Toeplitz Determinants in One and Higher Dimensions

Theorem 2.1. Let $g(z) = z + b_2 z^2 + b_3 z^3 + \cdots \in \Phi(\mathbb{U})$ , where $\Phi$ is same as given in Definition 1.3 and satisfy $$|\Phi''(0) + 2(\Phi'(0))^2| > 2\Phi'(0) > 0$$ If $g(z)/g'(z) \in \mathcal{M}_{\Phi}$ , then $$|T_{2,2}(g)| \le \frac{(\Phi'(0))^2}{4} \left(\frac{1}{2}\frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0)\right)^2 + (\Phi'(0))^2.$$ The bound is sharp. Proof. Since $g(z)/g'(z) \in \mathcal{M}_{\Phi}$ , therefore we have $$G(z) := \frac{zg'(z)}{g(z)} \in \Phi(\mathbb{U}),$$ which yields $G \prec \Phi$ . For $g(z) = z + b_2 z^2 + b_3 z^3 + \cdots$ , the Taylor series expansion of G(z) is given by $$G(z) = 1 + b_2 z + (2b_3 - b_2^2)z^2 + \cdots$$ Xu et al. [21] proved that <span id="page-4-0"></span> $$|b_3 - \lambda b_2^2| \le \frac{|\Phi'(0)|}{2} \max\left\{1, \left| \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + (1 - 2\lambda)\Phi'(0) \right| \right\}, \quad \lambda \in \mathbb{C}.$$ (2.1) Thus, whenever $|\Phi''(0) + 2(\Phi'(0))^2| \ge 2\Phi'(0)$ , the equation (2.1) yields $$|b_3| \le \frac{\Phi'(0)}{2} \left| \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right|. \tag{2.2}$$ Further, using the bound $|G'(0)| \leq \Phi'(0)$ , we obtain <span id="page-4-3"></span><span id="page-4-2"></span><span id="page-4-1"></span> $$|b_2| \le \Phi'(0). \tag{2.3}$$ From (1.1), we have $$|\det T_{2,2}(g)| = |b_3^2 - b_2^2| \le |b_3|^2 + |b_2|^2.$$ Clearly, the required bound follows directly from the above relation together with the bounds of $|b_3|$ and $|b_2|$ given in (2.2) and (2.3) respectively. To see the sharpness of the bound, consider the function $g_{\Phi}: \mathbb{U} \to \mathbb{C}$ given by $$g_{\Phi}(z) = z \exp \int_0^z \frac{(\Phi(it) - 1)}{t} dt = 1 + i\Phi'(0)z - \frac{1}{2} \left( (\Phi'(0))^2 + \frac{\Phi''(0)}{2} \right) z^2 + \cdots$$ (2.4) It can be easily seen that $g_{\Phi}(z)/g'_{\Phi}(z) \in \mathcal{M}_{\Phi}$ and $$|\det T_{2,2}(g_{\Phi})| = \frac{1}{4} \left( (\Phi'(0))^2 + \frac{\Phi''(0)}{2} \right)^2 + (\Phi'(0))^2,$$ which shows that the bound is sharp and completes the

Theorem 2.2. Let $g(z) = z + b_2 z^2 + b_3 z^3 + \cdots \in \Phi(\mathbb{U})$ , where $\Phi$ is same as given in Definition 1.3 and satisfy $$2\Phi'(0) - 2(\Phi'(0))^2 \le \Phi''(0) \le 6(\Phi'(0))^2 - 2\Phi'(0).$$ If $g(z)/g'(z) \in \mathcal{M}_{\Phi}$ , then $$|\det T_{3,1}(g)| \le 1 + 2(\Phi'(0))^2 + \frac{(\Phi'(0))^2}{4} \left(3\Phi'(0) - \frac{\Phi''(0)}{2\Phi'(0)}\right) \left(\frac{\Phi''(0)}{2\Phi'(0)} + \Phi'(0)\right).$$ The bound is sharp. Proof. Since $\Phi''(0) + 2(\Phi'(0))^2 > 2\Phi'(0)$ , by (2.2), we obtain <span id="page-5-1"></span><span id="page-5-0"></span> $$|b_3| \le \frac{\Phi'(0)}{2} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right). \tag{2.5}$$ Also, $6(\Phi'(0))^2 - \Phi''(0) \ge 2\Phi'(0)$ holds, hence (2.1) gives $$|b_3 - 2b_2^2| \le \frac{\Phi'(0)}{2} \left( 3\Phi'(0) - \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} \right). \tag{2.6}$$ From (1.2), we have $$|\det T_{3,1}(g)| = |2b_2^2b_3 - 2b_2^2 - b_3^2 + 1|$$ $$\leq 1 + 2|b_2|^2 + |b_3||b_3 - 2b_2^2|$$ Using the estimates for the second and third coefficients given in (2.5) and (2.2) together with the bound of $|b_3 - 2b_2^2|$ given in (2.6), required bound follows. The estimate is sharp for the function $g_{\Phi}(z) = z + \sum_{n=2}^{\infty} b_n z^n$ given by (2.4). For this function, we have $$1 - 2b_2^2 - b_3(b_3 - 2b_2^2) = 1 + 2(\Phi'(0))^2 + \frac{(\Phi'(0))^2}{4} \left(3\Phi'(0) - \frac{\Phi''(0)}{2\Phi'(0)}\right) \left(\frac{\Phi''(0)}{2\Phi'(0)} + \Phi'(0)\right),$$ which proves the sharpness of the bound. Remark 2.1. By taking $\Phi(z) = (1+z)/(1-z)$ , $\Phi(z) = (1+(1-2\alpha)z)/(1-z)$ and $\Phi(z) = (1+Dz)/(1+Ez)$ , Theorem 2.1 and 2.2 can be deduced to Theorem A, Theorem B and Theorem C, respectively. The bounds for other classes can also be obtained by changing the corresponding function $\Phi$ . For $\Phi(z) = ((1+z)/(1-z))^{\gamma}$ , the following result follows for the class $\mathcal{SS}^*(\gamma)$ .

