The Schwarzian norm estimates for Janowski convex functions

Lemma 2.1. (Dieudonné's lemma). [5, 6] Let $\omega \in \mathcal{B}_0$ and $z_0 \neq 0$ be a fixed point in $\mathbb{D}$ . The region of variability of $\omega'(z_0)$ is given by (2.1) $$\left|\omega'(z_0) - \frac{\omega(z_0)}{z_0}\right| \le \frac{|z_0|^2 - |\omega(z_0)|^2}{|z_0|(1 - |z_0|^2)}.$$ Moreover, the equality occurs in (2.1) if and only if $\omega \in \mathcal{B}_0$ is a Blaschke product of degree 2. The Dieudonné's lemma is an extension of the Schwarz's lemma as well as Schwarz-Pick lemma. Here, we note that a Blaschke product of degree $n \in \mathbb{N}$ is of the form <span id="page-3-0"></span> $$B(z) = e^{i\theta} \prod_{j=1}^{n} \frac{z - z_j}{1 - \bar{z}_j z}, \quad z, z_j \in \mathbb{D}, \ \theta \in \mathbb{R}.$$ The Dieudonné's lemma will be key in proving our main results. Before we state our main results, we introduce some sets which we will use throughout our next discussion. Let $E := \{(A, B) : -1 \le B < A \le 1\}$ and we consider following subsets of E. <span id="page-3-1"></span>(2.2) $$\begin{cases} E_1 & := \{(A,B) \in E : 1 - \sqrt{1 - B^2} < |A + B| < 1 + \sqrt{1 - B^2}\}, \\ E_2 & := \{(A,B) \in E \cup E_1^c : |A + B| \le |B|\} \\ & = \{(A,B) \in E : |A + B| \le 1 - \sqrt{1 - B^2}, |A + B| \le |B|\}, \\ E_3 & := \{(A,B) \in E \cup E_1^c : |A + B| > |B|\} \\ & = \{(A,B) \in E : |A + B| \ge 1 + \sqrt{1 - B^2}, |A + B| > |B|\}. \end{cases}$$ It evident that $E_1, E_2$ and $E_3$ are mutually disjoint and $E = E_1 \cup E_2 \cup E_3$ .

Theorem 2.1. For $-1 \le B < A \le 1$ , let $f \in C(A, B)$ be of the form (1.1) and $E_1$ , $E_2$ , $E_3$ are given by (2.2). Then the Schwarzian derivative $S_f(z)$ satisfies the inequality <span id="page-3-2"></span> $$(2.3) |S_f(z)| \le \frac{(A-B)(2-|A+B|(1-|z|^2))}{(1-|z|^2)(2-|A+B|(1-|z|^2)-2B^2|z|^2)}, z \in \mathbb{D}$$ for $(A, B) \in E_1 \cup E_2$ and <span id="page-3-3"></span> $$(2.4) |S_f(z)| \le \begin{cases} \frac{(A-B)(2-|A+B|(1-|z|^2))}{(1-|z|^2)(2-|A+B|(1-|z|^2)-2B^2|z|^2))} & \text{if } z \in S \cap \mathbb{D}, \\ \frac{|A^2-B^2|}{2(1-|B||z|)^2} & \text{if } z \in S^c \cap \mathbb{D} \end{cases}$$ for (A, B) ^∈ ^E3, where ^S ⁼ {^z ^∈ ^C : ^|z^| < δ¹ or, ^|z^| > δ2} with $$\delta_1 = \frac{|B| - \sqrt{B^2 - |A + B|(2 - |A + B|)}}{|A + B|}$$ and $$\delta_2 = \frac{|B| + \sqrt{B^2 - |A + B|(2 - |A + B|)}}{|A + B|}.$$ Proof. For −1 ≤ B < A ≤ 1, let f ∈ C(A, B) be of the form [\(1.1\)](#page-1-1). Then we have $$1 + \frac{zf''(z)}{f'(z)} \prec \frac{1 + Az}{1 + Bz}.$$ Thus, there exists an analytic function ^ω : ^D ^→ ^D with ^ω(0) = 0 such that $$1 + \frac{zf''(z)}{f'(z)} = \frac{1 + A\omega(z)}{1 + B\omega(z)}.$$ A simple computation gives $$\frac{f''(z)}{f'(z)} = \frac{(A-B)\omega(z)}{z(1+B\omega(z))},$$ and consequently, <span id="page-4-0"></span>(2.5) $$S_{f}(z) = \left[\frac{f''(z)}{f'(z)}\right]' - \frac{1}{2} \left[\frac{f''(z)}{f'(z)}\right]^{2}$$ $$= (A - B) \left[\frac{\omega'(z)}{z(1 + B\omega(z))^{2}} - \frac{2\omega(z) + (A + B)\omega^{2}(z)}{2z^{2}(1 + B\omega(z))^{2}}\right].