Growth, Distortion, Pre-Schwarzian and Schwarzian norm estimates for Generalized Robertson class

Theorem 2.1. For $-\pi/2 < \alpha < \pi/2$ and $0 \le \beta < 1$ the following are equivalent: (iii) $$f \in \mathcal{SP}^0_{\alpha}(\beta)$$ (ii) (2.4) $$\operatorname{Re}\left(1 + \frac{1}{2}\left(\left(e^{2i\alpha} - 2\beta e^{i\alpha}\cos\alpha\right) + 1\right)\frac{zf''(z)}{f'(z)}\right)$$ $$\geq 1 - (1 - \beta)^2\cos^2\alpha + \left(\frac{1 - |z|^2}{4}\right)\left|\frac{zf''(z)}{f'(z)}\right|^2$$ (iii) $$(2.5) \left| (1-|z|^2) \left( \frac{f''(z)}{f'(z)} \right) - 2(1-\beta) \cos \alpha \bar{z} \right| \le (1-\beta) \cos \alpha.$$ The inequalities (ii) and (iii) both are sharp for the function (2.6) $$f'_{\alpha,\beta}(z) = \frac{1}{(1-z)^{(1-\beta)\cos\alpha}} \text{ for } z \in \mathbb{D} \text{ with } \beta \in [0,1).$$ Proof of Theorem 2.1. For $-\pi/2 < \alpha < \pi/2$ and $0 \le \beta < 1$ let $f \in \mathcal{SP}^0_{\alpha}(\beta)$ be of the form (1.1). Then from (1.2), we have (2.7) $$1 + \frac{zf''(z)}{f'(z)} \prec \frac{1+Az}{1-z}, \text{ where } A = e^{-i\alpha} \left( e^{-i\alpha} - 2\beta \cos \alpha \right).$$ Consequently, there exists an analytic function $\omega: \mathbb{D} \to \mathbb{D}$ with $\omega(0) = 0$ such that $$1 + \frac{zf''(z)}{f'(z)} = \frac{1 + A\omega(z)}{1 - \omega(z)}.$$ Let $\omega(z) = z\phi(z)$ for some analytic function $\phi$ that satisfy $\phi(\mathbb{D}) \subseteq \mathbb{D}$ . From (2.7), we have (2.8) $$\frac{f''(z)}{f'(z)} = \frac{2G_1(\alpha, \beta)\omega(z)}{z(1 - \omega(z))}$$ which can be written as $$\frac{f''(z)}{f'(z)} = \frac{2G_1(\alpha, \beta) \phi(z)}{(1 - z\phi(z))}$$ Thus, we see that (2.9) $$\phi(z) = \frac{\frac{f''(z)}{f'(z)}}{2G_1(\alpha, \beta) + \frac{zf''(z)}{f'(z)}}.$$ Since $|\phi(z)|^2 \le 1$ , an easy computation shows that $$(2.10) \qquad \left| \frac{f''(z)}{f'(z)} \right|^2 \le \left( 2G_1(\alpha, \beta) + \frac{zf''(z)}{f'(z)} \right) \overline{\left( 2G_1(\alpha, \beta) + \frac{zf''(z)}{f'(z)} \right)}.$$ A simple computation shows that (2.11) $$(1 - |z|^2) \left| \frac{f''(z)}{f'(z)} \right|^2 \le (1 - \beta)^2 \cos^2 \alpha + 4 \operatorname{Re} \left( \frac{1}{2} \left( (e^{2i\alpha} - 2\beta e^{i\alpha} \cos \alpha) + 1 \right) \frac{z f''(z)}{f'(z)} \right)$$ which is equivalent to $$\operatorname{Re}\left(\frac{1}{2}\left(\left(e^{2i\alpha} - 2\beta e^{i\alpha}\cos\alpha\right) + 