Theory and Modern Applications

# Strong differential subordinations and superordinations obtained with some new integral operators

## Abstract

In this paper we study certain strong differential subordinations and superordinations obtained by using some new integral operators introduced in (Oros and Oros in Differential subordinations obtained with some new integral operator (to appear)).

MSC:30C80, 30C45, 30C20, 34A40.

## 1 Introduction and preliminaries

The concept of differential subordination was introduced in [1, 2] and developed in [3] by Miller and Mocanu. The concept of differential superordination was introduced in [4] like a dual problem of the differential subordination by Miller and Mocanu and developed in [5]. The concept of strong differential subordination was introduced in [6] by Antonino and Romaguera and developed in [7â€“17]. The concept of strong differential superordination was introduced in [18] like a dual concept of the strong differential subordination and developed in [19â€“21].

In [8] the author defines the following classes:

Denote by $\mathcal{H}\left(UÃ—\stackrel{Â¯}{U}\right)$ the class of analytic functions in $UÃ—\stackrel{Â¯}{U}$, where

$U=\left\{zâˆˆ\mathbb{C}:|z|<1\right\},\phantom{\rule{2em}{0ex}}\stackrel{Â¯}{U}=\left\{zâˆˆ\mathbb{C}:|z|â‰¤1\right\},\phantom{\rule{2em}{0ex}}\mathrm{âˆ‚}U=\left\{zâˆˆ\mathbb{C}:|z|=1\right\}.$

For $aâˆˆ\mathbb{C}$ and $nâˆˆ{\mathbb{N}}^{âˆ—}$, we denote by

$\mathcal{H}\mathrm{Î¾}\left[a,n\right]=\left\{f\left(z,\mathrm{Î¾}\right)âˆˆ\mathcal{H}\left(UÃ—\stackrel{Â¯}{U}\right):f\left(z,\mathrm{Î¾}\right)=a+{a}_{n}\left(\mathrm{Î¾}\right){z}^{n}+{a}_{n+1}\left(\mathrm{Î¾}\right){z}^{n+1}+â‹¯\right\},$

with $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, ${a}_{k}\left(\mathrm{Î¾}\right)$ holomorphic functions in $\stackrel{Â¯}{U}$, $kâ‰¥n$. Let

$A{\mathrm{Î¾}}_{n}=\left\{f\left(z,\mathrm{Î¾}\right)âˆˆ\mathcal{H}\left(UÃ—\stackrel{Â¯}{U}\right):f\left(z,\mathrm{Î¾}\right)=z+{a}_{n+1}\left(\mathrm{Î¾}\right){z}^{n+1}+â‹¯\right\},$

with $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, ${a}_{k}\left(\mathrm{Î¾}\right)$ holomorphic functions in $\stackrel{Â¯}{U}$, $kâ‰¥n+1$, and $A{\mathrm{Î¾}}_{1}=A\mathrm{Î¾}$,

denote the class of univalent functions in $\mathcal{H}\left(UÃ—\stackrel{Â¯}{U}\right)$,

denote the class of normalized starlike functions in $UÃ—\stackrel{Â¯}{U}$,

denote the class of normalized convex functions in $UÃ—\stackrel{Â¯}{U}$.

Let $A\left(p\right)\mathrm{Î¾}$ denote the subclass of the functions $f\left(z,\mathrm{Î¾}\right)âˆˆ\mathcal{H}\left(UÃ—\stackrel{Â¯}{U}\right)$ of the form

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

Definition 1.1 [7, 18]

Let $F\left(z,\mathrm{Î¾}\right)$ and $f\left(z,\mathrm{Î¾}\right)$ be members of $H\left(UÃ—\stackrel{Â¯}{U}\right)$. The function $f\left(z,\mathrm{Î¾}\right)$ is said to be strongly subordinate to $F\left(z,\mathrm{Î¾}\right)$, or $F\left(z,\mathrm{Î¾}\right)$ is said to be strongly superordinate to $f\left(z,\mathrm{Î¾}\right)$ if there exists a function w analytic in U, with $w\left(0\right)=0$ and $|w\left(z\right)|<1$ such that $f\left(z,\mathrm{Î¾}\right)=F\left(w\left(z\right),\mathrm{Î¾}\right)$ for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. In such a case, we write $f\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºF\left(z,\mathrm{Î¾}\right)$, $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. If $F\left(z,\mathrm{Î¾}\right)$ is univalent, then $f\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºF\left(z,\mathrm{Î¾}\right)$ if and only if $F\left(0,\mathrm{Î¾}\right)=f\left(0,\mathrm{Î¾}\right)$ and $f\left(UÃ—\stackrel{Â¯}{U}\right)âŠ‚F\left(UÃ—\stackrel{Â¯}{U}\right)$.

Remark 1.1 If $F\left(z,\mathrm{Î¾}\right)â‰¡F\left(z\right)$ and $f\left(z,\mathrm{Î¾}\right)â‰¡f\left(z\right)$, then the strong differential subordination or superordination becomes the usual notions of differential subordination and superordination, respectively.

