Some basic properties of certain subclasses of meromorphically starlike functions

Lemma 2.1. (See [16]) If the function $p \in \mathcal{P}$ is given by (1.2), then $$|p_k| \leq 2 \quad (k \in \mathbb{N}).$$

Lemma . Let β > and – γ – β > . Suppose also that the sequence {Ak}^∞ ^k= is defined by $$A_{1} = \frac{1 - \gamma - 2\beta}{1 - \beta} \quad and$$ $$A_{k+1} = \frac{2(1 - \gamma - 2\beta)}{1 - 2\beta + (\beta k + 1)(k + 1)} \left(1 + \sum_{l=1}^{k} A_{l}\right) \quad (k \in \mathbb{N}).$$ (2.1) Then $$A_{k} = \frac{1 - \gamma - 2\beta}{1 - \beta} \prod_{j=1}^{k-1} \frac{3 - 6\beta - 2\gamma + j(\beta j + 1 - \beta)}{1 - 2\beta + (\beta j + 1)(j + 1)} \quad (k \in \mathbb{N} \setminus \{1\}).$$ (2.2) Proof By virtue of [\(.\)](#page-2-0), we easily get $$[1 - 2\beta + (\beta k + 1)(k + 1)]A_{k+1} = 2(1 - \gamma - 2\beta) \left(1 + \sum_{l=1}^{k} A_l\right), \tag{2.3}$$ and $$\left[1 - 2\beta + (\beta k + 1 - \beta)k\right]A_k = 2(1 - \gamma - 2\beta)\left(1 + \sum_{l=1}^{k-1} A_l\right). \tag{2.4}$$ Combining [\(.](#page-2-1)) and [\(.\)](#page-2-2), we obtain $$\frac{A_{k+1}}{A_k} = \frac{3 - 6\beta - 2\gamma + k(\beta k + 1 - \beta)}{1 - 2\beta + (\beta k + 1)(k + 1)}.$$ (2.5) Thus, for k , we deduce from [\(.](#page-2-3)) that $$A_k = \frac{A_k}{A_{k-1}} \cdot \dots \cdot \frac{A_3}{A_2} \cdot \frac{A_2}{A_1} \cdot A_1 = \frac{1 - \gamma - 2\beta}{1 - \beta} \prod_{j=1}^{k-1} \frac{3 - 6\beta - 2\gamma + j(\beta j + 1 - \beta)}{1 - 2\beta + (\beta j + 1)(j + 1)}.$$ The proof of Lemma [.](#page-2-4) is evidently completed. The following two lemmas can be derived from [[,](#page-12-1) Theorem ] (see also [[\]](#page-12-2)), we here choose to omit the details of

Lemma . Let $$1 + \beta \lambda \left(\lambda + \frac{1}{2}\right) - \lambda - \frac{3}{2}\beta > 0. \tag{2.6}$$ Suppose also that f ∈ is given by [\(.\)](#page-0-1). If $$\sum_{k=1}^{\infty} \left[ k + \beta k(k-1) + \gamma \right] |a_k| \leq 1 - \gamma - 2\beta, \tag{2.7}$$ where (and throughout this paper unless otherwise mentioned) the parameter $\gamma$ is constrained as follows: $$\gamma := \lambda - \beta \lambda \left(\lambda + \frac{1}{2}\right) - \frac{\beta}{2},\tag{2.8}$$ then $f \in \mathcal{H}(\beta, \lambda)$ .

Lemma 2.4. Let $f \in \Sigma$ be given by (1.5). Suppose also that $\gamma$ is defined by (2.8) and the condition (2.6) holds true. Then $f \in \mathcal{H}^+(\beta, \lambda)$ if and only if $$\sum_{k=1}^{\infty} [k + \beta k(k-1) + \gamma] a_k \le 1 - \gamma - 2\beta.$$ (2.9)

