Differential subordinations using the Ruscheweyh derivative and the generalized Sălăgean operator

Abstract

In the present paper, we study the operator, using the Ruscheweyh derivative $R^{m}f(z)$ and the generalized Sălăgean operator $D_{\lambda }^{m}f(z)$, denote by $R{D}_{\lambda ,\alpha }^m:{\mathcal{A}}_n\to {\mathcal{A}}_n$, $R{D}_{\lambda ,\alpha }^{m}f(z)=(1-\alpha )R^{m}f(z)+\alpha D_{\lambda }^{m}f(z)$, $z\in U$, where ${\mathcal{A}}_n=\{f\in \mathcal{H}(U):f(z)=z+a_{n+1}{z}^{n+1}+\cdots ,z\in U\}$ 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, $U=\{z\in \mathbb{C}:|z|<1\}$ and $\mathcal{H}(U)$ the space of holomorphic functions in U.

Let ${\mathcal{A}}_n=\{f\in \mathcal{H}(U):f(z)=z+a_{n+1}{z}^{n+1}+\cdots ,z\in U\}$ and $\mathcal{H}[a,n]=\{f\in \mathcal{H}(U):f(z)=a+a_n{z}^n+a_{n+1}{z}^{n+1}+\cdots ,z\in U\}$ for $a\in \mathbb{C}$ and $n\in \mathbb{N}$.

Denote by $K=\{f\in {\mathcal{A}}_n:Re\frac{z{f}^{″}(z)}{f^{\prime }(z)}+1>0,z\in U\}$ 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 $w(0)=0$, $|w(z)|<1$, for all $z\in U$ such that $f(z)=g(w(z))$ for all $z\in U$. If g is univalent, then $f\prec g$ if and only if $f(0)=g(0)$ and $f(U)\subseteq g(U)$.

Let $\psi :{\mathbb{C}}^3\times U\to \mathbb{C}$, and let h be an univalent function in U. If p is analytic in U and satisfies the (second-order) differential subordination

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

(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 $\tilde{q}$ that satisfies $\tilde{q}\prec q$ 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 $D_{\lambda }^m:{\mathcal{A}}_n\to {\mathcal{A}}_n$,

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

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

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 $R^m:{\mathcal{A}}_n\to {\mathcal{A}}_n$,

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

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

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 $R{D}_{\lambda ,\alpha }^m:{\mathcal{A}}_n\to {\mathcal{A}}_n$,

$$R{D}_{\lambda ,\alpha }^{m}f(z)=(1-\alpha )R^{m}f(z)+\alpha D_{\lambda }^{m}f(z),z\in U.$$

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

This operator was studied also in [4–6] and [7].

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

For $\lambda =1$, we obtain $R{D}_{1,\alpha }^{m}f(z)=L_{\alpha }^{m}f(z)$, which was studied in [8–11].

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

Lemma 1.1 (Hallenbeck and Ruscheweyh [[12], Th. 3.1.6, p.71])

Let h be a convex function with $h(0)=a$, and let $\gamma \in \mathbb{C}\mathrm{\setminus }\{0\}$ be a complex number with $Re\gamma \ge 0$. If $p\in \mathcal{H}[a,n]$ and

$$p(z)+\frac{1}{\gamma }z{p}^{\prime }(z)\prec h(z),z\in U,$$

then

$$p(z)\prec g(z)\prec h(z),z\in U,$$

where $g(z)=\frac{\gamma }{n{z}^{\gamma /n}}{\int }_0^{z}h(t)t^{\gamma /n-1}dt$, $z\in U$.

Lemma 1.2 (Miller and Mocanu [12])

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

If $p(z)=g(0)+p_n{z}^n+p_{n+1}{z}^{n+1}+\cdots$, $z\in U$, is holomorphic in U and

$$p(z)+\alpha z{p}^{\prime }(z)\prec h(z),z\in U,$$

then

$$p(z)\prec g(z),z\in U,$$

and this result is sharp.

2 Main results

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

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

(2.1)

then

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

and this result is sharp.

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

$$R{D}_{\lambda ,\alpha }^{m}f(z)=z+\sum _{j=n+1}^{\mathrm{\infty }}\{\alpha {[1+(j-1)\lambda ]}^m+(1-\alpha )C_{m+j-1}^m\}{a}_j{z}^j,z\in U.$$

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

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

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

Then (2.1) becomes

$$p(z)+\frac{1}{\delta }z{p}^{\prime }(z)\prec h(z)=g(z)+\frac{nz}{\delta }{g}^{\prime }(z)\text{for }z\in U.$$

By using Lemma 1.2, we have

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

 □

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

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

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

(2.2)

then

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

where $q(z)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^{z}h(t)t^{\frac{\delta }{n}-1}dt$. The function q is convex, and it is the best dominant.

