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A new class of analytic functions defined by means of a generalization of the Srivastava-Attiya operator

Abstract

In this paper, we introduce a new class of analytic functions defined by a new convolution operator \(J_{(\lambda_{p}),(\mu _{q}),b}^{s,a,\lambda}\) which generalizes the well-known Srivastava-Attiya operator investigated by Srivastava and Attiya (Integral Transforms Spec. Funct. 18:207-216, 2007). We derive coefficient inequalities, distortion theorems, extreme points and the Fekete-Szegö problem for this new function class.

1 Introduction

Let \(\mathcal{A}\) denote the class of functions \(f(z)\) normalized by

$$ f(z)=z+\sum_{k=2}^{\infty}a_{k}z^{k}, $$
(1.1)

which are analytic in the open unit disk

$$\mathbb{U}=\bigl\{ z\mbox{: } z \in\mathbb{C} \mbox{ and } |z|< 1\bigr\} . $$

A function \(f(z)\) in the class \(\mathcal{A}\) is said to be in the class \(\mathcal{S}^{*}(\alpha)\) of starlike functions of order α in \(\mathbb{U}\) if it satisfies the following inequality:

$$ \Re \biggl(\frac{zf'(z)}{f(z)} \biggr)>\alpha \quad (z \in\mathbb{U}; 0\leqq\alpha< 1). $$
(1.2)

The largely investigated Srivastava-Attiya operator is defined as [1] (see also [24]):

$$ J_{s,a}(f) (z)=z+\sum_{k=2}^{\infty} \biggl(\frac{1+a}{k+a} \biggr)^{s}a_{k}z^{k}, $$
(1.3)

where \(z \in\mathbb{U}\), \(a \in\mathbb{C}\setminus\mathbb {Z}^{-}_{0}\), \(s \in\mathbb{C}\) and \(f \in\mathcal{A}\).

In fact, the linear operator \(J_{s,a}(f)\) can be written as

$$ J_{s,a}(f) (z):=G_{s,a}(z)*f(z) $$
(1.4)

in terms of the Hadamard product (or convolution), where \(G_{s,a}(z)\) is given by

$$ G_{s,a}(z):=(1+a)^{s} \bigl[ \Phi(z,s,a)-a^{-s} \bigr] \quad (z \in\mathbb{U}). $$
(1.5)

The function \(\Phi(z,s,a)\) involved in the right-hand side of (1.5) is the well-known Hurwitz-Lerch zeta function defined by (see, for example, [5, p.121 et seq.]; see also [6] and [7, p.194 et seq.])

$$\begin{aligned}& \Phi(z,s,a):=\sum_{n=0}^{\infty} \frac{z^{n}}{(n+a)^{s}} \\& \quad \bigl(a \in \mathbb{C}\setminus\mathbb{Z}_{0}^{-}; s \in\mathbb{C} \mbox{ when } |z|< 1; \Re(s)>1 \mbox{ when } |z|=1\bigr). \end{aligned}$$
(1.6)

Recently, a new family of λ-generalized Hurwitz-Lerch zeta functions was investigated by Srivastava [8] (see also [913]). Srivastava considered the following function:

$$\begin{aligned}& \Phi_{\lambda_{1},\ldots,\lambda_{p};\mu_{1},\ldots,\mu_{q}}^{(\rho _{1},\ldots,\rho_{p},\sigma_{1},\ldots,\sigma_{q})}(z,s,a;b,\lambda ) \\& \quad =\frac{1}{\lambda\Gamma(s)} \cdot\sum_{n=0}^{\infty} \frac{\prod_{j=1}^{p}(\lambda_{j})_{n\rho _{j}}}{(a+n)^{s}\cdot\prod_{j=1}^{q}(\mu_{j})_{n\sigma_{j}}} H_{0,2}^{2,0} \biggl[(a+n)b^{\frac{1}{\lambda}} \Bigm| \overline{ (s,1), \biggl(0,\frac{1}{\lambda} \biggr)} \biggr]\frac{z^{n}}{n!} \\& \qquad \bigl(\min\bigl\{ \Re(a),\Re(s)\bigr\} >0; \Re(b)>0; \lambda>0\bigr), \end{aligned}$$
(1.7)

where

$$\begin{aligned}& \Biggl(\lambda_{j} \in\mathbb{C}\ (j=1,\ldots,p) \mbox{ and } \mu_{j} \in \mathbb{C}\setminus\mathbb{Z}_{0}^{-}\ (j=1,\ldots,q); \rho_{j}>0\ (j=1,\ldots,p); \\& \quad \sigma_{j}>0\ (j=1,\ldots,q); 1+\sum _{j=1}^{q}\sigma_{j}-\sum _{j=1}^{p}\rho_{j}\geqq0\Biggr) \end{aligned}$$

and the equality in the convergence condition holds true for suitably bounded values of \(|z|\) given by

$$|z|< \nabla:= \Biggl(\prod_{j=1}^{p} \rho_{j}^{-\rho_{j}} \Biggr)\cdot \Biggl(\prod _{j=1}^{q}\sigma_{j}^{\sigma_{j}} \Biggr). $$

Here, and for the remainder of this paper, \((\lambda)_{\kappa}\) denotes the Pochhammer symbol defined, in terms of the gamma function, by

$$ (\lambda)_{\kappa}:=\frac{\Gamma(\lambda+\kappa)}{\Gamma(\lambda)}= \begin{cases} \lambda(\lambda+1)\cdots(\lambda+n-1) & (\kappa=n \in \mathbb{N}; \lambda\in\mathbb{C}), \\ 1 & (\kappa=0; \lambda\in\mathbb{C}\setminus\{0\}), \end{cases} $$
(1.8)

it being understood conventionally that \((0)_{0}:=1\) and assumed tacitly that the Γ-quotient exists (see, for details, [14, p.21 et seq.]).

