Schwarzian Norm Estimates for Analytic Functions Associated with Convex Functions

Theorem 2.1. For $c \in (0,3]$ , the following are equivalent: (i) $$f \in \mathcal{F}(c)$$ . (ii) $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge 1 - \frac{c}{2} + \left(\frac{1 - |z|^2}{2c}\right) \left|\frac{zf''(z)}{f'(z)}\right|^2$$ . (iii) $$\left| (1 - |z|^2) \left( \frac{f''(z)}{f'(z)} \right) - c\overline{z} \right| \le c.$$ The inequalities (ii) and (iii) both are sharp for the function (2.1) $$f_c(z) = \frac{(1-z)^{1-c} - 1}{c-1} \text{ for } z \in \mathbb{D} \text{ with } c \in (0,3].$$ As a consequence of Theorem 2.1, we obtain two results for functions in the class $\mathcal{F}(2)$ involving sharp inequalities. Corollary 2.1. If $f \in \mathcal{C} := \mathcal{F}(2)$ be of the form (1.1), then we have (2.2) $$\left| (1 - |z|^2) \left( \frac{f''(z)}{f'(z)} \right) - 2\bar{z} \right| \le 2.$$ The inequality (2.2) is sharp. Corollary 2.2. If $f \in \mathcal{C} := \mathcal{F}(2)$ be of the form (1.1), then we have (2.3) $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge \left(\frac{1 - |z|^2}{4}\right) \left|\frac{zf''(z)}{f'(z)}\right|^2.$$ The inequality (2.3) is sharp. Example 2.1. For the sharpness of the inequalities (2.2) and (2.3), we consider the function defined in (2.1) with c=2 as $$f_2(z) = \frac{1}{(1-z)} - 1$$ A simple computation using (2.1) shows that $$1 + \frac{zf_2''(z)}{f_2'(z)} = \frac{1+z}{1-z}.$$ Moreover, we have $$\operatorname{Re}\left(1 + \frac{zf_2''(z)}{f_2'(z)}\right) > 0$$ hence, it is clear that $f_2 \in \mathcal{F}(2)$ . To show the inequality (2.2) of Corollary 2.1 is sharp, we consider z = r < 1 and establish that $$\left| (1 - |z|^2) \left( \frac{f_2''(z)}{f_2'(z)} \right) - 2\bar{z} \right| = \left| (1 - r^2) \left( \frac{2}{1 - r} \right) - 2r \right| = |2 - 2r + 2r| = 2.$$ To show the inequality (2.3) in Corollary 2.2 is sharp, we see from (2.6) (Proof of Theorem 2.1) that (2.4) $$\phi(z) = \frac{\frac{f_2''(z)}{f_2'(z)}}{\frac{zf_2''(z)}{f_2'(z)} - 1} = 1.$$ Thus, it is clear that $|\phi(z)|^2 = 1$ which further leads to $$\operatorname{Re}\left(1 + \frac{zf_2''(z)}{f_2'(z)}\right) = \left(\frac{1 - |z|^2}{4}\right) \left|\frac{zf_2''(z)}{f_2'(z)}\right|^2.$$ Proof of Theorem 2.1. First, we will prove that (i) is equivalent to (ii). For $\beta > 0$ , let $f \in \mathcal{F}(c)$ be of the form (1.1). Then from (1.5), we have (2.5) $$1 + \frac{zf''(z)}{f'(z)} \prec 1 + \frac{cz}{1-z}.$$ Hence, then there exists an analytic function $\omega: \mathbb{D} \to \mathbb{D}$ with $\omega(0) = 0$ such that $$1 + \frac{zf''(z)}{f'(z)} = 1 + \frac{c\omega(z)}{1 - \omega(z)}.$$ Let $\omega(z) = z\phi(z)$ for some analytic function $\phi$ that satisfy $\phi(\mathbb{D}) \subseteq \mathbb{D}$ . Then it follows from (2.5) that $$\frac{f''(z)}{f'(z)} = \frac{c\omega(z)}{z(1 - \omega(z))}$$ which yields that (2.6) $$\phi(z) = \frac{\frac{f''(z)}{f'(z)}}{\frac{zf''(z)}{f'(z)} + c}.