Fekete-Szegő inequalities for certain class of analytic functions connected with q-analogue of Bessel function

Abstract

In this paper, we obtain Fekete-Szegő inequalities for a certain class of analytic functions f satisfying $1+\frac {1}{\zeta }\left [\frac {z\left (\mathcal {N}_{\nu,q}^{\lambda }f(z)\right)^{\prime }} {(1-\gamma)\mathcal {N} _{\nu,q}^{\lambda }f(z)+\gamma z\left (\mathcal {N}_{\nu,q}^{\lambda }f(z) \right)^{\prime }}-1\right ]\prec \Psi (z)$. Application of our results to certain functions defined by convolution products with a normalized analytic function is given, and in particular, Fekete-Szegő inequalities for certain subclasses of functions defined through Poisson distribution are obtained.

Introduction

Let $\mathcal {A}$ denote the class of analytic functions of the form:

$$ f(z)=z+\sum\limits_{k=2}^{\infty}a_{k}z^{k},\;z\in\mathbb{D}:=\{z\in\mathbb{C }:|z|<1\}, $$

(1)

and $\mathcal {S}$ be the subclass of $\mathcal {A}$ which are univalent functions in $\mathbb {D}$.

If $k\in \mathcal {A}$ is given by:

$$ k(z)=z+\sum\limits_{k=2}^{\infty}b_{k}z^{k},\;z\in\mathbb{D}, $$

(2)

then, the Hadamard (or convolution) product of f and k is defined by:

$$ (f\times k)(z):=z+\sum\limits_{k=2}^{\infty}a_{k}b_{k}z^{k},\;z\in\mathbb{D}. $$

(3)

If f and F are analytic functions in $\mathbb {D}$, we say that fis subordinate toF, written f≺F, if there exists a Schwarz functionw, which is analytic in $\mathbb {D}$, with w(0)=0, and |w(z)|<1 for all $z\in \mathbb {D}$, such that $f(z)=F(w(z)), z\in \mathbb {D}$. Furthermore, if the function F is univalent in $\mathbb {D}$, then we have the following equivalence (see [1] and [2]):

$$f(z)\prec F(z)\Leftrightarrow f(0)=F(0)\;\text{and}\;f(\mathbb{D})\subset F(\mathbb{D}). $$

The Bessel function of the first kind of order ν is defined by the infinite series:

$$J_{\nu}(z):=\sum\limits_{k\geq0}\frac{(-1)^{k}\left(\frac{z}{2} \right)^{2k+\nu}}{k!\Gamma\left(k+\nu+1\right)}, \;z\in\mathbb{C},\quad\left(\nu\in\mathbb{R}\right), $$

where Γ stands for the Gamma function. Recently, Szász and Kupán [3] investigated the univalence of the normalized Bessel function of the first kind $g_{\nu }:\mathbb {D}\rightarrow \mathbb {C}$ defined by (see also [4–6])

$$\begin{array}{*{20}l} g_{\nu}(z):=2^{\nu}\Gamma(\nu+1)z^{1-\frac{\nu}{2}}J_{\nu}(z^{\frac{1}{2}}) \\ =z+\sum\limits_{k=2}^{\infty}\frac{(-1)^{k-1}\Gamma(\nu+1)}{ 4^{k-1}(k-1)!\Gamma(k+\nu)} z^{k},\;z\in\mathbb{D},\quad\left(\nu\in\mathbb{R }\right). \end{array} $$

For 0<q<1, the q-derivative operator for g_ν is defined by:

$$\begin{array}{*{20}l} \partial_{q}g_{\nu}(z)= \partial_{q}\left[z+\sum\limits_{k=2}^{\infty}\frac{ (-1)^{k-1}\Gamma(\nu+1)}{4^{k-1}(k-1)! \Gamma(k+\nu)}z^{k}\right]:=\frac{ g_{\nu}(qz)-g_{\nu}(z)}{z(q-1)}= \notag \\ 1+\sum\limits_{k=2}^{\infty}\frac{(-1)^{k-1}\Gamma(\nu+1)}{4^{k-1}(k-1)! \Gamma(k+\nu)}[k,q]z^{k-1},\;z\in\mathbb{D}, \end{array} $$

where

$$ [k,q]:=\frac{1-q^{k}}{1-q}=1+\sum\limits_{j=1}^{k-1}q^{j},\qquad\left[0,q \right]:=0. $$

(4)

Using definition formula (4), we will define the next two products:

(i) For any non-negative integer k, the q-shifted factorial is given by:

$$[k,q]!:=\left\{ \begin{array}{lll} 1, & \text{if} & k=0, \\ \left[1,q\right]\left[2,q\right]\left[3,q\right]\dots[k,q], & \text{if} & k\in\mathbb{N}. \end{array} \right. $$

