Differential Sandwich-Type Results for Symmetric Functions Connected with a Q-Analog Integral Operator

Abstract

In this paper, we obtain some applications of the theory of differential subordination, differential superordination, and sandwich-type results for some subclasses of symmetric functions connected with a q-analog integral operator.

1. Introduction

The theory of q-analysis has an important role in many areas of mathematics and physics. Jackson [1,2] was the first that gave some application of q-calculus and introduced the q-analog of derivative and integral operator (see also [3]). Let $\mathcal{H}\left(\mathbb{U}\right)$ denote the class of analytic functions in the open unit disk $\mathbb{U}:=\{z\in \mathbb{C}:|z|<1\}$, and $\mathcal{H}[a,m]$ denote the subclass of functions $f\in \mathcal{H}\left(\mathbb{U}\right)$ of the form

$$f\left(z\right)=a+a_m{z}^m+a_{m+1}{z}^{m+1}+\dots ,z\in \mathbb{U},$$

with $a\in \mathbb{C}$ and $m\in \mathbb{N}:=\left\{1,2,\dots \right\}$.

In addition, let $\mathcal{A}\left(m\right)$ denote the subclass of functions $f\in \mathcal{H}\left(\mathbb{U}\right)$ of the form

$$f\left(z\right)=z+\sum _{k=m+1}^{\infty }{a}_k{z}^k,z\in \mathbb{U},$$

(1)

with $m\in \mathbb{N}$, and let $\mathcal{A}:=\mathcal{A}\left(1\right)$.

We define the integral operator ${\mathcal{K}}_{n,m}^{\alpha }:\mathcal{A}\left(m\right)\to \mathcal{A}\left(m\right)$, with $\alpha >0$ and $n\ge 0$, as follows:

$${\mathcal{K}}_{n,m}^{0}f:=f,$$

and

$${\mathcal{K}}_{n,m}^{\alpha }f\left(z\right):=\frac{{(n+1)}^{\alpha }}{\mathsf{\Gamma }\left(\alpha \right)z^n}\underset{0}{\overset{z}{\int }}{t}^{n-1}{\left(log\frac{z}{t}\right)}^{\alpha -1}f\left(t\right)dt,$$

where all the powers are the principal ones, and $log1=0$.

If $f\in \mathcal{A}\left(m\right)$ has the power expansion of the form in Equation (1), it can be easily verified that

$${\mathcal{K}}_{n,m}^{\alpha }f\left(z\right)=z+\sum _{k=m+1}^{\infty }{\left(\frac{n+1}{n+k}\right)}^{\alpha }{a}_k{z}^k,z\in \mathbb{U}.$$

For $0<q<1$, the q-derivative of the operator ${\mathcal{K}}_{n,m}^{\alpha }$ is defined by

$${\partial }_q{\mathcal{K}}_{n,m}^{\alpha }f\left(z\right):=\frac{{\mathcal{K}}_{n,m}^{\alpha }f\left(qz\right)-{\mathcal{K}}_{n,m}^{\alpha }f\left(z\right)}{z(q-1)},z\in \mathbb{U},$$

that is

$${\partial }_q\left[z+\sum _{k=m+1}^{\infty }{\left(\frac{n+1}{n+k}\right)}^{\alpha }{a}_k{z}^k\right]=1+\sum _{k=m+1}^{\infty }{\left(\frac{n+1}{n+k}\right)}^{\alpha }[k,q]a_k{z}^{k-1},z\in \mathbb{U},$$

(2)

where

$$[k,q]=\frac{1-q^k}{1-q}=1+\sum _{i=1}^{k-1}{q}^i,[0,q]=0,$$

It is easily to verify from Equation (2) that

$$z{\partial }_q{\mathcal{K}}_{n,m}^{\alpha }f\left(z\right)=z+\sum _{k=m+1}^{\infty }{\left(\frac{n+1}{n+k}\right)}^{\alpha }[k,q]a_k{z}^k,z\in \mathbb{U}.$$

For any non negative integer k, the q-number shift factorial is given by

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

while the q-generalized Pochhammer symbol for $r>0$ is defined by

$${\left[r,q\right]}_k=\left\{\begin{array}{ccc}1,\hfill & \mathrm{if}\hfill & k=0,\hfill \\ \left[r,q\right]\left[r+1,q\right]\dots \left[r+k-1,q\right],\hfill & \mathrm{if}\hfill & k\in \mathbb{N}.\hfill \end{array}\right.$$

For $\lambda >-1$, we define the operator ${\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }:\mathcal{A}\left(m\right)\to \mathcal{A}\left(m\right)$ by

$${\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)\ast {\mathcal{M}}_{q,\lambda +1}\left(z\right)=z{\partial }_q{\mathcal{K}}_{n,m}^{\alpha }f\left(z\right),$$

where

$${\mathcal{M}}_{q,\lambda +1}\left(z\right):=z+\sum _{k=m+1}^{\infty }\frac{{[\lambda +1,q]}_{k-1}}{\left[k-1,q\right]!}{z}^k,z\in \mathbb{U}.$$

From the above definition, we obtain

$$\begin{array}{c}{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)=z+\sum _{k=m+1}^{\infty }{\left(\frac{n+1}{n+k}\right)}^{\alpha }\frac{[k,q]\left[k-1,q\right]!}{{[\lambda +1,q]}_{k-1}}{a}_k{z}^k\\ =z+\sum _{k=m+1}^{\infty }\frac{[k,q]!}{{[\lambda +1,q]}_{k-1}}{\left(\frac{n+1}{n+k}\right)}^{\alpha }{a}_k{z}^k,z\in \mathbb{U},\\ (\alpha >0,\lambda >-1,m\ge 0,0<q<1)\end{array}$$

(3)

and from Equation (3) we can easily verify that

$$[\lambda +1,q]{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)=\left[\lambda ,q\right]{\mathcal{N}}_{n,m,q}^{\lambda +1,\alpha }f\left(z\right)+q^{\lambda }z{\partial }_q{\mathcal{N}}_{n,m,q}^{\lambda +1,\alpha }f\left(z\right),z\in \mathbb{U}.$$

We note that

$$\underset{q\to 1^{-}}{lim}{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)=:{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)=z+\sum _{k=m+1}^{\infty }\frac{k!}{{\left(\lambda +1\right)}_{k-1}}{\left(\frac{n+1}{n+k}\right)}^{\alpha }{a}_k{z}^k,z\in \mathbb{U}.$$

(4)

Definition 1.

