Abstract
Let ${\mathcal A}$ be the class of functions that are analytic in the unit disc ${\mathbb D}$, normalized such that $f(z)=z+\sum_{n=2}^\infty a_nz^n$, and let class ${\mathcal U}(λ)$, $0<λ\le1$, consists of functions $f\in{\mathcal A}$, such that \[ \left |\left (\frac{z}{f(z)} \right )^{2}f'(z)-1\right | < λ\quad (z\in {\mathbb D}). \] In this paper we determine the sharp upper bounds for the Hankel determinants of second and third order for the inverse functions of functions from the class ${\
Results & Lemmas (3)
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Lemma 1 · coeff
Lemma 1. For each function f in,, there exists function, analytic in, such that, and, for all, with (4) <span…
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)
$$\frac{z}{f(z)} = 1 - a_2 z - \lambda z \omega_1(z).$$
<span id="page-1-2"></span>Additionally, for
(5)
$$\omega_1(z) = c_1 z + c_2 z^2 + \cdots,$$
<span id="page-2-4"></span>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
<span id="page-2-3"></span>
$$z = [1 - a_2 z - \lambda z \omega_1(z)] f(z),$$
and after equating the coefficients,
<span id="page-2-0"></span>(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 · coeff
Lemma 2. Let for, and be given by. Then If, then f must be of the form (10) for some. <span id="page-2-1"></span>In the same paper ([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 $\phi \in [0, 2\pi]$ .
<span id="page-2-1"></span>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:
$$(11) |a_3| < 1 + \lambda + \lambda^2.$$
<span id="page-2-2"></span>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 · coeff
Theorem 1. Let,, and let its inverse is. Then (i) if the third coefficient of f satisfies inequality (11); Both results are sharp.
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.
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