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# Some differential subordinations using Ruscheweyh derivative and Sălăgean operator

Advances in Difference Equations20132013:150

https://doi.org/10.1186/1687-1847-2013-150

• Received: 9 April 2013
• Accepted: 10 May 2013
• Published:

## Abstract

In the present paper we study the operator defined by using the Ruscheweyh derivative ${R}^{n}f\left(z\right)$ and the Sălăgean operator ${S}^{n}f\left(z\right)$, denoted by ${L}_{\alpha }^{n}:\mathcal{A}\to \mathcal{A}$, ${L}_{\alpha }^{n}f\left(z\right)=\left(1-\alpha \right){R}^{n}f\left(z\right)+\alpha {S}^{n}f\left(z\right)$, $z\in U$, where ${\mathcal{A}}_{n}=\left\{f\in \mathcal{H}\left(U\right):f\left(z\right)=z+{a}_{n+1}{z}^{n+1}+\cdots ,z\in U\right\}$ is the class of normalized analytic functions with ${\mathcal{A}}_{1}=\mathcal{A}$. We obtain several differential subordinations regarding the operator ${L}_{\alpha }^{n}$.

MSC:30C45, 30A20, 34A40.

## Keywords

• differential subordination
• convex function
• best dominant
• differential operator
• Sălăgean operator
• Ruscheweyh derivative

## 1 Introduction

Denote by U the unit disc of the complex plane, $U=\left\{z\in \mathbb{C}:|z|<1\right\}$, and by $\mathcal{H}\left(U\right)$ the space of holomorphic functions in U. Let ${\mathcal{A}}_{n}=\left\{f\in \mathcal{H}\left(U\right):f\left(z\right)=z+{a}_{n+1}{z}^{n+1}+\cdots ,z\in U\right\}$ with ${\mathcal{A}}_{1}=\mathcal{A}$ and $\mathcal{H}\left[a,n\right]=\left\{f\in \mathcal{H}\left(U\right):f\left(z\right)=a+{a}_{n}{z}^{n}+{a}_{n+1}{z}^{n+1}+\cdots ,z\in U\right\}$ for $a\in \mathbb{C}$ and $n\in \mathbb{N}$. Denote by $K=\left\{f\in \mathcal{A}:Re\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{f}^{\mathrm{\prime }}\left(z\right)}+1>0,z\in U\right\}$ the class of normalized convex functions in U.

If f and g are analytic functions in U, we say that f is subordinate to g, written $f\prec g$, if there is a function w analytic in U, with $w\left(0\right)=0$, $|w\left(z\right)|<1$, for all $z\in U$, such that $f\left(z\right)=g\left(w\left(z\right)\right)$ for all $z\in U$. If g is univalent, then $f\prec g$ if and only if $f\left(0\right)=g\left(0\right)$ and $f\left(U\right)\subseteq g\left(U\right)$.

Let $\psi :{\mathbb{C}}^{3}×U\to \mathbb{C}$ and let h be a univalent function in U. If p is analytic in U and satisfies the (second-order) differential subordination
$\psi \left(p\left(z\right),z{p}^{\mathrm{\prime }}\left(z\right),{z}^{2}{p}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right);z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(1.1)

then p is called a solution of the differential subordination. The univalent function q is called a dominant of the solutions of the differential subordination, or more simply a dominant, if $p\prec q$ for all p satisfying (1.1).

A dominant $\stackrel{˜}{q}$ that satisfies $\stackrel{˜}{q}\prec q$ for all dominants q of (1.1) is said to be the best dominant of (1.1). The best dominant is unique up to a rotation of U.

Definition 1.1 (Sălăgean )

For $f\in \mathcal{A}$, $n\in \mathbb{N}$, the operator ${S}^{n}$ is defined by ${S}^{n}:\mathcal{A}\to \mathcal{A}$,
$\begin{array}{r}{S}^{0}f\left(z\right)=f\left(z\right),\\ {S}^{1}f\left(z\right)=z{f}^{\mathrm{\prime }}\left(z\right),\\ \cdots \\ {S}^{n+1}f\left(z\right)=z{\left({S}^{n}f\left(z\right)\right)}^{\mathrm{\prime }},\phantom{\rule{1em}{0ex}}z\in U.\end{array}$

Remark 1.1 If $f\in \mathcal{A}$, $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then ${S}^{n}f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{j}^{n}{a}_{j}{z}^{j}$, $z\in U$.

