Third Order Hankel Determinant for Inverse Functions of a Class of Starlike Functions of Order alpha

Lemma 2.1. Let $p \in P$ . Then, for some $\rho, \sigma, x \in \mathbb{D}$ , we have $$2c_{2} = c_{1}^{2} + x(4 - c_{1}^{2}),$$ $$4c_{3} = c_{1}^{3} + 2(4 - c_{1}^{2})c_{1}x - c_{1}(4 - c_{1}^{2})x^{2} + 2(4 - c_{1}^{2})(1 - |x|^{2})\sigma,$$ $$8c_{4} = c_{1}^{4} + (4 - c_{1}^{2})x[c_{1}^{2}(x^{2} - 3x + 3) + 4x]$$ $$-4(4 - c_{1}^{2})(1 - |x|^{2})[c_{1}(x - 1)\sigma + \overline{x}\sigma^{2} - (1 - |x|^{2})\rho].$$ $$(2.1)$$

Theorem 3.1. If $f(z) \in S^*(\alpha) (0 \le \alpha \le \frac{1}{2})$ , then $$|H_3(1)(f^{-1})| \le \frac{143}{18}(1-\alpha)^2.$$ The result is sharp for the function $f \in S^*(\alpha)$ , $(0 \le \alpha \le \frac{1}{2})$ given by $f(z) = z + \frac{\sqrt{286}}{6}(1-\alpha)z^4 + \cdots$ . Proof. From $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots$ there exists an analytic function $p \in P$ in the unit disk $\mathbb{D}$ with p(0) = 1 and $Re\{p(z)\} > 0$ such that $$\left\{ \frac{zf'(z) - \alpha f(z)}{(1 - \alpha)f(z)} \right\} = p(z) \Leftrightarrow zf'(z) - \alpha f(z) = (1 - \alpha)f(z)p(z). \tag{3.1}$$ We calculate $f'(z) = 1 + 2a_2z + 3a_3z^2 + 4a_4z^3 + \cdots$ and than replacing f(z), f'(z) and p(z) in 3.1 we get $$[z\{1+2a_2z+3a_3z^2+4a_4z^3+\cdots\}-\alpha\{z+a_2z^2+a_3z^3+a_4z^4+\cdots\}]$$ $$=[(1-\alpha)[\{z+a_2z^2+a_3z^3+a_4z^4+\cdots\}\times\{1+c_1z+c_2z^2+c_3z^3+\cdots\}].$$ Equating the coefficients of powers of $z, z^2, z^3$ and $z^4$ after simplifying, we get: $$a_{3} = \frac{(1-\alpha)}{2}(c_{2} + (1-\alpha)c_{1}^{2})$$ $$a_{4} = \frac{(1-\alpha)}{6}(2c_{3} + 3(1-\alpha)c_{1}c_{2} + (1-\alpha)^{2}c_{1}^{3}) \qquad (3.2)$$ $$a_{5} = \frac{(1-\alpha)}{24}(6c_{4} + 6c_{1}c_{3} - 6c_{1}c_{3}\alpha + 3(1-\alpha)(c_{2} + (1-\alpha)c_{1}^{2})c_{2} + (1-\alpha)(2c_{3} + 3c_{1}c_{2}(1-\alpha) + (1-\alpha)^{2}c_{1}^{3})c_{1}).