^{-1})$ are rotationally invariant, by Carathéodory Theorem we may assume, $c := c_1 \in [0, 2]$ (see, [2, 3]) and using the Lemma 1.1 and simplification of (2.8), we obtain (2.9) $$H_3(1)(f^{-1}) = \frac{1}{9216}(g_1(c,\delta) + g_2(c,\delta)\eta + g_3(c,\delta)\eta^2) + v(c,\delta,\eta)\rho),$$ where $\delta, \eta, \rho \in \overline{\mathbb{D}}$ and $$\begin{cases} g_1(c,\delta) := & \delta^2 (4-c^2)^2 (2c^2 - (36-13c^2)\delta) + 2c^2 \delta^2; \\ g_2(c,\delta) := & -8c\delta (4-c^2)^2 (1+\delta)(1-|\delta|^2); \\ g_3(c,\delta) := & -8(4-c^2)^2 (8+|\delta|^2)(1-|\delta|^2); \\ v(c,\delta,\eta) := & 72\delta (4-c^2)^2 (1-|\delta|^2)(1-|\eta|^2). \end{cases}$$ <span id="page-4-0"></span>Next, using $|\delta| = x$ , $|\eta| = y$ and the fact that $|\rho| \le 1$ , we easily obtain (2.10) $9216|H_3(1)(f^{-1})| \le (|g_1(c,\delta)| + |g_2(c,\delta)| + |g_3(c,\delta)| + |v(c,\delta,\eta)|) \le M(c,x,y),$ where M(c,x,y) is defined by $$M(c, x, y) := (h_1(c, x) + h_2(c, x)y + h_3(c, x)y^2 + h_4(c, x)(1 - y^2))$$ and $$\begin{cases} h_1(c,x) := & x^2(4-c^2)^2(2c^2+(36-13c^2)x) + 2c^2x^2; \\ h_2(c,x) := & 8cx(4-c^2)^2(1+x)(1-x^2); \\ h_3(c,x) := & 8(4-c^2)^2(8+x^2)(1-x^2); \\ h_4(c,x) := & 72x(4-c^2)^2(1-x^2). \end{cases}$$ The goal is to find the maximum values of M(c, x, y) on the closed cuboid $\Omega = [0, 2] \times [0, 1] \times [0, 1]$ . This involves determining the maximum values in the interior of $\Omega$ , the interior of the six faces, as well as at the vertices and edges (twelve in total). Case 1. In the interior of $\Omega$ : A simple computation shows that $$\frac{\partial M(c, x, y)}{\partial y} = 0 \text{ implies that } 8(4 - c^2)^2 (1 - x^2) [cx(1 + x) + 2(1 - x)(8 - x)y] = 0$$ only for $$y = -\frac{cx(1+x)}{2(8-x)(1-x)} = y_0(c,x)$$ since $(4-c^2)^2 \neq 0$ and $(1-x^2) \neq 0$ for all $c \in (0,2)$ and $x \in (0,1)$ . Moreover, we see that $y_0(c,x) < 0$ for all $c \in (0,2)$ and $x \in (0,1)$ . Hence, we deduce that M has no critical points in interior of $\Omega$ . Case 2. On the edges of $\Omega$ . Case 2.1. On x = 1, y = 0 or x = 1, y = 1, we see that $$M(c, 1, 0) = M(c, 1, 1) = (4 - c^2)^2 (36 - 13c^2) \le 576, c \in (0, 2).$$ Case 2.1. On x = 0, y = 1, we have $$M(c, 0, 1) = 64(4 - c^2)^2 < 1024, c \in (0, 2),$$ which is equivalent to $$64c^4 - 512c^2 \le 0$$ for $c \in (0, 2)$ which is true. Case 2.3. If c = 0, y = 0, then an easy computation shows that $$M(0, x, 0) = -576x^3 + 1152x \le M(0, x_1, 0) \le 256\sqrt{6} \approx 627.069, \ x \in (0, 1)$$ where $$x_1 = \sqrt{\frac{2}{3}} \approx 0.816497.$$ Case 2.4. If c = 0, y = 1, then we see that $$M(0, x, 1) = 1024 - 896x^2 + 576x^3 - 128x^4 \le 1024, \ x \in (0, 1)$$ which is equivalent to $$-128x^2 + 576x - 896 \le 0, \ x \in (0,1)$$ which is true. Case 2.5. If c = 0, x = 0, then we see that $$M(0,0,y) = 1024y^2 \le 1024, y \in (0,1).$$ Case 2.6. If c = 0, x = 1, then $$M(0,1,y) = 576, y \in (0,1).