Definition 1.2 (Ruscheweyh )

For $f\in \mathcal{A}$, $n\in \mathbb{N}$, the operator ${R}^{n}$ is defined by ${R}^{n}:\mathcal{A}\to \mathcal{A}$,
$\begin{array}{c}{R}^{0}f\left(z\right)=f\left(z\right),\hfill \\ {R}^{1}f\left(z\right)=z{f}^{\mathrm{\prime }}\left(z\right),\hfill \\ \cdots \hfill \\ \left(n+1\right){R}^{n+1}f\left(z\right)=z{\left({R}^{n}f\left(z\right)\right)}^{\mathrm{\prime }}+n{R}^{n}f\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.\hfill \end{array}$

Remark 1.2 If $f\in \mathcal{A}$, $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then ${R}^{n}f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{C}_{n+j-1}^{n}{a}_{j}{z}^{j}$, $z\in U$.

Definition 1.3 ()

Let $\alpha \ge 0$, $n\in \mathbb{N}$. Denote by ${L}_{\alpha }^{n}$ the operator given by ${L}_{\alpha }^{n}:\mathcal{A}\to \mathcal{A}$,
${L}_{\alpha }^{n}f\left(z\right)=\left(1-\alpha \right){R}^{n}f\left(z\right)+\alpha {S}^{n}f\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

Remark 1.3 If $f\in \mathcal{A}$, $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then ${L}_{\alpha }^{n}f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}\left(\alpha {j}^{n}+\left(1-\alpha \right){C}_{n+j-1}^{n}\right){a}_{j}{z}^{j}$, $z\in U$.

This operator was studied also in .

Lemma 1.1 (Hallenbeck and Ruscheweyh [, Th. 3.1.6, p.71])

Let h be a convex function with $h\left(0\right)=a$, and let $\gamma \in \mathbb{C}\mathrm{\setminus }\left\{0\right\}$ be a complex number with $Re\gamma \ge 0$. If $p\in \mathcal{H}\left[a,n\right]$ and
$p\left(z\right)+\frac{1}{\gamma }z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
then
$p\left(z\right)\prec g\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where $g\left(z\right)=\frac{\gamma }{n{z}^{\gamma /n}}{\int }_{0}^{z}h\left(t\right){t}^{\gamma /n-1}\phantom{\rule{0.2em}{0ex}}dt$, $z\in U$.

Lemma 1.2 (Miller and Mocanu )

Let g be a convex function in U and let $h\left(z\right)=g\left(z\right)+n\alpha z{g}^{\mathrm{\prime }}\left(z\right)$, for $z\in U$, where $\alpha >0$ and n is a positive integer.

If $p\left(z\right)=g\left(0\right)+{p}_{n}{z}^{n}+{p}_{n+1}{z}^{n+1}+\cdots$ , $z\in U$, is holomorphic in U and
$p\left(z\right)+\alpha z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
then
$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp.

## 2 Main results

Theorem 2.1 Let g be a convex function, $g\left(0\right)=1$ and let h be the function $h\left(z\right)=g\left(z\right)+\frac{z}{\delta }{g}^{\mathrm{\prime }}\left(z\right)$, $z\in U$.

If $\alpha ,\delta \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination
${\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta -1}{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.1)
then
${\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta }\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp.

Proof By using the properties of the operator ${L}_{\alpha }^{n}$, we have
${L}_{\alpha }^{n}f\left(z\right)=z+\sum _{j=2}^{\mathrm{\infty }}\left(\alpha {j}^{n}+\left(1-\alpha \right){C}_{n+j-1}^{n}\right){a}_{j}{z}^{j},\phantom{\rule{1em}{0ex}}z\in U.$

Consider $p\left(z\right)={\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta }={\left(\frac{z+{\sum }_{j=2}^{\mathrm{\infty }}\left(\alpha {j}^{n}+\left(1-\alpha \right){C}_{n+j-1}^{n}\right){a}_{j}{z}^{j}}{z}\right)}^{\delta }=1+{p}_{\delta }{z}^{\delta }+{p}_{\delta +1}{z}^{\delta +1}+\cdots$ , $z\in U$.

We deduce that $p\in \mathcal{H}\left[1,\delta \right]$.

Differentiating, we obtain ${\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta -1}{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}=p\left(z\right)+\frac{1}{\delta }z{p}^{\mathrm{\prime }}\left(z\right)$, $z\in U$.

