Sharp bounds of Hankel determinants of second and third order for inverse functions of certain class of univalent functions

Lemma 1. For each function f in $U(\lambda)$ , $0 < \lambda \le 1$ , there exists function $\omega_1$ , analytic in $\mathbb{D}$ , such that $|\omega_1(z)| \le |z| < 1$ , and $|\omega_1'(z)| \le 1$ , for all $z \in \mathbb{D}$ , with (4) $$\frac{z}{f(z)} = 1 - a_2 z - \lambda z \omega_1(z).$$ Additionally, for (5) $$\omega_1(z) = c_1 z + c_2 z^2 + \cdots,$$ we have (6) $$|c_1| \le 1$$ , $|c_2| \le \frac{1}{2}(1 - |c_1|^2)$ and $|c_3| \le \frac{1}{3} \left[ 1 - |c_1|^2 - \frac{4|c_2|^2}{1 + |c_1|} \right]$ . Using (4) and (5) we have $$z = [1 - a_2 z - \lambda z \omega_1(z)] f(z),$$ and after equating the coefficients, (7) $$a_{3} = \lambda c_{1} + a_{2}^{2},$$ $$a_{4} = \lambda c_{2} + 2\lambda a_{2}c_{1} + a_{2}^{3},$$ $$a_{5} = \lambda c_{3} + 2\lambda a_{2}c_{2} + \lambda^{2}c_{1}^{2} + 3\lambda a_{2}^{2}c_{1} + a_{2}^{4},$$ that we will use later on. From (3) and (7), after some calculations, we derive (8) $$A_{2} = -a_{2},$$ $$A_{3} = -\lambda c_{1} + a_{2}^{2},$$ $$A_{4} = -\lambda c_{2} + 3\lambda a_{2}c_{1} - a_{2}^{3},$$ $$A_{5} = -\lambda c_{3} + 4\lambda a_{2}c_{2} - 6\lambda a_{2}^{2}c_{1} + 2\lambda^{2}c_{1}^{2} + a_{2}^{4}.$$ We also need the next results from [16].

Lemma 2. Let $f \in \mathcal{U}(\lambda)$ for $0 < \lambda \le 1$ , and be given by $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ . Then $$(9) |a_2| \le 1 + \lambda.$$ If $|a_2| = 1 + \lambda$ , then f must be of the form (10) $$f(z) = \frac{z}{1 - (1 + \lambda)e^{i\phi}z + \lambda e^{2i\phi}z^2}$$ for some $\phi \in [0, 2\pi]$ . In the same paper ([16]) it was conjectured that for functions in $\mathcal{U}(\lambda)$ , $|a_n| \leq \sum_{i=0}^{n-1} \lambda^i$ holds sharply, and was claimed to be proven in the case n=3: $$(11) |a_3| < 1 + \lambda + \lambda^2.$$ The proof rely on another claim, that for all functions f from $\mathcal{U}(\lambda)$ , (12) $$\frac{f(z)}{z} \prec \frac{1}{(1+z)(1+\lambda z)}.$$ Recently, in [8], the second claim, and consequently the first one also, was proven to be wrong by giving a counterexample. Still, the subset of $\mathcal{U}(\lambda)$ when the inequality (11) and subordination (12) hold is nonempty, as the function $$f_{\lambda}(z) = \frac{z}{(1-z)(1-\lambda z)} = \sum_{n=1}^{\infty} \frac{1-\lambda^n}{1-\lambda} z^n = z + (1+\lambda)z^2 + (1+\lambda+\lambda^2)z^2 + \cdots$$ shows. Here $\frac{1-\lambda^n}{1-\lambda}\Big|_{\lambda=1}=n$ for all $n=1,2,3,\ldots$ Now we will give the sharp upper bound of the modulus of the second and the third Hankel determinant for the inverse functions of the functions from the class $\mathcal{U}(\lambda)$ .

Theorem 1. Let $f \in \mathcal{U}(\lambda)$ , $0 < \lambda \le 1$ , and let its inverse is $f^{-1}$ . Then (i) $|H_2(2)(f^{-1})| \le \lambda(1+\lambda+\lambda^2)$ if the third coefficient of f satisfies inequality (11); $$(ii) |H_3(1)(f^{-1})| \le \begin{cases} \frac{\lambda^2}{4}, & 0 < \lambda \le \frac{1}{4}, \\ \lambda^3, & \frac{1}{4} \le \lambda \le 1. \end{cases}$$ Both results are sharp.