Toeplitz Determinants for a Class of Holomorphic Mappings in Higher Dimensions

Theorem 2.1. Let $g \in \mathcal{H}(\mathbb{B},\mathbb{C})$ with g(0) = 1, $g(z) \neq 0$ , $z \in \mathbb{B}$ and suppose that G(z) = zg(z). If $(DG(z))^{-1}(D^2G(z)(z^2) + DG(z)(z)) \in \mathcal{M}_{\Phi}$ such that $\Phi$ satisfies $$|\Phi''(0) + 2(\Phi'(0))^2| > 2\Phi'(0) > 0.$$ then $$\left| \left( \frac{l_z(D^2G(0)(z^2))}{2!||z||^2} \right)^2 - \left( \frac{l_z(D^3G(0)(z^3))}{3!||z||^3} \right)^2 \right| \leq \frac{(\Phi'(0))^2}{4} + \frac{(\Phi'(0))^2}{36} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^2.$$ The bound is sharp. Proof. Fix $z \in X \setminus \{0\}$ and let $h : \mathbb{U} \to \mathbb{C}$ be defined by $$h(\zeta) = \begin{cases} \frac{l_z((DG(\zeta z_0))^{-1}(D^2G(\zeta z_0)((\zeta z_0)^2) + DG(\zeta z_0)\zeta z_0))}{\zeta}, & \zeta \neq 0, \\ 1, & \zeta = 0, \end{cases}$$ where $z_0 = \frac{z}{\|z\|}$ . Then $h \in \mathcal{H}(\mathbb{U})$ and $h(0) = \Phi(0) = 1$ . Since $(DG(z))^{-1}(D^2G(z)(z^2) + DG(z)(z)) \in \mathcal{M}_{\Phi}$ , therefore, we have $$\begin{split} h(\zeta) &= \frac{l_z((DG(\zeta z_0))^{-1}(D^2G(\zeta z_0)((\zeta z_0)^2) + DG(\zeta z_0)\zeta z_0))}{\zeta} \\ &= \frac{l_{z_0}((DG(\zeta z_0))^{-1}(D^2G(\zeta z_0)((\zeta z_0)^2) + DG(\zeta z_0)\zeta z_0))}{\zeta} \\ &= \frac{l_{\zeta z_0}((DG(\zeta z_0))^{-1}(D^2G(\zeta z_0)((\zeta z_0)^2) + DG(\zeta z_0)\zeta z_0))}{\|\zeta z_0\|} \in \Phi(\mathbb{U}), \;\; \zeta \in \mathbb{U}. \end{split}$$ Applying a similar method used in [7, Theorem 7.1.14], we get <span id="page-3-1"></span><span id="page-3-0"></span> $$(DG(z))^{-1} = \frac{1}{g(z)} \left( I - \frac{\frac{zDg(z)}{g(z)}}{1 + \frac{Dg(z)z}{g(z)}} \right). \tag{2.1}$$ A simple computation using the fact G(z) = zg(z) yields <span id="page-3-3"></span> $$D^{2}G(z)(z^{2}) + DG(z)(z) = (D^{2}g(z)(z^{2}) + 3Dg(z)(z) + g(z))z.$$ (2.2) By using (2.1) and (2.2), it follows $$(DG(z))^{-1}(D^2G(z)(z^2) + DG(z)(z)) = \frac{D^2g(z)(z^2) + 3Dg(z)(z) + g(z)}{g(z) + Dg(z)(z)}z.$$ (2.3) Consequently <span id="page-3-2"></span> $$l_z((DG(z))^{-1}(D^2G(z)(z^2) + DG(z)(z))) = \frac{D^2g(z)(z^2) + 3Dg(z)(z) + g(z)}{g(z) + Dg(z)(z)} ||z||.$$ (2.4) Using (2.4), we obtain $$h(\zeta) = \frac{l_{\zeta z_0}((DG(\zeta z_0))^{-1}(D^2G(\zeta z_0)((\zeta z_0)^2) + DG(\zeta z_0)\zeta z_0))}{\|\zeta z_0\|}$$ $$= \frac{D^2g(\zeta z_0)((\zeta z_0)^2) + 3Dg(\zeta z_0)(\zeta z_0) + g(\zeta z_0)}{g(\zeta z_0) + Dg(\zeta z_0)(\zeta z_0)}.$$ Equivalently, we can write $$h(\zeta)(g(\zeta z_0) + Dg(\zeta z_0)(\zeta z_0)) = D^2 g(\zeta z_0)((\zeta z_0)^2) + 3Dg(\zeta z_0)(\zeta z_0) + g(\zeta z_0).