$$ Since $|\phi(z)|^2 \le 1$ , an easy computation leads to (2.7) $$\left| \frac{f''(z)}{f'(z)} \right|^2 \le c^2 + 2c \operatorname{Re} \left( \frac{zf''(z)}{f'(z)} \right) + |z|^2 \left| \frac{f''(z)}{f'(z)} \right|^2.$$ By factorizing, we obtain $$c\left[c + 2\operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right)\right] \ge (1 - |z|^2) \left|\frac{f''(z)}{f'(z)}\right|^2$$ which implies that (2.8) $$\operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) \ge -\frac{c}{2} + \left(\frac{1-|z|^2}{2c}\right) \left|\frac{f''(z)}{f'(z)}\right|^2.$$ Consequently, we have the desired inequality $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \le 1 - \frac{c}{2} + \left(\frac{1 - |z|^2}{2c}\right) \left|\frac{zf''(z)}{f'(z)}\right|^2.$$ Next, we will prove that (ii) is equivalent to (iii). Multiplying (2.7) by $(1-|z|^2)$ both side, we have $$(1-|z|^2)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 \le c^2 (1-|z|^2) + 2c(1-|z|^2) \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right)$$ which implies that $$(1-|z|^2)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 - 2c(1-|z|^2) \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) + c^2|z|^2 \le c^2.$$ Hence, the required inequality is established. $$\left| (1 - |z|^2) \left( \frac{f''(z)}{f'(z)} \right) - c\bar{z} \right| \le c.$$ Considering the function $f_c$ defined in (2.1), we establish the sharpness of the above two inequalities, as confirmed in Corollary 2.1 and Corollary 2.2 for the spacial case c = 2. This completes the proof. In the next result, with the additional condition that f''(0) = 0, $\phi$ satisfies that $\phi(z) = z\psi(z)$ . for some analytic function $\psi$ with $|\psi(z)| < 1$ . We say that $f \in \mathcal{F}^0(c)$ if $f \in \mathcal{F}(c)$ and f''(0) = 0. In this way, we give our results as follows.

Theorem 2.2. For $c \in (0,3]$ , let $f \in \mathcal{F}^0(c)$ be of the form (1.1). Then we have (2.9) $$\frac{1}{(1+|z|^2)^{-\frac{c}{2}}} \le |f'(z)| \le \frac{1}{(1-|z|^2)^{\frac{c}{2}}}$$ and (2.10) $$\int_0^{|z|} \frac{1}{(1+\xi^2)^{-\frac{c}{2}}} d|\xi| \le |f(z)| \le \int_0^{|z|} \frac{1}{(1-\xi^2)^{\frac{c}{2}}} d|\xi|.$$ All of these estimates are sharp. Equality holds for the function $$f_{c,\lambda}(z) = \int_0^{|z|} \frac{1}{(1 - \lambda \zeta^2)^{c/2}} d|\zeta|$$ for some $\lambda \in \mathbb{C}$ with $|\lambda| = 1$ . As a consequence of Theorem 2.2, we obtain the following result. Corollary 2.3. If $f \in \mathcal{C} := \mathcal{F}^0(2)$ be of the form (1.1), then the sharp inequality (2.11) $$\frac{1}{(1+|z|^2)} \le |f'(z)| \le \frac{1}{(1-|z|^2)}$$ and (2.12) $$\int_0^{|z|} \frac{1}{(1+\xi^2)} d|\xi| \le |f(z)| \le \int_0^{|z|} \frac{1}{(1-\xi^2)} d|\xi|.