On $H_3(1)$ Hankel determinant for some classes of univalent functions

Lemma 2.1. ([2]) Let $p \in P$ , then $|c_k| \le 2$ , $k = 1, 2, \dots$ , and the inequality is sharp.

Lemma 2.2. ([7, 8]) Let $p \in P$ , then $$2c_2 = c_1^2 + x(4 - c_1^2) (2.1)$$ and $$4c_3 = c_1^3 + 2xc_1(4 - c_1^2) - x^2c_1(4 - c_1^2) + 2z(1 - |x|^2)(4 - c_1^2)$$ (2.2) for some x, z such that $|x| \le 1$ and $|z| \le 1$ .

Theorem 3.1. Let $f \in R$ . Then $$|a_2 a_3 - a_4| \le \frac{1}{2}.$$ The inequality is sharp. Equality is attained by $$f(z) = \int_0^z \frac{1+t^3}{1-t^3} dt.$$

Corollary 3.2. Let $f \in R$ . Then $$|H_3(1)| \le \frac{993}{1620}.$$

Theorem 3.3. Let $f \in S^*$ . Then $$|a_2a_3 - a_4| \le 2.$$ The inequality is sharp. Equality is attained by the Koebe function $k(z) = z/(1-z)^2$ . Proof. Let $f \in S^*$ . Then there exists a $p \in P$ such that zf'(z) = f(z)p(z). Equating coefficients we find that $a_2 = c_1$ , $2a_3 = c_2 + c_1^2$ and $6a_4 = 2c_3 + 3c_1c_2 + c_1^3$ . Thus we have $$|a_2a_3 - a_4| = \frac{1}{3}|c_1^3 - c_3|. (3.3)$$ Substituting for $c_3$ from Lemma 2, we obtain $$|a_2a_3 - a_4| = \frac{1}{12}|3c_1^3 - 2c_1(4 - c_1^2)x + c_1(4 - c_1^2)x^2 - 2(4 - c_1^2)(1 - |x|^2)z|.$$ (3.4) Since $|c_1| \leq 2$ by Lemma 1, let $c_1 = c$ and assume without restriction that $c \in [0, 2]$ . Applying the triangle inequality on (3.4), with $\rho = |x|$ , we obtain $$|a_2a_3 - a_4| \le \frac{1}{12}[3c^3 + 2(4 - c^2) + 2c(4 - c^2)\rho + (c - 2)(4 - c^2)\rho^2]$$ = $F(\rho)$ . Differentiating $F(\rho)$ , we have $$F'(\rho) = \frac{1}{12} [2c(4-c^2) + 2(c-2)(4-c^2)] > 0.$$ This implies that $F(\rho)$ is an increasing function of $\rho$ on [0,1] if $c \in [1,2]$ . In this case $F(\rho) \leq F(1) = c \leq 2$ for all $\rho \in [0,1]$ . It follows therefore that $F(\rho) \leq 2$ . On the other hand suppose $c \in [0,1)$ , then $F(\rho)$ is decreasing on [0,1] so that $F(\rho) \leq F(0)$ . That is $$F(\rho) \le \frac{3c^3 - 2c^2 + 8}{12}$$ = $G(c)$ . Hence we have $G(c) \leq G(0) = 2/3$ , $c \in [0,1)$ . This is less than 2, which is the case when $c \in [1,2]$ . Thus the maximum of the functional $|a_2a_3 - a_4|$ corresponds to $\rho = 1$ and c = 2. If $c_1 = c = 2$ in (2.1) and (2.2), then we have $c_2 = c_3 = 2$ . Using these in (3.3) we see that equality is attained which shows that our result is sharp. Furthermore, it is easily seen that the extremal function in this case is the well known Koebe function $k(z) = z/(1-z)^2$ . For $f \in S^*$ , using the known inequalities $|a_k| \le k$ , $k = 2, 3, \cdots$ [2], $|a_2 a_4 - a_3^2| \le 1$ [5] and $|a_3 - a_2^2| \le 1$ [6] together with Theorem 2 we have the next corollary.

Corollary 3.4. Let $f \in S^*$ . Then $$|H_3(1)| \le 16.$$ The inequality is sharp. Equality is attained by a rotation, $k_1(z) = z/(1+z)^2$ , of the Koebe function.

Theorem 3.5. Let $f \in C$ . Then $$|a_2 a_3 - a_4| \le \frac{1}{6}.$$ The inequality is sharp. Equality is attained by $$f(z) = \int_0^z \left\{ s. \exp\left(\int_0^s \frac{2t^3}{1 - t^3} dt\right) \right\} ds.$$

Corollary 3.6. Let $f \in C$ . Then $$|H_3(1)| \le \frac{15}{24}.$$ Acknowledgements. This work was carried out at the Centre for Advanced Studies in Mathematics, CASM, Lahore University of Management Sciences, Lahore, Pakistan during the author's postdoctoral fellowship at the Centre. The author is indebted to all staff of CASM for their hospitality, most especially Prof. Ismat Beg.