Theory and Modern Applications

# Differential subordinations using the Ruscheweyh derivative and the generalized Sălăgean operator

## Abstract

In the present paper, we study the operator, using the Ruscheweyh derivative ${R}^{m}f\left(z\right)$ and the generalized Sălăgean operator ${D}_{\lambda }^{m}f\left(z\right)$, denote by $R{D}_{\lambda ,\alpha }^{m}:{\mathcal{A}}_{n}\to {\mathcal{A}}_{n}$, $R{D}_{\lambda ,\alpha }^{m}f\left(z\right)=\left(1-\alpha \right){R}^{m}f\left(z\right)+\alpha {D}_{\lambda }^{m}f\left(z\right)$, $z\in U$, 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. We obtain several differential subordinations regarding the operator $R{D}_{\lambda ,\alpha }^{m}$.

MSC:30C45, 30A20, 34A40.

## 1 Introduction

Denote by U the unit disc of the complex plane, $U=\left\{z\in \mathbb{C}:|z|<1\right\}$ and $\mathcal{H}\left(U\right)$ the space of holomorphic functions in U.

Let ${\mathcal{A}}_{n}=\left\{f\in \mathcal{H}\left(U\right):f\left(z\right)=z+{a}_{n+1}{z}^{n+1}+\cdots ,z\in U\right\}$ 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 $K=\left\{f\in {\mathcal{A}}_{n}:Re\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}+1>0,z\in U\right\}$ 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 an univalent function in U. If p is analytic in U and satisfies the (second-order) differential subordination

$\psi \left(p\left(z\right),z{p}^{\prime }\left(z\right),{z}^{2}{p}^{″}\left(z\right);z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(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 (Al-Oboudi [1])

For $f\in {\mathcal{A}}_{n}$, $\lambda \ge 0$ and $n,m\in \mathbb{N}$, the operator ${D}_{\lambda }^{m}$ is defined by ${D}_{\lambda }^{m}:{\mathcal{A}}_{n}\to {\mathcal{A}}_{n}$,

$\begin{array}{c}{D}_{\lambda }^{0}f\left(z\right)=f\left(z\right),\hfill \\ {D}_{\lambda }^{1}f\left(z\right)=\left(1-\lambda \right)f\left(z\right)+\lambda z{f}^{\prime }\left(z\right)={D}_{\lambda }f\left(z\right),\hfill \\ \dots ,\hfill \\ {D}_{\lambda }^{m+1}f\left(z\right)=\left(1-\lambda \right){D}_{\lambda }^{m}f\left(z\right)+\lambda z{\left({D}_{\lambda }^{m}f\left(z\right)\right)}^{\prime }={D}_{\lambda }\left({D}_{\lambda }^{m}f\left(z\right)\right),\phantom{\rule{1em}{0ex}}z\in U.\hfill \end{array}$

Remark 1.1 If $f\in {\mathcal{A}}_{n}$ and $f\left(z\right)=z+{\sum }_{j=n+1}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then ${D}_{\lambda }^{m}f\left(z\right)=z+{\sum }_{j=n+1}^{\mathrm{\infty }}{\left[1+\left(j-1\right)\lambda \right]}^{m}{a}_{j}{z}^{j}$, $z\in U$.

Remark 1.2 For $\lambda =1$, in the definition above, we obtain the Sălăgean differential operator [2].

Definition 1.2 (Ruscheweyh [3])

For $f\in {\mathcal{A}}_{n}$, $n,m\in \mathbb{N}$, the operator ${R}^{m}$ is defined by ${R}^{m}:{\mathcal{A}}_{n}\to {\mathcal{A}}_{n}$,

$\begin{array}{c}{R}^{0}f\left(z\right)=f\left(z\right),\hfill \\ {R}^{1}f\left(z\right)=z{f}^{\prime }\left(z\right),\hfill \\ \dots ,\hfill \\ \left(m+1\right){R}^{m+1}f\left(z\right)=z{\left({R}^{m}f\left(z\right)\right)}^{\prime }+m{R}^{m}f\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.\hfill \end{array}$