For $f,g\in \mathcal{H}\left(\mathbb{U}\right)$, we say that f is subordinate to g, written $f\left(z\right)\prec g\left(z\right)$, if there exists a Schwarz function w, which is analytic in $\mathbb{U}$, with $w\left(0\right)=0$ and $\left|w\left(z\right)\right|<1$ for all $z\in \mathbb{U}$, such that $f\left(z\right)=g\left(w\right(z\left)\right)$, $z\in \mathbb{U}$. Furthermore, if the function g is univalent in $\mathbb{U}$, then we have the following equivalence (see [4,5]):

$$f\left(z\right)\prec g\left(z\right)\iff f\left(0\right)=g\left(0\right)andf\left(\mathbb{U}\right)\subset g\left(\mathbb{U}\right).$$

Let $k,h\in \mathcal{H}\left(\mathbb{U}\right)$, and let $\phi (r,s;z):{\mathbb{C}}^2\times \mathbb{U}\to \mathbb{C}$.

(i) If k satisfies the first order differential subordination

$$\phi (k\left(z\right),z{k}^{\prime }\left(z\right);z)\prec h\left(z\right),$$

(5)

then k is said to be a solution of the differential subordination in Equation (5). The function q is called a dominant of the solutions of the differential subordination in Equation (5) if $k\left(z\right)\prec q\left(z\right)$ for all the functions k satisfying Equation (5). A dominant $\tilde{q}$ is said to be the best dominant of Equation (5) if $\tilde{q}\left(z\right)\prec q\left(z\right)$ for all the dominants q.

(ii) If k satisfies the first order differential superordination

$$h\left(z\right)\prec \phi (k\left(z\right),z{k}^{\prime }\left(z\right);z),$$

(6)

then k is called to be a solution of the differential superordination in Equation (6). The function q is called a subordinant of the solutions of the differential superordination in Equation (6) if $q\left(z\right)\prec k\left(z\right)$ for all the functions k satisfying Equation (6). A subordinant $\tilde{q}$ is said to be the best subordinant of Equation (6) if $q\left(z\right)\prec \tilde{q}\left(z\right)$ for all the subordinants q.

Miller and Mocanu [6] obtained conditions on the functions h, q and $\phi$ for which the following implication holds:

$$h\left(z\right)\prec \phi (k\left(z\right),z{k}^{\prime }\left(z\right);z)\Rightarrow q\left(z\right)\prec k\left(z\right).$$

Applying these methods, in [7,8], the author studied general classes of first order differential superordinations and superordination-preserving integral operators. Using the results of Bulboacă [4] (see also [9,10]), the authors of [11] obtained sufficient conditions for functions $f\in \mathcal{A}$ to satisfy the double subordination

$$q_1\left(z\right)\prec \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec q_2\left(z\right),$$

where $q_1$ and $q_2$ are univalent functions in $\mathbb{U}$, normalized with $q_1\left(0\right)=q_2\left(0\right)=1$.

Sakaguchi [12] introduced a class $S_s^{*}$ of functions starlike with respect to symmetric points, which consists of functions $f\in \mathcal{A}$ satisfying the inequality

$$Re\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)-f(-z)}>0,z\in \mathbb{U},$$

that represents a subclass of close-to-convex functions, and hence univalent in $\mathbb{U}$. Moreover, this class includes the class of convex functions and odd starlike functions with respect to the origin (see [12,13]).

In addition, Aouf et al. [14] introduced and studied the class $S_{s,n}^{*}T(1,1)$ of functions n-starlike with respect to symmetric points, which consists of functions $f\in \mathcal{A}$, with $a_k\le 0$ for $k\ge 2$, and satisfying the inequality

$$Re\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)-D^{n}f(-z)}>0,z\in \mathbb{U},$$

where $D^n$ is the Sălăgean operator [15].

The classes defined in [12,13] could be generalized by introducing the next class of functions, defined with the aid of the ${\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }$ operator defined as follows:

Definition 2.

A function $f\in \mathcal{A}\left(m\right)$ with

$${\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\ne 0,z\in \dot{\mathbb{U}}:=\mathbb{U}\setminus \left\{0\right\},$$

(7)

is said to be in the class ${\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(\gamma ,\mu ,A,B)$ if it satisfies the subordination condition

$$\begin{array}{c}(1+\gamma ){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)\right)}^{\prime }-z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\prec \frac{1+Az}{1+Bz},\\ \left(\gamma \in \mathbb{C},0<\mu <1,-1\le B<A\le 1,m\in \mathbb{N},\alpha >0,n\ge 0,0<q<1,\lambda >-1\right).\end{array}$$

(8)

By specializing the parameters $\alpha$, $\lambda$ and q, we obtain the following subclasses:

(i) For $q\to 1^{-}$, we define the class ${\mathcal{W}}_{n,m}^{\lambda ,\alpha }(\gamma ,\mu ,A,B)$ as follows:

$$\begin{array}{c}{\mathcal{W}}_{n,m}^{\lambda ,\alpha }(\gamma ,\mu ,A,B):=\left\{f\in \mathcal{A}\left(m\right):(1+\gamma ){\left(\frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\right.\\ \left.-\gamma \left(\frac{z{\left({\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)\right)}^{\prime }-z{\left({\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\prec \frac{1+Az}{1+Bz}\right\},\end{array}$$

where the operator ${\mathcal{I}}_{n,m}^{\lambda ,\alpha }$ is defined by Equation (4);

(ii) For $q\to 1^{-}$, $\alpha =0$ and $\lambda =1$, we define the class ${\mathcal{N}}^{\gamma ,\mu }(m,A,B)$ that corrects the class defined by Muhammad and Marwan [16] as follows:

$$\begin{array}{c}{\mathcal{N}}^{\gamma ,\mu }(m,A,B):=\left\{f\in \mathcal{A}\left(m\right):(1+\gamma ){\left(\frac{2z}{f\left(z\right)-f(-z)}\right)}^{\mu }\right.\\ \left.-\gamma \left(\frac{z\left(f^{{}^{\prime }}\left(z\right)-f^{{}^{\prime }}(-z)\right)}{f\left(z\right)-f(-z)}\right){\left(\frac{2z}{f\left(z\right)-f(-z)}\right)}^{\mu }\prec \frac{1+Az}{1+Bz}\right\}.\end{array}$$

In this paper, we obtain some sharp differential subordination and superordination results for the functions belonging to the class ${\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(\gamma ,\mu ,A,B)$ to try to make a connection between a special subclass of analytic functions whose coefficients are given by the q-analog of integral operator and the differential subordination theory.