$$ The series expansion in terms of $\zeta$ gives $$\left(1 + h'(0)\zeta + \frac{h''(0)}{2}\zeta^2 + \cdots\right) \left(1 + 2Dg(0)(z_0)\zeta + \frac{3Dg(0)(z_0^2)}{2}\zeta^2 + \cdots\right) = 1 + 4Dg(0)(z_0)\zeta + \frac{9Dg(0)(z_0^2)}{2}\zeta^2 + \cdots$$ <span id="page-4-0"></span>Comparison of the homogenous expansions of either sides of the above equality provide $h'(0) = 2Dg(0)(z_0)$ . That is $$h'(0)||z|| = 2Dg(0)(z). (2.5)$$ Also, we have $$\frac{D^2G(0)(z^2)}{2!} = Dg(0)(z)z,$$ which gives $$\frac{l_z(D^2G(0)(z^2))}{2!} = Dg(0)(z)||z||.$$ Now, using $|h'(0)| \leq \Phi'(0)$ with (2.5), we obtain <span id="page-4-3"></span><span id="page-4-2"></span><span id="page-4-1"></span> $$\left| \frac{l_z(D^2G(0)(z^2))}{2!||z||^2} \right| \le \frac{\Phi'(0)}{2}.$$ (2.6) Moreover, for $\lambda \in \mathbb{C}$ , Xu et al. [17, Theorem 3.1] proved that $$\left| \frac{l_{z}(D^{3}G(0)(z^{3}))}{3!||z||^{3}} - \lambda \left( \frac{l_{z}(D^{2}G(0)(z^{2}))}{2!||z||^{2}} \right)^{2} \right| \\ \leq \frac{|\Phi'(0)|}{6} \max \left\{ 1, \left| \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \left( 1 - \frac{3}{2}\lambda \right) \Phi'(0) \right| \right\}, \quad z \in \mathbb{B} \setminus \{0\}.$$ (2.7) Since $|\Phi''(0) + 2(\Phi'(0))^2| \ge 2\Phi'(0)$ , therefore the above inequality gives $$\left| \frac{l_z(D^3 G(0)(z^3))}{3! ||z||^3} \right| \le \frac{\Phi'(0)}{6} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right). \tag{2.8}$$ Also, note that $$\begin{split} \left| \left( \frac{l_z(D^3 G(0)(z^3))}{3! ||z||^3} \right)^2 - \left( \frac{l_z(D^2 G(0)(z^2))}{2! ||z||^2} \right)^2 \right| \\ & \leq \left| \frac{l_z(D^3 G(0)(z^3))}{3! ||z||^3} \right|^2 + \left| \frac{l_z(D^2 G(0)(z^2))}{2! ||z||^2} \right|^2. \end{split}$$ The required bound follows from the above inequality together with the bounds given in (2.6) and (2.8). The result is sharp for the function G given by <span id="page-4-4"></span> $$DG(z) = I \exp \int_{0}^{T_u(z)} \frac{\Phi(it) - 1}{t} dt, \quad z \in \mathbb{B}, \quad ||u|| = 1.$$ (2.9) Clearly, $(DG(z))^{-1}(D^2G(z)(z^2) + DG(z)(z)) \in \mathcal{M}_{\Phi}$ and $$\frac{D^3G(0)(z^3)}{3!} = -\frac{1}{6} \left( \frac{\Phi''(0)}{2} + (\Phi'(0))^2 \right) (l_u(z))^2 z \text{ and } \frac{D^2G(0)(z^2)}{2!} = \frac{i\Phi'(0)}{2} l_u(z) z,$$ which immediately gives $$\frac{l_z(D^2G(0)(z^2))}{2!} = \frac{i\Phi'(0)}{2}l_u(z)||z||$$ and $$\frac{l_z(D^3G(0)(z^3))\|z\|}{3!} = -\frac{1}{6} \left( \frac{\Phi''(0)}{2} + (\Phi'(0))^2 \right) (l_u(z))^2 \|z\|^2.