$$ $a_2 = (1 - \alpha)c_1$ So, from (1.4) and (3.2), we obtain: $$H_3(1)(f^{-1}) = \frac{1}{144}(1-\alpha)^2(17(1-\alpha)^4c_1^6 - 51(1-\alpha)^3c_1^4c_2 + 8(1-\alpha)^2c_1^3c_3 + 9(1-\alpha)c_1^2\left(5(1-\alpha)c_2^2 - 2c_4\right) + 24(1-\alpha)c_1c_2c_3 - 27(1-\alpha)c_2^3 + 18c_2c_4 - 16c_3^2),$$ i.e. $$|H_3(1)(f^{-1})| = \left| \frac{1}{144} (1 - \alpha)^2 (17(1 - \alpha)^4 c_1^6 - 51(1 - \alpha)^3 c_1^4 c_2 + 8(1 - \alpha)^2 c_1^3 c_3 + 9(1 - \alpha) c_1^2 \left( 5(1 - \alpha) c_2^2 - 2c_4 \right) + 24(1 - \alpha) c_1 c_2 c_3 - 27(1 - \alpha) c_2^3 + 18c_2 c_4 - 16c_3^2 \right)|.$$ Substituting the equation of $c_2, c_3$ and $c_4$ from the given Lemma 2.1, in the right side of $|H_3(1)(f^{-1})|$ , we have $$|H_3(1)(f^{-1})| = \frac{1}{144}(1-\alpha)^2|17(1-\alpha)^4c_1^6 - \frac{51}{2}(1-\alpha)^3c_1^6$$ $$+ \frac{51}{2}(1-\alpha)^3c_1^4x(4-c_1^2) + 2(1-\alpha)^2c_1^6 + 4(1-\alpha)^2c_1^4x(4-c_1^2)$$ $$-2(1-\alpha)^2c_1^4(4-c_1^2)x^2 + 4(1-\alpha)^2c_1^3(4-c_1^2)(1-|x|^2)\sigma$$ $$+9(1-\alpha)c_1^25(1-\alpha)\frac{1}{4}(c_1^2+x(4-c_1^2))^2 - \frac{9}{4}(1-\alpha)c_1^6$$ $$-9(1-\alpha)c_1^2(4-c_1^2)xc_1^2(x^2-3x+3) - \frac{9}{4}(1-\alpha)c_1^2(4-c_1^2)x^24$$ $$\begin{split} &+\frac{9}{4}(1-\alpha)c_1^24(4-c_1^2)(1-|x|^2)c_1(x-1)\sigma\\ &-\frac{9}{4}(1-\alpha)c_1^2(-4)(4-c_1^2)(1-|x|^2)\overline{x}\sigma^2\\ &+\frac{9}{4}(1-\alpha)c_1^2(-4)(4-c_1^2)(1-|x|^2)(1-|\sigma|^2)\rho\\ &+24(1-\alpha)c_1^6\frac{1}{8}+24(1-\alpha)c_1\frac{1}{2}x(4-c_1^2)\frac{1}{4}c_1^3\\ &+24(1-\alpha)c_1\frac{1}{2}c_1^3x2(4-c_1^2)+24(1-\alpha)c_1\frac{1}{2}x(4-c_1^2)^22c_1x\\ &-24(1-\alpha)c_1\frac{1}{2}c_1^3x^2(4-c_1^2)-24(1-\alpha)c_1\frac{1}{2}x(4-c_1^2)^2x^2\\ &+24(1-\alpha)c_1\frac{1}{2}c_1^3x^2(4-c_1^2)-24(1-\alpha)c_1\frac{1}{2}x(4-c_1^2)^22(1-|x|^2)\sigma)\\ &-\frac{27}{8}(1-\alpha)c_1^6-\frac{27}{8}(1-\alpha)12xc_1^4+\frac{27}{8}(1-\alpha)3xc_1^6\\ &-\frac{27}{8}(1-\alpha)c_1^2(x^2(4-c_1^2)^2)-\frac{27}{8}(1-\alpha)x^3(4-c_1^2)^3+9(c_1^2+x(4-c_1^2))c_1^4\frac{1}{8}\\ &+9(c_1^2+x(4-c_1^2))\frac{1}{8}(4-c_1^2)x[c_1^2(x^2-3x+3)+4x]\\ &-9(c_1^2+x(4-c_1^2))\frac{1}{2}(4-c_1^2)(1-|x|^2)(c_1(x+1)\sigma+\overline{x}\sigma^2-(1-|\sigma|^2)\rho]\\ &-c_1^6-4c_1^4(4-c_1^2)x+2c_1^4(4-c_1^2)x^2-4(4-c_1^2)\sigma(1-|x|^2)c_1^3-4c_1^2x^2(4-c_1^2)^2x^4\\ &+2c_1(4-c_1^2)^2x^2\sigma(1-|x|^2)+2c_1^2(4-c_1^2)^2x^3-c_1^2(4-c_1^2)^2x^4\\ &+2c_1(4-c_1^2)^2x^2\sigma(1-|x|^2)+2(4-c_1^2)^2\sigma(1-|x|^2)c_1x^2\\ &-4(4-c_1^2)^2\sigma^2(1-|x|^2)^2]. \end{split}$$ For $c_1 = c \in [0, 2]$ , and sorting the expression we obtain $$H_3(1)(f^{-1}) = \frac{1}{144}(1-\alpha)^2(v_1(c,x) + v_2(c,x)\sigma + v_3(c,x)\sigma^2 + \psi(c,x,\sigma)\rho),$$ where $|x| \le 1, |\sigma| \le 1, |\rho| \le 1$ , and $$v_{1}(c,x) = 17(1-\alpha)^{4}c^{6} - \frac{51}{2}(1-\alpha)^{3}c^{6} + 2(1-\alpha)^{2}c^{6} - c^{6}$$ $$+ \frac{9}{8}c^{6} + \frac{9}{4}(1-\alpha)^{2}c^{6} - \frac{45}{8}(1-\alpha)c^{6} + 3(1-\alpha)c^{6}$$ $$- \frac{81}{2}(1-\alpha)xc^{4} + \frac{81}{8}(1-\alpha)xc^{6} + (4-c^{2})[(4-c^{2})(\frac{9}{4}(1-\alpha)^{2}c^{2}x^{2})]$$ $$+ 24(1-\alpha)c^{2}x^{2} - 12(1-\alpha)c^{2}x^{3} - \frac{27}{8}(1-\alpha)x^{2}c^{2} + \frac{9}{8}x^{2}c^{2}(x^{2} - 3x + 3)$$ $$+ \frac{9}{2}x^{3} - 4c^{2}x^{2} + 4c^{2}x^{3} - c^{2}x^{4} + \frac{51}{2}(1-\alpha)^{3}xc^{4} + 4(1-\alpha)^{2}c^{4}x$$ $$-2(1-\alpha)^{2}c^{4}x^{2} + \frac{9}{2}(1-\alpha)^{2}c^{4}x - 9(1-\alpha)c^{4}x(x^{2} - 3x + 3) - 9(1-\alpha)c^{2}x^{2}$$ $$+3(1-\alpha)c^{4}x + 24(1-\alpha)c^{4}x - 12(1-\alpha)c^{4}x^{2} + \frac{9}{8}c^{4}x + \frac{9}{8}c^{4}x(x^{2} - 3x + 3)$$ $$+\frac{9}{2}c^{2}x^{2} - 4c^{4}x + 2c^{4}x^{2}],$$ $$v_{2}(c, x) = (4 - c^{2})(1 - |x|^{2})[(4 - c^{2})(-\frac{9}{2}x(x + 1)c - 8cx + 4cx^{2})$$ $$+4(1-\alpha)^{2}c^{3} + 9(1-\alpha)c^{3}(x - 1) + 24(1-\alpha)c^{3} + 24(1-\alpha)cx$$ $$-\frac{9}{2}c^{3}(x + 1)] - 4(4-c^{2}),$$ $$v_{3}(c, x) = (4 - c^{2})(1 - |x|^{2})[9(1-\alpha)c^{2}\overline{x} - \frac{9}{2}c^{2}\overline{x}$$ $$-\frac{9}{2}|x|^{2}(4-c^{2}) - 4(4-c^{2})(1-|x|^{2})]$$ $$\psi(c, x, \sigma) = (4 - c^{2})(1 - |x|^{2})(1 - |\sigma|^{2})(-9(1-\alpha)c^{2} + \frac{9}{2}c^{2}).