On Sandwich-Type Results for a Subclass of Certain Univalent Functions Using a New Hadamard Product Operator

Abstract

This paper is concerned with studying some results by introducing a new Hadamard product $M_{\gamma ,b,v,\delta }^{C,\eta }$ operator for differential subordination and superordination for certain univalent functions in the open unit disc U. Firstly, we state some basic definitions and required theorems. Furthermore. Some sandwich theorems are derived. The differential subordination theory’s features and outcomes are symmetric to those derived using the differential superordination theory.

1. Introduction

Let $H=H\left(U\right)$ be the class of analytic functions in the open unit disk $U=\{z\in C:\left|z\right|<1\}$.

For n a positive integer and $a\in C.$ Let $H\left[a,n\right]$ be the subclass of $H$ of the form:

Assume that $A$ be a subclass of $H$ of functions $f$ of the form:

$$f\left(z\right)=a+a_n{z}^n+a_{n+1}{z}^{n+1}+\cdots \left(a\in C, N=\left\{1,2,3,\dots \right\}\right).$$

$$f\left(z\right)=z+{\displaystyle \sum }_{n=2}^{\infty }{a}_n{z}^n.$$

(1)

If $f\in A$ is given by (1) and $g\in A$ given by $g\left(z\right)=z+{\displaystyle \sum }_{n=2}^{\infty }{b}_n{z}^n.$

For f and g the Hadamard product (or convolution) is defined by

$$\left(f\ast g\right)\left(z\right)=z+{\displaystyle \sum }_{n=2}^{\infty }{a}_n{b}_n{z}^n=\left(g\ast f\right)\left(z\right).$$

Assume that both f and g are analytic defined in $U$, $f$ is called subordinate to $g$, in $U$ and denoted as $f\prec g$, if there is function, $w$, which is Schwarz analytic in $U$, and $w\left(0\right)=0,$ $\left|w\left(z\right)\right|<1 \left(z\in U\right)$, such that $f\left(z\right)=g\left(w\left(z\right)\right)$,$\left(z\in U\right)$. Moreover, i The equivalence connection exists if the function g is univalent in U. ([1,2,3]) is given as follows:

$$f\left(z\right)\prec g\left(z\right)\leftrightarrow f\left(0\right)=g\left(0\right) \mathrm{and} f\left(U\right)\subset g\left(U\right), z\in U.$$

Definition 1 

([1]). Let $\varphi \left(r,s,t;z\right):C^3\times U\to C$ and the analytic function is  $h\left(z\right)$ in$U$, such that$p$ and$\varphi \left(p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right);z\right)$ are univalent in$U$ and  $p$ satisfies the second–order differential superordination,

$$h\left(z\right)\prec \varphi \left(p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right);z\right), \left(z\in U\right).$$

(2)

Then we say that p is a differential superordination solution (2). A subordinate of the solutions of the differential superordination (2), or simply a subordinate, is an analytic function q(z), if p$\prec$q for all p satisfying (2). A univalent subordinate $\tilde{q}$(z) that satisfies q ≺ $\tilde{q}$ for all subordinants q of (2) is called the best subordinant.

Definition 2 

([3]). Let $\varphi \left(r,s,t;z\right):C^3\times U\to C$ and let$h\left(z\right)$ be univalent in$U$, and$p$ is analytic in$U$ satisfying the second−order differential subordination,

$$\varphi \left(p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right);z\right)\prec h\left(z\right), \left(z\in U\right),$$

(3)

Then we say that $p$ a solution of the differential subordination (3). The univalent function $q$ is called a dominant of the solution of the differential subordination (3) or more simply a dominant if $p\prec q$ for all $p$ satisfying (3). A dominant $\tilde{q}\left(z\right)$ that satisfies $q\prec \tilde{q}$ for every dominant $q$ of (3) is called the best dominant.

Sufficient requirements for the functions h, q and ϕ that satisfy the following condition, were obtained by many authors, see [4,5,6,7,8,9,10,11,12,13,14,15,16].

$$h\left(z\right)\prec \varphi \left(p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right);z\right)\Rightarrow q\left(z\right)\prec p\left(z\right), \left(z\in U\right).$$

(4)

By using the results (see [4,5,6,7,8,9,14,15,16]), we obtain sufficient conditions for normalized analytic functions satisfying:

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

where $q_1$ and $q_2$ are given univalent functions in$U$ with$q_1\left(0\right)=q_2\left(0\right)=1$. Also, many authors (see [2,4,5,6,7,8,9,10,11,12,13,14,15,16,17]) derived some differential-subordination and superordination results with some sandwich theorems.