Definition 1.2 [7]

We denote by ${Q}_{\mathrm{Î¾}}$ the set of functions $q\left(â‹\dots ,\mathrm{Î¾}\right)$ that are analytic and injective, as functions of z on $\stackrel{Â¯}{U}âˆ–E\left(q\left(z,\mathrm{Î¾}\right)\right)$, where $E\left(q\left(z,\mathrm{Î¾}\right)\right)=\left\{\mathrm{Î¶}âˆˆ\mathrm{âˆ‚}U:{lim}_{zâ†’\mathrm{Î¶}}q\left(z,\mathrm{Î¾}\right)=\mathrm{âˆž}\right\}$, and are such that for $\mathrm{Î¶}âˆˆ\mathrm{âˆ‚}Uâˆ–E\left(q\left(z,\mathrm{Î¾}\right)\right)$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. The subclass of ${Q}_{\mathrm{Î¾}}$ for which $q\left(0,\mathrm{Î¾}\right)=a$ is denoted by ${Q}_{\mathrm{Î¾}}\left(a\right)$.

Let $\mathrm{Î¨}:{\mathbb{C}}^{3}Ã—UÃ—\stackrel{Â¯}{U}â†’\mathbb{C}$, and let $h\left(z,\mathrm{Î¾}\right)$ be univalent in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. If $p\left(z,\mathrm{Î¾}\right)$ is analytic in $UÃ—\stackrel{Â¯}{U}$ and satisfies the (second-order) strong differential subordination

$\mathrm{Î¨}\left(p\left(z,\mathrm{Î¾}\right),z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right),{z}^{2}{p}_{{z}^{2}}^{â€²â€²}\left(z,\mathrm{Î¾}\right);z,\mathrm{Î¾}\right)â‰ºâ‰ºh\left(z,\mathrm{Î¾}\right),\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$
(1.1)

then $p\left(z,\mathrm{Î¾}\right)$ is called a solution of the strong differential subordination.

The univalent function $q\left(z,\mathrm{Î¾}\right)$ is called a dominant of the solutions of the strong differential subordination, or simply a dominant, if $p\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right)$ for all $p\left(z,\mathrm{Î¾}\right)$ satisfying (1.1). A dominant $\stackrel{Ëœ}{q}\left(z,\mathrm{Î¾}\right)$ that satisfies $\stackrel{Ëœ}{q}\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right)$ for all dominants $q\left(z,\mathrm{Î¾}\right)$ of (1.1) is said to be the best dominant of (1.1). Note that the best dominant is unique up to a rotation of $UÃ—\stackrel{Â¯}{U}$.

Let $\mathrm{Ï†}:{\mathbb{C}}^{3}Ã—UÃ—\stackrel{Â¯}{U}â†’\mathbb{C}$, and let $h\left(z,\mathrm{Î¾}\right)$ be analytic in $UÃ—\stackrel{Â¯}{U}$. If $p\left(z,\mathrm{Î¾}\right)$ and $\mathrm{Ï†}\left(p\left(z,\mathrm{Î¾}\right),z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right),{z}^{2}{p}_{{z}^{2}}^{â€²â€²}\left(z,\mathrm{Î¾}\right);z,\mathrm{Î¾}\right)$ are univalent in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$ and satisfy the (second-order) strong differential superordination,

then $p\left(z,\mathrm{Î¾}\right)$ is called a solution of the strong differential superordination. An analytic function $q\left(z,\mathrm{Î¾}\right)$ is called a subordinant of the solutions of the strong differential superordination, or more simple a subordinant if $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºp\left(z,\mathrm{Î¾}\right)$ for all $p\left(z,\mathrm{Î¾}\right)$ satisfying (1.1â€²). A univalent subordinant $\stackrel{Ëœ}{q}\left(z,\mathrm{Î¾}\right)$ that satisfies $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰º\stackrel{Ëœ}{q}\left(z,\mathrm{Î¾}\right)$ for all subordinants $q\left(z,\mathrm{Î¾}\right)$ of (1.1â€²) is said to be the best subordinant. Note that the best subordinant is unique up to a rotation of $UÃ—\stackrel{Â¯}{U}$.

We rewrite the operators defined in [22] for the classes presented earlier as follows.

Definition 1.3 [22]

For $f\left(z,\mathrm{Î¾}\right)âˆˆA{\mathrm{Î¾}}_{n}$, $nâˆˆ{N}^{âˆ—}$, $mâˆˆN$, $\mathrm{Î³}âˆˆC$, let ${L}_{\mathrm{Î³}}$ be the integral operator given by ${L}_{\mathrm{Î³}}:A{\mathrm{Î¾}}_{n}â†’A{\mathrm{Î¾}}_{n}$,

