Pre-Schwarzian and Schwarzian norm Estimates for class of Ozaki Close-to-Convex functions

Theorem 2.1. For $\frac{1}{2} \leq \lambda \leq 1$ , the following are equivalent: (i). $$f \in \mathcal{F}_0(\lambda)$$ . (ii). Re $$\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge \frac{1 - 2\lambda}{2} + \frac{1}{2} \left(\frac{1 - |z|^2}{1 + 2\lambda}\right) \left|\frac{f''(z)}{f'(z)}\right|^2$$ for $\lambda \in [1/2, 1]$ , (iii). $$\left| (1 - |z|^2) \frac{f''(z)}{f'(z)} - (1 + 2\lambda)\bar{z} \right| \le 1 + 2\lambda.$$ Proof of Theorem 2.1. Firstly, we will prove (i) is equivalent to (ii). If $f \in \mathcal{F}_0(\lambda)$ , then there exists a Schwarz function $\omega : \mathbb{D} \to \mathbb{D}$ such that <span id="page-4-1"></span>(2.1) $$1 + \frac{zf''(z)}{f'(z)} = \frac{1 + 2\lambda\omega(z)}{1 - \omega(z)}$$ holds with $\omega(0) = 0$ . Let $\omega(z) = z\phi(z)$ for some analytic function $\phi$ that satisfies $\phi(\mathbb{D}) \subseteq \mathbb{D}$ . Then it follows from (2.1) that $$\frac{f''(z)}{f'(z)} = \frac{(1+2\lambda)\omega(z)}{z(1-\omega(z))}$$ which shows that <span id="page-4-2"></span>(2.2) $$\frac{f''(z)}{f'(z)} = \frac{(1+2\lambda)\phi(z)}{(1-z\phi(z))}.$$ Consequently, we have (2.3) $$\phi(z) = \frac{\frac{f''(z)}{f'(z)}}{(1+2\lambda) + \frac{zf''(z)}{f'(z)}}.$$ Since $|\phi(z)|^2 \le 1$ , we have the inequality $$\left| \frac{f''(z)}{f'(z)} \right|^2 \le (1 + 2\lambda)^2 + 2(1 + 2\lambda) \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) + |z|^2 \left| \frac{f''(z)}{f'(z)} \right|^2.$$ By factorizing, an easy computation gives that <span id="page-5-0"></span>(2.4) $$(1+2\lambda) \left[ (1+2\lambda) + 2\operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) \right] \ge (1-|z|^2) \left| \frac{f''(z)}{f'(z)} \right|^2$$ which implies that $$2\operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) \ge -(1+2\lambda) + \left(\frac{1-|z|^2}{1+2\lambda}\right) \left|\frac{f''(z)}{f'(z)}\right|^2.$$ When $\lambda \in [\frac{1}{2}, 1]$ , it is easy to see that <span id="page-5-1"></span>(2.5) $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge \frac{1 - 2\lambda}{2} + \frac{1}{2}\left(\frac{1 - |z|^2}{1 + 2\lambda}\right) \left|\frac{f''(z)}{f'(z)}\right|^2.$$ Next, we will prove that (ii) is equivalent to (iii). By multiplying (2.4) by $(1 - |z|^2)$ both side, we have $$(1-|z|^2)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 \le (1+2\lambda)^2 (1-|z|^2) + 2(1+2\lambda)(1-|z|^2) \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right)$$ which turns out that $$(1-|z|^2)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 - 2(1+2\lambda)(1-|z|^2) \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) + (1+2\lambda)^2 |z|^2 \le (1+2\lambda)^2.$$ Thus, we have the desired inequality <span id="page-5-2"></span>(2.6) $$\left| (1 - |z|^2) \left( \frac{f''(z)}{f'(z)} \right) - (1 + 2\lambda) \bar{z} \right| \le (1 + 2\lambda).$$ Hence, complete the proofs. Remark 2.1. We have the following results for the class of convex functions which are immediate from Theorem 2.1. Corollary 2.1. If $f \in \mathcal{F}_0(1/2) = \mathcal{C}$ , then from (2.5), we have $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge \frac{1 - 2\lambda}{2} + \frac{1}{2}\left(\frac{1 - |z|^2}{1 + 2\lambda}\right) \left|\frac{h''(z)}{h'(z)}\right|^2$$ which implies that $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge \frac{1}{4}(1 - |z|^2) \left|\frac{f''(z)}{f'(z)}\right|^2.$$ Corollary 2.2. If $f \in \mathcal{F}_0(1/2) = \mathcal{C}$ , then we have from (2.6) $$\left| (1 - |z|^2) \left( \frac{f''(z)}{f'(z)} \right) - (1 + 2\lambda)\overline{z} \right| \le (1 + 2\lambda)$$ which implies that $$\left| (1 - |z|^2) \frac{f''(z)}{f'(z)} - 2\bar{z} \right| \le 2.$$ We obtain the following result which is distortion theorem and growth theorem for the class $\mathcal{F}_0(\lambda)$ .

