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Results & Lemmas (4)

Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.

Lemma 2.1 Lemma 2.1. If is of the form (2.1), then and for some For, there is a unique function with as in (2.2), namely For and, there is a unique…
Lemma 2.1. If $p \in \mathcal{P}$ is of the form (2.1), then $$(2.2) c_1 = 2p_1$$ $$(2.3) c_2 = 2p_1^2 + 2(1 - p_1^2)p_2$$ and $$(2.4) c_3 = 2p_1^3 + 4\left(1 - p_1^2\right)p_1p_2 - 2\left(1 - p_1^2\right)p_1p_2^2 + 2\left(1 - p_1^2\right)\left(1 - |p_2|^2\right)p_3$$ for some $p_1, p_2, p_3 \in \overline{\mathbb{D}} := \{z \in \mathbb{C} : |z| \le 1\}.$ For $p_1 \in \mathbb{T} := \{z \in \mathbb{C} : |z| = 1\}$ , there is a unique function $p \in \mathcal{P}$ with $c_1$ as in (2.2), namely $$p(z) = \frac{1 + p_1 z}{1 - p_1 z}, \quad z \in \mathbb{D}.$$ For $p_1 \in \mathbb{D}$ and $p_2 \in \mathbb{T}$ , there is a unique function $p \in \mathcal{P}$ with $c_1$ and $c_2$ as in (2.2) and (2.3), namely (2.5) $$p(z) = \frac{1 + (p_1 + \overline{p_1}p_2)z + p_2z^2}{1 - (p_1 - \overline{p_1}p_2)z - p_2z^2}.$$ For $p_1, p_2 \in \mathbb{D}$ and $p_3 \in \mathbb{T}$ , there is unique function $p \in \mathcal{P}$ with $c_1, c_2$ , and $c_3$ as in (2.2)-(2.3), namely $$p(z) = \frac{1 + (\overline{p_2}p_3 + \overline{p_1}p_2 + p_1)z + (\overline{p_1}p_3 + p_1\overline{p_2}p_3 + p_2)z^2 + p_3z^3}{1 + (\overline{p_2}p_3 + \overline{p_1}p_2 - p_1)z + (\overline{p_1}p_3 - p_1\overline{p_2}p_3 - p_2)z^2 - p_3z^3}, \quad z \in \mathbb{D}.$$ Next we recall the following well-known result due to Choi et al. [9]. Lemma 2.2 plays an important role in the proof of our main results.
Lemma 2.2 Lemma 2.2. Let A, B, C be real numbers and (i) If, then (ii) If AC < 0, then where
Lemma 2.2. Let A, B, C be real numbers and $$Y(A, B, C) := \max_{z \in \overline{\mathbb{D}}} (|A + Bz + Cz^2| + 1 - |z|^2).$$ (i) If $AC \geq 0$ , then $$Y(A, B, C) = \begin{cases} |A| + |B| + |C|, & |B| \ge 2(1 - |C|), \\ 1 + |A| + \frac{B^2}{4(1 - |C|)}, & |B| < 2(1 - |C|). \end{cases}$$ (ii) If AC < 0, then $$Y(A,B,C) = \begin{cases} 1 - |A| + \frac{B^2}{4(1-|C|)}, & -4AC\left(C^{-2} - 1\right) \le B^2 \land |B| < 2(1-|C|), \\ 1 + |A| + \frac{B^2}{4(1+|C|)}, & B^2 < \min\left\{4(1+|C|)^2, -4AC\left(C^{-2} - 1\right)\right\}, \\ R(A,B,C), & otherwise, \end{cases}$$ where $$R(A, B, C) = \begin{cases} |A| + |B| + |C|, & |C|(|B| + 4|A|) \le |AB|, \\ -|A| + |B| + |C|, & |AB| \le |C|(|B| - 4|A|), \\ (|A| + |C|)\sqrt{1 - \frac{B^2}{4AC}}, & otherwise. \end{cases}$$
Theorem 3.1 Theorem 3.1. Let given by (1.1) then The inequality is sharp.
Theorem 3.1. Let $f \in \mathcal{C}$ given by (1.1) then $$(3.1) |H_{2,1}(F_{f^{-1}}/2)| \le \frac{1}{33}.$$ The inequality is sharp.
Theorem 3.2 Theorem 3.2. Let given by (1.1) then The inequality is sharp.
Theorem 3.2. Let $f \in \mathcal{S}$ given by (1.1) then $$(3.10) |H_{2,1}(F_{f^{-1}}/2)| \le \frac{13}{12}.$$ The inequality is sharp.
Function classes studied:

Coefficient bounds & claims (4)

Machine-extracted from the paper text - useful for cross-referencing, not a verified fact.
coefficient_bound
H_{2,1}(F_{f^{-1}}/2) ≤ 1/33 for class C (sharp) [Theorem 3.1]
coefficient_bound
H_{2,1}(F_{f^{-1}}/2) ≤ 13/12 for class S* (sharp) [Theorem 3.2]
function_family
Class S*: f in A with Re(zf'/f) > 0
function_family
Class C: f in A with Re(1 + zf''/f') > 0

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