$$ Taking z = ru (0 < r < 1), we get $$\frac{l_z(D^3G(0)(z^3))}{3!\|z\|^3} = -\frac{1}{6} \left( \frac{\Phi''(0)}{2} + (\Phi'(0))^2 \right) \text{ and } \frac{l_z(D^2G(0)(z^2))}{2!\|z\|^2} = \frac{i\Phi'(0)}{2}.$$ (2.10) According to the above equations, we have $$\left| \left( \frac{l_z(D^3 G(0)(z^3))}{3! \|z\|^3} \right)^2 - \left( \frac{l_z(D^2 G(0)(z^2))}{2! \|z\|^2} \right)^2 \right| = \frac{(\Phi'(0))^2}{4} + \frac{1}{36} \left( \frac{\Phi''(0)}{2} + (\Phi'(0))^2 \right)^2,$$ which establishes the sharpness of the bound and completes the

Theorem 2.2. Let $g \in \mathcal{H}(\mathbb{B},\mathbb{C})$ with g(0) = 1, $g(z) \neq 0$ , $z \in \mathbb{B}$ and suppose that G(z) = zg(z). If $DG(z))^{-1}(D^2G(z)(z^2) + DG(z)(z) \in \mathcal{M}_{\Phi}$ such that $\Phi$ satisfy $$2\Phi'(0) - 2(\Phi'(0))^2 \le \Phi''(0) \le 4(\Phi'(0))^2 - 2\Phi'(0).$$ then $$|2a_2^2a_3 - a_3^2 - 2a_2^2 + 1| \le 1 + 2(\Phi'(0))^2 + \frac{(\Phi'(0))^2}{4} \left(\frac{\Phi''(0)}{2\Phi'(0)} - 3\Phi'(0)\right) \left(\frac{\Phi''(0)}{2\Phi'(0)} + \Phi'(0)\right),$$ where <span id="page-5-1"></span> $$a_3 = \frac{l_z(D^3G(0)(z^3))}{3!||z||^3}$$ and $a_2 = \frac{l_z(D^2G(0)(z^2))}{2!||z||^2}$ . The bound is sharp. Proof. Since $2\Phi'(0) \leq 4(\Phi'(0))^2 - \Phi''(0)$ , inequality (2.7) gives $$\left| \frac{l_z(D^3G(0)(z^3))}{3!||z||^3} - 2\left(\frac{l_z(D^2G(0)(z^2))}{2!||z||^2}\right)^2 \right| \le \frac{\Phi'(0)}{6} \left(2\Phi'(0) - \frac{1}{2}\frac{\Phi''(0)}{\Phi'(0)}\right) \tag{2.11}$$ for $z \in \mathbb{B} \setminus \{0\}$ . Also, since $\Phi$ satisfy $\Phi''(0) + 2(\Phi'(0))^2 \ge 2\Phi'(0)$ , therefore by (2.7), we have $$\left| \frac{l_z(D^3 G(0)(z^3))}{3!||z||^3} \right| \le \frac{\Phi'(0)}{6} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right), \quad z \in \mathbb{B} \setminus \{0\}.$$ (2.12) Also, we have <span id="page-5-0"></span> $$|2a_2^2a_3 - 2a_2^2 - a_3^2 + 1| \le 1 + 2|a_2|^2 + |a_3||a_3 - 2a_2^2|$$ The required bound follows directly from the above inequality along with the bounds given in (2.6) and (2.12), and the bound of $|a_3 - 2a_2^2|$ given by (2.11). Equality case holds for the function G(z) defined in (2.9) as for this function, we have $a_2 = \frac{i\Phi'(0)}{2}$ , $a_3 = -\frac{1}{6} \left( \frac{\Phi''(0)}{2} + (\Phi'(0))^2 \right)$ and hence $$2a_2^2a_3 - 2a_2^2 - a_3^2 + 1 = 2(\Phi'(0))^2 + \frac{(\Phi'(0))^2}{12} \left(\frac{\Phi''(0)}{2\Phi'(0)} - 3\Phi'(0)\right) \left(\frac{\Phi''(0)}{2\Phi'(0)} + \Phi'(0)\right) + 1,$$ which establish the sharpness of the result.