1\right)\frac{zf''(z)}{f'(z)}\right)$$ $$\geq -(1-\beta)^2\cos^2\alpha + \left(\frac{1-|z|^2}{4}\right)\left|\frac{zf''(z)}{f'(z)}\right|^2$$ We rewrite the last expression as (2.12) $$\operatorname{Re}\left(1 + \frac{1}{2}\left(\left(e^{2i\alpha} - 2\beta e^{i\alpha}\cos\alpha\right) + 1\right)\frac{zf''(z)}{f'(z)}\right)$$ $$\geq 1 - (1 - \beta)^2\cos^2\alpha + \left(\frac{1 - |z|^2}{4}\right)\left|\frac{zf''(z)}{f'(z)}\right|^2.$$ Multiplying both sides of equation (2.11) by $(1 - |z|^2)$ , we obtain (2.13) $$(1 - |z|^2)^2 \left| \frac{f''(z)}{f'(z)} \right|^2$$ $$\leq (1 - |z|^2)(1 - \beta)^2 \cos^2 \alpha + 4(1 - |z|^2) \operatorname{Re} \left( \frac{1}{2} \left( (e^{2i\alpha} - 2\beta e^{i\alpha} \cos \alpha) + 1 \right) \frac{zf''(z)}{f'(z)} \right)$$ which implies that $$(1 - |z|^2)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 - 4(1 - |z|^2) \operatorname{Re} \left( \frac{1}{2} \left( (e^{2i\alpha} - 2\beta e^{i\alpha} \cos \alpha) + 1 \right) \frac{z f''(z)}{f'(z)} \right) + |z|^2 (1 - \beta)^2 \cos^2 \alpha \le (1 - \beta)^2 \cos^2 \alpha.$$ Thus, we have $$(2.14) \qquad \left| (1-|z|^2) \left( \frac{f''(z)}{f'(z)} \right) - 2(1-\beta) \cos \alpha \bar{z} \right| \le (1-\beta) \cos \alpha.$$ This completes the proof. Example 2.1. For the sharpness of the inequalities (2.4) and (2.5), we consider the function defined in (2.6) with $\beta = 0$ as $$f_0'(z) = \frac{1}{(1-z)}$$ A simple computation using (2.6) shows that $$1 + \frac{zf_0''(z)}{f_0'(z)} = 1 + \frac{z}{1 - z}.$$ Moreover, it is easy to see that $$\operatorname{Re}\left(1 + \frac{zf_0''(z)}{f_0'(z)}\right) > 0,$$ hence, it is clear that $f_0 \in \mathcal{SP}_0^0(0)$ . To show the inequality (2.16) of Corollary 2.2 is sharp, we consider z=r<1 and establish that $$\left| (1 - |z|^2) \left( \frac{f_0''(z)}{f_0'(z)} \right) - \bar{z} \right| = \left| (1 - r^2) \left( \frac{1}{1 - r} \right) - r \right| = |1 - r + r| = 1.$$ To show the inequality (2.16) in Corollary 2.1 is sharp, we see from (2.9) (Proof of Theorem 2.1) that (2.15) $$\phi(z) = \frac{\frac{f_0''(z)}{f_0'(z)}}{\frac{zf_0''(z)}{f_0'(z)} + 2} = 1.$$ Thus, it is clear that $|\phi(z)|^2 = 1$ which further leads to $$\operatorname{Re}\left(1 + \frac{zf_{1/2}''(z)}{f_{1/2}'(z)}\right) = \left(\frac{1 - |z|^2}{4}\right) \left|\frac{zf_{1}''(z)}{f_{1}'(z)}\right|^2.$$ We have the following immediate results form Theorem 2.1.