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

$$\begin{aligned} H_{\mathfrak{p},\mathfrak{q}}^{m,n}(z)&=H_{\mathfrak{p},\mathfrak{q}}^{m,n} \left [z\Bigm| \begin{array}{c} (a_{1},A_{1}),\ldots,(a_{\mathfrak{p}},A_{\mathfrak{p}}) \\ (b_{1},B_{1}),\ldots,(b_{\mathfrak{q}},B_{\mathfrak{q}}) \end{array} \right ] \\ &=\frac{1}{2\pi\mathrm{i}}\int_{\mathcal{L}}\Xi(s)z^{-s} \, \mathrm{d}s \quad \bigl(z \in\mathbb{C}\setminus\{0\}; \bigl\vert \arg(z)\bigr\vert < \pi\bigr), \end{aligned}$$
(1.9)

where

$$ \Xi(s)=\frac{\prod_{j=1}^{m}\Gamma(b_{j}+B_{j}s)\cdot\prod_{j=1}^{n}\Gamma(1-a_{j}-A_{j}s)}{\prod_{j=n+1}^{\mathfrak{p}}\Gamma (a_{j}+A_{j}s)\cdot\prod_{j=m+1}^{\mathfrak{q}}\Gamma(1-b_{j}-B_{j}s)}, $$

an empty product is interpreted as 1, m, n, \(\mathfrak{p}\) and \(\mathfrak{q}\) are integers such that \(1\leqq m \leqq\mathfrak{q}\), \(0\leqq n \leqq\mathfrak{p}\), \(A_{j}>0\) (\(j=1,\ldots,\mathfrak{p}\)), \(B_{j}>0\) (\(j=1,\ldots,\mathfrak{q}\)), \(a_{j} \in\mathbb{C}\) (\(j=1,\ldots,\mathfrak{p}\)), \(b_{j} \in\mathbb{C}\) (\(j=1,\ldots ,\mathfrak{q}\)) and is a suitable Mellin-Barnes type contour separating the poles of the gamma functions

$$\bigl\{ \Gamma(b_{j}+B_{j}s)\bigr\} _{j=1}^{m} $$

from the poles of the gamma functions

$$\bigl\{ \Gamma(1-a_{j}+A_{j}s)\bigr\} _{j=1}^{n}. $$

It is worthy to mention that using the fact that [8, p.1496, Remark 7]

$$ \lim_{b\rightarrow0} \biggl\{ H_{0,2}^{2,0} \biggl[(a+n)b^{\frac{1}{\lambda }}\Bigm| \overline{ (s,1), \biggl(0, \frac{1}{\lambda} \biggr)} \biggr] \biggr\} =\lambda\Gamma(s)\quad (\lambda>0), $$
(1.10)

equation (1.7) reduces to

$$\begin{aligned} \Phi_{\lambda_{1},\ldots,\lambda_{p};\mu_{1},\ldots,\mu_{q}}^{(\rho _{1},\ldots,\rho_{p},\sigma_{1},\ldots,\sigma_{q})}(z,s,a;0,\lambda )&:= \Phi_{\lambda_{1},\ldots,\lambda_{p};\mu_{1},\ldots,\mu_{q}}^{(\rho _{1},\ldots,\rho_{p},\sigma_{1},\ldots,\sigma_{q})}(z,s,a) \\ & =\sum_{n=0}^{\infty}\frac{\prod_{j=1}^{p}(\lambda_{j})_{n\rho _{j}}}{(a+n)^{s}\cdot\prod_{j=1}^{q}(\mu_{j})_{n\sigma_{j}}} \frac{z^{n}}{n!}. \end{aligned}$$
(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

$$\begin{aligned}& \Phi_{\lambda_{1},\ldots,\lambda_{p};\mu_{1},\ldots,\mu_{q}}^{(\rho _{1},\ldots,\rho_{p},\sigma_{1},\ldots,\sigma_{q})}(z,s,a):=\sum _{n=0}^{\infty}\frac{\prod_{j=1}^{p}(\lambda_{j})_{n\rho _{j}}}{(a+n)^{s}\cdot\prod_{j=1}^{q}(\mu_{j})_{n\sigma_{j}}}\frac{z^{n}}{n!} \\& \quad \biggl(p,q \in\mathbb{N}_{0}; \lambda_{j} \in \mathbb{C}\ (j=1,\ldots ,p); a,\mu_{j} \in\mathbb{C}\setminus \mathbb{Z}_{0}^{-}\ (j=1,\ldots,q); \\& \quad \rho_{j},\sigma_{k} \in\mathbb{R}^{+}\ (j=1,\ldots,p; k=1,\ldots,q); \\& \quad \Delta>-1 \mbox{ when } s,z \in\mathbb{C}; \\& \quad \Delta=-1 \mbox{ and } s \in\mathbb{C} \mbox{ when } |z|< \nabla^{*}; \\& \quad \Delta=-1 \mbox{ and } \Re(\Xi)>\frac{1}{2} \mbox{ when } |z|= \nabla^{*}\biggr) \end{aligned}$$
(1.12)

with

$$\begin{aligned}& \nabla^{*}:= \Biggl(\prod_{j=1}^{p} \rho_{j}^{-\rho_{j}} \Biggr)\cdot \Biggl(\prod _{j=1}^{q}\sigma_{j}^{\sigma_{j}} \Biggr), \end{aligned}$$
(1.13)
$$\begin{aligned}& \Delta:=\sum_{j=1}^{q} \sigma_{j}-\sum_{j=1}^{p} \rho_{j}\quad \mbox{and}\quad \Xi :=s+\sum _{j=1}^{q}\mu_{j}-\sum _{j=1}^{p}\lambda_{j}+\frac{p-q}{2}. \end{aligned}$$
(1.14)

We propose to consider the following linear operator

$$ J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda}(f):\mathcal{A}\rightarrow \mathcal{A}, $$

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 \(G_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda }(z)\) is given by

$$\begin{aligned}& G_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda}(z) \\& \quad :=\frac{\lambda\prod_{j=1}^{q}(\mu_{j}) \Gamma(s)(a+1)^{s}}{\prod_{j=1}^{p}(\lambda_{j})}\cdot\Lambda (a+1,b,s,\lambda )^{-1} \\& \qquad {} \cdot \biggl[\Phi_{\lambda_{1},\ldots,\lambda_{p};\mu_{1},\ldots,\mu _{q}}^{(1,\ldots,1,1,\ldots,1)}(z,s,a;b,\lambda)- \frac{a^{-s}}{\lambda \Gamma(s)}\Lambda (a,b,s,\lambda ) \biggr] \\& \quad =z+\sum_{k=2}^{\infty}\frac{\prod_{j=1}^{p}(\lambda _{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}} \biggl(\frac {a+1}{a+k} \biggr)^{s} \biggl(\frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr) \frac{z^{k}}{k!} \end{aligned}$$
(1.16)

with

$$\Lambda (a,b,s,\lambda ):=H_{0,2}^{2,0} \biggl[ab^{\frac {1}{\lambda}}\Bigm| \overline{ (s,1), \biggl(0,\frac{1}{\lambda} \biggr)} \biggr]. $$