Theorem 2.3. Let $g \in \mathcal{H}(\mathbb{U}^n, \mathbb{C})$ with g(0) = 1, $g(z) \neq 0$ , $z \in \mathbb{U}^n$ and suppose that G(z) = zg(z). If $(DG(z))^{-1}(D^2G(z)(z^2) + DG(z)(z) \in \mathcal{M}_{\Phi}$ such that $\Phi$ satisfies $$|\Phi''(0) + 2(\Phi'(0))^2| \ge 2\Phi'(0),$$ <span id="page-6-3"></span>then $$\left\| \left( \frac{D^{3}G(0)(z^{3})}{3!} \right)^{2} - \left( \frac{D^{2}G(0)(z^{2})}{2!} \right)^{2} \right\| \\ \leq \frac{(\Phi'(0))^{2} \|z\|^{6}}{36} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^{2} + \frac{(\Phi'(0))^{2} \|z\|^{4}}{4}, \quad z \in \mathbb{U}^{n}. \right\}$$ (2.13) The bound is sharp. Proof. For $z \in \mathbb{U}^n \setminus \{0\}$ and $z_0 = \frac{z}{\|z\|}$ , define $h_k : \mathbb{U} \to \mathbb{C}$ such that <span id="page-6-4"></span> $$h_k(\zeta) = \begin{cases} \frac{\zeta z_k}{p_k(\zeta z_0) \|z_0\|}, & \zeta \neq 0, \\ 1, & \zeta = 0, \end{cases}$$ (2.14) where $p(z) = (DG(z))^{-1}(D^2G(z)(z^2) + DG(z)(z))$ and k satisfies $|z_k| = ||z|| = \max_{1 \le j \le n} \{z_j\}$ . By (2.3), we have $$h_k(\zeta) = \frac{D^2 g(\zeta z_0)((\zeta z_0)^2) + 3Dg(\zeta z_0)(\zeta z_0) + g(\zeta z_0)}{g(\zeta z_0) + Dg(\zeta z_0)(\zeta z_0)},$$ or, equivalently $$h_k(\zeta)(g(\zeta z_0) + Dg(\zeta z_0)(\zeta z_0)) = D^2 g(\zeta z_0)((\zeta z_0)^2) + 3Dg(\zeta z_0)(\zeta z_0) + g(\zeta z_0).$$ Comparison of same homogeneous expansions in the Taylor series expansions in terms of $\zeta$ yield <span id="page-6-1"></span><span id="page-6-0"></span> $$h'_k(0) = 2Dg(0)(z_0). (2.15)$$ Furthermore, from $G(z_0) = z_0 g(z_0)$ , we have $$\frac{D^2 G_k(0)(z_0^2)}{2!} = Dg(0)(z_0) \frac{z_j}{\|z\|}.$$ (2.16) Combining (2.15) and (2.16) with the fact $|h'_k(0)| \leq \Phi'(0)$ gives $$\left| \frac{D^2 G_k(0)(z_0^2)}{2!} \frac{\|z\|}{z_j} \right| \le \frac{\Phi'(0)}{2}.$$ If $z_0 \in \partial \mathbb{U}^n$ , then we get $$\left| \frac{D^2 G_k(0)(z_0^2)}{2!} \right| \le \frac{\Phi'(0)}{2}.$$ Since $$\frac{D^2 G_k(0)(z_0^2)}{2!}$$ , $k = 1, 2, \dots n$ are holomorphic functions on $\overline{\mathbb{U}}^n$ , therefore by the maximum modulus theorem of holomorphic functions on the unit polydisc, we have $$\left| \frac{D^2 G_k(0)(z_0^2)}{2!} \right| \le \frac{\Phi'(0)}{2}, \quad z_0 \in \mathbb{U}^n, k = 1, 2, \dots n.$$ <span id="page-6-2"></span>That is $$\left| \frac{D^2 G_k(0)(z^2)}{2!} \right| \le \frac{\Phi'(0) ||z||^2}{2}, \quad z \in \partial \mathbb{U}^n, k = 1, 2, \dots n.$$ (2.17) For $\lambda \in \mathbb{C}$ , Xu et al. [17, Theorem 3.2] established that $$\begin{vmatrix} \frac{D^{3}G_{k}(0)(z^{3})}{3!} - \lambda \frac{1}{2}D^{2}G_{k}(0)\left(z, \frac{D^{2}G(0)(z^{2})}{2!}\right) \\ \leq \frac{|\Phi'(0)|||z||^{3}}{6} \max\left\{1, \left|\frac{1}{2}\frac{\Phi''(0)}{\Phi'(0)} + \left(1 - \frac{3}{2}\lambda\right)\Phi'(0)\right|\right\}. \end{cases}$$ (2.18) Since $|\Phi''(0) + 2(\Phi'(0))^2| \ge 2\Phi'(0)$ , therefore, from (2.18), we get <span id="page-7-1"></span><span id="page-7-0"></span> $$\left| \frac{D^3 G_k(0)(z^3)}{3!