Corollary 2.3. If $g \in SS^*(\gamma)$ , then for $\gamma \in [1/3, 1]$ , the followings sharp inequalities hold: $$|\det T_{2,2}(q)| \le 9\gamma^4 + 4\gamma^2$$ and $|\det T_{3,1}(q)| \le 15\gamma^4 + 8\gamma^2 + 1$ . Next, we extend the above results on the unit ball $\mathbb{B}$ and on the unit polydisc $\mathbb{U}^n$ .

Theorem 2.4. Let $g \in \mathcal{H}(\mathbb{B},\mathbb{C})$ with g(0) = 1 and suppose that G(z) = zg(z). If $(DG(z))^{-1}G(z) \in \mathcal{M}_{\Phi}$ such that $\Phi$ satisfy $$|\Phi''(0) + 2(\Phi'(0))^2| > 2\Phi'(0)$$ . then $$\left| \left( \frac{l_z(D^2G(0)(z^2))}{2!||z||^2} \right)^2 - \left( \frac{l_z(D^3G(0)(z^3))}{3!||z||^3} \right)^2 \right| \leq \frac{(\Phi'(0))^2}{4} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^2 + (\Phi'(0))^2.$$ The bound is sharp. Proof. Xu et al. [21, Theorem 3.2] proved that $$\left| \frac{l_z(D^3 G(0)(z^3))}{3!||z||^3} - \lambda \left( \frac{l_z(D^2 G(0)(z^2))}{2!||z||^2} \right)^2 \right| \\ \leq \frac{|\Phi'(0)|}{2} \max \left\{ 1, \left| \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + (1 - 2\lambda)\Phi'(0) \right| \right\}, \quad \lambda \in \mathbb{C}, z \in \mathbb{B} \setminus \{0\}.$$ (2.7) Since $|\Phi''(0) + 2(\Phi'(0))^2| \ge 2\Phi'(0)$ , the above inequality gives <span id="page-6-4"></span><span id="page-6-3"></span> $$\left| \frac{l_z(D^3 G(0)(z^3))}{3!||z||^3} \right| \le \frac{\Phi'(0)}{2} \left| \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right|. \tag{2.8}$$ On the other hand, applying a similar method as in [9, Theorem 7.1.14] (also see [21, Theorem 3.2]), we obtain $$(DG(z))^{-1} = \frac{1}{g(z)} \left( I - \frac{\frac{zDg(z)}{g(z)}}{1 + \frac{Dg(z)z}{g(z)}} \right).$$ Therefore <span id="page-6-0"></span> $$(DG(z))^{-1}G(z) = z\left(\frac{zg(z)}{g(z) + Dg(z)z}\right), \quad z \in \mathbb{B},$$ which directly gives $$\frac{\|z\|}{l_z((DG(z))^{-1}G(z))} = 1 + \frac{Dg(z)z}{g(z)}.$$ (2.9) For fix $z \in X \setminus \{0\}$ and $z_0 = \frac{z}{\|z\|}$ , define the function $h : \mathbb{U} \to \mathbb{C}$ such that $$h(\zeta) = \begin{cases} \frac{\zeta}{l_z((DG(\zeta z_0))^{-1}G(\zeta z_0))}, & \zeta \neq 0, \\ 1, & \zeta = 0. \end{cases}$$ Then $h \in \mathcal{H}(\mathbb{U})$ and $h(0) = 1 = \Phi(0)$ . Further, since $(DG(z))^{-1}G(z) \in \mathcal{M}_{\Phi}$ , we find that $$\begin{split} h(\zeta) = & \frac{\zeta}{l_z((DG(\zeta z_0))^{-1}G(\zeta z_0))} = \frac{\zeta}{l_{z_0}((DG(\zeta z_0))^{-1}G(\zeta z_0))} \\ = & \frac{\|\zeta z_0\|}{l_{\zeta z_0}((DG(\zeta z_0))^{-1}G(\zeta z_0))} \in \Phi(\mathbb{U}), \quad \zeta \in \mathbb{U}. \end{split}$$ Taking (2.9) into consideration, we obtain <span id="page-6-5"></span> $$h(\zeta) = \frac{\|\zeta z_0\|}{l_{\zeta z_0}((Dg(\zeta z_0))^{-1}g(\zeta z_0))} = 1 + \frac{Dg(\zeta z_0)\zeta z_0}{g(\zeta z_0)}.$$ (2.10) In view of the Taylor series expansions of $h(\zeta)$ and $g(\zeta z_0)$ , the above equation gives $$\left(1+h'(0)\zeta + \frac{h''(0)}{2}\zeta^2 + \cdots\right) \left(1 + Dg(0)(z_0)\zeta + \frac{D^2g(0)(z_0^2)}{2}\zeta^2 + \cdots\right) = \left(1 + Dg(0)(z_0)\zeta + \frac{D^2g(0)(z_0^2)}{2}\zeta^2 + \cdots\right) \left(Dg(0)(z_0)\zeta + D^2g(0)(z_0^2)\zeta^2 + \cdots\right).$$ Comparison of homogeneous expansions yield that $h'(0) = Dg(0)(z_0)$ . That is <span id="page-6-2"></span> $$h'(0)||z|| = Dg(0)(z). (2.11)$$ Since G(z) = zg(z), therefore, we have <span id="page-6-1"></span> $$\frac{D^2G(0)(z^2)}{2!} = Dg(0)(z)z. (2.12)$$ Moreover, from (2.12), we conclude that <span id="page-7-0"></span> $$\frac{l_z(D^2G(0)(z^2))}{2!} = Dg(0)(z)||z||. \tag{2.13}$$ Thus, equation (2.13) together with (2.11) gives $$\left| \frac{l_z(D^2G(0)(z^2))}{2!} \right| = |Dg(0)(z)||z||| = |h'(0)||z||^2|.$$ Since $h \prec \Phi$ , therefore $|h'(0)| \leq \Phi'(0)$ . Consequently, we obtain <span id="page-7-2"></span><span id="page-7-1"></span> $$\left| \frac{l_z(D^2 G(0)(z^2))}{2! \|z\|^2} \right| \le \Phi'(0). \tag{2.14}$$ Using the bounds given in (2.14) and (2.8) together with $$\left| \left( \frac{l_z(D^2G(0)(z^2))}{2! \|z\|^2} \right)^2 - \left( \frac{l_z(D^3G(0)(z^3))}{3! \|z\|^3} \right)^2 \right| \le \left| \frac{l_z(D^3G(0)(z^3))}{3! \|z\|^3} \right|^2 + \left| \frac{l_z(D^2G(0)(z^2))}{2! \|z\|^2} \right|^2$$ the required bound follows. To see the sharpness, consider the function G given by $$G(z) = z \exp \int_0^{l_u(z)} \frac{(\Phi(it) - 1)}{t} dt, \quad z \in \mathbb{B}, \quad ||u|| = 1.$$ (2.15) It is a simple exercise to see that $(DG(z))^{-1}G(z) \in \mathcal{M}_{\Phi}$ and a quick calculation reveals that $$\frac{D^2G(0)(z^2)}{2!} = i\Phi'(0)l_u(z)z \text{ and } \frac{D^3G(0)(z^3)}{3!} = -\frac{1}{2}\left(\frac{\Phi''(0)}{2} + (\Phi'(0))^2\right)(l_u(z))^2z.$$ In view of the above equations, we have $$\frac{l_z(D^2G(0)(z^2))}{2!} = i\Phi'(0)l_u(z)||z||$$ and <span id="page-7-3"></span> $$\frac{l_z(D^3G(0)(z^3))\|z\|}{3!} = -\frac{1}{2} \left( \frac{\Phi''(0)}{2} + (\Phi'(0))^2 \right) (l_u(z))^2 \|z\|^2.$$ Setting z = ru (0 < r < 1), we get $$\frac{l_z(D^2G(0)(z^2))}{2!\|z\|^2} = i\Phi'(0) \text{ and } \frac{l_z(D^3G(0)(z^3))}{3!\|z\|^3} = -\frac{1}{2} \left(\frac{\Phi''(0)}{2} + (\Phi'(0))^2\right). \tag{2.16}$$ Thus for the function G, we have $$\left| \left( \frac{l_z(D^2G(0)(z^2))}{2!||z||^2} \right)^2 - \left( \frac{l_z(D^3G(0)(z^3))}{3!||z||^3} \right)^2 \right| = \frac{(\Phi'(0))^2}{4} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^2 + |\Phi'(0)|^2,$$ which proves the sharpness of the bound.