$$ Let us consider the transformation ζ(z) = ω ′ (z) − ω(z) z . By Dieudonné's lemma, the function ζ varies over the closed disk $$|\zeta(z)| \le \frac{|z|^2 - |\omega(z)|^2}{|z|(1 - |z|^2)},$$ for fixed |z| < 1 and z 6= 0. Using the transformation of ζ in [\(2.5\)](#page-4-0), we obtain $$S_f(z) = (A - B) \left[ -\frac{(A + B)\omega^2(z)}{2z^2(1 + B\omega(z))^2} + \frac{\zeta(z)}{z(1 + B\omega(z))^2} \right].$$ Thus, $$|S_f(z)| \le (A - B) \left[ \frac{|A + B||\omega(z)|^2}{2|z|^2|1 + B\omega(z)|^2} + \frac{|\zeta(z)|}{|z||1 + B\omega(z)|^2} \right]$$ $$\le (A - B) \left[ \frac{|A + B||\omega(z)|^2}{2|z|^2(1 - |B||\omega(z)|)^2} + \frac{|z|^2 - |\omega^2(z)|}{|z|^2(1 - |z|^2)(1 - |B||\omega(z)|)^2} \right].$$ For 0 ≤ s := |ω(z)| ≤ |z| < 1, we have <span id="page-4-1"></span> $$(2.6) |S_f(z)| \le (A-B) \left[ \frac{|A+B|s^2}{2|z|^2 (1-|B|s)^2} + \frac{|z|^2 - s^2}{|z|^2 (1-|z|^2) (1-|B|s)^2} \right]$$ $$= (A-B) \frac{2|z|^2 - s^2 (2-|A+B|(1-|z|^2))}{2|z|^2 (1-|z|^2) (1-|B|s)^2}$$ $$= (A-B)g(s),$$ where <span id="page-5-0"></span>(2.7) $$g(s) = \frac{2|z|^2 - s^2(2 - |A + B|(1 - |z|^2))}{2|z|^2(1 - |z|^2)(1 - |B|s)^2}, \quad 0 \le s \le |z| < 1.$$ Now, we wish to find the maximum value of g(s) over the closed interval [0, |z|]. To do this, we first find the critical points of g(s) in (0, |z|). A simple computation gives <span id="page-5-3"></span> $$g'(s) = \frac{2|B||z|^2 - s(2 - |A + B|(1 - |z|^2))}{|z|^2(1 - |z|^2)(1 - |B|s)^3},$$ and so g'(s) = 0 yields (2.8) $$s = \frac{2|B||z|^2}{2 - |A + B|(1 - |z|^2)} =: s_0(|z|).$$ Now, we have to check for what values of A and B, the point $s = s_0(|z|)$ lies in (0, |z|). We first note that the inequality (2.9) $$s_0(|z|) = \frac{2|B||z|^2}{2 - |A + B|(1 - |z|^2)} < |z|,$$ holds true if and only if k(|z|) > 0, where (2.10) $$k(|z|) = 2 - |A + B| - 2|B||z| + |A + B||z|^{2}.$$ We also note that k(0) = 2 - |A + B| > 0 and the discriminant of k(|z|) is given by <span id="page-5-4"></span> $$\Delta = 4((|A+B|-1)^2 + B^2 - 1).$$ Let $(A, B) \in E_1$ . If B = 0 then g'(s) < 0 in (0, |z|) and so, g(s) is a decreasing function in (0, |z|). Thus, the maximum of g(s) attains at s = 0. Thus, for B = 0, the desired result (2.3) follows from (2.6) and (2.7). If $B \neq 0$ then $s_0(|z|) > 0$ and $\Delta < 0$ . Hence, in this case k(|z|) has no real zero and so k(|z|) > 0. Consequently, $s_0(|z|)$ lies in (0, |z|). Let $(A, B) \in E_2$ . Then clearly $B \neq 0$ , $s_0(|z|) > 0$ and $\Delta \geq 0$ . If A + B = 0, then clearly k(|z|) > 0 and consequently, $s_0(|z|)$ lies in (0, |z|). If $A + B \neq 0$ then the zeros of k(|z|) in the real line are given by <span id="page-5-1"></span>(2.11) $$\delta_1 = \frac{|B| - \sqrt{B^2 - |A + B|(2 - |A + B|)}}{|A + B|}$$ and (2.12) $$\delta_2 = \frac{|B| + \sqrt{B^2 - |A + B|(2 - |A + B|)}}{|A + B|}.