Theorem 2.2. If $f \in \mathcal{F}_0(\lambda)$ , then for all $z \in \mathbb{D}$ and $\frac{1}{2} \leq \lambda \leq 1$ , $$\frac{1}{(1+|z|^2)^{\frac{1+2\lambda}{2}}} \le |f'(z)| \le \frac{1}{(1-|z|^2)^{\frac{1+2\lambda}{2}}}$$ and $$\int_0^{|z|} \frac{1}{(1+\xi^2)^{\frac{1+2\lambda}{2}}} d|\xi| \le |f(z)| \le \int_0^{|z|} \frac{1}{(1-\xi^2)^{\frac{1+2\lambda}{2}}} d|\xi|.$$ Proof of Theorem 2.2. Let $f \in \mathcal{F}_0(\lambda)$ . In view of (2.5), we obtain $\phi(0) = 0$ . By using the Schwarz lemma, we have $$\left| \frac{\frac{f''(z)}{f'(z)}}{(1+2\lambda) + \frac{zf''(z)}{f'(z)}} \right|^2 \le |z|^2$$ which shows that $$\left| \frac{f''(z)}{f'(z)} \right|^2 \le (1+2\lambda)^2 |z|^2 + 2(1+2\lambda)|z|^2 \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) + |z|^2 \left| \frac{zf''(z)}{f'(z)} \right|^2.$$ Thus, we have <span id="page-6-1"></span> $$(2.7) \qquad (1-|z|^4) \left| \frac{f''(z)}{f'(z)} \right|^2 \le (1+2\lambda)^2 |z|^2 + 2(1+2\lambda)|z|^2 \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right).$$ <span id="page-6-2"></span>Multiplying both side of (2.7) by $(1 - |z|^4)$ , we obtain (2.8) $$(1-|z|^4)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 - 2(1+2\lambda)|z|^2 (1-|z|^4) \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) \le (1+2\lambda)^2 |z|^2 (1-|z|^4).$$ Adding both side $((1+2\lambda)|z|^2\bar{z})^2$ in (2.8), we see that <span id="page-6-3"></span> $$(2.9) \quad (1-|z|^4)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 - 2(1+2\lambda)|z|^2 (1-|z|^4) \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) + (1+2\lambda)^2 |z|^4 \bar{z}$$ $$< (1+2\lambda)^2 |z|^2 (1-|z|^4) + (1+2\lambda)^2 |z|^4 \bar{z}.$$ Multiplying both side of (2.9) by |z|, a simple calculation shows that (2.10) $$\left| (1 - |z|^4) \frac{zf''(z)}{f'(z)} - (1 + 2\lambda)|z|^4 \right| \le (1 + 2\lambda)|z|^2$$ which implies that (2.11) $$\frac{-(1+2\lambda)|z|^2}{1+|z|^2} \le \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) \le \frac{(1+2\lambda)|z|^2}{1-|z|^2}.$$ Let $z = re^{i\theta}$ . Hence, <span id="page-7-0"></span> $$(2.12) \qquad \frac{-(1+2\lambda)r}{1+r^2} \le \frac{\partial}{\partial r} \left( \log|f'(re^{i\theta})| \right) \le \frac{(1+2\lambda)r}{1-r^2}.$$ When $1/2 \le \lambda \le 1$ integrating (2.12) w.r.t. r, we obtain (2.13) $$\frac{1}{(1+|z|^2)^{\frac{1+2\lambda}{2}}} \le |f'(z)| \le \frac{1}{(1-|z|^2)^{\frac{1+2\lambda}{2}}}$$ Next, for the growth part of the theorem, from the upper bound it follows that $$(2.14) |f'(re^{i\theta})| = \left| \int_0^r f'(re^{i\theta})e^{i\theta}dt \right| \le \int_0^r |f'(re^{i\theta})|dt \le \int_0^r \frac{1}{(1-t^2)^{\frac{1+2\lambda}{2}}}dt$$ which implies $$(2.15) |f(z)| \le \int_0^{|z|} \frac{1}{(1 - \xi^2)^{\frac{1+2\lambda}{2}}} d|\xi|$$ 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 $$(2.16) |f(z)| \ge |f(z_0)| \ge \int_0^{|z|} \frac{1}{(1+\xi^2)^{\frac{1+2\lambda}{2}}} d|\xi|.