Certain univalent functions, Bi-Bazilevic functions and subordination results have been recently studied in the following papers [18,19].

Now, for complex parameters $c_1,\dots ..c_t$ and $b_1,\dots ..b_r (b_j\in C\backslash \left\{0,-1,-2,\dots \right\},j=1,\dots ,r, \left|\gamma \right|<1)$, the $\gamma$-hypergeometric:

$$t{\Psi }_r^{} ={\displaystyle \sum }_{n=0}^{\infty }\frac{{\left(c_1,\gamma \right)}_n\dots .{\left(c_t,\gamma \right)}_n}{{\left(\gamma ,\gamma \right)}_n{\left(b_1,\gamma \right)}_n\dots .{\left(b_r,\gamma \right)}_n}{z}^n,$$

(5)

$$(t=r+1 such that t,r\in N_0=\left\{0,1,2,3,4,\dots \right\}; z\in U).$$

The γ-shifted factorial is involving by ${\left(c,\gamma \right)}_0=1 and {\left(c,\gamma \right)}_n=\left(1-c\right)\left(1-c\gamma \right)\left(1-c{\gamma }^2\right)\dots \left(1-c{\gamma }^{n-1}\right)$,$n\in N$, where $c$ any complex number and in terms of the Gamma function ${\left({\gamma }^{\mu },\gamma \right)}_n=\frac{{\Gamma }_{\gamma }\left(\mu +n\right){\left(1-\gamma \right)}^n}{{\Gamma }_{\gamma }\left(\mu \right)},$ such that ${\Gamma }_{\gamma }\left(y\right)=\frac{{\left(\gamma ,\gamma \right)}_{\infty }\left(1-\gamma \right)\left(1-y\right)}{{\left({\gamma }^y,\gamma \right)}_{\infty }}, 0<\gamma <1$.

The study suggests that note that and by utilizing ratio test, the series (5) converges absolutely in open unit disk $U$,$\left|\gamma \right|<1$.

$$2{\Psi }_1^{}={\displaystyle \sum }_{n=0}^{\infty }\frac{{\left(c_1,\gamma \right)}_n{\left(c_2,\gamma \right)}_n}{{\left(\gamma ,\gamma \right)}_n{\left(b_1,\gamma \right)}_n}{z}^n, (\left|\gamma \right|<1, z\in U)$$

is the$\gamma$-Gauss hypergeometric function see ([20,21]).

Recently Mohammed and Darus [22] defined the following: $D\left(c_i;b_j;\gamma \right)f:A\to A$.

$$D\left(c_i;b_j;\gamma \right)f\left(z\right)=z+{\displaystyle \sum }_{n=2}^{\infty }\frac{{\left(c_1,\gamma \right)}_{n-1}\dots .{\left(c_t,\gamma \right)}_{n-1}}{{\left(\gamma ,\gamma \right)}_{n-1}{\left(b_1,\gamma \right)}_{n-1}\dots .{\left(b_r,\gamma \right)}_{n-1}}{a}_n{z}^n.$$

Patel [23] defined an integral operator $I_{\nu ,\delta }^{\eta }$ on $A$ as follows:

For $\eta \in {{\rm N}}_0={\rm N}{{\displaystyle \cup }}^{ }\left\{0\right\}, \delta >0$ with $\nu +\delta >0$ and $\nu$ a real number.

Then for $f\in A$, we define the operator $I_{\nu ,\delta }^{\eta }$ by

$$\begin{gathered} I_{\nu ,\delta }^0 f\left(z\right)=f\left(z\right) \\ {I}_{\nu ,\delta }^1 f\left(z\right)=\left(\frac{\nu +\delta }{\delta }\right)z^{1-\left(\frac{\nu +\delta }{\delta }\right)} {{\displaystyle \int }}_0^z{t}^{\left(\frac{\nu +\delta }{\delta }\right)-2} f\left(t\right)dt, z\in U. \\ {I}_{\nu ,\delta }^2 f\left(z\right)=\left(\frac{\nu +\delta }{\delta }\right)z^{1-\left(\frac{\nu +\delta }{\delta }\right)} {{\displaystyle \int }}_0^z{t}^{\left(\frac{\nu +\delta }{\delta }\right)-2} I_{\nu ,\delta }^{1}f\left(t\right)dt, z\in U. \\ {I}_{\nu ,\delta }^{\eta } f\left(z\right)=\left(\frac{\nu +\delta }{\delta }\right)z^{1-\left(\frac{\nu +\delta }{\delta }\right)} {{\displaystyle \int }}_0^z{t}^{\left(\frac{\nu +\delta }{\delta }\right)-2} I_{\nu ,\delta }^{\eta -1}f\left(t\right)dt, z\in U. \\ =I_{\nu ,\delta }^1 \left(\frac{z}{1-z}\right)\ast I_{\nu ,\delta }^1 \left(\frac{z}{1-z}\right)\ast \dots \ast I_{\nu ,\delta }^1 \left(\frac{z}{1-z}\right)\ast f\left(z\right). \end{gathered}$$

