Srivastava and Gaboury Journal of Inequalities and Applications (2015) 2015:39

Theorem 1. Let $\alpha \in [0,1)$ . If $f(z) \in A$ satisfies the following equality <span id="page-5-0"></span> $$\sum_{k=2}^{\infty} \frac{(k-\alpha)}{k!} \left| \frac{\prod_{j=1}^{p} (\lambda_j + 1)_{k-1}}{\prod_{j=1}^{q} (\mu_j + 1)_{k-1}} \right| \left| \left( \frac{a+1}{a+k} \right)^s \right| \left| \left( \frac{\Lambda(a+k,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)} \right) \right| |a_k| \le 1 - \alpha, \tag{2.1}$$ then $$f \in \mathcal{S}^{s,a,\lambda,*}_{(\lambda_p),(\mu_q),b}(\alpha).$$ Proof Suppose that inequality (2.1) holds for $\alpha \in [0,1)$ . Let us define the function F(z) by $$F(z) := \frac{z(J_{(\lambda_p),(\mu_q),b}^{s,a,\lambda})'(f)(z)}{J_{(\lambda_p),(\mu_q),b}^{s,a,\lambda}(f)(z)} - \alpha \quad (f(z) \in \mathcal{A}).$$ $$(2.2)$$ It is sufficient to prove that $$\left| \frac{F(z) - 1}{F(z) + 1} \right| < 1 \quad (z \in \mathbb{U}) \tag{2.3}$$ to prove that $f(z) \in \mathcal{S}^{s,a,\lambda,*}_{(\lambda_p),(\mu_q),b}(\alpha)$ . In fact, we have that $$\begin{split} \left| \mathcal{F}(z) \right| &:= \left| \frac{F(z) - 1}{F(z) + 1} \right| = \left| \frac{\frac{z(J_{\lambda p}^{S,A,\lambda}(\mu_q),b}^{S,A,\lambda}(f)(z)}{J_{(\lambda p),(\mu_q),b}^{S,a,\lambda}(f)(z)} - \alpha - 1}{\frac{z(J_{\lambda p}^{S,A,\lambda}(\mu_q),b}^{S,a,\lambda}(f)(z))}{J_{(\lambda p),(\mu_q),b}^{S,a,\lambda}(f)(z)} - \alpha + 1} \right| \\ &= \left| \frac{\alpha z + \sum_{k=2}^{\infty} \frac{\prod_{j=1}^{p} (\lambda_j + 1)_{k-1}}{\prod_{j=1}^{q} (\mu_j + 1)_{k-1}} (\frac{a + 1}{a + k})^s (\frac{\Lambda(a + k, b, s, \lambda)}{\Lambda(a + 1, b, s, \lambda)}) \frac{(\alpha + 1 - k)}{k!} a_k z^k}{(2 - \alpha) z - \sum_{k=2}^{\infty} \frac{\prod_{j=1}^{p} (\lambda_j + 1)_{k-1}}{\prod_{j=1}^{q} (\mu_j + 1)_{k-1}} (\frac{a + 1}{a + k})^s (\frac{\Lambda(a + k, b, s, \lambda)}{\Lambda(a + 1, b, s, \lambda)}) \frac{(\alpha - 1 - k)}{k!} a_k z^k} \right|, \end{split}$$ and thus $$\begin{split} \left| \mathcal{F}(z) \right| & \leq \frac{\alpha |z| + \sum_{k=2}^{\infty} |\frac{\prod_{j=1}^{p} (\lambda_{j} + 1)_{k-1}}{\prod_{j=1}^{p} (\mu_{j} + 1)_{k-1}} ||(\frac{a+1}{a+k})^{s}||(\frac{\Lambda(a+k,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)})||\frac{(\alpha+1-k)}{k!}||a_{k}| \cdot |z|^{k}}{(2-\alpha)|z| - \sum_{k=2}^{\infty} |\frac{\prod_{j=1}^{p} (\lambda_{j} + 1)_{k-1}}{\prod_{j=1}^{q} (\mu_{j} + 1)_{k-1}} ||(\frac{a+1}{a+k})^{s}||(\frac{\Lambda(a+k,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)})||\frac{(\alpha-1-k)}{k!