$$ Setting |x| = x, $|\sigma| = y$ and utilizing the assumption $|\rho| \le 1$ , we obtain $$|H_3(1)(f^{-1})| \le \frac{1}{144} (1 - \alpha)^2 (|v_1(c, x)| + |v_2(c, x)|y + |v_3(c, x)|y^2 + |\psi(c, x, \sigma)|)$$ $$\le \frac{1}{144} (1 - \alpha)^2 Q(c, x, y),$$ (3.3) where $$Q(c, x, y) = q_1(c, x) + q_2(c, x)y + q_3(c, x)y^2 + q_4(c, x)(1 - y^2),$$ and $$q_{1}(c,x) = 17(1-\alpha)^{4}c^{6} + \frac{51}{2}(1-\alpha)^{3}c^{6} + \frac{17}{4}(1-\alpha)^{2}c^{6} + c^{6}$$ $$+ \frac{1}{8}c^{6} + \frac{21}{8}(1-\alpha)c^{6} + \frac{81}{2}(1-\alpha)xc^{4} + \frac{81}{8}(1-\alpha)xc^{6}$$ $$+ (4-c^{2})[(4-c^{2})(\frac{9}{4}(1-\alpha)^{2}c^{2}x^{2}$$ $$+ \frac{165}{8}(1-\alpha)c^{2}x^{2} - 12(1-\alpha)c^{2}x^{3} + \frac{9}{8}x^{2}c^{2}(x^{2}-3x+3) + \frac{9}{2}x^{3}$$ $$+ 4c^{2}x^{2} + 4c^{2}x^{3} + c^{2}x^{4}) + \frac{51}{2}(1-\alpha)^{3}xc^{4} + \frac{17}{2}(1-\alpha)^{2}c^{4}x + 2(1-\alpha)^{2}c^{4}x^{2}$$ $$+ 9(1-\alpha)c^{4}x(x^{2}-3x+3) + 9(1-\alpha)c^{2}x^{2} + 27(1-\alpha)c^{4}x$$ $$+ 12(1-\alpha)c^{4}x^{2} + \frac{23}{8}c^{4}x + \frac{9}{8}c^{4}x(x^{2}-3x+3) + \frac{9}{2}c^{2}x^{2} + 2c^{4}x^{2}]$$ $$q_{2}(c,x) = (4-c^{2})(1-x^{2})[(4-c^{2})(\frac{9}{2}x(x+1)c + 8cx + 4cx^{2})$$ $$+ 4(1-\alpha)^{2}c^{3} + 9(1-\alpha)c^{3}(x-1) + 24(1-\alpha)c^{3}$$ $$+ 24(1-\alpha)cx + \frac{9}{3}c^{3}(x+1)] + 4(4-c^{2})$$ $$q_3(c,x) = (4-c^2)(1-x^2)[9(1-\alpha)c^2x + \frac{9}{2}c^2x + \frac{9}{2}x^2(4-c^2) + 4(4-c^2)(1-x^2)]$$ $$q_4(c,x) = (4-c^2)(1-x^2)(9(1-\alpha)c^2 + \frac{9}{2}c^2).$$ Next, to find the upper bound of $|H_3(1)(f^{-1})|$ , we have to maximize Q(c,x,y) in the closed cuboid $\Omega:[0,2]\times[0,1]\times[0,1]$ . So, we need to consider the following cases: 1) In the interior of $\Omega$ : $$\frac{\partial Q(c,x,y)}{\partial y} = 0 \text{ imples that } q_2(c,x) + 2y(q_3(c,x) - q_4(c,x)) = 0,$$ SO $$y = -\frac{q_2(c, x)}{2(q_3(c, x) - q_4(c, x))} < 0$$ for $c \in (0,2)$ and $x \in (0,1)$ . Hence, we deduce that Q has no critical point in the interior of $\Omega$ . 2) On the edges of $\Omega$ : For x = 1, y = 0, we get $$Q(c,1,0) = q_1(c,1) + q_4(c,1)$$ $$= 17(1-\alpha)^4 c^6 + \frac{51}{2}(1-\alpha)^3 c^6 + \frac{17}{4}(1-\alpha)^2 c^6$$ $$+ \frac{21}{8}(1-\alpha)c^6 + \frac{81}{2}(1-\alpha)c^4 + \frac{81}{8}(1-\alpha)c^6 + \frac{9}{8}c^6$$ $$+ (4-c^2)[(4-c^2)(\frac{9}{4}(1-\alpha)^2c^2$$ $$+ \frac{165}{8}(1-\alpha)c^2 - 12(1-\alpha)c^2 + \frac{9}{8}c^2 + 9c^2 + \frac{9}{2})$$ $$+ \frac{51}{2}(1-\alpha)^3 c^4 + \frac{17}{2}(1-\alpha)^2 c^4 + 2(1-\alpha)^2 c^4$$ $$+ 9(1-\alpha)c^4 + 9(1-\alpha)c^2 + 27(1-\alpha)c^4$$ $$+ 12(1-\alpha)c^4 + \frac{23}{8}c^4 + \frac{25}{8}c^4 + \frac{9}{2}c^2] \le 1144,$$ (3.4) for 0 < c < 2 and $0 < \alpha < \frac{1}{2}$ . For x = 1, y = 1, we get $$Q(c, 1, 1) = q_1(c, 1) + q_2(c, 1) + q_3(c, 1) \le 1144,$$ for 0 < c < 2. 2.1) On $$x = 0, y = 1$$ , we get $$Q(c, 0, 1) = q_1(c, 0) + q_2(c, 0) + q_3(c, 0)$$ $$= 17(1-\alpha)^{4}c^{6} + \frac{51}{2}(1-\alpha)^{3}c^{6} + \frac{17}{4}(1-\alpha)^{2}c^{6} + c^{6} + \frac{1}{8}c^{6} + \frac{21}{8}(1-\alpha)c^{6} + (4-c^{2})[4(1-\alpha)^{2}c^{3} - 9(1-\alpha)c^{3} + 24(1-\alpha)c^{3} + \frac{9}{2}c^{3} + 4 + 4(4-c^{2})] \le 496$$ $$(3.5)$$ for 0 < c < 2 and $0 < \alpha < \frac{1}{2}$ . 2.2) For $$c = 0, y = 0,$$ $$Q(0, x, 0) = q_1(0, x) + q_4(0, x) = 72x^3 \le 72$$ which is equivalent to $x^3 \leq 1$ , which is true. 2.3) If $$c = 0, y = 1$$ we can see $$Q(0, x, 1) = q_1(0, x) + q_2(0, x) + q_3(0, x)$$ $$= 72x^3 + 16 + 4(1 - x^2) + 18x^2 + 16(1 - x^2) = 72x^3 - 2x^2 + 36 \le 106$$ which is equivalent to $72x^3 - 2x^2 - 70 \le 0$ which is true. 2.4) If $$c = 0, x = 0$$ we can see that $Q(0,0,y) = q_2(0,0)y + q_3(0,0)y^2 = 16y + 16y^2 \le 32, 0 < y < 1.$ 2.5) If $$c = 0, x = 1$$ we can see that $$Q(0, 1, y) = q_1(0, 1) + q_2(0, 1)y + q_3(0, 1)y^2 + q_4(0, 1)(1 - y^2)$$ $$= 72x^3 + 16y \le 88, 0 < y < 1.$$ 2.6) If $$c = 2, x = 0$$ we can see that $$Q(2,0,y) = q_1(2,0) + q_2(2,0)y + q_3(2,0)y^2$$ $$= 17(1-\alpha)^4c^6 + \frac{51}{2}(1-\alpha)^3c^6 + \frac{17}{4}(1-\alpha)^2c^6$$ $$+ \frac{1}{8}c^6 + \frac{21}{8}(1-\alpha)c^6 \le 496,$$ $$0 < y < 1, 0 < \alpha < \frac{1}{2}.$$ 2.7) If c = 2, x = 1 we can see that $Q(2,1,y) = q_1(2,1) + q_2(2,1)y + q_3(2,1)y^2$ $= 17(1-\alpha)^4c^6 + \frac{51}{2}(1-\alpha)^3c^6 + \frac{17}{4}(1-\alpha)^2c^6$ $+c^6 + \frac{1}{8}c^6 + \frac{21}{8}(1-\alpha)c^6 + \frac{81}{2}(1-\alpha)c^4 + \frac{81}{8}(1-\alpha)c^6 \le 1128,$ $0 < y < 1, 0 < \alpha < \frac{1}{2}.