Theorem 3.1. Let $\gamma$ be defined by (2.8). If $f \in \mathcal{H}(\beta, \lambda)$ with $0 < \beta < 2/5$ , then $$|a_1| \leq \frac{1-\gamma-2\beta}{1-\beta}$$ , and $$|a_k| \leq \frac{1-\gamma-2\beta}{1-\beta} \prod_{j=1}^{k-1} \frac{3-6\beta-2\gamma+j(\beta j+1-\beta)}{1-2\beta+(\beta j+1)(j+1)} \quad (k \in \mathbb{N} \setminus \{1\}).$$ Proof Suppose that $$q(z) := -\frac{zf'(z)}{f(z)} - \beta \frac{z^2 f''(z)}{f(z)} + \beta \lambda \left(\lambda + \frac{1}{2}\right) + \frac{\beta}{2} - \lambda. \tag{3.1}$$ Then, by the definition of the function class $\mathcal{H}(\beta,\lambda)$ , we know that q is analytic in $\mathbb{U}$ and $$\Re(q(z)) > 0 \quad (z \in \mathbb{U})$$ with $$q(0) = 1 - \gamma - 2\beta > 0.$$ It follows from (2.8) and (3.1) that $$q(z)f(z) = -zf'(z) - \beta z^{2}f''(z) - \gamma f(z).$$ (3.2) By noting that $$h(z) = \frac{q(z)}{1 - \gamma - 2\beta} \in \mathcal{P},$$ if we put $$q(z) = c_0 + \sum_{k=1}^{\infty} c_k z^k \quad (c_0 = 1 - \gamma - 2\beta),$$ by Lemma 2.1, we know that $$|c_k| \leq 2(1-\gamma-2\beta) \quad (k \in \mathbb{N}).$$ It follows from (3.2) that $$\left(c_0 + \sum_{k=1}^{\infty} c_k z^k\right) \left(\frac{1}{z} + \sum_{k=1}^{\infty} a_k z^k\right) = \left(\frac{1}{z} - \sum_{k=1}^{\infty} k a_k z^k\right) - \left(\frac{2\beta}{z} + \beta \sum_{k=1}^{\infty} k(k-1)a_k z^k\right) - \gamma \left(\frac{1}{z} + \sum_{k=1}^{\infty} a_k z^k\right).$$ (3.3) In view of (3.3), we get $$(1 - \gamma - 2\beta)a_1 + c_2 = -a_1 - \gamma a_1 \tag{3.4}$$ and $$c_{k+2} + (1 - \gamma - 2\beta)a_{k+1} + \sum_{l=1}^{k} a_l c_{k+1-l}$$ $$= -(k+1)a_{k+1} - \beta k(k+1)a_{k+1} - \gamma a_{k+1} \quad (k \in \mathbb{N}).$$ (3.5) From (3.4), we obtain $$|a_1| \le \frac{1 - \gamma - 2\beta}{1 - \beta}.\tag{3.6}$$ Moreover, we deduce from (3.5) that $$|a_{k+1}| \le \frac{2(1-\gamma-2\beta)}{1-2\beta+(\beta k+1)(k+1)} \left(1+\sum_{l=1}^{k} |a_l|\right) \quad (k \in \mathbb{N}).$$ (3.7) Next, we define the sequence $\{A_k\}_{k=1}^{\infty}$ as follows: $$A_1 = \frac{1 - \gamma - 2\beta}{1 - \beta} \quad \text{and} \quad A_{k+1} = \frac{2(1 - \gamma - 2\beta)}{1 - 2\beta + (\beta k + 1)(k + 1)} \left(1 + \sum_{l=1}^{k} A_l\right) \quad (k \in \mathbb{N}). \quad (3.8)$$ In order to prove that $$|a_k| \leq A_k \quad (k \in \mathbb{N}),$$ we make use of the principle of mathematical induction. By noting that $$|a_1| \leq A_1 = \frac{1 - \gamma - 2\beta}{1 - \beta}.$$ Therefore, assuming that $$|a_l| \leq A_l \quad (l = 1, 2, 3, ..., k; k \in \mathbb{N}).$$ Combining (3.7) and (3.8), we get $$|a_{k+1}| \leq \frac{2(1-\gamma-2\beta)}{1-2\beta+(\beta k+1)(k+1)} \left(1+\sum_{l=1}^{k} |a_l|\right)$$ $$\leq \frac{2(1-\gamma-2\beta)}{1-2\beta+(\beta k+1)(k+1)} \left(1+\sum_{l=1}^{k} A_l\right) = A_{k+1} \quad (k \in \mathbb{N}).$$ Hence, by the principle of mathematical induction, we have $$|a_k| \le A_k \quad (k \in \mathbb{N}) \tag{3.9}$$ as desired. By means of Lemma 2.2 and (3.8), we know that (2.2) holds true. Combining (3.9) and (2.2), we readily get the coefficient estimates asserted by Theorem 3.1. $\Box$ Following the earlier works (based upon the familiar concept of neighborhood of analytic functions) by Goodman [19] and Ruscheweyh [20], and (more recently) by Altintaş et~al.~[21-24], Cătaş [25], Cho et~al.~[26], Liu and Srivastava [27-29], Frasin [30], Keerthi et~al.~[31], Srivastava et~al.~[32] and Wang et~al.~[33]. Assuming that $\gamma$ is given by (2.8) and the condition (2.6) of Lemma 2.3 holds true, we here introduce the $\delta$ -neighborhood of a function $f \in \Sigma$ of the form (1.1) by means of the following definition: $$\mathcal{N}_{\delta}(f) := \left\{ g \in \Sigma : g(z) = \frac{1}{z} + \sum_{k=1}^{\infty} b_k z^k \text{ and} \right.$$ $$\left. \sum_{k=1}^{\infty} \frac{k + \beta k(k-1) + \gamma}{1 - \gamma - 2\beta} |a_k - b_k| \le \delta \left( \delta \ge 0 \right) \right\}. \tag{3.10}$$ By making use of the definition (3.10), we now derive the following result.