(ii) For any positive number r, the q-generalized Pochhammer symbol is defined by:

$$\left[r,q\right]_{k}:=\left\{ \begin{array}{lll} 1, & \text{if} & k=0, \\ \left[r,q\right]\left[r+1,q\right]\dots\left[r+k-1,q\right], & \text{if} & k\in\mathbb{N}. \end{array} \right. $$

For ν>0,λ>−1, and 0<q<1, define the function $\mathcal {I} _{\nu,q}^{\lambda }:\mathbb {D}\rightarrow \mathbb {C}$ by:

$$\mathcal{I}_{\nu,q}^{\lambda}(z):= z+\sum\limits_{k=2}^{\infty}\frac{ (-1)^{k-1}\Gamma(\nu+1)}{4^{k-1}(k-1)!\Gamma(k+\nu)} \frac{[k,q]!}{ [\lambda+1,q]_{k-1}}z^{k},\;z\in\mathbb{D}. $$

Remark 1

A simple computation shows that:

$$\mathcal{I}_{\nu,q}^{\lambda}(z)\times\mathcal{M}_{q,\lambda+1}(z)=z\, \partial_{q}g_{\nu}(z),\;z\in\mathbb{D}, $$

where the function $\mathcal {M}_{q,\lambda +1}$ is given by:

$$\mathcal{M}_{q,\lambda+1}(z):=z+\sum\limits_{k=2}^{\infty}\frac{ [\lambda+1,q]_{k-1}}{[k-1,q]!}z^{k},\;z\in\mathbb{D}. $$

Using the definition of q-derivative along with the idea of convolutions, we introduce the linear operator $\mathcal {N}_{\nu,q}^{\lambda }:\mathcal {A} \rightarrow \mathcal {A}$ defined by:

$$\begin{array}{*{20}l} \mathcal{N}_{\nu,q}^{\lambda}f(z):=\mathcal{I}_{\nu,q}^{\lambda}(z)\times f(z)= z+\sum\limits_{k=2}^{\infty}\psi_{k}a_{k}z^{k},\;z\in\mathbb{D}, \\ (\nu>0,\;\lambda>-1,\;0< q<1), \notag \end{array} $$

(5)

where

$$ \psi_{k}:=\frac{(-1)^{k-1}\Gamma(\nu+1)}{4^{k-1}(k-1)!\Gamma(k+\nu)}\cdot \frac{[k,q]!}{[\lambda+1,q]_{k-1}}. $$

(6)

Remark 2

From definition relation (5), we can easily verify that the next relations hold for all $f\in \mathcal {A}$:

(i) $[\lambda +1,q]\mathcal {N}_{\nu,q}^{\lambda }f(z)=[\lambda,q]\mathcal {N} _{\nu,q}^{\lambda +1}f(z) +q^{\lambda }z\partial _{q}\left (\mathcal {N} _{\nu,q}^{\lambda +1}f(z)\right), z\in \mathbb {D}$;

(ii) $\lim \limits _{q\to 1^{-}}\mathcal {N}_{\nu,q}^{\lambda }f(z)= \mathcal {I} _{\nu,1}^{\lambda }\times f(z)=:\mathcal {I}_{\nu }^{\lambda }f(z)=$

$z+\sum \limits _{k=2}^{\infty }\frac {k!}{(\lambda +1)_{k-1}}\frac {(-1)^{k-1} \Gamma (\nu +1)}{4^{k-1}(k-1)!\Gamma (k+\nu)}\,a_{k}z^{k}, z\in \mathbb {D}$.

Now, we define the class of functions $\mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)$ as follows:

Definition 1

Let $\Psi (z):=1+B_{1}z+B_{2}z^{2}+\dots, z\in \mathbb {D }$, with B₁>0, be a starlike (univalent) function with respect to 1, which maps the unit disk $\mathbb {D}$ onto a region included in the right half plane which is symmetric with respect to the real axis. For $\zeta \in \mathbb {C}^{\ast }$, and 0≤γ<1, the function $f\in \mathcal {A}$ is said to be in the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)$if the function

$$1+\frac{1}{\zeta}\left[\frac{z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z) \right)^{\prime}} {(1-\gamma)\mathcal{N}_{\nu,q}^{\lambda}f(z)+\gamma z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z)\right)^{\prime}}-1\right] $$

is analytic in $\mathbb {D}$ and satisfies:

$$\begin{array}{*{20}l} 1+\frac{1}{\zeta}\left[\frac{z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z) \right)^{\prime}} {(1-\gamma)\mathcal{N}_{\nu,q}^{\lambda}f(z)+\gamma z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z)\right)^{\prime}}-1\right] \prec\Psi(z) \\ \left(\nu>0,\;\lambda>-1,\;0< q<1,\;\zeta\in\mathbb{C}^{\ast},\;0\leq\gamma<1 \right). \end{array} $$