$$ For $(A,B) \in E_2$ with $A+B \neq 0$ , we have $|B|/|A+B| \geq 1$ and so $\delta_2 \notin (0,1)$ . Further, k(0) > 0 and k(1) = 2(1-|B|) > 0 for $B \neq -1$ . Therefore if $B \neq -1$ then k(|z|) has either two zeros or no zero in (0,1). Consequently, for $B \neq -1$ , k(|z|) has no zero in (0,1) and so, $s_0(|z|)$ lies in (0,|z|). Again, if B = -1 then $\delta_1 = 1 \notin (0,1)$ and <span id="page-5-2"></span> $$\delta_2 = \frac{2}{|A-1|} - 1 \ge 1.$$ Hence, for B = -1, k(|z|) has no zero in (0,1) and so, $s_0(|z|)$ lies in (0,|z|). Therefore, $s_0(|z|)$ lies in (0,|z|) if $(A,B) \in E_1 \cup E_2 \setminus \{(A,0)\}$ . Since the numerator of g'(s) is a linear function of s, and $g'(0) = 2|B|/(1-|z|^2) > 0$ , $g'(s_0(|z|)) = 0$ , it follows that g'(|z|) < 0. Hence the function g(s) is increasing in $(0, s_0(|z|))$ and decreasing in $(s_0(|z|), |z|)$ and consequently, the maximum of g(s) attains at $s_0(|z|)$ . Thus the desired result (2.3) follows from (2.6) and (2.7). Let $(A, B) \in E_3$ . Again, if B = -1 then k(|z|) have two zeros $\delta_1 = \frac{2}{|A-1|} - 1 \in (0, 1)$ and $\delta_2 = 1 \notin (0, 1)$ . If $B \neq -1$ then the function k(|z|) have two zeros $\delta_1$ and $\delta_2$ in (0, 1) which are given by (2.11) and (2.12), respectively. In any case, by Rolle's theorem, the function k'(|z|) has exactly one zero, say $\alpha$ , in $(\delta_1, \delta_2)$ . Since, k'(0) = -2|B| < 0 and k'(1) = 2(|A+B|-|B|) > 0, the function k(|z|) is strictly decreasing in $(0, \alpha)$ and strictly increasing in $(\alpha, 1)$ . Thus, we conclude that k(|z|) > 0 if $z \in S \cap \mathbb{D}$ , where $S = \{z \in \mathbb{C} : |z| < \delta_1$ or, $|z| > \delta_2\}$ ; and k(|z|) < 0 if $\delta_1 < |z| < \delta_2$ . Therefore, $s_0(|z|)$ lies in (0, |z|) if $z \in S \cap \mathbb{D}$ , and $s_0(|z|)$ does not lie in (0, |z|) if $z \in S^c \cap \mathbb{D}$ . Therefore, for $z \in S \cap \mathbb{D}$ , following the same argument as before, the function g(s) is increasing in $(0, s_0(|z|))$ and decreasing in $(s_0(|z|), |z|)$ and consequently, the maximum of g(s) attains at $s_0(|z|)$ . Thus the desired result (2.4) follows from (2.6) and (2.7). Again, for $z \in S^c \cap \mathbb{D}$ , from (2.7), we have $$g(|z|) = \frac{|A+B|}{2(1-|B||z|)^2} = \frac{|A+B|(1-|z|^2)}{2(1-|B||z|)^2}g(0) \ge g(0)$$ as $k(|z|) \leq 0$ for $z \in S^c \cap \mathbb{D}$ . Thus, the desired result (2.4) follows immediately. This completes the proof. Before we proceed further, let us discuss the sharpness of the estimate of $|S_f(z)|$ obtained in Theorem 2.1. Let $(A, B) \in E_3$ . We consider the function $K_{A,B}(z) \in \mathcal{C}(A, B)$ defined by (1.2). The Schwarzian derivative of $K_{A,B}$ is given by $$S_{K_{A,B}}(z) = -\frac{A^2 - B^2}{2(1 + Bz)^2}.