}||a_{k}| \cdot |z|^{k}} \\ & < \frac{\alpha + \sum_{k=2}^{\infty} |\frac{\prod_{j=1}^{p} (\lambda_{j} + 1)_{k-1}}{\prod_{j=1}^{q} (\mu_{j} + 1)_{k-1}} ||(\frac{a+1}{a+k})^{s}||(\frac{\Lambda(a+k,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)})||\frac{(k-\alpha-1)}{k!}|a_{k}|}{(2-\alpha) - \sum_{k=2}^{\infty} |\frac{\prod_{j=1}^{p} (\lambda_{j} + 1)_{k-1}}{\prod_{j=1}^{q} (\mu_{j} + 1)_{k-1}} ||(\frac{a+1}{a+k})^{s}||(\frac{\Lambda(a+k,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)})||\frac{(k-\alpha+1)}{k!}|a_{k}|} \leq 1, \end{split}$$ provided that (2.1) is satisfied. <span id="page-6-2"></span><span id="page-6-0"></span>The next theorem aims to provide coefficient inequalities for functions f(z) belonging to the class $\mathcal{S}^{s,a,\lambda,*}_{(\lambda_p),(\mu_q),b}(\alpha)$ .

Theorem 2. Let $\alpha \in [0,1)$ . If $f(z) \in S^{s,a,\lambda,*}_{(\lambda_n),(\mu_a),b}(\alpha)$ , then $$|a_{k}| \leq k! \left(\frac{2(1-\alpha)}{k-1}\right) \left| \left(\frac{a+k}{a+1}\right)^{s} \right| \left| \left(\frac{\Lambda(a+1,b,s,\lambda)}{\Lambda(a+k,b,s,\lambda)}\right) \right|$$ $$\cdot \left| \frac{\prod_{j=1}^{q} (\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p} (\lambda_{j}+1)_{k-1}} \right| \prod_{j=2}^{k-1} \left(1 + \frac{2(1-\alpha)}{j-1}\right) \quad (k \in \mathbb{N} \setminus \{1\}).$$ $$(2.4)$$ The result is sharp. Proof Let $$p(z) := \frac{\frac{z(f_{(\lambda_p),(\mu_q),b}^{s,a,\lambda})'(f)(z)}{f_{(\lambda_p),(\mu_q),b}^{s,a,\lambda}(f)(z)} - \alpha}{1 - \alpha} = 1 + c_1 z + c_2 z^2 + \cdots$$ Then p(z) is analytic and $$p(0) = 1$$ and $\Re(p(z)) > 0$ $(z \in \mathbb{U})$ . We note easily that <span id="page-6-3"></span> $$z\left(J_{(\lambda_p),(\mu_q),b}^{s,a,\lambda}\right)'(f)(z) = \left[(1-\alpha)p(z) + \alpha\right]J_{(\lambda_p),(\mu_q),b}^{s,a,\lambda}(f)(z).$$ With the help of (1.17), we find $$\frac{(k-1)}{k!} \frac{\prod_{j=1}^{p} (\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q} (\mu_{j}+1)_{k-1}} \left(\frac{a+1}{a+k}\right)^{s} \left(\frac{\Lambda(a+k,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)}\right) a_{k}$$ $$= (1-\alpha) \cdot \left[c_{k-1} + \sum_{m=2}^{k-1} \frac{\prod_{j=1}^{p} (\lambda_{j}+1)_{m-1}}{\prod_{j=1}^{q} (\mu_{j}+1)_{m-1}} \left(\frac{a+1}{a+m}\right)^{s} \left(\frac{\Lambda(a+m,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)}\right) \frac{a_{m}c_{k-m}}{m!}\right] (2.5)$$ for $k \in \mathbb{N} \setminus \{1\}$ . By making use of the Carathéodory lemma [29, p.41], we have <span id="page-6-1"></span> $$\frac{(k-1)}{k!