Putting q→1⁻, we obtain that $\lim \limits _{q\to 1^{-}}\mathcal {M} _{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)=:\mathcal {G}_{\nu }^{\lambda,\gamma }(\zeta ;\Psi)$, where

$$\begin{array}{*{20}l} \mathcal{G}_{\nu}^{\lambda,\gamma}(\zeta;\Psi):= \left\{1+\frac{1}{\zeta} \left[\frac{z\left(\mathcal{I}_{\nu}^{\lambda}f(z)\right)^{\prime}} { (1-\gamma)\mathcal{I}_{\nu}^{\lambda}f(z)+\gamma z\left(\mathcal{I} _{\nu}^{\lambda}f(z)\right)^{\prime}}-1\right] \prec\Psi(z)\right\} \\ \left(\nu>0,\;\lambda>-1,\;\zeta\in\mathbb{C}^{\ast},\;0\leq\gamma<1\right). \end{array} $$

In this paper, we obtain the Fekete-Szegő inequalities for the functions of the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)$. We give some application of our results to certain functions defined by convolution products with a normalized analytic function. In particular, Fekete-Szegő inequalities for certain subclasses of functions defined through Poisson distribution are obtained.

Fekete-Szegő problem

Denoted by $\mathcal {P}$, the well-known Carathéodory’s class of analytic functions in $\mathbb {D}$, normalized with P(0)=1, and having positive real part in $\mathbb {D}$, that is ReP(z)>0 for all $z\in \mathbb {D}$ (see [7]).

To prove our results, we need the following two lemmas.

Lemma 1

[8, Lemma 3] If $p(z)=1+c_{1}z+c_{2}z^{2}+\dots \in \mathcal {P}$, and α is a complex number, then

$$\max\left|c_{2}-\alpha c_{1}^{2}\right|=2\max\{1;\left|2\alpha-1\right|\}. $$

Lemma 2

[9, Lemma 1] If $p(z)=1+c_{1}z+c_{2}z^{2}+\dots \in \mathcal {P}$, then

$$\left|c_{2}-\alpha c_{1}^{2}\right|\leq\left\{ \begin{array}{lll} -4\alpha+2, & \text{if} & \alpha\leq 0, \\ 2, & \text{if} & 0\leq\alpha\leq 1, \\ 4\alpha-2, & \text{if} & \alpha\geq 1. \end{array} \right. $$

When α<0 or α>1, the equality holds if and only if $p(z)= \frac {1+z}{1-z}$ or one of its rotations.

If 0<α<1, then the equality holds if and only if $p(z)=\frac {1+z^{2} }{1-z^{2}}$ or one of its rotations.

If α=0, the equality holds if and only if:

$$p(z)=\left(\frac{1}{2}+\frac{\lambda}{2}\right)\frac{1+z}{1-z}+\left(\frac{1 }{2}-\frac{\lambda}{2}\right)\frac{1-z}{1+z}, \quad\text{with}\quad 0\leq\lambda\leq 1, $$

or one of its rotations.

If α=1, the equality holds if and only if:

$$\frac{1}{p(z)}=\left(\frac{1}{2}+\frac{\lambda}{2}\right)\frac{1+z}{1-z} +\left(\frac{1}{2}-\frac{\lambda}{2}\right)\frac{1-z}{1+z}, \quad\text{with} \quad 0\leq\lambda\leq 1. $$

Like it was mentioned in [9, pages 162–163], although the above upper bound is sharp, it can be improved as follows when 0<α<1:

$$ \left|c_{2}-\alpha c_{1}^{2}\right|+\alpha\left|c_{1}\right|^{2}\leq 2,\quad \text{if}\quad 0<\alpha\leq\frac{1}{2}, $$

(7)

and

$$ \left|c_{2}-\alpha c_{1}^{2}\right|+(1-\alpha)\left|c_{1}\right|^{2}\leq 2,\quad\text{if}\quad\frac{1}{2}\leq\alpha<1. $$

(8)

Theorem 1

If the function f given by (1) belongs to the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)$, with Ψ(z)=1+B₁z+B₂z²+… satisfying the conditions of Definition 1, and μ is a complex number, then:

$$\left|a_{3}-\mu a_{2}^{2}\right|\leq\frac{|\zeta|B_{1}}{2(1-\gamma)\psi_{3}} \cdot \max\left\{1;\left|\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{ 1-\gamma} -\frac{2\mu\zeta B_{1}\psi_{3}}{(1-\gamma)\psi_{2}^{2}} \right|\right\}, $$

where ψ_k,k∈{2,3}, are given by (6).