Combining (1.15) and (1.16), we obtain

$$\begin{aligned}& J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda}(f) (z) \\& \quad =z+\sum_{k=2}^{\infty}\frac{\prod_{j=1}^{p}(\lambda _{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}} \biggl(\frac {a+1}{a+k} \biggr)^{s} \biggl(\frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr)a_{k}\frac{z^{k}}{k!} \\& \qquad \bigl(\lambda_{j} \in\mathbb{C}\ (j=1,\ldots,p) \mbox{ and } \mu_{j} \in \mathbb{C}\setminus\mathbb{Z}_{0}^{-}\ (j=1,\ldots,q); p\leqq q+1; z \in \mathbb{U}\bigr), \end{aligned}$$
(1.17)

with

$$\min\bigl\{ \Re(a),\Re(s)\bigr\} >0;\qquad \lambda>0\quad \text{if } \Re(b)>0 $$

and

$$s \in\mathbb{C}; \qquad a \in\mathbb{C}\setminus\mathbb{Z}_{0}^{-} \quad \text{if } b=0. $$

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:

$$ J_{(\lambda_{p}),(\mu_{q}),0}^{s,0,\lambda}(f) (z):=\lim_{a\rightarrow 0} \bigl\{ J_{(\lambda_{p}),(\mu_{q}),0}^{s,a,\lambda}(f) (z) \bigr\} . $$
(1.18)

We can see that the operator \(J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda }\) generalizes several recently investigated operators such as:

  1. (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)].

  2. (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)].

  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].

  4. (iv)

    \(J_{(\gamma,1),(\gamma),0}^{s,a,\lambda}= J_{s,a}\), where \(J_{s,a}\) is the Srivastava-Attiya operator [1].

  5. (v)

    \(J_{(\gamma,1),(\gamma),0}^{-r,a,\lambda}= I(r,a)\) (\(a\geqq 0\), \(r \in\mathbb{Z}\)), where the operator \(I(r,a)\) is the one introduced by Cho and Srivastava [20].

  6. (vi)

    \(J_{(\beta,1),(\alpha+\beta),0}^{0,a,\lambda}= \mathcal {Q}_{\beta}^{\alpha}\) (\(\alpha\geqq0\), \(\beta>-1\)), where the operator \(\mathcal{Q}_{\beta}^{\alpha}\) was studied by Jung et al. [21].

  7. (vii)

    \(J_{(\gamma,1),(\gamma),0}^{1,a,\lambda}= J_{a}\) (\(a\geqq -1\)), where \(J_{a}\) denotes the Bernardi operator [22].

  8. (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].

  9. (ix)

    \(J_{(2,1),(2-\gamma),0}^{0,0,\lambda}= \Omega_{z}^{\gamma}\) (\(0\leqq\gamma<1\)), where \(\Omega_{z}^{\gamma}\) is the fractional integral operator investigated by Owa and Srivastava [24].

  10. (x)

    \(J_{(\lambda_{1}-1,\ldots,\lambda_{p}-1,1),(\mu_{1}-1,\ldots,\mu _{q}-1,0),0}^{0,a,\lambda}=H_{1}(\lambda_{1},\ldots,\lambda_{p}; \mu _{1},\ldots,\mu_{q})\) (\(p\leqq q+1\)), 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 \(\mathcal {S}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda,*}(\alpha)\) if \(J_{(\lambda _{p}),(\mu_{q}),b}^{s,a,\lambda}(f)\) is in the class \(\mathcal {S}^{*}(\alpha)\), that is, if

$$\begin{aligned}& \Re \biggl(\frac{z (J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda }(f) )'}{J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda}(f)} \biggr)>\alpha \\ & \quad \bigl(\lambda_{j} \in\mathbb{C}\ (j=1,\ldots,p) \mbox{ and } \mu_{j} \in \mathbb{C}\setminus\mathbb{Z}_{0}^{-}\ (j=1,\ldots,q); \\ & \quad z \in\mathbb{U}; 0\leqq\alpha< 1; p\leqq q+1 \bigr), \end{aligned}$$
(1.19)

with

$$\min\bigl\{ \Re(a),\Re(s)\bigr\} >0; \qquad \lambda>0 \quad \text{if } \Re(b)>0 $$

and

$$s \in\mathbb{C};\qquad a \in\mathbb{C}\setminus\mathbb{Z}^{-}\quad \text{if } b=0. $$

In this paper, we systematically investigate the class \(\mathcal {S}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda,*}(\alpha)\) of analytic functions defined above by means of the new generalized Srivastava-Attiya convolution operator \(J_{(\lambda_{p}),(\mu _{q}),b}^{s,a,\lambda}\). Especially, we derive coefficient inequalities, distortion theorems, extreme points and the Fekete-Szegö problem for this new function class.

2 Coefficient inequalities

Theorem 1

Let \(\alpha\in[0,1)\). If \(f(z) \in\mathcal{A}\) satisfies the following equality

$$ \sum_{k=2}^{\infty} \frac{(k-\alpha)}{k!}\biggl\vert \frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu _{j}+1)_{k-1}}\biggr\vert \biggl\vert \biggl(\frac{a+1}{a+k} \biggr)^{s}\biggr\vert \biggl\vert \biggl( \frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr)\biggr\vert |a_{k}|\leqq1-\alpha, $$
(2.1)

then

$$f \in\mathcal{S}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda,*}(\alpha). $$

Proof

Suppose that inequality (2.1) holds for \(\alpha\in[0,1)\). Let us define the function \(F(z)\) by

$$ F(z):=\frac{z (J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda} )'(f)(z)}{J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda}(f)(z)}-\alpha\quad \bigl(f(z) \in\mathcal{A} \bigr). $$
(2.2)

It is sufficient to prove that

$$ \biggl\vert \frac{F(z)-1}{F(z)+1}\biggr\vert < 1 \quad (z \in \mathbb{U}) $$
(2.3)

to prove that \(f(z) \in\mathcal{S}_{(\lambda_{p}),(\mu _{q}),b}^{s,a,\lambda,*}(\alpha)\).