$\begin{array}{r}{L}_{\mathrm{Î³}}^{0}f\left(z,\mathrm{Î¾}\right)=f\left(z,\mathrm{Î¾}\right),\\ {L}_{\mathrm{Î³}}^{1}f\left(z,\mathrm{Î¾}\right)=\frac{\mathrm{Î³}+1}{{z}^{\mathrm{Î³}}}{âˆ«}_{0}^{z}{L}_{\mathrm{Î³}}^{0}f\left(t,\mathrm{Î¾}\right){t}^{\mathrm{Î³}âˆ’1}\phantom{\rule{0.2em}{0ex}}dt,\\ {L}_{\mathrm{Î³}}^{2}f\left(z,\mathrm{Î¾}\right)=\frac{\mathrm{Î³}+1}{{z}^{\mathrm{Î³}}}{âˆ«}_{0}^{z}{L}_{\mathrm{Î³}}^{1}f\left(t,\mathrm{Î¾}\right){t}^{\mathrm{Î³}âˆ’1}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{2em}{0ex}}â€¦,\\ {L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)=\frac{\mathrm{Î³}+1}{{z}^{\mathrm{Î³}}}{âˆ«}_{0}^{z}{L}_{\mathrm{Î³}}^{mâˆ’1}f\left(t,\mathrm{Î¾}\right)â‹\dots {t}^{\mathrm{Î³}âˆ’1}\phantom{\rule{0.2em}{0ex}}dt.\end{array}$

By using Definition 1.3, we can prove the following properties for this integral operator:

For $f\left(z,\mathrm{Î¾}\right)âˆˆA{\mathrm{Î¾}}_{n}$, $nâˆˆ{\mathbb{N}}^{âˆ—}$, $mâˆˆ\mathbb{N}$, $\mathrm{Î³}âˆˆ\mathbb{C}$, we have

${L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)=z+\underset{k=n+1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(\mathrm{Î³}+1\right)}^{m}}{{\left(\mathrm{Î³}+k\right)}^{m}}â‹\dots {a}_{k}\left(\mathrm{Î¾}\right){z}^{k},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$
(1.2)

and

$zâ‹\dots {\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}=\left(\mathrm{Î³}+1\right){L}_{\mathrm{Î³}}^{mâˆ’1}f\left(z,\mathrm{Î¾}\right)âˆ’\mathrm{Î³}{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right),\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}.$
(1.3)

Definition 1.4 [22]

For $pâˆˆN$, $mâˆˆN$, $f\left(z,\mathrm{Î¾}\right)âˆˆA\left(p\right)\mathrm{Î¾}$, let H be the integral operator given by $H:A\left(p\right)\mathrm{Î¾}â†’A\left(p\right)\mathrm{Î¾}$,

$\begin{array}{r}{H}^{0}f\left(z,\mathrm{Î¾}\right)=f\left(z,\mathrm{Î¾}\right),\\ {H}^{1}f\left(z,\mathrm{Î¾}\right)=\frac{p+1}{z}{âˆ«}_{0}^{z}{H}^{0}f\left(t,\mathrm{Î¾}\right)\phantom{\rule{0.2em}{0ex}}dt,\\ {H}^{2}f\left(z,\mathrm{Î¾}\right)=\frac{p+1}{z}{âˆ«}_{0}^{z}{H}^{1}f\left(t,\mathrm{Î¾}\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{2em}{0ex}}â€¦,\\ {H}^{m}f\left(z,\mathrm{Î¾}\right)=\frac{p+1}{z}{âˆ«}_{0}^{z}{H}^{mâˆ’1}f\left(t,\mathrm{Î¾}\right)\phantom{\rule{0.2em}{0ex}}dt.\end{array}$

By using Definition 1.4, we can prove the following properties for this integral operator:

For $f\left(z,\mathrm{Î¾}\right)âˆˆA\left(p\right)\mathrm{Î¾}$, $mâˆˆ\mathbb{N}$, $pâˆˆ\mathbb{N}$, we have

${H}^{m}f\left(z,\mathrm{Î¾}\right)={z}^{p}+\underset{k=p+1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(p+1\right)}^{m}}{{\left(p+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right){z}^{k},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$
(1.4)

and

$z{\left[â‹\dots {H}^{m}f\left(z,\mathrm{Î¾}\right)\right]}^{â€²}=\left(p+1\right){H}^{mâˆ’1}f\left(z,\mathrm{Î¾}\right)âˆ’{H}^{m}f\left(z,\mathrm{Î¾}\right),\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}.$
(1.5)

We rewrite the following lemmas for the classes presented earlier (the proofs are similar to those found in [6]).

Lemma 1.1 [[3], Th. 3.4, p.132]

Let the function $q\left(z,\mathrm{Î¾}\right)$ be univalent in U for all $\mathrm{Î¶}âˆˆ\stackrel{Â¯}{U}$, and let Î¸ and Ï† be analytic in a domain D containing $q\left(UÃ—\stackrel{Â¯}{U}\right)$ with when $\mathrm{Ï‰}âˆˆq\left(UÃ—\stackrel{Â¯}{U}\right)$.

Set $Q\left(z,\mathrm{Î¾}\right)=z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)â‹\dots \mathrm{Ï†}\left(q\left(z,\mathrm{Î¾}\right)\right)$ and $h\left(z,\mathrm{Î¾}\right)=\mathrm{Î¸}\left(q\left(z,\mathrm{Î¾}\right)\right)+Q\left(z,\mathrm{Î¾}\right)$. Suppose that

1. (i)

$Q\left(z,\mathrm{Î¾}\right)$ is starlike univalent in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$,

2. (ii)

$Re\frac{z{h}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{Q\left(z,\mathrm{Î¾}\right)}>0$, $zâˆˆU$ for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$.