Theorem 2.5. Let $g \in \mathcal{H}(\mathbb{B},\mathbb{C})$ with g(0) = 1 and suppose that G(z) = zg(z). If $(DG(z))^{-1}G(z) \in \mathcal{M}_{\Phi}$ such that $\Phi$ satisfy $$2\Phi'(0) - 2(\Phi'(0))^2 \le \Phi''(0) \le 6(\Phi'(0))^2 - 2\Phi'(0),$$ then $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| \le 1 + 2(\Phi'(0))^2 + \frac{(\Phi'(0))^2}{4} \left(3\Phi'(0) - \frac{\Phi''(0)}{2\Phi'(0)}\right) \left(\frac{\Phi''(0)}{2\Phi'(0)} + \Phi'(0)\right),$$ where $$b_3 = \frac{l_z(D^3G(0)(z^3))}{3!||z||^3}$$ and $b_2 = \frac{l_z(D^2G(0)(z^2))}{2!||z||^2}$ . The bound is sharp. Proof. Since $2\Phi'(0) < \Phi''(0) + 2(\Phi'(0))^2$ , therefore from (2.7), we have <span id="page-8-1"></span><span id="page-8-0"></span> $$\left| \frac{l_z(D^3 G(0)(z^3))}{3!||z||^3} \right| \le \frac{\Phi'(0)}{2} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right). \tag{2.17}$$ Again, since $2\Phi'(0) + \Phi''(0) \leq 6(\Phi'(0))^2$ , the inequality (2.7) directly gives $$\left| \frac{l_z(D^3 G(0)(z^3))}{3!||z||^3} - 2\left(\frac{l_z(D^2 G(0)(z^2))}{2!||z||^2}\right)^2 \right| \le \frac{\Phi'(0)}{2} \left(3\Phi'(0) - \frac{1}{2}\frac{\Phi''(0)}{\Phi'(0)}\right). \tag{2.18}$$ Also, we have $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| \le 1 + 2|b_2|^2 + |b_3||b_3 - 2b_2^2|. \tag{2.19}$$ The required bound is derived by using the estimates given in (2.14) and (2.17), and the bound given by (2.18) in the above inequality. The equality case holds for the function G(z) defined by (2.15). It follows from (2.16) that for this function, we have $b_2 = i\Phi'(0)$ , $b_3 = -(\Phi''(0) + 2(\Phi'(0))^2)/4$ and hence $$1 - b_3(b_3 - 2b_2^2) - 2b_2^2 = 1 + 2(\Phi'(0))^2 + \frac{(\Phi'(0))^2}{4} \left(3\Phi'(0) - \frac{\Phi''(0)}{2\Phi'(0)}\right) \left(\frac{\Phi''(0)}{2\Phi'(0)} + \Phi'(0)\right),$$ which shows the sharpness of the bound.