$$ This completes the proof. Now, we will find the sharp bound of the pre-Schwarzian and Schwarzian norms in terms of value f''(0) in the class $\mathcal{F}_0(\lambda)$ , under the assumption that f''(0) = 0. In order to prove our result, we need the following lemma, which we can find in the proof of [16, Theorem 6]. Lemma A. [16] If $\phi(z): \mathbb{D} \to \mathbb{D}$ be analytic function, then (2.17) $$\frac{|\phi(z)|^2}{1 - |\phi(z)|^2} \le \frac{(\phi(0) + |z|)^2}{(1 - |\phi(0)|)^2 (1 - |z|^2)|)}$$ We derive results that provide sharp bounds for the pre-Schwarzian norm in terms of the value f''(0) for the function $f \in \mathcal{F}_0(\lambda)$ , particularly under the assumption f''(0) = 0.

Theorem 2.3. If $h \in \mathcal{F}_0(\lambda)$ , then for all $z \in \mathbb{D}$ and $\frac{1}{2} \leq \lambda \leq 1$ , then $$(2.18) ||Pf|| \le (1+2\lambda).$$ The inequality is sharp. Proof of Theorem 2.3. Since $\phi_1(z) = z\xi_1(z)$ , with $|\xi_1(z)| < 1$ , then in (2.2), we obtain $$\sup_{z \in \mathbb{D}} (1 - |z|^2) \left| \frac{f''(z)}{f'(z)} \right| \le \sup_{z \in \mathbb{D}} (1 - |z|^2) \frac{(1 + 2\lambda)|z\xi_1(z)|}{1 - |z|^2|\xi_1(z)|}$$ $$\le (1 + 2\lambda) \sup_{0 \le r \le 1} \frac{r(1 - r^2)}{(1 - r^2)}$$ $$= (1 + 2\lambda).$$ Thus we desired bound is established. The next part of the proof is to show that the bound is sharp. Henceforth, we consider the extremal function $f^*(z)$ is given by $$f^*(z) = \int_0^z \frac{1}{(1-\xi^2)^{\frac{1+2\lambda}{2}}} d\xi$$ for all $\frac{1}{2} \le \lambda \le 1$ . It can be easily shown that $||Pf^*|| = (1 + 2\lambda)$ . From Theorem 2.3, we obtain the following immediate result for the class C of convex functions. Corollary 2.3. If $f \in \mathcal{F}_0(1/2) = \mathcal{C}$ then for all $z \in \mathbb{D}$ , we have The bound is sharp. Sharpness of Corolary 4.3: For $\lambda = 1/2$ , it follows from (2.30) that $$\frac{f_{\frac{1}{2}}''}{f_{\frac{1}{2}}'}(z) = \frac{2z}{1-z^2}$$ and $Pf_{\frac{1}{2}} = \frac{2z}{1-z^2}$ . A simple computation thus yields that $$||Pf_{\frac{1}{2}}|| = \sup_{z \in \mathbb{D}} (1 - z^2) |Pf_{\frac{1}{2}}| = \sup_{z \in \mathbb{D}} (1 - |z|^2) \frac{2|z|}{1 - |z|^2} = 2$$ and we see the constant 2 is sharp. Remark 2.2. If $f \in \mathcal{F}_0(\frac{1}{2}) = \mathcal{C}$ , then for all $z \in \mathbb{D}$ , we have $$(1 - |z|^2) \left| \frac{h_1''(z)}{h_1'(z)} \right| \le 2|z|.$$ Our next result gives a sharp bound for the norm of the Schwarzian derivative when $f \in \mathcal{F}_0(\lambda)$ by a direct application of the Schwarz lemma.

