Hermitian-Toeplitz determinants for certain univalent functions

Lemma 1. [17, Lemma 3] If $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathcal{P}$ , then $$2p_2 = p_1^2 + \left(4 - p_1^2\right)\zeta,$$ for some $\zeta \in \bar{\mathbb{D}}$ .

Theorem 2. Let $f \in S^*(\varphi)$ and $\varphi(z) = 1 + B_1 z + B_2 z^2 + B_3 z^3 + \cdots$ , then the following holds: 1. $$1 - B_1^2 \le \det T_{2,1}(f) \le 1$$ . 2. If $$3B_1^4 - 8B_1^2 + 2B_1^2B_2 - B_2^2 < 0$$ and $B_1 \le |B_2 + B_1^2|$ , then <span id="page-2-0"></span> $$\det T_{3,1}(f) \le 1. \tag{4}$$ 3. If $3B_1^4 - 8B_1^2 + 2B_1^2B_2 - B_2^2 \ge 0$ and $B_1 \le |B_2 + B_1^2|$ , then <span id="page-2-1"></span> $$\det T_{3,1}(f) \le B_1^2(B_1^2 + B_2) - \frac{1}{4} (B_1^2 + B_2)^2 - 2B_1^2 + 1.$$ (5) All these inequalities are sharp.

Corollary 2.1. 1. If $f \in S_{sin}^*$ , then $0 \le \det T_{2,1}(f) \le 1$ and $\det T_{3,1}(f) \le 1$ . 2. If $$f \in \mathcal{S}_{o}^{*}$$ , then $0 \leq \det T_{2,1}(f) \leq 1$ and $\det T_{3,1}(f) \leq 1$ . 3. If $f \in S_P$ , then $1 - (64/\pi^4) \le \det T_{2,1}(f) \le 1$ and $\det T_{3,1}(f) \le 1$ . Remark. Theorem 2 yields some already known results for different subclasses of $S^*$ , obtained by an appropriate choice of $\varphi$ . - 1. If $f \in \mathcal{S}_{SC}^*$ , then $3/4 \leq \det T_{2,1}(f) \leq 1$ and $\det T_{3,1}(f) \leq 1$ [3, Theorem 2.1]. - 2. If $f \in \mathcal{S}_{R}^{*}$ , then $0 \leq \det T_{2,1}(f) \leq 1$ and $\det T_{3,1}(f) \leq 1$ [3, Theorem 2.2]. - 3. If $f \in \Delta^*$ , then $\det T_{3,1}(f) \le 1$ [15, Theorem 2]. - 4. If $f \in S^*$ , then $\det T_{3,1}(f) \le 8$ [4, Corollary 3], [11, Corollary 2]. - 5. If $f \in S^*(1/2)$ , then $\det T_{3,1}(f) \le 1$ [4, Corollary 4]. - 6. If $f \in \mathcal{S}_{N_e}^*$ , then $\det T_{3,1}(f) \leq 1$ [13, Theorem 4.2]. - 7. If we take $\varphi(z) = (1 + (1 2\alpha)z)/(1 z)$ , $\alpha \in (0, 1]$ , then we obtain the upper bound of $\det T_{3,1}(f)$ for $f \in \mathcal{S}^*(\alpha)$ [4, Theorem 3]. - 8. For $\varphi(z) = ((1+z)/(1-z))^{\alpha}$ , $\alpha \in [1/3,1]$ , we get the bound of $T_{3,1}(f)$ for function f belonging to the class of strongly starlike function $SS^*(\alpha)$ [15, Theorem 1], [11, Theorem 3]. - 9. For $\varphi(z) = (1 + Az)/(1 + Bz)$ , where $-1 \le B < A \le 1$ with $A B \le |A^2 3AB + 2B^2|$ , we get the bound of $T_{3,1}(f)$ for the class $S^*[A, B]$ [16, Theorem 2]. In a similar fashion, the bounds of $\det T_{2,1}(f)$ can also be found for all the above mentioned classes.

Theorem 3. If $f \in \mathcal{C}(\varphi)$ and $\varphi(z) = 1 + B_1 z + B_2 z^2 + B_3 z^3 + \cdots$ , then the following hold: - 1. $1 \frac{B_1^2}{4} \le \det T_{2,1}(f) \le 1$ . - 2. If $B_1 \leq |B_2 + B_1^2|$ , then <span id="page-4-1"></span> $$\det T_{3,1}(f) \le 1. \tag{7}$$ All these estimates are sharp.