We observe that $I_{\nu ,\delta }^{\eta }:A\to A$ is an integral operator and for $f$ given by (1), we have

$$I_{\nu ,\delta }^{\eta } f\left(z\right)=z+{\displaystyle \sum }_{n=2}^{\infty }{\left(\frac{\nu +\delta }{\nu +n\delta }\right)}^{\eta }{a}_n{z}^n.$$

(6)

It follows from (6) that $I_{\nu ,0}^{\eta } f\left(z\right)=f\left(z\right).$ By specializing the parameters, we get

(1)

$I_{1,1}^{\eta } f\left(z\right)={{\rm T}}^{\eta }f\left(z\right)$ (See [23,24]).

(2)

$I_{1-\delta ,1}^{\eta } f\left(z\right)={{\rm T}}_{\delta }^{\eta }f\left(z\right), \delta >0$ (See [23]).

(3)

$I_{\nu ,1}^{\eta } f\left(z\right)={{\rm T}}_{\nu }^{\eta }f\left(z\right), \nu >0$ (See [23]).

Definition 3.

Let$f\in A, z\in U,$we define a new operator$M_{\gamma ,b,v,\delta }^{C,\eta } f:A\to A$, where

$$\begin{array}{c}{M}_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)=D\left(c_i;b_j;\gamma \right)f\left(z\right)\ast I_{\nu ,\delta }^{\eta } f\left(z\right)\hfill \\ =z+{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\frac{{\left(c_1,\gamma \right)}_{n-1}\dots .{\left(c_t,\gamma \right)}_{n-1}}{{\left(\gamma ,\gamma \right)}_{n-1}{\left(b_1,\gamma \right)}_{n-1}\dots .{\left(b_r,\gamma \right)}_{n-1}}{\left(\frac{\nu +\delta }{\nu +n\delta }\right)}^{\eta }{a}_n{z}^n.\hfill \end{array}$$

(7)

We note from (7) that

$$z{\left(M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)\right)}^{\prime }+\frac{\nu }{\delta }{M}_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)=\frac{\nu +\delta }{\delta }{M}_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right).$$

(8)

The specific aim of this investigation is to find sufficient conditions for certain normalized analytic functions $f$ to satisfy:

$$q_1\left(z\right)\prec {\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\prec q_2\left(z\right),$$

and

$$q_1\left(z\right)\prec {\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right)}^{\lambda }\prec q_2\left(z\right),$$

where $q_1\left(z\right)$ and $q_2\left(z\right)$ are given univalent functions in $U$ with $q_1\left(0\right)=q_2\left(0\right)=1$.

In our paper, we derive some sandwich–type results involving the operator $M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)$.

Certain univalent and multivalent functions and differential subordination results for higher-order have been recently studied in the following papers [10,12,17,25,26,27,28,29,30,31,32].

2. Preliminaries

In order to establish our subordination and superordination results, we need the following lemmas and definitions.

Definition 4 

([2]). Denote by $Q$ the class of all functions$q$ that are analytic and injective on$\underset{\_}{U}\backslash E\left(q\right)$, where$\underset{\_}{U}=U{{\displaystyle \cup }}^{ }\left\{z\in \partial U\right\}$, and$E\left(q\right)=\left\{\zeta \in \partial U:q\left(z\right)=\infty \right\}$ and are such that$q^{\prime }\left(\zeta \right)\ne 0$ for$\zeta \in \partial U\backslash E\left(q\right)$. Further, let the subclass of$Q$ for which$q\left(0\right)=a$ be denoted by$Q\left(a\right)$,$Q\left(0\right)=Q_0$ and$Q\left(1\right)=Q_1=\left\{q\in Q:q\left(0\right)=1\right\}.$

Lemma 1 

([6]). Let $q\left(z\right)$ be a convex univalent function in$U$, let$\alpha \in C, \beta \in C\backslash \left\{0\right\}$ and suppose that$Re\left\{1+\frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}\right\}>max\left\{0,-Re\left(\frac{\alpha }{\beta }\right)\right\}$. If$p\left(z\right)$ is analytic in$U$ and$\alpha p\left(z\right)+\beta z{p}^{\prime }\left(z\right)\prec \alpha q\left(z\right)+\beta z{q}^{\prime }\left(z\right)$, then$p\left(z\right)\prec q\left(z\right)$ and$q$ is called the best dominant.