$$ All of these estimates are sharp. Equality holds for the function $$f_{2,\lambda}(z) = \int_0^{|z|} \frac{1}{(1 - \lambda \zeta^2)} d|\zeta|$$ for some $\lambda \in \mathbb{C}$ with $|\lambda| = 1$ . Proof of Theorem 2.2. Let $f \in \mathcal{F}^0(c)$ , and from (2.6), we obtain that $\phi(0) = 0$ . Then by using the Schwarz lemma, we get $$\left| \frac{\frac{f''(z)}{f'(z)}}{\frac{zf''(z)}{f'(z)} + c} \right|^2 \le |z|^2$$ which is equivalent to the inequality $$\left| \frac{f''(z)}{f'(z)} \right|^2 \le c^2 |z|^2 + 2c|z|^2 \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) + |z|^2 \left| \frac{zf''(z)}{f'(z)} \right|^2.$$ Thus, we have the estimate (2.13) $$(1 - |z|^4) \left| \frac{f''(z)}{f'(z)} \right|^2 \le c^2 |z|^2 + 2c|z|^2 \operatorname{Re} \left( \frac{zf''(z)}{f'(z)} \right).$$ Multiplying both sides the above inequality by $(1-|z|^4)$ , we obtain $$(1 - |z|^4)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 - 2c|z|^2 (1 - |z|^4) \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right)$$ $$\leq c^2 |z|^2 (1 - |z|^4).$$ Adding $(c|z|^2\bar{z})^2$ to both sides of the above inequality leads to $$(1 - |z|^4)^2 \left| \frac{f''(z)}{f'(z)} \right|^2 - 2c|z|^2 (1 - |z|^4) \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) + c^2|z|^4|\bar{z}|^2$$ $$< c^2|z|^2 (1 - |z|^4) + c^2|z|^4|\bar{z}|^2.$$ Multiplying both side by |z|, then by simple calculation, we obtain $$\left| (1 - |z|^4) \frac{zf''(z)}{f'(z)} - c|z|^4 \right| \le c|z|^2$$ which implies that $$\frac{c|z|^2}{1+|z|^2} \le \operatorname{Re}\left(\frac{zf''(z)}{f'(z)}\right) \le \frac{c|z|^2}{1-|z|^2}.$$ Let $z = re^{i\theta}$ . Then it is easy to see that $$\frac{cr}{1+r^2} \le \frac{\partial}{\partial r} \left( \log |f'(re^{i\theta})| \right) \le \frac{cr}{1-r^2}.$$ Integrating the above estimate w.r.t. r, we obtain $$\frac{1}{(1+|z|^2)^{-\frac{c}{2}}} \le |f'(z)| \le \frac{1}{(1-|z|^2)^{\frac{c}{2}}}.$$ Next, the growth part of the theorem follows from the upper bound $$|f'(re^{i\theta})| = \left| \int_0^r f'(re^{i\theta})e^{i\theta}dt \right| \le \int_0^r |f'(re^{i\theta})|dt \le \int_0^r \frac{1}{(1-t^2)^{\frac{c}{2}}}dt$$ which implies that $$|f(z)| \le \int_0^{|z|} \frac{1}{(1-\xi^2)^{\frac{c}{2}}} d|\xi|$$ 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 $$|f(z)| \ge |f(z_0)| \ge \int_0^{|z|} \frac{1}{(1+\xi^2)^{\frac{c}{2}}} d|\xi|.$$ Thus all the desired inequalities are established. For the function $f_{c,\lambda}(z)$ in the main result, the sharpness part can be shown easily, hence we omit the details. We will find the sharp bound of the pre-Schwarzian and Schwarzian norm for the function f in the class $\mathcal{F}(c)$ , under the assumption that f''(0) = 0. The following lemma will paly a key role to prove the result. Lemma A. [38] If $\phi(z) : \mathbb{D} \to \mathbb{D}$ be analytic function, then $$\frac{|\phi(z)|^2}{1 - |\phi(z)|^2} \le \frac{(\phi(0) + |z|)^2}{(1 - |\phi(0)|)^2 (1 - |z|^2)|)}$$ Our result is sharp bound of the pre-Schwarzian norm for $f \in \mathcal{F}^0(c)$ .