Then (2.1) becomes
$p\left(z\right)+\frac{1}{\delta }z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right)=g\left(z\right)+\frac{z}{\delta }{g}^{\mathrm{\prime }}\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$
By using Lemma 1.2, we have
$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}{\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta }\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

□

Theorem 2.2 Let h be a holomorphic function which satisfies the inequality $Re\left(1+\frac{z{h}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{h}^{\mathrm{\prime }}\left(z\right)}\right)>-\frac{1}{2}$, $z\in U$, and $h\left(0\right)=1$.

If $\alpha ,\delta \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination
${\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta -1}{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.2)
then
${\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta }\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where $q\left(z\right)=\frac{\delta }{{z}^{\delta }}{\int }_{0}^{z}h\left(t\right){t}^{\delta -1}\phantom{\rule{0.2em}{0ex}}dt$. The function q is convex and it is the best dominant.

Proof Let
$\begin{array}{rcl}p\left(z\right)& =& {\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta }={\left(\frac{z+{\sum }_{j=2}^{\mathrm{\infty }}\left(\alpha {j}^{n}+\left(1-\alpha \right){C}_{n+j-1}^{n}\right){a}_{j}{z}^{j}}{z}\right)}^{\delta }\\ =& {\left(1+\sum _{j=2}^{\mathrm{\infty }}\left(\alpha {j}^{n}+\left(1-\alpha \right){C}_{n+j-1}^{n}\right){a}_{j}{z}^{j-1}\right)}^{\delta }=1+\sum _{j=\delta +1}^{\mathrm{\infty }}{p}_{j}{z}^{j-1}\end{array}$

for $z\in U$, $p\in \mathcal{H}\left[1,\delta \right]$.

Differentiating, we obtain ${\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta -1}{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}=p\left(z\right)+\frac{1}{\delta }z{p}^{\mathrm{\prime }}\left(z\right)$, $z\in U$, and (2.2) becomes
$p\left(z\right)+\frac{1}{\delta }z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$
Using Lemma 1.1, we have
$p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}{\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta }\prec q\left(z\right)=\frac{\delta }{{z}^{\delta }}{\int }_{0}^{z}h\left(t\right){t}^{\delta -1}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,$

and q is the best dominant. □

Corollary 2.3 Let $h\left(z\right)=\frac{1+\left(2\beta -1\right)z}{1+z}$ be a convex function in U, where $0\le \beta <1$.

If $\alpha ,\delta \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination
${\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta -1}{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.3)
then
${\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta }\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where q is given by $q\left(z\right)=\left(2\beta -1\right)+\frac{2\left(1-\beta \right)\delta }{{z}^{\delta }}{\int }_{0}^{z}\frac{{t}^{\delta -1}}{1+t}\phantom{\rule{0.2em}{0ex}}dt$, $z\in U$. The function q is convex and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.2 and considering $p\left(z\right)={\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta }$, the differential subordination (2.3) becomes
$p\left(z\right)+\frac{z}{\delta }{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right)=\frac{1+\left(2\beta -1\right)z}{1+z},\phantom{\rule{1em}{0ex}}z\in U.$
By using Lemma 1.1 for $\gamma =\delta$, we have $p\left(z\right)\prec q\left(z\right)$, i.e.,
$\begin{array}{rl}{\left(\frac{{L}_{\alpha }^{n}f\left(z\right)}{z}\right)}^{\delta }& \prec q\left(z\right)=\frac{\delta }{{z}^{\delta }}{\int }_{0}^{z}h\left(t\right){t}^{\delta -1}\phantom{\rule{0.2em}{0ex}}dt\\ =\frac{\delta }{{z}^{\delta }}{\int }_{0}^{z}{t}^{\delta -1}\frac{1+\left(2\beta -1\right)t}{1+t}\phantom{\rule{0.2em}{0ex}}dt=\frac{\delta }{{z}^{\delta }}{\int }_{0}^{z}\left[\left(2\beta -1\right){t}^{\delta -1}+2\left(1-\beta \right)\frac{{t}^{\delta -1}}{1+t}\right]\phantom{\rule{0.2em}{0ex}}dt\\ =\left(2\beta -1\right)+\frac{2\left(1-\beta \right)\delta }{{z}^{\delta }}{\int }_{0}^{z}\frac{{t}^{\delta -1}}{1+t}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U.\end{array}$

□

Remark 2.1 For $n=1$, $\alpha =2$, $\delta =1$, we obtain the same example as in [, Example 2.2.1, p.26].

Theorem 2.4 Let g be a convex function such that $g\left(0\right)=1$ and let h be the function $h\left(z\right)=g\left(z\right)+\frac{z}{\gamma }{g}^{\mathrm{\prime }}\left(z\right)$, $z\in U$, where $\gamma >0$.