In fact, we have that

$$\begin{aligned} \bigl\vert \mathcal{F}(z)\bigr\vert &:=\biggl\vert \frac{F(z)-1}{F(z)+1} \biggr\vert =\biggl\vert {\frac {\frac{z (J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda} )'(f)(z)}{J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda}(f)(z)}-\alpha -1}{\frac{z (J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda} )'(f)(z)}{J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda}(f)(z)}-\alpha +1}}\biggr\vert \\ &=\biggl\vert \frac{\alpha z+ {\sum_{k=2}^{\infty}\frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}} (\frac{a+1}{a+k} )^{s} (\frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} )\frac{(\alpha +1-k)}{k!} a_{k} z^{k}}}{(2-\alpha) z- {\sum_{k=2}^{\infty}\frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}} (\frac{a+1}{a+k} )^{s} (\frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} )\frac{(\alpha -1-k)}{k!} a_{k} z^{k}}}\biggr\vert , \end{aligned}$$

and thus

$$\begin{aligned} \begin{aligned} \bigl\vert \mathcal{F}(z)\bigr\vert &\leqq\frac{\alpha|z|+ {\sum_{k=2}^{\infty} \vert \frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu _{j}+1)_{k-1}}\vert \vert (\frac{a+1}{a+k} )^{s}\vert \vert (\frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} )\vert \vert \frac{(\alpha +1-k)}{k!}\vert |a_{k}|\cdot|z|^{k}}}{(2-\alpha) |z|- {\sum_{k=2}^{\infty} \vert \frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}\vert \vert (\frac{a+1}{a+k} )^{s}\vert \vert (\frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} )\vert \vert \frac{(\alpha -1-k)}{k!}\vert |a_{k}|\cdot|z|^{k}}} \\ &< \frac{\alpha+ {\sum_{k=2}^{\infty} \vert \frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu _{j}+1)_{k-1}}\vert \vert (\frac{a+1}{a+k} )^{s}\vert \vert (\frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} )\vert \frac{(k-\alpha-1)}{k!} |a_{k}|}}{(2-\alpha) - {\sum_{k=2}^{\infty} \vert \frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu _{j}+1)_{k-1}}\vert \vert (\frac{a+1}{a+k} )^{s}\vert \vert (\frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} )\vert \frac{(k-\alpha+1)}{k!} |a_{k}|}} \leqq1, \end{aligned} \end{aligned}$$

provided that (2.1) is satisfied. □

The next theorem aims to provide coefficient inequalities for functions \(f(z)\) belonging to the class \(\mathcal{S}_{(\lambda_{p}),(\mu _{q}),b}^{s,a,\lambda,*}(\alpha)\).

Theorem 2

Let \(\alpha\in[0,1)\). If \(f(z) \in\mathcal{S}_{(\lambda_{p}),(\mu _{q}),b}^{s,a,\lambda,*}(\alpha)\), then

$$\begin{aligned} |a_{k}| \leqq& k! \biggl(\frac{2(1-\alpha)}{k-1} \biggr) \biggl\vert \biggl(\frac {a+k}{a+1} \biggr)^{s}\biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr)\biggr\vert \\ &{} \cdot\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}\biggr\vert \prod _{j=2}^{k-1} \biggl(1+\frac {2(1-\alpha)}{j-1} \biggr) \quad \bigl(k \in\mathbb{N}\setminus\{1\}\bigr). \end{aligned}$$
(2.4)

The result is sharp.

Proof

Let

$$p(z):=\frac{ {\frac{z (J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda } )'(f)(z)}{J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda }(f)(z)}}-\alpha}{1-\alpha}=1+c_{1}z+c_{2}z^{2}+ \cdots. $$

Then \(p(z)\) is analytic and

$$p(0)=1\quad \mbox{and}\quad \Re\bigl(p(z)\bigr)>0\quad (z \in\mathbb{U}). $$

We note easily that

$$z \bigl(J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda} \bigr)'(f) (z)= \bigl[(1- \alpha)p(z)+\alpha \bigr]J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda}(f) (z). $$

With the help of (1.17), we find

$$\begin{aligned}& \frac{(k-1)}{k!}\frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}} \biggl(\frac{a+1}{a+k} \biggr)^{s} \biggl(\frac {\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr) a_{k} \\& \quad =(1-\alpha) \cdot \Biggl[c_{k-1}+\sum _{m=2}^{k-1}\frac{\prod_{j=1}^{p}(\lambda _{j}+1)_{m-1}}{\prod_{j=1}^{q}(\mu_{j}+1)_{m-1}} \biggl( \frac {a+1}{a+m} \biggr)^{s} \biggl(\frac{\Lambda (a+m,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr) \frac{a_{m} c_{k-m}}{m!} \Biggr] \end{aligned}$$
(2.5)

for \(k \in\mathbb{N}\setminus\{1\}\).

By making use of the Carathéodory lemma [29, p.41], we have

$$\begin{aligned}& \frac{(k-1)}{k!}\biggl\vert \frac{\prod_{j=1}^{p}(\lambda _{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}\biggr\vert \biggl\vert \biggl(\frac{a+1}{a+k} \biggr)^{s}\biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr)\biggr\vert \cdot|a_{k}| \\& \quad \leqq2(1-\alpha) \\& \qquad {}\cdot \Biggl[1+\sum_{m=2}^{k-1} \biggl\vert \frac{\prod_{j=1}^{p}(\lambda _{j}+1)_{m-1}}{\prod_{j=1}^{q}(\mu_{j}+1)_{m-1}}\biggr\vert \biggl\vert \biggl( \frac{a+1}{a+m} \biggr)^{s}\biggr\vert \biggl\vert \biggl( \frac{\Lambda (a+m,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr)\biggr\vert \frac{|a_{m}|}{m!} \Biggr]. \end{aligned}$$
(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

$$ |a_{2}|\leqq4(1-\alpha)\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)}{\prod_{j=1}^{p}(\lambda_{j}+1)} \biggr\vert \biggl\vert \biggl(\frac{a+2}{a+1} \biggr)^{s}\biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+2,b,s,\lambda )} \biggr)\biggr\vert . $$
(2.7)