Proof

If $f\in \mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)$, then there exists a Schwarz function w, that is w is analytic in $\mathbb {D}$, with w(0)=0 and $\left |w(z)\right |<1, z\in \mathbb {D}$, such that:

$$ 1+\frac{1}{\zeta}\left[\frac{z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z) \right)^{\prime}} {(1-\gamma)\mathcal{N}_{\nu,q}^{\lambda}f(z)+\gamma z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z)\right)^{\prime}}-1\right]=\Psi(w(z)),\;z\in \mathbb{D}. $$

(9)

Since w is a Schwarz function, it follows that the function p₁ defined by:

$$ p_{1}(z):=\frac{1+w(z)}{1-w(z)}=1+c_{1}z+c_{2}z^{2}+\dots,\;z\in\mathbb{D}, $$

(10)

belongs to $\mathcal {P}$. Defining the function p by:

$$ p(z):=1+\frac{1}{\zeta}\left[\frac{z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z) \right)^{\prime}} {(1-\gamma)\mathcal{N}_{\nu,q}^{\lambda}f(z)+\gamma z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z)\right)^{\prime}}-1\right]= 1+d_{1}z+d_{2}z^{2}+\dots,\;z\in\mathbb{D}, $$

(11)

in view of (9) and (10), we have:

$$ p(z)=\Psi\left(\frac{p_{1}(z)-1}{p_{1}(z)+1}\right),\;z\in\mathbb{D}. $$

(12)

From (10), we easily get:

$$\frac{p_{1}(z)-1}{p_{1}(z)+1}=\frac{1}{2}\left[c_{1}z+\left(c_{2}-\frac{ c_{1}^{2}}{2}\right)z^{2}+\left(c_{3}+\frac{c_{1}^{3}}{4} -c_{1}c_{2} \right)z^{3}+\dots\right],\;z\in\mathbb{D}; $$

therefore,

$$\Psi\left(\frac{p_{1}(z)-1}{p_{1}(z)+1}\right)=1+\frac{1}{2}B_{1}c_{1}z + \left[\frac{1}{2}B_{1}\left(c_{2}-\frac{c_{1}^{2}}{2}\right)+\frac{1}{4} B_{2}c_{1}^{2}\right]z^{2}+\dots,\;z\in\mathbb{D}, $$

and from (12), we obtain:

$$ d_{1}=\frac{1}{2}B_{1}c_{1}\qquad\text{and}\qquad d_{2}=\frac{1}{2} B_{1}\left(c_{2}-\frac{c_{1}^{2}}{2}\right)+\frac{1}{4}B_{2}c_{1}^{2}. $$

(13)

On the other hand, from (11), according to (5), it follows that

$$ d_{1}=\frac{(1-\gamma)a_{2}\psi_{2}}{\zeta}\qquad\text{and}\qquad d_{2}= \frac{2(1-\gamma)a_{3}\psi_{3}}{\zeta}-\frac{(1-\gamma)(1+\gamma)a_{2}^{2} \psi_{2}^{2}}{\zeta}, $$

(14)

and combining (13) with (14), we have:

$$ a_{2}=\frac{\zeta B_{1}c_{1}}{2(1-\gamma)\psi_{2}}, $$

(15)

and

$$a_{3}=\frac{\zeta B_{1}}{4(1-\gamma)\psi_{3}}\left[c_{2}-\frac{c_{1}^{2}}{2} +\frac{1}{2}\frac{B_{2}}{B_{1}}c_{1}^{2}+\frac{\zeta B_{1}(1+\gamma) c_{1}^{2}}{2(1-\gamma)}\right]. $$

Therefore,

$$ a_{3}-\mu a_{2}^{2}=\frac{\zeta B_{1}}{4(1-\gamma)\psi_{3}} \left(c_{2}-\alpha c_{1}^{2}\right), $$

(16)

where

$$ \alpha=\frac{1}{2}\left[1-\frac{B_{2}}{B_{1}}-\frac{\zeta B_{1}(1+\gamma)}{ 1-\gamma}+\frac{2\mu\zeta B_{1}\psi_{3}} {(1-\gamma)\psi_{2}^{2}}\right], $$

(17)

and from Lemma 1, our result follows immediately. □

Putting q→1⁻ in Theorem 1, we obtain the next corollary:

Corollary 1

If the function f given by (1) belongs to the class $\mathcal {G}_{\nu }^{\lambda,\gamma }(\zeta ;\Psi)$, with Ψ(z)=1+B₁z+B₂z²+… satisfying the conditions of Definition 1, and μ is a complex number, then

$$\begin{array}{*{20}l} \left|a_{3}-\mu a_{2}^{2}\right|\leq \\ \frac{8|\zeta|B_{1}(\lambda+1)_{2}(\nu+1)_{2}}{3(1-\gamma)}\cdot \max\left\{1;\left|\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{1-\gamma} -\frac{3\mu\zeta B_{1}(\lambda+1)(\nu+1)}{2(1-\gamma) (\lambda+2)(\nu+2)} \right|\right\}. \end{array} $$