If $\alpha \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and the differential subordination
$\frac{\left(\gamma +1\right)z}{\gamma }\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}+\frac{{z}^{2}}{\gamma }\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}\left[\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}-2\frac{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n+1}f\left(z\right)}\right]\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.4)
holds, then
$z\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp.

Proof For $f\in \mathcal{A}$, $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, we have ${L}_{\alpha }^{n}f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}\left(\alpha {j}^{n}+\left(1-\alpha \right){C}_{n+j-1}^{n}\right){a}_{j}{z}^{j}$, $z\in U$.

Consider $p\left(z\right)=z\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}$ and we obtain $p\left(z\right)+\frac{z}{\gamma }{p}^{\mathrm{\prime }}\left(z\right)=\frac{\left(\gamma +1\right)z}{\gamma }\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}+\frac{{z}^{2}}{\gamma }\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}\left[\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}-2\frac{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n+1}f\left(z\right)}\right]$.

Relation (2.4) becomes
$p\left(z\right)+\frac{z}{\gamma }{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right)=g\left(z\right)+\gamma {g}^{\mathrm{\prime }}\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$
By using Lemma 1.2, we have
$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}z\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

□

Theorem 2.5 Let h be a holomorphic function which satisfies the inequality $Re\left(1+\frac{z{h}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{h}^{\mathrm{\prime }}\left(z\right)}\right)>-\frac{1}{2}$, $z\in U$, and $h\left(0\right)=1$.

If $\alpha \ge 0$, $\gamma \in \mathbb{C}\mathrm{\setminus }\left\{0\right\}$ is a complex number with $Re\gamma \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination
$\frac{\left(\gamma +1\right)z}{\gamma }\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}+\frac{{z}^{2}}{\gamma }\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}\left[\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}-2\frac{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n+1}f\left(z\right)}\right]\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.5)
then
$z\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where $q\left(z\right)=\frac{\gamma }{{z}^{\gamma }}{\int }_{0}^{z}h\left(t\right){t}^{\gamma -1}\phantom{\rule{0.2em}{0ex}}dt$. The function q is convex and it is the best dominant.

Proof Let $p\left(z\right)=z\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}$, $z\in U$, $p\in \mathcal{H}\left[1,1\right]$. Differentiating, we obtain $p\left(z\right)+\frac{z}{\gamma }{p}^{\mathrm{\prime }}\left(z\right)=\frac{\left(\gamma +1\right)z}{\gamma }\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}+\frac{{z}^{2}}{\gamma }\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}\left[\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}-2\frac{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n+1}f\left(z\right)}\right]$, $z\in U$, and (2.5) becomes
$p\left(z\right)+\frac{z}{\gamma }{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$
Using Lemma 1.1, we have
$p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}z\frac{{L}_{\alpha }^{n}f\left(z\right)}{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{2}}\prec q\left(z\right)=\frac{\gamma }{{z}^{\gamma }}{\int }_{0}^{z}h\left(t\right){t}^{\gamma -1}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,$

and q is the best dominant. □

Theorem 2.6 Let g be a convex function such that $g\left(0\right)=1$ and let h be the function $h\left(z\right)=g\left(z\right)+\frac{z}{\gamma }{g}^{\mathrm{\prime }}\left(z\right)$, $z\in U$, where $\gamma >0$.

If $\alpha \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and the differential subordination
$\frac{\left(\gamma +2\right){z}^{2}}{\gamma }\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}+\frac{{z}^{3}}{\gamma }\left[\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}-{\left(\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{2}\right]\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.6)
holds, then
${z}^{2}\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

This result is sharp.

Proof Let $p\left(z\right)={z}^{2}\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}$. We deduce that $p\in \mathcal{H}\left[0,1\right]$.

Differentiating, we obtain $p\left(z\right)+\frac{z}{\gamma }{p}^{\mathrm{\prime }}\left(z\right)=\frac{\left(\gamma +2\right){z}^{2}}{\gamma }\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}+\frac{{z}^{3}}{\gamma }\left[\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}-{\left(\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{2}\right]$, $z\in U$.

Using the notation in (2.6), the differential subordination becomes
$p\left(z\right)+\frac{1}{\gamma }z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right)=g\left(z\right)+\frac{z}{\gamma }{g}^{\mathrm{\prime }}\left(z\right).$
By using Lemma 1.2, we have
$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}{z}^{2}\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp. □

Theorem 2.7 Let h be a holomorphic function which satisfies the inequality $Re\left(1+\frac{z{h}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{h}^{\mathrm{\prime }}\left(z\right)}\right)>-\frac{1}{2}$, $z\in U$, and $h\left(0\right)=1$.