} \left| \frac{\prod_{j=1}^{p} (\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q} (\mu_{j}+1)_{k-1}} \right| \left| \left( \frac{a+1}{a+k} \right)^{s} \right| \left| \left( \frac{\Lambda(a+k,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)} \right) \right| \cdot |a_{k}| \\ \leq 2(1-\alpha) \\ \cdot \left[ 1 + \sum_{m=2}^{k-1} \left| \frac{\prod_{j=1}^{p} (\lambda_{j}+1)_{m-1}}{\prod_{j=1}^{q} (\mu_{j}+1)_{m-1}} \right| \left| \left( \frac{a+1}{a+m} \right)^{s} \right| \left| \left( \frac{\Lambda(a+m,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)} \right) \left| \frac{|a_{m}|}{m!} \right| \right|. \tag{2.6}$$ We have to prove that inequality (2.4) holds true for $k \in \mathbb{N} \setminus \{1\}$ . We will proceed by the principle of mathematical induction. If k = 2 in (2.6), we obtain $$|a_{2}| \leq 4(1-\alpha) \left| \frac{\prod_{j=1}^{q} (\mu_{j}+1)}{\prod_{j=1}^{p} (\lambda_{j}+1)} \right| \left| \left( \frac{a+2}{a+1} \right)^{s} \right| \left| \left( \frac{\Lambda(a+1,b,s,\lambda)}{\Lambda(a+2,b,s,\lambda)} \right) \right|.$$ (2.7) Now suppose that (2.4) is satisfied for $k \le n$ . Then, from (2.4) and (2.6), we have that $$\frac{n}{(n+1)!} \left| \frac{\prod_{j=1}^{p} (\lambda_{j}+1)_{n}}{\prod_{j=1}^{q} (\mu_{j}+1)_{n}} \right| \left( \frac{a+1}{a+n+1} \right)^{s} \left| \left( \frac{\Lambda(a+n+1,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)} \right) \right| \cdot |a_{n+1}| \\ \leq 2(1-\alpha) \left[ 1 + \sum_{m=2}^{n} \left| \frac{\prod_{j=1}^{p} (\lambda_{j}+1)_{m-1}}{\prod_{j=1}^{q} (\mu_{j}+1)_{m-1}} \right| \left( \frac{a+1}{a+m} \right)^{s} \right| \left( \frac{\Lambda(a+m,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)} \right) \left| \frac{|a_{m}|}{m!} \right] \\ \leq 2(1-\alpha) \left[ 1 + \sum_{m=2}^{n} \frac{2(1-\alpha)}{m-1} \prod_{j=2}^{m-1} \left( 1 + \frac{2(1-\alpha)}{j-1} \right) \right] \\ \leq 2(1-\alpha) \prod_{j=2}^{n} \left( 1 + \frac{2(1-\alpha)}{j-1} \right), \tag{2.8}$$ whence $$|a_{k}| \leq k! \left(\frac{2(1-\alpha)}{k-1}\right) \left| \left(\frac{a+k}{a+1}\right)^{s} \right| \left| \left(\frac{\Lambda(a+1,b,s,\lambda)}{\Lambda(a+k,b,s,\lambda)}\right) \right|$$ $$\cdot \left| \frac{\prod_{j=1}^{q} (\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p} (\lambda_{j}+1)_{k-1}} \right| \prod_{j=2}^{k-1} \left(1 + \frac{2(1-\alpha)}{j-1}\right) \quad (k \in \mathbb{N} \setminus \{1\}).$$ The result is sharp for the function f(z) given by $$f(z) = z + \frac{2(1-\alpha)}{k-1} \left(\frac{a+k}{a+1}\right)^{s} \left(\frac{\Lambda(a+1,b,s,\lambda)}{\Lambda(a+k,b,s,\lambda)}\right) \cdot \frac{\prod_{j=1}^{q} (\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p} (\lambda_{j}+1)_{k-1}} \prod_{j=2}^{k-1} \left(1 + \frac{2(1-\alpha)}{j-1}\right) z^{k} \quad (k \in \mathbb{N} \setminus \{1\}).$$