Theorem 2.6. Let $g \in \mathcal{H}(\mathbb{U}^n, \mathbb{C})$ with g(0) = 1 and suppose that G(z) = zg(z). If $(DG(z))^{-1}G(z) \in \mathcal{M}_{\Phi}$ such that $\Phi$ satisfies $$|\Phi''(0) + 2(\Phi'(0))^2| \ge 2\Phi'(0),$$ <span id="page-8-4"></span>then $$\left\| \left( \frac{D^{3}G(0)(z^{3})}{3!} \right)^{2} - \left( \frac{D^{2}G(0)(z^{2})}{2!} \right)^{2} \right\| \\ \leq \frac{(\Phi'(0))^{2} \|z\|^{6}}{4} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^{2} + \left( |\Phi'(0)| \|z\|^{2} \right)^{2}, \quad z \in \mathbb{U}^{n}. \right\}$$ (2.20) The bound is sharp. Proof. For $z \in \mathbb{U}^n \setminus \{0\}$ , let $z_0 = \frac{z}{\|z\|}$ . Define $h_k : \mathbb{U} \to \mathbb{C}$ such that <span id="page-8-5"></span> $$h_k(\zeta) = \begin{cases} \frac{\zeta z_k}{p_k(\zeta z_0) \|z_0\|}, & \zeta \neq 0, \\ 1, & \zeta = 0, \end{cases}$$ (2.21) where $p(z) = (DG(z))^{-1}G(z)$ and k satisfies $|z_k| = ||z|| = \max_{1 \le j \le n} \{|z_j|\}$ . Since $(D(G(z)))^{-1}G(z) \in \mathcal{M}_{\Phi}$ , we have $h_k(\zeta) \in \Phi(\mathbb{U})$ . Further, using (2.10), we have <span id="page-8-2"></span> $$h_k(\zeta) = 1 + \frac{Dg(\zeta z_0)\zeta z_0}{g(\zeta z_0)}$$ or equivalently, $$h_k(\zeta)g(\zeta z_0) = g(\zeta z_0) + Dg(\zeta z_0)\zeta z_0.$$ A comparison of homogeneous expansions obtained by the Taylor series expansions of g and $h_k$ about $\zeta$ gives <span id="page-8-3"></span> $$h'_k(0) = Dg(0)(z_0), \quad \frac{h''_k(0)}{2} = D^2g(0)(z_0^2) - (Dg(0)(z_0))^2.$$ (2.22) Also, using $G(z_0) = z_0 g(z_0)$ , we have $$\frac{D^3 G_k(0)(z_0^3)}{3!} = \frac{D^2 g(0)(z_0^2)}{2!} \frac{z_k}{\|z\|} \quad \text{and} \quad \frac{D^2 G_k(0)(z_0^2)}{2!} = Dg(0)(z_0) \frac{z_k}{\|z\|}. \tag{2.23}$$ Thus, from (2.22) and (2.23), we obtain $$\frac{D^2 G_k(0)(z_0^2)}{2!} \frac{\|z\|}{z_k} = h_k'(0),$$ which gives <span id="page-9-3"></span> $$\left| \frac{D^2 G_k(0)(z_0^2)}{2!} \frac{\|z\|}{z_k} \right| = |h'_k(0)| \le \Phi'(0).$$ If $z_0 \in \partial_0 \mathbb{U}^n$ , then $$\left| \frac{D^2 G_k(0)(z_0^2)}{2!} \right| \le \Phi'(0). \tag{2.24}$$ Since $$\frac{D^2 G_k(0)(z_0^2)}{2!}, \quad k = 1, 2, \dots n$$ are holomorphic on $\overline{\mathbb{U}}^n$ , by virtue of the maximum modulus theorem of holomorphic functions on the unit polydisc, we obtain $$\left| \frac{D^2 G_k(0)(z_0^2)}{2!} \right| \le \Phi'(0), \quad z_0 \in \partial \mathbb{U}^n, k = 1, 2, 3, \dots n.$$ <span id="page-9-1"></span>That is $$\left| \frac{D^2 G_k(0)(z^2)}{2!} \right| \le \Phi'(0) ||z||^2, \quad z \in \mathbb{U}^n, k = 1, 2, 3, \dots n.$$ (2.25) Therefore <span id="page-9-2"></span><span id="page-9-0"></span> $$\left\| \frac{D^2 G(0)(z^2)}{2!} \right\| \le \Phi'(0) \|z\|^2, \quad z \in \mathbb{U}^n.$$ According to the result established by Xu et al. [21, Theorem 3.3], we have $$\left| \frac{D^{3}G_{k}(0)(z^{3})}{3!} - \lambda \frac{1}{2}D^{2}G_{k}(0)\left(z, \frac{D^{2}G(0)(z^{2})}{2!}\right) \right| \\ \leq \frac{|\Phi'(0)|||z||^{3}}{2} \max\left\{1, \left| \frac{1}{2}\frac{\Phi''(0)}{\Phi'(0)} + (1 - 2\lambda)\Phi'(0) \right| \right\}, \quad k = 1, 2, \dots n.$$ (2.26) Since $|\Phi''(0) + 2(\Phi'(0))^2| \ge 2\Phi'(0)$ , therefore from (2.26), it follows that $$\left| \frac{D^3 G_k(0)(z^3)}{3!} \right| \le \frac{|\Phi'(0)| \|z\|^3}{2} \left| \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right|, \quad z \in \mathbb{U}^n, k = 1, 2, \dots n.$$ (2.27) Now, using the bounds given in (2.25) and (2.27), we obtain $$\left| \left( \frac{D^3 G_k(0)(z^3)}{3!} \right)^2 - \left( \frac{D^2 G_k(0)(z^2)}{2!} \right)^2 \right|$$ $$\leq \frac{(\Phi'(0))^2 ||z||^6}{4} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^2 + (\Phi'(0))^2 ||z||^4, \quad k = 1, 2, \dots n.$$ Therefore, $$\begin{split} \left\| \left( \frac{D^3 G(0)(z^3)}{3!} \right)^2 - \left( \frac{D^2 G(0)(z^2)}{2!} \right)^2 \right\| \\ & \leq \frac{(\Phi'(0))^2 \|z\|^6}{4} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^2 + (\Phi'(0))^2 \|z\|^4, \end{split}$$ which is the required bound. To prove the sharpness, consider the function <span id="page-10-2"></span> $$G(z) = z \exp \int_0^{z_1} \frac{\Phi(it) - 1}{t} dt, \quad z \in \mathbb{U}^n.$$ $$(2.28)$$ It is a simple exercise to check that $D(G(z))^{-1}G(z) \in \mathcal{M}_{\Phi}$ . From the above relation, we deduce that $$\frac{D^2G(0)(z^2)}{2!} = i\Phi'(0)z_1z, \quad \frac{D^3G(0)(z^3)}{3!} = -\frac{1}{2}\left(\frac{\Phi''(0)}{2} + (\Phi'(0))^2\right)(z_1)^2z.$$ By taking $z=(r,0,\cdots,0)$ , the equality in (2.20) holds.

