An estimation of the pre-Schwarzian norm for certain classes of analytic functions

Theorem 3.1. Let f ∈ S^∗ hyp. Then the pre-Schwarzian norm satisfies the following sharp inequality $$||P_f|| \le \begin{cases} \frac{st_s(1+t_s) + (1+t_s)(1-t_s)^{1-s} - (1-t_s^2)}{t_s} & \text{for } s \in (0,1) \\ 4 & \text{for } s = 1, \end{cases}$$ where t^s ∈ (0, 1) is the unique root of the equation $$\frac{(1-t)^{-s}\left(st^2(1-t)^s + t^2(1-t)^s + st^2 + (1-t)^s + st - t^2 - 1\right)}{t^2} = 0.$$ Proof. Let f ∈ S^∗ hyp. By the definition of the class S ∗ hyp, we have $$\frac{zf'(z)}{f(z)} \prec \frac{1}{(1-z)^s}.$$ Thus, there exist a Schwarz function ω ∈ B⁰ such that $$\frac{zf'(z)}{f(z)} = \frac{1}{(1 - \omega(z))^s}.$$ Taking logarithmic derivative on both sides with respect to z, we obtain $$P_f(z) = \frac{f''(z)}{f'(z)} = \frac{s\omega'(z)}{1 - \omega(z)} + \frac{1}{z} \left( \frac{1}{(1 - \omega(z))^s} - 1 \right).$$ Since |ω(z)| ≤ |z| < 1 and the branch of the logarithm is determined by log(1) = 0, thus, we have $$\frac{1}{(1 - \omega(z))^s} = \exp(-s\log(1 - \omega(z))) = 1 + \sum_{n=1}^{\infty} \frac{s(s+1)\cdots(s+n-1)}{n!} \omega^n(z) \quad (z \in \mathbb{D}).$$ In view of the Schwarz-Pick lemma, we have $$(1-|z|^{2})|P_{f}(z)| = (1-|z|^{2}) \left| \frac{s\omega'(z)}{1-\omega(z)} + \frac{1}{z} \left( \frac{1}{(1-\omega(z))^{s}} - 1 \right) \right|$$ $$\leq (1-|z|^{2}) \left( \frac{s|\omega'(z)|}{1-|\omega(z)|} + \frac{1}{|z|} \left( \frac{1}{(1-|\omega(z)|)^{s}} - 1 \right) \right)$$ $$\leq \frac{s\left(1-|\omega(z)|^{2}\right)}{1-|\omega(z)|} + \frac{(1-|z|^{2})}{|z|} \left( \frac{1}{(1-|\omega(z)|)^{s}} - 1 \right).$$ For $0 \le t := |\omega(z)| \le |z| < 1$ , we have $$(1-|z|^2)|P_f(z)| \le s(1+t) + \frac{(1-|z|^2)}{|z|} \left(\frac{1}{(1-t)^s} - 1\right).$$ Therefore, we have <span id="page-5-0"></span> $$||P_f|| = \sup_{z \in \mathbb{D}} (1 - |z|^2) |P_f(z)| \le \sup_{0 \le t \le |z| < 1} F_1(|z|, t), \tag{3.1}$$ where $$F_1(r,t) = s(1+t) + \frac{(1-r^2)}{r} \left(\frac{1}{(1-t)^s} - 1\right)$$ for $r = |z|$ . Now the objective is to determine the supremum of $F_1(r,t)$ on $\Omega = \{(r,t) : 0 < t \le r < 1\}$ . Differentiating partially $F_1(r,t)$ with respect to r, we obtain $$\frac{\partial}{\partial r}F_1(r,t) = -\left(\frac{1}{r^2} + 1\right)\left(\frac{1}{(1-t)^s} - 1\right) < 0.$$ Therefore, $F_1(r,t)$ is a monotonically decreasing function of $r \in [t,1)$ and it follows that $F_1(r,t) \leq F_1(t,t) = F_2(t)$ , where $$F_2(t) = s(1+t) + \frac{(1-t^2)}{t} \left(\frac{1}{(1-t)^s} - 1\right).$$ It is evident that $F_2(t) = 2(1+t)$ for s = 1. Hence, we have $||P_f|| \le 4$ . We consider the case, where 0 < s < 1. Differentiating $F_2(t)$ with respect to t, we obtain $$F_2'(t) = \frac{(1-t)^{-s} \left( st^2 (1-t)^s + t^2 (1-t)^s + st^2 + (1-t)^s + st - t^2 - 1 \right)}{t^2}$$ and $$F_2''(t) = \frac{s^2 t^3 + s^2 t^2 - st^3 + st^2 + 2t(1-t)^s - 2(1-t)^s - 2st - 2t + 2}{(1-t)^{s+1} t^3}.