If $p\left(z,\mathrm{Î¾}\right)$ is analytic in $UÃ—\stackrel{Â¯}{U}$ with $p\left(0,\mathrm{Î¾}\right)=q\left(0,\mathrm{Î¾}\right)$, $p\left(UÃ—\stackrel{Â¯}{U}\right)âŠ†D$ and

$\mathrm{Î¸}\left(p\left(z,\mathrm{Î¾}\right)\right)+z{p}^{â€²}\left(z,\mathrm{Î¾}\right)â‹\dots \mathrm{Ï†}\left(p\left(z,\mathrm{Î¾}\right)\right)â‰ºâ‰º\mathrm{Î¸}\left(q\left(z,\mathrm{Î¾}\right)\right)+z{q}^{â€²}\left(z,\mathrm{Î¾}\right)â‹\dots \mathrm{Ï†}\left(q\left(z,\mathrm{Î¾}\right)\right),$

then $p\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right)$, $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$ and $q\left(z,\mathrm{Î¾}\right)$ is the best strong dominant.

Lemma 1.2 [[5], Corollary 1.1]

Let $\mathrm{Î±},\mathrm{Î²},\mathrm{Î³}âˆˆ\mathbb{C}$, and let $h\left(z,\mathrm{Î¾}\right)$ be convex in U for all $\mathrm{Î¶}âˆˆ\stackrel{Â¯}{U}$, with $h\left(0,\mathrm{Î¾}\right)=a$ and $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºh\left(z,\mathrm{Î¾}\right)$, $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. Suppose that the differential equation $q\left(z,\mathrm{Î¾}\right)+\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{q\left(z,\mathrm{Î¾}\right)}=h\left(z,\mathrm{Î¾}\right)$ has a univalent solution $q\left(z,\mathrm{Î¾}\right)$ that satisfies $q\left(0,\mathrm{Î¾}\right)=a$.

If $p\left(z,\mathrm{Î¾}\right)âˆˆ\left[a,1\right]âˆ©{Q}_{\mathrm{Î¾}}$ and $p\left(z,\mathrm{Î¾}\right)+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{\mathrm{Î²}p\left(z,\mathrm{Î¾}\right)+\mathrm{Î³}}$ is univalent in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, then $h\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºp\left(z,\mathrm{Î¾}\right)+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{\mathrm{Î²}p\left(z,\mathrm{Î¾}\right)+\mathrm{Î³}}$ implies $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºp\left(z,\mathrm{Î¾}\right)$, $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. The function $q\left(z,\mathrm{Î¾}\right)$ is the best subordinant.

## 2 Main results

We first give results related to strong differential subordinations.

Theorem 2.1 Let $q\left(z,\mathrm{Î¾}\right)$ be univalent in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, with $q\left(0,\mathrm{Î¾}\right)=1$ and , and suppose that

1. (j)

$Req\left(z,\mathrm{Î¾}\right)>0$,

(jj) $Re\left(1+\frac{z{q}_{{z}^{2}}^{â€²â€²}\left(z,\mathrm{Î¾}\right)}{{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}âˆ’\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{q\left(z,\mathrm{Î¾}\right)}\right)>0$.

Let $nâˆˆ{\mathbb{N}}^{âˆ—}$, $\mathrm{Î³}âˆˆ\mathbb{C}$, $f\left(z,\mathrm{Î¾}\right)âˆˆA{\mathrm{Î¾}}_{n}$ and

$\begin{array}{r}\frac{{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)â‹\dots {\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}}{z}+\frac{z{\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{{z}^{2}}^{â€²â€²}}{{\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}}+\frac{{\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}}{{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}âˆ’1\\ \phantom{\rule{1em}{0ex}}â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right)+\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{q\left(z,\mathrm{Î¾}\right)},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},\end{array}$
(2.1)

then

$\frac{{\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}â‹\dots {L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}{z}â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right),\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$

and $q\left(z,\mathrm{Î¾}\right)$ is the best dominant.

Proof We let

$p\left(z,\mathrm{Î¾}\right)=\frac{{\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}â‹\dots {L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}{z},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}.$
(2.2)

Using (1.2) in (2.2), we have

$\begin{array}{rl}p\left(z,\mathrm{Î¾}\right)& =\frac{{\left[z+{âˆ‘}_{k=n+1}^{\mathrm{âˆž}}\frac{{\left(\mathrm{Î³}+1\right)}^{m}}{{\left(\mathrm{Î³}+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right){z}^{k}\right]}^{â€²}\left[z+{âˆ‘}_{k=n+1}^{\mathrm{âˆž}}\frac{{\left(\mathrm{Î³}+1\right)}^{m}}{{\left(\mathrm{Î³}+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right){z}^{k}\right]}{z}\\ =1+{A}_{n}\left(\mathrm{Î¾}\right){z}^{n}+â‹¯,\end{array}$

and since $p\left(0,\mathrm{Î¾}\right)=1$, we obtain that $p\left(z,\mathrm{Î¾}\right)âˆˆ\mathrm{Î¾}\left[1,n\right]$.