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

Definition 1.3 [4]

Let $\alpha ,\lambda \ge 0$, $n,m\in \mathbb{N}$. Denote by $R{D}_{\lambda ,\alpha }^{m}$ the operator given by $R{D}_{\lambda ,\alpha }^{m}:{\mathcal{A}}_{n}\to {\mathcal{A}}_{n}$,

$R{D}_{\lambda ,\alpha }^{m}f\left(z\right)=\left(1-\alpha \right){R}^{m}f\left(z\right)+\alpha {D}_{\lambda }^{m}f\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

Remark 1.4 If $f\in {\mathcal{A}}_{n}$, $f\left(z\right)=z+{\sum }_{j=n+1}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then $R{D}_{\lambda ,\alpha }^{m}f\left(z\right)=z+{\sum }_{j=n+1}^{\mathrm{\infty }}\left\{\alpha {\left[1+\left(j-1\right)\lambda \right]}^{m}+\left(1-\alpha \right){C}_{m+j-1}^{m}\right\}{a}_{j}{z}^{j}$, $z\in U$.

This operator was studied also in [46] and [7].

Remark 1.5 For $\alpha =0$, $R{D}_{\lambda ,0}^{m}f\left(z\right)={R}^{m}f\left(z\right)$, where $z\in U$ and for $\alpha =1$, $R{D}_{\lambda ,1}^{m}f\left(z\right)={D}_{\lambda }^{m}f\left(z\right)$, where $z\in U$.

For $\lambda =1$, we obtain $R{D}_{1,\alpha }^{m}f\left(z\right)={L}_{\alpha }^{m}f\left(z\right)$, which was studied in [811].

For $m=0$, $R{D}_{\lambda ,\alpha }^{0}f\left(z\right)=\left(1-\alpha \right){R}^{0}f\left(z\right)+\alpha {D}_{\lambda }^{0}f\left(z\right)=f\left(z\right)={R}^{0}f\left(z\right)={D}_{\lambda }^{0}f\left(z\right)$, where $z\in U$.

Lemma 1.1 (Hallenbeck and Ruscheweyh [[12], 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

$p\left(z\right)+\frac{1}{\gamma }z{p}^{\prime }\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

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 [12])

Let g be a convex function in U, and let $h\left(z\right)=g\left(z\right)+n\alpha z{g}^{\prime }\left(z\right)$, 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

$p\left(z\right)+\alpha z{p}^{\prime }\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

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 $h\left(z\right)=g\left(z\right)+\frac{nz}{\delta }{g}^{\prime }\left(z\right)$, for $z\in U$.

If $\alpha ,\lambda ,\delta \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and satisfies the differential subordination

${\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta -1}{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.1)

then

${\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta }\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp.

Proof By using the properties of operator $R{D}_{\lambda ,\alpha }^{m}$, we have

$R{D}_{\lambda ,\alpha }^{m}f\left(z\right)=z+\sum _{j=n+1}^{\mathrm{\infty }}\left\{\alpha {\left[1+\left(j-1\right)\lambda \right]}^{m}+\left(1-\alpha \right){C}_{m+j-1}^{m}\right\}{a}_{j}{z}^{j},\phantom{\rule{1em}{0ex}}z\in U.$

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

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

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

Then (2.1) becomes

By using Lemma 1.2, we have

$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\phantom{\rule{1em}{0ex}}{\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta }\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

□

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

If $\alpha ,\lambda ,\delta \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and satisfies the differential subordination

${\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta -1}{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.2)

then

${\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta }\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

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

Proof Let

$\begin{array}{rcl}p\left(z\right)& =& {\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta }={\left(\frac{z+{\sum }_{j=n+1}^{\mathrm{\infty }}\left\{\alpha {\left[1+\left(j-1\right)\lambda \right]}^{m}+\left(1-\alpha \right){C}_{m+j-1}^{m}\right\}{a}_{j}{z}^{j}}{z}\right)}^{\delta }\\ =& {\left(1+\sum _{j=n+1}^{\mathrm{\infty }}\left\{\alpha {\left[1+\left(j-1\right)\lambda \right]}^{m}+\left(1-\alpha \right){C}_{m+j-1}^{m}\right\}{a}_{j}{z}^{j-1}\right)}^{\delta }=1+\sum _{j=n\delta }^{\mathrm{\infty }}{p}_{j}{z}^{j}\end{array}$