$ 2.8) If c = 2, y = 0 we can see that $$Q(2, x, 0) = q_1(2, x)$$ $$= 17(1 - \alpha)^4 c^6 + \frac{51}{2} (1 - \alpha)^3 c^6 + \frac{17}{4} (1 - \alpha)^2 c^6$$ $$+ c^6 + \frac{1}{8} c^6 + \frac{21}{8} (1 - \alpha) c^6 + \frac{81}{2} (1 - \alpha) x c^4 + \frac{81}{8} (1 - \alpha) x c^6 \le 1128,$$ $$0 < x < 1, 0 < \alpha < \frac{1}{2}.$$ We also have $$Q(2, x, 1) = q_1(2, x) = Q(2, 1, y),$$ $Q(2, 0, 0) = Q(2, 0, y).$ - 3) On the faces of $\Omega$ : - 3.1) On x = 0 we have $$\begin{split} Q(c,0,y) &= 17(1-\alpha)^4c^6 + \frac{51}{2}(1-\alpha)^3c^6 + \frac{17}{4}(1-\alpha)^2c^6 \\ &+ c^6 + \frac{1}{8}c^6 + \frac{21}{8}(1-\alpha)c^6 + 4(1-\alpha)^2c^3(4-c^2)y - 9(1-\alpha)c^3(4-c^2) \\ &+ 24(1-\alpha)c^3(4-c^2)y + \frac{9}{2}c^3(4-c^2)y + 4(4-c^2)y + (4-c^2)4(4-c^2)y^2 \\ &+ (4-c^2)(9((1-\alpha)c^2 + \frac{9}{2}c^2)(1-y^2) \\ &= 17(1-\alpha)^4c^6 + \frac{51}{2}(1-\alpha)^3c^6 + \frac{17}{4}(1-\alpha)^2c^6 \\ &+ c^6 + \frac{1}{8}c^6 + \frac{21}{8}(1-\alpha)c^6 + (4-c^2)(4(1-\alpha)^2c^3y - 9(1-\alpha)c^3 + 24(1-\alpha)c^3y + \frac{9}{2}c^3y \\ &+ 4y + 4y^2 + (1-y^2)9(1-\alpha)c^2 + \frac{9}{2}c^2(1-y^2)) \leq 32, \\ 0 &< y < 1, 0 < \alpha < \frac{1}{2}. \\ \frac{\partial Q(c,0,y)}{\partial y} &= (4-c^2)(4(1-\alpha)^2c^3 + 24(1-\alpha)c^3 \\ &+ \frac{9}{2}c^3 + 4 + 8y(4-c^2) - 18y(1-\alpha)c^2 - 9yc^2) = 0 \\ \frac{\partial Q(c,0,y)}{\partial c} &= 102(1-\alpha)^4c^5 + 153(1-\alpha)^3c^5 \\ &+ \frac{51}{2}(1-\alpha)^2c^5 + 6c^5 + \frac{3}{4}c^5 + \frac{63}{4}(1-\alpha)c^5 \\ + (4-c^2)(4(1-\alpha)^2c^3 + 24(1-\alpha)c^3 + \frac{9}{2}c^3 + 4 + 8y - 18y(1-\alpha)c^2 - 9c^2y = 0. \end{split}$$ 3.2) On c = 0 we have $$Q(0, x, y) = 72x^{3} + 16y + 4(1 - x^{2})(18x^{2} + 16(1 - x^{2}))y^{2}$$ $$= 72x^{3} + 16y + 8(1 - x^{2})(8 + x^{2})y^{2} \le 72x^{3} + 16 + 8(1 - x^{2})(8 + x^{2})$$ $$= 72x^{3} + 16 + 64 + 8x^{2} - 64x^{2} - 8x^{4}$$ $$= 80 + 72x^{3} - 8x^{4} - 56x^{2} \le 88.