Differentiating (2.2), and after a short calculus, we obtain

$p\left(z,\mathrm{Î¾}\right)+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{p\left(z,\mathrm{Î¾}\right)}=\frac{{\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}â‹\dots {L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}{z}+\frac{z{\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{{z}^{2}}^{â€²â€²}}{{\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}}+\frac{z{\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}}{{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}âˆ’1.$
(2.3)

Using (2.3) in (2.1), the strong differential subordination (2.1) becomes

$p\left(z,\mathrm{Î¾}\right)+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{p\left(z,\mathrm{Î¾}\right)}â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right)+\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{q\left(z,\mathrm{Î¾}\right)},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}.$
(2.4)

In order to prove the theorem, we shall use Lemma 1.1. For that, we show that the necessary conditions are satisfied. Let the functions $\mathrm{Î˜}:\mathbb{C}â†’\mathbb{C}$ and $\mathrm{Ï†}:\mathbb{C}â†’\mathbb{C}$, with

$\mathrm{Î˜}\left(\mathrm{Ï‰}\right)=\mathrm{Ï‰}$
(2.5)

and

(2.6)

We check the conditions from the hypothesis of Lemma 1.1. Using (2.6), we have

$Q\left(z,\mathrm{Î¾}\right)=\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{q\left(z,\mathrm{Î¾}\right)}.$
(2.7)

Differentiating (2.7), and after a short calculus, we obtain

$\frac{z{Q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{Q\left(z,\mathrm{Î¾}\right)}=1+\frac{z{q}_{{z}^{2}}^{â€²â€²}\left(z,\mathrm{Î¾}\right)}{{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}âˆ’\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{{q}_{z}\left(z,\mathrm{Î¾}\right)}.$
(2.8)

Using (jj) in (2.8), we have

$Re\frac{z{Q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{Q\left(z,\mathrm{Î¾}\right)}>0,\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$
(2.9)

hence the function $Q\left(z,\mathrm{Î¾}\right)$ is starlike in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. Using (2.5) we have

$h\left(z,\mathrm{Î¾}\right)=\mathrm{Î¸}\left(q\left(z,\mathrm{Î¾}\right)\right)+Q\left(z,\mathrm{Î¾}\right)=q\left(z,\mathrm{Î¾}\right)+Q\left(z,\mathrm{Î¾}\right).$
(2.10)

Differentiating (2.10) and using (2.7), after a short calculus, we obtain

$Re\frac{z{h}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{Q\left(z,\mathrm{Î¾}\right)}=Re\left[q\left(z,\mathrm{Î¾}\right)+\frac{z{Q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{Q\left(z,\mathrm{Î¾}\right)}\right].$
(2.11)

Using (j) and (2.9) in (2.11), we have $Re\frac{z{h}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{Q\left(z,\mathrm{Î¾}\right)}>0$, $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. Using (2.5) and (2.6), we get

$\begin{array}{c}\mathrm{Î¸}\left(p\left(z,\mathrm{Î¾}\right)\right)=p\left(z,\mathrm{Î¾}\right),\phantom{\rule{2em}{0ex}}\mathrm{Ï†}\left(p\left(z,\mathrm{Î¾}\right)\right)=\frac{1}{p\left(z,\mathrm{Î¾}\right)},\hfill \\ Q\left(q\left(z,\mathrm{Î¾}\right)\right)=q\left(z,\mathrm{Î¾}\right),\phantom{\rule{2em}{0ex}}\mathrm{Ï†}\left(q\left(z,\mathrm{Î¾}\right)\right)=\frac{1}{q\left(z,\mathrm{Î¾}\right)},\hfill \end{array}$

and the strong differential subordination (2.1) becomes

$Q\left(p\left(z,\mathrm{Î¾}\right)\right)+z{p}^{â€²}\left(z,\mathrm{Î¾}\right)â‹\dots \mathrm{Ï†}\left(p\left(z,\mathrm{Î¾}\right)\right)â‰ºâ‰º\mathrm{Î¸}\left(q\left(z,\mathrm{Î¾}\right)\right)+z{q}^{â€²}\left(z,\mathrm{Î¾}\right)â‹\dots \mathrm{Ï†}\left(q\left(z,\mathrm{Î¾}\right)\right).$

Using Lemma 1.1, we obtain

$p\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right),\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\phantom{\rule{1em}{0ex}}\frac{{\left[{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}â‹\dots {L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}{z}â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right),\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$

and $q\left(z,\mathrm{Î¾}\right)$ is the best dominant.â€ƒâ–¡

Theorem 2.2 Let $q\left(z,\mathrm{Î¾}\right)$ be univalent in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, with $q\left(0,\mathrm{Î¾}\right)=pâˆ’1$ and , $zâˆˆU$, for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, and suppose that

1. (l)

$Req\left(z,\mathrm{Î¾}\right)>âˆ’1$,

(ll) $Re\left[1+\frac{z{q}_{{z}^{2}}^{â€²â€²}\left(z,\mathrm{Î¾}\right)}{{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}âˆ’\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{1+q\left(z,\mathrm{Î¾}\right)}\right]>0$, $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$.