Theorem 4. Let $f \in S^*(\varphi)$ and $B_1^2 \geq B_2$ , then the following inequalities hold: 1. If $\mu \notin [0, 4]$ , then <span id="page-5-1"></span> $$\det T_{3,1}(f) \ge \min \left\{ 1 - \frac{B_1^2}{4}, 1 - 2B_1^2 + \frac{3B_1^4}{4} + \frac{B_1^2 B_2}{2} - \frac{B_2^2}{4} \right\}. \tag{8}$$ 2. If $\mu = 4$ , then <span id="page-5-2"></span> $$\det T_{3,1}(f) \ge 1 - 2B_1^2 + \frac{3B_1^4}{4} + \frac{B_1^2 B_2}{2} - \frac{B_2^2}{4}. \tag{9}$$ 3. If $\mu \in (0,4)$ , then <span id="page-5-3"></span> $$\det T_{3,1}(f) \ge 1 - \frac{B_1^2}{4} - \frac{B_1^2(B_1^2 + 3B_1 - B_2)^2}{4(B_1(2B_1^2 - B_1 - 2B_2) + (3B_1^2 - B_2)(B_1^2 + B_2))},\tag{10}$$ where $$\mu = \frac{4B_1(B_1^2 + 3B_1 - B_2)}{(3B_1^2 - B_2)(B_1^2 + B_2) + B_1(2B_1^2 - 2B_2 - B_1)}.$$ The first two inequalities are sharp.

Theorem 5. If $f \in \mathcal{C}(\varphi)$ such that $B_1^2 \geq 2B_2$ , then $$T_{3,1}(f) \geq \begin{cases} \min\left\{1 - \frac{B_1^2}{36}, 1 - \frac{B_1^2}{2} + \frac{B_1^4}{18} + \frac{B_1^2 B_2}{36} - \frac{B_2^2}{36}\right\}, & \sigma \notin [0, 4], \end{cases}$$ $$T_{3,1}(f) \geq \begin{cases} 1 - \frac{B_1^2}{2} + \frac{B_1^4}{18} + \frac{B_1^2 B_2}{36} - \frac{B_2^2}{36}, & \sigma = 4, \end{cases}$$ $$1 - \frac{B_1^3 (B_1^3 + 4B_1^2 + 28B_1 - 8B_2)}{16(2B_1^4 + B_1^3 - B_1^2 - 2B_1B_2 + B_1^2B_2 - B_2^2)}, & \sigma \in (0, 4), \end{cases}$$ where $$\sigma = \frac{2B_1(B_1^2 + 16B_1 - 2B_2)}{2B_1^4 + B_1^3 - B_1^2 - 2B_1B_2 + B_1^2B_2 - B_2^2}.$$ The first two inequalities are sharp.

Theorem 6. Let $f \in \mathcal{K}$ for some function $g(z) = z + b_2 z^2 + b_3 z^3 + \cdots \in \mathcal{S}^*$ , such that <span id="page-10-0"></span> $$6|b_2|^3 - 4|b_2|(|b_3| - 1) - 4(|b_3| - 1)^2 + |b_2|^2(3|b_3| + 5) \ge 0, (13)$$ then 1. $$1 - \left(1 + \frac{|b_2|}{2}\right)^2 \le \det T_{2,1}(f) \le 1$$ . 2. If $$6|b_2|^3 + |b_2|^2(3|b_3| + 13) + 4|b_2|(|b_3| - 1) - 2(1 - |b_3|)^2 - 18 \le 0$$ , then <span id="page-10-2"></span> $$\det T_{3,1}(f) \le 1. \tag{14}$$ 3. If $6|b_2|^3 + |b_2|^2(3|b_3| + 13) + 4|b_2|(|b_3| - 1) - 2(1 - |b_3|)^2 - 18 > 0$ , then <span id="page-10-1"></span> $$\det T_{3,1}(f) \ge \frac{1}{18} \left( 6|b_2|^3 + |b_2|^2 (3|b_3| + 13) + 4|b_2|(|b_3| - 1) - 2(1 - |b_3|)^2 \right). \tag{15}$$ All these bounds are sharp.

Theorem 7. Let $f \in \mathcal{K}$ and $g(z) = z + \sum_{n=2}^{\infty} b_n z^n \in \mathcal{S}$ . If $\tilde{g}(z) = z + \sum_{n=2}^{\infty} i^{n-1} b_n z^n \in \mathcal{S}$ , then the following sharp bound hold: $$|\det T_{2,2}(f)| \le \frac{1}{4} (2 + |b_2|)^2 + \frac{1}{9} (|b_3| + 2|b_2| + 2)^2.$$