Theorem 3. Let $f(z) \in \mathcal{S}^{s,a,\lambda,*}_{(\lambda_p),(\mu_q),b}(\alpha)$ and $0 \leq \alpha < 1$ . Then <span id="page-7-0"></span> $$r - 2(1 - \alpha)r^{2} \sum_{k=2}^{\infty} \frac{k!}{k-1} \left| \left( \frac{a+k}{a+1} \right)^{s} \right| \left| \left( \frac{\Lambda(a+1,b,s,\lambda)}{\Lambda(a+k,b,s,\lambda)} \right) \right|$$ $$\cdot \left| \frac{\prod_{j=1}^{q} (\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p} (\lambda_{j}+1)_{k-1}} \right| \prod_{j=2}^{k-1} \left( 1 + \frac{2(1-\alpha)}{j-1} \right)$$ $$\leq |f(z)|$$ $$\leq r + 2(1-\alpha)r^{2} \sum_{k=2}^{\infty} \frac{k!}{k-1} \left| \left( \frac{a+k}{a+1} \right)^{s} \right| \left| \left( \frac{\Lambda(a+1,b,s,\lambda)}{\Lambda(a+k,b,s,\lambda)} \right) \right|$$ $$\cdot \left| \frac{\prod_{j=1}^{q} (\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p} (\lambda_{j}+1)_{k-1}} \right| \prod_{j=2}^{k-1} \left( 1 + \frac{2(1-\alpha)}{j-1} \right) \quad (|z| = r < 1)$$ $$(3.1)$$ and <span id="page-8-0"></span> $$1 - 2(1 - \alpha)r \sum_{k=2}^{\infty} \frac{k \cdot k!}{k - 1} \left| \left( \frac{a + k}{a + 1} \right)^{s} \right| \left| \left( \frac{\Lambda(a + 1, b, s, \lambda)}{\Lambda(a + k, b, s, \lambda)} \right) \right|$$ $$\cdot \left| \frac{\prod_{j=1}^{q} (\mu_{j} + 1)_{k-1}}{\prod_{j=1}^{p} (\lambda_{j} + 1)_{k-1}} \right| \prod_{j=2}^{k-1} \left( 1 + \frac{2(1 - \alpha)}{j - 1} \right)$$ $$\leq \left| f'(z) \right|$$ $$\leq 1 + 2(1 - \alpha)r \sum_{k=2}^{\infty} \frac{k \cdot k!}{k - 1} \left| \left( \frac{a + k}{a + 1} \right)^{s} \right| \left| \left( \frac{\Lambda(a + 1, b, s, \lambda)}{\Lambda(a + k, b, s, \lambda)} \right) \right|$$ $$\cdot \left| \frac{\prod_{j=1}^{q} (\mu_{j} + 1)_{k-1}}{\prod_{j=1}^{p} (\lambda_{j} + 1)_{k-1}} \right| \prod_{j=2}^{k-1} \left( 1 + \frac{2(1 - \alpha)}{j - 1} \right) \quad (|z| = r < 1).$$ (3.2) Proof Let $f(z) \in A$ be given by (1.1). Then, making use of Theorem 2, we find $$|f(z)| \leq |z| + \sum_{k=2}^{\infty} |a_k| \cdot |z^k|$$ $$\leq r + 2(1 - \alpha)r^2 \sum_{k=2}^{\infty} \frac{k!}{k - 1} \left| \left( \frac{a + k}{a + 1} \right)^s \right| \left| \left( \frac{\Lambda(a + 1, b, s, \lambda)}{\Lambda(a + k, b, s, \lambda)} \right) \right|$$ $$\cdot \left| \frac{\prod_{j=1}^q (\mu_j + 1)_{k-1}}{\prod_{j=1}^p (\lambda_j + 1)_{k-1}} \right| \prod_{j=2}^{k-1} \left( 1 + \frac{2(1 - \alpha)}{j - 1} \right) \quad (|z| = r < 1)$$ (3.3) and $$|f(z)| \geq |z| - \sum_{k=2}^{\infty} |a_{k}| \cdot |z^{k}|$$ $$\geq r - 2(1 - \alpha)r^{2} \sum_{k=2}^{\infty} \frac{k!