Lemma 2 

([3]). Let $q$ be univalent in$U$ and let$\varnothing$ and$\theta$ be analytic in the domain$D$ containing$q\left(U\right)$ with$\varnothing \left(w\right)\ne 0$, when$w\in q\left(U\right)$. Set$Q\left(z\right)=z{q}^{\prime }\left(z\right)\varnothing \left(q\left(z\right)\right)$ and$h\left(z\right)=\theta \left(q\left(z\right)\right)+Q\left(z\right)$, suppose that

(1) 

$Q$is starlike univalent in$U$,

(2) 

$Re \left(\frac{z{h}^{\prime }\left(z\right)}{Q\left(z\right)}\right)>0, z\in U$.

If $p$ is analytic in $U$ with $q\left(0\right)=p\left(0\right),p\left(U\right)\subseteq D$ and $\varnothing \left(p\left(z\right)\right)+z{p}^{\prime }\left(z\right)\varnothing \left(p\left(z\right)\right)\prec \varnothing \left(q\left(z\right)\right)+z{q}^{\prime }\left(z\right)\varnothing \left(q\left(z\right)\right)$, then $p\left(z\right)\prec q\left(z\right)$, and $q$ is the best dominant.

Lemma 3 

([2]). Let $q\left(z\right)$ be a convex univalent in the unit disk$U$ and let$\theta$ and$\varnothing$ be analytic in a domain$D$ containing$q\left(U\right)$. Suppose that

(1) 

$Re \left\{\frac{{\theta }^{\prime }\left(q\left(z\right)\right)}{\varnothing \left(q\left(z\right)\right)}\right\}>0$for$z\in U$,

(2) 

$Q\left(z\right)=z{q}^{\prime }\left(z\right)\varnothing \left(q\left(z\right)\right)$is starlike univalent in$z\in U$.

If $p\in H\left[q\left(0\right),1\right]{{\displaystyle \cap }}^{ }Q$, with $p\left(U\right)\subseteq D$, and $\theta \left(p\left(z\right)\right)+z{p}^{\prime }\left(z\right)\varnothing \left(p\left(z\right)\right)$ is univalent in $U$ and $\theta \left(q\left(z\right)\right)+z{q}^{\prime }\left(z\right)\varnothing \left(q\left(z\right)\right)\prec \theta \left(p\left(z\right)\right)+z{p}^{\prime }\left(z\right)\varnothing \left(p\left(z\right)\right)$, then $q\left(z\right)\prec p\left(z\right)$, and $q$ is the best subordinant.

Lemma 4 

([2]). Let $q\left(z\right)$ be a convex univalent in$U$ and$q\left(0\right)=1$. Let$\beta \in C$, that$Re\left\{\beta \right\}>0$. If$p\left(z\right)\in H\left[q\left(0\right),1\right]{{\displaystyle \cap }}^{ }Q$ and$p\left(z\right)+\beta z{p}^{\prime }\left(z\right)$ is univalent in$U$, then$q\left(z\right)+\beta z{q}^{\prime }\left(z\right)\prec p\left(z\right)+\beta z{p}^{\prime }\left(z\right),$ which leads to$p\left(z\right)\prec q\left(z\right)$ and$p\left(z\right)$ is called the best subordinant.

3. Differential Subordination Results

Here, some differential subordination results are introduced using the operator $M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)$.

Theorem 1.

Let$q\left(z\right)$be a convex univalent function in$U$with$q\left(0\right)=1$,$0\ne \alpha \in C\backslash \left\{0\right\}, \lambda >0$and assume that$q\left(z\right)$satisfies:

$$Re\left\{1-\frac{z{q}^{″}\left(z\right)}{q^{\prime }\left(z\right)}\right\}>max\left\{0,-Re\left(\frac{\lambda }{\alpha }\right)\right\}$$

(9)

If $f\in A$ satisfies the subordination

$$\lambda \left(\frac{-\nu }{\delta }+\left(\frac{\nu }{\delta }+1\right)\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}-1\right)+p\left(z\right)\prec q\left(z\right)+\frac{\alpha }{\lambda }z{q}^{\prime }\left(z\right),$$

(10)

then

$${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\prec q\left(z\right),$$

(11)

and $q\left(z\right)$ is the best dominant.

Proof. 