Using a similar proof like for Theorem 1 combined with Lemma 2, we can obtain the following theorem:

Theorem 2

If the function f given by (1) belongs to the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)$, with Ψ(z)=1+B₁z+B₂z²+… satisfying the conditions of Definition 1 and $\mu,B_{2}\in \mathbb {R}$, and ζ>0, then

$$\left|a_{3}-\mu a_{2}^{2}\right|\leq\left\{ \begin{array}{lll} \frac{\zeta B_{1}}{2(1-\gamma)\psi_{3}}\left[\frac{B_{2}}{B_{1}}+\frac{ \zeta B_{1}(1+\gamma)}{1-\gamma}- \frac{2\mu\zeta B_{1}\psi_{3}}{ (1-\gamma)\psi_{2}^{2}}\right], & \text{if} & \mu\leq\sigma_{1}, \\[1em] \frac{\zeta B_{1}}{2(1-\gamma)\psi_{3}}, & \text{if} & \sigma_{1}\leq\mu \leq\sigma_{2}, \\[1em] \frac{-\zeta B_{1}}{2(1-\gamma)\psi_{3}}\left[\frac{B_{2}}{B_{1}} +\frac{ \zeta B_{1}(1+\gamma)}{1-\gamma}-\frac{2\mu\zeta B_{1}\psi_{3}}{ (1-\gamma)\psi_{2}^{2}}\right], & \text{if} & \mu\geq\sigma_{2}, \end{array} \right. $$

with

$$ \sigma_{1}=\frac{(1-\gamma)\psi_{2}^{2}}{2\zeta B_{1}\psi_{3}}\left[-1+ \frac{B_{2}}{B_{1}} +\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}\right], $$

(18)

and

$$ \sigma_{2}=\frac{(1-\gamma)\psi_{2}^{2}}{2\zeta B_{1}\psi_{3}}\left[1+\frac{ B_{2}}{B_{1}} +\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}\right], $$

(19)

where ψ_k,k∈{2,3}, are given by (6).

Proof

With the same proof like those of Theorem 1, we obtain the equalities (16) and (17) hold.

(i) According to the first part of Lemma 2, we have:

$$\left|c_{2}-\alpha c_{1}^{2}\right|\leq-4\alpha+2,\;\text{if}\;\alpha\leq0. $$

Using (17), simple computation shows that the inequality α≤0 is equivalent to μ≤σ₁, and from (16) combined with the inequality $\left |c_{2}-\alpha c_{1}^{2}\right |\leq -4\alpha +2$, the first of our theorem is proved.

(ii) The second part of Lemma 2 shows that:

$$\left|c_{2}-\alpha c_{1}^{2}\right|\leq2,\;\text{if}\;0\leq\alpha\leq1. $$

From (17), it is easy to check that the inequality 0≤α≤1 is equivalent to σ₁≤μ≤σ₂. From the relation (16), the inequality $\left |c_{2}-\alpha c_{1}^{2}\right |\leq 2$ proves the second part of our result.

(iii) Finally, form the third part of Lemma 2, we have:

$$\left|c_{2}-\alpha c_{1}^{2}\right|\leq4\alpha-2,\;\text{if}\;\alpha\geq1. $$

The relation (17) shows immediately that α≥1 is equivalent to μ≥σ₂, while (16) combined with the inequality $\left |c_{2}-\alpha c_{1}^{2}\right |\leq 4\alpha -2$ proves the last part of our result. □

Taking q→1⁻ in Theorem 2, we get the next special case:

Corollary 2

If the function f given by (1) belongs to the class $\mathcal {G}_{\nu }^{\lambda,\gamma }(\zeta ;\Psi)$, with Ψ(z)=1+B₁z+B₂z²+… satisfying the conditions of Definition 1 and $\mu,B_{2}\in \mathbb {R}$, and ζ>0, then

$$\left|a_{3}-\mu a_{2}^{2}\right|\leq\left\{ \begin{array}{l} \frac{8\zeta B_{1}(\lambda+1)_{2}(\nu+1)_{2}} {3(1-\gamma)}\left[\frac{ B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{1-\gamma} -\frac{3\mu\zeta B_{1}(\lambda+1)(\nu+1)}{2(1-\gamma)(\lambda+2)(\nu+2)}\right], \\[1em] \hfill\text{if}\quad\mu\leq\eta_{1}, \\[1em] \frac{8\zeta B_{1}(\lambda+1)_{2}(\nu+1)_{2}} {3(1-\gamma)},\hfill\text{if} \quad \eta_{1}\leq\mu\leq\eta_{2}, \\[1em] \frac{-8\zeta B_{1}(\lambda+1)_{2}(\nu+1)_{2}} {3(1-\gamma)}\left[\frac{ B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{1-\gamma} -\frac{3\mu\zeta B_{1}(\lambda+1)(\nu+1)}{2(1-\gamma)(\lambda+2)(\nu+2)}\right], \\[1em] \hfill\text{if}\quad\mu\geq\eta_{2}, \end{array} \right. $$

with

$$ \eta_{1}=\frac{2(1-\gamma)(\lambda+2)(\nu+2)}{3\zeta B_{1}(\lambda+1)(\nu+1)} \left[-1+\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{(1-\gamma)} \right], $$

