Wang et al. Journal of Inequalities and Applications 2014, 2014:29

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 <span id="page-2-6"></span><span id="page-2-0"></span> $$A_{1} = \frac{1 - \gamma - 2\beta}{1 - \beta} \quad and$$ $$A_{k+1} = \frac{2(1 - \gamma - 2\beta)}{1 - 2\beta + (\beta k + 1)(k + 1)} \left(1 + \sum_{l=1}^{k} A_{l}\right) \quad (k \in \mathbb{N}).$$ (2.1) Then <span id="page-2-1"></span> $$A_{k} = \frac{1 - \gamma - 2\beta}{1 - \beta} \prod_{j=1}^{k-1} \frac{3 - 6\beta - 2\gamma + j(\beta j + 1 - \beta)}{1 - 2\beta + (\beta j + 1)(j + 1)} \quad (k \in \mathbb{N} \setminus \{1\}).$$ (2.2) Proof By virtue of [\(.\)](#page-2-0), we easily get <span id="page-2-2"></span> $$[1 - 2\beta + (\beta k + 1)(k + 1)]A_{k+1} = 2(1 - \gamma - 2\beta) \left(1 + \sum_{l=1}^{k} A_l\right), \tag{2.3}$$ and <span id="page-2-3"></span> $$\left[1 - 2\beta + (\beta k + 1 - \beta)k\right]A_k = 2(1 - \gamma - 2\beta)\left(1 + \sum_{l=1}^{k-1} A_l\right). \tag{2.4}$$ 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 <span id="page-2-5"></span> $$A_k = \frac{A_k}{A_{k-1}} \cdot \dots \cdot \frac{A_3}{A_2} \cdot \frac{A_2}{A_1} \cdot A_1 = \frac{1 - \gamma - 2\beta}{1 - \beta} \prod_{j=1}^{k-1} \frac{3 - 6\beta - 2\gamma + j(\beta j + 1 - \beta)}{1 - 2\beta + (\beta j + 1)(j + 1)}.$$ <span id="page-2-7"></span>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 $$\sum_{k=1}^{\infty} \left[ k + \beta k(k-1) + \gamma \right] |a_k| \leq 1 - \gamma - 2\beta, \tag{2.7}$$ <span id="page-3-4"></span>where (and throughout this paper unless otherwise mentioned) the parameter $\gamma$ is constrained as follows: <span id="page-3-0"></span> $$\gamma := \lambda - \beta \lambda \left(\lambda + \frac{1}{2}\right) - \frac{\beta}{2},\tag{2.8}$$ 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 <span id="page-3-3"></span> $$\sum_{k=1}^{\infty} [k + \beta k(k-1) + \gamma] a_k \le 1 - \gamma - 2\beta.$$ (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 <span id="page-3-1"></span> $$|a_k| \leq \frac{1-\gamma-2\beta}{1-\beta} \prod_{j=1}^{k-1} \frac{3-6\beta-2\gamma+j(\beta j+1-\beta)}{1-2\beta+(\beta j+1)(j+1)} \quad (k \in \mathbb{N} \setminus \{1\}).$$ 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 <span id="page-3-2"></span> $$\Re(q(z)) > 0 \quad (z \in \mathbb{U})$$ 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 <span id="page-4-0"></span> $$|c_k| \leq 2(1-\gamma-2\beta) \quad (k \in \mathbb{N}).$$ It follows from (3.2) that <span id="page-4-1"></span> $$\left(c_0 + \sum_{k=1}^{\infty} c_k z^k\right) \left(\frac{1}{z} + \sum_{k=1}^{\infty} a_k z^k\right) = \left(\frac{1}{z} - \sum_{k=1}^{\infty} k a_k z^k\right) - \left(\frac{2\beta}{z} + \beta \sum_{k=1}^{\infty} k(k-1)a_k z^k\right) - \gamma \left(\frac{1}{z} + \sum_{k=1}^{\infty} a_k z^k\right).$$ (3.3) In view of (3.3), we get <span id="page-4-2"></span> $$(1 - \gamma - 2\beta)a_1 + c_2 = -a_1 - \gamma a_1 \tag{3.4}$$ 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 <span id="page-4-3"></span> $$|a_1| \le \frac{1 - \gamma - 2\beta}{1 - \beta}.\tag{3.6}$$ Moreover, we deduce from (3.5) that <span id="page-4-4"></span> $$|a_{k+1}| \le \frac{2(1-\gamma-2\beta)}{1-2\beta+(\beta k+1)(k+1)} \left(1+\sum_{l=1}^{k} |a_l|\right) \quad (k \in \mathbb{N}).$$ (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 <span id="page-5-0"></span> $$|a_{k+1}| \leq \frac{2(1-\gamma-2\beta)}{1-2\beta+(\beta k+1)(k+1)} \left(1+\sum_{l=1}^{k} |a_l|\right)$$ $$\leq \frac{2(1-\gamma-2\beta)}{1-2\beta+(\beta k+1)(k+1)} \left(1+\sum_{l=1}^{k} A_l\right) = A_{k+1} \quad (k \in \mathbb{N}).