Let

$$p\left(z\right)={\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda },$$

(12)

then the function $p\left(z\right)$ is analytic in $U$ and $p\left(0\right)=1$. Therefore, if we differentiate (12) with respect to $z$ and by (8), in the last equation, it follows that

$$\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}=\lambda \left(\frac{z{\left(M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)\right)}^{\prime }}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}-1\right),$$

then

$$\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}=\lambda \left(\frac{-\nu }{\delta }+\left(\frac{\nu }{\delta }+1\right)\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}-1\right).$$

The subordination (10) follows from the hypothesis becomes $p\left(z\right)+\frac{\alpha }{\lambda }z{p}^{\prime }\left(z\right)\prec q\left(z\right)+\frac{\alpha }{\lambda }z{q}^{\prime }\left(z\right)$. □

An application of Lemma (1) with $\beta =\frac{\alpha }{\lambda }$ and $\alpha =1$, we obtain (11).

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

Corollary 1.

Let$\lambda >0, \alpha \in C\backslash \left\{0\right\}$and$\left(-1\le B<A\le 1\right)$.

Assume that$Re\left(\frac{1+Az}{1+Bz}\right)>max\left\{0,-Re\left(\frac{\lambda }{\alpha }\right)\right\}$. If$f\in A$is satisfies the following subordination condition:

$$p\left(z\right)\prec \frac{1+Az}{1+Bz}+\frac{\lambda }{\alpha }\frac{\left(A-B\right)z}{{\left(1+Bz\right)}^2},$$

where$p\left(z\right)$given by (12), then${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\prec \frac{1+Az}{1+Bz}$, and$\frac{1+Az}{1+Bz}$is called the best dominant.

Taking$A=1$and$B=-1$in corollary (1), we get following result.

Corollary 2.

Let$\lambda >0, \alpha \in C\backslash \left\{0\right\}$and suppose that$Re\left(\frac{1+z}{1-z}\right)>max\left\{0,-Re\left(\frac{\lambda }{\alpha }\right)\right\}$. If$f\in A$is satisfies the following subordination

$$p\left(z\right)\prec \frac{1+z}{1-z}+\frac{\lambda }{\alpha }\frac{2z}{{\left(1-z\right)}^2}$$

where$p\left(z\right)$given by (12), then${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\prec \frac{1+z}{1-z}$, and$\frac{1+z}{1-z}$is called the best dominant.

Theorem 2.

Let$q\left(z\right)$be a convex univalent function in$U$with$q\left(0\right)=1$,$q^{\prime }\left(z\right)\ne 0,\alpha \in C\backslash \left\{0\right\},\lambda >0$and suppose that$q\left(z\right)$satisfies:

$$Re\left\{1+\frac{m}{\epsilon }\left(q^m\left(z\right)+q^{m+1}\left(z\right)\right)+\frac{q^{m+1}\left(z\right)}{\epsilon }-z\frac{q^{\prime }\left(z\right)}{q\left(z\right)}+z\frac{q^{″}\left(z\right)}{q^{\prime }\left(z\right)}\right\}>0,$$

(13)

where $m\in C, \epsilon \in C\backslash \left\{0\right\}$ and $z\in U.$

Suppose that $\frac{z{q}^{\prime }\left(z\right)}{q\left(z\right)}$ is starlike univalent in $U$. If $f\in A$ satisfies:

$$G\left(z\right)={\varphi }_{\left(c,\eta ,\gamma ,b,\nu ,\delta ,\alpha ,m,\epsilon ;z\right)}\prec \left(1+q\left(z\right)\right){\left(q\left(z\right)\right)}^m+z\epsilon \frac{q^{\prime }\left(z\right)}{q\left(z\right)},$$

(14)

where

$$G\left(z\right)={\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right)}^{\lambda m}+{\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right)}^{\lambda \left(m+1\right)}+\epsilon \lambda \left(\frac{\nu +\delta }{\delta }\right)\left[\frac{M_{\gamma ,b,v,\delta }^{C,\eta -2} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}-\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right],$$

(15)

then ${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right)}^{\lambda }\prec p\left(z\right)$, and $p\left(z\right)$ is called the best dominant.

Proof. 

Let

$$p\left(z\right)={\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right)}^{\lambda }.$$

(16)

Then the function $p\left(z\right)$ is analytic in $U$ and $p\left(0\right)=1$. Therefore, differentiating (13) with respect to $z$ and by (8), it follows that

$$\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}=\lambda \left(\frac{\nu +\delta }{\delta }\right)\left[\frac{M_{\gamma ,b,v,\delta }^{C,\eta -2} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}-\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right].$$

(17)

$$\mathrm{Set} \theta \left(w\right)=\left(1+w\right)w^m \mathrm{and} \varphi \left(w\right)=\frac{\epsilon }{w}, w\ne 0,$$