$$ Let $$F_3(t) = \frac{s^2t^3 + s^2t^2 - st^3 + st^2 + 2t(1-t)^s - 2(1-t)^s - 2st - 2t + 2}{(1-t)^{s+1}}.$$ It is evident that $F_2''(t) = F_3(t)/t^3$ and $$F_3'(t) = \frac{-s(1-s)t^2(st+s-2t+4)}{(1-t)^{s+2}} < 0 \quad \text{for} \quad 0 < t < 1, \ 0 < s < 1.$$ Therefore, $F_3(t)$ is a monotonically decreasing function of $t \in (0,1)$ and it follows that $F_3(t) \leq \lim_{t \to 0^+} F_3(t) = 0$ , i.e., $F_2''(t) \leq 0$ for 0 < t < 1. Thus, $F_2'(t)$ is a monotonically decreasing function in t with $\lim_{t \to 0^+} F_2'(t) = s(s+3)/2$ and $\lim_{t \to 1^-} F_2'(t) = -\infty$ . Therefore, the equation $F'_2(t) = 0$ has the unique root $t_s$ in (0,1), as illustrated in Figure 3. Thus, $F_2(t)$ attains its maximum value at $t = t_s$ . From (3.1), we have $$||P_f|| \le F_2(t_s) = \frac{st_s(1+t_s) + (1+t_s)(1-t_s)^{1-s} - (1-t_s^2)}{t_s},$$ where $t_s \in (0,1)$ is the unique positive root of the equation <span id="page-6-0"></span> $$F_s(t) := \frac{(1-t)^{-s} \left( st^2 (1-t)^s + t^2 (1-t)^s + st^2 + (1-t)^s + st - t^2 - 1 \right)}{t^2} = 0. \quad (3.2)$$ To show that the estimate is sharp, we consider the function $f_1$ given by $$f_1(z) = z \exp\left(\int_0^z \frac{(1-t)^{-s} - 1}{t} dt\right).$$ The pre-Schwarzian norm of $f_1$ is given by $$||P_{f_1}|| = \sup_{z \in \mathbb{D}} (1 - |z|^2) |P_{f_1}(z)| = \sup_{z \in \mathbb{D}} (1 - |z|^2) \left| \frac{zs(1-z)^{-1} + (1-z)^{-s} - 1}{z} \right|.$$ On the positive real axis, we note that $$\sup_{0 \le r < 1} (1 - r^2) \frac{rs(1 - r)^{-1} + (1 - r)^{-s} - 1}{r}$$ $$= \begin{cases} \frac{sr_s(1 + r_s) + (1 + r_s)(1 - r_s)^{1 - s} - (1 - r_s^2)}{r_s} & \text{for } s \in (0, 1) \\ 4 & \text{for } s = 1, \end{cases}$$ where $r_s \in (0,1)$ is the unique root of the equation (3.2). Therefore, $$||P_{f_1}|| = \begin{cases} sr_s(1+r_s) + (1+r_s)(1-r_s)^{1-s} - (1-r_s^2) & \text{for } s \in (0,1) \\ \frac{r_s}{4} & \text{for } s = 1. \end{cases}$$ This completes the proof. In Table 1 and Figure 3, we obtain the values of $t_s$ and $||P_f||$ for certain values of $s \in (0,1)$ . We observe that, whenever $s \to 1^-$ , then $t_s \to 1^-$ and $||P_f|| \to 4^-$ . | s | 1/2 | 1/3 | 2/3 | 3/4 | 4/5 | 9/10 | 99/100 | |-----------|----------|----------|----------|----------|----------|----------|----------| | $t_s$ | 0.765186 | 0.721166 | 0.819069 | 0.851565 | 0.873603 | 0.926039 | 0.990582 | | $ P_f $ | 1.45876 | 0.926878 | 2.06701 | 2.41553 | 2.64591 | 3.18262 | 3.86967 | <span id="page-6-1"></span>TABLE 1. $t_s$ is the unique positive root of the equation (3.2) in (0,1)  <span id="page-7-0"></span>Figure 3. Graph of Fs(t) for different values of s in (0, 1) In the following result, we obtain the estimate of the pre-Schwarzian norm for functions in the class S ∗ L .