2. Preliminaries

To prove our results, we need the following definition and lemmas.

Definition 3

([5]). (Definition 2.2b., p. 21) Let $\mathcal{Q}$ be the set of all functions f that are analytic and injective on $\overline{\mathbb{U}}\setminus E\left(f\right)$, where $E\left(f\right):=\left\{\zeta \in \partial \mathbb{U}:\underset{z\to \zeta }{lim}f\left(z\right)=\infty \right\}$ and are such that $f^{\prime }\left(\zeta \right)\ne 0$ for $\zeta \in \partial \mathbb{U}\setminus E\left(f\right)$.

Lemma 1

([5]). (Theorem 3.1b., p. 71) Let the function H be convex in $\mathbb{U}$, with $H\left(0\right)=a$, and $\zeta \ne 0$ with $Re\zeta \ge 0$. If $\mathsf{\Phi }\in \mathcal{H}[a,m]$ and

$$\mathsf{\Phi }\left(z\right)+\frac{z{\mathsf{\Phi }}^{\prime }\left(z\right)}{\zeta }\prec H\left(z\right),$$

(9)

then

$$\mathsf{\Phi }\left(z\right)\prec \mathsf{\Psi }\left(z\right):=\frac{\zeta }{m{z}^{\frac{\zeta }{m}}}\underset{0}{\overset{z}{\int }}{t}^{\frac{\zeta }{m}-1}H\left(t\right)dt\prec H\left(z\right),$$

and the function Ψ is convex, $\mathsf{\Psi }\in \mathcal{H}[a,m]$, and is the best dominant of Equation (9).

Lemma 2

([17]). (Lemma 2.2., p. 3) Let q be univalent in $\mathbb{U}$, with $q\left(0\right)=1$. Let $\xi ,\phi \in \mathbb{C}$ with $\phi \ne 0$, and assume that

$$Re\left(1+\frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}\right)>max\left\{0;-Re\frac{\xi }{\phi }\right\},z\in \mathbb{U}.$$

If k is analytic in $\mathbb{U}$ and

$$\xi k\left(z\right)+\phi z{k}^{\prime }\left(z\right)\prec \xi q\left(z\right)+\phi z{q}^{\prime }\left(z\right),$$

(10)

then $k\left(z\right)\prec q\left(z\right)$, and q is the best dominant of Equation (10).

From [6] (Theorem 6, p. 820), we could easily obtain the following lemma:

Lemma 3.

Let q be convex in $\mathbb{U}$, and $k\ne 0$ with $Rek\ge 0$. If $g\in \mathcal{H}\left[q\right(0),1]\cap \mathcal{Q}$, such that $g\left(z\right)+kz{g}^{\prime }\left(z\right)$ is univalent in $\mathbb{U}$, then

$$q\left(z\right)+kz{q}^{\prime }\left(z\right)\prec g\left(z\right)+kz{g}^{\prime }\left(z\right),$$

(11)

implies that $q\left(z\right)\prec g\left(z\right)$, and q is the best subordinant of Equation (11).

Lemma 4

([18]). Let F be analytic and convex in $\mathbb{U}$, and $0\le \lambda \le 1$. If $f,g\in \mathcal{A}$, such that $f\left(z\right)\prec F\left(z\right)$ and $g\left(z\right)\prec F\left(z\right)$, then

$$\lambda f\left(z\right)+(1-\lambda )g\left(z\right)\prec F\left(z\right).$$

3. Main Results

Unless otherwise mentioned, we assume in the remainder of this paper that $\gamma \in \mathbb{C}$, $0<\mu <1$, $-1\le B<A\le 1$, $m\in \mathbb{N}$, $\alpha >0$, $n\ge 0$, $0<q<1$, $\lambda >-1$, and all the powers are understood as principle values.

Theorem 1.

If $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(\gamma ,\mu ,A,B)$ and $\gamma \in {\mathbb{C}}^{*}:=\mathbb{C}\setminus \left\{0\right\}$ with $Re\gamma \ge 0$, then

$${\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\prec \mathsf{\Psi }\left(z\right):=\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Azu}{1+Bzu}{u}^{\frac{\mu }{\gamma m}-1}du\prec \frac{1+Az}{1+Bz},$$

and Ψ is convex, $\mathsf{\Psi }\in \mathcal{H}[1,m]$, and is the best dominant.

Proof. 

If we define the function h by

$$h\left(z\right):={\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu },z\in \mathbb{U},$$

(12)

from Equation (7), it follows that h is an analytic function in $\mathbb{U}$, with $h\left(0\right)=1$. Differentiating Equation (12) with respect to z, we obtain that

$$\begin{array}{c}(1+\gamma ){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)\right)}^{{}^{\prime }}-z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ =h\left(z\right)+\frac{\gamma }{\mu }z{h}^{\prime }\left(z\right)\prec \frac{1+Az}{1+Bz}.\end{array}$$

(13)

Since

$${\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)=z+\sum _{k=m+1}^{\infty }{\alpha }_k{z}^k,\mathrm{and}{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)=-z+\sum _{k=m+1}^{\infty }{\alpha }_k{(-1)}^k{z}^k,$$

where

$${\alpha }_k=\frac{[k,q]!}{{[\lambda +1,q]}_{k-1}}{\left(\frac{n+1}{n+k}\right)}^{\alpha }{a}_k,k\ge m+1,$$

we have

$$U\left(z\right):={\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}={\displaystyle \frac{2z}{2z+{\displaystyle \sum _{k=m+1}^{\infty }}{\alpha }_k\left[1+{(-1)}^{k+1}\right]z^k}}={\displaystyle \frac{1}{1+{\displaystyle \sum _{s=m}^{\infty }}{\beta }_s{z}^s}},$$