Theorem 2.3. For $c \in (0,3]$ , let $f \in \mathcal{F}^0(c)$ be of the form (1.1), then the preschwarzian norm satisfies the inequality The inequality is sharp. Proof of Theorem 2.3. Since $\phi(z) = z\xi(z)$ , with $|\xi(z)| < 1$ , then in (2.6) we obtain $$\sup_{z \in \mathbb{D}} (1 - |z|^2) \left| \frac{f''(z)}{f'(z)} \right| \le \sup_{z \in \mathbb{D}} (1 - |z|^2) \frac{c|z\xi(z)|}{1 - |z|^2|\xi(z)|}$$ $$\le c \sup_{0 \le r \le 1} \frac{r(1 - r^2)}{(1 - r^2)}$$ $$= c$$ Thus, we have the required inequality $||Pf|| \le c$ . To show that the inequality is sharp, we consider the function $f_c^*$ given by $$f_c^*(z) = \int_0^z \frac{1}{(1-\xi^2)^{\frac{c}{2}}} d\xi.$$ It can be easily shown that $||Pf_c^*|| = c$ . This completes the proof. For the class C of convex functions, we obtain the following sharp result, derived from Theorem 2.3. Corollary 2.4. Let $f \in \mathcal{C} := \mathcal{F}^0(2)$ be of the form (1.1), then the pre-schwarzian norm satisfies the inequality $$||Pf|| \le 2.$$ The inequality is sharp. Sharpness of Corollary 2.4. For c=2, it follows from that $$\frac{f_2''}{f_2'}(z) = \frac{2z}{(1-z^2)}$$ and $Pf_2 = \frac{2}{1-z^2}$ . A simple computation thus yields that $$||Pf_2|| = \sup_{z \in \mathbb{D}} (1 - z^2) |Pf_2| = \sup_{z \in \mathbb{D}} (1 - |z|^2) \frac{2}{1 - |z|^2} = 2$$ and we see the constant 2 is best possible. Using the Schwarz lemma, we obtain a sharp bound for the Schwarzian derivative norm when $f \in \mathcal{F}(c)$ .

Theorem 2.4. For $c \in (0,3]$ , let $f \in \mathcal{F}^0(c)$ be of the form (1.1). Then the Schwarzian norm satisfies the inequality $$||Sf|| = \sup_{z \in \mathbb{D}} (1 - |z|^2)^2 |Sf(z)| \le \frac{c(4-c)}{2}.$$ The inequality is sharp. For the class of convex functions, we obtain the following immediate result from Theorem 2.4, showing that the sharp bound of ||Sf|| is 2. Corollary 2.5. If $f \in \mathcal{C} := \mathcal{F}^0(2)$ be of the form (1.1) with f''(0) = 0, then Schwarzian norm satisfies the inequality $$||Sf|| = \sup_{z \in \mathbb{D}} (1 - |z|^2)^2 |Sf(z)| \le 2.$$ The estimate is sharp. Proof of Theorem 2.4. Let $f \in \mathcal{F}^0(c)$ be of the form (1.1). Then from (2.6), we have $$\frac{f''(z)}{f'(z)} = \frac{c\phi(z)}{(1 - z\phi(z))}.$$ A simple calculation gives that $$Sf(z) = c \left( \frac{\phi'(z) + \left(1 - \frac{c}{2}\right)\phi^2(z)}{(1 - z\phi(z))^2} \right).$$ By using triangle inequality and Schwarz pick lemma, we obtain $$(2.14) (1-|z|^2)^2|Sf| \le c \left|\phi'(z) + \left(1 - \frac{c}{2}\right)\phi^2(z)\right| \frac{(1-|z|^2)^2}{|1 - z\phi(z)|^2}$$ $$= \frac{c(1-|z|^2)^2}{|1 - z\phi(z)|^2} \left(\frac{1-|\phi(z)|^2}{1-|z|^2} + \left(1 - \frac{c}{2}\right)|\phi(z)|^2\right).