} \right| \le \frac{\Phi'(0) ||z||^3}{6} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right) \tag{2.19}$$ for $z \in \mathbb{U}^n$ and $k = 1, 2, \dots n$ . Using the bounds from (2.17) and (2.19), we have $$\left| \left( \frac{D^3 G_k(0)(z^3)}{3!} \right)^2 - \left( \frac{D^2 G_k(0)(z^2)}{2!} \right)^2 \right|$$ $$\leq \frac{(\Phi'(0))^2 \|z\|^6}{36} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^2 + \frac{(\Phi'(0))^2 \|z\|^4}{4}$$ for $z \in \mathbb{U}^n$ and $k = 1, 2, \dots n$ . Therefore, $$\left\| \left( \frac{D^3 G(0)(z^3)}{3!} \right)^2 - \left( \frac{D^2 G(0)(z^2)}{2!} \right)^2 \right\|$$ $$\leq \frac{(\Phi'(0))^2 \|z\|^6}{36} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^2 + \frac{(\Phi'(0))^2 \|z\|^4}{4}, \quad z \in \mathbb{U}^n,$$ which is the required bound. To prove the sharpness of the bound, consider the function G given by <span id="page-7-2"></span> $$DG(z) = I \exp \int_{0}^{z_1} \frac{\Phi(it) - 1}{t} dt.$$ (2.20) It can be showed that $(DG(z))^{-1}(D^2G(z)(z^2) + DG(z)(z)) \in \mathcal{M}_{\Phi}$ and for $z = (r, 0, \dots, 0)'$ in (2.20), the equality case holds in (2.13).

Theorem 2.4. Let $g \in \mathcal{H}(\mathbb{U}^n,\mathbb{C}), \ g(0) = 1, \ g(z) \neq 0, \ z \in \mathbb{U}^n$ and suppose that G(z) = zg(z). If $(DG(z))^{-1}(D^2G(z)(z^2) + DG(z)(z)) \in \mathcal{M}_{\Phi}$ and $\Phi$ satisfy $$2\Phi'(0) - 2(\Phi'(0))^2 \le \Phi''(0) \le 4(\Phi'(0))^2 - 2\Phi'(0),$$ then $$\begin{aligned} \|2a_2^2a_3 - a_3^2 - 2a_2^2 + 1\| \\ &\leq 1 + \frac{(\Phi'(0))^2 \|z\|^6}{36} \left(2\Phi'(0) - \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)}\right) \left(\frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0)\right) + \frac{(\Phi'(0))^2 \|z\|^4}{2}, \end{aligned}$$ where $$a_3 = \frac{D^3 G(0)(z^3)}{3!}$$ and $a_2^2 = \frac{1}{2} D^2 G(0) \left(z, \frac{D^2 G(0)(z^2)}{2!}\right)$ . The bound is sharp. Proof. Since G(z) = zg(z), we have $$\frac{1}{2}D^2G_k(0)\left(z_0, \frac{D^2G(0)(z_0^2)}{2!}\right)\frac{z_k}{\|z\|} = \left(\frac{D^2G_k(0)(z_0^2)}{2!}\right)^2, \quad k = 1, 2, \dots, n,$$ where $z_0 = \frac{z_k}{\|z\|}$ and k satisfies $|z_k| = \|z\| = \max_{1 \le j \le n} \{|z_j|\}$ (see [16]). If $z_0 \in \partial \mathbb{U}^n$ , then $$\left| \frac{1}{2} D^2 G_k(0) \left( z_0, \frac{D^2 G(0)(z_0^2)}{2!} \right) \right| = \left| \frac{D^2 G_k(0)(z_0^2)}{2!} \right|^2.$$ Considering the function $h_k(\zeta)$ given in (2.14) and following the same methodology as in the proof of Theorem 2.3, we obtain (2.17), which together with the above relation yields <span id="page-8-1"></span><span id="page-8-0"></span> $$\left| \frac{1}{2} D^2 G_k(0) \left( z_0, \frac{D^2 G(0)(z_0^2)}{2!} \right) \right| \le \frac{(\Phi'(0))^2 ||z||^4}{4}. \tag{2.21}$$ Also, since $\Phi$ satisfy $2\Phi'(0) \leq 4(\Phi'(0))^2 - \Phi''(0)$ , therefore from (2.18), we obtain $$\left| \frac{D^{3}G_{k}(0)(z^{3})}{3!} - D^{2}G_{k}(0)\left(z, \frac{D^{2}G(0)(z^{2})}{2!}\right) \right| \\ \leq \frac{|\Phi'(0)|||z||^{3}}{6} \left(2\Phi'(0) - \frac{1}{2}\frac{\Phi''(0)}{\Phi'(0)}\right), \quad k = 1, 2, \dots n.