Theorem 2.4. If $f \in \mathcal{F}_0(\lambda)$ , for all $z \in \mathbb{D}$ and $\frac{1}{2} \leq \lambda \leq 1$ , then the Schwarzian norm $$||Sf_{\lambda}|| = (1 - |z|^2)^2 |Sh(z)| \le \frac{(1 + 2\lambda)(3 - 2\lambda)}{2}.$$ The inequality is sharp. Proof of Theorem 2.4. From (2.2), we have $$\left(\frac{f''(z)}{f'(z)}\right) = \frac{(1+2\lambda)\phi(z)}{(1-z\phi(z))}$$ A simple calculation shows that (2.19) $$Sf(z) = (1+2\lambda)\frac{\phi'(z) + (\frac{1-2\lambda}{2})\phi^2(z)}{(1-z\phi(z))^2}.$$ By using triangle inequality and Schwarz pick lemma, we obtain <span id="page-9-1"></span> $$(2.20) \quad (1-|z|^2)^2|Sf| \le (1+2\lambda) \left| \phi'(z) + \left(\frac{1-2\lambda}{2}\right) \phi^2(z) \right| \frac{(1-|z|^2)^2}{|1-z\phi(z)|^2}$$ $$= \frac{(1+2\lambda)(1-|z|^2)^2}{|1-z\phi(z)|^2} \left(\frac{1-|\phi(z)|^2}{1-|z|^2} + \left(\frac{1-2\lambda}{2}\right) |\phi(z)|^2\right).$$ We define the function $\Phi(z): \mathbb{D} \to \mathbb{D}$ such that (2.21) $$\Phi(z) = \frac{\bar{z} - \phi(z)}{1 - z\phi(z)}.$$ Since $\phi(\mathbb{D}) \subseteq \mathbb{D}$ , then we have $(1-|z|^2)(1-|z\phi(z)|^2)>0$ , hence it follows that $$(2.22) |\bar{z} - \phi(z)|^2 < |1 - z\phi(z)|^2.$$ Thus, we conclude that $|\Phi(z)|^2 < 1$ . Consequently, we have the estimates (2.23) $$1 - |\Phi(z)|^2 = \frac{(1 - |\phi(z)|^2)(1 - |z|^2)}{|1 - z\phi(z)|^2}$$ and <span id="page-9-0"></span>(2.24) $$\frac{(1-|z|^2)^2}{|1-z\phi(z)|^2} = \frac{(1-|\Phi(z)|^2)(1-|z|^2)}{(1-|\phi(z)|^2)}.$$ If we replace the expression (2.24) in (2.20), we have <span id="page-9-3"></span> $$(2.25) (1-|z|^2)^2|Sf(z)| \le (1+2\lambda)(1-|\Phi(z)|^2)\left(1+(\frac{1-2\lambda}{2})\frac{|\phi(z)|^2(1-|z|^2)}{(1-|\phi(z)|^2)}\right).$$ Since f''(0) = 0 implies that $\phi(0) = 0$ , using Lemma A, we easily obtain that <span id="page-9-2"></span>(2.26) $$\frac{|\phi(z)|^2}{1 - |\phi(z)|^2} \le \frac{|z|^2}{1 - |z|^2}.