(20)

and

$$ \eta_{2}=\frac{2(1-\gamma)(\lambda+2)(\nu+2)}{3\zeta B_{1}(\lambda+1)(\nu+1)} \left[1+\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{(1-\gamma)}\right]. $$

(21)

With a similar proof like for Theorem 1 and using the inequalities (7) and (8), we obtained the next result.

Theorem 3

If the function f given by (1) belongs to the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)$, with Ψ(z)=1+B₁z+B₂z²+… satisfying the conditions of Definition 1 and $\mu,B_{2}\in \mathbb {R}$, and ζ>0, then the next inequalities hold:

(i) for σ₁<μ≤σ₃, we have

$$ \left|a_{3}-\mu a_{2}^{2}\right|+\frac{(1-\gamma)\psi_{2}^{2}} {2\zeta B_{1}\psi_{3}}\left[1-\frac{B_{2}}{B_{1}}-\frac{\zeta B_{1}(1+\gamma)}{ 1-\gamma}+\frac{2\mu\zeta B_{1}\psi_{3}} {(1-\gamma)\psi_{2}^{2}}\right] \left|a_{2}\right|^{2}\leq\frac{\zeta B_{1}}{2(1-\gamma)\psi_{3}}; $$

(22)

(ii) for σ₃≤μ≤σ₂, we have

$$ \left|a_{3}\!-\mu a_{2}^{2}\right|+\frac{(1-\gamma)\psi_{2}^{2}} {2\zeta B_{1}\psi_{3}}\left[1+\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{ 1-\gamma}-\frac{2\mu\zeta B_{1}\psi_{3}} {(1-\gamma)\psi_{2}^{2}}\right] \left|a_{2}\right|^{2}\leq\frac{\zeta B_{1}}{2(1-\gamma)\psi_{3}}, $$

(23)

where σ₁ and σ₂ are defined by (18) and (19), respectively,

$$\sigma_{3}=\frac{(1-\gamma)\psi_{2}^{2}}{2\zeta B_{1}\psi_{3}} \left[\frac{ B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{(1-\gamma)}\right], $$

and ψ_k,k∈{2,3}, are given by (6).

Proof

With the same computations like in the proof of Theorem 1, we obtain the relations (16) and (17), while (15) is equivalent to:

$$ c_{1}=\frac{2(1-\gamma)\psi_{2}}{\zeta B_{1}}. $$

(24)

(i) To prove the first part of our theorem, we will use the inequality (7). Thus, according to (16), (17), and the above relation, it is easy to check that (7) could be written in the equivalent form (22), while the assumption $0<\alpha \leq \frac {1 }{2}$ is equivalent to σ₁<μ≤σ₃.

(ii) For the proof of the second part of our result, we will use the inequality (8). From (16), (17), and (24), it follows that (8) could be written in the form (23), and the assumption $\frac {1}{2}\leq \alpha <1$ is equivalent to σ₃<μ≤σ₂. □

Putting q→1⁻ in Theorem 3, we obtain the following result:

Corollary 3

If the function f given by (1) belongs to the class $\mathcal {G}_{\nu }^{\lambda,\gamma }(\zeta ;\Psi)$, with Ψ(z)=1+B₁z+B₂z²+… satisfying the conditions of Definition 1 and $\mu,B_{2}\in \mathbb {R}$, and ζ>0, then the next inequalities hold:

(i) for η₁<μ≤η₃, we have

$$\begin{array}{*{20}l} \left|a_{3}-\mu a_{2}^{2}\right| \\ +\frac{2(1-\gamma)(\lambda+2)(\nu+2)} {3\zeta B_{1}(\lambda+1)(\nu+1)}\left[ 1-\frac{B_{2}}{B_{1}} -\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}+\frac{ 3\mu\zeta B_{1}(\lambda+1)(\nu+1)}{2(1-\gamma) (\lambda+2)(\nu+2)}\right] \left|a_{2}\right|^{2} \\ \leq\frac{8\zeta B_{1}(\lambda+1)_{2}(\nu+1)_{2}}{3(1-\gamma)}; \end{array} $$