$$ 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. $\Box$ <span id="page-5-1"></span>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: <span id="page-5-3"></span><span id="page-5-2"></span> $$\mathcal{N}_{\delta}(f) := \left\{ g \in \Sigma : g(z) = \frac{1}{z} + \sum_{k=1}^{\infty} b_k z^k \text{ and} \right.$$ $$\left. \sum_{k=1}^{\infty} \frac{k + \beta k(k-1) + \gamma}{1 - \gamma - 2\beta} |a_k - b_k| \le \delta \left( \delta \ge 0 \right) \right\}. \tag{3.10}$$ 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 <span id="page-6-0"></span> $$\left| \frac{\frac{zf'(z)}{f(z)} + \beta \frac{z^2 f''(z)}{f(z)} + 1}{\frac{zf'(z)}{f(z)} + \beta \frac{z^2 f''(z)}{f(z)} + 2\gamma - 1} \right| < 1 \quad (z \in \mathbb{U}), \tag{3.13}$$ we easily find from (3.13) that a function $g \in \mathcal{H}(\beta, \lambda)$ if and only if <span id="page-6-2"></span> $$\frac{zg'(z)+\beta z^2g''(z)+g(z)}{zg'(z)+\beta z^2g''(z)+(2\gamma-1)g(z)}\neq\sigma \quad \big(z\in\mathbb{U};\sigma\in\mathbb{C};|\sigma|=1\big),$$ which is equivalent to <span id="page-6-1"></span> $$\frac{(g * \mathfrak{h})(z)}{z^{-1}} \neq 0 \quad (z \in \mathbb{U}), \tag{3.14}$$ 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 <span id="page-6-3"></span> $$\frac{(f * \mathfrak{h})(z)}{z^{-1}} \neq -\varepsilon \quad (|\varepsilon| < \delta; \delta > 0),$$ 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 <span id="page-6-4"></span> $$q(z) = \frac{1}{z} + \sum_{k=1}^{\infty} d_k z^k \in \mathcal{N}_{\delta}(f).$$ It follows from (3.10) that $$\left| \frac{((q-f)*\mathfrak{h})(z)}{z^{-1}} \right| = \left| \sum_{k=1}^{\infty} (d_k - a_k) c_k z^{k+1} \right| \le |z| \sum_{k=1}^{\infty} \frac{k + \beta k(k-1) + \gamma}{1 - \gamma - 2\beta} |d_k - a_k| < \delta. \quad (3.17)$$ <span id="page-7-4"></span><span id="page-7-2"></span><span id="page-7-1"></span><span id="page-7-0"></span> 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 $$q(z) \in \mathcal{N}_{\delta}(f) \subset \mathcal{H}(\beta, \lambda).$$ <span id="page-7-3"></span>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 <span id="page-8-0"></span> $$\frac{n+1+\beta n(n+1)+\gamma}{1-\gamma-2\beta} > \frac{n+\beta n(n-1)+\gamma}{1-\gamma-2\beta} > 1 \quad (n \in \mathbb{N}).$$ <span id="page-8-1"></span>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 <span id="page-8-2"></span> $$\left|\frac{h_1(z)-1}{h_1(z)+1}\right| \leq \frac{\frac{n+\beta n(n+1)+1+\gamma}{1-\gamma-2\beta} \sum_{k=n+1}^{\infty} |a_k|}{2-2\sum_{k=1}^{n} |a_k| - \frac{n+\beta n(n+1)+1+\gamma}{1-\gamma-2\beta} \sum_{k=n+1}^{\infty} |a_k|} \leq 1 \quad (z \in \mathbb{U}),$$ which shows that <span id="page-8-3"></span> $$\Re(h_1(z)) \ge 0 \quad (z \in \mathbb{U}). \tag{3.24}$$ 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 $n \in \mathbb{N}$ . <span id="page-9-0"></span>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 <span id="page-9-1"></span> $$\left|\frac{h_2(z)-1}{h_2(z)+1}\right| \leq \frac{\frac{n+\beta n(n+1)+2-2\beta}{1-\gamma-2\beta} \sum_{k=n+1}^{\infty} |a_k|}{2-2\sum_{k=1}^{n} |a_k| - \frac{n+\beta n(n+1)+2\beta+2\gamma}{1-\gamma-2\beta} \sum_{k=n+1}^{\infty} |a_k|} \leq 1 \quad (z \in \mathbb{U}),$$ which implies that <span id="page-9-3"></span> $$\Re(h_2(z)) \ge 0 \quad (z \in \mathbb{U}). \tag{3.27}$$ <span id="page-9-2"></span>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 <span id="page-9-4"></span> $$\Re\left(\frac{f'(z)}{f'_n(z)}\right) \ge \frac{(n+2)\gamma + (n+1)(n+2)\beta}{n+\beta n(n+1)+1+\gamma} \quad (n \in \mathbb{N}; z \in \mathbb{U}),$$ (3.28) <span id="page-9-7"></span>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 <span id="page-11-2"></span> $$f_k(z) = \frac{1}{z} - \frac{1-\gamma-2\beta}{k+\beta k(k-1)+\gamma} z^k.$$