Now suppose that (2.4) is satisfied for \(k\leqq n\). Then, from (2.4) and (2.6), we have that

$$\begin{aligned}& \frac{n}{(n+1)!}\biggl\vert \frac{\prod_{j=1}^{p}(\lambda _{j}+1)_{n}}{\prod_{j=1}^{q}(\mu_{j}+1)_{n}}\biggr\vert \biggl\vert \biggl(\frac {a+1}{a+n+1} \biggr)^{s}\biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+n+1,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr)\biggr\vert \cdot|a_{n+1}| \\& \quad \leqq2(1-\alpha) \Biggl[1+\sum_{m=2}^{n} \biggl\vert \frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{m-1}}{\prod_{j=1}^{q}(\mu _{j}+1)_{m-1}}\biggr\vert \biggl\vert \biggl( \frac{a+1}{a+m} \biggr)^{s}\biggr\vert \biggl\vert \biggl( \frac{\Lambda (a+m,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr)\biggr\vert \frac{|a_{m}|}{m!} \Biggr] \\& \quad \leqq2(1-\alpha) \Biggl[1+\sum_{m=2}^{n} \frac{2(1-\alpha)}{m-1}\prod_{j=2}^{m-1} \biggl(1+ \frac{2(1-\alpha)}{j-1} \biggr) \Biggr] \\& \quad \leqq2(1-\alpha)\prod_{j=2}^{n} \biggl(1+\frac{2(1-\alpha)}{j-1} \biggr), \end{aligned}$$
(2.8)

whence

$$\begin{aligned} |a_{k}| \leqq& k! \biggl(\frac{2(1-\alpha)}{k-1} \biggr)\biggl\vert \biggl(\frac {a+k}{a+1} \biggr)^{s}\biggr\vert \biggl\vert \biggl( \frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr)\biggr\vert \\ &{} \cdot\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}\biggr\vert \prod _{j=2}^{k-1} \biggl(1+\frac {2(1-\alpha)}{j-1} \biggr)\quad \bigl(k \in\mathbb{N}\setminus\{1\}\bigr). \end{aligned}$$

The result is sharp for the function \(f(z)\) given by

$$\begin{aligned} f(z) =& z+ \frac{2(1-\alpha)}{k-1} \biggl(\frac{a+k}{a+1} \biggr)^{s} \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr) \\ &{} \cdot\frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda _{j}+1)_{k-1}}\prod_{j=2}^{k-1} \biggl(1+\frac{2(1-\alpha)}{j-1} \biggr)z^{k} \quad \bigl(k \in\mathbb{N} \setminus\{1\}\bigr). \end{aligned}$$
(2.9)

 □

3 Distortion inequalities for the function class \(\mathcal{S}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda,*}(\alpha)\)

In this section, we establish distortion inequalities for functions belonging to the class \(\mathcal{S}_{(\lambda_{p}),(\mu _{q}),b}^{s,a,\lambda,*}(\alpha)\). These inequalities are given in the following theorem.

Theorem 3

Let \(f(z) \in\mathcal{S}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda ,*}(\alpha)\) and \(0\leqq\alpha<1\). Then

$$\begin{aligned}& r-2(1-\alpha)r^{2}\sum_{k=2}^{\infty} \frac{k!}{k-1}\biggl\vert \biggl(\frac {a+k}{a+1} \biggr)^{s} \biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr)\biggr\vert \\& \qquad {} \cdot\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}\biggr\vert \prod _{j=2}^{k-1} \biggl(1+\frac {2(1-\alpha)}{j-1} \biggr) \\& \quad \leqq\bigl\vert f(z)\bigr\vert \\& \quad \leqq r+2(1-\alpha)r^{2}\sum _{k=2}^{\infty}\frac{k!}{k-1}\biggl\vert \biggl(\frac{a+k}{a+1} \biggr)^{s}\biggr\vert \biggl\vert \biggl( \frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr)\biggr\vert \\& \qquad {} \cdot\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}\biggr\vert \prod _{j=2}^{k-1} \biggl(1+\frac {2(1-\alpha)}{j-1} \biggr)\quad \bigl(\vert z\vert =r< 1\bigr) \end{aligned}$$
(3.1)

and

$$\begin{aligned}& 1-2(1-\alpha)r\sum_{k=2}^{\infty} \frac{k\cdot k!}{k-1}\biggl\vert \biggl(\frac {a+k}{a+1} \biggr)^{s} \biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr)\biggr\vert \\& \qquad {} \cdot\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}\biggr\vert \prod _{j=2}^{k-1} \biggl(1+\frac {2(1-\alpha)}{j-1} \biggr) \\& \quad \leqq\bigl\vert f'(z)\bigr\vert \\& \quad \leqq1+2(1-\alpha)r\sum _{k=2}^{\infty}\frac{k\cdot k!}{k-1} \biggl\vert \biggl(\frac{a+k}{a+1} \biggr)^{s}\biggr\vert \biggl\vert \biggl( \frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr)\biggr\vert \\& \qquad {} \cdot\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}\biggr\vert \prod _{j=2}^{k-1} \biggl(1+\frac {2(1-\alpha)}{j-1} \biggr) \quad \bigl(\vert z\vert =r< 1\bigr). \end{aligned}$$
(3.2)

Proof

Let \(f(z) \in\mathcal{A}\) be given by (1.1). Then, making use of Theorem 2, we find

$$\begin{aligned} \bigl\vert f(z)\bigr\vert \leqq&|z|+\sum _{k=2}^{\infty}|a_{k}|\cdot\bigl\vert z^{k}\bigr\vert \\ \leqq& r+2(1-\alpha)r^{2}\sum_{k=2}^{\infty} \frac{k!}{k-1}\biggl\vert \biggl(\frac{a+k}{a+1} \biggr)^{s} \biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr)\biggr\vert \\ &{} \cdot\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}\biggr\vert \prod _{j=2}^{k-1} \biggl(1+\frac {2(1-\alpha)}{j-1} \biggr) \quad \bigl( \vert z\vert =r< 1\bigr) \end{aligned}$$
(3.3)

and

$$\begin{aligned} \bigl\vert f(z)\bigr\vert \geqq&|z|-\sum _{k=2}^{\infty}|a_{k}|\cdot\bigl\vert z^{k}\bigr\vert \\ \geqq& r-2(1-\alpha)r^{2}\sum_{k=2}^{\infty} \frac{k!}{k-1}\biggl\vert \biggl(\frac{a+k}{a+1} \biggr)^{s} \biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr)\biggr\vert \\ &{} \cdot\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}\biggr\vert \prod _{j=2}^{k-1} \biggl(1+\frac {2(1-\alpha)}{j-1} \biggr)\quad \bigl(\vert z\vert =r< 1\bigr). \end{aligned}$$
(3.4)