}{k-1} \left| \left( \frac{a+k}{a+1} \right)^{s} \right| \left| \left( \frac{\Lambda(a+1,b,s,\lambda)}{\Lambda(a+k,b,s,\lambda)} \right) \right|$$ $$\cdot \left| \frac{\prod_{j=1}^{q} (\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p} (\lambda_{j}+1)_{k-1}} \right| \prod_{j=2}^{k-1} \left( 1 + \frac{2(1-\alpha)}{j-1} \right) \quad (|z| = r < 1).$$ (3.4) From (1.1), we also have that $$|f'(z)| \leq 1 + \sum_{k=2}^{\infty} k \cdot |a_{k}| \cdot |z^{k-1}|$$ $$\leq 1 + 2(1 - \alpha)r \sum_{k=2}^{\infty} \frac{k \cdot k!}{k - 1} \left| \left( \frac{a + k}{a + 1} \right)^{s} \right| \left| \left( \frac{\Lambda(a + 1, b, s, \lambda)}{\Lambda(a + k, b, s, \lambda)} \right) \right|$$ $$\cdot \left| \frac{\prod_{j=1}^{q} (\mu_{j} + 1)_{k-1}}{\prod_{j=1}^{p} (\lambda_{j} + 1)_{k-1}} \right| \prod_{i=2}^{k-1} \left( 1 + \frac{2(1 - \alpha)}{j - 1} \right) \quad (|z| = r < 1)$$ (3.5) and $$|f'(z)| \ge 1 - \sum_{k=2}^{\infty} k \cdot |a_{k}| \cdot |z^{k}|$$ $$\ge 1 - 2(1 - \alpha)r \sum_{k=2}^{\infty} \frac{k \cdot k!}{k - 1} \left| \left( \frac{a + k}{a + 1} \right)^{s} \right| \left| \left( \frac{\Lambda(a + 1, b, s, \lambda)}{\Lambda(a + k, b, s, \lambda)} \right) \right|$$ $$\cdot \left| \frac{\prod_{j=1}^{q} (\mu_{j} + 1)_{k-1}}{\prod_{j=1}^{p} (\lambda_{j} + 1)_{k-1}} \right| \prod_{i=2}^{k-1} \left( 1 + \frac{2(1 - \alpha)}{j - 1} \right) \quad (|z| = r < 1).$$ (3.6) We thus obtain the results (3.1) and (3.2) asserted by Theorem 3.

Definition 1. The H-function involved in the right-hand side of (1.7) is the well-known Fox's H-function [15, Definition 1.1] (see also [14, 16]) defined by $$H_{\mathfrak{p},\mathfrak{q}}^{m,n}(z) = H_{\mathfrak{p},\mathfrak{q}}^{m,n} \left[ z \mid (a_{1},A_{1}),\dots,(a_{\mathfrak{p}},A_{\mathfrak{p}}) \right]$$ $$= \frac{1}{2\pi i} \int_{C} \Xi(s)z^{-s} \, ds \quad (z \in \mathbb{C} \setminus \{0\}; |\arg(z)| < \pi), \tag{1.9}$$ where $$\Xi(s) = \frac{\prod_{j=1}^{m} \Gamma(b_j + B_j s) \cdot \prod_{j=1}^{n} \Gamma(1 - a_j - A_j s)}{\prod_{j=n+1}^{p} \Gamma(a_j + A_j s) \cdot \prod_{j=m+1}^{q} \Gamma(1 - b_j - B_j s)},$$ an empty product is interpreted as 1, m, n, $\mathfrak{p}$ and $\mathfrak{q}$ are integers such that $1 \le m \le \mathfrak{q}$ , $0 \le n \le \mathfrak{p}$ , $A_j > 0$ $(j = 1, ..., \mathfrak{p})$ , $B_j > 0$ $(j = 1, ..., \mathfrak{q})$ , $a_j \in \mathbb{C}$ $(j = 1, ..., \mathfrak{p})$ , $b_j \in \mathbb{C}$ $(j = 1, ..., \mathfrak{q})$ and $\mathcal{L}$ is a suitable Mellin-Barnes type contour separating the poles of the gamma functions $$\left\{\Gamma(b_j+B_js)\right\}_{j=1}^m$$ from the poles of the gamma functions $$\left\{\Gamma(1-a_j+A_js)\right\}_{j=1}^n.