Abstract
This paper establishes sharp bounds for the second and third-order Toeplitz determinants associated with starlike functions $f$ in the unit disk such that $f(z)-z$ has a zero of order $k+1$ at $z=0$. These bounds are further extended to starlike mappings defined on the unit ball in a complex Banach space and on bounded starlike circular domains in $\mathbb{C}^n$. The derived results generalize several known bounds as special cases.
Results & Lemmas (8)
Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.
Lemma 1
Lemma 1. [13] Let be a normalized locally biholomorphic mapping. The mapping f is said to be starlike on if and only if The class of all…
Lemma 1. [13] Let $f : \mathbb{B} \to X$ be a normalized locally biholomorphic mapping. The mapping f is said to be starlike on $\mathbb{B}$ if and only if
$$\operatorname{Re}(l_z[Df(z)]^{-1}f(z)) > 0, \quad x \in \mathbb{B} \setminus \{0\}, \quad l_z \in T_z.$$
The class of all starlike mappings on $\mathbb{B}$ is denoted by $\mathcal{S}^*(\mathbb{B})$ .
Lemma 2
Lemma 2. [11] is said to be a bounded starlike circular domain if and only if there exists a unique real-valued continuous function, called…
Lemma 2. [11] $\Omega \subset \mathbb{C}^n$ is said to be a bounded starlike circular domain if and only if there exists a unique real-valued continuous function $\rho : \mathbb{C}^n \to \mathbb{R}$ , called the Minkowski functional of $\Omega$ , such that
1.
$$\rho(z) \ge 0$$
, $z \in \mathbb{C}^n$ ; $\rho(z) = 0 \Leftrightarrow z = 0$ ,
2.
$$\rho(tz) = |t|, \rho(z), t \in \mathbb{C}, z \in \mathbb{C}^n,$$
3.
$$\Omega = \{ z \in \mathbb{C}^n : \rho(z) < 1 \}$$
.
Furthermore, if $\rho(z)$ , $z \in \Omega$ , belongs to $\mathcal{C}^1$ except on some lower-dimensional manifold $E \subset \mathbb{C}^n$ , then $\rho(z)$ satisfies the following properties:
<span id="page-2-0"></span>
$$2\frac{\partial \rho(z)}{\partial z}z = \rho(z), \quad z \in \mathbb{C}^n \setminus E,$$
$$2\frac{\partial \rho(z)}{\partial z}\Big|_{z=z_0} = 1, \quad z_0 \in \partial\Omega \setminus E,$$
$$\frac{\partial \rho(\lambda z)}{\partial z} = \frac{\partial \rho(z)}{\partial z}, \quad \lambda \in (0, \infty), \ z \in \mathbb{C}^n \setminus E,$$
$$\frac{\partial \rho(e^{i\theta}z)}{\partial z} = e^{-i\theta}\frac{\partial \rho(z)}{\partial z}, \quad \theta \in \mathbb{R}, \ z \in \mathbb{C}^n \setminus E,$$
where
$$\frac{\partial \rho(z)}{\partial z} = \left(\frac{\partial \rho(z)}{\partial z_1}, \dots, \frac{\partial \rho(z)}{\partial z_n}\right)$$
.
Lemma 3
Lemma 3. [11] Let be a bounded starlike circular domain containing 0, whose Minkowski functional belongs to except on some…
Lemma 3. [11] Let $\Omega \subset \mathbb{C}^n$ be a bounded starlike circular domain containing 0, whose Minkowski functional $\rho(z)$ belongs to $C^1$ except on some lower-dimensional manifolds $E \subset \mathbb{C}^n$ . Let $f: \Omega \to \mathbb{C}^n$ be a normalized locally biholomorphic mapping. Then f is starlike on $\Omega$ if and only if
$$\operatorname{Re}\left(\frac{\partial \rho(z)}{\partial z} (Df(z))^{-1} f(z)\right) > 0, \quad z \in \Omega \setminus E.$$
The class of all starlike mappings on $\Omega$ is denoted by $S^*(\Omega)$ .
Theorem 5
Theorem 5. If, then The estimate is sharp k = 1, 2, 3.
Theorem 5. If $f(z) = z + \sum_{n=1}^{\infty} a_{nk+1} z^{nk+1} \in \mathcal{S}^*$ , then
$$\left|1 - a_{2k+1}^2 - 2a_{k+1}^2 + 2a_{k+1}^2 a_{2k+1}^2\right| \le \begin{cases} 1 + \frac{8}{k^2} + \frac{(k+2)(6-k)}{k^4}; & 1 \le k \le 3, \\ 1 + \frac{8}{k^2} + \frac{2+k}{k^3}; & k \ge 3, \end{cases}$$
The estimate is sharp k = 1, 2, 3.