From (1.1), we also have that

$$\begin{aligned} \bigl\vert f'(z)\bigr\vert \leqq&1+\sum _{k=2}^{\infty}k\cdot|a_{k}|\cdot\bigl\vert z^{k-1}\bigr\vert \\ \leqq&1+2(1-\alpha)r\sum_{k=2}^{\infty} \frac{k\cdot k!}{k-1}\biggl\vert \biggl(\frac{a+k}{a+1} \biggr)^{s} \biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr)\biggr\vert \\ &{} \cdot\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}\biggr\vert \prod _{j=2}^{k-1} \biggl(1+\frac {2(1-\alpha)}{j-1} \biggr) \quad \bigl(\vert z\vert =r< 1\bigr) \end{aligned}$$
(3.5)

and

$$\begin{aligned} \bigl\vert f'(z)\bigr\vert \geqq&1-\sum _{k=2}^{\infty}k\cdot|a_{k}|\cdot\bigl\vert z^{k}\bigr\vert \\ \geqq&1-2(1-\alpha)r\sum_{k=2}^{\infty} \frac{k\cdot k!}{k-1}\biggl\vert \biggl(\frac{a+k}{a+1} \biggr)^{s} \biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr)\biggr\vert \\ &{} \cdot\biggl\vert \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}\biggr\vert \prod _{j=2}^{k-1} \biggl(1+\frac {2(1-\alpha)}{j-1} \biggr)\quad \bigl(\vert z\vert =r< 1\bigr). \end{aligned}$$
(3.6)

We thus obtain the results (3.1) and (3.2) asserted by Theorem 3. □

4 Extreme points

This section is devoted to presenting the extreme points of the function class \(\mathcal{S}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda ,*}(\alpha)\). Let \(\mathcal{\widetilde{S}}_{(\lambda_{p}),(\mu _{q}),b}^{s,a,\lambda,*}(\alpha)\) be the subclass of \(\mathcal {S}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda,*}(\alpha)\) that consists in all functions \(f(z) \in\mathcal{A}\), which satisfy inequality (2.1). Then the extreme points of \(\mathcal{\widetilde{S}}_{(\lambda _{p}),(\mu_{q}),b}^{s,a,\lambda,*}(\alpha)\) are given by the following theorem.

Theorem 4

Let

$$ f_{1}(z):=z $$
(4.1)

and

$$\begin{aligned}& f_{k}(z):=z+\frac{k! (1-\alpha)}{(k-\alpha)}\biggl\vert \frac{\prod_{j=1}^{p}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\lambda _{j}+1)_{k-1}}\biggr\vert \biggl\vert \biggl(\frac{a+k}{a+1} \biggr)^{s}\biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )} \biggr) \biggr\vert z^{k} \\& \quad \bigl(k \in\mathbb{N}\setminus\{1\}\bigr). \end{aligned}$$
(4.2)

Then

$$f(z)\in\mathcal{\widetilde{S}}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda ,*}(\alpha)\quad (0 \leqq\alpha< 1) $$

if and only if it can be expressed in the following form:

$$ f(z)=\sum_{k=1}^{\infty} \gamma_{k} f_{k}(z) \quad \Biggl(\gamma_{k}>0; \sum _{k=1}^{\infty}\gamma_{k}=1\Biggr). $$
(4.3)

Proof

Suppose that

$$\begin{aligned} f(z) =&\sum_{k=1}^{\infty} \gamma_{k} f_{k}(z) \\ =&z+\sum_{k=2}^{\infty}\gamma_{k} \frac{k! (1-\alpha)}{(k-\alpha)} \biggl\vert \frac{\prod_{j=1}^{p}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\lambda _{j}+1)_{k-1}}\biggr\vert \\ &{}\cdot\biggl\vert \biggl(\frac{a+k}{a+1} \biggr)^{s}\biggr\vert \biggl\vert \frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )}\biggr\vert z^{k}. \end{aligned}$$
(4.4)

Then

$$\begin{aligned}& \sum_{k=2}^{\infty} \frac{(k-\alpha)}{k!}\biggl\vert \frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu _{j}+1)_{k-1}}\biggr\vert \biggl\vert \biggl(\frac{a+1}{a+k} \biggr)^{s}\biggr\vert \biggl\vert \biggl( \frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr)\biggr\vert \\& \qquad {} \cdot\gamma_{k} \frac{k! (1-\alpha)}{(k-\alpha)}\biggl\vert \frac{\prod_{j=1}^{p}(\mu_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\lambda _{j}+1)_{k-1}}\biggr\vert \biggl\vert \biggl(\frac{a+k}{a+1} \biggr)^{s}\biggr\vert \biggl\vert \frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+k,b,s,\lambda )}\biggr\vert \\& \quad =(1-\alpha)\sum_{k=2}^{\infty} \gamma_{k}=(1-\alpha) (1-\gamma _{1}) \\& \quad \leqq1-\alpha. \end{aligned}$$
(4.5)

Thus, by the definition of the function class \(\mathcal{\widetilde {S}}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda,*}(\alpha)\), we have

$$f \in\mathcal{\widetilde{S}}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda ,*}(\alpha)\quad (0\leqq \alpha< 1). $$

Conversely, if

$$f \in\mathcal{\widetilde{S}}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda ,*}(\alpha)\quad (0\leqq \alpha< 1), $$

then, by using (2.1), we may set

$$\begin{aligned}& \gamma_{k}=\frac{(k-\alpha)}{(1-\alpha) k!}\biggl\vert \frac{\prod_{j=1}^{p}(\lambda_{j}+1)_{k-1}}{\prod_{j=1}^{q}(\mu _{j}+1)_{k-1}}\biggr\vert \biggl\vert \biggl(\frac{a+1}{a+k} \biggr)^{s}\biggr\vert \biggl\vert \biggl(\frac{\Lambda (a+k,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr) \biggr\vert |a_{k}| \\& \quad \bigl(k \in\mathbb{N}\setminus\{1\}\bigr), \end{aligned}$$
(4.6)

which implies that

$$ f(z)=\sum_{k=1}^{\infty}\gamma_{k} f_{k}. $$

The proof of Theorem 4 is thus completed. □

5 The Fekete-Szegö problem

In this section, we shall obtain the Fekete-Szegö inequality for functions in the class \(\mathcal{S}_{(\lambda_{p}),(\mu _{q}),b}^{s,a,\lambda,*}(\alpha)\) when

$$s>0,\qquad a>0\quad \text{and}\quad 0\leqq\alpha< 1. $$

We need to recall an important lemma due to Ma and Minda [30] in order to prove our result involving Fekete-Szegö inequality.