Corollary 2.1. If $\alpha = \beta = 0$ , $f \in \mathcal{SP}_0^0(0) \subset \mathcal{C}$ , then from (2.12) we have (2.16) $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge \frac{1}{4}(1 - |z|^2) \left|\frac{zf''(z)}{f'(z)}\right|^2.$$ We obtain the following corollary for the class $\mathcal{C}$ .

Corollary 2.2. If $\alpha = \beta = 0$ , $f \in \mathcal{SP}_0^0(0) \subset \mathcal{C}$ , then from (2.14) we have $$\left| (1 - |z|^2) \frac{f''(z)}{f'(z)} - 2\bar{z} \right| \le 1.$$ Remark 2.1. The inequality $$\left| (1 - |z|^2) \left( \frac{f''(z)}{f'(z)} \right) - 2(1 - \beta) \cos \alpha \bar{z} \right| \le (1 - \beta) \cos \alpha.$$ is instrumental in the definition or characterization of the radius of concavity. In the next result, we establish the distortion theorem and growth theorem for the functions in the class $\mathcal{SP}_{\alpha}^{0}(\beta) = \{f \in \mathcal{SP}_{\alpha}(\beta) : f''(0) = 0\}.$

Theorem 2.2. For $-\pi/2 < \alpha < \pi/2$ and $0 \le \beta < 1$ , let $f \in \mathcal{SP}^0_{\alpha}(\beta)$ be of the form (1.1) for all $z \in \mathbb{D}$ , then the inequality $$\frac{1}{(1+|z|^2)^{(1-\beta)\cos\alpha}} \le |f'(z)| \le \frac{1}{(1-|z|^2)^{(1-\beta)\cos\alpha}}$$ and $$\int_0^{|z|} \frac{1}{(1+\xi^2)^{(1-\beta)\cos\alpha}} d|\xi| \le |f(z)| \le \int_0^{|z|} \frac{1}{(1-\xi^2)^{(1-\beta)\cos\alpha}} d|\xi|.$$ All of these estimates are sharp. Equality holds at a given point other than 0 for $$f(z) = \int_0^{|z|} \frac{1}{(1 - \lambda \zeta^2)^{(1-\beta)\cos\alpha}} d|\zeta|$$ for some $\lambda \in \mathbb{C}$ and $|\lambda| = 1$ . We have the following immediate result for a subclass of $\mathcal{C}$ of convex functions.