Let $pâˆˆ\mathbb{N}$, $f\left(z,\mathrm{Î¾}\right)âˆˆA\left(p\right)\mathrm{Î¾}$ and

$\frac{z{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{{z}^{2}}^{â€²â€²}}{{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}+\frac{{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}}{{z}^{pâˆ’1}}âˆ’p+1â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right)+1+\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{q\left(z,\mathrm{Î¾}\right)+1},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$
(2.12)

then $\frac{z{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}{{z}^{pâˆ’1}}âˆ’1â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right)$, and $q\left(z,\mathrm{Î¾}\right)$ is the best dominant.

Proof We let

$1+p\left(z,\mathrm{Î¾}\right)=\frac{{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}{{z}^{pâˆ’1}},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}.$
(2.13)

From (1.4), we have

$p\left(z,\mathrm{Î¾}\right)+1=\frac{{\left({z}^{p}+{âˆ‘}_{k=p+1}^{\mathrm{âˆž}}\frac{{\left(p+1\right)}^{m}}{{\left(p+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right){z}^{k}\right)}^{â€²}}{{z}^{pâˆ’1}}=p+\underset{k=p+1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(p+1\right)}^{m}}{{\left(p+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right)kz.$

Since $p\left(0,\mathrm{Î¾}\right)=pâˆ’1$, we obtain that $p\left(z,\mathrm{Î¾}\right)âˆˆ\mathcal{H}\mathrm{Î¾}\left[pâˆ’1,1\right]$. Differentiating (2.13), and after a short calculus, we obtain

$\frac{z{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{{z}^{2}}^{â€²â€²}}{{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}+\frac{{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}}{{z}^{pâˆ’1}}âˆ’p+1=p\left(z,\mathrm{Î¾}\right)+1+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{p\left(z,\mathrm{Î¾}\right)+1}.$
(2.14)

Using (2.14) in (2.12), the strong differential subordination becomes

$p\left(z,\mathrm{Î¾}\right)+1+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{p\left(z,\mathrm{Î¾}\right)+1}â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right)+1+\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{q\left(z,\mathrm{Î¾}\right)+1},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}.$
(2.15)

In order to prove the theorem, we shall use Lemma 1.1. For that, we show that the necessary conditions are satisfied. Let the functions $\mathrm{Î˜}:\mathbb{C}â†’\mathbb{C}$ and $\mathrm{Ï†}:\mathbb{C}â†’\mathbb{C}$, with

$\mathrm{Î˜}\left(\mathrm{Ï‰}\right)=\mathrm{Ï‰}+1$
(2.16)

and

(2.17)

We check the conditions from the hypothesis of Lemma 1.1. Using (2.17), we have

$Q\left(z,\mathrm{Î¾}\right)=z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)â‹\dots \mathrm{Ï†}\left(q\left(z,\mathrm{Î¾}\right)\right)=z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)â‹\dots \frac{1}{q\left(z,\mathrm{Î¾}\right)+1}.$
(2.18)

Differentiating (2.18), and after a short calculus, we obtain

$\frac{z{Q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{Q\left(z,\mathrm{Î¾}\right)}=1+\frac{z{q}_{{z}^{2}}^{â€²â€²}\left(z,\mathrm{Î¾}\right)}{{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}âˆ’\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{{q}_{z}\left(z,\mathrm{Î¾}\right)+1}.$
(2.19)

Using (ll) in (2.19), we have

$Re\frac{z{Q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{Q\left(z,\mathrm{Î¾}\right)}=Re\left(1+\frac{z{q}_{{z}^{2}}^{â€²â€²}\left(z,\mathrm{Î¾}\right)}{{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}âˆ’\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{{q}_{z}\left(z,\mathrm{Î¾}\right)+1}\right)>0,\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$
(2.20)

hence the function $Q\left(z,\mathrm{Î¾}\right)$ is starlike in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. Using (2.16) we have

$h\left(z,\mathrm{Î¾}\right)=\mathrm{Î¸}\left(q\left(z,\mathrm{Î¾}\right)\right)+Q\left(z,\mathrm{Î¾}\right)=q\left(z,\mathrm{Î¾}\right)+1+Q\left(z,\mathrm{Î¾}\right).$
(2.21)

Differentiating (2.21) and using (2.18), (2.20) and (l), after a short calculus, we obtain

$Re\frac{z{h}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{Q\left(z,\mathrm{Î¾}\right)}=Re\left[q\left(z,\mathrm{Î¾}\right)+1+\frac{z{Q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{Q\left(z,\mathrm{Î¾}\right)}\right]>0.$
(2.22)

Using (2.16) and (2.17), we get

$\begin{array}{c}\mathrm{Î¸}\left(p\left(z,\mathrm{Î¾}\right)\right)=p\left(z,\mathrm{Î¾}\right)+1,\phantom{\rule{2em}{0ex}}\mathrm{Ï†}\left(p\left(z,\mathrm{Î¾}\right)\right)=\frac{1}{p\left(z,\mathrm{Î¾}\right)+1},\hfill \\ \mathrm{Î¸}\left(q\left(z,\mathrm{Î¾}\right)\right)=q\left(z,\mathrm{Î¾}\right)+1,\phantom{\rule{2em}{0ex}}\mathrm{Ï†}\left(q\left(z,\mathrm{Î¾}\right)\right)=\frac{1}{q\left(z,\mathrm{Î¾}\right)+1},\hfill \end{array}$

and the strong differential subordination (2.12) becomes

$\mathrm{Î¸}\left(p\left(z,\mathrm{Î¾}\right)\right)+z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)â‹\dots \mathrm{Ï†}\left(p\left(z,\mathrm{Î¾}\right)\right)â‰ºâ‰º\mathrm{Î¸}\left(q\left(z,\mathrm{Î¾}\right)\right)+z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)â‹\dots \mathrm{Ï†}\left(q\left(z,\mathrm{Î¾}\right)\right).$