Theorem 3.2. Let the condition (2.6) hold true. If $f \in \Sigma$ satisfies the condition $$\frac{f(z) + \varepsilon z^{-1}}{1 + \varepsilon} \in \mathcal{H}(\beta, \lambda) \quad (\varepsilon \in \mathbb{C}; |\varepsilon| < \delta; \delta > 0), \tag{3.11}$$ then $$\mathcal{N}_{\delta}(f) \subset \mathcal{H}(\beta, \lambda). \tag{3.12}$$ Proof By noting that the condition (1.3) can be written as $$\left| \frac{\frac{zf'(z)}{f(z)} + \beta \frac{z^2 f''(z)}{f(z)} + 1}{\frac{zf'(z)}{f(z)} + \beta \frac{z^2 f''(z)}{f(z)} + 2\gamma - 1} \right| < 1 \quad (z \in \mathbb{U}), \tag{3.13}$$ we easily find from (3.13) that a function $g \in \mathcal{H}(\beta, \lambda)$ if and only if $$\frac{zg'(z)+\beta z^2g''(z)+g(z)}{zg'(z)+\beta z^2g''(z)+(2\gamma-1)g(z)}\neq\sigma \quad \big(z\in\mathbb{U};\sigma\in\mathbb{C};|\sigma|=1\big),$$ which is equivalent to $$\frac{(g * \mathfrak{h})(z)}{z^{-1}} \neq 0 \quad (z \in \mathbb{U}), \tag{3.14}$$ where $$\mathfrak{h}(z) = \frac{1}{z} + \sum_{k=1}^{\infty} c_k z^k \quad \left( c_k := \frac{k + \beta k(k-1) + 1 - [k + \beta k(k-1) + (2\gamma - 1)]\sigma}{2[\beta + (1 - \gamma - \beta)\sigma]} \right). \tag{3.15}$$ It follows from (3.15) that $$\begin{aligned} |c_k| &= \left| \frac{k + \beta k(k-1) + 1 - [k + \beta k(k-1) + (2\gamma - 1)]\sigma}{2[\beta + (1 - \gamma - \beta)\sigma]} \right| \\ &\leq \frac{k + \beta k(k-1) + 1 + [k + \beta k(k-1) + (2\gamma - 1)]|\sigma|}{2(1 - \gamma - 2\beta)|\sigma|} \\ &= \frac{k + \beta k(k-1) + \gamma}{1 - \gamma - 2\beta} \quad (|\sigma| = 1). \end{aligned}$$ If $f \in \Sigma$ given by (1.1) satisfies the condition (3.11), we deduce from (3.14) that $$\frac{(f * \mathfrak{h})(z)}{z^{-1}} \neq -\varepsilon \quad (|\varepsilon| < \delta; \delta > 0),$$ or equivalently, $$\left| \frac{(f * \mathfrak{h})(z)}{z^{-1}} \right| \ge \delta \quad (z \in \mathbb{U}; \delta > 0). \tag{3.16}$$ We now suppose that $$q(z) = \frac{1}{z} + \sum_{k=1}^{\infty} d_k z^k \in \mathcal{N}_{\delta}(f).$$ It follows from (3.10) that $$\left| \frac{((q-f)*\mathfrak{h})(z)}{z^{-1}} \right| = \left| \sum_{k=1}^{\infty} (d_k - a_k) c_k z^{k+1} \right| \le |z| \sum_{k=1}^{\infty} \frac{k + \beta k(k-1) + \gamma}{1 - \gamma - 2\beta} |d_k - a_k| < \delta. \quad (3.17)$$ Combining (3.16) and (3.17), we easily find that $$\left|\frac{(q\mathfrak{h})(z)}{z^{-1}}\right| = \left|\frac{([f+(q-f)]\mathfrak{h})(z)}{z^{-1}}\right| \ge \left|\frac{(f\mathfrak{h})(z)}{z^{-1}}\right| - \left|\frac{((q-f)\mathfrak{h})(z)}{z^{-1}}\right| > 0,$$ which implies that $$\frac{(q * \mathfrak{h})(z)}{z^{-1}} \neq 0 \quad (z \in \mathbb{U}).$$ Therefore, we have $$q(z) \in \mathcal{N}_{\delta}(f) \subset \mathcal{H}(\beta, \lambda).$$ The proof of Theorem 3.2 is thus completed. Next, we derive the partial sums of the class $\mathcal{H}(\beta,\lambda)$ . For some recent investigations involving the partial sums in analytic function theory, one can find in [28, 29, 34, 35] and the references cited therein.