Corollary 2.3. For $\alpha = \beta = 0$ , let $f \in \mathcal{SP}_0^0(0) \subset \mathcal{C}$ be of the form (1.1), then the inequality $$\frac{1}{(1+|z|^2)} \le |f'(z)| \le \frac{1}{(1-|z|^2)}$$ and $$\int_0^{|z|} \frac{1}{(1+\xi^2)} d|\xi| \le |f(z)| \le \int_0^{|z|} \frac{1}{(1-\xi^2)} d|\xi|.$$ All of these estimates are sharp. Equality holds at a given point other than 0 for $$f(z) = \int_0^{|z|} \frac{1}{(1 - \lambda \zeta^2)} d|\zeta|$$ for some $\lambda \in \mathbb{C}$ and $|\lambda| = 1$ . Proof of Theorem 2.2. Let $f \in \mathcal{SP}^0_{\alpha}(\beta)$ be of the form (1.1) and from (2.9), we obtain $\phi(0) = 0$ . Then by using the Schwarz lemma, we get (2.17) $$\left| \frac{\frac{f''(z)}{f'(z)}}{2G_1(\alpha,\beta) + \frac{zf''(z)}{f'(z)}} \right|^2 \le |z|^2$$ which implies that $$\left| \frac{f''(z)}{f'(z)} \right|^2 \le 4|z|^2 (1-\beta)^2 \cos^2 \alpha + 4|z|^2 \operatorname{Re} \left( \frac{1}{2} \left( (e^{2i\alpha} - 2\beta e^{i\alpha} \cos \alpha) + 1 \right) \frac{zf''(z)}{f'(z)} \right) + |z|^4 \left| \frac{f''(z)}{f'(z)} \right|^2.$$ Thus, we have $$(2.18) (1 - |z|^4) \left| \frac{f''(z)}{f'(z)} \right|^2$$ $$\leq 4|z|^2 (1 - \beta)^2 \cos^2 \alpha + 4|z|^2 \operatorname{Re} \left( \frac{1}{2} \left( (e^{2i\alpha} - 2\beta e^{i\alpha} \cos \alpha) + 1 \right) \frac{zf''(z)}{f'(z)} \right).$$ Multiplying both sides of (2.18) by $(1 - |z|^4)$ , we obtain $$(1 - |z|^4)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 - 4|z|^2 (1 - |z|^4) \operatorname{Re} \left( \frac{1}{2} \left( (e^{2i\alpha} - 2\beta e^{i\alpha} \cos \alpha) + 1 \right) \frac{zf''(z)}{f'(z)} \right)$$ $$\leq 4|z|^2 (1 - |z|^4) (1 - \beta)^2 \cos^2 \alpha.$$ Adding $(2(1-\beta)\cos\alpha|z|^2|z|)^2$ both side of the above inequality, we get (2.19) $$(1 - |z|^4)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 - 4|z|^2 (1 - |z|^4) \operatorname{Re} \left( \frac{1}{2} \left( (e^{2i\alpha} - 2\beta e^{i\alpha} \cos \alpha) + 1 \right) \frac{zf''(z)}{f'(z)} \right) + 4(1 - \beta)^2 \cos^2 \alpha |z|^4 |\bar{z}|^2$$ $$\leq 4|z|^2 (1 - |z|^4) (1 - \beta)^2 \cos^2 \alpha + 4(1 - \beta)^2 \cos^2 \alpha |z|^4 |\bar{z}|^2.$$ Multiplying both side by |z|, then by simple calculation $$(2.20) \left| (1 - |z|^4) \frac{zf''(z)}{f'(z)} - 2(1 - \beta)\cos\alpha |z|^4 \right| \le 2(1 - \beta)\cos\alpha |z|^2$$ which implies (2.21) $$\frac{-2(1-\beta)\cos\alpha|z|^2}{1+|z|^2} \le \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) \le \frac{2(1-\beta)\cos\alpha|z|^2}{1-|z|^2}.$$ Let $z = re^{i\theta}$ . Then, we obtain $$\frac{-2r(1-\beta)\cos\alpha}{1+r^2} \le \frac{\partial}{\partial r} \left(\log|f'(re^{i\theta})|\right) \le \frac{2r(1-\beta)\cos\alpha}{1-r^2}.$$ Case A. When $\alpha = 0$ and $0 \le \beta < 1$ if we integrate respect to r, we obtain (2.22) $$\frac{1}{(1+|z|^2)^{1-\beta}} \le |f'(z)| \le \frac{1}{(1-|z|^2)^{1-\beta}}$$ Case B. When $\alpha \neq 0$ , if we integrate respect to r, we obtain (2.23) $$\frac{1}{(1+|z|^2)^{(1-\beta)\cos\alpha}} \le |f'(z)| \le \frac{1}{(1-|z|^2)^{(1-\beta)\cos\alpha}}.$$ Next, for the growth part of the theorem, from the upper bound it follows that $$(2.24) |f'(re^{i\theta})| = \left| \int_0^r f'(re^{i\theta})e^{i\theta}dt \right| \le \int_0^r |f'(re^{i\theta})|dt \le \int_0^r \frac{1}{(1-t^2)^{(1-\beta)\cos\alpha}}dt$$ which implies $$(2.25) |f(z)| \le \int_0^{|z|} \frac{1}{(1-\xi^2)^{(1-\beta)\cos\alpha}} d|\xi|$$ for all $z \in \mathbb{D}$ . It is well-known that if $f(z_0)$ is a point of minimum modulus on the image of the circle |z| = r and $\gamma = f^{-1}(\Gamma)$ , where $\Gamma$ is the line segment from 0 to $f(z_0)$ , then $$(2.26) |f(z)| \ge |f(z_0)| \ge \int_0^{|z|} \frac{1}{(1+\xi^2)^{(1-\beta)\cos\alpha}} d|\xi|.$$ Thus, the inequalities are established. Now, we will find the sharp bound of the pre-Schwarzian and Schwarzian norms for functions in the class $\mathcal{SP}_{\alpha}^{0}(\beta) = \{f \in \mathcal{SP}_{\alpha}(\beta) : f''(0) = 0\}$ . The following lemma will play a key role to prove the result. Lemma A. [37] If $\phi(z) : \mathbb{D} \to \mathbb{D}$ be analytic function, then (2.27) $$\frac{|\phi(z)|^2}{1 - |\phi(z)|^2} \le \frac{(\phi(0) + |z|)^2}{(1 - |\phi(0)|)^2 (1 - |z|^2)|)}$$ We obtain the following result establishing a sharp bound of the pre-Schwarzian norm for $f \in \mathcal{SP}^0_{\alpha}(\beta)$ .