$$ In view of (2.26) and (2.25), we obtain $$(2.27) (1-|z|^2)^2|Sf(z)| \le (1+2\lambda)(1-|\Phi(z)|^2)\left(1+(\frac{1-2\lambda}{2})|z|^2\right).$$ Also, becouse $1 - |\Phi(z)|^2 \le 1$ , we see that (2.28) $$(1 - |z|^2)^2 |Sf(z)| \le (1 + 2\lambda) \left( 1 + \frac{1 - 2\lambda}{2} \right)$$ $$= \frac{(1 + 2\lambda)(3 - 2\lambda)}{2}.$$ Thus our desired inequality is established. Next part of the proof is to show the sharpness of the inequality. Henceforth, we consider the function $f_{\lambda}(z)$ is given by (2.29) $$f_{\lambda}(z) = \int_{0}^{z} \frac{1}{(1 - \xi^{2})^{\frac{1+2\lambda}{2}}} d\xi, \quad \text{for all } \frac{1}{2} \le \lambda \le 1$$ 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 $\frac{1}{2} \leq \lambda \leq 1$ . Note that <span id="page-10-0"></span>(2.30) $$\frac{f_{\lambda}''}{f_{\lambda}'}(z) = \frac{(1+2\lambda)z}{1-z^2} \text{ and } Sf_{\lambda} = \frac{(1+2\lambda)}{(1-z^2)^2} \left[ 1 + \left(\frac{1-2\lambda}{2}\right)|z|^2 \right]$$ which calculates $$||Sf_{\lambda}|| = \sup_{z \in \mathbb{D}} (1 - |z|^2)^2 |Sf_{\lambda}| = (1 + 2\lambda) \frac{(3 - 2\lambda)}{2}.$$ In general, the integral formula for $f_{\lambda}$ given in above does not give primitives in terms of elementary functions, however when $\lambda = 1/2$ , we see that $$f_{\frac{1}{2}}(z) = \int_0^z \frac{1}{(1-\xi^2)} d\xi = \frac{1}{2} \log\left(\frac{1+z}{1-z}\right),$$ where $||Sf_{\frac{1}{2}}|| = 2$ . For the class C, we have the an immediate result from Theorem 2.4. Corollary 2.4. If $f \in \mathcal{F}_0(\frac{1}{2}) = \mathcal{C}$ , then for all $z \in \mathbb{D}$ , we have $$||Sf_{\frac{1}{2}}|| \le 2.$$ The inequality is sharp. For the value |f''(0)| which is not necessarily zero, we will give a bound for the quantity $(1-|z|^2)^2|Sf(z)|$ with $f \in \mathcal{F}_0(\lambda)$ in the following.