Proof Let

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

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

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

$$p(z)+\frac{1}{\delta }z{p}^{\prime }(z)\prec h(z),z\in U.$$

Using Lemma 1.1, we have

$$p(z)\prec q(z),z\in U,\mathit{\text{i.e.}},{\left(\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{z}\right)}^{\delta }\prec q(z)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^{z}h(t)t^{\frac{\delta }{n}-1}dt,z\in U,$$

and q is the best dominant. □

Corollary 2.3 Let $h(z)=\frac{1+(2\beta -1)z}{1+z}$ 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

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

(2.3)

then

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

where q is given by $q(z)=(2\beta -1)+\frac{2(1-\beta )\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^z\frac{t^{\frac{\delta }{n}-1}}{1+t}dt$, $z\in U$. 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 $p(z)={(\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{z})}^{\delta }$, the differential subordination (2.3) becomes

$$p(z)+\frac{z}{\delta }{p}^{\prime }(z)\prec h(z)=\frac{1+(2\beta -1)z}{1+z},z\in U.$$

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

$$\begin{array}{rl}{\left(\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{z}\right)}^{\delta }& \prec q(z)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^{z}h(t)t^{\frac{\delta }{n}-1}dt\\ =\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^z{t}^{\frac{\delta }{n}-1}\frac{1+(2\beta -1)t}{1+t}dt\\ =\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^z[(2\beta -1)t^{\frac{\delta }{n}-1}+2(1-\beta )\frac{t^{\frac{\delta }{n}-1}}{1+t}]dt\\ =(2\beta -1)+\frac{2(1-\beta )\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^z\frac{t^{\frac{\delta }{n}-1}}{1+t}dt,z\in U.\end{array}$$

 □

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 $g(0)=1$, and let h be the function $h(z)=g(z)+\frac{nz}{\delta }{g}^{\prime }(z)$, $z\in U$.

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

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

(2.4)

holds, then

$$z\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{{(R{D}_{\lambda ,\alpha }^{m+1}f(z))}^2}\prec g(z),z\in U,$$

and this result is sharp.

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

$$R{D}_{\lambda ,\alpha }^{m}f(z)=z+\sum _{j=n+1}^{\mathrm{\infty }}\{\alpha {[1+(j-1)\lambda ]}^m+(1-\alpha )C_{m+j-1}^m\}{a}_j{z}^j,z\in U.$$

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

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

Relation (2.4) becomes

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

By using Lemma 1.2, we have

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

 □

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

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

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

(2.5)

then

$$z\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{{(R{D}_{\lambda ,\alpha }^{m+1}f(z))}^2}\prec q(z),z\in U,$$

where $q(z)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^{z}h(t)t^{\frac{\delta }{n}-1}dt$. The function q is convex, and it is the best dominant.

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

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

$$p(z)+\frac{z}{\delta }{p}^{\prime }(z)\prec h(z),z\in U.$$

Using Lemma 1.1, we have

$$\begin{array}{r}p(z)\prec q(z),z\in U,\mathit{\text{i.e.}},\\ z\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{{(R{D}_{\lambda ,\alpha }^{m+1}f(z))}^2}\prec q(z)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^{z}h(t)t^{\frac{\delta }{n}-1}dt,z\in U,\end{array}$$

and q is the best dominant. □

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

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

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

(2.6)

holds, then

$$z^2\frac{{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f(z)}\prec g(z),z\in U.$$

This result is sharp.

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

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

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

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

By using Lemma 1.2, we have

$$p(z)\prec g(z),z\in U,\mathit{\text{i.e.}},z^2\frac{{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f(z)}\prec g(z),z\in U,$$

and this result is sharp. □

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

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

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

(2.7)

then

$$z^2\frac{{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f(z)}\prec q(z),z\in U,$$

where $q(z)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^{z}h(t)t^{\frac{\delta }{n}-1}dt$. The function q is convex, and it is the best dominant.