Theorem 6
Theorem 6. Let and. If F(z) - z has a zero of order k + 1 at z = 0, then The bound is sharp.
Theorem 6. Let $f \in \mathcal{H}(\mathbb{B}, \mathbb{C})$ and $F(z) = zf(z) \in \mathcal{S}^*(\mathbb{B})$ . If F(z) - z has a zero of order k + 1 at z = 0, then
$$\left| \left( \frac{l_z(D^{2k+1}F(0)(z^{2k+1}))}{(2k+1)!||z||^{2k+1}} \right)^2 - \left( \frac{l_z(D^{k+1}F(0)(z^{k+1}))}{(k+1)!||z||^{k+1}} \right)^2 \right| \le \frac{(k+2)^2}{k^4} + \frac{4}{k^2}.$$
The bound is sharp.
Theorem 7
Theorem 7. Let and. If F(z) - z has a zero of order k + 1 at z = 0, then where <span id="page-7-0"></span> (16) The bound is sharp for k =…
Theorem 7. Let $f \in \mathcal{H}(\mathbb{B}, \mathbb{C})$ and $F(z) = zf(z) \in \mathcal{S}^*(\mathbb{B})$ . If F(z) - z has a zero of order k + 1 at z = 0, then
$$\left|1 - a_{2k+1}^2 - 2a_{k+1}^2 + 2a_{k+1}^2 a_{2k+1}^2\right| \le \begin{cases} 1 + \frac{8}{k^2} + \frac{(k+2)(6-k)}{k^4}; & 1 \le k \le 3, \\ 1 + \frac{8}{k^2} + \frac{2+k}{k^3}; & k \ge 3, \end{cases}$$
where
<span id="page-7-0"></span>
$$a_{2k+1} = \frac{l_z(D^{2k+1}F(0)(z^{2k+1}))}{(2k+1)!||z||^{2k+1}} \quad and \quad a_{k+1} = \frac{l_z(D^{k+1}F(0)(z^{k+1}))}{(k+1)!||z||^{k+1}}.$$
(16)
The bound is sharp for k = 1, 2, 3.
Theorem 8
Theorem 8. Let and. If F(z) - z has a zero of order k + 1 at z = 0, then The estimate is sharp.
Theorem 8. Let $f \in \mathcal{H}(\Omega, \mathbb{C})$ and $F(z) = zf(z) \in \mathcal{S}^*(\Omega)$ . If F(z) - z has a zero of order k + 1 at z = 0, then
$$\left| \left( 2 \frac{\partial \rho(z)}{\partial z} \frac{D^{2k+1} F(0)(z^{2k+1})}{(2k+1)! \rho^{2k+1}(z)} \right)^2 - \left( 2 \frac{\partial \rho(z)}{\partial z} \frac{D^{k+1} F(0)(z^{k+1})}{(k+1)! \rho^{k+1}(z)} \right)^2 \right| \le \frac{(k+2)^2}{k^4} + \frac{4}{k^2}.$$
The estimate is sharp.
Theorem 9
Theorem 9. Let and. If F(z) - z has a zero of order k + 1 at z = 0, then where <span id="page-10-1"></span> The bound is sharp for k = 1,…
Theorem 9. Let $f \in \mathcal{H}(\Omega, \mathbb{C})$ and $F(z) = zf(z) \in \mathcal{S}^*(\Omega)$ . If F(z) - z has a zero of order k + 1 at z = 0, then
$$\left|1 - a_{2k+1}^2 - 2a_{k+1}^2 + 2a_{k+1}^2 a_{2k+1}^2\right| \le \begin{cases} 1 + \frac{8}{k^2} + \frac{(k+2)(6-k)}{k^4}; & 1 \le k \le 3, \\ 1 + \frac{8}{k^2} + \frac{2+k}{k^3}; & k \ge 3, \end{cases}$$
where
<span id="page-10-1"></span>
$$a_{2k+1} = 2\frac{\partial \rho(z)}{\partial z} \frac{D^{2k+1} F(0)(z^{2k+1})}{(2k+1)! \rho^{2k+1}(z)} \quad and \quad a_{k+1} = 2\frac{\partial \rho(z)}{\partial z} \frac{D^{k+1} F(0)(z^{k+1})}{(k+1)! \rho^{k+1}(z)}. \tag{28}$$
The bound is sharp for k = 1, 2, 3
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