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

Differentiating, we obtain ${\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta -1}{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }=p\left(z\right)+\frac{1}{\delta }z{p}^{\prime }\left(z\right)$, $z\in U$, and (2.2) becomes

$p\left(z\right)+\frac{1}{\delta }z{p}^{\prime }\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

Using Lemma 1.1, we have

$p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\phantom{\rule{1em}{0ex}}{\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta }\prec q\left(z\right)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_{0}^{z}h\left(t\right){t}^{\frac{\delta }{n}-1}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,$

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 $\alpha ,\delta ,\lambda \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and satisfies the differential subordination

${\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta -1}{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.3)

then

${\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta }\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

where q is given by $q\left(z\right)=\left(2\beta -1\right)+\frac{2\left(1-\beta \right)\delta }{n{z}^{\frac{\delta }{n}}}{\int }_{0}^{z}\frac{{t}^{\frac{\delta }{n}-1}}{1+t}\phantom{\rule{0.2em}{0ex}}dt$, $z\in U$. 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 $p\left(z\right)={\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta }$, the differential subordination (2.3) becomes

$p\left(z\right)+\frac{z}{\delta }{p}^{\prime }\left(z\right)\prec h\left(z\right)=\frac{1+\left(2\beta -1\right)z}{1+z},\phantom{\rule{1em}{0ex}}z\in U.$

By using Lemma 1.1, for $\gamma =\delta$, we have $p\left(z\right)\prec q\left(z\right)$, i.e.,

$\begin{array}{rl}{\left(\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z}\right)}^{\delta }& \prec q\left(z\right)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_{0}^{z}h\left(t\right){t}^{\frac{\delta }{n}-1}\phantom{\rule{0.2em}{0ex}}dt\\ =\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_{0}^{z}{t}^{\frac{\delta }{n}-1}\frac{1+\left(2\beta -1\right)t}{1+t}\phantom{\rule{0.2em}{0ex}}dt\\ =\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_{0}^{z}\left[\left(2\beta -1\right){t}^{\frac{\delta }{n}-1}+2\left(1-\beta \right)\frac{{t}^{\frac{\delta }{n}-1}}{1+t}\right]\phantom{\rule{0.2em}{0ex}}dt\\ =\left(2\beta -1\right)+\frac{2\left(1-\beta \right)\delta }{n{z}^{\frac{\delta }{n}}}{\int }_{0}^{z}\frac{{t}^{\frac{\delta }{n}-1}}{1+t}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U.\end{array}$

□

Remark 2.1 For $n=1$, $\lambda =\frac{1}{2}$, $\alpha =2$, $\delta =1$, we obtain the same example as in [[13], Example 4.2.1, p.125].

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

If $\alpha ,\lambda ,\delta \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and the differential subordination

$\begin{array}{r}z\frac{\delta +1}{\delta }\frac{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{2}}+\frac{{z}^{2}}{\delta }\frac{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{2}}\left[\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}-2\frac{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)}\right]\\ \phantom{\rule{1em}{0ex}}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U\end{array}$
(2.4)

holds, then

$z\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)\right)}^{2}}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp.