$$ For B > 0 and any $z_0 \in S^c \cap \mathbb{D}$ with $-1 < z_0 \le 0$ , we have $|S_{K_{A,B}}(z_0)| = |A^2 - B^2|/2(1 - |B||z_0|)^2$ . Again, for B < 0 and any $z_0 \in S^c \cap \mathbb{D}$ with $0 \le z_0 < 1$ , we have $|S_{K_{A,B}}(z_0)| = |A^2 - B^2|/2(1 - |B||z_0|)^2$ . This shows that the second inequality in (2.4) is sharp in such cases. Next, suppose that A, B and $z_0$ satisfy either of the following two conditions: - (C1) $(A, B) \in E_1 \cup E_2 \text{ and } -1 < z_0 < 1,$ - (C2) $(A, B) \in E_3 \text{ and } -1 < z_0 < 1 \text{ with } z_0 \in S \cap \mathbb{D}.$ Depending on A and B, we choose a pair of unimodular real numbers (p,q) as follows: <span id="page-7-1"></span>(2.13) $$\begin{cases} (p,q) = (1,1) & \text{when } A+B \leq 0, \\ (p,q) = (-1,1) & \text{when } A+B > 0, B \geq 0, \\ (p,q) = (-1,-1) & \text{when } A+B > 0, B < 0. \end{cases}$$ For given A, B and $z_0$ satisfying (C1) or (C2), we choose (p, q) as above and consider the function $f_{z_0,p,q}$ defined by (2.14) $$1 + \frac{zf_{z_0,p,q}''(z)}{f_{z_0,p,q}'(z)} = \frac{1 + A\phi(z)}{1 + B\phi(z)},$$ where <span id="page-7-2"></span> $$\phi(z) = \frac{pz(z-b)}{1-bz},$$ and b is a solution of the equation (2.15) $$\frac{z_0(z_0 - b)}{1 - bz_0} = qs_0(|z_0|) = \frac{2q|B|z_0^2}{2 - |A + B|(1 - z_0^2)},$$ $$\ni.e., b = \frac{z_0(2 - 2q|B| - |A + B|(1 - z_0^2))}{2 - 2q|B|z_0^2 - |A + B|(1 - z_0^2)}.$$ <span id="page-7-0"></span>We know that whenever A, B and $z_0$ satisfy (C1) or (C2), the point $s_0(|z_0|)$ lies in $[0, |z_0|)$ , where $s_0(|z|)$ is given by (2.8). This ensures that $b \in (-1, 1)$ and $\phi$ is a Blaschke product of degree 2 with $\phi(0) = 0$ . Hence, the function $f_{z_0,p,q}$ belong to the class $\mathcal{C}(A, B)$ . For the function $f_{z_0,p,q}$ , the Schwarzian derivative is given by $$S_{f_{z_0,p,q}}(z) = \left[\frac{f_{z_0,p,q}''(z)}{f_{z_0,p,q}'(z)}\right]' - \frac{1}{2} \left[\frac{f_{z_0,p,q}''(z)}{f_{z_0,p,q}'(z)}\right]^2$$ $$= (A - B) \left[ -\frac{(A+B)\phi^2(z)}{2z^2(1+B\phi(z))^2} + \frac{\phi'(z) - \frac{\phi(z)}{z}}{z(1+B\phi(z))^2} \right]$$ $$= (A - B) \frac{-(A+B)(z-b)^2 + 2p(1-b^2)}{2(1-bz+Bpz(z-b))^2}.$$ Substituting the value of b, given in (2.15), and then evaluating the Schwarzian derivative $S_{f_{z_0,p,q}}(z)$ at $z_0$ , we obtain $$S_{f_{z_0,p,q}}(z_0) = -(A-B)\frac{2q^2|B|^2z_0^2(2p+(A+B)(1-z_0^2)) - p(2-|A+B|(1-z_0^2))^2}{(1-z_0^2)(2-|A+B|(1-z_0^2) + 2pqB|B|z_0^2)^2}.$$ Therefore, for any pair of (p,q), given in (2.13), we have <span id="page-7-3"></span> $$(2.16) |S_{f_{z_0,p,q}}(z_0)| = \frac{(A-B)(2-|A+B|(1-|z_0|^2))}{(1-|z_0|^2)(2-|A+B|(1-|z_0|^2)-2B^2|z_0|^2)}.$$ This shows that the inequality (2.3) and the first inequality in (2.4) are sharp for real z. The above discussion shows that the estimate of the Schwarzian derivative $|S_f(z)|$ obtained in Theorem 2.1 is sharp for certain real values of z. This also helps us to obtain the sharp estimate of the Schwarzian norm $||S_f(z)||$ for functions in C(A, B) which is given below.

