A Conjecture on $H_3(1)$ For Certain Starlike Functions

Lemma 1.1. Let $p(z) = 1 + \sum_{n=1}^{\infty} a_n z^n \in \mathcal{P}$ . Then, <span id="page-1-3"></span> $$2p_2 = p_1^2 + \gamma(4 - p_1^2), \tag{1.3}$$ $$4p_3 = p_1^3 + 2p_1(4 - p_1^2)\gamma - p_1(4 - p_1^2)\gamma^2 + 2(4 - p_1^2)(1 - |\gamma|^2)\eta$$ (1.4) and <span id="page-1-4"></span> $$8p_4 = p_1^4 + (4 - p_1^2)\gamma(p_1^2(\gamma^2 - 3\gamma + 3) + 4\gamma) - 4(4 - p_1^2)(1 - |\gamma|^2)(p_1(\gamma - 1)\eta + \bar{\gamma}\eta^2 - (1 - |\eta|^2)\rho), \quad (1.5)$$ for some $\gamma$ , $\eta$ and $\rho$ such that $|\gamma| \le 1$ , $|\eta| \le 1$ and $|\rho| \le 1$ . In 2021, Kumar and Gangania [10], introduced a new class $\mathcal{S}_{\wp}^*$ by choosing $\varphi(z) = 1 + ze^z$ in (1.2) defined as $$\mathcal{S}_{\wp}^* = \left\{ f \in \mathcal{A} : \frac{zf'(z)}{f(z)} \prec 1 + ze^z \right\}.$$ They used the similar strategy and obtained the bound as, $|H_3(1)| \leq 0.150627$ for $\mathcal{S}_{\wp}^*$ and also proposed a conjecture which is stated as follows: <span id="page-1-0"></span>Conjecture 1.2. [10, Page no. 33] If $f \in \mathcal{S}_{\wp}^*$ , then the sharp bound for the third Hankel determinant is given by $$|H_3(1)| \le \frac{1}{9} \approx 0.1111...,$$ with the extremal function $f(z) = z \exp\left(\frac{1}{3}(e^{z^3} - 1)\right) = z + \frac{1}{3}z^4 + \frac{2}{9}z^7 + \cdots$ In this article, together with the proof of conjecture, the estimates for $a_6$ and $a_7$ in association with the fourth order Hankel determinant for the functions in the class $\mathcal{S}_{\wp}$ are obtained. For $\mathcal{S}^(\varphi)$ , the general third Hankel determinant for second and third fold symmetric functions are also estimated.

Theorem 2.1. Let $f \in \mathcal{S}_{\wp}^*$ . Then, <span id="page-1-5"></span> $$|H_3(1)| \le \frac{1}{9}. (2.5)$$ The result is sharp.

Lemma 2.2. Let $p = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathcal{P}$ . Then <span id="page-5-1"></span> $$|p_n| \le 2, \quad n \ge 1,\tag{2.15}$$ <span id="page-5-2"></span> $$|p_{n+k} - \mu p_n p_k| \le \begin{cases} 2, & 0 \le \mu \le 1; \\ 2|2\mu - 1|, & elsewhere, \end{cases}$$ (2.16) and $$|p_1^3 - \mu p_3| \le \begin{cases} 2|\mu - 4|, & \mu \le 4/3; \\ 2\mu\sqrt{\frac{\mu}{\mu - 1}}, & 4/3 < \mu. \end{cases}$$ (2.17)

Lemma 2.3. Let $f \in \mathcal{S}_{\wp}^*$ . Then $|a_6| \le 47/60 \approx 0.7833$ and $|a_7| \le 503/480 \approx 1.0479$ .

Theorem 3.1. Let $f \in \mathcal{S}_{\wp}^*$ . Then $|H_4(1)| \leq 2.54589$ . 3.2. Third Hankel determinant for 2&3 fold symmetric functions for $S^*(\varphi)$ . In the recent times, it has been observed that finding the sharp estimates of third Hankel determinant for general Ma-Minda class is not feasible till now. But for some classes, sharp estimates have been obtained, for instance, see [1, 8, 20] and now including Theorem 2.1 as well which motivated us to settle the Conjecture 1.2. Further looking at the difficulty of the general class, we restrict ourselves to answer the problem for the n-fold symmetric functions.

Theorem 3.3. Let $f \in \mathcal{S}^*(\varphi)$ . Then - (1) $\widehat{f} \in \mathcal{S}^{(3)}(\varphi)$ implies that $|H_3(1)| \leq |B_1|^2/9$ . (2) $\widehat{f} \in \mathcal{S}^{(2)}(\varphi)$ implies that $$|H_3(1)| \le \frac{1}{4}|B_1| \times \begin{cases} \frac{1}{6}(B_2 - \frac{9}{8}B_1^2 + B_1^2), & \frac{9}{4}B_1^2 \le 2(B_2 + B_1^2 - B_1); \\ \frac{1}{6}B_1, & 2(B_2 + B_1^2 - B_1) \le \frac{9}{4}B_1^2 \le 2(B_2 + B_1^2 + B_1); \\ \frac{1}{6}(-B_2 + \frac{9}{8}B_1^2 - B_1^2), & 2(B_2 + B_1^2 + B_1) \le \frac{9}{4}B_1^2. \end{cases}$$ The estimate in (1) is sharp.

Definition 3.2. [3] A function $f \in \mathcal{A}$ is called n-fold symmetric if $f(e^{2\pi i/n}z) = e^{2\pi i/n}f(z)$ which holds for all $z \in \mathbb{D}$ and n is a natural number. We denote the set of n-fold symmetric functions by $\mathcal{A}^{(n)}$ . Let $f \in \mathcal{A}^{(n)}$ , then f has power series expansion $$f(z) = a_1 z + a_{n+1} z^{n+1} + a_{2n+1} z^{2n+1} + \cdots$$ Therefore, for $f \in \mathcal{A}^{(3)}$ and $f \in \mathcal{A}^{(2)}$ respectively, we have <span id="page-7-0"></span> $$H_3(1) = -a_4^2$$ and $H_3(1) = a_3(a_5 - a_3^2)$ . (3.8) Now we conclude this paper with the following result: