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Ma-Minda φ-classes studied in this paper:
Abstract
For $-1\leq B<A\leq 1$, let $\mathcal{S}^*(A,B)$ denote the class of Janowski starlike functions which satisfy the subordination relation $zf'(z)/f(z)\prec (1+Az)/(1+Bz)$. In the present article, we determine the sharp of pre-Schwarzian norm for the functions in the class $\mathcal{S}^*(A,B)$.
Results & Lemmas (7)
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Lemma 1.1
Lemma 1.1. [1, Lemma 2.2] Suppose that is an analytic function with. If, then
Lemma 1.1. [1, Lemma 2.2] Suppose that $\omega : \mathbb{D} \to \mathbb{D}$ is an analytic function with $\omega(0) = 0$ . If $-1 < A \le 1$ , then
$$M(z) := \frac{(|z|^2 - |\omega(z)|^2) + |\omega(z)|(1 - |z|^2)|2 + A\omega(z)|}{|z|(1 - |\omega(z)|^2)|2 + A\omega(z)|} \le 1, \quad z \in \mathbb{D}.$$
Lemma 2.1
Lemma 2.1. For, let (2.1) Then satisfies the following sharp inequalities. (i) If then for every, we have (2.2) (ii) If AB < 0 then for…
Lemma 2.1. For $-1 \le B < A \le 1$ , let
(2.1)
$$\phi_{A,B}(z) = \frac{2 + Az}{(1 + Az)(1 + Bz)}.$$
Then $\phi_{A,B}(z)$ satisfies the following sharp inequalities.
(i) If $AB \geq 0$ then for every $z \in \mathbb{D}$ , we have
(2.2)
$$|\phi_{A,B}(z)| \le \begin{cases} \phi_{A,B}(-|z|) & \text{for } 0 \le B < A \le 1\\ \phi_{A,B}(|z|) & \text{for } -1 \le B < A \le 0. \end{cases}$$
(ii) If AB < 0 then for every $z \in \mathbb{D}$ , we have
(2.3)
$$|\phi_{A,B}(z)| \le \begin{cases} \phi_{A,B}(|z|) & \text{for } |z|^2 \le \frac{A+2B}{A^2B} \\ \phi_{A,B}(-|z|) & \text{for } |z|^2 \ge \frac{A+2B}{A^2B}. \end{cases}$$
Theorem 2.1
Theorem 2.1. For, let be of the form (1.1). Then we have the sharp inequality, where is given by (1.3). (i) If, then where and are given by…
Theorem 2.1. For $-1 \le B < A \le 1$ , let $f \in \mathcal{S}^*(A, B)$ be of the form (1.1). Then we have the sharp inequality $||T_f|| \le ||T_{K_{A,B}}||$ , where $K_{A,B}$ is given by (1.3).
(i) If $AB \geq 0$ , then
$$||T_{K_{A,B}}|| = \begin{cases} (A-B)\gamma_1(\alpha_1), & \text{for } 0 \le B < A < 1, \\ (A-B)\gamma_1(\alpha_1), & \text{for } A = 1, B < \frac{1}{3}, \\ 2, & \text{for } A = 1, B \ge \frac{1}{3}, \\ (A-B)\gamma_2(\alpha_2), & \text{for } -1 < B < A \le 0, \\ 2(2+A), & \text{for } B = -1, -1 < A \le 0, \end{cases}$$
where $\gamma_1$ and $\gamma_2$ are given by (2.8) and $\alpha_1$ , $\alpha_2$ are the unique roots in (0,1) of the equations $h_1(x) = 0$ , $h_2(x) = 0$ respectively and $h_1$ , $h_2$ are given by (2.9), (2.10) respectively.
(ii) If AB < 0 with $\beta = \sqrt{(A+2B)/A^2B}$ , then
$$||T_{K_{A,B}}|| = \begin{cases} (A-B)\gamma_1(\alpha_1), & \text{for } A+2B>0, \\ 2(A-B), & \text{for } A+2B=0, \\ (A-B)\max\{\gamma_2(\alpha_2), \gamma_1(\beta)\}, & \text{for } A+2B<0, \ \beta<1, \\ (A-B)\gamma_2(\alpha_2), & \text{for } A+2B<0, \ \beta\geq1, \ B\neq-1, \\ 2(2+A), & \text{for } A>0, \ B=-1, \end{cases}$$
where $\gamma_1$ and $\gamma_2$ are given by (2.8) and $\alpha_1$ , $\alpha_2$ are the unique roots in (0,1) of the equations $h_1(x) = 0$ , $h_2(x) = 0$ respectively and $h_1$ , $h_2$ are given by (2.9), (2.10) respectively.
Corollary 2.1
Corollary 2.1. [1, Theorem 2] Let, be of the form (1.1). Then the sharp inequality, holds and equality occurs for the function.
Corollary 2.1. [1, Theorem 2] Let $f \in \mathcal{S}^*(\alpha)$ , $0 \le \alpha < 1$ be of the form (1.1). Then the sharp inequality $||T_f|| \le 6 - 4\alpha$ , holds and equality occurs for the function $k_{\alpha}(z) = z/(1-z)^{2(1-\alpha)}$ .
Corollary 2.2
Corollary 2.2. Let be of the form (1.1). Then the sharp inequality, holds and equality occurs for the Koebe function. Taking A = 1 and B =…
Corollary 2.2. Let $f \in \mathcal{S}^*$ be of the form (1.1). Then the sharp inequality $||T_f|| \le 6$ , holds and equality occurs for the Koebe function $k(z) = z/(1-z)^2$ .
Taking A = 1 and B = 0, we get the following result.
Corollary 2.3
Corollary 2.3. Let be of the form (1.1). Then the sharp inequality, holds and equality occurs for the function. Taking and,, we get the…
Corollary 2.3. Let $f \in \mathcal{S}^*(1,0)$ be of the form (1.1). Then the sharp inequality $||T_f|| \leq 9/4$ , holds and equality occurs for the function $f(z) = ze^z$ .
Taking $A = \alpha$ and $B = -\alpha$ , $0 < \alpha < 1$ , we get the following result.
Corollary 2.4
Corollary 2.4. Let, be of the form (1.1). Then the sharp inequality holds, where is an unique root in (0,1) of the equation: Moreover, the…
Corollary 2.4. Let $f \in \mathcal{S}^*(\alpha, -\alpha)$ , $0 < \alpha \le 1$ be of the form (1.1). Then the sharp inequality
$$||T_f|| \le \begin{cases} 2\alpha(1-x_0^2)(2+\alpha x_0)/(1-\alpha^2 x_0^2), & \text{for } 0 < \alpha < 1, \\ 6, & \text{for } \alpha = 1 \end{cases}$$
holds, where $x_0$ is an unique root in (0,1) of the equation:
$$\alpha^{3}x^{4} + (\alpha^{3} - 3\alpha)x^{2} + (4\alpha^{2} - 4)x + \alpha = 0.$$
Moreover, the equality occurs for the function $f(z) = z/(1 - \alpha z)^2$ .
Function classes studied:
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