$$ which is equivalent to $$72x^3 - 8x^4 - 56x^2 \le 8, 0 < x < 1,$$ which is true. 3.3) On c = 2 we have $$Q(2, x, y) = q_1(2, x) = 17(1 - \alpha)^4 c^6 + \frac{51}{2}(1 - \alpha)^3 c^6$$ $$+ \frac{17}{4}(1 - \alpha)^2 c^6 + c^6 + \frac{1}{8}c^6 + \frac{21}{8}(1 - \alpha)c^6$$ $$+ \frac{81}{2}(1 - \alpha)xc^4 + \frac{81}{8}(1 - \alpha)xc^6 = 17(1 - \alpha)^4 64$$ $$+ \frac{51}{2}(1 - \alpha)^3 64 + \frac{17}{4}(1 - \alpha)^2 64 + 64 + \frac{1}{8}64$$ $$+ \frac{21}{8}(1 - \alpha)64 + \frac{81}{2}(1 - \alpha)x16 + \frac{81}{8}(1 - \alpha)x64 \le 1144$$ $$0 < \alpha < \frac{1}{2}, 0 < x, y < 1.$$ 3.4) On x = 1 we have $$Q(c, 1, 1) = q_1(c, 1) + q_2(c, 1) + q_3(c, 1) \le 1144$$ for $0 < c < 2, 0 < \alpha < \frac{1}{2}$ . 3.5) On y = 0 we have $$Q(c, x, 0) = q_1(c, x) + q_4(c, x) = 17(1 - \alpha)^4 c^6 + \frac{51}{2}(1 - \alpha)^3 c^6$$ $$+ \frac{17}{4}(1 - \alpha)^2 c^6 + c^6 + \frac{1}{8}c^6 + \frac{21}{8}(1 - \alpha)c^6$$ $$+ \frac{81}{2}(1 - \alpha)xc^4 + \frac{81}{8}(1 - \alpha)xc^6$$ $$+ (4 - c^2)[(4 - c^2)(\frac{9}{4}(1 - \alpha)^2 c^2 x^2$$ $$+ \frac{165}{8}(1 - \alpha)c^2 x^2 - 12(1 - \alpha)c^2 x^3 + \frac{9}{8}x^2 c^2 (x^2 - 3x + 3) + \frac{9}{2}x^3$$ $$+ 4c^2 x^2 + 4c^2 x^3 + c^2 x^4) + \frac{51}{2}(1 - \alpha)^3 xc^4 + \frac{17}{2}(1 - \alpha)^2 c^4 x + 2(1 - \alpha)^2 c^4 x^2$$ $$+ 9(1 - \alpha)c^4 x(x^2 - 3x + 3) + 9(1 - \alpha)c^2 x^2 + 27(1 - \alpha)c^4 x$$ $$+12(1-\alpha)c^{4}x^{2} + \frac{23}{8}c^{4}x + \frac{9}{8}c^{4}x(x^{2} - 3x + 3) + \frac{9}{2}c^{2}x^{2} + 2c^{4}x^{2}]$$ $$+(4-c^{2})(1-x^{2})(9(1-\alpha)c^{2} + \frac{9}{2}c^{2}).$$ Then $$\frac{\partial Q(c,x,0)}{\partial x} = \frac{81}{2}(1-\alpha)c^4 + \frac{81}{8}(1-\alpha)c^6$$ $$+(4-c^2)[(4-c^2)(\frac{9}{2}(1-\alpha)^2c^2x + \frac{165}{4}(1-\alpha)c^2x - 36(1-\alpha)c^2x^2 + \frac{9}{8}c^2(4x^3 - 9x^2 + 6x) + \frac{27}{2}x^2$$ $$+8c^2x + 12c^2x^2 + 4c^2x^3) + \frac{51}{2}(1-\alpha)^3c^4 + \frac{17}{2}(1-\alpha)^2c^4 + 4(1-\alpha)^2c^4x$$ $$+9(1-\alpha)c^4(3x^2 - 6x + 3) + 18(1-\alpha)c^2x + 27(1-\alpha)c^4$$ $$+24(1-\alpha)c^4x + \frac{23}{8}c^4 + \frac{9}{8}c^4(3x^2 - 6x + 3) + 9c^2x + 4c^4x]$$ $$+(4-c^2)(-2x)(9(1-\alpha)c^2 + \frac{9}{2}c^2) = 0$$ Also, if we calculate $$\frac{\partial Q(c, x, 0)}{\partial c} = 0,$$ we get system of equations that has no critical point on $(0,2)\times(0,1)$ , and $$\max Q(c, x, 0) \le 80$$ for c = 0, x = 0. 