with

$${\beta }_s=\frac{{\alpha }_{s+1}\left[1+{(-1)}^s\right]}{2},s\ge m.$$

Moreover,

$$U\left(z\right)=\frac{1}{1+{\displaystyle \sum _{s=m}^{\infty }}{\beta }_s{z}^s}=1+\sum _{j=1}^{\infty }{\gamma }_j{z}^j,z\in \mathbb{U},$$

with unknowns ${\gamma }_j$, $j\ge 1$, we have

$$1=\left(1+{\beta }_m{z}^m+{\beta }_{m+1}{z}^{m+1}+\dots \right)\left(1+{\gamma }_{1}z+{\gamma }_2{z}^2+\cdots +{\gamma }_m{z}^m+{\gamma }_{m+1}{z}^{m+1}+\dots \right),$$

and equating the corresponding coefficients it follows that

$${\gamma }_1={\gamma }_2=\cdots ={\gamma }_{m-1}=0,{\gamma }_m=-{\beta }_m,{\gamma }_{m+1}=-{\beta }_{m+1},\dots ,$$

hence

$$U\left(z\right)=1+\sum _{j=m}^{\infty }{\gamma }_j{z}^j\in \mathcal{H}[1,m].$$

According to Equation (12), we have

$$h=U^{\mu },\mathrm{with}U\in \mathcal{H}[1,m],$$

and using the binomial power expansion formula, we get

$$h=U^{\mu }\in \mathcal{H}[1,m].$$

Now, from the subordination in Equation (13), using Lemma 1 for $\zeta ={\displaystyle \frac{\mu }{\gamma }}$, we obtain our result. □

Taking $q\to 1^{-}$ in Theorem 1, we obtain the following corollary:

Corollary 1.

If $f\in {\mathcal{W}}_{n,m}^{\lambda ,\alpha }(\gamma ,\mu ,A,B)$ and $\gamma \in {\mathbb{C}}^{*}:=\mathbb{C}\setminus \left\{0\right\}$ with $Re\gamma \ge 0$, then

$${\left({\displaystyle \frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\prec \mathsf{\Psi }\left(z\right):=\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Azu}{1+Bzu}{u}^{\frac{\mu }{\gamma m}-1}du\prec \frac{1+Az}{1+Bz},$$

and Ψ is convex, $\mathsf{\Psi }\in \mathcal{H}[1,m]$, and is the best dominant.

Remark 1.

The above theorem shows that

$${\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(\gamma ,\mu ,A,B)\subset {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(0,\mu ,A,B),$$

for all $\gamma \in \mathbb{C}$ with $Re\gamma \ge 0$.

Moreover, the next inclusion result for the classes ${\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(\gamma ,\mu ,A,B)$ holds:

Theorem 2.

If ${\gamma }_1,{\gamma }_2\in \mathbb{R}$ such that $0\le {\gamma }_1\le {\gamma }_2$, and $-1\le B_1\le B_2<A_2\le A_1\le 1$, then

$${\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }({\gamma }_2,\mu ,A_2,B_2)\subset {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }({\gamma }_1,\mu ,A_1,B_1).$$

(14)

Proof. 

If $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }({\gamma }_2,\mu ,A_2,B_2)$, since $-1\le B_1\le B_2<A_2\le A_1\le 1$, it is easy to check that

$$\begin{array}{c}\left(1+{\gamma }_2\right){\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\\ -{\gamma }_2\left({\displaystyle \frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right){\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\\ \prec \frac{1+A_{2}z}{1+B_{2}z}\prec \frac{1+A_{1}z}{1+B_{1}z},\end{array}$$

(15)

that is $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }({\gamma }_1,\mu ,A_1,B_1)$, hence the assertion in Equation (14) holds for ${\gamma }_1={\gamma }_2$.

If $0\le {\gamma }_1<{\gamma }_2$, from Remark 1 and Equation (15), it follows $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(0,\mu ,A_1,B_1)$, that is

$${\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\prec \frac{1+A_{1}z}{1+B_{1}z}.$$

(16)

A simple computation shows that

$$\begin{array}{c}\left(1+{\gamma }_1\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -{\gamma }_1\left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ =\left(1-\frac{{\gamma }_1}{{\gamma }_2}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ +\frac{{\gamma }_1}{{\gamma }_2}\left[\left(1+{\gamma }_2\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\right.\\ \left.-{\gamma }_2\left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\right],z\in \mathbb{U}.\end{array}$$

(17)

Moreover,

$$0\le \frac{{\gamma }_1}{{\gamma }_2}<1,$$

and the function ${\displaystyle \frac{1+A_{1}z}{1+B_{1}z}}$, with $-1\le B_1<A_1\le 1$, is analytic and convex in $\mathbb{U}$. According to Equation (17), using the subordinations in Equations (15) and (16), from Lemma 4, we deduce that

$$\begin{array}{c}\left(1+{\gamma }_1\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -{\gamma }_1\left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\prec \frac{1+A_{1}z}{1+B_{1}z},\end{array}$$

that is $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }({\gamma }_1,\mu ,A_1,B_1)$. □

Taking $q\to 1^{-}$ in Theorem 2, we obtain the following corollary:

Corollary 2.

If ${\gamma }_1,{\gamma }_2\in \mathbb{R}$ such that $0\le {\gamma }_1\le {\gamma }_2$, and $-1\le B_1\le B_2<A_2\le A_1\le 1$, then

$${\mathcal{W}}_{n,m}^{\lambda ,\alpha }({\gamma }_2,\mu ,A_2,B_2)\subset {\mathcal{W}}_{n,m}^{\lambda ,\alpha }({\gamma }_1,\mu ,A_1,B_1).$$

Example 1.

For the special case $A_1=1$ and $B_1=-1$, Theorem 2 and Corollary 2 reduce to the next examples, respectively:

Suppose that ${\gamma }_1,{\gamma }_2\in \mathbb{R}$ such that $0\le {\gamma }_1\le {\gamma }_2$, and $-1\le B_2<A_2\le 1$.

1. If $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }({\gamma }_2,\mu ,A_2,B_2)$, then

$$\begin{array}{c}Re\left\{\left(1+{\gamma }_1\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\right.\\ \left.-{\gamma }_1\left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\right\}>0,z\in \mathbb{U};\end{array}$$

2. If $f\in {\mathcal{W}}_{n,m}^{\lambda ,\alpha }({\gamma }_2,\mu ,A_2,B_2)$, then

$$\begin{array}{c}Re\left\{\left(1+{\gamma }_1\right){\left(\frac{2z}{{\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\right.\\ \left.-{\gamma }_1\left(\frac{z{\left({\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\right\}>0,z\in \mathbb{U};\end{array}$$

Theorem 3.