Using Lemma 1.1, we have $p\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right)$, i.e., $\frac{{\left[{H}^{m}f\left(z,\mathrm{Î¾}\right)\right]}_{z}^{â€²}}{{z}^{pâˆ’1}}âˆ’1â‰ºâ‰ºq\left(z,\mathrm{Î¾}\right)$ and $q\left(z,\mathrm{Î¾}\right)$ is the best dominant.â€ƒâ–¡

Next we give results related to strong differential superordinations.

Theorem 2.3 Let $h\left(z,\mathrm{Î¾}\right)$ be convex in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, with $h\left(0,\mathrm{Î¾}\right)=a$. Suppose that the differential equation

$q\left(z,\mathrm{Î¾}\right)+\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{q\left(z,\mathrm{Î¾}\right)}=h\left(z,\mathrm{Î¾}\right),\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$
(2.23)

has a univalent solution $q\left(z,\mathrm{Î¾}\right)$ that satisfies $q\left(0,\mathrm{Î¾}\right)=a$ and $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºh\left(z,\mathrm{Î¾}\right)$.

If $p\left(z,\mathrm{Î¾}\right)âˆˆ\mathcal{H}\left[a,1\right]âˆ©{Q}_{\mathrm{Î¾}}$ and $p\left(z,\mathrm{Î¾}\right)+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{p\left(z,\mathrm{Î¾}\right)}$ is univalent in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, $f\left(z,\mathrm{Î¾}\right)âˆˆA\mathrm{Î¾}$, then

$h\left(z,\mathrm{Î¾}\right)â‰ºâ‰º\frac{{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}{z}+\frac{z{\left({L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}{{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}âˆ’1$
(2.24)

implies $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰º\frac{{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}{z}$, $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. The function $q\left(z,\mathrm{Î¾}\right)$ is the best subordinant.

Proof We let

$p\left(z,\mathrm{Î¾}\right)=\frac{{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}{z},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}.$
(2.25)

From (1.2), we have

$p\left(z,\mathrm{Î¾}\right)=\frac{z+{âˆ‘}_{k=2}^{\mathrm{âˆž}}\frac{{\left(\mathrm{Î³}+1\right)}^{m}}{{\left(\mathrm{Î³}+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right){z}^{k}}{z}=1+\underset{k=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(\mathrm{Î³}+1\right)}^{m}}{{\left(\mathrm{Î³}+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right){z}^{kâˆ’1}$
(2.26)

and since $p\left(0,\mathrm{Î¾}\right)=1$, we obtain that $p\left(z,\mathrm{Î¾}\right)âˆˆ\mathcal{H}\left[1,1\right]âˆ©{Q}_{\mathrm{Î¾}}$.

Differentiating (2.26), and after a short calculus, we obtain

$p\left(z,\mathrm{Î¾}\right)+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{p\left(z,\mathrm{Î¾}\right)}=\frac{z{\left({L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}{{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}+\frac{{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}{z}âˆ’1.$
(2.27)

Using (2.27) in (2.24), the strong differential superordination becomes

$q\left(z,\mathrm{Î¾}\right)+\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{q\left(z,\mathrm{Î¾}\right)}â‰ºâ‰ºp\left(z,\mathrm{Î¾}\right)+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{p\left(z,\mathrm{Î¾}\right)},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}.$
(2.28)

Using Lemma 1.2, we obtain $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºp\left(z,\mathrm{Î¾}\right)$, i.e., $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰º\frac{{L}_{\mathrm{Î³}}^{m}f\left(z,\mathrm{Î¾}\right)}{z}$, $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$.â€ƒâ–¡

Example 2.1 Let $h\left(z,\mathrm{Î¾}\right)=\frac{1âˆ’z\mathrm{Î¾}}{1+z\mathrm{Î¾}}$, $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, with $Re\left(1+\frac{z{h}_{{z}^{2}}^{â€²â€²}\left(z,\mathrm{Î¾}\right)}{{h}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}\right)=Re\frac{1âˆ’z\mathrm{Î¾}}{1+z\mathrm{Î¾}}>0$, $zâˆˆU$, $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$. From Theorem 2.3 we have that if $m=1$, $n=1$, $\mathrm{Î³}=âˆ’1+i$, $f\left(z,\mathrm{Î¾}\right)=z+\frac{\mathrm{Î¾}}{2}{z}^{2}$, and $\frac{1+i}{2}z+\frac{1+\mathrm{Î¾}\frac{1+i}{2}z}{1+\frac{1+i}{2}\frac{\mathrm{Î¾}}{2}z}$ is univalent in U for $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, then