It follows that$\theta \left(w\right)$ and $\varphi \left(w\right)$ are analytic in$C\backslash \left\{0\right\}$, and $\varphi \left(w\right)\ne 0,w\in C\backslash \left\{0\right\}$. In addition,

$$Q\left(z\right)=z{q}^{\prime }\left(z\right) \varphi \left(q\left(z\right)\right)=z\epsilon \frac{q^{\prime }\left(z\right)}{q\left(z\right)} \mathrm{and} h\left(z\right)=\theta \left(q\left(z\right)\right)+Q\left(z\right)=\left(1+q\left(z\right)\right){\left(q\left(z\right)\right)}^m+z\epsilon \frac{q^{\prime }\left(z\right)}{q\left(z\right)}.$$

It is clear that $Q\left(z\right)$ is starlike univalent in $U$, we have

$$Re\left\{\frac{z{h}^{\prime }\left(z\right)}{Q\left(z\right)}\right\}=Re\left\{1+\frac{m}{\epsilon }\left(q^m\left(z\right)+q^{m+1}\left(z\right)\right)+\frac{q^{m+1}\left(z\right)}{\epsilon }-z\frac{q^{\prime }\left(z\right)}{q\left(z\right)}+z\frac{q^{″}\left(z\right)}{q^{\prime }\left(z\right)}\right\}>0.$$

By a straightforward computation, we obtain

$$G\left(z\right)=\left(1+p\left(z\right)\right){\left(p\left(z\right)\right)}^m+z\epsilon \frac{p^{\prime }\left(z\right)}{p\left(z\right)},$$

(18)

where $G\left(z\right)$ is given by (15). From (14) and (18), we have

$$\left(1+p\left(z\right)\right){\left(p\left(z\right)\right)}^m+z\epsilon \frac{p^{\prime }\left(z\right)}{p\left(z\right)}\prec \left(1+q\left(z\right)\right){\left(q\left(z\right)\right)}^m+z\epsilon \frac{q^{\prime }\left(z\right)}{q\left(z\right)}.$$

(19)

Thus, from Lemma (2), it follows that $p\left(z\right)\prec q\left(z\right)$.

By (16), we get the result. □

Taking $q\left(z\right)=\frac{1+Az}{1+Bz} \left(-1\le B<A\le 1\right),$ in Theorem (2), for every $\lambda >0, \epsilon \in C\backslash \left\{0\right\}$ the condition (13) becomes

$$Re\left\{1+\frac{m}{\epsilon }\left({\left(\frac{1+Az}{1+Bz}\right)}^m+{\left(\frac{1+Az}{1+Bz}\right)}^{m+1}\right)+\frac{1}{\epsilon }{\left(\frac{1+Az}{1+Bz}\right)}^{m+1}-\frac{\left(A-B\right)z}{\left(1+Bz\right)\left(1+Az\right)}-\frac{2zB}{1+Bz}\right\}>0$$

(20)

Thus, we obtain the following corollary.

Corollary 3.

Let$\lambda >0, \alpha \in C\backslash \left\{0\right\}, m\in C$and$-1\le B<A\le 1$. Assume that (20) hold. If$f\in A$and$G\left(z\right)\prec \left(1+\frac{1+Az}{1+Bz}\right){\left(\frac{1+Az}{1+Bz}\right)}^m+\frac{\left(A-B\right)z\epsilon }{\left(1+Bz\right)\left(1+Az\right)}$, where$G\left(z\right)$given by (15), then${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\prec \frac{1+Az}{1+Bz}$, and$\frac{1+Az}{1+Bz}$is called the best dominant.

Taking$p\left(z\right)={\left(\frac{1+z}{1-z}\right)}^{\rho } (0<\rho \le 1)$, in Theorem (2), the condition (20) becomes

$$Re\left\{1+\frac{m}{\epsilon }\left({\left(\frac{1+z}{1-z}\right)}^{\rho m}+{\left(\frac{1+z}{1-z}\right)}^{\rho \left(m+1\right)}\right)+\frac{1}{\epsilon }{\left(\frac{1+z}{1-z}\right)}^{\rho \left(m+1\right)}-\frac{2z\rho }{1-z^2}+\frac{2z\left(z+\rho \right)}{1-z^2}\right\}>0.$$

(21)

Thus, we obtain the following corollary.

Corollary 4.

Let$\lambda >0, \alpha \in C\backslash \left\{0\right\}, m\in C$and$\left(0<\rho \le 1\right)$. Assume that (21) holds. If$f\in A$and$G\left(z\right)\prec \left(1+\frac{1+z}{1-z}\right){\left(\frac{1+z}{1-z}\right)}^{\rho m}+\frac{2z\epsilon \rho }{1-z^2}$, where$G\left(z\right)$given by (15), then${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\prec {\left(\frac{1+z}{1-z}\right)}^{\rho }$, and${\left(\frac{1+z}{1-z}\right)}^{\rho }$is the best dominant.