If $\alpha \ge 0$, $\gamma \in \mathbb{C}\mathrm{\setminus }\left\{0\right\}$ is a complex number with $Re\gamma \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination
$\frac{\left(\gamma +2\right){z}^{2}}{\gamma }\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}+\frac{{z}^{3}}{\gamma }\left[\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}-{\left(\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{2}\right]\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.7)
then
${z}^{2}\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where $q\left(z\right)=\frac{\gamma }{{z}^{\gamma }}{\int }_{0}^{z}h\left(t\right){t}^{\gamma -1}\phantom{\rule{0.2em}{0ex}}dt$. The function q is convex and it is the best dominant.

Proof Let $p\left(z\right)={z}^{2}\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}$, $z\in U$, $p\in \mathcal{H}\left[0,1\right]$.

Differentiating, we obtain $p\left(z\right)+\frac{z}{\gamma }{p}^{\mathrm{\prime }}\left(z\right)=\frac{\left(\gamma +2\right){z}^{2}}{\gamma }\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}+\frac{{z}^{3}}{\gamma }\left[\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}-{\left(\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{2}\right]$, $z\in U$, and (2.7) becomes
$p\left(z\right)+\frac{1}{\gamma }z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$
Using Lemma 1.1, we have
$p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}{z}^{2}\frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\prec q\left(z\right)=\frac{\gamma }{{z}^{\gamma }}{\int }_{0}^{z}h\left(t\right){t}^{\gamma -1}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,$

and q is the best dominant. □

Theorem 2.8 Let g be a convex function such that $g\left(0\right)=1$ and let h be the function $h\left(z\right)=g\left(z\right)+z{g}^{\mathrm{\prime }}\left(z\right)$, $z\in U$.

If $\alpha \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and the differential subordination
$1-\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}}{{\left[{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.8)
holds, then
$\frac{{L}_{\alpha }^{n}f\left(z\right)}{z{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

This result is sharp.

Proof Let $p\left(z\right)=\frac{{L}_{\alpha }^{n}f\left(z\right)}{z{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}$. We deduce that $p\in \mathcal{H}\left[1,1\right]$.

Differentiating, we obtain $1-\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}}{{\left[{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}}=p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)$, $z\in U$.

Using the notation in (2.8), the differential subordination becomes
$p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right)=g\left(z\right)+z{g}^{\mathrm{\prime }}\left(z\right).$
By using Lemma 1.2, we have
$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}\frac{{L}_{\alpha }^{n}f\left(z\right)}{z{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp. □

Theorem 2.9 Let h be a holomorphic function which satisfies the inequality $Re\left(1+\frac{z{h}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{h}^{\mathrm{\prime }}\left(z\right)}\right)>-\frac{1}{2}$, $z\in U$, and $h\left(0\right)=1$.

If $\alpha \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination
$1-\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}}{{\left[{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.9)
then
$\frac{{L}_{\alpha }^{n}f\left(z\right)}{z{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where $q\left(z\right)=\frac{1}{z}{\int }_{0}^{z}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt$. The function q is convex and it is the best dominant.

Proof Let $p\left(z\right)=\frac{{L}_{\alpha }^{n}f\left(z\right)}{z{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}$, $z\in U$, $p\in \mathcal{H}\left[0,1\right]$.

Differentiating, we obtain $1-\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}}{{\left[{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}}=p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)$, $z\in U$, and (2.9) becomes
$p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$
Using Lemma 1.1, we have
$p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}\frac{{L}_{\alpha }^{n}f\left(z\right)}{z{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}\prec q\left(z\right)=\frac{1}{z}{\int }_{0}^{z}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,$

and q is the best dominant. □

Corollary 2.10 Let $h\left(z\right)=\frac{1+\left(2\beta -1\right)z}{1+z}$ be a convex function in U, where $0\le \beta <1$.