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

$R{D}_{\lambda ,\alpha }^{m}f\left(z\right)=z+\sum _{j=n+1}^{\mathrm{\infty }}\left\{\alpha {\left[1+\left(j-1\right)\lambda \right]}^{m}+\left(1-\alpha \right){C}_{m+j-1}^{m}\right\}{a}_{j}{z}^{j},\phantom{\rule{1em}{0ex}}z\in U.$

Consider $p\left(z\right)=z\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)\right)}^{2}}$, and we obtain

$p\left(z\right)+\frac{z}{\delta }{p}^{\prime }\left(z\right)=z\frac{\delta +1}{\delta }\frac{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{2}}+\frac{{z}^{2}}{\delta }\frac{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{2}}\left[\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}-2\frac{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)}\right].$

Relation (2.4) becomes

$p\left(z\right)+\frac{z}{\delta }{p}^{\prime }\left(z\right)\prec h\left(z\right)=g\left(z\right)+\frac{nz}{\delta }{g}^{\prime }\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

By using Lemma 1.2, we have

$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\phantom{\rule{1em}{0ex}}z\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)\right)}^{2}}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

□

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

If $\alpha ,\lambda ,\delta \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and satisfies the differential subordination

$\begin{array}{r}z\frac{\delta +1}{\delta }\frac{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{2}}+\frac{{z}^{2}}{\delta }\frac{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{2}}\left[\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}-2\frac{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)}\right]\\ \phantom{\rule{1em}{0ex}}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\end{array}$
(2.5)

then

$z\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)\right)}^{2}}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

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

Proof Let $p\left(z\right)=z\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)\right)}^{2}}$, $z\in U$, $p\in \mathcal{H}\left[1,n\right]$.

Differentiating, we obtain $p\left(z\right)+\frac{z}{\delta }{p}^{\prime }\left(z\right)=z\frac{\delta +1}{\delta }\frac{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{2}}+\frac{{z}^{2}}{\delta }\frac{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{2}}\left[\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}-2\frac{{\left(R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n+1}f\left(z\right)}\right]$, $z\in U$, and (2.5) becomes

$p\left(z\right)+\frac{z}{\delta }{p}^{\prime }\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

Using Lemma 1.1, we have

$\begin{array}{r}p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\\ z\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{{\left(R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)\right)}^{2}}\prec q\left(z\right)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_{0}^{z}h\left(t\right){t}^{\frac{\delta }{n}-1}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,\end{array}$

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 $h\left(z\right)=g\left(z\right)+\frac{nz}{\delta }{g}^{\prime }\left(z\right)$, $z\in U$.

If $\alpha ,\lambda ,\delta \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and the differential subordination

${z}^{2}\frac{\delta +2}{\delta }\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}+\frac{{z}^{3}}{\delta }\left[\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{″}}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}-{\left(\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}\right)}^{2}\right]\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U$
(2.6)

holds, then

${z}^{2}\frac{{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

This result is sharp.

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

Differentiating, we obtain $p\left(z\right)+\frac{z}{\delta }{p}^{\prime }\left(z\right)={z}^{2}\frac{\delta +2}{\delta }\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}+\frac{{z}^{3}}{\delta }\left[\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{″}}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}-{\left(\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}\right)}^{2}\right]$, $z\in U$.

Using the notation in (2.6), the differential subordination becomes

$p\left(z\right)+\frac{1}{\delta }z{p}^{\prime }\left(z\right)\prec h\left(z\right)=g\left(z\right)+\frac{nz}{\delta }{g}^{\prime }\left(z\right).$

By using Lemma 1.2, we have

$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\phantom{\rule{1em}{0ex}}{z}^{2}\frac{{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp. □

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

If $\alpha ,\lambda ,\delta \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and satisfies the differential subordination

${z}^{2}\frac{\delta +2}{\delta }\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}+\frac{{z}^{3}}{\delta }\left[\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{″}}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}-{\left(\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}\right)}^{2}\right]\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.7)

then

${z}^{2}\frac{{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

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

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

Differentiating, we obtain $p\left(z\right)+\frac{z}{\delta }{p}^{\prime }\left(z\right)={z}^{2}\frac{\delta +2}{\delta }\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}+\frac{{z}^{3}}{\delta }\left[\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{″}}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}-{\left(\frac{{\left(R{D}_{\lambda ,\alpha }^{n}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{n}f\left(z\right)}\right)}^{2}\right]$, $z\in U$, and (2.7) becomes

$p\left(z\right)+\frac{1}{\delta }z{p}^{\prime }\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

Using Lemma 1.1, we have

$p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\phantom{\rule{1em}{0ex}}{z}^{2}\frac{{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\prec q\left(z\right)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_{0}^{z}h\left(t\right){t}^{\frac{\delta }{n}-1}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,$

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 $h\left(z\right)=g\left(z\right)+nz{g}^{\prime }\left(z\right)$, $z\in U$.