4. Differential Superordination Results

Theorem 3.

Let$q\left(z\right)$be a convex univalent function in$U$with$q\left(0\right)=1$,$\alpha \in C, \lambda >0$with$Re\left(\alpha \right)>0$. If$f\in A$such that

$${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\ne 0,$$

and suppose that $f$ satisfies the condition:

$${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\in H\left[q\left(0\right),1\right]{{\displaystyle \cap }}^{ }Q.$$

(22)

If the function $p\left(z\right)$ given by (12) is univalent and the superordination condition:

$$q\left(z\right)+\frac{\alpha }{\lambda }z{q}^{\prime }\left(z\right)\prec p\left(z\right)+\lambda \left(\frac{-\nu }{\delta }+\left(\frac{\nu }{\delta }+1\right)\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}-1\right), \mathrm{holds},$$

(23)

then$q\left(z\right)\prec {\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda },$ and$q\left(z\right)$ is called the best subordinant.

Proof. 

Let

$$p\left(z\right)={\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }.$$

(24)

Differentiating (12) with respect to $z$, we get

$$\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}=\lambda \left(\frac{-\nu }{\delta }+\left(\frac{\nu }{\delta }+1\right)\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}-1\right)$$

(25)

$$\lambda \left(\frac{-\nu }{\delta }+\left(\frac{\nu }{\delta }+1\right)\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}-1\right)+p\left(z\right)\prec q\left(z\right)+\frac{\alpha }{\lambda }z{q}^{\prime }\left(z\right).$$

A simple computation and using (8), from (25), we get $\lambda \left(\frac{-\nu }{\delta }+\left(\frac{\nu }{\delta }+1\right)\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}-1\right)=\frac{\alpha }{\lambda }z{p}^{\prime }\left(z\right)$.

By Lemma (4), we obtain the desired result.

Taking $q\left(z\right)=\frac{1+Az}{1+Bz} \left(-1\le B<A\le 1\right),$ in Theorem (3), we obtain the following corollary. □

Corollary 5.

Let$\lambda >0, \alpha \in C\backslash \left\{0\right\}$and$\left(-1\le B<A\le 1\right)$, such that

$${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\in H\left[q\left(0\right),1\right]{{\displaystyle \cap }}^{ }Q.$$

If the function$p\left(z\right)$given by (12) is univalent in$U$and$f\in A$satisfies the superordination condition:

$$\frac{1+Az}{1+Bz}+\frac{\alpha }{\lambda }\frac{\left(1+Bz\right)Az-\left(1+Az\right)Bz}{{\left(1+Bz\right)}^2}\prec p\left(z\right),$$

(26)

then$\frac{1+Az}{1+Bz}\prec {\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda },$and$\frac{1+Az}{1+Bz}$is called the best subordinant.

Theorem 4.

Let$q\left(z\right)$be a convex univalent function in$U$,$q^{\prime }\left(z\right)\ne 0$,$\alpha ,\epsilon \in C\backslash \left\{0\right\}, \lambda >0$,$m\in C$and suppose that$q\left(z\right)$satisfies:

$Re\left\{\frac{1}{\epsilon }\left(\left(m\right)q^m\left(z\right)+\left(m+1\right)q^{m+1}\left(z\right)\right)\right\}>0.$ Let $f\left(z\right)\in A$ and suppose that satisfies the next condition:

$${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right)}^{\lambda }\in H\left[q\left(0\right),1\right]{{\displaystyle \cap }}^{ }Q,$$

and

$${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right)}^{\lambda }\ne 0.$$

If the function $p\left(z\right)$ given by (16) is univalent in $U$ and

$${\left(q\left(z\right)\right)}^m+{\left(q\left(z\right)\right)}^{m+1}+\epsilon \frac{z{q}^{\prime }\left(z\right)}{q\left(z\right)} \prec G\left(z\right),$$

(27)

then, $q\left(z\right)\prec {\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right)}^{\lambda },$ and$q\left(z\right)$ is called the best subordinant.

Proof. 