If $\alpha \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination
$1-\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}}{{\left[{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.10)
then
$\frac{{L}_{\alpha }^{n}f\left(z\right)}{z{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where q is given by $q\left(z\right)=\left(2\beta -1\right)+2\left(1-\beta \right)\frac{ln\left(1+z\right)}{z}$, $z\in U$. The function q is convex and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.9 and considering $p\left(z\right)=\frac{{L}_{\alpha }^{n}f\left(z\right)}{z{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}$, the differential subordination (2.10) becomes
$p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right)=\frac{1+\left(2\beta -1\right)z}{1+z},\phantom{\rule{1em}{0ex}}z\in U.$
By using Lemma 1.1 for $\gamma =1$, we have $p\left(z\right)\prec q\left(z\right)$, i.e.,
$\begin{array}{rl}\frac{{L}_{\alpha }^{n}f\left(z\right)}{z{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}& \prec q\left(z\right)=\frac{1}{z}{\int }_{0}^{z}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=\frac{1}{z}{\int }_{0}^{z}\frac{1+\left(2\beta -1\right)t}{1+t}\phantom{\rule{0.2em}{0ex}}dt\\ =\frac{1}{z}{\int }_{0}^{z}\left[\left(2\beta -1\right)+\frac{2\left(1-\beta \right)}{1+t}\right]\phantom{\rule{0.2em}{0ex}}dt=\left(2\beta -1\right)+2\left(1-\beta \right)\frac{ln\left(1+z\right)}{z},\phantom{\rule{1em}{0ex}}z\in U.\end{array}$

□

Example 2.1 Let $h\left(z\right)=\frac{1-z}{1+z}$ be a convex function in U with $h\left(0\right)=1$ and $Re\left(\frac{z{h}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{h}^{\mathrm{\prime }}\left(z\right)}+1\right)>-\frac{1}{2}$.

Let $f\left(z\right)=z+{z}^{2}$, $z\in U$. For $n=1$, $\alpha =2$, we obtain ${L}_{2}^{1}f\left(z\right)=-{R}^{1}f\left(z\right)+2{S}^{1}f\left(z\right)=-z{f}^{\mathrm{\prime }}\left(z\right)+2z{f}^{\mathrm{\prime }}\left(z\right)=z{f}^{\mathrm{\prime }}\left(z\right)=z+2{z}^{2}$.

Then ${\left({L}_{2}^{1}f\left(z\right)\right)}^{\mathrm{\prime }}=1+4z$,
$\begin{array}{c}\frac{{L}_{2}^{1}f\left(z\right)}{z{\left({L}_{2}^{1}f\left(z\right)\right)}^{\mathrm{\prime }}}=\frac{z+2{z}^{2}}{z\left(1+4z\right)}=\frac{1+2z}{1+4z},\hfill \\ 1-\frac{{L}_{2}^{1}f\left(z\right)\cdot {\left({L}_{2}^{1}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}}{{\left[{\left({L}_{2}^{1}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}}=1-\frac{\left(z+2{z}^{2}\right)\cdot 4}{{\left(1+4z\right)}^{2}}=\frac{8{z}^{2}+4z+1}{{\left(1+4z\right)}^{2}}.\hfill \end{array}$

We have $q\left(z\right)=\frac{1}{z}{\int }_{0}^{z}\frac{1-t}{1+t}\phantom{\rule{0.2em}{0ex}}dt=-1+\frac{2ln\left(1+z\right)}{z}$.

Using Theorem 2.9, we obtain
$\frac{8{z}^{2}+4z+1}{{\left(1+4z\right)}^{2}}\prec \frac{1-z}{1+z},\phantom{\rule{1em}{0ex}}z\in U,$
induce
$\frac{1+2z}{1+4z}\prec -1+\frac{2ln\left(1+z\right)}{z},\phantom{\rule{1em}{0ex}}z\in U.$

Theorem 2.11 Let g be a convex function such that $g\left(0\right)=0$ and let h be the function $h\left(z\right)=g\left(z\right)+z{g}^{\mathrm{\prime }}\left(z\right)$, $z\in U$.

If $\alpha \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and the differential subordination
${\left[{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}+{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.11)
holds, then
$\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{z}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

This result is sharp.

Proof Let $p\left(z\right)=\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{z}$. We deduce that $p\in \mathcal{H}\left[0,1\right]$.

Differentiating, we obtain ${\left[{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}+{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}=p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)$, $z\in U$.

Using the notation in (2.11), the differential subordination becomes
$p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right)=g\left(z\right)+z{g}^{\mathrm{\prime }}\left(z\right).$
By using Lemma 1.2, we have
$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{z}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp. □

Theorem 2.12 Let h be a holomorphic function which satisfies the inequality $Re\left(1+\frac{z{h}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{h}^{\mathrm{\prime }}\left(z\right)}\right)>-\frac{1}{2}$, $z\in U$, and $h\left(0\right)=0$.

If $\alpha \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination
${\left[{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}+{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.12)
then
$\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{z}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where $q\left(z\right)=\frac{1}{z}{\int }_{0}^{z}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt$. The function q is convex and it is the best dominant.