Theorem 2.7. Let $g \in \mathcal{H}(\mathbb{U}^n, \mathbb{C})$ with g(0) = 1 and suppose that G(z) = zg(z). If $(DG(z))^{-1}G(z) \in \mathcal{M}_{\Phi}$ such that $\Phi$ satisfy $$2\Phi'(0) - 2(\Phi'(0))^2 \le \Phi''(0) \le 6(\Phi'(0))^2 - 2\Phi'(0),$$ then $$||2b_2^2b_3 - b_3^2 - 2b_2^2 + 1||$$ $$\leq 1 + \frac{(\Phi'(0))^2||z||^6}{4} \left(3\Phi'(0) - \frac{1}{2}\frac{\Phi''(0)}{\Phi'(0)}\right) \left(\frac{1}{2}\frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0)\right) + 2(\Phi'(0))^2||z||^4,$$ where $$b_3 = \frac{D^3 G(0)(z^3)}{3!}$$ and $b_2^2 = \frac{1}{2} D^2 G(0) \left(z, \frac{D^2 G(0)(z^2)}{2!}\right)$ . 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 [19]). If $z_0 \in \partial \mathbb{U}^n$ , then $$\left|\frac{1}{2}D^2G_k(0)\bigg(z_0,\frac{D^2G(0)(z_0^2)}{2!}\bigg)\right| = \left|\frac{D^2G_k(0)(z_0^2)}{2!}\right|^2.$$ Consider the function $h_k(\zeta)$ defined in (2.21). Following the same method as in the proof of Theorem 2.6, we get (2.24), which together with the above relation yield $$\left| \frac{1}{2} D^2 G_k(0) \left( z_0, \frac{D^2 G(0)(z_0^2)}{2!} \right) \right| \le |\Phi'(0)|^2.$$ Since <span id="page-10-1"></span> $$\left| \frac{1}{2} D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \right|, \quad k = 1, 2, \dots n$$ are holomorphic functions on $\overline{\mathbb{U}}^n$ . By virtue of the maximum modulus theorem of holomorphic functions on the unit polydisc, we obtain <span id="page-10-0"></span> $$\left| \frac{1}{2} D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \right| \le |\Phi'(0)|^2 ||z||^4, \quad z \in \mathbb{U}^n, k = 1, 2, \dots n.$$ (2.29) Furthermore, since $2\Phi'(0) < \Phi''(0) + 2(\Phi'(0))^2$ , therefore by (2.26), we have $$\left| \frac{D^3 G_k(0)(z^3)}{3!} \right| \le \frac{\Phi'(0) \|z\|^3}{2} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right), \quad z \in \mathbb{U}^n, k = 1, 2, \dots n.$$ (2.30) Again, since $\Phi$ satisfies $2\Phi'(0) + \Phi''(0) < 6(\Phi'(0))^2$ , the inequality (2.26) gives <span id="page-11-0"></span> $$\left| \frac{D^3 G_k(0)(z^3)}{3!} - D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \right| \le \frac{\Phi'(0) \|z\|^3}{2} \left( 3\Phi'(0) - \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} \right) \tag{2.31}$$ for $z \in \mathbb{U}^n$ and $k = 1, 2 \cdots n$ . Now, using (2.29), (2.30) and (2.31), we have $$\begin{split} & \left| 1 + D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \left( \frac{D^3 G_k(0)(z^3)}{3!} \right) - D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \right. \\ & \left. - \left( \frac{D^3 G_k(0)(z^3)}{3!} \right)^2 \right| \\ & \leq 1 + \left| \frac{D^3 G_k(0)(z^3)}{3!} \right| \left| \frac{D^3 G_k(0)(z^3)}{3!} - D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \right| \\ & \left. + \left| D^2 G_k(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \right| \\ & \leq 1 + \frac{(\Phi'(0))^2 \|z\|^6}{4} \left( 3\Phi'(0) - \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} \right) \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right) + 2(\Phi'(0))^2 \|z\|^4 \end{split}$$ for $z \in \mathbb{U}^n$ and $k = 1, 2, \dots n$ . Therefore $$\begin{split} & \left\| 1 + D^2 G(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \left( \frac{D^3 G(0)(z^3)}{3!} \right) - D^2 G(0) \left( z, \frac{D^2 G(0)(z^2)}{2!} \right) \\ & - \left( \frac{D^3 G(0)(z^3)}{3!} \right)^2 \right\| \\ & \leq 1 + \frac{(\Phi'(0))^2 \|z\|^6}{4} \left( 3\Phi'(0) - \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} \right) \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right) + 2(\Phi'(0))^2 \|z\|^4, \end{split}$$ which is the required bound. Using the same argument as in Theorem 2.6, the function G provided in (2.28) leads to the sharpness of the bound, which completes the proof. Remark 2.2. (i) When $\mathbb{B} = \mathbb{U}$ and $X = \mathbb{C}$ , Theorem 2.4 and Theorem 2.6 are equivalent to Theorem 2.1. (ii) In case of $\mathbb{B} = \mathbb{U}$ and $X = \mathbb{C}$ , Theorem 2.5 and 2.7 are equivalent to Theorem 2.2.