$$ It is worthy to mention that using the fact that [8, p.1496, Remark 7] <span id="page-2-0"></span> $$\lim_{b \to 0} \left\{ H_{0,2}^{2,0} \left[ (a+n)b^{\frac{1}{\lambda}} \mid \overline{(s,1), \left(0, \frac{1}{\lambda}\right)} \right] \right\} = \lambda \Gamma(s) \quad (\lambda > 0), \tag{1.10}$$ equation (1.7) reduces to $$\Phi_{\lambda_{1},...,\lambda_{p};\mu_{1},...,\mu_{q}}^{(\rho_{1},...,\rho_{p},\sigma_{1},...,\sigma_{q})}(z,s,a;0,\lambda) := \Phi_{\lambda_{1},...,\lambda_{p};\mu_{1},...,\mu_{q}}^{(\rho_{1},...,\rho_{p},\sigma_{1},...,\sigma_{q})}(z,s,a)$$ $$= \sum_{n=0}^{\infty} \frac{\prod_{j=1}^{p} (\lambda_{j})_{n\rho_{j}}}{(a+n)^{s} \cdot \prod_{i=1}^{q} (\mu_{j})_{n\sigma_{i}}} \frac{z^{n}}{n!}.$$ (1.11)

Definition 2. The function $\Phi_{\lambda_1,\ldots,\lambda_p;\mu_1,\ldots,\mu_q}^{(\rho_1,\ldots,\rho_p,\sigma_1,\ldots,\sigma_q)}(z,s,a)$ involved in (1.11) is the multiparameter extension and generalization of the Hurwitz-Lerch zeta function $\Phi(z,s,a)$ introduced by Srivastava et al. [13, p.503, Eq. (6.2)] defined by $$\begin{split} &\Phi_{\lambda_{1},\ldots,\lambda_{p};\mu_{1},\ldots,\mu_{q}}^{(\rho_{1},\ldots,\rho_{p},\sigma_{1},\ldots,\sigma_{q})}(z,s,a):=\sum_{n=0}^{\infty}\frac{\prod_{j=1}^{p}(\lambda_{j})_{n\rho_{j}}}{(a+n)^{s}\cdot\prod_{j=1}^{q}(\mu_{j})_{n\sigma_{j}}}\frac{z^{n}}{n!}\\ &\left(p,q\in\mathbb{N}_{0};\lambda_{j}\in\mathbb{C}\;(j=1,\ldots,p);a,\mu_{j}\in\mathbb{C}\setminus\mathbb{Z}_{0}^{-}\;(j=1,\ldots,q);\\ &\rho_{j},\sigma_{k}\in\mathbb{R}^{+}\;(j=1,\ldots,p;k=1,\ldots,q); \end{split}$$ $\Delta > -1$ when $s, z \in \mathbb{C}$ ; $\Delta = -1$ and $s \in \mathbb{C}$ when $|z| < \nabla^*$ ; $$\Delta = -1 \text{ and } \Re(\Xi) > \frac{1}{2} \text{ when } |z| = \nabla^*$$ (1.12) with $$\nabla^* := \left(\prod_{j=1}^p \rho_j^{-\rho_j}\right) \cdot \left(\prod_{j=1}^q \sigma_j^{\sigma_j}\right),\tag{1.13}$$ $$\Delta := \sum_{j=1}^{q} \sigma_j - \sum_{j=1}^{p} \rho_j \quad \text{and} \quad \Xi := s + \sum_{j=1}^{q} \mu_j - \sum_{j=1}^{p} \lambda_j + \frac{p-q}{2}. \tag{1.14}$$ We propose to consider the following linear operator <span id="page-3-0"></span> $$J^{s,a,\lambda}_{(\lambda_p),(\mu_q),b}(f): \mathcal{A} \to \mathcal{A},$$ defined by $$J_{(\lambda_p),(\mu_q),b}^{s,a,\lambda}(f)(z) = G_{(\lambda_p),(\mu_q),b}^{s,a,\lambda}(z) * f(z),$$ (1.15) where \* denotes the Hadamard