Proof Let $p\left(z\right)=\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{z}$, $z\in U$, $p\in \mathcal{H}\left[0,1\right]$.

Differentiating, we obtain ${\left[{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}+{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}=p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)$, $z\in U$, and (2.12) becomes
$p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$
Using Lemma 1.1, we have
$p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{z}\prec q\left(z\right)=\frac{1}{z}{\int }_{0}^{z}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,$

and q is the best dominant. □

Corollary 2.13 Let $h\left(z\right)=\frac{1+\left(2\beta -1\right)z}{1+z}$ be a convex function in U, where $0\le \beta <1$.

If $\alpha \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination
${\left[{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}+{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.13)
then
$\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{z}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where q is given by $q\left(z\right)=\left(2\beta -1\right)+2\left(1-\beta \right)\frac{ln\left(1+z\right)}{z}$, $z\in U$. The function q is convex and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.12 and considering $p\left(z\right)=\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{z}$, the differential subordination (2.13) becomes
$p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right)=\frac{1+\left(2\beta -1\right)z}{1+z},\phantom{\rule{1em}{0ex}}z\in U.$
By using Lemma 1.1 for $\gamma =1$, we have $p\left(z\right)\prec q\left(z\right)$, i.e.,
$\begin{array}{rl}\frac{{L}_{\alpha }^{n}f\left(z\right)\cdot {\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{z}& \prec q\left(z\right)=\frac{1}{z}{\int }_{0}^{z}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=\frac{1}{z}{\int }_{0}^{z}\frac{1+\left(2\beta -1\right)t}{1+t}\phantom{\rule{0.2em}{0ex}}dt\\ =\frac{1}{z}{\int }_{0}^{z}\left[\left(2\beta -1\right)+\frac{2\left(1-\beta \right)}{1+t}\right]\phantom{\rule{0.2em}{0ex}}dt\\ =\left(2\beta -1\right)+2\left(1-\beta \right)\frac{ln\left(1+z\right)}{z},\phantom{\rule{1em}{0ex}}z\in U.\end{array}$

□

Example 2.2 Let $h\left(z\right)=\frac{1-z}{1+z}$ be a convex function in U with $h\left(0\right)=1$ and $Re\left(\frac{z{h}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{h}^{\mathrm{\prime }}\left(z\right)}+1\right)>-\frac{1}{2}$.

Let $f\left(z\right)=z+{z}^{2}$, $z\in U$. For $n=1$, $\alpha =2$, we obtain ${L}_{2}^{1}f\left(z\right)=-{R}^{1}f\left(z\right)+2{S}^{1}f\left(z\right)=-z{f}^{\mathrm{\prime }}\left(z\right)+2z{f}^{\mathrm{\prime }}\left(z\right)=z{f}^{\mathrm{\prime }}\left(z\right)=z+2{z}^{2}$, $z\in U$.

Then ${\left({L}_{2}^{1}f\left(z\right)\right)}^{\mathrm{\prime }}=1+4z$,
$\begin{array}{c}\frac{{L}_{2}^{1}f\left(z\right)\cdot {\left({L}_{2}^{1}f\left(z\right)\right)}^{\mathrm{\prime }}}{z}=\frac{\left(z+2{z}^{2}\right)\left(1+4z\right)}{z}=8{z}^{2}+6z+1,\hfill \\ {\left[{\left({L}_{2}^{1}f\left(z\right)\right)}^{\mathrm{\prime }}\right]}^{2}+{L}_{2}^{1}f\left(z\right)\cdot {\left({L}_{2}^{1}f\left(z\right)\right)}^{\mathrm{\prime }\mathrm{\prime }}={\left(1+4z\right)}^{2}+\left(z+2{z}^{2}\right)\cdot 4=24{z}^{2}+12z+1.\hfill \end{array}$

We have $q\left(z\right)=\frac{1}{z}{\int }_{0}^{z}\frac{1-t}{1+t}\phantom{\rule{0.2em}{0ex}}dt=-1+\frac{2ln\left(1+z\right)}{z}$.

Using Theorem 2.12, we obtain
$24{z}^{2}+12z+1\prec \frac{1-z}{1+z},\phantom{\rule{1em}{0ex}}z\in U,$
induce
$8{z}^{2}+6z+1\prec -1+\frac{2ln\left(1+z\right)}{z},\phantom{\rule{1em}{0ex}}z\in U.$

Theorem 2.14 Let g be a convex function such that $g\left(0\right)=0$ and let h be the function $h\left(z\right)=g\left(z\right)+\frac{z}{1-\delta }{g}^{\mathrm{\prime }}\left(z\right)$, $z\in U$.