Theorem 2.2. For $-1 \le B < A \le 1$ , let $f \in \mathcal{C}(A, B)$ be of the form (1.1) and $E_1$ , $E_2$ , $E_3$ are given by (2.2). (i) If $(A, B) \in E_1 \cup E_2$ , then (2.17) $$||S_f|| \le \begin{cases} 2, & \text{for } B = -1, \\ A - B, & \text{for } B \neq -1, |A + B| \le 2(1 - B^2), \\ (A - B)\gamma(\alpha), & \text{for } B \neq -1, |A + B| > 2(1 - B^2), \end{cases}$$ <span id="page-8-3"></span><span id="page-8-0"></span>where $\gamma$ is given by (2.18) $$\gamma(t) = \frac{(1-t^2)(2-|A+B|(1-t^2))}{2-|A+B|(1-t^2)-2B^2t^2},$$ and $\alpha$ is the unique root in (0,1) of the equation h(t)=0 with <span id="page-8-2"></span>(2.19) $$h(t) = (2 - |A + B|)(|A + B| + 2B^{2} - 2) - 2|A + B|(2 - |A + B|)t^{2} + |A + B|(2B^{2} - |A + B|)t^{4}.$$ <span id="page-8-5"></span>(ii) If $(A, B) \in E_3$ , then (2.20) $$||S_f|| \le \frac{2|A^2 - B^2|(1 - \sqrt{1 - B^2})^2}{B^4}.$$ Moreover, all the estimates are sharp.

Corollary 2.1. Let $f \in C(1,-1) =: C$ be of the form (1.1). Then the Schwarzian norm satisfies the sharp inequality $$||S_f|| < 2.$$ If we choose $A = 1 - 2\alpha$ , $0 \le \alpha \le 1$ and B = -1 in Theorem 2.2 then the first inequality in (2.17) and (2.20) provides Schwarzian norm estimate for the class of convex functions of order $\alpha$ , which was obtained by Suita [28]. Also, the inequality (2.20) in Theorem 2.2 in conjunction with Theorem A gives the quasiconformal extension. Corollary 2.2. Let $f \in C(1-2\alpha,-1) =: C(\alpha), 0 \le \alpha \le 1$ be of the form (1.1). Then the Schwarzian norm satisfies the sharp inequality $$||S_f|| \le \begin{cases} 2, & \text{for } 0 \le \alpha \le 1/2, \\ 8\alpha(1-\alpha), & \text{for } 1/2 \le \alpha \le 1. \end{cases}$$ Further, f can be extended to a $4\alpha(1-\alpha)$ -quasiconformal mapping of the Riemann sphere $\widehat{\mathbb{C}}$ when $1/2 < \alpha \leq 1$ . If we choose $A = \alpha$ , $0 < \alpha \le 1$ and B = 0 in Theorem 2.2 then the second inequality in (2.17) in conjunction with Theorem A gives the following result. Corollary 2.3. Let $f \in \mathcal{C}(\alpha,0)$ , $0 < \alpha \leq 1$ be of the form (1.1). Then the Schwarzian norm satisfies the sharp inequality $||S_f|| \leq \alpha$ and equality occurs for the function $f_0$ which is given by (2.23). Further, f can be extended to a $\alpha/2$ -quasiconformal mapping of the Riemann sphere $\widehat{\mathbb{C}}$ . If we choose $A = \alpha$ and $B = -\alpha$ , $0 < \alpha \le 1$ in Theorem 2.2 then the second inequality in (2.17) in conjunction with Theorem A gives the following result. Corollary 2.4. Let $f \in \mathcal{C}(\alpha, -\alpha)$ , $0 < \alpha \le 1$ be of the form (1.1). Then the Schwarzian norm satisfies the sharp inequality $||S_f|| \le 2\alpha$ and equality occurs for the function $f_0$ which is given by (2.23). Further, f can be extended to a $\alpha$ -quasiconformal mapping of the Riemann sphere $\widehat{\mathbb{C}}$ when $0 < \alpha < 1$ . Data availability: Data sharing not applicable to this article as no data sets were generated or analyzed during the current study. Authors contributions: All authors contributed equally to the investigation of the problem and the order of the authors is given alphabetically according to their surname. All authors read and approved the final manuscript. Acknowledgement: The second named author thanks the University Grants Commission for the financial support through UGC Fellowship (Grant No. MAY2018-429303).