$$ (2.22) Thus, from (2.19), (2.21) and (2.22), we have $$\begin{split} & \left| 1 + D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \left( \frac{D^3 G_k(0)(z^3)}{3!} \right) - D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \right. \\ & \left. - \left( \frac{D^3 G_k(0)(z^3)}{3!} \right)^2 \right| \\ & \leq 1 + \left| \frac{D^3 G_k(0)(z^3)}{3!} \right| \left| \frac{D^3 G_k(0)(z^3)}{3!} - D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \right| \\ & \left. + \left| D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \right| \\ & \leq 1 + \frac{(\Phi'(0))^2 \|z\|^6}{36} \left( 2\Phi'(0) - \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} \right) \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right) + \frac{(\Phi'(0))^2 \|z\|^4}{2} \end{split}$$ for $z \in \mathbb{U}^n$ and $k = 1, 2, \dots n$ . Therefore $$\begin{split} & \left\| 1 + D^2 G(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \left( \frac{D^3 G(0)(z^3)}{3!} \right) - D^2 G(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \\ & - \left( \frac{D^3 G(0)(z^3)}{3!} \right)^2 \right\| \\ & \leq 1 + \frac{(\Phi'(0))^2 \|z\|^6}{36} \left( 2\Phi'(0) - \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} \right) \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right) + \frac{(\Phi'(0))^2 \|z\|^4}{2}, \end{split}$$ which is the required bound. Sharpness of the bound can be seen from the function G(z) defined in (2.20) by taking $z = (r, 0, \dots, 0)$ , which completes the proof. Note that if $g \in \mathcal{H}(\mathbb{B})$ and $(Dg(z))^{-1}(D^2g(z)(z^2) + Dg(z)(z)) \in \mathcal{M}_{\Phi}$ , then various choices of $\Phi$ give different subclasses of holomorphic mappings. For instance, when $\Phi(z) = (1 + (1 - 2\alpha)z)/(1 - z)$ and $\Phi(z) = (1 + z)/(1 - z)$ , we easily obtain $g \in \mathcal{K}_{\alpha}(\mathbb{B})$ and $g \in \mathcal{K}(\mathbb{B})$ , respectively. For these classes, Theorem 2.1 to Theorem 2.4 yield the following results.

Corollary 2.5. Let $g \in \mathcal{H}(\mathbb{B},\mathbb{C})$ and $G(z) = zg(z) \in \mathcal{C}_{\alpha}(\mathbb{B})$ . Then the following inequality holds: $$\begin{split} \left| \left( \frac{l_z(D^2 G(0)(z^2))}{2! ||z||^2} \right)^2 - \left( \frac{l_z(D^3 G(0)(z^3))}{3! ||z||^3} \right)^2 \right| \\ & \leq \frac{2(1-\alpha)^2 (2\alpha^2 - 6\alpha + 9)}{\alpha}, \quad l_z \in T_z, \ z \in \mathbb{B} \setminus \{0\}. \end{split}$$ If $\mathbb{B} = \mathbb{U}^n$ and $X = \mathbb{C}^n$ , then <span id="page-9-0"></span> $$\left\| \left( \frac{D^{3}G(0)(z^{3})}{3!} \right)^{2} - \left( \frac{D^{2}G(0)(z^{2})}{2!} \right)^{2} \right\| \\ \leq (1 - \alpha)^{2} \|z\|^{4} + \frac{(2\alpha^{2} - 5\alpha + 3)^{2} \|z\|^{6}}{9}, \ z \in \mathbb{U}^{n}. \right\}$$ (2.23) All these bounds are sharp. Remark 2.1. In case of n=1, $\mathbb{B}=\mathbb{U}$ and (2.23) reduces to the following: $$\left| \left( \frac{G^{(3)}(0)}{3!} \right)^2 - \left( \frac{G''(0)}{2!} \right)^2 \right| \le \frac{2(1-\alpha)^2 (2\alpha^2 - 6\alpha + 9)}{9},$$ which is equivalent to Theorem C.