Theorem 2.5. If $f \in \mathcal{F}_0(\lambda)$ , for all $z \in \mathbb{D}$ and $\frac{1}{2} \leq \lambda \leq 1$ and $\gamma = |\phi(0)| = \frac{|f''(0)|}{(1+2\lambda)}$ , then $$(1-|z|^2)^2|Sf(z)| \le (1+2\lambda)\left(1+\frac{1-2\lambda}{2}\frac{1+\gamma}{1-\gamma}\right).$$ Proof of Theorem 2.5. Let $\gamma = |\phi(0)|$ . In view of the Lemma A, we see that <span id="page-11-0"></span>(2.31) $$\frac{|\phi(z)|^2}{1 - |\phi(z)|^2} \le \frac{(\gamma + |z|)^2}{(1 - \gamma^2)(1 - |z|^2)}.$$ If we substitute in (2.31) in (2.25), then we obtain $$(1-|z|^2)^2|Sf(z)| \le (1+2\lambda)(1-|\Phi(z)|^2)\left(1+(\frac{1-2\lambda}{2})\frac{(\gamma+|z|)^2}{(1-\gamma^2)}\right).$$ From the fact that |z| < 1 and $1 - |\Phi(z)|^2 \le 1$ , we easily calculate that $$(1-|z|^2)^2|Sf(z)| \le (1+2\lambda)\left(1+\frac{1-2\lambda}{2}\frac{1+\gamma}{1-\gamma}\right)$$ which is the required inequality. This completes the

Theorem 3.1. If $f = h + \bar{g} \in \mathcal{G}(\beta)$ , then $||P_f|| \leq 2\beta + 1$ . The estimate is best possible. Proof of Theorem 3.1. Let, $f = h + \bar{g} \in \mathcal{G}(\beta)$ . Then f holds the subordination relation $$1 + \frac{zf''(z)}{f'(z)} \prec \frac{1 - (1+\beta)z}{1 - z},$$ which gives us <span id="page-11-2"></span> $$\left|\frac{f''(z)}{f'(z)}\right| \le \frac{\beta}{1-|z|}.$$ By the Schwarz-Pick lemma and (3.1), we have $$\begin{aligned} ||P_f|| &= \sup_{z \in \mathbb{D}} \left(1 - |z|^2\right) |P_f| \\ &= \sup_{z \in \mathbb{D}} \left(1 - |z|^2\right) \left| \frac{f''(z)}{f'(z)} - \frac{\bar{\omega}(z)\omega'(z)}{1 - |\omega(z)|^2} \right| \\ &\leq \sup_{z \in \mathbb{D}} \left(1 - |z|^2\right) \left( \left| \frac{f''(z)}{f'(z)} \right| + \left| \frac{\bar{\omega}(z)\omega'(z)}{1 - |\omega(z)|^2} \right| \right) \\ &\leq \sup_{z \in \mathbb{D}} \left(1 - |z|^2\right) \left( \frac{\beta}{1 - |z|} + \frac{|\bar{\omega}(z)|}{1 - |z|^2} \right) \\ &\leq \sup_{z \in \mathbb{D}} \left(\beta(1 + |z|) + |\omega(z)|\right) \\ &= 2\beta + 1. \end{aligned}$$ We now show that the estimate is best possible. For $$\begin{cases} \sqrt{1-\beta} \le t < 1 & \text{when } \beta \in [0,1) \\ \frac{1-2\alpha}{3-2\alpha} \le t < 1 & \text{when } \beta \in [1,\infty), \end{cases}$$ we consider the function $f = h + \bar{g} \in \mathcal{G}(\beta)$ with the second complex dilatation $\omega_t(z) = \frac{z-t}{1-tz}$ and h(z) be such that $$1 + \frac{zh_1''(z)}{h_1'(z)} = \frac{1 - (1+\beta)z}{1 - z}.$$ Then clearly $f \in \mathcal{G}(\beta)$ for all t in $[0, \infty)$ . A simple calculation gives that $$\frac{\bar{\omega}_t(z)\omega_t(z)}{1 - |\omega_t(z)|^2} = \frac{\bar{z} - t}{(1 - tz)(1 - |z|^2)}.$$ Consequently, we have $$||P_{f_t}|| = \sup_{z \in \mathbb{D}} (1 - |z|^2) \left| \frac{h_1''(z)}{h_1'(z)} - \frac{\bar{\omega}(z)\omega'(z)}{1 - |\omega(z)|^2} \right|$$ $$= \sup_{z \in \mathbb{D}} (1 - |z|^2) \left| \frac{\beta}{1 - |z|} - \frac{\bar{z} - t}{(1 - tz)(1 - |z|^2)} \right|.$$ We set $$M_t = \sup_{z \in \mathbb{D}} (1 - |z|^2) \left| \frac{\beta}{1 - |z|} - \frac{\bar{z} - t}{(1 - tz)(1 - |z|^2)} \right|$$ <span id="page-12-0"></span>(3.2) $$= \sup_{z \in [0,1)} \left| \beta(1+r) - \frac{r-t}{1-tr} \right| = \sup_{z \in [0,1)} \phi(r),$$ where $$\phi(r) = \beta(1+r) - \frac{r-t}{1-tr}.