Corollary 3.1. Let $g \in \mathcal{H}(\mathbb{B}, \mathbb{C})$ with g(0) = 1 and $G(z) = zg(z) \in \mathcal{S}^*(\mathbb{B})$ . Then the following holds: $$\left| \left( \frac{l_z(D^2G(0)(z^2))}{2!||z||^2} \right)^2 - \left( \frac{l_z(D^3G(0)(z^3))}{3!||z||^3} \right)^2 \right| \le 13, \quad z \in \mathbb{B} \setminus \{0\}, l_z \in T_z.$$ If $\mathbb{B} = \mathbb{U}^n$ and $X = \mathbb{C}^n$ , then $$\left\| \left( \frac{D^3 G(0)(z^3)}{3!} \right)^2 - \left( \frac{D^2 G(0)(z^2)}{2!} \right)^2 \right\| \le 9 \|z\|^6 + 4 \|z\|^4, \ z \in \mathbb{U}^n.$$ All the estimates are sharp.

Corollary 3.2. Let $g \in \mathcal{H}(\mathbb{B}, \mathbb{C})$ with g(0) = 1 and $G(z) = zg(z) \in \mathcal{S}_{\alpha}^*(\mathbb{B})$ . Then the following holds: $$\left| \left( \frac{l_z(D^2G(0)(z^2))}{2!||z||^2} \right)^2 - \left( \frac{l_z(D^3G(0)(z^3))}{3!||z||^3} \right)^2 \right| \le (1 - \alpha)^2 (4\alpha^2 - 12\alpha + 13), \ z \in \mathbb{B} \setminus \{0\}.$$ If $\mathbb{B} = \mathbb{U}^n$ and $X = \mathbb{C}^n$ , then $$\left\| \left( \frac{D^3 G(0)(z^3)}{3!} \right)^2 - \left( \frac{D^2 G(0)(z^2)}{2!} \right)^2 \right\| \le (1 - \alpha)^2 ((3 - 2\alpha)^2 \|z\|^6 + 4\|z\|^4), \ z \in \mathbb{U}^n.$$ All the estimates are sharp.

Corollary 3.3. Let $g \in \mathcal{H}(\mathbb{B}, \mathbb{C})$ with g(0) = 1 and $G(z) = zg(z) \in \mathcal{SS}^*_{\gamma}(\mathbb{B})$ . Then for $\gamma \in [1/3, 1]$ , the following holds: $$\left| \left( \frac{l_z(D^2 G(0)(z^2))}{2! ||z||^2} \right)^2 - \left( \frac{l_z(D^3 G(0)(z^3))}{3! ||z||^3} \right)^2 \right| \le 9\gamma^4 + 4\gamma^2, \quad z \in \mathbb{B} \setminus \{0\}, l_z \in T_z.$$ If $\mathbb{B} = \mathbb{U}^n$ bnd $X = \mathbb{C}^n$ , then $$\left\| \left( \frac{D^3 G(0)(z^3)}{3!} \right)^2 - \left( \frac{D^2 G(0)(z^2)}{2!} \right)^2 \right\| \le 9\|z\|^6 \gamma^4 + 4\|z\|^4 \gamma^2, \quad z \in \mathbb{U}^n.$$ All the estimates are sharp.