Theorem 2.3. For $0 \le \beta < 1$ and $-\pi/2 < \alpha < \pi/2$ , let $f \in \mathcal{SP}^0_{\alpha}(\beta)$ be of the form (1.1) for all $z \in \mathbb{D}$ , then the pre-Schwarzian norm satisfies the inequality $$(2.28) ||Pf|| \le 2(1-\beta)\cos\alpha.$$ The inequality is sharp. Proof of Theorem 2.3. Since $\phi(z) = z\xi(z)$ , with $|\xi(z)| < 1$ , then in (2.8) we obtain $$\sup_{z \in \mathbb{D}} (1 - |z|^2) \left| \frac{f''(z)}{f'(z)} \right| \le \sup_{z \in \mathbb{D}} (1 - |z|^2) \frac{2(1 - \beta) \cos \alpha |z\xi(z)|}{1 - |z|^2 |\xi(z)|}$$ $$\le 2(1 - \beta) \cos \alpha \sup_{0 \le r \le 1} \frac{r(1 - r^2)}{(1 - r^2)}$$ $$= 2(1 - \beta) \cos \alpha.$$ The extremal function is given by (2.29) $$f^*(z) = \int_0^z \frac{1}{(1 - \xi^2)^{(1-\beta)\cos\alpha}} d\xi.$$ It can be easily shown that $||Pf^*|| = 2(1-\beta)\cos\alpha$ . This completes the proof. $\square$ We have the following immediate result from Theorem 2.3.

Corollary 2.4. For $\alpha = \beta = 0$ , let $f \in \mathcal{SP}_0^0(0) \subset \mathcal{C}$ be of the form (1.1) for all $z \in \mathbb{D}$ , then the pre-Schawarzian norm $$||Pf|| \le 2.$$ The inequality is sharp. Sharpness of Corollary 2.4. For $\alpha = 0, \beta = 0$ , it follows from that $$\frac{f_0''(z)}{f_0'(z)} = \frac{2z}{1-z^2}$$ and $Pf_0 = \frac{2}{1-z^2}$ . A simple computation thus yields that $$||Pf_0|| = \sup_{z \in \mathbb{D}} (1 - z^2) |Pf_0| = \sup_{z \in \mathbb{D}} (1 - |z|^2) \frac{2}{1 - |z|^2} = 2$$ and we see the constant 2 is sharp. In our next result, we give a sharp bound for the norm of the Schwarzian derivative when $f \in \mathcal{SP}_{\alpha}^{0}(\beta) = \{f \in \mathcal{SP}_{\alpha}(\beta) : f''(0) = 0\}$ by a direct application of the Schwarz lemma.