If $\alpha ,\lambda \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and the differential subordination

$1-\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{″}}{{\left[{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\right]}^{2}}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U$
(2.8)

holds, then

$\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

This result is sharp.

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

Differentiating, we obtain $1-\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{″}}{{\left[{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\right]}^{2}}=p\left(z\right)+z{p}^{\prime }\left(z\right)$, $z\in U$.

Using the notation in (2.8), the differential subordination becomes

$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec h\left(z\right)=g\left(z\right)+nz{g}^{\prime }\left(z\right).$

By using Lemma 1.2, we have

$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\phantom{\rule{1em}{0ex}}\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp. □

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

If $\alpha ,\lambda \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and satisfies the differential subordination

$1-\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{″}}{{\left[{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\right]}^{2}}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.9)

then

$\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

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

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

Differentiating, we obtain $1-\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{″}}{{\left[{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\right]}^{2}}=p\left(z\right)+z{p}^{\prime }\left(z\right)$, $z\in U$, and (2.9) becomes

$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

Using Lemma 1.1, we have

$p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\phantom{\rule{1em}{0ex}}\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}\prec q\left(z\right)=\frac{1}{n{z}^{\frac{1}{n}}}{\int }_{0}^{z}h\left(t\right){t}^{\frac{1}{n}-1}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,$

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 $\alpha ,\lambda \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and satisfies the differential subordination

$1-\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{″}}{{\left[{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\right]}^{2}}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.10)

then

$\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

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

Proof Following the same steps as in the proof of Theorem 2.9 and considering $p\left(z\right)=\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}$, the differential subordination (2.10) becomes

$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec h\left(z\right)=\frac{1+\left(2\beta -1\right)z}{1+z},\phantom{\rule{1em}{0ex}}z\in U.$

By using Lemma 1.1 for $\gamma =1$, we have $p\left(z\right)\prec q\left(z\right)$, i.e.,

$\begin{array}{rl}\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}{z{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}& \prec q\left(z\right)=\frac{1}{n{z}^{\frac{1}{n}}}{\int }_{0}^{z}h\left(t\right){t}^{\frac{1}{n}-1}\phantom{\rule{0.2em}{0ex}}dt\\ =\frac{1}{n{z}^{\frac{1}{n}}}{\int }_{0}^{z}\frac{1+\left(2\beta -1\right)t}{1+t}{t}^{\frac{1}{n}-1}\phantom{\rule{0.2em}{0ex}}dt\\ =\frac{1}{n{z}^{\frac{1}{n}}}{\int }_{0}^{z}{t}^{\frac{1}{n}-1}\left[\left(2\beta -1\right)+\frac{2\left(1-\beta \right)}{1+t}\right]\phantom{\rule{0.2em}{0ex}}dt\\ =\left(2\beta -1\right)+\frac{2\left(1-\beta \right)}{n{z}^{\frac{1}{n}}}{\int }_{0}^{z}\frac{{t}^{\frac{1}{n}-1}}{1+t}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U.\end{array}$

□

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 $Re\left(\frac{z{h}^{″}\left(z\right)}{{h}^{\prime }\left(z\right)}+1\right)>-\frac{1}{2}$.