(ii) for η₃≤μ≤η₂, we have

$$\begin{array}{*{20}l} \left|a_{3}-\mu a_{2}^{2}\right| \\ +\frac{2(1-\gamma)(\lambda+2)(\nu+2)}{3\zeta B_{1} (\lambda+1)(\nu+1)}\left[ 1+\frac{B_{2}}{B_{1}} +\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}-\frac{ 3\mu\zeta B_{1}(\lambda+1)(\nu+1)}{2(1-\gamma) (\lambda+2)(\nu+2)}\right] \left|a_{2}\right|^{2} \\ \leq\frac{8\zeta B_{1}(\lambda+1)_{2}(\nu+1)_{2}}{3(1-\gamma)}, \end{array} $$

where η₁ and η₂ are defined by (20) and (21), respectively, and

$$\eta_{3}=\frac{2(1-\gamma)(\lambda+2)(\nu+2)} {3\zeta B_{1}(\lambda+1)(\nu+1) }\left[\frac{B_{2}}{B_{1}} +\frac{\zeta B_{1}(1+\gamma)}{(1-\gamma)}\right]. $$

Applications to functions defined by poisson distribution

In [10], Porwal studied a power series whose coefficients are probabilities of the Poisson distribution, that is:

$$\mathrm{I}_{m}(z)=z+\sum\limits_{k=2}^{\infty}\frac{m^{k-1}}{(k-1)!} e^{-m}z^{k},\;z\in\mathbb{D},\quad(m>0), $$

and motivated by this investigation Srivastava and Porwal [11] introduced the linear operator $\mathcal {I}^{m}:\mathcal {A}\rightarrow \mathcal {A}$ defined by:

$$\mathcal{I}^{m}f(z):=\mathrm{I}_{m}(z)\times f(z)=z+\sum\limits_{k=2}^{\infty} \frac{m^{k-1}}{(k-1)!}e^{-m}a_{k}z^{k},\;z\in\mathbb{D}, $$

where $f\in \mathcal {A}$ has the form (1).

Definition 2

Let the function Ψ satisfying the conditions of Definition 1. For $\zeta \in \mathbb {C}^{\ast }, 0\leq \gamma <1$, and $k\in \mathcal {A}$, the function $f\in \mathcal {A}$ is said to be in the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }\left (\zeta ;k; \Psi \right)$ if $f\times k\in \mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)$, that is"

$$1+\frac{1}{\zeta}\left[\frac{z\left(\mathcal{N}_{\nu,q}^{\lambda}(f\times k)(z)\right)^{\prime}} {(1-\gamma)\mathcal{N}_{\nu,q}^{\lambda}(f\times k)(z) +\gamma z\left(\mathcal{N}_{\nu,q}^{\lambda}(f\times k)(z)\right)^{\prime}}-1 \right] $$

is analytic in $\mathbb {D}$ and satisfies

$$\begin{array}{*{20}l} 1+\frac{1}{\zeta}\left[\frac{z\left(\mathcal{N}_{\nu,q}^{\lambda}(f\times k)(z)\right)^{\prime}} {(1-\gamma)\mathcal{N}_{\nu,q}^{\lambda}(f\times k)(z) +\gamma z\left(\mathcal{N}_{\nu,q}^{\lambda}(f\times k)(z)\right)^{\prime}}-1 \right]\prec\Psi(z) \\ \left(\nu>0,\;\lambda>-1,\;0< q<1,\;\zeta\in\mathbb{C}^{\ast},\;0\leq\gamma<1 \right). \end{array} $$

A special case of the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }\left (\zeta ;k;\Psi \right)$ is obtained for k=I_m; hence, $f\in \mathcal { M}_{\nu,q}^{\lambda,\gamma }\left (\zeta ;\mathrm {I}_{m};\Psi \right)$ if and only if $\mathcal {I}^{m}f\in \mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)$.

Applying Theorems 1 and 2 for the function f×k given by (3), we get the following results, respectively:

Theorem 4

If the function f given by (1) belongs to the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }\left (\zeta ;k;\Psi \right)$, with $\Psi (z)=1+B_{1}z+B_{2}z^{2}+\dots, k\in \mathcal {A}$ is given by (2) with b₂b₃≠0, and μ is a complex number, then

$$\left|a_{3}-\mu a_{2}^{2}\right|\leq\frac{|\zeta|B_{1}}{2(1-\gamma)|b_{3}| \psi_{3}}\cdot \max\left\{1,\left|\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)} {1-\gamma}-\frac{2\mu\zeta B_{1}b_{3}\psi_{3}}{(1-\gamma) b_{2}^{2}\psi_{2}^{2}}\right|\right\}, $$

where ψ_k and k∈{2,3} are given by (6).