Theorem 3.2. Let f ∈ S^∗ L . Then the pre-Schwarzian norm satisfies the following inequality $$||P_f|| \le \frac{2s(1-t_s^2)}{1-st_s} + \frac{(1-t_s^2)\left((1+st_s)^2-1\right)}{t_s},$$ where t^s ∈ (0, 1) is the unique positive root of the equation $$\frac{-3s^4t^4 + 2s^3t^3 + (s^4 + 7s^2)t^2 - (2s^3 + 8s)t + 3s^2}{(1 - st)^2} = 0.$$ Proof. Let f ∈ S^∗ L . By the definition of the class S ∗ L , we have $$\frac{zf'(z)}{f(z)} \prec (1+sz)^2.$$ Thus, there exists a Schwarz function ω(z) ∈ B⁰ such that $$\frac{zf'(z)}{f(z)} = (1 + s\omega(z))^2.$$ Taking logarithmic derivative on both sides with respect to z, we obtain $$P_f(z) = \frac{f''(z)}{f'(z)} = \frac{2s\omega'(z)}{1 + s\omega(z)} + \frac{1}{z}\left((1 + s\omega(z))^2 - 1\right),$$ In view of the Schwarz-Pick lemma, we have $$(1 - |z|^{2})|P_{f}(z)| \leq (1 - |z|^{2}) \left( \frac{2s|\omega'(z)|}{|1 + s\omega(z)|} + \frac{|(1 + s\omega(z))^{2} - 1|}{|z|} \right)$$ $$\leq (1 - |z|^{2}) \left( \frac{2s|\omega'(z)|}{1 - s|\omega(z)|} + \frac{(1 + s|\omega(z)|)^{2} - 1}{|z|} \right)$$ $$\leq \frac{2s \left( 1 - |\omega(z)|^{2} \right)}{1 - s|\omega(z)|} + \frac{(1 - |z|^{2}) \left( (1 + s|\omega(z)|)^{2} - 1 \right)}{|z|}.$$ For 0 ≤ t := |ω(z)| ≤ |z| < 1, we obtain $$(1-|z|^2)|P_f(z)| \le \frac{2s(1-t^2)}{1-st} + \frac{(1-|z|^2)\left((1+st)^2-1\right)}{|z|}.$$ Therefore, we have <span id="page-8-0"></span> $$||P_f|| = \sup_{z \in \mathbb{D}} (1 - |z|^2) |P_f(z)| \le \sup_{0 \le t \le |z| \le 1} F_4(|z|, t), \tag{3.3}$$ where $$F_4(r,t) = \frac{2s(1-t^2)}{1-st} + \frac{(1-r^2)((1+st)^2-1)}{r}$$ for $|z|=r$ . Now our objective is to determine the supremum of $F_4(r,t)$ on $\Omega = \{(r,t) : 0 < t \le r < 1\}$ . Differentiating partially $F_4(r,t)$ with respect to r, we obtain $$\frac{\partial}{\partial r}F_4(r,t) = -\left(\frac{1}{r^2} + 1\right)\left((1+st)^2 - 1\right) < 0.$$ Therefore, $F_4(r,t)$ is a monotonically decreasing function of $r \in [t,1)$ and it follows that $F_4(r,t) \leq F_4(t,t) = F_5(t)$ , where <span id="page-8-1"></span> $$F_5(t) = \frac{2s(1-t^2)}{1-st} + \frac{(1-t^2)\left((1+st)^2 - 1\right)}{t}.$$ (3.4) Differentiate $F_5(t)$ twice with respect to t, we obtain $$F_5'(t) = \frac{-3s^4t^4 + 2s^3t^3 + \left(s^4 + 7s^2\right)t^2 - \left(2s^3 + 8s\right)t + 3s^2}{(1 - st)^2},$$ $$F_5''(t) = \frac{2\left(3s^5t^4 - 7s^4t^3 + 3s^3t^2 + 3s^2t + 2s^3 - 4s\right)}{(1 - st)^3}.