Corollary 2.6. Let $g \in \mathcal{H}(\mathbb{B}, \mathbb{C})$ and $G(z) = zg(z) \in \mathcal{C}_{\alpha}(\mathbb{B})$ . Then for $\alpha \in [0, 1/2]$ , the following sharp bound holds: $$|2a_2^2a_3 - a_3^2 - 2a_2^2 + 1| \le \frac{8\alpha^4 - 34\alpha^3 + 71\alpha^2 - 72\alpha + 36}{9}, \quad z \in \mathbb{B} \setminus \{0\},\$$ where $$a_3 = \frac{l_z(D^3G(0)(z^3))}{3!||z||^3}$$ and $a_2 = \frac{l_z(D^2G(0)(z^2))}{2!||z||^2}$ . <span id="page-9-1"></span>Corollary 2.7. Let $g \in \mathcal{H}(\mathbb{U}^n, \mathbb{C})$ and $G(z) = zg(z) \in \mathcal{C}_{\alpha}(\mathbb{U}^n)$ . Then for $\alpha \in [0, 1/2]$ , the following sharp inequality holds: $$||2a_2^2a_3 - a_3^2 - 2a_2^2 + 1|| \le 1 + 2||z||^4 (1 - \alpha)^2 + \frac{(1 - \alpha)^2 (9 - 18\alpha + 8\alpha^2)||z||^6}{9}$$ (2.24) for $z \in \mathbb{U}^n$ , where $$a_3 = \frac{D^3 G(0)(z^3)}{3!}$$ and $a_2^2 = \frac{1}{2} D^2 G(0) \left(z, \frac{D^2 G(0)(z^2)}{2!}\right)$ . Remark 2.2. When n = 1, Corollary 2.7 is equivalent to Theorem D. In particular, for $\alpha = 0$ , we obtain the following results for the class $\mathcal{C}$ in higher dimensions.

Corollary 2.8. Let $g \in \mathcal{H}(\mathbb{B}, \mathbb{C})$ and $G(z) = zg(z) \in \mathcal{C}(\mathbb{B})$ . Then the following holds: $$\left| \left( \frac{l_z(D^2G(0)(z^2))}{2!||z||^2} \right)^2 - \left( \frac{l_z(D^3G(0)(z^3))}{3!||z||^3} \right)^2 \right| \le 2, \quad l_z \in T_z, \ z \in \mathbb{B} \setminus \{0\}.$$ If $\mathbb{B} = \mathbb{U}^n$ and $X = \mathbb{C}^n$ , then <span id="page-9-2"></span> $$\left\| \left( \frac{D^3 G(0)(z^3)}{3!} \right)^2 - \left( \frac{D^2 G(0)(z^2)}{2!} \right)^2 \right\| \le \|z\|^4 + \|z\|^6, \quad z \in \mathbb{U}^n.$$ (2.25) All these bounds are sharp. Remark 2.3. For n = 1, (2.25) is equivalent to Theorem A.

Corollary 2.9. Let $g \in \mathcal{H}(\mathbb{B},\mathbb{C})$ and $G(z) = zg(z) \in \mathcal{C}(\mathbb{B})$ . Then the following sharp bound holds: $$|2a_2^2a_3 - a_3^2 - 2a_2^2 + 1| \le 4, \quad z \in \mathbb{B} \setminus \{0\},$$ where $$a_3 = \frac{l_z(D^3G(0)(z^3))}{3!||z||^3}$$ and $a_2 = \frac{l_z(D^2G(0)(z^2))}{2!||z||^2}$ .

Corollary 2.10. Let $g \in \mathcal{H}(\mathbb{U}^n, \mathbb{C})$ and $G(z) = zg(z) \in \mathcal{C}(\mathbb{U}^n)$ . Then for $\alpha \in [0, 1/2]$ , the following sharp estimation holds: $$|2a_2^2a_3 - a_3^2 - 2a_2^2 + 1| \le 1 + 2||z||^4 + ||z||^6$$ for $z \in \mathbb{U}^n$ , where $$a_3 = \frac{D^3 G(0)(z^3)}{3!}$$ and $a_2^2 = \frac{1}{2} D^2 G(0) \left(z, \frac{D^2 G(0)(z^2)}{2!}\right)$ . Remark 2.4. when n = 1, Corollary 2.10 is equivalent to Theorem B.