Lemma 1

If

$$p(z)=1+c_{1}z+c_{2}z^{2}+\cdots $$

is an analytic function in \(\mathbb{U}\) such that

$$\Re\bigl(p(z)\bigr)>0 \quad (z \in\mathbb{U}), $$

then

$$ \bigl\vert c_{2}-\nu c_{1}^{2}\bigr\vert \leqq \begin{cases} -4\nu+2 & (\nu\leqq0), \\ 2 & (0\leqq\nu\leqq1), \\ 4\nu-2 & (\nu\geqq1). \end{cases} $$

When \(\nu<0\) or \(\nu>1\), the equality holds true if and only if

$$ p(z)=\frac{1+z}{1-z} $$
(5.1)

or one of its rotations. If \(0<\nu<1\), then the equality holds true if and only if

$$ p(z)=\frac{1+z^{2}}{1-z^{2}} $$
(5.2)

or one of its rotations. If \(\nu=0\), then the equality holds true if and only if

$$ p(z)= \biggl(\frac{1+\omega}{2} \biggr) \biggl(\frac{1+z}{1-z} \biggr)+ \biggl(\frac{1-\omega}{2} \biggr) \biggl(\frac{1-z}{1+z} \biggr)\quad (0\leqq\omega \leqq1) $$
(5.3)

or one of its rotations. If \(\nu=1\), then the equality holds true if and only if \(p(z)\) is the reciprocal of one of the functions such that the equality holds true in the case \(\nu=0\).

Theorem 5

Let

$$s>0, \qquad a>0,\qquad 0\leqq\alpha< 1 $$

and

$$\lambda_{j} >-1 \quad (j=1,\ldots,p),\qquad \mu_{j} >-1 \quad (j=1,\ldots,q). $$

If \(f \in\mathcal{S}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda,*}(\alpha )\), then

$$ \bigl\vert a_{3}-\tau a_{2}^{2}\bigr\vert \leqq \begin{cases} 3(1-\alpha) \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{2}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{2}} \\ \quad {}\cdot (\frac{a+3}{a+1} )^{s} (\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+3,b,s,\lambda )} )(-4\nu+2) & (\tau\leqq\sigma_{1}), \\ 6(1-\alpha) \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{2}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{2}} \\ \quad {}\cdot (\frac{a+3}{a+1} )^{s} (\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+3,b,s,\lambda )} ) & (\sigma_{1}\leqq\tau\leqq\sigma_{2}), \\ 3(1-\alpha) \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{2}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{2}} \\ \quad {}\cdot (\frac{a+3}{a+1} )^{s} (\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+3,b,s,\lambda )} )(4\nu-2) & (\tau\geqq\sigma_{2}), \end{cases} $$

where

$$\begin{aligned}& \nu: =(1-\alpha) \Biggl(\frac{4 \tau}{3}\prod _{j=1}^{p} \biggl(\frac {\lambda_{j}+2}{\lambda_{j}+1} \biggr)\prod _{j=1}^{q} \biggl(\frac{\mu _{j}+1}{\mu_{j}+2} \biggr) \biggl(\frac{a+2}{a+3} \biggr)^{s} \biggl(\frac {a+2}{a+1} \biggr)^{s} \\& \hphantom{\nu: =}{} \cdot \biggl(\frac{\Lambda (a+3,b,s,\lambda )}{\Lambda (a+2,b,s,\lambda )} \biggr) \biggl( \frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+2,b,s,\lambda )} \biggr)-1\Biggr), \end{aligned}$$
(5.4)
$$\begin{aligned}& \sigma_{1} = \frac{3}{4}\prod _{j=1}^{p} \biggl(\frac{\lambda _{j}+1}{\lambda_{j}+2} \biggr)\prod _{j=1}^{q} \biggl(\frac{\mu_{j}+2}{\mu _{j}+1} \biggr) \biggl(\frac{a+1}{a+2} \biggr)^{s} \\& \hphantom{\sigma_{1} =}{} \cdot \biggl(\frac{\Lambda (a+2,b,s,\lambda )}{\Lambda (a+3,b,s,\lambda )} \biggr) \biggl( \frac{\Lambda (a+2,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr) \end{aligned}$$
(5.5)

and

$$\begin{aligned} \sigma_{2} =& \frac{3(2-\alpha)}{4(1-\alpha)}\prod _{j=1}^{p} \biggl(\frac {\lambda_{j}+1}{\lambda_{j}+2} \biggr)\prod _{j=1}^{q} \biggl(\frac{\mu _{j}+2}{\mu_{j}+1} \biggr) \biggl(\frac{a+1}{a+2} \biggr)^{s} \\ &{} \cdot \biggl(\frac{\Lambda (a+2,b,s,\lambda )}{\Lambda (a+3,b,s,\lambda )} \biggr) \biggl(\frac{\Lambda (a+2,b,s,\lambda )}{\Lambda (a+1,b,s,\lambda )} \biggr). \end{aligned}$$
(5.6)

The result is sharp.

Proof

For \(f \in\mathcal{S}_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda ,*}(\alpha)\), let

$$ p(z)=\frac{ {\frac{z (J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda } )'(f)(z)}{J_{(\lambda_{p}),(\mu_{q}),b}^{s,a,\lambda }(f)(z)}}-\alpha}{1-\alpha}=1+c_{1}z+c_{2}z^{2}+ \cdots. $$
(5.7)

Then, with the help of (2.5), we have

$$ a_{2}=2(1-\alpha)c_{1} \frac{\prod_{j=1}^{q}(\mu_{j}+1)}{\prod_{j=1}^{p}(\lambda_{j}+1)} \biggl(\frac{a+2}{a+1} \biggr)^{s} \biggl(\frac {\Lambda (a+1,b,s,\lambda )}{\Lambda (a+2,b,s,\lambda )} \biggr) $$
(5.8)

and

$$ a_{3}=3(1-\alpha) \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{2}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{2}} \biggl( \frac{a+3}{a+1} \biggr)^{s} \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+3,b,s,\lambda )} \biggr) \bigl[{c_{2}}+(1-\alpha)c_{1}^{2} \bigr]. $$
(5.9)