Theorem 5

If the function f given by (1) belongs to the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }\left (\zeta ;k;\Psi \right)$, with Ψ(z)=1+B₁z+B₂z²+… satisfying the conditions of Definition 1 and $\mu,B_{2}\in \mathbb {R}, k\in \mathcal {A}$ is given by (2) with b₂b₃≠0, and ζ>0, then

$$\left|a_{3}-\mu a_{2}^{2}\right|\leq\left\{ \begin{array}{lll} \frac{\zeta B_{1}}{2(1-\gamma)|b_{3}|\psi_{3}}\left[\frac{B_{2}}{B_{1}} + \frac{\zeta B_{1}(1+\gamma)}{(1-\gamma)} -\frac{2\mu\zeta B_{1}b_{3}\psi_{3}}{(1-\gamma)b_{2}^{2}\psi_{2}^{2}}\right], & \text{if} &\mu\leq\sigma_{1}, \\[1em] \frac{\zeta B_{1}}{2(1-\gamma)|b_{3}|\psi_{3}}, & \text{if} &\sigma_{1}\leq\mu\leq\sigma_{2}, \\[1em] \frac{-\zeta B_{1}}{2(1-\gamma)|b_{3}|\psi_{3}}\left[\frac{B_{2}}{B_{1}} + \frac{\zeta B_{1}(1+\gamma)}{(1-\gamma)} -\frac{2\mu\zeta B_{1}b_{3}\psi_{3}}{(1-\gamma)b_{2}^{2}\psi_{2}^{2}}\right], & \text{if} &\mu\geq\sigma_{2}, \end{array} \right. $$

with

$$\sigma_{1}=\frac{(1-\gamma) b_{2}^{2}\psi_{2}^{2}}{2\zeta B_{1}b_{3}\psi_{3}} \left[-1+\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}\right], $$

and

$$\sigma_{2}=\frac{(1-\gamma) b_{2}^{2}\psi_{2}^{2}}{2\zeta B_{1}b_{3}\psi_{3}} \left[1+\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}\right], $$

and ψ_k,k∈{2,3}, are given by (6).

For k:=I_m, we have

$$b_{2}=me^{-m}\quad\text{and}\quad b_{3}=\frac{m^{2}}{2}e^{-m}, $$

and for this special case from Theorems 4 and 5, we deduce to the following results, respectively:

Theorem 6

If the function f given by (1) belongs to the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }\left (\zeta ;\mathrm {I}_{m};\Psi \right)$, with Ψ(z)=1+B₁z+B₂z²+…, and μ is a complex number, then

$$\left|a_{3}-\mu a_{2}^{2}\right|\leq\frac{|\zeta|B_{1}}{(1- \gamma)m^{2}e^{-m}\psi_{3}}\cdot \max\left\{1;\left|\frac{B_{2}}{B_{1}}+ \frac{\zeta B_{1}(1+\gamma)}{1-\gamma} -\frac{\mu\zeta B_{1}\psi_{3}}{ (1-\gamma)e^{-m}\psi_{2}^{2}}\right|\right\}, $$

where ψ_k and k∈{2,3} are given by (6).

Theorem 7

If the function f given by (1) belongs to the class $\mathcal {M}_{\nu,q}^{\lambda,\gamma }\left (\zeta ;\mathrm {I}_{m};\Psi \right)$, with Ψ(z)=1+B₁z+B₂z²+… satisfying the conditions of Definition 1 and $\mu,B_{2}\in \mathbb {R}$, and ζ>0, then

$$\left|a_{3}-\mu a_{2}^{2}\right|\leq\left\{ \begin{array}{l} \frac{\zeta B_{1}}{(1-\gamma)m^{2}e^{-m}\psi_{3}} \left[\frac{B_{2}}{B_{1}} +\frac{\zeta B_{1}(1+\gamma)} {(1-\gamma)}-\frac{\mu\zeta B_{1}\psi_{3}}{ (1-\gamma) e^{-m}\psi_{2}^{2}}\right],\;\text{if}\quad\mu\leq\sigma_{1}^{ \ast}, \\[1em] \frac{\zeta B_{1}}{(1-\gamma)m^{2}e^{-m}\psi_{3}},\hfill\text{if} \quad\sigma_{1}^{\ast}\leq\mu\leq\sigma_{2}^{\ast}, \\[1em] \frac{-\zeta B_{1}}{(1-\gamma)m^{2}e^{-m}\psi_{3}}\left[\frac{B_{2}}{B_{1}} +\frac{\zeta B_{1}(1+\gamma)} {(1-\gamma)}-\frac{\mu\zeta B_{1}\psi_{3}}{ (1-\gamma)e^{-m}\psi_{2}^{2}}\right],\;\text{if}\quad\mu\geq\sigma_{2}^{ \ast}, \end{array} \right. $$

with

$$\sigma_{1}^{\ast}=\frac{(1-\gamma)e^{-m}\psi_{2}^{2}}{\zeta B_{1}\psi_{3}} \left[-1+\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}\right], $$

and

$$\sigma_{2}^{\ast}=\frac{(1-\gamma)e^{-m}\psi_{2}^{2}}{\zeta B_{1}\psi_{3}} \left[1+\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}\right], $$

where ψ_k and k∈{2,3} are given by (6).