Let the function $p\left(z\right)$ defined on $U$ by (16). Then a computation show that

$$\frac{z{q}^{\prime }\left(z\right)}{q\left(z\right)}=\lambda \left(\frac{\nu +\delta }{\delta }\right)\left[\frac{M_{\gamma ,b,v,\delta }^{C,\eta -2} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}-\frac{M_{\gamma ,b,v,\delta }^{C,\eta -1} f\left(z\right)}{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}\right].$$

(28)

By setting $\theta \left(w\right)=\left(1+w\right)w^m$ and $\varphi \left(w\right)=\frac{\epsilon }{w}, w\ne 0$, we see that $\theta \left(w\right)$ and $\varphi \left(w\right)$ are analytic in $C\backslash \left\{0\right\}$, and that $\varphi \left(w\right)\ne 0,w\in C\backslash \left\{0\right\}$. In addition, $Q\left(z\right)=z{q}^{\prime }\left(z\right) \varphi \left(q\left(z\right)\right)=z\epsilon \frac{q^{\prime }\left(z\right)}{q\left(z\right)}$.

Clearly, $Q\left(z\right)$ is starlike univalent in $U$, we have $Re\left\{\frac{{\theta }^{\prime }\left(q\left(z\right)\right)}{\varphi \left(q\left(z\right)\right)}\right\}=Re\left\{\frac{1}{\epsilon }\left(m{q}^m\left(z\right)+\left(m+1\right)q^{m+1}\left(z\right)\right)\right\}>0.$ By a straightforward computation, it follows that:

$$G\left(z\right)=\theta \left(p\left(z\right)\right)+p\left(z\right)=\left(1+p\left(z\right)\right){\left(p\left(z\right)\right)}^m+z\epsilon \frac{p^{\prime }\left(z\right)}{p\left(z\right)},$$

(29)

where $G\left(z\right)$ is given by (15).

From (27) and (29), it follows that:

$$\left(1+q\left(z\right)\right){\left(q\left(z\right)\right)}^m+z\epsilon \frac{q^{\prime }\left(z\right)}{q\left(z\right)}\prec \left(1+p\left(z\right)\right){\left(p\left(z\right)\right)}^m+z\epsilon \frac{p^{\prime }\left(z\right)}{p\left(z\right)}.$$

(30)

Thus by applying Lemma (3), we get $q\left(z\right)\prec p\left(z\right)$.

The proof is complete. □

5. Sandwich Results

If we combine Theorem (1) with Theorem (3), we get the sandwich Theorem:

Theorem 5.

Let$q_1$and$q_2$be a convex univalent functions in$U$with$q_1\left(0\right)=q_2\left(0\right)=1$and$q_2$satisfies (9). Suppose that$Re\left\{\alpha \right\}>0, \lambda >0, \alpha \in C\backslash \left\{0\right\}.$

If $f\in A$, such that

$${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\in H\left[q\left(0\right),1\right]{{\displaystyle \cap }}^{ }Q,$$

and the function $p\left(z\right)$ defined by (12) is univalent and satisfies,

$$q_1\left(z\right)+\frac{\alpha }{\lambda }z{q}_1^{’}\left(z\right)\prec p\left(z\right)\prec q_2\left(z\right)+\frac{\alpha }{\lambda }z{q}_2^{’}\left(z\right),$$

(31)

implies that

$$q_1\left(z\right)\prec {\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\prec q_2\left(z\right),$$

where$q_1$ and$q_2$ are respectively, the best subordinant and the best dominant of (31).

If we combine Theorem (2) with Theorem (4), we get the sandwich Theorem:

Theorem 6.

Let$q_i$be two convex univalent functions in$U$, such that$q_i\left(0\right)=1, q_i^{\prime }\left(z\right)\ne 0 \left(i=1,2\right)$. Suppose that$q_1$and$q_2$satisfies (24) and (14), respectively.

If $f\in A$ and assume that $f$ satisfies the next conditions:

$${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\ne 0.$$

and

$${\left(\frac{M_{\gamma ,b,v,\delta }^{C,\eta } f\left(z\right)}{z}\right)}^{\lambda }\in H\left[q\left(0\right),1\right]{{\displaystyle \cap }}^{ }Q,$$

and $\varphi \left(z\right)$ is univalent in $U$, then

$$\left(1+q_1\left(z\right)\right){\left(q_1\left(z\right)\right)}^m+z\epsilon \frac{q_1^{\prime }\left(z\right)}{q_1\left(z\right)}\prec \varphi \left(z\right)\prec \left(1+q_2\left(z\right)\right){\left(q_2\left(z\right)\right)}^m+z\epsilon \frac{q_2^{\prime }\left(z\right)}{q_2\left(z\right)}$$

(32)

Implies $q_2\left(z\right)\prec \varphi \left(z\right)\prec q_2\left(z\right)$, and $q_1$ and $q_2$ are the best subordinant and the best dominant respectively and $\varphi \left(z\right)$ takes the form as in (15).