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

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

$$p(z)+\frac{1}{\delta }z{p}^{\prime }(z)\prec h(z),z\in U.$$

Using Lemma 1.1, we have

$$p(z)\prec q(z),z\in U,\mathit{\text{i.e.}},z^2\frac{{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{R{D}_{\lambda ,\alpha }^{m}f(z)}\prec q(z)=\frac{\delta }{n{z}^{\frac{\delta }{n}}}{\int }_0^{z}h(t)t^{\frac{\delta }{n}-1}dt,z\in U,$$

and q is the best dominant. □

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

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

$$1-\frac{R{D}_{\lambda ,\alpha }^{m}f(z)\cdot {(R{D}_{\lambda ,\alpha }^{m}f(z))}^{″}}{{[{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }]}^2}\prec h(z),z\in U$$

(2.8)

holds, then

$$\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{z{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}\prec g(z),z\in U.$$

This result is sharp.

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

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

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

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

By using Lemma 1.2, we have

$$p(z)\prec g(z),z\in U,\mathit{\text{i.e.}},\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{z{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}\prec g(z),z\in U,$$

and this result is sharp. □

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

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

$$1-\frac{R{D}_{\lambda ,\alpha }^{m}f(z)\cdot {(R{D}_{\lambda ,\alpha }^{m}f(z))}^{″}}{{[{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }]}^2}\prec h(z),z\in U,$$

(2.9)

then

$$\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{z{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}\prec q(z),z\in U,$$

where $q(z)=\frac{1}{n{z}^{\frac{1}{n}}}{\int }_0^{z}h(t)t^{\frac{1}{n}-1}dt$. The function q is convex, and it is the best dominant.

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

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

$$p(z)+z{p}^{\prime }(z)\prec h(z),z\in U.$$

Using Lemma 1.1, we have

$$p(z)\prec q(z),z\in U,\mathit{\text{i.e.}},\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{z{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}\prec q(z)=\frac{1}{n{z}^{\frac{1}{n}}}{\int }_0^{z}h(t)t^{\frac{1}{n}-1}dt,z\in U,$$

and q is the best dominant. □

Corollary 2.10 Let $h(z)=\frac{1+(2\beta -1)z}{1+z}$ 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

$$1-\frac{R{D}_{\lambda ,\alpha }^{m}f(z)\cdot {(R{D}_{\lambda ,\alpha }^{m}f(z))}^{″}}{{[{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }]}^2}\prec h(z),z\in U,$$

(2.10)

then

$$\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{z{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}\prec q(z),z\in U,$$

where q is given by $q(z)=(2\beta -1)+\frac{2(1-\beta )}{n{z}^{\frac{1}{n}}}{\int }_0^z\frac{t^{\frac{1}{n}-1}}{1+t}dt$, $z\in U$. 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 $p(z)=\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{z{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}$, the differential subordination (2.10) becomes

$$p(z)+z{p}^{\prime }(z)\prec h(z)=\frac{1+(2\beta -1)z}{1+z},z\in U.$$

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

$$\begin{array}{rl}\frac{R{D}_{\lambda ,\alpha }^{m}f(z)}{z{(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}& \prec q(z)=\frac{1}{n{z}^{\frac{1}{n}}}{\int }_0^{z}h(t)t^{\frac{1}{n}-1}dt\\ =\frac{1}{n{z}^{\frac{1}{n}}}{\int }_0^z\frac{1+(2\beta -1)t}{1+t}{t}^{\frac{1}{n}-1}dt\\ =\frac{1}{n{z}^{\frac{1}{n}}}{\int }_0^z{t}^{\frac{1}{n}-1}[(2\beta -1)+\frac{2(1-\beta )}{1+t}]dt\\ =(2\beta -1)+\frac{2(1-\beta )}{n{z}^{\frac{1}{n}}}{\int }_0^z\frac{t^{\frac{1}{n}-1}}{1+t}dt,z\in U.\end{array}$$

 □

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

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

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

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

We have $q(z)=\frac{1}{z}{\int }_0^z\frac{1-t}{1+t}dt=-1+\frac{2ln(1+z)}{z}$.

Using Theorem 2.9, we obtain

$$\frac{2z^2+2z+1}{{(1+2z)}^2}\prec \frac{1-z}{1+z},z\in U,$$

induce

$$\frac{1+z}{1+2z}\prec -1+\frac{2ln(1+z)}{z},z\in U.$$

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

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

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

(2.11)

holds, then

$$\frac{R{D}_{\lambda ,\alpha }^{m}f(z)\cdot {(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{z}\prec g(z),z\in U.$$

This result is sharp.