3.6) On $$y = 1$$ we have $$Q(c, x, 1) = q_1(c, x) + q_2(c, x) + q_3(c, x)$$ $$\frac{\partial Q(c, x, 1)}{\partial x} = \frac{81}{2}(1 - \alpha)c^4 + \frac{81}{8}(1 - \alpha)c^6$$ $$+ (4 - c^2)[(4 - c^2)(\frac{9}{2}(1 - \alpha)^2c^2x$$ $$+ \frac{165}{4}(1 - \alpha)c^2x - 36(1 - \alpha)c^2x^2 + \frac{9}{8}c^2(4x^3 - 9x^2 + 6x) + \frac{27}{2}x^2$$ $$+ 8c^2x + 12c^2x^2 + 4c^2x^3)$$ $$+ \frac{51}{2}(1 - \alpha)^3c^4 + \frac{17}{2}(1 - \alpha)^2c^4 + 4(1 - \alpha)^2c^4x$$ $$+ 9(1 - \alpha)c^4(3x^2 - 6x + 3) + 18(1 - \alpha)c^2x + 27(1 - \alpha)c^4$$ $$+ 24(1 - \alpha)c^4x + \frac{23}{8}c^4 + \frac{9}{8}c^4(3x^2 - 6x + 3) + 9c^2x + 4c^4x]$$ $$+ (4 - c^2)(-2x)[(4 - c^2)(\frac{9}{2}x(x + 1)c + 8cx + 4cx^2)$$ $$+ 4(1 - \alpha)^2c^3 + 9(1 - \alpha)c^3(x - 1) + 24(1 - \alpha)c^3$$ $$\begin{aligned} +24(1-\alpha)c^2x + \frac{9}{2}c^3(x+1)] \\ +(4-c^2)(1-x^2)[(4-c^2)(\frac{9}{2}(2x+1)c + 8c + 8cx) \\ +9(1-\alpha)c^3 + 24(1-\alpha)c^2 + \frac{9}{2}c^3] \\ +(4-c^2)(-2x)[9(1-\alpha)c^2x + \frac{9}{2}c^2x + \frac{9}{2}x^2(4-c^2) \\ +4(4-c^2)(1-x^2)] + (4-c^2)(1-x^2)[9(1-\alpha)c^2 + \frac{9}{2}c^2 + 9x(4-c^2) \\ +4(4-c^2)(-2x)] = 0. \end{aligned}$$ If we calculate $$\frac{\partial Q(c, x, 1)}{\partial c} = 0,$$ we will got a system which has no solutions on $(0,2) \times (0,1)$ . So Q(c,x,1) has no critical point on $(0,2) \times (0,1)$ and hance, $$\max Q(c, x, 1) = 80$$ for c = 0, x = 0. - 4) On the vertices of $\Omega$ : - 4.1) We note that $$Q(0,0,1) = 80,$$ $$Q(0,0,0) = 80.$$ Summerizing 1) to 4), we get $$\max_{(c,x,y)\in\Omega} \{Q(c,x,y)\} \le 1144.$$ Consequently, from inequality (3.3) we have that $$|H_3(1)(f^{-1})| \le \frac{143}{18}(1-\alpha)^2.$$ In the end, we will show that the upper bound is sharp. We consider the function f obtained by choosing $a_2 = a_3 = a_5 = 0$ and $a_4 = \frac{\sqrt{286}}{6}(1 - \alpha)$ . So, it follows from $H_3(1)(f)$ that $$|H_3(1)(f^{-1})| = \frac{143}{18}(1-\alpha)^2,$$ which prove that the bound is sharp.