Theorem 2.4. For $0 \le \beta < 1$ and $-\pi/2 < \alpha < \pi/2$ let $f \in \mathcal{SP}^0_{\alpha}(\beta)$ be of the form (1.1) for all $z \in \mathbb{D}$ , then the Schwarzian norm $$||Sf|| = (1 - |z|^2)^2 |Sf(z)| < 2(1 - \beta)\cos\alpha (2 - (1 - \beta)\cos\alpha).$$ The inequality is sharp. Proof of Theorem 2.4. From (2.8), we have $$\frac{f''(z)}{f'(z)} = \frac{2G_1(\alpha, \beta)\phi(z)}{(1 - z\phi(z))}.$$ A simple calculation shows that $$Sf(z) = 2G_1(\alpha, \beta) \left( \frac{2\phi'(z) + (2 - 2G_1(\alpha, \beta)) \phi^2(z)}{2(1 - z\phi(z))^2} \right).$$ By using triangle inequality and Schwarz pick lemma, we obtain (2.30) $$(1 - |z|^2)^2 |Sf| \le 2|G_1(\alpha, \beta)| \left| 2\phi'(z) + (2 - 2G_1(\alpha, \beta)) \phi^2(z) \right| \frac{(1 - |z|^2)^2}{2|1 - z\phi(z)|^2}$$ $$= \frac{2|G_1(\alpha, \beta)|(1 - |z|^2)^2}{|1 - z\phi(z)|^2} \left( \frac{1 - |\phi(z)|^2}{1 - |z|^2} + (1 - |G_1(\alpha, \beta)|) |\phi(z)|^2 \right).$$ We define the function $\Psi(z): \mathbb{D} \to \mathbb{D}$ such that $$\Psi(z) := \frac{\bar{z} - \phi(z)}{1 - z\phi(z)}.$$ Since $\phi(\mathbb{D}) \subseteq \mathbb{D}$ then $(1-|z|^2)(1-|z\phi(z)|^2) > 0$ , it follows that $$|\bar{z} - \phi(z)|^2 < |1 - z\phi(z)|^2$$ Hence, we can conclude that $|\Psi(z)|^2 < 1$ . A simple computation leads to $$1 - |\Psi(z)|^2 = \frac{(1 - |\phi(z)|^2)(1 - |z|^2)}{|1 - z\phi(z)|^2}$$ and (2.31) $$\frac{(1-|z|^2)^2}{|1-z\phi(z)|^2} = \frac{(1-|\Psi(z)|^2)(1-|z|^2)}{(1-|\phi(z)|^2)}.$$ If we replace the expression (2.31) in (2.30), we have (2.32) $$(1-|z|^2)^2|Sf(z)| \le 2|G_1(\alpha,\beta)|(1-|\Psi(z)|^2)\left(1+(1-|G_1(\alpha,\beta)|)\frac{|\phi(z)|^2(1-|z|^2)}{(1-|\phi(z)|^2)}\right).$$ Since h''(0) = 0 implies that $\phi(0) = 0$ , using Lemma A, we obtain (2.33) $$\frac{|\phi(z)|^2}{1 - |\phi(z)|^2} \le \frac{|z|^2}{1 - |z|^2}.$$ Using (2.33) in (2.32), we obtain $$(1-|z|^2)^2|Sf(z)| \le 2|G_1(\alpha,\beta)|(1-|\Psi(z)|^2)\left(1+(1-|G_1(\alpha,\beta)|)|z|^2\right).$$ Again, since $1 - |\Psi(z)|^2 \le 1$ , then $$\sup_{z \in \mathbb{D}} (1 - |z|^2)^2 |Sf(z)| \le \sup_{z \in \mathbb{D}} 2(1 - \beta) \cos \alpha \left( 1 + (1 - (1 - \beta) \cos \alpha) |z|^2 \right)$$ $$= 2(1 - \beta) \cos \alpha \left( 2 - (1 - \beta) \cos \alpha \right).$$ Next part of the proof is to show that the inequalities are sharp. The family of parameterized functions defined as: $$(2.34) f_{\alpha,\beta}(z) = \int_0^z \frac{1}{(1-\xi^2)^{(1-\beta)\cos\alpha}} d\xi, \text{for } -\pi/2 < \alpha < \pi/2, \ 0 \le \beta < 1$$ maximizes the Schwarzian norm defined as: $$||Sf|| = \sup_{z \in \mathbb{D}} (1 - |z|^2)^2 |Sf|$$ and from this, the sharpness of the inequality holds for $-\pi/2 < \alpha < \pi/2, \ 0 \le \beta < 1$ . Note that (2.35) $$\begin{cases} f''_{\alpha,\beta}(z) = \frac{2z(1-\beta)\cos\alpha}{1-z^2}, \\ Sf_{\alpha,\beta} = \frac{2(1-\beta)\cos\alpha}{(1-z^2)^2} \left(1 + (1-(1-\beta)\cos\alpha)|z|^2\right) \end{cases}$$ which calculates $$||Sf_{\alpha,\beta}|| = \sup_{z \in \mathbb{D}} (1 - |z|^2)^2 |Sf_{\alpha,\beta}|$$ = $\sup_{z \in \mathbb{D}} 2(1 - \beta) \cos \alpha \left( 1 + (1 - (1 - \beta) \cos \alpha) |z|^2 \right)$ \(\le 2(1 - \beta) \cos \alpha (2 - (1 - \beta) \cos \alpha). In general, the integral formula for $f_{\alpha}$ given in above does not give primitives in terms of elementary functions, however when $\alpha = 0$ and also $\beta = 0$ , we have $$f_{0,0}(z) = \int_0^z \frac{1}{(1-\xi^2)} d\xi = \frac{1}{2} \log\left(\frac{1+z}{1-z}\right),$$ where $||Sf_{0,0}|| = 2$ .