$\frac{1âˆ’z\mathrm{Î¾}}{1+z\mathrm{Î¾}}â‰ºâ‰º\frac{1+i}{2}z+\frac{1+\mathrm{Î¾}\frac{1+i}{2}z}{1+\frac{1+i}{2}\frac{\mathrm{Î¾}}{2}z}$

implies

$\frac{1}{1+z\mathrm{Î¾}}â‰ºâ‰ºz+\frac{\mathrm{Î¾}}{2}\frac{1+i}{2}{z}^{2},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}.$

Theorem 2.4 Let $h\left(z,\mathrm{Î¾}\right)$ be convex in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, with $h\left(0,\mathrm{Î¾}\right)=p$. Suppose that the differential equation

$q\left(z,\mathrm{Î¾}\right)+\frac{z{q}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{q\left(z,\mathrm{Î¾}\right)}=h\left(z,\mathrm{Î¾}\right),\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$

has a univalent solution $q\left(z,\mathrm{Î¾}\right)$ that satisfies $q\left(0,\mathrm{Î¾}\right)=p$ and $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºh\left(z,\mathrm{Î¾}\right)$.

If $p\left(z,\mathrm{Î¾}\right)âˆˆ\mathcal{H}\left[p,1\right]âˆ©{Q}_{\mathrm{Î¾}}$ and $p\left(z,\mathrm{Î¾}\right)+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{p\left(z,\mathrm{Î¾}\right)}$ is univalent in U for all $\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}$, $f\left(z,\mathrm{Î¾}\right)âˆˆA\left(p\right)\mathrm{Î¾}$, then

$h\left(z,\mathrm{Î¾}\right)â‰ºâ‰º1+\frac{{z}^{2}{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{{z}^{2}}^{â€²â€²}}{{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U},$
(2.29)

implies $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰º\frac{z{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}{{H}^{m}f\left(z,\mathrm{Î¾}\right)}$. The function $q\left(z,\mathrm{Î¾}\right)$ is the best subordinant.

Proof Using (1.5) in (2.29), the strong differential superordination becomes

$h\left(z,\mathrm{Î¾}\right)â‰ºâ‰º1+\frac{z{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{{z}^{2}}^{â€²â€²}}{{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}.$
(2.30)

We let

$p\left(z,\mathrm{Î¾}\right)=\frac{z{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}{{H}^{m}f\left(z,\mathrm{Î¾}\right)}.$
(2.31)

From (1.4), we have

$p\left(z,\mathrm{Î¾}\right)=\frac{z\left(p{z}^{pâˆ’1}+{âˆ‘}_{k=p+1}^{\mathrm{âˆž}}\frac{{\left(p+1\right)}^{m}}{{\left(p+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right)k{z}^{kâˆ’1}\right)}{{z}^{p}+{âˆ‘}_{k=p+1}^{\mathrm{âˆž}}\frac{{\left(p+1\right)}^{m}}{{\left(p+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right){z}^{k}}=\frac{p+{âˆ‘}_{k=p+1}^{\mathrm{âˆž}}\frac{{\left(p+1\right)}^{m}}{{\left(p+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right){z}^{k}}{1+{âˆ‘}_{k=p+1}^{\mathrm{âˆž}}\frac{{\left(p+1\right)}^{m}}{{\left(p+k\right)}^{m}}{a}_{k}\left(\mathrm{Î¾}\right){z}^{kâˆ’p}}.$

Since $p\left(0,\mathrm{Î¾}\right)=p$, we obtain that $p\left(z,\mathrm{Î¾}\right)âˆˆ\mathcal{H}\left[p,1\right]âˆ©{Q}_{\mathrm{Î¾}}$.

Differentiating (2.31), and after a short calculus, we obtain

$p\left(z,\mathrm{Î¾}\right)+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{p\left(z,\mathrm{Î¾}\right)}=1+\frac{z{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{{z}^{2}}^{â€²â€²}}{{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}.$
(2.32)

Using (2.32) in (2.30), the strong differential superordination becomes

$h\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºp\left(z,\mathrm{Î¾}\right)+\frac{z{p}_{z}^{â€²}\left(z,\mathrm{Î¾}\right)}{p\left(z,\mathrm{Î¾}\right)},\phantom{\rule{1em}{0ex}}zâˆˆU,\mathrm{Î¾}âˆˆ\stackrel{Â¯}{U}.$
(2.33)

Using Lemma 1.2, we obtain $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰ºp\left(z,\mathrm{Î¾}\right)$, i.e., $q\left(z,\mathrm{Î¾}\right)â‰ºâ‰º\frac{z{\left({H}^{m}f\left(z,\mathrm{Î¾}\right)\right)}_{z}^{â€²}}{{H}^{m}f\left(z,\mathrm{Î¾}\right)}$.â€ƒâ–¡

Remark 2.1 Using another integral operator, the author finds interesting results in strong differential subordinations and superordinations in [14].

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## Acknowledgements

The author thanks the referee for his/her valuable suggestions to improve the present article.

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Correspondence to Georgia Irina Oros.

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Oros, G.I. Strong differential subordinations and superordinations obtained with some new integral operators. Adv Differ Equ 2013, 317 (2013). https://doi.org/10.1186/1687-1847-2013-317