Suppose that q is univalent in $\mathbb{U}$, with $q\left(0\right)=1$, and let $\gamma \in {\mathbb{C}}^{*}$ such that

$$Re\left(1+{\displaystyle \frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}}\right)>max\left\{0;-Re\frac{\mu }{\gamma }\right\},z\in \mathbb{U}.$$

(18)

If $f\in \mathcal{A}\left(m\right)$ such that Equation (7) holds, and satisfies the subordination

$$\begin{array}{c}(1+\gamma ){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ \prec q\left(z\right)+\frac{\gamma }{\mu }z{q}^{\prime }\left(z\right),\end{array}$$

(19)

then

$${\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\prec q\left(z\right),$$

and q is the best dominant of Equation (19).

Proof. 

Since $f\in \mathcal{A}\left(m\right)$ such that Equation (7) holds, it follows that the function h defined by Equation (12) is analytic in $\mathbb{U}$, and $h\left(0\right)=1$. As in the proof of Theorem 1, differentiating Equation (12) with respect to z, we obtain that Equation (19) is equivalent to

$$h\left(z\right)+\frac{\gamma }{\mu }z{h}^{\prime }\left(z\right)\prec q\left(z\right)+\frac{\gamma }{\mu }z{q}^{\prime }\left(z\right).$$

Using Lemma 2 for $\xi :=1$ and $\phi :={\displaystyle \frac{\gamma }{\mu }}$, we get that the above subordination implies $h\left(z\right)\prec q\left(z\right)$, and q is the best dominant of Equation (19). □

For the special case $q\left(z\right)={\displaystyle \frac{1+Az}{1+Bz}}$, with $-1\le B<A\le 1$, Theorem 3 reduces to the following corollary:

Corollary 3.

Let $\gamma \in {\mathbb{C}}^{*}$ and $-1\le B<A\le 1$, such that

$$max\left\{-1;-{\displaystyle \frac{1+Re\frac{\mu }{\gamma }}{1-Re\frac{\mu }{\gamma }}}\right\}\le B\le 0,\mathit{or}0\le B\le min\left\{1;{\displaystyle \frac{1+Re\frac{\mu }{\gamma }}{1-Re\frac{\mu }{\gamma }}}\right\}.$$

(20)

If $f\in \mathcal{A}\left(m\right)$ such that Equation (7) holds, and satisfies the subordination

$$\begin{array}{c}(1+\gamma ){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ \prec \frac{1+Az}{1+Bz}+\frac{\gamma }{\mu }\frac{(A-B)z}{{(1+Bz)}^2},\end{array}$$

(21)

then

$${\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\prec \frac{1+Az}{1+Bz},$$

and ${\displaystyle \frac{1+Az}{1+Bz}}$ is the best dominant of Equation (21).

Proof. 

For $q\left(z\right)={\displaystyle \frac{1+Az}{1+Bz}}$, the condition in Equation (18) reduces to

$$Re\frac{1-Bz}{1+Bz}>max\left\{0;-Re\frac{\mu }{\gamma }\right\},z\in \mathbb{U}.$$

(22)

Since

$$inf\left\{Re\frac{1-Bz}{1+Bz}:z\in \mathbb{U}\right\}=\left\{\begin{array}{ccc}{\displaystyle \frac{1+B}{1-B}},\hfill & \mathrm{if}\hfill & -1\le B\le 0,\hfill \\ {\displaystyle \frac{1-B}{1+B}},\hfill & \mathrm{if}\hfill & 0\le B<1,\hfill \end{array}\right.$$

we easily check that Equation (22) holds if and only if the assumption in Equation (20) is satisfied, whenever $-1\le B<1$. □

Taking $q\to 1^{-}$ in Theorem 3, we obtain the following corollary:

Corollary 4.

Suppose that q is univalent in $\mathbb{U}$, with $q\left(0\right)=1$, and let $\gamma \in {\mathbb{C}}^{*}$ such that

$$Re\left(1+{\displaystyle \frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}}\right)>max\left\{0;-Re\frac{\mu }{\gamma }\right\},z\in \mathbb{U}.$$

If $f\in \mathcal{A}\left(m\right)$ such that Equation (7) holds, and satisfies the subordination

$$\begin{array}{c}(1+\gamma ){\left(\frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\prec q\left(z\right)+\frac{\gamma }{\mu }z{q}^{\prime }\left(z\right),\end{array}$$

then

$${\left({\displaystyle \frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\prec q\left(z\right),$$

and q is the best dominant of Equation (19).

Theorem 4.

Let q be convex in $\mathbb{U}$, with $q\left(0\right)=1$, and $\gamma \in {\mathbb{C}}^{*}$, with $Re\gamma \ge 0$. In addition, let $f\in \mathcal{A}\left(m\right)$ such that

$${\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\in \mathcal{H}[q\left(0\right),1]\cap \mathcal{Q},$$

(23)

and assume that the function

$$\begin{array}{c}(1+\gamma ){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ \mathit{is}\mathit{univalent}\mathit{in}\mathbb{U}.\end{array}$$

(24)

If

$$\begin{array}{c}q\left(z\right)+\frac{\gamma }{\mu }z{q}^{\prime }\left(z\right)\prec (1+\gamma ){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu },\end{array}$$

(25)

then

$$q\left(z\right)\prec {\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu },$$

and q is the best subordinant of Equation (25).

Proof. 

Letting the function h be defined by Equation (12), then $h\in \mathcal{H}\left[q\right(0),m]$, and from Equation (23) we have that $h\in \mathcal{H}\left[q\right(0),1]\cap \mathcal{Q}$. As in the proof of Theorem 1, differentiating Equation (12) with respect to z, we obtain that

$$q\left(z\right)+\frac{\gamma }{\mu }z{q}^{\prime }\left(z\right)\prec h\left(z\right)+\frac{\gamma }{\mu }z{h}^{\prime }\left(z\right).$$

Now, according to Lemma 3 for $k:={\displaystyle \frac{\gamma }{\mu }}$ we obtain the desired result. □

Taking $q\left(z\right)={\displaystyle \frac{1+Az}{1+Bz}}$, with $-1\le B<A\le 1$, in Theorem 4, we obtain the following corollary:

Corollary 5.