$$ 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 $$|\bar{z} - \phi(z)|^2 < |1 - z\phi(z)|^2$$ Thus, we conclude that $|\Psi(z)|^2 < 1$ . It is easy to see that $$1 - |\Psi(z)|^2 = \frac{(1 - |\phi(z)|^2)(1 - |z|^2)}{|1 - z\phi(z)|^2}$$ and (2.15) $$\frac{(1-|z|^2)^2}{|1-z\phi(z)|^2} = \frac{(1-|\Psi(z)|^2)(1-|z|^2)}{(1-|\phi(z)|^2)}.$$ If we replace the expression (2.15) in (2.14), then we have $$(2.16) \qquad (1-|z|^2)^2|Sf(z)| \le c(1-|\Psi_1(z)|^2)\left(1+\left(1-\frac{c}{2}\right)\frac{|\phi(z)|^2(1-|z|^2)}{(1-|\phi(z)|^2)}\right).$$ Since f''(0) = 0 implies that $\phi(0) = 0$ , using Lemma A, we obtain (2.17) $$\frac{|\phi(z)|^2}{1 - |\phi(z)|^2} \le \frac{|z|^2}{1 - |z|^2}.$$ Using (2.17) in (2.16), we obtain $$(1 - |z|^2)^2 |Sf(z)| \le c(1 - |\Psi(z)|^2) \left(1 + \left(1 - \frac{c}{2}\right) |z|^2\right).$$ Again, considering that 1 − |Ψ(z)| ² ≤ 1, one can readily observe that $$\sup_{z \in \mathbb{D}} (1 - |z|^2)^2 |Sh(z)| \le c \left( 1 + \left( 1 - \frac{c}{2} \right) \right) = \frac{c(4 - c)}{2}.$$ Thus the desired inequality is obtained. The sharpness of the inequality follows from the Example 2.2. □ Example 2.2. The family of parameterized functions defined as: $$f_c(z) = \int_0^z \frac{1}{(1-\xi^2)^{c/2}} d\xi$$ , for $c \in (0,3]$ maximizes the Schwarzian norm defined as: $$||Sf_c|| = \sup_{z \in \mathbb{D}} (1 - |z|^2)^2 |Sf_c|$$ and from this, the sharpness of the inequality holds for c > 0. Note that $$\frac{f_c''}{f_c'}(z) = \frac{cz}{1-z^2}$$ and $Sf_c = \frac{c}{(1-z^2)^2} \left[ 1 + \left(1 - \frac{c}{2}\right) |z|^2 \right]$ which shows that $$||Sf_c|| = \sup_{z \in \mathbb{D}} (1 - |z|^2)^2 |Sf_c|$$ $$= c \left( 1 + \left( 1 - \frac{c}{2} \right) \right)$$ $$= \frac{c(4 - c)}{2}.$$ Given that the value |f ′′(0)| is not necessarily zero, we provide a bound for the quantity (1 − |z| 2 ) 2 |Sf(z)| where f ∈ F(c). Theorem 2.5. If f ∈ F(c), for all z ∈ D with c ∈ (0, 3] and $$\gamma = |\phi(0)| = \frac{|f''(0)|}{c},$$ then $$(1-|z|^2)^2|Sf(z)| \le c\left(1+\left(1-\frac{c}{2}\right)\frac{1+\gamma}{1-\gamma}\right).$$ We have the following immediate result from Theorem 2.5 for the class C of convex functions. Corollary 2.6. If f ∈ C := F(2) be of the form (1.1), then the inequality for c = 2, with $$\gamma = |\phi(0)| = \frac{|f''(0)|}{2}$$ the following inequality holds: $$(1 - |z|^2)^2 |Sf(z)| \le 2.$$ Proof of Theorem 2.5. Let γ = |ϕ(0)|. Applying the Lemma A, we calculate $$\frac{|\phi(z)|^2}{1 - |\phi(z)|^2} \le \frac{(\gamma + |z|)^2}{(1 - \gamma^2)(1 - |z|^2)}.$$ If we substitute the above inequality into (2.14), we get $$(1-|z|^2)^2|Sf(z)| \le c(1-|\Phi_1(z)|^2)\left(1+\left(1-\frac{c}{2}\right)\frac{(\gamma+|z|)^2}{(1-\gamma^2)}\right).$$ From the fact that |z| < 1 and 1 − |Φ1(z)| ² ≤ 1, we can easily calculate $$(1-|z|^2)^2|Sf(z)| \le c\left(1+\left(1-\frac{c}{2}\right)\frac{1+\gamma}{1-\gamma}\right) = 2.$$ This is the desired bound. □