$$ Case 2.7. For the other edges, c = 2, x = 0 or c = 2, x = 1 or c = 2, y = 0 or c = 2, y = 1 or c = 2, y = 0, we have $$M(2,0,y) = M(2,1,y) = M(2,x,0) = M(2,x,1) = M(c,0,0) = 0$$ for all $c \in (0, 2)$ , $x \in (0, 1)$ and $y \in (0, 1)$ . Case 3. On the faces of $\Omega$ . Case 3.1. On c=2, we have $$M(2, x, y) = 0, x \in (0, 1), y \in (0, 1).$$ Case 3.2. On c=0, we see that $$M(0, x, y) = 1152 - 576x^{3} + (1024 - 1152x - 896x^{2} + 1152x^{3} - 128x^{4})y^{2}$$ $$= 576(2 - x^{2}) + 128(8 - x)(1 - x)^{2}(1 + x)y^{2}$$ $$\leq 576(2 - x^{2}) + 128(8 - x)(1 - x)^{2}(1 + x)$$ $$= 1024 - 896x^{2} + 576x^{3} - 128x^{4} \leq 1024, (x, y) \in (0, 1) \times (0, 1)$$ which is equivalent to $$-896x^2 + 576x^3 - 128x^4 \le 0$$ for all $x \in (0, 1)$ which is true. Case 3.3. On x = 0, we have $$M(c, 0, y) = 64(4 - c^2)^2 y^2 \le 1024, c \in (0, 2), y \in (0, 1).$$ Case 3.4. On x = 1, we have $$M(c, 1, y) = (4 - c^2)^2 (36 - c^2) \le 576, c \in (0, 2), y \in (0, 1).$$ Case 3.5. On y = 0, we have $$M(c, x, 0) = (4 - c^{2})^{2} \left(72x(1 - x^{2}) + x^{2}(2c^{2} + (36 - 13c^{2})x + 2c^{2}x^{2})\right).$$ Then a simple computation shows that $$\frac{\partial}{\partial c}M(c,x,0) = 2c(4-c^2)((8-6c^2)x^4 + (39c^2+20)x^3 + (8-6c^2)x^2 - 144x) = 0$$ and $$\frac{\partial}{\partial x}M(c,x,0) = (4-c^2)^2(72+4c^2x-(108+39c^2)x^2+8c^2x^3) = 0.$$ The above system of equation has no critical point on $(0,2) \times (0,1)$ , and $$\max\{M(c,x,0)\} \le 256\sqrt{6} \text{ for } c=0, x=\sqrt{2/3}.$$ Case 3.6. On y = 1, we have $$M(c, x, 1) = (4 - c^{2})^{2} \left[x^{2} (2c^{2}x^{2} + (36 - 13c^{2})x + 2c^{2}) + 8cx(1 + x)(1 - x^{2}) + 8(8 + x^{2})(1 - x^{2})\right].$$ Furthermore, we see that $$\frac{\partial}{\partial c}M(c,x,1) = (4-c^2)[32x(1-x)(1+x)^2 - 40c^2x(1-x)(1+x)^2 - 6c^3x^2(2-13x+2x^2) + 8c(-32+30x^2-31x^3+6x^4)] = 0$$ and $$\frac{\partial}{\partial x}M(c,x,1) = (4-c^2)^2[c^2x(4-39x+8x^2) - 4x(28-27x+8x^2) - 8c(-1-2x+3x^2+4x^3)] = 0.$$ However, it can be shown that the above system has no solutions on $(0,2) \times (0,1)$ . Thus, it follows that M(c,x,1) has no critical point on $(0,2) \times (0,1)$ and hence, $$\max M(c, x, 1) = 1024$$ attained at c = 0, x = 0. Case 4. On the vertices of $\Omega$ . A simple computation gives the following estimates $$M(0,0,0) = M(2,0,0) = M(2,1,0) = M(2,1,1) = M(2,0,1) = 0,$$ $M(0,1,0) = M(0,1,1) = 576, M(0,0,1) = 1024.$ Summerizing Case 1 to Case 4, we easily obtain that $$\max\{M(c, x, y) : (c, x, y) \in \Omega\} \le 1024$$ Consequently, in view of inequality (2.10), we obtain that $$|H_3(1)(f^{-1})| \le \frac{1}{9}.$$ Thus the bound of the theorem is established. To complete the proof, it is sufficient to show that this bound is sharp. In order to show that we consider the function given by (2.1), which is equivalent to choosing $a_2 = a_3 = a_5 = 0$ and $a_4 = 1/3$ . By (1.7), a simple computation shows that $A_2 = A_3 = A_5 = 0$ and $A_4 = 1/3$ . Thus it follows from (1.5) that $|H_3(1)(f^{-1})| = 1/9$ establishes that the bound is sharp. This completes the