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

Then ${\left(R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\right)}^{\prime }={f}^{\prime }\left(z\right)=1+2z$,

$\begin{array}{c}\frac{R{D}_{\frac{1}{2},2}^{1}f\left(z\right)}{z{\left(R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\right)}^{\prime }}=\frac{z+{z}^{2}}{z\left(1+2z\right)}=\frac{1+z}{1+2z},\hfill \\ 1-\frac{R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\cdot {\left(R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\right)}^{″}}{{\left[{\left(R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\right)}^{\prime }\right]}^{2}}=1-\frac{\left(z+{z}^{2}\right)\cdot 2}{{\left(1+2z\right)}^{2}}=\frac{2{z}^{2}+2z+1}{{\left(1+2z\right)}^{2}}.\hfill \end{array}$

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

$\frac{2{z}^{2}+2z+1}{{\left(1+2z\right)}^{2}}\prec \frac{1-z}{1+z},\phantom{\rule{1em}{0ex}}z\in U,$

induce

$\frac{1+z}{1+2z}\prec -1+\frac{2ln\left(1+z\right)}{z},\phantom{\rule{1em}{0ex}}z\in U.$

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

If $\alpha ,\lambda \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and the differential subordination

${\left[{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\right]}^{2}+R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{″}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U$
(2.11)

holds, then

$\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{z}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

This result is sharp.

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

Differentiating, we obtain ${\left[{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\right]}^{2}+R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{″}=p\left(z\right)+z{p}^{\prime }\left(z\right)$, $z\in U$.

Using the notation in (2.11), the differential subordination becomes

$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec h\left(z\right)=g\left(z\right)+nz{g}^{\prime }\left(z\right).$

By using Lemma 1.2, we have

$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\phantom{\rule{1em}{0ex}}\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{z}\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp. □

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

If $\alpha ,\lambda \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and satisfies the differential subordination

${\left[{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\right]}^{2}+R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{″}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.12)

then

$\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{z}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

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

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

Differentiating, we obtain ${\left[{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\right]}^{2}+R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{″}=p\left(z\right)+z{p}^{\prime }\left(z\right)$, $z\in U$, and (2.12) becomes

$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

Using Lemma 1.1, we have

$\begin{array}{r}p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\\ \frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{z}\prec q\left(z\right)=\frac{1}{n{z}^{\frac{1}{n}}}{\int }_{0}^{z}h\left(t\right){t}^{\frac{1}{n}-1}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,\end{array}$

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 $\alpha ,\lambda \ge 0$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and satisfies the differential subordination

${\left[{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }\right]}^{2}+R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{″}\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.13)

then

$\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{z}\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

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

Proof Following the same steps as in the proof of Theorem 2.12 and considering $p\left(z\right)=\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{z}$, the differential subordination (2.13) becomes

$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec h\left(z\right)=\frac{1+\left(2\beta -1\right)z}{1+z},\phantom{\rule{1em}{0ex}}z\in U.$

By using Lemma 1.1 for $\gamma =1$, we have $p\left(z\right)\prec q\left(z\right)$, i.e.,

$\begin{array}{rl}\frac{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\cdot {\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{z}& \prec q\left(z\right)=\frac{1}{n{z}^{\frac{1}{n}}}{\int }_{0}^{z}h\left(t\right){t}^{\frac{1}{n}-1}\phantom{\rule{0.2em}{0ex}}dt\\ =\frac{1}{n{z}^{\frac{1}{n}}}{\int }_{0}^{z}\frac{1+\left(2\beta -1\right)t}{1+t}{t}^{\frac{1}{n}-1}\phantom{\rule{0.2em}{0ex}}dt\\ =\frac{1}{n{z}^{\frac{1}{n}}}{\int }_{0}^{z}{t}^{\frac{1}{n}-1}\left[\left(2\beta -1\right)+\frac{2\left(1-\beta \right)}{1+t}\right]\phantom{\rule{0.2em}{0ex}}dt\\ =\left(2\beta -1\right)+\frac{2\left(1-\beta \right)}{n{z}^{\frac{1}{n}}}{\int }_{0}^{z}\frac{{t}^{\frac{1}{n}-1}}{1+t}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U.\end{array}$

□

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 $Re\left(\frac{z{h}^{″}\left(z\right)}{{h}^{\prime }\left(z\right)}+1\right)>-\frac{1}{2}$.