$$ A simple computation shows that $$\phi'(r) = \beta - \frac{1 - t^2}{(1 - rt)^2}$$ and $\phi''(r) = -\frac{2t(1 - t^2)(1 - rt)}{(1 - rt)^4} < 0$ for all $r \in [0, 1)$ . Now, $\phi'(r) = 0$ gives $$r = r_0 := \frac{1}{t} - \frac{1}{t} \sqrt{\frac{1 - t^2}{\beta}}.$$ Thus the maximum value of $\phi$ is attained at $r_0$ . Hence, we have <span id="page-13-8"></span>(3.3) $$M_t = \phi(r_0) = \frac{1 + 2\beta - 2\beta\sqrt{\frac{1 - t^2}{\beta}}}{t}.$$ In view of (3.2) and (3.3), it is clear that $M_t \leq ||P_{f_t}|| \leq 2\lambda + 1$ . We note that $M_t$ is increasing function for t and $M_t \to 1 + 2\beta$ as $t \to 1$ . This shows that estimate is the best possible.

Definition 1.1. (see [9, Definition 1.1]) Let $f \in \mathcal{A}$ be a locally univalent for $z \in \mathbb{D}$ and let $-1/2 < \lambda \le 1$ . Then $f \in \mathcal{F}(\lambda)$ if, and only if <span id="page-2-1"></span>(1.4) $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) > \frac{1}{2} - \lambda \text{ for } z \in \mathbb{D}.$$ It is clear that, when $-1/2 \le \lambda \le 1/2$ , functions defined by (1.4) provide a subset of $\mathcal{C}$ , with $\mathcal{F}(1/2) = \mathcal{C}$ , and, since $1/2 - \lambda \ge -1/2$ when $\lambda \le 1$ , functions in $\mathcal{F}(\lambda)$ are close-to-convex when $1/2 \le \lambda \le 1$ . We shall cal members f of $\mathcal{F}(\lambda)$ when $1/2 \le \lambda \le 1$ Ozaki close-to-convex functions and denote this class by $\mathcal{F}_0(\lambda)$ . One important note that in contrast to the definition of the class $\mathcal{K}$ , the definition of $\mathcal{F}(\lambda)$ does not involve an independent starlike function g, but, as was shown in [38], members of $\mathcal{F}(1)$ have coefficients which grow at the same rate as those in $\mathcal{K}$ , that is, $\mathcal{O}(n)$ as $n \to \infty$ . 1.1. Pre-Schwarzian and Schwarzian derivatives of analytic functions. The pre-Schwarzian and Schwarzian derivatives are important research tools in geometric function theory, especially to characterize Teichumüller space by using the pre-Schwarzian and Schwarzian derivatives embedding models. They are also powerful tools in the research of the inner radius of univalency for planer domains and quasiconformal extensions, (see [30,31]). The research of pre Schwarzian and Schwarzian derivatives has a long history and is closely related to other disciplines. As early as 1836, Kummer firstly introduced the Schwarzian derivative when studying hypergeometric partial differential equations. Later, people have done a lot of research on the relationship between Schwarzian derivative and Univalent function. Several sufficient conditions for univalent analytic functions are obtained by using the notions of Pre-Schwarzian and Schwarzian derivatives. It is well known that the pre-Schwarzian norm $||Ph|| \leq 6$ for the Univalent analytic function h is defined in $\mathbb{D}$ . In 1972, Becker [13] used the pre-Schwarzian derivative to obtain the sufficient condition that the function in $\mathbb{D}$ is univalent, in other words, if $||Ph|| \le 1$ , then the function h is univalent in $\mathbb{D}$ . For different aspect of Schwarzian derivatives, we refer to the articles [2, 12, 15] and references therein. 