Definition 1.1. [13] Suppose $\alpha \in [0,1)$ and $g: \mathbb{B} \to X$ is a normalized locally biholomorphic mapping. If Re $$\{l_z[(Dg(z))^{-1}(D^2g(z)(z^2) + Dg(z)(z))]\} \ge \alpha ||z||, l_z \in T_z, z \in \mathbb{B} \setminus \{0\},$$ then f is called a quasi convex mapping of type B and order $\alpha$ on $\mathbb{B}$ . If $\mathbb{B} = \mathbb{U}^n$ and $X = \mathbb{C}^n$ , then the above condition reduces to $$\left| \frac{q_k(z)}{z_k} - \frac{1}{2\alpha} \right| < \frac{1}{2\alpha}, \quad \forall z \in \mathbb{U}^n \setminus \{0\},$$ where $$q(z) = (q_1(z), q_2(z), \dots, q_n(z))' = (Dg(z))^{-1}(D^2g(z)(z^2) + Dg(z)(z))$$ is a column vector in $\mathbb{C}^n$ and k satisfies $$|z_k| = ||z|| = \max_{1 \le j \le n} \{|z_j|\}.$$ For $\mathbb{B} = \mathbb{U}$ and $X = \mathbb{C}$ , the relation is equivalent to $$\operatorname{Re}\left(1 + \frac{zg''(z)}{g'(z)}\right) > \alpha, \quad z \in \mathbb{U}.$$ Let $\mathcal{K}_{\alpha}(\mathbb{B})$ denote the class of quasi convex mappings of type B and order $\alpha$ . When $\alpha = 0$ , Definition 1.1 is the definition of quasi convex mapping of type B introduced by Roper and Suffridge [15].

Definition 1.2. Let $\Phi: \mathbb{U} \to \mathbb{C}$ be a biholomorphic function such that $\operatorname{Re} \Phi(z) > 0$ , $\Phi(0) = 1$ , $\Phi'(0) > 0$ and $\Phi''(0) \in \mathbb{R}$ . Let $\mathcal{M}_{\Phi}$ be the class of mappings given by $$\mathcal{M}_{\Phi} = \left\{ p \in \mathcal{H}(\mathbb{B}) : p(0) = Dp(0) = I, \ \frac{l_z(p(z))}{\|z\|} \in \Phi(\mathbb{U}), \ z \in \mathbb{B} \setminus \{0\}, l_z \in T_z \right\}.$$ In case of $\mathbb{B} = \mathbb{U}^n$ and $X = \mathbb{C}^n$ , we have $$\mathcal{M}_{\Phi} = \left\{ p \in \mathcal{H}(\mathbb{B}) : p(0) = Dp(0) = I, \ \frac{p_j(z)}{z_j} \in \Phi(\mathbb{U}), \ z \in \mathbb{U}^n \setminus \{0\} \right\},$$ where $p(z) = (p_1(z), p_2(z), \dots, p_n(z))'$ is a column vector in $\mathbb{C}^n$ and j satisfies $|z_j| = ||z|| = \max_{1 \le k \le n} \{|z_k|\}$ . In 1999, Roper and Suffridge [15] gave a sufficient condition for a normalized biholomorphic convex mapping on the Euclidean unit ball in $\mathbb{C}^n$ . Later, Zhu [20] provided a brief proof of this theorem. Xu et al. [17] obtained the sharp bounds of Fekete-Szegö inequality for the class of quasi-convex mappings of type B and order $\alpha$ defined on the unit ball $\mathbb{B}$ and on the unit polydisc in $\mathbb{C}^n$ . Liu and Liu [14] derived the sharp estimates of all homogenous expansions for a subclass of holomorphic mappings of quasi-convex mappings of type B and order $\alpha$ in higher dimensions. Contrary to the coefficient inequalities for many subclasses of S, only few are known for homogeneous expansions for subclasses of biholomorphic mappings in the case of several complex variables [5, 9, 8, 10, 6, 19]. In case of one complex variable, many coefficient problems are studied for the class $\mathcal{K}$ such as Theorems A-D. A natural question arises that how to retain these results in higher dimensions. Providing an answer to this question is the aim of this study.