Corollary 2.5. If $f \in \mathcal{SP}^0_{\alpha}(\beta)$ , then for all $z \in \mathbb{D}$ and $\alpha = 0, \beta = 0$ , we have $||Sf|| \leq 2$ . The inequality is sharp. Without requiring |f''(0)| to be zero, we derive a bound for $(1-|z|^2)^2|Sf(z)|$ for functions in $\mathcal{SP}_{\alpha}(\beta)$ .

Theorem 2.5. If $f \in \mathcal{SP}_{\alpha}(\beta)$ , for all $z \in \mathbb{D}$ and $-\pi/2 < \alpha < \pi/2$ , $0 \le \beta < 1$ and (2.36) $$\xi = |\phi(0)| = \frac{|f''(0)|}{2(1-\beta)\cos\alpha},$$ then $$(1 - |z|^2)^2 |Sf(z)| \le 2(1 - \beta) \cos \alpha \left( 2 + (1 - \beta) \cos \alpha \frac{(\xi + |z|)^2}{(1 - \xi^2)} \right).$$

Corollary 2.6. If $f \in \mathcal{SP}_0(0) := \mathcal{C}$ , for all $z \in \mathbb{D}$ with (2.37) $$\xi = |\phi(0)| = \frac{|f''(0)|}{2},$$ then inequality $$(1 - |z|^2)^2 |Sf(z)| \le 2.$$ Proof of Theorem 2.5. Let $\xi = |\phi(0)|$ . Applying the Lemma A, we calculate (2.38) $$\frac{|\phi(z)|^2}{1 - |\phi(z)|^2} \le \frac{(\xi + |z|)^2}{(1 - \xi^2)(1 - |z|^2)}.$$ If we substitute (2.38) in (2.32), we obtain (2.39) $$(1-|z|^2)^2|Sf(z)| \le 2(1-\beta)\cos\alpha(1-|\Phi_1(z)|^2)\left(2+(1-\beta)\cos\alpha\frac{(\xi+|z|)^2}{(1-\xi^2)}\right).$$ From the fact that |z| < 1 and $1 - |\Phi_1(z)|^2 \le 1$ , we can easily calculate $$(2.40) (1-|z|^2)^2|Sf(z)| \le 2(1-\beta)\cos\alpha\left(2+(1-\beta)\cos\alpha\frac{(\xi+|z|)^2}{(1-\xi^2)}\right).$$ This completes the

Theorem 3.1. If $f \in \mathcal{SP}_{\alpha}(\beta)$ , for all $z \in \mathbb{D}$ and $-\pi/2 < \alpha < \pi/2$ , $0 \le \beta < 1$ then Re $(T_{f(z)}) > 0$ for $|z| < R_{\alpha,\beta,Co(A)}$ , where $R_{\alpha,\beta,Co(A)}$ is the least value of $r \in (0,1)$ satisfying $\Phi_A(r) = 0$ with $$\Phi_A(r) = (A + 1 - 2(1 - \beta)\cos\alpha)r^2 - 2(A + 1 + (1 - \beta)\cos\alpha)r + A - 1.$$ The radius $R_{\alpha,\beta,Co(A)}$ is best possible.

Theorem 3.2. The radius of convexity for the class of function $SP_{\alpha}(\beta)$ is at least $\frac{1}{(1-\beta)\cos\alpha-1}$ .

Definition 3.1. The radius of concavity (w.r.t $\mathcal{F}$ ), a subclass of $\mathcal{A}$ is the largest number $R_{\mathcal{F}} \in (0,1]$ such that for each function $f \in \mathcal{F}$ , $Re(T_f(z)) > 0$ for all $|z| < R_{\mathcal{F}}$ , where $T_f(z)$ is defined in (2.1).

Definition 3.2. The number $r \in [0, 1]$ is called the radius of convexity of a particular subclass $\mathcal{F}_{\beta}$ of the class $\mathcal{A}$ of normalized analytic functions (where $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ ) in the unit disk $\mathbb{D}$ if r is the largest number such that the function f is convex in the disk |z| < r. A function f analytic in a region $\Omega$ is convex in $\Omega$ if it maps $\Omega$ onto a convex region. For an analytic function f in the unit disk $\mathbb{D}$ , the condition for f to be locally univalent and convex in a disk |z| < r is given by the inequality you provided: (3.3) $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) > 0, \quad \text{for } |z| < r.$$