Therefore, we find

$$\begin{aligned} a_{3}-\tau a_{2}^{2} =&3(1- \alpha) \frac{\prod_{j=1}^{q}(\mu _{j}+1)_{2}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{2}} \biggl(\frac {a+3}{a+1} \biggr)^{s} \biggl( \frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+3,b,s,\lambda )} \biggr) \bigl[{c_{2}}+(1-\alpha )c_{1}^{2} \bigr] \\ &{} -4\tau(1-\alpha)^{2}c_{1}^{2} \frac{\prod_{j=1}^{q}(\mu _{j}+1)^{2}}{\prod_{j=1}^{p}(\lambda_{j}+1)^{2}} \biggl(\frac {a+2}{a+1} \biggr)^{2s} \biggl( \frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+2,b,s,\lambda )} \biggr)^{2} \\ =&3(1-\alpha) \frac{\prod_{j=1}^{q}(\mu_{j}+1)_{2}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{2}} \biggl(\frac{a+3}{a+1} \biggr)^{s} \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+3,b,s,\lambda )} \biggr)\Biggl[c_{2}-c_{1}^{2}(1- \alpha) \\ &{} \cdot\Biggl(\frac{4 \tau}{3}\prod_{j=1}^{p} \biggl(\frac{\lambda _{j}+2}{\lambda_{j}+1} \biggr)\prod_{j=1}^{q} \biggl(\frac{\mu_{j}+1}{\mu _{j}+2} \biggr) \biggl(\frac{a+2}{a+3} \biggr)^{s} \biggl(\frac {a+2}{a+1} \biggr)^{s} \\ &{} \cdot \biggl(\frac{\Lambda (a+3,b,s,\lambda )}{\Lambda (a+2,b,s,\lambda )} \biggr) \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+2,b,s,\lambda )} \biggr)-1 \Biggr)\Biggr]. \end{aligned}$$
(5.10)

We thus write

$$ a_{3}-\tau a_{2}^{2}=3(1-\alpha) \frac{\prod_{j=1}^{q}(\mu _{j}+1)_{2}}{\prod_{j=1}^{p}(\lambda_{j}+1)_{2}} \biggl(\frac {a+3}{a+1} \biggr)^{s} \biggl( \frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+3,b,s,\lambda )} \biggr) \bigl(c_{2}-\nu c_{1}^{2} \bigr), $$
(5.11)

where

$$\begin{aligned} \nu: =&(1-\alpha) \Biggl(\frac{4 \tau}{3}\prod _{j=1}^{p} \biggl(\frac {\lambda_{j}+2}{\lambda_{j}+1} \biggr)\prod _{j=1}^{q} \biggl(\frac{\mu _{j}+1}{\mu_{j}+2} \biggr) \biggl(\frac{a+2}{a+3} \biggr)^{s} \biggl(\frac {a+2}{a+1} \biggr)^{s} \\ &{} \cdot \biggl(\frac{\Lambda (a+3,b,s,\lambda )}{\Lambda (a+2,b,s,\lambda )} \biggr) \biggl(\frac{\Lambda (a+1,b,s,\lambda )}{\Lambda (a+2,b,s,\lambda )} \biggr)-1 \Biggr). \end{aligned}$$
(5.12)

The result asserted by Theorem 5 follows by applying Lemma 1.

Moreover, if \(\tau<\sigma_{1}\) or \(\tau>\sigma_{2}\), then the equality holds true if and only if

$$ J_{(\lambda_{p}),(\tau_{q}),b}^{s,a,\lambda}(f) (z)=\frac{z}{ (1-\mathrm{e}^{\mathrm{i}\theta}z )^{2(1-\alpha)}}\quad ( \theta\in\mathbb {R}). $$
(5.13)

For \(\sigma_{1}<\tau<\sigma_{2}\), the equality holds true if and only if

$$ J_{(\lambda_{p}),(\tau_{q}),b}^{s,a,\lambda}(f) (z)=\frac{z}{ (1-\mathrm{e}^{\mathrm{i}\theta}z )^{1-\alpha}}\quad ( \theta\in\mathbb {R}). $$
(5.14)

If \(\tau=\sigma_{1}\), then the equality holds true if and only if

$$\begin{aligned} J_{(\lambda_{p}),(\tau_{q}),b}^{s,a,\lambda}(f) (z)&= \biggl(\frac {z}{ (1-\mathrm{e}^{\mathrm{i}\theta}z )^{2(1-\alpha)}} \biggr)^{[(1+\omega )/2]} \biggl(\frac{z}{ (1+\mathrm{e}^{\mathrm{i}\theta}z )^{2(1-\alpha )}} \biggr)^{[(1-\omega)/2]} \\ &=\frac{z}{ [ (1-\mathrm{e}^{\mathrm{i}\theta}z )^{1+\omega} (1+\mathrm{e}^{\mathrm{i}\theta}z )^{1-\omega} ]^{1-\alpha}}\quad (\theta\in\mathbb{R}; 0\leqq\omega\leqq1). \end{aligned}$$
(5.15)

Finally, when \(\tau=\sigma_{2}\), the equality holds true if and only if \(J_{(\lambda_{p}),(\tau_{q}),b}^{s,a,\lambda}(f)(z)\) satisfies the following condition:

$$ \frac{z (J_{(\lambda_{p}),(\tau_{q}),b}^{s,a,\lambda} )'(f)(z)}{J_{(\lambda_{p}),(\tau_{q}),b}^{s,a,\lambda}(f)(z)}=(1-\alpha )p(z)+\alpha, $$
(5.16)

where

$$ \frac{1}{p(z)}= \biggl(\frac{1+\omega}{2} \biggr) \biggl( \frac {1+z}{1-z} \biggr)+ \biggl(\frac{1-\omega}{2} \biggr) \biggl( \frac {1-z}{1+z} \biggr)\quad (0< \omega<1). $$
(5.17)

 □

We conclude this paper by mentioning that, by suitably specializing the parameters involved, our main results (Theorem 1 to Theorem 5) would yield a number of (known or new) results for much simpler function classes, which were investigated in several earlier works by employing many special cases of the new generalized Srivastava-Attiya operator.

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Srivastava, H.M., Gaboury, S. A new class of analytic functions defined by means of a generalization of the Srivastava-Attiya operator. J Inequal Appl 2015, 39 (2015). https://doi.org/10.1186/s13660-015-0573-z

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