Theorem 3.3. Let $f \in \Sigma$ be given by (1.1) and define the partial sums $f_n(z)$ of f by $$f_n(z) = \frac{1}{z} + \sum_{k=1}^{n} a_k z^k \quad (n \in \mathbb{N}).$$ (3.18) If $$\sum_{k=1}^{\infty} \frac{k + \beta k(k-1) + \gamma}{1 - \gamma - 2\beta} |a_k| \le 1, \tag{3.19}$$ where $\gamma$ is given by (2.8) and the condition (2.6) holds true, then 1. $$f \in \mathcal{H}(\beta, \lambda)$$ ; 2. $$\Re\left(\frac{f(z)}{f_n(z)}\right) \ge \frac{n + \beta n(n+1) + 2\beta + 2\gamma}{n + \beta n(n+1) + 1 + \gamma} \quad (n \in \mathbb{N}; z \in \mathbb{U}),\tag{3.20}$$ and $$\Re\left(\frac{f_n(z)}{f(z)}\right) \ge \frac{n + \beta n(n+1) + 1 + \gamma}{n + \beta n(n+1) + 2 - 2\beta} \quad (n \in \mathbb{N}; z \in \mathbb{U}).$$ (3.21) The bounds in (3.20) and (3.21) are sharp. Proof First of all, we suppose that $$f_1(z)=\frac{1}{z}.$$ We know that $$\frac{f_1(z) + \varepsilon z^{-1}}{1 + \varepsilon} = \frac{1}{z} \in \mathcal{H}(\beta, \lambda).$$ From (3.19), we easily find that $$\sum_{k=1}^{\infty} \frac{k + \beta k(k-1) + \gamma}{1 - \gamma - 2\beta} |a_k - 0| \le 1,$$ which implies that $f \in \mathcal{N}_1(z^{-1})$ . By virtue of Theorem 3.2, we deduce that $$f \in \mathcal{N}_1(z^{-1}) \subset \mathcal{H}(\beta, \lambda).$$ Next, it is easy to see that $$\frac{n+1+\beta n(n+1)+\gamma}{1-\gamma-2\beta} > \frac{n+\beta n(n-1)+\gamma}{1-\gamma-2\beta} > 1 \quad (n \in \mathbb{N}).$$ Therefore, we have $$\sum_{k=1}^{n} |a_k| + \frac{n + \beta n(n+1) + 1 + \gamma}{1 - \gamma - 2\beta} \sum_{k=n+1}^{\infty} |a_k| \le \sum_{k=1}^{\infty} \frac{k + \beta k(k-1) + \gamma}{1 - \gamma - 2\beta} |a_k| \le 1.$$ (3.22) We now suppose that $$h_{1}(z) = \frac{n + \beta n(n+1) + 1 + \gamma}{1 - \gamma - 2\beta} \left( \frac{f(z)}{f_{n}(z)} - \frac{n + \beta n(n+1) + 2\beta + 2\gamma}{n + \beta n(n+1) + 1 + \gamma} \right)$$ $$= 1 + \frac{\frac{n + \beta n(n+1) + 1 + \gamma}{1 - \gamma - 2\beta} \sum_{k=n+1}^{\infty} a_{k} z^{k+1}}{1 + \sum_{k=1}^{n} a_{k} z^{k+1}}.$$ (3.23) It follows from (3.22) and (3.23) that $$\left|\frac{h_1(z)-1}{h_1(z)+1}\right| \leq \frac{\frac{n+\beta n(n+1)+1+\gamma}{1-\gamma-2\beta} \sum_{k=n+1}^{\infty} |a_k|}{2-2\sum_{k=1}^{n} |a_k| - \frac{n+\beta n(n+1)+1+\gamma}{1-\gamma-2\beta} \sum_{k=n+1}^{\infty} |a_k|} \leq 1 \quad (z \in \mathbb{U}),$$ which shows that $$\Re(h_1(z)) \ge 0 \quad (z \in \mathbb{U}). \tag{3.24}$$ Combining (3.23) and (3.24), we deduce that the assertion (3.20) holds true. Furthermore, if we put $$f(z) = \frac{1}{z} - \frac{1 - \gamma - 2\beta}{n + \beta n(n+1) + 1 + \gamma} z^{n+1},$$ (3.25) then $$\frac{f(z)}{f_n(z)}=1-\frac{1-\gamma-2\beta}{n+\beta n(n+1)+1+\gamma}z^{n+2}\rightarrow \frac{n+\beta n(n+1)+2\beta+2\gamma}{n+\beta n(n+1)+1+\gamma} \quad (z\rightarrow 1^-),$$ which implies that the bound in (3.20) is the best possible for each $n \in \mathbb{N}$ . Similarly, we suppose that $$h_2(z) = \frac{n + \beta n(n+1) + 2 - 2\beta}{1 - \gamma - 2\beta} \left( \frac{f_n(z)}{f(z)} - \frac{n + \beta n(n+1) + 1 + \gamma}{n + \beta n(n+1) + 2 - 2\beta} \right)$$ $$= 1 - \frac{\frac{n + \beta n(n+1) + 2 - 2\beta}{1 - \gamma - 2\beta} \sum_{k=n+1}^{\infty} a_k z^{k+1}}{1 + \sum_{k=1}^{\infty} a_k z^{k+1}}.$$ (3.26) In view of (3.22) and (3.26), we conclude that $$\left|\frac{h_2(z)-1}{h_2(z)+1}\right| \leq \frac{\frac{n+\beta n(n+1)+2-2\beta}{1-\gamma-2\beta} \sum_{k=n+1}^{\infty} |a_k|}{2-2\sum_{k=1}^{n} |a_k| - \frac{n+\beta n(n+1)+2\beta+2\gamma}{1-\gamma-2\beta} \sum_{k=n+1}^{\infty} |a_k|} \leq 1 \quad (z \in \mathbb{U}),$$ which implies that $$\Re(h_2(z)) \ge 0 \quad (z \in \mathbb{U}). \tag{3.27}$$ Combining (3.26) and (3.27), we readily get the assertion (3.21) of Theorem 3.3. The bound in (3.21) is sharp with the extremal function f given by (3.25). We thus complete the proof of Theorem 3.3. In what follows, we turn to quotients involving derivatives. The proof of Theorem 3.4 below is similar to that of Theorem 3.3, we here choose to omit the analogous details.