Let $\gamma \in {\mathbb{C}}^{*}$, with $Re\gamma \ge 0$, and $-1\le B<A\le 1$. If $f\in \mathcal{A}\left(m\right)$ such that the assumptions in Equations (23) and (24) hold, and satisfies the subordination

$$\begin{array}{c}\frac{1+Az}{1+Bz}+\frac{\gamma }{\mu }\frac{(A-B)z}{{(1+Bz)}^2}\prec (1+\gamma ){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu },\end{array}$$

(26)

then

$$\frac{1+Az}{1+Bz}\prec {\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu },$$

and ${\displaystyle \frac{1+Az}{1+Bz}}$ is the best subordinant of Equation (26).

Taking $q\to 1^{-}$ in Theorem 4, we obtain the following corollary:

Corollary 6.

Let q be convex in $\mathbb{U}$, with $q\left(0\right)=1$, and $\gamma \in {\mathbb{C}}^{*}$, with $Re\gamma \ge 0$. In addition, let $f\in \mathcal{A}\left(m\right)$ such that

$${\left({\displaystyle \frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\in \mathcal{H}[q\left(0\right),1]\cap \mathcal{Q},$$

and assume that the function

$$\begin{array}{c}(1+\gamma ){\left(\frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\mathit{is}\mathit{univalent}\mathit{in}\mathbb{U}.\end{array}$$

If

$$\begin{array}{c}q\left(z\right)+\frac{\gamma }{\mu }z{q}^{\prime }\left(z\right)\prec (1+\gamma ){\left(\frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right)}^{\mu },\end{array}$$

then

$$q\left(z\right)\prec {\left({\displaystyle \frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu },$$

and q is the best subordinant of Equation (25).

Combining Theorems 3 and 4, we obtain the following sandwich-type theorem:

Theorem 5.

Let $q_1$ and $q_2$ be two convex functions in $\mathbb{U}$, with $q_1\left(0\right)=q_2\left(0\right)=1$, and let $\gamma \in {\mathbb{C}}^{*}$, with $Re\gamma \ge 0$. If $f\in \mathcal{A}\left(m\right)$ such that the assumptions in Equations (23) and (24) hold, then

$$\begin{array}{c}{q}_1\left(z\right)+\frac{\gamma }{\mu }z{q}_1^{\prime }\left(z\right)\prec \mathsf{\Theta }\left(z\right):=(1+\gamma ){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ \prec q_2\left(z\right)+\frac{\gamma }{\mu }z{q}_2^{\prime }\left(z\right),\end{array}$$

(27)

implies that

$$q_1\left(z\right)\prec \mathsf{\Phi }\left(z\right):={\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\prec q_2\left(z\right),$$

and $q_1$ and $q_2$ are, respectively, the best subordinant and the best dominant of Equation(27).

Combining Corollaries 4 and 6, we obtain the following sandwich-type theorem:

Corollary 7.

Let $q_1$ and $q_2$ be two convex functions in $\mathbb{U}$, with $q_1\left(0\right)=q_2\left(0\right)=1$, and let $\gamma \in {\mathbb{C}}^{*}$, with $Re\gamma \ge 0$. If $f\in \mathcal{A}\left(m\right)$ such that the assumptions in Equations (23) and (24) hold for the operator ${\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }$ replaced by ${\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }$, then

$$\begin{array}{c}{q}_1\left(z\right)+\frac{\gamma }{\mu }z{q}_1^{\prime }\left(z\right)\prec \widehat{\mathsf{\Theta }}\left(z\right):=(1+\gamma ){\left(\frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\\ -\gamma \left(\frac{z{\left({\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\prec q_2\left(z\right)+\frac{\gamma }{\mu }z{q}_2^{\prime }\left(z\right),\end{array}$$

(28)

implies that

$$q_1\left(z\right)\prec \widehat{\mathsf{\Phi }}\left(z\right):={\left({\displaystyle \frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\prec q_2\left(z\right),$$

and $q_1$ and $q_2$ are, respectively, the best subordinant and the best dominant of Equation (27).

Example 2.

Taking $q_j=1+r_{j}z$, with $0<r_1<r_2$, $j=1,2$ in Theorem 5 and Corollary 7, we obtain the next examples, respectively:

Let $\gamma \in {\mathbb{C}}^{*}$, with $Re\gamma \ge 0$.

1. If $f\in \mathcal{A}\left(m\right)$ such that the assumptions in Equations (23) and (24) hold, then

$$r_1\left|1+\frac{\gamma }{\mu }\right|<|\mathsf{\Theta }\left(z\right)-1|<r_2\left|1+\frac{\gamma }{\mu }\right|,z\in \mathbb{U}\Rightarrow r_1<|\mathsf{\Phi }\left(z\right)-1|<r_2,z\in \mathbb{U},(0<r_1<r_2)$$

where Θ and Φ are given in Theorem 5, and the obtained bounds $r_1$ and $r_2$ are the best possible.

2. If $f\in \mathcal{A}\left(m\right)$ such that the assumptions in Equations (23) and (24) hold for the operator ${\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }$ replaced by ${\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }$, then

$$r_1\left|1+\frac{\gamma }{\mu }\right|<|\widehat{\mathsf{\Theta }}\left(z\right)-1|<r_2\left|1+\frac{\gamma }{\mu }\right|,z\in \mathbb{U}\Rightarrow r_1<|\widehat{\mathsf{\Phi }}\left(z\right)-1|<r_2,z\in \mathbb{U},(0<r_1<r_2)$$

where $\widehat{\mathsf{\Theta }}$ and $\widehat{\mathsf{\Phi }}$ are given in Corollary 7, and the obtained bounds $r_1$ and $r_2$ are the best possible.

Example 3.

Putting $q_j=e^{r_{j}z}$, with $0<r_1<r_2\le 1$, $j=1,2$ in Theorem 5 and Corollary 7, we obtain the next examples, respectively:

Let $\gamma \in {\mathbb{C}}^{*}$, with $Re\gamma \ge 0$.