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

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

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

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

By using Lemma 1.2, we have

$$p(z)\prec g(z),z\in U,\mathit{\text{i.e.}},\frac{R{D}_{\lambda ,\alpha }^{m}f(z)\cdot {(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{z}\prec g(z),z\in U,$$

and this result is sharp. □

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

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

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

(2.12)

then

$$\frac{R{D}_{\lambda ,\alpha }^{m}f(z)\cdot {(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{z}\prec q(z),z\in U,$$

where $q(z)=\frac{1}{n{z}^{\frac{1}{n}}}{\int }_0^{z}h(t)t^{\frac{1}{n}-1}dt$. The function q is convex, and it is the best dominant.

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

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

$$p(z)+z{p}^{\prime }(z)\prec h(z),z\in U.$$

Using Lemma 1.1, we have

$$\begin{array}{r}p(z)\prec q(z),z\in U,\mathit{\text{i.e.}},\\ \frac{R{D}_{\lambda ,\alpha }^{m}f(z)\cdot {(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{z}\prec q(z)=\frac{1}{n{z}^{\frac{1}{n}}}{\int }_0^{z}h(t)t^{\frac{1}{n}-1}dt,z\in U,\end{array}$$

and q is the best dominant. □

Corollary 2.13 Let $h(z)=\frac{1+(2\beta -1)z}{1+z}$ 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

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

(2.13)

then

$$\frac{R{D}_{\lambda ,\alpha }^{m}f(z)\cdot {(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{z}\prec q(z),z\in U,$$

where q is given by $q(z)=(2\beta -1)+\frac{2(1-\beta )}{n{z}^{\frac{1}{n}}}{\int }_0^z\frac{t^{\frac{1}{n}-1}}{1+t}dt$, $z\in U$. 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 $p(z)=\frac{R{D}_{\lambda ,\alpha }^{m}f(z)\cdot {(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{z}$, the differential subordination (2.13) becomes

$$p(z)+z{p}^{\prime }(z)\prec h(z)=\frac{1+(2\beta -1)z}{1+z},z\in U.$$

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

$$\begin{array}{rl}\frac{R{D}_{\lambda ,\alpha }^{m}f(z)\cdot {(R{D}_{\lambda ,\alpha }^{m}f(z))}^{\prime }}{z}& \prec q(z)=\frac{1}{n{z}^{\frac{1}{n}}}{\int }_0^{z}h(t)t^{\frac{1}{n}-1}dt\\ =\frac{1}{n{z}^{\frac{1}{n}}}{\int }_0^z\frac{1+(2\beta -1)t}{1+t}{t}^{\frac{1}{n}-1}dt\\ =\frac{1}{n{z}^{\frac{1}{n}}}{\int }_0^z{t}^{\frac{1}{n}-1}[(2\beta -1)+\frac{2(1-\beta )}{1+t}]dt\\ =(2\beta -1)+\frac{2(1-\beta )}{n{z}^{\frac{1}{n}}}{\int }_0^z\frac{t^{\frac{1}{n}-1}}{1+t}dt,z\in U.\end{array}$$

 □

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

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

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

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

We have $q(z)=\frac{1}{z}{\int }_0^z\frac{1-t}{1+t}dt=-1+\frac{2ln(1+z)}{z}$.

Using Theorem 2.12, we obtain

$$6z^2+6z+1\prec \frac{1-z}{1+z},z\in U,$$

induce

$$2z^2+3z+1\prec -1+\frac{2ln(1+z)}{z},z\in U.$$

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

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

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

(2.14)

holds, then

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

This result is sharp.

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

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

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

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

By using Lemma 1.2, we have

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

and this result is sharp. □

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

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

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

(2.15)

then

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

where $q(z)=\frac{1-\delta }{n{z}^{\frac{1-\delta }{n}}}{\int }_0^{z}h(t)t^{\frac{1-\delta }{n}-1}dt$. The function q is convex, and it is the best dominant.

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

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

$$p(z)+\frac{1}{1-\delta }z{p}^{\prime }(z)\prec h(z),z\in U.$$

Using Lemma 1.1, we have

$$\begin{array}{r}p(z)\prec q(z),z\in U,\mathit{\text{i.e.}},\\ \frac{R{D}_{\lambda ,\alpha }^{m+1}f(z)}{z}\cdot {\left(\frac{z}{R{D}_{\lambda ,\alpha }^{m}f(z)}\right)}^{\delta }\prec q(z)=\frac{1-\delta }{n{z}^{\frac{1-\delta }{n}}}{\int }_0^{z}h(t)t^{\frac{1-\delta }{n}-1}dt,z\in U,\end{array}$$

and q is the best dominant. □

Author’s contributions

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