1.2. Pre-Schwarzian and derivatives of Harmonic functions. A twice continuously differentiable complex-valued function f = u + iv in a domain $\Omega$ is called harmonic if u and v both are harmonic in $\Omega$ or equivalently if it satisfies the Laplace equation $\Delta f = 4f_{z\bar{z}} = 0$ . Every harmonic mapping f has a canonical representation of the form $f = h + \bar{g}$ , where h and g are analytic functions in $\Omega$ called the analytic and co-analytic part of f, respectively. The jacobian of $f = h + \bar{g}$ is given by $J_f(z) = |f_z|^2 - |f_{\bar{z}}|^2 = |h'(z)|^2 - |g'(z)|^2$ . The harmonic mapping f is called orientation-preserving or sense-preserving mapping if $J_f(z) > 0$ and is called orientation-reversing or sense-reversing mapping if $J_f(z) < 0$ . For a sense-preserving harmonic mapping $f = h + \bar{g}$ , its second complex dilatation is $\Omega$ with $\omega = g'/h'$ and $|\omega(z)| < 1$ in $\Omega_1$ . According to Lewy's theorem, see [32], a harmonic mapping $f = h + \bar{g}$ is locally univalent in $\Omega$ if its Jacobian $J_f(z) \neq 0$ . Stuyding the sufficient conditions for a locally univalent harmonic mapping to be univalent , as well as the necessary conditions for a univalent harmonic mapping is an interesting research area. For harmonic pre-Schwarzian and its applications, we refer to the articles [16, 20, 21, 34, 44]. Harmonic mappings play the natural role in parameterizing minimal surfaces in the context of differential geometry. Planner harmonic mappings have application not only in the differential geometry but also in the various field of engineering, physics, operations research and other intriguing aspects of applied mathematics. The theory of harmonic functions has been used to study and solve fluid flow problems (see [1]. The theory of univalent harmonic functions having prominent geometric properties like starlikeness, convexity and close-to-convexity appear naturally while dealing with planner fluid dynamical problems. For instance, the fluid flow problem on a convex domain satisfying an interesting geometric property has been extensively studied by Aleman and Constantin [1]. With the help of geometric properties of harmonic mappings, Constantin and Martin [14] have obtained a complete solution of classifying all two dimensional fluid flows. Let $\mathcal{H}$ denotes the class of all harmonic mappings $f = h + \bar{g}$ in $\mathbb{D}$ with the normalization h(0) = h'(0) - 1 = g(0) = 0. The function $f = h + \bar{g}$ is of the form $$h(z) = z + \sum_{n=2}^{\infty} a_n z^n$$ and $g(z) = \sum_{n=1}^{\infty} b_n z^n$ . As harmonic mapping is the generalization of analytic function, there are many open problem and interesting fact in planer harmonic mappings, (see, e.g. [18, 19, 42]). In 2003, Chuaqui et al. [17] defined the Schwarzian derivative of harmonic mappings, if $\omega$ equals the square of an analytic functions. Without assuming any additional condition on $\omega$ , in [20], Hernández and Martín gave another definition of Schwarzian derivative for the harmonic mappings. For locally univalent harmonic mapping $f = h + \bar{g}$ , Sf is given by $$Sf = (\log J_f)_{zz} - \frac{1}{2} (\log J_f)_z$$ $$= Sh + \frac{\bar{\omega}}{1 - |\omega|^2} \left( \frac{h''}{h'} \omega - \omega'' \right) - \frac{3}{2} \left( \frac{\omega' \bar{\omega}}{1 - |\omega|^2} \right)^2,$$ where Sh is the classical Schwarzian derivative of the analytic function h. The pre-Schwarzian derivative of harmonic mapping $f = h + \bar{g}$ defined as $$Pf(z) = (\log J_f)_z = \frac{h''(z)}{h'(z)} - \frac{\omega'(z)\bar{\omega}(z)}{1 - |\omega(z)|^2}.$$