1. If $f\in \mathcal{A}\left(m\right)$ such that the assumptions in Equations (23) and (24) hold, then

$$\left(1+\frac{\gamma }{\mu }z\right)e^{r_{1}z}\prec \mathsf{\Theta }\left(z\right)\prec \left(1+\frac{\gamma }{\mu }z\right)e^{r_{2}z}\Rightarrow e^{r_{1}z}\prec \mathsf{\Phi }\left(z\right)\prec e^{r_{2}z},(0<r_1<r_2\le 1)$$

where Θ and Φ are given in Theorem 5, and $e^{r_{1}z}$ and $e^{r_{2}z}$ are, respectively, the best subordinant and the best dominant.

2. If $f\in \mathcal{A}\left(m\right)$ such that the assumptions in Equations (23) and (24) hold for the operator ${\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }$ replaced by ${\mathcal{I}}_{n,m,q}^{\lambda ,\alpha }$, then

$$\left(1+\frac{\gamma }{\mu }z\right)e^{r_{1}z}\prec \widehat{\mathsf{\Theta }}\left(z\right)\prec \left(1+\frac{\gamma }{\mu }z\right)e^{r_{2}z}\Rightarrow e^{r_{1}z}\prec \widehat{\mathsf{\Phi }}\left(z\right)\prec e^{r_{2}z},(0<r_1<r_2\le 1)$$

where $\widehat{\mathsf{\Theta }}$ and $\widehat{\mathsf{\Phi }}$ are given in Corollary 7, and $e^{r_{1}z}$ and $e^{r_{2}z}$ are, respectively, the best subordinant and the best dominant.

Theorem 6.

If $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(0,\mu ,1-2\rho ,-1)$, with $0\le \rho <1$, then $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(\gamma ,\mu ,1-2\rho ,-1)$ for $\left|z\right|<R$, where

$$R={\left(\sqrt{\frac{{\left|\gamma \right|}^2{m}^2}{{\mu }^2}+1}-\frac{\left|\gamma \right|m}{\mu }\right)}^{\frac{1}{m}}.$$

(29)

Proof. 

For $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(0,\mu ,1-2\rho ,-1)$, with $0\le \rho <1$, let the function h be defined by

$${\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }=(1-\rho )h\left(z\right)+\rho ,z\in \mathbb{U}.$$

(30)

Hence, the function h is analytic in $\mathbb{U}$, with $h\left(0\right)=1$, and since $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(0,\mu ,1-2\rho ,-1)$ is equivalent to,

$${\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\prec \frac{1+(1-2\rho )z}{1-z},$$

it follows that $Reh\left(z\right)>0$, $z\in \mathbb{U}$.

As in the proof of Theorem 1, since $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(0,\mu ,1-2\rho ,-1)$, with $0\le \rho <1$, we deduce that

$${\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\in \mathcal{H}[1,m],$$

and from the relation in Equation (30), we get $h\in \mathcal{H}[1,m]$. Therefore, the following estimate holds

$$\left|z{h}^{\prime }\left(z\right)\right|\le \frac{2m{r}^{m}Reh\left(z\right)}{1-r^{2m}},\left|z\right|=r<1,$$

that represents the result of Shah [19] (the inequality (6), p. 240, for $\alpha =0$), which generalize Lemma 2 of [20].

A simple computation shows that

$$\begin{array}{c}\frac{1}{1-\rho }\left\{(1+\gamma ){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\right.\\ \left.-\gamma \left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }-\rho \right\}\\ =h\left(z\right)+\frac{\gamma }{\mu }z{h}^{\prime }\left(z\right),z\in \mathbb{U},\end{array}$$

hence, we obtain

$$\begin{array}{c}Re\left\{\frac{1}{1-\rho }\left[(1+\gamma ){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }\right.\right.\\ \left.\left.-\gamma \left(\frac{z{\left({\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right)}^{\prime }}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right){\left(\frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}\right)}^{\mu }-\rho \right]\right\}\\ \ge Reh\left(z\right)\left[1-\frac{2\left|\gamma \right|m{r}^m}{\mu \left(1-r^{2m)}\right)}\right],\left|z\right|=r<1,\end{array}$$

(31)

and the right-hand side of Equation (31) is positive provided that $r<R$, where R is given by Equation (29). □

Theorem 7.

Let $f\in {\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(\gamma ,\mu ,A,B)$, let $\gamma \in {\mathbb{C}}^{*}$ with $Re\gamma \ge 0$, and $-1\le B<A\le 1$.

1. Then,

$$\begin{array}{c}\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Au}{1-Bu}{u}^{\frac{\mu }{\gamma m}-1}du<Re{\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\\ <\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Au}{1+Bu}{u}^{\frac{\mu }{\gamma m}-1}du,z\in \mathbb{U}.\end{array}$$

(32)

2. For $\left|z\right|=r<1$, we have

$$\begin{array}{c}2r{\left(\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Aur}{1+Bur}{u}^{\frac{\mu }{\gamma m}-1}du\right)}^{-\frac{1}{\mu }}<\left|{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right|\\ <2r{\left(\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Aur}{1-Bur}{u}^{\frac{\mu }{\gamma m}-1}du\right)}^{-\frac{1}{\mu }}.\end{array}$$

(33)

All these inequalities are the best possible.

Proof. 

From the assumptions, using Theorem 1, we obtain that

$${\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\prec \mathsf{\Psi }\left(z\right):=\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Azu}{1+Bzu}{u}^{\frac{\mu }{\gamma m}-1}du,$$

(34)

and the convex function $\mathsf{\Psi }\in \mathcal{H}[1,m]$ is the best dominant. Therefore,

$$\begin{array}{c}Re{\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }<\underset{z\in \mathbb{U}}{sup}Re\left(\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Azu}{1+Bzu}{u}^{\frac{\mu }{\gamma m}-1}du\right)\\ =\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\underset{z\in \mathbb{U}}{sup}Re\left(\frac{1+Azu}{1+Bzu}\right)u^{\frac{\mu }{\gamma m}-1}du=\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Au}{1+Bu}{u}^{\frac{\mu }{\gamma m}-1}du,z\in \mathbb{U},\end{array}$$

and

$$\begin{array}{c}Re{\left({\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }>\underset{z\in \mathbb{U}}{inf}Re\left(\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Azu}{1-Bzu}{u}^{\frac{\mu }{\gamma m}-1}du\right)\\ =\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\underset{z\in \mathbb{U}}{inf}Re\left(\frac{1-Azu}{1-Bzu}\right)u^{\frac{\mu }{\gamma m}-1}du=\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Au}{1-Bu}{u}^{\frac{\mu }{\gamma m}-1}du,z\in \mathbb{U}.\end{array}$$