If $\alpha \ge 0$, $\delta \in \left(0,1\right)$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and the differential subordination
${\left(\frac{z}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{\delta }\frac{{L}_{\alpha }^{n+1}f\left(z\right)}{1-\delta }\left(\frac{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n+1}f\left(z\right)}-\delta \frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.14)
holds, then
$\frac{{L}_{\alpha }^{n+1}f\left(z\right)}{z}\cdot {\left(\frac{z}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{\delta }\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

This result is sharp.

Proof Let $p\left(z\right)=\frac{{L}_{\alpha }^{n+1}f\left(z\right)}{z}\cdot {\left(\frac{z}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{\delta }$. We deduce that $p\in \mathcal{H}\left[1,1\right]$.

Differentiating, we obtain ${\left(\frac{z}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{\delta }\frac{{L}_{\alpha }^{n+1}f\left(z\right)}{1-\delta }\left(\frac{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n+1}f\left(z\right)}-\delta \frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\right)=p\left(z\right)+\frac{1}{1-\delta }z{p}^{\mathrm{\prime }}\left(z\right)$, $z\in U$.

Using the notation in (2.14), the differential subordination becomes
$p\left(z\right)+\frac{1}{1-\delta }z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right)=g\left(z\right)+\frac{z}{1-\delta }{g}^{\mathrm{\prime }}\left(z\right).$
By using Lemma 1.2, we have
$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\phantom{\rule{1em}{0ex}}\frac{{L}_{\alpha }^{n+1}f\left(z\right)}{z}\cdot {\left(\frac{z}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{\delta }\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp. □

Theorem 2.15 Let h be a holomorphic function which satisfies the inequality $Re\left(1+\frac{z{h}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{h}^{\mathrm{\prime }}\left(z\right)}\right)>-\frac{1}{2}$, $z\in U$, and $h\left(0\right)=1$.

If $\alpha \ge 0$, $\delta \in \left(0,1\right)$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination
${\left(\frac{z}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{\delta }\frac{{L}_{\alpha }^{n+1}f\left(z\right)}{1-\delta }\left(\frac{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n+1}f\left(z\right)}-\delta \frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.15)
then
$\frac{{L}_{\alpha }^{n+1}f\left(z\right)}{z}\cdot {\left(\frac{z}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{\delta }\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where $q\left(z\right)=\frac{1-\delta }{{z}^{1-\delta }}{\int }_{0}^{z}h\left(t\right){t}^{-\delta }\phantom{\rule{0.2em}{0ex}}dt$. The function q is convex and it is the best dominant.

Proof Let $p\left(z\right)=\frac{{L}_{\alpha }^{n+1}f\left(z\right)}{z}\cdot {\left(\frac{z}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{\delta }$, $z\in U$, $p\in \mathcal{H}\left[0,1\right]$.

Differentiating, we obtain ${\left(\frac{z}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{\delta }\frac{{L}_{\alpha }^{n+1}f\left(z\right)}{1-\delta }\left(\frac{{\left({L}_{\alpha }^{n+1}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n+1}f\left(z\right)}-\delta \frac{{\left({L}_{\alpha }^{n}f\left(z\right)\right)}^{\mathrm{\prime }}}{{L}_{\alpha }^{n}f\left(z\right)}\right)=p\left(z\right)+\frac{1}{1-\delta }z{p}^{\mathrm{\prime }}\left(z\right)$, $z\in U$, and (2.15) becomes
$p\left(z\right)+\frac{1}{1-\delta }z{p}^{\mathrm{\prime }}\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$
Using Lemma 1.1, we have
$\begin{array}{c}p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}}\text{,}\hfill \\ \frac{{L}_{\alpha }^{n+1}f\left(z\right)}{z}\cdot {\left(\frac{z}{{L}_{\alpha }^{n}f\left(z\right)}\right)}^{\delta }\prec q\left(z\right)=\frac{1-\delta }{{z}^{1-\delta }}{\int }_{0}^{z}h\left(t\right){t}^{-\delta }\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,\hfill \end{array}$

and q is the best dominant. □

## Declarations

### Acknowledgements

Dedicated to Professor Hari M Srivastava.

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

## Authors’ Affiliations

(1)
Department of Mathematics and Computer Science, University of Oradea, str. Universitatii nr. 1, Oradea, 410087, Romania

## References

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