Theorem 3.4. Let $f \in \Sigma$ be given by (1.1) and define the partial sums $f_n(z)$ of f by (3.18). If the conditions (2.6) and (3.19) hold, where $\gamma$ is given by (2.8), then $$\Re\left(\frac{f'(z)}{f'_n(z)}\right) \ge \frac{(n+2)\gamma + (n+1)(n+2)\beta}{n+\beta n(n+1)+1+\gamma} \quad (n \in \mathbb{N}; z \in \mathbb{U}),$$ (3.28) and $$\Re\left(\frac{f_n'(z)}{f'(z)}\right) \ge \frac{n + \beta n(n+1) + 1 + \gamma}{(n-2)(n+1)\beta + 2(n+1) - n\gamma} \quad (n \in \mathbb{N}; z \in \mathbb{U}). \tag{3.29}$$ The bounds in (3.28) and (3.29) are sharp with the extremal function given by (3.25). Finally, we prove the following inclusion relationship for the function class $\mathcal{H}(\beta,\lambda)$ .

Corollary 3.7. Let $f \in \mathcal{H}^+(\beta, \lambda)$ . Suppose also that $\gamma$ is defined by (2.8) and the condition (2.6) holds true. Then $$a_k \leq \frac{1-\gamma-2\beta}{k+\beta k(k-1)+\gamma}.$$ Each of these inequalities is sharp, with the extremal function given by $$f_k(z) = \frac{1}{z} - \frac{1-\gamma-2\beta}{k+\beta k(k-1)+\gamma} z^k.$$