$$ Let $$F_6(t) = 3s^5t^4 - 7s^4t^3 + 3s^3t^2 + 3s^2t + 2s^3 - 4s.$$ Differentiating $F_6(t)$ with respect to t, we obtain $$F_6'(t) = 3s^2 (4s^3t^3 - 7s^2t^2 + 2st + 1) = 3s^2(1 - st)^2(1 + 4st) > 0.$$ Therefore, $F_6(t)$ is a monotonically increasing function of $t \in [0,1)$ and it follows that $$F_6(t) \le F_6(1) = 3s^5 - 7s^4 + 3s^3 + 3s^2 + 2s^3 - 4s = -s(1-s)(3s^3 - 4s^2 + s + 4) < 0.$$ Therefore, $F_5''(t) < 0$ and hence, we have $F_5'(t)$ is a monotonically decreasing function of t with $F_5'(0) = 3s^2$ and $\lim_{t\to 1^-} F_5'(t) = (-2s^4 + 10s^2 - 8s)/(1-s)^2 = 2s\left(s^2+s-4\right)/(1-s) < 0$ . This leads us to conclude that the equation $F_5'(t) = 0$ has the unique root $t_s$ in (0,1). This shows that $F_5(t)$ attains its maximum at $t_s$ . From (3.3) and (3.4), we have $$||P_f|| \le F_5(t_s) = \frac{2s(1-t_s^2)}{1-st_s} + \frac{(1-t_s^2)\left((1+st_s)^2-1\right)}{t_s},$$ where $t_s \in (0,1)$ is the unique positive root of the equation <span id="page-8-2"></span> $$G_s(t) := \frac{-3s^4t^4 + 2s^3t^3 + (s^4 + 7s^2)t^2 - (2s^3 + 8s)t + 3s^2}{(1 - st)^2} = 0.$$ (3.5) This completes the proof. In Table 2 and Figure 4, we obtain the values of $t_s$ and $||P_f||$ for certain values of $s \in (0, 1/\sqrt{2}]$ . <span id="page-9-0"></span> | s | 1/2 | 11/20 | 2/3 | 3/5 | $1/\sqrt{2}$ | |----------------------------|---------|----------|----------|----------|--------------| | $t_s$ | 0.19266 | 0.213611 | 0.265836 | 0.235311 | 0.285555 | | $\overline{P_f \parallel}$ | 2.07478 | 2.30104 | 2.85492 | 2.53348 | 3.05755 | Table 2. $t_s$ is the unique positive root of the equation (3.5) in (0,1)  <span id="page-9-1"></span>FIGURE 4. Graph of $G_s(t)$ for different values of s in $(0, 1/\sqrt{2})$ In the following result, we establish the sharp estimate of the pre-Schwarzian norm for the functions in the class $C_{hyp}$ .

Theorem 3.3. For any $g \in C_{hyp}$ , the pre-Schwarzian norm satisfies the following sharp inequality $$||P_g|| \le \begin{cases} \frac{(1+r_s)(1-r_s)^{1-s} - (1-r_s^2)}{r_s} & \text{for } 0 < s < 1\\ 2 & \text{for } s = 1, \end{cases}$$ where $r_s \in (0,1)$ is the unique root of the equation $$\frac{(1-r)^{-s}\left(r^2(1-r)^s+r^2s-r^2+(1-r)^s+rs-1\right)}{r^2}=0.$$

Theorem 3.4. For any $g \in C_L$ , the pre-Schwarzian norm satisfies the following sharp inequality $$||P_g|| \le \frac{2\left(\sqrt{3s^2+4}+4\right)\left(3s^2+2\sqrt{3s^2+4}-4\right)}{27s}.$$