Corollary 3.4. Let $g \in \mathcal{H}(\mathbb{B}, \mathbb{C})$ with g(0) = 1 and $G(z) = zg(z) \in \mathcal{S}^*(\mathbb{B})$ . Then the following holds: $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| \le 24,$$ where $$b_3 = \frac{l_z(D^3G(0)(z^3))}{3!||z||^3}, \quad b_2 = \frac{l_z(D^2G(0)(z^2))}{2!||z||^2}, \quad l_z \in T_z.$$ The estimate is sharp.

Corollary 3.5. Let $g \in \mathcal{H}(\mathbb{B}, \mathbb{C})$ with g(0) = 1 and $G(z) = zg(z) \in \mathcal{S}_{\alpha}^*(\mathbb{B})$ . Then for $\alpha \in [0, 2/3]$ , the following holds: $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| \le 12\alpha^4 - 52\alpha^3 + 91\alpha^2 - 74\alpha + 24$$ where $$b_3 = \frac{l_z(D^3G(0)(z^3))}{3!||z||^3}$$ and $b_2 = \frac{l_z(D^2G(0)(z^2))}{2!||z||^2}$ , $l_z \in T_z$ . The estimate is sharp.

Corollary 3.6. Let $g \in \mathcal{H}(\mathbb{B},\mathbb{C})$ with g(0) = 1 and $G(z) = zg(z) \in \mathcal{SS}^*_{\gamma}(\mathbb{B})$ . Then for $\gamma \in [1/3,1]$ , the following holds: $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| < 15\gamma^4 + 8\gamma^2 + 1,$$ where $$b_3 = \frac{l_z(D^3G(0)(z^3))}{3!||z||^3}$$ and $b_2 = \frac{l_z(D^2G(0)(z^2))}{2!||z||^2}$ , $l_z \in T_z$ . The estimate is sharp. When $\mathbb{B} = \mathbb{U}^n$ and $X = \mathbb{C}^n$ , we obtain the following bounds:

Corollary 3.7. Let $g \in \mathcal{H}(\mathbb{U}^n, \mathbb{C})$ with g(0) = 1 and $G(z) = zg(z) \in \mathcal{S}^*(\mathbb{B})$ . Then the following holds: $$||2b_2^2b_3 - b_3^2 - 2b_2^2 + 1|| \le 15||z||^6 + 8||z||^4 + 1, \ z \in \mathbb{U}^n,$$ where $$b_3 = \frac{D^3 G(0)(z^3)}{3!}$$ and $b_2^2 = \frac{1}{2} D^2 G(0) \left(z, \frac{D^2 G(0)(z^2)}{2!}\right)$ . The estimate is sharp.

Corollary 3.8. Let $g \in \mathcal{H}(\mathbb{U}^n, \mathbb{C})$ with g(0) = 1 and $G(z) = zg(z) \in \mathcal{S}^*_{\alpha}(\mathbb{B})$ . Then for $\alpha \in [0, 2/3]$ , the following holds: $$||2b_2^2b_3 - b_3^2 - 2b_2^2 + 1|| \le (1 - \alpha)^2(2\alpha - 3)(6\alpha - 5)||z||^6 + 8(1 - \alpha)^2||z||^4 + 1, \quad z \in \mathbb{U}^n,$$ where $$b_3 = \frac{D^3 G(0)(z^3)}{3!}$$ and $b_2^2 = \frac{1}{2} D^2 G(0) \left(z, \frac{D^2 G(0)(z^2)}{2!}\right)$ . The estimate is sharp.

Corollary 3.9. Let $g \in \mathcal{H}(\mathbb{U}^n, \mathbb{C})$ with g(0) = 1 and $G(z) = zg(z) \in \mathcal{SS}^*_{\gamma}(\mathbb{B})$ . Then for $\gamma \in [1/3, 1]$ , the following holds: $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| \le 15\gamma^4||z||^6 + 8\gamma^2||z||^4 + 1,$$ where $$b_3 = \frac{D^3 G(0)(z^3)}{3!}$$ and $b_2^2 = \frac{1}{2} D^2 G(0) \left(z, \frac{D^2 G(0)(z^2)}{2!}\right)$ . The estimate is sharp.

Definition 1.1. Let $g: \mathbb{B} \to X$ be a normalized locally biholomorphic mapping and $\alpha \in (0,1)$ . We say that g is starlike of order $\alpha$ if $$\left| \frac{1}{\|z\|} l_z([Dg(z)]^{-1}g(z)) - \frac{1}{2\alpha} \right| < \frac{1}{2\alpha}, \quad \forall z \in \mathbb{B} \setminus \{0\}, \ l_z \in T(z).$$ In case of $X = \mathbb{C}^n$ and $\mathbb{B} = \mathbb{U}^n$ , the above condition is equivalent 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 $$q(z) = (q_1(z), q_2(z), \dots, q_n(z))' = (D(g(z)))^{-1}g(z)$$ 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 $$\operatorname{Re} \frac{zg'(z)}{g(z)} > \alpha, \quad z \in \mathbb{U}.$$ Let $\mathcal{S}_{\alpha}^{}(\mathbb{B})$ denote the class of starlike mappings of order $\alpha$ on $\mathbb{B}$ . When $X = \mathbb{C}$ and $\mathbb{B} = \mathbb{U}$ , the class $\mathcal{S}_{\alpha}^{}(\mathbb{U})$ is denoted by $\mathcal{S}^{*}(\alpha)$ .