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

Then ${\left(R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\right)}^{\prime }={f}^{\prime }\left(z\right)=1+2z$,

$\begin{array}{c}\frac{R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\cdot {\left(R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\right)}^{\prime }}{z}=\frac{\left(z+{z}^{2}\right)\left(1+2z\right)}{z}=2{z}^{2}+3z+1,\hfill \\ {\left[{\left(R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\right)}^{\prime }\right]}^{2}+R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\cdot {\left(R{D}_{\frac{1}{2},2}^{1}f\left(z\right)\right)}^{″}={\left(1+2z\right)}^{2}+\left(z+{z}^{2}\right)\cdot 2=6{z}^{2}+6z+1.\hfill \end{array}$

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

$6{z}^{2}+6z+1\prec \frac{1-z}{1+z},\phantom{\rule{1em}{0ex}}z\in U,$

induce

$2{z}^{2}+3z+1\prec -1+\frac{2ln\left(1+z\right)}{z},\phantom{\rule{1em}{0ex}}z\in U.$

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

If $\alpha ,\lambda \ge 0$, $\delta \in \left(0,1\right)$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and the differential subordination

${\left(\frac{z}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\right)}^{\delta }\frac{R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)}{1-\delta }\left(\frac{{\left(R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)}-\delta \frac{{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U$
(2.14)

holds, then

$\frac{R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)}{z}\cdot {\left(\frac{z}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\right)}^{\delta }\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

This result is sharp.

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

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

Using the notation in (2.14), the differential subordination becomes

$p\left(z\right)+\frac{1}{1-\delta }z{p}^{\prime }\left(z\right)\prec h\left(z\right)=g\left(z\right)+\frac{nz}{1-\delta }{g}^{\prime }\left(z\right).$

By using Lemma 1.2, we have

$p\left(z\right)\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\phantom{\rule{1em}{0ex}}\frac{R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)}{z}\cdot {\left(\frac{z}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\right)}^{\delta }\prec g\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

and this result is sharp. □

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

If $\alpha ,\lambda \ge 0$, $\delta \in \left(0,1\right)$, $n,m\in \mathbb{N}$, $f\in {\mathcal{A}}_{n}$ and satisfies the differential subordination

${\left(\frac{z}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\right)}^{\delta }\frac{R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)}{1-\delta }\left(\frac{{\left(R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)}-\delta \frac{{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$
(2.15)

then

$\frac{R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)}{z}\cdot {\left(\frac{z}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\right)}^{\delta }\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

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

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

Differentiating, we obtain ${\left(\frac{z}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\right)}^{\delta }\frac{R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)}{1-\delta }\left(\frac{{\left(R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)}-\delta \frac{{\left(R{D}_{\lambda ,\alpha }^{m}f\left(z\right)\right)}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\right)=p\left(z\right)+\frac{1}{1-\delta }z{p}^{\prime }\left(z\right)$, $z\in U$, and (2.15) becomes

$p\left(z\right)+\frac{1}{1-\delta }z{p}^{\prime }\left(z\right)\prec h\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.$

Using Lemma 1.1, we have

$\begin{array}{r}p\left(z\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in U,\phantom{\rule{1em}{0ex}}\mathit{\text{i.e.}},\\ \frac{R{D}_{\lambda ,\alpha }^{m+1}f\left(z\right)}{z}\cdot {\left(\frac{z}{R{D}_{\lambda ,\alpha }^{m}f\left(z\right)}\right)}^{\delta }\prec q\left(z\right)=\frac{1-\delta }{n{z}^{\frac{1-\delta }{n}}}{\int }_{0}^{z}h\left(t\right){t}^{\frac{1-\delta }{n}-1}\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}z\in U,\end{array}$

and q is the best dominant. □

## Author’s contributions

The author drafted the manuscript, read and approved the final manuscript.

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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 Loriana Andrei.

### Competing interests

The author declares that she has no competing interests.

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Andrei, L. Differential subordinations using the Ruscheweyh derivative and the generalized Sălăgean operator. Adv Differ Equ 2013, 252 (2013). https://doi.org/10.1186/1687-1847-2013-252

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### Keywords

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