product (or convolution) of analytic functions, and the function $G^{s,a,\lambda}_{(\lambda_p),(\mu_q),b}(z)$ is given by <span id="page-3-1"></span> $$G_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda}(z) := \frac{\lambda \prod_{j=1}^{q} (\mu_{j}) \Gamma(s)(a+1)^{s}}{\prod_{j=1}^{p} (\lambda_{j})} \cdot \Lambda(a+1,b,s,\lambda)^{-1} \cdot \left[ \Phi_{\lambda_{1},\dots,\lambda_{p};\mu_{1},\dots,\mu_{q}}^{(1,\dots,1,1,\dots,1)}(z,s,a;b,\lambda) - \frac{a^{-s}}{\lambda \Gamma(s)} \Lambda(a,b,s,\lambda) \right]$$ $$= z + \sum_{k=2}^{\infty} \frac{\prod_{j=1}^{p} (\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q} (\mu_{j}+1)_{k-1}} \left( \frac{a+1}{a+k} \right)^{s} \left( \frac{\Lambda(a+k,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)} \right) \frac{z^{k}}{k!}$$ $$(1.16)$$ with $$\Lambda(a,b,s,\lambda) := H_{0,2}^{2,0} \left[ ab^{\frac{1}{\lambda}} \mid \overline{(s,1), \left(0,\frac{1}{\lambda}\right)} \right].$$ Combining (1.15) and (1.16), we obtain <span id="page-3-2"></span> $$J_{(\lambda_p),(\mu_q),b}^{s,a,\lambda}(f)(z)$$ $$= z + \sum_{k=2}^{\infty} \frac{\prod_{j=1}^{p} (\lambda_j + 1)_{k-1}}{\prod_{j=1}^{q} (\mu_j + 1)_{k-1}} \left(\frac{a+1}{a+k}\right)^s \left(\frac{\Lambda(a+k,b,s,\lambda)}{\Lambda(a+1,b,s,\lambda)}\right) a_k \frac{z^k}{k!}$$ $$(\lambda_j \in \mathbb{C} \ (j=1,\ldots,p) \text{ and } \mu_j \in \mathbb{C} \setminus \mathbb{Z}_0^- \ (j=1,\ldots,q); p \leq q+1; z \in \mathbb{U}), \tag{1.17}$$ with $$\min\{\Re(a),\Re(s)\}>0;$$ $\lambda>0$ if $\Re(b)>0$ and $$s \in \mathbb{C}$$ ; $a \in \mathbb{C} \setminus \mathbb{Z}_0^-$ if $b = 0$ . Remark 1 It follows from (1.15) and (1.17) that the operator $J_{(\lambda_p),(\mu_q),0}^{s,a,\lambda}(f)$ (special case of (1.17) when b=0) can be defined for $a \in \mathbb{C} \setminus \mathbb{Z}^-$ by the following limit relationship: $$J_{(\lambda_p),(\mu_q),0}^{s,0,\lambda}(f)(z) := \lim_{a \to 0} \left\{ J_{(\lambda_p),(\mu_q),0}^{s,a,\lambda}(f)(z) \right\}. \tag{1.18}$$ We can see that the operator $J^{s,a,\lambda}_{(\lambda_p),(\mu_q),b}$ generalizes several recently investigated operators such as: - (i) If p = 2, q = 1 and b = 0, then $J_{(\lambda_1, \lambda_2), (\mu_1), 0}^{s,a,\lambda} = J_{\lambda_1, \lambda_2; \mu_1}^{s,a}$ , where $J_{\lambda_1, \lambda_2; \mu_1}^{s,a}$ is the linear operator introduced by Prajapat and Bulboacă [17, p.571, Eq. (1.8)]. - (ii) $J_{(\gamma-1,1),(\nu),0}^{s,a,\lambda} = I_{a,\nu,\gamma}^{s}$ , where $I_{a,\nu,\gamma}^{s}$ is the generalized operator recently studied by Noor and Bukhari [18, p.2, Eq. (1.3)]. - (iii) $J_{(\gamma-1,1),(\nu),0}^{0,0,\lambda} = I_{\nu,\gamma}^s$ , where $I_{\nu,\gamma}^s$ is the Choi-Saigo-Srivastava operator [19]. - (iv) $J_{(\gamma,1),(\gamma),0}^{s,a,\lambda} = J_{s,a}$ , where $J_{s,a}$ is the Srivastava-Attiya operator [1]. - (v) $J_{(\gamma,1),(\gamma),0}^{-r,a,\lambda} = I(r,a)$ ( $a \ge 0$ , $r \in \mathbb{Z}$ ), where the operator I(r,a) is the one introduced by Cho and Srivastava [20]. - (vi) $J_{(\beta,1),(\alpha+\beta),0}^{0,a,\lambda} = \mathcal{Q}^{\alpha}_{\beta}$ ( $\alpha \geq 0$ , $\beta > -1$ ), where the operator $\mathcal{Q}^{\alpha}_{\beta}$ was studied by Jung et al. [21]. - (vii) $J_{(\gamma,1),(\gamma),0}^{1,a,\lambda} = J_a$ ( $a \ge -1$ ), where $J_a$ denotes the Bernardi operator [22]. - (viii) $J_{(\gamma,1),(\nu),0}^{0,0,\lambda} = \mathcal{L}(\gamma,\nu)$ , where $\mathcal{L}(\gamma,\nu)$ is the well-known Carlson-Shaffer operator [23]. - (ix) $J_{(2,1),(2-\gamma),0}^{0,0,\lambda} = \Omega_z^{\gamma}$ ( $0 \le \gamma < 1$ ), where $\Omega_z^{\gamma}$ is the fractional integral operator investigated by Owa and Srivastava [24]. - (x) $J_{(\lambda_1-1,\ldots,\lambda_p-1,1),(\mu_1-1,\ldots,\mu_q-1,0),0}^{0,a,\lambda}=H_1(\lambda_1,\ldots,\lambda_p;\mu_1,\ldots,\mu_q)$ ( $p\leq q+1$ ), where the operator $H_1(\lambda_1,\ldots,\lambda_p;\mu_1,\ldots,\mu_q)$ is the Dziok-Srivastava operator [25, 26] which contains as special cases the Hohlov operator [27] and the Ruscheweyh operator [28]. We say that a function $f \in \mathcal{A}$ is in the class $\mathcal{S}^{s,a,\lambda,}_{(\lambda_p),(\mu_q),b}(\alpha)$ if $J^{s,a,\lambda}_{(\lambda_p),(\mu_q),b}(f)$ is in the class $\mathcal{S}^(\alpha)$ , that is, if $$\Re\left(\frac{z(J_{(\lambda_p),(\mu_q),b}^{s,a,\lambda}(f))'}{J_{(\lambda_p),(\mu_q),b}^{s,a,\lambda}(f)}\right) > \alpha$$ $$\left(\lambda_j \in \mathbb{C} \ (j=1,\ldots,p) \text{ and } \mu_j \in \mathbb{C} \setminus \mathbb{Z}_0^- \ (j=1,\ldots,q);$$ $$z \in \mathbb{U}; 0 \leq \alpha < 1; p \leq q+1\right), \tag{1.19}$$ with $$\min\{\Re(a),\Re(s)\}>0;$$ $\lambda>0$ if $\Re(b)>0$ and $$s \in \mathbb{C}$$ ; $a \in \mathbb{C} \setminus \mathbb{Z}^-$ if $b = 0$ . <span id="page-5-1"></span>In this paper, we systematically investigate the class $\mathcal{S}^{s,a,\lambda,*}_{(\lambda_p),(\mu_q),b}(\alpha)$ of analytic functions defined above by means of the new generalized Srivastava-Attiya convolution operator $J^{s,a,\lambda}_{(\lambda_p),(\mu_q),b}$ . Especially, we derive coefficient inequalities, distortion theorems, extreme points and the Fekete-Szegö problem for this new function class.