Definition 1.2. [16] Let $g: \mathbb{B} \to X$ be a normalized locally biholomorphic mapping and $\gamma \in (0,1]$ . We say that f is strongly starlike mapping of order $\gamma$ if $$\left|\arg l_z([Dg(z)]^{-1}g(z))\right| < \frac{\pi}{2}\gamma, \quad \forall z \in \mathbb{B} \setminus \{0\}, \ l_z \in T(z).$$ In case of $\mathbb{B} = \mathbb{U}^n$ and $X = \mathbb{C}^n$ , the above condition is equivalent to $$\left|\arg\frac{q_j(z)}{z_j}\right| < \frac{\pi}{2}\gamma, \quad z \in \mathbb{U}^n \setminus \{0\}.$$ where $$q(z) = (q_1(z), q_2(z), \dots, q_n(z))' = (D(g(z)))^{-1}g(z)$$ is a column vector in $\mathbb{C}^n$ and j satisfies $$|z_j| = ||z|| = \max_{1 \le k \le n} \{|z_k|\}.$$ In case of $\mathbb{B} = \mathbb{U}$ and $X = \mathbb{C}$ , the relation is equivalent to $$\left|\arg \frac{zg'(z)}{g(z)}\right| < \frac{\pi}{2}\gamma, \quad z \in \mathbb{U}.$$ Let $\mathcal{SS}^_{\gamma}(\mathbb{B})$ denote the class of starlike mappings of order $\gamma$ on $\mathbb{B}$ . When $X = \mathbb{C}$ and $\mathbb{B} = \mathbb{U}$ , the class $\mathcal{SS}^_{\gamma}(\mathbb{U})$ is denoted by $\mathcal{SS}^*(\gamma)$ . For a biholomorphic function $\Phi: \mathbb{U} \to \mathbb{C}$ , which satisfies $\Phi(0) = 1$ and $\operatorname{Re} \Phi(z) > 0$ , Kohr [15] introduced the class $\mathcal{M}_{\Phi}$ containing the functions $p \in \mathcal{H}(\mathbb{B})$ such that Dp(0) = I and $||z||/l_z(p(z)) \in \Phi(\mathbb{U})$ . Here, we additionally take $\Phi'(0) > 0$ , $\Phi''(0) \in \mathbb{R}$ and define the following:

Definition 1.3. Let $\Phi: \mathbb{U} \to \mathbb{C}$ be a biholomorphic function such that $\Phi(0) = 1$ , $\operatorname{Re} \Phi(z) > 0$ , $\Phi'(0) > 0$ and $\Phi''(0) \in \mathbb{R}$ . We define $\mathcal{M}_{\Phi}$ to be the class of mappings given by $$\mathcal{M}_{\Phi} = \left\{ p \in \mathcal{H}(\mathbb{B}) : p(0) = 0, D(p(0)) = I, \frac{\|z\|}{l_z(p(z))} \in \Phi(\mathbb{U}), z \in \mathbb{B} \setminus \{0\}, \ l_z \in T(z) \right\}.$$ For $\mathbb{B} = \mathbb{U}^n$ and $X = \mathbb{C}^n$ , the above relation is equivalent to $$\mathcal{M}_{\Phi} = \left\{ p \in \mathcal{H}(\mathbb{U}^n) : p(0) = 0, D(p(0)) = I, \frac{z_k}{p_k(z)} \in \Phi(\mathbb{U}), z \in \mathbb{U}^n \setminus \{0\} \right\},\,$$ where $p(z) = (p_1(z), p_2(z), \dots, p_n(z))'$ is a column vector in $\mathbb{C}^n$ and 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 $$\mathcal{M}_{\Phi} = \left\{ p \in \mathcal{H}(\mathbb{U}) : p(0) = 0, p'(0) = 1, \frac{z}{p(z)} \in \Phi(\mathbb{U}), z \in \mathbb{U} \right\}.$$ Also, note that, if $g \in \mathcal{H}(\mathbb{B})$ and $D(g(z))^{-1}g(z) \in \mathcal{M}_{\Phi}$ , then suitable choices of $\Phi$ in Definition 1.3 provide different subclasses of holomorphic mappings. For instance, when $\Phi(z) = (1+z)/(1-z)$ , $\Phi(z) = (1+(1-2\alpha)z)/(1-z)$ and $\Phi(z) = ((1+z)/(1-z))^{\gamma}$ , we easily obtain that $g \in \mathcal{S}^(\mathbb{B})$ , $g \in \mathcal{S}^_{\alpha}(\mathbb{B})$ and $g \in \mathcal{S}^*_{\gamma}(\mathbb{B})$ respectively. In this paper, we obtain the sharp bounds of $|\det T_{2,2}(g)|$ and $|\det T_{3,1}(g)|$ for a class of holomorphic functions in the unit disk, which contain the above results as special cases. Further, these results are generalized in higher dimensions.