In addition, since

$$\begin{array}{c}{\left|{\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right|}^{\mu }<\underset{z\in \mathbb{U}}{sup}\left|\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Azu}{1+Bzu}{u}^{\frac{\mu }{\gamma m}-1}du\right|\\ =\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\underset{z\in \mathbb{U}}{sup}\left|\frac{1+Azu}{1+Bzu}\right|u^{\frac{\mu }{\gamma m}-1}du=\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Aur}{1+Bur}{u}^{\frac{\mu }{\gamma m}-1}du,\left|z\right|=r<1,\end{array}$$

we get

$$\left|{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right|>2r{\left(\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Aur}{1+Bur}{u}^{\frac{\mu }{\gamma m}-1}du\right)}^{-\frac{1}{\mu }},$$

while

$$\begin{array}{c}{\left|{\displaystyle \frac{2z}{{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)}}\right|}^{\mu }>\underset{z\in \mathbb{U}}{inf}\left|\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Azu}{1-Bzu}{u}^{\frac{\mu }{\gamma m}-1}du\right|\\ =\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\underset{z\in \mathbb{U}}{inf}\left|\frac{1-Azu}{1-Bzu}\right|u^{\frac{\mu }{\gamma m}-1}du=\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Aur}{1-Bur}{u}^{\frac{\mu }{\gamma m}-1}du,\left|z\right|=r<1,\end{array}$$

implies

$$\left|{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }f(-z)\right|<2r{\left(\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Aur}{1-Bur}{u}^{\frac{\mu }{\gamma m}-1}du\right)}^{-\frac{1}{\mu }}.$$

The inequalities of Equations (32) and (33) are the best possible because the subordination in Equation (34) is sharp. □

Taking $q\to 1^{-}$ in Theorem 7, we obtain the following corollary:

Corollary 8.

Let $f\in {\mathcal{W}}_{n,m}^{\lambda ,\alpha }(\gamma ,\mu ,A,B)$, let $\gamma \in {\mathbb{C}}^{*}$ with $Re\gamma \ge 0$, and $-1\le B<A\le 1$.

1. Then,

$$\begin{array}{c}\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Au}{1-Bu}{u}^{\frac{\mu }{\gamma m}-1}du<Re{\left({\displaystyle \frac{2z}{{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)}}\right)}^{\mu }\\ <\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Au}{1+Bu}{u}^{\frac{\mu }{\gamma m}-1}du,z\in \mathbb{U}.\end{array}$$

2. For $\left|z\right|=r<1$, we have

$$\begin{array}{c}2r{\left(\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Aur}{1+Bur}{u}^{\frac{\mu }{\gamma m}-1}du\right)}^{-\frac{1}{\mu }}<\left|{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f\left(z\right)-{\mathcal{I}}_{n,m}^{\lambda ,\alpha }f(-z)\right|\\ <2r{\left(\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Aur}{1-Bur}{u}^{\frac{\mu }{\gamma m}-1}du\right)}^{-\frac{1}{\mu }}.\end{array}$$

All these inequalities are the best possible.

Taking $q\to 1^{-}$, $\alpha =0$ and $\lambda =1$ in Theorem 7, we obtain the following corollary:

Corollary 9.

Let $f\in {\mathcal{N}}^{\gamma ,\mu }(m,A,B)$, let $\gamma \in {\mathbb{C}}^{*}$ with $Re\gamma \ge 0$, and $-1\le B<A\le 1$.

1. Then,

$$\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Au}{1-Bu}{u}^{\frac{\mu }{\gamma m}-1}du<Re{\left({\displaystyle \frac{2z}{f\left(z\right)-f(-z)}}\right)}^{\mu }<\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Au}{1+Bu}{u}^{\frac{\mu }{\gamma m}-1}du,z\in \mathbb{U}.$$

2. For $\left|z\right|=r<1$, we have

$$\begin{array}{c}2r{\left(\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1+Aur}{1+Bur}{u}^{\frac{\mu }{\gamma m}-1}du\right)}^{-\frac{1}{\mu }}<\left|f\left(z\right)-f(-z)\right|\\ <2r{\left(\frac{\mu }{\gamma m}\underset{0}{\overset{1}{\int }}\frac{1-Aur}{1-Bur}{u}^{\frac{\mu }{\gamma m}-1}du\right)}^{-\frac{1}{\mu }}.\end{array}$$

All these inequalities are the best possible.

Example 4.

Putting $\mu =\gamma =m=1$, $A=1-2\beta$ ($0\le \beta <1$), and $B=-1$ in Corollary 9, we get the next special case.

If $f\in {\mathcal{N}}^{1,1}(1,1-2\beta ,-1)$ with $0\le \beta <1$, then:

1. The next inequality holds:

$$Re{\displaystyle \frac{2z}{f\left(z\right)-f(-z)}}>2\beta -1+2(1-\beta )ln2,z\in \mathbb{U}.$$

2. For $\left|z\right|=r:=0.9$, we have

$$\frac{1.8}{1+3.116855762\beta }<|f\left(z\right)-f(-z)|<\frac{1.8}{1-0.5736580307\beta }.$$

Remark 2.

Part (ii) of Corollary 9 corrects the Corollary (3.10) studied by Muhammad and Marwan [16].

Concluding, all the above results give us information about subordination and superordination properties, inclusion results, radius problem, and sharp estimations for the classes ${\mathcal{M}}_{n,m,q}^{\lambda ,\alpha }(\gamma ,\mu ,A,B)$, together general sharp subordination and superordination for the operator ${\mathcal{N}}_{n,m,q}^{\lambda ,\alpha }$. For special choices of the parameters $\gamma \in \mathbb{C}$, $0<\mu <1$, $-1\le B<A\le 1$, $m\in \mathbb{N}$, $\alpha >0$, $n\ge 0$, $0<q<1$, and $\lambda >-1$, we may obtain several simple applications connected with the above-mentioned classes and operator.